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R such that s2 lim inf . > 0 s-»+oo e > 0 and g is bounded then the same holds true for f + g, which can be deduced from the fact that for some k > 0 R which satisfies V(t,u,v)£E, 6 Lp(a,b) and R > r such that fRR S1/" € Lp(a,b), r > 0 and R> r that satisfy (4.18). Define E C [a,b] x M2 from (4.7) (with a — a, b — b, a = a and f3 = 0). Then, for every Lp- Caratheodory function f : E —> E such that for a.e. t £ [a,b] and all (u,v) £ M2, with (t,u,v) £ E, € LP(a,b), r>0andR>r that satisfy (4.18). Define E C [5,6] x M2 from (4.7) (with a = a, b = b, a = & and (5 = 0). Then, for every LP-Caratheodory function f : E —> R such that for a.e. t € [a,b] and all (u,v) G R 2 , with (t,u,v) € E, sgn(v)f(t,u,v) < ${t)(p{\v\), (resp. sgn(v)f(t,u,v) > -tp(t) R be a positive continuous function satisfying (5.4) and f : E — M a continuous function which satisfies (4.14) (with (p = 0 such that for all t £ A (resp. for all t G B), f(t,a(t),a'(t))>-N R be a positive continuous function satisfying — r 0 satisfy (1.9) and i/iat i/ie function f satisfies one of the one-sided Nagumo conditions (a), (b), (c), or (d) in Theorem 1-6.1. 0 such that (1.9) holds and f satisfies one of the one-sided Nagumo conditions (a), (b), (c), or (d) in Theorem 1-6.1, with a = cti and 0 = /32Then the problem (1.5) has at least three solutions ui,u2,ua G W2'p(a, b) such that for all t € [a, b] «i(t) < «i(t) < ft(*). «2(t) < Mt) < fo{t), m(t) < u3(t) < u2{t) and there exist t\, t2 £ [a, b] with {u + tv) - cf>(u) t ~ t for all t > 0 and all v G H. The idea is to pass to the limit when t —* 0, for some appropriate choice of v. To prove (a), choose v = w and note that by (4.2) (with v = w), the function T(u + tw) is strictly increasing for t G M + . We compute then 0, ), v) > 0 and s+ f+(t) uniformly in t and u—»+oo l) with support in IQ SO that / F(t,(p(t))dt>0. Jo It follows that for s large enough $(y) < 0. Hence, for those values of s, $(aa) < $(0) = 0, i.e. a% ^ ai = 0. By Proposition III-2.7, 0:2 is a strict lower solution of (3.1) and we deduce from (3.6) that a.- 0. Step 3 - Construction of a strict upper solution /3i of (3.1) so iftai a i = 0 < /9i < (h o,nd «2 ^ /?i- Assumption (HQ) in Theorem 1.5 holds with <7o = £ > 0 small enough. We deduce then as in Claim 1 of this theorem the existence of a strict W^^-upper solution 0i > 0 which we can choose small enough so that «2 j£ 0iConclusion - The proof follows now from Theorem III-2.11. D Remark Such a result can be obtained using variational arguments as in Example IV-1.1. 2, ds = Jo
and V{t,u,v)eE, \f{t,u,v)\
4. A PRIORI BOUND ON THE DERIVATIVE
47
We have / h
-7-^ ds > 7-7- ds > > —: ~JA ¥>(*)
5- = >
- = 00.
On the other hand, if 5 = 1, f + g does not satisfy a Nagumo condition. One computes n+ 1 1 and /
ds
JB IB s + *
which shows that
ds < oo.
It is now easy to see that we can regularize / — ip to obtain a continuous function with the same property. In many cases however, f + g still satisfies a Nagumo condition with (4.9). For example, if / satisfies such conditions with
A similar result holds if / and g satisfy a Nagumo condition with the same function
I
00
sds
ip(s) > max/?(i) - mina(t),
(4-10)
where r = max{^h)-_l{a), ^ f ® } > 0. Then there exists R > 0 such that for every continuous function f : E —> R which satisfies ,i;)|<¥J(M)
(4.11)
48
I. THE PERIODIC PROBLEM
and every solution u of (4.6) on [a, b] such that a < u < 0, we have Hu'lloo < RProof. Define R to be such that R
sds
> max/3(£) - mma(t). m
(4.12)
Let u be a solution of (4.6) such that a
\u(r)\
\
\ < b —a
r
.
Now consider an interval / = [to,ti] or [ti,t0] such that u'(i) > 0 on /, io) = r, u'(^i) > r. Then we have
f'(tl) iff! = T1 "'(O^W M = f
Mto) V{s)
Jt0 V{u'{t))
Jt0
u'(t)f(t,u(t),u'(t)) ({>{t))
,
u'{t)dt\ = \u{f,\) — u(to)\ < max/3(i) — mina(t). From (4.12) we deduce that u'(ti) < R. In the same way, we prove that for any t € [a,b], u'(t) > —R and the result follows. A fundamental generalization of this proposition concerns one-sided Nagumo conditions. This applies to solutions u such that some a priori bound on the derivative is known at the end points of the interval of definition, which is clearly the case for the Neumann problem u" = f{t,u,u'), u'(a) = A, u'(b) = B.
,
(
. >
Notice also that if u solves the periodic problem (4.1), there exists a £ [a, b] such that u'(a) = 0 and the periodic extension of u(t) solves (4.13) with a, b = a + b - a and A = B - 0. Our first result concerns solutions such that a one-sided a priori bound on the derivative is known at both end points.
4. A PRIORI BOUND ON THE DERIVATIVE
49
Proposition 4.5. Let a and f3 € C([a,6]) be such that a < (3. Define E C [a, b] x M2 from (4.7) (with a = a, b = b, a — a and (3 = (3). Let r > 0, R > r and
f(t,u,v)<
(4.14)
and for every solution u of (4.6) on [a, b] which is such that a < u < f3, u'(a) < r and u'(b) > —r we have
Similarly, for every continuous function f : E —> R which satisfies V(t, u, v) e E,
f(t, u, v) > -
(4.15)
and for every solution u of (4.6) on [a, b] which is such that a < u < {3, u'(a) > —r and u'(b) < r we have
IKHoo < RProof. Let u be a solution as above and assume by contradiction that mintu'(t) < — R. Hence, there exists an interval / = [to,ii] such that u'(t) < 0 on /, u'(to) < —R and u'{t\) — —r. We deduce then the contradiction sds /
f~u ' °' sds
¥>($) ~ 7- U '(ti)
__ [
tl
(-u'(t))(-u"(t))
M
_ fH
(-u'(t))f(t,u(t),u'(t))
/
s ds cp[s)
A similar argument holds if maxt u'(t) > R. The second part of the proposition is proved in the same way. Example 4.2. As an example, notice that condition (4.14) applies for a problem such as u" + u'32 + g(t, u) = 0, u(a) = u(b), u (a) — u'(b).
50
I. THE PERIODIC PROBLEM
Remark In the previous proposition, the condition f(t,u,v)<(p(\v\) uses the same function ip for v > 0 and v < 0. A trivial generalization would be f{t,u,v) <
sgn(v)f(t, u, v) <
(4.16)
and for every solution u of (4.6) on [5,5], which is such that a < u < [3 and \u'(a)\ < r, we have
llu'lU < RSimilarly, for every continuous function f : E —» R which satisfies V(t, u, v) e E,
sgn(v) f(t, u, v) > -Cp{\v\)
(4.17)
and for every solution u of (4.6) on [a,b], which is such that a < u < J3 and \u'(b)\ < r, we have llu'Hoo < RThe reader will easily find examples of functions f(t,u, v) which satisfy one and only one of these one-sided conditions. Exercise 4.1. Prove Proposition 4.6. Each of the conditions (4.14), (4.15), (4.16) and (4.17) is called a onesided Nagumo condition. Remark Nagumo conditions and more specifically one-sided Nagumo conditions have a nice geometric interpretation. Assume for simplicity a and ft are constant and f(t,a,0) < 0 < f(t,(3,O). Consider then the assumption f{t,u,v) <
4. A PRIORI BOUND ON THE DERIVATIVE
51
In the phase space (u, u') we define the curve u' = v(u), where u is the solution of the problem
,
( ) V o
V
If we assume (4.10) and choose vo £ [0, r] it is easy to see that v is defined on [a,/3]. Also, if u £ [a,0], the vector field corresponding to the differential equation (4.6) points downward along that curve. An assumption such as f(t, u, v) >
n(u) < v! < v{u)}.
This geometry is essential to obtain existence of solutions from a degree argument or a modification method as we did in this chapter. Notice that it is not essential for the vector field to point upward along the curve u' — fj,(u) and downward along the curve u' = u(u). Other geometries are possible. These underlie the various one-sided Nagumo conditions we mentioned above. 4.4
A Generalization: the Caratheodory Case
Observe that if
J
^ds>\mLP(maX/3(t)-mma(t))1^.
(4.18)
Define E C [a, b] x R2 from (4.7) (with a - a and 0 — ft). Then, for every IP -Caratheodory function f : E —» E such that for a.e. t € [a,b\ and all (u,v) £ M.2, with (t,u,v) £ E, \f(t,u,v)\<$(t)
52
I. THE PERIODIC PROBLEM
Proof. Let u be a solution of (4.6) and t £ [a,b] be such that u'(t) > R. We can choose, as in the proof of Proposition 4.4, to < h (or t0 > *i) such that u'(t0) — r, u'(ti) = R and r < u'(s)
Jt0 f 1 ip(s)un/ll(s) ds < ll^Hip f * u'(s) ds ha \Jt0 1/g ~mina(t)) .
1/9
Prom (4.18) we obtain a contradiction and deduce that u'(t) < R. In the same way we prove that u'(t) > —R. D As a first remark, notice that (4.18) is obviously implied by
L o
°
sl/9
—T-T ds = CO.
Also one-sided Nagumo conditions can be worked out for the Caratheodory case just as in Proposition 4.5. Proposition 4.8. Let a and (3 £ C([a,5]) be such that a < J3 and p, q € [l,oo] such that i + i = 1. Assume there exist
/(*,«,«) <$(t)£(M), and for every solution u of (4.6) on [a,b] such that a < u < J3, u'(a) < r and u'(b) > —r, we have Hu'llao < R. p Similarly, for every L -Caratheodory function f : E —» M such that for a.e. t € [a,b] and all (u,v) £ M2, with (t,u,v) £ E, f(t,u,v)>-$(t)
5. DERIVATIVE DEPENDENCE : C2-SOLUTIONS
53
Proposition 4.9. Let a and 0 € C([a,b]) be such that a < 0 and p, q 6 [l,oo] such that - + - — 1. Assume there exist
Exercise 4.2. Prove Propositions 4.8 and 4.9.
5 5.1
BVP with Derivative Dependence : C2-Solutions Definitions
Consider the periodic boundary value problem u" = /(*,«, u'), u(a) = u(b), u'(a) = u'(b).
,-,, ^° '
The dependence of / in the derivative u' does not really change the definitions of lower and upper solutions. These are straightforward extensions of Definitions 2.1. Definitions 5.1. A function a 6 C([a, &]) such that a(a) = a(b) is a C2lower solution of (5.1) if its periodic extension on R, defined by ct(t) = a(t + b — a) is such that for any to S R either D-a(t0) < D+a(tQ) or there exist an open interval IQ with to G -^o and a function a® e C ' ( / o ) such that (a) a(to) = cto(to) and a(t) > ao(t) for all t € IQ,' (b) a'0'(t0) exists and a'0'(t0) > f(to,a0{to),a'o(to)). A function (i £ C([a,b\) such that /3(a) = f3(b) is a C2-upper solution of (5.1) if its periodic extension on R is such that for any to € R either D~0(to) > D+0(to) or there exist an open interval Io with to £ Jo and a function 0Q £ C1(/o) such that (a) P(t0) = f30{t0) and /?(*) < 0o(t) for all t € J o ; (b) $ (to) exists andj30'(t0) < f(t0,/3o(t0), Potto))-
54
I. THE PERIODIC PROBLEM
Remark Let a be a lower solution and assume to is such that D_a(t 0 ) > D+a(t0). Prom the definition, there exists then ao S C1(/o) such that a(to) = ao(*o) and a(t) > cto(t) on Jo. It follows that D-a(t0) < D-a(tQ) < a'0{t0) < D+a(t0) < D+a(t0) < D-a(t0), which implies a has a derivative at to and a'(t0) = a'o(to). As in Section 2, we can build lower and upper solutions from maxima of lower solutions and minima of upper solutions. Proposition 5.1. Let a* S C([a, b]) (i = 1, ...,n) be C2-lower solutions of (5.1). Then the function a(t) = max cti(t),
t G [a,b]
is a C2-lower solution of (5.1). Proposition 5.2. Let (3j £ C([a, b]) (j — 1, ...,m) be C2'-upper solutions of (5.1). Then the function = ^ m ^ i ) , te[a,b] is a C2-upper solution of (5.1). Exercise 5.1. Prove the above propositions. 5.2
Existence of Solutions
For a first and simple approach, we state a theorem which uses a two-sided Nagumo condition, which as we will see later, can be generalized in various directions. Theorem 5.3. Let a and j3 be C2 -lower and upper solutions of the problem (5.1) such that a < (3, E be defined in (4.7), if : IR+ —> 1R be a positive continuous function satisfying (4.9) and f : E —» M be a continuous function which satisfies (4.8). Then the problem (5.1) has at least one solution u € C2([a, b]) such that for all t € [a, b] a(t) < u(t) <
5. DERIVATIVE DEPENDENCE : C2-SOLUTIONS
55
Proof. Consider the modified problem u" = A/(t, 7 (*.«),«') +
p on \a, b]. Assume there exists f 0 such that u(to) = mint u(t) < —p. This assumption leads to a contradiction since in such a case 0 < u"(*0) = A[/(to, a(to), 0) +
Let us define the
operators L : D o m t C Cx{[a,b\) -> C([a,b]),u i-» u" -u, N*:C ([a,b])-*C{[a,b]), u ^ Xf(t, 7 (t, u),«') + ¥>(|u'|)(u - A7(t,«)) - u, 1
where DomI = {u£ C2([a,6]) | u(a) = u(b), u'(a) - u'(b)}. Observe that L has a compact inverse. Hence, we can define the completely continuous operator Tx(u) = L-'Nxiu). From degree theory, we have that
where O = { « e ^([0,6]) I Hloo < p, IKHoo < R}Using the Odd Mapping Theorem A-1.5, we compute that
and the problem (5.2) with A = 1 has a solution u.
56
I. T H E P E R I O D I C PROBLEM
Conclusion - Using the argument in Theorem 2.3, we prove that this solution u is such that for all t € [a,b]
a(t) < u(t) < 0(t). Hence u is also a solution of (5.1). Notice that we can try to prove this result using the modified problem
u" - u = f{t,f{t,u),8{u'))-i{t,u), u(a)=u(b), u'(a)=u'(b),
,
.
[p a)
-
where j(t,u) is defined from (1.3) and S(v) = min{.R,max{—R,v}}. The proof follows as in Theorem 2.3 provided a and /? are in WltO°(a, b). On the other hand, with this last assumption we can weaken the condition (4.9) on (p and assume
L
T > max/3(i) - mina(t). (5.4) o ¥>(«) * * As always for periodic problems, we can also replace the two-sided Nagumo condition by a one-sided one. Theorem 5.4. Let a, f) 6 W1'°°(a, b) be C2-lower and upper solutions of the problem (5.1) such that a<(3,Ebe defined in (4.7),
and
f
sds —^ > max/3(t) — min a(t). lo V>{s) t t Jo
, p - a l U , ||2>-j8|U Halloo} < H
Consider then the modified problem (5.3). Step 1 - Using Schauder's Theorem, we prove that problem (5.3) has a solution u. Step 2 - The solution u is such that a(t) < u(t) < /3(t) on [a,b]. This result follows from the argument in the proof of Theorem 2.3. As a consequence, u satisfies f(t,u,d(u>)), u" = u{a) = u{b), u'(a) = u'{b).
K
>
5. DERIVATIVE DEPENDENCE : C2-SOLUTIONS
57
Step 3 - The solution u is such that ||u'||oo < fl. As u solves a periodic problem, there exists a G [a, b] such that u'(a) = 0. Let b — a + b — a and consider the periodic extension of u on [a,b\. Observe that for all (t,u,v) G E (with a — a and b = b), f(t,u,6(v))<
(resp. f(t,/3(t),P'(t))
(5.6)
Then the problem (5.1) has at least one solution u G C2([a, b]) such that for all t G [a, b] a(t) < u(t) < /?(*). Proof. Let R be large enough so that
i
rR
s d s
an\
-
- r T > max/?(*)-: /o V( Increasing the value of N if necessary, we can assume N > max{\ f(t,u,v)\
\ t G [a,b], a(t)
0(t), \v\ < R}.
Consider then the modified problem u" -u = f(t,u,u')-j(t,u), u(a)=u(b), u'(a)=u'(b), where f(t,u,v) := max{min{/(£,7(£,u),v),N}, —N} and j(t,u) is denned in (1.3).
58
I. THE PERIODIC PROBLEM
Step 1 - Using Schauder's Theorem we prove that problem (5.7) has a solution u. Step 2 - The solution u is such that a < u < /?. periodicity. Let to S [a, b[ be such that
Extend a and u by
u(t0) - a(t 0 ) = min(u(t) - a{t)) < 0, which implies £>_a(t0) > D+a{t0) x
and there exists ao G C {Io) as in Definition 5.1. It follows that to is a minimum of u — ao, (u — ao)'(to) = 0 and (u - ao)"(*o) > 0. Prom the remark that follows Definition 5.1, we have a'(to) = a^to). Hence to € A and /(to,u(to),u'(t o )) < f(to,a(to),a'(tQ))
=
f(t0,a0(to),a'0(to)).
We compute now 0 < u"{to) - a'o'ito) = /(to, u(t0), u'(t0)) + "(to) - ao(*o) - "o (*o) < /(t 0 , a o (t o ), a o (t o )) - ao'(to) + u(t0) - ao(to) < 0 which is a contradiction. In a similar way we prove that u < j3. As a consequence, u satisfies u" = max{min{/(t,u,n'),iV},-iV}, «(a) = u(6), u'(o) = u'(b).
,_ , '
K
Step 3 - The solution u is such that \u'(t)\ < R on [a,b\. Observe that, for all (t, u, v) £ E, max{min{/(t,u,u),iV},-JV} <tp(\v\). Prom Proposition 4.5 (with r = 0 and
IKIloo < RHence |/(t, u(t), u'(t))\ < N and the function u is a solution of (5.1). Notice the idea of the modification (5.7). Here / is truncated rather than u and u'. Remark also that condition (5.6) is satisfied if a, (3 G W1'°°(a, b) or if / does not depend on u'.
6. DERIVATIVE DEPENDENCE : W^-SOLUTIONS
59
Remark The Theorems 5.3, 5.4, and 5.5 are still valid if we replace the Nagumo condition (4.8) or the one-sided Nagumo condition (4.14) by this last one-sided condition or any of the following one: (a) for all (t,u,v) G E, f{t,u,v) > -
Structure of the Set of Solutions
The following results concern existence of maximal and minimal solutions between a and @ as well as existence of a continuum of solutions. Theorem 5.6. Let a and /? be C2-lower and upper solutions of the problem (5.1) such that a < /3, E be defined in (4.7), ip : M+ —> R be a positive continuous function satisfying (4.9) and f : E —+ R be a continuous function which satisfies (4.8). Then the problem (5.1) has a minimal solution umin G C2([a,b]) and a maximal solution umax G C2([o, b}) in [a,(3], i.e. 01 < Umin < Umax < P,
and any other solution u of (5.1) such that a < u < j3 satisfies Umin
Umax-
Theorem 5.7. Assume the hypothesis of Theorem 5.6 hold and f is nondecreasing with respect to u. Then for any to G [a,b] and u* G K with umin(to) < u* < umax(t0), there exists a solution u G C2([a,b]) of (5.1) such that umin < u < umax and u(to) = u*. Exercise 5.2. Prove the above results adapting the proofs of Theorems 2.4 and 2.5. Exercise 5.3. State and prove the counterpart of Theorem 5.6 under the assumptions of Theorem 5.5.
6
BVP with Derivative Dependence : W2'x-Solutions
For problems with a nonlinearity that depends on the derivative and satisfies Z^-Caratheodory conditions, the definitions of lower and upper solutions are the obvious extensions of Definitions 3.1.
60
I. THE PERIODIC PROBLEM
Definitions 6.1. A function a £ C({a,b}) such that a(a) = a(b) is a Wr2'1-lower solution of (5.1) if its periodic extension on R, defined by a(t) — a(t + b — a), is such that for any to &M. either D-.a(t0) < D+a(t0) or there exists an open interval Io such that to € Jo, a € W2'l(Io) and for a.e. t € IQ, a"(t)>f(t,a(t),a'(t)). A function (5 £ C([a, b]) such that j3{a) = /3(6) is a W^-upper solution of (5.1) if its periodic extension on R, defined by P(t) — f3{t + b — a), is such that for any to £ R either D~0(to) > D+/3(t0) or there exists an open interval Io such that to € Io, 0 € W2il(Io) and for a.e. t G Jo, Our first result uses lower and upper solutions which are in W1'00. This assumption is not essential but it allows a method of proof which is interesting by itself. Theorem 6.1. Let a and 0 € Wx'°°(a, b) be W2'1-lower and upper solutions of (5.1) such that a < (3. Let E be defined from (4.7) and p, q £ [l,oo] be such that | + | = 1. Assume f : E —> R satisfies Lp-Caratheodory conditions. Assume moreover there exist
/ —r^r ds > |M|LI,(max/3(i) - mina(t)) 1 ^ s Jo V( ) * *
(6.1)
and that the function f satisfies one of the following one-sided Nagumo conditions (a) for a.e. t £ [a, b] and all {u,v) £ R 2 such that a(t)
f3(t),
(d) for a.e. t£ [a,b] and all (u,v) € R 2 such that a(t)
6. DERIVATIVE DEPENDENCE : W2>1-SOLUTIONS
61
A natural idea to prove this theorem is to use a modified problem. For example, we could consider (5.3) as in Theorem 5.4. To prove then that solutions of the modified problem lie between a and /3 requires however that / satisfies an extra assumption such as Condition (A) in Proposition III1.5. To avoid such an unnecessary condition, we will use another modified problem, which will need several auxiliary results. Lemma 6.2. Let u e W1'oo(a,b). and
Then u+ = max{u,0} G W1'oo{a,b)
= jtu(t),
ifu(t)>0,
^u+(t)=0,
ifu(t)<0.
at Proof. See [188, Theorem Al (p.50)].
D
Lemma 6.3. Let u G Whoo(a,b) and (un)n C W 1 ' 0 0 ^^) be such that un —> u in W1><x(a,b). Then u+ —+ u+ in C([a,b]) and for almost every t G [a, b]
i Proof. The first part of the proof comes from the continuity of the application P : u h-> u+ as follows from
To prove the second part, notice that for almost every t, the derivatives
&u+(t), £tu{t), ftu+(t) and ftun(t)
exist and lim n ^ oo ftun{t)
=
ftu{i).
Let t be such a point. If u{t) > 0, we have Jjtt + (i) = ^u(t) and, for n large enough, un{i) > 0, which implies that for such an n,
=
n(t)
and ^ ^ ( t ) = !«+(«).
Similarly, if u(t) < 0, we have Jju + (t) = 0 and for n large enough,
At last if u(t) = 0 and as at point t both derivatives -^u+(t) and Jju(t) exist, we have ^u+(t) = 0 and also ^u(t) = 0. As -jftU+(t) is ^un(t) or 0, this implies ^
|«+(t) « 0 = | « ( t ) .
D
62
I. THE PERIODIC PROBLEM
Corollary 6.4. Let a and 0 € W1>o°(a, b) be such that a < /3, and define j(t,u) from (1.3). Then, for any u e W1>0o(a,6), we /taw 7(.,u) € W1'°°{a,b) and ifu(t)
lim ^7(*,«n(*)) n—nx at
4 at
Proof. One can notice that 7 (i,
u(t)) = «(t) + (a - «)+(*) - (« - /?)+(*)
and use Lemmas 6.2 and 6.3. Lemma 6.5. Let a and /3 S M^1'0O(a, 6) 6e s«c/i that a < f3, define 7(t, u) from (1.3) and E from (4.7), and assume / : £ - » ! satisfies Ll-Caratheodory conditions. Then, for any u G C1([a, 6]), #ie function (6.2)
where G(t,s) is the Green's function corresponding to the problem (1.4), is mC 1 ([a, 6]). Furthermore, the operator
defined by (6.2) is completely continuous. Proof. Notice that given u e C1([a,6]), it follows from Corollary 6.4 that /(s,7(s,w(s)), j^j(s,u(s))) is in L1(a,b). It is then straightforward to see that Tu is in C 1 (fa, 6]).
6. DERIVATIVE DEPENDENCE : W^-SOLUTIONS
63
Next, let un —> u in Cx([a, 6]). Prom \T(un)(t)
-T(u)(t)\ fb
= | / Ja
G(t,s)f(s,j(s,un(s)),4-j(s,un(s)))ds b
- f Ja
G(t,s)f(s,1(S,u(s)),£j(s,u(s)))ds\ b
<max|G(t,s)| / |/(s,7(a,un(s)),* Ja -f(s, Lebesgue's Dominated Convergence Theorem, and Corollary 6.4, we have T{un) -> T(u) in C([a, b]). Similarly, one proves -^T(un) —> - -^T(u) in C([a,6]) as u n — u in Cx{[a, b)). The continuity of T follows. At last, the complete continuity of T follows as usually from ArzelaAscoli Theorem. Proof of Theorem 6.1 : Let R > maxdla'Hoo, ||/3'||oo} be large enough so that fR sx/q ds > \\ip\\Lp(maxP(t) —mi Jo f(s) Consider the modified problem u" -u = f{t, y(t, u), ^7(-, u)) - 7(i, u), where f(t,u,v) := f(t, u, max{min{u, R], — R}) and f(t,u) is defined by (1.3). Notice that (6.3) is a functional equation. Claim 1 - The problem (6.3) has at least one solution. We can write (6.3) as a fixed point problem u = Tu, where
(Tu)(t)= f Ja
This operator is clearly bounded and from Lemma 6.5 we know it is completely continuous. Hence by Schauder's Theorem, T has a fixed point which is a solution of (6.3).
64
I. THE PERIODIC PROBLEM
Claim 2 - The solution u of (6.3) satisfies on [a, b] a(t) < u(t) < j3{t). This claim follows from the argument used to prove Theorem 3.1. Claim 3 - The solution u of (6.3) is such that ||«'||oo 5= R- This result is proved from the argument used in Theorem 5.4 with Proposition 4.5 replaced by Proposition 4.8 or 4.9. Conclusion - It follows now from Claims 2 and 3 that the solution u of (6.3) solves (5.1). The following theorem extends Theorem 6.1 to lower and upper solutions whose derivative is not in L°°(a,b). As in Theorem 5.5, we have to assume a one-sided bound on f(t,a(t),a'(t)) and f(t,/3(t),/3'(t)). Notice that such a condition holds if a and /? £ W1'°°(a,b). Theorem 6.6. Let a and 0 £ C([a,b]) be W2'1-lower and upper solutions of (5.1) such that a < /?. Define A C [a,b] (resp. B c [a,b]) to be the set of points where a (resp. f3) is derivable. Let E be defined from (4.7) and p, q € [1, oo] be such that - + - = 1. Assume f : E —> K satisfies Lv-Caratheodory conditions and there exists N g L1(a, b), N > 0 such that for a.e. t e A (resp. for a.e. t e B) f{t,a{t), a'{t)) > -N(t)
(resp. f(t,P(t),/3'(t)) < N(t)).
Assume moreover there exist y? 6 C(M+,Rj) and tp € Lp{a,b) satisfying (6.1) and that the function f satisfies one of the one-sided Nagumo conditions (a), (b), (c), or (d) in Theorem 6.1. Then the problem (5.1) has at least one solution u € W2>p(a, b) such that for all t € [a, 6] a(t) < u(t) < Proof. The proof proceeds in several steps. Step 1: The modified problem. Let R > 0 be large enough so that
4 Increasing N if necessary, we can assume N(t) > \f(t, u, v)\ if t e [a,b], a{t)
a'(t))\,
\v\<S
=
XB(t)
m a x \f(t,0(t),/3'(t) \v\<8
+ v ) - f(t,0{t),
f3'(t))\,
6. DERIVATIVE DEPENDENCE : W^-SOLUTIONS
65
where 7 is defined from (1.3), XA and XB are the characteristic functions of the sets A and B. It is clear that w^ are L1-Caratheodory functions, non-decreasing in 5, such that u>i(t,0) = 0 and |wj(i,5)| < 2N(t). We consider now the modified problem u"-u = f(t,u,u')-w(t,u), u(a)=u{b), u'(a)=u'{b),
, , '
{0
where uj(t,u) = 0(t) - wa(t,u - /?(*)), if u > /?(*), = «, if a(t) < u < /3(t), = a(t) +un(t, a(t) -u), ii Step 2: Existence of a solution of (6.4). Notice that the operator
defined by
= f
G(t,S)[f(S,u(s),u'(s))-w(s,u(s))}ds,
Ja
where G(t,s) is the Green's function corresponding to (1.4), is completely continuous and bounded. Hence, by Schauder's Theorem, T has a fixed point u which is a solution of (6.4). Step 3: The solution u of (6.4) is such that a < u < 0. Let us assume on the contrary that for some to € M in(w(t) - a(t)) = u(t0) - a(tQ) < 0. Then, as in Theorem 3.1, there exists an open interval Jo with to S ^o> a e VF2>1(70) and, for a.e. t e Jo, a"(t)>f(tMt),a'(t)). Furthermore u'(*o) — ct'(to) = 0 and for t > to near enough to \u'(t)-a'(t)\
As wi is non-decreasing and f(t, a(t),a'(t)) < f(t,a(t),a'(t)), we have for t > to near enough to
u'(t) - a'(t) = [ (u"(s) - a"(a)) ds Jtn
< / [f(s,a(s),u'(s)) - f(s,a(sW(s)) Jt0
+u(s) — a(s) — u>i(s,a(s) — u(s))] ds < 0.
66
I. THE PERIODIC PROBLEM
This inequality proves u(to) — a (to) is not a minimum of u — a which is a contradiction. A similar argument holds to prove u < (5. Step 4: The solution u of (6.4) is such that ||u'||oo < R- Observe that for all (t,u,v) £ E, one of the conditions (a), (b), (c), or (d) in Theorem 6.1 is satisfied with f(t, u, v) replaced by f(t, u, v). Extending u by periodicity and using Propositions 4.8 or 4.9, we prove then that every solution u £ [a, /?] of (6.4) satisfies
IKHoo < R. Conclusion. It follows from Step 3 and 4 that the solution u of (6.4) solves
(5.1). We can improve Theorems 6.1 and 6.6 by assuming that / is an LrCaratheodory function and that conditions (a) to (d) are satisfied with tp £ Lp(a, b), where r is not necessarily equal to p. Observe that a Nagumo condition \f(t, u,v)\ < ip(t)
where 1 < a < 3/2. Existence of a solution in W2'2~e(0,1), for e > 0 small enough, follows from Theorem 6.1 with a(t) — —1, (3{t) = 0, p = ^r^ < 2, ^ and
—r—r ds = I
sa + l
ds — OO.
In Theorem 6.6, the limit cases p = oo, q — 1, and p = 1, q — oo are of interest. If p = oo, q = 1, it is natural to choose ijj = 1 and (6.1) reads
I
°° sds —y-r > max * B(t) — mina(i). In this case, Theorems 6.6 and 5.5 differ from the regularity assumptions. If p — 1, q = oo, condition (6.1) becomes /0
rK")
and we obtain the following corollary.
6. DERIVATIVE DEPENDENCE : W^-SOLUTIONS
67
Corollary 6.7. Let a and (3 G C([a, b}) be W2'1-lower and upper solutions of (5.1) such that a < j3. Define A C [a,b] (resp. B C [a, b]) to be the set of points where a (resp. /3) is derivable. Let E be defined by (4.7). Assume f : E —> R satisfies L1 -Caratheodory conditions and there exists N G Lx{a,b), N > 0 such that for a.e. t G A (resp. for a.e. t £ B) f(t,a(t),a'(t))>-N(t)
(resp.
Assume moreover there exist
f
ds
Jo
and that the function f satisfies one of the one-sided Nagumo conditions (a), (b), (c), or (d) in Theorem 6.1. Then the problem (5.1) has at least one solution u G W2'1(a,b) such that for all t € [a, b] a(t) < u(t) < The conditions of the corollary are satisfied if / : E —> R satisfies a L1-Lipschitz condition in v, i.e. there exists L £ Lx{a,b) such that for a.e. t G [a,b] and all (u,vi), (u,v2) G R2 (t,u,v1),(t,u,v2)£E
=»
\f(t,u,vi)-f(t,u,v2)\
Exercise 6.1. Prove a result similar to Corollary 6.7 assuming there exist ip G C(R + ,Rj) and ip G L1(a,b) with
I
00
ds
= CO
and that the function / satisfies one of the one-sided Nagumo conditions (a) for a.e. t G [a,b] and all (u,v) G R 2 such that a(t)
(3(t),
(b) for a.e. t G [a,b] and all (u,v) G R 2 such that a{t)
p(t),
(c) for a.e. t G [a,b] and all (u,v) G R2 such that a(t)
< (if>{t) + M M M ) ;
(d) for a.e. t G [a,b] and all (u,v) G R 2 such that a(t)
sgn(v)f(t,u,v)
> -(
p(t),
68
I. THE PERIODIC PROBLEM
Hint : See [187]. Notice that the Example 4.1 shows we cannot replace the Nagumo condition by the assumption that solutions of the given problem (5.1) between a and /3 are a priori bounded in C1([a,6]). We can however replace the Nagumo condition by any condition so that an a priori bound in C([a, b]) on solutions of the corresponding modified problem implies an a priori bound in (^([a, b}). In particular, we have the following result for the Rayleigh equation. Theorem 6.8. Let a and/3 £ C([o, 6]) be W2>1-lower and upper solutions of (4.4) such that aRbe a Caratheodory function such that for some H £ L2(a,b), for a.e. t £ [a,b] and all (u, v) £ E 2 with (t, u, v) £ E, \h(t,u,v)\
Then the problem (4.4) has at least one solution u £ W2'2(a, b) such that for all t £ [a, b] a{t) < u(t) < Proof. Define A C [a, b] (resp. B C [a, b\) to be the set of points where a (resp. (3) is derivable and
, 6) = XA(t) max \h(t, a(t),a'(t) +v)~ h(t, a(t),a'(t))\, \v\<8
) = XBit) max \h(t,p{t),0\t) +v)~ h(t,p(t),p'(t))\, \v\<S
where XA and XB are the characteristic functions of the sets A and B. It is clear that Wj are Caratheodory functions, non-decreasing in 5, such that uJi(t,O) = 0 and \ui{t,6)\ < 2H(t). Consider the family of modified problems u" - C{t)u = -[Xg(u') u(a)=u(b),
u'{a)=u'(b),
(6.5)
where C £ Lx{a,b) is chosen such that C(t) > \g(0)\ + 1 + 3H(t) on [a,b], A G [0,1], ~f(t,u) is defined from (1.3), and w(t, u) = C{t)P(t) - U2(t, u - 0(t)), if u > 0(t), = C(t)u, i£a(t)
6. DERIVATIVE DEPENDENCE : W^-SOLUTIONS
69
Claim 1 - Define p = max-fllaHoo, ||/?||oo} + 2. Then every solution u of
(6.5) satisfies ||u||oo < P- Let us assume on the contrary that for some t0 GM minu(t) — u(t0) < —p. Hence, for t > to close enough to to, we have \u'(t)\
— h(s,a(s),u'(s)) —uji(s,a(s) — u(s))}ds
<- I \C{s) - \g(u'(s))\ - m(s)} ds < 0. Jto
This inequality proves that u(to) is not a minimum of u which is a contradiction. A similar argument holds to prove that u < p. Claim 2 - There exists R > 0 such that every solution u of (6.5) satisfies ||w'||oo < R- The proof follows the argument in Proposition 4.1. Claim 3 - There exists a solution u of (6.5) with A = 1. Define the operator
Tx(u) = - / G(t, s)[Xg(u'(s)) + h(s, j(s,u(s)), u'(s)) + w(s,u(s))\ ds, Ja where G(t, s) is the Green's function of u" - C{t)u = f{t), u(a)=u(b), u'{a) = u'{b). Observe that there exists Ro > 0 such that TQ(C1{[a,b])) C Hence,
B(0,RQ).
and by the properties of the degree, we prove easily that (6.5) with A = 1 has a solution. Claim 4 - The solution u of (6.5) with A = 1 is such that a(t) < u{t) < on [a,b]. Extend a and u by periodicity. Let to £ [a, b[ be such that u(t0) - a{t0) = min(u(t) - a(t)) < 0.
70
I. T H E P E R I O D I C PROBLEM
Prom the definition of lower solution there exists an open interval IQ such that t 0 G Jo, a G W2'l{IQ) and for a.e. t G IQ,
a"(t) + g{a'{t)) + h(t, a(t), a'(t)) > 0. Further, for t > to near enough to, \u'{t)-a'(t)\
Proof. Consider the family of modified problems u" - C(t)u = u(a) = u(b), u'(a) = u'(b), where C G L1(a,b) is chosen such that for a.e. t G [a,b], \g(a(t))\ < C(t), lff(/?(t))| < C(t) and for every (t,u) G E, \h{t,u)\ < C(t). Claim 1 - Let p = maxdaHoo, ||/3||oo} + 3. Then every solution u of (6.6) satisfies ||u||oo < P- This result follows as Claim 1 in Theorem 6.8. Claim 2 - There exists R > 0 such that any solution u of (6.6) is such that Hu'lloo < R. Use the argument of Proposition 4.3.
6. DERIVATIVE DEPENDENCE : W^-SOLUTIONS
71
Claim 3 - There exists a solution u o/(6.6) with A = 1. The problem (6.6) is equivalent to the fixed point problem u = Txu, where Tx : ^([a.b]) -» C^K&D is defined by
(Txu)(t) = - f Ja
with G(t, s) the Green's function corresponding to u" - C{t)u = /(£), u(a) = u(b), u'(a) = u'(6). Observe that for some _R0 > 0 we have To(C1([a,6])) C £?(0,-R0)- Hence, by the properties of the degree, we prove easily that (6.6) with A = 1 has a solution u. Claim 4 ~ The solution u of (6.6) with A = 1 is such that a(t) < u(t) < j3{t) on [a, b]. The proof repeats the argument of Claim 4 in Theorem 6.8. Existence of solutions between the maximum of lower solutions and the minimum of upper solutions can be proved along the lines of Theorem 3.2. We state such a result for (5.1) with a Nagumo condition. A similar statement is valid for (4.4) or (4.5) under the assumptions of Theorems 6.8 or 6.9. Theorem 6.10. Let an € C([a,b]) (i = l , . . . , n ) and Bj € C([a,b]) (j = 1 , . . . ,m) be respectively W2'1 -lower solutions and W2'1 -upper solutions of (5.1). Define Ai C [a, 6] (resp. Bj c [a, b]) to be the set of points where «j fresp. 8j) is derivable. Assume a := max a; < 3 := min 8j. 3
l
l<j<m
Define E = {(t,u,v) € [a,b] x R2 | minjaj < u < maxj/3j} and let p, q e [l,oo] be such that - + - = 1. Assume f : E —> K satisfies VCaratheodory conditions and there exists N G Ll(a,b), N > 0 such that for all i and a.e. t 6 Ai (resp. for all j and a.e. t £ Bj) f(t,ai(t),afi(t))
> ~N(t)
(resp. / ( i , ^ ( t ) , ^ ( t ) ) < N(t)).
Assume moreover there exist tp € C(R+,lRo~) and ip € Lp(a,b) that satisfy (6.1) and assume that the function / satisfies one of the one-sided Nagumo conditions (a), (b), (c), or (d) in Theorem 6.1. Then the problem (5.1) has at least one solution u £ W2'p(a, b) such that for all t £ [a, b] a(t) < u(t) <
72
I. THE PERIODIC PROBLEM
Proof. Let R and JV be chosen as in the proof of Theorem 6.6. Define then f(t,u,v) = m&x{min{f(t,u,v), N(t)},
-N(t)},
gi(t,u,v) = f_(t,ai(t),v) - ai(t) - wu(t,ai(t) - u), if u < «*(£), = f(t,u,v) — u, iiu>ai(t), hj(t, U, V) = f_(t, Pj(t),v) f(t,u,v)-u, =
) + LJ2j(t, U - Pj(t)), if U > Pj(t),
where w u ^ ^ ^ X ^ W max
j the characteristic functions of the sets Ai and Bj, We consider now the modified problem u" -u = F(t,u,u'), u(o) = «(6), «'(a)=u / (6),
, , ^ j
where i,u,u) = min gdt,u,v), l
if u
= max hj(t,u,v),iif3(t)
l<j<m
Adapting the argument in Theorem 3.2 and 6.6, we prove (6.7) has a solution such that a(t) < u(t) < j3(t) and ||u'||oo < R- The proof D follows. The structure of the set of W2>1-solutions is similar to the structure of C2-solutions. In particular, the following result concerns existence of extremal solutions. Theorem 6.11. Let a and fi £ C([a,b\) be respectively W2'1-lower and upper solutions of (5.1) such that a < p, Define A C [a, b] (resp. B C [a, b}) to be the set of points where a (resp. 0) is derivable. Let E be defined from (4.7) and p, q G [l,oo] fee such that | + | = 1. Assume / : E —» R satisfies Lp-Caratheodory conditions and there exists N e L1(a, b), N > 0 such that for a.e. t G A (resp. for a.e. t G B) f(t,a(t),a'(t))
> -N(t)
(rvsp. f(t,/?(*),/?(*)) < N(t)).
Assume moreover there exist G Lp(a,b) that satisfy (6.1) and assume that the function f satisfies one of the one-sided Nagumo conditions (a), (b), (c), or (d) in Theorem 6.1.
6. DERIVATIVE DEPENDENCE : W^-SOLUTIONS
73
Then the problem (5.1) has a minimal solution umin 6 W2'1(a,b) and a maximal solution umax £ W2tl(a, b) in [a,0\, i.e. ot < umin < umax < fi and any other solution u of (5.1) with a < u < j3 satisfies
Exercise 6.2. Prove the above theorem. Exercise 6.3. As in Theorem 3.5, prove that if / is non-decreasing with respect to u we have a continuum of solutions. Exercise 6.4. As an application of Theorem 6.6, we can study the influence of a parameter on the number of solutions. Consider the problem u" + g(u) sgn(ti') \u'\r + /(*, u) - s(t), u(0) = u(27r), «'(0) = «'(27r)
,
{
'
where / is an Lp-Caratheodory function, g is continuous, and s G Lp(0, 2K). It is easy to observe that the Nagumo condition is satisfied if 0 < r < 2 — 1/p. Assume further lim /(£, u) = — oo, U—S-+OO
lim /(t,u) = +oo, U—» —OO
uniformly in t. (a) Prove that for all s € Z/°°(0,27r), the problem (6.8) has at least one solution. (b) Suppose moreover, there exist real numbers v < u such that f(t, v) < f(t, u). Prove then that for any s € L°°(0, 2TT) with f(t, v) < s(t) < f(t,u) problem (6.8) has at least two solutions. Hint : Use constant lower and upper solutions : ai < /3i = v < cxi — u < Combining the lower and upper solution method with other methods, we can prove the existence of a third solution of (6.8) for every s with f(t,v) < s(t) < f(t,u). We refer the interested reader to Exercise III-1.4. Observe that using one-sided Nagumo conditions, we can replace the restriction on r by a sign condition on g.
74
I. THE PERIODIC PROBLEM
Exercise 6.5, Consider the singular problem u" + g(u)u' + -= h{t), u u(a) = u(b), u'(a) = u'(b), where g is continuous and h £ L1 (a, b) is such that Ja h(t) dt > 0 and has an essential upper bound. Prove existence of a positive solution. Hint : We can choose a > 0 constant small enough and (3{t) = c + w(t), where c is a large enough constant and w is solution of u" + g(u + c)u' = h{t) - 7 - i - / h(t) dt, f u(a) = u(b), u'(a) = u'{b), / u{t) dt = 0. Ja The idea is to prove an a priori bound on w which is independent of c and then to pick c so that /3 is an upper solution and 0 > a. For more details see [149].
Chapter II
The Separated BVP 1 1.1
C2-Solutions The method of lower and upper solutions
Consider the problem u" = f(t,u,u'), aiu(a) — O2u'(o) = AQ, bxu{b) + b2u'{b) = B o ,
(1.1)
where Ao, Bo € E, oi, h € R, a2, b2 G M+, a\ + o| > 0 and b\ + b\ > 0. This problem is called the separated boundary value problem and contains as special cases the Dirichlet problem u" = f(t,u,u'), u{a) = Ao, u(b) = B o , the Neumann problem u" = f(t,u,u'), u'(a) = A,, u'{b) - Bo, and the Robin problem u" = f(t,u,u'), u'(a) = Ao, u(b) = B o . If we are looking for C2-solutions the Definitions 1-5.1 can be adapted to problem (1.1) as follows. 75
76
II. T H E SEPARATED BVP
Definitions 1.1. A function a G C([a,b}) is a C2-lower solution of (1.1) if (a) for any to G]a, b[, either D-a(to) < D+a(to), or there exist an open interval IQ C }a, b[ with to G Jo and a function ao 6 C 1 (/o) such that (i) a(to) = cto(to) and a(t) > ao(t) for all t G IQ; (ii)a'0'(t0) exists and ao'(to) > f(to,ao{to),a'o{to)); (b) aia(a) - a2D+a(a) < Ao, ha(b) + & 2 D-a(6) < Bo. A function /3 G C([a,b]) is a C2-upper solution o/(l.l) if (a) for any t0 G ]o, b[, either D~(3(t0) > D+f3(t0), or there exist an open interval IQ C }a, b[ with to € IQ and a function A) G Cl{IQ) such that (i) /3(t0) = /?o(to) and j3(t) < /30(t) for all t e h; (ii)/30'(t0) exists and/3'0'(t0) < f(tQ,/30(t0),/30(t0)); (b) ax/3(a) - a2D+/3{a) > Ao, bi0(b) + b2D~P(b) > BoRemark Let a be a lower solution and assume to is such that -D-a(to) >
D+a(t0).
As in Chapter I, we can observe that in that case a has a derivative at io and a'(to) = ao(*o)As in Chapter I, we can build lower and upper solutions from maxima of lower solutions and minima of upper solutions. The proofs of these propositions are left as exercises. Proposition 1.1. Let on G C([a, b]) (i — l,...,n) be C2-lower solutions of (1.1). Then the function a(t) = max aAt),
t G [a, b]
l
is a C2-lower solution
of (1.1).
P r o p o s i t i o n 1.2. Let f3j G C([a, b]) (j = 1, (1.1). Then the function /?(*)=
min 0j(t),
, rn) be C 2 -upper solutions
of
t£[a,b]
l<j<m
is a C2-upper solution of (1.1). The main existence results for solutions of (1.1) parallel the corresponding ones for the periodic problem.
1. C 2 -SOLUTIONS
77
T h e o r e m 1.3. Let Ao, Bo G R, ax, b\ G R, a 2 , 62 G K + with af+a^
> 0,
6? + b\ > 0. Assume a and /3 G C([a, b]) are C2-lower and upper solutions of the problem (1.1) SMC/I t/iai a < /3. Lei £ := {(£, it, u) G [a, b] x R 2 | a(t) < u < /3(«)}, (1.2) y : R + —» R 6e a positive continuous function satisfying
r i£=oo,
(i.3)
and assume f : E —> M is a continuous function which satisfies (1.4) V(t,u,v)€E, \f(t,u,v)\
n" = Xf(t,7(t,«),«') + v(l«'l) (« (o) = A[A0 - 017(0, u(a)) + a 2 u'(o) + 7(0, u(a))}, ftau'(6) + 7(6,
(1.5)
where 7 : [a, 6] x E —> R is defined by (), (), = u, if a(i)
(1.6)
Let p > 0 be such that, for all t € [a, 6], - P < a(t) < P(t) < p, f(t,a{t),O)+
Bo - bia(b) + a(b) > -p, So - fti/9(ft) + 0(b) < p.
Claim 1 - Every solution u of (1.5) with A G [0,1] is such that ||u||oo < /" Assume there exists to G ]o, 6[ such that mint u(t) = u(*o) ^ ~P- We have u'(to) = 0 and obtain the contradiction 0 < u"(t0) = A[/(t 0 , a(*o), 0) + ¥>(0) (u(t 0 ) - a(«o))] + (1 - A)v(0)«(t o ) < 0.
78
II. THE SEPARATED BVP
If the minimum is achieved for to = a, we have the contradiction —p > u(a) = A[AQ — aia(a) + a,2u'(a) + a(a)] > \[AQ — aia(a)+a(a)} > —p. In case the minimum is for to = b, we come to the same contradiction. At last, we prove in a similar way that u(t) < p. Claim 2 - There exists R > 0 such that every solution u of (1.5) with X G [0,1] satisfies ||«'||oo < R- The claim follows choosing R > 0 from Proposition 1-4.4, where a = —p, J3 = p and (p(v) = (1 + 2p)
ATi:C1([o,6])->C([a,6])x]R2,
i = 1,2,
where Lu = (u", u(a), u(b)), JV o (u)
Ni(u) = (f{t,7(t,u),it') ^o - ai7(a,"(«)) + a2«'(o) + 7(a. Bo - fc17(h, «(&)) - hu'ib) + 7(6, Observe that L has a compact inverse. Hence, we can define the completely continuous operator T A : C1([o,6])->C1([a,&]) by T\(w) = L~1(XN\ + (1 — A)iVo)(ii). From degree theory, we have that deg(T 0 ,n)=deg(ri,n), where 0 = {u £ ^([a^]) | HuHoo < /?, ||u'||oo < R}- Using the Odd Mapping Theorem A-1.5, we compute that and the problem (1.5) with A = 1 has a solution u. Claim 4 - The solution u of (1.5) with X = 1 is such that for all t G [a,b], a(t) < u(t) < P(t). If u(to) — a(to) = mmt(u(t) — a(t)) < 0 for to e]o,b[, then this result follows from the argument in the proof of Theorem 1-2.3. If to — a, we have u'(a) — D+a(a) > 0. This fact together with (1.5) and the definition of a C2-lower solution provides the contradiction AQ > a\a(a) — a,2D+a(a) > a\a{a) — a,2u'(a) = AQ — u(a) + a(a) > Ao. A similar argument holds if to = b and the rest of the claim follows as well.
1. C2-SOLUTIONS
79
Conclusion - It follows from the preceding claims that u is also a solution of (1.1). As in Theorem 1-5.4, we can improve condition (1.3) in case a and 0 are in Wl'°°(a,b). More generally, we can prove a result similar to Theorem 1-5.5 which uses one-sided bounds on f(t,a(t),a'(t)) and f(t,0(t),0'(t)). Theorem 1.4. Let Ao, Bo G R, a\, b% G R, a2, b2 G E + with a? + a\ > 0, b\ + b\> 0. Let a and 0 G C([a, b\) be C2-lower and upper solutions of the problem (1.1) such that a < 0. Define A c [a, b] (resp. B C [a, b]) to be the set of points where a (resp. 0) is derivable. Let E be defined by (1.2),
f
—p- > max0(t) - mina(t),
(1.7)
where r = m a x { g ( ^Jr: f > , ^ I ° ( 6 ) } , ond fe< / : £ -> M 6e a confinuow function which satisfies (1.4). Assume there exists N > 0 SMC/J that for all t £ A (resp. for all t £ B) f(t,a(t), a'(t)) > -N
(resp. f(t,0{t),0'(t))
< N).
Then the problem (1.1) ftas at feast one solution u G C2([a, b]) such that for all t G [a, b] a(t) < u(t) < 0{t). Proof. The proof of this theorem follows the proof of Theorem 1-5.5. Let R be large enough so that
I
R
sds
> max/3(i) — mina(t).
Increasing the value of N if necessary, we can assume N > max[0,H] v( u )Consider then the modified problem u" = f(t, u, u') + arctan(u — j(t, u)), u(a) - a2u'(a) = Ao - axj(a, u(a)) + 7(0, u(a)), u(b) + b2u'(b) - Bo - 617(6, «(6)) + 7(6, «(6)),
(1.8)
where /(i, u, v) :— max{min{/(t, 7(4, u), u), AT}, —AT} and 7(i, u) is defined in (1.6). Step 1 - Taking the inverse of the operator Lu = (u" ,u(a)—a%u'(a),u{b) + b2u'(b)), we transform (1.8) into a fixed point problem in C1. Next, using Schauder's Theorem we prove existence of a fixed point u.
80
II. THE SEPARATED BVP
Step 2 - The solution u is such that on [a,b], a(t) < u(t) < /3(i). The claim follows from the argument in the proof of Theorem 1.3 and 1-5.5. As a consequence, u satisfies
u" = max{min{/(*, u, u'),N}, —N}, aiu(a) — O2u'(a) = Ao, blU{b) + b2u'(b) = Bo.
(1.9)
Step 3 - The solution u is such that ||w'||oo < R(t,u,v)£E, \max.{mm{f(t,u,v),N},-N}\
Observe that for all
<
From Proposition 1-4.4, every solution u £ [a,0\ of (1.9) satisfies
II"' Hence \f(t,u(t),u'(t))\
< N and the function u is a solution of (1.1).
In the proof of this theorem, we could have used the modified problem u" — u = f(t, u,u') — 7(4, u), u(a) — a2u'(a) = AQ — 017(0, u(a)) + j(a, u(a)), u(b) + b2ii'(b) — Bo — 617(6, u(b)) + 7(6, u(b)), equally for the periodic problem. In this last problem, the aim of the modification was two-fold. First, we needed the linear problem to be invertible which is why we replaced u" by u" — u. Second, we needed to confine solutions between a and 0 which was obtained by the modification of / outside a and /3 adding a term as u — j(t, u). For the separated boundary value problem, we do not need to modify the linear problem. However, we still need to modify / outside a and /? which has been done using the term arctan(u — j(t,u)); however, the alternative u — j(t,u) works as well. Example 1.1. Consider the Dirichlet boundary value problem ( ) u(0) = 0, u(l) = 1. It is easy to see that a{t) = 0 is a lower solution and that /3(f) = 2t is an upper solution. Notice that \f(t,u,v)\ and
r i
< ( u 2 + l ) l n ( i / + l ) if v2 > e - l x s as
(s2 + 1) ln(s2 + 1)
1 .
,,
j
o
vv
=
00.
1. C 2 -SOLUTIONS
81
Hence f(t,u,v) satisfies a Nagumo condition. Prom Theorem 1.4, this problem has a solution u such that for alH £ [0,1], 0 < u(t) < 2t. As in the periodic case, if the nonlinearity depends on u' the existence of a well ordered pair of lower and upper solutions alone, i.e. without a Nagumo condition does not guarantee the existence of a solution. Such a situation holds true modifying Example 1-4.1. A simpler time independent example is the following. Example 1.2. Consider the problem u" = g{u') + u + 1, u(0) = 0, u(T) = 0, where g is continuous, g(0) = 0, g(v) > 0 if v ^ 0 and \/g € L1(K). Observe that —1 and 0 are lower and upper solutions of this problem. Further every solution is a convex function such that u e [—1,0]. Hence, u" > g(u') and JQ -rjpy dt > T. As l/g € L 1 (E) we have a contradiction for T large enough. As for the periodic case, such examples prove we cannot replace the Nagumo condition assuming that solutions between a and /? are a priori bounded in C1([a,6]). In case the nonlinearity is independent of v!, as lower and upper solutions are continuous and / is continuous, we automatically have the existence of N such that
f(t,a(t))>-N,
f(t,P(t))
Hence, for the problem u" = /(«,«), aiu(a) — azu'ia) = AQ, hu(b) + b2u'(b) = Bo,
(1-10)
the following result is a consequence of Theorem 1.4. Theorem 1.5. Let Ao, Bo e R, ai, h e R, a2, b2 £ M+ -with a\-\-a\> 0, b\+b\ > 0. Assume a and 0 G C([a,b]) are C2-lower and upper solutions of the problem (1.10) such that a < (5. Define E = {(t,u) € [a,b] x R | a(t)
82
II. T H E SEPARATED B V P
As for the periodic problem, we have the existence of maximal and minimal solutions between lower and upper solutions. T h e o r e m 1.6. Let Ao, Bo G R, a i , 61 G R, 02, &2 € R + with of +a\ > 0, b\ + b\> 0. Assume a and (3 G C([a,b]) are C2-lower and upper solutions of the problem (1.1) such that a < (3. Define A C [a,b] (resp. B C [a,b]) to be the set of points where a (resp. p) is derivable. Define E from (1.2), let ip : R + —> R be a positive continuous function satisfying (1.7) (with r defined as in Theorem I.4) and suppose f : E —> R is a continuous function which satisfies (1.4). Assume further there exists N > 0 such that for all t G A (resp. for all t G B), f(t,a(t),a'(t))
> -N,
(resp. /(*,£(*),^(i)) < N).
Then the problem (1.1) has a minimal solution umin and a maximal solution umax G C2([a,b]) in [a,0\, i.e. < umin < umax < and any other solution u of (1.1) with a
Exercise 1.1. Prove this theorem adapting the proof of Theorem 1-2.4. 1.2
A First Application
The following proposition is an example of a possible application. Proposition 1.7. Let f € C([0,1] xR 2 ) be such that §£ G C([0,1] xR 2 ) and assume there exists
L
sds ip(s)
=
00
such that for any (t,u,v) G [0,1] x R 2 , \f(t,u,v)\ <
fu{t,u,v)>k. Then the problem u" = f(t,u,u>),
u'(0) = r, u'(l) = s has at least one solution.
. C2-SOLUTIONS
83
Proof. Let M = max(r,s) and m = min(r,s), define v(t) = m^- — M^ 2 ' and choose A > 0 large enough so that v(t) — A < 0 and u" — f(t,0,v') — fc(u(£) - A) > 0. It is easy to see that a(t) = v(t) — A is a lower solution. Similarly, taking B large enough, we can build an upper solution /3(i) = B + w(t) > a(t), where w(t) = M^- — r?v ~f' . The proposition follows then from Theorem 1.3. 1.3
A Model Example in Singular Perturbations
Another field of applications concerns singular perturbation problems. The problem eu" + uu' — u = 0, (1.11) it(0) = A, u(l) = B, where e > 0, A, B € 1R is a model example whose solutions exhibit a large variety of interesting behaviours as e goes to zero. These can be shown off through lower%and upper solutions. To this end we shall assume here that e > 0 is as small as needed. Observe also that f(t,u,v) = (u — uv)/e satisfies a Nagumo condition. Case 1 : 0 < B — 1 < A - There exists a solution o/(l.ll) with a boundary layer att = O. The functions a(t)=t + B-l
and (3(t) = t + B - 1 + (A - B + l)exp((l - B)-)
are lower and upper solutions of (1.11). Hence, we have a solution ue of (1.11) between a and (3.
Fig. 1 :
II. THE SEPARATED BVP
84
Further, as the solution ue lies between the lower and upper solutions, it is easy to see that the solution uo(t) = t + B - 1 of the reduced problem uv! -u — 0, u{\) = B approximates ue as e —> 0 uniformly on compact subsets of ]0,1] but not on [0,1]. In such a case we say that there is a boundary layer att = 0. The case B — I > A> 0 is similar. Case 2:0
1 - B. The function a{t) = 0, = t-X,
if t € [0, A[, ifie[A,l],
is a lower solution and if * € [0, X[,
/?(*) = Aexp(-i/Ve) +
where K = Aexp(—X/i/e) + ^/e, is an upper solution.
Fig. 2 : 0 < A, 0 < B < 1 To verify that 3 is an upper solution we check that 0{t) < /30(t) :=t-X
for all t € Jo = [0,1] and
+ Kexp(-(t
-
1. C2-SOLUTIONS
85
Notice that we use the full generality of Definitions 1.1 for t = A, since /3 ^ /3o in a neighbourhood of this point. It follows that we have a solution ut of (1-11) between a and j3 and we see that lim uJt) = 0,
ift€]O,A[,
= t-X,
ifte[\,l],
uniformly on compact subsets of ]0,1]. In this problem, two solutions of the reduced problem uv! — u = 0,
are needed to approximate ue, i.e. UQ = 0 if t E]0, A[ and uo = t — A if t G [A, 1], There is a boundary layer at t = 0 and it is clear that a nonuniformity in the limit of the derivative u'e takes place at the interior point t = A. In such a case, we speak of a transition or interior layer in the derivative at t = A. Case 3:0
equation are needed to approximate the solution, i.e. lim ue(t) =t + A, E—»0 +
on [0, -A[,
on[-A,l-B[, = 0, = t + B-l, on [1-5,1].
B
a(t)
Fig. 3 : 0 < B < A For e small, a lower solution reads a(t) = \{t + A - \/(* + -
+ 4e1/a),
if t < t0 = 1 - B iit>t0,
86
II. THE SEPARATED BVP
and an upper solution .1/2
= \{t - 1 + B + y/(t - 1 + B) 2 + 4CV2), if t > tiCase 4-'l
— B
layer att-O.
there exists a solution of (1.11) with a boundary
Define A = t a n h - 1 ( - ^ / ( B - 1)). The functions and
are lower and upper solutions. It follows there exists a solution ue of (1.11) between a and /? and lim ue(t) =t
if
0,1].
0+
The convergence is uniform on compact subsets of ]0,1] and we have a boundary layer at t = 0. u
Fig. 4 : 1-B
Case 5 : A < 0, 0 < B, A + 1 < B, \A + B\ < 1 - There exists a solution of (1.11) with a transition or interior layer att = 1 ~'^~ B . Let e > 0, '0
and t\ > to be such that
l-A-B §
1. C2-SOLUTIONS
87
It can be seen that t i - £ o - » O a s e - » 0 and that for e small enough, a(t) =t + A, = t-to
if£e[0,£ o [, if£e[£0,l],
B-A-l
is a lower solution. In the same way, we consider e > 0, £2 = satisfy i-__V, B—A—1 (4.
4. \
1-
^ ~ ^ — e and £3 < £2 that
o ta+B—1
t a n n — ^ — ( £ 2 — £3) = I j _ ^ _ 1 . We prove t h e n t h a t £3 — £ 2 — > 0 a s e — > 0 and t h a t i(£-£3), @(t) = £ - £2 + B ~ ^ - 1 t a n h 2-
if£e[0,£2[, if t e [t 3 ,1],
is an upper solution.
Fig. 5 : A < 0, 0 < B, A + 1 < J5, \A + B\ < 1 Hence, we have a solution ue of (1.11) between a and (3 and we see that lim^ ue(t) = 14- A, = £+ 5 - 1 ,
on
on on]i=4=fi,l].
Here, we have a transition layer at £ = 1 ~4~ j B
88
2
II. THE SEPARATED BVP
W 21 -Solutions
2.1 The Method of Lower and Upper Solutions In this section we consider nonlinearities / which are L1-Caratheodory. As in Sections 1-3 and 1-6, we adapt accordingly the definitions of lower and upper solutions. Definitions 2.1. A function a (zC([a, b}) is a W2il-lower solutiono/(1.1) if (a) for any t0 e]a,b[, either D-a(t0) < D+a(to), or there exists an open interval Jo C ]a, b[ such that to G Jo, a £ W2:1(I0) and, for a.e. te IQ, a"(t)>f(t,a(t),a'(t)); (b) oia(a) - a2D+a{a) < Ao, haib) + b2D-a(b) < Bo. A function 0 G C({a,b\) is a W2il-upper solution o/(l.l) if (a) for any t0 e]a,b[, either D~f3(t0) > JJ>+/3(io), or there exists an open interval IQ C]a,b[ such that to E Jo, /? G W2'l{Io) and, for a.e. te Io,
(b) aip(a) - a2D+p(a) > Ao, bx(3{b) + b2D-/3{b) > Bo. Theorem 2.1. Let Ao, Bo e R, a\, b\ € R, a2, b2 e K+ with a\ + a?, > 0 and b\ + b\ > 0. Assume a and /3 G C([a, b]) are W2'1-lower and upper solutions of problem (1.1) such that a < j3. Define A C [a, b] (resp. B C [a, b)) to be the set of points where a (resp. /3) is derivable. Let E be defined in (1.2), p G [1, oo] and f : E -> R be an W-Caratheodory function. Suppose there exists N G L:(a, b), N > 0 such that for a.e. t e A (resp. for a.e. t € B), f(t,a(t),a'(t))
> -N{t),
(resp. /(*,/?(*),0'W) < N(t)).
Assume moreover there exist ip G C(R + ,Rj) g l/9
ds > ||^||z,p(max/9(t)
an
d ^ ^ Lp(a,b) satisfying
-mina(t))1/q,
Jr
whelL r n a x i ^ l , ffl^fel} and a = fr; (b) for a.e. t G [a, b] and all (u,v) with (t,u,v) G E,
2. W 2il -SOLUTIONS
89
Then the problem (1.1) has at least one solution u £ W2'p(a, b) such that for all t £ [a, b] a{t) < u(t) < Proof. The proof proceeds in several steps. Step 1 - The modified problem. Let R be large enough so that
Increasing N if necessary, we can assume N(t) > \f(t,u,v)\ if t £ [a,b], a(t)
-N(t)},
\v\<5
wa(t, S) = XB(t) max \f(t,0(t), ^'(*) + «) - /(*, Ptt), 0'(t))\, \v\<8
where 7 is defined from (1.6), XA and XB are the characteristic functions of the sets A and B. It is clear that Wj are L1-Caratheodory functions, non-decreasing in 6, such that u>i(t, 0) = 0 and joji(t,
(2.1)
where w(t1u) = -w 2 (t,u-)9(0), = 0, = Wi(t, a(t) — u),
if if if u
Step 2 - Existence of a solution u of (2.1). We argue as in Theorem 1.4. Step 3 - The solution u of (2.1) satisfies on [a,b] a(t) < u(t) < Assume u — a has a negative minimum and let to = sup{t € [a, b] | u(t) — a(t) = min (u(s) - a(s))}. s6[a,b]
If £0 — 0,, we have u'{d) — D+a(a) > 0. This observation together with (2.1) and the definition of W2'1-lowex solution provides the contradiction Ao > a.ia(a) — a,2D+a(a) > aia(a) — a2u'(a) = Ao — u(a) + a(a) > Ao.
90
II. T H E SEPARATED BVP
A similar argument holds if to = b. If to G ]a, b[, there exist an open interval IQ and a ty G /o, with ii > to, such that a G W2tl(I0), t0 G J o , "'(*i) - a'(*i) > 0 and for a.e. t G Io a"(t)>f(t,a(t),a'(t)). Further u'(to) — ot'{to) = 0 and, for t near enough £o> |u'(i)-V(i)|
As Wi is non-decreasing and f(t, a(t),a'(t)) < f(t,a(t),a'(t)), the contradiction
we come to
rti
0 < u'(*i) - a'(h) = /
(u"(s) - a"(s))ds
t 1
[f{s,a(s),u'(s)) -f(s,a{s),a'(s))
-wi(s,a(s) - «(s))]d« < 0.
- T/ie solution u of (2.1) is SMC/I iftat ||u'||oo < -R- Observe that for any (t,u,v) G E, \f(t,u,v) — u(t,u)\ = \f(t,u,v)\ < tp(t)tp(\v\) so that the claim follows from Proposition 1-4.7. Conclusion. A consequence of Steps 3 and 4 is that along the solution u of (2.1), \f(t,u(t),u'(t))\ < N(t). Hence, f(t,u(t),u'(t)) = f(t,u(t),u'(t)) and u solves (1.1). Remark 2.1. Consider the Dirichlet problem u" = f(t,u,u'), u(a) — 0, u(b) = 0. As u G W2'1(a,b) C C 1 ([a,6]), the assumptions of Theorem 2.1 imply some one-sided bounds on a and j3,
for some C > 0. In the applications, we can use the following simplification of Theorem 2.1. Corollary 2.2. Let AQ, Bo G E, a\, 6i G R, o 2 , 62 G M+ wii/i of + o\ > 0 anrf b\ + b\> 0. Assume a and (3 G W liOO (a, 6) are W 2ll -fower and upper solutions of problem (1.1) such that a < j3.
2. W^-SOLUTIONS
91
Let E be defined in (1.2), p £ [l,oo] and f : E -> R be an D>Caratheodory function. Assume moreover there exist p £ C(R + ,RQ") and ip £ Lp(a,b) such that o
gl/q
(a) / —rr ds — +oo, where a = -2-? and p l Jo V>(s) ~ (b) for a.e. t£[a,b] and all (u,v) with (t,u,v)£E, \f(t,u,v)\
f Jr
where r = fr^> Q — ~?y an^> for a-e- * G [a>^] {t,u,v)£E, \f(t,u,v)\(t)
an
d a^ (u>v) such that
Then the problem (1.1) has at least one solution u £ W2'p(a,b) such that for all t £ [a, b] a < u(t) < /3. If / does not depend on u' as in problem (1.10), Theorem 2.1 reduces to the following one. Theorem 2.4. Let Ao, Bo £ R, a\, bt £ R, a2, b2 £ R + with a\ + a\> 0, b\ + b22> 0. Assume a and /3 £ C([a,b)) are W2'1-lower and upper solutions of the problem (1.10) such that a < (3. Let E = {(t,u) £ [a,b] x R | a(t)
92
II. T H E SEPARATED B V P
As for the periodic problem, we have solutions of (1.1) between maxima of lower solutions and minima of upper solutions. Theorem 2.5. Let Ao, Bo G R, ai, bi G R, o 2 , b2 G R+ with a\ + a% > 0, b\ + b\ > 0. Let cti G C([a,b]) (i = l , . . . , n ) and Bj G C([a,b]) (j = l , . . . , m ) fee respectively W2'1 -lower solutions and W2'1 -upper solutions of (1.1). Define Ai c [a, 6] (Vesp. I?,- C \a,b]) to be the set of points where a» (resp. 3j) is derivable. Assume a := max ati < 3 := min /3,-. \
~~
l<j<m
J
Define E = {(t,u,v) G [a,b] x R 2 | minjaj < u < maxj/3^} and let p, q € [l,oo] be such that - + - = 1. Assume / : £ - » ! satisfies 1PCaratheodory conditions and there exists N G L1(a,b), N > 0 such that for all i and a.e. t G Ai (resp. for all j and a.e. t £ Bj) /(t,<*(*),oj(t)) > -N(t)
(resp. /(t.&W.^W) < N(t)).
Assume moreover there exist tp G C ( R + , R j ) and ip G Lp(a,b) that satisfy conditions (a) and (b) in Theorem 2.1. Then the problem (1.1) has at least one solution u G W2'p(a, b) such that for all t G [a, 6] a(t) < u{t) < f3{t). Exercise 2.1. Prove the previous theorem using as a guideline the proof of Theorem 1-6.10. As previously we have the existence of maximal and minimal solutions between a and 3 in the following sense. Theorem 2.6. Let Ao, Bo G R, a\, h G E, a2, h G R+ with a\ + a\ > 0 and b\ + b\ > 0. Assume a and 3 G C([a, b]) are W2'1 -lower and upper solutions of problem (1.1) such that a < 8. Define A C [a,b] (resp. B C [a,b]) to be the set of points where a (resp. 3) is derivable. Let E be defined in (1.2), p G [1, oo] and f : E ->Rbean Lp'-Caratheo1 dory function. Suppose there exists N G L (a,b), N > 0, such that for a.e. t G A (resp. for a.e. t G B) f(t,a(t),a'(t))
> -N(t),
(resp. f(t,P(t),3'(t)) +
<
Assume moreover there exist ip G C(R ,Ro") and tp G Lp(a,b) satisfying conditions (a) and (b) in Theorem 2.1. Then the problem (1.1) has a minimal solution umin G W2'p(a,b) and a maximal solution umax G W2'p(a,b) in [a,0\, i.e. m i n
< Umax <
2. W^-SOLUTIONS
93
and any other solution u such that a
Exercise 2.2. Prove Theorem 2.6 adapting the argument used to prove Theorem 1-2.4. Exercise 2.3. Suppose the assumptions of Theorem 2.6 and that there exist (a) an upper solution $ such that for all t £ [a,b], a(t) < /?(£) < fi(t) and (b) a continuum of lower solutions S in C([a, b]) such that a £ S and for some a £ S, a /3. Prove there exist then two solutions of (1.1). Hint : See [22]. 2.2
A Multiplicity Result for the Neumann Problem
Consider the problem u" + f(t,u)=0, u'{a) = 0, u'{b) = 0,
.
[
, '
where f(t,u) is L1-Caratheodory. Assume that for some sequences (an)n and (bn)n of positive numbers such that ai < b\ < ... < an < bn < ... we have f{t,an)>0 and f(t,bn)<0. It is clear then that an and bn are W2'l-lowei and upper solutions of (2.2). Existence of infinitely many solutions follows. The problem is more difficult if we add some forcing h(t) which might be unbounded. Consider for example the problem
f(t,u) = h(t), u'(0) = 0, u'{l) = 0.
(2-3)
To deal with this problem, let us recall that the operator L : DomL C C([0,1]) -> X C L^O, l),u H+ U",
(2.4)
where X = { u 6 L^O, 1) | /„ u(t) dt = 0} and DomL = {u € W2'1^, 1) Jo u(t) dt = 0 and u'(0) = u'(l) = 0}, has a continuous inverse.
94
II. T H E SEPARATED BVP We can obtain lower solutions from the following result.
Proposition 2.7. Let c be the norm of the inverse of L defined in (2.4), / be an L1 -Caratheodory function and let h £ Ll(0,1) satisfy Jo h(t) dt = 0. Assume that there exists a > 0 such that for a.e. t G [0,1] and allu € [a-c\\h\\Li,a +c\\h\\Li], f(t,u)>0. 21 Then there exists a W ' -lower solution a of (2.3) with ||/i|| L i] on [0,1]. Proof. The function a = a + L~l(h) is a lower solution which is in [a — U c\\h\\Lx,a + c\\h\\Li]. Upper solutions are obtained along the same lines. Proposition 2.8. Let c be the norm of the inverse of L defined in (2.4), / be an L1-Caratheodory function and let h € i 1 (0,1) satisfy Jo h(t) dt = 0. Assume that there exists b > 0 such that for a.e. i € [0,1] and allu e [b - c\\h\\Li,b + c\\h\\Li), f(t,u) <0. 2>1 Then there exists a W -upper solution (3 of (2.3) with 0{t)e[b-c\\h\\Li,b + c\\hUi] on [0,1]. Exercise 2.4. Prove the existence of an infinite number of solutions for the Neumann problem u" + sin sfu — h(t), u'(0) = 0, u'{l) = 0,
r1
in case he L1 (0,1) and | /
h(t)dt\
Jo In Propositions 2.7 and 2.8, lower and upper solutions are obtained from a sign condition on f(t,u), which holds on intervals long enough. A similar result can be obtained if the intervals are small but \f(t,u)\ is large enough. Proposition 2.9. Let / : [0,1] x R —» R be an L2-Caratheodory function and h G L 2 (0,1). Assume there exist sequences (an)n, [bn)n o,nd (c n ) n of numbers such that (a) lim an = lim bn - Cn = +oo; n—>oc
n—>oo
(b) for some e > 0 and any n, anCn > e; (c) for any u£[bncn, bn + cn], f(t, u) > an. Then, for any R > 0, there exists a function a(t) > R, which is a W2'1-lower solution for (2.3).
2. W 2>1 -SOLUTIONS
95
Proof. Let R > 0 be fixed. Consider the problem w" -dw = h(t), u>'(0) = 0, tu'(l) = 0, where d > 0. Any solution w of this problem is such that / w'2dt + d w2dt = hw dt. Jo Jo Jo Hence, we have the a priori estimates 1
1
~~ d
~ \fd
Also, for some to,
Hence, we obtain
w2{t) = w2(t0) +2 I w(s)w'(s)ds < " J^ 2 (1 + 2d*). Jt0
"
Let us choose do such that IIMI2
,
n so that
and next d > d0 so that a n = ||/i||£,2(l + 2d3)5. For this choice of d and n, and all £ G [0,1], we have On > d|w(*)l and
Now, it is easy to see that a(t) = bn + w(t) > it is a lower solution since a" + f(t, a) - h(t) >dw + h(t) +anA similar result holds for upper solutions.
h(t) > 0.
96
II. THE SEPARATED BVP
Proposition 2.10. Let f : [0,1] x R —* R be an L2-Caratheodory function and h € L2(0,1). Assume there exist sequences (an)n, (bn)n and {cn)n of positive numbers such that (a) lim an = lim bn — Cn = +co; n—>oo
n—>oo
(b) for some e > 0 and any n, anCn > e; (c) for any u £ [bn - Cn, bn + d ] , f{t, u) < -an. Then, for any R > 0, there exists a function P(t) > R, which is a W2'1 -upper solution for (2.3). Exercise 2.5. Prove that the problem u" + it sin u2 = h(t), «'(0) = 0, «'(1) = 0, where h S L 2 (0,1), has an infinite number of solutions. 2.3
A Multiplicity Result for the Dirichlet Problem
For the Neumann problem, existence of an infinite number of solutions was deduced from sequences of lower and upper solutions (ctn)n and (/3n)n which are well-ordered " i < Pi <
< an < /3n < ...
Such a geometry might seem difficult to find for a Dirichlet problem since we cannot enclose the lower and upper solutions in disjoint intervals. It occurs however if the nonlinearity / oscillates enough around the first eigenvalue. To build such an example, consider the Dirichlet problem
u + f(t,u)=0, = 0, u{ir) = 0,
l
0J
where f(t, u) is an L1-Caratheodory function defined as follows. First, let h e ] 0 , 1 [be such that
define and let /i, v satisfy
3. ONE-SIDED NAGUMO CONDITION
97
Next, we choose positive sequences (An)n and (Bn)n such that kAn < (1 - ^)An < kBn < (1 + £)Bn < kAn+1
and An+1 >
f Bn.
At last, we choose / so that f(t, u) > fiu,
on [0, TT] x y [kAn, (1 - £)An],
/(t, u) < -MU,
on [0, IT] x ( J [fcBn, (1 + *J)-Bn],
oo
n=l
\f{t,u)\ < vu,
on [O,TT] x R+.
We shall then consider nested lower and upper solutions
and Pn(t) = Bn{sint + t-^-) G [an,an+1]. To verify that an is a lower solution, notice that in case an(t) < kAn, i.e. t < ti or t > 7r — t\, we have < + a n + / ( t , a n ) > An[\ - i ^ ^ - ku] > 0 and if an(t) > kAn, we compute < + a« + /(t,On) > A n [i - £ +fe/i]> 0. Similarly, we can check that j3n is an upper solution. As a consequence the problem (2.5) has an infinite number of ordered positive solutions.
3
One-Sided Nagumo Condition
One-sided Nagumo conditions were introduced in Sections 1-4.3 and 1-4.4. These ideas can be applied for the separated boundary value problem (1.1). Theorem 3.1. Let Ao, Bo e M, Oi, &i € R, a2, b2 £ M+ with aj+a%> 0, b\ + b\ > 0. Suppose a and /? € C([a,b]) are W2:1-lower and upper solutions of the problem (1.1) such that a < (3. Define A C [a,b] (resp. B C [a, b]) to be the set of points where a (resp. 0) is derivable.
98
II. T H E SEPARATED BVP
Let E be defined in (1.2) and, for some p G [l,oo], / : E —* R. be an IP-Caratheodory function. Assume there exists N G L1(a,b), N > 0 suc/i £/iai /or a.e. i G A (resp. /or a.e. t £ B), f(t, a(t), a'(t)) > -N(t),
(resp. f(t,0(t),0'(t))
< N(t)).
If moreover there exist r > 0,
(resp. u'(a) > —r and u'(b) < r);
-rj- ds > ||V»||i,p(max)8(t) - xmna{t))l>q, where q =
^ ;
(c) for a.e. t G [a,b] and all (u,v) G M2 such that (t,u,v) G E, we have f(t, u, v) < V(*MM)
(resp. f(t, u, v) > -1/(t)ip{\v\)).
Then the problem (1.1) has at least one solution u G W2'p(a, b) such that for all t G [a, 6] a(t) < u{t) < Proof. We proceed as in the proof of Theorem 2.1 except for Step 4 where we use Proposition 1-4.8 rather than Proposition 1-4.7. T h e o r e m 3.2. Let Ao, Bo G R, ai, b\ G E, a2, b2 G R + with a\+a%> 0, b\ + b\ > 0. Suppose a and f3 G C([a, b]) are W2'1 -lower and upper solutions of the problem (1.1) such that a < 0. Define A C [a,b] (resp. B C [a, b}) to be the set of points where a (resp. /3) is derivable. Let E be defined in (1.2) and for some p G [l,oo], / : E —> M be an Lp-Caratheodory function. Assume there exists N G L ^ a , 6), N > 0 such that for a.e. t G A (resp. for a.e. t G B), f(t,a(t),a'(t))
> -N(t),
(resp. f(t,0{t),0'(t))
< N(t)).
If moreover there exist r > 0, ip G C(R + , Rj}") and ip G Lp(a, b) such that (a) for all u solution of (1.1) with a
i
s
P{ )
(resp. \u'(b)\ < r);
ds > HV'lU^max^t) - mina(i)) 1 / 9 , where q = - ^ - ; t
t
3. ONE-SIDED NAGUMO CONDITION
99
(c) for a.e. t G [a,b] and all (u,v) G R2 such that (t,u,v) G E, we have sgn(v)f(t,u, v) < tp(t)ip(\v\) (resp. sgn(v)f(t,u,v) > -ip(t)tp(\v\)). Then the problem (1.1) has at least one solution u G W2'p(a, b) such that for all t G [a, b] a(t) < u(t) < Proof. We proceed as in the proof of Theorem 2.1 except for Step 4 where we use Proposition 1-4.9 rather than Proposition 1-4.7. The conditions |it'(a)| < r or |u'(&)| < r are automatically satisfied in case of the Neumann problem or of the Robin problem. Example 3.1. Consider the boundary value problem u" = -u'2n+1 + k sin27rt, u'(0) - 0, «(1) = 0, with n > 0 and k G R. It is easy to see that Theorem 3.2 applies with lower and upper solutions
Assumption (c) is satisfied since f(t,u,v) = — v2n+1+k sin27rf < k sin27ri, for v > 0, and f(t,u,v) > k sin27rf for v < 0. Notice that Theorem 2.1 does not apply if n is large enough. For other problems, the estimate r can sometimes be deduced from the upper and lower solutions if they coincide at t = a or t = b. Example 3.2. Consider the boundary value problem u" = — u'2n sinvrt + k sin27rf, u(0) = 0, u(l) = 0, with n > 0 and k G [—1,1]. Theorem 3.1 applies now with
Assumption (a) is satisfied since the conditions a < u < (3 and a(0) = 0(0) = 0 imply a'(Q) = - 1 < u'(0) < /3'(0) = 1.
100
II. THE SEPARATED BVP
We mentioned in Section 1-4.3 that the one-sided Nagumo conditions impose that along curves u' — /x(u) and u' = v(u) > /i(u) the vector field is not tangent from above along u' = fi(u) and not from below along u' = v{u). The following proposition describes such a situation for the Dirichlet problem. Proposition 3.3. Let R > 0, let a, 0 G C([a, b}) be such that a < (3 and define E from (1.2). Then for every Lx -Caratheodory function f : E —> R which satisfies uf(t,u,v)>0
if(t,u,v)eEand\v\>R,
(3.1)
and every solution u of u" = f(t,u,u'), u(o) = 0, u(b) = 0, such that a < u < P, we have
Proof. Let u be a solution of (3.2) such that a < u < 0. Claim 1 - The function u is such that (u(t),u'(t)) £ Fi := {(u, v) | u > 0, v > R} if t e [a, b[. Assume on the contrary that for some t\ £ [a,b[, (u(ti),u'(ti)) G F\. Notice first that we can find a £2 > ^1 such that ^(£2) > 0 and u'fa) > R. As u(b) — 0, we can find a t3 £]t2,b] such that (u(t),u'(t)) £ F\ for any t G [^2i^3[ and v!(tz) = R. This result gives the contradiction R = u'(t3) = u'(t a ) + Jt*3 /(*,u(s),u'(s))ds > R. Claim 2 - The function u is such that {u(t),u'(t)) £ F2 := {(u, v) I u < 0, v > R} if t G [a, 6]. This claim is proved reversing the time and using the arguments of Claim 1. Conclusion - For t G [a, b], we deduce from the above claims and the continuity of u' that (u(t),u'(t)) <£ FXV) F2, i.e. u'{t) < R. We prove in a similar way that u'(t) > —R. The assumptions of this proposition impose some conditions on the vector field (u(t),u'(t)) associated with (3.2); see figure 6.
3. ONE-SIDED NAGUMO CONDITION
101
\R
0
a
-R
\ Fig. 6 : Vector field in Proposition 3.3 The above proposition can be improved in two ways. We can replace the sign conditions (3.1) by Nagumo conditions (see the remark after Proposition 1-4.6). This procedure amounts to replacing the segments v! = R and u' = — R by curves deduced from these Nagumo conditions. Another improvement concerns the inversion of the vector field on the segments u' = R and u' = —R which can take place at some point u = /Ui(i) and u = ^%{t). Such a generalization is worked out in the next proposition. Proposition 3.4. Let R > 0, let a, f3 G C([a,b]) be such that a < f3 and define E from (1.2). Assume there exist n\ and ^2 € W1]1(o, b) such that -R < fi[, R > / 4 , /ii(a) > 0 > fj,2(a) and px(6) < 0 < /i2(&).
Then for every L1 -Caratheodory function f : E —> M which satisfies (u-m(t))f(t,u,v)>Q (u-n2(t))fit,u,v)>0
ifv<-R, ifv>R,
on E and every solution u of (3.2) such that a < u < /3, we have IKIloo < -RFunctions diagonals.
and ^2 that satisfies such assumptions are often called
Exercise 3.1. Prove the above result. Hint : Let u be a solution of (3.2) such that a /x2(i), v > R} for some t € [a,b[, the condition u(b) — 0 cannot be satisfied; (ii) if (u(t),u'(t)) € F2{t) := {(u,v) \ u < ^(t), v > R} for some* e [a,b], the condition u(a) = 0 cannot be satisfied.
102
II. THE SEPARATED BVP
Exercise 3.2. State and prove the counterpart of Theorems 3.1 and 3.2 using Propositions 3.3 or 3.4. 4 4.1
Dirichlet Problem Derivative Independent Problems
Dirichlet boundary value problems u" = f(t,u), u(o) = 0, u(b) = 0
.
k
,
;
can be studied for more general nonlinearities than -L1-Caratheodory functions. This remark follows from the analysis of the linear problem u" = h(t), u(a) = 0, u(b) = 0.
{
,
}
Its solution makes sense as a function in Co([a, b]) n Wjo'c (o, b) provided
(cfr. Proposition 4.1). For example, the problem t(l - t)u" = 1, u(0) = 0, u(l) = 0, has the solution u(t) = tint + (1 - t)ln(l - t) which is in Co([0,1]) f~l 14^^(0,1) but not in C1([0,1]). Notice that in some cases, problems with such singularities turn out to have solutions in C°°([a, b\). This is the case in the example u(0) = 0, u(l) = 0, whose solution is u(t) = t(l — t). The following proposition makes precise the main result we need concerning the linear problem (4.2). Proposition 4.1. // heA:={h\(s-a)(b-
s)h(s) G L\a, b)},
the problem (4.2) has one and only one solution u e W2'A(a,b) := {u e Wl'l{a,b) \ u" G A) C C([a,b]) nC'Qa.lD
4. DIRICHLET PROBLEM
103
such that u(t) = /
G(t,s)h{s)ds,
Ja where G(t, s) is the Green's function corresponding to (4.2). Further, we have ,6
where \\h\\ji — / (s — a)(b — s)\h(s)\ds. Ja Proof. Define ()
-(t-a)[ feh{a)ds, forte]o,6[, Jt
= 0,
for t € {a, b}.
Claim 1 - t i e Wf^{a,b) n W^^a.ft) and w" = h. Let c G]a,6[ be fixed. As h e A, we have ( — a)h € /^(a,!:) and (6 — -)h € L 1 (c,6). It follows that
[ ^h(s)dseW1'1(a,c)
and
Ja
/
feh{s)ds
eW1'1^).
Jt
Hence u £ W|o'c (a, 6) and we compute
Jt
Ja
Using Hake's Theorem and Lebesgue's Dominated Convergence Theorem we obtain u' £ Ll(a,b). At last we deduce from a similar argument that v! S W^(a,b) and u"(t) = h(t). Claim 2 - The function u is continuous and u(a) = u(b) = 0. Prom Claim rt
1, u G CQa, b[). Next, as / f5f ft(s) ds € W 1 ' 1 ^,*:), we have
On the other hand, we can write rb
< f X[t,b]^){t~tba-S) Ja
\Hs)\ds,
104
II. THE SEPARATED BVP
where X[t,b] is the characteristic function of the interval [t,b\. Further, as X[t,b](sy
b-a
\h(S)\ ^ ' 6-o
\h(S)\
€ L
( f l ' 6 )'
the Lebesgue's Dominated Convergence Theorem implies rb
lim(t-a) / |5f h(s) ds = 0. a
*~*
Jt
it lim u(t) = 0. Similarly we obtain ob lim u(t) — 0. This result proves that t—^a
Oaim 3 - \\u\\oo < -^z^Wh^A- We deduce from (4.3) that
~ Ja and the claim follows. Remark If ft € L}Oc(a> &)> a n y solution u of (4.2) reads
u(t)=u(to)+
h(s)(t-s)da,
where to is chosen such that u'(t0) = 0. If further h is one-signed we can deduce from Levi's Theorem that the integrals o
I
rb
(s — a)h(s)ds
and
Ja
/ (b — s)h(s) Jto
exist so that (s — a)(b — s)h(s) £ Ll{a, b). Notice that this result does not hold anymore if h changes sign. Consider for example the problem u" = - £ s i n i ,
u(0) = 0, u ( i ) = 0,
which has a continuous solution u(t) = i s i n j although s ( - — s)h(s) $ The Proposition 4.1 suggests how to generalize the Caratheodory conditions in case of the Dirichlet problem. Let us write the boundary value problem (4.1) as an operator equation
Lu = Nu.
4. DIRICHLET PROBLEM
105
To inverse the linear operator L, we deduce from Proposition 4.1 that the function Nu should be in A A natural assumption to ensure this property is captured with the following definition. A function / : J ) c [ a , 6 ] x l - * R is said to be an A-Caratheodory function or to satisfy A-Caratheodory conditions if it is a Caratheodory function such that for any r > 0 there exists h € A and for all (t, u) £ D with |u| < r \f(t,u)\
(4.4)
Notice that in the introductory examples, the functions 1 fit.u) = —. J v
' '
u r
and
—1
t(-\ — t)
;
r
t(-\ _ t)
are A-Caratheodory functions. Observe also that L1(a, b) C A so that this definition generalizes the classical L1-Caratheodory conditions on / . For Dirichlet problems, the notion of M^2'v-lower and upper solutions particularizes as follows. Definitions 4.1. A function a £ C([a, b]) is said to be a W2il-lower solution of (4.1) if (a) for any to £ ]a, b\, either D-a(to) < D+a(to), or there exists an open interval Io c]a, b[ such that t0 £ Io, a £ W2'l{I0), and for a.e. t £ Io, a"(t)>f(t,a{t)); (b) a{a) < 0, a(b) < 0. A function j3 € C([a, b]) is a W2>1-upper solution o/(4.1) if (a) for any t0 €}a,b[, either £>~/3(t0) > D+/3(t0), or there exists an open interval IQ C]a, b[ such that to £ IQ, /3 £ W / 2 ' 1 (/Q), and for a.e. t € Jo,
(b) 0(a) > 0, P{b) > 0. We can now state the main result for the Dirichlet problem (4.1). Theorem 4.2. Assume that a and j3 are W2'1 -lower and upper solutions of (4.1) such that a < f3. Let E - {(t,u) £ [a,b] x R | a{t) < u < p(t)} and f : E —> R be an A-Caratheodory function. Then the problem (4.1) has at least one solution u £ W2'~A(a,b) such that for all t £ [a, b] a(t) < u(t) < f3(t).
106
II. THE SEPARATED BVP
Proof. We consider the modified problem u{a) = 0, u(b) = 0, where 7 is defined by (1.6). Claim 1 - The problem (4.5) has at least one solution u € W2>A(a, b). Define the operator T : C([a, b]) —» C([a, b]) given by
where G(t,s) is the Green's function associated with (4.2). We can prove that T is completely continuous and bounded. Hence by Schauder's Fixed Point Theorem, T has a fixed point u which is a solution of (4.5) in W2'A(a,b). Claim 2 - Any solution u of (4.5) satisfies a
fb
|| / G{t,s)h(s)ds\\00 <+oo and / |§f(i,s)| h(s)ds G Ll{a,b). (4.6) Ja
Ja
We can see that in the periodic case, the set of functions h : }a, b[—* M+ measurable that satisfy (4.6) is in fact L1(a,6). The same holds true for the separated boundary value problem (1.10) if 02^2 ¥" 0- I n t n a t sense, Theorem 4.2 has the same generality as Theorem 1-3.1 and Theorem 2.4. Example 4.1. Consider the boundary value problem u" + \u\V2 - } = 0, u(0) = 0, u(n) = 0. It is easy to see that j3(t) = 0 is an upper solution and a(t) = t In £ — t is a lower solution. Hence we have a solution u such that for all t £ [0, n] tln--t<
u(t) < 0.
Notice that in this example the function f(t, u) is not I^-Caratheodory.
4. DIRICHLET PROBLEM
107
If we want more regularity, we need more restrictive conditions on / . For example, we have the following theorem. Theorem 4.3. Assume that a and f3 are W2'1 -lower and upper solutions of (4.1) such that a < (3. Let E = {(t,u) G [a,b] x R | a(t) < u < (3{t)} and f : E —» R satisfy Caratheodory conditions. Assume there exists a measurable function h such that Ja (s-a)h(s) ds < oo and for a.e. t £ [a, b] and allucR with (t,u) G E, \f(t,u)\
a{t) < u(t) < /3(t). Proof. Existence of a solution u G W2'~A(a,b) follows from Theorem 4.2. Further
u(t)= f
G(t,s)f{s,u(s))ds,
Ja
where G(t, s) is the Green's function corresponding to (4,2). It is now standard to see that u G C1(]o, b]). In the same way, we can consider the problem / ( )
u(a) = 0, blU(b) + b2u'(b) = 0.
(4.7)
Theorem 4.4. Let b\ G R, b 0 and assume that a and fi are W2'1 -lower and upper solutions of (4.7) such that a < /?. Let E = {(t,u) e [a,b] x R \ a{i)
a(t) < u{t) < Exercise 4.1. Prove the above theorem.
108
II. THE SEPARATED BVP
4.2
Derivative Dependent Problems
Consider now the Dirichlet problem with derivative dependence u" = f(t,u,u'), u(a) = 0, u(6) = 0.
,._. '
{
A natural idea would be to impose a Nagumo condition \f{t,u,v)\
t
$~ 1 (2| / ,
* ,
i
ds
*
= / where io —TrThen there exists h € L1(o, 6) nC(]a, b[) such that, for every a < a% < < ^ < h < b, every Ljoc-Caratheodory function f : E — R which satisfies for a.e. t e [01,61] and all (u,v) G R2 with (t,u,v) G E
4. DIRICHLET PROBLEM
109
and every solution u of u" = f(t,u,u')
(4.9)
defined on [a\,bi] and such that a < u < 0, we have WeKA],
\u'(t)\
Remark 4.1. Let us comment assumption (b). (i) If ip £ Ll{a, b), condition (b) is implied by condition (a); (ii) In case
If $~1(2(6 - a)rp(-)) € A, we have
I*
(
and it follows that
In the same way, we prove
it Proo/. 5tep i - Existence of a function h £ C(]a,b[) that satisfies the assertions of the proposition. Let c, d be such that ^J* < c < £ < d < £ j and M > g^(max t /3 — mint a). Define hi to be the solution of h[ = ip{t)ip{hi), t e [c,6[, fri(c) = M, the solution of ) , t€]a,d],
h2{d) = M,
110
II. THE SEPARATED BVP
and h(t) = h2(t), = max{hi(£),/i2(t)}, = hi(t),
on }a,c[, on [c, d[, on [d,b[.
Let Oi, 61 be such that a < a\ < ^ip < ^ ^
f* u"(s) j
f*
7*0 V ( u ( s ))
Ao
i.e. u'(t) < /ii(i) on / and hence on [r, 61]. In the same way, we prove u'(t) < h,2{t) on [oi,r]. It follows that u'(t) < h(t) on [ai,&i]. At last, we obtain from the same argument that u'(t) > -h(t) on [ai,bi]. Step 2 - ft G Ll{a,b). We compute
JM
V{r)
Jc
Hence we write - / ip(s)ds) - / ip(s)ds+ / ip(s)ds) Jc
J£
and as $~ x is convex
from which we deduce h\ G Ll{c, b). In the same way, we have / Jt and deduce /12 G L1(a,rf).
4. DIRICHLET PROBLEM
111
Remark The condition (p non-decreasing is not essential. If it is not satisfied, we replace condition (b) by rt
/ ij){a)da) £ L\c,b) Jc
and $~ 1 ($(M) + / il)(s)ds) £ Lx(a, d), Jt where M, c and d are defined in the proof of Proposition 4.5. Theorem 4.6. Let a, f3 £ C([a,b]) be W2'1-lower and upper solutions of the problem (4.8) such that a < f3. Define A C [a,b] (resp. B C [a,b]) to be the set of points where a (resp. (3) is derivable. Let E be defined by (1.2) and f : E —> R be an Ljoc-Caratheodory function. Suppose there exists N £ Ljoc(a,b), N > 0 such that for a.e. t £ A (resp. for a.e. t £ B), f(t,a{t),a'{t))
> -N{t),
(resp. / (
Assume moreover there exist tp £ Ljoc(a,b) and a non-decreasing function ip £ C(M+,Rj) such that (a) I
Jo
= oo;
(b) for some £ £ }a, b[,
/ iP(s)ds\)£L1(a,b), where
« (u) =
/
ds —r-r
Jo
Then the problem (4.8) has at least one solution u£C([a,b})nW^c(a,b) such that for all t £ [a, b], a(t) < u{t) <
112
II. THE SEPARATED BVP
Proof. Step 1 - The modified problem. Let (o n ) n , (&„)„ C]o, b[, (An)n, (Bn)n C K be such that lim an — a, lim bn = b, lim An = 0, lim Bn — 0, n—>oo
n—>oo
n—>oo
n—*oo
a(a n ) < A n < /?(an), a(6 n ) < B n < /3(6n). Consider the modified problem
We can assume that for any n, a n < ^ ^ < ^ < bn. Hence by Theorem 2.1 and Proposition 4.5 problem (4.10) has, for every n, a solution un satisfying on [an, bn] a(t) < un(t) < p(t),
\u'n(t)\ < h(t),
with h given by Proposition 4.5. Step 2 - Existence of a solution u of (4.8). Using Arzela-Ascoli Theorem, we can find a subsequence (u^)n C (un)n that converges in C1([oi,6i]). Proceeding by induction, for any k £ N, we build {vfyn which is a subsequence of (w^~1)n that converges in Cl({a,k, 6fe]). It follows that the diagonal sequence (u™)n converges pointwise to some function u and that for any compact K C ]a, b[, the convergence takes place in CX{K). Hence, u satisfies on ]a,b[ u" = f(t,u,u') and a{t)
P{ S
M is defined in the proof of Proposition 4.5. Hence, by the remark following Proposition 4.5, this result generalizes Theorem 2.1 for the Dirichlet case as well as Theorem 4.2.
4. DIRICHLET PROBLEM
113
As a first illustration consider the following example where Theorem 2.1 does not apply. Example 4.2. Consider the boundary value problem a
u" =
+u + t,
u(0) = 0, u(l) = 0, where 0 < o < 1 and 1 < n < 2 — a. Existence of a solution follows from Theorem 4.6 with a(t) = - 1 , 0(t) = 0, f = \, ip(t) = i + 1, rt
(f(y) = max{l,y°} and explicit computation of $ - 1 (2| /
ip(s)ds\).
il/2
In Theorem 4.6 we do not use the full power of condition (c) so that we can generalize it in the following way. Theorem 4.7. Let the assumptions of Theorem 4-6 be satisfied with (c) replaced by (c') there exist a
> —ip(t)ip(\v\) < 4>(t)
ift
G]a,d[, ifte]c,b[.
Then the problem (4.8) has at least one solution
ueC([a,b])nW?£(a,b) such that for all t S [a, b], a(t) < u(t) < P{t). Exercise 4.2. Prove Theorem 4.7. Remark In Theorem 4.6 and 4.7, we can assume a control on only for |w| > R.
f(t,u,v)
Example 4.3. Consider the following modification of Example 4.1 u" + \u\1'2 - jtu11/* - \ = 0, u(0) = 0, U(TT) = 0, where n > 0. We can apply Theorem 4.7 to prove the existence of a solution choosing a(t) = t In £ - t, f3{t) = 0, £ = TT/2, C = 7r/3, d = 2TT/3, ip(v) = max{l,u1//3} and = !+TT1/2,
= 7+7r 1 / 2 + A-,
on[0,ir/3], on]7r/3,7r].
114
II. THE SEPARATED BVP
Hence, there is a solution u such that for all t G [0, n] t In | - 1 < u{t) < 0.
5
Other Boundary Value Problems
In the Chapters VI to X we will present applications of the method of lower and upper solutions to a variety of problems in the theory of differential equations. The ideas of these two first chapters can also be extended to solve other problems. This section presents a selection of such extensions. The first one concerns radial solutions of partial differential equations. Here, we have to rewrite the method so as to include systems with singularities. In the second one, we investigate differential equations with deviating arguments. These two first applications concern second order problems. Therefore, we can use the maximum principle on which the theory was grounded and we just have to adapt the method to the problem at hand. The third application concerns a fourth order problem which imposes a more fundamental generalization of the underlying maximum principle. In the fourth application we consider the existence of bounded solutions. Here the lower and upper solutions method applies as such to solve an auxiliary problem. A more elaborate application is the study of the Ginzburg-Landau model which is a vector problem where the use of lower and upper solutions technique is not straightforward. Reading these applications, the reader will realize how large the scope is of the method. To end this introduction, let us notice that the existence of lower and upper solutions does not necessarily imply existence of solutions. For example, it is proved in [242] that there exists some c > 0 and some function f(t,u), which is continuous and almost periodic so that the equation u" + cu' = f{t,u) has no almost periodic solution although there exist constant lower and upper solutions, i.e. /(*,<*) < 0
5. OTHER BOUNDARY VALUE PROBLEMS 5.1
115
Radial Solutions
Consider the problem -Av = f(\x\,v), v = 0,
in BR, on dBR,
.
. '
K
where BR denotes the ball of center 0 and radius R in W1 (with n > 1) and \x\ the euclidian norm of x G Mn. It is well known that v is a classical radial solution of (5.1) if and only if v(x) = u(|x|), where u is a solution of -{tn-lu')'=tn-'f{t,u), u'(0) = 0, u(R) = 0.
(b
-}
More generally, we consider the problem )u')' = q(t)f(t,u), (pu')(0) = 0, u(U) = 0,
.
,
{b d)
-
with / continuous, p, q continuous and p, q > 0 on ]0, R]. Here the condition (pu')(0) = 0 means limt_>o(P'"')(i) = 0. The difficulty is that we must allow the equation to be singular at t — 0 since in the model example p(0) = 0. However the zero of p is balanced by a zero of q. This is clear from the following lemma which is fundamental in our approach. Lemma 5.1. Let p, q € C([0, R]), p,q>0 on ]0, R], define w(t) = jf ^f and assume qw 6 L 1 (0,i?). Then for any h € C([0, R}), there exists a unique function u € C([0, R\)f\ Cl(]Q,R]), withpu' GC^io.iJ]), solution of )u')' = q(t)h(t),
= °. u(R) = 0.
Further, we have
Proof. The proof follows from the explicit solution
rR u(t) = / G(t,s)q(s)h(s)ds, Jo where G(t, s) = w(t) if s < t and G(t, s) = w(s) if s > t.
D
Remark 5.1. The condition
{s)ds =0
l!
(54)
-
116
II. THE SEPARATED BVP
implies u'(0) = 0 and also qw € Ll(0,R). Hence, if we reinforce the assumptions of Lemma 5.1 with condition (5.4), we obtain existence of a solution of -(p(t)u')' = q(t)h(t), u'(0) = 0, u(R) = 0. The main existence result is as follows. Theorem 5.2. Let f € C([0,R] x R), p, q e C{[0,R]), p, q > 0 on }0,R], define w(t) = Jt -4^y and assume qw € i 1 (0, R). Consider a and 0 G C([0,i?])nC1(]0,i2]) withpa' andp/3' G C1([0, i?]), a < f3 and suppose -(pa')'(t) < q(t)f(t,a(t)), (pa')(Q) > 0, a(i?) < 0,
-(pp')'(t) > q(t)f(t,f3(t)), (pp')(0) < 0, P(R) > 0.
T/ien i/ie problem (5.3) /ias at ieasi one solution u such that a to, near enough to, p(t)(u'(t) — a'(t)) < p(to)(w'(to) — a'(to)) = 0 which contradicts the fact that to is a minimum of u(t) — a(t).
5. OTHER BOUNDARY VALUE PROBLEMS
117
If to = 0, we deduce from the boundary conditions (pw')(0) — (pa')(0) < 0. On the other hand, we can write for all i S ]0, R] small enough -(pa')'(t)
w(0) = 1, tu'(O) - 0,
and ZQ(A) to be the first positive zero of w\. Assume there exists A > o such that zo(A) > R- With such assumptions, it is easy to see that if B > 0 is large enough, the functions a(t) — 0 and /?(t) = Bw\(t) are lower and upper solutions of (5.2). Notice that J3 is an eigenfunction of an eigenvalue problem (t"~ V ) ' + At"-1™ = 0,
u/(0) = 0, w{zQ) - 0,
with zo > R- Such eigenfunctions are often the keys to the construction of lower and upper solutions. 5.2
Differential Equations with Deviating Arguments
Consider the following boundary value problem u" = f(t,u,u(h(t))), aiu-a2u' — Ao(t), bxu + b2u' = B Q ( t ) ,
te[o,i], te[a,0], te[l,b],
(5.6)
where a < 0 and 1 < b. The functions / , ylo, Bo and h are continuous but no restriction on the sign of h(t) — t is imposed so that both retarded
118
II. T H E SEPARATED BVP
and advanced shifts are allowed. A solution of (5.6) is supposed to be in C([a, b}) n Cl{I) n C2([0,1]), where / = [a, b], I=[a,l), 7 = [0,6], I = [0,1],
if if if if
a2 ^ 0 and 62 o2/0and62 a2 = 0 and b2 a 2 = 0 and 62
^ 0, =0, ^ 0, = 0.
A first result uses a monotonicity assumption on / . T h e o r e m 5.3. Let a < 0, b > 1, / : [0, l ] x K 2 ^ R, AQ : [o,0] -+ R, Bo : [1,6] — R and ft : [0,1] > [a, b] be continuous functions, ai, a 2 , &i, 62 6e nonnegative numbers such that a\ + a 2 > 0 and 61+62 > 0. Assume f(t, u, v) is non-increasing in v and there exist a and j3 £ C([a,b]) n C ^ J ) nC 2 ([0,l]) such that for any t € [a,b], a(t) < /3(t) and a"(t)>f(t,a(t),a(h(t))), oia(t) - O2Q'(t) < A)(*)» b2a'(t) < B0(t),
j3"(t) < /(t,/3(t),/3(ft(t))), oi)9(*) - a 2/3'W > i4o(t), biP(t) + b20'(t) > Bo(t),
V i e [0,1], Vi e [a, 0], Vt € [1,6].
Tften i/ie problem (5.6) ftas ai feasi one solution u such that for all [a,b], a(t)
t G [0,1], t G [a, 0], t£[l,b],
(5.7)
where "f{t,u) is defined in (1.6). Claim 1 - Problem (5.7) has at least one solution. We write (5.7) as an integral equation u = Tu. (5.8) The operator T : C([a, b]) —> C([a, b}) is defined as follows, (i) For te [0,1], we set (Tu){t) =
/
where G(t, s) is the Green's function associated with u" - u = /(<), aiu(0) - a2u'{0) = 0, 6iu(l) + 62w'(l) = 0, and uo(t) is the unique solution of u" -u = 0, aiu(0) - O2tt'(0) = A o (0), 6 l U (l) + b2u'{l) = Bo(
5. OTHER BOUNDARY VALUE PROBLEMS
119
(ii) For t G [a, 0[ we write (Tu)(t) = (Tu)(0) exp(^-t) + / exp( 2i (t - s)) ^& ds,
if a2 / 0, if a2 = 0.
(iii) Similarly, for i G]l,6] we define
if
The operator T is completely continuous and bounded. It follows from Schauder's Theorem that T has a fixed point which is a solution of the modified problem (5.7). Claim 2 - Solutions u of the modified problem (5.7) are such that a < u < p. Let u be a solution of the modified problem (5.7) and assume minj(u(i) — a(t)) = «(to) - «(*o) < 0. Assume first a^ ^ 0 and notice that for t € [a, 0], u(t) < a(t) implies u'if) < a'(t). If it were not the case, we come to the contradiction A0(t) = a\u{t) - a2u'(t) < a\a{t) - a2a'(£) < A0(t). As a consequence, io > 0. Similarly, if 62 7^ 0, to < 1. In case a2 = 0, we also have for t £ [a, 0], Ao(t) = alU(t) > oia(t) and u(t) > a(t). Hence, io > 0- Similarly, if 62 = 0, we deduce to < 1In all cases, a global negative minimum of u — a takes place at to € int/. It is such that u'(to) — a'(to) = 0 and we can compute 0 < u"(t0) - a"(to) < /(*o,a(*o),7C»(to),«(ft(to)))) +«(*o) - a(to) - /(*o,o(*o),o(h(*o))) < u(t0) - a(t0) < 0, which is a contradiction. It follows from these arguments that u — a has no negative minimum, i.e. that u > a. Similarly, we prove u < /3.
120
II. THE SEPARATED BVP
Conclusion - It follows from Claim 2 that the solution u of (5.7) solves (5.6). Exercise 5.1. Prove that we can improve Theorem 5.3 by replacing the non-increasing assumption on / by the following hypothesis on the lower and the upper solutions : for all t G [0,1] and y G [a(h(t)),(3{h(t))\, a"(t) > f(t,a(t),y) and 0"(t) < /(t,/?(£),y). If we reinforce the notion of lower and upper solution, we can replace the monotonicity condition by a bound on the shifts h(t) — t. Theorem 5.4. Let a < 0, b > 1, f : [0,1] x E 2 -> R, Ao : [a,0] -> R, Bo [1,6] —> R and h : [0,1] —> [a,b] be continuous functions, a\, a2, b\, hi be nonnegative numbers such that a\ + 0,2 > 0 and b\ + 62 > 0. Assume there exist a and /? > a such that a"(t)>f(t,a(t),a(h(t))), axa{t) - a2a'(t) < A0(t), halt) + b2a'(t) < B0(t),
p"(t) < f(t,p(t),p(h(t))), Vt € [0,1], ax(3{t) - a2p'(t) > A0(t), Vt G [a, 0], bx(3{t) + b2f3'(t) > B0(t), Vt € [1,6].
Then there exists S such that if maxt \h(t) — t\<8, the problem (5.6) has at least one solution u such that for all t G [a,b], a(t) < u(t) < /3(i). Exercise 5.2. Prove the above theorem. Hint : See [171] for the Dirichlet problem and [85] for the general case. 5.3
Fourth Order BVP
Consider the fourth order problem / ( ) u(0) = 0, u(l) = 0, «'(0) = 0, u'(l) = 0.
(5.9)
The technique of lower and upper solution can be extended to such boundary value problems provided we can make sure that any solution of the modified problem is between the lower and the upper solution. This last property follows from the maximum principle worked out in Claim 2 of the proof below.
5. OTHER BOUNDARY VALUE PROBLEMS
121
Theorem 5.5. Let a, 0 € C4([0,l]) be such that a < 0, define E = {(t,u) e [0,1] x R | a{t)
a(0) < 0, a(l) < 0, a'(0) < 0, a'(l) > 0,
TTien the problem (5.9) /ias at least one solution u such that a
(5.10)
where j(t,u) is defined from (1.6). Claim 1 - The problem (5.10) has at least one solution. If / € C([0,1]), the linear problem
u^ = f(t), u(0) = 0, u(l) = 0, «'(0) = 0, u'(l) = 0, has a unique solution
u(t)= f
Jo
G(t,s)f(s)ds,
where the Green's function G(t,s), which can be computed explicitly, is continuous and nonnegative. We can then write (5.10) as an integral equation 10
and prove existence of solutions from Schauder's Theorem. Claim 2 - The solution u of (5.10) is such that a 0, w(l) = B > 0, w'(0) = C > 0, io'(l) = -D < 0, where /(*) := u' 4 '(t) - a<4>(i) = f(t,-y(t,u(t))) - a^(t) reads
> 0. Its solution
122
II. THE SEPARATED BVP
w(t)= f
G{t,s)f{s)ds
°+A(l - i) 2 (l + 2t) + Bt2(l + 2(1 - t)) + Ct(l - t)2 + Dt2(l -1), which is clearly positive. In a similar way, we prove that (3(t) — u(t) > 0. Conclusion - It follows from Claim 2 that the solution u of (5.10) solves (5.9). D Remark If / is non-decreasing in u, the differential inequalities in Theorem 5.5 are satisfied provided 0(4)(t)>f(t,/3(t)) 5.4
and
a^(t)
Bounded Solutions
We consider the second order differential equation «" = /(*,«,«'),
(5.H)
where / is a continuous function defined on R3. The following theorem gives sufficient conditions for the existence of a bounded solution u £ C2(R) of (5.11). Theorem 5.6. Let a, /3 G C2(R) be such that a<(3, E = {(t,u,v) G R3 | a(t) < u < 0(t)} and f : E —> R be continuous. Assume that a and j3 are such that for alltER. and 0"(t) <
a"(t)>f(t,a(t),a'(t))
Assume also that for any bounded interval I, there eodsts a positive continuous function ipi : M+ —> R that satisfies
L
00
sds = 00
and for all te I, (u, v) € R2 with a(t)
5. OTHER BOUNDARY VALUE PROBLEMS
123
Proof. Using Theorem 1.3 the problems u" = f(t,u,u'), u(—n) — a(—n), u{n) = a(n), have a solution un such that a
\u'(t)\ < R.
Proof. Consider the modified equation u" = f(t,u,u'),
(5.12)
where f(t,u,v) — f(t, R, min(u, R)))- If / is a bounded interval, the function f(t, u, v) is bounded on D = I x [A, B) x 1R. The constant functions a(t) — A, j5{t)=B and (pi(v)=max.D \f(t,u,v)\ satisfy the assumption of Theorem 5.6. Hence there exists a solution u(t) € [A,B}oi (5.12). Assume f(t,u,R) > 0 if u G [-^j-S] a n d suppose that for some to, u'(to) > R. From Lagrange's Theorem, there exists some t e [to, to + (B — A)/R] such that \u'{t)\ < R. Let ij = max{£ > t0 \ u'(s) > Ron [to,t[}. We have then tx e [to,to + (B - A)/R], u'{ti) = R and u"(h) < 0. This implies f(t,u(ti),u'(ti)) < 0 which contradicts the assumption. We obtain similar contradictions in the other cases. Hence, |u'(i)| < R and u solves (5.11).
124
II. THE SEPARATED BVP
Example 5.2. Consider the differential equation u" = g(u) + \u'\s+p(t),
(5.13)
where g and p are some continuous functions, p is lower bounded and s > 0. If there exist A < B such that g(A)+p(t)<0
Ginzburg-Landau Model
Symmetric solutions of the Ginzburg-Landau model for superconductivity in a slab with a parallel external magnetic field are solutions of the boundary value problem u" = ci(u 2 + h2v2 - l)u, u'(0) = 0, «'(1) = 0,
v" = c2u2v, v(0) = 0, w'(l) = 1.
,
,
[
}
The theory of lower and upper solution can be used for such systems. More generally, we consider in this section the problem u" = f(u, hv)u, u'(0) = 0, «'(1) = 0,
v" = g(u)v, w(0) = 0, v'(l) = 1.
, °
{
, '
Theorem 5.8. Assume / : E x R+ —* R and g : R -* R + are two continuous functions such that (a) for some a > 0 and all v > 0, f{a,v) > 0; (b) f(0, ) is increasing; (c) there exist ho > 0 and UQ solution of u" = /(0, hot)u,
u'(0) = 0, u'(l) = 0,
wi£/i wo(i) > 0 for any t € [0,1]. Then, for every h < ho, the problem (5.15) has at least one solution (u, v) with u > 0 on [0,1]. Proof. Part 1 - The operator
where S(u) is the solution of v" = g(u(t))v,
v(0) = 0, v'(l) = 1,
5. OTHER BOUNDARY VALUE PROBLEMS
125
is well-defined and continuous. For any u G C([0,1], R), we define v(t; u) to be the solution of the Cauchy problem i/(0) = 0, t/(0) = 1.
v"=g{u(t))v,
Notice that v(-;u) € C([0,1],R+) and v'(l;u) > v'(0;u) = 1. It follows that S is defined from
and is continuous. Part 2 - Construction of a lower solution a. Let h G [0, hQ[. By continuous dependence, there exists hi G [h, ho[ such that the solution u\ of u" = /(0,
u(0) = uo(0), u'(0) = 0,
satisfies u\(t) > 0 on [0,1], Hence we have f11dd — I -j:{u'iuo - u\v!Q)(t) dt Joi at
= [ Jo
{f(0,h1t)-f{0,hot))u1{t)uo(t)dt<0,
which proves u'i(l) < 0. Prom continuous dependence, we can find next ao(t) > 0 and e > 0 such that = 1, aJ,(O) = 0, oj,(l) < 0. Choose 7] > 0 such that for any t G [0,1]
and 6 G]0, a] so that for any |w| < 5 and v G [0, hi], \f(0,v)-f(U!v)\
, , °- j
l
126
II. THE SEPARATED BVP
Using Green's function, we write (5.16) as a fixed point problem and existence of a solution u follows from Schauder's Theorem. If maxt u(t) = u(to) > a we have the contradiction 0 > u"(t0) = f{a, hS(j(-, u))(to))a + u(tQ) - a > 0. Suppose now that mmt(u(t) — a(t)) = u(to) — a(to) < 0. This assumption leads to the contradiction 0 < u"(t0) - a"(t0) < u(*o) + [/(a(*o), hS(-y(; «))(*<>)) - l]«(*o) - /(0, Mo + e)a(t0) f(0,hS(j(;u))(to))\a(to) < u(t0) - a(to) + |/(a(*o),ft5(7(.,u))(to)) < u(t0) - a(t0) + r]a(t0) - [/(0,Mo + e) - /(0,fti*o)]o(*o)< 0. As a conclusion 7(£, w(i)) = it(i) and (u,S(u)) solves problem (5.15). Exercise 5.3. Prove that for c-i large enough, there exists 0 < h* < oo such that (i) for h £ [0, /i*], the problem (5.14) has at least one solution (u,v) with u> Oon [0,1]; (ii) for /i > ft*, the problem (5.14) has no solution. Hint : See [90].
6
PDE Problems
6.1 The Dirichlet Problem The method of lower and upper solutions is easy to extend to partial differential equations. A natural framework for this method is the Lp-theory of elliptic equations. Consider for example the Dirichlet problem Au + f(x,u) = 0, u = o,
infi, on an,
, . * ;
where Q, is a bounded domain in R^ and / : O x R —> R. To deal with such a problem, we need to recall basic facts on elliptic equations. A first one concerns the invertibility of the Laplacian operator. Proposition 6.1. Let Q be a bounded domain in RN with boundary dQ, of class C1'1 and let 2 < p < oo. Then, there exists XQ € 1R such that for all A > Ao and f € LP(Q), the problem —Au + Xu = /, u = 0,
in fi, on d£l,
6. PDE PROBLEMS
127
has a unique solution u G W2'P(Q). Further there exists a constant C > 0 such that
Proof. See [302, Lemmas 3.21 and 3.22]. This result means that the operator L : W02lP(O) -» Lp(Cl),
(6.2)
defined by Lu = — Au + Xu has a continuous inverse. Another fundamental tool is the strong maximum principle. Proposition 6.2. Let Q be a bounded domain in M.N with boundary dfl of class C1'1, p > N and A > 0. If u £ W2'P(Q) satisfies —Au + Xu < 0,
in O,
then either u does not have a positive maximum in fi or u is constant. Proof. See [302, Theorem 3.27]. In this section, we consider nonlinearities / : il x M.k —> E, with k = 1 or 1 + N. We shall say that / satisfies Caratheodory conditions if (a) for a.e. x £ 0, the function f(x, is continuous, ) is measurable. (b) for all « e R k , the function If further we have (c) for all r > 0, there exists h G LP(O) such that for a.e. x £ Q, and all v G Rk with ||w|| < r, \f(x,v)\ < h(x), we say that / satisfies IP-Caratheodory conditions. We can now state the main result of the method which paraphrases Theorem 4.2. Theorem 6.3. Let O be a bounded domain in RN with boundary dQ of class C1'1, p > N and let f : O x I —> R satisfy IP -Caratheodory conditions. Consider a and [3 G W2'P(Q) with a < (3 on Q, and suppose Aa(x) + f(x, a(x)) > 0 in fi, A/3(x) + f(x, P{x)) < 0 intt,
a < 0 on d£l, /3 > 0 on 90.
Then the problem (6.1) has at least one solution u such that a
128
II. T H E SEPARATED B V P
Proof. The proof is a remake of the proof of Theorem 4.2. We consider the modified problem u),
= o,
in H,
.
on an,
[ 6)
.
where j(x,u) = min{max{a(a;),u},/3(x)}, A > Ao and Ao is denned in Proposition 6.1. Claim 1 - The problem (6.3) has at least one solution u G W2>P(Q). Define the operator T : C(fi) -> C(fi) given by
Tu = L^Nu, where L~l is the inverse of the operator Lu — —Au + Xu, defined by (6.2), and N : C(Q) - L p (fi) is denned from (Nu)(x)_= f(x,y(x,u(x))) + Xj(x, u(x)). As W 2 ' p (fi) is compactly embedded in C(O), we can prove that T is completely continuous and bounded. Hence by Schauder's Fixed Point Theorem, T has a fixed point u which is a solution of (6.3) in W2'P(Q,). Claim 2 - Any solution u of (6.3) satisfies a < u < 0. Observe first that a(x) < u(x) < P{x) on dQ. Assume a — u has a positive maximum at some point i o £ f ! , We can find QQ regular enough and x\ S fio such that x0 G O 0 ,
a(x) > u{x) on Q.o,
and
a(x\) — u(x\) < a(x0) - u(x0). We compute then that on fio, - A ( a — u) + X(a — u)< f(x, a) - f(x, j(x, u)) + X(a - j(x, u)) = 0 and deduce a contradiction from the strong maximum principle (Proposition 6.2). In a similar way, we prove /3 — u does not have a negative minimum. Conclusion - Prom Claim 2, the function u, solution of (6.3), solves (6.2).
6.2
The Gradient Dependent Dirichlet Problem
Consider the problem Au + f(x, u, Vu) = 0, in
u = o,
on an,
fi,
, {
'
6. PDE PROBLEMS
129
where the nonlinearity / depends on Vw. To deal with such a problem, we need a priori bounds on derivatives. These bounds can be obtained as in the ODE case using a Bernstein condition though the proof is somewhat more elaborate. Proposition 6.4. Let Q be a bounded domain in RN with boundary dQ of class C1'1, p > N and a and (3 € W2'p(£l) with a < 0. Assume also K > 0 and h G LP(Q) are given. Then there exists R > 0 so that given any function f : Q. xR x RN —> K that satisfies Lp-Caratheodory conditions and is such that for a.e. x £ tt, all u € [a(x),/3(x)] and all v € WLN, \f(x,u,v)\
(6,5)
and given any solution u of
0, u = 0,
inQ., on dfl,
satisfying a(x) < u(x) < P(x) on Q,, we have ||Vu||oo < R. Proof. See (302, Lemma 5.10]. Theorem 6.5. Let O be a bounded domain in M.N with boundary dft of class C1'1, p > N and let a and j3 € W2'P(Q) with a < /? on fl. Assume f : Q, x M x M.N —> E satisfies Lp-Caratheodory conditions and (6.5) for some K > 0 and h 6 L P (O). At last, suppose Aa(ar) + f{x, a(x), Va(x)) > 0 in Q, a < 0 on dQ, A0(x) + f(x, f3(x), V/3(x)) < 0 in to, (3 > 0 on dQ,. Then the problem (6.4) has at least one solution u such that a < u < j3 on fi. Proof. Let Ao be given by Proposition 6.1, X > XQ and R > 0 as denned in Proposition 6.4. Increasing R if necessary, we can suppose that R > max{||Va||oo, l i V / ^ } . Let N e LP(O) be such that N(x) > \f(x,u,v)\ if i e S ] , a(x)
= min{max{-AT(x),/(a;,7(a;,u),u)},iV(a;)},
ui(x,S)= mpc \f(x,a(x),Wa{x) + v) - f(x,a(x),Wa(x))\, \\v\\<6
u>2(x,6) = max \f(x,/3(x), V0(x) +v) - f(x,f3(x), lbll<«
130
II. THE SEPARATED BVP
where j(x,u) — min{max{a(x),u},P(x)}. It is clear that Wj are I/P-Caratheodory functions, non-decreasing in 5 and such that iJi(x, 0) = 0 and \u)i{x,6)\ <2N(x). Consider then the modified problem -Au + Xu = f(x,u, Vu) u(x) = 0,
+LJ(X,U),
infi, on da,
. K
. '
where ui(x, u) = \(3{x) - wa(x, u - 0(x)), = Xu, = Xa(x) +uii{t, a(x) — u),
if u > 0(x), iia(x)
Claim 1 - The problem (6.6) has at least one solution u 6 W2'P(Q). This repeats the argument of the corresponding step in the proof of Theorem 6.3. Claim 2 - Any solution u of (6.6) satisfies a < u < /3. Assume that a — u has a positive maximum at some point XQ £ 0. In this case Va(xo) = )- Hence we can find QQ regular enough and x\ e OQ such that io € flo,
||Vu(a;) - Va(a;)|| < a(x) - u(x) on fi0,
and a(xi) - u{xi) < a(x0) - u(x0). We compute then that on fioi —A(a — u) + X(a — u) < f(x, a, Va) — f(x, u, Vu) — w\ (x, a(x) - u) < 0 and deduce a contradiction from Proposition 6.2. In a similar way, we prove that 0 — u does not have a negative minimum. Conclusion - Let u be a solution of (6.6) given by Claim 1. From Claim 2, a < u < f3 on Cl and u is a solution of —Au — f(x, u, Vu), in fi, u(x) — 0, on dfl. Notice now that for a.e. x G f2, all u € [a(x),/3(x)], and all v e RN,
\f(x,u,v)\
6. PDE PROBLEMS 6.3
131
The General Problem
The method extends to the very general semilinear elliptic problem Lu = f(x,u,Vu), u{x) = 0,
Bu(x) — c(x)i
in 0, on dtt \ F,
(6.7)
° n r,
where fi is a bounded domain in WLN, F is an open and closed subset of the boundary dQ, L is a linear second order elliptic operator, B is a linear first order boundary operator, and C, is a given function. Theorem 6.6. Let Q be a bounded domain in RN with boundary dQ, of class C1'1 and T an open and closed subset of 9 0 . Let p > N and L : W2'P(Q) —> LP(Q) be a strongly elliptic operator defined by , , d2u
^
,,
with a,ij € C(Q) such that for some a > 0, Y^fj=i ai,j(x)€i£j ^ a lC| 2 for a ^ x e H , Let also C G W 1 ~p' p (r) and B : W2*(Q) -* W 1 "p' p (r) be defined by N
„
Bu(x) = J2bi(x)^(x)+b(x)u(x), where bi and b e C Oll (r) are such that, if v(x) denotes the unit outward normal atxeT, for all xeT, J^iLi bi(x)ui(x) > K > 0. Let a and 0 e W2>P(Q) with a < 0 on 0 . Assume f i f i x M x R ^ - + E satisfies Lp-Caratheodory conditions and (6.5) for some K > 0 and h G ). yli last, suppose
a < 0, Ba < C,
)>0,
m 0, ondQ,\T, on T,
> 0, > C,
ondn\T, on T.
the problem (6.4) /ias ai least one solution u such that a < u < f3 in Q. Proof. See [93].
132 6.4
II. THE SEPARATED BVP Weak Lower and Upper Solutions
The method of lower and upper solutions can be worked out for weak solutions. Let fi be a bounded domain in WLN and / : Q. x K —> R. We consider the problem Find u £ HQ(Q.) such that for any v £ HQ (0.),
f[(Vu\ Vv)-f(x,u)v}dx = O. Jn
(6.8)
A standard result using weak lower and upper solutions is as follows. Theorem 6.7. Let Q be a bounded domain in M.N and let f : fi x R —v K satisfy Caratheodory conditions. Consider a and /? G H1 (fl) with a < (3 on Q,, a+ £ HQ(Q,), 0" £ HQ(Q.) and suppose that for any v G HQ(Q), v>0,
[
f{x, a(x))v(x)] dx < 0,
(6.9)
a:) | Vv(x)) - f(x, 0(x))v(x)} dx > 0.
(6.10)
Jn
Assume also there exists h £ LP(Q) withp > 2N/(N + 2) such that for a.e. x £ Q and all a(x)
= 0, V«Gfl3(n),
(6.11)
where j(x,u) — min{max{a(a;),u},/0(a;)}. Claim 1 - The problem (6.11) has at least one solution. Define p' = p/{p—l) and the operator T : IP (f2) —» Lp (Q) where Tu = w is the solution of / [(Viu | Vw) - f{x,j(x,u))v]dx Jn
= 0, Vu G H£(Q).
Prom Lax-Milgram theorem, such a solution exists, is unique, and ||w||^i < C||/i||ip. It follows that T is a bounded compact operator and the claim follows from Schauder's Theorem.
6. PDE PROBLEMS
133
Claim 2 - Any solution u of (6.11) satisfies a 0,
L
[(Va - Vu | Vu) - {f{x, a) - fix, 7(2;, u)))v{x)\ dx < 0.
/n
For v = {a — u)+ £ HQ(Q), we obtain
L n
|V(a-u)+|2<0
so that (a — u)+ = 0. In a similar way, we prove /3 > u. Conclusion - From Claim 2, the function u, solution of (6.11), solves (6.8). A function a (resp. /?) that satisfies (6.9) (resp. (6.10)) is called a weak lower solution (resp. a weak upper solution) of (6.8). Such functions can also be used in case the nonlinearity is gradient dependent. For generalizations in the framework of Theorem 6.5, we refer to [38, 39].
This Page is Intentionally Left Blank
Chapter III
Relation with Degree Theory 1
The Periodic Problem
It has been seen that the periodic problem «" = /(«,«), «(o) = u(6), u'(a) = u'(b)
{
n
n
'
is equivalent to the fixed point problem u = Tu,
(1.2)
where T : C([a, b]) -> C{{a, b]) is denned from (Tu)(t) := / G(t,a)[/(3,tt(a)) -u(a)]d5 Ja
(1.3)
and G(i, s) is the Green's function corresponding to (1-1.4). In this chapter, we associate the set Q = {ue C([a,b\) | Vt e [o,fe], a(t) < u(t) < /3(t)}
(1.4)
to the pair of lower and upper solutions (a, (3) and wish to consider the degree deg(J-T,fi). To this end we have to reinforce the concepts of lower and upper solutions so that the boundary of D, does not contain solutions of (1.1). This modification amounts to requiring that the solutions cannot be tangent to the curves u = a{t) and u = 135
136
III. RELATION WITH DEGREE THEORY
respectively from above or from below and motivates the definitions we introduce in the following section. For problems with a nonlinearity / which depends on the derivative u" = /(«,«,«'), u(a) = u(b), u'(a) = u'(b),
n
^^
the corresponding fixed point problem reads i-b
(t) = (Tu)(t) := [ G(t,s)[f(sMs),u'(s))-u(s)}ds, Ja 1
1
where T : C ([a, 6]) —> C ([a,6]). Given lower and upper solutions a and /3, and some R > 0, we shall compute the degree deg(/-T,fi), where a={ue
^([a.fe]) | Vi G [a, b], a(t) < u(t) < /3(t), \u'(t)\ < R}.
As above, this requirement will force us to reinforce the concepts of lower and upper solutions and furthermore to choose R to be an a priori bound on the derivative v!. 1.1
The Strict Lower and Upper Solutions
Definitions 1.1. A C2 or W2>1-lower solution a of (1.5) (resp. (l-l)j is said to be strict if every solution u of (1.5) (resp. (1.1),) with u > a is such that u(t) > a(t) on [a,b]. Similarly, a C2 or W2'1 -upper solution J3 of (1.5) (resp. (1.1) j is said to be strict if every solution u of (1.5) (resp. (1.1)) with u < 0 is such that u{t)
1. THE PERIODIC PROBLEM
137
Proof. Let u be a solution of (1.5) such that u> a and assume by contradiction that min (u(t) - a(t)) = u(t0) - a(tQ) = 0. te[a,6]
We have u'(to) — ct'(to) = 0 which follows from assumption (b) in case t0 = a or b. Hence, we obtain the contradiction 0 < u"{t0) - a"(t0) = /(t 0 , a(tQ),a'(t0)) - a"(t0) < 0. Using the same argument we obtain the corresponding result for upper solutions. Proposition 1.2. Let f : [a, b] x R2 —> R be continuous and p € C2([a, b}) be such that (a) for all t G [a,b], ff'(t) < f(t,P(t),P'(t)); (b) fta) = f3(b), F(a) < ff(b). Then f3 is a strict C2-upper solution of (1.5). If / is not continuous but Z/P-Caratheodory, these last results no longer hold. In fact, even the stronger condition for a.e. t G [a,b], a"{t) > f(t,a(t),a'(t))
+1
(1.6)
does not prevent solutions u of (1.5) to be tangent to the curve u = a(t) from above. Such a situation occurs, for example, for the bounded function /(*,«):=-1,
2
+sint := ; , 1 + sm t := — sint,
if«<-l, . if — 1 < u < sins, ~ if sini < u,
if we consider a(t) = —1, u(t) = sint, a — 0 and b = 2TT. This remark motivates the following proposition. Proposition 1.3. Let / : [ a , t ] x i 2 - > i be an L1 -Caratheodory function. Let a £ C([a, b]) be such that a(a) = a(b) and consider its periodic extension on R defined by a(t) = a(t + b — a). Assume that a is not a solution of (1.5) and for any t0 G K, either (a) £)_a(t 0 ) < D+a{t0) or (b) there exist an open interval IQ and eo > 0 such that and t o e Jo, aeW2'1^) for a.e. t G Io, allu G [a(t),a(t)+e 0 ] and all v G [a'(t)—eo,a'(t)+eo], a"(t) >/(*,«,«). a is a strict W2'1-lower solution of (1.5).
138
III. RELATION WITH DEGREE THEORY
Proof. The function a is a H^2>1-lower solution since clearly it satisfies Definition 1-6.1. Let u be a solution of (1.5) such that u > a. As a is not a solution, there exists t* such that u(t*) > a(t*). Extend u and a by periodicity and assume by contradiction that t0 = inf{t > t* | u(t) = a(t)} exists. As a — u is maxima at to, we have D-a{to) — u'(to) > D+a(to) — u'(to). Therefore, assumption (b) applies. We have that u'(to) ~&'(to) = 0 and there exist IQ and eo > 0 according to (b). It follows that we can choose t\ e Jo with tx < to such that u'(ti) — a'{t{) < 0 and for every t£]h,to[ u(t) < a(t) + c0,
\u'(t) - a'(t)\ < e0.
Hence, for almost every t G]£i,£o[> we can write
a"(t)>f(t,u(t),u'(t)), which leads to the contradiction 0 < («' - a')(t0) - («' - a')(*i) = / "[/(*. «(*).«'(*)) " a"(t)]dt < 0. D In the same way we can prove the following result on strict upper solutions. Proposition 1.4. Let f : [o,i]xM 2 ->R be an Ll-Caratheodory function. Let 0 € C([a, b}) be such that /3(o) = /?(&) and consider its periodic extension on R defined by /?(£) = f3(t + b — a). Assume that /3 is not a solution of (1.5) and for any to G K, either (a) D-0(to) > D+0{to) or (b) there exist an open interval Jo and eo > 0 such that to£ la, peW2>l(I0) and fora.e. t e Io, alluE [/3(t)-eo,/3(t)] andallv £ [/3'(t)-eo,P'{t)+eo}, (3"(t) < f(t,u,v). Then f3 is a strict W2'1 -upper solution of (1.5). Observe that in the continuous case, if a satisfies the conditions of Proposition 1.1, then it satisfies the conditions of Proposition 1.3. The converse, however, does not hold since Proposition 1.3 does not require strict inequalities. To be able to say that a function a which satisfies (1.6) is a strict lower solution, we need some more regularity on / . In particular we have the following result.
1. THE PERIODIC PROBLEM
139
Proposition 1.5. Let f : [a, b] x R2 —> E be an Lx-Caratheodory function that satisfies the assumption (A) for all to G [a, b], (UO,VQ) G R2 and e > 0, there exists 6 > 0 such that \t-to\ < 6,\u-uo\ < S, \v-vo\ < S => \f(t,u,v)-f(t,uo,vo)\<e. 2>l Let A > 0 and a G W (a, b) be such that
a"(t)>f(tMt),<*'(t)) + A, a(o) = a(b), a'{a) > a'(b). Then a is a strict W2'1 -lower solution of (1.5). Remark Notice that in this proposition, Condition (A) does not imply that / is continuous. For example, we can take f(t,u,v) = g(u,v) + h(t) with g continuous and h € L1(a,b) or f(t,u,v) = g(u,v)h(t) with g continuous and h G L°°(a,b). Proof. Let us deduce this proposition from Proposition 1.3. To this end, we will prove that for any to G R, there exist an open interval IQ and eo > 0 such that to G /o and for a.e. t G Jo, for all u, v with a(t) < u < a(t) + eo, oc'(t) — eo < v < a'(t) + eo, we have a"(t)>f(t,u,v). Let to G R be fixed. Define u0 = a(t0), v0 = a'(t0) and e - A/2. We deduce then from Assumption (A) a 5 > 0 such that t-to\<
5, \u - a(to)\ < 8, \v - a'(tQ)\ < S => \f(t, u, v) - f(t, a(t 0 ), a'(*o))| < A/2. Let now rj < 6 be such that for any £ G [to — ?7, io + ?7]> |a(t) - a(t o )| < S/2, \a'{t) - a'(£0)| < 5/2. Choose Jo =]£o — Wito + ?7[ and eo = S/2. The result follows now since a"(t) - f(t, u, v) = a"(t) - f(t, a(t), a'(t))
+f(t,a(t),a'(t)) - f(t,a(to),a'(to)) +f(t,a(to),a'(tQ)) - f(t,u,v) >A-A/2~ A/2 = 0. In this last proposition, we can weaken the assumption a G W2'1(a,b) and allow angles as previously. Also, we can make Condition (A) one-sided and impose (A') for all t 0 G fai b], ("0; ^o) € R2, and e > 0, there exists S > 0 such that t - t o | < 6, \ui -uo\ < 5, u2 G [ui.wi +5[, \vi-vo\<5,\v2-vo\
140
III. RELATION WITH DEGREE THEORY
For strict upper solutions we have a similar result. Proposition 1.6. Let f : [a, b] x R2 —> R be an L1 -Caratheodory function that satisfies Assumption (A) in Proposition 1.5. Let B > 0 and fi £ W2'1(a,b) be such that P"(t)
- f(t,uuv)
<
(b) for a.e. t G [a, b] and all u, v\, V2 € R,
Then every W2>1 -lower solution a (resp. every W2'1-upper solution j3) of (1.5) which is not a solution is a strict W2'1 -lower solution (resp. a strict W2'1 -upper solution). Proof. Let u be a solution of (1.5) such that u > a. As in the proof of Proposition 1.3, we find an open interval /o, to £ Io and ti S 7o with £1 < to such that u(to) = a(t0), u'(t0) = a'(tQ), u'(t\) — a'(ti) < 0 and for a.e. tel0 a"(t)>f(t,a(t),a'(t)). Let w = u — a and observe that, on [ti,to[i -w" +l(t)sgn(w'(t))w' + k(t)w > -f(t,u{t),u'{t)) + f(t,a(t),a'(t)) +l(t)\u'(t) - a'(t)\ + k(t)(u(t) - a(t))
1. THE PERIODIC PROBLEM
141
Defining P(t) = - f i-t
Q(t) = -
l(s)sgn(w'(s))ds,
Jti
ep(-s)k(s)ds,
ft
R(t) = - / Q(s) e - p ( s ) ds,
Jti
Jti
we compute
l[(w'(t) - k(t)w{t)) Hence, we have the contradiction
0 > K(t)e p W + Q(t)w(t))eRW |£ = -w'(h) > 0.
D
Exercise 1.2. Assume / satisfies the assumptions of Proposition 1.7 and a is a VF2'Mower solution (resp. (3 is a W2il-upper solution) of (1.5) which is not a solution. Prove that any W2]1-upper solution /? > a (resp. W2'1lower solution a < 0) is such that /3 > a. Remark 1.1. In Proposition 1.7, we can assume (a) and (b) to hold only in a neighbourhood of {(£, a(t),a'(t)) \ t £ [a, 6] such that a'(t) exists} (resp. {(t,/3(t),f3'(t)) 11 € [a, b] such that /?'(£) exists}). However this proposition does not hold without the Lipschitz conditions. Even a one-sided Holder condition is not enough as shown by the following example
Here /3(t) = tA is an upper solution which is not a solution of (1.7). On the other hand, u(t) = 0 is a solution such that u(0) = /3(0). Hence, /3 is not strict. 1.2
Existence and Multiplicity Results
Now we can prove the key result of this section. Theorem 1.8. Let a and 0 € C([a,b]) be strict W2'1 -lower and upper solutions of problem (1.5) such that on [a,b], a(t) < /3(t). Define A C [a,b] (resp. B C [a,b]) to be the set of points where a (resp. (i) is derivable. Let E be defined by E := {(*, u, v) G [a, 6 ) x K 2 | a(t)
(1.8)
142
III. RELATION WITH DEGREE THEORY
and p, q £ [l,oo] be such that h + h — 1- Assume f : E —> R satisfies Lp-Caratheodory conditions and there exists N £ L 1 (a,6), N > 0 such that for a.e. t £ A (resp. for a.e. t £ B) f(t,a(t),a'(t))
> -N(t)
(resp. f(t,p(t),/?'(*))
< N(t)).
Assume moreover there exist tp £ C(R + ,Rj), ip £ Lp(a, b) and R > 0 such that f ^rds> \\ip\\Lp(ma,x(3(t)-mina(t))1/g (1.9) < S t l Jo P\ ) and that the function f satisfies one of the one-sided Nagumo conditions (a), (b), (c), or (d) in Theorem 1-6.1. Then deg(/-T,fi) = l, (1.10) l l where T : C {[a,b]) -> C {[a,b\) is defined by fb
(Tu)(t):= I G(t,s)[f(s,u(s),u'(s))-u(s)]ds, Ja
(1.11)
G(t,s) is the Green's function corresponding to (1-1.4) and 0 is given by n = {u£ Cl([a,b\) | Vt £ [a,6], a(t) < u(t) < /3(t), \u'(t)\ < R}. (1.12) In particular, the problem (1.5) has at least one solution u £ W2'p(a, b) such that for all t £ [a, 6] a(£) < u(t) Proof. Increasing N if necessary, we can assume N(t) > \f(t,u,v)\ [a, 6], a(t) < u < /3(i) and |u| < R. Define then
if t G
/ ( i , u, v) - max{min{/(i, 7 (t,u), v), N(t)}, -N(t)}, max |/(t,a(t),a'(t) +v)- f(t,a(t), a'(t))\, \v\<5
max \f(t, /3(i), /3'(i) + v) - /(t, /?(i),^(i))|, H
where 7 is defined from (1-1.3), XAa n d Xs a r e the characteristic functions of the sets A and B. It is clear that w, are L1-Caratheodory functions, non-decreasing in 5 such that Wj(t,0) = 0 and |wj(t,5)| < 2N(t). We consider now the modified problem u" - u = Jit, u, u') - w(t, u), «(o) = u(6), u'(a) = u'{b), where w(t, u) - /3(i) - w2(t, u - j8(*)), = u, — a(t)+u>i(t,a(t)-u),
if « iia( iiu
,
x
K
;
1. THE PERIODIC PROBLEM
143
This problem is equivalent to the fixed point problem u = Tu, where T : Cx{[a,b)) -+ Cl{\a,b]) is defined by (fu)(t) = /
G(t,S)[f(s,u(s),u'(s))-u;(s,u(s))}ds.
J
Observe that T is completely continuous and there exists R large enough so that 0 C B(0,R) and f(Cl([a,b])) C B(0,R). Hence we have, by the properties of the degree, deg(I-f,B(O,R)) = l. We know that every fixed point u of T is a solution of (1.13). Arguing as in the proof of Theorem 1-6.6, we see that a < u < 0 and ||w'||oo < RAs a and (3 are strict, a < u < (3. Hence, every fixed point of T is in Q and by the excision property we obtain deg(J - T,fi)= deg(J - f, Q) = deg(J - T, B(0,5)) = 1. Existence of a solution u such that for all t G [a,b], a{t) < u{t) follows now from the properties of the degree. A generalization concerns the use of several lower and upper solutions. T h e o r e m 1.9. Let at € C([a,b]) (i = l , . . . , n ) and fa G C([a,b}) {j = 1 , . . . , m ) be respectively W2>1 -lower and upper solutions of (1.5). Assume
a :— max a.; l
and 3 := min /37l<j<m
satisfy a(t) < 0(t) on [a,b] and are strict, i.e. any solution u of (1.5) with a
(resp. /(*,&(*),$(t)) < JV(t)).
Assume moreover there exist
144
III. RELATION WITH DEGREE THEORY
Then deg(/-T>n) = l, where T : C1([a,6]) —* (^([a, &]) is defined by (1.11) and Q is given by (1.12). /n particular, the problem (1.5) /i«s at /east one solution u G W 2 ' p (a, &) suc/i i/iai for all t G [a, 6] a(i) < w(f) < /?(£). Exercise 1.3. Prove Theorem 1.9 paraphrasing the argument of the previous proof and using the idea of Theorem 1-6.10. In case / is independent of u', this result reduces to Theorem 1.10. Let on G C([a,b]) (i = l , . . . , n ) and /3j G C([a,b]) (j = 1,... ,m)feerespectively W2'1 -lower and upper solutions of. (1.1). Assume a := max a, and B := min /3,!<*<«
1<J<"»
satisfy a(t) < (3(t) on [a,b] and are strict (see Theorem 1.9). Let E := {(t, u) £ [a,ii] x R | minjaj(t) < u < max j ftj(t)} and assume f : E —> M satisfies L1-Caratheodory conditions. Then d e g ( / - T , Q ) = l, where T : C([a, b\) — C([a, 6]) is defined by (1.3) and 0 is given 6y (1.4). /n particular, the problem (1.1) ftos a£ teasi one solution u G VF2ll(a,5) SMC/I i/iai for all t € [a, b]
a(t) < u(t) < /?(*). Observe that we can replace the Nagumo conditions by any condition so that an a priori bound in the space C([a,6]) on solutions u of the corresponding modified problem implies an a priori bound on HU'HQO which is the case for the Rayleigh and the Lienard equation. Consider first the Rayleigh equation u + g(u) + h(t,u,u')=Q, u(a) = u(b), u'(a)=u'{b). Theorem 1.11. Let a solutions of (1.14) such (1.8), g G C(E) and h for some H G L2(a,b), {t,u,v)€E,
{
'
and j3 £ C([a, b]) be strict W2'1 -lower and upper that on [a,b], a(t) < fi(t). Let E be defined by : E — R be a Garatheodory function such that for a.e. t G [a,b], and for all (u,v) G M2 with \h(t,u,v)\
1. THE PERIODIC PROBLEM
145
Then deg(7-T,n) = l, where O is defined by (1-12) (with R > 0 large enough), T : C1([a, 6]) —> C1([a, 6]) is defined by ,&
(r«)(t) := - / G(t,s)[g(u'(s)) + h{s,u(s),u'{s)) +C(s)u{s)]ds, Ja G(t,s) is the Green's function of u"-C(t)u = f(t), u(a) =«(&), «'(o) = «'(&),
U
,
j
and C € Lx(a, 6) is c/iosen such that C(t) > \g(0)\ + 1 + ZH(t) on [a, b]. In particular, the problem (1.14) has at least one solution u £ W2'2(a, b) such that for all t £ [a, 6] a(t) < u(t) < f3(t). Proof. The proof repeats the arguments of Theorem 1-6.8. Using the notations therein, observing that a and (3 are strict, and denoting fii = {u e C^IM]) I IMIoo < p, \\u'\\oo < R}, we have deg(/ - T,fi) = deg(J - TuSI) = deg(J - T,,^) = deg(J - To, Ox) = deg(J - To, This concludes the proof. The next result concerns the Lienard equation u"+ g(u)u' + h(t, u) = 0, u(a) = u(b), u'(a) = u'(b).
.
[
.
b}
Theorem 1.12. Let a and j3 G C([a,b]) be strict W2'1-lower and upper solutions o/(1.16) such that on [a,b], a(t) < (3{t). LetE = {{t,u) e [a,b] x R | a(t) < u < p(t)}, g € C(R) and h : E -+R be an L1 -Caratheodory function. Then d e g ( / - T , n ) = l, w/iere fi is defined by (1.12) fwii/i i? > 0 Zan/e enough), T : Cl([a, b]) —> C1([o,6]) is defined by (Tu)(t) :=- f G(t, 3)\g(u(8))u'(8) + f(s, u(s)) + C(s)u(s)} ds, Ja
146
III. RELATION WITH DEGREE THEORY
G(t,s) is the Green's function of (1.15) and C € Ll{a,b) is chosen such that for a.e. t G [a,b], \g(ot{t))\ < C(t), \g{0(t))\ < C{t) and for every (t,u)EE, \f(t,u)\
Assume further 0i anda2 are strict W2'1 -upper and lower solutions. Define Ai C [a, b] (resp. Bi C [o, b]) to be the set of points where a.i (resp. Pi) is derivable. Let E be defined by E := {(t,u,v) S [a,b] x R2 | min at(t)
(1.17)
and p, q £ [l,oo] be such that - + - = 1. Assume f : E —> M satisfies V-Caratheodory conditions and there exists N € Ll(a, b), N > 0 such that for i = l,2 and a.e. t € Ai (resp. a.e. t G Bi) f(t,a,(t),aj(t))
> -N(t)
(resp. f(t,&(*),#(*))
< N(t)).
Assume moreover there exist
1. THE PERIODIC PROBLEM
147
Notice that the conditions 113(^1) > 0i(ti) and u^t-z) < 012(^2) is a localization condition that implies that 113 ^ u\ and U3 ^ u^. Proof. Define gi(t, u, v) = f(t, cti(t),v) - ai(t) - wii(t, oti(t) - u), if u < at(t), itu>ai(t), — f(t,u,v) -u, hi(t, u, v) = f(t, fa(t),v) - 0i( = f(t,u,v)-u, where w«(t, *) = XA*(*) max \f(t, o<(t), |f|<«
w«(M) = Xfli(t) max \f(t, AW, X/ii and XBj are the characteristic functions of the sets Ai and Bj. We consider now the modified problem u" -u = F{t,u,u'), «(a)=u(6), u'(a) = u'(b),
,
{
> '
where F(t,u,v) =g1(t,u,v), = f(t,u,v)-u,
ifu
Let us choose fc so that 0\ < /?2 + k and aj — A; < a2) a n d define T : Cl([a,b\)-* C\[a,b}) by (Tu)(i)= / where G(t,s) is the Green's function corresponding to (1-1.4). Step 1 - Computation of deg(I — T, fii,i), where fiM = {u G C^Io.b]) I Vt G [a,6],ai(t) - A; < u(t) < fa(t), \u'(t)\ < R}. Define the alternative modified problem u" -u = F(t,u,u'), «(o)=u(6), u'(fl)=«'(6), where F(t,u,u') = gi(t,u,v), ifu
h 1Q . V
148
III. RELATION WITH DEGREE THEORY
Define next f : Cl([a, b]) -> ^([a, b}) by fb {Tu)(t)= / G(t,s)F(s,u(s),u'(s))ds. Ja For any A £ [0,1], we consider then the homotopy T\ = AT + (1 — X)T. Claim 1 : If A £ [0,1] and u is a fixed point ofT\, we have at\ < u < fa. This result follows from the usual maximum principle argument as in Step 3 of the proof of Theorem 1-6.6. Claim 2 : If X £ [0,1] and u £ fliti is a fixed point ofT\, we have u < j3\. Assume there exists to £ [a,b] such that u(to) = /3i(to)- We deduce from Claim 1 that cti(t) < u(t) < fait) for aU t £ [a,b\. Hence we prove, as in Step 4 of the proof of Theorem 1-6.6, that ||w'||oo < -R so that u solves (1.5). As further f3\ is a strict upper solution, the claim follows. Claim 3 : deg(J — T, Oi,i) = 1. It follows from the above claims that a.\ — k and Pi are strict lower and upper solutions of (1.19) and we deduce from Theorem 1.8 and the properties of the degree that deg(J - T, O u ) = deg(7 - TA, fiM) = deg(J - f, O M ) = 1. Step 2 - deg(7 - T, Q2,2) = 1, where «2,2 = {«£ C^fa.t]) | Vt G [a,b],a2(t) < u(t) < fa{t) + k, \u'[t)\ < R}. The proof of this result parallels the proof of Step 1. Step 3 - There exist three solutions Sj (i = 1,2,3) o/(1.18) such that a\ — k <
< Pi, a2 < v.2 < fa + k, a\ — k <
< fa + k
and there exist t\, t% £ [a, b] with
Mh) > Pi(h), Mb) < Mh)The two first solutions are obtained from the fact that deg(/-T,n 1 , 1 ) = l
and deg(J - T,O2]2) = 1.
Define n 1)2 = {u£ Cl([a,b}) | Vt £ [a,b], ax(t)-k
< u(t) < fa(t)+k, \u'{t)\ < R}.
We have
i = deg(/-r,n 1 , 2 ) = deg(J - T, ni,i) + deg(J - T, na,a) + deg(J - T, ni >2 \(fti,i U O2>2)) which implies and the existence of u3 £ fiii2 \ (fli,! U ^2,2) follows.
1. THE PERIODIC PROBLEM
149
Step 4 - There exist solutions Ui (i = 1,2,3) of (1.5) such that
iAere exist t\, ti e [a, 6], with U3(ti) > /?l(tl) We know that solutions u of (1.18) are such that i.e. they are solutions of (1.5). Next, from Theorem 1-6.11, we know there exist extremal solutions umin and umax of (1.5) in [cti,/^]. The claim follows then with ui = umin, ui = umax and 1*3 = W3. Observe that in this theorem ui < min{/31,/32} and u-x > Example 1.1. In Example 1-1.4, we have proved that if h € C([0, 2TT]) is such that \\h\\Lt < 3, \h\ < cos(f H/iH/,1), the problem u" + sinu = h(t), u(Q) = u(2n), u'{0) = u'{27r),
,
K
, >
has at least one solution. With Theorem 1.13, it is now easy to extend this example to functions h € L^O, 2n) and to complement it as follows. Let w to be the solution of w" = h(t),
w(0) = w(2ir), w'(0) = W'(2TT), w=0,
and Using Propositions 1.5, 1.6 and Theorem 1.13, we find three solutions of (1.20), i.e. ct\ < Ui < fii, oi2 < v>2 < 02 and a.\ < uz < 02 with U3(£i) >
0\{t\) and 1^3(^2) < ^2(^2) for some t\ and ^2 6 [0,2TT]. Notice that ui might be U2 — 2TT but 113 ^= u\ mod 2vr. Hence, this problem has at least two geometrically different solutions. Example 1.2. Consider the equation (1 + e cos t)u" — (2e sin t)u' + asinu = 4esint
(1-21)
which describes the motions of a satellite in the plane of its orbit around its center of mass. Here u is twice the angle between the radius vector of the satellite and one of its principal axes of inertia lying in the plane of the orbit, t is the true anomaly, and a and e are parameters such that \a\ < 3 and 0 < e < 1.
150
III. RELATION W I T H D E G R E E THEORY
Let us prove that for all a and e e ] — 1,1[ with |e| < |a|/4, the equation (1.21) has at least two 27r-periodic solutions which do not differ by a multiple of 2TT. TO this end we apply, in case a > 0, Theorem 1.13 with
and obtain three solutions a\ < ui < pi, cti < u-2 < fa and a\
These conditions improve the preceding ones if a is small. To prove this result in case a > 0, let
be a 2?r-periodic solution of (1 + ecos£)z" — (2esin£)z' = 4esint. If we can find a constant c such that on [0,2ir] = z{t)+ce]0,n[, this function is a strict lower solution as follows from Proposition 1.1. The result follows then from Theorem 1.13 with c*i < A = a\ + n < a-i = a.\ + 2?r < fa = ot\ + 3TT. The extrema of ati are difficult to compute. However if we choose c so that ai — -^ JQ^ ai dt = IT/2, we can compute (using |e| < eo and Proposition A-4.1) that 6t\ — OL\ — a.\ satisfies
The function a% = d i + 7r/2 works.
1. THE PERIODIC PROBLEM
151
Example 1.3. Consider the problem u" + cu' - u3 + 3u = p(t), u(0) =W(2TT), U ' ( 0 ) = U'(2TT),
(1.22)
where pG I°°(0,27r). Claim 1 - For all p G L°°(0,2TT), (1.22) has at least one solution. This result follows from Theorem 1-6.9, where a,\ < —1 and j3% > 1 are constants such that a\ - 3ax < -|blloo and /?f - 3/32 > ||p||ooClaim. 2 - For all p € L°°(0,27r) such that ||p||oo < 2, (1.22) has at least three solutions. This result follows from Theorem 1.13, where a\ and 02 are chosen as in Claim 1, /?i = —1 and «2 = 1- The functions f3\ and 0:2 are strict upper and lower solutions as follows from Propositions 1.6 and 1.5. Claim 3 - For all p e L°°(0,2n) such that \\p\\oo = 2, (1.22) has at least two solutions. Let a, and /3j (i = 1,2) be chosen as in Claim 2. The existence of solutions Uj € [CKJ, /3J] follows then from Theorem 1-6.9. This result is best possible as follows from the case c ^ 0 and p constant. Here, we can give an exact count of the number of solutions. Multiplying the equation (1.22) by u' and integrating we obtain ||ti'||z,2 = 0. Hence, in this case, the problem admits only constant solutions. It is then easy to see that it has exactly (a) three solutions for p 6 ] — 2,2[, (b) two solutions for p — — 2 or 2, (c) one solution for p G ] — 00, —2[ or p G ]2, +oo[. This situation is represented at figure 1.
Fig. 1 : Diagram of the solutions of (1.22)
152
III. RELATION WITH DEGREE THEORY
We can generalize this example and complement Exercice 1-6.4. Exercise 1.4. Consider the problem u" + 9 («)8gn(u')|«T + /(*, u) = s(t), «(0) = u(2ir), «'(0) = U'(2TT),
^
}
where / is an Lp-Caratheodory function that satisfies Condition (A) in Proposition 1.5, g is continuous, and s G L°°(0,2ir). Assume 0 < r < 2 - 1/p, lim f(t, u) ——oo, lim fit,u) =+oo, u—*+oo
u—*—oo
uniformly in t and there exist real numbers v < u such that f(t, v) < f(t, u). Prove that for any s G i°°(0,27r) with f(t,v) < s(t) < f(t,u), problem (1.23) has at least three solutions. Hint : Use constant upper and lower solutions : a\ < /3\ = v < a% — u < Exercise 1.5. Adapt Theorem 1.13 to the Rayleigh and the Lienard equations (1.14) and (1.16). Another way to obtain multiplicity results is to exhibit domains fii D Q such that deg(/ - T, n x ) = 0 and deg(7 - T, 0) = 1, which is the basic idea of the following theorem. Theorem 1.14. Let k > 0 and a, 0 G C([a, b]) be respectively a W2'1lower solution and a strict W2'1 -upper solution of the problem (1.5) with a < f3 < k. Define A C [a, b] (resp. B C [a, b]) to be the set of points where a (resp. (3) is derivable. Let p, q G [l,oo] with - + - = 1, and E = {(t,u,v) G [a,6] x R2 a(t) < u}. Assume f : E —* R satisfies Lp-Caratheodory conditions and there exists N G Ll(a, b), N > 0 such that for a.e. t G A (resp. for a.e. teB) f(t,a(t),a'(t))>-N(t) (resP.f( LetR>O,
C(R+, Rj) and ip G Lp(a, b)
r
o V{s)
ds >
anrf suppose that for a.e. t G [a, 6] and aH (u, v) G M2 w^/i a(t) < u < k,
1. THE PERIODIC PROBLEM
153
Assume at last that for every solution u e W2'1(a,b) of u" < f(t,u,u'), u{a) = u(b), u'(a) = u'(b),
(1.24)
with u > a we have u < k on [a, b]. Then the problem (1.5) has at least two solutions u\, u2 G W2'p(a,b) such that for all t € [a, b] a(t) < ui(t) < (3{t), ux{t) < u2(t) < k and for some ti € [a, b], Proof. Define f(t,u,v)
= f(
and wi(t, 5) = XA(t) max \f(t, a(t), a'(t) +v)- f(t, a(t),a'(t))\, \v\<5
where 7i(i, u) = max{a(f),u} and XA is the characteristic function of the set A. We consider now the modified problem u" -u = f{t,u,u')-u){t,u)-s, u(a)=u(6), u'(a) = «'(&),
.
.
l
;
where s > 0 and uj(t, u) = a(t) + u)\(t,a(t) —u), if u < a(t), = u, if u > a(t). Solutions of (1.25) solve the fixed point problem u(t) = (Tu)(t) + s, where rb a
and G(t,s) is the Green's function corresponding to (1-1.4). Ckim i - Solutions u of (1.25), with s > 0, are such that a 0, are such that ||tt'||oo < R- FVom Claim 1, a < u < k. Arguing then as in Theorem 1-6.6, Claim 2 follows from Proposition 1-4.8.
154
III. RELATION WITH DEGREE THEORY
Claim 3 - deg(/ - T, fi) = 1, where O = {u € C x (h &]) I Vt G [a, b], a{t) 1 < u(i) < /3(i), |u'(i)| < -R}. Notice that a — 1 is a lower solution of (1.25) and from Claim 1, this lower solution is strict. The claim follows then from Theorem 1.8. Claim 4 - d e g ( J - r , f i i ) = 0 , where Ox = {u € frfab]) | Vt G [a,b],a(t)1 < «(i) < fc, |u'(i)| < i?}. Solutions u of (1.25) are such that
As further they are in fix, it is clear that ||w—Tudoo = s is a priori bounded. Hence for s = s<j large enough there is no solution of (1.25). It follows that deg(J - T, fiO - deg(/ - T - sOl fii) = 0. Conclusion - Prom Claim 3, there exists a solution u\ e fi. By Theorem 1-6.11, we can choose ui to be the minimal solution in O. Next, by excision, we deduce from the previous claims that deg(J-T,n1\fi) = - l , and there exists a second solution u^ S fii \ Cl. If ui ^ u\, ui and U2 are upper solution and we deduce from 1-6.10 existence of a solution u^ with a < us < min{wi,xt2} which contradicts ui to be minimal. D Notice that we cannot use a one-sided Nagumo condition such as f(ttu,v)>-il>(t)
— k < ui < u\.
Exercise 1.7. Paraphrase Theorem 1.14 for the Rayleigh and the Lienard equations (1.14) and (1.16). If / does not depend on the derivative u', Theorem 1.14 reduces to the following.
1. THE PERIODIC PROBLEM
155
Theorem 1.15. Let k > 0, a, f3 e C([a,b]) be respectively a W2'1 -lower solution and a strict W2>1 -upper solution of (1.1) such that a < 0 < k. Let E = {(t,u) G [a, 6] x R | a(t) < u} and assume f : E —> R is an Ll -Caratheodory function. Suppose moreover that for all solutions u £ W2'1 (a, b) of u"
Ul{t)
and for some t\ 6 [a, b], Example 1.4. In Example 1-1.2, we have proved that if h € C([a, b}) is such that for all t € [o, b], -1/4 < h(t) < 0, then ax{t) = - 1 and a2(t) = 0 are lower solutions and /3i(t) = —1/\/2 and /?2(£) = l/%/2 are upper solutions of u" + u — u2 = h(t), (1.26) u(a) = u(b), u'(a) = ix'( If we reinforce the condition on h assuming —1/4 < h(t) < 0 for all t e [a, b], we can use Theorem 1.15 and prove the existence of a third solution U3 such that for some t\ € [a, b], U3(t\) > l/\/2- To this end, we apply this theorem with a = 0 and j3 — l/%/2- The upper solution f3 is strict from Proposition 1.2. It remains to prove an a priori bound on the positive upper solutions of (1.26). Observe that u4 - u2 > —\ on R + . Hence, if u is an upper solutions of (1.26), u" < h(t) - (u4 - u2) < h(t) +
\<\-
Moreover mintu(t) = u(to) < 1 as otherwise u"(t) < 0 on [a,b] which contradicts the periodicity of u. Extending u and h by periodicity and as u'(to) = 0 we have, for alH £ [to, to + b — a[, rt
u(t) = u(t0) + / u"(s)(t - s) ds< 1 + §(& - a)2, Jto
which proves the required a priori bound. If we assume further —1/4 < h{t) < 0 for all t £ [a, b], the lower solution a = 0 is also strict as follows from Proposition 1.1 and we conclude, using
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III. RELATION WITH DEGREE THEORY
Theorem 1.13, that this problem has a fourth solution u^ such that for all tG[o,6], Ml < U4 < U2-
Notice at last that if h is constant, the four solutions are easy to exhibit. 2
The Dirichlet Problem
2.1
The Strict Lower and Upper Solutions
Consider the Dirichlet problem u" = f(t,u,u'),
, s [
u{a) = 0, u{b) = 0,
'
where / is an Lp-Caratheodory function. It has been seen that this problem is equivalent to the fixed point problem u(t) = (Tu)(t) := f G(t,s)f(s,u(s),u'(S))ds, Ja
(2.2)
where G(t, s) is the Green's function corresponding to «" = /(*), u(a) = 0, u(b) = 0.
[
}
In this chapter, we consider the degree of I — T for an open set O of functions u that lie between the lower and upper solutions a and (3. However, we want to allow a and /3 to satisfy the boundary conditions. In that case, the set fi = { u e $([a,b]) I IKHoo < R,Vte]a,bl
a(t) < u{t) < /?(*)}
is not open in Co([a,&]). Similarly, if / does not depend on u', we cannot use the set Q. = {ue C0([a,b}) I Vt e]o,6[, a(t) < u(t) < /?(*)}, which might not be open in Co ([a, b}). A way out is to impose some additional condition on the function at the boundary points t = a and t — b. For example, the set {u 6 C&([a,b]) I Hu'lloo < J2,Vt e]a,&[, a(t) < u(t)
D+a(a) < u'{a) < D+f3(a), D^a{b) > u'(b) > D~(3(b)}
2. THE DIRICHLET PROBLEM
157
is open in Co([a, b}). To work in a space of continuous function, we need to generalize the conditions a(t) < u(t) on ]a,b[, D+a(a) < u'(a) and D~a{b) > u'(b) for such functions. Let us first introduce the following definition. Definition 2.1. Let u, v E C([a, b]). We write u >- v or v - v(t), on }a,b[, + D+u(a) > D v(a), ifu(a) = v(a), D~u{b) < £>_u(6), ifu(b) = v(b), or, which is equivalent, if there exists e > 0 such that for any t G [a, b] u(i)-v{t)
>ee(t),
5
where e(t) := sin(7r|5^) * the eigenfunction corresponding to the first eigenvalue of the problem u" + Xu = 0,
u(a) = 0, u(b) = 0.
The next step is to define the concept of strict lower and upper solutions. Definitions 2.2. A lower solution a of (2.1) is said strict if every solution u of (2.1) with a 0 such that for any t€ [a,b] u(t) - a(t) > e 1. The connection is obvious if we notice that sin(7r|5|) and 1 are the first eigenfunctions of the corresponding eigenvalue problems u" + Xu = 0,
u(a) = 0, u(b) = 0,
and u" + \u = 0,
u(a) = u(b), u'(a) = u'(b).
More generally, a lower solution a is strict if for every solution u with a < u, there exists e > 0 such that a(t) + epi(t) < u(t), where tpi is the first eigenfunction of the corresponding eigenvalue problem. Most of the results we present for the Dirichlet problem extend to the separated boundary conditions. Such generalizations are left to the reader as exercises.
158
III. RELATION WITH DEGREE THEORY
A first result concerns lower and upper solutions which are C2. Proposition 2.1. Let / : [ o , 6 ] x R 2 - > R be continuous and a e C([a, b]) n C2{]a,b[) be such that (a) for all t e}a,b[, a"(t) > f(t,a(t),a'(t)); (b) for t0 e {a, b}, either a(t0) < 0 ora(t0) = 0, aeC 2 (]a,6[U{i 0 }), and a"(t0) > f(to,a(to),a'(to)). Then a is a strict C2'-lower solution of (2.1). Proof. From the assumptions, a is a C2-lower solution of (2.1). Let then u be a solution of (2.1) such that a
«(<„)-a(t B )
It follows there exists a subsequence of (£„)„ that converges to a point t0 such that u(to) = a(to)- If io G ]a, b[, we have u'(io) = a'(to)- On the other hand, if to — a, we know that a £ C2([a, b[). Further, we deduce from (2.4) that u(tn) - u(a) < a(tn) + ^ sin(Tr^) - a(a) tfi
a
tyi
Ot
This result implies that u'(a) < a'(a). As further u—a is minimum at t = a, we also have u'{a) > a'(a). Hence, u'(a) = a'(a). A similar reasoning applies if t0 = & so that in all cases u'(t0) — a'(t0) = 0. At last, we obtain the contradiction 0 < u"(to)-a"(to) = /(to, a(t 0 ), a'(tQ))-a"(t0) < 0. In a similar way, we can write the following. Proposition 2.2. Let / : [ a , 5 ] x I 2 - > R be continuous and f3 € C([a, b}) D C2(]a,b[) be such that (a) for all t €}a,b{, 0"{t) < f(t,0(t),0'(t)); (b) for t0 G {a, b}, either 0(to) > 0 or 0(tQ) = 0, /3 e C2Qa,b[U{t0}) and f3"{t0) < f(to,/3(to),l3'(to)). Then 0 is a strict C2-upper solution of (2.1). Notice that an upper solution (3 such that 13"{t) < f(t, 0(t),P(t)) on ]a, b[, (3(a) > 0,
0(b) > 0
is not necessarily strict. Consider for example the problem (2.1) defined on [a, b) = [0,1] with /(*,«,«') = 0, = 7u/t2, = 7t,
ifu<0, ifO
2. THE DIRICHLET PROBLEM
159
The function u(t) = 0 is a solution and (5{t) — t3 > u(t) is an upper solution which is not strict (as 0(0) = w(0) and /3'(0) = u'(Q)) but satisfies 0"(t) < f(t,P(t),(3'(t)) on ]0,l], 0(0) = 0 and (3(1) > 0. In the Caratheodory case, we can use the following propositions. Proposition 2.3. Let f : [a,b]xR2 —>R be an Ll-Caratheodory function. Assume that a £ C([a,b}) is not a solution of (2.1) and that (a) for all t0 £ ]a, b[, either £>_a(£0) < D+a(to) or there exist an open interval Io C [a, b] and e > 0 such that tQ £ IQ, a e W 2il (/ 0 ) and for a.e. t£ Io, all u £ [a(t),a(t) + esin(7r|5^)] and alive [a'(t)-e,a'(t) + e], a"(t)>f(t}u,v); (b) either a(a) < 0 or a(a) — 0 and there exist e > 0 such that a £ W2>1(a,a + e) and for a.e. t e [a,a + e], all u G [a(t),a(t) + esin(Tr^r)] and all v £ [a'(t)-e,a'(t) + e), a"(t)>f(t,u,v); (c) either a(b) < 0 or a(b) = 0 and there exist e > 0 such that a £ W2tl(b — e,b) and for a.e. t £ [b — e,b], all u £ [a(t),a(t) + esin(7r|52-)] and all v £ [ ' ( ) ' ( ) ] a"(t)>f(t,u,v). 21 Then a is a strict W ' -lower solution of (2.1). Proof. Notice first that a satisfies Definition II-2.1 and therefore is a W2'1lower solution. Let u be a solution of (2.1) such that u > a. Arguing by contradiction as in Proposition 2.1 there exists a sequence (£„)„ that satisfies (2.4) and that converges to a point io such that u(£o) = a(£o) and u'(to) = a'(to). As a is not a solution, we can find t* such that u(t*) > a(t*). Assume io < t* and define t\ = max{£ < t* \ u(t) = a(t)}. We prove then that u(ti) = ct(ti), u'(t\) = a'(ti) and fix e > 0 from the assumptions. Next, for t > t\ near enough t\, u(t) £\a(t),a(t)+esin(n£t)]
and u'(t) £ [a'(t) - e,a'(t) + e].
Hence, we compute u'(t) - a'(t) = / [f(s, u(s), u'(s)) - a"(s)] ds<0 for t >ti, near enough t\ which contradicts the definition of t\. A similar argument holds if to > t*. D
160
III. RELATION WITH DEGREE THEORY
Strict W2il-upper solutions can be obtained from a similar proposition. Proposition 2.4. Let / : [ a , 6 ] x l 2 - » R be an L1-Caratheodory function. Assume that ft G C([a,6]) is not a solution of (2.1) and that (a) for all tQ G }a, b{, either D~ft(t0) > D+(3(to) or there exist an open interval 1$ C [a, b] and e > 0 such that to G IQ, ft G W2'1^) and for a.e. t € Io, all u e [ft(t) - esin(7r|5f ),ft(t)} and alive [ft'(t) -e,F(t) + e}, ft"(t) < f(t,u,v); (b) either ft(a) > 0 or ft(a) = 0 and there exist e > 0 such that ft G W2'1(a,a + e) and for a.e. t G [a,a + e], all u G \fl{t) - esin^f^2-),/?(£)] and all v G [ ' ( ) ' ( )
]
P"(t)
or
ft"(t)
with A > 0, are strict lower or upper solution.
2. THE DIRICHLET PROBLEM
161
Proposition 2.5. Let f : [a, 6] x R2 —> R be an L1 -Caratheodory function that satisfies the assumption (B) for all to £ [a,b], (uo,«o) £ R2 ond e > 0, £/iere exists 5 > 0 such that \t — to\ < 5, |ui - u o | < (5, U2 6 [«i,«i +^ s in( 7r |rt)[> |ui - « o | <<5, |u2-vo| <<5 => f(t,u2,v2) - f{t,ui, vi) < e. Let A > 0 and a G W 2il (a, b) be such that a"(t)>f(t,a(t),a'(t)) + A, a(a) < 0, a{b) < 0. Then a is a strict lower solution of (2.1). Proof. The proof follows as for Proposition 1.5. Notice that Condition (B) holds if f(t,u,v) = g{u,v) + h(t) with g continuous and h € L1(a, b) or if f(t,u,v) = g(u,v)h(t) with R be an L1 -Caratheodory function that satisfies Condition (B) of Proposition 2.5. Let B > 0 and f3 € W2'l{a,b) be such that /?(a) > 0, 0(b) > 0. Then {3 is a strict upper solutions of (2.1). Notice that it is easy to generalize these results using lower and upper solutions with corners. As for the periodic problem, we can study cases where / satisfies a Lipschitz condition in v and a one-sided Lipschitz condition in u. Proposition 2.7. Let f : [a, 6] x R2 —> R be an L1-Caratheodory function such that for some k, I e Ll{a, 6;R + ), (a) for a.e. t € [a, b], all u\, v,2 €M. with ui < u% and » e R , f(t,u2,v)
- f(t,ui,v)
< k(t) (u2 - ui);
(b) for a.e. t £ [a, b] and all u, v\, v2 G R, \f(t,u,v2) - f(t,u,vi)\
< l(t)\v2 -
Vl\.
Let a (resp. ft) be a lower (resp. upper) solution of (2.1) which is not a solution and assume
162
III. RELATION WITH D E G R E E THEORY
(c) either a(a) < 0 (resp. (3(a) > 0) or a(a) = 0 (resp. 0(a) = 0) and there exists an interval IQ = [a, c[c [a, 6] such that a G W2'l(IQ) (resp. 0 G W2
(resp. 0"(t) <
f(t,0(t),0'(t)));
(d) either a(b) < 0 (resp. 0(b) > 0) or a(b) = 0 (resp. 0(b) = 0) and there exists an interval Jo =]c, b] C [a, 6] SKC/I i/iat a e W 2 ' 1 ^ ) fresp. ,8 G W2ll(-^o)j a"d/or a.e. t G J o , a"(f) > /(«,a(t),a'(*))
(resp. /?"(*) < /(*,/?(*),/3'(t))).
T/ien Q: fresp. /3j is a strict lower solution (resp. a strict upper solution) of (2.1). Proof. Let u be a solution of (2.1) such that u > a. We argue as in Proposition 2.1 to exhibit a point to G [a, b] such that u(to) = &(ta) and «'(*o) = a'(to)- Next, we conclude as in Proposition 1.7. Remark 2.1. As for periodic problem, we can see from the problem (1.7) that this proposition does not hold without the Lipschitz conditions on / . We can however assume (a) and (b) to hold only in a neighbourhood of {(t,a(t),a'(t)) 11 G [a,b] such that a'(t) exists} (resp. {(t,f3(t),/3'(t)) 11 G [a,b] such that (3'(t) exists}). Exercise 2 . 1 . Assume / satisfies the assumptions of Proposition 2.7 and a, /3 > a are VF2>1-lower and upper solutions of (2.1). Assume further a or /3 is not a solution and satisfies conditions (c) and (d) of Proposition 2.7. Prove then that a -< 0. 2.2
Existence and Multiplicity Results
The main result of this section relates strict lower and upper solutions with degree theory. Theorem 2.8. Let a and 0 G C([a,6]) be strict lower and upper solutions of the problem (2.1) such that a -<, 0. Define A C [a, 6] (resp. B c [a, b}) to be the set of points where a (resp. 0) is derivable. Let E be defined by
b—a and let p, q G [1, oo] be such that i + i = 1.
b- a
2. THE DIRICHLET PROBLEM
163
Assume f : E —» R satisfies Lp -Caratheodory conditions and there exists N € Lx(a, b), N > 0 such that for a.e. t E A (resp. for a.e. t € B) f(t,a(t),a'(t))
> -N(t)
{resp. f(t,p(t),/3'(t))
< N(t)).
Assume moreover there exist tp e C(R+,Ro"), ip G Lp(a,b) and R> r such that —T-T- ds > ||i/)||ip(max/3(t) - mina(£)) 1/9 t
ip(s)
t
and that the function f satisfies for a.e. t € [a,6] and all (u,v) with (t,u,v)eE, Then where T : C%([a, b\) -> C$([a, b}) is defined by (T«)(t):= / G(t,s)f(s,u(s),u'(s))ds, Ja with G(t, s) the Green's function corresponding to (2.3) and fl = { « e C%([a,b]) \a^u<(3,
IKH^ < R}.
(2.5)
In particular, the problem (2.1) has at least one solution u e W2'v(a,b) such that Proof. Increasing N if necessary, we can assume N(t) > \f(t,u,v)\ if t € [a,6], a(t)
max \f(t, /3(t),$'(t) +v)- f(t, P(t),0'(t))\, \v\<5
where 7 is defined from (1-1.3), XA and XB a r e the characteristic functions of the sets A and B. We consider now the modified problem u" =f(t,u,u')-u(t,u), u{a) = 0, u{b) = 0, where = 0, = u>i(t,a(t) — u),
Ua(t)
, > K
}
164
III. RELATION WITH DEGREE THEORY
This problem is equivalent to the fixed point problem u = Tu, where T : C<J([a, &]) - » C^([a, b}) is defined by
(Tu)(t)= f
G(t,s)[f{s,u(s),u'(s))-u;(s,u(S))]ds.
Ja
Observe that T is completely continuous and we can find R large enough so that fi C B(0,R) and T(C&([a,b})) C B(0,R). Hence we have, by the properties of the degree, deg(I-T,B(O,R)) = l. We know that every fixed point of f is a solution of (2.6) and arguing as in Step 3 and 4 of the proof of Theorem II-2.1 we prove a < u < (3 and llu'Hoo < R. As a and /3 are strict, a -< u -< /3. Hence, every fixed point of T is in 0 and by the excision property we obtain deg(J -T,Q) = deg(J - f,Q) = deg(J - f,5(0,-R))
= 1.
Existence of a solution u such that a < u -< /3 follows now from the properties of the degree. Exercise 2.2. Consider Theorem 2.8 with a one-sided Nagumo condition. Exercise 2.3. Extend Theorem 2.8 to the separated boundary value problem u" = f(t,u,u'), a\u(a) — a.2u'(a) = AQ, b1u(b)+b2u'(b) = B0, in case AQ, BO e R, oi, 6i G R, a2, 62 € R + , of + a| > 0 and 6f + 63 > 0. Hint : See [31]. In the case where / does not depend on the derivative u', this result reduces to the following theorem.
2. THE DIRICHLET PROBLEM
165
Theorem 2.9. Let a and (3 £ C([a,b]) be strict W2>1-lower and upper solutions of the problem «" = /(*.«), u(a) = 0, u(6) = 0, SMC/I i/iat a -< /3. De/ine J5 : = {(t,u)
€ [a, 6] x R | a(t)
/?(*)} and
assume f : 2? —> R is an Ll -Garatheodory function. Then, for R > 0 Za7i0e enough, d e g ( / - r , n ) = i, w/iere fi is de/med 6y (2.5), T : C&([a,b]) -* C^{[a,b]) by rb
(Tu)(t):= f
G(t,s)f(s,u(s))ds
Ja
and G(t,s) is the Green's function corresponding to the problem (2.3). In particular, the problem (2.7) has at least one solution u e W2>1(a, 6) such that a -'A(a, b) and it is not meaningful to define the corresponding fixed point problem on ^([a, 6]). This remark forces us to make stronger assumptions on strict lower and upper solutions. Theorem 2.10. Let a and (3 € C([a, b]) be strict W2<1 -lower and upper solutions of the problem (2.7) such that a(a) < 0 < /3(a),
a(b) < 0 < /3(b)
and
Define E := {(t,u) G [a, 6] x M | a(t)
Vt € ]a, b[, a{t) < /?(*).
p(t)} and assume f : E -> R
deg(/-T,n) = l, where 0 = {u e C0([a,6]) | Vt € [a, 6], a(t) < u(t) < /?(*)},
166
III. RELATION WITH DEGREE THEORY
T : C0([a,b}) -> C0([a,b}) is defined by fb
(Tu)(t):= /
G(t,s)f(s,u(s))ds
and G(t,s) is the Green's function corresponding to the problem (2.3). In particular, the problem (2.7) has at least one solution u G. W2'-A(a, b) such that for all t £ [a, b], a{t) < u(t) < /3(i). Exercise 2.4. Prove Theorem 2.10. The simplest multiplicity result that we can deduce from Theorem 2.8 is obtained when we have two pairs of strict lower and upper solutions. Theorem 2.11. (The Three Solutions Theorem) Let cti, ft and a2, 02 G C([a,6]) be two pairs of W2'1-lower and upper solutions of (2.1) such that on [a,b] ai(t) < 0i(t), ai(t) < fa(t), a2(t) < and there exists to 6 [a, 6] with > Pi(to). Assume further ft anda2 are strict W2'1-upper and lower solutions. Define Ai C [a, b] (resp. Bi C [a, b}) to be the set of points where a>i (resp. Pi) is derivable. Let E be defined by (1.17), p, Q £ [1, oo] be such that i + - = 1 and r = m a x { f t ^ I ° ' < f c ) , Mb)bZT{a)}Suppose f : E -> R is an U>-Caratheodory function and there exists N £ L1(a,b), N > 0 such that for i — 1,2 and
a.e. ( 6 Aj fresp. a.e. t £ Bi) > (i
f \' I I I
f \ (I I J J ^ — /V 1 / 1
I "/"V Q T l
i l l
(i 'i I I
/ * i f f I 1 ""^ / v f / I 1
Assume moreover there exist ip £ C(M+,Ro"), %j> £ Lp(a,b) and R> r such that
jf
ds > \\il)\\iJp(xnax.P2(i) — m i n a i ( t ) ) 1 ' 9
holds and for a.e. t 6 [a, b] and all (u,v) with (t,u,v) S E, \f(t,utv)\
2. THE DIRICHLET PROBLEM
167
and there exist t\, t2 € [a,b] with u3(ti) > /?i(ti), us(t2) < a2(t2). Exercise 2.5. Prove the preceding result adapting the argument of Theorem 1.13. As in the periodic case, another way to obtain multiplicity result is to exhibit domains fii D O such that deg(/-T,Oi) = 0 and deg(7 - T,Q) = 1. Such a result can be obtained if we assume that we have an a priori bound on all the upper solutions which satisfies the boundary conditions. Here we assume / does not depend on the derivative u'. Theorem 2.12. Let k > 0 and a, j3 € C([a, b)) be respectively a W2'1 -lower solution and a strict W2>1 -upper solution of (2.7) with a < /3 < k. Let E = {(t,u) 6 [ o , t ] x R | a(t) < u} and f : E -> R be an L1Caratheodory function. Assume moreover that for all solutions u € W2>1(a, b) of u{a) = 0, u{b) = 0, with u > a we have u
u(a) = 0, u(b) = 0, where ji(t,u) problem
= max.{a(t),u}.
Solutions of (2.8) solve the fixed point
where
(Tu)(t) = f G(t,s)f(t,'y1(t,u))ds Ja
168
III. RELATION WITH DEGREE THEORY
and G(t, s) is the Green's function corresponding to (II-4.2). As in the proof of Theorem 1.14, we prove that solutions of (2.8) with is a priori s > 0 are such that a < u < k. Hence \\Tu — u||oo = s j j bounded, which implies that for s = SQ large enough, (2.8) has no solution. Now we deduce from the equation that there exists some R > 0 so that solutions u of (2.8) with 0 < s < SQ satisfy ||u'||oo < R- The rest of the proof follows as in Theorem 1.14. Again we can consider the case where / satisfies ,4-Caratheodory conditions. Theorem 2.13. Let k > 0 and a, j3 € C([a, b}) be respectively a W2'1-lower solution and a strict W2'1-upper solution of (2.7) with a < 0 < k, /3(a) > 0 and (3(b) > 0. Let E = {{t,u) G [a,b] x R | a(t) < u} and f : E -+ R be an ACaratheodory function. Assume moreover that for all solutions u £ W2<1(a,b) of u"
< ft(t), ui(t) < U2(t)
and there exists to £ [a, b) with U2(to) > (3(t0). Exercise 2.6. Prove the above theorem.
3
Non Well-ordered Lower and Upper Solutions
3.1 Introduction We noticed already in Chapter I that the method of lower and upper solutions depends strongly on the ordering a < (3. Consider for example the periodic problem u + u = smt, CQ 1 "\ u(0) = U{2IT), U'(0) = u'(2ir).
K
}
Clearly this problem has no solution although a(t) = 1 is a lower solution and (3(t) = — 1 < a(t) is an upper solution. On the other hand, lower and
3. THE NON WELL-ORDERED CASE
169
upper solutions a, (3 satisfying the reversed ordering condition (3 < a arise naturally in situations where the corresponding problem has a solution. As a very simple example, we can consider the linear problem u(0) =
u" + §u = sint, U(2TT), U'(0) = u'(2vr).
The functions a = § and /? = — § are lower and upper solutions such that a > 0. Notice, however, that the unique solution u(t) — —3sin£ does not lie between the lower and the upper solution. More generally, we can consider the problem u(0) = «(2TT), «'(0)
== u'(2ir),
K
'
where A lies between the two first eigenvalues of the corresponding periodic problem, i.e. 0 < A < 1. It is easy to see that, in such a case, there exist constant lower and upper solutions of (3.2) which are in the reversed order. Further, the solution u of (3.2) lies between all constant upper and lower solution, /? < u < a, if and only if 0 < A < h. We shall see in Section 3.5 that if we reinforce this condition, i.e. A < | , such a result holds true for any problem u" + Xu = f(t), u(0) = u(2n), u'(0) = u'(2ir), where / is continuous. More precisely, the solution lies between every pair of lower and upper solutions with a > 0. This limitation is optimal since we can find, for A > | , forcings / and lower and upper solutions a > /3 such that the corresponding solution does not lie between /3 and a. The use of the method of lower and upper solutions is strongly related to the interaction of the nonlinearity with the eigenvalues of the corresponding linear problem. Consider the Dirichlet problem u" + / ( t , u ) = 0 , u(a) = 0, u(b) = 0,
,„„, '
K
together with lower and upper solutions a and /3 such that a"{t) + f{t, a(t)) > 0, 0"(t) + f(t, /?(*)) < 0,
a(a) < 0, a(b) < 0, /3(o) > 0, (3(b) > 0.
Let us assume a < /? and ) > A. u—v
(3.4)
170
III. RELATION WITH DEGREE THEORY
It follows that -(/?"(*) - a"{t)) > f(t,(3(t)) - /(*,<*(*)) > A(/3(t) - a(t)) and integrating twice by part, we obtain 2
A / 03(t)-a(i))mn(7rt=a)dt< ( ^ ) Jo
^
/
/ (£(*) - a(t)) sin ( * £ * ) * . Ja
If furthermore A > Ai, where Ai = (jr^) 2 is the first eigenvalue of the Dirichlet problem, then this inequality implies a = f3 which is then a solution of (3.3). This observation means that it is not possible to work the method of lower and upper solutions, with a < /?, if the nonlinearity satisfies (3.4) with A > Aj, i.e. if it is "above the first eigenvalue". This is a fundamental limitation of the method. A dual conclusion holds if we assume a and (3 satisfy the reversed order a > (3 and ) < A. (3.5) u—v In this case, we deduce a(a) = (3{a) = 0, a(b) = /3(6) = 0, -(a"(t) - P"(t)) < f(t, a(t)) - f(t, 0(t)) < X(a(t) and
f
rb
Ja
rb
(a1 - /3')2 ds-X
I (a- (3)2 ds
= - f (a- 13)" {a -0)ds-\ Ja
rb
f (a- f3)2 ds < 0.
Ja
If A < Ai, the above inequality implies a = f3 and this function is a solution of (3.3). It follows now that it is not possible to work a method of lower and upper solutions, with (3 < a, if the nonlinearity satisfies (3.5) with A < Xi, i.e. if it is "under the first eigenvalue". It is easy to see that the same conclusions hold for other boundary value problems such as the periodic one u" + f(t,u)=0, u(0) = u(27r), u'(0) = U'(2TT). Here the first eigenvalue is Ai = 0 and the method does not work with a < /?, if the nonlinearity satisfies (3.4) with A > 0 or with /3 < a, if the nonlinearity satisfies (3.5) with A < 0. For the periodic problem we can complete this remark. If we assume the lower and upper solutions, a and (3, to be not "well-ordered", i. e. a(io) > /3(*o) for some t0 £ [0,2TT], and u-v
3. THE NON WELL-ORDERED CASE
171
we can prove a > /3. Assume by contradiction a(ti) < fl(ti) for some t\. Extending u := a — /? by periodicity, we define then Si := inf{t < to \ u > 0 on [t, to]} and S2 := sup{t > to | u > 0 on [to, t]} < s\ + 2ix. The function u satisfies the Dirichlet problem u" + q(t)u = a{t), u(si) = 0, u(s3) = 0, where q{t) < | and a > 0. It is easy to see that the first eigenvalue A of the problem u" + q(t)u + Xu = 0, u(si) = 0, u(s2) = 0, is positive. It follows then from the maximum principle (Theorem A-5.2) that u < 0 on ]si, S2[ which is a contradiction. The existence of lower an upper solutions a, /3 such that a > /? is not sufficient to guarantee the solvability of this problem which is clear from Problem (3.1). The difficulty there comes from the interference of the nonlinearity with the second eigenvalue of the problem A2 = 1. In the theorems we present in this section, we will put assumptions so that the nonlinearity remains "below this second eigenvalue". 3.2
A Periodic Non-resonance Problem
Consider the problem u" + Au = /(i,u,u'), B(u(a),u(b),u'(a),u'(b)) = 0, where / is bounded and B represents the boundary conditions. This problem can be thought of as a perturbation of the linear eigenvalue problem u" + Xu = 0, B{u{a),u{b),u'(a),u'(b)) = 0. In this section, we shall apply the method of lower and upper solutions to such problems when A is the first eigenvalue. Our first result concerns the periodic problem u" = f(t,u,u'), u(o) = u(6), «'(a)=u'(6)
, „,
{
}
with one-sided boundedness of the nonlinearity. Here the first eigenvalue is Ai = 0.
172
III. RELATION WITH DEGREE THEORY
Theorem 3.1. Let a and /3 G C([a, b}) be W2<1 -lower and upper solutions of (3.6) such that a /3. Define A C [a, b) (resp. B C [a, b}) to be the set of points where a (resp. 0) is derivable. Assume f : [a, b] x R 2 —> R is an L1 -Caratheodory function and there exists N G Ll{a, b), N > 0 such that for a.e. t G A (resp. a.e. t G B) f(t,a(t),a'(t))
> -N(t)
(resp. f(t,/?(t),/?(*))
< #(*))
Assume further that for some h G L 1 (a, b) either f(t,u,v)
or f(t, u, v) > h(t) on [a, b] x R2. Then there exists a solution u of (3.6) in S:={u€
C H M ) | 3*i,*a G [a,b], u(h) > /3(tx), u{t2) < a(t2)}.
(3.7)
Proof. For each r > 2(6 — a) 2 , we define / r (t, u, v) = f(t, u, v), = (1 + r - \u\)f(t, u, v) + (|u| - r)», = 7,
if |w| < r, if r < |u| < r + 1,
and consider the problem u" = fr(t,u,u'), u(a) = u{b), u'(a) = u'(b).
, {
, }
Claim - There exists k > 0 SMC/I tftaf for any r > 2(b — a)2, solutions u of (3.8), which are in S, are such that ||u||oo < ^- Consider the case f(t,u,v) < h(t) on [a, b] x R 2 . Let u £ S be a solution of (3.8) and let t0, t\ and £2 € [a, 6] be such that tt'(to) = 0,
u(ti) > /?(*!) > —H^Hoo
and
u{t2) < a(t2) < ||a||oo.
Extending u by periodicity, we can write for t £ [to, to + 6 — a] rto+b—a
>
-\\h\\Ll
It follows that for t G [ti,t! + & — a]
u(t) = u(h) + f u'{s) ds > -||/3|U - ||h||L1 (b - a) - M
3. THE NON WELL-ORDERED CASE
173
and for t £ [£2 — b + a, £2] u{t) = u(t a ) -
2
Hence, we have ||«||oo < 2(101100 + ||/3||oc + ||ft|Ui(6 - a)) =: k. A similar argument holds if f(t,u,v)
> h(t).
Conclusion - Consider the problem (3.8), with r > max{fc, 2(b — a) 2 }. It is easy to see that ot\ — —r — 2 and /?2 = r + 2 are lower and upper solutions. Assume j3 is not a strict upper solution. There exists then a solution u of (3.8) such that u < f3 and for some t\ e [a, b\, u(t\) = /3(ti). As further a ^ /3, there exists (2 G [ai^] such that 01(^2) > Pfa)- It follows that a(*2) > ^(£2), u & S, and we deduce from the claim that ||u||oo < k. Hence, u is a solution of (3.6) in 5. We come to the same conclusion if a is not a strict lower solution. Suppose now that /?i = /? and a 2 = a are strict upper and lower solutions. We deduce then from Theorem 1.13 the existence of three solutions of (3.8) one of them, u, being such that for some t\, t /?(£i) and u{ti) < a{t2). Hence, u £ S and from the claim ||u||oo < k. This implies that u solves (3.6) and proves the theorem. Example 3.1. The problem
where f(u) = 12|u|1/2, if u<4, = 28 - u, if 4 < u < 28, = 0, if 28 < u, is such that 0(t) = i 4 and a(t) — 28 > /3(t) are respectively upper and lower solutions. The solutions in the corresponding set S are the constant functions 0 and 28 but are not in the interior of «S. A possible generalization concerns the use of several lower and upper solutions. Exercise 3.1. Let a» 6 C([a,b\) (i - l , . . . , n ) and fa £ C([a,b}) (j = 1 , . . . , m) be respectively W2'1-lower and upper solutions of (3.6) such that a := max a. ^ 3 := min BA,
174
III. RELATION W I T H D E G R E E THEORY
Assume / : [a, 6] x K2 —> R is an Caratheodory function such that for some h £ L 1 (o, 6) either on[a,6]xR2,
f(t,u,v)
< h(t)
f(t,u,v)
> h{t) on [a, 6] x R 2 .
or
Prove there exists a solution u £ <S of (3.6), where S is defined in (3.7). In a similar way, we can consider the Lienard equation u" + g{u)u' + f(t, u) = 0, u(a) = u(b), u'{a)=u'(b).
, {
. '
Theorem 3.2. Let a and /? £ C([a, b}) be W2<1-lower and upper solutions of (3.9) such that a (3. Assume g £ C(R) and f : [a, 6] x R —> R is an L1 -Caratheodory function such that for some h £ L1(a, b) either f{t, u) < h(t) on [a, b] x R, or f(t,u)
>h(t)
on [a,b] x R.
Then there exists a solution u £ S of (3.9), where S is defined in (3.7). Proof. For each r > 8(6 — a) 2 , we define
fr(t,u) = f(t,u), if \u\
Claim - There exists k > 0 suc/i i/iai /or any r > 8(6 — a) 2 , solutions u of (3.10), which are in S, are such that ||u||c» < k. Consider the case f(t,u) > h(t) on [a, 6] x R. Let u £ 5 be a solution of (3.10). Integrating (3.10), or multiplying this equation by u and integrating, we obtain respectively
/ fr{t,u(t))dt Ja
=0
and
/ u'2{t)dt= Ja
J Jo,
fr(t,u{t))u(t)dt.
3. THE NON WELL-ORDERED CASE
175
It follows that rb
«'ll£»= / J
fr(t,u(t))(u(t)-\\u\\00)dt<2(\\h\\Lr+^(b-a))\\U\\00,
Define now ti and £2 G [o, &] to be such that u(h) > /?(tO > —H/^Hoo
and
u(t a ) < a ( t 2 ) < ||a||oo-
Extending u by periodicity, we can write for t £ [ii, t\ + b — o]
u(t) = u(ti) + J u'(s) ds > -H/3IU - IKIM6 - a)1/2
and for £ E [*2 — & + a, ^2] «(*) = «(*2) - / h
2
u'{s) ds < Halloo + (2(6 - a)||ft||Li \\u\Ufl2 + M a .
Hence, we obtain for some k > 0 Moo < k. A similar argument holds if f(t,u) < h(t). Conclusion - The rest of the proof repeats the argument used in Theorem 3.1. 3.3
Interaction with the First Fucik Curve
The reason of imposing in Theorem 3.1 a boundedness assumption, such as f(t,u,v) < h(t), is to make sure the nonlinearity does not interfere with the second eigenvalue of the corresponding linear problem. An alternative way to prevent such an interference is to assume some asymptotic control on the quotient f(t, it, v)/u as |u| goes to infinity. We can generalize further assuming different behaviours of this quotient as it goes to plus or minus infinity. The following theorem concerns such a generalization. To simplify, we consider the problem u" + / ( t , u ) = 0 , u{a) = u(b), u'(a) = u'(fe).
,
[
s '
176
III. RELATION WITH DEGREE THEORY
Theorem 3.3. Let a and /3 6 C([a,6]) be W2'1 -lower and upper solutions 0/(3.11) such that a £(3. Let / : [a, b] x R —> E be an L1 -Caratheodory function such that for some functions a < 0, 6 > 0 in L1(a, b), < liminf ^ ^ < limsup ^ ^ < b {t), U—
U
u—
(3.12)
U
uniformly in t 6 [a, 6]. Assume further that for any p, q s L1(a, 6), wit/i a+ < p < 6+ and a_ < q < 6_, i/ie nontrivial solutions of u"+p(t)u+-q(t)u=0, u(a) = u(6), «'(o) = u'(6),
, . ^ °;
where u+ {t) = max{u(t), 0} andu~(i) = max{—u(i),0}, do not have zeros. Then the problem (3.11) has at least one solution u £ S, where S is defined in (3.7). Proof. Step 1 - Claim : There exists e> 0 so that for any p, q G L1(a, b), with a+ — e < p
un -» u in C1([a,b}),
tn-*tQ.
It follows that a+ < p < b+, a_ < q < 6_, u is a solution of (3.13) and u(to) = 0 which contradicts the assumption. S'iep 3 - The modified problem. Let us choose R > 0 large enough so that a+ - e < g + (t, u) = ILLHI
= -^f'^ <6_+e,
for u > R, for u < - f l
and extend these functions on [a,fe] x l so that these inequalities remain valid. As / is L1-Caratheodory, there exists k € L1(a,b) such that f(t, u) = g+(t, u)u+ - g- (t, u)u~ + h(t, u) and \h(t,u)\
3. THE NON WELL-ORDERED CASE
177
Next, for each r > 1, we define
= (1 + r — = 0, hr(t,u) = h(t,u), = (l+r-\u\)h(t,u), = 0,
if |u| < r, if r < |u| < r + 1 , ifr + l < | u | , if |u| < r, if r < |u| < r + l, if r + 1 < \u
and consider the modified problem u" + g+(t,u)u+ - g-(t,u)u- + hr{t,u) = 0, u(o) = u(b), u'{a) = u'(b). Step 3 - Claim : There exists k > 0 such that, for any r > k, solutions u of (3.14), which are in S, are such that ||u||oo < k. Assume by contradiction, there exist sequences (rn)n and (un)n C S, where rn > n and un is a solution of (3.14) with r = rn such that ||un||oo ^ n- As un € <S, there exist sequences (£i n ) n and (t2n)n C \a,b] such that un(t\n) > /3(tin) and Un(t2n) < a(t2n)Consider now the functions vn = wn/||un||oo which solve the problems < + g+n(t, un)v+ - g~n (t, un)v f vn(a) = vn(b), v'n(a) = v'n(b).
= 0,
Going to subsequence, we can assume as above g+n (;un) - p , g~n (;un) - q in
0 in L\a,b),
It follows that v satisfies (3.13) and by assumption has no zeros. Hence, we come to a contradiction since v(ti) > 0 and vitz) < 0 which implies v has a zero. Conclusion - We deduce from Theorem 3.1 that problem (3.14) with r > maxfUalloo, ||/?||oo>fe} has a solution u£ S and conclude from Step 3. D In this theorem we control asymptotically the nonlinearity using the functions , . Next we impose some admissibility condition on the box [a+,&+] x [a_,6_] which is to assume that for any (p,q) S [a+,6+] x [a_,6_], the nontrivial solutions of problem (3.13) do not have zeros. Such a condition implies that the nonlinearity does not interfere with the second eigenvalue A2 of the periodic problem, i.e. (A2,A2) £ [o+,6+] x [a_,6_].
178
III. RELATION WITH DEGREE THEORY
This remark can be completed considering the second curve of the Fucik spectrum. The Fucik spectrum is the set T of points (/i, v) e M2 such that the problem u" + y,u+ — vu~ = 0, u(a) — u(b), u'{a) = u'(b), has nontrivial solutions. From explicit computations of the solution it is easy to see that n=l
where and
The following proposition relates the admissibility of the box [a+,&+] x [a_,6_] with the Fucik spectrum. Proposition 3.4. Let (n,v) G F% and p, q € L1(a, 6). Assume that p{t) < M> i(t) < vi
for a.e. t € [a,b]
and for some set I C [a, b] of positive measure p(t) < n,
q(t) < v,
for a.e. t£ I.
Then the nontrivial solutions of problem (3.13) have no zeros. Proof. Assume there exists a nontrivial solution u which has a zero. Extend u by periodicity and let to and t\ be consecutive zeros such that u is positive on ]*o»*i[- Define v(t) = sin(y/ju(i — to)) and compute (uv' - vu')
= / to
(p(t) - n)u(t)v(t) dt.
Jta
If t\ —to < -j=, we come to a contradiction 0 < -u(ti)u'(ii) < 0. Hence t\ — to > -j= and we only have equality in case p(t) = \i on [io,ti]. Similarly, we prove the distance between two consecutive zeros t\ and t -y^ with equality if and only if q(t) = v on [ti, t^. It follows that 7T
7T
b — a > to — to > —= H—:= = 6 — a.
3. THE NON WELL-ORDERED CASE
179
These inequalities imply that £2 — *o = b—a, p(t) =/jon [i0, £1] and q(t) = v on [ti, ^2] which contradicts the assumptions. Remark Assumption (3.12) can be replaced by the following: For every e > 0, there is jE € Lx(a, b) such that, for a.e. t G [a, b] and every ueR, f(t,u) = g+(t,u)u+ - g-(t,u)u~ +h(t,u), with
and h, L1-Caratheodory functions satisfying +e
and There is a vast literature concerning system which are asymptotic to positively homogeneous problem as in the above theorems. Most of these results can be adapted to a situation where there exist non well-ordered lower and upper solutions. We present here some extensions as exercises. Exercise 3.2. Let a and /? G C([a,b}) be W2'1-lower and upper solutions o/(3.11) such that a £ (3. Let / : [a, b] x R —> E be an L1-Caratheodory function such that for some functions a < 0, 6_ > 0 in L1(a, b), a-(t) < liminf ^ ^ - < limsup ^ ^ < 6_(t), a+(t)< liminf ^ ^ , uniformly int£ [a,b]. Assume further that for any q e L1 (a, b), withq
andanyie [a,b[,
has only the trivial solution (where q(t) — q(t — (b — a)) fort £]fe,i+b — a]). Prove then that the problem (3.11) has at least one solution u e S, where S is defined by (3.7). Hint : See [95].
180
III. RELATION WITH DEGREE THEORY
Consider the problem u"+g(u) = h(t), u(a) = u(b), u'{a) = u'(b).
^
'
For such a problem, we can obtain similar result using conditions on the potential G{u) = f£ g(s)ds. In this case lower bounds on 2 ^ are not necessary anymore but we cannot prove the solution is in S. The following problem concerns such a result. Exercise 3.3. Let a and f3 G C([a, b]) be W2>1-lower and upper solutions of (3.15). Let h G L°°(a, b), g G C(K) be such that for some (/i, v) G T2 g(u) , ,. q(u) lim sup -1-L < fx and lim sup -^-^ < v, u->+oo
U
«-+-oo
U
.. . ,2G(u) ,. . 2G(u) lim mi 5—- < fi or lim inft —5-^- < v. u—*+oo
u
u—
u
Prove then that the problem (3.15) has at least one solution. Hint : In case a > /3 see [147]. The proof without this condition can be built using the ideas in [79]. 3.4
Multiplicity Results
Existence of several solutions are studied in Sections 1.2 and 2.2. Some additional results can be obtained using non well-ordered lower and upper solutions. The following result complements Theorem 3.3. Theorem 3.5. Let a\ and a2 € C([a, b]) be W2'l-lower solutions of (3.11) and fi G C([a, b]) be a strict W2'1-upper solution such that a.\ < fi and
a2£P.
Assume f : [a, 6] x K —» R is an L1 -Caratheodory function such that for some function a+ G L1(a,b), ^ ^
> a+(t),
(3.16)
uniformly in t G [a, b]. Then the problem (3.11) has at least two solutions u\ and u2 such that oti < u\ < j3, i*2 G S
and u\ < u2,
where S is defined in (3.7) with a = max{ai, a2}.
3. THE NON WELL-ORDERED CASE
181
Proof. For any r > max{j|ai||OO) Halloo ||/?||oo}> we consider the modified problem u(a)=u(b),
{
u'(a) = «'(6),
'
where fr(t,u) = f(t,ai(t)) -u + ai(t), = /(t,u), = ( 1 + r - « ) / ( £ , u), = 0,
if u < ax(t), ifai(t)
CZaim 1 - .Every solution of (3.17) is suc/i tfia£ u > a\. This result follows from the usual maximum principle argument as it is used, for example, in the proof of Theorem 1-3.1. Claim 2 - There exists k > Oso that for anyr > maxdlaiHoo, ||<22||oo> Halloo}
and any solution u 6 S of (3.17), we have ||u||oo < fc. As u £ S, there exist to and t\ such that u(to) = min u(t) < u(t\) < ot{ti) < ||o!||oote[a,6]
Further, we deduce from the asymptotic character of / that there exists a+ and h € Lx(a,b) such that for a.e. t E [a, b] and all u > ot\{t), fr(t,u)
>-[a+(t)u
Hence, we have for t £ [to,to + b — a), rt
u(t)=u(t0)-
fr(s,u(s))(t-s)ds o
t
a+(s)u(s) ds to
and the claim follows from Gronwall's Lemma. Conclusion - Consider the problem (3.17) for r > maxfllaiHoo, ||a2||oo, ||/?||oo, k}, where k is given in Claim 2. It follows from Theorem 1-3.4 that there exists a solution ui of (3.17) which is minimal in [a\,0\. Hence, a\(t) < ui(t) < r which implies this solution solves also (3.11). Next, we can apply Exercise 3.1 to obtain a solution u2 G S of (3.17). From Claim 1, 1x2 > «i, and from Claim 2, U2 < k < r. Therefore u-i is a solution of (3.11).
182
III. RELATION WITH DEGREE THEORY
At last, notice that if u2 J* u\, u\ and u2 are upper solutions of (3.17) and we deduce from Theorem 1-3.2 the existence of a solution. U3 with Q!i < W3 < min{ui,U2} which contradicts u\ to be minimal. Assumptions such as (a) f(t, u) is nonnegative or (b) f(t,Q) > 0 and f(t,ui) - f(t,u2) > -M{ux-u2), if ux > u2 > 0, imply the asymptotic behaviour (3.16) and we can choose a% = 0. Hence, existence of nonnegative upper and lower solutions implies existence of two nonnegative solutions one of them being nontrivial. Observe that the existence of the third solution of u(a)=u(b),
u'(a)=u'(b),
if —1/4 < h(t) < 0 obtained in Example 1.4 can be deduced from Theorem 3.5 in a quite obvious way. We can extend Theorem 3.5 to the Lienard equation (3.9). Theorem 3.6. Let a\ and a2 e C([a,b]) be W2'1-lower solutions o/(3.9) and (3 € C([a, b]) be a strict W2tl-upper solution such that aj < (3 and Assume g E C(R) and f : [a, b] x R —» R is an L1 -Caratheodory function such that for some function h £ Ll{a, b), liminf/(t,M) > hit), u—*+ca
uniformly in t S [a, b]. Then the problem (3.9) has at least two solutions u\ and u2 such that di
and u\ < u2,
where S is defined in (3.7) with a = rmQi.{ai,a2}. Proof. The proof is exactly the same as for Theorem 3.5 except Claim 2 which repeats the argument used to prove the claim in Theorem 3.2. Exercise 3.4. Consider the Lienard equation u" + f(u)u' + g(u) + h(t, u, u') = 0, u{a) = u(b), u'(a) = u'(b),
,
f
^ 2
Lti)
where / , g : R —> R are continuous functions and h : \a, b] x R —> R is continuous and bounded. Prove that if infs \f(s)\ > 0, the existence of a lower and an upper solution implies the solvability of (3.18). Hint : See [235].
3. THE NON WELL-ORDERED CASE 3.5
183
Lower and Upper Solutions in the Reversed Order
In case a > j3 on [a,b], we can ask whether the solution of (3.11) satisfies the "strong localization" /3 < u < a on [a, b] as in the well ordered case. This is not always the case under the assumptions of Theorem 3.3 as shown by problem (3.2). But it is true if
f(t,u)-f(t,v) u—v
<
/_jr ~ \b — a,
This analysis is the content of this section. Theorem 3.7. Let a and (3 G W2>1(a, 6) be lower and upper solutions of (3.11) such that a > /3, E = {(t,u) G [a,b] x R \ /3(t) < u < a(t)} and f : E —> M be an Ll -Caratheodory function. Assume there exists k G]0, (jrs) 2 ] such that for a.e. t G [a, b] and allu G [(3(t),a(t)], f(t,a(t))
- ka{t) < f(t,u) -ku< f(t,P(t)) - k/3(t).
Then the problem (3.11) has at least one solution u G W2'1(a,b) such that for all t G [a, b] < u(t) < a(t). Proof. We can assume that a and (3 are not solutions; otherwise, the result is trivial. Let us consider the modified problem
u" + f(t,u)=0, u(a) = u(b), u'{a) = u'(b),
{
}
where f(t, u) = f(t, a{t)) + k(u - a{t)), = /(*,«),
iiu> a(t), if 0(t) < u < a(t), if «
By Theorem 3.3 and the Remark after Proposition 3.4, problem (3.19) has a solution u G S. It remains to prove the announced localization. Let us prove u < a. Observe that v = a — u satisfies
v" + kv = (a" + f{t, a)) + {f(t, u) - ku) - (/(*, a) - ka) =: h{t), v(a) - v(b) = 0, v'(a) - v'(b) > 0 where h G L^o.b) is such that h > 0, h ^ 0. By Corollary A-6.3, v > 0, i.e. u < a. In the same way, we prove u > @. O
184
III. RELATION WITH DEGREE THEORY
Remark Observe that the limitation on k in Theorem 3.7 is optimal if we want the strong localization. Consider the problem u" + ku = f(t), u(a) = u(b), u'(a) = u'(b), with k > ( E ^ ) 2 . Let c, d be such that ( t ^ ) 2
and f + g = b - a,
and define = ^
sinc(t-a),
if te[a,a + f],
= * ^ £ s i n d ( t - i + 5 - a ) , if t
This problem has /3 = 0 and a = ^ max{fc~c, d ^"fc} as upper and lower solutions satisfying the reversed ordering p < a. But its solution is given by u{t):=\smc(t-a),
ifie[a,a+f],
: = i s i n d ( i - f + 5 - a ) , ifte]a + f,6]
which is not one-signed, i.e. it is not between the upper solution /3 = 0 and the lower solution. Exercise 3.5. Prove that under the assumptions of Theorem 3.7, if for some k e]0, (gr^) 2 ], for a.e. t € [a,b], and for all u, v € [/3(i),a(t)], u ^ u
f(t,u)-f(t,v) u —v
~~
then the solutions of (3.11) in [0,a] are ordered. Prove also there are solutions umin, umax € W2'l{a,b) such that (3 < umin < umax < a, and any other solution u of (3.11) such that (3 < u < a satisfies min
Umax.
3. THE NON WELL-ORDERED CASE 3.6
185
The Dirichlet Problem
In this Section 3, we have worked out the periodic boundary value problem. These results can be extended to deal with other boundary conditions. We present here such an extension to the Dirichlet problem. The theorems we state are the counterpart of previously obtained results and their proof repeats previous arguments. Our first concern is a non-resonance result for the problem u{a) = 0, u(b) = 0, 2
where ($r^) is the first eigenvalue of the corresponding eigenvalue problem u{a) = 0, u(b) = 0.
^3'21^
Theorem 3.1 can then be modified as follows. Theorem 3.8. Assume a and 0 € C([a, b]) are W2'1 -lower and upper solutions of (3.20) such that a % fi. Suppose f : [a, b] x M2 -» R is a Caratheodory function such that for some h S L1 (a, b) \f(t,u,v)\<
h(t), on [a,b) x K2.
Then there exist a solution u of (3.20) in 3ti,t2 G [a,b], u(tx) > /9(ti), u(t2) < a(t2)}. Proof. For each r > 1, we define fr(t,u,v) = f(t,u,v), if|u|
V-M)
Claim 1 - For some C > 0, a(t) -< Csm{-K^) and (3{t) y Let r > maxfUalloo, ||/?||oc}- Define w to be the solution of w= f,
w(a) = r + 2, w(b) = r + 2,
e(t) — sin(7r|5|), choose k > 0 such that ke + w > r + 2 on [a, b] and define /32 = —a.\ = ke + w. Observe that fii and a\ are upper and lower solutions of (3.23). Hence the claim follows from Remark II-2.1.
186
III. RELATION WITH DEGREE THEORY
Claim 2 - There exists k > 0 such that for any r > k, solutions u of (3.23), which are in S, are such that ||u||oo < k. Assume by contradiction there exist sequences (r n ) n and (un)n C <S, where rn>n and un is a solution of (3.23) with r = r n , such that ||«n||oo > n. Let us write
un(t) =un(t) + une{i). ,6
where / un{t)e(t) dt — 0. We deduce from Proposition A-4.5 that for Ja K>0 some K
< K ["Kit) + {^fun{t)\e{t)dt Ja fb
rb
< K / \frn(t,un(t),u'n(t))\e(t)dt < K / (h(t) + tesW)e{t)dt. Hence, for some C\ > 0 and n large enough, we can write ||fin||ao
(3-24)
Let to De such that u'n(to) = 0. It follows that
\u'n(t)\ < | A ( ^ ) 2 M*)| + \frn(s,Un(s),u'n Jta
for some Ci > 0. Assume now that for some subsequence lim unk = +00. This assumpk—>oo
tion implies Unk(t) > (unk - C3\\unk\\Ci)e(t) > (unk - C 4 ( l + ^ ) ) e ( t ) , for some C3 and C4 > 0. Hence, for rife large enough, unk >- a which contradicts unk e S. In a similar way, we prove that no subsequence of (un)n goes to —00. It follows that un and also, using (3.24), Hunloo a r e bounded which contradicts the assumption. Conclusion - Let r > maxJUaHoo, ||/3||oo>fc}- As in Claim 1, we define fj2 = — aj = ke + w and conclude as in the proof of Theorem 3.1. Next we consider the problem of interaction with Fucik spectrum. A result similar to Theorem 3.3 holds for the Dirichlet problem () u(a) - 0, u(b) - 0.
3. THE NON WELL-ORDERED CASE
187
Theorem 3.9. Let a, f3 € C([a,b]) be W2'1-lower and upper solutions of (3.25) such that a £ 0. Let f : [a, b] x R - R be an Ll-Caratheodory function such that for some functions a < (^rj) 2 , > (jr^) 2 in L1(a,b), a (t) < liminf ^ ^
< limsup ^ ^
< b (t),
uniformly in t G [a, 6]. Assume further that for any p, q G L1(a, 6), witft a+ < p < 6+ and a_ < g < 6_, i/ie nontrivial solutions of u"+p(t)u+-q(t)u-=0, u(a) = 0, u(6) = 0,
. ^
,
0)
where u + (t) = max{u(i),0} and u~(t) — max{-u(t),0}, do not have interior zeros. Then the problem (3.25) has at least one solution u £ S, where S is defined by (3.22). Exercise 3.6. Prove the above theorem. The condition on p, q, can be related with the Fucik spectrum as in Proposition 3.4. Roughly speaking, the nontrivial solutions of the problem (3.26) do not have zeros if the box [a+, 6+] x [a_, 6_] lies under the second Fucik curve T2 of the Dirichlet problem. At last, we can write multiplicity theorems for (3.25) which paraphrase results worked out in Section 3.4. We state here a situation with two lower solutions and a strict upper one. Theorem 3.10. Let at\, a% G C([a, b}) be W^-lower solutions of (3.25) and {3 e C([a, b}) be a strict W2'1-upper solution such that a\ < p and &2 £ @- Assume f : [a,b] x R —» R is an L1 -Caratheodory function such that for some function a+ £ L1(a, b), (3.16) holds uniformly int G [a, b). Then the problem (3.25) has at least two solutions u\ and u-i such that a\ < u\ -< f3, U2 G S
and u\ < 1x2,
where S is defined in (3.22) with a = max{ai,a2}Exercise 3.7. Prove Theorem 3.10.
188
III. RELATION WITH DEGREE THEORY
Hint : Adapt the proof of Theorem 3.5. In Claim 2, notice that u G S implies the existence of sequences (un)n and (tn)n such that un —> u in C^d^b}) and un{tn) < a{tn) < Ce(tn). Here e(t) = sin(Tr^f) and C is such that a < Ce as in Remark II-2.1. We deduce then the existence of to such that u(to) < Ce(to) and u'(to) = Ce'(io) and compute /"* u(t) < Ce(t0) + Ce'(io)(i - *o) - / fr(s, u(s))(t - s) ds. The proof follows then as in Theorem 3.5. In very much the same way, we can state a dual result with two upper solutions and a strict lower one. Theorem 3.11. Let a £ C([a,b\) be a strict W2'1-lower solution of (3.25) andP\, (3% € C([a,&]) be W2'1-uppersolutions such thata < 02 anda ^ j3i. Assume f : [a, b) x E —> R is an L1 -Caratheodory function such that for some function a_ G i 1 (a, b),
uniformly intE
[a, b].
Then the problem (3.25) has at least two solutions u-i and u^ such that u\ e S,
a < U2 -< P2 and u\ < U2,
where S is defined in (3.22) with /3 = min{/?i,/?2}-
Exercise 3.8. Extend the results of this section to the separated boundary value problem u" = f(t,u), a\u(a) — a2u'(a) = 0, blU{b) + b2u'(b) = 0, in case oi, h G E, a2, b2 € M+, a\ + a\ > 0 and b\ + b\ > 0.
Chapter IV
Variational Methods 1
The Minimization Method
Another approach in working with lower and upper solutions is to relate them with variational methods. Consider for example the problem «" + /(*,«) = 0 , u(a) = 0, u{b) = 0,
n U
,x >
where / is an L1-Caratheodory function. It is well known that the related functional rb ,ti2(f\
J
t
(1.2)
with F{t,u) = J™ f(t,s)ds is of class C1 and its critical points are the solutions of (1.1). A first link between the two methods is that existence of a well ordered pair of lower and upper solutions a and /3 implies the functional
[a, b}xR\ a(t)
(1.3)
Assume f : E —> R is an L -Caratheodory function. Then the functional
min (f>(v). veH%(a,b) a
Further, u is a solution of (1.1). 189
190
IV. VARIATIONAL METHODS
Proof. Let us consider the modified problem «" + /(t,7(*,«))=0, u(a) = 0, u(b) = 0,
,
{
. '
where j(t,u) = max{a(t),min{u,/3(£)}}. Define the functional /
M - F(t,«(*))] dt,
Ja
where F(t,u) — J™ f(t,f(t,s))ds. As f{t,^{t,u)) is L1-Caratheodory, (j> 1 is of class C and its critical points are precisely the solutions of (1.4). Moreover is weakly lower semi-continuous and coercive. Hence, (f> has a global minimum u (see Theorem A-2.2) which is a solution of (1.4). As in Theorem II-2.1, we prove that u satisfies a < u < 0, and hence is a solution of (1.1). Notice that on {u e Hg(a,b) \ a < u < /?}, the function <[>(u) —
,
K
, *
where / : [a, b] x R —> K is an L1-Garatheodory function that satisfies /(t,0) = 0 , f(t,R) < 0 for some R > 0 and there exists fi £ ^ with 0 < n(t)
m i n cf>{v) <
This last inequality implies u ^ 0. The method applies to other boundary value problems such as the periodic problem u" + /(t,u) = 0, n fix { ' u(a) = u(b), u'(a) = u'{b).
1. T H E MINIMIZATION METHOD
191
Here, the associated functional reads rb
J
11-F(t,u(t))]dt,
(1.7)
with F(t,u) = / 0 " f{t, s) ds and H^r{a, b) = {u £ Hl{a, b) | u(a) = u{b)}. We can also consider the Neumann problem u" + f(t,u)=0, u'(a) = 0, u'(6) = 0,
, {
, 3
with whom we associate the functional
- F(t,u(t))} dt.
Ja For both these problems, we can state the corresponding result.
(1.9)
Theorem 1.2. Let a and f3 be W2'1 -lower and upper solutions of (1.6) (resp. (1-8),) with a < j3 on [a,b] and E be defined from (1.3). Assume f : E —> R is an L1-Caratheodory function. Then the functional
min
min <^(v)J.
uEH' (o,6)
v€H1(a,b)
a
OL
Further, u is a solution of (1.6) (resp. (1-8)J. Exercise 1.1. Prove Theorem 1.2. As an application, consider the problem + h(t) = 0, t , ( 0 ) = 0 , «(*) = (>.
(L10)
T h e o r e m 1.3. Let /J,, h € L°°(0, IT) with /x0 = essinf /x(£) > 0 and g : K —> R 6e a continuous function. Let us denote G(u) = JQ g(s) ds. Assume that —oo < liminf — \ 1 < 0 and lim sup —— = +oo. o
u*
o W
T/ien i/ie problem (1.10) /tas <wo infinite sequences of solutions (un)n and (vn)n satisfying . . . < Vn+\
Un+i < .. .
and lim (maxtin(t)) = +oo, n—>oo
t
lim (mini; n (i)) = —oo. n—>oo
t
192
IV. VARIATIONAL METHODS
Proof, Step 1 - Claim: For every M > 0 there exists 0, upper solution of (1.10), with fi(t) > M on [0,7r]. First observe that if g is unbounded from below on [0, +oo[, we have a sequence of constant upper solutions /?„ —> +oo. In the contrary, we can assume there exists K > 0 such that g{u) > -K for u > 0. Given M > 0, we can choose d so that d>2M.
We define then /3 to be a solution of the Cauchy problem
«" + NI«»07(«) + ^ ) + Woo = o> u(0) = d, t*'(O)=O. Assume there exists to 6 ]0,7r] such that /3(t) > M on [0, to[ and P(tg) = M. Notice that on [0,t 0 ], £'(*) < Oand ||M|U(G ; (/3(t))+^/3(t))+||/i||oo/3W > 0. From the conservation of energy for (1.11), we have
0<-^)
which leads to the contradiction /3(£o) ^ f > Af Hence, the claim follows. In a similar way, we prove the following. Claim - For every M > 0 i/iere exists a lower solution a of (1.10) such that a(t) < -M on [O,TT]. Step 2 - Claim; There exists a sequence of positive real numbers (sn)n with sn —> +oo and z >- 0 such that
Define 2 to be a C1-function such that 0 < z(t) < 1 on ]0, ir[, z(0) — 0, z(ir) — 0, z'(0) > 0, Z'(TT) < 0 and z(t) = 1 on [e,ir - e] for some e > 0. Choose (sn)n a sequence of positive real numbers with sn —> +00 and
1. THE MINIMIZATION METHOD
193
G n
^ ' —> +00. Recall that the assumptions imply G(u) > —L(u2 + 1), for some L > 0. We compute then s2z'2(
= / i-^T2 l Jo f
=
S
- li(t)G{snz(t)) - h(t)snZ(t)} dt q27i2/f\
^LM
rv-e dt
./[O,7r]\[e,7r-e]
_ G(Sn) /
^
- / < s2n\\z'\\le
M ( t )dt
Je
^
n{t)G{snz(t))dt-sn
J[O,7r]\[e,7r-e]
I h(t)z(t)dt JO
- G(s n )Mo(7r - 2c) + L ( s ^
It follows t h a t
5iep 5 - Claim: There exists a sequence of negative real numbers (tn)n with tn —> —00 and z >- 0 such that 4>(tnz) —+ —00. The argument is similar to Step 2. S'tep 4 - Conclusion. By Step 1, we have a\,/3i lower and upper solutions of (1.10) with cti < f3i. Hence, we obtain from Theorem 1.1 a solution u\ of (1.10) such that ot\ < ui < 0i. From Step 2, we have z and s% such that s\z > u\ and 4>{s\z) < (f>{ui). Moreover, Step 1 provides the existence of an upper solution /?2 with u\ < s\z < /?2- Now, by Theorem 1.1, we have a solution ui of (1.10) satisfying
and <^>(w2) =
min
It follows that U2 ^u\. Iterating this argument and reproducing it in the negative part, we prove the result. Remark 1.1. The condition on G(u)/u2 cannot be replaced by analogous conditions on g(u)/u as observed in [230]. Problem 1.2. Let / : [0, +oo[—* R be a continuous function. Let us denote F(u) = Jo" f(s) ds. Assume that liminf — ^ = 0 and u-»0+
U2
limsup u_*0+
Prove then that the problem u" + /(«) = 0, u(0) = 0, u(n) = 0
= +00. Ul
194
IV. VARIATIONAL METHODS
has a sequence (un)n of positive solutions satisfying lim (maxun(i)) = 0. n—>oo
t
Hint ; See [240]. As a next problem consider the following prescribed mean curvature equation (
«(0) = 0, u(l) = 0. This equation is equivalent to u" = A(l + i/ 2 ) 3 / 2 /(i,«),
«(0) - 0, «(1) - 0.
As in Example 1.1, the following result provides a nontrivial nonnegative solution. Notice however that the equation does not satisfy a Nagumo condition which will force us to modify not only the dependence in u but also in the derivative u'. Problem 1.3. Let / : [ 0 , l ] x R - » R b e a continuous function such that f(t, 0) < 0 and define F(t, u) = /0" f(t, s) ds. Assume that for some [a, 6] C ][ lim
max — „
u-»0+ t£[a,6]
UZ
= — oo.
Prove then that there exists a number A* > 0 such that the problem (1.12) has for each A e]0,A*[, a nontrivial nonnegative solution. Hint : See [92] or [146]. Define p(s) = (1 + s)- 1 / 2 ,
if 0 < s < 1,
a(t) = 0 and /3(t) = t ( l - i ) . Consider then the functional 4>: H&(0,1) defined by
f
where P(v) — f^p^ds, F(t, u) = Jo" f(t, j(t, s))ds and as usually, "f(t, u) max{a(i),min{u,/3(i)}}. Critical points of
«(0) - 0, u(l) = 0,
2. LOCAL MINIMUM OF THE FUNCTIONAL
195
which can also be written u" = A(p(«'2) + 2p'(u'2)u'2r1f(tMt,
u)),
u(0) = 0, u(l) = 0.
The proof follows then if we notice that for A > 0 small enough, a and 0 are respectively lower and upper solutions and
2
Local Minimum of the Functional
Theorem 1.1 defines the solution u of (1.1) as a minimizer of the functional <j>, defined by (1.2), on the set C = {u £ Hl(a,b) \ a < u < j3}, which does not imply that u is a local minimizer of that functional. Indeed, if a and /3 coincide at some point, the set C has empty interior and there is no evidence that the function u minimizes
J
(2.1)
The main result of this section proves that such a minimizer, minimizes locally 0 on HQ(G, b). Theorem 2.1. Let f : [a, b) x M. —» R be an Ll-Carathe'odory function. Assume UQ £ Cg([a,&]) is a local minimizer of 4> defined by (2.1) i.e.
3r > 0, \/v E CcUMD,
IMIci < r =* 4>{u0) < j
then UQ is a local minimizer of
\\v\\Hi < r =}> 4>(u0) <
R e m a r k 2.1. This result is somewhat surprising since a fig-neighbourhood is much bigger than a Cg-neighbourhood. It is false in general, but here we can use the special structure of the functional. Notice also that the above theorem still holds if u 0 G HQ (a, b). Proof. Notice first that, as uo minimizes locally $ in Cg([a, b\), we have
t Ja ° for all £ G C^([a, b}) and by density for all £ G H^(a, b).
196
IV. VARIATIONAL METHODS
Suppose the conclusion does not hold. Then Vr > 0, 3vr € H$(a, b),
\\vr\\Hi < r and cp(u0 + vr) < cf>(u0),
(2.2)
where ||ii||^i = J u'2 dt. By a standard lower semi-continuity argument, we prove that the problem min^(u), where V = {u e H^(a, b) \ ip(u) = f*(u' - u'o)2 dt-r2
££(b)( )
< 0} has a solution
Prom Lagrange multiplier theory (see [69]), there exists (iT < 0 so that
J
(wU'-f(t,Wr)t;)dt = # K ) ( 0 = ^ # K ) ( O = Mr J (W'r-U'o){;'dt,
for all £ e H&{a, b). It follows that
(1-Mr) / {W'r-U'Q)S! dt= f Ja
(f{t,Wr)-f(t,
Ja
for all £ € Ho(a,b), i.e. wr - u0 € W 2 l l (a,6) and
-(1 -
M
r ) « - «o) = f{t,wr) - f(t,u0).
Further, as wr -* uQ in HQ (a, 6) for r —» 0, the wr are bounded in C([a, b\) and there is a function h € L 1 (a, 6) such that for any r G [0,1], \w"(t) — u d'WI — MO o n [°i^]- We deduce now from Arzela-Ascoli's Theorem that up to a subsequence, the wr converge in Co([a, &]) as r goes to zero. Next as wr —» UQ in HQ(a,b) we conclude wr —» «o in Cg([a,6]). As 4>(wr) = 4>{wr) < 4>(uo), this result proves that UQ does not minimize cf> and concludes the proof. Theorem 2.2. Let a and (3 be strict W2'1 -lower and upper solutions of (1.1) with a -< P and E be defined by (1.3). Assume f : E —> M is an L1 -Caratheodory function. Then there exists a local minimizer u of the functional
2. LOCAL MINIMUM OF THE FUNCTIONAL
197 H1
(PS) // («„)„ C i?o(o,6) is such that (j){un) is bounded and W
u(0) = 0, u(n) = 0,
(2.3)
has a nontrivial nonnegative solution. Proof. Critical points of the functional
r
Jo
are solutions of u" + {u+f = 0,
«(0) = 0, u{n) = 0
(2.4)
and, as these solutions are nonnegative, they are solutions of (2.3). Observe then that <j> has a mountain pass geometry. First, for e > 0 small enough a(t) = —e and /3(t) = esint are strict lower and upper solution of (2.4). Hence by Theorem 2.2, the problem (2.4) has a solution UQ which is a local minimizer of <j>. Next, there exists R > 0 such that
u(0) = 0, u(ir) = 0,
(2.5)
198
IV. VARIATIONAL METHODS
where g £ C((0, A]) is such that g(0) = 0, g(A) = 0, and there exists r > 0 such that g(u) < u on [0,r]. Assume there exists a strict lower solution a with 0 -< a < A. Then the problem (2.5) has at least two solutions ui, «2, with 0 ^ u\ <* U2 < A and 1*2 >- a. To prove this property we first extend g to the whole line g(u) — g(u), HO
Z
Jo
U.-G(u(t))}dt,
where G(u) = /Q1* g(s)
2. LOCAL MINIMUM OF THE FUNCTIONAL
199
Theorem 2.4. Let f : [a,b] x R —> K be an L1 -Caratheodory function. Assume (a) f(t,O) >0o.e, on [a,b]; such that 0 < B\ < ... < Bn and for i € (b) there exist B\,...,Bn
{l,...,n}, f(t,Bi) < 0 a.e. on [a,b]; (c) there exist Ai,..., An^i and vi,..., vn~\ > 0 such that Bi < Ai < B2 <
andforie{l,...,n-l}
< ^ n - i < Bn
and all (t,u) £ [a,b] x [0,Bi[
where F(t, u) = / " f(t, s) ds; (d) there exists k € I/(a,b) such that for a.e. t 6 [a, b] and allu € [0,S n ], f{t,u) + k(t)u is non-decreasing. Then there exists A such that for all A > A, the problem (2.6) has at least In — 1 ordered solutions 0 < « l -< U-2 -< . . . -< M2n-1-
Remark 2.2. (a) Roughly speaking the above hypotheses imply that for fixed t, the graph of f(t,.) has n positive humps and n negative ones, each positive hump having greater area than the previous negative one. (b) In case F(t, Ai) > F(t,u) for all (t,u) £ [a,b] x [0,5j[ is not satisfied, the theorem is no more valid. Consider for example the problem u" + A cos u = 0, u(a) = 0, u(b) = 0. By elementary methods, it is easy to see that for A > 0 this problem has a unique positive solution u with ||u||oo < f (c) More qualitative properties of the set of solutions can be found in [98] and [202]. Proof. Define first J(t,u) = f(t,u), ifu>0, = /(i,0),
ifu<0
and
/
^ i - XF(t,u(t))]dt,
Ja
where F(t, s) = JQS f{t, x) dx. Notice that critical points of (f> are solutions of u" + Xf(t,u)=0, . , u(a) = 0 «(6) = 0
V
;
200
IV. VARIATIONAL METHODS
Such solutions are nonnegative and therefore solve (2.6). Let AQ be a negative constant, and observe that it is a strict lower solution of (2.7). Next we deduce from assumptions (b), (d), and Proposition III-2.7 that for all i £ {1,... , n}, Bj is a strict upper solution. Step 1 - By Theorem 2.2, there exists a local minimizer u\ of cf> in HQ(CI, b) which is a solution of (2.7) such that AQ -< ui -< B\. Solutions of (2.7) are nonnegative so that
m i n
Step 2 - There exists w £ HQ(O>, b) and Ai > 0 such that wj < w < B
h{t)dt
J[a,b]\{a+S,b-S]
Choose w £ HQ(a,b) to be such that w(t) = A\ on [a + 5,b — S] and u\ < w < A\. We compute, for A large enough,
[F(t,A1)-F(t,ui(t))}dt
+X I
[FfoAJ-Fit
J[a,b]\[a+8,b-5]
rb wl2(t) < / Ja
f vx{b-a)-
2
f /
h(t)Axdt
J[a,b]\la+S,b-S]
2 Step 3 - By Step 2, min
Hence, by Theorem 1.1, there exists 113 solution of (2.7) such that u\ 113 < B(u3) —
min
<j)(v).
H^(b)
Arguing as in Proposition III-2.7 and using assumption (d), we prove «i -< U3. Hence using Theorem 2.1, we prove u% is a local minimizer of
2. LOCAL MINIMUM OF THE FUNCTIONAL
201
Step 4 - Consider now the modified function /(*,«) =7(*,«i(*)) = /(*,«i(*))> = /(*,«)=/(*,«), = f{t, U3(t)) = f(t, U3(t)),
if u
< «i(0. ifui(t)
and the corresponding functional <j>: FQ1 (a, 6) -> R , « H+ /
- XF(t, u(t))\ dt,
where F(t, s) = JQS f(t, x) dx. Let us prove that u\ and U3 are local minimum of (j> in /fo(a, b). To this end, observe that =7(tl«1(t)K _ _ ifu
— / [u'i(u — ui)~ — Xf(t,ui)(u — ui)~]dt Ja 2 b ,..
..
\-'
dt and hence, by Theorem 2.1, u\ is a local minimum of
u{a) = 0, u(b) - 0. As in Theorem II-2.1, we deduce from the definition of f(t,u) that ui < 1*2 < W3. Using again assumption (d) we obtain u\ -< u
202
IV. VARIATIONAL METHODS
Conclusion - Iterating this process, we obtain the required solutions. We can now write a variational version of Amann's Three Solutions Theorem. Exercise 2.2. Let ot\, /3i and CK2, 02 £ C([a, b}) be two pairs of W2>1-lower and upper solutions of (1.1) such that on [o, 6] #,(*), ai(t) < /3i(t), a2(t) < and there exists to € [a, b] with
Assume further /?x and a 2 are strict upper and lower solutions. Let E be defined by (1.3) (with a — min{ai, 0:2} and /3 = max{/3i, #2}) and suppose /:£?—> M is an L1-Caratheodory function. Prove that the problem (1.1) has at least three solutions u\, u^, 113 e W2>l{a,b) such that 011 < ui < min{/9i,/32}, max{ai,a2}
3
The Minimax Method
3.1 The Minus Gradient Flow One of the main techniques in variational methods uses the deformation of paths or surfaces along the minus gradient (or pseudo-gradient) flow. In this section, we study the dynamical system associated with this flow. We shall define the minus gradient flow using the following assumptions : {H) Let / : [ o , 5 j x R - » K , (t,u) H-> /(£,u) be an I^-Caratheodory function, locally Lipschitz in u. Let also m £ Ll{a, b) be such that m > 0 a.e. in [a, b] and f(t, u) + m(t) u is increasing in u. Now, let us define on HQ(a,b) the scalar product (u,v)Hi = I [u'{t)v'{t) + m{t)u{t)v{t)\dt Ja and let
F(t,u)= f f(t,s)ds. Jo
3. THE MINIMAX METHOD
203
It is then easy to see that the functional
/
- F(t,u(t))]dt
is of class C1 and V
(3.1)
N : H&(a, b) -> Ll(a, b), u <-> f(t, u) + m(t)u, ^Kh and Kh is defined to be the unique solution of -u" + m{t)u = h{t),
u(a) = 0, u(b) = 0.
Let us notice at last that if Assumptions (H) are satisfied, the function
defined from (3.1) is locally lipschitzian. Next, we define a (^-function tpr : R -> [0,1] such that ipr(s) - 1 if s > r and Vr(s) = 0 if s < r - 1. We consider then the Cauchy problem u(0) = u0,
{6
j
where u0 G Cg([a,6]). Prom the theory of ordinary differential equations, we know that the solution u( ; UQ) of (3.3) exists, is unique, and is defined in the future on a maximal interval [0,o;(uo)[- We also know that for any t € [0,w(uo)[, the function u(t; ) : Co([a,&]) —> Co([a,6]) is continuous. We call the minus gradient flow the local dynamical system defined on e^([a,b})byu(t;u0). We could have defined the minus gradient flow in X = Co ([a, 6]) or HQ(O,,b). However, such choices are not suitable in our context. We have to work with sets such as {u € X \ u -< 0} and {u € X \ u >- a}, where a and j3 are lower and upper solutions that can satisfy the boundary conditions. With the Co ([a, b}) or the H^(a, b)-topology, these sets have in such a case empty interior which creates major difficulties. A first result shows that the solutions of (3.3) are defined for all t > 0. Proposition 3.1. Let Assumptions (H) be satisfied and u(t;uo) be the minus gradient flow defined for some r e R . Then for any UQ G Cg([a, 6]) we have W(UQ) = +oo.
204
IV. VARIATIONAL METHODS
Proof. Notice that
which implies that for all £ € [O,w(«o)[
(3.4)
Observe also that (j)(u(t; uo)) > min{r — l,
< /
i>r(^(u{s;uo)))\\V
J t\
H i
Hence, if U(UQ) < +oo, there exists it* 6 ffo(&5&) such that u(t;uo) -5 u* as t —> w(uo)- It follows that the function u( ;uo) : [0, w(ito)] —* C([o,6]), where U(U>(UQ);UO) = u*, is continuous and
Let a(t) = i(jr((j)(u(t;uo)))- For all t € [0,w(uo)], we can write /j.
\
u(t;uo)=v,Qe Hence, U(-;UQ)
rt —I Jo a(r)dr
,
rt I
ft —I a(r)dr
+ I e J' Jo
t \TS-\T
and contradicts the maximality of ui(uo). Invariant S e t s
An important property of the cones
and
\ J
^ n\i\
ti\
6Cn(la,oJ).
[0,w(uo)] —* CQ([a,6]) is continuous, which implies c1 it(£; Uo) —> u* as i —» w(uo)i
3.2
i
a(s)KlVu(s;ua)as
3. THE MINIMAX METHOD
205
which are associated to lower and upper solutions a and /? of (1.1) is that they are positively invariant. To make this precise, let us introduce the following definitions. Definition 3.1. Let u(t;uo) be the minus gradient flow defined for some r £ M. A nonempty set M C CQ([<J, b]) is called a positively invariant set if Vu0 € M, Vt > 0, u(t; u0) E M.
As a first example, notice that (3.4) implies that the set 4>c = {u € Cl([a,b\) | 4>(u) < c}
is positively invariant. Also, unions and intersections of positively invariant sets are positively invariant. To investigate the cones Ca and C® we need the following lemma. Lemma 3.2. Let Assumptions (H) be satisfied. Assume a E W2>1(a, 6) is a lower solution of (1.1). Then for all u> a, we have KNu > a, where K and N are defined by (3.2). Proof. Let u > a, set w = KNu — a and observe that w satisfies -w" + m(t)w = f(t, u(t)) + m{t)u(t) - (~a"(t) + m(t)a(t)) > f(t, u(t)) + m{t)u{t) - (f(t, a{t)) + m(t)a(t)) > 0, w(a) > 0, w(b) > 0. It follows from the maximum principle that w > 0. Proposition 3.3. Let Assumptions (H) be satisfied and u(t; UQ) be the minus gradient flow defined for some r £ R. Assume a E W2tl(a,b) is a lower solution of (1.1). Then the set Ca defined from (3.5) is positively invariant. Similarly, if 0 E W2:1(o, b) is an upper solution of (1.1), the set C13 defined from (3.6) is positively invariant. Proof. If the claim is wrong, we can find UQ £ Ca and t\ > 0 so that for all t £ [0,*i[, u(t;u0) £ Ca and u(ti;u 0 ) £ dCa. Let w(t) = u(t;u0) — a, define a(t) = ipr(^>(u(t;uo))) and observe that for alHE [0,ti[ ftw(t) = ftu{t;u0)
= -a(t)(u(t,uQ) - KN(u(t;u0))) = -a(t)w(t) + h(t),
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IV. VARIATIONAL METHODS
where from Lemma 3.2, h(t) = a(t)(KN(u(t;u0)) As w(0) — UQ — a >- 0 we have
- a) > 0 for t G [0,*i[.
Jo which contradicts u(ti;uo) G dCa. 3.3
D
Non Well-ordered Lower and Upper Solutions
The first result of this section provides Palais-Smale type sequences from non well-ordered lower and upper solutions. As usual, this proposition gives a solution of (1.1) with the help of the Palais-Smale condition. Proposition 3.4. Let Assumptions (H) be satisfied. Suppose a and f3 G W2'l(a, b) are lower and upper solutions of (1.1) and a jt /?. Define Ca and Cp from (3.5) and (3.6), r = { 7 G C([0, l],C5([a,&])) I 7(0) G &,
7 (1) 6
Ca},
r 7 = {s G [o, i] 17(«) G c^ia, b\) \ (cf> u ca)} and assume c := inf max^>(7(s)) G M, where
M(i;uo)G^1([c-5,c
and there exists an increasing unbounded sequence (tn)n C M+ such that -5 0 as n -+ oo. Proof. Let us fix 5 G]0,1[ and define E = <^)c-'5UCaUC^. Observe that £ is positively invariant. Define A(E) = {UQ G Cg([a,b]) \ 3t > 0, u(i;wo) 6 .E}. Obviously, this set is open and positively invariant. Consider a path 7 G T so that for all s G T 7 , <j>(rr(s))
3. THE MINIMAX METHOD
207
such that u(ts;*f(s)) £ E. Assume next that for every u £ N , there exists sn £ [0,1] such that u(n, 7(s n )) £ E. Going to a subsequence, we can assume sn —> s* £ [0,1] and using the contradiction assumption, there exists ta* > 0 such that u(ts*;y(s*)) £ E, As E is open, for all n large enough u(ts*;'y(sn)) £ E which leads to a contradiction as E is positively invariant and u(n; 7(s n )) ^ E. Notice now that 7 = u(T; 7(-)) is in T and such that ^(7(-)) < c — d on T^ which contradicts the definition of c. Conclusion - By construction of A(E) and as (fi(u(-\ uo)) is non-increasing, we have for all t > 0, u(t\uQ) £
and there exists an increasing unbounded sequence (tn)n which verifies ^
D
In order to obtain existence of solutions of (1.1), we need to prove that the sequence (u(tn;uo))n converges towards such a solution, which holds true in case we assume the Palais-Smale Condition (PS). Theorem 3.5. Let Assumptions (H) be satisfied. Suppose a and f3 £ W2tl(a, b) are lower and upper solutions of (1.1) and a j£ j3. Define T and T7 from (3.7) and assume c := inf max<^(7(s)) £ R, where
1 < 0(Ufc)
and V(j)(vk) = 0.
Let us fix k £ N. Prom Proposition 3.4, there exist Uk £ CQ([O, b]) such that Vt > 0,
u(t; uk) £ r i([c -l,c+i})\
(C» U Ca).
and there exists an increasing unbounded sequence (tn)n C M+ such that -?0
as n —» 00.
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IV. VARIATIONAL METHODS
Notice that (f>(u(t; uk)) > c — 1 so that u(t; uk) solves
f
(u), u(0) = ufc.
Using the Palais-Smale Condition (PS), there exists a subsequence that we still write (u(tn;uk))n and vk £ fig (a, 6) s u c n H1
u(tn\Uk)
—? Wfc
as n —> oo,
c - £ < (ufe) < c + i and V^(u fe ) = 0. CZaim 1 : There exists R > 0 swc/i that for all s e [0,+oo[, l|w(*;Mfc)||iyi > -R impiies ||V^(u(s;ufe))||ffi > ^ . If not, there exists a sequence (sm)m C [0, +oo[ such that \\u(sm;uk)\\Hi >m
(3.8)
and ||V0(u(sm;iife))||#i < ^ . As (j>{u{t;uk)) is bounded, by Palais-Smale Condition, there exists a subsequence {smi)j so that u(smi;uk) converges in i?o(a, b) which contradicts (3.8). Claim 2 : \\u(t;v,k)\\Hl *s bounded on M.+ . Assume that for some t > 0, > RQ = max{||wfe||fl-i,-R}. Then there exists t\ £ [0,i] so that #i = RQ and for any s G [h,t], ||u(s;tifc)||jji > i?o > R- It follows that
On the other hand, we have (t;uk) -u(ti;uk)\\Hi
= || / ^7(f>(u(s;uk))ds\\Hi
J < |^(«(i; «fc)) - ^(u(ti;
As 4>(u{t;uk)) is bounded, the claim follows.
3. THE MINIMAX METHOD
209
Claim 3 : Vk & CQ([O,,b)) \ (C@ U Ca). To prove this claim let us show c1 that for some subsequence u(tn;uk) -^ vk. Consider the sequence {wn)n C C£([a,b]), defined by e-{tn~s)(KNu(s;uk))(r)ds,
wn(r) = [ " Jo
with K and N denned from (3.2). As ||u(t; Ufc)||#i is bounded, there exists h G I}(a, b) so that U ( S ; Wfe ))
+ m(-)(«(s; Wfe )
-KNu(s;uk))](r)ds\
e o
Using Arzela-Ascoli Theorem, we can find a subsequence (wni)i converging in Co ([a, &]). The same holds true for u(tn; itfe) =
'" + / e~(tn~s)KNu(s; uk) ds, Jo
i.e.
for some % and as u(tn;uk)
c1 u(tni;uk) -Avk —? Vfe the claim follows.
Conclusion - Prom Palais-Smale Condition, a subsequence of (vk)k converges in JF/QC^I^) to some function v. As vk = KNvk, the convergence also holds in Co([a, 6]) so that Q),
<£(«) = c and V0(u) = 0.
D
Theorem 3.6. Le£ Assumptions (H) be satisfied. Suppose the functions a i , O!2 G -W 2il (o,6) are iower solutions of (1.1) and Pi, fa G W 2>1 (a, b) are upper solutions of (1.1), none of them being solution. Assume they satisfy
Assume at last the Palais-Smale Condition (PS) is satisfied. Then the problem (1.1) has at least 7 solutions such that a i -< U3, u3 -fc Pi, u3 ^ a2, a 2 -< ui} Ui -A Pi, v-i / " l , u 5 -K /3 2 , u5 -fi Pi, u5 i- a 2 ) u6 -< Pi, UQ -fi Pi, u6 / ai, u-j is not comparable with a* and Pi for i = 1, 2.
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IV. VARIATIONAL METHODS
Proof. Notice that from Proposition III-2.7 and Assumptions (H), the lower and upper solutions oti and ft are strict. The two first solutions u\ and U2 are then obtained by Theorem 2.2, while the solutions Ui for i = 3,4,5,6 are obtained repeating the argument of Theorem 3.5 with, for example, for T = { 7 € C([0,l],C ai ) | 7(0) € C?\ 7(1) € Ca2). For that which concerns the last solution, let us define
and observe that c € K. Let u(t; uo) be the minus gradient flow defined with r = c — 1. We will prove that for any 5 G ]0,1[, there exists wo £ ^([a, b]) such that Vt € [0, +oo[,
«(t, uo) £(l>-1([c-d,c + 6}) \ (C A U Cai U
Fix S e ]0,1[, let a G E be such that c<
max 4>{o-(r, s)) < c + 5
and define E — ^c~5 U (C3* U Cai U C 32 U C a J . Observe that E is an open positively invariant set. Define A(E) = {uo G Cg([a, 6]) j 3t > 0, u(i;tio) G E}. Let us prove there exists UQ £ cr(Tcr) \ A(E). Otherwise, as in Proposition 3.4, we prove there exists T so that for all (r, s) G [0,1]2, u(T; a(r, s)) G E. Notice that &(r, s) = u(T; a{r, s)) G E, and m.ax.
3. THE MINIMAX METHOD 3.4
211
A Four Solutions Theorem
This section deals with a problem (1.1) which has the trivial solution u — 0. We consider assumptions which imply existence of lower and upper solutions cti, 0i so that ai < 0i < 0 < a2 < 02The Three Solutions Theorem (Theorem III-2.11) provides three solutions, two one-sign solutions u\ e [ai,/?i] and u-i € [a.2,02] and a third one that can be the zero solution. The difficulty is to obtain a third nontrivial solution. Here such a result is obtained assuming the slope ^'"^ crosses the two first eigenvalues. Theorem 3.7. Let Assumptions (H) be satisfied and assume: (i) there exist X > A2 = /£Ha\i and <> > 0 such that for a.e. t £ [a, b] and allue [-6,6], u (ii) there exist p, < \\ = (f,la)i allu 6 M with \u\ > R,
an
d R> 0 such that for a.e. t € [a, b] and
u Then the problem (1.1) has at least three nontrivial solutions Ui such that U\ ~< 0, U2 >- 0 and U3 changes sign. Proof. Claim - There exists a.\ -< 0 which is a strict lower solution of (1.1). Let h G Lx(a, b) be such that h > 0 and for a.e. t € [a, b] and all u < 0, f(t,u)> fiu Define then ai to be the solution of u" + /iu = h(t),
u(a) = 0, u(b) = 0.
As fj, < Ai and h > 0 we have from the maximum principle a\ -< 0 and a'{{t) + f(t,ai(t))
> a'{(t) + mi{t) - h(t) = 0, ai(o) = 0, ai(b) = 0,
i.e. a\ -< 0 is a lower solution which is not a solution. Using Assumptions (H), it follows from Proposition III-2.7 that ai is a strict lower solution. Claim - There exists 0% >- 0 which is a strict upper solution of (1.1). We
construct 02 as we did for a i .
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IV. VARIATIONAL METHODS
The modified problem. Consider the modified problem
u" + f(t,u) = 0, u(a) = 0, u(b) = 0,
{
}
where
f(t,u)=f(t,a1{t)), = /(t,«), = f(jt,fo(t)),
ifu
and the corresponding functional
with F{t,u)
= / f(t,s)ds.
As usual (see Theorem II-2.1) it is easy to
Jo
see that every solution it of (3.9) satisfies a\ < u < fo and is a solution of (1.1). Existence of the solutions u\ andu2- Define
f(t, Using Assumptions (H) and Proposition III-2.7, these are strict upper and lower solutions. Theorem 1.1 applies then with a = on and /? = /3j, which implies the existence of solutions u\ G C&1 and U2 € C a 2 . Existence of a third nontrivial solution. Observe that as <j> is coercive, it satisfies the Palais-Smale Condition and we can apply Theorem 3.5 with a — a2 and /3 = /3j. This observation proves the existence of a solution U3 G Cg([a, b]) \ (C^ 1 U Ca2), i.e. M3 ^ «i and U3 ^U2. The main problem is to see that u% is not the trivial solution. To this aim, we prove that c=<j>(u3) < 0 = 0(0). Define T from (3.7), with a = a2, 13 = fa, and 7 G T by 7 (a)
= 2e[(2S - 1 ) ^ + (1 - \2s -
where y>2(*) — sin(27r|^). Observe that 7(0) = -2ev3i -< fa, 7(1) = 2ev?i >- o a , « i -< -tetpi
< 7(s) < Aetpi -< fa for all s G [0,1].
3. THE MINIMAX METHOD
213
Moreover, b
(
[(
- l)Vl(t)
+ (1 - |2a -
< j [2e2[(2s -A2e2[(2s - l)^i(t) + (1 - |2s - l|)^ 2 (t)] 2 ] dt < e2(6 - o)[(2s - l) 2 (Ai - A) + (1 - |2s - 1|)2(A2 - A)] < e2(6 - a)[(2s - I ) 2 + (1 - |2s - 1|)2](A2 - A)
<-£(&-a). Hence c < — \{b — a) < 0, which implies the third solution «3 is nontrivial. Claim - The function U3 changes sign. Assume U3 > 0 and define 77 = max{r > 0 | U3 — T
> (A - Ai)(i - to)r] // ^i(s) rfs > 0, Jto J which contradicts the minimality of U3 — 77^1 for t — to. It follows that 3 > 5(pi >- Q!2 which contradicts the localization of 113. We prove in a similar way that U3 cannot be negative. Therefore U3 changes sign. An additional solution can be obtained combining variational method and degree theory. Here we impose that the slope ~ ^ lies between two consecutive eigenvalues for small values of u. Theorem 3.8. Assume that f € Cl{{a,b] x R) satisfies Assumptions (H) together with
214
IV. VARIATIONAL METHODS
(i) there exist p, q, k > 2 (k E N) and 8 > 0 such that for a.e. t G [a, b] and all u € [—5,5],
fwj t/tere exist fj. < Ai = /b^a..2 and i? > 0 swc/i i/iai for a.e. t £ [a, b] and allueR with \u\ > R, u Then the problem (1.1) has at least four nontrivial solutions Uj such that u\ -< 0, u^ >~ 0 and U3, U4 change sign. Proof. Part 1 - Existence of solutions ui -< 0 -< «2- As in the proof of Theorem 3.7, we choose strict lower solutions «j and strict upper solutions Pi such that a\ ~< —Stpi -< /3i = — £<^i
and 02 = eyi -< 6
where £ € ]0,5[. R-om Theorem II-2.6, the problem (1.1) has two solutions ui -< /3i and U2 >- «2 such that u\ is the maximum solution in [ai,/3i] and «2 is the minimum solution in [a^i/^]- Moreover we can prove as in the proof of Theorem 3.7 that Mi -< —bipi
a n d U2 >- 8ip\.
Part 2 - Existence of a changing sign solution 113. Consider now the modified problem , . «" + /(t,«) = 0, l j «(o) = 0, u(6)=0, where /(*,«) = /(«,«! (*)), = /(*,«), = f(t,U2(t)),
ifu<m(t), ifui(t)
and the corresponding functional
with F(t, u(t)) = J£ f(t, s) ds. As usual (see Theorem II-2.1), it is easy to see that every solution of (3.10) satisfies ttj < u < u2 and is a solution of (1.1). As in the proof of Theorem 3.7, we see that the problem (3.10) and
3. THE MINIMAX METHOD
215
hence (1.1) has a third solution U3 ^ 0, which changes sign and is such that
and deg (I - KN, Q2) = 1
where, for R large enough, ||u'||oo
fi2 = {u € Co ([a, 6]) \ a2 ^ u ^ U2 + 1, |K||oo < -R}, and /f, iV are defined from N :C^{[a,b]) -> L^a,/)),^1-> /(*,«) +m(t)u, K:L 1 (o.6)-»C 0 1 ([o,6]),/i M iir/i where Kh is the unique solution of -u" + m{t)u = /i(i),
u(a) = 0, ti(6) = 0
and m is defined in Assumption (H). Suppose 4>(ui) >
Let us prove next that, for r small enough, \deg(I-KN,B(0,r))\ = l.
216
IV. VARIATIONAL METHODS
Consider the homotopy % = 0, «" + s /(£, u ) + ( l - s u(a) - 0, u(b) = 0.
^
;
Notice that for a.e. t £ [a,b] and all u G [max(—8,ui(t)),mm(u2(t),6)] k
p
(
u Any solution u of (3.11) is such that u" + A(t)u = 0,
) 2
q
u(a) = 0, u{b) = 0,
with A(t) := s&jffiL + (1 - s ) 2 ! 2 . We can find r > 0 small enough such that if u e 9 5 ( 0 , r ) we have A(i) G [p,g] C]Afc,Afe+i[ for t G [a,6]. By eigenvalue comparison, we conclude that u = 0. Hence, using Theorem A-1.6, we can write |deg(/-/S-JV.BfO.r))! = |deg(/ - K(^I),B{0,r))\
= 1.
We come now to the contradiction deg (I-KN,B(0tR)) = deg(I-KN,ni)_+deg(I-KN,n2) +deg (/ - KN, JS(u3, e)) + deg (7 - KN, B(0, r)) = 2-1 , which proves existence of an additional nontrivial solution u$ of (3.10). and Recall that such a solution lies in [1*1,1x2] and from the definition of 31 U2, "4 ^ C'' U C a a . Arguing as in Theorem 3.7, we prove then that this solution changes sign. 3.5
A Dual Application
The result we consider in this section is somewhat dual to Theorem 3.7. Here we assume the slope ^ " ' crosses the two first eigenvalues the other way round. It is lower than the first eigenvalue for small values of |u| and between two consecutive ones for large |u|. Theorem 3.9. Let Assumptions (H) be satisfied together with (i) there exist fj, < Xi = /{,^o\a and 5 > 0 such that for a.e. t € [a, b] and allu G R with \u\ < S,
3. THE MINIMAX METHOD
217
(ii) there exist p, q, k > 2 (k £ N) and R > 0 such that for a.e. t 6 [a,6] and all M 6 R with \u\ > R, ^n
(fc + l)27T2
TTien i/ie problem (1.1) /ias a£ 2eas£ t/iree nontrivial solutions Ui such that u\ -< 0, W2 >- 0 and U3 changes sign. Proof. As in the proof of Theorem 3.7 we introduce a, 0:2, strict lower solutions and /?i, /? strict upper solutions of (1.1) such that Pi -< a -< 0 -< 13 -< a2. Part 1 - Existence of a solution «i -< 0 of (1.1). Define the sequence of modified problems u" + fp(t,u)=0, . . [6 u(o) = 0, u{b) = 0, ' where fp(t,u) = f(t,u), — 0,
H-p
Notice first that solutions of (3.12) are such that u < 0. Next arguing as in Theorem III-3.3 there exists K > 0 such that for all p > 0 and all u e W2'l{a,b), solution of (3.12) with u $ C01 U Ca, we have ||u|| c i < K. Let p > K be fixed and consider the modified functional
with Fp(t, u) = / / p (i, s) ds. Since
218
IV. VARIATIONAL METHODS
Part 3 - Existence of a solution u3 of (1.1) which changes sign. Let
sin(ir|Es), fi{t) = sin(27r|Ef) and define
with F(t, u)=
Jo
f{t, s) ds.
Claim 1 : 4>(R((2r — l)y>i+(l — |2r —l|) -~o° o,s \R\ —+ oo, uniformly inr £ [0,1]. Let ft e L ^ a , 6), h(t) > 0, be such that
f(t,u) >pu-h(t), ifu>0, f(t,u)
f
\u'2{t) -p(t)u2(t)}dt+
I
h(t)\u(t)\dt.
Ja
Hence, we compute + (1 - |2r -
- l)V?(t) + (1 - |2r - 1|)V?W) -p((2r - l)Vf(t) + (1 - |2r - 1|) W ) ) ] dt / h(t)\(2r - l)Vl(f) + (1 - |2r - A 2 )f / ((2r - 1) V(«) + (1 - |2r / h(t)\(2r - l)(pi(t) + (1 - |2r Ja and the claim follows. Notations - Introduce the following notations : m = inf &(u) > -co, C0nc S = {(reC([O,l]2)Co1([a,6]))| a(r, 0) = 0,
K
' "
Observe first that if R is large enough, a(r, s) = Rs((2r - l ) V l + (1 - |2r defines a surface element in E. Also, it is easy to see that c> m.
3. THE MINIMAX METHOD
219
Claim 2 : Let u(t;uo) be the minus gradient flow defined with r = m — 1. Then, for any 5 G]0,1[, there exists UQ G Co([a,b]) such that
Vt e [0, +oo[,
u(t, «o) G 4T\[c -&,c + 5[)\ (Cfi U Ca).
Fix 5 G ]0,1[, let a G £ be such that c<
max 6(o~(r,s)) < c + S
and define E = ^)c~'5 U C 3 U C a . Observe that E is an open positively invariant set. Define A(E) = {u0 G C^{[a,b\) | 3t > 0, u(t;u0) € E}. Let us prove there exists u0 € a(To-) \ A(E). Otherwise, as in Proposition 3.4, we prove there exists T so that for all (r, s) G [0,1]2, u(T; <x(r, s)) G £. Notice that ff(r, s) = u(T;
which contradicts the definition of c and proves the claim. Hence, by construction of A(E) and as <^>(u(-;wo)) is non-increasing, for alHG [0,+oo[,
Claim 3 : The Palais-Smale Condition (PS) holds. Let us decompose the space ffo(o,&) into orthogonal components HQ(a,b) = H~ © ff+, where H~ = span{
* ^ * + 1 and consider
u),u+ - u_) ff i - J \(u+ + u_)(u+ - u_ ds
— I (f(s,u) — Xu)(u+ —u-)ds. Ja We deduce from the assumptions that rb
f / {f(s,u) - Xu)(u+ - u _ ) d s < <J||u|||a +Cb|MU 2 < 5||u||£,2 +Ci||u|| ff i Ja for some S <
fc+1
2~
k
and some CQ, C\ > 0. On the other hand, we have
/ {((u'+)2 - (u'_)2)-X((u+)2
- (u_) 2 )]rfs > S\\u\\2L2 +e\\u\\2H>,
220
IV. VARIATIONAL METHODS
for some e > 0. It follows then fT7/Affl ^
n
\ "^> e*ll
ll^
Z*"' limit
('Q 1 Q \
H1
Let now ( u n ) n be such that (f>(un) is bounded and V(f>(un) —§ 0. We have We deduce from (3.13) that ||un||#i dition follows from Theorem A-2.1.
ls
bounded and the Palais-Smale Con-
Conclusion - For every 5 €]0,1[, we have proved the existence of UQ such that for all t e [0, +oo[, u(t;u0) G
5,c + S}) \ (Ca U
Hence, there exists an increasing unbounded sequence (tn)n so that = -\\V(l>(u(tn;u0))\\2H,
^
-» 0.
As the Palais-Smale Condition holds, we prove as in Theorem 3.5 the existence of u3 e Co([a, b}) \ (Ca U C 3 ) solution of (1.1) such that ^(^3) = c. We deduce then from the localization U3 € CQ([O, b}) \ (Ca U C 3 ) that U3 changes sign. 3.6
A Seven Solutions Theorem
Combining the results of the previous sections we obtain the existence of at least seven solutions in case the slope "^ is bigger than the second eigenvalue for small and large values of |u| and there exist lower and upper solutions. Theorem 3.10. Let f € C1([a, 6] x R) satisfy Assumptions (H) together with (i) there exist fj,, X, k > 2 (k £ N) and 5 > 0 such that for a.e. t € [a, b) and all u S [—S, 6], f(tu) (fc + l) 2 7r 2 k2w2 A 71 \2 < A* ^ ^ < h z T T (0 — a)2 u (0 — ay (ii) there exist p, q, I > 2 (I 6 N) and R > 0 such that for a.e. t € [a, 6] and all u £ R with \u\ > R, s^l
z==
71
—Q ^
v? ^ P —
(6 — ay
u
77
To
— ^i+1)
(o — a)^
3. THE MINIMAX METHOD
221
(Hi) there exist a and (3 lower and upper solutions of (1.1) which are not solutions of (1.1) such that a -< 0 -< /?. Then the problem (1.1) has at least seven nontrivial solutions such that two of them are negative, two are positive, and three change sign. Proof. We now observe that arguing as in Theorem 3.8, we obtain four solutions in Ca l~l C 3 , one negative, one positive and two changing sign, and using the arguments of Theorem 3.9 we prove the existence of three solutions outside Ca l"l C13, one negative, one positive, and one changing sign.
3.7
Crossing of the Second Eigenvalue
Using the arguments of the preceding sections, we can also consider the case where the nonlinearity crosses only the second eigenvalue and prove the existence of a solution which changes sign. This use of the lower and upper solution method in a case of interaction only with the second eigenvalue is the main interest of this section. Theorem 3.11. Let Assumptions (H) be satisfied together with (i) there exist /u, A and 6 > 0 such that for a.e. t £ [a, b] and all u € [—5, 6], Al = 77
rx < \X <
(0 — a)
< A < 77
2
u
rn = A2;
(0 — ay
(ii) there exist p, q, k > 2 (k 6 N) and R > 0 such that for a.e. t e [a,b] and allu € R with \u\ > R,
(b — ay
u
{b — ay
Then the problem (1.1) has at least one nontrivial changing sign solution. Proof. Let ipi(t) = sin(7r|5^), ip2(t) — sin(27r|5^) and define (f>(u) =
fb ,u>2
r
, ^
/ [—£-* — F(t, u(t))\dt, where F(t, u) = I f{t,s)ds. Ja Jo Claim 1 - For every p € [—8,5] \ {0} we have
222
IV. VARIATIONAL METHODS
which implies that for p G [5,6} \ {0},
2 / P2V?(t)dt- I F{t,pyx{t))dt Ja
Ja
fb
0(tt) = | / u^tjctt- / F(t,u(t))dt J
Ja
}b 2
>\\
fb
u' {t) d t - \ l Ja
u2{t) dt > \{1 -j;)
Ja
fb
u'2{t) dt > e.
Ja
Claim 3 -4>{R{{2r — l)y>i + (l — |2r — lj)y>2)) —* — °o &s \R\ ~* °o, uniformly inr € [0,1]. The proof is similar to the one in Theorem 3.9. Notations - Introduce the following quantities :
^ such that max0(i?((2r-l)v?i+(l-|2r-l|)¥? 2 )) < - 1 ,
a(t) = fpxft) and
r£[0l]
S = {a G C([0,l]a,C01([ol6])) | ff(r,0) = 5(2r - l)
3. T H E MINIMAX M E T H O D
223
Using a Generalized Intermediate Value Theorem or a degree argument, it is easy to see that given a G £, the function (JQ <x(r, s)y>\ dt, \\cr(r, s)||#i — rj) has a zero (ro,So) G [0,1] 2 , i.e. any surface a G S intersects the set {u € £ | ||u||#i = *?} It follows then from Claim 2 that for such a a we have max <j)(cr(r, s)) > e, which implies c > e > 0. Claim 4 - Let u(t; uo) be the minus gradient flow defined by
u(0) = uo, where_ u0 G C&([a, b]) and tp G CX(R, [0,1]) is such that ^(s) = 1 if s>e/2 and ^J(S) = 0 i/ s < 0. T/ien /or any <$ G ]0, e[, ^e?r exists a uo G CQ([O, 6])
Vte[0,+oo[,
u^uojG^-Hlc-tf.c
Fix 5 G ]0, e[, let a G S be such that c<
max
and define E = 4>c~s I) C13 U Ca- Observe that E is an open positively invariant set. Define A{E) = {u0 G C^({a,b}) | 3t > 0, u(t;w 0 ) G S } . We prove as in Theorem 3.9 there exists uo G o{Ta) \ A(E) such that Vf G [0,+oo[,
i t ( t ; « o ) G ^ 1 ( [ c - 5 , c + (5])\(C /3 UC < a ).
Claim 5 : The Palais-Smale Condition (PS) holds. The argument is the same as in Theorem 3.9. Conclusion - For every S G]0,e[, we have proved the existence of uo such that for all t G [0, +oo[, u(t,u0)
G ^- x ([c - 6,c + 6}) \ (Ca U C"3).
Hence, there exists an increasing unbounded sequence ( t n ) n so that -(j}(u{tn;u0))
= -1|V
-> 0.
As the Palais-Smale Condition holds, we prove as in Theorem 3.5 the existence of v G C^{[a, b}) \ (Ca U C3) solution of (1.1) such that (j)(v) = c. As c > 0 we know that v ^ 0, and as in Theorem 3.7, we prove that v changes sign.
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IV. VARIATIONAL METHODS
We can also consider the case where / crosses the second eigenvalue the other way on. Theorem 3.12. Let Assumptions (H) be satisfied and assume : (i) there exist \> X2 — r**a\i and 5 > 0 such that for a.e. t £ [a, b] and all u G [-£,5], u
>
(ii) there exist p, q and R > 0 such that for a.e. t 6 [a, 6] and all u 6 with \u\ > R, 7T 2
f(t,U)
.
47T2
Then the problem (1.1) /ias at least one nontrivial solution u which changes sign. Exercise 3.1. Prove Theorem 3.12.
4 A Reduction Method The main difficulty in using the lower and upper solutions method is to find these functions in concrete situations. Consider a variational problem associated to a functional (j): H —> R, where H is a Hilbert space. The idea of this section is to consider a foliation described by an equation T(u) — £, where F : H —> M. and £ is a real parameter. Assume <j>(u) has a minimum on each of the sets r - 1 ( ^ ) . We can then define the reduction function
In this section, we show how this function can help to find lower and upper solutions so that we can apply the previous theory. 4.1
Lower and Upper Solutions in a Weak Sense
The lower and upper solutions we obtain using the reduction have to be considered in a weak sense which is related to the notion of weak solutions i.e. of the derivative of the functional.
4, A REDUCTION METHOD
225
Definition 4.1. A function a G H 1 (a, b) is a weak lower solution of (1.1) if a(a) < 0, a(b) < 0 and for every v G HQ (a, b), v > 0, ,6
/ \a'(t)v'(t) - f{t, a(t))v(t)] dt < 0. Ja In a similar way, a function ft G Hl{a,b) is a weak upper solution of (1.1) ifP{a) > 0, P(b) > 0 and for every v G H%(a, b), v>0,
f
\P'{t)v'{t)-f{t,i3(t))v{t)\dt>Q.
Ja
Definition of strict weak lower and upper solutions can be made as in Definition III-2.2. Remark 4 . 1 . Notice that if a G W2>1(a, b) is a W 2il -lower solution of (1.1), it is a weak lower solution. Similarly a W2'1 -upper solution of (1.1), which is in W2<1(a, b), is a weak upper solution. R e m a r k 4.2. If (j>(u) is defined from (1.2), the function a E HQ(O,,b) is a weak lower solution of (1.1) if for every v G HQ (a, b), v > 0,
Similarly the function /? G H$ (a, b) is a weak upper solution of (1.1) if for every v G HQ (a, b), v > 0,
This remark turns out to be a key to prove existence of such lower and upper solutions. Theorem 4.1. Assume f : [a, b] x M —> R is L1 -Caratheodory and there exist a, /3 G Hl{a, b) weak lower and upper solutions of (1.1) with a < f3. Then the problem (1.1) has at least one solution u G W2'l(a,b) such that for allt£ [a,b], a(t) < u(t) < P(t), If moreover a andj3 are strict, then for R > 0 large enough, deg(I—T, O) = 1, where T and Cl are defined by
/ G(t,s)f{s,u(s))ds Ja
226
IV. VARIATIONAL METHODS
with G(t, s) the Green's function of -u" = h(t), u{a) = 0, u(b) = 0 and Sl = {i*eC 0 1 ([a,i]) | a ^ u ^ / 3 , Hu'lloo < R}. Proof. Consider t h e modified problem
u" + f(t,1(t,u))=0, u{a) = 0, u(b)=0,
.
{
. '
where ^y(t,u) = max{a:(£),min{u,/3(£)}}. As f(t,^f(t,u)) is bounded, this problem has a solution. Let us prove that every solution u of (4.1) is such that a < u < /?. Assume on the contrary that maxt(a — u) > 0. Observe that (a — u)+ £ Ho (a, b), (a — u)+ ^ 0 and we have the contradiction 0= / [u'((a-u)+)'-/(t,a)(a-u)+]dt Ja <*- u)'((a - u)+)' dt = - [ ((a- u)+)n dt < 0. J
Hence u > a. In a similar way, we prove u< /3. The degree result is proved as in Theorem III-2.8. In a similar way we can consider for example the periodic problem (1.6). Definition 4.2. A function a € Hper(a,b) is a weak lower solution of (1.6) if for every v e H^er(a, b), v>0, f b
/ a'(t)v'(t) - f(t, a(t))v{t)] dt < 0. Ja
In a similar way, a function j3 £ H^er{a,b) is a weak upper solution of (1.6) if for every v 6 H^er{a,b), v>0,
i
/ 3 W ( i ) - f(t,(3(t))v(t)]dt>0.
Ja
Definition of strict weak lower and upper solutions can be made as in Definition III-l.l.
4. A REDUCTION METHOD
227
Remark 4.3. Notice that if a € W 2>1 (a,6) is a W/2'1-lower solution of (1.6), it is a weak lower solution. Similarly a T¥ 2il -upper solution of (1.6), which is in W2'l{a, b), is a weak upper solution. Remark 4.4. If
{V
every v £ H^er(a,b), v>0, Theorem 4.2. Assume f : [a, b] x E —» R is L1 -Caratheodory and there exist a, P £ H^er(a,b) weak lower and upper solutions o/(1.6) with a < fl. Then the problem (1.6) has at least one solution u G W2
T:C{[a,b])->C{[a,b]):u~
/ G(t,s)[f(s,u(s))
+u(s)}ds
Ja
with G(t, s) the Green's function of -u" + u = h(t), u(a) = u(b), u'(a) = u'(b) and Q = {u£ C([a,b}) | a(t) < u(t) < j3(t) on [a,b]}. Proof. The first part of the proof follows as in Theorem I-1.1 together with Theorem 4.1 and the second part follows as in Theorem III-1.8.
4.2
Foliations and Reductions
In this section we present the framework of the reduction method in some abstract Hilbert space. D e f i n i t i o n 4 . 3 . Given a Hilbert space H (with scalar product norm || \\), we say that F:H - > R
) and
is a foliation of H if it is weakly continuous, surjective, and convex. The sets r - 1 ( ^ ) , where { 6 M , are called the leaves of the foliation.
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IV. VARIATIONAL METHODS
Let us first recall some well known facts of standard convex analysis. For a convex, continuous function F a notion of subgradient dT(u) can be defined (see for instance [248]). The subgradient can be equivalently characterized by means of the right Gateaux derivative DrT{u)(v) := hm t—*'U"'"
C
which exists for all v G H due to the convexity of F. Precisely, it is well known that w e dT(u) if and only if (w,v)
VDEF.
(4.2)
With this notion we can state a necessary condition for solutions of a constraint minimization problem. Proposition 4.3. Let H be a Hilbert space, (j> : H —> E be of class C1, T : H - » R be a foliation and $ e E . Assume u is a solution of min
\T(u)\ < C2
implies
\\u\\ < Cs.
Proposition 4.4. Let H be a Hilbert space, $ : H —> W. be a C1, weakly lower semicontinuous function and letT : H —> M. be a foliation admissible for 4>. Then the reduction function = min 6(u)
r(«)=c
is well defined, continuous, and satisfies f(T(u))
< 0(u),
Vw € H.
(4.3)
4. A REDUCTION METHOD
229
Proof. To see that tp is well defined, for any given ( £ l , take a sequence (un)n such that r(«n) = £,
lim (j)(un) = inf
r(u)=J
Due to the admissibility condition on F, (un)n is bounded in H. Denote again by (un)n a subsequence which is weakly convergent and let u be its weak limit. Since F is weakly continuous, T(u) = £ and, since
r(u)=£
Hence the infimum is a minimum and
Then, taking un € H such that 4>{un) =
m i n 4>{u)
the admissibility condition says that [un)n is bounded in H. If u is the weak limit of a convergent subsequence, which will be denoted again by (un)n, once more T(u) = £ and we have the contradiction
n-*oo
To prove the upper semicontinuity of
f{)
^()
First observe that, as F is surjective, 0 ^ dT(u) which means that for some veH DrT{u)(v) < 0. Also if w 6 dT(v), we have w ^ 0 and 0 < IMI2 < DrT{u){w).
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IV. VARIATIONAL METHODS
We use w to prove that ip is upper semicontinuous at £ from the right. By contradiction assume that for some e > 0 and £„ —> £, £n > £, we have
Since DrT(u)(w) > 0, we can choose tn —> 0 + such that F(u + tnw) — £„. Then we should have
and
which is a contradiction. The same argument, where w is replaced by v, leads to the upper semicontinuity from the left. The set of the constrained minimizers on the leaves of the associated foliation will be denoted by Mi := {u G H | r(«) = £, 4>{u) = v»(0}. Going back to the problems (1.1) and (1.6), remember that a lower solution (resp. an upper solution) is such that for all v > 0 (V(j)(u),v) < 0
(resp. (V^(u).t;) > 0).
On the other hand, any solution u of the corresponding constraint problem min
r(«)=c
is such that for some A £ R and w £ dT(u), V<j)(u) = Xw.
Hence, if X(w, v) is one-signed for all v > 0, the corresponding solution u turns out to be a lower or an upper solution. A first step in this direction requires that the Hilbert space H is endowed with a partial ordering and that F is monotone. Definition 4.5. Let H be a Hilbert space with partial ordering <. We say that F is an increasing function on H when u
implies T(u) < F(u).
Lemma 4.5. Let H be a Hilbert space with partial ordering < and T : H —> R be convex and increasing. Then for any w £ dT(u) and any v G H with v > 0 we have (w,v) > 0 .
4. A R E D U C T I O N METHOD
231
Proof. For all v > 0, we can write (w,v) > F(u) — T(u — v)>0. Consider a situation where
> 0,
!>-¥>(&) < 0
(4.4)
(see for instance [220]). The question is now how to connect the sign of the Dini derivative of ip at a given £ to the sign of the Lagrange multipliers of a given u G Mj. The next statement provides a satisfactory answer. Proposition 4.6. Let H be a Hilbert space, <j> : H —> R be a C1, weakly lower semicontinuous function and letT : H —> K be a foliation admissible for
= D+
t-+o+
t — 4>{u) < ii m 4>(u r i _ i+—tw)I — rL2
we conclude that A > 0, the strict inequality arising as soon as D+(p(£) > 0.
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IV. VARIATIONAL METHODS
To prove (b), let us choose a v G H such that DTT(u)(v) < 0. Recall that this is possible as 0 ^ dT(u). It follows then that T(u + tv) is a strictly decreasing function for t € K + small and
t-» o+
Also,
< ^ flu + fa)-flu) As (w,v) < DrT(u)(v) < 0, we conclude that A < 0, the strict inequality arising as soon as £)_(£) < 0. Remark 4.5. It is not difficult to realize that everything works unchanged when the convex foliation F is replaced by a smooth function, which is moreover a regular constraint, which means that VF(u) ^ 0 holds for all u G H. In that case, we can even sharpen the previous result, by proving that holds for the (unique) Lagrange multiplier A of any u G Mj. As a consequence, Proposition 4.6 and Lemma 4.5 trivially imply the following statement. Proposition 4.7. Let H be a Hilbert space with partial ordering, (f>: H —> ]R be a C1, weakly lower semicontinuous function and let T : H —> R be an increasing foliation admissible for 4>. Let further £ £ R, u S M$. Then
> 0 implies (V<£(u),u) > 0 for all v>0, D-
(V
The main limitation in applying these results is that we have no information on the ordering of the lower and upper solutions. In fact, in case we have a maximum of the reduction ip, we have £i < £2 satisfying
> 0,
D-ipfa) < 0
4. A REDUCTION METHOD
233
(see for instance [220]) and u\ e M^ and uz G Mf2. This means that u\, u
Application to the Periodic Problem
In this section, we use the previous ideas to deal with a nonlinearity which is non-resonant with the first Fucik curve. The basic condition (4.5) is a variant of the so-called Ahmad, Lazer and Paul condition (see [4]). Theorem 4.8. Let f : [a,b] xR —> R be an Ll-Caratheodory function such that b limsup/ F(t,u)dt =+oo, (4.5) o Ja
where F(t, u) = /0" /(£, s) ds and for some
< 0,
> 0 in L1(a, b),
a (t) < liminf f&$- < limsup f-^- < b (t), u—
u
o
U
uniformly in t. Assume moreover there exist \i > u > 0 with _1_ _1_ b-a and a set I C [a, b] of positive measure such that b+(t) < [i, b-(t) < v, for a.e. t £ [a,6], b+(t) < fi, b-(t) < v, for a.e. te I. Then the periodic problem (1.6) has at least one solution u 6 W2>1(a, b). Proof. We just have to prove that (f,(u) = f [Ja
2
F{t, u)\ dt
and
T(u) = r-^— [ [/xu+ - vvT\ dt b — a Ja
defined on Hper(a, b) verify the conditions of Proposition 4.7 and that ip, defined in Proposition 4.4, is not monotone. In that case, we apply Proposition 4.7, Theorem III-3.3 (with the modification given by Theorem 4.2), and Proposition III-3.4 to conclude. It is well known that cf> : Hper(a,b) — M is C1 and weakly lower semicontinuous and it is obvious that F is weakly continuous, convex, and increasing. To see that T is surjective, we just have to observe that if £ > 0, = € anc^ if C < 0, r(£/z/) = £. Let us prove that T is admissible.
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IV. VARIATIONAL METHODS
If it is not the case, there exists a sequence (un)n such that |r(tt n )| < C2, Ci a^d ilun||ffi,,r —» oo- First observe that this result implies ~* 00 If n °t, we have for some D\,
Ci > 4>(un) = J\^f - F(t,
un
and ||Tin||/fi is bounded which contradicts the assumption. For each n, set vn = «n/||wn||oo- Clearly vn satisfies
+ - w-] dt\ < -£\-,
'6-i
(4.6)
and, for every e > 0, there exists 7e € -L^a, b) such that
I
2
(4'?)
2
o &
n
O &
|L, II ' ' ll^nlloo
In particular, for some D2 > 0 and every n, ||u n ||#i er < D^. Up to a subsequence, (vn)n converges weakly in Hper and strongly in C([a,6]) to some function v with ||i>||oo = 1- As e is arbitrary, passing to the limit in (4.6) and (4.7), we have
rb / [fiv+ - vv~) dt = 0, ,2 2 6 Ja
+2
2
V / V V
By the variational characterization of the first Fucik curve given in [78,120],
\-{»—
+ v—)\dt = ja[\-{b+V—
+ b-.V—)\dt = O
and using [78, Lemma 4.3], we know that v satisfies v" + [piv+ - vv~] = 0, v(a) = v(b), v'(a) = v'(b). Hence either v = 0 or v ^ 0 a.e. on [a, b\. The second possibility is excluded as then r v12
v+2
v-2
0 > J j ^ - ( b + ^ - + b_V-Y)]dt
fb v12
J
We conclude that v = 0 which contradicts IMIoo
v+2
V =
/ !
4. A R E D U C T I O N METHOD
235
As liminf ip(£) < — limsup / F(t,^/fi)dt
= —oo,
liminf y>(£) < —limsup / F(t,£/i>)dt = —oo, we conclude that (p has a maximum. The result follows. Our second result concerns the case / is one-sided superlinear. T h e o r e m 4.9. Let f : [a, b] x R —> R be an L1 -Caratheodory function such that ,6
limsup/ F(t,u)dt = +oo, u—
Ja
w/iere F(i, u) = Jo" / ( i , s) ds anrf /or some o_(t) < liminf ^ u » o o
^ u
< 0, 6_ > 0 m £ 1 (a, 6),
u->-oo
< l i m s u p ^ ^ < 6_(t), liminf ^ W
uniformly in t. Assume moreover b- < (jz^) 2 o,.e. on [a,6] and b- < ( j ^ ) 2 on a subset of positive measure of [a, b}. Then the problem (1.6) has at least one solution u € W 2 l l (o, 6). Proof. Again we just have to prove that b
=
/
Ja
u12 [
F(t,u)]dt
and
T(u) = maxu(t) t
defined on Hper(a,b) verify the conditions of Proposition 4.7, and that tp, defined in Proposition 4.4, is not monotone. In that case, we apply Proposition 4.7 and Exercise III-3.2 (with the modification given by Theorem 4.2) to conclude. As in the previous case, the only nontrivial assumption is the admissibility. If it is not the case, there exists a sequence (un)n such that |r(tt n )| < C2, <j>(v-n) < C\ and Hunlltf1 —> 00. Arguing as in the proof of Theorem 4.8, we see also that ||u^||oo —* +00. For each n, set vn = -rr-^Ph Clearly vn satisfies IN H (4.8)
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IV. VARIATIONAL METHODS
and for every e > 0, there exists 7e € Ll{a,b) such that b a -2 2
v2 2
In particular, there exists Di such that for every n, ||un||ffi < A2Hence, up to a subsequence, (vn)n converges weakly in Hper and strongly in C([a, b]) to some function v with ||u~||c» = 1- Passing to the limit in (4.8) and (4.9) and as e is arbitrary, we have
maxu(i)=O and / — f — t As ( j r ^ ) 2 is the first eigenvalue of the Dirichlet problem on an interval of lenght b — a and for some t\ £ [a, b], v € HQ{t\, t\ + b — a), we have t\-\-b—a
/
!
I'i,
2
2 (?L_(^) ?L] 2 b a 2
~
Hence u(i) = — s i n ( j ^ ( t — t{)) and the above equality contradicts 6_ < (fe^r)2 o n a s u bset of positive measure of [a, b], which proves the admissibility and concludes the proof.
4.4
Application to the Dirichlet Problem
Using Theorem III-3.9 and Theorem 4.1, we can obtain the equivalent of Theorem 4.8 for the Dirichlet problem, which we leave as an exercise. Instead we concentrate on other types of results in this section. These results have also their counterpart in the periodic case. Our first result concerns a strong resonance with the first eigenvalue. Theorem 4.10. Let h 6 Lx(0,7r) with J* h(t)sintdt = 0, / : [0,7r] x R —* K be an L1 -Caratheodory function, define F(t,u) = Jo f(t,s)ds and assume lim /(£, u) ~ 0 and lim F(t, u) = 0, \u\—too
|u|too
4. A REDUCTION METHOD
237
uniformly in t G [0,7r]. Then the problem u" + u + f(t,u) u(0)
= 0,
= h(t), ) =
0,
has at least one solution. Proof. We apply Theorem 4.1 or Theorem III-3.9 (with the modification given by Theorem 4.1) and Proposition 4.7 with Jo
and
l
l fir
F : Fo(°>7r) - * R , u - * / Jo Jo It is easy to see that
w" + w = h(t),
w(0) = 0, W{K) = 0,
/ w(t)smtdt = Q, Jo
and
Coo
- / o ' t 2 ^ - 2 ^ + /i(*)^oo(t)] dt. We shall prove that lira
Let (^n)n be an unbounded sequence and u^n G M^ n . We write £„ sin i + it>£n with J o ty^n (t) sin t di = 0. Observe that w%n solves w
'( + ^^n + /(*» Cn sin t + iwSn) = /i(t) + y,n sin t, wc,(0) = 0, t« €n (7r)=O,
. ^
U;
for some /i n G R. Multiplying (4.10) by ioj n and by sint and integrating, we have r«J?
/ o
(t)
[-^
to? (t)
- - % ^ - /(*,CnSint + «fc,(i)HB(*) + fcW^nt*)] * = 0,
2 r
Mn = - / /(*. £n sin t + tf$n (t)) sin i dt. 1" Jo We deduce then from the above equalities that (tO{n)n C i?o(0,7r) and ) n are bounded. As w^n is a solution of (4.10), these results imply the
238
IV. VARIATIONAL METHODS
existence of a function k G LX{Q,"K) such that for all n, \w'J(t)\ < k(t) on [0,7r]. Hence, up to a subsequence, w^n —> w in HQ(0,IT). Passing to the limit in (4.10) we see that w = Woo. Moreover (n
J^
2
-
2
€n
and passing to the limit, we observe that \ -h(t)Woo(t)]dt
— Coo.
As the sequence (£ n ) n is arbitrary, we conclude that lim tp((i) = Coo, |C|-»oo
and tp is not strictly monotone. Our second application concerns oscillating potentials. Theorem 4.11. Let f : [Q,n] x E —* R be an L1 -Caratheodory function such that, for a.e. t G [0, n] and all « e i , \f(t,u)\
p / = — oo and limsup
( j ^ )
= +oo.
|€| |€|-»oo JO
ICI— o Jo
where F(t,u) = J^ f(t, s) ds. Then the problem u" +u + f{t,u)=0, u(0) = 0, u(vr) = 0,
has infinitely many solutions. Proof. We apply Proposition 4.7 with
and /
It is easy to see that cf> is C1 and weakly lower semicontinuous, and that F is weakly continuous, convex, increasing, and surjective. Moreover the admissibility of F can be proved as in Theorem 4.8. It remains to study the reduction function tp denned by Proposition 4.4.
4. A R E D U C T I O N METHOD
239
For every £ and u% E M%, denote u^ = £ sin t+w% with JQ w^(i) sin tdt — 0. As in the proof of Theorem 4.10, we prove that ||u^|| #1 and, using (4.10), are bounded independently of £. Hence we have
ip{£) = (f>{v,£) = I [—
=-rlt
F(t, ^sint +
di
sJdt+oi
Jo as |^| —> oo and liminf (p(£) = — oo and
lim sup
These results imply that we have two unbounded sequences (£,), and {Vi)i with D+ip(£i) > 0 and D-ip^i) < 0. By Proposition 4.7, we have an infinite number of lower and upper solutions c^ = Vi sin t + wVi and ft = £isin£ + ui&. These functions can be ordered cti -< (Hi < on+i because ||ttfM||ci is bounded independently of/x. We conclude by applying Theorem 4.1.
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Chapter V
Monotone Iterative Methods 1
An Abstract Formulation
Let Z b e a Banach space. An order cone K C Z is a closed set such that for all u and v G K, u + v G K, for all t G 1R+ and u G K, tu G K, if u G if and —u G i£ then rx = 0. Such a cone K induces an order on Z: u < v if and only if v — u & K. We write equivalently u < v or v > u. The cone is said to be normal if there exists c > 0 such that 0 < u < v implies ||u|| < c||u||. The following theorem gives conditions for an increasing sequence (an)n to converge to a fixed point of T. Theorem 1.1. Let X C Z be continuously included Banach spaces so that Z has a normal order cone. Let a and 0 G X, a < /3, £ = {u£X\a
(1.1)
and let T : £ —» X be completely continuous in X. Assume that the sequence (an)n defined by a0 = a,
an- Tan-i
(1.2)
is bounded in X and for all n G N a n < an+i < 0Then the sequence (an)n converges monotonically in X to a fixed point uofT such that a
242
V. MONOTONE ITERATIVE METHODS
Proof. Claim - The sequence (ctn)n converges in X. The sequence («„)„ is increasing and included in £. As the set A — {an \ n £ N} is bounded in X, T(A) is relatively compact in X. Hence, any sequence (ank)k C (a n ) n has a converging subsequence in X and therefore in Z. As the order cone is normal and the sequence is monotone, the sequence itself converges in Z, i.e. there exists u € Z so that a u. It follows that all such subsequence converging in X have the same limit u, which implies x an ->u. Next, we deduce from the continuity of T that u is a fixed point of T. D A similar result holds to prove the convergence of the sequence (/3n)nTheorem 1.2. Let X C Z be continuously included Banach spaces so that Z has a normal order cone. Let a and fi € X, a < /3, £ be defined by (1.1) andT : £ —> X be completely continuous in X. Assume the sequence (/3n)n defined by (1.3) 0o=0, 0n=T0n-i, is bounded in X and for all n € N 0n > 0n+i > a. Then the sequence (/3n)n converges monotonically in X to a fixed point vofT such that As a corollary we can write the following result which deals with maps T that are monotone increasing, i.e. u < v implies Tu < Tv. Theorem 1.3. Let X C Z be continuously included Banach spaces so that Z has a normal order cone. Let a and f3£X,a
2. THE WELL-ORDERED CASE
243
Proof. Claim 1 - The sequence (an)n converges in X to a fixed point umin of T such that a < umin < fi. As T is monotone increasing, we prove by recurrence that for any n £ N, an < an+i < /?. Hence, (an)n C £ and since T{£) is relatively compact in X the sequence (an)n is bounded in X. The claim follows now from Theorem 1.1. Claim 2 - The sequence (/3n)n converges in X to a fixed point umax of T such that umin < umax < /3- As for Claim 1 we prove existence of a fixed point umax such that a < umax < (3. Next, as T is monotone increasing and a < f3 we deduce by recurrence that an < (3n. At last, going to the limit we obtain umin < umax. Claim 3 - Any fixed point u G £ ofT verifies umin
1 2(n
TV
M
l
\ —
l 2n+l"
The assumptions of the theorem are satisfied for a = —1/2 and /3 = 1/2. The sequence (an)n — {—^)n converges to the fixed point u = 0. However u = 0 is not minimal since the points un = — 2n 1 +1 € [—|, ^] are fixed points ofT.
2
Well-ordered Lower and Upper Solutions
2.1 The Periodic Problem Consider the periodic boundary value problem u" = f(t,u), u(a)=u(b), u'{a)=u'{b),
{
,„. >
where / is a continuous function. Our aim is to build an approximation scheme, easy to compute, that converges to solutions of (2.1). To this end, given continuous functions a and P, and M > 0, we consider the sequences (ccn)n and (f3n)n defined by a0 = a, < - Man = f(t, on-i) - Man-U an(a) = an(b), a'n(a) = a'n(b)
(2-2)
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V. MONOTONE ITERATIVE METHODS
and /%' - M(3n = f(t, /3n_i) The approximations an and /3n are "easy to compute", in the sense that for every n, the problems (2.2) and (2.3) are linear and have unique solutions which read explicitely ,6 ,6
/
where G(t, s) is the Green function of the problem u" -Mu = f(t), u(a) = u(b), u'(o) = u'(6).
.
K
'
Clearly this method does not avoid numerical difficulties such as those related to stiff systems. The next theorem proves the convergence of the an and /3 n . T h e o r e m 2 . 1 . L e t a and {3 € C2([a, b}), a < @ and
E := {(t, u) e [a, 6] x E | a(t)
(2.5)
Assume f : E —» R is a continuous function, there exists M > 0 such that for all (t,ui), (t,U2) S E, u\ < 162 implies f(t,U2) — f(t,ui) < M(u2 — ui) and for all t € [a, b] a"(t) > f(t,a(t)), a(a) = a(b), a'(a) > a'(b), Then the sequences (an)n and {j3n)n defined by (2.2) and (2.3) converge monotonically inCl(\a,b\) to solutions umin andumax of (2.1) such that a < umin < umax < p. Further, any solution u of (2.1) with graph in E verifies I^min ™ ^ — ^ m o i
Proof. Let X = C ^ M ] ) , Z = C([a,6]), K = {u <E Z \ u(t) > 0 on [a, 6]} be the ordered cone in Z and £ be defined from (1.1). Define the operator T : S -> X by Tu(t) = / G(t,«)(/(«, u(s)) - Mu(s)) ds, Ja
2. THE WELL-ORDERED CASE
245
where G(t, s) is the Green function of (2.4). This operator is continuous in X and monotone increasing. Further, T{£) is relatively compact in X, a
Remark Recall that existence of the minimal and maximal solutions umin and umax follows from Theorem 1-2.4. Next, we consider a derivative dependent problem u" = f(t,u,u'), u(a) = u(b), u'(a) = u'(b).
,
.
K 0)
As above, given a, (3 € Cl{\a, b]) and L > 0, we consider the approximation schemes a0 = a, < - Lan = f{t, « „ _ ! , < _ ! ) - Lan-i, an(a) = an(b), a'n(a) = a'n(b) and
0o=0, 0" - L0n - /(t^-i.iflj,.!) - L/3n_i, A.(o)=A,(6), f3'n(a) = &(b). Such problems lead to a major difficulty. A straightforward application of the previous ideas would be to assume that for any u\, U2, v\ and v%, ui < U2 implies f(t,u2, V2) — f{t,U\, v{) < L(u2 — u\). This would mean that / does not depend on derivatives. The following result turns out the difficulty. Theorem 2.2. Let a and j3 e C2([a, b]), a < 0 and E : = {(t, u, v) G [a, 6] x M2 | a{t)
/3(t)}.
(2.9)
Assume > R is a continuous function, there exists M > 0 such that for all (t,ui,v), {t,U2,v) G E, ui
implies f(t,U2,v) — f{t,u\,v)
< M(u2 — ui);
(2-10)
there exists N > 0 such that for all (t,u,v\), (t,u,V2) € E, \f(t,u,v2) - f(t,u,t»i)| < N\v2 - wi|;
(2.11)
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V. MONOTONE ITERATIVE METHODS
and for all t £ [a, b] a"{t) > fit, a(t), a'(t)),
a(a) = a(b), a'[a) > a'(b),
At last, let L > 0 be such that L>M+
N2 N i — + — v / iV 2 + 4M
(2.12)
and for all t £ [a,b] f(t,a(t),a'(t))
- f{t,P(t),0'(t))
+ L(0{t) - a(t)) > O.
(2.13)
Then, the sequences (an)n and (0n)n defined by (2.7) and (2.8) converge monotonically in C1([a, 6]) to solutions u andv of (2.6) such that a 0 satisfies -w" + N\w'\ + (M + l)w = h{t) > 0, w(a) - w{b), w'(b) > w'(a). Thus, using the maximum principle (Theorem A-5.3 with p(t) = —Nsignw' and q(t) — —(M + 1)) we can prove that if a ^ j3, our assumptions imply a < f3 on [a,b]. Also if u is a solution of (2.6) such that a ^ u ^ /3, we have a < u < j5. (b) It is clear from Remark (a) that the assumptions on L are satisfied if L is large enough so that the theorem applies for any values of M and N which satisfy the assumptions (2.10) and (2.11). (c) The conditions on L are immediately satisfied with L — M if the function / does not depend on the derivative v! (i.e. N = 0). (d) If a (resp. 0) is a solution of (2.6), we have an = a (resp. j3n — /3) for all n e N. Proof of Theorem 2.2 : The proof uses Theorems 1.1 and 1.2 with X = CHK&D, Z = C([a,b\) and K = {u £ Z | u(t) > 0 on [a,b]} as the ordered cone in Z. Let £ be defined from (1.1). The operator T : £ —+ X, defined by = f G(t,s)(f(s,u(s),u'(s)) - Lu(s))ds, Ja where G(t,s) is the Green function of (2.4) with M = L, is completely continuous in X. With these notations, the approximation schemes (2.7) and (2.8) are equivalent to (1.2) and (1.3).
2. THE WELL-ORDERED CASE
247
A : Claim - Let L > 0 satisfy (2.12). Then the functions an defined recursively by (2.7) are such that for all n s N , (a) an is a lower solution, i.e. aZ(t)>f(t,an(t),a>n(t)), ^14> an(a) = an(b), a'n(a) > a'n(b), (b) an+i > an. The proof is by recurrence. Initial step : n = 0. The condition (2.14) for n = 0 is an assumption. Next, w = a i — ao is a solution of -w" + Lw = a%(t) - f(t, ao(t), a'0(t)) > 0, w(a) =w{b), w'(a) <w'{b). Hence, we deduce (b) from the maximum principle (Theorem A-5.3). Recurrence step - 1st part : assume (a) and (b) hold for some n and let us prove that <+l(*)>/(*.On+l(*)X+l(t)). an+i(a) = an+1(b), a'n+1(a) > a'n+1(b). Let w = an+i — an. We have = /(*>ain+i.a4+i) ~ /(*) a n,a{,) - L(an+i -an) < M(an+i - an) + N\a'n+1 - a'n\ - L(an+i - an) = (M - L)w + N\w'\. On the other hand, w satisfies -w" + Lw = h(t), w(a) = w(b), w'(b) - w'(a) = A,
(2.15)
with h(t) := o£(£) - /(t,a n (t),a^(i)) > 0 and .4 > 0. Its solution iu reads [ /"* r- hw(t) = k \ h(s) cosh v i ( ^ + s — t) ds 1
fb
+ / ft(s) cosh V i ( ^ p + 1 - s) ds + A cosh \/L(t where Hence, to pprove a n +i is a lower solution, we only have to verify
f +NvT| sinh y/Z( ^
+ s -1) |] h{s) ds < 0,
) ,
248
V. MONOTONE ITERATIVE METHODS rb rb
/ \(M Jt L
-L) cos
and (M - X) cosh VI(i Since h is nonpositive and
) + 7VvT| sinh y/L(t - ^ ) | < 0.
(M - L) coshx + iVVT| sinhx| < (M - L + NVZ)\sinhx\ for all x G R, we obtain (M - L)w + JV|u/| < 0 if M - L + Ny/Z < 0, which follows from (2.12). Recurrence step - 2d part: assume (a) and (b) hold for some n and let us prove that a n +2 > c*n-i_i. The function w = cx-n+2 — &n+i satisfies (2.15), where
Ht)
<+i(t) - f(t,an+1(t),a'n+1(t))
and
A = 0.
Erom the previous step h(t) > 0 and the claim follows from the maximum principle (Theorem A-5.3). B : Claim - Let L > 0 satisfy (2.12). Then the functions f3n defined recursively by (2.8) are such that for all n £ N , (a) (3n is an upper solution, i.e. 0n(a) = /?„(&), Fn(a) < Pn(b), (b) A,+i < PnThe proof of this claim parallels the proof of Claim A. C : Claim - an < j3n. Define, for all i e N, Wi = fa — on and h{(t) := / ( i , a i ( t ) , ^ ( i ) ) - /(t,A(*)./3*(*)) + L (AW - a iW)The proof of the claim is by recurrence. Initial step : a.\ < j3\. The function IUJ is a solution of (2.15) with h = ho > 0 and A = 0. Using the maximum principle (Theorem A-5.3), we deduce that wi > 0, i.e. «i < f5\. Recurrence step : Let n > 2. Ifhn-2 > 0 and a n _i < /3n_i, t/ien /in_i > 0 and ara < /9n. First, let us prove that for all t € [a, b], the function hn-\ is nonnegative. Indeed, we have
2. THE WELL-ORDERED CASE
249
Notice that wn-i is a solution of (2.15) with h(t) = 7in_2(£) > 0 and A ~ 0. Hence, we can proceed as in the proof of Claim A to show that /i n _i > 0. It follows then from the maximum principle (Theorem A-5.3) that wn is nonnegative, i.e. an < /?„. D : Claim - There exists R > 0 such that any solution u of u" > f(t, u, u'), u(a) = u(b), u'{a) = u'(b), with a < u < (3 satisfies ||u'||oo < R- We deduce from the assumptions that u" = f(t,u,u') + h(t), where h(t) > 0, f(t, u, v!) + h(t) > - max \f(t, u, 0)| - iV|u'|
(2.16)
and F — {(t,u) \ t € [o, b],a(t)
E : Claim - There exists R > 0 such that any solution u of u" < fit, u, u'), u(a) = u(b), u'(a) = u'(b), with a < u < f3 satisfies ||w'||oo < R- The proof repeats the argument of Claim D. F : Conclusion -We deduce from Theorems 1.1 and 1.2 that the sequences (an)n and (/3n)n converge monotonically in C1([a, 6]) to functions u and v which are solutions of (2.6) such that a
Further, any solution u of (2.6), such that a
250
V. MONOTONE ITERATIVE METHODS
Proof. Existence of extremal solutions umin and umax follows from Theorem 1-5.6. If a is not a solution, we deduce from the above Remark (a) that Umin > a- Hence, choosing L > 0 large enough so that (2.12) is satisfied and f{t, a(t),a'(t)) - f{t, umin(t), u'min{t)) + Huminit) - a{t)) > 0 on [a,b], we can apply Theorem 2.2 with /3 = umin and obtain that u — lim an is such that a < u < umin, whence u = uminn—*oo
Similarly, we prove lim j3n = umax. n—too
Remark Observe that the problem ')2 = sint, u(0) = U(2TT), U'(0) = u'(27r), cannot be worked out from Theorem 2.2 as it does not satisfy (2.11). However, we can proceed as follows. Notice first that such a problem satisfies a Nagumo condition. Next, we know that lower and upper solutions, a and P € [—1,1], of problems that satisfy this Nagumo condition have a priori bounded derivatives: ||o:'||oo and ||/3'||oo < R- We can modify then the equation for |u'| > R so that the same a priori bound on the derivatives can be obtained for the modified problem together with (2.11). It follows then that the approximations defined from (2.7) and (2.8) are the corresponding approximations for the modified problem so that convergence follows from Theorem 2.2. The following theorem uses an approximation scheme which, from a computational point of view, is more involved since it uses piecewise linear equations. On the other hand, the analysis is simplified since the coefficients in the approximation scheme are the Lipschitz constants on the nonlinearity. Notice also that in this theorem we use different approximation schemes for the an and for the /3n. Theorem 2.4. Let a and /3 € C2i[a, b]), a < f3 and E be defined by (2.9). Assume f : E —> M is a continuous function, there exists M > 0 such that for all (t,ui,v), (t,u2,v) £ E, u\
implies fit,U2,v) —
fit,ui,v)<Miu2—ui);
there exists N > 0 such that for all it,u,v\), it, u, v^) € E, \fit,u,v2) - fit,u,vi)\
and for all t G [a, b]
a"it) > fit, ait), a'it)),
a(o) = a(6), a'(a) > a'(6),
2. THE WELL-ORDERED CASE
251
Let ao = a, and /?o = /? Then the problems
a'n - N\a'n - a'n_x\ - Man = /(*,«„_!, aj,^) - Ma n _i, «nW = «n(&). < ( a ) = a n( & ) and #,' + N\(3'n - ^ _ j (3n(a) = 0n(b), p>n(a) = (3>n(b), define sequences (an)n and (/3n)n that converge monotonically in Cx([a, to solutions umin and umax of (2.6) such that
Furthermore, any solution u of (2.6), such that a < u < f3, verifies I^min _ ^ _ V"max
Proof. Let X = Cl{[a,b\), Z = C([a,b}), K = {u G Z \ u(t) > 0 on [a,b}} be the ordered cone in Z and £ be defined from (1.1). A : Definition of a completely continuous operator T. u £ X the problem
v" - N\v' - u'(t)\ ~Mv = f(t,u(t),u'(t)) v{a) = v(b), v'(a) = v'(b),
Claim - For any
- Mu(t),
.
K
. '
has a unique solution v. Notice that if k > 0 is large enough, — k and k are well-ordered lower and upper solutions of (2.17). Existence of a solution of this problem follows then from Theorem 1-5.3 with
= max w(t) > 0. te[a,b]
We compute w'(t0) — v'2(to) — t>i(£o) = 0 and obtain the contradiction w"(t0) = Mw(t0) + N(\v'2(t0) - u'(to)\ - \v[(t0) - «'(«o)|) > 0. Hence w < 0. Similarly we prove w > 0, which implies w = 0, and the solution of (2.17) is unique. Now, we can define the operator
T :S -» X,ui->Tu, where £ is defined from (1.1) and Tu is the solution of (2.17). This operator is completely continuous.
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V. MONOTONE ITERATIVE METHODS
B : Claim - For all n € N, an+\ > an. Initial step : w = a\ — ao = Ta — a > 0. Notice that w solves the problem w" - N\w'\ - Mw = f(t, a(t),a'(t)) - a"(t) < 0, w(b) - w(a) = 0, w'{b) - w'(a) = a'(a) - a'(b) > 0. The claim follows from the maximum principle argument used in Claim A. Recurrence step - Assume an+i —an>0 and prove w = OJn+2 — an+l > 0. The function w satisfies w" - N\w'\ - Mw = f(t,an+1,a'n+1) - f(t,an,a'n) - M(an+1 - an) - N\a'n+1 - a'n\ <0 and w(a) = w(b), w'(a) = w'(b). The claim follows from the same maximum principle argument. C : Claim - an < /?. The proof repeats the argument of the previous claims. D : Claim - The sequence (a n ) n is bounded in X. The functions an solve problems u" + N\u'\ ~Mu = gn(t), u{a) = u(b), u'{a) = u'(b), where gn = N(\a'n\ + \a'n - a'n_x\ - K . J )
+iV|<_11 + f(t, <*„_!, c£-l) - /(*. «n-I, 0) +/(t,an_i,0)-Man_i > /(t,a n _i,0) - M o n _ i > f(t,p,O)-M0. As the gn are uniformly lower bounded, the boundedness of the an follows now from Proposition 1-4.5. E : Conclusion -The convergence of the sequence (an)n to a solution umin of (2.6) follows from Theorem 1.1. To prove um%n is a minimum solution, let u be any solution of (2.6) such that a
Similarly, we prove the convergence of the sequence (/3n)n to a maximal solution using Theorem 1.2. 2.2
The Neumann Problem
The Neumann problem
u" = f(t,u), u'(a) = 0, u'(b) = 0,
.
.
(
'
2. THE WELL-ORDERED CASE
253
where / is a continuous function, can be investigated along the lines of the periodic problem. Theorem 2.5. Let a and 0 e C2([a, b]), a < 0 and let E be defined from (2.5). Assume f : E —> R is a continuous function, there exists M > 0 such that for all (t,ui), (t,U2) € E, u\ < U2 implies f{t,U2) — f(t,ui) < M(u2 — u\) and for all t € [a, b] a"(t)>f(t,a(t)), P"(t)
a'(a)>0, a'(b)<0, / 3 » < 0 , /3'(6) > 0.
Then the sequences (an)n and (@n)n defined by a0 = a , o£ - Man = /(i,a n _i) - Man-i, a'n(a) = 0, a'n(b) = 0, and W - M/3n = f(t,/3n^) - Mpn-x, 0>n(a) = 0, 0'n{b) = 0, converge monotonically in C1([a, b]) to solutions umin and umax of (2.18) such that a < umin < umax < 0Further, any solution u of (2.18) with graph in E verifies Umin
Exercise 2.1. Prove the above theorem. Hint : See the proof of Theorem 2.1. To deal with derivative dependent equations u" = /(*,«, u'), u'(a) = 0, u'{b) = 0,
r2 {
,^ }
we can use the ideas of Theorem 2.2. Theorem 2.6. Let a and 0 G C2([a,b}), a < 0 and let E be defined from (2.9). Assume f : E —> E is a continuous function, there exists M > 0 such that for all (t,u\,v), (t,U2,v) G E, Mi < U2 implies f(t,u2,v)
— f{t,u\,v)
< M(u2 — wi);
254
V. MONOTONE ITERATIVE METHODS
there exists N > 0 such that for all {t,u,v\), (t,u,V2) G E, \f(t,u,v2)-f(t,u,v1)\ and for all t £ \a, b] a"(t) > f{t,a{t), a'(t)),
P"{t) < f(t,/?(«),/?'(*)), 13'{a) < 0, /?'(&) > 0. Then, if L is large enough, the sequences (an)n and (f3n)n defined by a0 -a,
a'n - Lan = / ( ^ a n - i , ^ ! ) - Lan_i, a'n(a) = 0, a'n(b) = 0 and
&{a) = 0, ^(6) = 0 converge monotonically in Cx{[a, b\) to solutions umin such that a < umin < umax < p.
and umax
of (2.19)
Further, any solution u of (2.19), such that a < u < /?, verifies Umin
Umax-
Exercise 2.2. Prove the above theorem. Hint : See the proofs of Theorem 2.2 and Corollary 2.3, or [65]. We can investigate problem (2.19) along the lines of Theorem 2.4.
Theorem 2.7. Let a and /3 € C2([a, b]), a < /3 and let E be defined from (2.9). Assume f : E —> R is a continuous function, there exists M > 0 such that for all (t,ui,v), (t,U2,v) £ E, wi < us implies f(t,U2,v) — f(t,ui,v)<M(u2
— ux);
there exists N > 0 such that for all (t,u,vi), (t,u,V2) S E, \f(t,u,v2) - f(t,u,vi)\ and for all t G [a, 6] a"(t) > f(t, a(t),a'(t)), P"(t) < f(t,P(t),p'(t)),
< N\v2-vi\;
a'(a) > 0, a'(b) < 0, f3'(a) < 0, F{b) > 0.
2. THE WELL-ORDERED CASE
255
Let CXQ = a and (3o = P- Then the problems
< - AT|o4 - < _ i | - Man = /(*,a n _i,<-i) a'n(a) = 0, a'n{b) = 0 and
/ 3 » = o, #,(&) = o define sequences (an)n and (/?„)„ t/iai converge monotonically in C1([a, 6]) io solutions umin andumax of (2.19) SMC/I ttat a < Wmin < "max < /?
Further, any solution u of (2.19), SKC/J i/iai ct
Exercise 2.3. Prove the above theorem. Hint: See the proof of Theorem 2.4 or [234]. 2.3
The Dirichlet Problem
Consider the Dirichlet problem u" = f(t,u), u(a) = 0, u(b) = 0,
.
l
,
Uj
where / is a continuous function. Theorem 2.8. Let a and 0 G C2([a, b]), a < fi and let E be defined from (2.5). Assume f : E —> R is a continuous function, there exists M > 0 such that for all (t,ui), (tjU?) E E, u\ < W2 implies fit, u^) — f(t, ui) < M{u% — «i), and for all t E [a, b] a"(t)>f(t,a(t)),
a(a) < 0, a(b) < 0,
P"(t)
P{a) > 0, 0(b) > 0.
T/ien i/ie sequences (an)n and (/?„)„ defined by a0 = a, Man-i, n = /(t,a n _i) an(a) = 0, a n (6) = 0
256
V. MONOTONE ITERATIVE METHODS
and /%' - M/3n = /(*,&_!) - MPH-I, Pn(a) = 0, /?„(&) = 0 converge monotonically in C1([a, 6]) to solutions umin and umax such that a < umin < umax < p.
of (2.20)
Further, any solution u of (2.20) with graph in E verifies
Exercise 2.4. Prove the above theorem. Hint: See the proof of Theorem 2.1. In case of derivative dependent equations u" = f{t,u,u'), u(a) = 0, u(b) = 0,
.
{
,
}
approximation schemes similar to (2.7) and (2.8) do not work. Indeed if we try to repeat the argument of Theorem 2.2, we have to prove a priori bounds on the derivatives of lower and upper solutions (see Parts D and E). As noted in Remark 2.1, it implies we use one-sided Nagumo condition which, for the Dirichlet problem, imposes the lower and upper solutions to satisfy the boundary conditions. We can think this condition is not very restrictive since the first iterates a\ and /?i already satisfy such conditions. However L must also verify (2.13) and this might not be the case even for large values of L. We have then to consider alternative approximation schemes. Theorem 2.9. Let a and f3 G C2([a,b\), a < 0 and let E be defined from (2.9). Assume f : E —> M is a continuous function, there exists M > 0 such that for all (t,ui,v), (t,U2,v) £ E, ui
implies f(t,U2,v)-f(t,ui,v)<M(u2
— ui);
there exists N > 0 such that for all (t,u,vi), (t,u,V2) € E, f{t,U,V2)-f(t,U,V1)\
and for all t £ [a, b] a"(t) > f{t,a(t), a'(t)),
a(a) = 0, a(b) = 0,
0"{t) < f{t, 0it),TO).
/3(o) = 0- W) = 0-
2. THE WELL-ORDERED CASE
257
At last, let KQ G C([a,6]) be such that KQ(O) > 0 and for all t £ [a, b], K0(t) = -K0(b + a-t). Then, for L large enough, the sequences (an)n and (/?„)„ defined by a0 = a, 0O = 0, - Lan+1 = f(t,an,a'n) - ^LK0(t)a'n - Lan, an+i(a) = 0, an+i(b) = 0 1
and (3Z+1 - VLK0(t)p'n+1
- Lpn+1 = f(t,pn,f3'n) - yiK0{t)p'n
- L/3n,
/ 3 n + 1 ( a ) = 0, /3 n + 1 (6) = O
converge monotonically in C1([a,6]) to solutions umin and umax of (2.21) such that Oi < Umin < Umax < 0-
Further, any solution u of {2.21) with graph in E verifies
Exercise 2.5. Prove the above theorem. Hint : See the proofs of Theorem 2.2 and Corollary 2.3 or [63]. 2.4
Bounded Solutions
In this section we consider bounded solutions of the differential equation u" + cu' = f(t,u).
(2.22)
To this end, we define the set BC(R) - {u £ C(R) | u £ L°°(R)}. Theorem 2.10. Let a and P £ C2(R) D LOC(M), a < 0 and let E := {(t,u) £ R2 | a(t)
allt£R
a"(t)+ca'(t)>f(t,a(t)),
0"(t)+c0>(t)
258
V. MONOTONE ITERATIVE METHODS
define sequences (an)n and (/3n)n C BC(M) that converge monotonically and uniformly on bounded intervals of M. to solutions umin and umax of (2.22) such that a < umin < umax < (3. Further, any solution u of (2.22) with graph in E verifies Proof. First recall that given p G SC(IR), the problem y" + cy'-Xy=p{t) with A > 0 has a unique solution in BC(M) given by = /
G(t,s)p(s)ds,
where G(t, s) =
exp(f (S - 0) exp(-i/|t - s\),
with v = yX + ^-. This result implies the (an)n and (/3n)n are uniquely defined. Claim 1 - If an is such that for all t g R aZ(t)+ca>n(t)>f(t,an(t)), an{t) < then the function ccn+i G BC(M) defined by a'n+1 + ca'n+1 - Man+1 = f(t, an) - Man, satisfies for all t G R, ) + ca'n+l(t) >
f(t,an+l(t)),
It is enough to observe that an and (3 satisfy for alH G M. < ( 0 + ca'n(t) - Man(t) > f(t,an(t)) - Man(t), p"(t) + cf3'{t) - M/9(0 < f(t,0(t)) ~ M0{t) < f(t,an(t)) - Man(t). Hence, by Theorem II-5.6, the solution a n +i of u" + cu' — Mu — f(t, an) — Man
3. THE REVERSED ORDER CASE
259
is such that an < a n +i < f3 on R and then satisfies also for all t € R < + 1 ( t ) +ca'n+1(t) - Man+1(t) = f(t,an(t))
- Man(t)
>f(t,an+1(t))-Man+1(t), Claim 2 - If (3n is such that for allt 6 R
ft(t) + cl3'n{t) u. Similarly we have umin
3
Lower and Upper Solutions in Reversed Order
The monotone approximation method can be used in case the lower and upper solutions are in the reversed order /? < a. This method works for any boundary value problem such that a uniform anti-maximum principle holds. This is the case for the periodic and the Neumann problems. We can also work bounded solutions on R, However the method does not apply to the Dirichlet problem since, in this case, we only have a non-uniform anti-maximum principle. 3.1
The Periodic Problem
Consider the periodic boundary value problem u" = f(t,u), u(a) - u(b), u'{a) = u'(b),
.
K
, '
260
V. MONOTONE ITERATIVE METHODS
where / is a continuous function. In Section 2, we have built an approximation scheme for solutions of (3.1) based on the maximum principle. Here, we consider a similar approach based on the anti-maximum principle. Given continuous functions a and /?, and M > 0, we consider the sequences (an)n and (/3n)n defined by cto = a,
a n (a) = an(&), < ( a ) = a^(6) and ft + M0n = f(t, /?„_!) + M/?n_!,
(3-3)
pn(a)=pn{b), &(a)=P'n(b). If M is not an eigenvalue of the periodic problem, i.e. M ^ (fz^) 2 with n G N, the functions a n and /3n, solutions of (3.2) and (3.3) can be written explicitly =
,6 / b
Pn(t)=
f Ja
where G(t, s) is the Green function of the problem u" + Mu = f(t), u(a) = u(b), u'(a)=u'(b).
{
, ° '
The next theorem indicates a framework to obtain convergence of the an and f3n to extremal solutions of (3.1). Theorem 3.1. Let a and (3 e C2{\a, b}), /3
(3.5)
Assume f : E —> 1R is a continuous function, there exists M £]0, (^r^)2] such that for all (t,ui), (t,U2) 6 E, u\ < u% implies f(t,U2) — f(t,u\) > —M(u2 — u\) and for all t € [a, b] a"(t) > f(t,a(t)), a(a) = a(b), a'(a) > a'(b), P"{t) < f(t, P{t)), 0(a) = P(b), f3'(a) < f3'(b).
3. THE REVERSED ORDER CASE
261
Then the sequences (an)n and (/3n)n defined by (3.2) and (3.3) converge monotonically in ^([a, b]) to solutions umax and umin of (3.1) such that P < umin < umax < a. Further, any solution u of (3.1) with graph in E verifies
Proof. Let X = Cx([a,&]), Z = C([a,b}), K = {u & Z \ u(t) > 0 on [a,b]} be the ordered cone in Z and S = {ueX\(3
(3.6)
The operator T : £ —* X, defined by = / Ja
G(t,s)(f{s,u(s))
where G(t,s) is the Green function of (3.4), is continuous in X and monotone increasing (see Corollary A-6.3). Further, T(S) is relatively compact in X, /3 < T/3 and a > Ta. The proof follows now from Theorem 1.3, where a and (3 have to be interchanged. Next, we consider the derivative dependent problem u" = f(t,u,u'), u(a)=u(b), u'{a)=u'{b).
. > °'
K
As above, given a, /3 € (^([a, b]) and L > 0, we consider the approximation schemes o^ + Lan = f(t,an-1,o^_1) + Lan-1, an(a) = an(b), <(o) = a'n(b)
(3-8)
/?£ + L0n = /(*,&-!, fl,-i) + ^A»-i. (3n(a) = pn(b), pn(a) = Fn{b).
(3 9)
and
"
The following result paraphrases Theorem 2.2 in a case where the antimaximum principle applies. Theorem 3.2. Let a and /3 € C2([a, b]), /3 < a and E := {(*, u, v) e [a, b] x R 2 | (3{t)
(3.10)
262
V. MONOTONE ITERATIVE METHODS
Assume f : E —> R is a continuous function, there exists M G]0, (J^ such that for all (t,u\,v), (t,U2,v) G E, ui < u2
implies f(t,u2,v)
- f(t,ui,v)
> -M(u2 - u i ) ;
there exists N > 0 such that for all (t,u, v\), (t,u,V2) G E, \f(t,u,v2)-f(t,u,vi)\
<JV|«a-«i|;
and for all t G [a, b] a"(t) > f(t,a(t),a'(t)), (3"(t) < f(t,P(t),(3'(t)),
a(a) = a(b), a'(a) > a'(b), /3(o)
At last, let L G [M, ( ^ ) 2 ] be such that (L - M) cos \ / I ( ^ ) - i V \ f l s i n VZ(^)>0 and f(t, a(t), a'(t)) - f(t, 0(t), ff{t)) + L(a(t) - /?(£)) > 0.
(3.11)
Then, the sequences (an)n and (/3n)n defined by (3.8) and (3.9) converge monotonically in (^([a, b}) to solutions u and v of (3.7) such that (3
Tu(t)= /
G(t,s)(f(s,u(S),u'(s))
Ja
where G(t, s) is the Green function of (3.4) with M replaced by L, is completely continuous in X. With these notations, the approximation schemes (3.8) and (3.9) are equivalent to (1.3) and (1.2). A : Claim - Let L G [M, (jr^) 2 ] satisfy (3.11). Then the functions an defined recursively by (3.8) are such that for all n G N, (a) an is a lower solution, i.e. an(a) = an(b),a'n(a)>a'n(b), (bj an+x < an. The proof is by recurrence.
[6
^>
3. THE REVERSED ORDER CASE
263
Initial step : n = 0. The condition (3.12) for n = 0 is an assumption. Next, w — c*o — a.\ is a solution of w" + Lw = < ( t ) - f(t,ao(t),a'o(t)) > 0, w(a) = w(b), w'(a) > w'(b). Hence, we deduce (b) from the anti-maximum principle (Corollary A-6.3). Recurrence step - 1st part : assume (a) and (b) hold for some n and let us prove that a'Ji+1(t)>f(t,an+1(t),a'n+1(t)), an+1(a) = an+1(b), a'n+l(a) > a'n+1{b). Let w = an — an+i > 0. We have n+i
f(,n+u
n+i)
= -f(t, an,a'n) + f{t,an+i,a'n+1) - L(an - an+1) < M(an - an+i) + N\a'n+1 - a'n\ - L(an - an+i) = (M - L)w + N\w'\. On the other hand, w satisfies w" + Lw = h(t),
w(a) = w(b), w'(a) - w'(b) = C,
with h(t) := a£{t) - f{t,an{t),a'n(t))
(3.13)
> 0 and C > 0. Observe that
w(t) = /
/
io
it
Hence, using (3.11) and denoting D = 2v / Xsin\/£(^ a ), we compute {M -L)w(t) + N\w'(t)\ < — [C[(M - L) cos \ / £ ( ^ -t) + N^/Z\ sin y/Zi^
-t)\]
rt
+ / h(s)[(M - L) cos VL(^ Ja
+ s~t)
+ NVL\ sin VZ{^
+ s -1)\] ds
+ I h(s)[(M - L) cos V T ( ^ +1 - s)
V\ sin>/!(*=* +1 - s)\] ds] Hence a n +i is a lower solution.
264
V. MONOTONE ITERATIVE METHODS
Recurrence step - 2d part: assume (a) and (b) hold for some n and let us prove that an+2 < oin+i- The function w = an+i — an+2 satisfies (3.13), where h(t) := a£+1(t) - f(t,an+1(t),a'n+1(t))
and C = 0.
From the previous step h(t) > 0 and the claim follows from the antimaximum principle (Corollary A-6.3). B : Claim - Let L e [M, (jf^) 2 ] satisfy (3.11). Then the functions (3n defined recursively by (3.9) are such that for all n £ N, (a) fin is an upper solution, i. e. f3n(a) = &(&), p'n(a) < (3'n(b), (b) 0n+1 > Pn-
The proof of this claim parallels the proof of Claim A. C : Claim - an> /3n. Define, for all t € N, w; = Qj - ft and hi(t) := f(t, <*(«),£*&)) - f{t, pi(t), fl(t)) + L(ai(t) - ft(t)). The proof of the claim is by recurrence. Initial step : ot\ > ft. The function u>i is a solution of (3.13) with h = h0 > 0 and C = 0. Using the anti-maximum principle (Corollary A-6.3), we deduce that ui\ > 0, i.e. a i > ft. Recurrence step : Let n > 2. ///i n _2 > 0 anrf a n _i > fti-i, i/ien /in_i > 0 and a n > /?„. First, let us prove that for all i £ [a, b], the function /i n _i is nonnegative. Indeed, we have ftn_! = /(*, On-i, < _ i ) - f(t, pn-U ^ _ x ) + L K _ ! - /? n _0
> -M(a n _ x - /?„_!) - JVK_! - pn_x\ + L(an^ - (3^) =
(L-M)wn-1-N\w'n_1\.
Recall that wn-i is a solution of (3.13) with h(t) = hn-2{t) > 0 and C = 0. Hence, we can proceed as in the proof of Claim A to show that /i n _i > 0. It follows then from the anti-maximum principle (Corollary A-6.3) that wn is nonnegative, i.e. an > j3n. D : Claim - There exists R > 0 such that any solution u of u" > f(t,u,u'),
u(a) = u(b), u'{a) = u'(b),
with P < u < a satisfies ||u'||oo < R. We deduce from the assumptions that u" = f(t,u,u') + h(t),
3. THE REVERSED ORDER CASE
265
where h(t) > 0, f(t,u,u') + h{t) > - m a x F |/(i,u,0)| - N\u'\ and JF = {(t,u) | t G [a, b],/3(t)
u{a) = u(b), u'{a) = u'(b),
with f3 < u < a satisfies WU'WOQ < R. The proof repeats the argument of Claim D. F : Conclusion - We deduce from Theorems 1.2 and 1.1, where a and /3 have to be interchanged, that the sequences (an)n and {/3n)n converge monotonically in Cl([a, b}) to functions u and v such that 0 < v < a and {3 < u < a. Further Claim C implies v < u. Example 3.1. Consider the problem u" — k arctan v! + cu3 — sin t, u(0) = u{2n), u'(Q) = u'(2n), where c > 0 and k > 0. This problem has a = c~1//3 and /3 = —c"1/3 as lower and upper solutions. To satisfy the assumptions of Theorem 3.2, let M = 3c1/3 and N — k, choose L e]3c1/(3, l/4[ and assume c < 12~3
and
k < L=^=p- cotanVX7r.
As in Section 2, we can modify the analysis in case we agree to compute the approximations an and /3n from piecewise linear problems. The following result adapts Theorem 2.4 to lower and upper solutions in reversed order. Theorem 3.3. Let a and j3 € C2([a,b]), /3 < a and let E be defined by (3.10). Assume f : E —» R is a continuous function, there exists M > 0 such that for all (t,ui,v), (t,U2,v) E E, u\
implies f(t,U2,v) — f(t,ui,v)
> — M(u2 — u\),
there exists N > 0 such that for all (t,u,vi), (t,u,V2) € E, \f(t,u,v2) - /(t,u,t»i)| < N\v2 - vi\ and for all t £ [a, b]
a"(t) > f(t,a(t),a'(t))t P"{t) < f(
a(a) = a(b), a'(a) > a'(
266
V. MONOTONE ITERATIVE METHODS
Assume also b — a < 20(M, N/2), where l_arctanh^A£, N FN*-M
6{M, N) =
ifO<M
I
77' I
N
£ — arctan
,
ifQ
Let do — ot and 00 = 0- Then the problems
al - N\a'n - < _ ! | + Man = f(t,On-i, aj,-i an{a) = an(b), a'n{a) = a'n(b) and
define sequences (an)n and (/3n)n that converge monotonically in C1([o, b]) to solutions umax and umin of (3.7) such that
<
Further, any solution u of (3.7), such that j3
Umax-
Proof. Let X = ^([a^]), Z = C([a,b}), K = {u e Z \ u{t) > 0 on [a,6]} be the ordered cone in Z and £ be denned from (1.1). A : Definition of a completely continuous operator T. Claim 1 - For any u G X the problem
v" - N\v' - u'(t)\ +Mv = f(t,u(t),u'(t)) v(a) = v(b), v'(a) — v'(b),
+ Mu{t),
(3.15)
has a solution v. Consider the problem v" - s\v' - u'{t)\ + Mv = a(t),
v(a) = v{b), v'(a) = v'
(3.16)
with s e [0, N] as an homotopy parameter and a e C([a,b]). Assume that solutions of (3.16) are not a priori bounded in X. Hence, there exists sequences (sn)n C [0,N] and (vn)n C X such that vn solves (3.16) with s = sn and limn-^ Hfnllc1 — + 00 - We define then vn — vnj\vn\c^ and going to subsequences we can assume _ C1 _ vn —+ v,
_ " L2 „ , vn —i v a n d sn —» s.
3. THE REVERSED ORDER CASE
267
Going to the limit in (3.16), we obtain that v solves v" - s\v'\ +Mv = 0,
v(a) = v(b), v'(a) = v'(b).
If M < s 2 /4, Theorem A-6.2 implies v = 0. On the other hand, if M > s 2 / 4 . we have b - a < 26(M, JV/2) < 20 (M, s/2) < 2\{M, -s/2) (see (A6.3) and (A-6.4)) and the conclusion iJ = 0 follows from the same theorem. Hence, in all cases, this contradicts lim
H->
Tu,
where £ is defined from (1.1) and Tu is the solution of (3.15). This operator is completely continuous. B : Claim - For all n S N, an > an+\. Initial step : w = ao — a.\ = a — Ta > 0. Notice that w solves the problem w" + N\w'\ + M m = a"{t) - /(*, a(t), a'(t)) > 0, w{a) - w(b) = 0, w'{a) - w'{b) = a'(a) - a'(b) > 0. The claim follows from the anti-maximum principle (Theorem A-6.2). Recurrence step - Assume an — an+i > 0 and prove w = an+\ — an+2 > 0. The function w satisfies w" + N\w'\ + Mw = f(t,an,a'n) - f(t,an+i,a'n+1) >0
+ N\a'n+1 -a'n\ + M{an - an+1)
and the claim follows from the anti-maximum principle (Theorem A-6.2). C : Claim - an > (3. The proof repeats the argument of the previous claims.
268
V. MONOTONE ITERATIVE METHODS
D : Claim - The sequence (an)n is bounded in X. The functions an, n > 1, solve the problems u" + N\u'\ + Mu = gn(t), u{a) = u(b), u'(a) = u'(b), where
n _i > f(t,0,Q) As the p n are uniformly lower bounded, the boundedness of the an follows now from Proposition 1-4.5.
E : Conclusion - The convergence of the sequence (an)n to a solution umax of (3.7) follows from Theorem 1.2, where a and 0 have to be interchanged. To prove umax is the maximum solution, let u be any solution of (3.7) such that 0 < u < a. We apply then Theorem 1.2 with the a and /3 of this theorem replaced respectively by u and a and obtain uma,x = lim an > u. n—>oo
Similarly, we prove the convergence of the sequence ({3n)n to a mimimal solution using Theorem 1.1. 3.2
The Neumann Problem
The Neumann problem u"=f(t,u), u'(a) = 0, u'(b) = 0,
{A
l)
where / is a continuous function, can be investigated along the lines of the periodic problem. Theorem 3.4. Let a and @ 6 C2([a,6]), 0 < a and let E be defined from (3.5). Assume f : E —* M is a continuous function, there exists M € ]0, (2(^y) 2 ] such that for all (t, m), (t, u2) G E, ui < u^ implies f{t,U2) — f{t,u\) > —M(u2 — «i) and for all t € [a, b] a"{t) > f(t, a(t)), 0"(t)
a'(o) > 0, a'(b) < 0, 0>{a) < 0, 0'(b) > 0.
Then the sequences (an)n and (Pn)n defined by ao = a, < + Man = f(t, a n _!) a'n(a) = 0, a'Jb) = 0,
3. THE REVERSED ORDER CASE
269
and
Po = 0 , /%' + Mpn = f(t, /3n_0 + MPn-x,
# » = o, /?;(&) = o, converge monotonically in Cl{[a,b\) to solutions umax such that
and umin of (3.17)
P < Umin < Umax < a. Further, any solution u of (3.17) with graph in E verifies Umin
Umax-
Exercise 3.1. Deduce the above theorem from Theorem 1.3. To deal with derivative dependent equations u" = f(t,u,u'), u'(a) = 0, u'(b) - 0,
,
„.
W
' *>
we can use the ideas of Theorem 3.2. T h e o r e m 3.5. Let a and 0 e C2([a,b\), jS < a and let E be defined from (3.10). Assume f : E —» R is a continuous function, there exists ] such that for all {t,ux,v), (t,u2,v) S E,
m
implies f(t,U2,v) —
f(t,ui,v)>
there exists N > 0 such that for all (t,u, i>i), (t,u,V2) £ E, \f(t,u,v2) - f(t,u,Vl)\
< N\v2 - « i | ;
and for all t € [a, 6] a"(t) > f{t,a(t),a'(t)), P"{t) < f{t,/3(t),P'(t)),
a'(a) > 0, a'(b) < 0, Pi") < 0, f3'(b) > 0.
At last, let L e [M, (2,£_a) )2] be such that
(L - M) cos \/I(6 - a) - iVvTsin VZ(b - a) > 0 and 0.
270
V. MONOTONE ITERATIVE METHODS
Then, the sequences (an)n and (/?„)„ defined by ao = a,
a'^+Lan
= /(t,a n _i,a4_ 1 ) + La!n_i, < ( o ) = 0, a'n(b) = 0,
and 0% + LPn = f(t,0n-i,0'n_l) + L&-i, P'n(a) = 0, p'n(b) = 0, converge monotonically in ^([a, b]) to solutions u and v of (3.18) such that (3 < v R is a continuous function, there exists M > 0 such that for all (t,ui,v), (t,U2,v) € E, ui
implies f(t,U2,v) — f(t,u\,v)
> — M(u2 — ui)\
there exists N > 0 such that for all (t,u,vi), (t,u,V2) € E, \f{t,u,v2) - f{t,u,vi)\ <
N\v2-Vl\;
and for all t £ [a, b] a"{t) > f(t, a{t), a'(t)), a'{a) > 0, a'(b) < 0, 0"(t) < f(t,p(t),P'(t)), 0>(a) < 0, /3'(6) > 0. Assume also b — a < 8(M, N/2), where 8 is defined from (3.14). Let ao = a and 0o = 0. Then the problems a£ - N\a'n - a^_j| + Man = /(^an-i.a^j) + Man_i, a'n(a) = 0, a'n{b) = 0, and 0Z + N\0>n - ^ _ i | + M0n = f(t,/3n-l,P'n-l) 0>n(a) = O,l3'n(b) = O,
+ ^ - 1 ,
define sequences (an)n and ({3n)n that converge monotonically inCl{[a, b}) to solutions umax and umin of (3.18) such that P < Umin < Umax < a.
3. THE REVERSED ORDER CASE
271
Further, any solution u of (3.18), such that 0
Exercise 3.3. Prove the above theorem along the lines of the proof of Theorem 3.3. Hint : See [51].
3.3
Bounded Solutions
Existence of bounded solutions of (2.22) can be deduced from lower and upper solutions in the reversed order. The following result parallels Theorem 2.10. Here we use BC2(R) = {u e C2(R) | u,u',u" e L°°(R)}. Theorem 3.7. Let a and 0 € BC2(R), a>0 and let E := {(t,u) € M.2 | j3(t) < u < a(t)}. Assume c > 0, f : E —> 1 is B continuous bounded function such that for all (t, u{), (t, U2) € E, Ml < U2 =^ f{t,U2) - f(t,Ul)
> -\{U2
-U\).
Assume further that for alltEM a"(t) +
ca'(t)>f(t,a(t)),
p"(t) + cP'(t)
c2 a
f (J. a \ _i c2 a
-jPn+l — J\t,Pn) + -^-Pn,
define sequences (an)n and (0n)n C SC(K.) that converge monotonically and uniformly on bounded intervals of K to solutions umax and umin of (2.22) such that P < Umin £ Umax < Oi.
Further, any solution u of (2.22), such that (3
Proof. First observe that given p G BC(M.), the unique solution in $C(R) of
y" + cy'+£y= p(t) is given by y(t)=
f+00 y+oo / G(t,s)p(s)ds, J—oo
272
V. MONOTONE ITERATIVE METHODS
where
Claim - / / an
G(t,s) = ( t - s ) e x p ( - f ( t - s ) ) , = 0, is such that for alltGM.
if s < t, if s > t.
On(t)
then the function an+\ defined by
satisfies for all t € M.,
n an(t) > an+1(t) > It is enough to observe that w = an — an+i solves w" + cw' + ^w = o£ + ca'n - f(t, an). As a'n + ca'n — f(t, an(t)) > 0 we deduce from the form of the solution that w > 0, i.e. an > an+\ and also, for all t £ R , <+i(<) +ca'n+1(t) = f(t,an(t)) + £(an{t)-an+1(t))
> f(t,an+1(t)).
The inequality /3 < a n +i can be proved in the same way. The rest of the proof follows as in Theorem 2.10.
4
A Mixed Approximation Scheme
In the previous section, we considered sequences (an)n and (Pn)n which converge to solutions of a boundary value problem such as (2.1). The reader might have noticed that computing an and /?„ can be a difficult problem. In this section, we give a method of approximations which is simple but provides only bounds on the solutions. We also give assumptions which imply these bounds to be equal, in which case they are solutions. Consider a Dirichlet problem u" = f{t,u, u),
u(a) = 0, u(b) = 0.
(4.1)
To simplify the argument, we assume here / is independent of the derivative u', which is not essential and could be developed as in the previous sections. We shall use the following definition.
4. A MIXED APPROXIMATION SCHEME
273
Definition 4.1. Functions a and j3 G C([a, b}) are coupled lower and upper quasi-solutions of (4.1) if (a) for any t G [a, b), a(t) < /3(t); (b) for any t0 G }a, b[, either D_a(to) < D+a(t0), or there exists an open interval Jo c]a, b[ such that to G Jo, a £ W2'1^), and for a.e. t G J o
(c) for any t0 e ]o, b[, either D~P(t0) > D+/3(t0), or there exists an open interval IQ C ]a, b[ such that to G IQ, /3 G W2'x{h), andfora.e. t G / 0 P"(t)
u{a) - 0, u(b) = 0,
,
,
«" = /(*,«,u),
v(a) = 0, u(6) = 0.
K
}
Proposition 4.1. Lei ao, /?o S C([a, b]), E := {(t,«,«) 11 e [a, 6], u, t; G [a o (t), A)(*)]} (4-3) 1 anrf f : E —>R be an L -Caratheodory function such that f(t,u,v) is nonincreasing in u and non-decreasing in v. Assume ao and (3Q are coupled lower and upper quasi-solutions of (4.1). Then, the sequences (an)n and ((3n)n, defined for n > 1 by
a£ = /(t,an_i,/3n_:i), /%' = / ( t ^ - i . a n - i ) ,
an(a) = 0, an(b) = 0, /3n(a) = 0, &(&) = 0,
converge monotonically in Cl([a,b}) to functions umin and umax. {uminiumax) is a solution of (4.2) such that
The pair
ao < umin < umax < (30. Moreover, any solution (u,v) of (4.2) with ao < u < (3Q, ag < v < (3o is such that
Umax>
Umin
274
V. MONOTONE ITERATIVE METHODS
Proof. LetX =C1([a,b])xC1([a,b]), Z = C([a,b}) xC([a,b}), K = {(u,v) G Z I u > 0,v < 0} and £ = {(u,v) € X | a0 < u < fa, a0 < v < (30}. We define T : £ —> X, (u, v) — t > T(u,v), where T(ii, i>) is the solution (a;,t/) of x" = f(t, u,v), y" = f(t,v,u),
x(a) = 0, x(b) = 0, y(a) = 0,y(b)=0
and verify T is continuous, monotone increasing, T(£) is relatively compact in X and (ao,/?o) < T(oo,/%), (/3o,ao) > Theorem 1.3 applies with a = (ao,/3o) and /3 = (y3o,ao)> and the claims follow. Notice that if u is a solution of the given problem (4.1), then (u, u) is a solution of the auxiliary problem (4.2), thus umm and umax are bounds on solutions of (4.1). If the bounds umin and umax given in Proposition 4.1 are equal, umin is a solution of the initial problem (4.1), and this theorem provides an approximation scheme to a solution of (4.1). In particular, this will be the case if solutions of the auxiliary problem (4.2) are unique. Unfortunately, these solutions are in general not unique as follows from the problem u" + seu = 0, v v" + se = 0,
u(0) = u(l) = 0, v(0) = v(l) = 0
and Theorem VIII-3.6, if we choose s > 0 small enough. The next proposition proves, under appropriate assumptions, the uniqueness of solutions of the auxiliary problem (4.2). Hence, it also proves convergence of the sequences defined in Proposition 4.1 to the unique solution of the given problem (4.1). Theorem 4.2. Suppose the assumptions of Proposition 4-1 hold. Assume moreover (i) there exists e > 0 such that ao > e/?o/ (ii) for every s G [e, 1[, almost every t € [a, b] and every u, v € [ao, @o] with sv < u < v, sf(t,*,sv)>f(t,u,v). Then, the functions umin andumax defined in Proposition 4-1 are equal, i.e. are solutions of (4.1). Proof. From assumption (i), we deduce ' r a n i _i " m m _;
4. A MIXED APPROXIMATION SCHEME
275
Let so = sup{s | sumax < umin}. It is obvious that s0 € [e, 1] and that s u o max < umin. From the definition of SQ, we deduce the existence of to € [a, b] such that Umin(*0) - S0Umax{t0)
= 0,
u'min(t0)
- SOu'max(tO)
= 0.
If not we have u
«min(*) - soumax{t) > 0, for t e }a, b[, 'min(a) ~ sou.'max(a) > 0, u'min(b) - sQu'max{b)
< 0.
Hence there exists e > 0 so that umin(t) — Soumax(t) > eumax(t) on [a, &]. This contradicts the definition of SQ. If t0 ^ b, we also have t\ > to such that u'min(ti) — sou'max(t\) > 0. Assume now that SQ < 1. Hence, we can write s
OU'max ~ sof(t,Umax,Umin)
> SQf(t, j^U
-^ J\y} I^mini "Umax)
^min^
which leads to the contradiction 0 < (u'min - sou'max)\h
= f 1*0
X
« i n ( t ) - soiCaxM) dt < 0.
JtQ
A similar argument holds if to = b. Hence so = 1 and umax = umin. D Conditions of Theorem 4,2 can be checked on any pair of lower and upper quasi-solutions. In some cases, it is useful to use some iterate (an, /?„) from the sequence defined in (4.4). Theorem 4.3. Let a0, 0O G C([o, b}) and E be defined from (4.3). Suppose f : E —> M is an Ll-Caratheodory function such that for some K > — (g3^) 2 , f(t,u,v) — Kv is non-increasing in u, non-decreasing in v and for almost every t and all u, v with ao(t) < u < v < Po(t),
[f(t,u,v) - f(t,v,u)](v -u)<
£
Assume ao and j30 are coupled lower and upper quasi-solutions o/(4.1). Then, both sequences (an)n and (0n)n, defined for n > 1 by o£ - Kan = / ( t . o n - i , A,_i) - K0n-i, /?;' - K0n = f(t,/?„_!, a n _ i ) - Kan^, converge to the same solution u of (4.1).
on (a) = 0, an(b) = 0, /3n(a) = 0, f3n{b) = 0,
276
V. MONOTONE ITERATIVE METHODS
Proof. As in Proposition 4.1, we can prove that the sequences (an)n and (Pn)n converge respectively to functions u = umin and v = umax > u solutions of u" -Ku = f(t, u, v) - Kv, v" -Kv = f(t, v, u) - Ku,
u(a) = 0, u(b) = 0, v(a) = 0, v(b) = 0.
If u ^ v, we compute / [(«" - u") - K(v - u)](u - v) dt = / [f(t, v, u) - f(t, u, v) - K(u - v)](u - v) dt Ja
b
(u~v)2dt
'/.
and also
rbb
- / [(?;" - u") - K(v - u)](v - u) dt Ja
= f {(v' - u'f + K{v - uf] dt b
which is a contradiction. Another result in this direction uses a one-sided Lipschitz condition on the function / . Theorem 4.4. Let a0, (3Q e C([a,6]) and E be defined from (4.3). Assume g : E —> M. is an L1 -Caratheodory function such that g(t,u,v) is 2
non-increasing in u and non-decreasing in v and for some L > ff,Za)» > g(t, u, v) — Lv is non-increasing in v. Assume OCQ and PQ are coupled lower and upper quasi-solutions of (4.1) with f(t,u,v)
=
-L(u-v)].
Then, both sequences (an)n and (f3n)n, defined for n > 1 by a'n = /(t,Q n _i,/? n _i), Pn = /(*, A.-1. On-l), converge to the same solution u of u" = g(t,u,u),
an(a) = 0, an(b) = 0, 0n{a) = 0, pn(b) = 0, u(a) = u(b) = 0.
4. A MIXED APPROXIMATION SCHEME
277
Proof. By Proposition 4.1, the limit functions u = lim an
and v = lim (3n > u
n—*oo
n—»oo
exist and are such that u" =
u(a) = 0, u{b)=O,
v" =
v(a) = 0, v(b) = 0.
Hence, w = v — u is a nonnegative solution of w" + Lw = h(t), w(a)=w{b) = 0, where h{t) = \[g{t,v,v) - g(t,u,v) + g(t,v,u) - g(t,u,u)] < 0. Such a nontrivial nonnegative solution does not exists if L > ,fe^ .^ as otherwise we have the contradiction b
=
F b b / (w"(s)
r / /
w(s) sin J S J ( S - a) ds
+ Lw(s)) sin ^{s
- a) ds
b
= / h(s) sin-^(s
- a) ds < 0.
D
This Page is Intentionally Left Blank
Chapter VI
Parametric Multiplicity Problems 1
Periodic Solutions of the Lienard Equation
Consider the periodic problem for a Lienard equation u" + g{u)u'+ f(t,v) = a, u(a) = u(b), u'(o) = u'(b),
n {
n j
where a < 6, s G R, g : R —» R is continuous and / : [ a , ! i j x l - » I is £1-Caratheodory. This chapter is concerned with the Ambrosetti-Prodi problem. This reduces here to study the dependence of the number of solutions on the parameter s. We first state conditions such that problem (1.1) has at least one solution for s in an interval ]SQ,SI] and no solution for s < soTheorem 1,1. Let g : R —» R be continuoiis, f : [a,b] X R —> R be L1Garatheodory and s € R. Assume there exist si 6 I , sa > si and r > 0 such that for all u < —r and a.e. t € [a,b] /(«,0)
T/ien tftere exists SQ E [—oo, sj] s«c/i that (i) for s < so,feeproblem (1.1) ftas no solution; (ii) for s G ]«o> 82] U {si}, the problem (1.1) ftas ai tosi one solution. Proof. Notice first that for s € [si,S2], a(t) = — r is a lower solution of (1.1) and f3(t) = 0 an upper one. It follows then from Theorem 1-6.9 that problem (1.1) with s € [si,S2] has a solution u\ € W2'l{a, b). 279
280
VI. PARAMETRIC MULTIPLICITY PROBLEMS
Define next SQ 6 [—00, si] to be the infimum of those s such that (1.1) has a solution. Hence, Claim (i) holds. To prove Claim (ii), let us fix s G]SQ,S2]. There exists s* G]SO,S] such that (1.1) with s — s* has a solution u*. Observe that /?(£) = u*(t) is an upper solution of (1.1). Further a(t) = r* < —r is a lower one and we can choose r* small enough so that a(t) = r* < /3(£). The claim follows now from Theorem 1-6.9. Remark It follows from the proof of the above theorem that if ua is a solution corresponding to sa and tt& is a solution corresponding to sj, > sa, ua is an upper solution of (1.1) with s = sj, and we can choose Ub < ua. The following examples show that the maximal interval of existence of solutions can be the closed interval [so,oo[ or the open one ]SQ,OO[. Dependence of solutions on s is represented in figure 1. Example 1.1. Consider the problem u" + u~ = s, u(a)=u(h), u'(a) = u'(b). It is easy to see from a phase plane analysis that this problem has a unique solution u(t) = — s if s > 0, no solution if s < 0, and an infinite number of solutions, the nonnegative constants, for s = 0.
Fig. 1 : Examples 1.1 and 1.2 Example 1.2. The problem u" + exp(—u) = s, u(a) = u(b), u'(a) = u'(b), has a unique constant solution if s > 0 and no solution if s < 0. Uniqueness of the solution follows easily from the decreasingness of the nonh'nearity exp(—u).
1. THE LIENARD EQUATION
281
The next example shows that the maximal interval of existence of solutions can be bounded. Example 1.3. The problem u" — arctanu = s, «(o) = u(b), u'(o) = u'(b), has a unique constant solution for any s e ] — vr/2, ic/2[. Further, any solution is such that fb = — / & & Ja i.e. there is no solution if s &] — w/2, w/2[. 1
Notice that the uniqueness of solutions in the above examples shows that we have to put additional assumptions to obtain multiplicity results, which is the case in the following theorem. Theorem 1.2. Let s G l , j : i - » I 6 e continuous and / : [ a , & ] x R - > I be an L1-Caratheodory function such that (A) for all tg £ [a, b], UQ € M and e > 0, there exists 6 > 0 such thai =» \f(t,u)-f{t,uo)\<€. \t-tQ\<6,\u-Uo\<5 Assume there exist si e R, s% > s% and r > 0 such that for all \u\ > r and a.e. t £ [a,b] /(*,0)<Sl<«2
Then there exists so €] — oo,sx] such that (i) for s < so, the problem (1.1) has no solution; (ii) for s = SQ, the problem (1.1) has at least one solution; (in) for s £]SQ,S2], the problem (1.1) has at least two ordered solutions. Remark As we already noticed after Proposition III-1.5, assumption (A) does not imply that / is continuous. Proof. Let so be defined as in Theorem 1.1. Claim I - so G K. Let hr G Ll{a, b) be such that for any |u| < r, \f(t,u)\ < hr(t). It follows then that for any t i £ l and a.e. t 6 [a, b], f(t,u) "> min{/(t,0),— hr(t)} = —hr(t). Hence, direct integration of (1.1) proves that S = T ^ / f(t,u(t))dt>f(t,u(t))dt> b-aja b-a Ja i.e. there is no solution for s < s*. The claim follows.
282
VI. PARAMETRIC MULTIPLICITY PROBLEMS
Claim 2 - For s €}SQ,S2], the problem (1.1) has at least two solutions. First observe that a solution u* of (1.1) with s = s* is a strict W2'1upper solution of (1.1) with s > s*, which follows from Proposition III-1.6. Moreover, for every A large enough, the constants a = A and a = —A axe lower solutions. Increasing A if necessary, we can assume a < u* and a jt u*. The claim follows now from Theorem III-3.6. Claim 3 - For s = SQ, the problem (1.1) has a solution. By definition of so, we can find a sequence (sn)n and corresponding solutions un of (1.1) with s = sn, such that sn —> SQ. Notice first that for any n and some £«i |wn(in)| < T. Next, arguing as in Theorem 111-3,2, we prove that the ||^n||oo &re a priori bounded. This result implies, as in Proposition 1-4.3, an a priori bound on ||«n||ci and the claim follows from Arzela-Ascoli Theorem. D This theorem has the obvious following corollary. Corollary 1.3. Let s £ R, g : E -> E be continuous and / : [a, b] x E -> E be an L1-Caratheodory function such that for all to € [a, 6], UQ G E and e > 0, there exists 5 > 0 that verifies \t-to\ <S, \u-uQ\ <5 =* \f(t,u)-f(t,uo)\<e. Assume that for some si £ R and a.e. t £ [a, 6], /(£, 0) < si, and tftat lim f(t,u) = +oo |«|-»oo
uniformly int £ [a, &]. Tften t/iere exists SQ € E SMG/I i/iai fij /or s < so, t/te problem (1.1) feas no solution; (ii) for s = so, i/ie problem (1.1) feas at feast one solution; (Hi) for s > SQ, the problem (1.1) /ias at feast t«/o ordered solutions. The following examples show that the count of solution can be exact, see Example 1.4, or that the number of solutions can be larger as in Example 1.5. Example 1.4. Consider the problem u" + u' + u2 = s, u(a) = «(&), u'(a) = u'(b). Multiplying the equation by u' and integrating gives ||u'||f,2 = 0. Hence this problem has only the constant solutions u = /
1. THE LIENARD EQUATION
283
Example 1.5. In a similar way, we prove that the solutions of problem u" + u' + (u2 - I) 2 = s, u(a) =«(&), u'(a)=u'(b), are the constants u =
l
s1/2.
u,
Fig. 2 : Examples 1.4 and 1.5 Exact count of solutions can be given for the frictionless problem u" + f{t,u) = s, u(a) = u(b), u'(a) = u'(b),
(1.2)
in case the nonlinearity is strictly convex. Theorem 1.4. Let s G K and f : [a, b] x I convex function such that
R be a continuous, strictly
lim f(t,u) — +oo |u|—>oo
uniformly in t G [a, b]. Assume moreover that for all t G [a, b] and u ^ v f(t,u)-f{t,v) U — V
2TT
b-a
Then there exists s o S K such (i) for s < so, the problem (1.2) has no solution; (ii) for s = so, the problem (1.2) has exactly one solution; (Hi) for s > so, the problem (1.2) has exactly two solutions which are ordered. Proof. From Corollary 1.3, we only have to prove that for s > so the problem (1.2) has at most two solutions and for s = SQ at most one.
284
VI. PARAMETRIC MULTIPLICITY PROBLEMS
Claim 1 - Solutions of (1.2) are strictly ordered. Let u and v be two solutions which axe not ordered. Extending these solutions by periodicity, we can find t\ £ [a,b] and t2 > ti, with ^2 — ii < (b — a)/2, such that w — u — v is one sign on ]t\, t2[, say w > 0, and w(ti) — iu(ta) = 0. Notice that
"+*<,)«,=o,
(())MW)
A Sturm-Liouville argument leads then to the contradiction 0=
Notice at last that the convexity assumption implies / is locally lipschitzian. Hence, we deduce from uniqueness of solutions of the Cauchy problem, that solutions of (1.2) which are ordered are strictly ordered. Claim 2 - Problem (1.2) has at most two solutions. Let u$ < u^ < ui be three solutions of (1.2). The functions vi = u% — u^ and «2 = "a — U3 are such that v" + ki{t)vi = 0 and v% + fa(t)v2 = 0, where ,_ f(t,Uj(t)) - f(t,U3(t))
,_
f(t,U2(t))-f(t,U3(t))
A Sturm-Liouville argument gives then the contradiction
(
i
(
= /
(ki{t)-k2(t))vi(t)v2(t)dt>0.
5 - If problem (1.2) ««i/i s = s* ftas too solutions, then s* ^ SQ. Let «i < « 2 be two solutions corresponding to s*. Choose an s < s* near enough s* so that
It follows then that 0 = (u\ + u2)l2 is a C2-upper solution of (1.2) with s — s. Also, for r large enough, a = —r is a C2-lower solution such that a < 0. Hence, it follows from Theorem 1-2.3 that problem (1.2), with s = §, has a solution and so < § < s*.
2. THE RAYLEIGH EQUATION 2
285
Periodic Solutions of the Rayleigh Equation
Parametric multiplicity can occur in other problems. In this section, we consider the periodic problem for a Rayleigh equation u" + g{u') + f(t,u,u') = s, u(a) = u(b), u'(a) = u'(b).
, ( {A l) -
A first result concerns existence of one solution. In this section, for u G L2(a, b), we write u(t) = u(t) + % fb fb where u — / u(t) dt and / u(t) dt — 0. Also, we shall use the o-a>Ja Ja following spaces : 1
Lp{a, b) = {% G Lp(a, b)\u = 0 } , &]) | t»(o) = U(b), «'(<*> = U'(b), U = 0 } .
^
Theorem 2.1. Let s G M., g : WL —> 1 5e a continuous function and / : [a, 6] x K2 —> K 6e a Camtheodory function such that for some h G L2(a, 6), a.e. t € [a,6] and all (u,v) G K2,
Then there exists a nonempty, bounded interval I such that (i) if s £ I, the problem (2.1) has no solution; (ii) if s G J, the problem (2.1) has at least one solution. Proof. Let M. be the set of s such that (2.1) has a solution. Claim 1 : LetR > s/b — a||/i||/,2. Then any solutionu of (2.1) is SMC/I that Utt'Hoo < B.
(2.2)
Multiplying (2.1) by it" and integrating on [a, b], we get
ll«"lli» = - t f(t,U,u')U"dt < ||fc|M|u"|Ua. Hence, < \\h\\L* and if we choose to such that u'(to) = 0, we have
(2.3)
||«"||L»
ft
,
,
\u'(t)\ = | / u"(s)ds\ < Vb-a\\u"\\Li < y/b- a\\h\\Li < R. Jt0
286
VI. PARAMETRIC MULTIPLICITY PROBLEMS
A modified problem. Consider the function
with S(y) = max{min{y, R}, — R}. Repeating the proof of Claim 1, it is clear that any solution u of
) + f(t,u,u) = s,
. . {
u(o) = u(b), u'(a) = u'(b).
'
satisfies (2.2). Hence u is a solution of (2.1) if and only if it is a solution of (2.4). Claim 2 : M is nonempty. defined by
Consider the projector P : L2(a,b) —> It
rb Pu=z / u(t)dt, b-a Ja i
denote by M the compact inverse of the linear operator
L : C^K b}) -* L2(a, 6 ) , « H -U", with domain DomL = H2(a,b) nCper([a,6]) and let be defined from Nu = g{v!) + /(-, u, u'). The operator f = M(I - P)N is completely continuous and bounded so that Schauder's Fixed Point Theorem provides a solution of the equation u = Tu. This last equation is equivalent to -u" = g(u') + f(t,u,u') - P(g(u>) + u(a) = u(b), u'(a) = u'(b), Pu = 0,
,
which proves that s := P(g(u') + f(.,u,u')) G A4. Claim 8 : M. is bounded. Direct integration of (2.4) shows that
\\h\\v y/b-a
Claim 4 M. is an interval. Consider ri,t2&M. with r\ < r% and let «i, «2 be the corresponding solutions of (2.1). For s G]ri,r2[, the functions «i and U2 are respectively PF2>1-upper and lower solutions of (2.4). Recall that g(v) + f(t, u, v) is bounded by a L2-function. Hence, we can deduce from Theorems 1-6.8 or III-3.1 (according as the lower and upper solutions are ordered or not) that problem (2.4), and therefore (2.1), has a solution.
2. THE RAYLEIGH EQUATION
287
Example 2.1. Consider the problem u" + u'3 - f(u) = s, u(a) = u(b), u'(a) = u'(b), where / is a continuous, bounded function. Multiplying the equation by u' and integrating, we prove that all solutions are constant. Particularizing / ( « ) , it is easy to find examples with different structures of the solution set. Consider first f(u) = arctanu. In this case there is a solution, which is unique, if and only if s G ] — TT/2, TT/2[. If we choose f(u) = ue"'"', we have no solution if s $ [—e"1, e" 1 ]; one solution if s = - e " 1 , 0 or e~1\ and two if s £ j — e~ 1 ,e~ 1 [\{0}. Another example is f{u) = sinu. Here there is no solution if s £ [—1, +1] and an infinite number of them for s £ [—1, +1]. The next theorem gives conditions to have at least two solutions for some values of s. It describes a solution set such as in figure 3, where the vertical axis represents functions mod 2TT.
Fig. 3 : Solution set in Theorem 2.2 Theorem 2.2. Let s £ 1 , j : R -» K k a continuous function and f : [a, b] x K2 —> M. be a Caratheodory function such that (a) for a.e. t £ [a, b] and all (u,v) £ R 2 , f(t,u,v) — f(t,u + 27r,v); (b) for all to £ \a,b], (UQ,VQ) £ R 2 and e > 0, there exists 5 > 0 such that \t - t01 < 5, \u- uo\ < 5, \v - vo\ < 6 => \f(t,u,v) - f{t,uo,vo)\ < e; (c) for some h £ L2(a,b), a.e. t £ [a,b] and all (u,v) £ R 2 , \f(t,u,v)\
288
VI. P A R A M E T R I C MULTIPLICITY PROBLEMS
Proof. Prom Theorem 2.1, the set of s such that (2.1) has at least one solution is a nonempty, bounded interval I, i.e. cl/ = [so,si]. Step 1 - Claim : I is closed. Let (sn)n be a sequence in I converging to a s and (un)n be the corresponding solutions of (2.1). Prom the periodicity, we can assume (adding a multiple of 2n to un if necessary) that un(a) G [0,2n]. It follows now from (2.2) that \un{t)\ = \un{a) + I u'n(r)dr\ < 2TTwhich, using (2.3), gives that the sequence (un)n is bounded in H2(a,b). As H2(a,b) is compactly embedded in C1([a,6]), a subsequence converges to some function u G (^([a, b\) and going to the limit in (2.1) (with s = sn), it follows that u is a solution of (2.1) with the given s. Step 2 - Existence of strict W2'1 -lower and upper solutions of (2.1) for s (E]SQ, SI[. Consider the modified problem (2.4), where R > 0 is defined from Claim 1 of Theorem 2.1. Let uSo and uSl be the solutions of (2.1) with s = SQ and s = si respectively. As / is periodic, we can choose k G Z such that « i := uSl < Pi := uso + 2kn and «i(i«) + 2TT > /?i(i») for some Pi + 2?r £* G [a, b]. By Proposition III-1.6, the functions /?i and are strict W 2>1 -upper solutions of (2.4) for s > so. Similarly, we deduce from Proposition III-1.5 that ot\ and ai := ot\ + 2TT are strict W 2ll -lower solutions if s < s\. Step 3 - Claim : If s G]s o ,si[, problem (2.1) has at least two ordered solutions that do not differ by a multiple of 2TV. Recall that g(v) + f(t,u,v) is bounded by a L2 function. It follows then from Theorem III-1.13 that the problems (2.4), and also (2.1), have at least two solutions ui and u2 which are such that a.\ < u\ < f3\, u\ < U2, ^2(^1) > Pi(ti) for some ti G [a,b] and 1*2(^2) < 02(^2) for some t2 G [a,b]. Hence ui — u\ cannot be a multiple of 2?r. Example 2.2. In Theorem 2.2 the solution set can reduce to one point as in the problem
u" + sin(t + u) = s, u(0) = U(2TT), U'(0) = U'(2TT).
Let u be a solution, multiply the equation by 1 + u' and integrate, which gives s = 0 and so = s i = 0. A classical example of differential equation with periodic nonlinearity is the pendulum equation u" + g(u') + Asinu = s+p(i), u{a) = u(b), u'(a) = u'(b).
. K
-, '
2. THE RAYLEIGH EQUATION
289
The next result gives some estimates on the interval / of admissible s. Proposition 2.3. Let g : M —> R be a continuous function, A f l and there K e]0,7r[. Then for all p £ L2(a,b) with \\p\\Lt < K{b_j^b_a, exists s c G R (independent of A) with (2.6) such that (2,5) has at least one solution for any s with
|s-sc|<|A|sin^.
(2.7)
Proof. Let u* £ H2(a,b) be a solution of u" = p(t)-g(u') + Pg(u% t*(o) = u(b), u'(o) = u'(6), P« = 0,
,
.
l
j
where P / = -^ fa f(t) dt. Multiplying by u*" and integrating we get 2 - Further, we obtain from Propositions A-4.3 and A-4.1
ll«1l<x, < ^ T ^ l K i b < ^g^llplU' < f and Now we can modify (2.8) as in Claim 1 of Theorem 2.1 and obtain from Schauder's fixed point Theorem a solution u* of (2.8). Let sc := Pg(u*'). Notice that (2.6) holds from the estimate on ||ii**[|oo. Assume A > 0 and observe that a(t) := u*(t) + § € [f - § , § + §] and /?(*) := u*(t) + ^ € [3f - f, ^ + f ] . We also fix s according to (2.7). It is then easy to check that a is a W 'Mower solution for (2.5) since a" + g(a') + Asina — s — p(t) = Asin(w* + | ) — (s — sc) > ^4(sin(u* + f) - sin ^ ^ ) > 0. Similarly, we see that 0 is a W^2>1-upper solution so that the existence of a solution follows from Theorem 1-6.8. A similar argument holds if A < 0. If the friction force is linear, we can give an explicit estimate of the interval I of admissible s. Corollary 2.4. Let e G R, A € R and K 6 ]0, JT[. Then, for allp £ L2(a, b) with \\p\\2 < K.. _Syji and for any s such that
290
VI. PARAMETRIC MULTIPLICITY PROBLEMS
the problem u" + cu' + A sin u = s + p(t), u{a) = u{b), u'(a) = u'(b) has at least one solution. Proof. One has to see that in the proof of the above proposition sc =
^jbau*'(t)dt
= o.
a
Many results can be worked out for the pendulum equations. A variant of the previous result is the following. Proposition 2.5. Let g : R —> R be a continuous function, A £ R and p 6 L2(a,b). Let u* £ Cper([a,b]) be a solution of u"+g(u')=p(t), u(a) = u(b), u'(a) = u'(b). Assume A = ifmax u*(t) — min u*(t)) < 52 and s\ < 2 \e[a,b]
te[a,b]
~
~
IAICOSA.
Then problem (2.5) has at least one solution. Exercise 2.1. Prove the above proposition. Example 2.3. As an example of application of the above proposition, consider the problem u" + sin u = s + n sin nt, U(0) = U(2TT), U'(0) = U'(2TT),
which has a solution if \s\ < cos(^). Our next result concerns the Ambrosetti-Prodi problem for the Rayleigh equation u" + g(u') + f{t, u, u') = s + h(t, u, u'), . . K } u(a) = u{b), u'(a) =u'(b). Here the nonlinearity can be unbounded. More precisely, h(t,u,v) is supposed to be uniformly bounded by a L2-function and f(t,u,v) satisfies a uniform limit as u —> oo. Such a structure is modelled from a physical system with a restoring force —f(t,u,v) and a forcing h £ L2(a,b). Theorem 2.6. Let s G R, g : R -> R be continuous and f, h : [a, b] x K2 —» R be Caratheodory functions. Assume that for all to £ [a,b], (uo,^o) & R2 and e > 0, there exists a 5 > 0 such that \t — to\ < S, \u- uo\ < 5, \v - VQ\ < 6 => \f{t, u, v) - f(t, uo,vo)\ < e and \h(t, u, v) - hit, u0, vo)\ < e. Assume further
2. THE RAYLEIGH EQUATION
291
(a) there exist d > c > 0 such that for all v e R
(b) there exists a function k\ € I?(a, b) such that for a.e. t e [a, b] and all (u,v) &M.2 \h(t,u,v)\
Tiften ^/iere exists SQ € K sucft &a£ fij t / s < so, t/ie problem (2.9) ft&s no solution; (ii) if s = so, the problem (2.9) has at least one solution; (Hi) if s > SQ, the problem (2.9) has at least two ordered solutions. Proof. Claim 1 : Given s* there exist RQ > 0 and R% > 0 so that any solution u of (2.9) with s < s* is such that \\u\\ao
\\u'\\oo
Let us pick R > 0 large enough so that for a.e. t E [a, b] and all (u, v) G R2 with \u\ > R, f(t,u,v) > 0 and choose &2 according to assumption (c). Hence, we have f(t,u,v) > —k2(t) for a.e. t and all u, v. If we multiply the equation in (2.9) by (u') + an^ integrate, we obtain c / (u')+2 dt<
I [u" + g(u') + (f(t,u,v!) + k2(t)))(u')+ dt
Ja
u
- f [s + h(t,u,u') + k2(t)]{u')+ dt Ja This gives a bound K\ on ]|(w')+IU2 Moreover, we have
&n
& therefore ||(ti')+||x,i < K\y/b — a.
0= a
and \\u'\\Lt =2\\(u>)+\\Li
<
292
VI. PARAMETRIC MULTIPLICITY PROBLEMS
Take now K% := s* + (H^iUi3 + 2dKi)/y/b~—H. If u is a solution of (2,9), direct integration gives fb
,&
/ f(t,u,u')dt = s(b-a)+
[h(t,u,u') - g(u')]dt
Ja
J Q>
< s*(b-a) + \\ki\\iJ2^/b-a + d / \u'\dt Ja < s*(b-a) + \\ki\\La,/b^a + idKis/b^
= K2(b - a).
Next, by assumption (d), we can pick r > 0 such that for a.e. t G [a, b], all i ) 6 l and every \u\ > r, f(t,u,v) > K%. Hence, given u solution of (2.9), we can choose to such that |tt(to)| < r and we have
/ Ja
The bound on ||ii'||cio follows now from Proposition 1-4.1. Claim 2 : Equation (2.9) has a solution for some s large enough. Prom Theorem 2.1, there exists some si such that t \
m
>, \
' ,)l\
( 2 - 10 )
v
u(a) = u(b), u'(a) = u'(b) ' has a solution u, and therefore a family of solutions u+C, with C € R. Let «o be the element of this family with mean value zero. Multiplying (2.10) by U'Q and integrating, we obtain ||«olU2 < ll^illi2 ^d from Propositions A-4.3 and A-4,1 ||«o||oo < Co : = i
Fix s > si + sup{/(t, u,w) | t € [a, 6], |u| < Co, |«| < Ci}. Then, we claim that UQ is an upper solution of (2.9). Indeed, "o
/(> o, d) ( i , o) < ug + g(u'Q) + f{t, u0, u'o) + h (t) = si + /(*, uo, «i) < s.
On the other hand, it is possible to get an ordered lower solution by considering the equation u" + g(u') - h(t) = 82, u(o) = u(b), u'(o) = u'(b).
/911>
(
}
2. THE RAYLEIGH EQUATION
293
As above, it is clear there exists some S2 such that equation (2.11) has a family of solutions u + C, with C € M.. Let u% be an element of this family small enough so that for every t 6 [a, b], ui{t) < uo{t)
and
s% + f(t,ui{t),u[(t))
> s,
which is possible from (d). Then, < + fl(«i) + /(*i«ii«i)-M*»«i.«i) > < + g(u[) + f(t, ui, «i) - fci(t) = s2 + /(t, MI, «i) > s. Claim 2 follows then from Theorem 1-6.8, using the fact that wi and UQ are ordered lower and upper solutions of (2.9). Claim 3 : The set M of all the s such that (2.9) has a solution is bounded below. From Claim 2, M is not empty. Let so € M and u be a solution of (2.9) with s < SQ. Rrom Claim 1, a direct integration of (2.9) leads to
s > - ^ 3 ^ (J(k2 + h) dt J where &2 is associated with RQ in assumption (c). Claim 4 : M is closed. Let (sn)n be a sequence in M converging to s* and («n)n be the corresponding solutions of (2.9). Prom Claim 1, the sequence (un)n is bounded in Cl([a, b]) and from Arzela^Ascoli Theorem it converges to a solution of (2.9) with s = s*. Claim 5 : M = [SQ, OO[. Let si G M, u\ be the solution of (2.9) with s — Si and let s > sj. The function ui is a W2>1-upper solutions of (2.9) since u" + g(u[) + f(t,ui, u'x) - h(t, ui,u[) - s = si - s < 0. An ordered tF2il-lower solution a can be obtained by the argument in Claim 2. It follows now from Theorem 1-6.8 that s £ M. Claim 6 : If s € int^M, equation (2,9) has two ordered solutions. Let si € M be such that < s and u\ be the corresponding solution. It follows from Proposition III-1.6 that «i is a strict W2|1-upper solution. As in Claim 2 we can find a lower solution below «x. The proof follows now from the argument in Claim 1 and Theorem III-1.14.
294
3
VI. PARAMETRIC MULTIPLICITY PROBLEMS
A Two Parameters Dirichlet Problem
We can write the problem studied in the first section as u" + g{u)u' + f{t,u) = p{t) + s, u(a) = u(b), u'{a) =u'(b), and think of it as studying the range of the operator 1Z : L1(o, b) —» Ll{a, b), defined by VJU, = u" + g{u)u' + f(t,u) and with domain DomTZ = {u G W2'1(a,b) | u(a) = u(b), u'(a) = u'(b)}. In that section, we gave conditions so that given p £ Ll(a,b), the set of functions h(t) = p(t) + s, with s G R, intersects the range of Tl along a closed half-line and that on the corresponding open half-line there are at least two solutions. Given p and q G L}{a,b), a similar problem could be worked out for the line h(t) = p(t) + sq(t), where s G R. Consequently, we study the problem u" + g(u)u' + f(t, u) = p(t) + sq(t), u(a) = u(b), u'(a) = u'(b). The choice q{t) = 1 is somewhat natural since this function is the first eigenfunction of the corresponding linear problem. However this is not fundamental. To compare two such approaches, we consider in this section a two parameters problem, i.e. we fix a plane in Ll{a,b) and look for the intersection of the range of the operator with this plane. More precisely, we work the Dirichlet problem u" + u + f(t,u) =r + s(p(t), u(0)=0, u(7r)=0,
. [
. '
where ip(t) = J-smt and r, s are parameters. This can be seen as a generalized Ambrosetti-Prodi problem. Throughout the section, for u € L 1 (0,TT), we write u(t) = u(t)+u
r Jo
h(t)tp(t) dt > s
3. A TWO PARAMETERS DIRICHLET PROBLEM
295
and ,u) >h(t) uniformly in t. Then there exists a non-increasing Lipschitz function ro : R RU {—cx>} with 2-J- TQ{S) + s < s such that (i) ifr< ra(s), the problem (3.1) has no solution; (ii) ifro(s) < r < ^\/^(s — s), the problem (3.1) has at least one solution.
)vT
Fig. 4 : Illustration of Theorem 3.1 To prove this theorem we need the following lemma which provides lower solutions Lemma 3.2. Let f satisfy L1-Caratheadory conditions and assume (A-l) holds. Then, for any z € CQ([0, IT}) and each (r,s) such that
the problem (3.1) has a W2'1 -lower solution a such that a < z. Proof. Let us choose e > 0 so that
Prom (A-l), we can pick R > \\z\\oo such that for a.e. t and all u < —R, f(t,u) > h(t) — e. Also, from the Carath&dory conditions there exists k € L^O.-n-) such that for a.e. t and all u e] - oo,R], f(t,u) > k(t). Next, we choose S > 0 small enough so that
/ (h(t) - k(i)Mt)dt < hh-2 Jlr - a],
296
VI. PARAMETRIC MULTIPLICITY PROBLEMS
where F$ := [0, S]u[ir- 6, w]. The function
g(t):=k(t)-r, if t£Fs, := h(t) - (r + e), if t e h
[$, v-S[,
is such that g = [h- 2 i / | r] - / (h(t) - k(t))
on Fs, on Is.
Hence a < z and since
a" + a + f(t, a)-r-
stp(t) >w" + w + g(t) - stp(t) = (g - s)tp(t) > 0,
a is a W^2>1-lower solution, We can now proceed to the proof of Theorem 3.1. Proof of Theorem 3.1 : Step 1 - Definition of ro satisfying (i) and (ii). Define ro(s) 6 l U {-oo} by I~Q(S) :=
min{|-y/^(s — s),inf{r | (3.1) has a solution for (r, s)}}.
Notice that condition (i) holds by construction of ro. Let now (r,s) be given such that By definition, there exists ri £ [ro(s), r [ and a solution u\ of (3.1) for {r\, s). The function u\ is a PF^-upper solution of (3.1) for (r,s). Prom Lemma 3.2, we obtain a lower solution a < u\ and the existence of a solution of (3.1) for the given (r, s) follows from Theorem II-2.4. Step 2 - The function ro(s) is non-increasing and Lipschitz. Let sg be such that TQ(S2) < f \ / f ( ^ ~ S2 )" ^ o r B;ny ^ -*" 0, small enough, there exist rj\ C
3. A TWO PARAMETERS DIRICHLET PROBLEM
297
and a solution «2 of (3.1) for {r<2,,s%). Next, for any s\ < s2 close enough tos2,
ri=r 2 + yf (s2-si) is such that 2 y ^ ri + si < s. The function «2 is a W2>1-upper solution of (3.1) for (ri, Si) and from Lemma 3.2 there exists a W2il-lower solution a < «2- This result together with Theorem II-2.4 implies ro(si) < r\. It follows that
ro(si) < ri < ro(s2) + y f (s2 - s and as ?? is arbitrary,
< ro(s2) + y f (s2 - sx). On the other hand, let si be such that ro(si) < \-\f^{$ — s\). Then, for any rj > 0, we can find r\ £}ro(si),ra(si) +rj\ so that there exists a solution u\ of (3.1) for (ri,si). If s^ > si, this function is a W-^-upper solution of (3.1) for (ri, s 2 ). Using the same arguments as above, it follows that ro(s2) < r o (si). Hence, the claim follows. Example 3.1. Consider the linear problem sin* u(0) = 0, U(TT) = 0, which has a solution if and only if 2-J ^r + s = 0. In this case I is any negative number and TQ(S) = \yf (s — s). Theorem 3.1 states that there is no solution if r < ro(s) and as s is any negative number, we have that there is no solution if r < — j - ^ j s , which is the best one-sided condition we can give. Example 3.2. The function ra(s) can take the value —oo as follows from the problem u(0) = 0, u(ir) = 0, which has solutions for all values of r and s. Here f(t,u) = — u, s is any positive constant and ro(s) = — oo.
298
VI. PARAMETRIC MULTIPLICITY PROBLEMS
To rule out the case where ro(s) = —oo, we shall impose some lower bound on / . Theorem 3.3. Let f satisfy L1 -Caratheodory conditions, (A-l) and (A-2) there exists k £ L1(O,TT) such that for a.e. t £ [0,7r] and all « £ R , f{t,u)>k(t). Then there exists a non-increasing Lipschitz function ro : R —> R with f ro(s) + s < s such that (i) if r < ro(s), the problem (3.1) has no solution; (ii) ifro(s) < r < \y/^{s — s), the problem (3.1) has at least one solution. Proof. Let u be a solution of (3.1), multiply the equation by
o
k{t)
Jo
The rest of the proof follows from Theorem 3.1. To make sure the region
where there is at least one solution is nonempty, we reinforce the assumption on the nonlinearity for large negative values of u. Theorem 3.4. Let f satisfy L1 -Caratheodory conditions, (A-2) and (A-l*) lim f(t,u) = +00 uniformly in t. u—»-oo
Then there exists a non-increasing Lipschitz function r^ : R —> R such that (i) if r < ro(s), the problem (3.1) has no solution; (ii) if ro(s) < r, the problem (3.1) has at least one solution. Proof. Claim 1 - For every s e t , problem (3.1) has an upper solution if r is large enough. Let m £ L1(0,7r) be such that for a.e. t £ [O,TT] and all u £ [—1,1], \f(t, u)\ < m(t). By Proposition A-4.4, there exists K > 0 such that for all u £ W2
\u" +u\ipdt.
Next, we choose p £ C([0,n]) such that \\m — PIIL1 < 57f VlF anc ^ define (3 as the solution of u" + u + m — p — (fh — p)
3. A TWO PARAMETERS DIRICHLET PROBLEM
209
We have Halloo
p)ip\
- p\\Li < 1
and therefore
/?"(i)+/3(t)+/(i,/3(t)) < 0"{t)+0(t)+m(t)
=p(t)+(m-p)ip(t)
< r+«p(t)
if T is large enough. Claim 2 - For every s £ R, problem (3.1) has a solution if r is large enough. Let s € R and, according to Claim 1, choose f(s) large enough so that problem (3.1) for (f(s), s) has an upper solution /?. Let s > 2y | r («)+ s. According to (A-l*), the assumptions of Lemma 3.2 are satisfied and problem (3.1) for (f(s),s) has a lower solution a < j3. We deduce then from Theorem II-2.4 the existence of a solution of (3.1) for (f(s),s). Conclusion. Prom Theorem 3.3, the function ro(s) exists such that ro(s) < f(s) and (i) if r < ro(s), the problem (3.1) has no solution; (ii) if ro(s) < r < f \ / J ( s —s), the problem (3.1) has at least one solution. As s is arbitrarily large, the second conclusion holds for all r > ro(s). Example 3.3. Multiplicity results are not straightforward as follows from the problem u" + u+ = r, u(0) = 0, U{-K) = 0, which corresponds to f(t,u) = u~. Here, there is no solution if r < 0, an infinite number if r = 0 and only one if r > 0. To obtain multiplicity results, we will have to reinforce (A-l) assuming a control of the nonlinearity for large values of \u\. Also, we shall use degree theory, i.e. strict lower and upper solutions, which forces us to assume additional assumptions as in Section III-2. Here we consider condition (B), which is satisfied in particular if f(t,u) — g(t,u) + h(t), where g is continuous and h € L1(a, 6). Theorem 3.5. Let f satisfy Ll-Caratheodory conditions together with (A-l**) lim f(t, u) = +oo uniformly in t and |tt|-»OO
(B) given to 6 [0, IT], « o € l and e > 0, there exists 6 > 0 such that for a.e. t £]to~S,to+8[n[0,w] andallui £ {UQ—S,UO+S], U-Z G [ui,ui+Ssm,t], f{t,u2)-f(t,ui)<e.
300
VI. PARAMETRIC MULTIPLICITY PROBLEMS
Then, there exists a non-increasing Lipschitz function ro : K —+ K such that (i) if r < ro(s), the problem (3.1) has no solution; (ii) if r = ro(s), the problem (3.1) has at least one solution; (Hi) ifro(s) < r, the problem (3.1) has at least two ordered solutions. To prove this result, we use the following lemma which provides the necessary a priori bounds to apply Theorem III-2.12. Lemma 3.6. Suppose the assumptions of Theorem 3.5 are verified. Then, given s and r*, there exists an R > 0 such that for all solutions u of u" + u + f(t,u)
{
,
}
we have Proof. Let u b e a solution of (3.2). Write u(t) =u
r
\u"(t) + u(t)\tp(t) dt. Jo Using (A-l**) and the Caratheodory conditions, we can find k G Ll(0, it) such that for a.e. t 6 [O,TT] and all n e t ,
f{t,u) > k(t).
(3.3)
Hence, we compute \u"{t) + u(t)\ < \u"(t) + u(t) + f(t,u(t)) - r* - sip{t)\ /(<,u(i))| + |r*| + |s|¥>(i) u{t) + f(t, u{t)) - r* - s
/
\u"{t) + u(t)\ip(t) dt <
r Jo
(r*+s
r Jo
- \r*
s\.
3. A TWO PARAMETERS DIRICHLET PROBLEM
301
Hence, we obtain
||fi|U < K[2 Ji (r* + |r*|) + (a + |»|) + 2 f \k(t)\
(2 J | r * + s) > llminf / f(t,u f(t, k(t))ip(t)dt = +oo.
D
Proof of Theorem 3.5 : Let TQ(S) be defined from Theorem 3.4 and (r, s) be such that ro(s) < r. By definition, there exist ri G [ro(s),r[ and a solution «i of (3.1) for (ri,s). By Proposition III-2.6 the function u\ is a strict upper solution of (3.1) for (r,s). As (A-l) is satisfied for any s, we can assume 2y ^ r+s < s and using Lemma 3.2 we obtain a lower solution a -< tii. Prom Lemma 3.6, there exists an R > 0, such that for all solutions u of u(0) = 0, U(TT) = 0, we have ||«||oo < R- It follows then from Theorem III-2.12 that the problem (3.1) has at least two ordered solutions. Let now (r,s) be such that r = ro(s). We can find rn > ro(s) such that linin^oo rn = ro(s) and solutions un of (3.1) for (rn,s). From Lemma 3.6, there exists an R > 0 such that ||Mn||oo < -R and from Caratheodory conditions, we have k £ Lx(0,7r) so that for all n € N, |«JJ(t)| < k(t). Using Arzela-Ascoli Theorem, we find a subsequence (unk)k that converges in C1([0,7r]) to some function u € C1([O,TT]). Going to the limit in (3.1), it is easy to see that u solves (3.1) for (r, s) — (ro(s), s). D The next theorem gives conditions for the function ro(s) to be decreasing. Theorem 3.7. Let f satisfy L1 -Caratheodory conditions together with assumptions (A-l**) and (B). Assume one of the functions g(t,x) = f(t,x) or g(t,x) — —f(t,x) is such that (A-3) given R > 0, we have Ve > 0,38 > 0, (Vu0 : |«o| < Rsp(t)),(W : \u\ < R
302
VI. PARAMETRIC MULTIPLICITY PROBLEMS
Proof. Let s0 e K. Step 1 - Claim: For all s* > so, there exists r* < ro(so) and a W2'1 -upper solution (i of (3.1) with r = r* and s = s*. Assume g{t,x) = f(t,x) in (A-3). The proof is similar in case g(t,x) = —f(t,x). Prom Theorem 3.5, there is a solution UQ of u'o +uo + f(t, uo) = r o (s o ) + soip(t), u o (0)
= 0, WO(TT) = 0.
Let w be the solution of w" + w = - ( l - : w(0) = 0, w(ir) = 0 , w = 0.
Choose then K > 0 and R > 0 such that for all t G [0, n], \w(t)\ < Ktp(t),
\uo(t)\ < -f
and
Define e > 0 such that so + e < s*
and pick 5 > 0 from assumption (A-3). Next, we choose m G]0,1] small enough so that
2mK <S,
so +
and let r* := ro(s o ) — m. Define
and observe that 0 < uo(t) - P(t) < 2mK(p(i) < 8
f{t, (3(t)) = < ( i ) + «o(i) + m(w"(t) + w{t)) + f(t, so
+ f{t,P(t))
< (ro(so) - m) + (so + Hence /3 is a W 2 i l -upper solution of (3.1) with r = r* and s = s*. Step 2 - As (A-l) is satisfied for any s, we can assume 2w |- r* + s* < s and it follows from Lemma 3.2 that there exists a M/2'1-lower solution a < ft. We obtain then from Theorem II-2.4 the existence of a solution of (3.1) for the given (r*,s*). This result implies that ro(s*)
3. A TWO PARAMETERS DIRICHLET PROBLEM
303
Remark The condition (A-3) is a one-sided continuity condition for the norm ||u|| = sup te[o,7r]
Such a condition can be obtained from a continuity assumption together with a one-sided Lipschitz condition near zero. Proposition 3.8. Let f be a continuous function and (A-3*) there exist L > 0 and R > 0 such that f(t,u) + Lu or —f(t,u) + Lu is non-decreasing for a.e. t G [0, TT] and u £ [—R,R]. Then assumption (A-3) holds. Proof. Assume f(t,u) + Lu is non-decreasing. Let RQ > 0 be given and choose r e ]0, §] so that RQV(T) < R. For t £ FT := [0, r] U [TT - r, TT] and all it0, U in [—/20v(*)) -Rov(*)]> we have 0 < uo — u => f(t, u) — f{t, u0) < L(u0 — u). If t G [r, 7T — r], we compute from the continuity of / Ve' > 0,352 > 0,
|uo-u| < S2 =* / ( t , u ) - / ( t , u 0 ) < e' < 4 T ^ O '
(3-4)
Now, given e, we choose Si and e' small enough so that L5x < e,
—r-r < e, ¥>(T)
we choose 52 from (3.4) and we put 6 := min(5i,52 V ^). It is then easy to see that if |uo|, |u| < Ro
and /(«, u) - f(t, u0) < ^ j v ( t ) < e^(t).
D
The decreasing character of r 0 can also be deduced from a Holder condition.
304
VI. PARAMETRIC MULTIPLICITY PROBLEMS
Problem 3.1. Let / be a continuous function that satisfies (A-1**) and (A-4) either g(t,u) = f(t,u) or g(t,u) = —f(t,u) is such that for some L > 0, a e]l/3,l[ and r > 0, for a.e. t and all u, v S [—r,r] with 0 < u — v < 1, we have g(t,u)-g(t,v)
4
Strong Resonance Problems
In this section, we study strong resonance problems. These are problems which avoid resonance using a nonlinearity that can vanish at infinity. As in the previous section, for u € L^O,^), we write u(t) = u{t) + uip(t), where
u— u(t)ip(t)dt and / u{t)
Our first result concerns the problem u" + u + f(t,u) = s
.
[
. '
where / is an Z^-Caratheodory function. Theorem 4.1. Let s £ R andp £ L 1 (0,TT). Assume that f is an L}-Caratheodory function such that for some k € Ll(0,ir), for a.e. t G [0,7r] and all u£R, \f(t,u)\
4. STRONG RESONANCE PROBLEMS
305
Step 1 - The set M is nonempty. Consider the projector Q : L1 (0,7r) —» L^O.TT) defined by Qu(t) = u(t) = u(t) - ( / u(r)ip(r) dr)ip(t) Jo and denote by M the inverse of the linear operator L : CQ([0, IT]) —> 1/(0, n), u H-> w" + «, with domain Dom L = W2'1 (0, ?r) n C^ ([0, w]). The problem u(0) = 0, u(w) = 0, u = 0 can be written u = fu:=M(p-Qf{-,u)). The operator f : CQ([0,n]) —» CQ([Q,it)) is completely continuous and bounded. Hence, using Schauder's Theorem, it has a fixed point u which is a solution of (4.1) with s— I Jo
f(r,u(r))tp(r)dr.
Step 2 - The set M. is bounded. Observe that if u is a solution of (4.1), \s = I ; f(t,u(t))(p(t)dt Jo
< / K(i)<^(£)cre. ^o
Step 3 - The set M is an interval. Let us define the real numbers so := inf{s | (4.1) has a solution} and si := sup{s | (4.1) has a solution}, which are such that M C [so,si]. Fix s E]so,si[, There exists a € [SQ,S] and ua solution of (4.1) with s — a. Also, there exists b 6 [s, s{\ and Ub solution of (4.1) with s = b. It is easy to see that ua and tt& are upper and lower solutions of (4.1). The claim follows then from Theorems II-2.4 or III-3.8 according as Ub < ua or u& ^ ua. D Remark 4.1. For any a £ i , the problem u(0) = 0, U(TT) = 0, u = 0,'
^4'2^
can be written u = fau := M(p - Qf(; atp + u)), where M and Q are defined in Step 1 of the above proof. The operator "]) —» Co([0,7r]) is completely continuous and bounded thus
306
VI. PARAMETRIC MULTIPLICITY PROBLEMS
Schauder's Theorem applies. As further the bound is uniform in o, there is a K > 0 so that for any given a G M, the problem (4.2) has a solution ua € l an£ CQ([O,TT]) and any such solution verifies ||fio||c < K i |«aOOI ^ Ktp{t). Hence, for any value of a € M the function u — ua + atp solves (4.1) with = / f(r,ua{r) + acp(r))
4. STRONG RESONANCE PROBLEMS
307
Theorem 4.2. Let s g l andp € £ 1 (0, IT). Assume that f is an L^-Caratheodory function such that (a) for some k € Ll{Q,ir), for a.e. t € [O,TT] and all u € R, \f(t,u)\
|u|—»oo
XTien iftere easst so < 0 < sj such
fij if s $ [SQ,SI], the problem (4.1) tes no solution; (ii) if s E [so,si], t/ie problem (4.1) ftas ai teasi one solution. Proof. Let J be given from Theorem 4.1, so = inf/ and sj = sup/. Step 1 - so < 0 < &i. Let us first prove there exists r > 0 and u = u + wp solution of (4.1) with s = r. Recall from Remark 4.1 that the solutions u of (4.2) are such that ||5||oo £ K and 1^(^)1 ^ K
uf(t,u)>2f
and
Next we pick a <5 > 0 small enough so that
/" k(t)dt+ f a u large enough such that
>K + R, and finally a solution u of (4.2). Observe that these choices imply that u(t) = u(t) + utp(t) > Rtp(t) if t G [0,IT] and u(t) > Riit £[8,n - §], Let A = {t e [0,ir] | |«(t)| < R} C [0,iS] U [TT - 6,ir] and 5 = [0,7r] \ A. We compute then
/ f(t,u{t))u(t)dt= JO
/ /(<,u(*))u(t)dt + I f(t,u(t))u(t)dt JA
JB
_
Jo
I f(t,u(t))u(t)dt= Jo
f f(t,u(t))u(t)dt+
f
JA
f(t,u(t))u(t)dt
JB r§
/«tr
k(t)dt+ /
1
k(t)dt\ +fn J
308
VI. PARAMETRIC MULTIPLICITY PROBLEMS
and / f{t,u{t))u{t)dt-
f(t,u(t))ip(t)dt=
/ Jo
Jo
f k(t)dt+ [ Jo
f(t,u{t))u{t)dt
k(t)dt
-fir
+ ^
(-K-25)
Jir-5
r Hence, r :—
f(t, u(t))tp(t) dt > 0 and u = u+wp solves (4.1) with s = r. Jo In a similar way, we prove there exists r < 0 and u — u + wp solution of (4.1) with s = r. Step 2 - so £ I, s\ £ I. Let us prove there exists a solution of (4.1) for s = so- Let sn £}so,s\[ be such that rimn_»0O sn = SQ and let un be the corresponding solution of (4.1). Multiplying (4.1) by tp and integrating we see that sn = J^ f(t, un(t))
f(t,un(t))tp(t)dt
= Q,
n—too JQ
which contradicts SQ < 0. The claim follows then from Arzela-Ascoli Theorem. In a similar way, we prove S\ £ I. Example 4.3. A typical example where Theorem 4.2 applies is
" +u + l^2=Mt)
u
+ P(t),
u(0) = 0, U(TT) = 0. To obtain a multiplicity result we use strict lower and upper solutions, which forces to assume additional assumptions as in Section III-2. Theorem 4.3. Let s £ R and p £ L^O, 7r). Assume that f is an L1Caratheodory function such that for some L £ L1(a,b;W+) and for a.e. t £ [0,7r], f(t,u) + L(t)u is non-decreasing in u and (a) for some k £ L 1 (0,TT), for a.e. t £ [0, TT] and all « £ l , \f{t,u)\
4. STRONG RESONANCE PROBLEMS
309
(b) lim f(t,u) = 0 and liminf «/(£,«) > m > 0 uniformly in t, \u\—»oo
|ti|—>oo
Then there exist SQ < 0 < si such that (i) if s £ [so, si], the problem (4.1) has no solution; (ii) if s G [so,si], the problem (4.1) has at least one solution; (Hi) if s €]so,si[\{0}, the problem (4.1) has at least two ordered solutions. Proof, Define so and s\ from Theorem 4.2. Let s* £]0, si[. A similar argument holds if s* 6 ]SQ, 0[. Part 1 - Construction of a strict lower solution. The problem (4.1) with s = a G]s*, SI[ has a solution ua which is such that < + «o + /(*, ua) - s*v(t) - p{t) = (a- s*)
s
*
Next, we pick S > 0 small enough so that
y^f f k(t) dt+ f
k(t) dt
2
Recall that there exists if > 0 so that for any u, the solution u of (4.2) satisfies |fi(t)| < Kip(t), Hence we can find u large enough so that uip(t) + u(t) > R on [5, tr — 5] and Sy + fi >^ « o , where u is the corresponding solution of (4.2). Then, we have r=
Jo
f(t,u(t))ip(t)dt <s*.
The function iti = u + tup is a solution of (4.1) with s = r, i.e. it is a W2il-upper solution of (4.1) with s = s* such that ua -< ui. Erom the same argument but with « small enough, we obtain a W2'1upper solution «2 such that U2 -< ua.
310
VI. PARAMETRIC MULTIPLICITY PROBLEMS
Conclusion - The result follows from Theorem III-3.11.
O
Problem 4,1. Prove the following result. Let s € K andp € Ll(0, n). Assume that f is an L1 -Caratheodary function such that for some k € L°°(0, IT), for a.e. t £ [O,TT], and all u £ K,
l/(t, «)!<*(*) Assume further that for some /i £ [1,2[
lim {u^fit,
u) = 0 and liminf sgn(u)|u|" f(t,u) > m > 0
uniformly in t. Then there exist SQ < 0 < s\ such that (i) if s $ [SQ,SI], the problem (4.1) has no solution; (ii) if s G [«o,si], the problem (4.1) has at least one solution. If moreover, for some L £ L^OjTnR"1") and for a.e. t G [O,TT], f(t,u) + L(t)u is non-decreasing in u then, (in) if s e]so,si[\{0}, the problem (4.1) has at least two solutions. Hint: Combine the ideas of Theorems 4.2, 4.3 and VII-5.3. In the preceding theorems, resonance is avoided with a restoring force f(t, u) depending on the position u. A similar result holds for a system with a nonlinearity that depends only on the derivative. More precisely, we consider the boundary value problem u(0) = 0, u(ir) - 0,
{
'
where g is a continuous function in M and p e L1 (0, w). The following theorem gives conditions so that the structure of the solution set is described by figure 5, where I is 0. Theorem 4.4. Let g : M. —* R be locally Lipschitz continuous and such thai the limits g(—oo) and g(+oo) exist and are finite. Let s 6 K and
l
Then there exists real numbers so = so(p,g) <8i = Sx(fi,g) such that I
y/$ig{-oo) + s(+oo)) e [so,si]
(4.4)
and problem (4.3) has (i) no solution if s $ [so,sj]; (ii) at least one solution if s € ]SQ, S\[, S £ {so, Si} \ {1} or s = SQ — si; (Hi) at least two ordered solutions «i and « 2 if s £}SQ,SI[\{1}.
4. STRONG RESONANCE PROBLEMS
311
Fig. 5 : generic examples of Theorem 4.4 To prove this theorem, we shall use the following lemma which shows that v takes negative values arbitrarily near t = 0. Lemma 4.5. Let m £ L 1 (0,a) and v e W2>1(0,a) satisfy (a) v" + m(t)v' + v < 0 in ]0, o[, (b) t/(0) = 0, v'(0) = 0. Then there exists oi € ]0, a[ sweft ftai w(aj) < 0. Proof. Suppose that v > 0 in ]0, o[. Let M(t) = Jo m(s) ds. Assumption (a) can be written (v'(t) exp M(t))' + v(t) exp M(t) < 0
in ]0, o[
and v'(t)ex.pM(t) is strictly decreasing in [0, a]. From (b) we infer that v < 0 in ]0, o[, a contradiction. Proof of Theorem 4-4 " First we note that without loss of generality we can suppose I — 0, i.e. g{—oo) + g(+oo) = 0. In fact, replacing g, p and s by h(v) :=g(v) - \ , / f /, q(t) = p(t) - | , / f f(t) and r = s - | y ^ F , where i = [(£) + r^(t), we obtain an equivalent problem that satisfies the same assumptions as (4.3) with I — 0. Henceforth, from now on we assume this condition. Claim 1 - There exists an s* such that problem (4.3), with s = s*, has at least one solution. We proceed as in Remark 4.1. The equivalent of (4.2) is u" + u = p — Qg(atp' + u'), (4.5) u(0) = 0, u(ir) = 0 , u = 0 which can be written as a fixed point problem u = faii := M(p - Qg(aip' to which Schauder's Theorem applies. There exists a K > 0 so that given any a 6 M, the problem (4.5) has a solution ua with ||wa||ci <
312
VI. PARAMETRIC MULTIPLICITY PROBLEMS
the function u — ua + ay> solves (4.3), with s = s* :— f£ g(a(p'(r) + u'a(r))(P(r) dr. Claim 2 - (i) holds. If u is a solution of (4.3), \s\ —
g{u'{t))v{t)dt\ o
is bounded and we can define SQ := inf{s | (4.3) has a solution} and si := sup{s | (4.3) has a solution}. Hence, assertion (i) is satisfied. Claim 5 - 1 = 0 £ [so>si]. Using Lebesgue's Theorem and the uniform bound on Hfiallc1! we have
F
alim too
/ g(atp'(r) + u'a[r))(p(r) dr — 0. ~ Jo Hence, choosing a sufficiently large, we obtain numbers
F Jo
arbitrarily small such that (4.3) has a solution, i.e. s £ [SQ,BI\. Going to the limit we have 1 = 0 6 [«o, «i]. - If s €]so,si[, the problem (4.1) has at least one solution. Let Claim s G]SO,SI[. AS in Step 3 of the proof of Theorem 4.1 it is easy to build lower and upper solutions of (4.3) so that the claim follows from Corollary II-2.2 or Theorem III-3.8 according as the lower and upper solutions are well-ordered or not. Claim 5 - For any e > 0 there exists K$ > 0 such that if u is a solution of (4.3) for some s with \s\ > e, then \\u\\ci < KQ- Assume that there exists a sequence (sn)n with \sn\ > e and corresponding solutions un with ||un||c» —* oo. Write un(t) = Un
ft*
= 0, la
i.e. lim sn — 0, which is a contradiction. Claim 6 - If s £ {SQ, SI}\ {I}, the problem (4.3) has at least one solution. This claim follows from Claim 5 and Arzela-Ascoli Theorem.
4. STRONG RESONANCE PROBLEMS
313
Claim 7 - (in) holds. Let s G]0,SI[. We build a strict lower solution a as in Part 1 of Theorem 4.3, using Proposition III-2.7 together with Remark III-2.1. Then we choose u large enough so that if u is the corresponding solution of (4.5), we have for all t G [0, TT] \a{t)-u(t)\<wp(t) and
f
+ u'(t))tp(t) dt < s.
Jo
In that way, + Jo u are upper solutions of (4.3). We conclude applying Corollary II-2.2 and Theorem III-3.8. Example 4.4. A model example is u" + u + arctanu' = sip(t), u(0) = 0, u(n) = 0, which is quite different from Example 4.2 as we have here a set ]SQ, SI[\{0} so that the problem has two solutions. Here So < 0 < s\ as follows from Proposition 4.6 below. Example 4.5. Notice also that Theorem 4.4 can be applied to the resonant problem u" + u = sip(t), ) = 0, U(TT) = 0, which is such that So = S\ — I = 0. The proposition that follows gives conditions that rule out the situation of the last example. Proposition 4.6. Assume that g : R —> R is of class C1, g(0) = 0, the limits g(—oo) and g(+oo) exist and g'(0) ^ 0. Let s G R and p G C([0,TT]). Then there exists e > 0 such that, if \\p\\oo < £, we have so(p,g) < 0 < si(P, 9)Proof. The mapping ^':C2([0,7r])nC01([0,7r]) -> C([O,TT]), U ^ u"+u+g{u') is differentiable at u = 0. Since the problem v" + v + cv' = 0, v(0) = 0, v(ir) = 0, where c is a nonzero constant, has only the trivial solution, it follows that .F'(O) is an isomorphism. Hence, by the Inverse Mapping Theorem, the range of T contains an open ball centered at the origin and we can conclude.
314
VI. PARAMETRIC MULTIPLICITY PROBLEMS
We can also have that I coincides with one of the values SQ or si as in the following proposition. Proposition 4.7. Let p £ Z1(0,TT) and g : E —» K be locally Lipschitz continuous such that the limits g(—oo) and g(+oo) exist and are finite. Assume that for allv£R g{v) < g(-oo) = g(+oo). Then we have sQ(p,g) < si(p,g) = I = 2J~ g(+oo). Proof Notice that any solution u of (4.3) is such that
s= f JO
g(uf(t)Mt)dt<2x-g{+oo). V 7T
Erom Claim 1 of the proof of Theorem 4.4, we know there exists a solution u of (4.3) for some value of s and the claim follows.
Chapter VII
Resonance and Non-resonance 1
A Periodic Non-resonant Problem
Consider the problem u" + f(t,u) = h(t), u(a)=u(b), «'(o)=u'(6).
. . {
}
Our first result considers a nonlineaxity with an asymptotic slope under the first eigenvalue. Theorem 1.1. Let h £ Ll{a,b). conditions and
Assume f satisfies
limsup ——— < 7(4) ^ 0 u
|u|-»oo
uniformly int£
L1-Caratheodory
[a, b],
where 7 G L 1 (a,6). Then the periodic problem (1.1) has at least one solution. Proof. Let Ai be the first eigenvalue of the problem u(a) = u(b), u'(a) — u'(b). It is known (see Proposition A-3.9) that the corresponding eigenfunction
=
min
J£Li_^
L _ J _ > 0, b 2
UGH\{O}
[ u dt
315
316
VII. RESONANCE AND NON-RESONANCE
where H = {u e Hl(a,b) | u(a) = u(b)} (see Theorem A-3.7). If Ai = 0, we compute
0= / (V'{(t) + 1(t)iPl(t))dt= Ja
[ 7Wvi(t)d«. Ja
This equation contradicts the fact that j(t)(pi(t) < 0 on a set of positive measure. Step 2 - Construction of an upper solution. Let ip be the solution of the boundary value problem
We deduce from the maximum principle, Theorem A-5.3, that ip(t) > 0 on \a,b}. By hypothesis, there exists r G Ll{a, b) such that for all u > 0 and a.e. t€[a,b],
/(*,«)< (7(0+ y ) « + r(t). Let w be the solution of w" + (7(0 + %) w + r(t) - h{t) = 0, w(a) = w(b), w'(a) = w'(b). If k is large enough, the function /3(t) — w(t) + kip(t) is a positive upper solution of (1.1). Step 3 - Conclusion, In the same way we construct a negative lower solution a < 0 and the claim follows from Theorem 1-3.1. Notice that we can always choose 7 G L°°(a, b) since max{7, —1} satisfies also the assumptions. Moreover, we can generalize this result using functions 7 that can take positive values provided these are not too large. For such functions, we define 7 + (0 = max(7(t),0) and 7~(0 = max(-7(t),0). Theorem 1.2. Let h G Ll{a,b). conditions and limsup ———- < 7(£) |u|-«x>
u
Assume f satisfies Ll-Caratheodory uniformly int€ [a, b],
where 7 6 L1(a, 6) is such that ||7+||^ < 5 ^
and
|| 7 +|| L1 - (1 - ^ | | 7 + | | } / 2 ) 2 | | 7 - | U l < 0 .
Then the periodic problem (1.1) has at least one solution.
1. A PERIODIC NON-RESONANT PROBLEM
317
Proof. Define Ai to be the first eigenvalue of the eigenvalue problem u(a) — u(b), u'(a) = u'(b), and if i the corresponding eigenfunction which can be chosen positive (see Proposition A-3.9). Step 1 - Ai > 0. Let us prove that for all u e Hl{a, b), u^Q, u(a) = u(b), / [u'2(t)->y(t)u2(t)}dt>Q. Ja Otherwise, there exists u =£ 0 such that
f u'2(t)dt< ff j(t)u2(t)dt Ja
(1.2)
Ja
and
ll«'ll£a
with c = J^jf- (see Proposition A-4.1) and \u{t)\ > \u\ - \u(t)\ > \\u\\oo - INloo - \u(t)\ x — 9C > l 1 / 2 ' \/ i IU/H ll u l|oo _ QIIT/II ^||"||oo > ^ lUill |i"||oov(~\ ^ r l| l| '7v + l||£,i
It follows now that f -y(t)u2(t)dt = f -y+(t)u2(t)dt-
f
Ja
Ja
Ja
j-(t)u2(t)dt
< N I L [ll7 IU> - (1 - 2c|| 7 + ||l, / 1 2 ) a ||7-|M < 0 +
and we have a contradiction with (1.2). Step 2 - Construction of an upper solution. Notice first that there exists an r > 0 such that for all u > r and a.e. t € [a, 6],
/(*,«) <(7(*) + y ) « . Define w to be the solution of %)w - h(t) = 0, w(a) — w(b), w'(a) — w'(b). Then for k > 0 large enough, /3(t) = w(t) + kipi(t) > r is an upper solution.
318
VII. RESONANCE AND NON-RESONANCE
Step 3 - Conclusion. In the same way, we construct a negative lower solution and we conclude from Theorem 1-3.1. We can obtain similar results for a nonlinearity with asymptotic slope between the first and the second eigenvalue. Theorem 1.3. Let h G L1(a,b). conditions and
Assume f satisfies L1 -Caratheodory
7i(i) < liminf ———- < limsup ———- < 72 (t) |u|-»oo
U
uniformly in t € [a, b],
u
|u|_»oo
where 71 and 72 S L1(a,b) are such that rb
1
li(t)dt>0,
7l^0
and
^
Ja
Then the periodic problem (1.1) has at least one solution. Proof. Let A* be the i-th eigenvalue of u(a) = u(b), u'(a) = u'(b), and let
min u£H\{0}
where
6
$(u), X
'
f
la
u2 dt
and
H = {u& H^a,b) \ u(a) =
It follows that Ai < $(1) = -^5lli{t)dt < 0. If Ai = 0, we have $(1) = 0 which implies we can choose
2. RESONANCE FOR THE PERIODIC PROBLEM
319
Step 3 - Conclusion. Proposition III-3.4 proves that nontrivial solutions of the problem u" + p{t)u+ - q(t)u~ = 0, u(a)=u{b), u'(a)=u'(b), do not have zeros if 71 < p < 72 and 71 < q < 72. The claim follows now from Theorem III-3.3.
2
Resonance Conditions for the Periodic Problem
Consider the periodic problem (1.1) and notice that by adding a constant to both the functions f(t,u) and h(t), we can suppose that
f.
b
h(t) dt = 0.
A first result, which is of Landesman-Lazer type concerns a nonlinearity with asymptotic slope on the left of the first eigenvalue. Theorem 2.1. Let h G Ll(a,b) be such that Ja h(t)dt = 0. Assume f satisfies L1 -Caratheodory conditions and (a) there exist s+ € R and f+ £ Lx(a,b) such that if u> s+ and
[
f+(t)dt<0;
Ja (b) there exist s _ £ R and / _ G L1 (a, b) such that
/(t,u)>/_(t) and
I
ifu<s^
b
f-(t)dt>0.
Then the periodic problem (1.1) has at least one solution. Proof. To build a lower solution a, we consider the solution w of w" + (/_ - h)(t) = 0, w(a) — w(b), w (a) = w'(b), with Ja w(t) dt = 0. Then for k large enough, a(t) = —k + w(t) is a lower solution. In the same way we construct an upper solution (3 > a and the theorem follows from Theorem 1-3.1.
320
VII. RESONANCE AND NON-RESONANCE
The next result uses the Landesman-Lazer conditions to consider a nonlinearity with asymptotic slope between the two first eigenvalues. Theorem 2.2. Let h € L1(a,b) be such that J*h(t)dt satisfies L1 -Caratheodory conditions and
- 0. Assume f
(a) there exist s+ G R, /+ G L1 (a, b) such that ifu>s+
we have
f(t,u)>f+{t) and
h
L
f+(t)dt>0;
(b) there exist s_ G M, / _ G L 1 (a, 6) such that ifu<s~
we have
/(t,u)
I
f-(t)dt<0\
Ja
(c) limsup tuHoo
—- ^ I u
I uniformly in t. \o-aj
Then the periodic problem (1.1) has at least one solution. Proof. The argument of the proof of Theorem 2.1 gives lower and upper solutions a and /3 such that a > 0. We conclude then as in Step 3 of Theorem 1.3. Another type of result uses a monotone nonlinearity. Theorem 2.3. Let h G L1(a,b) be such that J^h(t)dt continuous, non-increasing function. Then the problem
— 0 and f be a
u" + f{u) = h(t), u(o) = u(b), u'(a) = u'{b) has at least one solution if and only i/0 G Im(/). Proof. The sufficient part of the claim can be deduced from Theorem 2.1. The necessary condition follows by direct integration of the equation.
3. A NON-RESONANT DIRICHLET PROBLEM
3
321
A Non-resonant Dirichlet Problem
Consider the Dirichlet problem "+u + f(t,u)=h{t), u{0) = 0, «(TT) = 0.
{6
'
The first results consider a nonlinearity with asymptotic slope on the left of the first eigenvalue. Theorem 3.1. Let h € L1(0,TT). Assume f satisfies Ll-Caratheodory conditions and there exists 7 € Ll(0,ir) such that lim sup ———- < 7(4) ^ 0 uniformly in t. M-+00
u
Then the Dirichlet problem (3.1) has at least one solution. Proof. Let Ai be the first eigenvalue of the problem u(0) = 0, u(n) = 0. Arguing as in the proof of Theorem 1.1 and using Theorem A-3.2 we see that Ai > 0. Hence, we can find a function ip(t) > 0 on [0, n] such that
By hypothesis, there exists a € Ll(0,n) such that for all u > 0 and a.e. te
[O,TT],
/(t,«)<(7(*) + y ) « + «(*) Let w be the solution of w" + (1 + 7(i) + ^ ) w + a{t) - h(t) = 0, w(0) = 0, W(TT) = 0. If B is large enough, the function @{t) = w{t) + Bip(t) is a positive upper solution. In the same way we construct a lower solution a < 0 and the theorem follows from Theorem II-2.4. If the variables u and t separate as in {6
«(0) = 0, u(ir) = 0,
we can use conditions on the potential F(u
Jo
>
322
VII. RESONANCE AND NON-RESONANCE
Theorem 3.2. Let h € L°°(0,7r). Assume / e C(E) is such that liminf ^ «—»+oo
< 0 and liminf ^ H l < 0,
ti^
u—>—oo
U^
where F(u) — I f(s) ds. Jo Then the problem (3.2) has at least one solution. Proof. Claim - There exists a nonnegative upper solution of (3.2). In case u + f(u) is unbounded from below on M+ we choose /3 € R.+ such that P + f(fi) ^ —Halloo- The constant function /3 verifies the claim. Assume next u + f(u) is bounded from below on E + and consider the equation u" + u + f(u) + M = 0, (3.3) where M > \\h\\oo and u + f(u) + M > 1 on M+. We deduce from the assumptions that there exist a sequence (vn)n and an e > 0 such that vn —> +oo and ^(fn) < — ev^- Define V(u) = | u 2 + F(u) + Mu and notice that the sequence ( 2 ^n ~ ^( v n))n goes to infinity with n since
Therefore, we can find a UQ > 0 such that V(uo) = u0 + /(wo) + M > 1 and for all u € [0, uo] ii^M 2 - V(U) <
{
V(UO).
Consider now the solution /? of (3.3) such that /3(TT/2) = u0, /3'(TT/2) = 0, whose energy is E = V(UQ). This solution cannot take the value zero on the interval ]f - T, f + T[, where T--1
The claim follows. Conclusion - In a similar way we build a negative lower solution and the theorem follows from Theorem II-2.4. Example 3.1. Consider the problem (3.2) where u3 f(u) = usin(m(u + 1)) + -5—r cos(ln(u2 + 1)). 2
(3.4)
3. A NON-RESONANT DIRICHLET PROBLEM
323
Theorem 3.1 does not apply since |u|-+oo
u
However, we compute F(u) = \ sin(ln(u2 + 1)) and F(u) F(u) liminf —r-^- = — 4 and liminf —5— = — 4, so that existence follows from Theorem 3.2. The next result considers a nonlinearity with asymptotic slope between the first and the second eigenvalue. Theorem 3.3. Let h G L1(0,7r). Assume f satisfies L1 -Caratheodory conditions and li(t)
< l i m i n f ———- < l i m s u p ———- < 7 2 ( 0 uniformly |u|-»oo
U
|u|-.oo
u
int£
[0,TT],
where 71 and 72 G L1 (0, TT) are such that /"
t) sin21 dt > 0, 71 / 0 and 72(t) ^ 3.
JO
T/ien i/ie Dirichlet problem (3.1) /ias
324
VII. RESONANCE AND NON-RESONANCE
We can write an analog of Theorem 3.2 in case the nonlinearity has an asymptotic slope between the first and the second eigenvalue. Theorem 3.4. Let h G L°°(0,7r). Assume f <E C(R) is such that F(u) , , F(u) limsup —— > 0 and limsup —^— > 0, where F(u) = / f(s) ds and Jo limsup
u
|u|-*oo
< 3.
(3-5)
Then the problem (3.2) has at least one solution. Proof. We deduce from the assumptions that there exist a sequence (vn)n and an e > 0 such that vn —> +oo and F(vn) > ev%. Define V(u) = ^u2 + F(u) — H/illooU and notice that the sequence ( % v% — V(vn))n goes to - c o with n since
Therefore, we can find u§ > 0 which is such that for all u £ [0, >
£
Observe that V(u) < V(uo) for all u £ [0,wo[ and then f(ua) — \\h\\oo > 0. Consider now the solution w of
V'{UQ)
=
f(w)-\\h\\oo = Q, tu(0) = u 0 , w'(0) = 0. There exists T > 0 such that w(t) > 0 on [0,T[, w(T) = 0 and w'(t) < 0 on ]0, T[. We compute then
Wo
) - V(u)
Hence, the function a{t)=w{T-t), iite[0,T[, = «o, if*G [T,ir-T[, = w(t-n + T), if t € [n - T, n] is a nonnegative lower solution of (3.2).
3. A NON-RESONANT DIRIGHLET PROBLEM
325
In the same way, we build a nonpositive upper solution /?. If u + f(u) is unbounded from below for u —> +00, the problem (3.2) has an upper solution ft > a and the result follows from Theorem II-2.4. We work in the same way the case u + f(u) unbounded from above for u —> —00. Hence we can assume
N |u|->+oo
0
U
and we conclude by Theorem III-3.9 and the equivalent of Proposition III3.4 for the Dirichlet problem. Example 3.2. Consider again the problem (3.2) with / defined in (3.4). We know that existence of a solution follows from Theorem 3.2. We can also apply Theorem 3.4 so that existence follows both from a non-resonance result on the left and on the right of the first eigenvalue. Exercise 3.1. Prove, using the Three Solutions Theorem (Theorem III2.11), that problem (3.2) with / defined in (3.4)and h e L°°(0,7r) has at least three solutions. We can generalize the above theorem, replacing (3.5) by a condition on the potential. Exercise 3.2. Let h E L°°(0,7r). Assume / 6 C(R) is such that limsup —Tjp > 0 and limsup —^— > 0, where F(u) = / /(s) ds, and for some /i, v such that , 1
t
0
lim sup
< n and lim sup
< u,
2F(u) , , 2F(u) lim mir —)r-i- < fi and lim mi r—^- 1 < u. u—>+oo
u
u—
u
Prove then that the problem (3.2) has at least one solution. Hint : See Exercise III-3.3 or [79]. Our last result concerns the problem u(0) = 0, «(ir) = 0, where the nonlinearity f(t, u) is monotone.
+ J-
= 1,
326
VII. RESONANCE AND NON-RESONANCE
Theorem 3.5. Let h £ Ll(Q,ir) and 7 £ L 1 (°i 7r ) &e such that f(t) $ 0. Assume / : [O,TT] X R —> R satisfies L1-Caratheodory conditions, f(t,O) £ L°°(0,7r) and for each t £ [0,7r], /(£,.) is o non-increasing function. Then the Dirichlet problem (3.6) has a unique solution. Proof. Compute first p
u
|u|—oo
|u|-»oo
U
uniformly in f. Hence by Theorem 3.1 we have the existence of a solution u. We only have to prove its uniqueness. Assume that u and v are two distinct solutions of (3.6) and let w = u — v. Prom the proof of Theorem 3.1, we know Ai > 0 and we compute 2 2 0 = - f w'2(s)ds+ [ (l + -y(s))w ( (s)ds
+ J (f(s, u(s)) - f(s, v(s)))(u(s) - v(s)) ds <-Ai rw2(s)ds+
r(f(s,u(s)) - f(s,v(s)))(u(s) - v(s))ds < 0,
Jo
Jo
which is a contradiction.
4
Landesman-Lazer conditions for the Dirichlet problem
Consider the Dirichlet problem (3.1) and define ip(t) = J^sint. Section 2, we can assume
As in
r I h(t)tp(t) dt = 0.
Jo
The first results consider a nonlinearity with asymptotic slope on the left of the first eigenvalue. Proposition 4.1. Let h £ L^O,?!") be such that f^ h(t) sint dt = 0. 4ssume f is an L1-Caratheodory function and (a) there exist s+ £ R and f+ £ L1(0, ?r) such that f(t,u) < f+(t)
ifu>s+
and Jo
/+(*) sinidi < 0;
sint
4. LANDESMAN-LAZER CONDITIONS
327
(b) there exist s_ G R and /_ G L1 (0, IT) such that f(t,u)> /_(*)
i/u<s_sin£
and /o Jo
/_ (t) sin t dt > 0.
Then the Dirichlet problem (3.1) has at least one solution. Proof. Let us build a lower solution a(t) < s_ sin t. To this end, define first w to be the solution of w" + w + /_(t) - h(t) - ( r /_(a)
and u > s+ > 0 => /(t,u) < f+(t)
with /+(*) sinidi < 0, / /_(*) sintdt > 0, Jo as in Theorem 2.1, are not sufficient to prove the result in Proposition 4.1. Consider for example the problem (3.1) with h = 0, f(t,u) = g(u)sint, g(u) = cosu if u £ [—7r/2,7r/2] and #(u) = 0 if u £ [-ir/2,ir/2]. This problem has no solution since such a solution u verifies /J7 g{u{t)) sin21 dt — 0, which is impossible. The following is a result in this direction. Theorem 4.2. Let h G L1(0,7r) be such that J* h(t)sintdt = 0. Assume f is an L1-Caratheodory function, and there exist /+, /_ € L1(0, TT) such that
328
VII. RESONANCE AND NON-RESONANCE
(a) limsup/(i,u) < f+(t) uniformly in t and
I f+{t) sintdt < 0 ; Jo (b) liminf f(t,u) > /_(i) uniformly in t and u*oo
/ Jo
_(£) sintdt >0.
Then the Dirichlet problem (3.1) has at least one solution. Proof. We prove that assumption (a) implies the corresponding assumption in Proposition 4.1. Observe first that (a) implies that there exist r and rj > 0 such that for a.e. t £ [0, TT] and all u > r, we have
f(t,u)
k+(t) sintdt+
(\k+(t)\ + ar(t)) sintdt < 0 Jit—e
and s+ such that s+ sini > r for all t € [e, TT — e]. Then the assumption (a) of Proposition 4.1 is fulfilled with ifte[e,7r-e], f+{t) = k+(t), = \k+(t)\+ar(t), ift$[e,n-e]. D Example 4.1. A simple example of application of the above theorem is u" + u — arctanu = k(t), u{0) = 0, U(TT) = 0, where k = J ^ k(t)ip(t) dt e ] - \/27r, >/27r[. Here we take ft(t) = fc(f) and / ( i , u) — — arctanu — kip{t).
4. LANDESMAN-LAZER CONDITIONS
329
Remark 4.2. Theorem 4.2 is no more true if either JQ f+(t) sintdt = 0 or J* f-(t) sintdt = 0. Consider for example u" + u + exp(-u 2 ) = 0 , u(0) = 0, u(n) = 0. This problem has no solution (This fact can be proved multiplying the equation by sint and integrating.) but liminf f(t,u) > f-(t) = 0, limsup f(t,u) < f+(t) = 0 and
/ /+(*) sintdt = 0, / f-(t) sintdt = 0. Jo Jo Notice also that assumptions such as (a) and (b) in Proposition 4.1 or Theorem 4.2 cannot be necessary as they are essentially asymptotic. To build counter-examples, we can take any problem which has a solution and modify the equation for |u| > s sin t, where s is large enough, so that the assumptions (a) and (b) do not hold. Similar results can be obtained for systems with asymptotic slope between the first and the second eigenvalues. Here we use lower and upper solutions in the reversed order. Proposition 4.3. Let h € L1(0,7r) be such that f£ h(t) sintdt — 0. Assume / is an Ll-Caratheodory function and (a) there exist s+ £ R, /+ 6 L 1 (0 ) TT) such that f(t,u) > f+(t)
if u> s+ sint
and Jo Jo
f+{t) sintdt >0;
(b) there exist s_ e i , / _ 6 L^O.TT) such that f(t,u)
(t)
ifu<s-
sin t
and
r I /_ (t) sin t dt < 0; Jo (c) limsup ' ^ 3 uniformly in t. Then the Dirichlet problem (3.1) has at least one solution.
330
VII. RESONANCE AND NON-RESONANCE
Proof. Lower and upper solutions a and j3 are build as in the proof of Proposition 4,1 except that a = a
I
o
f+(t) sintdt > 0 ;
(b) limsup/(t,u) < f~(t) uniformly in t and
f
f-(t)
sintdt < 0 ;
Jo
(c) limsup ———- ^ 3 uniformly in t. M-»oo
u
Then the Dirichlet problem (3.1) has at least one solution. Proof. This proof repeats the argument in the proof of Theorem 4.2.
5
Other Resonance Conditions
An alternative to the Landesman-Lazer conditions given in Theorems 4.2 and 4.4 is to control the resonance using repulsive forces in case the asymptotic slope is on the left of the first eigenvalue and attracting ones if it is on the right. Here, the assumption f£ h(t) sin t dt = 0 is a real restriction since we cannot add a constant to f{t,u). Theorem 5.1. Leth G L^OjTr) and f satisfy L1-Caratheodory conditions. Assume si (a) / h(t) sintdt = 0; Jo (b) uf(t,u) < 0 for a.e. t € [O,TT] and all u G R. Then the Dirichlet problem (3.1) has at least one solution.
5. OTHER RESONANCE CONDITIONS
331
Proof. Let w be the solution of w" + w = h(t), w(0) = 0, w(ir) - 0,
and define a(t) = w(t) — asint, where a is large enough so that a < 0. It is easy to see that a is a lower solution. Similarly, we define (3(t) = w(t) + bsint which is nonnegative if b is large enough and therefore is an upper solution. The proof follows then from Theorem II-2.4. Remark An alternative proof would be to deduce this theorem from Proposition 4.1. Notice however that we cannot apply Theorem 4.2. Consider for example the problem u" + u — uexp(—u2) = sin2i, u(0) = 0, U(TT) = 0. Existence of at least one solution follows from Theorem 5.1. On the other hand, any function / + (t) such that f(t,u) = — uexp(—u2) < /+(t) if u > s+sint has to be nonnegative and Jo /+(£) sin tdt > 0. Hence, we cannot use Theorem 4.2 and Proposition 4.1 can only apply with /+(£) = 0. If the asymptotic slope is on the right of the first eigenvalue we can prove a dual result. Theorem 5.2. Leth € L1(0,TT) and f satisfy L1 -Caratheodory conditions. Assume
r (a) I h{t) sin t eft = 0; (b) u f(t, u)>0 for a.e. t€ [0, n) and all u e E; (c) limsup ———- ^ 3 uniformly in t. |u|—oo
u
Then the Dirichlet problem (3.1) has at least one solution. Proof. Lower and upper solutions a and (3 are built as in the proof of Theorem 5.1 except that here a > 0 and (3 < 0. We conclude then using Theorem III-3.9. Remark Here also, we can deduce this result from Proposition 4.3.
332
VII. RESONANCE AND NON-RESONANCE
An alternative concerns strong resonance. Here f(t, u) goes to zero as u —> 0 but we control the rate of decrease of f(t,u). Theorem 5.3. Let h G L1(0, TT) and f be a Caratheodory function. Assume fir
(a) / h{t) sin £ eft = 0; Jo (b)for some k G L°°(0,7r), for a.e. t G [O,ir] and allue R, \f{t,u)\ < k(t); (c) for some /u G [1,2[ and liminf sgn(it)|u|'i/(i,M) > m > 0 lim \u\*t~1f(t,u)=Q |ti|—>OO
|«|—>0O
uniformly in t. Then the problem (3.1) has at least one solution. Proof. The proof follows the line of Step 1 in the proof of Theorem VI-4.2. Let ip(t) = yj^sint and define K as in Remark VI-4.1. Hence, for every s e R , the solution u of (VI-4.2) is such that ||u||oo < K and |u(i)| < K
and
«"/(*,«) > 3 f .
Pick next 5 > 0 small enough so that
where x 1S the characteristic function of [0,5] U [ir — 5, TT]. Then we choose a > 0 large enough such that a sin 5 > K+R and define a(t) — u{t)+aip{t), where u is the corresponding solution of (VI-4.2). Notice that a(t) > R
I Ht,c)a"vdt = j KUcfi—^it-
JJ(t,a)
oM-i
L
2
io
lit
V-H*)-
6. FREDHOLM ALTERNATIVE RESULTS
333
We deduce, as in Remark VI-4.1, that a solves u" + u + f{t, u) = sotp(t) + h(t), u(0) = 0, U(TT) = 0, with so = Jg f(t,a(t))tp(t)dt > 0 and therefore a is a lower solution of (3.1). The construction of an upper solution is similar and we conclude from Theorem III-3.8. D Remark The condition lim | « | " - 1 / ( t , « ) = 0 alone does not ensure existence of a solution as follows from the example in Remark 4.2. Example 5.1. An example of application of Theorem 5.3 is the following u" + u+ j^p - e-W cosu = h{t), u(0) = 0, u(n) = 0, where f£ h(t) sin tdt — 0. Here Landesman-Lazer's conditions (Proposition 4.1) do not hold since 1+w
— e~'"' cosu < /+(*),
foru>s+sini
implies f+(t) > 0 and f£ f+(t)sintdt > 0. Proposition 4.3 does not apply either since 1+u
— e~'"' cosu > f+(t),
for u > s+ sint
implies f+(t) < 0, f+(t) < —1/2 in a neighbourhood of t — 0, and therefore J* f+(t) sin tdt < 0. Also, Theorems 5.1 and 5.2 cannot be used since uf(t, u) is not one sign.
6
Fredholm Alternative Results
The next theorem considers the inverse of the operator Fu = u" + u — qu~ defined on W 2il (0, n) (1 CQ([0, TT]). More precisely, we consider the problem
334
VII. RESONANCE AND NON-RESONANCE =h(t), u" + u-qu w(0) = 0, U(TT) = 0,
.
[
'
where q ^ 0 and h £ Ll(Q,ir). This corresponds to a Predholm alternative for the problem u" + fiu+ — wuT = h(t), u(0) = 0, «(TT) = 0, with (/i, v) = (1,1 + q) on the first curve of the Fucfk spectrum. Theorem 6.1. Let h E /^(0,7r) and q =£0- Then the problem (6.1) has a solution if and only if q / h(t)smtdt<0. (6.2) Jo Proof. Claim 1 - The condition (6.2) is sufficient. Let w be the solution of T
w" + w = h(t) — (^ / h(s) sin sds) sin t, * Jo w(0) = 0, w(w) = 0, such that J£ w(t) sintdt = 0. Assume g > 0 and define a = w + A sin t, with A large enough so that a > 0. This function is a lower solution since a" + a = h(t)-(-
/ h(s) sinsds) sint > h(t) = h{t) + qa~. Jo
In a similar way, we build a nonpositive upper solution /3 = w — Bsint, where B is large enough, and the claim follows from Theorem III-3.9. In case q < 0, a = w — Asixit, with A large enough so that a is a nonpositive lower solution and /? = w + Bsint, with B large enough so that /3 is a nonnegative upper solution. The claim follows then from Theorem II-2.4. Claim 2 - The condition (6.2) is necessary. Let u be a solution of (6.1). Multiplying this equation by sin t and integrating, we have PIT
P-K
i-ir
q I h(t) sintdt = q / (u"+u—qu~)sintdt = — g2 / u~sin£d£<0. ./o
JO
JO
Some results can be obtained from a monotonicity assumption. Theorem 6.2. Let h 6 L1^,*) be such that $* h(t)sintdt = 0 and f : [0,7r] x ]R > R. be an L1 -Caratheodory function. Assume that for each t g [0,7r], f(t,.) is a non-increasing function.
6. FREDHOLM ALTERNATIVE RESULTS
335
Then the Dirichlet problem (3.1) has at least one solution if and only if there exists ( £ 8 such that
I
f(t,£smt)sintdt
= O.
(6.3)
o
Proof. Claim 1 - The condition (6.3) is sufficient. Let to be a solution of w" + w = h- f(t,£sint), to(0) = 0, w(n) = 0. Choose then A and B large enough so that a := w — A sin t < £ sin t < w + B sin t =: /3. We can see that a is a lower solution and 8 an upper one so that the claim follows from Theorem II-2.4. Claim 2 - The condition (6.3) is necessary. Let u be a solution of (3.1). Multiplying this equation by sin t and integrating, we obtain / f(t,u(t))sintdt = O. Jo On the other hand, there exist £i and £2 such that for all t E [0,7r], £1 sin t < u(t) < £2s m tHence and as / is non-increasing we obtain /
f(t,&Bmt)s,mtdt
< 0 < /
f (t, & sin t) sin tdt
Jo Jo and conclude by an application of the Intermediate Value Theorem. Example 6.1. Notice that the function f(t, u) = — exp u + a verifies the assumptions of Theorem 6.2 for any a > 0. Remark 6.1. If / is decreasing, we can prove that this solution is unique. Let u and v be two distinct solutions of (3.1) and let w = u — v. We compute
0 = - f w'2(s)ds+ f w2(s)ds +/
(f(sMs))-f(s,v(s)))(u(s)-v(S))ds
< / (f(s,u(s))-f(s,v(8)))(u(s)-v(S))ds<0, Jo which is a contradiction.
336
VII. RESONANCE AND NON-RESONANCE
If / is only non-increasing, uniqueness does not occur in general. For example, consider the case where f(t,u) = 0 if (t, u) G [0, TT] X [—1,1] and h{t) = 0, then for all a G [—1,1], u(t) = asint is a solution. This example is generic as follows from the next result which completes the equivalent for Dirichlet problem of Theorem 1-3.5. Theorem 6.3. Under the hypothesis of Theorem 6.2, the setS of solutions of (3.1) is convex. Moreover, there exists u G W^2'1(0,7r) with J^ u(t) sin t dt = 0 and an interval J c l such that S = {u = u + csin t: c G J } . Proof. Let u and v be two solutions of (3.1). The function w — u — v is such that T
PIT
0= - /
w'2(t)dt+
<-
w'2(t) dt+
Jo
r7t
w2(t)dt + Jo
(f(t,u(t))-f(t,v(t)))w(t)dt
w2(t) dt < 0.
Hence, w(t) — asint and we can find a set J c E such that S — {u = u + csin 11 c G J } , where u(t) = u(t) — ^(JQ U(S)sinsds) sint and u is any solution of (3.1). We only have to prove that J is an interval. To this aim, let c\ < 0% be in J and c G]ci,C2[. Observe that h(t) = u"(t) + u(t) + f(t,u(t) + c2sint) < (u(t) + csint)" + {u(t) + csint) + f(t,u{t) + csint) < u"{t) + u(t) + f(t, u(t) + cx sint) = h{t). This relation shows that v(t) = u(t)+csint is also a solution and completes the proof. Remark 6.2. Notice that without the monotonicity assumption on f(t,.) the set of solutions might be non-convex. Consider for example the problem u" + u + (u2 - sin21)2 = 0, u(0) = 0, U(TT) = 0. It is easy to see that u(t) — — sin t and u(t) = sin t are solutions. Further, if u is another solution then / (u2(t) - sin21)2 sintdt = O.
Jo
7. DERIVATIVE DEPENDENT PROBLEM
337
This equation implies that u(t) = — sin t and u(t) = sin t are the only two solutions. Exercise 6.1. Prove the same kind of results for the Neumann boundary value problem.
7
Derivative Dependent Dirichlet Problem
Consider the problem u(0) = 0, U(TT) = 0. First, we consider a non-resonance result which considers a one-sided linear control of the nonlinearity f(t, u, v). Theorem 7.1. Let f satisfy L1 -Caratheodory conditions and assume that for anyr0 ro > 0 there exist + = l,
Assume also there eocist b G Ll(0, T T ; R + ) , 5, 7 G K + , and r% > 0
/or a.e. t € [0,7r] and a// (w,^) G R2, with \u\ > r\, we have sgn(u) f(t,u, v) < b(t) + 7|u| + 5\v\. If further IT < T(5,7), w/iere
if 5* - 4 7 > 0,
f/ien t/ie problem (7.1) /ias a solution.
338
VII. RESONANCE AND NON-RESONANCE
Proof. Let y be a solution of
on [0, TT] . Define v to be the solution of v" - Sv' + jv = 0, «(TT/2) = 1, «'(7r/2) = 0,
on [7r/2,7r] and extend v by symmetry on [0,7r] SO that v(t) — V(TT — t). Next, verify that for all values of <5 and 7, the condition n < T(5,7) implies v(t) > 0 and v'(t)(t — TT/2) < 0, whence v satisfies the equation v" + 5\v'\ +7?; = 0. It follows that if k > 0 is large enough f3 = y + kv > r\ and we can compute < y"(t) + kv"(t) + b(t) + 8(\y'(t)\ + k\v'(t)\) + j(y{t) + kv{t)) = 0. Hence, 0 is an upper solution. We build similarly a lower solution and conclude from Corollary II-2.2. Remark 7.1. If 5 = 0, the condition becomes 7 < 1, which is best possible since 1 is the first eigenvalue of the corresponding linear problem u" + Xu = 0,
u(0) = 0, u(n) = 0.
Next, we consider a non-resonance result with a quadratic control of the nonlinearity f(t,u,v). Theorem 7.2. Let f satisfy L1 -Caratheodory conditions and assume that for any r§ exist p, q £ [1, oo] such that - + - = 1, tp £ r^ > > 0, there ea and ip £ Lp(0, TT) with ds = oo, and for a.e. t £ [0,7r], aH tt G [—ro,ro] and all u € E
|/(t,«,«)|
1 -7 . 7
7. DERIVATIVE DEPENDENT PROBLEM
339
Proof. Construction of an upper solution. First, let us choose a 7 a € J7,1[ and an R > r\ such that Vu>R,
h ( u ) < ^ ^ la
Next, we choose 7& €]7,7 a [ and e > 0 such that (
+
)
la
1b
Define then to to be a solution of w" + iw + b{t) = 0,
w > R,
and Y(u) = ulb. This function solves
y
76
At last, we choose k such that ^ < k2 < 1, and define v (t) = Vo cos k (t—f). If Vb is large enough, this function verifies v"+ —v + Kw'2(t)vl-"lb <0,
v>0,
76
i z ^ ( 1 + 1)A. It is now easy to check that the function @(t) = w(t) + Y(v(t)) > R is a positive upper solution since
where K =
p" + f(t, $, 13') < w" + Y"(v)v'2 + Y'{v)v" +b + i(w + Y(v)) + i ^ [ ( l + e)^v'2 + (1 + 1)J£] < 0. Conclusion. We obtain in a similar way a lower solution and the theorem follows from Corollary II-2.2. Remark 7.2. In case h(u) = 0, we recover the non-resonance condition
H-.OO
uniformly in t.
340
VII. RESONANCE AND NON-RESONANCE
Example 7.1. A non-strict inequality limsup|u|/i(u) <
1—7
|«|—>oo
does not
7
imply existence as follows from the problem u
~rTv "" \u\) >7 (i+|«|) u U(0) = 0, U{7t) = 0,
—u'
(7 2\
which has no solution. To verify this claim, let us introduce the solution Y(u) of the Cauchy problem
and verify that 7 (l+«) If u solves the problem (7.2), it is easy to see that u > 0 and the function v(t) = Y(u(t)) verifies v" + v + 7 = 0,
v(0) = 0, v(n) = 0.
But as JQ 7 sin t dt =^ 0, it is clear from Fredholm alternative that this last equation has no solution. The same ideas can be used to study a resonance problem with a quadratic control. In case ft, = 0, it reduces to Landesman-Lazer conditions. Theorem 7.3. Let f be an L1 -Caratheodory function and assume that for any ro > 0, there exist p, q G [1, oo] such that | + | == 1,
—rr ds = OO)
and for a . e . t G [0, TT], all u G [—ro, r o ] and a l l v e R
Assume also that there exist an L1 -Caratheodory function g(t,u) and a continuous function h(u) > 0 such that for a.e. t G [0,?r] and all (u,v) G t, u, v) <
g(t,u)+h(u)v2.
Define H{y) = \ f h(s)dsl Jo
Q(y)= f Jo
expH(r)dr
7. DERIVATIVE DEPENDENT PROBLEM
341
and F(t,y) = sgn(y)g(t,y)expH(y)
- Q(y).
Assume at last there exists 7+ and')- 6 L1(0,TT) such that 7+(t) > lim sup F(t,y),
7-W < liminf F(*,y),
uniformly in t and / 7+(t)sintdt<0, Jo
/ 7- (£) sin t dt > 0. Jo
Then the problem (7.1) has a solution. Proof. Notice first that the function Q : E + -+ M+ is an increasing homeomorphism whose inverse Y(x) = Q~1(x) is the solution of the problem y" + h(y)yn = 0,
y(0) = 0, t/'(0) = 1.
Define then f3(t) — Y(v(t)), where the function v(t) > 0 will be chosen later. We compute /?" + f(t, 0, /3') < [Y"(v) + h(Y(v))Y'2(v)]v12 + Y'(v)v" + g(t, Y{v)) = Y'(v)[v" + g(t, Y(v)) exp H(Y(v))], which proves that 0 is an upper solution if v" + g(t, Y(«)) exp H(Y(v)) = v" + v + F(t, Y(v)) < 0. Notice then that limsupF(t,y(t;)) = limsupF(f,i/) < 7+(t). v—»+oo
y—>+oo
Hence, following the proof of Theorem 4.2 and Proposition 4.1, we can find a function v(t) > 0 such that v" + v + F(t, Y(v)) < 0. A lower solution is obtained in a dual way and we conclude from Corollary II-2.2. D Example 7.2. The simplest example to consider is a problem with a linear restoring force and a quadratic friction u" + b(t) + 7it + u'2 — 0, u(0) = 0,
U(TT)
= 0.
342
VII. RESONANCE AND NON-RESONANCE
Here Theorem 7.2 does not apply since h(u) — 1 and limsup |u|/i(u) = +oo. On the other hand, we can apply Theorem 7.3. We compute F{t,y) = 8gn(y)[(|6(t)| + f\y\ - l)e |y| + !] If 7 < 0, we have lim F(t, y) = Too. y—
If 7 = 0 and \b(t)\ ^ 1, we obtain
r lim sup F(t, y) < 7+ (t) y—»+oo
with / 7+ (t) sin t dt < 0 JO
and ATT
liminfF(t,i/) >7_(t) w i t h / 7_(f)sinidt > 0, where 1+(t) = — 7-(i) = (|&(0l ~ 1)-^ + 1 an<^ ^ ^s l^S 6 enough. In both cases, existence follows from Theorem 7.3. Problem 7.1. Theorem 7.3 extends Theorem 4.2 to derivative dependent problems. This theorem is a resonance result at the left of the first eigenvalue. Consider a similar extension of Theorem 4.4 and obtain a resonance result at the right of the first eigenvalue for derivative dependent problems. Hint : See [93]. In the following theorem, the existence result is obtained from the dependence in the derivative. Theorem 7.4. Letp, q € [l,oo] be such that - + - = 1. Assume f : [0,7r] x ffi2 —» K satisfies U'-Caratheodory conditions and that for anyro > 0, there exist y € C(R+,Rj) and ip € LP(O,TT) with
I
as = oo,
and for a.e. t € [O,TT], all u € [—ro,ro] and
Assume moreover there exist r\ > 0, h E L1([0, TT],]R+) and g G such that
f Jo
>IWU
7. DERIVATIVE DEPENDENT PROBLEM
343
and, for a.e. t e [0, TT] and all (u,v) € M.2, with \u\ > r\, sgn(u)f(t,u,v)
z(w) = 0.
It is easy to see that p(t) = T\ + / z(s) ds > r\ is an upper solution. In a Jo similar way, we compute a lower solution and existence of a solution follows from Corollary II-2.2. Example 7.3. Consider the boundary value problem u" = eu(u'2 + 1), u(0) = 0, u(ir) = 0. It is easy to observe that the Nagumo condition in Theorem 7.4 is satisfied. The other conditions of this theorem follow with g(v) = v2 + 1 and h(t) = e~ n for ri large enough.
This Page is Intentionally Left Blank
Chapter VIII
Positive Solutions In this chapter, we consider positive solutions of the Dirichlet problem ti" + / ( t , u ) = 0 , u(a) = 0, u{b) = 0,
( K
.
}
where / : [a, b] x M —> R is an L1-Caratheodory function. Most of these results extend to other boundary value problems such as the separated boundary value problem or the periodic problem. Such extensions are left as exercises. Let us recall that given u, v G C([a, b]), we write u >- v or v -< u if u{t) > v{t) on ]a,b[, D+u(a) > D+v(a) if u(a) = v(a), D~u(b) < D-v(b) if u(6) - «(6). Here, positive solutions u : [a, b] —> R are such that u y 0.
1
Existence of One Solution
A necessary condition for a positive solution of (0.1) to exist is that inf (t,u)6[o,6]xR+
/(*,")?(*)"
sup
(t,u)e[o,6]xK+
f(t,u) u
where q € Ll(a,b) and Ai(g) is the first eigenvalue of the spectral problem u" + q(t)u + Xu = 0, «(o) = 0, u{b) = 0. 345
,
U
.
;
346
VIII. POSITIVE SOLUTIONS
This condition can be obtained multiplying the equation (0.1) by
The Sublinear Case
Our first concern is a sublinear case, where the "slope" of / — qu is greater than a first eigenvalue near zero and smaller near infinity. Theorem 1.1. Define
and for all u> p and a.e. t S [a, b],
Then the boundary value problem (0.1) has at least one solution u y- 0. Proof. Claim 1 - There exists a 6 W2:1(a, b), a W2'1-lower solution of (0.1) such that ay-0, a(o) = 0, a{b) = 0. Define v
'
llvi(-;ato)Hoo-
The function a G W2>l(a, b) is such that a{t) X 0, a(a) = 0, a(b) = 0 and a" + f{t, a) > a" + qo{t)a = -Ai(g o )a > 0. Claim 2 - The problem (0.1) has a W2'1-upper solution 0 G W2tl(a, b) such that for all t G]a,6[, (3(t) > a(t). Let k G Ll(a, b) be such that for a.e.
1. E X I S T E N C E O F ONE SOLUTION
347
t e [a,b] and all u € [Q,p], |/(«,u)| + |goo(*)|P+l/oo(*)l < &(*) Choose next e > 0 small enough so that fb (/oo + kx) := / (foo(s) + fc(s)x(s))
D function
and U-.+00
u
hold uniformly in t. Then the problem (0.1) has at least one solution u >~ 0. Proof. The assumptions imply Ai(7o) < 0 and Ai(70O) > 0. The proof follows then from Theorem 1.1 with QQ — ~fo — e, qoo = 7e» + e, /oo = — 1 and e > 0 small enough, As an exercice the reader can prove the following result which extends Theorem 1.1 to separated boundary value problems. Here we define \i(q) to be the first eigenvalue of the spectral problem u" + q(t)u + Xu = 0, a\u(a) — a,2u'(a) — 0,
hu(b) + b2u'(b) = 0.
(1-2)
348
VIII. POSITIVE SOLUTIONS
Theorem 1.3. Letai, b\ G E, az, b2 € E + witha\+a\ > 0 andb\+b\ > 0. Assume f : [a, b] x R —+ R is an L1-Caratheodory function that satisfies (HQ~) and (H^), where
(1.3)
has at least one solution u >~ 0. An alternative uses the comparison spectral problem = 0, u(a) = 0, u ( 6 ) = 0 .
K
, '
Let us write fii(r) and ipi(t;r), the corresponding eigenvalues and eigenfunctions. The following result parallels Theorem 1.1. Theorem 1.4. Define il>i(-;r) to be a positive eigenfunction corresponding to /ii(r), the first eigenvalue of the spectral problem (1.4), and assume f : [a, b] x E — M is on L1 -Caratheodory function that satisfies (HQ~) there exist 6 > 0 and ro € Z/1(a, 6), ro(i) > 0 a.e. on [a, 6], such that Mi(ro) < 1 and for allu €]0,5] and a.e. t € [a, 6], /(*,«)-ro(t)«>0; (^^,) i/iere erzsi p > 0, /«, and r ^ £ ^ ( a , 6), r ^ ^ ) > 0 a.e. on [a, 6], swc/i iftai /Ui(roo) > 1,
if
Q,
and for all u> p and a.e. i s [a, 6], f(t,u)~roo(t)u
TTten t/ie boundary value problem (0.1) /ias ai least one solution u >- 0. Proof. Verify that
is a lower solution and build an upper solution as in Claim 2 of the proof of Theorem 1.1.
1. EXISTENCE OF ONE SOLUTION
349
Remark 1.1. We must notice that Theorem 1.1 is equivalent to Theorem 1.4 if qo(t) > 0 and q<x>(t) > 0 a.e. on [a,b]. The first result is however somewhat more general since it allows cases where qo or goo change sign. Recall first that (see Theorem A-3.2), for q G ^(ttjb), q(t) > 0 a.e. on [a, b], we can write uUq) =
rbu'2dt min -^h
€^\{0} \
? d t
and Xi(q) =
min
Ja
GH»\{
fb(u'2-qu2)dt —-—-—
Hence it is easy to see that 1 <S> Xi(q) < 0 and m(q) > 1 <* Ax(g) > 0. This result proves in particular the equivalence of (HQ~) and (HQ), in case qo(t) > 0 a.e. on [a,b}. Next, if / satisfies (H^) with qoo(t) > 0 a.e. on [a, b], we have for all u > p and a.e. t £ [a, b],
with roo(t) = goo(t) + Ai(q-oo)- Observe that ^1(^00) = 1, ^1(^00) =
We prove in the same way that (.ff^,) implies (H^). 1.2
The Superlinear Case
The superlinear case is a "reversed" case, where the nonlinearity is over the first eigenvalue near +00 and under it near zero. The next theorem presents an existence result under such assumptions that refer to the spectral problem (1.1). Theorem 1.5. Let / : [a,b] x K —> E be an L1 -Caratheodory function such that f(t,O) = 0 on [a,b], define tfi(-;q) to be a positive eigenfunction corresponding to \i(q), the first eigenvalue of the spectral problem (1.1) and assume (HQ) there exist S > 0 andqo £ Ll(a,b), qo(t) > 0 a.e. on [a,b], such that Ai( 0 and for all u G]0,<5] and a.e. t G [a, b], f(t,u)-qo(t)u
350
VIII. POSITIVE SOLUTIONS
(#+,) there exist p > 0, qx and foe £ ^fab),
such that Ai(goo) < 0,
rb
f
and for all u> p and a.e. t £ [a, b), f(t,U)-qoo{t)u
>/«,(«).
Then the boundary value problem (0.1) has a nontrivial solution u > 0. //, moreover, there exists h £ Lx{a,b) such that for all u £ [0,S] and a.e. t £ [a,b], \f(t,u)\ < h(t)u, (1.5) then u y 0. Notice that assumption (HQ) implies that (1.5) follows from the condition f(t,u) > -h(t)u. Proof. If for some B > 0, B
1. E X I S T E N C E O F ONE SOLUTION
351
Remark Notice that assumptions (HQ) and (H+,) alone do not imply the existence of a solution. Consider for example the problem u" + 4 u - 2 - s i n 2 < = 0, u(0) = 0, u(n) = 0. From Fredholm alternative, one proves this problem has no solution although f(t, u) = 4w — 2 — sin 2t satisfies (HQ ) and (H^) with qo = Qoo — 1 and /oo = 1. As in Corollary 1.2, we can write the following consequence of Theorem 1.5. Corollary 1.6. Let f : [a, b] x R —> R be an L1-Caratheodory function such that f(t,O) = 0 on[a,b). Assume that for some 70 and 700 £ £ 1 (a, b),
and —a
uniformly in t. Then the problem (0.1) /ias at least one nontrivial solution u>0. If moreover, there exists h € L1(a,6) such that for some 8 > 0, all u € [0,6] and a.e. t G [a,b], \f(t,u)\
Theorem 1.5 can be extended to the separated boundary value problem. Theorem 1.7. Let 01, &i € R, a2, b2 € K+ with a\ + a\ > 0 and 6? + &2 > 0. Assume f : [a, 6] x R —> M is an L1 -Caratheodory function that satisfies f(t,O) — 0 on [a,b], (HQ) and (H^), where ipi(-;q) is a positive eigenfunction corresponding to Xi(q), the first eigenvalue of the spectral problem (1.2). Then the boundary value problem (1.3) has at least one nontrivial solution u>0. If, moreover, there exists h € L1(a,b) such that for all u G [0,8] and a.e. t e [a,b], \f(t,u)\
352
VIII. POSITIVE SOLUTIONS
As for Theorem 1.5 we can use assumptions that refer to the spectral problem (1.4). Such assumptions give a result which is equivalent to Theorem 1.5 if (?oo > 0 as follows from Remark 1.1. Theorem 1.8. Let f : [a, 6] x R —» R be an L1-Caratheodory junction such that f(t,O) = 0 on [a,b], define ipi{-;r) to be a positive eigenfunction corresponding to Mi(r)> ^ e firs^ eigenvalue of the spectral problem (1.4), and assume (HQ) there exist 5 > 0 and r^ £ L1(a, 6), ro(t) > 0 a.e. on [a, b], such that f*i(ro) > 1; and for all u S ]0, $} and a.e. t € [a, 6], f(t,u)-ro(t)u
r^t)
> 0 a.e. on [a,b],
b
I
Ja
and for all u> p and a.e. t E [a, b],
Then the boundary value problem (0.1) has at least one nontrivial solution u>0. If, moreover, there exists h € Ll{a,b) such that for all u G [0,5] and a.e. t £ [a,b],
\f(t,u)\
J
at infinity
f(t,s)ds.
Theorem 1.9. Let f : [a, b] x R —> R be an L1-Caratheodory function that satisfies for some a+ € L1(a,b), limin^^+jx, ^ > a+(t) uniformly in t €L [a,b]. Assume f(t,O) = 0 on [a,b], (HQ) and there exist p > 0 and ^^ £ L1(a,b) such that for a.e. t £ [a,b] and all u > p —a where F(t,u) — / Jo
f(t,s)ds.
1. EXISTENCE OF ONE SOLUTION
353
Then the boundary value problem (0.1) has at least one nontrivial solution u>0. If, moreover, there exists h G Lx{a,b) such that for all u G [0,5] and o.e. t G [a,b], \f(t,u)\
Hence for any u G HQ (a, b), u > 0, we have
and we can find K large enough so that
f [| JO
Let r > X||v?i(-;7oo)||oo
an
d define
/(*,«) =0, ifu<0, = f(t,u), ifO
F(t,u)= and
f f(t,a)da
Jo
It is easy to see that -F(t,u)=0, > -e(t)u,
ifu<0, if u > r,
for some £ G L1(o,6). This implies $(u) is coercive on H(j(a,b). Hence, it has a minimum a which is such that a" + f(t,a)=0, a(a) = 0, a(b) = 0.
354
VIII. POSITIVE SOLUTIONS
As min$(it) < $(Kipi(-;'y00))_= $(l
The Semipositone Problem
In this section, we consider positive solutions of the parametric problem u" + sf(u) = 0,
,
i
u(a) = 0, u(b) = 0,
K
'
where /(0) can be negative. This problem is called the semipositone problem. Theorem 1.10. Let s € ] ( ^ ) 2 , ( ^ ) 2 [ - Assume / : R -> R is a continuous function such that
iimsupZM
(i,7)
and f(u)>u m > 0 and a — 1 H
—m 1
on [0,
Then the problem (1.6) /ifls a solution us y 0. Proo/. C/oim 1 - There exists a C2-lower solution a y 0 of (1.6). The function
is such that a >- 0, a(o) = a(6) = 0, a(t) < am and a"(t) + af{a(t)) > a"(t) + s(a(t) - m) = 0. Claim 2 - The problem (1.6) /ias a C2-upper solution /3 > a. We deduce from (1.7) that assumption (H^) is satisfied with qoo = | + e, /oo = ~1 and e > 0 small enough. The upper solution /? is obtained then as in Claim 2 of the proof of Theorem 1.1.
1. EXISTENCE OF ONE SOLUTION Conclusion-
355
The proof follows from Theorem II-1.5.
Exercise 1.2. Extend Theorem 1.10 to deal with a function / such that f(u) > pu — m on [0, am], for some p > 0. Remark With the assumptions of Theorem 1.10 and if /(0) < 0, the problem (1.6) has no positive solution for s > 0 small. More precisely, if f(u) < ku on R+ and u y 0 is a solution of (1.6), we compute 0= / (u" + sf(u))Sin(^t)dt b
b
2
< f («" + sku)sin(^-t)dt = (sk-(^) )
f
Ja
uSin(^
Ja
which is a contradiction if s > 0 is small enough. For large values of s, we can write the following result which gives large solutions. Theorem 1.11. Let f : K —> R be a continuous function that satisfies ^ - <0 and /or some r > 0 and rn > 0 f(u)>m
on [r, +oo[.
Then, for s large enough, (1.6) /ias a solution us y 0 suc/i i/iat lim lluglloo = +oo. s—»oo
Proof. Claim 1 - For eac/i s Zarye enough the problem (1.6) ftas a C2-lower solution ag y 0 swc/i i/iai ||as||oo —» oo as s —> oo. Let k > 0 be such that for any u € [0,r], /(tt) > —k. Define
The claim is then verified with as(t) = ?j(t — a)2, = r + 2djEsE.(t
if a
Claim 2 - For s as in Claim 1, the problem (1.6) has a C2-upper solution Ps >®s- We proceed as in Claim 2 of the proof of Theorem 1.10.
356
VIII. POSITIVE SOLUTIONS
Conclusion - The proof follows from Theorem II-1.5. 1.4
A Problem with Indefinite Weight
Consider now the problem u" + sq(t)g(u) = 0, u(a) = 0, u(b) = 0,
,
.
(1 0}
'
where g is nonnegative and q can change sign. Theorem 1.12. Assume g : K.+ —* R + is a continuous function such that g(0) > 0 and q G Ll{a, b) satisfies (i) / Ja
G(t,r)q(r)dr>Qfort£]a,b{,
where G(t,r) = fr-^ft-*) if r G [a,t[ and G{t,r) = (*-gK6-r) if r f«j / (6 - r)q(r) dr > 0, Ja
e
/ (r - a)q(r)dr > 0. Ja
Then for s > 0 small enough, (1.8) /ias a solution us >- 0 swc/i limlluslloo = 0 . Proo/. 5iep 1 - Construction of a W2'1-lower solution of (1.8) /or values of s. Let b
fb
«o(t) = / G(t,r)|qf(r)|dr
and tto(i) = /
Ja
We have
G(t,r)q(r)dr.
Ja
a. 0 — a
and Hence, we can fix e > 0 small enough so that a(t) = s(fl(0)«o(*) - euo(i)) >- 0. Next, we choose s > 0 small enough, i.e. a small enough, so that for
te]a,b[ a" + sq(t)g(a) = %(£)( 0. It follows that a is a W2il-lower solution of (1.8).
1. EXISTENCE OF ONE SOLUTION
357
Step 2 - Construction of a W2'1 -upper solution of (1.8). The function (3(t) — srv0(t) > 0, with r e]0,1[, is an upper solution for small values of s since in this case 0" + sq(t)g{/3) < sr(-\q(t)\ + s^q^gip))
< 0.
Conclusion - Notice at last that, for small values of s, a(t) < sg(0)uo(t) < sg(O)vo(t) < srvo(t) = f3(t). The theorem follows then from Theorem II-2.4.
D
Remark The hypothesis of Theorem 1.12 are almost optimal. If there exists a family of positive solutions us(t) of (1-8) that tends to zero with s, we have
I
S
Ja as s goes to zero, we obtain and going to the limit rb
1 Ja
It follows that UQ > 0 which implies -Un 0) =
and
r—- <7( r ) dr ^ °h— a Hence, the conditions (i) and (ii) with non-strict inequalities are necessary. Another case of interest concerns radial solutions of the Laplacian. Here we have to consider a mixed problem (tnuj + stnq(t)g(u) = 0, u'(0) = 0, u(l) = 0.
, , ^ ;
Exercise 1.3. Let g : K + —> R+ be a continuous function such that g(0) > 0, n e N and assume q £ L1(0,1) satisfies
f1 i
r
(i) / — ( / vnq(v) da) dr > 0 for t £ [0, If; Jt rn Jo (ii) / rnq{r) dr > 0. Jo
358
VIII. POSITIVE SOLUTIONS
Then prove that for s > 0 small enough, (1.9) has a solution us y 0 such that lirojujoo = 0 .
2
Some Multiplicity Results
2.1 The Sub-superlinear Case Multiplicity results can be obtained if the nonlinearity crosses the first eigenvalue twice. The sub-superlinear case concerns a nonlinearity with a "slope" which is greater than the first eigenvalue both at 0 and at +oo. Such an assumption alone does not imply the existence of solutions as can be seen from the example u(0)
= 0,
) =
0.
We have to add assumptions that force a "double crossing" of the first eigenvalue. Theorem 2.1. Let / : [a, 6] x R —» E satisfy L1-Caratheodory conditions, (HQ) and (#+,), where
Define
with 0 < A < S small enough so that a < 0. We know that ^i(S9o) >- 0, i.e. ay 0, and a" + f(t, a) > a" + qo(t)a = -Xi(qo)a > 0. Claim 2 - There exists k > /3 such that for all solution u > a of u" +
f(t,u)<0,
u(o) = 0, u(6) = 0,
{
'
we have u(t) < k on [a,b]. Suppose that, for any n, there exists a solution un > a of (2.1) such that max^] un(t) > n. Let us write un(t) —
2. SOME MULTIPLICITY RESULTS
359
Un(t)+untpi(t; goo), where fa ( ^ ( i ) ^ * ; qoo)-Qoo('t)un(t)'Pi{t; goo)) dt = 0. Rom Proposition A-4.5, we can write, for some K > 0, <-Kr / | ^ + b
b
< K[2 f
qooun
«
b b
u
<2K
- I
( ^
II {u'n + QooUn + \l{qoo)Un) +Vl(-) 9oo) dt. Ja
As we obtain < + goo(tK + /oo(t)
ifu n (i)>p, if «»(*) e [a(t),p[,
and ||«n||oo < 2X / [ftp(t) + |goo(t)|p+ |/ooWI]Vl(t;«oo)
> - M * ) - koo(«)|p- IM /oo(*) - Ai^ooKW > /oo(*),
if «n(t) e [a(t),/>], if «„(«) > A
we can use Fatou's Lemma and obtain the contradiction b
ff b
0 = lim /
-K(
> lim
b
> I
foo(t)
Ja
Conclusion - From Theorem III-2.12, problem (0.1) has at least two soluO tions ui ^. U2 > a y 0.
360
VIII. POSITIVE SOLUTIONS
For systems controlled by nonnegative coefficients go and goo, we can reformulate Theorem 2.1 using the spectral problem (1.4). Theorem 2.2. Let f : [a, b] x E —> E satisfy L1 -Carathiodory conditions, (HQ) and {&+,), where tpi(-;q) is a positive eigenfunction corresponding to Hi(q), the first eigenvalue of the spectral problem (1.4). Assume there exists a strict W 2 ' 1 -upper solution (5 >- 0 of (0.1). Then the problem (0.1) has at least two solutions u\ >- 0 and ui^u\. The next result gives conditions under which we have the required strict upper solution. Proposition 2.3. Let f : [a, b] x K —> R satisfy L1-Caratheodory conditions. Let r 6 L1(a,b), r(t) > 0 a.e. on [a,b], and assume that H\(r), the first eigenvalue of (1.4) satisfies Hi{r) > 1. Suppose that there exist M > rn > 0 such that for a.e. t € ]a, b[ and all u € }m, M[, f(t,u)
Then the problem (0.1) has a strict W2'1-upper solution /3 such that j3(t) £ [m, M] for every t € [a, b]. Proof. Let ipi(-;r) be a positive eigenfunction of (1.4) corresponding to /ii(r) such that HV'iGsOlloo < M — m. Then 0 = ipi(-;r) +m is a strict upper solution. Indeed, for a.e. t £ [a, 6] and all u E [m, P" + f(t,u) < V4'(-; r) + r(t)(u - m) < -r(t)[/n(r)i&i(-;r) - (u - m)] < 0 and /3(a) = (3(b) = m > 0. The proof follows now from Proposition III2.4. D The next result gives a simple alternative condition that implies the existence of a strict upper solution. Proposition 2.4. Let f : [a,bjxM. —> E satisfy L1 -Caratheodory conditions and assume there exist R > 0, D > 0 such that for a.e. t and all s £ [0, it], we have f{t,s)
2. SOME MULTIPLICITY RESULTS
361
Proof. The proof follows immediately from Proposition III-2.4. As above, the reader can prove as an exercice the following extension to the separate boundary value problem. Theorem 2.5. Leta\, bx € E, a2, b2 € R+ witha\+a\ > 0 andb\+b\ > 0. Assume f : [a, b] x R —> R satisfies L1-Caratheodory conditions, (HQ) and (H^), where ¥>i(-;
The Super-sublinear Case
We can work out a situation symmetric to the preceeding one, the supersublinear case, where the "slope" of / is smaller than the first eigenvalue both at 0 and at +oo. In this case also we have to impose assumptions that force a real double crossing of the eigenvalue if we want a multiplicity result. This remark is clear from the example u" + \u = 0, u(0) = 0, U(TT) = 0. Using the idea of Section 2.1, we shall assume existence of a strict lower solution. Such an assumption will not be sufficient to ensure multiple solutions as follows from the following example u" - 2\u -21 = - 2 , it(0) = 0, u{T) = 0. It can be seen from a phase plane analysis that there exists To < \f2ir so that for T 6]Tb,\/27r[ this problem has two positive solutions and for T > \f2ir only one. Nevertheless, there exists a strict lower solution. For example, if \/2V < T < 2\/27r, we can use a(t) — max{l — cos V%t, 1 — cos\/2(T-t)}. To obtain multiple positive solutions, we will assume further f(t, 0) = 0. Notice that (HQ) already implies f(t,O) < 0. Theorem 2.6. Let f : [a, b] x K. —* R be an L1-Caratheodory function such that f(t,O) = 0. Assume (HQ), (H^) hold with X\(q) and fi(-;q) the first eigenvalue and the corresponding positive eigenfunction of the spectral problem (1.1). Assume further there exists a strict W2'1 -lower solution a^0 of (0.1). Then the problem (0.1) has two solutions u\ and u2 such that 0 S u\ S U2
and u2 >- 0.
362
VIII. POSITIVE SOLUTIONS
Proof. Step 1 - Lower solutions. Notice that a\ = 0 and a2 = a are Ty2il-lower solutions, ct2 being strict, and a\ < a2. Step 2 - Upper solutions. As in Claim 1 of the proof of Theorem 1.5, we deduce from (HQ) the existence of a strict Ty2il-upper solution of (0.1) Pi e W2'l{a, b) such that /?i > a : = 0 and ft ^ a2. Also, as in Claim 2 of the proof of Theorem 1.1, we deduce from (H^) the existence of a W2jl-upper solution /32 G W2'l(a, 6) such that fa > «2
and fa > ftConclusion - Prom Theorem III-2.11, problem (0.1) has three solutions UQ, U\ and Ui such that 0 <
UQ
U\ % U2.
Notice that ito can be the zero solution so that we only proved the existence of two nontrivial solutions. Corollary 2.7. Let f : [a, b] x R —> R be an L1 -Caratheodory function such that ton M = 0 and lim ^ = 0 (2 . 2 ) U-+0+
U
u-*+oa
u
uniformly in t. Assume moreover there exist M* > M > m > 0 such that for a.e. t € [a,b] and all u £ [0,m] f(t,u)>0 and for a.e. t G [a, 6] and all u € [m, M*]
Then the problem (0.1) has at least two solutions uj X 0 and u2 ^ uiProof. The assumptions /(t,0) = 0, (H^) and ( F ^ ) follow from (2.2). Next, let r = ^- 2 J J^ m , a = 0 + r and 6 = b — r. We deduce then from Proposition III-2.3 that
is a strict W^-lower solution. The existence of two solutions 0 ^ u\ ^ u2 follows now from Theorem 2.6. Clearly, u = 0 is a solution. However, (2.2) implies uniqueness of solutions of the Cauchy problem with initial conditions u(to) — u'(t0) = 0 so that Ui(t) y 0.
2. SOME MULTIPLICITY RESULTS
363
As an exercise, we can prove the following extension of Theorem 2.6. Theorem 2.8. Let ai,bx£ R, a2, b2 € K + with a\ + a\ > 0 and b\ + b\ > 0. Assume f : [a, b] x M. —> R is on 1} -Caratheodory function such that f(t,O) = 0 and(HQ), (H^) hold with \\{q) andtpi(-;q) the first eigenvalue and the corresponding positive eigenfunction of the spectral problem (1.2). Assume further there exists a strict W2'1 -lower solution a^.0 of (1.3). Then the problem (1.3) has two solutions u\ and «2 such that 0 % u\ ^ U2
2.3
and
u2 >~ 0.
Applications of Variational Equations
Using the variational equation of (0.1) along a solution u we can obtain other types of multiplicity results. Theorem 2.9. Let f : [a, b] x E —> K be a continuous function and assume there exist a and /? > a which are C2-lower and upper solutions of (0.1) but not solutions of (0.1). Assume also there exists a solution UQ of (0.1), with a < UQ < /?, so that -7fe(t,u) exists and is continuous in a neighbourhood of {(t,uo(t)) \ t € [a, 6]}. Assume at last that the solution of
y" + |£(t,«o(«)) V = 0, y(a) = 0, y\a) = 1,
(2.3)
has a zero on }a,b[. Then the boundary value problem (0.1) has, aside from UQ, at least two additional solutions ui and u2 with a -< u\ -< UQ -< u% -< /3'. R e m a r k Define M to be such that | £ ( i , «o(*)) + M > 0. To require that the solution of (2.3) has a zero on ]a, b[ is equivalent to require that the first eigenvalue Ai of
&t,uo{t)) + M)u, ou
u{a) = 0, u(b) = 0,
satisfies Ai < 1. Let ip\ be the eigenfunction corresponding to Ai and y be the solution of (2.3). For any t € ]a, b], we have ft y(t)
= ( A i - l ) / ( ^ ( s > U o W ) + M)y( S M(a)ds. 7a OU (2.4)
364
VIII. P O S I T I V E SOLUTIONS
If y is positive on ]a, b{, evaluating (2.4) for t = b we obtain Ai > 1 and if y has a first zero t\ € ]a, b[, an evaluation of (2.4) for t = t\ leads to Ai < 1. The result follows. Proof of Theorem 2.9 : As a and /? are not solutions, arguing as in Proposition III-2.7, we can prove that a -< u0 -< /?. Next, using the above remark, we can find e > 0 and 5 > 0 such that §L(t,uo(t)) + M - e > 0, the first eigenvalue Ai(e) of
y" + My = AxtcXffti, «o(t)) + M »(o) = 0, y(6) - 0, satisfies Aj(e) < 1 and for all v with \v\ < S and all t € [a, 6], we have
Denote by ^ >- 0 the eigenfunction of (2.5) corresponding to Ai(e) with H^i lloo — 1- It is then easy to see, using Propositions III-2.3 and III-2.4, that a\ — UQ + | ^ i and /3i = UQ — |^>i are strict lower and upper solutions of (0.1). Decreasing S if necessary, we have a -< f3\ -< ai -< j3 and we conclude by Theorem III-2.11. D When we have a lower and an upper solution a < p, another way to "cross" the first eigenvalue is to ask that the minimal solution wm;n and the maximal solution ii ma x satisfy umm ~< u m a x . We deduce from Theorem VII-3.5 that problem (0.1) has a unique solution if the slope of / is always strictly smaller than Ai. On the other hand, if the slope of / is always strictly greater than Ai, we have a = 0 and hence izmin = a = 0 = umax. Theorem 2.10. Let f : [a,b] x R —+ R be continuous and a and (3 be lower and upper solutions of (0.1) such that a < /3. Assume the minimal and maximal solutions um-m and u m a x are such that a -< umin -< umax -< 0 and let §£(£, u) be continuous in a neighbourhood of {(t,um-m(t)) | t € [o, b}} U {(t,Umax(i)) I t £ [ai&]}- Then the boundary value problem (0.1) has at least three solutions it m i n -< u -< uma,x provided the boundary value problem y(a) = 0, y(b) = 0, has only the trivial solution for v = u m j n and u m a x . Proof. Let M be such that &(t,umin(t)) + M > 0, ^(t,umax(t)) By assumption, the eigenvalue problem
-y" + My = X(^(t,umin(t)) y(a) - 0,
W(6) = 0,
+ M)y,
+ M > 0.
3. PARAMETRIC PROBLEMS
365
is such that Ai(umin) 7^ 1. If Ai(umin) < 1, we prove arguing as in Theorem 2.9 that the problem (0.1) has a solution u such that a
umin.
This result contradicts the definition of um\n and hence Ai(Mmjn) > 1. In the same way we prove that Ai(u max ) > 1. As Ai(um;n) > 1 and Ai(u max ) > 1, we construct as in Theorem 2.9, ot\ and Pi strict lower and upper solutions of (0.1) such that a < Umin -< 01 -< at -< "Umax < P-
Existence of a third solution u follows now from Theorem III-2.11. Further Uniin ~< u since otherwise, there exists to G ]a, b[ so that u(t0) = umin(£o) and u'(to) = u'm[n(to). Prom uniqueness of solutions of the Cauchy problem, we come to the contradiction u — u m i n . Similarly, we have u -< umax. D Remark This result is also true for two ordered solutions ui -< u?, (not necessarily the maximum and minimum ones) but then we loose the localization U\ < U -< U2-
3
Parametric Problems
In this section, we study the parametric problem u" + s/(t,«) = 0, «(0) - 0, u(l) = 0,
,„
n {6 l)
-
in case / and s are nonnegative. With such an assumption, nontrivial solutions are positive, i.e. u >- 0. We noticed in Section II-4 that for such a Dirichlet problem it is natural to allow / to be singular, i.e. to satisfy ,4-Caratheodory conditions. This section is written with this setting. Our first result establishes existence of at least one solution for small values of the parameter s. Theorem 3.1. Let / : [0,1] x M —> IR+ satisfy A-Caratheodory conditions and assume f(t,O) ^ 0. Then there exists sj G]0, oo[U{oo} such that (a) for any s £]0, s\[, (3.1) has at least one positive solution; (b) for any s > s\, (3.1) has no solution.
366
VIII. POSITIVE SOLUTIONS
Proof. Claim 1 : for any s > 0 small enough, there is a positive solution us of (3.1). Let h E Abe such that for a.e.te[0,l], Vu€[O,l], Define j3\ to be the solution of
ffl + h(t) = 0, From Proposition II-4.1, we know that /3i(t)>0
and
||AI|oo
Next, we choose s > 0 small enough so that s||ft||yv < 1 and define for s e]0,s], (3 := s0i. We compute then
Hence, a = 0 and P > a are W^'Mower and upper solutions for (3.1) and the claim follows from Theorem II-4.2. Claim 2 : Define s\ — sup{s | (3.1) has a solution} e]0,oo[U{oo}. Then for any 5 e]0, si[ problem (3.1) has a positive solution. From Claim 1, it follows that si > 0. Let us fix s e]0,si[ and s* G [s,si] such that (3.1), with s — s*, has a solution u*, which has to be positive. Notice that a = 0 and j3 — u* are W^2>1-lower and upper solutions for (3.1), with s = s, and the claim follows from Theorem II-4.2. Remarks (a) If in the proof of Claim 1 we let s go to zero, we have that ||j9||oo a nd hence ||ws||oo goes to zero. This fact shows that there is, in the space (u, s), a "branch" of solutions that emanates from the origin. Further, if we assume / G C1([0,1] x R), we can apply the implicit function theorem to the equation $(s, u) = u" + sf(t,
u)=0
and prove that this "branch" is actually a curve parametrized by s. (b) If s\ < 82, it follows from the proof of Claim 2 that there exist corresponding solutions u\ and U2 which are ordered : ui < U2- It can also be proved that for any r G]0, SI[ there exists a minimal positive solution ur which is an increasing function of r, i.e. r\ < ri implies that for all tG[0,l],urx(t)<«r.(t).
3. PARAMETRIC PROBLEMS
367
Example 3.1. If we consider the example
«" + f =0, u(0) = 0, u(l) = 0, whose solution is u(t) = stln(l/t), it is clear that solutions do not have, in general, bounded derivatives. Example 3.2. Notice that the condition f(t,O) ^ 0 is essential. The theorem does not hold for the problem u" + s\u\ = 0, u(0) = 0, u(l) = 0. Example 3.1 has a positive solution for any s > 0. Hence without additional assumptions we might have s\ — +oo. The next result gives conditions so that si £ R + . Theorem 3.2. Let f : [0,1] x R —> R + satisfy A-Caratheodory conditions. Assume f(t,O) ^ 0 and there exists k G L°°(0,1) such that k>0, k{t) =£ 0 and for a.e, t£ [0,1], Vtt > 0, f(t, u) > k(t)u. Then there exists Si E Mj such that (a) for any s €]0, si[, (3.1) has at least one positive solution; (b) for any s > si, (3.1) has no solution. Proof. From Theorem 3.1, it is enough to prove that for s > 0 large enough, there is no solution of (3.1). Consider the eigenvalue problem u" + Xk(t)u = 0, u(0) = 0, u(T) = 0, and let Ai > 0 and (pi be its first eigenvalue and eigenfunction. Assume there exists a solution u of (3.1) with s > \\. We compute 0 = / dt-}r(u'(i)¥>i(t) - u(*M(t)) dt<<
7
r1 JJo
(Xi -- a)fc(t)«(*)vi(*) * < 0,
which is impossible. Remarks (a) Notice there is no point to assume that k G A, since we can replace k(t) by min{fc(i), 1}, which is in L°°(0,1). (b) It follows from the proof that si < Aj. Hence, in case k is constant, Ai = n2fk and there is no solution if s £]7r2/fc, oo[.
VIII. POSITIVE SOLUTIONS
368
Example 3.3. The type of problem we consider can have unique solutions as it is clear from the example u" + s(u + 1)+ = 0, u(0) = 0, u(l) = 0, whose solution "w cos(V/5/2) is unique and defined for s £ ]0,TT2[. u
Si — 7T2
Fig. 1 : Example 3.3 To better understand the role of the lower bound on f(t,u), write the following modification of Theorem 3.2.
we can
Proposition 3.3. Let / : [ 0 , l ] x i - > E + satisfy A-Caratheodory conditions. Assume f{t, 0 ) ^ 0 and there exists a continuous function g : R —* R+ such that (i) sup /
— = = = = = = = < +oo, where G{u) := / g{v)dv;
"0>0 JO
(ii)
u-»0+
\/G{Uo) U
— (j{U)
Jo
> 0;
(Hi) for a.e, t£ [0,1], Vu > 0, f(t,u) > g(u). Then there exists s% E Kj such that (a) for any s s]0,si[, (3.1) has at least one positive solution; (b) for any s > s\, (3.1) has no solution. Proof. Assumption (ii) is such that for a small enough and s large enough, a(t) = asin?rt is a Wr2|1-lower solution of u" + sg(u) = 0, u(0) = 0, «(1) = 0.
(3.2)
3. PARAMETRIC PROBLEMS
369
Any solution u of (3.1) is a W2>1-upper solution for (3.2). Hence if s\ = oo, we deduce from Theorem II-4.2 that the problem (3.2) has solutions for any large value of s. But this is impossible since assumption (i) and a standard time-map argument imply that for s large enough, (3.2) has no solution. We can obtain more precise results for linearly bounded nonlinearities. Proposition 3.4. Let / : [0,1] x R —» M.+ satisfy L°° -Caratheodory conditions. Assume f(t, 0 ) ^ 0 and for some a > 0, b > 0 and allu>0 f{t,u) 0 and compute for B large enough 0"(t)+sf(t,(3(t)) < -sbB cos Vsb(t - 2) + s[a + bB(cos Vsb{t - | ) - cos ^ ) ] = s[a - bB cos ^ p ] < 0. The proof follows now from Theorem II-2.4 with a(t) = 0 . The situation is somewhat different if / is asymptotically linear near the origin as follows from the following result. Proposition 3.5. Let f : [0,1] x R —> R + satisfy A-Caratheodory conditions. Assume that for some c £ [0,oo[, d € [0,oo[U{+oo}, for a.e. t € [0,1] and allu>0 u Then, for s € ] 0 , j [ U ] ^ - , o o [ , (3-1) has no positive solution. Proof. Assume u is a positive solution of (3.1) and let v(t) = sinTrt. If sc> 7r2, we have the contradiction 0= / ^(u'(t)v{t)-u(t)v'(t))dt= Jo dt
f (V2 -sf^'^)u(t)v(t)dt Jo V u{t) 1
< 0.
Similarly, if sd < TTTT2, , w wee obtain i m i l a r l y , if obtain
0 = // ^( ^({)(){){)) f U'{t)v(t)-U{t)v'{t))dt= f Jo
dt
v
Jo
(TT2-Sf^'^)u(t)v ^)()()
«(*) ' U
370
VIII. POSITIVE SOLUTIONS
Notice that the nonexistence of positive solutions for s €]^r-,oo[ is essentially the remark (b) after Theorem 3.2. Example 3.4. Consider first the problem u" + s(u + h(t)e~u)+ = 0, u(Q) = 0, u(l) = 0, where h(t) G]0,l] and u+ — max(u,0). We have here, with the notations of Propositions 3.4 and 3.5, a = 1, b = 1, c = 1, d = oo so that these propositions imply there exists a solution if s e]0,7r2[ and no solution if s G ]7r 2 , oo[, i.e. s\ = 7T2.
Example 3.5. Consider next the problem u" + smax(l,u- 1) =0, M(0) = 0, u(l) = 0. Here a = 1, b = 1, c = 1/2, d = oo so that we can deduce from the same Propositions 3.4 and 3.5 there is a solution if s €]O,TT2[ and there is no solution if s G]27r2,oo[, i.e. s\ £ [7r2,27r2]. Example 3.6. Another example is u" + s(u - 1)+ = 0, u(0) = 0, u(l) - 0, Here c = 0, d = 1 and there is no solution if s £ ]0, TT2[. In the next theorem, we obtain more solutions. To this end, some additional assumptions on the nonlinearity / are necessary. Theorem 3.6. Let f : [0,1] x R —> ]R+ satisfy A-Caratheodory conditions and assume that given r > 0 and r)> 1, there exists some e > 0 such that Vui, u € [0,r] and for a.e. t G [0,1], ui
,„ „, ^' '
Suppose f(t, 0) ^ 0 and further that for some k € L°°(0,1) such that k > 0, k(t) ^ 0, and a > 0, we have for a.e. t <= [0,1], Vw > 0,
f(t,u) >
k(t)(l+ua)u.
Then there exists s\ G R j such that (a) for any s G]0,SI[, (3.1) has at least two positive solutions which are ordered; (b) for s = s\, (3.1) has at least one positive solution; (c) for any s> s\, (3.1) has no solution.
3. PARAMETRIC PROBLEMS
371
Proof. Define s\ from Theorem 3.2, fix s G]0,s\[ and choose r\ &\s,s\[. Let u\ be a solution of (3.1) with s = r\ and e G ]0,1] be small enough so that for all u, v £ [0, ||«i||oo + 1] a n d for a.e. t G [0,1] v
r
+ e =» fit,u) < s
Arguing as in Proposition III-2.3, we prove that /? = u\ + e is a strict W2il-upper solution of (3.1) with s = s. Claim 1 : There exists RQ > 0 such that for all positive solutions u of u" + sk(t)u1+a = 0, u(0) = 0, «(1) = 0,
,
{
.
]
we have ||u||oo < -Ro- Assume by contradiction there exists a sequence {un)n of solutions of (3.4) such that An := ||un||oo > n. Notice that un(t) > j4n(^(t) with y?(t) = min{i, 1 — t}. Hence, for n large enough, we have the contradiction f1 0 = - / (u'n(t) + n2un(t)) sin = / (sk^u^it) 0
> \s(f
l
dt -K2un(t)) sin irtdt
k(t)ip1+a{t)sinntdt)A^-
Claim 2 : There exists R > 0 such that solutions u of u" + sf(t,u)<0, u(0) - 0, u(l) = 0, satisfy ||u||oo < R- Assume the claim is wrong. Hence, there exist solutions un of (3.5), such that ||un||oo —* +oo as n —> +oo. Let Hi > 0 and
on [*0,ti].
372
VIII. POSITIVE SOLUTIONS
Hence, a is a W2'1-lower solution for (3.4). Choose now un such that un > a and notice that un is a W2il-upper solution for (3.4). If ||a||oo > Ro> we get, from Theorem II-2.4, a solution of (3.4) with maximum larger than RQ, which contradicts Claim 2. Claim 3 : Problem (3.1), with s = s, has at least two solutions. This result follows from Theorem III-2.13 using /? = u\ + e as strict upper solution and a = 0 as a lower one. Claim 4 for s = sj, (3.1) has at least one solution. Consider a sequence (sn)n C [0, si] such that limn_^oo sn = Si and corresponding solutions un of (3.1) with s = sn. Erom Claim 2 and the ,4-Caratheodory conditions on /(i, it), there exists h £ A such that
K(t)l=*n[- / rf(r,un{r))dr+ f (1 -r)f{r,un(r))dr t°
l
*
< sn [ ( rh(r) dr+ f (1 - r)h(r) dr] € L\0,1). Therefore, from Arzela-Ascoli Theorem, there exists a subsequence (uni)i which converges in C([0,1]) to some function VQ. Prom the closedness of D the derivative, VQ 6 W|o'c (0,1) and satisfies (3.1), with s —
Fig. 2 : Theorem 3.6 Remarks (a) Notice that condition (3.3) is satisfied if f(t, u) is continuous and f(t,u) > 0 on [0,1] x [0,+oo[. However condition (3.3) is somewhat restrictive if f(t, u) takes zero value as follows from the example /(£, u) = sin2 2irt + u(l + u2) which does not satisfy (3.3). (b) If / e C1([0,1] x R), we can arrange the two solutions us and vs such that ||ti«||oo — 0 and ||ua|(oo -* oo as s -+ 0. Indeed, if some subsequence of (||us||oo)s is bounded as s —» 0, then for some sub-subsequence vs —* v in
3. PARAMETRIC PROBLEMS
373
C([0,1]) and v satisfies v"(t) — 0, v{0) = v(l) = 0. Hence v = 0, but in a neighbourhood of the origin, we deduce from the implicit function Theorem that there exists a single branch of solutions which has to be us. (c) The Theorem 3.6 establishes a structure of the solution set that can be illustrated with figure 2. We must notice that this set can be more complicate as follows from the study of the model example u" + sh{t)eu = 0, u(0) = 0, u(l) = 0, (see [125]). (d) An exact count of solutions can be given in case the nonlinearity is strictly convex. Such a proposition with the relevant additional assumptions is in [125]. Notice that a related result was already obtained for another problem in Theorem VI-1.4. Another multiplicity result can be obtained from the type of assumptions used in Section 2.2. Here, by opposition with the first part of this section, the multiplicity takes place for large values of s and the solutions are bounded with s —> oo. Theorem 3.7, Let f : [0,1] x M —> M. be an Ll-Caratheodory function such that for some r > 0 and k G L^O, 1;
0
(3.6)
Assume moreover (a) f(t,O) = 0 for a.e. t £ [0,1]; (b) lim sup ———- < 0 holds uniformly in t; u u_>o+ (c) f{t,r) < 0 for a.e. t e [0,1]; (d) there exists an open interval IQ C [0,1], rj > 0 and uo £ [0, r] such that for all t G IQ / f(t,s)ds>r). Jo Then, for s > 0 large enough, there exist at least two solutions u\ y 0 and U2 ^ u\ of (3.1). Proof. Step 1 - Lower and upper solutions of (3.1). Notice that a% — 0
and 02 = r are respectively W2'1-lower and upper solutions of (3.1).
374
VIII. POSITIVE SOLUTIONS
Step 2 - Construction of a second lower solution oti >- a\ = 0. Let e > 0 and 7(14) = max{0,min{u,r}}. Consider then the modified problem u" + sf(t,j(u))-eu = Q, t*(0) = 0, u(l) = 0, together with the associated functional $ : HQ(0, 1)-»R defined by
where F(t,u) — JJJ* f(t,j(s))ds. As there exists a function h G L^O, 1) so that |/(i,7(it))| < M*)> ^ e functional $(u) is coercive. Hence, it has a minimum 0:2 such that oi{(t) + s/(t,7(aa(*))) - ea2(t) = 0, Q 2 (0) = 0, o a (l) = 0. It is now standard to see that 0 < ot2(t) < r so that > 0, a2(0) = 0, a 2 (l) = 0. Next, using assumption (d), we can find a function
Chapter IX
Problems with Singular Forces 1
A Periodic Problem
In this chapter, we axe mainly interested in problems which are singular as u goes to zero and, as we are dealing with scalar equations, this assumption implies solutions are one-signed, say positive. To begin with we shall consider the periodic problem
u+g(u)u
+ f{t,u) = h{t),
.
{ J u(a) = u{b), u'{a) = u'(b). The first result we examine deals with a nonlinearity f(t, u) which represents an attractive force. Here a singularity such as f(t, u) = ^ can take place and it helps in rinding lower solutions. However such a singularity is not necessary, we only need the nonlinearity to be large enough near the origin (see condition (b) in the following theorem).
Theorem 1.1. Let g G C(M+), h G Ll{a,b) and let / : [a,6] x R^ -» R satisfy Caratheodory conditions and (a) for any 0 < r < s, there exists a function k G L1 (a, b) such that for a.e. t G [a, 6] and all u G [r,s], we have \f(t,u)\
376
IX. PROBLEMS WITH SINGULAR FORCES and ,6
,&
/ fo(t)dt< Ja
/ h{t)dt. Ja
Then the problem (1.1) has at least one positive solution. Proof. Condition (b) implies a(t) — a is a lower solution of (1.1). To construct an upper solution, we introduce the function = fa(t) - h(t) - ^
/ (/o(t) - h(t)) dt e L\a, b) J
and prove that for all c £ l , the problem u"(t) + g(\u(t) + c\)u'(t) + 4>{t) = 0, u(a)=u(b), u'(a) =«'(&),
K
}
has a solution 0O G W2'\a,b) such that $0 := 5 ^ J*f30(t)dt = 0. To this aim consider the homotopy u"(t) + \{g(\u(t) + c\)u'(t) + W)] = 0, u{a)=u(b), u'(o)=u'(6).
(
^
,
]
Solutions of this problem with mean value zero are fixed points of the operator TA : CHK&J) -+ C\[a,b}),u ~ Txu, where Pfab])
= {u € ^([0,6]) | f*u(s)ds = 0}, Txu = X
G(t, s)[g(\u(s) + c\)u'(s) + cf>{s)} ds
Ja and G(t, s) is the corresponding Green function. Let us prove that the fixed points of T\ are a priori bounded in C([a, b}) and then, arguing as in Proposition 1-4.3, in C1([a, 6]). Multiply (1.3) by u and integrate, we obtain by Proposition A-4.1
1. A PERIODIC PROBLEM
377
Now it is easy to observe that if we choose c = T^||^||Z,I + R and (3Q the corresponding solution of (1.2), 0(t)=Po(t)
+ c>R
is the desired upper solution. We conclude then by application of Theorem 1-6.9. Examples Consider the problem u" + g{u)u' + ^ = h, u{a)=u(b), u'(a)=u'(b), where g : K+ - t R i s a positive continuous function, k and h are positive constants. This problem has a positive solution and it can be proved from Theorem 1.1. Let us change now the restoring force into a repulsive one and study the problem u ^ = h, v u(a)=u(b), u'(a) = u'(b). ' ' The above theorem does not apply though there exists a positive solution u = (k/h)2. Prom a phase plane analysis, it is clear that this feature is due to the fact that this solution is not a saddle point so that we cannot use the method of lower and upper solutions as in the above theorem. As we already pointed out, Theorem 1.1 does not really use the fact that f(t,u) is singular for u = 0. This observation is clear from the further modification of the example = h, u" + g(u)u' + u(o) = u(6), tt'(a) = u'(b). If k > h, this problem has a positive solution and it can be found from the above theorem. Exercise 1.1. Consider the problem u" + f(u,u')=s+p(t), u(a) = u(b), u'(a) = u'{b). Assume / :]0, +oo[xE —> R is a continuous function, Lipschitz in the second variable and such that lim f(u,0) = +oo. 0+
378
IX. PROBLEMS WITH SINGULAR FORCES
Assume furthermore that there exist an r > 0, R > r and a continuous function g : R —* R such that for any u G [r, R) and D £ R , f(u,v)
v OJ
Theorem 1.2. Let g G C(M+), /i G Ll{a, b) and assume f : [a, b] x R j -> E satisfies Caratheodory conditions together with (a) for any 0 < r < s, there exists a function k € L1(a,b) such that for a.e. t G [a, b] and all u G [r, s], we have \f(t,u)\
/(*,/?)-M*)<0; ('cj (strong force) i/iere exisi p G ]0, f}\, I G L 1 (a, 6) and / G C(]0, p]) ««£/i /
f~(u)du
=+oo
and /""(«)= max{—/(u),0},
JO
swc/i i/iai for all u G]0,p] and a.e. t G [a,6]
f(t,u)
and
/(*,«)<£(«);
fd,) #iere esisi R> /3 and fo G Ll{a,b) such that, for a.e. t G [a,6] and aH /(*,«) >/o(t) and
,6
/ Ja
,b
h{t)dt>
I h(t)dt; Ja
1. A PERIODIC PROBLEM
379
(e) there exists a function 7 € Ll{a,b) such that ^(t) < (jr^) 2 o,.e. on [a, b\ with strict inequality on a subset of positive measure and u—*+oo
«*
uniformly in t € [a, b]. Then the problem (1.1) has at least one positive solution. Remarks As will be clear from the proof, assumptions (b) and (d) provide upper and lower solutions. The strong force condition (c) gives a lower a priori bound on solutions. It is important to notice that here we do not impose that f(t, u) goes to —oo as u goes to zero, but that the bound / has a negative part whose primitive is unbounded as u goes to zero. Notice that some control of the singularity such as the strong force assumption (c) is necessary. This fact is clear from [199] where it is proved that the problem u(a) = u(b), u'(a) = u'(b), has no solution for some negative h G C([a, &]), At last, assumption (e) forces the nonlinearity to be, for large values of u, "under" the asymptote of the first nontrivial Fucik curve for the periodic problem, which is also the first eigenvalue of the Dirichlet problem u" + Xu = 0,
u(a) — 0, u(b) = 0.
Such a condition is somewhat natural if we realize that large solutions of the periodic problem (1.5) look like solutions of a Dirichlet problem. Also, such a condition cannot be avoided. This fact is clear from [42] where it is proved that for some h £ C([0, 2TT]) the problem u(0) = U(2TT), U'(0) = u'(2ir), has no solution. Example 1.1. It is easy to see from Theorem 1.2 that the model example (1.5) has a solution if 7 e Ll{a, b) is such that
and h € I<1(o, 6) is lower bounded.
380
IX. PROBLEMS WITH SINGULAR FORCES
Proof of Theorem 1.2 : The modified problem - Let 5 G]0, p], define the
truncated function fs(t,u) = f(t,u), ifu>6, = f(t,S), iiu<S, and consider the modified problem u" + g(\u\)u' + fs(t,u) = h(t), u{a) = u(b), u'(a) = u'(b).
(
;
Claim 1 - there exists a solution u of (1.6) in S = {ue C([a,b}) | 3tut2
e [a, 6], u(*i) > /3(h), u(t2) < a(t2)}.
We first deduce from (b) that the constant function j3(t) = (3 > S is an upper solution for (1.6). Next, we obtain a lower solution from condition (d) and the argument used in the proof of Theorem 1.1 to define the upper solution. Notice at last that for some k E L1(a,&), all u G M and a.e. t € [a, b], we have fs(t,u) > min{ min f(t,u)J0{t)} > k(t). 5
We deduce now the claim from Theorem III-3.2. Claim 2 - There exists an L+ > 0 so that any solution u € <S of (1.6) with 0 < S < p satisfies maxt u(t) < L+. Let u € S be a solution of (1.6) such that max t u(t) > R = maxfii, Haloo}. From the definition of <S, there exists t% G [a,b] so that ufa) < afo) < R- Hence, extending u by periodicity if necessary, we can find a' < b' so that b' — a' < b — a, u(a') = u(b') = R, u(t) > R on ]a',b'[ and max ult) — max u(t). a'
a
Let now e > 0 be given by Proposition A-4.6 and choose M > R such that for all u > M and a.e. t G [a, b]
Define then
p(t,u) = t&p-, forw>M, = I&M, for
u
<M ,
and
fs(t,u)-p(t,u)u.
These functions are such that for some k G L1(o, 6)
, max <<
1. A PERIODIC PROBLEM
381
The function v = u — R is nonnegative on [a', b'] and solves the problem v" + g(v + R)v' + p(t, u(t)){v + R)+ qs(t, u(t)) = h{t), v(a') = 0, v(b') = 0. Hence, we compute pb po
/ Ja'
v'2(t)dt-
rb pa
Ja' <- f v(t)(v"(t) + g(v(t) + R)v'(t) + p(t,u(t))(v(t) + R)) V = f (qs(t,u(t))-h(t))v(t)dt Ja' and using Proposition A-4.6 we obtain II2
\\H'i(a',b')
—£ l 2
b'
(q5(t, u{t)) - h(t) + ( 7 (t) + f )5)«(t) dt, Ja'
i.e.
llUllHl(a',b') ^ ^"llul|oo, for some K > 0 independent of <5. It follows that for all t £[a', b'\ «(*) =
so that \\v\\oc < K(b - a) and
maxu(t) < R + K(b - a) =: L+.
Claim 3 - There eocists an L~ so that any solution u € S of (1.6) with 0 < 6 < p satisfies mint u(t) > L~. Let L+ be given from Claim 2. We deduce from (a) and (c) that there exists k € L1 (o, 6) so that for all ue]0,L+{ and a.e. t G [a,b] /(*,«)< k(t).
382
IX. PROBLEMS WITH SINGULAR FORCES
Using Claim 2, it follows that we can write for a.e. t € [a', 6'] fs(t,u(t))
(f5(t,u(t))~h(t))dt = 0 and
f u'2(t)dt = f Ja
(fs(t,u(t))-h(t))u(t)dt
Ja
It follows then that pb Ja
b
< 2 / \k{t) - h(t)\ \\u\\oo dt < 2{\\h\\Li + HfclUOIMloo, Ja
i.e.
KIU=<(2(||^IU1 + ||%0)1/2IHI^2.
Define now ti € [a, 6] to be such that u{t{) > @(t\) > — ||/3||oo- Extending u by periodicity, we can write for t S [ti,t\ + b — a] *
u{t) = u(ti)+ / u'(s)ds> -||/3|| oo -(2(6-a It is now easy to obtain L~ so that u(t) > L~ on [a,b}. Claim 4 ~ There exists an N > 0 so that any solution u € S of (1.6) with 0 < S < p satisfies WU'WQQ < N. As in Claim 3, we can find k G L1(a,6) so that for all solution u G 5 of (1.6) with 0 < 5 < p, and a.e. t G [a,b], fs(t,u(t))
u'(t) = /V(») - /«(«,«(«)) -ff(|u(s)|)u'(s))ds -2
max |G(u)|), L-
V
^M
1. A PERIODIC PROBLEM
383
where G(u) = / g(\s\) ds. Similarly, we have for t G [t — b + a,t\ Jo u'(t) < \\h\\Lx + \\k\\Li + 2
max
\G(u)\ =: N.
L~
Claim 5 - There exists a £ G ]0, p] so that any solution u€ S of (1.6) with 0 < S < p, is such that u(t) > £ on [a, b]. Define £ > 0 to be so that
f / - ( u ) du > (||ft||Li + p|| L i)JV + (1 + (ft - a) JJ
max Z-
Let u € 5 be a solution of (1.6). Define the set A = {t G [a, 6] | u'(<) > 0} and suppose by contradiction that there exist t\, ti G A so that u(ti) = £, w(i2) = P and w(t) < p o n [*i,t2]. Multiplying (1.6) by u' and integrating on B = [ti,t2] (~l A, we get
I g(\u(S)\)u'2(s)ds+ f B
JB
From the a priori bounds on ||«||oo and ||u'||oo) we obtain then g{\u{s)\)u'\s) ds<(b-
a)N2
max
B
and it follows that - / fs(s, u(s))u'(s) ds < N\\h\\Li + ^ + (b - a)N2
max
On the other hand, let B~ = {t e B \ f(u(t)) < 0} and B+ We compute then B+
=B\B~.
f5(S,u(s))ul(s)dS
and -
fs(s,u(s))u'(s)ds>-
[
JB-
f(u(s))u'(s)ds= f f-(u(s))u'(s)ds
f-(u{s))u'{s)ds=
JB
f
f-(u)du.
J£
We deduce then the contradiction f-(u)
du< N\\e\\Li+N\\h\\Li+2£-
+ (b - a)N2
max L~
|
384
IX. PROBLEMS WITH SINGULAR FORCES
Conclusion - From Claim 1, we know that problem (1.6) with 8 — £ has a solution. According to Claim 5 this solution solves (1.1). D The next result deals with u" + f(t,u) = h(t), u(a) = u(b), u'(a) = u'(b).
(1.7)
Here the nonlinearity allows resonance with the first eigenvalue of the Dirichlet problem as u goes to infinity. The non-resonance condition (c) implies a lower bound on solutions so that the strong force condition is no more necessary. Theorem 1.3. Let h £ L1(a, 6) and assume f : [a, b] x E j —* K satisfies Caratheodory conditions together with (a) for any 0 < r < s, there exists a function k £ L1(a, 6) such that for a.e. t £ [a, b] and all u € [r, s], we have \f{t,u)\
f(t,u)>fo(t) and /
fa(t)dt>
Ja
/ h(t)dt; Ja
(c) there exists S > 0 such that for allu> 0 and a.e. t G [a, b]
Then the problem (1.7) has at least one positive solution. Proof. The modified problem - Let /3o G]0, (^p) 2 £[ be fixed. For r > we define the truncated function = /(*,«), = f(t,r),
iij30
and consider the modified problem u" + fr(t,u) = h(t), u(a) = u(b), u'(a)=u'{b).
{
}
1. A PERIODIC PROBLEM
385
Notice that /?(£) = (5Q is an upper solution of (1.8). Next, we deduce a lower solution as in the proof of Theorem 1.2. At last, we notice that for some k G L1(a,6), all u G]0,+oo[ and a.e. t G [a, 6], fr(t,u)<
sup
f(t,u)
0o
We deduce now from Theorem III-3.1 the existence of a solution u G S of (1.8). Claim - There exists an M > 0 so that for any r large enough, solutions u G S of (1.8) are such that j3$ < u < M. Notice first that if u is a solution of (1.8) u{a) = u(b), u'(a) = u'(b). It follows now from the anti-maximum principle (Corollary A-6.3) that
u>p0-
To obtain an upper bound on the solutions, let k £ i 1 (a, b) be so that for all r > R, u > flo and a.e. t € [a, 6] fr(t, u) < Ht) - min{ min /(*, u), fQ(t)} < k(t). It follows as in Theorem 1.2 that
Next, as u G <S, there exists some i G [o, b] so that w(f) < ||a||oo. Hence, we can write ft
u(t) =u(t)+
/ u'(s) ds < Halloo + ||fc||ii (b - a) =: M. Jt
Conclusion - It follows now from the previous steps that problem (1.8), with r > M large enough, has a solution u with /30 0. It is then easy to see from Theorem 1.3 that the problem u(a) =«(& has at least one positive solution.
386
IX. PROBLEMS WITH SINGULAR FORCES
Notice also that for k > 0 and h(t) > 0 small enough, the existence of the solution of u(a) = u{b), u'{a) = u'(b), can be deduced from Theorem 1.3. The following result is an alternative to Theorem 1.3 which present an explicit non-resonance condition on h. Theorem 1.4. Let h G ^(ttjb) and assume / : [a,b] x RQ" —» R satisfies Caratheodory conditions together with (a) for any 0 < r < s, there exists a function k G L1 (a, b) such that for a.e. t £ [o, b] and all u G [r,s], we have \f(t,u)\
fo{t)dt> / h(t)dt; Ja
(d) there exists 8 > 0 so that for any t G [a, b] rt+b-a
f it
h(s) sin7r#5^ ^ s ^ &
Jt has at least one positive solution. Then the problem (1.7)
Proof. The modified problem - For r > /30 — ^ f 8, we consider the modified problem (1.8). The function @(t) — w(t) — B is an upper solution of (1.8) if B > 0 is large enough and w solves the problem 1
fb
w" = h(t) - /(*,/30) - r=- / (h(s) - f(s,(30))ds, Ja w(a) = w(b), w'(a) = w'(b). As in Theorem 1.2, we deduce from (c) the existence of a lower solution of the form a(t) = v(t) + A. At last, we see as in Theorem 1.3 that fr(t,u) is upper bounded by a Z, ^function. Existence of a solution in S of the modified problem (1.8) follows then from Theorem III-3.1.
1. A PERIODIC PROBLEM
387
Claim - Solutions u of the modified problem (1.8) are such that u > ^ r 5 . Let to be such that u(to) = minte[ai,] u(t). Multiplying (1.8) by sir and integrating on [to, to + b — a], we obtain 5<
h(t) sin ir^-dt
=
Jto
[u"(t) +fr(t,u(t))}sinnj=%
dt
Jto i-ta+b—a Jt0
Conclusion - An upper bound on the solutions of (1.8) in 5 is obtained as in the proof of Theorem 1.3. Hence, for r > 0 large enough, the solutions of the modified problem solve (1.7). Example 1.3. Consider once again Example 1.2, with h(t) = - 1 , if t G [a, a + e], h(t) = 1, if t G [a + e, 6]. We cannot use Theorem 1.3 to prove existence of positive solutions but Theorem 1.4 applies if e > 0 is small enough. We can also replace the assumption (e) in Theorem 1.2 by a condition on the friction term g(u). Consider the problem u" + g(u)u'+ f(u) = h(t), u(a) u(b), u'(a) = u'{b).
{
.
}
Theorem 1.5. Let g G C0R+), h G L2{a,b) and f G C(Rj). Assume moreover that (a) for some {3 > 0 and a.e. t G [a, b], we have - h(t) < 0; (b) (strong force) there exist a p G ]0, /3[ and I G E suc/t
I
f {u) du - +oo, o where f (u) = max{—f(u), 0} and for all u G ]0, p] there exists an R> ft such that ifu> R, rb
>bh f h(t)dt; Ja
(d) there exists an A > 0 such that \g(u)\ > A for every u > 0. Then the problem (1.9) has at least one positive solution.
388
IX. PROBLEMS WITH SINGULAR FORCES
Proof. For 5 €]0,/>], we define fs(u) = /(u), if u > 6, = f(6), iiu<8, and consider the modified problem o) = u(b), u'(a) = u'(b).
v
'
As in the proof of Theorem 1.2, we have the existence of a lower solution a and an upper solution 0 < a, and we deduce from Theorem III-3.2 the existence of a solution of (1.10) in <S. Claim - There exists an L > 0 so that any solution u € 5 of (1.10) with 0 < S < p is such that maxt u(t) < L+. Assume g(u) > A. The proof is similar if g(u) < —A. Multiplying (1.10) by v! and integrating, we obtain pb
pb
Ah'Wh < / g{\u\)u'2dt= j Ja Ja
hu'dt<\\h\\L2\\u'\\L2,
i.e. \\u'h*<j\\h\\L*. As u € S, there exists a t% so that u(t
u(t) = u(t2) + / u'(s)ds< Halloo + *7p||/i||i,a =: L+. Conclusion - We conclude the proof as in Theorem 1.2.
2 A Sublinear Dirichlet Problem In Section II-4, we noticed that a natural frame to study the Dirichlet problem u(a) = 0, u(b) = 0 , considers nonlinearities which are .A-Caratheodory and defines solutions to be in W2'A(a,b). Here, we are interested in problem (2.1) with a singularity at u — 0. In this case, / cannot satisfy ,4-Caratheodory conditions. However, solutions can still be in W2vA(0, TX) as is clear from the example n «(0) = 0, u(ir) = 0.
K
'
2. A SUBLINEAR DIRICHLET PROBLEM
389
Its solution reads u(t) = | y/t{ir -1) and is in W2'A(0, TT). Notice that f(t, u) = iv]_t\t is only ,4-Caratheodory on sets of the form [0, IT] X [e, +oo[, with e > 0 and not on [0, IT] X ]0, +OO[. AS the boundary conditions impose the solution to go into the singularity u = 0, there is no hope to obtain a positive solution from a simple application of the theory in Section II-4. The next theorem considers a sublinear problem where the nonlinearity is over the first eigenvalue near the origin (see (a)) and under it near infinity (see (b)). Our first result is written in the continuous case so that solutions are C2. Theorem 2.1. Let f : ]a, 6[XRQ" - » l k a continuous function and assume that for every compact set L C Mj, there is an h^ G AnCQa, b[) such that for all t G ]a, 6[ and u € L, \f{t,u)\
<j2u
Then the problem (2.1) has at least one solution
ueC([a,b],R+)nC2Qa,blK)Remark Assumption (a) is equivalent to assume there exist k > j ^ and a function a\ G C2([a,b],~B.+) such that (i) t G]a, b[ implies ai(t) > 0; (ii) f(t, u) > k2u, for all t G ]a, b[ and 0 < u < oi(t); (iii) o'/(i) > 0, for all t G [a, (2a + 6)/3] U [(a + 26)/3, b]. Proof. Step 1 - Construction of lower solutions. Consider &2 such that < &2 < min{A;, ^ ^ } and the function a2(t) = A2cosk2(t
-—),
390
IX. PROBLEMS WITH SINGULAR FORCES
where A2 is chosen small enough so that
f(t,u) > k2u, for all t e ] 2 ^ -^fc^
+ mzl a n d
u
Next, we choose a\ from the remark and let ax(t) = Aia\{t), where A\ £]0,1] is small enough so that for some points t\ g]a, t2 e]^,b[, we have (a-1) ai(t) > a2(t), for all t e [Mi] U [t2,b]; (b-1) a 2 (t) > ai(t), for all t € [ti,t2]. Notice that for any /* : ]o, 6[xMj —> E such that
2a h
^[
/*(*,«) > /(«,«), for all (t,u) e]a,6[xEj, we have (a-2) a'({t) + /,(t,ai(t)) > a'/(i) + fc2ai(*) > 0, for all t € [a,ix] U [ta,6]; (b-2) aa'(t) + f*(t,a2(t)) > -k%a2(t) + k2a2(t) > 0, for all t € [h,t2]. Step 2 - Approximation problems. We define for each n € N, n > 1, 7/n(t) = max{a and set fn(t,u) = We have that for each index n, /„ : ]o, 6[XMQ" —> R is continuous and fn(t,u) = f{t,u), for all (t,u) £ Kn x R+, where Hence, the sequence of functions {fn} converges to / uniformly on any set K x MQ", where K is an arbitrary compact subset of ]a, b[. Next we define fn(t,u) —min{fi(t,u),---
,fn(t,u)}.
Each of the functions /j is a continuous function defined on ]a, 6[XMQ"; moreover fi(t,u) > f2{t,u)
> fn{t,u) > fn+i(t,u)
> f(t,u)
2. A SUBLINEAR DIRICHLET PROBLEM
391
and the sequence (fn)n converges to / uniformly on the compact subsets of ]a,6[xMj since /„(*, u) = f(t, u), for all t G Kn and u G Mj. Define now a decreasing sequence (en) C R j such that lim en = 0, n—+oo
/(*, u) > fe2w, for all t € ifn and u G ]0, en], and consider the sequence of approximation problems u" + fn(t,u) = 0 , u(a) = e n , u(6) = en. p 3 - en is a lower solution of (2.3). It is clear that for any c G ]0, en] fn{t,c) > f{Vn(t),c) > k2C> 0. As the sequence (en) is decreasing, we also have fn(t,en)
= min fi(t,en) > k2en > 0, l
which proves the claim. Step 4 ~ Existence of a solution u\ of (2.3), with n = 1, such that max{ai(f), a2(t), e j < Notice first that m.ax{ai(t),oi2(t).,ei} is a C2-lower solution of (2.3) with n = 1. Next, we can find from assumption (b), M > max{ai(t),o>2(t), e\} and h G ADC(]a,b[), h>Q, such that for alH G]o,b[ and it G [M,oo[
Hence, we can write /i(i,u) =max{/(ij1 (t),u)J(t,u)} < j2u +h{t)+ h{rn(t)). Choose /? G C([a, 6]) n C2Qa, b[) such that f3(a) = M, P(b) = M, i.e.
= M+
f Ja
G(t,s)[h(s) + h(r)i(s)) + 72M]ds > M,
392
IX. PROBLEMS WITH SINGULAR FORCES
where (~ilx
nN l
sin j(s—a) sin 7(6—t) ~
sin7(t—a) sin7(b—s) 7 sin 7 ( 6 - 0 ) '
-c
.
? XS
- ^ t — lm
N o t i c e t h a t P i sw e l l - d e f i n e d a n d b o u n d e d s i n c e h— h o r j i + h G A
and
t
sin7(s — a) sin 7(6 — t) h(s) ds < / sin "/(s — a) h(s) ds + I sin 7(6 — s)h(s)ds < 00, Ja i(a+6)/2 I sin j(t — a) sin 7(6 — s) h(s) ds Jt sin 7(s — a) h(s)ds + I sin 7(6 — s)h(s)ds < 00. < I Ja J(a+b)/2 It is easy to see now that P" + fi(t, P) < P" + i2p + h(t) + h(rji(t)) = 0. Consider now the change of variables x = u — ej and apply Theorem II-4.2 to obtain a solution u\ of (2.3), with n = 1, that verifies max{oi(t),a a (t),ci} < «i(t) < P(t). Step 5 - The problem (2.3) has at least one solution un such that ma,x(a1(t),a2(t),en)
< un(t) < un-\(t).
Let us notice that wn_i is an upper solution of (2.3) since 0 = < _ i ( t ) + /n-l(t,«n-l(*)) > <_i(*) + fn{t,Un-X{t)) and «n-l(o) = «n-l(&) =
e
n - l > £n-
The claim follows by Theorem II-4.2. Step 6 - Existence of a solution of (2.1). Consider now the pointwise limit u(t) = lim un(t). n—>oo
It is clear that for any n > 1, i(£), a2(t)} < u(t) < un(t),
Vt G ]o, b[.
2. A SUBLINEAR DIRICHLET PROBLEM
393
Let now K c]a,6[ be a compact interval. There is an index n* = n*(K) such that K C Kn for all n> n* and therefore for these n> n*, 0 = < ( i ) + fn(t,un(t))
= < ( t ) + /(*,u n (i)),
Vt G K.
Hence the function un is a solution of the equation in (2.1) for all t G K and n>n*. Moreover sup{|/(t,u)| \t£K,
max{ai(i),a 2 (t)} < u < un.(t)} < +oo.
Then by Arzela-Ascoli Theorem it is standard to conclude that u is a solution of (2.1) on the interval K. Since K was arbitrary, we find that u G C2(\a,b[,R%) and for all t &]a,b[,
Since u(a) — u(b) = lim en — 0, n
it remains only to check the continuity of u at £ = o and t = b. Let e > 0 be given. Take n£ such that une(0) < e. By the continuity of unc (t) at £ = a, we can find a constant 8 = Se > 0 such that for any t £]a,a + 5[ 0
C(]a, b[). If we assume / > 0 on ]a, 6[ and
/ (t - a)(6 - *)(/(*) + \h(t)\) dt < +oo, existence of a solution follows from Theorem 2.1. Notice that in this example h can change sign, / and h can be singular or zero at t = a and t = b.
394
IX. PROBLEMS WITH SINGULAR FORCES
The next result improves condition (b) in Theorem 2.1. It replaces the eigenvalue condition to build the upper solution by condition (2.4) which is reminiscent of the variational approach and allows a time-mapping argument. Theorem 2.2. Let f :]a,b[xM.Q —» R. be a continuous function. Assume that for every compact set L C Kj, there is hi 6 AnCQa,b[) such that for all t € ]a, b[ and u £ L, \f(t,u)\
£M<(_) 11-++00
U2
x
b-a>
'
where $(u) = Jo" ip(s) ds, and for some M > 0, allt e ]a, b[ and all u > M, f(t,u)<tp{u)
+ h(t).
(2.5)
Then the problem (2.1) has at least one solution ti€C([a,6] ) R + )nC 2 (]a,6[,Kj). Proof. The argument is the same as in Theorem 2.1 except for Step 4. Step 4 - Existence of a solution u\ of (2.3), with n = 1, SMC/I
We already know that max{ai(t), aa(i), ei} can be used as a lower solution. Hence, we only have to find an upper solution 0, which we will write =w We can assume M > vaax.{ai{t),a2{t),ex). Hence, using (2.5), we can write for t s ]a, b[ and u> M fi(t,v) = max{/(Tji(t),«),/(*,«)} < V(«) T/ie function w(t). Define u; e C([a, 6],R+) nC2(]a,b[,Rj) to be such that to(a) = 0, w(b) = 0.
2. A SUBLINEAR DIRICHLET PROBLEM
395
We have for all t £ [a, b] ,6
w(t)= / G(t,s)[h(s) + h(rn(s))]ds>0, Ja where G(t, s) is the corresponding Green function and h can be assumed to be nonnegative. An auxiliary problem. Consider the function + CQU+,
ip(u) = ip(M)
for all u < M — \\w
OO!
= ip(u + ||ti?||oo) + eou + , for all u > M — ||icl|oo! with u+ = max{0, u}, and choose eo > 0 small enough so that liminf^P< hJL.)
u-»+oo u2
\°~aJ
,
(2.6)
where 'J'(u) = Jo" ^(s) ds. The function xj) : R —> WQ is continuous and verify lim ip(u) — +00. (2.7) We consider now the auxiliary problem (2-8) where L € K + . The function fto. From the condition (2.6), we can find a sequence (it,)* going to infinity and such that for some 5 e ]0, jr^[, lim (2*(«i) - 52u?) = - c o . Hence, we can find a sequence (Li)i going to infinity and such that - *(«)) < 52(L? - w2),
V« e [0, Li[.
Next, we deduce from the conservation of energy that any solution u of (2.8) is such that du
/
2
\/2 Ju Let us choose then Li large enough so that du
,. , f1 S
ds
396
IX. PROBLEMS WITH SINGULAR FORCES
We choose /?o to be the solution of (2.8) for L = Li. It follows then that 2 1 - 2
^
Hence, /?o is defined on [o, 6] and /30 > M. TTie upper solution. Now, we have /?(£) = fio(t) + w(t) > M and P"(t) + /i(t,0(t)) = h(t,/3(t)) - 1>(Po(t)) - h(t) - ftfai(t)) < fi(t,0{t)) ~
3
A Derivative Dependent Dirichlet Problem
Next, we consider a problem with derivative dependent nonlinearity u" + /(t,u,u') = 0) u(a) = 0, u(b) = 0.
m<, {
'
In the previous section, we imposed the slope ^ to be larger than the first eigenvalue Ai as u goes to zero and smaller than Ai as u goes to infinity. To extend these sublinear assumptions to (3.1) we consider the following auxiliary piecewise linear problem u" + B\u'\ + Cu = 0, u(0) = 1, u'(Q) = 0, with B, C > 0. Its solution is positive on ]
%
2'
[
wnere
if B2 - 4C < 0, if B2 - AC = 0. We can now write the following theorem where we consider Caratheodory conditions in order to diversify the framework.
3. A DERIVATIVE DEPENDENT PROBLEM
397
Theorem 3.1. Let f : }a, b[ xMj x l - > R satisfy Caratheodory conditions and assume that for each compact set L C Rg" x R there exists h^ G Ljoc(a,b) such that for a.e. t e]a,b[ and all (u,v) G L, \f(t,u,v)\
>Bi|u|-
(b) there exists £ 2 , C2 > 0 with b-a < T(B%, C2), M > 0 and h€ A such that for a.e. t G [a, b] and all (u,v) G [M, oo[xR, f(t,u, v) < B2\v\ + C2u + h(t); (c) for any compact set L C R j , there exist a non-decreasing function
00,
t
Then the problem (3.1) has at least one solution u G C([a, b}) n C^a,6[,R+) n W^(a, b). Remark 3.1. Assumption (a) is equivalent to assume there exist B%, C\ > 0, with b - a > T(B1, Ci), and a function 01 G Co([a,6],R+) such that (i) t G]O, b[ implies ai(t) > 0; (ii) f(t,u,v) > Bi\v\ + Ciu, for alii e]a,b{, 0
398
IX. PROBLEMS WITH SINGULAR FORCES
Step 1 - Construction of lower solutions. Decreasing B\ and C\ if necessary, we can assume that r(Si,Ci) > (b — a)/3. We define then a2{t) = )) where a2 is the solution of a)/2)=0, and A2 is chosen small enough so that f(t,u,v) >Bi\v\+Ciu for a.e. t G [&£& - r ^ c 0 ; kta+ r ( y ) [ ; all 0 < u < Q2(t) and all v G M. Next, we choose ai from Remark 3.1 and let ai(t) = Aiax(t), where A% G ]0,1] is small enough as in the corresponding Step 1 in Theorem 2.1. Step 2 - Approximation problems. We define the sequence of approximation problems
u"+ /„(«,«,«') = 0,
(32)
u(a) = c n , u(6) = e n as in the proof of Theorem 2.1. The functions /» are defined on ]a, 6[xRg" x R, satisfy fl(t,U,v) >f2{t,U,v)
>fn(t,U,v)
>fn+i(t,U,v)
>f(t,U,v)
and the sequence (fn)n converges to / uniformly on compact subsets of ]a,b[ XRQ" x R. The sequence (e n ) n C RQ" is such that lim en = 0, n—>oo
f(t,u,v) >Bi\v\+Ciu,
for all t e [a+ 55^,6 - 5^1], u G]0,en], v G K.
Step 3 - a3(t) := en is a lower solution of (3.2). Step 4 ~ Existence of a solution u\ of (3.2), with n = 1, swcft i/iat max(ai(t),a a (t),ei) < «i(t). Notice first that max(a1(t), a2{t), ei) is a P72il-lower solution of (3.2) with n = l. Next, we prove that for some h £ A f1{t,u,v)=mnx{f(rn(t),u,v),f(t,u,v)}
4. A MULTIPLICITY RESULT
399
and define /3 such that 0" + B2\0'\ + C2l3 + h(t) = O, with R > 0 large enough to enforce (3 > M. It turns out that /? is an upper solution of (3.2), with n = 1. At last, we verify that
and that for e such that y^-j < C, we can write
$- 1 (2| f (iP(s)
< (1 - e)*-^!^ I ^ ^WdsD + e ^ d l / * TP(s)ds\) + e<S>-\l\ f Hence, $~ 1 (2| /
(V(s) + V(»7i(*)))d*l) G L\a,b).
The claim follows now from Theorem II-4.6. Step 5 - The problem (3.2) has at least one solution un such that max.(a1(t),a2(t),en) < un{t) < u n _i(t). Step 6 - Existence of a solution of (3.1). The proof of these steps repeats the corresponding argument used in the proof of Theorem 2.1 and uses Proposition II-4.5.
4
A Multiplicity Result
In this section, we consider a multiplicity result for the problem «" + /(t,«) = 0, u(a) = 0, u(b) = 0.
,
{
, '
The idea is, as in Section VIII-2.1, to impose conditions so that the nonlinearity crosses twice the first eigenvalue. A first crossing is ensured by the
400
IX. PROBLEMS WITH SINGULAR FORCES
existence of lower and upper solutions. The second one follows from conditions at infinity. Our first result assumes such sub-superlinear conditions and allows some resonance in the second crossing. Theorem 4.1. Let a and (3 G C([a, b]) be respectively a W2>1 -lower solution and a strict W2'1 -upper solution of (4.1) with 0 < a(t) < f3(t) for t G ]a, b{, /3(a) > 0 and /3(b) > 0. Let E = {(t,u) G [a,b] x R | a(t) < u} and / : E -> R be an ACaratheodory function. Assume further there exist p > 0 and c, d G A, d > 0 on [a, b] such that (a) for a.e. t G [a, b] and all u> p, f{t,u)
>d(t)u-c{t);
(b) the first eigenvalue HI of
u(a) - 0, u{b) = 0, is such that /ii < 1; ,6
(c) /
lixnmi[f(t,u)-fiid(t)u}ipi(t)dt>0,
where ipi is a positive eigenfunction of (4.2) associated with \i\. Then the problem (4.1) has at least two solutions u% andu^ G W2'^(a, b) such that for all t G [a, b] a(t)
Ul{t)
Proof. Claim - There exists R > 0 such that every solution u> a of u(a) = Q,'u{b)~=0,
(43)
satisfies max u(t) < R. te[a,b]
Suppose that for any n, un is a solution of un(a) = 0, un(b)=0, with un > a and maxte[oj,] un(t) > n. The function un can be written as u^t)
=un(t)+un(t),
4. A MULTIPLICITY RESULT
401
where un{t) = antpi{t), an G R, and f*u'n(t)ip[(t) dt = 0. From Proposition A-4.4, we know \u^ + /iidtin|V>l dt ||«n||oo
< K[2 / « + mdun)+il>idtJa
*b
« + pidun)ipidt]. Ja
Using Lemma A-3.15, we have fb
fb
/ « + mduntyi dt = / (^' + ^ i K < l ( = 0. Ja
J a.
As / is an .A-Caratheodory function, there exists an h 6 A so that for a.e. t G [a,b] and all u G [a(t),p], \f(t,u)\ < h(t). Hence, < + md(t)un < -f(t, un) < md{t)(p - un) + h(t) + \c(t)\ + fiid(t)un
and we obtain, using Proposition A-3.14, rb
<2K
(h(t) + \c(t)\ Ja b <1KKX f (h(t) + \e(t)\ + ind{t)p){t -a)(b- t) dt. Ja As ||un||oo —* oo> we have an —» oo and un(t) —> oo for all t £]a,b[. Notice next that for a.e. t G [a, b] and all u G E j , (f{t,u) - fud^u^it)
> (-h(t) - |c(i)| -
Ml d(t)p)^!(t)
Using Fatou's lemma, we obtain the contradiction rb
0 = nlim / - « + fj,idun)tpi dt -*°°Ja > Jirn^ / (f(t,un) - mdun)ipi dt > / liminf (f(t, un) — u,idun)il)i dt > 0, Ja n-*°° where we have used (c).
G L\
402
IX. PROBLEMS WITH SINGULAR FORCES
Conclusion - The proof follows now from Theorem III-2.13. In very much the same way we can prove the following which is a corollary of the preceding theorem. Here we do not allow resonance at infinity. Theorem 4.2. Let a and /3 € C([a, b}) be respectively a W2'1 -lower solution and a strict W2'1-upper solution of (4.1) with 0 < a(t) < (3{t) for t e ]a, b[, P(a) > 0 and /3(6) > 0. Let E = {(t,u) € [a,b] x R | a{t) < u} and f : E -» M be an ACaratheodory function. Assume further there exist a p > 0 and c, d G A, d > 0 on [a, b] such that (a) for a.e. t e [a, b] and all u> p, f(t,u)>d(t)u-c(t); (b) the first eigenvalue nx of (4.2) is such that fxi < 1. Then the problem (4.1) has at least two solutions u\ andu% g W2'A(a, b) such that for all t € [o, b] a(t) < wi(i) < (3{t), Ul(t) < u2{t). We can obtain lower and upper solutions from the following propositions. Proposition 4.3. Let d £ A, d(t) > 0 a.e. on \a,b] and assume that the first eigenvalue of (4.2) satisfies fi\ > 1. Suppose that there exist some M > m > 0 such that for a.e. t S [a, b] and all u £ ]m, M[, f(t,u)
Then the problem (4.1) has a strict W2>1 -upper solution {3 such that @(t) £ [m, M] for every t G [a, b}. Proof. Let (3\ be a positive eigenfunction of (4.2) that corresponds to /^i and such that maxt/?i(t) < M — m. Then /3(t) f3i(t) + m is a strict W2ll-upper solution. In fact, for to €}a,b[, let SQ = min{
4. A MULTIPLICITY RESULT
403
Proposition 4.4. Let d € A, d(t) > 0 a.e. on [a,b] and assume that the first eigenvalue of (4.2) satisfies /ix < 1- Suppose that there exists I > 0 such that for a.e. t £ [a,b] and all u E]0, l[, f(t, u) > d(t) u. Then the problem (4.1) has a W2'l-lower solution a such that a(t) £ [0,1] for every t £ [a, b]. Proof. Let a be a positive eigenfunction of (4.2) that corresponds to n\ and such that maxt a(t) < I. Then a(t) is a PF2il-lower solution. Remark 4.1. If I < m, Propositions 4.3 and 4.4 give the existence of ordered lower and strict upper solutions of problem (4.1). Remark 4.2. Notice that the lower solution a given by Proposition 4.4 is concave. Hence, there exists a k > 0 such that a(t) >k(t-a)(b~t). This result can be used to verify the regularity assumption in Theorem 4.1. For example, if we consider / : ]a, b[ XEQ" —> R defined by /(*,«) =
(t-a)(b-t)ua
with 0 < a < 1, then, for any t € ]a, b[ and any u € [cx(t), R], we have
Example 4.1. Consider the problem u(0) = 0, u{n) = 0. The main feature of this example is that each of the terms 4r, — ^, it provides one of the basic assumptions in Theorem 4.1. The function a can be build from the singularity ^r near the origin. The upper solution follows from the term — ^ which dominates on a bounded interval away from the origin. The term u implies the appropriate behaviour at infinity, which is clear if we check the following, (a) The function
404
IX. PROBLEMS WITH SINGULAR FORCES
is a lower solution if ao > 0 is small enough, (b) The constant function is a strict upper solution. (c) Given an R > 0, if u £ \ot{t), R] we have
"0 -R
— t(TT-t)
and /in G ^t. (d) Take d(t) = tr^_t\ of
Easy computations show that the first eigenvalue
u" + nd(t)u ~ 0, u(0) = 0,
U(IT) = 0
is HI = 1 with eigenfunction ipi{t) = £(TT — t). Then the assumptions of Theorem 4.1 are satisfied since 1 2 1 \-u) —2u] = +oo. ; r f 3 (- s2s U u-»+oo i(7T — t) U2 U U u»+oo
iminf , w) — u,id(t)u] = lliminf
Chapter X
Singular Perturbations 1
Boundary Layers at One End Point
This section deals with problems such as eu" + u' - 1 =0, u(0) = 0, u(l) = 0, where e > 0 is a small parameter. This problem has the solution
which tends to uo(t) = t — 1 as e —> 0, uniformly on compact subsets of ]0,1]. The convergence is however nonuniform at the left endpoint t — 0. In such a case, the solution is said to have a boundary layer at the end point t = 0. A symmetric situation holds for the problem M(0)
= 0, u(l) = 0.
Here, the convergence is uniform on compact subsets of [0,1[ and the boundary layer takes place at the right endpoint t = 1, At last, let us recall that a richer example, which exhibits several types of boundary layers, e u" + uu' — u = 0, u(0) = A, «(1) = B, has been worked out in Section II-1.3. 405
406
X. SINGULAR PERTURBATIONS
The following theorems consider the boundary value problem eu" + / ( t , u , u ' ) = 0 , u(o) = 0, «(6) = 0.
h {
n )
They describe cases where a boundary layer takes place at the left end point t = a. Symmetric results with a boundary layer at the right endpoint t — b can easily be deduced. Theorem 1.1. Let u0 € C2([a,b]) and let F= {(t,u) \a
ri(t)
where r\ and r% are two continuous functions on [a, b] such that for all t G [a, b]
and uo(a) +r1(a) < 0 < uQ(a) +r2(a). Consider a function / G CX(F x R) such that V(t,u,v)eFxM,
|/(t,u,v)|
(1.2)
+
where (f E C(R ) satisfies
f Assume
sds
oo.
(1.3)
Jo
(a) wo is a solution of the reduced problem f(t,uo(t),u'o(t))
= 0,
(b) there exists a \i > 0 such that for all (t, u, v) £ F x R,
Then for e small enough, there exists a solution us of (1.1) such that
Condition (b) can be interpreted as a stability assumption on the boundary layer equation Oil)
— + f(t,u,v) = O, where (t, u) € F are fixed parameters. This assumption is essential to have the boundary layer at the initial point t — a. If we reverse the inequalities, i.e. g£(£, u, v) < —fi < 0, the boundary layer takes place at the terminal point t = b.
1. BOUNDARY LAYERS AT ONE END POINT
407
Next, we show that an asymptotic result on the derivative can be obtained provided we reinforce the assumption on |^(t,w, v). Theorem 1.2. Under the assumptions of Theorem 1.1 and if |£(t, u, v) is bounded on F x l , we have
Proof of Theorem 1.1: Part A - The reduced -problem. Claim 1 - There exists a function h £ ^(F) such that (i) for any (t, u, v) G F x R, /(£, u,v) = 0 if and only if v = h(t, u);
(ii) §(t,u) = -§£(*,u,h(t,u)) For any (t, u) G F, assumption (b) implies f(t, u,.) is a strictly increasing function such that lim f(t,u,v)
= +oo and
lim f(t,u,v) = —oo.
It follows that we can define h(t, u) to be the unique v such that f(t, u, v) = 0. From implicit derivation, h £ CX(F) and (ii) is satisfied. Claim 2 - For some Q > 0, eo > 0 small enough and any z with \s\ < £o, the solution vc,(t,e) of the Cauchy problem u' = h{t,u),
u(6)=e
(1.4)
is defined on [a, b] and we have
\vo{t,e) - uo(t)\ < \e\ exp J(6 -t) = O(e), \v'0(t,e) - u'0(t)\ < Uf-exp jf (6 -t) = 0{e). Notice first that as u0 E ^([a^]) and / G ^{F x R), there exist R > 0 and Q > 0 such that for all (t, u) € F and all i> with |v — Uo(£)| < -R>
Hence, in a neighbourhood of uo(t), h(t,u) satisfies a Lipschitz condition with respect to u with constant Ql\x. The proof follows then from an application of Gronwall Lemma. Part B - Construction of C2-lower and upper solutions of (1.1) in case uo(a) > 0.
408
X. SINGULAR PERTURBATIONS
Choose K\ and K2 such that -n(a)
> Kx > uo(a) and K2 > % IKI
and define Ai and A2 to be the greater and the smaller roots of eX2 — Q = 0. We have
Notice that for e > 0 small enough the functions a(t) = vo(t, -e) - Kx exp(-Ai(t - a)) - e i ^ e x p ^ ^ - *)) - 1], (3(t) = vo(t, e) + eK2[exp(X2(b -1)) - 1], have their graph in F. Claim 1 - 13 > a. It is clear that (3{t) - a(t) > vo(t,e) - vo(t, -e). Also, as vo(b, e) > va(b, —e) we deduce from the uniqueness of the solutions of the Cauchy problem (1.4) that for all t £ [a,b], vo(t,e) > vo(t, —e), i.e. 13 > a. Claim 2 - For e > 0 small enough, ea"(t) + f(t,a(t),a'{t))>0
and ep"(t) + f{t,p{t),j3'{t)) < 0.
Let us write vo(t) for vo{t, —e). Observe then that for e small enough, {(*, vo(t) + e ( a ( « ) - vo(t)))
\te[a,
b], £ € [0,1]} C F
and
f(t,a, a') = f(t, v0,v'o) + §£(*, v0 + £i(a - vo),v'Q)(a - v0) for some 0 < & = &(t) < 1, i = 1,2. Hence, we have ea" + f{t, a, a') = e < - eKx\\exp(-Ax(i
- a)) - e2K2\\ exp(A2(6 - 1 ) )
+f(t,vo,v'o) -§{1{t,vQ+^(a-v0),v'Q)[K1exp(-X1(t~a))+eK2(exp(X2(b-t))-l)} +^(t,a,v'0+&(a'-v'0))[KlX1eM-X1(t-a))+eK2X2exp{X2(b-t))}
1. BOUNDARY LAYERS AT ONE END POINT
409
and we obtain e a"(t) + f(t, a(t), a'(t)) > e t#(i) + f{t, vo(t), v'0(t)) + eK2Q +Kr exp(-Ai(* - a))[-eA? + XifJ, - Q] +eK2 exp(A2(6 - t))[-e\% + A2/x - Q\By definition of Aj and as f(t,vo(t),v'o(t)) = 0, we can write ea"(t) + f(t, a(t), a'(t)) > e[«J(t) + K2Q] = e « ( t ) + K2Q + o(l)] > 0. In the same way we prove
Claim 3 - For e small enough, a(a) < 0 < /?(a), a(6) < 0 < /3(6). We compute o(a) = «b(a, - e ) - if i - £X2 < -uo(a) - Jfi < 0, and obtain similarly (3{a) > 0 and 0(b) > 0. Part C - 27ie case UQ(O) < 0. The same arguments hold with a(t) - vo(t, -e) - eK2[exp{X2(b -1)) - 1], /3(i) = vo(t,e) + Kx exp(-Ai(t - a)) + eif2[exp(A2(6 - t)) - 1], and appropriate K\ and i
+ £ize, u'0)ze - — (t, uE,u'o + &z'e)z's,
where £i = ^i(^), ^2 = &(t) G [0,1]. Hence, we can find M > 0 large enough so that e\z'J\ < mp\§L(t,u,u>0(t))\\ze\ F
FxK
(1.5)
410
X. SINGULAR PERTURBATIONS
Let No — 2r/(b — a), where ||ze||oo < >*. If we define NE to be such that ~"! lNo
s
,
irM
——as = l + S
£
,
we deduce from (1.5) that |4MI ^ -We- As the function ds is convex, we compute 2rM
o) > 1>'{N0)(Ne - N0) > j ^ i N , - No)
£
and it follows that I4(*)l < Ne < No + f (1 + N0)(b - o) = 0(1). Conclusion - We know that \ze(t)\ <S(t,£);=L(£
+ e-
for some L > 0 and we have also that //
df
,
9f
This result implies £Z£ ^ C^j{ty£) ~~ \IZ£ -\-£l\^
it Ze ^ 0,
EZ'I > -QS(t, e) - nz'e - EK,
if z'E < 0,
where Q is defined as in the proof of Theorem 1.1 and K > 0. It follows that
L 'I which implies using Claim 1
(
I4WI e < I4(a
J
1. BOUNDARY LAYERS AT ONE END P O I N T
411
More generally, we consider the boundary value problem eu" + f(t,u,u') = Q, a\u{a) — a2u'(a) — A, bxu{b) + b2u'(b) = B.
(1-6)
As above the assumption (b) will ensure that the boundary layer takes place at the initial point t — a. Further, we assume that at this point the boundary condition is not of Dirichlet type, i.e. a2 j= 0. This assumption implies that the boundary layer that appears in the function u£ is not of first order, i.e. Theorem 1.3. Let A, B G R, 01, h G R, a2 > 0, b2 > 0, bj + bj > 0. Let UQ G C2([a,&]), F be a closed neighbourhood of the graph of UQ and f G Cl{F x R). Assume (a) UQ is a solution of the reduced problem f(t, « 0 (t), u'0(t)) = 0,
blUo(b) + b2u'0(b) = B;
(b) there exists /J, > 0 such that for all (t, u, v) G F x M,
(c) A = h^(b,uo(b),u'o(b))
- b2§L(b,u0(b),u'Q(b))
> 0.
Then for e small enough, there exists 0 solution ue of (1.6) such that
Proof. Part A - The reduced problem. As in the proof of Theorem 1.1, we define the function h G CX(F) such that f(t, u, v) = 0 if and only if v = h(t, u). Next for 770 > 0 small enough and any r\ with |T/| < r]o, we can prove that the solution vo(t,r]) of the Cauchy problem u' = h(t,u), is denned on [a, b] and verifies
u{b) = uo(b) + rj
(1.7)
412
X. SINGULAR PERTURBATIONS
for some Q > 0. At last, we compute bivo(b,rj) + b2v'0{b,r,) -B = h{uo(b) + r?) + b2h(b,u0(b) + rj) - B = 6i»7 + &2|g(fc, «o(6) + €»7)»?. where £ e [0,1]. As h G C 1 ^ ) . we have
§(6,uo(&))]!? + ofa) Part B - Construction of C2-lower and upper solutions of (1.6). Choose K\, K2 and K3 > 0 such that K\ > -^\aiuo(a)
- a2u'0(a) - A\,
K2 > ^ and let M = ^ + 0(1) and A2 = — + O(e) be as in the proof of Theorem 1.1. Define now a{t) = vo(t, -eK3) - eKx exp(-Ai(t - a)) - e/£r2[exp(A2(6 - t)) - 1], 0(t) - vo(t, eK3) + EKX exp(-Ai(i - a)) + eiC2[exp(A2(6 - t)) - 1]. If e is small enough, these functions have their graph in F. First, we prove as previously that /3 > a. Next, repeating the argument in the proof of Theorem 1.1 we have for small values of e > 0 ea"(t) + f[t,a(t),a'(t))
> e[v%(t, -EK3) + K2Q) > 0
and £/?"(*) + /(i,/3(t),/3'(t))<0. At last we compute for e > 0 small enough aia(a) — ^ ( / ( a ) — A = a\Uo(a) — a2u'0(a) — A — K\a2yi, + O(e) < 0, ha(b) + b2a'(b)-B = blv0(b, -eK3) + b2v'0(b, -eK3) - B -eKx{bi - 62Ai)e-Al(6-a) + eK2b2X2 %(,o(b),u>Q(b))}-1 + eK2b2£+o(e) <0 and in a similar way we obtain ai(3(a) - a2p'(a) -A>0, b2/3'(b) - B > 0.
2. BOUNDARY LAYERS AT BOTH END POINTS
413
Condition (b) implies a one-sided Nagumo condition which together with the boundary conditions implies a bound on u'(a), so that existence of a solution uE € [a, 0] follows from Theorem II-3.2. Part C - Estimate on the derivative u'e. First we deduce from the boundary conditions that u'e(a)-u'0(a)=O(l). Second, we know that |«E(a) — wo(o)| < S(t,e) = Le for some L > 0, and we deduce as in the proof of Theorem 1.2
<(t)-«(,(*) = O(e + exp-f(t-a)). Example 1.1. Consider the problem eu" + | — u + | ( t + \){u' + u' ) = 0, u'(0) =: 0, u{\) = 2. The hypothesis of Theorem 1.3 are satisfied with uo(t) — t + 1, ^(t,u,v) | and A = 3.
2
>
Boundary Layers at Both End Points
Consider the problem u(o) = A, u(6) = B,
( 2-1 )
where e is a small positive parameter. The following theorem uses a stability assumption <-m<0 which produces boundary layers at both end points. Theorem 2.1. Let u0 € C2([a, b}) and let E = {(t,u) \a
n(t)
where r\ and r2 are two continuous functions on [a, b] such that for all t € [a, b] r1(t)<0
414
X. SINGULAR PERTURBATIONS
Consider a function f e ^(E) and assume (a) UQ is a solution of the reduced problem f(t,uo(t)) = 0; (b) there exists m > 0 such that for all (t, u) G E, < -m<0. Then for e small enough, there exists a solution u£ of (2.1) such that for any t 6 [a, b]
W{t) - uo(t)\ < |«0(o) - A\ e x p ( - y f (t - a)) +\uo(b) - B\ exp( Proo/. Consider for example the case where ua(a) < A, uo(b) > B, the other cases are treated in a similar way. It is easy to see that a(t) = uo(t) - (uo(b) - B) e x p ( - ^ f (6 -1)) is a C2-lower solution of (2.1),
= uo(t) + (A- uo(a)) e x p ( - y f (t - a)) is a C2-upper solution and a(t) < f3{t). Whence, we can apply Theorem II-1.5. D Remark Theorem 2.1 implies that ue goes to u0 with e > 0, uniformly on compact subsets of ]a, b[. Boundary layers at both end points can occur in systems with derivative dependence. Consider for example the problem eu" - tu'\u'\ - u = 0 u ( - l ) = l, u(l) = 2 . Its solution satisfies lim uE(t) = 0 if t ^ , with uniform convergence on compact subsets of ] — 1,1[. The limiting function ua(t) = 0 is the continuous solution of the reduced equation tu'\u'\+u~0. Exercise 2.1. Prove the above statements using lower and upper solutions.
3. ANGULAR SOLUTIONS
3
415
Angular Solutions
Here we consider a problem su" + f(t,u,u')=0, u(a) = A, u(b) - B,
.
,
{6A)
and work out conditions so that it has a solution that converges uniformly to a function UQ € C([a, b\) which is only piecewise C2 but satisfies both boundary conditions. These are the so-called angular solutions. In this case a boundary layer in the derivative u'£ takes place at points to where the derivative u'o is discontinuous. Notice that boundary layers also appear in the function w0 but only in terms of higher order with respect to e. A first result of this type is the following. Theorem 3.1. Let u0 e C([a, b}) be such that for some t0 € }a, b[, u0eC2{[a,t0}),
u0eC2([t0,b})
and Diuo(to) < DruQ(t0).
Let also E be a neighbourhood of {(t, u,v)\t£
[a, b] \ {t0}, u = uo(t), v = u'0(t) or t = t0, u = uo(tQ), Diuo(to)
Consider a function f £ Cl(E) and assume (a) UQ is a solution of the reduced problem f(t,uQ(t),u'0{t)) =0, ift€ [a,b] \ {t0}, uo(a) = A, uo(b) = B; W %(t,uQ(t),DiuQ(t))
and
4<5O < u'b(t0) - u'a(t0).
416
X. SINGULAR PERTURBATIONS
Next, we consider the sets TZa = {(t,u,v)
\a
+ 50,\u-
ua(t)\
Tlb = {(t, u,v)\to-6o
< So, \v - u'a{t)\ < SQ},
ub(t)\ < So, \v - u'b{t)\ < 50},
Ut0 = {(t,u,y)\to-6o
+ 6o, < So, u'a{t) + S0
u'b(t) - So}.
If So > 0 is small enough, it is clear from assumptions (b) and (c) that there exist constants fi > 0 and Q > 0 such that u,v)|
ii{t,u,v)e1laUilb;
{t,u,v)ena\ f(t,u,v)<-n,
%{t,u,v)>ix, if (t,u,v)e Kb; if {t,u,v)£llta.
We transform now / outside of 1Za U%bU7£t0 s ° that the modified function / satisfies a Nagumo condition and consider the modified problem eu" + f(t,u,u') = 0, u(a) = A, u{b) = B.
, ->
{4 l
Part 2 - Construction of C2-lower and upper solutions of (3.2). Let Ai be the root of the equation eA2 — /i\ + Q = 0 such that Ai = (/i - \ / y - 4 e Q ) / 2 £ = Ql\i + 0{e) > 0. and choose a K > 0 large enough so that QK > (exp(Ai(i0 - a)) - 1) max \u'0'(t)\, t€[a,to\
QK > (exp(Ai(6 -1 0 )) - 1) max \u'0'(t)\. t€[ta,6]
Let also A2 = (u'b(to) - _u'a(tQ) - 2SQ) which is positive for So small enough and such that for e > 0 small enough
Define now e
( )
_l
a(t) = uo(t) - eK e A i ( t o _ o ) _ x , )
i
,
if
a
3. ANGULAR SOLUTIONS
417
and
+eVV2(*-to),
0{t) = uo(t) + eK^^f^ = uo(t) + eK \
\ + ei/2eA2(to-t)i
if a < t < t0, if t < t < b
We can verify then that, for e > 0 small enough, these functions are respectively C2-lower and upper solutions of (3.2), which is a tedious but straightforward calculation. For example, we compute for t < t0, ea" + f{t,a,a') - Ea" + f(t,uo,u'o) + &(t,x,y)(a
- u0) + &(t,x,y)(a'
- u'o)
> ea" + Q(a - u0) - /J,{a' - u'o)
where (x,y) is on the segment joining (a,a') and (uo,u'o). Similarly, we have to check
sP" + f(t,0,P')>O separately for (t,/3(t),/3'(t)) in each of the regions TZa, Tib and TZt0- We also have to notice that D~(3(t0) > D+f3(to). At last, we deduce from Theorem II-1.3 that the modified problem of (3.2) has a solution u£ with a