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2p}, (2.13c) 2; (c) uIan <- 0. Then
x E G(p),
as is illustrated in Figure 2.2. This choice of cp is legitimate because kp(x - y) = 0 when Iy - xj >- p, so that supp kp(x -.) = 0, and certainly kp(x -.) is infinitely differentiable and non-negative in Q. (ii) Now let up(x) =
kp(x - y) u(y) dy, x E G(p),
(2.14)
J
where, equally well, the integral could be written as one over P'(x, p). Then up E C00(G(p)) by Exercise 1.23; the present boundary 30 plays no
part when x E G(p). The definition of distributional subsolution states that 0
<
Ala(kp(x - .), u; 0)
f
[k(x
N
a jj
-=1
N
- E bi [kp(x_Y)] N
a,j
E
f
n
i,1=1
u(y) + ckp(x - y) u(y)
dy
Yj
i=1
a2
[xxjk'(x - Y)] u(y) dy
N a
+ i
1
b1 f [ axe kp(x - y)] u(y) dy +c J kp(x - y) u(y) dy
= L1up(x) Thus up is a C2-subsolution relative to L1 and G(p); by the weak maximum principle for Lo and for L, maxaG(p) up
up(s)
maxac(p)(up)+
if c = 0, if c < 0.
2 Some Maximum Principles for Elliptic Equations
48
ag (x,P)
4
A2 Fig. 2.2.
Consequently, if
limp-o up(s) = u(),
(2.15)
lim supp-o { maxao(p) up } < maxan u,
(2.16a)
maxan u+,
(2.16b)
lim supp-0 { maxaG(p)(uP)+ }
then (2.12) will follow [because for c = 0 we shall be able to contradict u +,u, p > 0, by choosing p sufficiently small; similarly for c < 0].
2.2 The weak maximum principle
49
(iii) Consider in passing the statements limp-o { maxaG(p) up } = maxan u,
(2.17a)
limp...,o { maxaG(P)(uP)+ } = maxan u+.
(2.17b)
These may seem simpler than (2.16a,b) and, with (2.15), they certainly imply (2.12). Moreover, (2.17a,b) are true. However, their proof is longer, and slightly harder, than that of (2.16a,b) because a lower bound for maxaG(p) up emerges less easily than the upper bound that we shall find. (iv) Since u E C(S2) and S2 is compact, u is uniformly continuous: for every e > 0 there is a number 6, > 0 such that
y,z e f and ly - z1 < 6,
Iu(y) - u(z)I < E;
(2.18)
we reduce SE, if necessary, in order that 6, < dist(c, 892). To prove (2.15), we observe that, for every 3s > 0, If
lu(g) - up()1 =
<
kP(b -
if p<8E
J
=
u(y)} dy
E.
To prove (2.16a), we write
M := maxan u,
vP := uP I aG(P)'
Now, if x E aG(p) and y E R(x, p), then dist(y, 892) < 3p [because dist(x, 892) = 2p and ly - xI < p]; if also 3p < 5 , then u(y) < M + E [because dist(y, a92) < 8E and by (2.18)]. Accordingly, for all x E aG(p) and every s > 0, kp(x - y) u(y) dy
vP(x) mx,P)
f = M+E, <
kp(x - y) (M + E) dy
if 3p < 8E
(x,P)
(2.19)
which proves (2.16a).
It remains to prove (2.16b). If M < 0, then (2.19) shows that, for 3p < 8_M and for all x E aG(p), we have vp(x) < 0 and hence (vp)+(x) = 0; therefore, both sides of (2.16b) are zero. If M = 0, then (2.19) shows that vp(x) < E for every E > 0 and for all x E aG(p), if 3p < 5g; again
50
2 Some Maximum Principles for Elliptic Equations
Fig. 2.3.
both sides of (2.16b) are zero. If M > 0, then maxan u+ = M, and (2.19) implies (2.16b) once more. 2.3 The boundary-point lemma and the strong maximum principle Lemma 2.12 (the boundary-point lemma for balls). Suppose that
(a)Bcz0isaball, uEC(B); (b) u is a C2-subsolution relative to Lo or L and B, or a distributional subsolution relative to Ll and B; (c) u(x) < u(p) for all x E B and some p E 3B, with u(p) >- 0 when the coefficient c is not the zero function. Let m be an outward unit vector at p (m n > 0 and Imi = 1, where n denotes the outward unit normal to aB at p). Then lim in ftlo
u(p) - u p - tm) > 0 ,
l
( 2 . 20)
which implies that
a m (P) :
= li mto
u(p) - p - tm) > 0
( 2 . 21 )
t
whenever this one-sided directional derivative exists.
Proof (i) As in Figure 2.3, let B =: R(q, p) and A := R(q, p) \ .4(q, 2 p); it will suffice to consider the annular set A. Also, let M := u(p) = SUPB u.
2.3 The boundary-point lemma and the strong maximum principle 51
If we can find a function v E C2(A) such that v(p) = 0,
(I)
u+v < M on A,
(III)
then we can prove (2.20) as follows. Let w := u + v. For 0 < t < 2p,
w(p) - w(p - tm) = M - w(p - tm) t
t
>0
[by (I) and (III)],
whence lim inftlo
u(p) - u(p - tm) t
= lim infrl0
> liminftl0 8m
{w(p) - w(p - tm)} - {v(p) - v(p - tm)} t
-v(p) + v(p - tm) t
(p) > 0
[by (II) ]
(ii) Consider the function defined on A by
v(x) := 6 (e Kr2 - e Kp2) ,
r := Ix - qI,
and shown in Figure 2.4; both positive constants S and K are still to be chosen.
Certainly v E C2(A); also (I) and (II) hold, since
<0. r=p
For (III), we shall use the weak maximum principle, first considering the
values of u + v on A. For r = p we have u < M, v = 0 and hence u + v < M, with equality at p. For r = p, we have u < M by hypothesis (c); if M - a denotes the maximum of u2 for r = 12P [the supremum of a continuous function on a compact set is attained], then a > 0. Choose S = a; then u < M - a and v < a for r = 2p. Accordingly, maxaA (u + v) = M.
(iii) If we can choose K so that Lv >- 0 in A (hence so that L0v >- 0
in A, or Liv > 0 in A), then condition (III) will follow from one of
2 Some Maximum Principles for Elliptic Equations
52
<S
.r
Fig. 2.4.
our three versions of the weak maximum principle, applied to u + v and A. The condition u(p) >- 0 when c # 0 banishes one difference in these versions. To deal with a distributional subsolution u, we add to the given condition,
Ald(cp, u; A) > 0 whenever (p e C°(A) and cp > 0, the condition Ald((p,v;A) >- 0 for the same gyp; this property of v will be implied by integration by parts [as in (2.9) and (2.10)] once we have Llv > 0 in A. Then, with u+v E C(A) and Ald(cp,u+v;A) >- 0, condition (III) will follow from Theorem 2.11. (iv) It remains to calculate Lv and choose K. Since 2
ai e -Kr
aiaj eKr2
= e -Kr' (-K)2(xj - q1), = eKr2 {4K2(x, - gi)(xi - qi) - 2KS,i}
we have
Lo(x) = exr2
4K2 E affi(x) (xi - gi)(x1 - qj) i,)
-2K Eaffi(x)-2K i
i
+ C(x) (eKr2 - eKP2 }
bi(x) (xi - qj)
2.3 The boundary-point lemma and the strong maximum principle 53 By the condition (2.3) of uniform ellipticity, I Lv(x)
>
4K2Aor2 - 2K supA
e-Kr2
I aiiI + I bl p I - SUPA ICI
i
>0
for xEA
if we choose K sufficiently large, because r2 > (2 p)2.
Note a change of direction in the statement of the next theorem: there is no mention of 52 or of 852.
Theorem 2.13 (the strong maximum principle). Suppose that (a) 12 is a region (open and connected, possibly unbounded); (b) u is a C2-subsolution relative to Lo or L and 52, or a generalized subsolution relative to Ll and 52; (c) supn u >- 0 when the coefficient c is not the zero function. Under these hypotheses, if supn u is attained at a point of 0, then u is constant in 52.
Proof Let M := supn u, and assume that this supremum is attained at z E Q. Define
F := {x E 52
I
u(x) = M),
G := {x E 52
I
U(X) < M} ;
then F is closed in the metric space 52, and not empty because z E F; the set G is open in the metric space Q. If G is empty, the theorem is true. Suppose then that there is a point xo E G. We shall obtain a contradiction by means of Lemma 2.12, first using the result that, because 52 is open and connected in RN, it is pathwise connected (Burkill & Burkill 1970, p.44; Cartan 1971, p.42). This implies existence of a continuous arc
y:={fi(t) I 0
Let z be the first point of y at which u(x) = M; here `first' means `with smallest t'. Possibly z = z. Let q be any point of y that is strictly between xo and z, and is such that Iq - zI < dist(y, 852) when 52 has a boundary. Now consider the ball B := R(q, p) with p := dist(q, F). Then p < Iq - zI < dist(y, 812), so that B c 52; also, B c G by construction. There exists a point p E F n 8B because F is closed (possibly p = x). All
2 Some Maximum Principles for Elliptic Equations
54
xo = l; (0) e G
Fig. 2.5.
the hypotheses of Lemma 2.12 hold, so that at p the outward normal derivative an
(p) = n (Vu)(p) > 0.
But since p E F, it is an interior maximum point of u E C'(0). Hence (Vu)(p) = 0 and we have our contradiction.
There are many boundary-point lemmas for elliptic operators and sets
other than balls, but Lemma 2.12 is probably the heart of the matter. Theorem 2.15 is a consequence of that lemma, seasoned by a touch of the strong maximum principle. First, we need a definition.
Definition 2.14 A set fl has the interior-ball property at a point p E as2 if there exists a ball B0 c ) such that p E aBo; it has the exterior-ball property at p if there exists a ball B1 c RN j j such that p E aB1.
Figure 2.6 shows two cases of the interior-ball property for 92, and therefore two cases of the exterior-ball property for RN \ S2. Note that a unit vector m at p, outward from an interior ball B0, need not be outward from S2.
Theorem 2.15 (a boundary-point theorem for fl). Suppose that (a) 0 is a region; (b) u is a C2-subsolution relative to Lo or L and S2, or a generalized subsolution relative to L, and S2;
2.3 The boundary-point lemma and the strong maximum principle 55
aS2
Fig. 2.6.
(c) there is a point p E 00 such that u E C(f) U {p}) and u(p) = supu u, with u(p) > 0 when the coefficient c is not the zero function; (d) ) has the interior-ball property at p. Let m be a unit vector at p, outward from an interior ball Bo at p. Then either liminftlo
u(p) - u(p - tm) > 0 t
(2.22)
(which implies that (au/am)(p) > 0 whenever this derivative exists), or u is constant in Q.
Proof Let xo be the centre of the ball Bo and let po :_ lp - xol, so that Bo = P4(xo, po). Now consider the smaller ball B := M(q, po) with z hence q := 1(p + xo). Since B c Bo U {p}, we have B c C1 U {p} and u E C(B).
If u(x) < u(p) for all x E B, then Lemma 2.12 implies (2.22). If u(z) = u(p) for some z E B, then u(z) = supu u and the strong maximum principle implies that u is constant in n.
Suppose that p is what may be called an edge point ; for example, f2 = (0,1)2 =R2 and p= (0, 0), or S = (0,1)3 c 1[83 and p = (1, 0, 0). Then S2 lacks the interior-ball property at p, but something can still be said, for a subsolution, about an outward derivative or difference quotient at p. This is the subject of Appendix E.
Remark 2.16 (on the condition c < 0 in S2). For a subsolution u, if supu u = 0 in an application of the weak or strong maximum principle, or if SUPB u = 0 in an application of the boundary-point lemma for a ball B, then the condition c < 0 in S2 (imposed in Definition 2.3) can be omitted.
56
2 Some Maximum Principles for Elliptic Equations
Proof We use the decomposition c(x) = c+(x)+c (x) [defined in Chapter 0, (v)]. The foregoing theorems and lemma are valid for the operator Land bilinear form Ai defined by
L-u := Lu - c+(x)u (= Lou + c-(x)u Al (cp, u; S2) := Al (gyp, u; S2) - c+
Jn
cpu
(c+ = c > 0).
When supn u = 0, we can use L- in place of L because Lu 0 and u < 0 imply that L-u >- 0. Again, when supn u = 0, we can use Al in place of Al because A, (gyp, u; S2) >- 0, cp >_ 0 and u < 0 imply that Al (gyp, u; S2) >- 0.
2.4 A maximum principle for thin sets f
All our maximum principles so far have required that the coefficient c(x) < 0 for all x E S2, unless it happens to be known for a subsolution u that supn u = 0 (Remark 2.16), or for a supersolution v that infn v = 0. In this section we proceed to a weak maximum principle for thin sets f in which both c(x) and u(x) are unrestricted in sign a priori. By a thin set f) we mean one of specified diameter and small volume: 1921 < 6, where the positive number 6 depends only on diam S2 and on constants
independent of n. To derive this maximum principle, we need some form of the basic estimate for elliptic equations that is presented here as Theorem 2.18. This estimate, in turn, is a consequence of elementary
results in Appendix A for the Newtonian potential and of the weak maximum principle in Theorem 2.11. Given a bounded open subset G of RN, we define a modified Newtonian kernel K by
-ZIxj +1diamG 1
2 log
diam G Ixl
Ix11
KN
-2
ifN=1, ifN = 2,
(2.23)
ifN > 3,
where x 0 if N >- 2 and where KN is as in (A.18b) of Appendix A. This function differs from the Newtonian kernel K introduced by (A.18) only for N = 1 or 2, and then only by the addition of a constant which ensures that K(xo - x) >_ 0 whenever xo,x E G. The corresponding
2.4 A maximum principle for thin sets S)
57
modified Newtonian potential of a suitable density function g : G - R is defined by v(xo) :=
Jc
K(xo - x) g(x) dx,
xo E RN,
(2.24)
but here we restrict attention to field points xo E G.
Lemma 2.17 Let G be a bounded open subset of RN and let v be the modified Newtonian potential defined by (2.23) and (2.24). If g E Lp(G)
with 1 < p < oo for N = 1, or with N/2 < p < oo for N >- 2, then v E C(G) and v is a distributional solution (Definition A.7) of -A v = g in G. Moreover, Iv(xo)I < I(N, p) (diam G)2-(N/P)
for all xo E G,
II g I Lp(G) II
(2.25)
where, with the notation 1/p + 1/q = 1 and ?N := I0-IN(0,1)1, 1/q
2F
(I'
1
for N > 3,
(log p F(N, p) =
1/q
q
1
t(2, p) _
f or 1
q+1)
N1 2
I
p dp }
/
( N)
(1 < p < oo),
JJJ
1/q
1/P
(N - Nq + 2q )
N
2 (<).
(2.26)
Proof For N >- 2, it follows from Theorem A.6 that v E C(G), and from Theorem A.8 that v is a distributional solution of -Av = g not merely in G but in RN; for N = 1, the proofs are similar in strategy but much easier in detail. The bound (2.25) results from the Holder inequality; we integrate k(xo - x)q over the ball °(xo, diam G), using R Ix - xol as variable of integration.
Notation and terminology The next theorem involves both the constantcoefficient operator L1 (Definition 2.3) and the Lebesgue space L1(S2). Confusion will be avoided by unfailing display of S2 in the symbol Lp()). Extending slightly the terminology in Definition 2.10, we shall say that L1u+f >- 0 in fl in the distributional sense if u and f are locally integrable
2 Some Maximum Principles for Elliptic Equations
58
in S2 and
E aij (ajaig) u - E bj (ajcp) u + c(pu + (pf
J
>_ 0
j=1
i,j=1
whenever (p E Cc(Q) and (p > 0.
(2.27)
Theorem 2.18 (a basic estimate for the operator L1). Suppose that (a) 0 is bounded, u E C(ST); (b) L1u + f >_ 0 in fl in the distributional sense, where f E Lp(S2) with
1
maxi u < A II f I
Lp(K2) II
,
(2.28)
where A is independent of u and X521 (but depends on diam12). In fact, coarse inequalities give
A = I'( N, p) exp (11 diam S2) (di am 0)2-(N/p) ,
(2 . 29)
where I'(N, p) is as in Lemma 2.17, )o is the (positive) smallest eigenvalue of the matrix (aij) and It = (b1, ... , bN) is the vector of coefficients in the term It V of L1.
Proof (i) We make two co-ordinate transformations. First, let P be an orthogonal N x N matrix such that the transformation y = Px makes the yj-axes principal axes of the matrix a; say (PaP-'),j =: Aidij for i, j E { 1, ... , N}, where 20 := min j A j > 0. Second, we make a dilatation
z = Ey, where Eij = , 1/2Sij, in order to transform L1 to A + Writing
b* := EPb,
G := EP(S2),
u*(z) := u (P-1E-1z) = u(x),
and transforming f and (p like u, we obtain from hypothesis (b) that in G in the d.s.,
where A and V are with respect to z, and d.s. means 'distributional sense'. More explicitly,
{ (Acp*
- b* O(p* + c(p*) u* + (p* f * } dz > 0
whenever (p* E Cc(G) and (p* >- 0.
2.4 A maximum principle for thin sets S2
59
(ii) Next, first derivatives are removed by the transformation u*(z) =: rl(z) u(z),
f*
(z) =: n(z) f(z), where j(z) := exp
(P*(z) _: (-i2b*
1z)iP(z),
z).
1 (2.30)
Hypothesis (b) now becomes
(A - k2) u(z) + f (z) >- 0 in G in the d.s.,
(2.31a)
where -k2 := c - !I b* I2 < 0; more explicitly, a
(Dip - k2ip) + ipf } dz > 0 whenever ip E COO(G) and
>- 0.
Conditions (a) to (c) also imply that u E C(G), that f E Lp(G) for the same p as in (b), and that (2.31b)
uIaG<_0.
(iii) We compare the function u with the modified Newtonian potential
v of (f)+. (Note that v(z) :=
(f)+ _ (f+)^ .) In other words,
I
K(z - i;)
(f)+ (4) dt',
z E G,
G
whence v E C(G) and
(A - k2) v(z) + (f )+ (z) = -k2v(z) in G in the d.s., (2.32a) v(z)
>0
on
(2.32b)
by Lemma 2.17 and because k (z - l; ) > 0 and (f) (o >- 0. Let w
v; then (2.31a), (2.31b) and (2.32a), (2.32b) imply that
0 in G in the d.s.,
(L - k2) w >- k2v WIaG <_ 0,
and the weak maximum principle (Theorem 2.11) ensures that max-d w < maxaG W +
= 0.
The inequality (2.25) now yields
maxG u 5 maxi v < F(N, p) (diam G)2-(Nlp) III I L(G)I.
(2.33)
2 Some Maximum Principles for Elliptic Equations
60
Returning to 0, u and f, we note first that 21/2 Ibl,
b*
diam G < 21/2 diam Q.
We may suppose that 0 E S2, because the operator L1 and the norm II f I Lp(f) II are invariant under translation of co-ordinate axes; then (2.30) implies that C
1
maid u < exp 2 II f I LP(G) II
I b i diam fl 20
maxG u,
Ibl dia0m S21 AO N/2p
<- exp ( 2
II f I LP(1) II.
The result (2.28) now follows from (2.33) and these inequalities.
As was mentioned earlier, the virtue of the following maximum principle is that the signs of the coefficient y(x) and of the subsolution u are both unrestricted in 92. Theorem 2.19 (a maximum principle for thin sets fl). Suppose that (a) S2 is bounded, u E C(O); (b) L10u + y(x)u >_ 0 in 12 in the distributional sense, where L10 is the operator L1 with coefficient c = 0 and y E L,,,(L ); (c) Ulan <0. Then
u<0 on S2
whenever I0I < 6,
(2.34)
where the positive number 6 is independent of u and I0I (but depends on diam S2). In fact, we may take 61/N =
20
diamS
Ily
exp I
_ Ibl diam S2 20
)
(2.35)
where the notation is that explained after (2.29).
Proof (i) The first step is to write L10 + y(x) > 0 in a more tractable form. We introduce a constant c< 0 with Icl so large that g(x) := -c + y(x) >_ 0 almost everywhere in 12; this can be done with II g I L,,(0) II <- 211 y I L.(92) II. Then
L1u + g(x)u = L10u + y(x)u >- 0
in S2 in the d.s.,
where d.s. means 'distributional sense', as before.
2.5 Steps towards Phragmen-Lindelof theory
61
(ii) The second step is to decompose g(x)u:
Llu+g(x)u+ >- -g(x)u- >- 0 in 12 in the d.s., to recall that ul an < 0, and to apply Theorem 2.18 with f = gu+; the choice p = N is admissible for all N E N. This yields maxi u < A II gu+ I LN(G) 11 < A II g I L.(n) II maxj u+ 15211/N.
(2.36)
If maxi u < 0, then (2.34) holds. If maxis u >- 0, then (2.36) and our bound for 11 g I L,,,(52)11 imply that
(maxi u+) { 1- 2A 11 Y
I
L.(K2) 11 InI IIN } <- 0,
from which (2.34) and (2.35) follow if we choose 1521 to be so small that the expression in braces is positive.
2.5 Steps towards Phragmen-Lindelof theory
All three versions of the weak maximum principle in §2.2 require SZ to be bounded and u to be continuous on S2. If we relax one or other of these conditions, what other hypotheses will ensure that a subsolution, relative to L and 52, can be bounded above in terms of its values on 852? This is the theme of the remainder of this chapter, but, as it stands, the question is much too wide; we narrow it as follows.
(a) Among the many unbounded, proper subsets of RN that might be considered, our favourite will be the half-space D := { x E RN XN > I
01.
(b) In the rest of this chapter, the dimension N > 2. (c) The condition u c C(S2) will be relaxed at only one or two boundary points; typically to u E C(S2 \ {p}), where p is a specified point of M.
(d) Only the Laplace operator A will be considered. There is no disgrace in this restriction; good answers to our question are sensitive to details of the differential operator L, and each proof involves a comparison function tailored rather closely to the task in hand. To launch here into the more general theory initiated by Gilbarg (1952) and E. Hopf (1952a) would be a catastrophic attempt to run before we have learned to walk.
2 Some Maximum Principles for Elliptic Equations
62
Definition 2.20 We shall say that u is subharmonic in C1 if it is a distribu-
tional subsolution relative to A and fl; that is, if u is locally integrable in 0 and
r, (L p)u >- 0
whenever
(p e Cc(f2)
and
(p >- 0.
Then u is superharmonic in S2 if -u is subharmonic there; u is harmonic in 52 if it is both subharmonic and superharmonic in S2. This definition of `harmonic' (which follows inevitably from the useful definition of `subharmonic' that we have adopted) scarcely does justice to harmonic functions. Theorems B.6 and B.10 show that, if u is harmonic
in 92 according to Definition 2.20, then, after re-definition on a set of
measure zero, not only is u a C2-solution of Au = 0, but also u is real-analytic in 92.
Our opening question can now be replaced by the following. If u is continuous on b and subharmonic in D, to what rate of growth, as JxJ -+ oo, must u(x) be restricted in order that supD u = supaD u? If u is continuous merely on S2 \ {p} and subharmonic in S2, to what rate of growth, as x -> p, must u(x) be restricted in order that supn u = supan\{P} u?
We begin by inspecting some simple and explicit harmonic functions
that vanish on the boundary of the half-space D or on a punctured boundary 8S2 \ {p}; these functions indicate rates of growth that are too large in the context of our questions. Examples 1. The harmonic polynomials Pl(x) = xN, p2(x) = 2x1xx, p3(x) = 3xixN - $N,...,
pm(x) = Im (xl + ixN)m,...
(2.37)
all vanish on 8D; if N > 3, there are many more such polynomials. But, apart from the zero function, no function springs to mind that
is continuous on D, is harmonic in D, vanishes on 8D and is o(r) as r := JxI -> oo. This is significant: the critical rate of growth for the result supD u = supaD u will turn out to be close to growth like r as r -+ co.
2. If we seek functions that are continuous on D \ {0}, tend to zero at infinity, are harmonic in D and vanish on OD \ {0}, then the prototype is ql(x) = xN /rN
,
x E RN \ {0},
r := IxI.
(2.38)
This is the potential of a particular dipole (§A.4); more precisely, the
2.5 Steps towards Phragmen-Lindelof theory
63
potential of a multipole of type (0, ... , 0, 1). Differentiating this repeatedly
in horizontal directions (with respect to xj, j < N - 1), we generate multipole potentials like
q2(x) = algl(x) = -NxlxNr N-2 02g1(x) = NXN{ (N+2)x2 -r2 }r-N-4, q3(x) =
(2.39)
which retain the properties listed before (2.38), but have a stronger singularity at the origin, relative to ql, and a more rapid decay at infinity. However, no non-trivial function springs to mind that has the properties listed before (2.38) and is o(r-N+1) as x -> 0 with x E D. Again this is significant: the critical rate of growth for the result supra u = supan\{P} u will turn out to be close to growth like Ix - p1-N+1 as x --> p, when 892 is smooth.
3. We now allow a singularity at the south pole p := (0, .... 0, -a) of the ball B := _4(0, a) in RN. The Poisson kernel (§B.5) gives an example of a function continuous on {p}, harmonic in B and equal to zero on 8B \ {p}: P(x, p) = const. (a2 - r2)Ix - pI-N, x E B \ {p}, r := 1x4.
(2.40)
Again the singularity at p is of dipole type, and again appropriate differentiation generates a stronger singularity at p, while conserving the value zero on 8B \ {p}. Thus the harmonic function
Q(x) = (xNa1 -x1aN)P(x,p) = const. xl(a2 - r2)Ix - pl-N-2,
x E W\ {p},
(2.41)
has a quadrupole singularity at p. Rather as in Example 2, no non-trivial function springs to mind that is continuous on B \ {p}, harmonic in B, equal to zero on 8B \ {p} and is o(Ix - pj-N+1) as x -+ p; this is significant in the same way as before. The Phragmen-Lindelof theory that follows must be distinguished from Phragmen-Lindelof theory for holomorphic functions (complex-analytic functions). In that theory one supposes that, for example, supaD l u + iv I is known, where D is the upper- half of the complex plane C; the analogous situation for us, when N = 2, is that only supaD Iul (or only supaD lvi) is known. In the case of holomorphic functions (Hille 1973, Chapter 18; Titchmarsh 1932, §§5.6-5.8) much more can be inferred because much more is given. Definition 2.21 Let D := { x E RN I XN > 0 }, N >- 2; let Da = Dn-4(0, a)
2 Some Maximum Principles for Elliptic Equations
64
Fig. 2.7.
and l7a := D n 8-4(0, a) (Figure 2.7); denote the equator of R(0, a) by E := 8D n 0-4(0, a). A function with values V (x, a) will be called a comparison function of
the first kind if (a) for each a E (0, oo),
V(., a) E C(Da \ E) n C2(Da)
and
(AV)(.,a) = 0 in Da, V (. ,
a) > 0
on ra;
V(., a) >_ 0 on Da \ E, (2.42) (2.43) (2.44)
(b)inf{V(x,a) I xEfa}->ooas a function A : D --+ (0, oo) such that V (xo, a) < A(xo) whenever xo E D and a >_ 21xol. E
Functions having these properties will be displayed in due course; first,
we prove the lemma for which the definition has been designed. This lemma shows that, if a function u is continuous on D, is subharmonic in D and is smaller, in order of magnitude, than V (. , a(n)) on some sequence (raw) of hemispheres marching to infinity, then we retain the result SUPD U = supaD u.
2.5 Steps towards Phragmen-Lindelof theory
65
The proof will show that the important case of the growth condition (2.45) is that in which the limit inferior equals zero. Lemma 2.22 Let V be a comparison function of the first kind. If u E C(D), if u is subharmonic in D and if
lim info-,, sup
u(x)
xEra }50,
V (x, a)
(2.45)
then
SUPD U = SupaD u.
Proof (i) We may suppose that supaD u < oo, otherwise the result is trivial. Let u := u - supaD u. Then supaD u = 0 and Ii also satisfies the growth condition (2.45), because the definition of u and hypothesis (b) imply that sup
Iu(x) - u(xI
V (x a)
xEra }-->0 asa --goo.
Let both xo E D and s > 0 be given; we shall prove the lemma by showing that u(xo) < s.
(ii) If lim info-ao sup { u(x)
I
X E IF,, } < 0,
then no comparison function is needed. For, there is a sequence (a(n)) tending to infinity for which the supremum of u(x) over ra(n) tends to a non-positive limit. We choose a(k) so large that xo E Da(k) and so large that u(x) < s on ra(k). Then the weak maximum principle, Theorem D.11, applied to u on Da(k) shows that u(xo) < c as desired. (iii) It remains to consider the following case: there is a number A >- 0 such that
sup {u(x)
I
X E ra } > 0 whenever a > A,
and
lim infa-,,, sup
u(x) V (x, a)
= 0.
X E ra T
Let
s(a) := sup
u(x)
I
I
V(x,a)
l x E ra } for a > A; JJ
(2.46)
2 Some Maximum Principles for Elliptic Equations
66
XN
D
Fig. 2.8.
then s(a) > 0. We so choose a that, at the given point x0, s(a) V(xo,a) < 1E;
(2.47)
this can be done because (2.46) states that there is a sequence (a(n)) for which s(a(n)) -- 0 as n -+ oo and a(n) -> oo, while hypothesis (c) ensures that V (xo, a(n)) < ..(xo) whenever a(n) > 21xoI. With a now fixed at this value, define cp(x) Da,a
for x E Da \ E, {x E D I IxI < a, dist(x,E) > 6 > 0}
u(x) - s(a) V (x, a)
(see Figure 2.8). Choose 6 so small that xo E Da,g and so small that
dist(x,E) = 6 and x E
u(x) < ZE;
this choice is possible because uI E < 0 and u E C(D).
(iv) Finally, we apply the weak maximum principle (Theorem 2.11) to pp on Da,g. Certainly cp E C(Da,g), and qo is subharmonic in Da,g, because
u is subharmonic there and V(., a) is harmonic. The boundary values of q' are as follows. On BDa,b n 8D we have u(x) < 0 and V (x, a) >_ 0, hence ep(x) < 0. On the part of ODa,g distant 6 from E we have u(x) < e and V (x, a) >- 0, hence cp(x) < E. On
i
z
2.5 Steps towards Phragmen-Lindelof theory
67
3Da,b fl I'a we have cp(x) < 0 by the definition of s(a):
x E Fa
T(x) = V (x, a) V (xx a) - supYEra V (y, a)
0. JJJ
Therefore the weak maximum principle implies that cp(x) < 1c on Da,b; it follows from (2.47) that u(xo) = cp(xo) + s(a)V(xo, a) < s,
as desired.
The next item is a naive application of Lemma 2.22, based on a simple comparison function and intended to make Definition 2.21 less mysterious. In this example, V is independent of a, and does not have a discontinuity on the equator E. The full force of Lemma 2.22 will emerge only in §2.7, after more elaborate comparison functions have been constructed. Example 2.23 Let D :_ { x E R2
I
x2 > 0 }. If u E C(D), if u is subhar-
monic in D and if, for some constant /3 E (0, 1),
lim info-,,, sup { a-#u(x)
I
x E Fa } <_ 0
(in particular, if u(x) = o(rfl) for some / E (0, 1) as r
(2.48)
xj -> oo), then
SUPD U = supoD u.
Proof Denote points of D by x = (r cos 0, r sin 0), 0 < 0 < it. We claim that the formula k := (1 - /3) 2,
V(x) := r' sin(fl0 + k),
xE
defines a comparison function of the first kind. For, referring to Definition 2.21, we observe that V E C(D) n C2(D); that V >_ 0 on D, with V > 0 on {0}, because
sin(/30 + k) > sink
for 0< O< ir;
and that AV = 0 in D because V(x) = Ime'kzp
(z := x1 +ix2 = re'°).
Thus V satisfies condition (a) of Definition 2.21; it satisfies (b) because V(x) > a' sink when x E Ta; for (c), we may choose 2(xo) := V(xo) or 2(xo) := ro.
2 Some Maximum Principles for Elliptic Equations
68
G G a
F
p
F
Fig. 2.9.
If the growth condition (2.48) implies (2.45) for the present function V, then Lemma 2.22 implies the present result. Now,
sink <
a#
1
when x E Fa,
so that the two growth conditions are equivalent. Definition 2.24 Let G be a connected open set in RN, N > 2, let p E 8G be given, let Ga := G\R(p, a), let ya := Gfla (p, a) and let F := 8GnOR(p, a).
Here a E (0, ao) and ao is a positive constant depending only on G. (Three possible configurations are shown in Figure 2.9; in the third, G :_ R(p, ao) \ {p} and F is empty.) A function with values W (x, a) will be called a comparison function of the second kind if (a) for each a E (0, ao), W(., a) E C(Ga \ F) f1 C2(Ga)
and
W(., a) >_ 0
(AW)(. , a) = 0 in Ga, W(., a) > 0
on ya;
on Ga \ F,
(2.49)
(2.50) (2.51)
2.5 Steps towards Phragmen-Lindelbf theory
69
(b) inf{ W(x,a) I X E ya } - oo as a - 0; (c) there is a function A : G -> (0, oo) such that W (xo, a) < A(xo) whenever xo E G, a < 1 Ixo - pI and a < ao. Lemma 2.25 Let G be as in Definition 2.24 and let W be a comparison function of the second kind. Let S2 be a bounded open subset of G such that p E A2 n 8G. If U E C(S2 \ {p}), if u is subharmonic in S and if
lim infa.o sup
u(x) 1 W (x, a)
x E S2 n 8g(p, a) } < 0,
(2.52)
then
supn U = supan\{P} U.
Proof The proof resembles that of Lemma 2.22, but to condense it ruthlessly would be a false economy. (i) We may suppose that supao\{p} u < oo, otherwise the result is trivial. Let u := u - supa0\{P} u. Then supan\{p}u = 0 and u also satisfies (2.52),
because u - u is a (finite) constant and by condition (b) in Definition 2.24.
Let both xo E S and s > 0 be given; we shall prove the lemma by showing that u(xo) < E. To this end, write
fl,, := S2 \ V(p, a) and as := S2 n 8R(p, a). (ii) If
lim infa.o sup {u(x) I x E as } <- 0,
then no comparison function is needed. We argue as in the proof of Lemma 2.22, using small surfaces aa(n) with a(n) -> 0 instead of large hemispheres Fa(n) with a(n) -* oo.
(iii) It remains to consider the following case: there is a number a > 0 such that sup { u(x) I X E as } > 0 whenever a < a, and
lim infa.o sup
u(x) W (x, a)
Let r
s(a) := sup {
u(x)
l W(x,a)
x E as } = 0.
l x E as } J
for a < a;
(2.53)
2 Some Maximum Principles for Elliptic Equations
70
an
Fig. 2.10.
then s(a) > 0. Choose a to be such that, at the given point xo, s(a) W (xo, a) < 1 e;
(2.54)
this can be done because of (2.53) and hypothesis (c) in Definition 2.24. With a now fixed at this value, define ip(x) := i(x) - s(a) W (x, a)
for x E S2a \ F.
If 8S2 intersects F, define Qa,b
{ x E 0 I Ix - pI > a, dist(x, F) > S > 0 }
(Figure 2.10); choose 6 so small that xo E S2a,b and so small that
dist(x, F) = S and x E S2
u(x) < 18;
this choice is possible because ul ffinF 0 and u E C (S2 \ {p}). If 8S2 does not intersect F (in particular, if F is empty), define S2a,b := Sta. In either case, W(. , a) is continuous on S2a,b and u(x) < i s on 8S2a,b \ 6a (iv) Now apply the weak maximum principle, Theorem 2.11, to ip on S2a,b, observing that W E C(S2a,6) and that W is subharmonic in Stb. On af2a,b \ Ca we have u(x) < Zs and W(x,a) > 0, hence ip(x) < Zs. On
2.5 Steps towards Phragmen-Lindelof theory
71
af2a,b n as we have ip(x) < 0 by the definition of s(a). The maximum principle ensures that p(x) < E on a,b ; it follows from (2.54) that z
u(xo) = p(xo) + s(a) W (xo, a) < s, as desired.
O
Like Example 2.23, our first application of Lemma 2.25 will involve only simple comparison functions. But the result is better than that of Example 2.23; it is not restricted to R2 and it is best possible in a certain sense (Exercise 2.44) when no smoothness is demanded of 8S2 at p.
Theorem 2.26 Let SZ be bounded in RN, N > 2, let p E % and write as := 92 n 8-4(p, a). Let b be so large that f2 c R(p, b). If u E C(92 \ {p}), if u is subharmonic in S2 and if lim infa.o sup S
log(b/a)
lim info-o sup { aN-2u(x)
I
x E as } < 0 when N=2,
(2.55)
x E as } < 0
(2.56)
when N >_ 3,
then
supn u = supan\{P} U.
Proof The comparison functions are potentials of point sources (multiples of Newtonian kernels), discussed at some length in Appendix A.
(i) For N = 2, we choose G := .4(p, b) \ {p} for the set in Definition 2.24 and define
W(x) := log
x
for x E 4(p,b) \ {p}.
b pl
Then Ga = l(p, b) \ R(p, a) and ya = 0.4(p, a); condition (a) of Definition 2.24 is satisfied. Since W(x) = log(b/a) when x E ya, condition (b) holds. For (c), we choose , (xo) := W(xo). Thus W is a comparison function of the second kind. The growth conditions (2.52) and (2.55) coincide for this function W ; therefore Lemma 2.25 implies the present result. (ii) For N >_ 3, we choose G := RN \ {p} for the set in Definition 2.24 and define W(x) := Ix _ pl-N+2
for x E RN \ {p}.
One checks without difficulty, very much as in (i), that this function W is a comparison function of the second kind. The growth conditions (2.52)
72
2 Some Maximum Principles for Elliptic Equations
and (2.56) coincide for this W ; again Lemma 2.25 implies the present result.
2.6 Comparison functions of Siegel type
This section concerns functions g(. ; a), ge(. ; a) and $2(. ; a, b) with the property that ag(. ; a) is a useful comparison function of the first kind, while a N+1ge(. ; a) and a N+l$2(. ; a, b) are corresponding comparison functions of the second kind; ge(. ; a) and $2(. ; a, b) are defined on different domains. These functions will enable us to extend Example 2.23 to half-spaces in RN for all N > 2; to improve the rate of growth allowed in Example 2.23 from approximately o(rfl), where r := Ixl and # < 1, to approximately o(r2/xN) as r -* oo; and to improve the rate of growth allowed in Theorem 2.26 from approximately o(Ix - pl-N+2), for N >- 3
and x - p, to something slightly bigger than o(Ix - pl-N+i), provided that S has the exterior-ball property at p (Definition 2.14). The functions g, ge and 92 will be called of Siegel type because g for N = 2, displayed here in (2.59), was introduced into Phragmen-Lindelof theory by D. Siegel (1988). The construction of g for all N >- 2, from the Poisson integral formula for functions harmonic in a ball (§B.5), is the subject of Appendix C. The functions ge and $2 result from applications to g of the Kelvin transformation (§B.3). It will be convenient to use the signum function, defined by
-1 sgn t :=
0 1
if if if
t < 0, t = 0,
(2.57)
t>0.
Theorem 2.27 Let B := R(0, a) in RN, N > 2, and let E
x E aB
xN = 0 } denote the equator of B. (a) There exists a function g = g(. ; a), which we call the primary function of Siegel type, such that g E C(B \ E) fl C0O(B) and
Ag = 0
in B, g(x) = a/xN on aB \ E, g(x)I < const. IxNI/a if r := Ixl < a/2, where the constant depends only on N. Also, sgn g(x) = sgn xN on and g(x; a) depends only on x/a.
(2.58a) (2.58b) (2.58c)
E,
2.6 Comparison functions of Siegel type
73
(b) For N = 2, let (x, y) E R2 and z = x + iy E C. Then, on B \ E c R2, g(x, y; a) = Im (a
_
a z
ay
a
a+z
ay
(a - x)2 + y2 +(a + x)2 + y2
.
(2.59)
Proof See Appendix C.
Corollary 2.28 The exterior function ge = ge(. ; a) of Siegel type is defined by
ge(x; a) :=
(a) r
N-2
a2
g I r2 x; a I
,
x E l[8N \ (B U E),
(2.60)
where again r := Ixi and B, E are as in Theorem 2.27. It follows that ge E C (RN \ {B U E}) n C°°(RN \ B) and that
Age = 0
in RN \ B, ge(x) = a/xN on 8B \ E, Ige(x)I < const.aN-lr-NIxNI if r > 2a,
(2.61a)
(2.61b) (2.61c)
where the constant depends only on N. Also, sgn ge(x) = sgn xN on RN \ (B U E), and ge(x; a) depends only on x/a.
Proof A calculation, done fully in Theorem B.15, shows that (a)N+2(Ag)
(_x) a2= 0
Age(x) _ if r > a and hence Ia2x/r2I = a2/r < a. The remaining properties of Se are immediate consequences of the definition (2.60) and the corresponding properties of g.
Inspection of Corollary 2.28 and Definition 2.24 shows that a N+lgei restricted to D \ (B U E) (where D is our usual half-space), is a comparison
function of the second kind, with G = D and p = 0 in the notation of Definition 2.24. The restriction to D \ (B U E) is needed in order that ge >- 0. Therefore aN+lge can be used in Lemma 2.25 for sets 92 that are on one side of a hyperplane containing the point p of 8S2 at which u may be discontinuous; this is illustrated in Figure 2.11. In particular, a-N+lge can be used for convex sets S2. Suppose now that S2 has merely the exterior-ball property (Definition 2.14) at the specified point p E 852. Then a suitable comparison function
a N+1g2(. ; a, b) is found by inversion relative to a sphere as follows.
2 Some Maximum Principles for Elliptic Equations
74
XN
Fig. 2.11.
Choose co-ordinates so that p and an exterior ball Bq at p are given by
p = (0,...,0,2b), q = (0,...,0,b), Bq =-V(q,b)
(2.62)
for some b > 0, as shown in the right half of Figure 2.12. Define
V := {
E RN N < 2b I, VQ := V \ -4(p, a) with 0 < a < b;
we shall use ge(p - ; a) for l; E Va \ E, where E now denotes the equator of R(p, a). Under the transformation 4b2
equivalently
x=b
x
(r.=lx1>0), (2.63)
2
(p := I c I > 0),
which is inversion relative to the sphere 0-4(0,2b),
2.6 Comparison functions of Siegel type
75
Fig. 2.12.
2 Some Maximum Principles for Elliptic Equations
76
the punctured half-space V \ {0} has image G := RN \ Bq ; (2.64a) the ball R(p, a) has image B* := -4(p*, a* ), where
1
P*
1 - (a/2b)2 p'
1- (a/2b)2
a*
a
the equator E has image F := 8G n 8B*.
(2.64b) (2.64c)
Note that p is a fixed point of the map (2.63) and that, near p, this transformation is approximately reflection in the hyperplane V. Therefore 8B* is close to 7_4(p, a) for small radii a. A more precise statement is that 4b 2
Ix-p12 = 727 b ( xN -PN + 2b
zzN
PN - b
\
I .
(2.65)
Here the factor 4b2/r2 will be unimportant when $2 comes to be used in Theorem 2.36 [because 4b2/r2 -_+ 1 as x -> p], but we shall need Z(x) := XN - PN + I
x-
2
Zb
I
,
(2.66)
which is almost dist(x, OG) for points x near OG. In fact, a calculation shows that 1 z(x) < dist(x, 8G) z Iac
= 0, 1- 2
b -
<1 for all x E G,
z (x )
(2 . 67a, b)
and
maxXEaB. Z(x) = a (1
- 2ba
(2.67c)
Corollary 2.29 Let ge be as in Corollary 2.28. We adopt the notation (2.62), (2.64a,b,c) and (2.66) for any b > 0 and any a E (0, b). Then the two-ball function $2 = $2( ; a, b) of Siegel type is defined by
(2b)N $2 (x; a, b)
2
4b ge
(p-
x; a)
,
x E G \ (B* U F U {0})
////
and by $2(0)
(2.68)
limr1o 92(x) = 0. It follows that $2 E C (G \ {B* U F}) n
Cc0(G \ B *) and that
A92=0 in G \ B*, $2(x) = C 2b r / N-4 z(x) a
on G n 8B*,
(2.69a)
(2.69b)
2.7 Some Phragmen-Lindelof theory for subharmonic functions
if x E G \ 4(p', 2a'), where p' := then 1$2()I <_ const.
1
aN-1
1
1 - (a/b)2
N2
Ip
p, a' :=
- 4b x 2
rb
1
1 - (a/b)2
N I
pN
a,
77
l
2
_ 4b
xN
(2.69c)
where the constant depends only on N. Also, $2(x) = 0 on aG \ B*, and $2(x) > 0 in G \ B * .
Proof We have IL92 = 0 in G \ B* by Theorem B.15, already cited in the proof of Corollary 2.28. The remaining properties of $2 follow from those of ge by direct calculation.
2.7 Some Phragmen-Lindelof theory for subharmonic functions We return to the half-space D := { x E ][8N
I
XN > 0 }, N >_ 2.
Theorem 2.30 If u E C(D), if u is subharmonic in D and if lim inf
a
x
a } = 0,
(2.70)
then
SUPD U = SUP3D U.
Proof Let g continue to denote the primary function of Siegel type (Theorem 2.27), and let V (. , a) := ag(. ;a) on Da \ E, where Da :_ D n 9(0, a) and E := OD n 09(0, a). Theorem 2.27 shows that this V is a comparison function of the first kind (Definition 2.21); in particular, V (x, a) =
a2
>_ a
for x E I'a := D n 0-4(0, a),
XN
and
V(xo, a) < const. xoN
for x0 E D and a >- 2Ix0I,
where the constant depends only on N. Therefore the present theorem is implied by Lemma 2.22. The supremum over iTa in (2.45) of that lemma can now be written as a maximum over ra because the function with values XN u(x)/a2 is continuous on D; the maximum cannot be negative, because of values for XN = 0.
2 Some Maximum Principles for Elliptic Equations
78
Corollary 2.31 Let G be an unbounded open subset of the half-space D. If u E C(G), if u is subharmonic in G and if
lim infa, max
xN u(x) a2
x E G,
Ix! = a
< 0,
(2.71)
then
SUPG U = SUNG U.
Proof In the proof of Lemma 2.22 we replace D by G, keeping the same comparison function V of the first kind. Thus Da is replaced by GQ := G n -4(0, a) and Fa now means G n 8-4(0, a). We define s(a) and choose the radius a exactly as before. If 8G does not intersect the equator E, we need not remove a neighbourood of E from GQ. The shape of 8G is unimportant; what matters is that u(x) < 0 on 8G by the definition of u, and that cp(x) < 0 on I'a by the definition of s(a). After this extension of Lemma 2.22, the present corollary results from
the choice V = ag made in the proof of Theorem 2.30. The maximum in the growth condition (2.71) may be negative once more, because XN need not descend to zero when x E G and IxI = a. Remark 2.32 (i) If we add to Corollary 2.31 the hypotheses: G is connected and u c C1(G), then
u(x) < supaG u for all x E G,
unless u is constant on This is an immmediate consequence of the strong maximum principle (Theorem 2.13). (ii) Let G be as in Corollary 2.31. If u E C(G), if u is superharmonic in G and if
lim supa,,, min {
XN u(x) a2
xEG, Ixl=a }>_0,
(2.72)
then
infG u = infaG u.
This follows from Corollary 2.31 by an argument like that in Remark 2.7.
(iii) For a function u that is continuous on and harmonic in G (hence is in C00(G), by Theorem B.6), we wish to conclude that infaG u < u(x) < supaG u for all x E G
2.7 Some Phragmen-Lindelof theory for subharmonic functions
79
(with strict inequality for x E G if G is connected and u is not a constant). It may be worthwhile to retain both (2.71) and (2.72) as hypotheses, but the simpler condition r l liminfa, max { xNI (x) x E G, lxi = a 0 (2.73) a2
111
is sufficient.
When a subset of D is significantly narrower near infinity than is D itself, much larger rates of growth are permissible. Corollary 2.31 is then far from sharp. We illustrate this by two examples; observe that, just as Theorem 2.30 extends to unbounded open subsets of D, so Examples 2.33 and 2.34 extend to unbounded open subsets of S and of H, respectively. Example 2.33 Consider the sector S := { (r cos 0, r sin 0) E R2
I
r > 0, 0 < 0 < Q }
,
$ E (0,27r).
If u E C(S), if u is subharmonic in S and if lim infR-,, max
sin(7cd/R/l)Pu(x, y)
(x, y) E S,
l
I(x, y)I = R y = 0, )))
(2.74)
then
sups U = supas U.
Proof This statement is merely a transcription of Theorem 2.30, for N = 2, under the conformal map of S onto D. Write z = x + iy = re'B for points of S, and _ + irl = pe" for points of D; the appropriate mapping is equivalently p = r"/p,
t = nO/l3,
r >_ 0, 0 < 0 < p.
(2'75)
This is a homeomorphism of the closed sector S onto the closed halfplane D; it is also a C°° map, with C°° inverse, of S \ {0} onto D \ {0}. Let u(c, rl) := u(x(l;, il), under the mapping (2.75). Then u E C(D) because u E C(S). Also, a satisfies the growth condition (2.70) because u satisfies (2.74). We now show that u is subharmonic in D. y), rl(x, y)). Given 0 E CC°(D) satisfying 0, define 'p(x, y) := Then 'p c Cc°(S), cp 0 and coxx+cpyy
+ Pnn = I
dl; / dz 12'
di; drl =
dt' dz
2
dx dy,
80
2 Some Maximum Principles for Elliptic Equations
f
I G
b
H
x
Fig. 2.13.
so that
AD
ipnn) u d d'1 = .f
s((Px + (pyy) u dx dy > 0. S
The result 5UPD u = supaD u now implies that sups u = supas u.
If we allow /3 = 2n in the definition of S, then the foregoing result remains true even though S = R2 and (2.75) is no longer a homeomorphism of S onto D. Indeed, the weaker condition u E C(D) can replace u E C(S), provided that (2.74) is replaced by
lim infR-, sup
sin(0/2) u(x,.y) R1/2
1
(x, Y) E S>
I(x, Y)I = R T = 0 (2.76)
for /3 = 2ir. The condition u E C(D) is weaker for /3 = 2ir in that it allows limiting values u(x, 0+) as y 10 and u(x, 0-) as y T 0 such that u(x, 0+) * u(x, 0-) for x > 0. When contemplating supas u, we must then regard the upper and lower sides of 8S as distinct.
Example 2.34 Consider the half-strip H C(H), if u is subharmonic in H and if
lim inf,- max
sin(ny/b) u(c, y) exp(ne/b)
(0, oo) x (0, b) in ]R2. If u E
0 S y <_ b
th en
SUPH U = SUNH U.
I
= 0,
(2.77)
2.7 Some Phragmen-Lindelof theory for subharmonic functions
81
Proof This result is implied by Corollary 2.31 and the conformal map of H onto G := D \ -4(0,1); as in Theorem 2.30, the maximum in the growth condition cannot be negative [because the function in question is zero for y = 0, b]. Again let z = x + iy and + iq = pee; the relevant map is now (Figure 2.13) rcz
= exp T, equivalently p = exp
7rX
b
,
t=
71Y ,
} x>_0, 0
b
The rest is essentially as in Example 2.33.
Finally, we derive two more results for subharmonic functions that may be discontinuous at p E 852; as was promised earlier, these theorems
allow a rate of growth larger than that in Theorem 2.26, for certain boundaries 852.
Theorem 2.35 Let 52 be bounded in RN, N >_ 2, let p E 852 and assume that 52 is on one side of a hyperplane A containing p. (Figure 2.11 shows a case with p = 0 and A = IX I XN =01- ) Let dA(x) := dist(x,A). If u E C(S2 \ {p}), if u is subharmonic in 52 and if lim infa,o max { aN-2dA(x) u(x) I x E S2 n OR(p, a) } <_ 0,
(2.78)
then
supn u = supan\{P} U.
Proof Choose co-ordinates so that p = 0, A = { x
xN = 0 } and 52 lies in the half-space D; then dA(x) = XN. With ge denoting the exterior function of Siegel type (Corollary 2.28), define W(. , a) := a-N+lge( ; a) on D \ (B U E), where B := -4(0, a) and E := 8D n B. Then Corollary 2.28 shows W to be a comparison function of the second kind (Definition 2.24); in particular W (x, a) =
a
N+2
> aN+i
I
for X E ya := D f18B,
XN
and
W(xo,a) < const. IxoI-NxON
for x0 E D and a <
where the constant depends only on N.
Zjxol,
2 Some Maximum Principles for Elliptic Equations
82
Accordingly, the theorem follows from Lemma 2.25. The supremum over KI n 8R(p, a) in (2.52) of that lemma can now be written as a maximum over S2 n 8R(p, a) because the function with values aN-2dA(x) u(x) is continuous on S2 \ {p}.
In the next theorem, the growth condition (2.79) may seem absurd because of the elaborate p*, a* and because of the detailed knowledge of u that seems to be assumed. However, as with other growth conditions
that we have met, there are simpler statements that imply (2.79). For example, if u(x) = o(Ix - pl-'+') as x -> p, or if dB denotes distance to an exterior ball at p and u(x) = o (Ix - pI-N+2 /dB(x)) as x p, then (2.79) is amply satisfied.
Theorem 2.36 Let S2 be bounded in RN, N > 2, and assume that S2 has the exterior-ball property at p E 8SI (Definition 2.14). Let R(q, b), with b = l p - q I, be an exterior ball at p such that 2q - p S2 (Figure 2.14). Let dB (x) := dist (x, R(q, b)). If u E C(SZ \ {p}), if u is subharmonic in n and if
lim infa.o max { aN-2dB(x) u(x)
I
x E S1 n 8-4(p*, a*) } < 0,
(2.79)
where
p* :=P+
2
1
- q), a* 2 1 ((a/2b)2 (p
1 - (a/2b)2'
(2.80)
then
sup0 u = supan\{P} U.
Proof (i) We make two changes in Definition 2.24 and Lemma 2.25. (Readers who distrust such tinkering with previous results may prefer to prove the theorem by means of Exercise 2.45.) First, the ball _V(p, a) is replaced by 9(p*, a* ), where p* and a* are as in (2.80). Second, the
condition a < 1Ixo - p1, which accompanies the inequality W(xo, a) < 2(xo) in (c) of Definition 2.24, is replaced by
a'<1Ixo-p'I, where p'
(a/b)2
p + 21 - (alb)2 (P - q), a,
a
1 - (alb)2
(2.81)
2.7 Some Phragmen-Lindelof theory for subharmonic functions
83
a.)
Fig. 2.14.
Here b is fixed and, in the proof of Lemma 2.25, the radius a is always
chosen to be so small that certain inequalities hold. Such choices are not thwarted by the new perturbation terms in p,k, a*, p' and a', so that Lemma 2.25 remains valid [with 89(p*, a*) replacing 89(p, a) in (2.52)].
(ii) Choose co-ordinates so that p = (0,..., 0, 2b) and q = (0,..., 0, b), as in Figure 2.12; Let G := RN \ R(q, b) and B* := V(p*, a* ). Referring to Corollary 2.29, define
for x E G \ (B* U F) .
W* (x, a) :=a N+1$2(x; a, b)
Then W* is a comparison function of the second kind, according to the modified Definition 2.24; from (2.69b) we obtain
W(x,a) =
(2b)N_4
Z(x)
> aN+l J 1 - 0 (b) } for x E GnaB*,
l
and (2.69c) implies a bound W* (xo, a) < A(xo)
for xo E G and a' < z Ixo - p'I
The growth conditions
lim info-o sup
u(x) W* (x, a)
X E fl n aB* } < 0
(2.82)
2 Some Maximum Principles for Elliptic Equations
84
and (2.79) are equivalent because our choice of co-ordinates and (2.67) imply that x 2b
=1+ o(a) \b
zx =1+ o(a) for xEGnOB*, \b
and
dB(x)
and because the function with values aN-2dB(x)u(x) is continuous on SZ \ {p}. Therefore the modified Lemma 2.25 implies the present theorem.
2.8 Exercises
Exercise 2.37 Suppose that 1(p, a) c 0 c P1(q, b) in RN, and that f := S2 -> R satisfies 0 < k < f(x) < l for all x E fl. Prove that, if it exists, the solution u EC(?!) n C2(S2) of the Dirichiet problem -Du = f in .0, u I an = 0 is bounded by
(a2-Ix-p12)
N
If only the foregoing information is given, can this estimate be improved?
Exercise 2.38 Assume that 1 is a bounded region, that u is subharmonic in g (Definition 2.20) and v superharmonic in 0, that u, v E C(S2) and
that uk, < vlM. Prove that either u(x) < v(x) for all x E 0, or u = v.
Exercise 2.39 Let u be harmonic and continuous in a region S ; then u E C°°(fl), by Theorem B.6. Prove that if supn Ioul is attained in fl, then Vu is a constant vector. Show by an example that, if info foul is attained in 12, then Vu need not be a constant vector. Exercise 2.40 Prove that, if Oil is of class C2, then S2 has the interior-ball property (Definition 2.14) at every boundary point.
Exercise 2.41 Writing x = (r cos 0, r sin 0) for points of R2, consider the region fl := { (r cos 0, r sin 0) E ][i;2
I
r > 0, 0<0 <#I for # E (0, 2n],
and the function u defined by u(x)
-r'lfl sin 0
no
if x E SZ \ {0},
ifx=0.
2.8 Exercises
85
Show that (a) Du = 0 in fl; (b) for i E (0, it) the boundary-point lemma cannot be applied at the origin, and its conclusion does not hold there; (c) for f E [it, 2n] the boundary-point lemma does indeed describe the behaviour of u near the origin. What distinguishes the case it in (c)? Exercise 2.42 Let 0 be a bounded region with 3 of class C1 and with the interior-ball property (Definition 2.14) at every boundary point. The Neumann problem for L in S2 (where L is as in Definition 2.3) is to find v such that
Lv = f in S2,
8n an
= g,
v E C1(S2) n C2(S2),
(2.83)
where f,g are given functions and 8v/8n denotes the outward normal derivative.
Prove that, if they exist, any two solutions of (2.83) differ only by a constant, and that this constant is zero when the coefficient c is not the zero function. Exercise 2.43 The Lebesgue spine. Write x = (x1, x2, z) for points of 1183, let s (xi + x22)1/2 and define
52:={xER3 l 0
v(x) :=
x E SZ \ {0},
Cd {s1
belongs to C00(S2) and satisfies Av = 0 in 92; that v(x) = (z + 1Z 1) log 1 + cp(x) s
with T E C(S2) and T (O) = 1;
and that v has no extension in C(S2). Define g E C(8S2) by g(0) := 3 and g(x) := v(x) for x E 0 \ {0}. Prove that the Dirichlet problem of finding u such that
Du = 0 in 0,
ul an
= g,
u E C(52) n C2(S2)
has no solution. [Assume that u exists and apply Theorem 2.26 to u - v and to -u + v.]
86
2 Some Maximum Principles for Elliptic Equations
Exercise 2.44 (i) Let 12 be bounded in RN, N > 2, and let Ofl have an isolated point (for example, fl = R(O,1) \ {0}). Use Theorem 2.26 to prove that the Dirichlet problem of finding u such that
Du = 0 in
12,
ul an
= g,
u E C(12) n C2(12)
has no solution for certain functions g E C(a12). (ii) Prove that Theorem 2.26 is best possible in the following sense. If the hypotheses (2.55) and (2.56) are changed to
lim supa-o sup {
log(b/a) lim supa,o sup {aN-2u(x)
I x E oa } < oo l
when N=2,
x E Ua } <00 when N> 3,
and the other hypotheses remain unchanged, then the conclusion is false.
Exercise 2.45 Prove Theorem 2.36, for the case when supan\{P} u < co, by inversion relative to the sphere 0.(2q - p, 2b), by use of the corresponding Kelvin transform (§B.3) of the function u := u - supan\{P} u and by application of Theorem 2.35. [Use convenient co-ordinates, as in (2.63)] Exercise 2.46 Let 12 be an unbounded open subset of the half-space D :_ { x E RN l XN > 0 }, N > 2, and let s2 have the exterior-ball property (Definition 2.14) at each point of the set P := { pl,... , pk } = 812. Suppose that u E C(S2 \ P); that u is subharmonic in Q; that u satisfies both
lim infR-,, max {
XN u(x) R2
xES2, Ixj=R}<0
(cf. Corollary 2.31) and the growth condition (2.79) on small surfaces 0 n OR(p*, a*) for each m E { 1, ... , k}. (The radius b of exterior balls at the points pm can be chosen to be independent of m.) Prove that supra u = supan\p u.
3
Symmetry for a Non-linear Poisson Equation in a Symmetric Set 0
3.1 The simplest case
In this chapter we prove theorems about positive solutions u of Au + f (u) = 0 in a bounded symmetric set fl. Theorem 1.2 (which asserted spherical symmetry and monotonicity of u in a ball) appears as Corollary 3.5, with slightly less smoothness required off now than was demanded
before, and again as Corollary 3.9, (b), for a function f that may be discontinuous. Our main concern is with sets 0 and solutions u that are symmetrical only under reflection in a single, fixed hyperplane; results for this situation imply those for balls, as Lemma 1.8 may have indicated already.
Two definitions set the stage as regards f and fl.
Definition 3.1 Let W c R. A function f : W -+ R is locally Lipschitz continuous if, for each compact set E c W, there is a constant A(E) such that If
f(t)l
the number A(E) is called a Lipschitz constant of PIEThe function f is uniformly Lipschitz continuous if a single Lipschitz constant A(W) serves for all s, t c W.
For example, if W = (0, co) and f (t) = t-1 + It - 11 - t2, then for the compact subset [8, n] of (0, oo) we may use the Lipschitz constant A([S, n]) =
S_Z
+ 1 + 2n.
But no single Lipschitz constant will serve for this function and for all s and t in (0, oo). 87
3 Symmetry for a Non-linear Poisson Equation
88
Definition 3.2 Recall from Definition 1.7 that x''1 denotes the reflection
of any point x E RN in the hyperplane Tµ(k). A set S c W" will be called Steiner symmetric relative to T,(k) if, whenever x E S, the closed line segment from x to xµ°k is also in S; that is, if
xES={(1-O)x+ex"'k I0<e<1}
S.
0 Notation This will be as in Definition 1.7, but with the understanding that k = el := (1,0,... , 0) and with no display of k. We write x" (x2, ... , XN). Thus
T, :_ { E RN I 1 = t, },
xµ := (2µ - x1, x")
and g, (x) := g(xµ) for any function g : S2 -+ R whenever xµ E S2. Given a bounded region S that is Steiner symmetric relative to To (Figure 3.1), we also define (for all µ E I[8)
Z(µ):={xEnIx1>µ}, Y(µ):={xERNIXP EZ(p)},
(3.1)
and call these a cap and reflected cap, respectively. Note that Y(11) c S2
if µ > 0. Also, M := sup{ xl
I
x E E2 }, so that Z(µ) is empty if and
only if it >- M. Theorem 3.3 Let 0 be bounded, connected and Steiner symmetric relative to the hyperplane To. Assume that u := S2 --> R has the following properties. (a)
u EC(ii) fl C2(52),
(b)
Au + f (u) = 0 in S2,
u > 0 in S2,
uI an
= 0. (3.2)
where f : [0, oo) -> R is locally Lipschitz continuous. Then
u(z) < u(z'") if µ E (0,M) and z E Z(µ),
(3.3)
81u(x) < 0 if x1 > 0 and x E S2.
(3.4)
Proof (i) Let w(., p) := u - uµ on Z (µ) for µ E [0, M). Our main task is to prove (3.3), that
z E Z(µ) r w(z, µ) = u(z) - u(zµ) < 0 whenever p E (0, M). We begin by combining the equations
Au(z) + f (u(z)) = 0 for z E Z(µ), Au(y) + f (u(y)) = 0 for y E Y(µ),
(*)
3.1 The simplest case
89
Fig. 3.1.
where p E [0, M). In the second of these, set y = zµ. Then u(y) = u(zµ) _
uµ(z); also 8/8y1 = -8/8z1 and 8/8yj = 8/8zj for j = 2,...,N; hence 'Luµ(z) + f (uµ(z)) = 0 for z E Z(p).
Subtracting this from the equation for u(z), and defining f(uU
y(,p)
10
-f
uµuu)
at points z where u(z) uµ(z), at points z where u(z) = uµ(z),
we obtain Aw + y(z, p)w = 0 for z E Z(p),
(3.6)
where w = w(z, p). Also, the identity Z(p) = 52 n { x E RN I xl > p j implies that 8Z(µ) c 852 U Tµ; therefore w(z, p) < 0
for z E 8Z (p),
(3.7)
with strict inequality when z E 8Z(p) \ T. and z-u E 52 [because then z E 852, so that u(z) = 0, while u(zµ) > 0], and with equality when z E 8Z(µ) n T. Of course, there is a third possibility: that w(z,p) = 0 because z and zµ both belong to 00 \ T.
90
3 Symmetry for a Non-linear Poisson Equation
(ii) The next step is to show that (*) holds for all sufficiently small, positive values M - µ. (a) First, we apply the maximum principle for thin sets (Theorem 2.19) to the set Z(µ), the operator A + y(., µ) and the function w(. , µ). In order to have data independent of p, we use diam Z (0) for the diameter in Theorem 2.19. Since U E C(S2), there is a number U > 0 such that 0 < u(x) < U on S2; hence there is a Lipschitz constant A = A([0, U]) > 0
such that If (u) - f(u) I < Al u - u; accordingly Iy(z, µ)I < A for all z E Z (p) and all it E [0, M). Since w(. , p) is a C2-solution of (3.6), it is certainly a distributional solution. Therefore Theorem 2.19 implies existence of a number 6 > 0, independent of µ E [0, M), such that
IZ(µ)I <S=w(z,p)<0 forzEZ(µ).
(3.8)
(b) Second, we sharpen (3.8) to w(z,p) < 0 in Z(µ) by means of the strong maximum principle (Theorem 2.13). Now that we have supZ(p) w(. , p) = 0 [the value 0 being attained on Tµ], we may ignore the sign of y(z, µ), by Remark 2.16; we apply Theorem 2.13 to each component of Z (µ) and to the function w(., µ). Assume that w(zo, u) = 0 at some point zo E Z(µ); let Z0(µ) be the component of Z(µ) that contains zo. Then, by Theorem 2.13 and the continuity of u on n, we have
w(z, µ) = 0 at all z E Zo(µ). This is a contradiction because for µ > 0 there exist points p E OZo(y) \ Tµ such that pµ E S [see a hint to Exercise 3.13] ; then w(p, p) < 0, as was noted after (3.7). Consequently, IZ(,a)I < 6
w(z,p) < 0 for z E Z(µ).
(3.9)
(iii) Let (m, M) be the largest open interval of y in which (*) holds; assume (for contradiction) that m > 0.
(a) Fix Z E Z(m) and let y 1 m. Since u(z) < u(zµ) for µ E (m,zl), and since u(zµ) varies continuously for fixed z and varying p, we have w(z, m) < 0. This inequality can be sharpened to w(z, m) < 0 for z E Z(m),
(3.10)
by means of (3.6), (3.7) and the strong maximum principle; the argument is like that in (ii)(b).
(b) Let 6 be the small positive number, independent of it E [0, M), inferred from Theorem 2.19 and introduced before (3.8). Let F C Z(m) be a compact set satisfying I Z(m) \ FI <
ZS
(Figure 3.2); the existence of
3.1 The simplest case
91
F
F as2
T.
TM
Tm_E
Fig. 3.2.
such an F follows from the representation of Z(m) as a countable union of dyadic cubes (Exercise 3.14). Then
w(z,m) < -c for all z E F and for some c > 0,
(3.11)
by (3.10), by the continuity of w(. , m) and by the compactness of F.
(c) Now consider wE := w(., m - e) for 0 < e < so, and let GE Z (m - e) \ F. The number eo is to be so small that I GE I < 6, that m - e > 0 and that wE(z) < -Zc for z E F. This last is possible by the uniform continuity of u on 0. We apply the maximum principle for thin sets (Theorem 2.19) to the set GE, the operator A + y(., m - e) and the function wE; the data diamZ(0), Iy(z,µ)I < A and S used in (ii)(a) can be used again here. By (3.7) and because wE < -Zc on F we have wE < 0 on BGE; therefore Theorem 2.19 implies that wE < 0 on GE. Consequently WE < 0 in Z (m - e); then, by the strong maximum principle once more
[applied as in (ii)(b) to each component of Z(m - e)], wE(z) < 0 for z E Z (m - e). This contradicts the maximality of the interval (m, M) with m > 0, and thus completes the proof that (*) holds for all u E (0, M). (iv) That 01u < 0 in Z(0) will now be inferred from the boundary-point lemma for balls (Lemma 2.12). Given p E S2 with p1 > 0, we set p = p1
92
3 Symmetry for a Non-linear Poisson Equation
and observe that there is a ball B := M(p + pe1, p) in Z(p) [because S is open, hence contains a ball M(p, 2p)]. Now w(., p) is a C2-solution relative to A+ y(., p) and B; also, w(z, p) < 0 for z E B while w(p, p) = 0. It follows from Lemma 2.12 and Remark 2.16 that -(alw)(p, p) > 0. This proves that (alu)(p) < 0 because the identity
aiw(x, p) = as { u(x) -u(2µ - xl, x")
(al u) (x) + (alu) (2p - xl, x")
l1
(3.12)
shows that (alw) (p, p) = 2 (alu) (p) when pl = p. Corollary 3.4 Under the hypotheses of Theorem 3.3, u is an even function
of xl: u(-xl, x") = u(xl, x") for all x E S2.
(3.13)
Proof Fix z E Z(0) and let p j 0. Since u(z) < u(zµ) for p E (0,zl), and since u(zu) varies continuously for fixed z and varying p, we obtain u(z) < u(z°) for z E Z(0). Then, by continuity of u on S2,
u(xl, x") < u(-xi,x") for xl >- 0 and x E 1. But, by applying to points of Y (O) the various arguments that we have used for points of Z(0) (in other words, by moving T. from p = -M to p = 0 rather than from p = M to p = 0), we obtain
u(-xl, x") < u(xl, x") for xl >- 0 and x E S1 Together, these two inequalities imply (3.13).
Corollary 3.5 If in Theorem 3.3 the set S is a ball, say KI = R(0, a) c then u is spherically symmetric (depends only on r := IxI) and du/dr < 0
for 0 < r < a. Proof (i) Let k be a given unit vector (k E RN and Jkl = 1); let R(k) be an orthogonal N x N matrix such that the transformation z = R(k)x
yields zl = x k; let u(z) = u(x) under this transformation. Then u satisfies the hypotheses of Theorem 3.3, hence is an even function of zl by Corollary 3.4. In the notation of Definition 1.7 we have u(x°") = u(x) for all x E RN, provided that we set u(x) := 0 outside R(0, a). This holds for every unit vector k; therefore Lemma 1.8 establishes the spherical symmetry of u.
3.2 A discontinuous non-linearity f
93
(ii) That du/dr < 0 for 0 < r < a follows from (3.4); indeed, now that u is known to depend only on r, it suffices to have (3.4) on a single ray from the origin. 0
3.2 A discontinuous non-linearity f
There are two reasons for relaxing the condition in Theorem 3.3 that f be Lipschitz continuous. First, functions f with simple discontinuities arise naturally in problems of Newtonian potentials, steady vortex flows and magnetohydrodynamics (§4.4 and Exercises 3.16 and 4.21 to 4.24). Second, although Theorem 3.3 and its corollaries serve well as introduction to a strategy of proof, they do not show how robust and flexible the method is. The elliptic equation (3.6) satisfied by w was perhaps the most important ingredient of the proof of Theorem 3.3; in this section we shall see that Lemmas 3.7 and 3.8, which are mere shadows of equation (3.6), are quite enough to prove the results of the previous section for a less pleasant function f. Notation We continue to use the notation described after Definition 3.2. In particular, it is implicit that the unit vector k = el. Theorem 3.6 Let SZ be bounded, connected and Steiner symmetric relative to the hyperplane To. Assume that u : S2 -+ R has the following properties. (a) u E C(n) n C1(KI), u > 0 in 11, ul an = 0. (b) For all cp E CC°(S2f),
J
(3.14)
{
where f has a decomposition f = f 1 + f2 such that f 1
: [0, oo) --+ R is locally Lipschitz continuous, while f2 : [0, oo) -+ IR is non-decreasing and is identically 0 on [0, K] for some lc > 0.
Then the previous conclusions hold:
u(z) < u(z") 81u(x) < 0
if µ E (0, M) and z E Z(µ), if
x1 > O and x
J.
(3.15) (3.16)
Remarks 1. Equivalent hypotheses on f. Consider the conditions: f = 91 + 92, where g1 [0, oo) -+ R is locally Lipschitz continuous, while :
$2 : [0, oo) -> R is non-decreasing and is uniformly Lipschitz continuous
on [0, K] for some x > 0. These conditions are not more general than
3 Symmetry for a Non-linear Poisson Equation
94
those in the theorem: let $3(t)
I
Sl
g2(t)
if 0 < t < K,
$2(ac)
if t > K,
and define f, := 91 + g3, f2 := 92 - 932. Upward jumps and downward jumps of f. The function f in Theorem 3.6 may 'jump upwards' (as its argument increases) in that the hypotheses allow simple discontinuities
f (c+) - f (c-) > 0,
where f (c+) := limtlc f (t), f (c-) := limtTC f (t),
(3.17)
and where c > K [with f (K) = &-)]. If f jumps downwards, then the result need not hold (Exercise 3.19). Indeed, if f is merely Holder continuous [see remarks preceding Definition A.10] at a point of an interval in which it decreases, then (3.15) and (3.16) may be false (Exercise 3.18).
3. Notation and terminology. Let d(x) := dist(x, OQ) for points x E S2. The condition uI an = 0 and the uniform continuity of u on '2 imply that
3h > 0 such that d(x) < h
u(x) < x;
(3.18)
we define
Zh() := Z(p) n j x E 0 I d(x) < h }.
(3.19)
The phrase in the generalized sense will be a slight extension of the terminology in Definition 2.10; its meaning is displayed in Lemma 3.7. As before, w(. , p)
:= u - uµ on Z(µ) for µ E [0, M).
4. Method. Since equation (3.6), which was basic to the proof of Theorem 3.3, is no longer available, we must make do with the substitutes in the following two lemmas.
Lemma 3.7 If it E [0, M), then Aw + y(z, µ)w >- 0 in Zh(µ) in the generalized sense,
(3.20a)
where w = w(z,,u) and y is defined as before but with f 1 replacing f . That is,
fZh()
dz>0 whenever cp E C,°(Zh(p)) and q > 0,
(3.20b)
3.2 A discontinuous non-linearity f Y(., µ)
f1(uu
=
- u1(up)
0
95
at points z where u(z) uµ(z), at points z where u(z) = u, (z). (3.20c)
Proof (i) In order to derive an equation for w that will be useful both for this lemma and for the next, we abbreviate Z (p), Y (y) to Z, Y and proceed from for all cp E C,(Z ),
f{_vW Vu + Wf (u) } dy = 0
for all W E CC°(Y ).
Given cp E C°(Z), we choose W = cpµ [so that suppp c Y] and set y = zµ. Then a/ayl = -a/azl and a/aye = a/az; for j = 2,...,N; also, W(y) = (pp(y) = (p(yµ) = (p(z) and u(y) = u(zµ) = u,,(z), whence
f{_VVu,+f(u)} dz = 0 for all
E C°°(Z).
Subtract this equation from the equation for u(z); then { -VT Vw+4p[f1(u)-fi(uµ)] } dz
J
(3.21)
z
(p [f2(u) - f2(uN,)] dz
for all cp E CC°(Z).
(ii) Now restrict attention to cp E C,°(Zh) with cp >- 0, where Zh = Zh(µ),
and integrate only over Zh in (3.21). The definition (3.20c) of y shows that, on the left-hand side of (3.21), f i(u) - fi(uµ) = A. , µ)w.
On the right-hand side, f2(u) = 0 because u(z) < K in Zh, by (3.19) and (3.18), while f2(uµ) >- 0 because f2 >- 0 on [0, oo). Thus
- IZ, ( [ f2(u) - f2(u) ] dz > 0 whenever 4P E C(Zh) and cp >_ 0, from which (3.20b) follows.
Lemma 3.8 If u E [0, oo) and w(z, µ) < 0 for all z E Z(µ), then Ow - Aw >- 0
in Z(µ) in the generalized sense;
(3.22)
96
3 Symmetry for a Non-linear Poisson Equation
here w = w(z, p) and A is a Lipschitz constant for f 1 on the interval [0, supra u]. Thus A Z 0 and A is independent of p.
Proof In (3.21) we have
fl(u)-fl(uµ) <_ AIu-u4 = -Aw, by the definition of A and because w(.,p) < 0 in Z(p), and f2 (u) - f2(uµ) < 0,
because u < uµ in Z(p) [by hypothesis] and f2 is non-decreasing. Acwith cp Z 0, cordingly, for all ( E
- fz V9 - Vw dz
= >-
-f P{fl(u)-fl(uµ)+f2(u)-f2(uµ)} dz z
Jz
(3.23)
cpAw dz,
and this is (3.22).
Proof of Theorem 3.6 (i) As before, our main task is to prove that
z E Z(p)
w(z, p) = u(z) - u(zµ) < 0
(*)
whenever p E (0, M). Again we begin with small values of M - p, and proceed very much as in the proof of Theorem 3.3, step (ii). (a) In order to apply the maximum principle for thin sets (Theorem 2.19) to the set Zh(p), the operator A + y(z, p) and the function w(. , p), we choose data A and 6 independent of p E [0, M), as in the earlier proof. Of course, f, replaces f ; the constant A is as in Lemma 3.8, so that Iy(z,p)I <_ A; and diamZ(0) serves as the diameter in Theorem 2.19. If M - h < p < M, then Z (p) = Zh(p); therefore Lemma 3.7 applies to Z (p). We have boundary values w(z, p) < 0 for z E aZ(p); consequently, by Theorem 2.19,
M - h< p < M and IZ(p)I < 6
w(z, p) < 0 for z E Z(p).
(3.24)
(b) With p as in (3.24), let Zo(p) be an arbitrary component of Z(p). Exactly as before, there are points p E aZo(p) \ T, such that pµ E 0, so that w(p, p) < 0. In view of (3.24), Lemma 3.8 now applies. Then the strong maximum principle (Theorem 2.13), this time for a generalized subsolution relative to A - A and Zo(p), shows that
M - h :s; p < M and IZ(p)I < 6
w(z, p) < 0 for z E Z(p). (3.25)
3.2 A discontinuous non-linearity f
97
(ii) Let (m, M) be the largest open interval of p in which (*) holds; assume (for contradiction) that m > 0. (a) Fixing Z E Z(m) and letting p 1 m, we obtain [precisely as in the proof of Theorem 3.3, step (iii)(a)] that w(z, m) < 0. Then Lemma 3.8 applies for p = m and shows that w(. , m) is a generalized subsolution
relative to A - A and to each component of Z(m). As before, each component of Z(m) has boundary points p at which w(p,m) < 0; the strong maximum principle (Theorem 2.13) shows that
w(z, m) < 0 for z E Z (m).
(3.26)
(b) Next, we prove that
71u<0 on Z(m)\852.
(3.27)
Given p E Z(m) \ 752, we set y = pl ; there is a ball R(p + 9e1, 9) in Z(p)
for some 9 > 0 and w(z,p) < 0 in this ball because p = pi >- m. The boundary-point lemma for balls (Lemma 2.12), applied as in the proof of Theorem 3.3, step (iv) [but now for a generalized subsolution relative to A - A], shows that (01u)(p) = 11(01w)(p, p) < 0.
(c) Letting wE := w(.,m-s) for 0 < E < Eo, as before, we wish to obtain a contradiction by showing that wE < 0 in Z(m - e) if Eo is sufficiently small. We continue to imitate the proof of Theorem 3.3, but a further construction is required because Lemma 3.7 applies only to sets near 751, while the hypothesis w(., p) < 0 of Lemma 3.8 remains to be proved for p < m. The new step is first to observe that (3.27) extends to certain subsets of Z(m - p) for some p > 0, and then to introduce a cylindrical set
EE := (m - E, m + i] XS, T>0, S (--R N-1, EE c Z(m - E),
(3.28)
as well as a compact subset F of Z(m); see Figure 3.3. We can prove that wE(z) < 0 for z E EE by integrating 71u from z'-E to z, provided that {m} x S is a compact subset of f1 0 and that i and so are sufficiently small.
Let GE := Z (m - E) \ (EE U F). We shall prove in (iii) that EE and F can be so chosen that, for 0 < E <- Eo and for some Eo > 0,
we < 0 on EE,
GE cZh(m-E),
wE < 0 on F, IGEI <6.
(I), (II)
(III), (IV)
98
3 Symmetry for a Non-linear Poisson Equation
E
F
E
F
an TM
Tm_E
Fig. 3.3.
Here S is the positive constant, independent of u E [0, M), inferred from Theorem 2.19 and introduced in step (i)(a). We are now on a familiar path to the desired contradiction. Conditions (III) and (IV) allow application
to wE and GE first of Lemma 3.7 and then of Theorem 2.19; for the latter, we have wE < 0 on 3G, by (I), (II) and by the usual conditions on OZ (m - E). Therefore wE < 0 on GE, whence wE < 0 on Z(m - E), with strict inequality on EE and F. By Lemma 3.8, wE is now a generalized subsolution relative to A - A
and to each component of Z (m - E); provided that m - E > 0, each component of Z(m - E) has boundary points p at which wE(p) < 0; consequently, the strong maximum principle (Theorem 2.13) shows that wE < 0 in Z(m-E). This contradicts the maximality of the interval (m, M) with m > 0. (iii) Here is a detailed description of EE and F that leads to conditions (I) to (IV).
(a) Let Sm := {m} x S, where S is a closed subset of RN-1 such that Sm c (Tm n 0) and such that, in the hyperplane Tm, all points between Sm and an are sufficiently near an; say x E (Tm n n) \ Sm
dist(x, T. n On) < 1h.
(3.29)
3.2 A discontinuous non-linearity f
99
The existence of such a set Sm follows from the representation of Tm n 91
as a countable union of dyadic squares, that is, of (N - 1)-dimensional dyadic cubes (Exercise 3.14).
By (3.27) and the continuity of alu on the compact set Sm, there is a number c = c(S) > 0 such that 81u < -c on Sm. Moreover, 81u is uniformly continuous on each compact subset of 0. Therefore, there exists a cylindrical set [m - p, m + a] x S within fl, with p > 0 and a > 0, such that
81u < 0 on [m - p, m + a] XS.
(3.30)
In view of (3.27), the number a is restricted only by the need to have this set within fl. Now, for every z > 0,
m-Zx>-m-2z. Hence in the definition (3.28) of EE it suffices for condition (I) that 0 < i < a and Zr < p, and then that 0 < so < Zr; for it follows that, when z E E, the line segment from zm-E to z is in the set [m-p, m+a] x S, so that wE < 0 on EE, by (3.30).
(b) Let E' := EE n Z(m); the sets E' and F are to be independent of E. The complement of E' in Z(m) \ Z(m + i) is, or can be made, small in two senses. First, (3.29) and the fact (made precise in Exercise 3.13, (iii)) that Tµ n f2 cannot widen as y increases through positive values, imply that
z E { Z(m) \ Z(m + z) } \ E' - d(z) < 1h.
(3.31)
Second, if i is sufficiently small, then
{Z (m) \ Z (M + i)} \ El < 16.
(3.32)
The number i is now chosen so that the condition (I) and (3.32) hold; S and i are fixed henceforth, but so may be reduced further.
(c) Condition (II) will follow, as in the earlier proof, from the compactness of F and the uniform continuity of u(zµ) for fixed z and varying p, provided that so is sufficiently small. We now choose F so that (III) and (IV) hold for sufficiently small so.
Let F denote, for n E No, the union of all (closed) dyadic cubes of edge-length 2-" (Exercise 3.14) that are subsets of the open set Z(m). Then N1/22-n whenever x E Z(m) \ F,,; also, I Z(m) \ FnI -* 0 dist(x, 8Z(m)) <
3 Symmetry for a Non-linear Poisson Equation
100
as n -+ oo. Therefore we can choose k to be so large that
x E Z(m) \
min{ h,i },
(3.33)
and
I Z(m) \ FkI < 46.
(3.34)
Define F := Fk \ int E'. [The definition F := Fk would also serve and would not change GE, but the interiors of E' and F would intersect] We argue as follows to establish (III) and (IV) for sufficiently small 8o. (d) Consider (III). For Z E GE \ Z(m), contemplate the point (m, z"). If (m, z") E S2, then, by (3.29),
d(z) < I z - (m, z") I + dist ((m, z"), act) < s + 1 h. If (m, z") 91, then there is a point of 8Q on the closed line segment from z to (m, z"). In either case, d(z) < h if Eo < h. Z For Z E GE n { Z(m) \ Z(m + i) }, (3.31) states that d(z) < 1h.
For z E GE n Z(m + i), we have z E Z(m) \ Fk and zl > m + 2; by (3.33), a point of 7Z(m) that is nearest z must be in 0 rather than in Therefore d(z) < h, by (3.33). (e) To prove (IV), we observe that IGE \ Z(m)I < 16 if go is sufficiently small, and that
IGEnZ(m)I
(iv) That alu < 0 in Z(0) is proved as before [proof of Theorem 3.3, step (iv)], except that now w(. , µ) is a generalized subsolution relative to
0 - A and a suitable ball. Corollary 3.9 (a) Under the hypotheses of Theorem 3.6, u is an even function of xl :
u(-xl, x") = u(xl, x") for all x E S2.
(b) If in Theorem 3.6 the set 0 is a ball, say 0 = -4(0, a) c IRN, then u is spherically symmetric (depends only on r := lxi) and du / dr < 0 for
0
Proof The proofs of Corollaries 3.4 and 3.5 serve here also.
3.3 Exercises
101
Fig. 3.4.
3.3 Exercises
Exercise 3.10 In Corollary 3.5 the conclusion du / dr < 0 for 0 < r < a cannot be extended to r = a, even when the hypothesis u E Cl(S2) is added. Demonstrate this by means of the set 92 := R(0,1) c R2 and functions f and u as follows.
f(t)
a2(t-y) for all t > 0,
u(x) := Jo(/3r)+y
for r := lxi < 1,
where the Bessel function Jo is described by Exercise 1.19 and Figure 3.4, fi denotes the smallest positive zero of JO' and y := -Jo(/J).
Exercise 3.11 Let n := .4(0, 1) c RN for N = 1 or 3. Find analogues of the functions f and u in Exercise 3.10. [Fragments of Exercise 1.21 are relevant for N = 3] Exercise 3.12 Adopting the notation introduced after Definition 3.2, give an example of a set S2 c R2 that is as in Theorem 3.3, has 812 of class Coo and is such that the cap Z(µ) has infinitely many components for some value u E (0, M).
Exercise 3.13 Let S2 be bounded and Steiner symmetric relative to To = To(e'); here we use Definition 3.2 and the notation introduced after it. Prove the following. (i) 0 is Steiner symmetric relative to To. (ii) For an arbitrary component Zo(µ) of the cap Z(µ), where 0 < µ < M, there exist boundary points p E 8Zo(p) \ T. such that pµ E S2.
3 Symmetry for a Non-linear Poisson Equation
102
[Given any point zo E Zo(u), define p = zo + rlel, where q is the smallest positive number such that p E OS2.] (iii) If x, y E S2 and yi > x1 > 0, y" = x", then dist(y, 892) < dist(x, 892).
[Assume the contrary and consider the largest balls in 92 with centres x and y.] (iv) The boundary 892 need not be of class C.
Exercise 3.14 By a dyadic cube in RN we mean a set of form
Qn(P) = { x E RN I pj2-" < xj < (pj + 1)2-" for j = 1,...,N }, where p = (p1,...,pN), each pj E Z and n E No. [Thus a dyadic cube is closed, has edges of length 2-n for some n E No, and has faces on each of which a co-ordinate is constant and equal to an integral multiple of 2-n .]
Given a bounded open subset 0 of RN, define Fn to be the union of all dyadic cubes of edge-length 2' that are subsets of Q. Prove the following.
(i) If X E 92 and dist(x, 892) > N1/22-" (where n E No), then x E Fn. (ii) Fn c Fn+1 and S2 = Un o Fn. (iii) 92 is a countable union of dyadic cubes. (iv) In \ FnI 0 as n co. Exercise 3.15 We seek all functions u E C(ST) n C 1(92), other than the zero
function, that are generalized solutions (Definition A.7) of
u"+).fH(u-1)=0 in 92:=(-1,1)c]R,
Ulan = 0,
where A, is a positive constant and fH is the Heaviside function (see Chapter 0, (v)). Prove the following.
(a) For A. > 4 there are exactly two non-trivial, generalized solutions; they involve the points 2 (1 +
a1 i a2
1 - 4/2)
respectively,
and are, for j = 1, 2,
uj(x) =
1+x
if-1 <x<-aj,
1 + 22(a? - x2)
if -aj < x < aj,
1
-x
T_- aj
if aj<x<1.
3.3 Exercises
103
(b) For 2 = 4 there is exactly one non-trivial, generalized solution. (c) For 0 < 2 < 4 there is no non-trivial, generalized solution.
To obtain bonus marks, prove (a) to (c) without appeal to Theorem 3.6 or Corollary 3.9. [Theorem B.6, applied to u(x) wherever u(x) < 1 and to u(x) + 22x2 wherever u(x) > 1, implies a priori that u E C°°(G), where G xE
(-1,1)
1
u(x) 1 1 }.]
Exercise 3.16 The solution of this exercise becomes, under the transformation in Exercise A.25, the Stokes stream function of `Hill's vortex in a ball'. Exercise 4.23 will explain this phrase and will account for the strange notation adopted here. However, the reader needs no knowledge of all this in order to do the exercise. Let B := R(0, b) c R5. We seek all functions w E C(B) n C1(B), other than the zero function, that are generalized solutions (Definition A.7) of
Lw + 2f H (w -' U) = 0 in B,
wan=0,
where 2 and U are positive constants and f H is the Heaviside function (see Chapter 0, (v)). It will be convenient to define 15U 2b2a2(1 - a3)
AO = min o<«
_
75 6
2
-2/3 U
62.
(5)
Use Corollary 3.9 to prove the following.
(a) For 2 > 20 there are exactly two non-trivial, generalized solutions; they involve the two solutions al and a2 of the equation 2(
1 - a3) =
15U
2Ab2,
0
with the notation r :_ lxi, a := alb when a = al, or a := alb when a = a2, the two solutions are
W(X) _
Ua3 (2 - a3 - 2 r
\
2
-21 1
Ua3 (X3
-a33
2)\
if r < a,
ifa
(b) For 2 = 20 there is exactly one non-trivial, generalized solution. (c) For 0 < 2 < 20 there is no non-trivial, generalized solution. [A remark like the hint for Exercise 3.15 applies here.]
3 Symmetry for a Non-linear Poisson Equation
104
Exercise 3.17 Suppose that in Exercises 3.15 and 3.16 we do not prescribe A, but rather prescribe the `energy norms' r
1/2
1
f u'2 dx }
=n>0
{ '
(lull
l
(Exercise 3.15),
J
1
l 1/2
Iow12 dx }
11wII fB
=n>0
(Exercise 3.16),
JJJ
and calculate 2 a posteriori. Prove that each problem has exactly one non-trivial, generalized solution for prescribed q > 0. Exercise 3.18 Consider, for arbitrary N E N,
Lu+f(u)=0
inB :=-4 (0,r6-) cIIBN
(3.35)
0, U I OB =
where, for a constant p > 2, 4p(p-1)(1-t)1-(2/p)+p(N+2p-2)(1-t)1-(1/0
if
0
f(t) -4p(p -1)(t -1)1-(2/p) + 2p(N + 2p - 2)(t
-1)1-(1/P)
if t > 1. (3.36)
If p is large, does f fall far short of the local Lipschitz continuity demanded in Theorem 3.3?
Verify that there is a solution u(.,c) E C2(B) as follows for each c E M(0,1), so that the conclusions of Theorem 3.3 and its corollaries are false in this case. u(x, c) =
1+(1-p2)p
if p := Ix - cl < 1,
1
if p> 1 and r := lxl < 2,
1-(Zr2_2)p
if 2
Write further solutions of (3.35) with (3.36).
Exercise 3.19 Consider the problem (3.35) with the change that now
f (t) :=
8 + 2(N + 2)(1- t)1/2
if 0 < t < 1,
0
if t = 1,
-8 + 4(N + 2)(t -1)1/2
if t > 1.
Why does this function fail to satisfy the hypotheses on f in Theorem 3.6?
3.3 Exercises
105
Exhibit a generalized solution u(.,c) E C(B) n C'(B), for each c E R(O,1), that shows the conclusions of Theorem 3.6 and Corollary 3.9 to be false in this case. [Generalized solutions are introduced in Definition A.7.]
4
Symmetry for the Non-linear Poisson Equation in RN
4.1 Statement of the main result
This chapter continues exploration of symmetry for the equation Du + f (u) = 0; the function f is very much like that in Theorem 3.6 and may jump upwards as its argument increases. However, the equation now governs u on the whole of ][8N; relative to the situation when 0 is a ball, we lose both the boundary condition u = 0 on a and the knowledge that, if u is spherically symmetric, then the centre of symmetry of u must be the centre of the ball 91. It turns out that this loss of information is tolerable if the behaviour of u(x) as xJ -* oc is admissible in the sense of Definition 4.1. Such admissible asymptotic behaviour supplies
both a unique point as a possible centre of symmetry and a means of establishing, outside some large ball, the monotonicity condition (*) which was our main objective when proving Theorems 3.3 and 3.6. By the end of Lemma 4.6, appropriate behaviour outside a large ball
will have been demonstrated; two methods are then available for a proof of monotonicity and of spherical symmetry on all of RN. The first method, which we adopt in the main text of the chapter, makes no appeal to the maximum principle for thin sets. The second method, which is the subject of Exercise 4.27 and 4.28, imitates the proofs of Theorems 3.3 and 3.6; this is possible, under an additional condition on discontinuities of f, because the monotonicity condition has already been established outside a bounded set. Corollary 4.10 returns to a more modest symmetry: if f (u(x)) is replaced by f (u(x), x") in the non-linear Poisson equation, where f (., x")
resembles the previous f for all x" E ]RN-1, and if u has admissible asymptotic behaviour, then u enjoys the earlier monotonicity and sym106
4.1 Statement of the main result
metry relative to a hyperplane { x
I
107
xl = const. } that can be located a
priori.
Definition 4.1 Let u E C'(R'). We shall say that u has admissible asymptotic behaviour if one of the following conditions holds outside some ball, say for r :_ xI >- Ru. The numbers K, m, b and y are all [strictly] positive constants, with b E (0,1], and c = (c1,. .. , CN) is a constant vector. (A)
u(x) = -Kr"' + (c x)rri-2 + h(x),
where lVh(x)I = 0
(rni-2-b) and, if m-1-6 < 0, then h(x) -+ 0 as
r -+ co.
u(x) = K log Y + (c x)r-2 + h(x), r where IOh(x)I = O(r-2-6 ) and h(x) 0 as r -+ oo. (B)
(C)
u(x) =
Kr-"
+ (c
x)r-,,,-2
+ h(x),
where IOh(x)I = O(r-ni-2-6 ) and h(x) ---> 0 as r -* oo.
The essential features of Definition 4.1 are these. First, u(x) decreases as r -+ oo; that u(x) -+ -oo in cases (A) and (B), while u(x) -+ 0 in case (C), is less important. Second, if c * 0, then c x does not depend only on r. Therefore, if u should turn out to be spherically symmetric about some point q E RN, then u(x + q) will contain no term (c x)r" (where a E R is a constant). This enables us to find a priori the only point q that can be a centre of symmetry of the function u. Theorem 4.2 Suppose that a function u : RN -* R has the following properties.
(a) u E C1(]R^'); it has admissible asymptotic behaviour; and u > 0 on RN in case (C). (b) For all 'p E Cc (RN), J N {-Otp Vu + cpf (u)} = 0,
(4.1)
where f has a decomposition f = f 1+ f2 such that f, : R-+ R is locally Lipschitz continuous, while f2 : R -> R is non-decreasing. In case (C), f 1 and f2 need be defined and need have these properties only on [0, oo).
4 Symmetry for the Non-linear Poisson Equation in RN
108
Let
_ q:
1 Km
c
1
-c K
for cases (A) and (C),
v(x) := u(x + q). (4.2)
for case (B),
Then v is spherically symmetric (depends only on r) and dv/ dr < 0 for
r>0. The proof of Theorem 4.2 will take the form of five lemmas and a proposition. A first step is to verify that, under the transformation (4.2), the function v has every good property of u and has better asymptotic behaviour in that the second-order terms in the expansion of u are absent from the expansion of v. Lemma 4.3 The function v in (4.2) satisfies conditions (a) and (b) of Theorem 4.2, but with improved asymptotic behaviour in that one of the following conditions holds for r >- R + 2IgI _: R, . As before, K, m, S and y are positive constants, with S E (0, 1]. (A)
v(x) = -Krm + g(x),
where IOg(x)I =
O(r"i-2_6),
and, ifm-1-8<0,then g(x)-->0as r->co. (B)
v(x) = K log r + g(x),
where IOg(x)I = 0 (r-2-S)
,
and g(x)-* 0as r-+oo. (C)
v(x) = Kr-' + g(x),
where IVg(x)I = O(r m-2-s),
and g(x)->0as r->oo. Proof (i) It is clear from (4.2) that v E C1(R") and that v > 0 on RN in case (C).
(ii) Here is a merely formal calculation showing the removal of terms (c x)r" in Definition 4.1. For case (A), -K(r2 + 2q x + IgI2)mJ2 + C (x + q)Ix +
qlni-2
2+ so that terms of order rm-1 are cancelled by the choice q = c/Km. For case (B),
KlogIx+gl
2]}
4.1 Statement of the main result
=x
log
Y
r
- (q x)r-2 +
109
>
so that terms of order r-1 are cancelled by the choice q = c/K. For case (C), qI-m
iclx +
= xr-' { 1 - m(q x)r-2 + ... },
so that terms of order r-ni-1 are cancelled by the choice q = c/xm. (iii) To show that v satisfies (4.1), we suppose that any V in Cc(RN) is given. Choose cp(y) = yp(y - q) in (4.1); then { -(V1V)(Y JRN
- q) (Vu)(y) + 1V (y - q) f (u(y)) } dy = 0,
and the substitution y = x + q yields (4.1) for W and v.
(iv) For a rigorous justification of the claims made for Vg and g, we define functions F, G and H by Ix + qIa =: ra + a(q 1
+
log Ix
=: log gI
c (x + q) I x + q I
a-2
x)ra-2 + F(x,
a),
1
r - (q - x)r-2 + G(x),
=: (c . x)ra-2 + H(x, a),
where a E R is an arbitrary constant. Then, with h still denoting the remainder function in Definition 4.1, g is given by one of the formulae
g(x) _ -iF(x, m) + H(x, m) + h(x + q), g(x) = iG(x) + H(x, 0) + h(x + q),
g(x) = rcF(x, -m) + H(x, -m) + h(x + q). Accordingly, to prove the claims made for Vg it is sufficient to show that, for each k E { 1, ... , N}, a F(x, a) axk
< const. ra-3,
aak G(x) < const. r-3, a H(x, a) axk
< const. ra-3,
where the constants may depend on a, q, N and k, but not on x. To this
110
4 Symmetry for the Non-linear Poisson Equation in RN
end, let cp(t) := Ix + tq la, 0< t< 1; then the definition of F becomes
= (p(1) - (p(0) - cp'(0) = 1(1 - t) T "(t) dt
F(x, a)
0
N
1
f(1 - t)
2
jEqj7j lx+tql°
dt,
(4.5a)
j=1
0
where x and q are fixed and a; means, as always, differentiation with respect to the jth argument of the operand. Similarly, N
E qj8j
(1 - t)
G(x) _
2
dt,
(4.5b)
{ c (x + tq) lx + tq I a-2 } dt.
(4.5c)
log
(j=1
Ix + tql
N
H(x, a) =
E qjaj
J
1=1
It is sufficient to justify differentiation with respect to xk under the integral sign for 21gI < r < M and each M. We may suppose that IgI > 0 (otherwise, there is nothing to prove), and then the bounds
0<
IqI
< Zr < x+tqI <
Zr < 2M
(4.6)
imply uniform continuity of each differentiated integrand.
The desired result (4.4) is now implied by the following estimates, which are slight extensions of Lemma A.4 and are proved in the same way. For any y E RN \ {0} and any multi-index $ (Definition A.3), I&Iylal lap log IYI
Iap(c'
<
(4.7a)
const.lyla-I/tl
<
const. IYI-ICI
<-
const.
if I$1 >- 1,
(4.7b)
I
y)IYIa-2I
lyla-1-IBI;
(4.7c)
the Leibniz rule (Exercise A.23) can be used for derivation of (4.7c), and the constants may depend on everything except y. To prove (4.4), we first set y = x + tq and then use (4.6).
(v) Regarding the claims made for g itself: in view of (4.3) it is enough to prove that, if a < 1 + S, then F(x, a), G(x) and H(x, a) all tend to zero as r -> oo. This is the case; since 6 < 1, we have a < 2,
4.2 Four lemmas about reflection of v
111
and (4.5a) to (4.7c) show that F(x, a) = O(ra-2), G(x) = O(r-2) and H(x, a) = O(ra-2).
4.2 Four lemmas about reflection of v Definition 4.4 Let k be a given unit vector (k E RN and Iki = 1), and let N
ER,j (k) xj
x,
(i = 1,...,N),
j=1
where R(k) is an orthogonal N x N matrix such that x1 = x k. By variables aligned with k we mean
(a) the co-ordinates x1 and x" := (x2,-,5W; (b) the description Tµ :_ { c E RN I (Definition 1.7); (c) the description zµ
1 = p } of the hyperplane T, (k)
(2p - x1, x") of the reflected point xµ,k;
(d) the version gyp, such that i%p(x1,...,. ) = io(R(k)x) = cp(x), of any given function (p from RN into R;
(e) the open half-spaces k (p) := { x E RN
I
xl < p } and Z(p)
{ x E RN I xl > p } that are separated by the hyperplane Tµ.
Until the contrary is stated, k will be arbitrary but fixed ; we shall use variables aligned with k, but shall omit the tilde as long as there is no danger of confusion. Let v be the function in Lemma 4.3 (more precisely, the version of that function aligned with k); we wish to show that v(-xl, x") = v(xi, x") for all x E RN, and that 81v(x) < 0 whenever x1 > 0. If these conditions hold (Figure 4.1), then p>0
and
Y E Y(p)
v(yµ) < v(y),
and we begin by showing that this inequality holds whenever Iyi is sufficiently large. The four key results of this section will be labelled by Roman numerals.
Lemma 4.5 There is a number R(p), depending only on v and p, such that
p > 0,
y E Y(p)
and
IYI > R(p)
v(Y1) < v(Y)
The function R is non-increasing on (0, 1] and is constant on [1, oo).
(I)
112
4 Symmetry for the Non-linear Poisson Equation in RN
v (x)
x Fig. 4.1. x'
µ-h
µ+h
µ Fig. 4.2.
Proof (i) We consider only points outside the ball -4(0, as in Lemma 4.3, and adopt the notation (Figure 4.2)
z:=Y",
R is
h:=2(z1-Y1)>0,
s = Iy1= {(µ - h)2 + Iy"I2}112,
t := Izl = { (µ + h)2 + Iy"I2 }1/2,
so that
t2-s2=4µh>0,
t>s>--R,,.
(4.8)
The basis of the proof is the size of the gap, in the asymptotic description of v, between the leading term and the remainder function g.
4.2 Four lemmas about reflection of v
113
We shall demonstrate this for two of four cases: case (A) of Lemma 4.3 with m >- 2, and case (C) of that lemma. Case (A) with 0 < m < 2, and case (B), are left as exercises for the reader. The letters C1, C2,... will denote constants independent of y and µ, but possibly depending on v (and hence on N, m, ...). (ii) Case (A) with m >- 2 and N >- 2. By Lemma 4.3, m > 2,
v(Y) - v(z) = x(tm - sm) + g(Y) - g(z),
r >- R.
for
Cirm-2-s
IVg(x)I 5
We estimate separately for 4µh < s2 and for 4µh > s2.
(a) 4µh < s2. By convexity of the function r , Tm/2, 0 < T < oo, and by (4.8), tm - sm = (s2 + 4µh)m/2 - (s2)m/2 >
(52)m12-14µh = 2msm-2uh.
2 the two-dimensional plane To estimate g(y) - g(z), define PyZ to be containing the points 0, y and z (or any such plane if y" = 0), and define t to be the circular arc in PyZ from y to z, centred at (µ, 0). Then points x E IF have r = IxI E [s, t], and IF has length nh at most, so that Ig(Y) - g(z)I =
Jr
Vg(x) dxI <
C17Ctm-2-ah
if m - 2 - S >_ 0,
C1nsm`2-bh
if m - 2 - 8 < 0,
and t2 = s2 + 41th < 2s2. Hence for this case [(A) with m >- 2, N 4µh:5 s2], v
2 and
C2) > 0 if sa > 2km µ
V(Z) > sm-2-6h 2xm sa
(b) 4µh > s2. This condition implies that t2 > 252 and hence that f
tm - sm = tm { 1 -
r
()/2}
lm
\t/
> tm
1_
> 2 tm.
J
To bound g(y) - g(z), we observe that m - 1 - 8 >- 0 (because m >- 2 and 8 < 1). Reducing S slightly if necessary, we may suppose that m - 1 - 8 > 0 whenever m >- 2; then g can be estimated as follows by integrating radially outwards from 8-4(0, R(0,Re); we write ro := Ixol and g(xo) = g
-
( ro xo l + f 1olro R
Let xo be any point outside dS(Txo) dT
dT,
4 Symmetry for the Non-linear Poisson Equation in RN
114
where dg(Txo)
= Ixo - (Vg)(Txo)I <- Clro(iro)ni-2-b,
dT
so that Ig(xo)I
< maxr=R, Ig(x)I +
C1ro-1-61 I Tm-2-6 di Jo
< C3ro-1-a
(ro >_
if C3 is sufficiently large. Consequently, (g(Y) - g(z)I < 2C3t
and, for this case [(A) with m >_ 2, N >- 2 and 4µh > s2], v(Y) - v(z)
> tm-1-S (K t1+o
- 2C3)
2 tm-1-a J K (21/2s)1+a - 2C3
>0
J
23/2-6/2C3
if S1+6 >
C4-
K
(c) Collecting results, we define 1/a
C2
R(µ) := max R (2Kmmin{µ,1})
C1/(1+a) 4
'
(4.9)
and the lemma is proved for case (A) with m >_ 2 and N >- 2. (iii) Case (C) with N >- 2. By Lemma 4.3, v(Y) - v(z) = K(s m - t-') + g(y) - g(z),
log(x)I
C5r-m-2-1
<_
m > 0,
for r > R,.
Again we estimate separately for 41th < s2 and for 41th > s2.
(a) 4µh< s2. By strict convexity of the function T H T m/2, 0 < T < 00,
s m - tm =
(s2)-m/2
- (s2 +
4µh)-m/2
> 2 (s2 +
(252)-m/2-14µh = C6 s m-2µh.
2
Integration along F gives, essentially as before, Ig(Y) - g(z)I <- C57csn-2-6h.
4µh)-m/2-14µh
4.2 Four lemmas about reflection of v
115
Hence for this case [(C) with N > 2 and 4µh < s2],
v(Y) - v(z) > s m-2-S h (C6Kµs ° - CSn) > 0 if s S >
C5
C6 K/2
(b) 4µh > s2. Since now t2 > 2s2, s
-m
m/2
lm - t m =S m{1- (S) }>s-m1
=C7sm.
-(2)
Next, for any point xo outside R(0, R,) we obtain more easily than before dg(ixo)
Ig(xo)I =
C5ro m- 1-6
dT
m+l+S'
d2
1
so that 2C5
1 g(Y) - g(z)I <-
m+l}S
Sm -1-S
Then, for this case [(C) with N >- 2 and 4µh > s2], v(Y) - v(z) > s' 1-6 (C7Ks'+S
+
-
> 0
m + 1
ifs
1+S
2C5
Cg.
> C7K(m+ 1 + 6)
(c) Collecting results, we define
R(p) := max
Rn,
( C6K
1/S
C5
l} /J
Cg/(1+S)
(4.10)
and the lemma is proved for case (C) with N >- 2.
(iv) Modifications for N = 1. When the domain of v is R, we must modify certain estimates of Ig(y)-g(z)I because no path outside the ball (-R,,, connects y and z if y < -R,,. In other words, we have no substitute for the arc F when N = 1 and y < -R,,. However, if y >- R we can replace integration of Vg along F by integration along R from y to z. Also, integrals of Vg along radial lines from (R,/ro) xo to xo, or from xo to infinity, can be used as before. (For N = 1 and xO < -R0, the latter are from xo to -oo.) Since the arc F was used only for 41h < s2, we need change only estimates that are subject to both 41,&< s2,
equivalently,
t2 < 2s2,
and
y = p - h = -s < -R,,, equivalently, h = s + µ >- R + µ.
4 Symmetry for the Non-linear Poisson Equation in RN
116
Then, for case (C) with N = 1, by inequalities in (iii) that remain valid and by h > s, m- t m> C6 S ri-2µh > C6 s--n'-lµ, S g(y) - g(z) I
<-
SM-1-b'
2C5
m+1+6
whence v(Y) - v(z)
>S
`1_S (c6Ks -
2C5
m+1+6)
if sa >
>0
2C5
C6Kµ(m+1+8)
For case (A) with m >- 2 and N = 1, by inequalities in (ii) that remain valid and by h >s, t < 2112s, t' - S' >- 2msm-2µh > 2ms'-1µ, g(y) - g(z) I <_ 2C3
tni-1-S
(21/2s)m-1-S
< 2C3
C9 Sni-1-S,
whence
v(Y) - v(z) > sni-1-S (2xmµsa - C9) > 0
if S6 >
C9 2K µ
Lemma 4.6 There is a number µ* > 1 such that
µ > µ* and y E Y(µ)
v(y') < v(Y)
(II)
Proof (i) Let R(µ) be as in Lemma 4.5, so that R(µ) = R(1) for µ >- 1, and define R1 := max{ 1, R(1) },
c1 := min rsR, v(x).
[Here r := lxi, as before] (ii) For cases (A) and (B) of Lemma 4.3, choose µ* to be such that r >- µ*
v(x) < c1 - 1;
this is possible because v(x) --> -oo as r - oo. Then µ* > RI; Figure 4.3 illustrates the situation.
If µ >- µ*, y E Y(µ) and lyl > R1, then also lyl > R(µ) [because µ* > 1 and R1 >- R(1)]; Lemma 4.5 ensures that v(yµ) < v(y). If µ >- µ*,
4.2 Four lemmas about reflection of v
117
Fig. 4.3.
y E Y(µ) and IyI < RI, then v(y) >- cl [by definition of cl] and Iyµ) > µ* [because yl < Rl < p*] so that v(yµ) < cl - 1 < v(Y).
(iii) For case (C) of Lemma 4.3, choose y* to be such that
r > µ*
v(x) < 1 cl ;
this is possible because now cl > 0 and v(x) -+ 0 as r -+ oo. The rest is as in (ii), with Zc1 replacing cl - 1. O
Although all four results of this section will be needed to prove Theorem 4.2, it is the next lemma that is the heart of the matter. Observe that the generalized Poisson equation (4.1) is used only in Lemma 4.7,
and that the whole theorem rests on our ability to sharpen < to < in Lemma 4.7.
Lemma 4.7 Assume that for some y > 0 we have (a) v(yµ) < v(y) for all y E Y(µ), (b) v(Yo) * v(Yo)for some yo E Y(µ)
4 Symmetry for the Non-linear Poisson Equation in RN
118 Then
v(yµ) < v(y) for all y E Y(µ),
(III.a)
alv(x) < 0 for all x E Tµ.
(III.b)
and
Here (III.a) implies (III.b).
Proof Let Y := Y (µ) and Z := Z (p) for the given u, and let w := vµ - v [which means, according to Definition 1.7, that w(x) = v(xµ) - v(x) for all x E RN]. By hypothesis, w(y) < 0 for y E Y ; we shall prove strict inequality by means of the maximum principle. (i) A generalized Poisson equation for w. By (4.1) and Lemma 4.3, we have J N { -OW Vv + Wf (v) } = 0
for all W E C,°(RN).
(4.11)
Given cp E C,(Y), choose W = cpµ in this equation; then suppcpµ c Z and
-Ocpµ(z) . Vv(z) + cpµ(z) f (v(z)) } dz = 0.
Set z = yµ. Then y = zµ E Y, a/azi = -a/ay, and a/azj = a/ayj for j = 2,...,N; also, cpµ(z) = T(zµ) = cp(y) and v(z) = v(yµ) = vµ(y). Accordingly, jr { -V q (Y) - Vv (Y) + cp(Y) f (vµ(Y)) } dy = 0.
(4.12)
Now choose p = T in (4.11), and subtract the result from (4.12); then f { -Ocp Vw + (p [f (vµ) - f (v)] } = 0 for all
E C,00(Y).
(4.13)
(ii) Application of the strong maximum principle.
In order to control f, we consider balls in Y rather than the whole of Y. Let D :_ M(c, p),
c E Y,
p := min{ 1, dist(c, Tµ) },
so that D c Y. To estimate f (vµ(y)) - f (v(y)) at points y of D, we recall that f = f1 + f2 with f 1 locally Lipschitz continuous and f2 nondecreasing. By the continuity of v, there is a number M = M(c, v, µ)
4.2 Four lemmas about reflection of v
119
such that Iv(y)I < M and Iv(yµ)I < M whenever y E D; hence there is a Lipschitz constant A = A(M) > 0 such that, in D,
f1(vµ) - fl(v) < Alvµ - vj = -Aw
[since w < 0].
Also,
fz(vµ) - fz(v) < 0,
because vµ < v and f2 is non-decreasing. By addition of these bounds and by (4.13),
O=f {-V .Vw+w[f(vµ)-f(v)]} <_-f{-VVw-Aw} whenever p E Cc(D) and co >- 0; in other words, w is a generalized subsolution relative to A - A and D. Now suppose that w(c) = 0 = SUPD w at some c E Y. Then the strong maximum principle, Theorem 2.13, states that wID = 0, and this shows that the zero set of w, say X :_ { y e Y w(y) = 0 }, is open in Y. Since w is continuous, X is also closed in Y. Then, since Y is connected, I
either X = Y or X = 0; our hypothesis that w(c) = 0 at some c E Y implies that X = Y, which contradicts hypothesis (b) of the lemma. Consequently, X must be empty, and this proves (III.a). (iii) Application of the boundary-point lemma for balls. Let p E Tµ be given. Define
B :_ -4 ((pl - p, p"), p)
for any p E (0,1].
We know from (ii) that w is a generalized subsolution relative to A - A and B, and that w < 0 in B. In addition, w(p) = 0 because pµ = p. Then the boundary-point lemma for balls, Lemma 2.12, states that (al w)(p) > 0. Now, (01 w) (p) = -2 (01 v) (p) because
81 w(x) = xj { v(2µ - x1, x") - v(xi, x") } , so that (01v) (p) < 0. This proves (III.b).
Lemma 4.8 The set I ,u > 0
1
(Ill.a) holds : y E Y(µ)
v(yµ) < v(y)} is open in R. (IV)
120
4 Symmetry for the Non-linear Poisson Equation in RN
Proof Assume that the lemma is false. Then, for some value y = 2a > 0 at which (III.a) holds, there is a sequence (µn) in (a, 3a) such that (III.a) fails at each µn and µn 2a as n --+ oo. This failure of (III.a) must be for IyI < R(a), because (I) ensures that
it ? a, y E Y(µ) and IyI > R(a)
v(Yµ) < v(Y),
where R(µ) < R(a). Thus there are sequences (µn) and (yn) [here Yn is not a co-ordinate, but a point of RN] such that µn -* 2a, Yn E Y(µn), ynl < R(a)
and v (yn) >_ v(Yn)
By the relative compactness of (yn), and resort to a subsequence if need be, we may suppose that yn tends to some point q in . (0, R(oc)) as n - oo. Then q E Y(2oc) and v(g2a) > v(q). Since (III.a) holds at µ = 2a, we must have q E Tea. But (Ill.b) holds at µ = 2a [because (III.a) does]; hence (8iv)(q) < 0, and this contradicts v (yn^) >- v(yn) when n is sufficiently large.
4.3 Proof of Theorem 4.2 and a corollary
Proposition 4.9 The results (I) to (IV) imply that v is spherically symmetric
and that dv/dr < 0 for r > 0; that is, they imply Theorem 4.2.
Proof (i) Let (m, oo) be the largest open interval of µ in which (III.a) holds. This m exists, by (II) and (IV).
Assume that m > 0. We show that the hypotheses preceding (III.a) hold for y = m. First, yEY(m)=v(Y'n)<_v(Y),
by (Ill.a) for y > m and by continuity: fix y E Y(m) and let µ ] m. Second, by (I),
y E Y(m) and IyI > R(m)
v(Ym) < v(Y)
Therefore (III.a) holds at u = m. But then (IV) states that (III.a) holds in (m - e, oo) for some e > 0, so that (m, oo) is not maximal for (III.a). This contradiction shows that m = 0. (ii) We now have
v(xl, x") < v(-xl, x") whenever xl >- 0,
by (Ill.a) for µ > 0 and by continuity: fix -x1 < 0 and let p 10. [For xl = 0 we have equality]
4.3 Proof of Theorem 4.2 and a corollary
121
Keeping the co-ordinates xt = zt = x k and x" = z", which are aligned with k, we repeat the argument for the unit vector -k; this yields
v(xt, x") >- v(-xi,x")
whenever xl >- 0.
Thus v" is an even function of zt for arbitrary unit vector k, hence v is spherically symmetric, by Lemma 1.8.
That dv/dr < 0 for r > 0 is shown by (Ill.b).
The method also yields symmetry relative to a single co-ordinate, in place of spherical symmetry, for a wider class of functions f. In the following corollary there is only one system of co-ordinates (no alignment with an arbitrary unit vector). We still write "
The remark about f, (., x") means that, for each compact set E C R, there is a constant A(E) independent of x" such that If 1(s, x") - f 1(t, x") I < A(E)Is - tI whenever s, t E E and x" E RN-t Corollary 4.10 Suppose that u : RN -> R satisfies condition (a) of Theorem 4.2 and that, in condition (b), f (u(x)) is replaced by f (u(x), x"), where f now has a decomposition f = ft + f2 such that f, (., x") : IR --> R is locally Lipschitz continuous uniformly over x" E RN-1, while f2(.,x") : R -> R is non-decreasing for each x" E RN-t. In case (C), these conditions need hold only on [0, oo). Again define q and v by (4.2). Then v(-xl, x") = v(xt, x") for all x E RN, and 7tv(x) < 0 whenever x1 > 0.
Proof The only changes are these. (i) The function v satisfies not (4.1) but, for all ip E C,(RN), { -Vip(x) Vv(x) + W(x) F (v(x), x") } dx = 0,
(4.14)
LN
where
F(t, x") := f (t, x" + q")
for all t E I[8,
or for all t E [0, oo) in case (C). This follows from step (iii) of Lemma 4.3. Then the decomposition f = f t + f2 implies a decomposition F = F1 + F2 with the same properties. (ii) The results (I) to (IV) are derived not for an arbitrary unit vector k, but only for k = (1,0,. .. , 0) [relative to the original co-ordinate system]. Proposition 4.9 is restricted similarly.
4 Symmetry for the Non-linear Poisson Equation in RN
122
4.4 Application to some Newtonian potentials Notation As in (A.18) of Appendix A, the Newtonian kernel for RN is
-ZIxI
K(x) = K(x; N) :_
I
log
KN
Ixl
in
118,
in
1182 \ {0},
N > 3,
in RN \ {0},
IxIN_2
where KN =
Nx"2
1
(N - 2)1a-4N(0,1)1
(N
aMN(0,1)I
2),
=
(N12)!
The formula for the surface area of the unit sphere is a result of Exercise 1.14.
The characteristic function XA of any subset A of a universal set U is defined by 1 1 if x E A, XAx)
if x E U \ A.
0
We define the Heaviside function to be fH := X(oM, so that fH(t)
1
if t>0,
0
if t < 0.
(The value fH(O) = 0 is a little unorthodox.) Recall Theorem 1.1: if G is a bounded open subset of RN (N > 1), A is a positive constant, u(x) := A
f c
K(x - ) di;
for all x E R 1,
(4.15)
and
ul aG
= constant = /3,
(4.16)
then G is a ball. The truth of this will emerge from the more general Theorem 4.13, but it is worthwhile to anticipate here the reasons that (4.15) and (4.16) allow application of Theorem 4.2. First, results in Appendix A show that u E C'(RN) and that the asymptotic behaviour of u (as r -> co) is admissible. Second, we shall infer from the maximum
principle that u(x) > /3 if and only if x E G; therefore, in place of the classical result (Appendix A) that
Au =
-2 in G, 0
in RN
4.4 Application to some Newtonian potentials
123
we shall find that u is a generalized solution of the equation ,Lu + Afx(u - /3) = 0 in RN
(4.17)
In other words, u satisfies (4.1) with f = 2fH(.-f), and this f has a suitable decomposition: choose f l = 0. We consider, in the first instance, the Newtonian potential of a given density function F on a given set G; let
w(x) := fG K(x -) F(l;) d
for all x E RN, F E LP(G) with p > N.
where G is open and bounded in RN,
(4.18)
Lemma 4.11 The potential w in (4.18) enjoys the following properties. (i) w E C1(RN) (ii) Extend F to be zero outside G; then
for all cpEC`°(RN).
,RN
(4.19)
(iii) If fG F > 0, then the asymptotic behaviour of w is admissible in the sense of Definition 4.1; for N = 1 we have case (A) with m = 1, for N = 2 we have case (B), and for N 3 we have case (C) with m = N - 2.
Proof (i), (ii) Theorem A.11 establishes somewhat more than the C' property of w and also shows that w is a generalized solution of -Ow = F in RN; in other words, shows that (4.19) holds. (iii) Let R,v be so large that G c -4(0, -12R,,), and let r := IxI > R, For N = 1, we observe from (A.56), or else directly from the definition (4.18), that
w(x) = -Kr + cxr-1
(N = 1, r > Rw),
where K
1J 2
F( )
2
G
d
G
For N = 2, Theorem A.12 states that w(x) = K log
1
r
+ (c
x)r_2
+ h(x)
(N = 2, r it R,,),
where K
G
27r
d
(j = 1, 2),
124
4 Symmetry for the Non-linear Poisson Equation in RN
Ih(x)I < const. r 2,
IVh(x)I < const. r-3.
For N >- 3, Theorem A.12 states that w(x) = Kr N+2 + (c - x)r-N + h(x)
(N >- 3, r >-
where K = KN
f F( )
/
cj = (N - 2)KN J
(1 = 1,..., N),
d
IVh(x)l < const. r N-i
(h(x)I < const. r-N
Lemma 4.12 Suppose that in (4.18) we have F(I;) > 0 almost everywhere in G, and that wIaG = constant=#. Then w(x) > /3 if and only if x E G; equivalently,
XG(x) = fH(w(x) - /3) for all x E RN.
Proof It is enough to prove that x E G
(4.20)
w(x) > /3 and that x E
RN \G=> w(x)< f3. (i) Consider points of G. By Lemma 4.11, w E C'(-G) and
- J Vq Vw = <0 G
f
cpF
for all
(4.21a)
cp E
G
if also q >- 0.
(4.21b)
Thus w is a generalized supersolution relative to A and G. By the weak maximum principle, Theorem 2.11 with c = 0, we have w >- /3 in G. Now apply the strong maximum principle, Theorem 2.13, to any component [maximal connected subset] Go of G. If w(xo) = /3 at some point xo E Go, then w = /3 in Go, hence Vw = 0 in Go. This contradicts (4.21a) if we choose cp to be such that supp cp c Go, p >- 0 everywhere and T > 0 somewhere in Go.
(ii) Given any point y E RN \ G, we prove that w(y)
Let
0R(O,R)\G, with R so large that y E SZ and w < f3 on 09(0, R). This last is possible
because, if N = 1 or 2, then w(x) -+ -oo as r -> oo; if N > 3, then /3=wlaG>0andw(x)-->0asr--*oo. Now in (4.19)wehaveF=Oin 0; hence w E C 1(fi) and
-Jn
for all pEC10°(f)),
4.4 Application to some Newtonian potentials
125
so that w is a generalized solution relative to A and S2. [In fact, subsolution would suffice.] By the weak maximum principle, Theorem 2.11 with c = 0, maxjj w = maxan w = P. 11
Accordingly, the density function of the potential u in (4.15) and (4.16)
is F = .1xG = *fH(u(.) - f) on IRN, and, as explained earlier, Theorem 1.1 follows from Theorem 4.2. However, we need not restrict attention to density functions that are constant in G. In the next theorem we consider certain density functions F = g o u that are positive and may vary with
the potential u itself; if u is constant on the boundary 8G, then G is a ball once more. Whether such functions u exist, and whether, if they do, they are of interest, will be considered briefly after the theorem has been proved.
Theorem 4.13 Let G be a bounded open subset of IRN (N >- 1), and suppose that a measurable function u : RN -p R satisfies
u(x) = fK(x - ) g(u()) dfor all x E RN,
(4.22)
uIaG = constant = f,
(4.23)
where g : R - R is as follows. (a) 0 < g(t) < M for all t E R and for some constant M. (b) g is continuous except at finitely many points. (c) The restriction of g to [f, oo) has a decomposition g 91 + 92 such that gl(f) = 0 and gl is locally Lipschitz continuous, while $2(f) > 0 and $2 is non-decreasing. Then G is a ball, there is a point q E RN such that the definition v(x)
u(x + q) makes v spherically symmetric, and dv/dr < 0 for r > 0. Proof We verify that u satisfies the hypotheses of Theorem 4.2. (i) u E C1(1RN). This will follow from Lemma 4.11 if g o u E LP(G) with
p> N. In fact, g o u E L,,(G) if g o u is measurable, because hypothesis (a) ensures that g o u is bounded. To prove that g o u is measurable, we let F denote the finite set of points at which g is discontinuous and let h gIR\F. If W is open in R, then g-1(W) = E U h-1(W) for some set E F, and h-1(W) is open in R because it can be written U(hm)-1(W), where each restriction h. of h is continuous. Thus g-1(W) is a Borel set
126
4 Symmetry for the Non-linear Poisson Equation in RN
in R, which makes g Borel-measurable; by a standard result (Kingman & Taylor 1966, p.108; Rudin 1970, p.31), g o u is then measurable.
(ii) That u has admissible asymptotic behaviour also follows from Lemma 4.11. Case(C) of such behaviour occurs if and only if N >- 3; then the positivity of K and of g imply that u > 0 on RN. (iii) The generalized Poisson equation (4.1) is established as follows. By (4.19),
f
N
{
Vu + coF } =0 for all
E CC°(RN),
where
F(x) = g (u(x)) XG(x) for all x E ][8N [by Lemma 4.12] = g(u(x)) fH(u(x) - /3)
= f (u(x)) if we set f (t) := g(t) fH(t - /3) for all t E R. Also, let
fi(t) := ge(t) fH(t -13)
for j= 1,2 and all t E R;
then fl : R -+ R is locally Lipschitz continuous, f2 : R -> R is nondecreasing, and f = f, + f2. (iv) Accordingly, Theorem 4.2 applies to u; there is a point q E RN such that u(x + q) = v(x) =: V(r), and
say
(r := IxI),
V'(r) < 0 for r > 0.
(4.24)
Since V (O) = max u > /3, there is exactly one number b (for the function
V in question) such that V(b) = /3, and since x + q E G if and only if V (r) = u(x + q) > /3 (by Lemma 4.12), we have G = 3(q, b).
El
Apart from the case in which g is a constant, are there functions u that satisfy both (4.22) and (4.23)? The answer is that such functions exist in profusion; we shall see a few of them in Example 4.15 and in Exercises 4.25 and 4.26. However, equations (4.22) and (4.23) do not form the best starting point for an existence proof; it is more usual and profitable to proceed from (4.1), constructing solutions u in a Sobolev space by means of some variational principle or other. This allows functions g that are less restricted than those in Theorem 4.13. Further properties of u then follow from the theory of Sobolev spaces and from regularity theory for
4.4 Application to some Newtonian potentials
127
solutions of elliptic equations. We do not follow this course here because some of these steps are outside the range of the present book. The proof of Theorem 4.13 shows that it is essentially a particular case of Theorem 4.2. Whether functions u as in Theorem 4.2 are of interest or of importance is, of course, in the eye of the beholder. For N = 2 and N = 5 there are applications to steady vortex flows of an ideal fluid and to the problem in magnetohydrostatics of the confinement of a plasma in equilibrium; some of these are indicated by Exercises 4.18 to 4.24. Example 4.14 In Theorem 4.13, let
g(t) = const. = A > 0
for all t E R.
(This is a return to Theorem 1.1 and Exercise 1.12.) If N = 1, (4.22) implies that /3 < 0; if N >- 3, that /j > 0. In these cases there is a unique radius b = b(A, /3) such that G = R(q, b) for some q E RN. In fact, (A.23) shows that 11/2 (N# 2). b = { N(N - 2)/3/A. If N = 2, then (A.23) shows that the radius b of the ball G is determined by b2 log b2
=
4fl
(N = 2)
(4.25)
Consider the graph of b2log(1/b2) for b > 0. If /3 > 0, then (4.25) has no solution when 4/3/A > 1/e, one solution when 4/3/A = 1/e, and two solutions when 4/3/A < 1/e. If /3 < 0, then (4.25) has exactly one positive solution.
Note that, for a given solution u in Theorem 4.13, there is always a unique radius b such that G = JJ(q, b), as was mentioned in the proof. However, for given data g and l3, there may well be more than one solution V as in (4.24), and hence more than one radius b; this is the case in Example 4.14 if N = 2 and 0 < 4/3/A < 1/e. Example 4.15 In Theorem 4.13, let
N = 2, where
g(t) = A + (µ - A)fx(t - a) for all t E R, 0 < A < µ,
/3 < 0,
/3
and A, p, a are constants. The function f in Theorem 4.2 is then as in Figure 4.4. We propose to list all possible solutions V of (4.22) and (4.23) for the N and g in (4.26); here V is as in (4.24).
4 Symmetry for the Non-linear Poisson Equation in IRN
128
f (t) =µ
f(t)=A
f(t)=0
1
t
a
/3
Fig. 4.4.
(i) If V1(0) < a for a given solution V1, then Vl (r) < a for all r > 0 and we call V1 a one-step solution of the present problem. It is a particular one of the solutions in Example 4.14 (with N = 2 and /3 < 0), and, by (A.23), 2
=
Vl (b)
V1(0) =
-2 log b = Q, 1
(4.27)
?b2 <_
Clog b + 2
(4.28)
a.
Define a to be the inverse of the function s '-slogs, s >- 1 (Figure 4.5). Then (4.27) becomes b2 = a(-4/3/2); the condition (4.28) for existence of a one-step solution becomes + 4 a I - 4A
(4.29)
a.
(ii) If V(0) > a for a given solution V, then there is exactly one radius a for that solution such that V(a) = a, and exactly one radius b for that solution such that V(b) = /3; we note that 0 < a < b and call this V a two-step solution. It is easy to write V in terms of A,µ a and b: one adds the potentials of density 2 in R(0, b) and of density µ - 2 in 9(0, a), each of these being given by (A.23). However, our aim is to find V when A, µ, a and /3 are given. Knowing that
2
- µr2 4 + (µ -22)a2 (logo+211 + ,b2 (log 1o+2 V(r)
(µ -1)a2 2 (µ {
1
r < a,
log r-2r24 +2b2 2 /Clogb+2a
2A)a2
1
+
1
_22 }log r,
1
r > b,
(4.3G)
4.4 Application to some Newtonian potentials
129
e
s
Fig. 4.5.
we seek a and b such that (µ - .1)a2 2
.b2 (
.1a2
1
log a
+
4
_
2 2
2
(µ 2A)a
1
log b
2
+
} log
+
1
1) _
(4.31a)
(4.31b)
This may seem an unattractive task, but it is of a kind that must be faced in non-linear problems. The corresponding calculations for N = 1 (Exercise 4.25) and for N >_ 3 (Exercise 4.26) turn out to be easier.
130
4 Symmetry for the Non-linear Poisson Equation in RN
(iii) With the notation
:_ (/I
- 1) a2,
q :=b 2, (4.32)
4a µ A:=-A, B:=-A4/3 >0, k:=A-1>0,
we now seek
> 0 and n > 0 such that
(og
q - 1) = A,
+ log k + k +
(4.33a)
(4.33b)
( + r1) log q = B. The non-negative solution fi(n) :=
of (4.33b) is (Figure 4.6) 1
g - h,
1 < q < 112,
(4.34)
where 111 and f2 are the unique solutions of B
(4.35)
121og h2 = B,
h1109 h1 = k-+-, ,
whence 1 < n1 < r12. We are interested only in solutions for which 0 < a < b, hence 0 < < kn; therefore l;(r1) is acceptable only on ('11,112), but it is convenient to have the function defined on (1,112]. It follows from (4.34) that
(q) > 0
on (1, r12),
'('1) < 0 and "(j) > 0
on (1,'12]. (4.36)
In view of (4.33a), the next step is to define
h('1) :_ (r1)
+ q(log 11 - 1),
log (r1) + log k + k
1 < q < 112,
1
ll
(4.37)
and to contemplate all solutions of h(r1) = A. Here h('1) = h('1;k,B) and we have k > 0, B > 0 and A < B; the parameter A may have either sign or be zero.
Remark 1 For each k > 0 and B > 0, h'(r1i) < 0
and
limn1
2
h'(q) = co.
(4.38)
Proof The definition (4.37) implies that
l h'(q) = C'('1) { log 01) + log k + k + 1 } + log r1.
(4.39)
4.4 Application to some Newtonian potentials
131
11
Fig. 4.6.
For h = ril, we have (ql) = kris and [by (4.34) and (4.35)]
k+1 -1, -1=(log ril)2ril log rii B
whence h(hl)=-(k+1)- (TO
As nT rig,
+11
+1 <0.
9-11
-oo; then h'(n) -(logB
'(n) --> '(n2) _
(1k
112 )2n2
oo because
-1
- 1 < 0. 109 q2
Remark 2 There is exactly one point no E (ri1,1J2) such that h'(i o) = 0. At this stationary point rfo, the restriction of h to N1,112] has a strict local and global minimum.
Proof Remark 1 shows that hl and rig are not stationary points of h, and that there is at least one stationary point, say 1o, in 01, 0. Thus
4 Symmetry for the Non-linear Poisson Equation in RN
132
B
h(7)) .17 112
B - a(B)
Fig. 4.7.
h'(r1o) = 0; since '(r1o) < 0 by (4.36), and log no > 0, we infer from (4.39)
that log o(no) + log
1
1
+ + 1 > 0.
Also, "(rio) > 0 by (4.36), so that (
2
log x('10)+log+k+1}+
>0. +X10
Thus at any stationary point in ('j1,112) the function h has a strict local minimum. Hence there can be at most one such stationary point, and we have seen that there is at least one. Remark 3 The definitions of (r1), 11,112 and h(r1) imply that
h(nl) = (k+1)r1ilogi1 = B, h(112)
= 112(1og 112 - 1) = B - a(B),
where a is the function defined after (4.28). (iv) We can now collect ingredients for a list of solutions.
(a) Remarks 1 to 3 show that h is essentially as in Figure 4.7. The figure is misleading in that B - a(B) may have either sign or be zero (in fact, it is negative, zero or positive according as B < e, B = e or B > e),
4.5 Exercises
133
and in that negative values h"(q) occur only for large values of k, but the figure is sufficient for counting solutions of the equation h(r1) = A.
(b) A two-step solution V is determined completely by the solution n E (ni, n2) of h(r1) = A, because is then given by (4.34), the radii a and b are given by (4.32), and V is given by (4.30).
(c) The necessary and sufficient condition for existence of a one-step solution Vl is [by (4.29) and (4.32)] A 5 B - v(B). (v) Conclusions
The results in (iv) imply the following list of all
possible solutions. Recall that B > 0 and A < B by hypothesis, and that h(q) = h(r1;k,B).
(a) For B - Q(B) < A < B, there is one solution V; it is a two-step solution. (b) For A = B - a (B), there are two solutions: one has 7 E 01, u2) and
is a two-step solution, the other has 7 ='12, Vi (0) = a and is a one-step solution.
(c) For min, h(r1) < A < B - Q(B), there are two distinct two-step solutions and a one-step solution with Vl (0) < a.
(d) For A = min, h(r1), there is a two-step solution and a one-step solution with V1(0) < a. (e) For A < min, h(j), there is a one-step solution with V1(0) < a.
4.5 Exercises Exercise 4.16 Prove Lemma 4.5 for case (A) of Lemma 4.3 with 0 < in < 2.
[You may find it helpful to separate estimates of g for 4µh > s2,
m-1-6 >0from those for4ph>s2,m-1-8 <0.Ifm-1-6 =0, we may reduce S slightly.]
Exercise 4.17 Prove Lemma 4.5 for case (B) of Lemma 4.3.
Exercise 4.18 This exercise and the next are concerned only with the occurrence in hydrodynamics of the equation A1p + f (w) = 0. Let v :1[83 -+ 1[83 denote the steady (that is, time-independent) velocity, and p : R3 R the steady pressure, of an inviscid fluid having constant density p > 0. Because of the identity
(V V)v=(Vxv)xv+V(zwI2)I
4 Symmetry for the Non-linear Poisson Equation in JRN
134
the equations of motion, under an extraneous force field -V(F, are
divv := V v = 0,/ curly := V x\v (4.40)
xv=-V(P+(F+IIV12 P
//
one calls ( the vorticity and p/p + (F + !Iv I2 a Bernoulli function (see Batchelor 1967, pp.74 and 160; Lamb 1932, pp. 4-6). (i) Flow independent of x3, with stream function W(xl, X2). Show that,
for any given functions W E C1(R) and H E C'(R), the velocity field and Bernoulli function v = (-821p, 01 W, W (W))
and
P + (F +
1
2
Iv12
= H(1V )
satisfy equations (4.40) if lp E C2(R2) and OW(x) = H' (W(x)) - W (W (x)) W' (W(x))
(4.41)
at all points x = (xl, x2) E JR2.
(ii) Cylindrically symmetric flow, with Stokes stream function 'P(xl, s). Write x = (x1, s cos 9, s sin 9), s > 0, for points of 1R3, and let
el := (1, 0, 0), e' := (0, cos 9, sin 9), es := (0, - sin 9, COO).
Show that, for any given functions F E C1(R) and H E C'(R), the velocity field and Bernoulli function
v = s (_OTael +
e' + r('t`)es)
(s > 0)
and
PP +(D+2wj2=H('I') satisfy equations (4.40) for s > 0 if T E C2 (R x (0, oo)) and z
(axi +
2
as2
- sa
'F = szH'('t') - T('`)T'('`)
(4.42)
at all points (x1, s) E R x (0, co).
Exercise 4.19 Transformation of the Hicks equation (4.42). Show that under the transformation 'P(xl, s) = s2X(xl, s), in which X is Greek capital chi, (4.42) becomes 2
z 5X
2
= (8xi + aS2 + s as X = H'(S2X) -
s2 r(s2X)r1(S2X ),
(4.43)
4.5 Exercises
135
where s > 0 and A5X = { (8/402+ +(a/8y5)2 }X if we write xl = yi + y5)'/2 (cf. Exercises 1.18 and A.25). In the context of generalized solutions, H and F need not be in C'(R). Let and s = (y22 +
H(t) = -2tfH(t) + const.
and
IF(t) = ytfH(t)
(4.44)
for all t E IR; here A,,y are constants (A > 0, y > 0) and fH is the Heaviside function (Chapter 0, (v)). Also, write
U = const. > 0.
X(xi, s) = x(xl, s) - U, 2
This corresponds to `P = W - 2 Us2, where V will be required to `vanish at infinity' while -2 Us2 is the Stokes stream function of velocity (U, 0, 0). Show that (4.43) becomes
o5x=-{.1+Y2(x-IU) }.fH(x-2U) at all points (x 1, s) such that s > 0 and x(xl, s)
(4.45)
2 U.
Exercise 4.20 Let W be a C2-solution of (4.41) on 1
where w = const. > 0,
Vl = const.,
V2 = const.,
(this corresponds, for the fluid, to 'solid-body' rotation about the origin plus a uniform velocity plus something smaller), or K
X2
Vip(x) _ --- (X1, X2) - (ci,-C2)l r4 where
K = const. > 0,
x22
-
cl = const.,
2x
Y4 2 +O(r 3),
c2 = const.,
(this could be due to negative vorticity in a bounded set).
(i) Show that the level sets of w are the point q where W has its maximum, and circles with centre q. Find q in terms of the data for each of the two cases.
(ii) The streamlines in R3 are the solution curves of dx/dt = v(x), where now x e ][R3 and v is as in Exercise 4.18, (i). Show that a streamline is either the line A := { (q 1, q2, x3) -oo < x3 < oc }, or a circle in I
a plane of constant x3 and with centre q, or a helix contained in a cylindrical surface with axis A.
136
4 Symmetry for the Non-linear Poisson Equation in RN
Exercise 4.21 Consider the elliptic disk D := { (x, y) E R2 axe + fy2 < 1 }, where a and /3 are positive constants, and consider the function I
1 -axe _ fl y2 2(a + '
Wo(x, Y)
(x, Y) E D.
Q)
Note that /Wo = -1 in D, WO = 0 on 8D. (Thus WO is the stream function of flow in D with vorticity -1 and with velocity tangential to 8D.) We seek an extension W of Wo as follows: W E C1(R2) fl C2(R2 \ 8D),
W = Wo on D, /W = 0 in R2 \ D, W < 0 in R2 \ D and the asymptotic behaviour of W, as I(x,y)I -+ oo, is admissible (Definition 4.1). Prove that such a function W exists if and only if a = /3. If D is replaced by an arbitrary bounded open set S c R2, with smooth boundary 852, is there an analogous result?
Exercise 4.22 Let E be the Hilbert space formed by completion of the set in the norm I II defined by hull :=
(u,u),
where (u,v)
:=f
Vu- Vv. 5
The space E is embedded in Llo/3(R5). Referring to Exercise 4.19, we discard the cylindrical symmetry there; we call w :R5 -+ R a finite-energy solution of (4.45) if w E E \ {0} and
l
{A+y2(w-ZU)}fH(w-ZU)] =0
for all p E C'(R5). From these two properties of w it follows (by methods
beyond us here) that the equivalence class w E E has a representative w E C1(R5); that w has admissible asymptotic behaviour (Definition
4.1) of type (C), with m = 3 and 6 = 1; that w > 0 on R5 ; and that w E C2(R5 \ S), where S { y E R5 w(y) = 2 U I. Given these facts, deduce that every finite-energy solution w of (4.45) is as follows, for some point q E R5.
(i) If 2 > 0 and y = 0, define a := (15U/22)1/2; then
1U(5-3a2) if r
(4.46) 2Ua33
1
if r>a,
where r := IyI (cf. Example 4.14 with N = 5 and /1 = 2 U).
4.5 Exercises
137
(ii) If A > 0 and y > 0, a unique radius a is determined by
0 < ya < /Jl = 4.49...
and
yaJ512(ya)
J3/2(ya)
-
3 Uy2 2 2
where the Jv are Bessel functions (Exercises 1.19, 1.21) and IJ1 is the smallest positive zero of J3/2i then
3U AJ3/2(32)-B+3 2
w(Y + q) = xl(r)
(yr) /
if r
3
(4.47)
3
1
if r>a,
2 U a3
where r :=1Y1,
A :=
(ya) 1/2
J5/2(ya)
,
B :_
J312(ya)
yaJ512(ya)
Exercise 4.23 The uniqueness of Hill's vortex and of Hicks's vortex. Let F be the set of functions p : 1[8 x [0, oo) - II8 defined by OYn,s) = s2u(Yl,s) = s2u(Y),
s = (Y2 + ... + Ys )1/2,
where u is a cylindrically symmetric function in the space E (Exercise 4.22). Show that, if (p, u and gyp, v are such pairs of functions, then aW
ay)
aW
ff 1s2{ ayl ayl + 8s 8s
OW
} s dy1 ds = 1 J Du(Y) Vv(y) dY
Rx(o,oo)
2n2
R5
It follows that F can be made a Hilbert space; also, that a flow in R3 with Stokes stream function W E F (and with no velocity in the direction es) has finite kinetic energy. [Take (p = W.] Now consider all generalized solutions `P = W - 1 Use, with W E F, of (4.42) with H and IF as in (4.44). Prove that, for some constant c E JR and with r := (x1 + s2)1/2, ip(x1 + C, S) _
I s2xo(r) s2X1(r)
if 2 > 0 and y = 0, if 2 > 0 and > 0,
where Xo is as in (4.46) and xl as in (4.47). These functions were discovered, without reference to 1R5, by M.J.M. Hill (1894, for y = 0) and W.M. Hicks (1899, for y > 0). The streamlines of Hill's vortex are shown in Figure 4.8; for those of Hicks's vortex, see Moffatt (1969).
138
4 Symmetry for the Non-linear Poisson Equation in RN S
U `I' = const. < 0
D s xl
Fig. 4.8.
Exercise 4.24 The equations of magnetohydrostatics relate the magnetic induction B : R3 1R3 to the current density j : 1183 -> R3 and pressure p 1[83 --+ 118 of an electrically conducting fluid in equilibrium. Under an idealization that is frequently made (Ferraro & Plumpton 1966, p.35; Thompson 1962, pp.47 and 53), these equations are :
curl B=µj, divB = 0, j x B = V(p + p4)), where the constants µ > 0 and p > 0 are respectively the magnetic permeability and density of the fluid, and -VI is an extraneous force field.
Translate (as far as it interests you to do so) the results of Exercises 4.18 to 4.23 into the language of magnetohydrostatics. Exercise 4.25 In Theorem 4.13, let
N = 1, g(t) = A + (µ - WO - a) for all t e I[8, where
0<2<µ, #
List all possible solutions V for these data; here V is as in (4.24). [The strategy is as in Example 4.15; the tactics differ. For a two-step solution, defined to be one satisfying V(0) > a, a possible method is to let V(a) = a, V(b) )./p; to seek := alb E (0, 1) such that and 0 (I
Og2 + 110
(1 - 0)g + 0 and to prove that cps (. , 0) has exactly one zero o E (0, 1) for fixed 0.]
4.5 Exercises
139
Exercise 4.26 In Theorem 4.13, let
where
N>-3, g(t)=2+(p-2)fH(t-a) for all O<2
t E R,
List all possible solutions V for these data; again V is as in (4.24). [For a two-step solution, again defined to be one satisfying V(0) > a, a possible method is to let V (a) = a, V (b) and 0 :_ 2/p; to seek := alb E (0,1) such that V(
2
0) :=
2
a
> 1;
to prove that, if 0 E (0,21N) and is fixed, then ip (. , 0) has exactly one zero o E (0, 1); and to prove that, if 0 E [2/N, 1), then 0) < 0 for
Exercise 4.27 This exercise concerns an alternative to Lemmas 4.7 and 4.8 and to Proposition 4.9 in the case when f2 = 0, so that f is locally Lipschitz continuous. Let (m, oo) be the largest open interval of p such that Y E 1'(µ)
.
w(Y, µ) := v (Yµ) - v(Y) < 0;
(*)
assume (for contradiction) that m > 0. With R(p) as in Lemma 4.5, define
Ro:=R(2m)+1,
X(p):={xERN I -Ro<xl
Exercise 4.28 Here we extend the method in Exercise 4.27 to certain discontinuous functions f.
(a) Let f be as in Theorem 4.2 with the additional condition that f2(t) = 0 whenever t < lc, for some 1c E R when u satisfies (A) or (B) of Definition 4.1, or for some K > 0 when u satisfies (C). [Cf. Theorem 3.6 and Remark 1 following it.] (b) Let R2 be so large that v(x) < K whenever r > R2, and choose
Ro:=max{R(Zm)+1, R2+1}
140
4 Symmetry for the Non-linear Poisson Equation in RN
in order to construct a set X(µ) like that in Exercise 4.27. For functions
f as in (a), complete the proof of Theorem 4.2 beyond Lemma 4.6 by combining the method of Exercise 4.27 with that used to prove Theorem 3.6.
5
Monotonicity of Positive Solutions in a Bounded Set 1
5.1 Prospectus
This chapter concerns positive solutions of the equations
Du +f(U) = 0
(5.1)
and
(5.2)
in a bounded set 0 that need not be symmetric; in (5.2), the coefficient b is a constant vector. Although (5.1) is a particular case of (5.2), the two are treated separately because (5.2) will be examined under stronger restrictions on f and on the subset of n to be considered. These stronger restrictions are not necessary, but without them one must either work long and hard or use advanced techniques; in fact, equations far more general than (5.2) can be treated by advanced versions of the method. Our purpose here is merely to show, at an elementary level, that the techniques of this chapter are not restricted to the non-linear Poisson equation (5.1). Our first task (§5.2) is to explore the nature of caps and reflected caps
for a set 92 that lacks Steiner symmetry. For suitable sets S2 and unit vectors k, we shall introduce alpha caps Z(a(k),k) that generalize the caps Z (0,el) of the Steiner symmetric sets S2 in Chapter 3. We shall also define beta caps Z (/3(k), k) ; these form a smaller class of possibly smaller caps and have an additional property that simplifies certain proofs. The first monotonicity theorem in §5.3 is a variant of Theorem 3.6 for
alpha caps and for equation (5.1). A further result for (5.1) concerns the hyperplane T,,(k)(k) that bounds an alpha cap: if there is a point xo E TT(k)(k) n 52 at which the derivative of u normal to the hyperplane 141
142
5 Monotonicity of Positive Solutions in a Bounded Set n
vanishes [that is, (k Vu) (xo) = 0], then both u and ) must be symmetric relative to T (k)(k).
For positive solutions of (5.2), similar, but slightly more elaborate, results will appear.
The reader who dislikes the geometry of caps, and who is interested only in equation (5.1), need read §5.2 only as far as Theorem 5.3.
5.2 On the geometry of caps and reflected caps Definition 5.1 Let 92 be a bounded region. For every unit vector k E RN and every p E R, the hyperplane Tµ(k) and the reflection xµ,k of x are as in Definition 1.7. (a) The sets
Z(p,k):={xEf) I x E RN I xµ,k E Z(p,k) }
will be called a cap and reflected cap, respectively. The right-hand boundary of the cap Z(p, k) is rµ(k) := OZ(lp,k) \ TT(k). We also define
M(k) := sup{ x k I X E K21 = sup{ u I Z(p,k) is not empty }. (b) Iff, for a given unit vector k, there is a number a < M(k) such that Y (µ, k) c S2 whenever p e (a, M(k)), then S2 will be called admissible for the direction k. In that case, we define a(k) := inf { a
I
p E (a, M(k))
Y (µ, k) c S2 }
and call Z (a(k), k) the alpha cap for the direction k. Figure 5.1 shows examples of caps, reflected caps and alpha caps with
k = el := (1,0,.. . , 0) and with no display of k. Note that if, for the moment, we define y := inf { p I Y (p) c S2 }, then y can be very different from a. The following lemma and theorem are very simple, but have important consequences.
5.2 On the geometry of caps and reflected caps
143
Fig. 5.1.
Lemma 5.2 Let 92 be admissible for a given direction k. Let p E [a, M) and z E Z (q), the label k being omitted from the symbols in Definition 5.1. Denote the closed line segment from z to zµ by
(1-6)z+9zµ I 0
Proof Since S2 is open, there is a point z* := z+2ek in S2 for some s > 0. Let x E (. Then x is the reflection of z* in TQ [that is, x = (z*)-] if
a=
k) =
where
because xE(,
5 Monotonicity of Positive Solutions in a Bounded Set n
144
so that
ll.+E a, we have z* E Z(a) and hence x = (Z*)' E Y (a). Since a > a, the definition of a ensures that Y (a) c S2. Therefore X E S2, as desired.
The significance of the next little theorem is that the inclusion Y (p, k) c S2 extends from u > a(k) to y = a(k).
Theorem 5.3 If 92 is admissible for a given direction k, then for that direction the reflection in TT(k) of the alpha cap is in 0. In other words,
Y(a(k),k) c n. Proof Omitting the label k again, we wish to show that, if z E Z(a), then z" E S2. This follows from Lemma 5.2 with it = a. The remainder of this §5.2 leads to the result that a bounded region f2 with On of class C1 is admissible for every direction k. At the same time we establish properties of the outward unit normal n (to the boundary of such a set) that become useful in the context of Theorem 5.12, which concerns equation (5.2). This material may be of interest in its own right, but it is not needed for Theorems 5.10 and 5.11, which concern equation (5.1).
Definition 5.4 Let 8S2 be of class Cc for some I E No and let p e 852. As in Chapter 0, (viii), the set f2 is defined globally in terms of co-ordinates xj, while yj are `local' co-ordinates such that 8S2 has a representation YN = h(y'),
where y' := (Yl, ... , YN-1),
(5.3)
near the boundary point p. (a) The co-ordinate transformation will now be written y = Yp(x) := A(p)(x - p),
(5.4)
where A(p) is an orthogonal N x N matrix depending on p.
(b) Let Q'(0, p) :_ { y' E
RN-1 I
-p < yi < p, j = 1, ... , N - 1 }
denote a cube about the origin in RN-1, with edges of length 2p > 0. We define (Figure 5.2)
'rp,a(p):_{yERN Y EQ'(0,p), I
-a
5.2 On the geometry of caps and reflected caps
145
XN
xl
Fig. 5.2.
and its x-image %
(p)
Y, 1('r,,,Q(p)),
and call these sets rhomboid neighbourhoods, of the points 0 and p respectively. The yN-axis is to point into 92, and the positive lengths p and a are to be so small that x E 892 n GWp,,(p)
x E 9 n ()&p,,(p)
if if
y = Yp(x) E *-p,,(p) and YN = h(y'),
(5.5a)
y = Yp(x) E Vp,,(p) and YN > h(y').
(5.5b)
(The word rhomboid is used faute de mieux and because *-PAP) is approximately rhomboid when N = 2, e >- 1 and p is sufficiently small.) Exercise 5.5 Let 8S2 be of class Cc for some f >_ 1, let p E 892 and let the outward unit normal v(y') = A(p)n(x),
where y = A(p)(x - p),
146
5 Monotonicity of Positive Solutions in a Bounded Set S2
be as in equation (0.2). Abbreviating A(p) to A and writing Vx (a/ax,, ... , a/aXN), define new local co-ordinates z1, ... , zN by z = By = BA(x - p), where B is an orthogonal N x N matrix such that N
n(p) ' VXZN = V(0) ' V yZN = E vj(0)BNj * 0.
(5.6)
i=1
Prove the following.
(a) There is a set W, open in RN and containing p, such that aft fl W has a representation ZN = g(z'), z' E H, g E C'(H),
(5.7)
where H is open in JN-1 and convex. (b) There exist new rhomboid neighbourhoods, say Va,b(p) and its ximage Ua,b(p), characterized by z' E Q'(0, a) and by IZN - g(z')I < b, such that a statement analogous to (5.5a, b) holds when n(p) ' VXZN < 0(c) If the matrix B satisfies BN1 = -vj(0) for j = 1, ... , N, then ZN = -n(p) ' (x - p) for all x E ]RN,
(a1g)(0) = 0 for j = 1, ... , N - 1, (5.8)
so that the hyperplane { Z ZN = 0 } is tangent to aft at x = p (at z = 0), as is shown in Figure 5.3. I
[For a form of the implicit-function theorem that yields the Cc property of g in (5.7), and not merely C' smoothness, see Dieudonne 1969, p.272, or Fleming 1965, p.117]
Notation Until the contrary is stated, the unit vector k will be fixed; we adopt variables aligned with k (Definition 4.4). That is, we rotate the co-ordinate axes and use
xl =x'k, x" =(x2,...,xN)I Q:={5 EGRN
I xE1}, Z(µ):={zEsz z,>µ M:=sup{., I zE1 }
and so on. Again we omit the tilde of these aligned variables as long as there is no danger of confusion.
5.2 On the geometry of caps and reflected caps
147
x1
Fig. 5.3.
Lemma 5.6 Let n be a bounded region with M of class C'. Define
F:= { x E 3c
I
x1 =M1.
Let p E F and let l;j be the local co-ordinates called zj in Exercise 5.5, (c), so that the cN-axis is along the inward normal at p, and afl is described
near p by N = g('). Then
n(p) = el := (1,0,...,0) and
0
(5.9)
for all ' in the domain of g. Proof Observe first that M is the supremum of the continuous function x --> x1 not merely over fl, but also over the compact set S2 (Exercise 1.4); this supremum is attained in S2, but not in the open set 92. Thus F is not empty and is closed in RN. It is immediate that n(p) = e1 implies 0: if n(p) = e1, then
xl - M = -i;N = -g(l;') for x E ail and x near p; if g(l;') < 0 for some l;', then the definition of M is contradicted. That n(p) = el also seems self-evident: if n(p) * el (Figure 5.4), then there are points x in S2 near p for which x1 > M, so that the definition of
148
5 Monotonicity of Positive Solutions in a Bounded Set C1
TM
Fig. 5.4.
M is contradicted again. However, an analytical form of this argument is worthwhile.
= C(x - p), so that C is an
Write the co-ordinate transformation as orthogonal N x N matrix. Then
-(CNI, CN2, ... , CNN)
n(p) =
(5.10)
and
N-1
xl - M =
Cil j + CNIg( ')
and x near p. (5.11)
for x E 3f
i=1
Here g(l;') = o(Il;'I) as
- 0, because
g() - g(0) _ .f r
(t
l
dt = Jo
r
(Vg)(tr) dt
with g(0) = 0, (Vg)(0) = 0 by (5.8) and with Vg continuous by (5.7).
Assume that C1 0 for some m c {1,. .. , N - 11. Set l;' = (.... )in(5.11),with ,>0ifCl <0ifCnil <0. to be so small that x1 - M > 0. This is a contradiction, so that Ci1 = 0 for i = 1, ... , N - 1. The rows Since
we can choose
5.2 On the geometry of caps and reflected caps
149
and columns of (C,j) are unit vectors; therefore CN1 = ±1. If CN1 = 1, we have N
i=1
which is again a contradiction when ' = 0 and bN > 0. Accordingly, CN, = -1, and we see from (5.10), both sides of which are unit vectors, that n(p) = el.
Theorem 5.7 Let 0 be a bounded region with 00 of class C'; we use variables aligned with a given unit vector k. There is a number 6 > 0 such that, if µ E (M - 6, M), then (a) Y (p) =!Q, which implies that n is admissible for the arbitrary direction k,
(b) nl(x) > 0 for all x E I'µ.
Proof (i) Let F be as in Lemma 5.6. We shall use Exercise 5.5 again. The rhomboid neighbourhoods Ua,b(P) of all points p e F form an open cover of the compact set F. [That is, each set Ua,b(P) is open and F is a subset of the union of all of them.] We extract a finite subcover, say { Ua,b(pr)
I
r = 1,2,...,s },
where a = a(pr) and b = b(pr). [When F is a finite set, this extraction is unnecessary.] Figure 5.5 shows one set of the finite subcover; the local co-ordinates l;j are as in Lemma 5.6. Now define G
U Ua,b(Pr), r=1
M-s:=sup{x1 I xE J\G}=max{x, I xES2\G,}, 6 :=min{ s,2b(pl),...,Zb(ps) In describing M - e, we have used Exercise 1.4 once more. Note that e > 0, otherwise the definition of F would be contradicted, since G covers F.
(ii) Let p E (M - 6, M). We wish to show that Y (p) c S2; equivalently, that
z E 0 and z, > µ
z" E .Q.
150
5 Monotonicity of Positive Solutions in a Bounded Set 92
aUa b(Pr)
Fig. 5.5.
Now the conditions zl > It, u > M - 6 and 6 < e imply that zi > M - s, so that z ¢ 0 \ G. Therefore Z E S n G, whence z E S2 n Ua,b(pr) for some
r E {1,...,s}. Returning to the co-ordinate transformation = C(p)(x-p) used in Lemma 5.6, we define
( := C(pr)(z - pr),
j := C(pr)(z' - pr).
(5.12)
We shall show that g(0 < ?IN < b, where now g = g(. , pr) and b = b(pr) for the particular r introduced before (5.12); then zµ E f2. By Lemma 5.6, N = M - x1 when p E F; therefore reflection in Tµ is reflection in { I N = M - u }. Accordingly,
qN = 2(M - µ) - N,
(5.13)
in which 2(M - p) < 26 < b and SN > g(Z'') >- 0, so that 1N < b. Next, by (5.13) and repeated use of the inequality N = M-z1 < M-p, '1N > M - µ > SN > gW) = g(11'), since n' by (5.12). The results '1N < b and r1N > g('1') now show (as we noted above) that zµ E S2, and hence that Y (y) c S2.
(iii) It remains to consider x E Fµ for P E (M - 6, M) and to show that n1(x) > 0. [Recall that F. := 3Z(µ) \ Tµ and that the identity Z(p) = S2 n { x x1 > p } implies that 8Z(µ) (-- Of) U Tµ; therefore f1, c 8S2 and here x E Of).] Since 6 <- s, we have x1 > p. > M - s; then x 0 0 \ G, by the definition of M - s. Consequently x E 3) n G, I
5.2 On the geometry of caps and reflected caps
151
so that x E OS2 n Ua,b(p') for some r E {1,...,s}. Equivalently, under the transformation = C(p') (x - pr) we have E Va,b(p'), N = g = g(., p'), and
nl(x)=-VN()=
1
{ (alg)(')2 + ... + (aN-ig)(')2 +
>0.
111/2
The condition established in Theorem 5.7 that nI(x) > 0 for all x E I'µ is a useful property of the right-hand boundary of a cap. It prompts the following definition, which is phrased in terms of our original variables rather than in terms of those aligned with k. Definition 5.8 Let 92 be a bounded region with On of class C1. For every unit vector k E RN we define
1(k) := inf{ v I µ E (o, M(k))
n(x) k > 0 for all x E T'N,(k)
/3(k) := max{ a(k), 1(k) },
and call Z (/3(k), k) the beta cap for the direction k.
Figure 5.6 illustrates this definition for k = e1. The set in part (a) of the figure is 521 :=
2
x E 1182
I
(XI - cxi) + x2 < 1, xl < 1/2c }
,
c = const. E (0, 4) , (5.14)
and we define \/ l
4c
(5.15)
.
The reader should verify that, for the set 521 and the direction e1,
the critical values of p are a (e') = y > 0 and 1 (e') = 0, so that
Q(el)=a(e') The set in part (b) of Figure 5.6 is 922
x E 1[82
I
(xl + cx21)2 + x2 < 1, x1 > -1/2c
},
c = const. E (0, 4) ,
(5.16)
and now a (el) = 0 and A (e') = 0. In part (c) of Figure 5.6 a small protuberance has been added to 922 near the point (M,0), where M := sup{ xl X E 922 }. This can be done I
152
5 Monotonicity of Positive Solutions in a Bounded Set Q
Fig. 5.6.
in such a way that a (el) is unchanged, 2 (e') is greatly increased, 00 remains of class Coo and 00 is unchanged for x1 <
ZM.
Remark 5.9 If SZ is a bounded region with aft of class C', then, for each unit vector k, (i) n(x) k > 0 for all x E rA(k)(k), (ii) n(x) k >- 0 for all x E I'a(k)(k). Proof We revert to variables aligned with k, so that n(x) k and I'A(k)(k) become nl(x) and FA, respectively.
(i) If X E f'A, then x E aft and x1 > A. Hence x E I'µ for some y > 2 [for example, it = 1(x, + A)] ; the definition of 2 now ensures that n, (x) > 0.
(ii) Assume for contradiction that p E F,, and n,(p) < 0. By this last, z := p+Eel E 0 and y := p-eel V SZ for sufficiently small s > 0. Choose ,u = p1; then It > a, z E Z(µ) and zµ = y V S). This contradicts the definition of a.
5.3 Monotonicity in f2
153
Remark 5.9 implies that, when A(k) > a(k), we have n(x) k > 0 for all x E I'fiki(k) and n(x) k >- 0 for all x E I'a(k)(k). Although this distinction may seem slight, it makes certain proofs much easier for beta caps than for alpha caps.
5.3 Monotonicity in S2
Our first monotonicity theorem, for sets f2 that may lack symmetry, is an
extension of Theorem 3.6 that takes longer to state than to prove. The reader who finds the proof somewhat glib is invited (in Exercise 5.25) to spell out the corresponding extension of Theorem 3.3. This smaller task may illuminate the present proof. Theorem 5.10 Let 0 be a bounded region that is admissible for the direction
el (Definition 5.1); we abbreviate T. (e'), Z (p,el), a (et),... to Tµ, Z(p), a,.... Assume that u : S2 -+ l1 has the following properties. (a) u E C(92) n C1(f2), u > 0 in f2, u = 0 on I'a. (b) For all (p E C,"°(S2)),
{-V
Vu+cpf(u)} =0,
(5.17)
where f has a decomposition f = f 1 + f2 such that f 1
: [0, oo) -* R is locally Lipschitz continuous, while f2 : [0, oo) -+ d8 is non-decreasing and is identically 0 on [0, ic] for some 1c > 0.
Then the conclusions of Theorem 3.6 hold in the alpha cap:
u(z) < u (zµ) if 01u(x) < 0
if
p E (a, M) and z E Z(p),
(5.18)
xl > a and x E f2.
(5.19)
Proof Scrutiny of the proof of Theorem 3.6 shows that there we used only conditions that have counterparts here: rather than the full Steiner symmetry of SZ relative to To, we used only inclusions implied here by Lemma 5.2; rather than the full boundary condition u = 0 on 8f2, we used only that u = 0 on F0. (Of course, the full conditions were used for Corollary 3.9.) Therefore the proof of Theorem 3.6 extends to the present situation, provided that the earlier interval (0, M) is replaced by the present (a, M). Theorem 5.11 Let f2 and u be as in Theorem 5.10. Suppose that, in addition,
(01u)(xo) = 0 at some point xo E T,, ma.
(5.20)
154
5 Monotonicity of Positive Solutions in a Bounded Set C1
Let Zo(a) be the component of Z(a) that contains xo+Eel for sufficiently small e > 0, and let Yo(a) be the reflection of Zo(a) in T. Then u(xa) S2
= u(x) =
for all x E Zo(a),
Yo(a) U Zo(a).
(5.21) (5.22)
Proof (i) Once again let w(., u) := u-uµ on Z(µ), this time for µ E [a, M). Fix Z E Z(a) and let µ 1 a. Since u(z) < u(zµ) for µ E (a, z1), by (5.18), and since u(zµ) varies continuously for fixed z and varying p, we have w(z, a) < 0. Lemma 3.8 now states that Aw - Aw >- 0
in Z(a) in the generalized sense,
(5.23)
where w = w(z, a) and the constant A is described in Lemma 3.8.
(ii) Suppose for contradiction that w(zo, a) 0 at some point zo E Zo(a). Then w(., a) < 0 everywhere in Zo(a) [because, if the supremum 0 of w(., a) were attained in Zo(a), then w(., a) would be the zero function
in Zo(a), by (5.23) and the strong maximum principle, whereas our hypothesis is that w(zo, a) * 0]. Now consider the point xo in (5.20), observing that w(xo, a) = 0 because
xo E T. We apply the boundary-point lemma for balls (Lemma 2.12) to the function w(., a) and the ball B := -4 (xo + pet, p), with p so small that B c Zo(a). It follows that (01 w) (xo, a) < 0, which contradicts (5.20) because (81w)(xo,a) = 2(81u)(xo). Accordingly, w(z,a) = 0 first for all z E Zo(oc), and then, by continuity, for all z e Zo(a); thus we have proved (5.21). (iii) It remains to prove (5.22); let q E 8Yo(a) \ T. Then q" E 8Zo(a) \ Ta c F., so that u (q") = 0 by hypothesis (a) of Theorem 5.10. It now follows from (5.21), with x = q", that u(q) = u (q") = 0. But then q E Ofl [otherwise we would have u(q) > 0]; in other words,
8Yo(a) \ T,,
852.
(5.24)
This implies, since 0 is connected, that S2 = Yo(a) U Zo(a). Some readers may regard this implication of (5.24) as self-evident; others may consider our claim to be too intuitive or too condensed. A detailed proof is offered in §5.4.
We turn now to an analogue of Theorem 5.10 for the equation
Au+b181u+f(u)=0 in 0,
(5.25)
5.3 Monotonicity in 52
155
in which b1 = const. >- 0. For equation (5.2) with b * 0, we have chosen
variables aligned with b/jbj; in any case, the unit vector k = el. For the moment we suppose that f is locally Lipschitz continuous. The new feature is this: upon setting w(., p) := u - uµ on Z(p) for p E [a, M), we obtain the equation ,Lw - b181 w + y(z, p)w = -2b1(31u) (z)
for z E Z(p),
(5.26)
in place of (3.6); again w = w(z, p) and y is as in (3.5). When b1 > 0, we can infer that w(. , p) is a subsolution, relative to 0 - b181 + y and Z (P), only if we know that 81 u < 0 in Z (p). It is sufficient in the first instance to know that 81u < 0 in some neighbourhood of F,, but to prove even this is not easy if Fis not smooth, or if IF,, is smooth but contains points x at which n1(x) = 0, or if f (0) < 0. Therefore we restrict attention to boundaries slightly smoother than those of class C1, to beta caps and their right-hand boundaries Tp, and to functions f with f (0) >- 0. Theorem 5.12 Let SZ be a bounded region with 852 of class C' and with the interior-ball property (Definition 2.14) at every point of rp. Here I'p is the right-hand boundary of the beta cap Z(/3) for the direction e1 (Definitions 5.1 and 5.8) and T,, (e'), Z (µ,e1), /3 (e1) ,... are abbreviated to Tµ, Zµ, .... Assume that u : S2 --> R has the following properties.
(a)uEC1(4), u>0inn, u=0onF#. (b) For all cp E C'(52),
-VT - Vu+cpb181u+cpf(u)} =0, where b, = const.
(5.27)
0, f (0) >- 0 and f is otherwise as in Theorems 3.6 and
5.10.
Then the usual monotonicity conditions hold in the beta cap: u(z) < u (zµ)
81u(x) < 0
if if
p E (/i, M) and z E Z (p), x 1 > /3
and
x E 52.
(5.28) (5.29)
Setting w(. , p) := u - uµ on Z (p) for y E [/3, M), we wish to prove that
z E Z(p)
(01u) (z) < 0 and w(z,,u) := u(z) - u (zµ) < 0
(**)
whenever p E (/3, M). The proof of Theorem 3.6 involved a boundary strip Zh(p), a subset of Z(p) having width h independent of p, with the property that u(x) < iC for x E Zh(p). That h was independent of p was a luxury rather than an essential ingredient of the proof. This time, we shall use a boundary strip WI(p), still within Z(p), with the property that
156
5 Monotonicity of Positive Solutions in a Bounded Set 0
both a1u(x) < 0 and u(x) < lc whenever x E W,(/.t); as far as we know a priori, it might be that the width q(p) 10 asp . f3. This prospect may be uncomfortable, but we shall find that it is not catastrophic. In fact, Lemmas 5.13 and 5.14 will almost reduce the proof of Theorem 5.12 to that of Theorem 3.6. Notation Once again y(., p)
f1(u) - f 1(uµ)
at points z where u(z)
0
at points z where u(z) = uµ(z).
u - uµ
uµ(z),
(5.30)
Since u vanishes on I'p and is uniformly continuous on n,
3 h > 0 such that
dist (x,rp) < h and x E 12
u(x) < K. (5.31)
We write
w, (/,,) := Z(u) n { x I dist(x,r) < q(p) }
(5.32)
for the boundary strip, to be constructed in Lemma 5.13, with the additional property that a1u < 0 in W,(p). Lemma 5.13 For each p E (f, M) there is a number ri(p) > 0 such that
(atu) (z) < 0 and u(z) < K.
z E WW(p)
(5.33)
It follows that ,Lw - b181w + y(z, p)w >- 0
in W,, (,u) in the generalized sense,
(5.34)
where w = w(z, p) and y is defined by (5.30).
Proof (i) Let p E (fl, M) and p E I'µ. Then p E fp; since /1 >- A, Remark
5.9 shows that nt(p) > 0. We shall use the boundary-point lemma to prove that (0 1u) (p) < 0. By hypothesis there is a ball B c Q such that pEaB;then, forallcpEC,11°(B)with 'p 0,
-
f{-Vco-Vu+coblalul
= - f cp{[ft(u)-ft(0)]+.ft(0)+f2(u)} < fco{Au_fi(o)-f2(u)},
where A is our usual Lipschitz constant for f t on the interval [0, supra u], so that A >- 0. Since f 1(0) >- 0 and f 2(u) >- 0 by hypothesis,
<0, JB
5.3 Monotonicity in S2
157
so that u is a generalized supersolution relative to 0 + b1 1 -A and B. Moreover, u(x) > u(p) = 0 for all x E B, and el is an outward unit vector at p because ni(p) > 0. Thus Lemma 2.12 implies that (alu)(p) < 0. (ii) Let µ E [a, M). Calculating as in the proof of Lemma 3.7, step (i), and abbreviating Z(p) to Z, one finds that, for all (p E C,(Z),
J
{ -VVw-Wb1aiw+w[fi(u)-.fi(ua)] } dz =
+ [f2(u) - f2(uµ)] } dz, z where w = w(z, µ), p = lp(z), ai = alazi and so on. -Jip{2b,
(5.35)
(iii) Let µ E (/3,M). We have shown in (i) that aiu < 0 on the compact set t ; by hypothesis, aiu is (uniformly) continuous on S2. Hence a, u< -c(µ) on ,,, for some c(µ) > 0, and then there is a number
p(µ) > 0 such that (aiu)(z) < 0 if z E Z(µ) and dist(z,f) < p(µ). Choose ii(y) to be the smaller of p(µ) and the h in (5.31), observing that dist(z,I'f) < dist(z,r,) because f,, c I'p. With this choice of q(µ), condition (5.33) holds.
Now restrict (5.35) to test functions cp E C°°(W,(µ)) with 9 > 0. On
the right-hand side, (aiu)(z) < 0 and f2(u(z)) = 0 for z E WI(µ), by (5.33), while f2(ul,(z)) >- 0 because f2 >- 0 on [0,00). Thus the right-hand member of (5.35) is non-negative, which implies (5.34).
Lemma 5.14 If µ e [/3,M) and if, for all z E Z(µ), both (01u)(z) < 0 and w(z, µ) < 0, then
Lw - b1a1w - Aw >- 0
in Z(µ) in the generalized sense;
(5.36)
here w = w(z, µ) and A is as in Lemma 3.8.
Proof If cp > 0 in (5.35), then -cobiaiu >- 0 in Z(µ) by hypothesis. The rest follows from the proof of Lemma 3.8.
Proof of Theorem 5.12 (i) In order to prove that (**) holds for all sufficiently small, positive M - µ, we modify the proof of Theorem 3.6, step (i), as follows. First,
M-I
2M I< < M
dist(z,I'/) <
(fl+M)
for all z E Z(µ).
Second, we may and shall suppose that the width q(µ) of W,(µ) does
not decrease as µ increases [because, if µ < v and z E Z(v) and
5 Monotonicity of Positive Solutions in a Bounded Set 52
158
< q(µ), then dist(z,I'µ) < q(p) and so (81u)(z) < 0 and
dist(z,I'v
u(z) < K]. Conse/quently,\
I R 2M dist (z, Tµ) < q C /12 M ) < rl(µ)
for all z E Z (µ),
so that Z (µ) = W,(u) for such values of µ. Then Lemma 5.13 and the maximum principle for thin sets (Theorem 2.19, with 6 as in the proof of Theorem 3.6) ensure that
M-
I 11
\
R2M I < µ <M and
µ> _
M
an d
2
I
Z( p)I
<6
/(aiu)(z) <0 and w(z,µ) < 0 for all z E Z(µ).
That (**) holds for these values of p now follows from Lemma 5.14, from the strong maximum principle (Theorem 2.13) and from the existence of boundary points p, for each component of Z (p), at which w(p, µ) < 0.
(ii) Let (m, M) be the largest open interval of µ in which (**) holds; assume (for contradiction) that m > /i. (a) First, a1u < 0 in Z(m) because, if z E Z(m), then z E Z(µ) for some p > m. Second, we obtain w(z,m) < 0 for z E Z(m) by fixing z E Z(m) and letting y 1 m. Then Lemma 5.14 and the strong maximum principle imply that w(z, m) < 0
for z E Z (m).
(5.37)
The condition alu < 0 now extends to T,,, fl 52 by application of the boundary-point lemma to w(.,m) and to balls in Z(m) with boundary meeting Tm; therefore,
a1u < 0
on Z(m) \ M.
(5.38)
(b) Writing wE := w(., m - e) for 0 < E < Eo, we shall obtain a contradiction by showing that a1u < 0 and wE < 0 in Z(m - E) if Eo is sufficiently small. Provided that m - eo > fl, we already have that a1 u < 0 in W,1(m - E), by Lemma 5.13. The only changes from the proof of Theorem 3.6 are that Ow-b1a1w replaces Lw in the two lemmas and that now we must so choose E, and
F that
GE :=Z(m-s)\(EEUF) e W,,(m-s).
5.4 A little topology
159
For this last, the main step is a suitable choice of the cross-section S of the cylindrical set E. To make this choice, we refer to (3.29), replace the h there by rl (2 (/3 + m)), and demand that the conditions on so include
m-Eo>- 2(Q+m). Then rl( z) > rl (z (/3 + m)) for m -to < p < M [by the second observation in (i)].
We already have alu < 0 on F, by (5.38), and the earlier construction gives a1 u < 0 on EE ; that w, < 0 on EE U F is argued as before. Theorem 5.15 Let 91 and u be as in Theorem 5.12. Suppose that, in addition,
(a1u)(xo) = 0
at some point
xo E Tp n Q.
(5.39)
Let Zo(/3) be the component of Z(/3) that contains xo + sel for sufficiently small s > 0, and let Yo(/3) be the reflection of Zo(f3) in Tp. Then
u(x'3) = u(x) for all
x E Zo(/3),
K2 = Yo (fl) U Zo(/3),
(5.40) (5.41)
and the coefficient b1 = 0.
Proof The proof of Theorem 5.11 remains valid here if we replace the interval (a, M) by (fl, M), Theorem 5.10 by Theorem 5.12, and Lemma 3.8 by Lemma 5.14.
It remains to prove that b1 = 0. Setting µ = /3 in (5.35), we observe from (5.40) and (5.41) that u = uil on Z(/3) and hence w(.,/3) = 0 on Z(/3); therefore (5.35) with µ = /3 reduces to ipb,Oju = 0
for all
cp c=
z (a)
Since 31u < 0 in Z(/3), by (5.29), it follows from a result of Exercise 1.16 that b1 = 0.
5.4 A little topology Remark 5.16 This section is concerned only with a full proof of (5.22) in Theorem 5.11. In addition to the basic fact that S2 is open, bounded and connected, we have the following data. (A) Zo := Zo(a) is a component of the alpha cap Z := Z(a).
160
5 Monotonicity of Positive Solutions in a Bounded Set E2
(B) Yo := Yo(a) is the reflection in Ta of Zo (so that Yo c S) by Theorem 5.3).
(C)8Yo\Tj c8S2. The question is whether these conditions imply that 0 = Yo U Zo. We recall that, for sets A and B in a metric space,
AUB=AUB, AnBcAnB, 8(AUB) c 8AU3B, 8(AnB) c 8AU0B.
Lemma 5.17 Let C be a connected, non-empty set in a metric space (M, d).
If A c C and C n 8A is empty, then either A = 0 or A = C.
Proof The pair (C,dI c.c) is also a metric space, and C = into A U 8cA U extc A,
where intc, 8c and extc denote respectively the interior, boundary and exterior relative to C. Given the hint in Exercise 5.26, one checks without difficulty that 8cA c C n 8A (where 8A is the boundary of A relative to M), so that 8cA is empty. Since C is connected, it cannot be the union of two disjoint, non-empty sets that are open relative to C; one of intc A and extc A must be empty. If intc A is empty, then
AcC=extcA=intc(C\A) cC\A
A=0.
If extc A is empty, then
AcC=intcAcA
A=C.
Lemma 5.18 Let D and E be sets in a metric space, with D connected, non-empty and open. If D intersects E but not 8E, then D c E. Proof Let A := D n E. Then A c D and D n 8A is empty because
DnOA cDn (8DU8E) =DnBE =0. Accordingly, we may apply Lemma 5.17 with C = D : either D n E = 0 or D n E = D. The former is contrary to hypothesis; the latter implies
that D c E.
5.4 A little topology
161
Lemma 5.19 If { Gt t E T } is a family of disjoint open sets in a metric space and G := UtET Gt, then each aGt c aG. I
Proof Let x E 3Gs for some s E T. Then X E G, x
GS and, since
every open set containing x intersects Gs, we have x 0 G, if t * s. Thus x E G \ G. But G \ G = aG because G is open. O Proposition 5.20 With 0, Zo and Yo as in Remark 5.16, define
A := Yo uZou (S2naZo) u (S2n3Yo).
(5.42)
Then A = 0, which implies that S2 = A = Yo U Zo.
Proof (i) That A = Yo U Zo is immediate from the definition (5.42). We
shall prove that A = Q by means of Lemma 5.17 (with C = 0 and M = RN). First, A c 0 because Yo c n by (B) of Remark 5.16 and Zo Q by (A). Second, we shall prove in (ii) that A is open in RN ; therefore,
aA=A\A = Y0UZ0\{Y0UZ0U(nnaZ0)U(12naYo)} = a(YouZo) \ { (12nazo) u (On OYO) }
c
(aYO u aZo) \ { (s2 n azo) u (s2 n aYo) }
,
which shows that S2 n OA is empty. Thus Lemma 5.17 applies; since A is not empty, we have A = 0.
(ii) To prove that A is open in RN, we consider separately three types of point in A. (a) Let X E Yo U Zo. Then X E int A because Yo U Zo is open. (b) Let p E 0 n aZ0. Now aZo c OZ by Lemma 5.19, and OZ c OOUTa because Z = 0 n { x I xl > a 1; since 0 and Of) are disjoint, p E s2 n Ta.
Hence there exists a radius p = p(p) > 0 such that 9(p, p) c 0. To show that M(p, p) c A, we consider B+ S
B_
.(p,p)n{x xl>a}, M(P, p) n T«,
P(p,p)n{x x,
Observe that -'(p, p) intersects Zo because p E aZo, and that S and B_ do not intersect Zo; therefore B+ intersects Zo. Next, B+ does not intersect aZo because aZo c % U Ta. Therefore Lemma 5.18, with D = B+ and E = Z0, shows that B+ c Zo.
162
5 Monotonicity of Positive Solutions in a Bounded Set S2
Then B_ c Yo (by reflection in Ta) and S c 12 n aZo (because S c 0 and points of S, being limit points of B+, are also limit points of Zo). We have shown that B+, B_ and S are each contained in a subset of A; consequently, M(p, p) c A and so p E intA.
(c) Let q E 92 n aYo. Then q E T, because aYo \ Ta c as2 by (C) of Remark 5.16. In addition, there is a sequence in Yo such that y -+ q. But then yn (yn)a E Zo, and yn -+ qa = q. Hence q E 92 n aZo, and q E int A by step (b).
5.5 Exercises
Exercise 5.21 A boundary 00 is of class Co,' if, in each local representation YN = h(y', p) of as2 [Chapter 0, (viii)], the function h(. , p) is uniformly Lipschitz continuous, so that h(., p) E Cb"(G(p)) according to Definition A. 10.
Show by means of an example that, for part (a) of Theorem 5.7 it is not enough to demand that 1 be a bounded region with aQ of class Co,l
Exercise 5.22 The elliptic disk
E :={(x,y) ER2 I Axe-Bxy+Cy2<1}, where
A := 1 + a sine y,
B := a sin 2y,
C := 1 + a cos 2 y,
is characterized by the parameters a > 0 and y E [0, n/2]; the major axis has length 2 and slope tan y, while the minor axis has length 2(1 + Prove that, for the set E, a)-'12.
a(e')
(e') =
z(1+a)-112(asin2y)(1+asin2y)-1/2,
according to Definitions 5.1 and 5.8. Find the solution u E C(E) n C2(E) of
Au+1=0 in E,
UIM=0.
For the direction e1, how well does Theorem 5.10 describe the graph of u? Give separate answers for y = 0, 0 < y < it/2 and y = 7r/2. [The function u can be found from Exercise 4.21.] Exercise 5.23 Let 0 be the set called S 2 in (5.16); assume that a function u satisfies the hypotheses in Theorem 5.10, strengthened to u = 0 on as2.
5.5 Exercises
163
Show that u (xi, x2) < u (-xi, x2) whenever xi > 0 and x E 92, and that u (xl, x2) = u (xl, -x2) for all x E SZ. Establish the signs of 81 u and 02u throughout S2, except for the sign of alu in the open subset { x E 92 I -y < xl < 0 }, where y is as in (5.15). Infer that the maximum of u can occur only in the line segment (-y, 0) x {0}. Exercise 5.24 Let 0 and .1 = 2 (el) be as in Definition 5.8. Prove that there exists a point xo E TT n 00 such that nI(xo) < 0; moreover, such that nl(xo) = 0 or nl(xo) = -1. Exercise 5.25 State a result that extends Theorem 3.3 in the way that Theorem 5.10 extends Theorem 3.6. Prove your result by adjusting the notation in the proof of Theorem 3.3.
Exercise 5.26 Let A c B c M, where (M, d) is a metric space and B is not empty. Write 8A and BBA, respectively, for the boundary of A in M
and the boundary of A relative to B. Prove that 8BA c B n 8A, with equality if B is open in M. Give an example of inequality. [A set D c B is open relative to B if and only if D = B n G for some G open in M. For a set S in a metric space (X, d), a point p belongs to 7S if and only if every open set containing p intersects both S and X \ S.]
Exercise 5.27 Define an annulus or spherical shell by A
x E RN
a< IxJ 2, 0 < a < b. Suppose that u E C(A) n C2(A),
u > 0 in A and u = 0 on 8.(0, b), Du + f (u) = 0
where f
:
in
A,
[0, oo) -> l[8 is locally Lipschitz continuous.
(5.43a) (5.43b)
Let 8/8r
(x/IxJ) . V.
Prove that, if x, y c A with xJ E ((a + b)/2, b), y/lyl = x/IxI and JyI+IxI =a+b, then au
(x) < 0 and u(x) < u(y);
(5.44a)
also that Or ( x
)<0 if
1xI=a+b
(5 . 44b)
Exercise 5.28 Add to (5.43a, b) the condition
u=0
on
09(0, a).
(5.43c)
164
5 Monotonicity of Positive Solutions in a Bounded Set C1
Display a spherically symmetric solution (depending only on r := lxl) of (5.43a, b, c) for the case in which
)zt
N=3 and f(t)=( b n
(for all t >- 0).
Show that, for this case, (5.44a, b) are satisfied amply when alb is small, but are satisfied only by a hair's breadth when alb is very close to 1. [A small part of Exercise 1.21 is relevant.] Exercise 5.29 This exercise prepares for the description, in Exercise 5.30, of a solution of (5.43a, b, c) that is not spherically symmetric. It is to be hoped that the essence of the argument emerges, even though parts of it are beyond the range of this book. With A as in Exercise 5.27, let H be the real Hilbert space formed by completion of the set C,(A) in the norm II . 11 defined by IlvII :=
(v,w) :=
where
(v,v),
fVv
Vw.
(i) Derive, or improve on, the following inequalities. (a) If V E H and is spherically symmetric, say r := lxl and v(r) := v(x), then IIvIIZ
D(r)2
q(r),
where q(r)
r-a
if a < r < z(a+b),
b - r if 2(a+b)
(b) For all v E H,
fv2
N-1
<
I(b-a)2 8
(a2ab )
IIv112.
(ii) Elements of H that are not spherically symmetric may have unbounded values. Verify that the following functions w satisfy IlwII < 00. Given a point p E A, choose 6 so that 0 < 6 < dist(p, 3A), and define w(x) := hN I Ix
pl)
for all x E A \ {p},
where, if A c R2, we use v
h2(t) (t)
(log t) - 1 if 0 < t < 1, y = const. E (0, { 0
ift> 1,
,
5.5 Exercises
165
or, if A c RN with N > 3, we use hN(t)
t-7 -1 if0
(o_i)
0
fro l7U
= J a(r, 90) dr = -
f
b 8v(xo)
0
ar (r, 9o) dr,
in which v E Q°(A), 9:= x/Ixl,v(r, 9) := v(x), and from the BuniakowskySchwarz inequality. For each function w in (ii) one can show, by extensions of Exercises 1.23 and 1.25, that there is a sequence of functions w,, E Cc(A) such that II w - w II -+ 0 as n -+ oo; therefore each w is indeed in H.]
Exercise 5.30 Here we shall describe a solution u of (5.43a, b, c) with f : R --> R defined by
f (t) = eg(t) := t'(t - c)fH(t - c)
for all
t E R,
where fH denotes the Heaviside function [Chapter 0, (v)], c is a given positive constant, and the positive constant t will be calculated a posteriori because lull is prescribed (cf. Exercise 3.17). Let the Hilbert space H be as in Exercise 5.29 and let
G(t) := 1(t-c)ZfH(t-c)
for all
t E IR.
For each p > 0, the variational problem of maximizing fA G(v(x)) dx v E H Il v Il = p } has a solution u E S. such over the sphere Sp that
JG(u)>0
and
(u, 9) = 8
J
g(u)(p
for all
c: H,
A
where ,,
= Ilull2/ f g(u)u. A
Moreover, the equivalence class u (of functions equal almost everywhere
in A) has a representative uo that satisfies (5.43a, b, c), for f and e as above.
166
5 Monotonicity of Positive Solutions in a Bounded Set C1
Given all this, show that the representative uo of u is not spherically symmetric if p is sufficiently small; in fact, if b - a
0 < C2Iaa(0,a)I)
1/2
p < c.
Appendix A. On the Newtonian Potential
A.1 Point sources in ][83
(i) By the potential of a unit source at the point c E 1R3 we mean the function 4) defined by 1
(D(x)
.
Ix
1 c_,
x E 1[83 \ {c}.
(A.1)
The words `potential' or `potential function' imply that the vector field -V1 is the object of primary interest; the phrase `unit source' or source of unit strength is intended to convey the result (A.3) below. There is a good case for believing that God devised the function 1 on the first day of the Creation. Before considering a small part of the evidence for this belief, we note two properties of the function. (a)
satisfies the Laplace equation :
0q) :_ (a1 + az + a3)1 = 0 For, writing R := Ix - cI
aR
in R3 \ {c}.
(A.2)
(x1 - ci)2 + .. + (X3 - c3)2 }112, we have
x; - ci R 21
1
'R
'
3 (xi - ci)2
a`R=R4
R
1 x,-ci R2
R
1 R3'
summing this last over i from 1 to 3, we obtain (A.2). (b) If 92 is a bounded open set that contains the point c, and aQ is of class C 1, then
- fn a = 1. an
(A.3)
Here n denotes the outward unit normal on an (see Figure A.1), a/an n V and the element dS of surface area is implied. 167
Appendix A. On the Newtonian Potential
168 x3 A
an x2
Fig. A.1.
To prove (A.3), we introduce a ball R(c, p) with p so small that R(c, p) c f); on the sphere 8.(c, p), the unit normal n outward from 0' := S2 \ R(c, p) points towards c. Hence (with R = Ix - cl) atj
d
On l a_4(c,p)
1
dR 4nR
1
I
4R 2, R=p
V
so that 8(D
(A.4)
1.
an
Now apply the divergence theorem to the vector field V (D and the set Q':
0=/n M'=Jas 8n + I -O(D
04),
aa(c,p) On
(A.5)
then (A.3) follows from (A.4) and (A.5).
(ii) Electrostatics This may seem an absurd heading for the single rule
that is stated here. But this rule is the basis of the whole subject, and leads to the electric fields of arbitrary distributions of electric charge. Consider a point charge Q (a particle having electric charge Q) at the point c E R3. According to Coulomb's hypothesis of 1785, illustrated in Figure A.2, the force on a test particle, having unit electric charge and located at any point x * c, is X
-QOI(x),
(A.6)
C= K Ix - C13 where the positive constant K is the dielectric constant of whatever E(x)
4Qx
material is assumed to occupy R3. The force on a particle at x having
A.1 Point sources in R3
169
Q or
C
x3
x1
Fig. A.2.
charge q is qE(x), and the vector field E is the electric field of the point charge Q. The formula (A.6) is an example of an inverse-square law, because E(x) has direction along the line from c to x, and magnitude proportional to IX - CI-2.
(iii) Signs No single convention, for the sign of potential functions and of source strength, is satisfactory for all applications. One reason is that positive point charges repel each other, whereas point masses attract each other. Nevertheless, we retain the definition and terminology in (i) for all three cases of our brief flirtation with physics. (iv) Gravitation Here the basic hypothesis is traditionally attributed to Newton, who published it in his Principia of 1687 after perhaps twenty years' contemplation. However, it may be that Hooke has not had his share of the credit (Arnold 1990, Fauvel et al. 1988). Consider a point mass M (a particle having mass M) at the point c E R3. According to the Newtonian hypothesis, the force on a test particle, having unit mass and located at any point x * c, is F(x)
_
M x-c
_ M pcfi (x), 47rkg Ix - c13 - kg
(A.7)
where kg is a positive gravitational constant. The force on a particle at x having mass m is mF(x). (Our signs are unfortunate here in that the potential energy of this particle is -Mm0(x)/kg apart from an additive constant.) Again we have an inverse-square law.
170
Appendix A. On the Newtonian Potential
(v) Point sources in hydrodynamics In this subject it is the Laplace equation that is basic, and point sources can be used to construct explicit solutions. For the flow of a highly idealized fluid past a compact set (or body) E in R3, one seeks a velocity potential p such that -V(p represents the fluid velocity and such that cp = 0 in
][8
3
8cp
\ E,
8n
= 0 on 8E,
(A.8a)
-Vcp(x) --+ (U,0,0) as lxi - oo.
Here U is a given positive constant, so that the velocity `at infinity' is prescribed. Let Q := R3 \ E. By a solution of the problem (A.8) we mean a function cp that satisfies not only (A.8a) but also the conditions (p E C' (?i) n C2p,
as r := lxi --+ co,
Vgo(x) + (U, 0, 0) = O(r2) and qp(x) + Uxi -+ 0. (A.8b)
Theorem A.1 If 8Q is of class C' and SZ is connected, then the problem (A.8) has at most one solution.
Proof Let w := cpl - c02 be the difference of two solutions, so that w E C1(4) n C2(12) and
Lw = 0 in as r -* oo,
52,
8w
an
= 0 on 852,
Vw(x) = O(r-2) and w(x) -+ 0.
Then w(x) = O(r-1) as r -* co, by Remark 4 before Theorem A.2 below. Substituting V = wVw into the divergence theorem
Jn
v-
V=fn n V,
where V is to be in C1(S2',1R3) and a suitable approximation SZ' to 52 will be chosen presently, we obtain the energy identity 10' IvwI2
=
w
wLw,
Jasz 8n fo, in which the left-hand member is twice a kinetic energy, for a fluid of density 1. We cannot begin with the energy identity for the set 92, for two reasons: Q is unbounded and w need not be in C2(S2). (It would be artificial to demand in (A.8) that p E C2 (s2).)
A.1 Point sources in 1R3
171
To define SZ', we first set S A := S2 fl 9(0, A) with A so large that as2 = OE c -4(0,A), and then choose IT := I)A,,,,, the mth approximation to IT A in the sense of Theorem D.9 (Appendix D). Since 1 A,m c QA c Q, we have w E C2(SiA,,,,) and Aw = 0 in S2QA,,,,, so that IvwI2
J
=
LA w all'
Let m - co; by Theorem D.9 and because aw/an = 0 on OS2,
f InA
IVw12=
aw
famoA wan.
Now let A -+ oo; we have waw/an = O(A-3) for r = A, while the surface area I09(0,A)l = 4irA2, so that finally fn ww12 = 0. This implies that JVwj = 0 (the zero function) because IVwI is continuous; then w(x) is constant in I) because 92 is connected, and, in fact, w = 0 in n because
w(x)-*0as r-*oo. If S2 is not connected, we still obtain IVwI = 0 in the foregoing proof, so that the problem (A.8) admits at most one velocity field -v(p. For the construction of a solution of (A.8) by an elementary method, it is usual and profitable to cheat by writing
cp(x) = -Uxi +
47r
q = const. > 0,
1
1
Ix + cl
Ix - cl
(A.9)
'
c = (y, 0, 0) with y > 0,
and by then showing that this cp is the velocity potential of flow past a certain body E called a Rankine solid (Figure A.3). Note that the three terms in (A.9) are the potentials, respectively, of a uniform stream with velocity (U, 0, 0), of a point source at -c of strength q and of a point source at c of strength -q. Thus cp is defined on R3 \ {-c, c}, indeed, cp E C110(R3 \ {-c,c}), but we are interested mainly in the velocity -VT on and outside E. It is far from obvious that the formula (A.9) implies
the picture in Figure A.3; before taking up this question in (vi), we consider a limiting case. Set q = k/2y, and let y --> 0 with k fixed and positive, and with x * 0. Then q
1
1
47r
Ix + cl
Ix - cl
_
l
k ((al 1)
4n 2y 47
xl /
xj
2y + 0(72) }
47r IX13
J
(A.10)
172
Appendix A. On the Newtonian Potential s = (x2 + x3 )1'2
W- x1
tit --Ot-
a
Fig. A.3.
This is the potential of a dipole (or doublet) at the origin, of strength k and type (-1,0,0), or of strength -k and type (1,0,0); the type states the direction from negative source to positive source before the limit is taken. This function satisfies the Laplace equation in R3 \ {0}, because A and 01 commute when acting on C3 functions. Now define spherical co-ordinates r, 0, 2 by x = (xl, x2i x3) = (r cos 0, r sin 0 cos A, r sin 0 sin A),
0, 050Sn, -it
(A. 11)
Let k = 2nUa3; then the limiting form of (A.9) is
/
1a3
c00(x) _ -U cos 0 I r + 2 r2
(r > 0),
I
(A.12)
and the reason for our choice of k is that as°(x)=-Ucosel-a33)
which vanishes if r = a. Thus 4po is the unique solution of the problem (A.8) when the body is a closed ball: E = 9(0,a). (vi) The Stokes stream function In addition to the spherical co-ordinates r, 0, 2 in (A. 11), we introduce cylindrical co-ordinates xl, s, 2 (Figure A.4) : x = (x1, s cos A, s sinA) ,
s=
(x2 -}- x3)12
,
-z[ < 2 < n.
(A. 13)
Suppose that a potential function p satisfies the Laplace equation
A.1 Point sources in 1183
173
Fig. A.4.
°tp = 0 in SZ and is cylindrically symmetric about the xl-axis. (This last means that f2 is a figure of revolution about the xl-axis and that
=0
-X3acp
TA
in fl.)
If both these conditions hold, we write cp(x) = ip(xl, s) and, apart from an additive constant, define the Stokes stream function W(x) = ip(xl, s) corresponding to the potential cp by a;p
pp
axl(-, as)
-S
I
Ts
(_, a sax1
(A.14) .
A better definition will be given in Exercise A.24, but the present one is legitimate because (by a formula for ° in Exercise 1.18)
a a1p
a aip
ax, as
as ax,
(020 s
020
axi + as2 +
100 as)
s
- s°cp = 0.
We now show that the vector field V4 is tangential to the level sets W(x) = const. } of W ; it is this fact that makes W a splendid companion to cp. Here we adopt the convention that the zero vector is both perpendicular and parallel to every vector. First, the vector fields V9 and VW are orthogonal (are perpendicular to each other): {x
I
vcp 0W =
airy aip
ax, ax, +
LIP aip-
00 00
00 00
as as - s -axl as + as axi = 0.
Second, at any point xo E f1, the vector (VW)(xo) is normal to the level
Appendix A. On the Newtonian Potential
174
set { x E 0 I W(x) = const. = W(xo) }. Together, these two facts show that (Ocp)(xo) is tangential at xo to the level set of W containing xo.
Accordingly, a picture of the level sets of W (which are surfaces of revolution in R3 wherever IVW1 * 0) or of the level sets of ip- (which are curves in the half-plane R x [0, c) wherever I Vip I * 0) indicates the directions of the vector field VT. Such a picture also indicates, very approximately, the magnitudes V p1 = IOW I /s. For, let 8(x) be the distance from a point x on a level set to an adjacent level set; if 8(x) is small, then VW(x)1
IW2 - W1 1160,
where W1 and W2 are the constant values of W on the level sets in question.
For the velocity potentials in (A.9) and (A.12), the definition (A.14) yields
W(x) _ -1 Us2 + 2
4
x, +y
41r
{(x1 + y)2 + S21112
X1 -Y
/
- {(xi - y)2 + sz}1/2
,
(A.15)
3I
WOW = - 2 U sine e I r2 - a
.
(A.16)
To calculate Wo one first writes (A. 14) in terms of///spherical co-ordinates
(Exercise A.18). The boundary aE of the Rankine solid is a part of the level set {x W(x) = 0}. Detailed description of aE still requires I
work (Exercise A.26), but the Stokes stream function enables us to establish the main properties of this boundary before resorting to numerical calculation.
A.2 The Newtonian potential: first steps The Newtonian kernel This is another name for the potential of a unit source at the origin, but now the setting is RN with N E { 1 , 2, 3, ...}. [In (A.21) below, K is said to be the kernel of the integral operator acting on f.]
It is a practical outlook, not a hollow desire for generality, that prompts our interest in the Laplace operator a; + . + aN for all N. The case N = 1 has the virtue that easy arguments (as in Remark 1.5 and Exercise 1.17) may point the way to results that are true in higher dimensions. In R2, there is the golden result that the real and imaginary parts of a holomorphic or analytic function u+iv, of the complex variable z = x1 + ix2, satisfy the Laplace equation; this leads to many explicit solutions, some of which are even relevant to physical situations. (Such
A.2 The Newtonian potential: first steps
175
situations involve long cylindrical bodies with axis normal to the plane being considered, and of arbitrary cross-section.) The prevalence of the Laplace operator in JR3 has been illustrated, in a small way, in §A.1. The story does not end there; wholly concrete problems, originally set in R3, can lead to the Laplace equation in R5 (Exercise A.25). Let IN(c, p) denote, for the moment, the ball in JRN with centre c and radius p, and let ia.N(c, p)) denote the surface area of its spherical boundary. [Note that -41 (c, p) is the interval (c - p, c + p) of length or `volume' 2p; hence the `surface area' I8-41(c, p)I of the corresponding sphere is d(2p)/dp = 2.] Our extension K(x - c; N) of the D(x) in (A.1) is the result of three conditions: K(x; N) is to depend only on Ixl and N; we demand that dK(x;N) 1 whenever r := Ixl > 0; (A.17) dr = l a_4N(0, r) I
-
andK(x;N) is to vanishforr=Oif N= 1,forr=1 if N=2, and as r --> co if N > 3. Accordingly,
-ZIxI
K(x;N) .--
in
118,
in R2 \ {0},
2 log I
(A.18a)
1
where NJG "2
1
KN
(N - 2) Ia.N(O,1)I
(N
(N/2)!. (A.18b)
2),
The formula for the surface area of the unit sphere is a result of Exercise 1.14. Note that the formula stated for N >- 3, in (A.18a), is also valid for N = 1. The basic properties (A.2) and (A.3) of (D are unchanged, and are proved exactly as before.
AK(x - c; N) = 0
for x E RN \ {c}.
(A.19)
If 52 is a bounded open set in 1[8N that contains the point c, and a52 is of class C1, then
8K (x-c;N) = 1
(A.20)
here n denotes the outward unit normal on 852.
The Newtonian potential Let G be a bounded open set in RN and let
Appendix A. On the Newtonian Potential
176
f:G
R be a given function. Provided that the integral exists for all
x E RN, we call u(x)
:=.f K(x-l;;N) c
d
(x E RN, d = dg1... di;N) (A.21)
the Newtonian potential of the density function f . In classical physics the following hypothesis is made. The rule that
11:K(x-i;j;3)Qj, K
xE][83
J=1
is the electric potential at x of charges Q j at the points J, extends to the rule that (1/1c)u(x) is the potential, at all x E R3, of charge density f. Here f is measured in units of charge per volume. Essentially the same
hypothesis is made for the gravitational potential of a density f (now measured in units of mass per volume). One often calls cj or a source point, and x afield point. The new feature, that the potential is defined at points where there is charge or mass, prompts the question: is there a simple relationship between u and f at points of G? A short answer is that the Poisson equation -L u = f holds in G, provided that f is sufficiently smooth; this will be proved in §A.5. In this section we shall give an answer that is less direct but is both more general and easier to prove: if f has a certain integrability property, then a weak form of the Poisson equation holds in G.
Notation Henceforth we omit the label N when its absence causes no confusion. For some purposes the notation in u(xo) = jK(xo - x) f (x) dx,
xo E RN,
(A.22a)
is more convenient than that in (A.21); accompanying symbols are r := IxI,
ro := Ixol,
R := Ixo - xI,
(A.22b)
so that, for example, K(x - xo) = KNR-N+2 if N * 2.
There are a few cases for which the integral defining the Newtonian potential can be evaluated explicitly (for which the potential can be found `in closed form'). Of these, the case of constant density in a ball is the simplest and perhaps the most valuable; we state the result and then explain its derivation in a sequence of remarks.
A.2 The Newtonian potential: first steps
177
The potential of unit density in a ball Let B :_ 9(0, a); then
ux
JB
K x-
cN(a), r< a,
2N
d
(A.23a)
r >: a,
IBIK(x),
where IBI = ir N12aN/(N/2)! denotes the volume of B, and
- i1a 2 cN(a) =
a2
if N = 1, 1
if N = 2,
(log a + 2 I
(A.23b)
Remarks 1. For N = 1 the result (A.23) follows from any easy exercise
in integration. The same is true for N = 2 if we use the notation in (A.22) and the series log
1
1
Iz
Re log = log z - zo - zol =
1
ro
°°
1
+En
r
cos n(B - 80), (-)o
r < ro, ) (A.24)
where z = xl + ix2 = re'0. [If r > ro, we interchange z and zo.] For N = 3 we choose, for source points x, spherical co-ordinates r, 0, A such
that the axis {0 = 0} passes through the field point xo (Figure A.5). The Fubini theorem and a theorem about co-ordinate transformations (Apostol 1974, p.421; Weir 1973, p.158) allow us to write " 2 / n (r2 sin g de uB (xo) = 47r , r dr J 2 + ro - 2rro cos 0) 1/2 0 0
-
f
" rz
d. ,
and this is an elementary integral. These integrations yield (A.23) for N = 1, 2,3 (the properly sceptical reader is invited to verify this). The function UB for N = 3 is shown in Figure A.6. One easily checks that, for all N < 3, both UB and duB / dr are continuous at r = a; hence UB E C'(]RN) for N< 3.
2. Observe that for r >- a the potential UB is merely that of a point source at the origin of strength IBI, and that - ILuB =
(02+...+02N)x1+ 0
at least for N < 3.
+xN 2N
-1 ifra,
(A.25)
Appendix A. On the Newtonian Potential
178
Fig. A.S.
a2/2
a213
a
Fig. A.6.
We now proceed backwards. The result (A.25) can be regarded as an ordinary differential equation in terms of r, and one can use this equation to construct, with relatively little labour, a tentative formula for
all values of N. However, we shall exclude the case N = 1, because a third condition `at infinity' would be required in what follows. Using a formula in Exercise 1.18 for A acting on functions of r alone, we seek V(r) such that
LV=
d
rN 1 dr
(rN_1 dr
V=
-1, 0
r > a,
2
V(r) - r logy ->0 as r-3oo if N=2, V(r)-30 as r ->oo if N>-3,
A.2 The Newtonian potential: first steps
179
and such that V E C1 [0, oo) n C2([0, oo) \ {a}). This last condition fixes cer-
tain constants that are unknown initially and rules out singular behaviour at the origin. An elementary calculation now yields z
... AT-7 .f'_
Vlvlk, I -
l
//
J -4 +2a2( Aloga+2 I, r a;
r2
for N>-3:
2N+2a N-2' r
V(r)=
1
aN
N(N - 2)
2
1
1
rN-2
ra.
These are the functions on the right-hand side of (A.23a). 3.
It remains to prove that uB(x) = V(r) for N > 4, where uB(x)
denotes the integral in (A.23a). This follows from Theorem A.2 below, which gives a second characterization of the Newtonian potential when f and aG are exceptionally smooth. One easily verifies that, with G = B and v(x) = V(r), the hypotheses of Theorem A.2 are satisfied. The theorem has other applications, because one may well ask: when is a given function v the Newtonian potential of -AvIG? 4. We note the following in connection with condition (c) of Theorem A.2. If, as r := xI -> oo, IVw(x)I = 0(r-1-m) and w(x) -> 0, where the constant m > 0, then w(x) = 0(r-m) as r -+ oo. The reason is this: there are positive constants C1 and C2 such that, for r > C1, 00 dw(tx)
1w(X)1 = 1
dt
dt
r <J
00
co
x (Vw)(tx) dt C2r-m
IxIC2t1-mjxj-1-m
1
dt =
m
In Theorem A.2 the condition Av E Lp(G), p > N/2, is needed because condition (a) does not restrict sufficiently the behaviour of 5.
(Lv)(x) as x approaches
G.
Theorem A.2 Let G be a bounded open subset of RN with N > 2; suppose
that either 8G is of class C1 or G is listed in Remark D.4. Consider a function v : RN -+ R with the following properties.
Appendix A. On the Newtonian Potential
180
(a) v E C1(RN) n C2(1RN \ aG). (b) Lv E LP(G) for some p > N/2, and L v = 0 in RN \ G. (c) As r := lxi --* oo,
Vv(x) = O(r) and v(x) -> 0 if N
3,
V{ v(x) - cK(x) } = O(r2) and v(x) - cK(x) ---> 0 if N = 2, for some constant c. T h e n v is the Newtonian potential o f -Ov I G
Proof (i) In this first step, v need not be the function specified above. Let 0 be a bounded open set with 3f of class C1, or else a set listed in Remark D.4, and let v and w be in C2(52). Then the divergence theorem for the vector field vVw states that
j{Vv.Vw+vL.w}=fv. Interchanging v and w, then subtracting the result, we obtain the Green identity
fn
{vtw-wpv}=fn {van-wan}. a
(A.26)
JJ
(ii) Let xo be a given field point in RN \ aG, arbitrary apart from x0
aG, but fixed henceforth. If xo E G, we define (Figure A.7) GE := G \ °(xo, s),
HA := -4(0, A) \ G,
where E is so small that °R(xo, E) c G and A is so large that G c -4(0, A).
If xo E R' \ G, we define HA as before, but with A now so large that G U {xo} c -4(0, A), and use the sets G and HA,c := HA \ Mxo, e),
where a is so small that -V(xo, e) c HA. From now until step (v), we suppose that xo E G; for the other case, the argument is changed only slightly.
(iii) Let GE,,,, and HA,,,, be mth approximations to G. and HA, respectively, in the sense of Theorem D.9. We apply the Green identity (A.26) to GE,,,, and HA,,,,, choosing v to be as in the statement of the present theorem, and choosing w(x) = K(x - xo) =: Ko(x), say. [We cannot apply the Green identity immediately to GE and HA, because v need not
A.2 The Newtonian potential: first steps
181
a-4 (0, A)
Fig. A.7.
be in C2(Ge) and in C2(HA).] Since AK° = 0 in RN \ {x°} and Av = 0 in RN \ G, there results
-
/
K°
v
f I
Gem
0 =
OK°
8v
8n
- K 0 an
8K° 8n
- K°
v
V
HA,
(A . 27a)
'
8v 1
8n j
.
(A . 27b)
In order to apply Theorem D.9, we must verify that (a)
vajK° - K°a v E C(GE) n C(HA) for each j E { 1, ... , N},
(J3) K°w E Ll(GE). Now, K° is infinitely smooth outside -4(xo, e), so that the hypothesis v E C1(RN) implies (a), and hypothesis (b) implies more than (fl) [namely,
that K°Lv is in Lp(GE) with p > N/2, which is better than membership of L1(Ge) because GE is bounded; see Chapter 0, (xiv)].
Letting m -+ oo, using Theorem D.9, and adding the two identities which result from (A.27a,b), we find that the two contributions of 8G cancel each other, because v and K° are in C1(RN \ {x°}) and because the outward unit normals n to aGE and aHA are in opposite directions at each point of 8G where they exist. Accordingly,
avf, J KoAv=J +JA v 8n -K08n GE
aBE
s
where BE := 1(xo, 8), SA := a9(0, A) and n points towards x° on BBe.
(iv) Now let 6 --+ 0. Define Xe to be the characteristic function of GE
Appendix A. On the Newtonian Potential
182
[equal to 1 on GE, and to 0 on TRN \ GE] ; then
Jf KoAv
= J XEKoAv G
GE
JG
KoAv
by the Lebesgue dominated convergence theorem, since, with S dist(xo, aG), we have Ko integrable and Ov bounded on Bb, while Ko is bounded and Lv is integrable on G \ Bb. Next, c
I
aKo n, an
- v(xo) _
{v(x) - v(xo)} LBE
Ko(x) an
dS(x) --0,
because aKo/an = 1/I3BEl on aBE by the definitive property (A.17) of the Newtonian kernel and because v(x) -+ v(xo) when x E OBE. Finally, Koav -+ 0 fB, an because Koav/an = o(e N+1) uniformly on aBE, while IaBEI = Thus (A.28) yields
-
/ / (aKo av 1 JGKow = v(xo) + JsA tv an - Ko an J
const.cN-1
(A.29)
(v) We now show that the last integral in (A.29) tends to zero as A -* oo, for each fixed x0 E RN \ 0G. First, Ko(x) is close to K(x) when r is sufficiently large; in fact, for r >_ 2ro and for each j we have (ajK)(x - xo) - (a1K)(x)I <- xol maxo«
For N >- 3 and r = A -> oc, both terms vOKo/ar and Koav/ar are 0 (A-err+s), while the surface area ISA I is O (AN-1) ; then the integral over
SA is O (A-N+2) and tends to zero. For N = 2 and r = A
aK v
oo
- Ko ar
= {cK(x) + O (r-1) } {
oo,
d(x) + 0 (r-2) }
{dK(x) -{K(x)+O(r-1)} c+O(r 2) d
}
= 0 (A-2log A) , because the dominant terms cancel. Since ISAI = 21rA, the integral over SA is O (A-' log A) and tends to zero. Therefore (A.29) reduces to
v(xo) = -
JG
Koty
(xo E RN \ OG)
.
(A.30)
A.2 The Newtonian potential: first steps
183
aN
Fig. A.8.
(vi) Finally, we use continuity to extend the result (A.30) to points xo E 8G. The function v is more than continuous on RN, by hypothesis.
Theorem A.6 will show (independently of the present theorem) that the condition AV E LP(G), p > N/2, makes the integral in (A.30) a continuous function of xo on RN. Therefore the function v + fG KoLv, being zero outside aG and continuous on 1RN, must vanish also on OG.
To make further progress, we need (a) a notation that enables us to write partial derivatives of arbitrary order without undue labour, (b) bounds for all derivatives of the Newtonian kernel K. Definition A.3 We write as
:= a1l a2z ... aN,
xa := xal xaz 2 1
xaN . N
the N-tuple a :_ (a1, a2, ... , aN) of integers aj >_ 0 is called a multi-index of length N; its order is lal := a1 + 062 + + N. (Figure A.8 shows a set of multi-indices of fixed order, and also illustrates the partial ordering
a < fJ introduced in Exercise A.23.) It is to be understood that x° = 1
even when xi = 0. The label Jul = m, when appended to E, means summation over all multi-indices of order m, the length N being implied by the context. For example, if N = 2, Caxa = C(3,0)xl + C(2 1)xlx2 + C(1 2)xlx2 + C(0,3)x2. Ia1=3
Appendix A. On the Newtonian Potential
184
Lemma A.4 For all multi-indices /1 if N * 2, or for I#I > 1 if N = 2,
X# 0,
Cplxl-N+2-IRI,
(A.31)
where the constant C1q depends only on f (the length of which species N).
Proof We shall prove by induction that (for
as above and r := IxI)
aflK(x; N) = r N+2-21ftIPp(x),
x
0,
(A.32a)
where P p is a homogenous polynomial in x1, ... , xN of degree 1$1 (possibly the zero polynomial): cp,a x".
Pp(x) =
(A.32b)
Iai=1R1
[Many of the coefficients cp, a are zero; for example,
P(10) _ -2x1 = P(l,o)(x) = -xl/21C, 1
-12x,
P(o,1,o)(x) = -x2/47t,
P(2)(x) = 0, P(1,1)(x) = xlx2/1t,
P(o,0,2)(x) = (-xi - x2 + 2x2)/41.]
Assume that (A.32) is true for IP1 = m (it is true for 1P1 = 0 if N or for 1P1 = 1 if N = 2). Then
2,
a;a#K(x; N)
_ (-N + 2 -
2m)r-N+1-2m(xj
/r)P#(x) +
= r-N-2m{ (-N + 2 - 2m)xjPp(x) + (xi +
r-N+2-2majPp(x)
+ xN)aiPp(x) },
where ajPp is a homogenous polynomial in x1, ... , xN of degree m - 1. r-N-2m is a homogeneous polynomial in x1, ... , xN Hence the factor of of degree m + 1. This proves (A.32) for I Q 1 = m + 1 and hence for all f (except $ = 0 if N = 2). Finally, (A.32) implies (A.31) because IPp(x)l <
El
const.lxlIBI.
We now extend the density function f in (A.21) to have domain RN by setting
f(x) :=0 if xERN\G. The support of f, written supp f and defined fully in Chapter 0, (iv), is the smallest closed set outside which f equals zero. The next theorem concerns the set RN \ supp f, which always contains
RN \'G; it also contains any open subset of G on which f happens to
A.2 The Newtonian potential: first steps
185
be zero. For the moment, we relax the condition that the potential be defined at all points of RN.
Theorem A.5 If f E L1(G), then the Newtonian potential u off has the properties u E CGO(RN \ suppf)
and Au = 0 in RN \ supp f.
Proof (i) The essence of the matter is the formula
(a"u)(xo) = fG(a"K)(xo - x) f(x) dx for all xO E ][8N \ suppf and all a.
(A.33)
Since RN \ supp f is open, we have (for xo E RN \ supp f ) So := dist(xo, supp f) > 0, 80 :!:-:1
1
- xI < 80 + diam G
for all x E supp f,
so that nothing can go wrong in merely formal, repeated differentiation. However, as preparation for more dangerous cases, we shall justify the formula (A.33). The symbol (a"K)(xo-.) will denote the function with values (a"K)(xo - x); the point xO is a parameter, and is outside suppf throughout the proof.
(ii) First, does the integral in (A.33) always exist? Since f E L1(G), it
is sufficient that (a"K)(x0-.) be bounded and measurable on supp f, and this is the case because (A.32) shows [when we replace the x there by x0 - x] that (a"K)(xo-.) is continuous on the compact set suppf. (iii) Next, is the integral in (A.33) the result of operating on u with a"? Assume that (A.33) is true for lal = m (it is true for Jai = 0); we prove it for lal = m + 1 by calculating (a;a"u)(xo). Let
h := (hjbji)N1 = (0,...,0,hp 0,..., 0)
with hj * 0,
and define the difference kernel
k(xo,x,h)
:=
1
h
{ (a"K)(xo+h-x)- (a"K)(xo-x) - hj (a;a°`K)(xo - x) I,
(A.34)
Appendix A. On the Newtonian Potential
186
because then, if Ihl <-
280,
(a"u)(xo + h) - (8"u)(xo) hi
= <-
Jc
-
f
(aia"K)(xo - x) f (x) dx
k(xo, x, h) f (x) dx (A.35)
Il k(xo, . , h) I L.11 I if 1, G
where, in the present case, IIk(xo,. , h) I L. II = maxxESUppf Ik(xo, x, h)l.
If this norm of k tends to zero as hj -> 0, then the inequality (A.35) justifies the formula (A.33) for l aI = m + 1. Now define
0 < t < 1,
g(t) := (8"K)(xo + th - x), because then k(xo,x,h) = h { g(1) - g(0) - g'(0)} = h
Referring to (A.31), we let I' := max,p,
f I
(1 - t) g"(t) dt.
Cp, and bound k as follows.
"t Ih (i 02a"K)(xo+th-x)I < h2f l«I+2(,8)-N-I"I IgOI= 20 because Ix0 + th - xl > Il k(xo, , h)
28o;
also Jul = m. Accordingly,
I L.11 < Z hi I rm+2
(160)-N-. 2- 0 as h j -> 0.
(A.36)
This completes the inductive proof of (A.33).
(iv) The continuity of a"u in RN \ supp f follows from the continuity of 8"K away from its singular point. Again let h E RN, with magnitude IhI < 280; now let h have arbitrary direction. By application of (A.31) to 001K, or else by (A.32),
(a"K)(xo + h - x) - (8"K)(xo - x)I <
A"aON+t-I"I IhI,
where the constant A" depends only on a. Consequently I
(a"u)(xo+h)-(a"u)(xo)I <M(xo,a,f) IhI,
where
M (xo, a, f) := A"SDN+1-I«I fiii.
(v) That L u = 0 in RN \ suppf follows from (A.33) and the fact that (AK) (xo - x) = 0 whenever x0 * x.
A.2 The Newtonian potential: first steps
187
The small-ball technique This is a reliable, although slightly laborious,
method that we use repeatedly henceforth. Consider the function in (A.34) with lal = 0 and h := (h1,0,...,0), h1 # 0, so that 1
k(xo, x, h) := hJ { K(xo + h - x) - K(xo - x) - hl (81K)(xo - x) } , (A.37)
where now xo may be in the support of f. We bound some specified norm of k (or of a function like k) as follows. (a) For x E M(xo, 21hl), the three terms of k are treated separately; each makes a small contribution. (b) For x 0 M(xo,21h1), the function k as a whole is bounded, as it was
in the proof of Theorem A.5, by means of a formula for the remainder of a short Taylor series. In this context we write
R := Ix - xol,
Rh := Ix - xo - hl
(A.38)
(Figure A.9). Also, F will be used as a symbol for any positive constant independent of xo, x, h, f and G (but possibly depending on N and, in Theorem A.6, the exponent p). Thus F may have different values at successive appearances within a single line. The estimates of the method are simple applications of the following inequality, usually with c = xo or c = xo + h. Suppose that 91 c M(c, M) in IRN, that cp : 0 -> R is measurable and that Ipp(x)I < ip (Ix - cl)
almost everywhere in
SZ,
where ip is defined and non-negative on (0, M), and the function with values RN-11p(R) belongs to L1(0, M). Then
f
M
w(Ix - cl) dx = aa(0,1)
w(R)RN-1
dR.
(cM>
(A.39)
Equality of the last two integrals follows, as it did in Remark 1 after (A.23), from Fubini's theorem and a theorem about co-ordinate transformations (Apostol 1974, p.421; Weir 1973, p.158).
Theorem A.6 Let G be a bounded open subset of RN with N >- 2. If f E Lp(G) for some p > N/2, then the Newtonian potential u of f is uniformly continuous on IIBN.
Appendix A. On the Newtonian Potential
188
DR (x0,21hI
Fig. A.9.
Proof (i) Let q be the Holder conjugate of p [that is, 1/p + 1/q = 1]; since p > N/2, we have -Nq + 2q + N > 0. The integral in the definition K (xo - x) f (x) dx,
14(X0)
X o E 1[8N
G
exists because the integrand is measurable and the integral can be bounded by means of the Holder inequality; the calculations in step (ii) imply that K(xo-.) E Lq(G). (ii) To prove the continuity of u, we introduce the difference kernel
a(xo, x, h) := K(xo + h - x) - K(xo - x), so that Iu(xo + h) - u(xo)I
= <
a(xo, x, h) f (x) dx IG II
I
Lq(G)II 11f
Lp(G)II.
I
(A.40)
If, as I hl - 0, this norm of a tends to zero independently of xo E R', then the uniform continuity of u follows. To this end we apply the small-ball technique; in addition to the symbols R, Rh and r [defined by (A.38) and the remark following it], we use a notation that is abbreviated but self-explanatory.
(a) R < 2Ihl, N >- 3. Note that R < 21hl implies Rh < 31h1. Since
A.2 The Newtonian potential: first steps
189
K(xo + h - x) = FRh N+2, the definition of a and the triangle inequality for the norm of Lq give l 1/g
JR<21hj lalq dx }
1/q
3IhI
R Nq+2gR1N-1 dRh 0
1/q
p21hl
+rJ
R-Nq+2gRN-1 dR
o
1711h I -Nq+2q+N }1/q
(N >- 3);
(A.41)
the integrals converge at Rh = 0 and R = 0 because -Nq + 2q + N > 0, which we noted in (i). A similar integration, but over the ball { x I R < diam G + dist(xo, G) }
,
shows that K(xo -.) E Lq(G) if N >- 3. (a') R < 21hl, N = 2. Since now K(xo + h - x) = F log(1/Rh), 1/q
l JR<21hl
Ialg dx
1/q
3Ihl
I
Jo
IlogRhlgRh dRh
<- F { Ih12 I log Ih1
Iq
+ Ih12 } 1/q
+
similar
(N = 2).
(A.42)
(b) R >- 21hl, N > 2. Define b(t) := K(xo + th - x), 0 < t < 1, because then
a(xo, x, h) = b(1) - b(0) = I b'(t) dt. 1
Now,
Ixo+th-x1 > Ixo-x1-IhI > !R, 2
Ih1 <- 12 R
whence, by (A.31), N hj(8jK)(xo + th - x)
I b'(t) I =
<
rlhiR-N+1,
j=1
so that la(xo,x,h)I
<-
FIhIR-N+1,
We may suppose that N/2 < p < N, because the volume IGI is finite, whence a function in LP(G) is also in L,(G) for 1 <- r < p [as explained in Chapter 0, (xiv)]. The following constants t may then depend on IGI.
With p < N, we have -Nq + q + N < 0, so that the next integral will
Appendix A. On the Newtonian Potential
190
converge at infinity, thereby simplifying the estimate because we need not mention diam G and dist(xo, G). In fact, 1/q k!21hl
1/q
o0
<
jaIq dx
T'
l
JJJ
=
jhNq
I21h1
R-Nq+q RN-1 dR
IF J I hI -Nq+2q+N
JJ
111q,
(A.43)
and the exponent of Ihi is the same as the final one in (A.41) and (A.42).
It is legitimate to add the inequality (A.43) to (A.41) or to (A.42) [because ((x + f3)1/P < a1/P + 3u/P for real numbers a >- 0, $ > 0 and p > 1]. The uniform continuity of u follows from (A.40), either (A.41) or (A.42), and (A.43).
Theorem A.6 may seem unexciting, but the result is a good one in the following sense. If the condition p > N/2 is widened slightly to p >- N/2,
then not only may u fail to be continuous, but it may fail to exist on a set that is dense in G [that has closure equal to G] ; Exercise A.27 demonstrates this. We turn now to the question of how the Newtonian potential is related
to the Poisson equation -Au = f in G, even when the density function f is so bad that the potential has no derivative. One method is not to explore in detail the behaviour of u and f at each point of G, but to ask how u and f combine with arbitrary test functions lp E CC°(G). These are differentiable any number of times and have compact support within
G; the set Q°(i2) is defined fully in Chapter 0, (vii). In the following definition, fl need not be bounded and v need not be a Newtonian potential. Definition A.7 As always, let Sl be an open non-empty subset of ][81'. We shall say
(a) that v is a C2-solution of -Ov = g in S2 if v E C2(92) and the equation holds pointwise:
- Av(x) = g(x)
at each x E S2;
(A.44)
(b) that v is a generalized solution of -A v = g in 92 if v E C'(S2) and
in
V (P Vv =
J
Mpg
whenever cp E CC°(S2);
(A.45)
(c) that v is a distributional solution of -/v = g in 0 if v is locally
A.2 The Newtonian potential: first steps
191
integrable in S2 (integrable on each compact subset of S2) and
- J (AT)v = J cog n
whenever cp e C,(S2).
n
(A.46)
Regarding the condition qp >- 0, which appears in some statements resembling (A.45) and (A.46), see Remark 8 below.
Remarks 6. An explanation of Definition A.7 seems desirable. Note first that, when V E C2(S2) and g E C(S2), equation (A.44) is equivalent to the statement
Jn
co{ Av + g } = 0
whenever
E CC°(1).
(A.44')
For, to pass from (A.44) to (A.44'), we multiply (A.44) by (p and integrate over 52; to pass from (A.44') to (A.44), we use a result in Exercise 1.16. Second, (A.45), (A.46) and (A.44') are related by integration by parts:
Vv = - J
Jn
c Av
if
E CC°(S2)
and v E C2(S2), (A.47)
n
(A(p)v = fvVv if
E C(S2) and v E C); (A.48)
these formulae are derived from first principles in Remark 7. If v is a C2-solution of -iv = g in S [which implies that g E C(S2)], then v is also a generalized solution, by (A.44') and (A.47). Similarly, a generalized solution is also a distributional solution.
Conversely, if a generalized solution should happen to be in C2(92), then it is also a C2-solution when g E C(92); if a distributional solution should happen to be in C' (S2), then it is also a generalized solution. To sum up, we may say that (A.45) and (A.46) are as close to (A.44)
as is possible for functions in the classes that house generalized and distributional solutions.
7. To derive (A.47) and (A.48) we do not need a divergence theorem, because there is no boundary integral in these identities. Define T to be zero not merely in S2 \ supp cp but in RN \ supp cp; then 'p E C,(R') and it is sufficient to prove that, for each j e { 1, ... , N},
f f
8j((p8jv) = 0 if
E CC°(RN) and v E C2(supp
(A.49)
N
8j{(8j'p)v} = 0 if cp E Cc(RN) and v E C1(supp'p); (A.50)
N
summation over j yields (A.47) and (A.48). Fubini's theorem allows us to
Appendix A. On the Newtonian Potential
192
replace each volume integral in (A.49) and (A.50) by a repeated integral
and to integrate first with respect to xj. Then, if b is so large that the open cube (-b, b)N contains supp cp, 0o
a.i(gn3.iv) dxi = f ai(pa.iv) dxi = [Tajv] b;=-b = 0, b
00
and similarly for (A.50). 8.
Definition A.7 resembles a part of Definition 2.10, and in Remark
6 we noted the relevance of Exercise 1.16. But there seems to be an inconsistency: in Exercise 1.16 and Definition 2.10 only non-negative test functions are used, q' E C,°(0) and p > 0. The answer is that the condition cp 0 is needed for the inequalities in Exercise 1.16 and for the subsolutions in Definition 2.10, but not for the definition of generalized and distributional solutions. In fact, let (A.46)+ denote the statement which results from adding the condition cp >- 0 to (A.46); if v and g are locally integrable in n, then (A.46) and (A.46)+ are equivalent statements ; we sketch the proof. It is obvious that (A.46) implies (A.46)+; to prove that (A.46)+ implies
(A.46), one uses the decomposition (p = (p+ + cp- of a given function cp E Cc(K2) and applies the smoothing operation in Exercise 1.23 to form
approximations an to (p+, and -Nn to (p-, that allow use of (A.46)+ for an, v, g and for fn, v, g. Then a limiting procedure yields (A.46): if an and fn are formed with the same smoothing kernel, say a kernel s, with smoothing radius 2 ", we have A (an - Qn) (x)
=
f(5n)(X_y){+(y)+Q_(y)} dY f sn(x - y) (A(p)(Y) dy n
L p(x)
n
as
oo,
uniformly over x E 92. Thus L (an - fn) -> Aip in Lo°(92) and, similarly, an - fin -+ (p in LCo(S2); consequently, (A.46)+ for an, v, g and for fin, v, g implies (A.46) for (p, v, g. Similar remarks apply to (A.45).
9. The proof of the next theorem involves the identity
(p(xo) _ -
JRN
K (xo - x) L p(x) dx
if
cp E CGO (RN)
and xo E RN. (A.51)
This is implied by Theorem A.2 if the set G in that theorem is chosen to
A.2 The Newtonian potential: first steps
193
be a ball (or a cube) containing the support of cp; then the hypotheses of the theorem are satisfied with much room to spare. Indeed, we need only steps (i) and (iv) of the proof of Theorem A.2, and it may be worthwhile to demonstrate this.
Let B, := °(xo, s), let n be the unit normal on 3B, outward from RN \ B,, hence pointing towards x0, and let Ko(x) := K(x - xo). Since cp is infinitely smooth everywhere, and identically zero outside the compact set/supp (p,
IN
- x) i p(x) dx
RN
= lim,-o
f
Ko Lco
.
Il2N\BE
-(OKo) (Ko_)+f a(p
= lime-o aB,
8n
n
N\BE
= -T(xo) by step (iv) of the earlier proof and because L Ko = 0 in RN \ BE.
Theorem A.8 Let G be a bounded open subset of RN with N > 2. If f E LP(G) for some p > N/2, and f := 0 on IRN \ G, then the Newtonian potential u off is a distributional solution of -Au = f in RN. Proof In view of Definition A.7, we must prove that u is locally integrable
in RN, which is certainly true because u E C(RN) by Theorem A.6, and that
-.fN(oO)u=,fN(Pf
whenever p E C,'°(W'').
Now, by Fubini's theorem and (A.51),
-.f (D(P) u = RN _ _
j
A(P(x) G
N
K(x -
- f fO {fRNK_x(x) dx
ff()
dx
d JJJ
} dG
) d,
and this last equals fRN (p f because f vanishes outside G.
Of course, u is also a distributional solution of -Au = f in G [because local integrability in R ' implies local integrability in G, and because an
Appendix A. On the Newtonian Potential
194
identity that holds for all test functions cp e CC°(RN) certainly holds for all qp E Cc (G)].
A.3 Continuity of the force field Vu
In the remainder of this appendix it is to be understood, unless the contrary is stated, that G is bounded and open in RN, that u is the Newtonian potential of f : G
IR, and that formulae like (A.52) below
hold for all j in { 1, ... , N}.
The pointwise form of the Poisson equation -Au = f in G, which will be established in §A.5, is conceptually important and often useful. But in many contexts, such as that of §4.1, we need consider only first derivatives of the potential. It is remarkable that this can be done without reference to the boundary 8G; Theorem A.11 below remains true when 8G is pathological. Note also that our daily lives, as particles on the boundary of the earth, would be very different if the gravitational force field were discontinuous at G.
Lemma A.9 If f EL
(u)(xo) =
f
for some p > N, then
for all xo E RN.
(BK) (xo - x) f (x) dx
(A.52)
Proof (i) We proceed very much as in the proof of Theorem A.6. Since now p > N, the Holder conjugate q of p [such that 1/p + 1/q = 1] now satisfies -Nq + q + N > 0. The integral in (A.52) exists because the integrand is measurable and
because the Holder inequality provides a bound; this bound can be inferred from the calculations in step (ii).
(ii) To justify the formula (A.52), we may set j = 1 without any real loss of generality, as the proof of Theorem A.5 shows. As in (A.37), let h := (h1,0,...,0), h1 * 0, and define k by k(xo, x, h) :=
1
hl
{ K(xo + h - x) - K(xo - x) - h1(81K)(xo - x) }.
Then I u(xo + hh) - u(xo)
-
/'
(01K)(xo - x) .f (x) dx
fk(xo,x,h)f(x) dx
IIf LP(G)II, and the lemma will follow if this norm of k tends to zero as h1 -* 0. In Iik(xo,.,h)
I
Lq(G)II
I
A.3 Continuity of the force field Vu
195
fact, we shall prove more: that the norm of k(xo,.,h) in Lq(RN) tends to zero, uniformly over xO E RN. We use the small-ball technique. (a) R < 2IhI, N * 2. Since K(xo + h - x) = +FRh N+2 (minus only for N = 1), the definition of k and the triangle inequality for the norm of Lq yield 1/q
fR<21hl I kl q dx
1/9
3Ih
IhI-q f Rh Nq+2q Rh -1 dRh 0
1/q
21h1
IhI-q
+ IF
JR-Nq+2qRN-1 dR
1/q
2jhj
+I
}
R-Nq+q RN-1
dR
o
< I' {IhI -Nq+q+N } 1/q
where -Nq + q + N >0. (a') R < 2IhI, N = 2. Since K(xo + h - x) = Flog(1/Rh), the first integrand becomes IlogRhlgRh, the second becomes IlogRIgR, the third is unchanged. Consequently,
<
fdx
{
F { hi-log hI q +
12-}1/q
<21h1
where q < 2.
(b) R >- 2IhI, N >- 1. Let g(t) := K(xo + th - x), 0< t< 1; as before, IhI < 12R
Ixo+th - x) > Ixo - xI - IhI ? 1R,
so that, by (A.31), Ig"(t)I = Ihi (02K)(xo + th - x)I < hi { maxlfl1=2 CR } (IR) N , and
Ik(xo,x,h)I
=
hi{g(1)-g(0)-g'(0)} 1
hi
1
j(1_t)g"(t)dt
<
FIhiR-N.
Just as we assumed in the proof of Theorem A.6 that N/2 < p < N, so we may suppose here that N < p < oo; the following constants IF may
Appendix A. On the Newtonian Potential
196
then depend on IGI. With p < co, hence q > 1, the next integral will converge at infinity and yield a simple estimate. In fact, 1/q
1/q
f
J R>2Ihj
<
Iklq dx
0
IF
I hl q
J
R-NgRN-1 dR
(-Nq + N < 0)
21hl
F{ I h l q-Nq+N }J 1 /q '
and the exponent of Ihl is the same as the final one in (a) and (a'). Thus Il k(xo, . , h) I Lq (lRN) II -+ 0 independently of x0, as h1 --+ 0.
It is now appropriate to distinguish certain continuous functions on RN that are decidedly better than arbitrary members of C(RN). First, observing that functions in the set C(0) can be unbounded when .0 is unbounded, we shall define the normed linear space Cb (5i) of bounded continuous functions on S2. Second, we introduce Holder continuity. A function v : RN -> R is Holder continuous at the point c if and only if there are constants A E (0, 1], A and 6 > 0 such that Iv(x) - v(c)l < AIx - cI2
whenever x E B(c, 6).
The number A is a Holder exponent and A is a Holder constant. When A = 1, the name Lipschitz often replaces the name Holder. Here is an example. If, for some a E R,
1+ar 1 + I log rI'
W(X)
to,
r := Ixl > 0, r = 0,
then w E C(RN). But w E Cb(RN) only if a = 0, and w is not Holder continuous at the origin, whatever a may be. In the following definition, we demand uniform Holder continuity on Q. [We take suprema over S2, rather than S2, because this is sufficient in the definition and is traditional] Definition A.10 (a) The normed linear space Cb (i) consists of the norm defined by IIv
I CbMII := sup,En Iv(x)I
and of the set
{v:S2-*JRI vEC( ),
IIvICb())II
A.3 Continuity of the force field Vu
197
(b) For given A E (0, 1], the Holder constant of w : S2 -> R is [w]A :=sup
1 1w(X) _
x,y E S2, 0 < Ix - yl < 1
I
,
in which the final constant 1 may be replaced by any positive constant characteristic of 92 or of the problem in hand. (c) The normed linear space Cb'2 (S2) consists of the norm defined by v
Cb'1(n)
:= supxEsi Iv(x)I + [vIA
and of the set { v E Cb (S2)
I
[V]A<001-
0 It is seldom necessary to know the Holder constant [w], of w; usually one is content with a Holder constant A such that Iw(x)-w(y)I <- Alx-yl2
for all x and y as above. The spaces Cb (0) and C"2' (S2) are Banach spaces [are complete]. We shall use neither this fact nor the considerable
machinery to which the definition leads. But the definition gives us a convenient way of stating the next result and those in §A.5. Theorem All If f E LP(G) for some p E (N, oo) and f := 0 on IRN\G, then the first derivatives of u are bounded and uniformly Holder continuous: aju E Cb'2'(RN),
with A = 1 - N/p,
and u is a generalized solution of -Au = f in IRN.
Proof As in the proof of Lemma A.9, the Holder conjugate q of p satisfies
0 < -Nq +q + N< 1,
(A.53)
with equality on the right only if N = 1. Again we may take j = 1. (i) First we bound a1 u. By Lemma A.9, I(a1u)(xo)I
=
.f (a, K)(xo - x) f (x) dxl G
IIf Lp(G)II to bound this norm of 31K, we consider two cases. Let do := dist(xo, G) and D := diam G. II
(a1K)(xo-.)
I
Lq(G)II
I
Appendix A. On the Newtonian Potential
198
(a) do < 1. Then Ix - xol < D + 1 when x E G; by the left-hand inequality in (A.53), fD+1
J Ia1K
Iq < IT]
R-Nq+q RN-1 dR = F(D + 1)-Nq+q+N
1. Since do < Ix - xol < D + do when x E G, we have R >- 1 and use the right-hand inequality in (A.53): (b) do
IG
Ia1K
Iq
<
r
do+D
R-Nq+q RN-1 dR
(-Nq + q + N - 1 < 0)
do
fdo+D
<TI
dR = I'D.
do
(ii) Now we establish the uniform Holder continuity of a1u. Let h E RN have magnitude IhI < 1 and arbitrary direction. Define
a(xo,x,h) := (01K) (xo + h - x) - (01K)(xo-x); then
I(01u)(xo + h) - (71u)(xo)I =
L
a(xo, x, h) f (x) dx
IIa(xo,.,h)
I
Lq(G)II IIf
I
Lp(G)II. (A.54)
We apply the small-ball technique.
(a) R < 2IhI. Since I (81K)(xo + It - x)I < FRh N+1 by (A.31), we have 1/9
-
fR<21hl
1/q
3IhI
<
Ialq dx J
I'
Rh Nq+q Rh -1 dRh
+ similar
0
11/q
IF J IhI-Nq+q+N
(b) R >- 2IhI. Let b(t) := (71K)(xo + th - x), 0< t< 1. As before, we have Ixo + th - xl >- 1R, and, by (A.31), N
Ib'(t)I =
hi(8i81K)(xo + th - x)
< rIhIR-N,
i=1
a(xo, x, h)I = b(l) - b(0)I =
(jb'(t) 1
dt <- rIhIR-N.
A.4 Multipoles and the far field
199
Then, exactly as in the proof of Lemma A.9, step (b), 1/9
dx }J
l
< r S hJ l
-Nq RN-1 dR } 1/q (-Nq + N < 0) JJJ
= 1711h
I q-Nq+N }1/q
and
q Nq+N=1-N+N 1-1 p
q
Accordingly, (A.54) implies that, whenever xo E RN and lhl < 1, I(Olu)(xo + h)
- (aiu)(xo)I < F 11f
I
LP(G)II lhl1.
(iii) The potential u is certainly in C1(RN), by what we have just proved. Integration by parts, as in (A.48) or (A.50), gives
Vu = - f (AT)u N
for all
E C,(RN),
N
and the proof of Theorem A.8 shows that -fRN(A(P)u = IRN (pf.
Therefore the potential is a generalized solution of -Au = f in RN. A.4 Multipoles and the far field In (A.10) we encountered the potential of a dipole in R3; inevitably, this
is one of a hierarchy of point singularities. Let a be a multi-index of length N (Definition A.3); then cpa(x) := a(-1)1"18"K(x;N),
x E RN \ {0},
(A.55)
is the potential of a multipole at the origin, of strength a and type a. As before, App, = 0 in RN \ {0}, because A and a" commute when acting on functions with continuous derivatives of order 2 + lad. For N = 1, multipoles are not exciting: 9(1)(x) is constant in (-oo,O) and in (0,00); (p(m)=0if m - 2. A multipole is called a dipole if loci = 1, or a quadrupole if loci = 2. Figure A.10 shows the field lines of three such potentials; these are curves to which -Vgpa is tangential. In the case of R2, the field lines are drawn with the help of a stream function (Exercises A.19 and A.21); in the case of R3 and cylindrical symmetry, with the help of a Stokes stream function (§A.1, item (vi)).
Appendix A. On the Newtonian Potential
200
x2
xI
.xl
.x1
dipole of type (1,0,0)
quadrupole of type (2,0)
dipole of type (1,0) Fig. A.10.
Equation (A.23a) showed that, for constant density in a ball B, the potential at points outside B is that of a point source at the centre of B. One cannot expect so simple a result in general, but Theorem A.12 provides a corresponding result that is not complicated for an observer sufficiently far from G. This will be discussed after the proof, but it should
be noted now that the terms of the series in Theorem A. 12 decrease in order of magnitude for large IxI as jal increases. The theorem is stated only for N > 2 because its analogue for one dimension is immediate: if N = 1 and (a, b) G, then the definition of u implies that
-Ii x f f() -lx u(x) = l x 2
f
G
f f
2
2
x> b,
G
(A.56) a. x
G
Once again we relax the condition that the potential be defined at all points of RN.
Theorem A.12 Let a := sup { IxI I x E G }. If N >- 2, f E Li(G) and IxI > 2a, then
u(x) =
6"(-1)I"18"K(x) + Ek(x), 0<1al5k-1
(A.57a)
A.4 Multipoles and the far field
201
where
afO
Qa
... aN
(A.57b)
10flEk(x)l < const. IxI-N+2-k-IBI
(A.57c)
al! 062 !
G
and
for all multi-indices P. The constant depends on k, i, f and G, but not on X.
Proof We have u(x) =
f K(x -
Ixl > 2a,
ICI < a,
G
0< t< 1. Then
and set g(t) := K(x (m) %
(mt
= %() -i i
()t
m
K(x-t),
j=1
where x and are fixed, and a j means, as always, differentiation with respect to the jth argument of the operand. By the multinomial theorem (Exercise A.22), m
N
E j aj E
=
j=1
M
"a"'
IaI=m
whence g(m)(t)
= (-1)m E a 11 (811K)(x
-
Ial=m
Accordingly [Taylor's theorem or integration by parts], k-1
K(x -
g(1) _
M)
g(mi0)
+
m=0
J
1 (1
t1 )i1 g(k)(t) dt
(-1)1«1 a a"K (x) + Ak (x, ), 0
where Ak(x,
) :=(-1)k
f
I
dt.
(Ik lal=k
Appendix A. On the Newtonian Potential
202
into the integral defining u, we obtain (A.57a,b)
Substituting for with
Ek(x) _
c
d
Ak(x,
It remains to bound aflEk. The result (A.33) allows us to differentiate
K(x - ) with respect to x under the integral sign fc, and it is certainly legitimate to differentiate the part Ak(x, ) of K(x - ) under the integral sign fo , because that integrand is C°° by the condition Ix - tl; > a. Now (A.31) and the inequality Ix - tcI - 1Ixj imply that
a"+QK x- t
<
ck,fl l x-N+2-k-Ipl
where Ck
max - C"+fl;
then Ak(x,
)
< const. ak lxrN+2-k-IRI
and
Ia%(x)I < const. ak If
I
Li(G)II
IxI-N+2-k-I#I,
where the constant depends only on k and R.
Remarks 1. The theorem states that, as IxI oc, the dominant part of the potential, and of all its derivatives, is due to a source (Ial = 0) at the origin, of strength equal to the total charge or mass f f. (We assume for purposes of description that, at each order Ial < k - 1, at least one coefficient o" 0.) Next in order of magnitude come the dipole terms
(Ial = 1); their strengths are moments f & of the density function. Then, still smaller, come the quadrupole terms (Ial = 2); their strengths
are quadratic moments f
apart from a numerical factor.
The hierarchy of multipole terms, becoming smaller (for large Ixl) as 1061 increases, continues in this way.
2. Provided that fG f * 0, we can so choose co-ordinates that the dipole terms vanish. (This suggests the first step in the proof of Theorem 4.2.) Define the centre c of charge or mass by c..-
d
and translate the origin to c, in effect, by setting v(x) := u(x + c). Multi-indices are not always convenient; we now write Q := .7a...,O) =
fL
.f,
A.5 Second derivatives of u at points in G a1 := 6(1,0....,o) =
.f G
i
aN := 0(0,...,0,1) _
203
.f N f G
so that cj = aj/Q. Then for xl > 2(a + Icl), say, v(x)
T. 6"(-1)1"1(8"K)(x + c) + 0 (IXI-N)
=
0
N
= Q K(x) + E cjajK(x)+ O
(IxI_N)1
j=1
N
Y ajajK(x)+ 0
(IXI-N)
+ O (IXI-N)
j=1
= QK(x) + 0
(IXI-N)
(A.58)
because Qc j = a p
3. Let N = 2 and fix x with lxl > a. Then one can use (A.24) to prove that the series in the theorem, and the series resulting from it by repeated differentiation, all converge as k -+ oo. For N > 3, a corresponding result is true (but more difficult to prove). However, convergence of these series is not the object of our interest here; rather, we have sought an ordered picture of the field at large distances from G.
A.5 Second derivatives of u at points in G
For N > 2, the condition f E C(G) fails to imply that u E C2(G); this is shown by Exercise A.29. The theme of this section is that Holder
continuity of f on G ensures not merely that u E C2(G), but that second derivatives of u are Holder continuous in G with the same Holder exponent as f. In symbols, f E C6'µ(G)
a,aju E C6'µ()
whenever p E (0, 1) and f2 c G. }
The reason for the condition p < 1 will be mentioned after the condition
has been used. [Of course, if f E C6'1(G), then f E C6'µ (G) for each p E (0, 1).] Except when G is a ball, we shall not pursue the more difficult
question of regularity up to the boundary: if f E C6'µ(G), does 8'8juIG have an extension to G that is in C (G) or perhaps in C6'µ (G) ? Exercise A.30 shows that the answer can be No when f is constant on G and 8G has a corner; a positive answer requires a condition on the smoothness of 8G and more analysis than is appropriate here.
Appendix A. On the Newtonian Potential
204
For N = 1, the analogue of Theorems A.15 and A.16 is again very simple; the definition of u implies that
- u"(x) = f (x) whenever x E G and f E C (G).
(A.59)
Notation Following convention, we often omit the subscript b from Cb (S2)
and Cb'2 (S2) when fl is bounded, because then C (52) contains only bounded functions. The ambiguity, that C (S2) may or may not be not be normed, causes little difficulty. Lemma A.13 Assume that N >- 2. Let B := 9(0, a), let f E Co'u (B) for some p E (0, 1), and let
v(xo) :_ f K(xo - x) f (x) dx,
xo E IR' .
B
Extend f to be zero outside B. Then [for all i and j in { 1, ... , N}] (a,ajv)(xo)
= fs
(a,ajK)(xo - x) { f (x) - f (xo) } dx -S,JfN)
(A.60)
whenever x0 E RN \ aB.
Proof If xo E RN \ B, then f (xo) = 0 and (A.60) is a particular case of (A.33). Therefore we suppose in the remainder of this proof that xo E B. The existence of the integral in (A.60) will be implied once again by the calculations that follow. It is enough to prove the result for i = 1; let h = (h1, 0,..., 0),
0 < IhI < 1,
IhI < dist(xo, aB).
Note first that, by Lemma A.9 and (A.23a),
J
{ (ajK) (xo + h - x) - (ajK)(xo - x) } dx
_ (ajuB)(xo + h) - (ajuB)(xo) _ -N { (xo + h)j - xoj } h1
=-81;N. Hence, by a second application of Lemma A.9,
(ajv)(xo + h) - (ajv)(xo) =
J
{ (a;K)(xo + h - x) - (a;K)(xo - x) } x {.f (x) - f(xo)} dx - 6I j N f(xo)
205
A.5 Second derivatives of u at points in G
We are now on a familiar path. Fix j and define
hl { (a,K) (xo + h - x) - (8jK) (xo - x)
k(xo, x, h)
-hi (01ajK)(xo-x)}; then (aiv)(xo + h) - (aiv)(xo) h1
f (ai7iK)(xo-x){f(x)-f(xo)} dx+81jfN
f
B
_
)
k(xo, x, h) { f (x) - f (xo) } dx.
B
If this last integral tends to zero as h1 --+ 0, then the lemma will be IIf I COP(R)II and R = Ix - xol; the following coarse
proved. Let Ilf II
bound will be useful.
if R < 1,
[f]µRµ
If (W )-f(xo)I 2IIf
I
if R
c(B)IIR'`
21fI R.
(A.61)
1
In the inevitable application of the small-ball technique, constants F now depend on N and µ. We recall the notation Rh = Ix - xo - hl. (a) R < 2IhI. By the definition of k, by (A.31) and by (A.61), k{ f (x) - f (xo) } dx JR<2lhl 3IhI
<
FIhl-1
J +F f
RhN+1 If I I I
I
hI µ Rh-1 dRh + similar
0
21h1
,
R-N Il f II RJ' RN-1 dR
0
= rllf II IhI".
(b) R > 2IhI. Let g(t) := (aajK)(xo + th - x), Ixo + th - xl > ZR, we have from (A.31) IS "(t)I = Ihi (aia.iK)(xo + th - x)I < Ik(xo, x, h)I =
f(1 - t)g"(t) dt <
0 < t < 1. I'lh12R-N-1
I'IhiR-N-1;
Since
Appendix A. On the Newtonian Potential
206
accordingly,
f
< rlhl
k{ f (x) - f (xo) } dx z21hl
f
R-N-' Ill II R" RN-' dR lhl
= rllf II IhI"
Remarks 1. At the last step of the foregoing proof, the integrand was If p = 1 had been allowed, then for this case we would have IIfIIR-2+".
been driven to some estimate like
f
2a
k{ f (x) - f (xo) } dx < FIhl f
_
R !2lhl
If II R' dR = Fllf II hI log
lhl
.
IhI
(A.62)
At this stage, the logarithm would do no harm, but in Theorems A.15 and A. 16 the effect of µ = 1 and of a similar inequality would be that the Holder exponent 1 is not conserved. 2. Again let B :_ -4(0, a) in RN, N >- 2; denote the usual small ball by H := . (xo, 2lhl) with 0 < lhi < a/4, say. It will be desirable to bound the integral I3(xo, h) :_
J
(aiajK)(xo - x) dx,
where xo E B,
(A.63)
\H
independently of xo and h. Our standard method gives za
R-N RN-' dR = IF log
IIij(xo, h)I 5 IF
a ,
IhI
21hl
which is not good enough for our purpose. The logarithm can be avoided by a more careful calculation, which unfortunately runs to some length; this is the subject of Lemma A.14. 3. If Lemma A. 14 and Theorem A. 15 are thought to be of no interest,
then one can follow a shorter path to Theorem A.16. This requires a bound for Iii only when the balls B and H are concentric. Now, with the notation B0 := M(xo, p), p > 2Ihl, we have
f
(ai7jK)(xo - x) dx o\H
ni(x) (ajK) (x - xo) dS(x) fa(Bo\H)
I(VK)(x-xo)I dS(x)
aBouaH
2
in view of the definitive property (A.17) of the Newtonian kernel.
(A.64)
A.5 Second derivatives of u at points in G
207
Lemma A.14 Let Iij be as in (A.63). Then N
E Ii j(xo, h)2 < const., i,j=1
where the constant is independent of xo and h; indeed, it depends only on N.
(i) It will be convenient to rotate the co-ordinate frame: let x = T(xo)z and xo = T(xo). o, where T(xo) is an orthogonal N x N
Proof
matrix depending on xo. Now,
Kij(xo - x)
(aiajK)(xo - x) =
N(xoi - x,)(xoj - xj) - aijlxo - x12 6NIxo - xlN+2 (A.65)
where 6N := la°4N(0,1)1. Since the transformation of co-ordinates depends on xo, we use the explicit form of Kij, rather than the rule for transforming partial derivatives, to infer that N
Kij(xo - x) =
TipTjgKpq (zo - z),
where T = T(xo).
(A.66)
p,q=1
Let H := _4(zo, 2I hl) and let I pq (zo, h) denote the integral with respect to z of Kpq (zo - z) over B \ H; then N
Iij(xo,h) = E Tip Tiqlpq (x0, h), p,q=1
and, by the orthogonality of T, N
N
I ij(xo, h)2 = i,j=1
I pq (xo, h) 2.
(A.67)
p,q=1
(ii) Given xo E B, we so choose T that zo = (Ixol, 0, ... , 0); we shall use these special co-ordinates 5 j and (A.67) to prove the Lemma, but henceforth we omit the tilde for ease of writing. Two configurations are possible (Figure A.11): case I if H c B, or case II if H intersects 8B; we shall write formulae and inequalities that are valid for both. It is immaterial whether the integration is over B \ H or over B \ H; in the rest of this proof we shall use B \ H because it is open. The divergence theorem is valid for both configurations because 8 (B \ H) is of class C°° in case I, while the set B \ H of case II is listed in Remark D.4.
Appendix A. On the Newtonian Potential
208
x1
Fig. A.11.
(iii) If i # j, then Iij = 0. For, at least one of i and j must differ from 1; suppose that j * 1. Then xo1 = 0 and Sid = 0, so that Ki3(xo - x) is an odd function of xj, while B \ H is symmetrical about the hyperplane { x I xj= 0 }. Also, N
N
E Kli (xo -x) = 0 for x * xo
Ill (xo, h) _ - E I.i1(xo, h) ; i=2
f=1
therefore it is sufficient to bound 1122 I, ... , INN by cylindrical symmetry about the x1-axis.
and these are all equal,
(iv) Let n = n(x) be the unit normal on B (B \ H) outward from B \ H,
and let dS = dS(x). Then
Ili(xo, h)I = f\H (B?K)(x - xo)
dxl = J (B\H) nl (8jK)(x - xo) dS
nj (O jK) (x - xo) dS + 1
(A.68)
aB\H
because I(VK)(x - xo)I = 1/I0HI when x E BH. Now let j * 1. Then n1(x) = xj/a on BB and xj - xo1 = xj, so that nj (BjK)(x-xo) dS oB\H
x
Jas\H aaNIX - xo!N 1
/
(N - 1)aON JaB\H
dS
(j#1)
x2+...+x Ix - xolN
dS
by cylindrical symmetry about the x1-axis. Transforming to polar coordinates, we set x1
= acosO, (x EBB, 0<0 <7r);
A.5 Second derivatives of u at points in G
209
we also write xol = Ixol =: a(l - 6), so that 0 < 6 < 1. Then
Ix-xo12 = a2{1-2(1-6)cos0+(1-S)2} =
- 6) sine 0
62 + 4(1
a2
and, for some angle /I in [0, ir/4) that depends on Ixo1 and IhI,
Js\H nj (ajK) (x - xo) dS a2 sin2 0
1
(N - 1)aUN ffl" aN {g2 + 4(1
- 6) sin2(0/2) } N/2
/f' [ 4 cos2(0/2) sin2(6/2) 1 (N - 1)ax p S2 + 4(1 - 6) sin2(0/2) J
aN_1(a sin 9)N-2 a d6
N/2
6N_1
do.
(A.69a)
We replace fi by 0 in this integral, discard the 62 if 6 < 2, and discard the 4(1 - 6) sin2(0/2) if 6 > ; accordingly, 2
4 cos2(0/2) sin2(0/2) 1 62 + 4(1
N/2
- 6)sin2(0/2)J
dB
2N/2 fo (cos(0/2))N do
if 0 < 6
2N fo (sin 0)N do
if 2 < 6 < 1,
fn/2
< 2N+I
J0
(sin 0)N do
21
for 0<6< 1,
(A.69b)
because
fn
fn/2 0\ N I cos 2 I do = 2 (sin (p)N
J
LLLIIj3(xo,h)I,
dcp
20 = it2 - cpJ
.
j 1, that depends only on N; in view of (A.67) and remarks in (iii), this bound The results (A.68) /and (A.69) yield a bound for proves the lemma.
Remark 4. When proving uniform Holder continuity, with the supremum of the function already bounded, one need consider only points x and y satisfying 0 < Ix - yl < 6 for some 6 > 0, because Ix - YI > b
Iw(x) - w(Y)I < 2 sup. Iw(z)I Ix - YIN
b2
Appendix A. On the Newtonian Potential
210
Theorem A.15 As in Lemma A.13, let B := 9(0,a) in RN, N > 2, let f E C°' (B) for some it E (0, 1) and let v be the Newtonian potential of f I B. Then each second derivative "AB has an extension to B that is in C°'µ(B), and v is a C2-solution of -,Lv = f in B.
Proof In this proof (. )ij := ai8j(.) for functions (but not for the Kronecker delta), and Ilf II
Iif
I
Co'µ (B) II
(i) To bound vij(xo) independently of xo E B, we use the formula (A.60) for vij(xo) and the bound (A.61) for If(x) - f(xo)1; accordingly,
= I f K`3(xo-x){.f(x)-f(xo)}dx-8ijf
Ivi>(xo)I
F
<
J
Za
(x0)
R-N IfII R'` RN-1 dR + Il.fNII
o
<
(A.70)
rllfll(aµ+1).
(ii) To prove uniform Holder continuity of vii, we consider field points x0 and xo+h in B, with 0 < Ih1 < a/4 [recall Remark 4]. The corresponding small ball is H := . (xo, 21hl). This time we do not separate the two steps of the small-ball technique, but proceed from the decomposition vij(xo + h) - vij(xo)
j
f
Kij(xo - x){ f (x) - f (xo) } dx nB
+ J \H { Kij(xo + h - x) - Kij(xo - x) } { f (x) - f (xo + h) } dx +{ f (xo) - f (xo + h) }
-1I
fB\H
Kij(xo - x) dx
bij{f(xo+h)- f(xo)}.
(A.71)
The inequality (A.61), with x0 replaced by xo + h [hence with R replaced by Rh := Ix - xo - hl], yields If (x) - f (xo + h)I < 21If II Rh
(A.72)
for all x and x0 + h in B. Then the modulus of the first integral in (A.71) is bounded by
F
f 0
3Ihl
RhNI.f11RRRJ-1dRh=IIl.f11Ihi.
A.5 Second derivatives of u at points in G
211
The second integral is similar. The first factor of the third integrand is estimated by our usual procedure for R > 21h1; the second factor by means of (A.72), in which Rh < (3/2)R for R > 21h1. Then the modulus of the third integral is bounded by IF
R-N_1lhl 00
IIf 11 Rµ RN-' dR = rllf 11 Jhlµ.
J21h1
The fourth integral has been bounded in Lemma A.14. Hence the fourth and fifth terms on the right of (A.71) also contribute r11f 11 I h1µ, so that lv,j(xo + h) - vtj(xo)I < rl1f 11
l hjµ
(A.73)
for all xo and xo + h in B with 0 < Ihl < a/4. Therefore v,j has a unique continuous extension to B (Dieudonne 1969, pp. 56-57) and this extension belongs to C0"µ(B).
(iii) That v E C2(B) follows from what has just been proved; that -(Ov)(xo) = f (xo) at each xo E B is implied by (A.60), since xo. Therefore v is a C2-solution of the (LK)(xo - x) = 0 for x Poisson equation in B.
Theorem A.16 Let G be bounded and open in RN, N >- 2, and let f E C°'µ (G) for some p e (0, 1). Then each second derivative of the Newtonian
potential u off has the property whenever S2 c G,
8i8ju E C°'µ (52)
(A.74)
and u is a C2-solution of -Au = f in G.
Proof In this proof (. ),j := a,a j(.) for functions (but not for the Kronecker delta), and If 11 = IIf C°'µ (G) (I. (i) Let an (open) set fl, such that 12 = G, be given. Then dist(12, aG) > 0 because S2 and aG are compact and disjoint; let p := 1 dist(S2,UG). For each c E S2 and all xo E RN, define B := R(c, p) and I
v(xo)
:=
f K(xo - x) f (x) dx, a
K(xo - x) f (x) dx.
w(xo)
(A.75a) (A.75b)
G\B
[The symbols v(xo, c) and w(xo, c) would be more accurate but more clumsy.] Then B c G and u = v + w. Application of (A.60) to v and of
Appendix A. On the Newtonian Potential
212
(A.33) to w shows that, whenever c E S2 and xo E B, v1i( xo)
=
JB w (xo) = f
K ,j ( xo
- x ){ f (x ) - f (xo ) } dx - S,j fN
K1(xo - x) f (x) dx .
,
(A . 76 a) (A.76b)
\B
In particular, when xo E S2 is given, we may choose c = x0, so that the field point is at the centre of B. [This is legitimate because (A.76) holds for every c E S2 and every xo E -4(c, p), but it must be remembered that the differentiation with respect to xoi and xoi was for fixed c and before the choice c = xo was made.] (ii) To bound uij(xo) independently of xo E S2, we choose c = x0, adapt (A.70) to the present situation, obtaining Ivij(xo)I < rllf ll (Pµ + 1),
and infer from (A.76b) that, with D := diam G, Iwij(xo)I < IF
D
R-N
Ilf II
RN-1
dR = F IIf II log P
(iii) To prove uniform Holder continuity of uij on S2, we consider field points x0 and x0 + h in S2, with 0 < h i < p14 [recall Remark 4] ; again we choose c = xo. The identity (A.71) and estimates like those leading from (A.71) to (A.73), again with H := -V(xo,21hl) but now with B = -4(x0,p)1 with 0 < Ihl < p/4 and with (A.64) replacing Lemma A.14, yield Ivij(xo + h) - vij(xo)I < rllf II Ihlµ whenever xo, xo + h E S2 and 0 < IIt I < p14.
Next, our standard method of estimating for R > 21hl gives I wig (xo + h) - wil(xo) I =
J \B{Kid(xo+h-x) -Kij(xo-x)}f(x) dx
< r f R-N-1 IhI IIfll
RN-1 dR
P
= IIIf1IIhi/P whenever xo, xo + h E S2 and 0 < IhI < p/4. Thus uij E C°°µ(5l).
(iv) The result (A.74) implies that u E CZ(G) [because for each point xo E G there is an open set 0 = O(xo) such that xo E S and 52 c G]. To prove that -(L u) (xo) = f(xo) at a given point xo E G, choose any ball
A.6 Exercises
213
B such that xo E B and B c G; in (A.75) and (A.76), let B denote this new ball. It follows that u = v + w, that -(Av) (xo) = f(xo) and that (Aw) (xo) = 0. Thus u is a C2-solution of the Poisson equation in G. A.6 Exercises
Exercise A.17 Considering the Newtonian potential in one dimension (N = 1), (i) derive the formula (A.23) for the potential of unit density in (-a, a); (ii) prove that, if f E C (G), then -u" = f in G.
Exercise A.18 Show that in terms of spherical co-ordinates r, 0,A the definition (A.14) of the Stokes stream function becomes 1
CWr,
r Po) = r sin 0 C- r
(PrJ
Derive (modulo an additive constant) the Stokes stream function WO in (A.16) from the potential To in (A.12).
Exercise A.19 Let z = x + iy = re'B denote an arbitrary point of the complex plane C. [The point (x, y) E 1[82 and the point x + iy E C can be considered one and the same for many purposes, such as the definition of balls, of open sets and of continuity] If x is holomorphic (or analytic) in an open set n c C, if x = (p + iW, and if (p is regarded as a potential function, then x is called a complex potential, W a stream function, and - dx(z)/ dz a complex velocity (or complex electric field or ... ). (i) How is dx(z)/ dz related to V(p(x, y) and to VW(x, y)? Show that the vector field VT is tangential to the level sets of W. [Recall that the zero vector is both normal and parallel to every vector] (ii) The complex potential of a unit source at the origin is -(log z)/2n, after some suitable restriction of 0. Use this to find the analogues in the plane (JR2 or C) of the (po and WO in §A.1 for `flow past a closed ball'.
Exercise A.20 In the analogue for JR2 of the problem (A.8), one allows any bad behaviour of cp that is not passed on to V(p. (i) Referring to Exercise A.19, (ii), exhibit infinitely many solutions for flow past a closed ball in R2 under the condition V P(x, y) + (U, 0) _ 0 (r-1) as r -* oo. [Contemplate the complex potential i log z, z * 0.] (ii) State and prove an analogue, for flow in 1R2, of the uniqueness result in Theorem A.I.
Appendix A. On the Newtonian Potential
214
Exercise A.21 (i) Show that in the plane a multipole at the origin, of unit strength and of type a with lal >- 1, has complex potential iaz
Pa(z) =
(_)1
iaz
1
d
dzlog
27L
1)!
z
z-I«I
(z
0).
(ii) For a = (1,0), find equations describing the field lines; sketch these level sets of W(1,O).
(iii) Explain, and demonstrate by an example, how your picture in (ii)
can be combined with a picture of the conformal map ( = z", where n E { 2,3,4.... } and 0< arg z < 2ir/n, to yield a sketch of any multipole in the plane with dal > 2. [To draw a picture of the map C = z", write z = re'B and draw corresponding level sets of r and of 0 in the z-plane and in the C-plane.]
Exercise A.22 Derive the multinomial theorem: for any positive integer m,
(XI + ... + XN)m
m!
=
xa,
where a! := al ! ... aN
!
dal=m
[One method is induction on N; a second is a combinatorial argument; a third proceeds from 00
(1-x1-...-xN)E(x1+...+xN)m=1
(Ix1I+...+.IxNI<1).]
M=0
Exercise A.23 For multi-indices of length N, the partial ordering a < $ means that ai < Qj for j = 1, ... , N. As before, a ! := al ! ... OWL Derive the Leibniz rule for differentiating the product of sufficiently smooth functions f and g, namely, g) _
l1
a<
( - a) l a l (a af (aag
Exercise A.24 The Stokes stream function ip can exist (and be useful) without an accompanying potential function. Write x = (xl, x2, x3) = (x, s cos 2, s sin 1) for points of R3, with s > 0 and -it < .l < it. Suppose that we are given a cylindrically symmetric vector field V of the form V(x) = (V1 (x), V2(x), V3 (x)) = (u(x, s), v(x, s) cos A, v(x, s) sin A),
A.6 Exercises s = (x2 + x3)"2
215
P
f V
4
. (xo, so)
(a,b) .-
Fig. A.12.
Suppose also that V is solenoidal (or
and that V E C'(IR3,1R3) divergence-free):
in R3 V = ax + s as (sv) = 0 limiting values being taken for s = 0. Define, for all (xo, so) E R x [0, oc), (X so)
p(xo, so; P) :=
J(a,b)
{ sv dx -su ds },
where (a, b) is a fixed `reference point'; the path P of integration is in the half-plane R x [0, oo), is connected, and consists of finitely many straight-line segments, each parallel to the x-axis or to the s-axis (Figure A.12).
(i) Prove that for fixed (xo, so) the value W(xo, so ; P) is independent of the choice of P, so that ip(xo, so; P) = ip(xo, so). [The interior of the set bounded by two different paths is, or can be made, a finite union of open rectangles; the divergence theorem applies]
(ii) Show that
which is a generalization of (A.14). Deduce that the vector field (u, v) is tangential to the level sets of W. Exercise A.25 The equation 2
2
(a x2 + as2 - s a w(x's) = s2f(x,s)
(A.77)
Appendix A. On the Newtonian Potential
216
can be used to find a vector field V as in Exercise A.24 for which curl V is prescribed by av
au
ax
as =
sf(x,s).
(In hydrodynamics, the equations of motion allow f, but not sf, to be piecewise constant.) Show that, if we set yp(x, s) =: s2x(x, s) and interpret x as x = yl and s then (A.77) becomes as s = (y2 + + y5)1/2,
a a
(Ax) (Y)
yl)2
a 2 + ... + (aYs)
x(Y1, s) = f (Yi, s)
Exercise A.26 This question concerns the Stokes stream function in (A.15) for flow past Rankine's solid E. Write x for the previous x1 and consider F(x, s)
4n-2= Asx+y q
s)
As2 - cos 01 + cos 02,
Rl
+
x-y RZ
say,
(x + y)2 + s2 }1/2 respectively, R1 > 0, R2 > 0, s >- 0, and A, y are given positive constants. In order to describe OE, prove the where R1, R2 following.
(a) Ifx>0ands>0,then FX(x,s)>0. (b) If -y < x < y and s > 0, then FS(x, s) > 0. (c) For fixed x E [-y, y], there is exactly one number g(x) such that g(x) > 0 and F(x,g(x)) = 0; for fixed s E (0, g(0)], there is exactly one number h(s) such that h(s) 0 and F(h(s), s) = 0. (d) If s > 0 and F(x, s) = 0, then FF(x, s) 0. [Assume the contrary and observe that 3 cos 0 - cos3 0 decreases on (0, n).]
(e) There is exactly one number a such that a > y and F(a, s)/s2 -* 0
as s-*0. Deduce that aE has a representation s = g(x), -a < x < a. What can be said about the function g? Exercise A.27 In this exercise we relax the condition that the Newtonian potential is to exist at all points of IR'. Let B :_ -4(0,1) in R3; Theorem A.6 states that, if f E Lp(B) with p > 3/2, then the Newtonian potential of f is uniformly continuous on 1R3. (i) Prove that fo r-1 (log(3/r)) -a dr exists (is finite) if and only if a > 1.
A.6 Exercises (log(3/r))_l
(ii) Let g(x) := r-2
217
for x E R3 \ {0} and r := IxI < 2.
Show that gIB E L3/2(B) and that the Newtonian potential of gIB exists only in 1[83 \ {0}.
(iii) Let al, a2, ... , an, ... , with an = (anl, an2, an3), be an enumeration of all points in B that have rational co-ordinates. Define f : B -+ l1 by .f (x)
E00 2-ng(x - an),
IxI < 1,
n=1
where g is as in (ii). Prove that f E L312(B), that the Newtonian potential of f fails to exist on the set A := (an), and that A = B.
Exercise A.28 Suppose that N >- 3, that G = -4(0, b) \ R(0, a) (with b > a > 0) and that f (x) := rm for some constant m E R. Then the Newtonian potential of f : G --+ R is bm+2 - am+2
0
(m + 2)(N - 2)
(m + 2)(m + N)
am+Nr-N+2
(m
N)(N - 2)'
a
_N+2
(m + N)(N - 2) r
,
r >_ b,
provided that m * -2 and m * -N. (i) Verify the foregoing claim.
(ii) How can corresponding formulae for m = -2 and m = -N be obtained economically? (iii) For this density function f, many multipole strengths 6a in Theorem A. 12 differ from zero; for example, all those with a = (2n, 0, ... , 0)
and n E N. Why are the terms in (A.57a) with 1 < Jul < k - 1 absent from the present representation of u for r >- b?
Exercise A.29 Write x = (x, y) = (r cos 0, r sin 0) for points of R2, and contemplate the function u defined by
r = 0, u(x) =
r2 cos 20 ( 4 + log (loge ) 4
r-2 cos 20,
0
Appendix A. On the Newtonian Potential
218
Let B :_ -4(0,1); define f := -Au on B \ {0}, and define f on aB and at 0 in such a way that f E C(BA. (i) Show that 02U 0 C(B), so that u C2(B), even though f E CM. (ii) Prove that u is the Newtonian potential of f. [The proof of Theorem A.2 implies that, for the present exercise and
forIxol> 5,0<8<1, uaKo
u(xo) = f6
r-S
-
Koau
ar
where Ko(x) := K(x - xo).]
Exercise A.30 Write x = (x, y) = (r cos 0, r sin 0) for points of R2, and consider the sector
Sp :={xER2 I 0
$E(0,it).
The Newtonian potential of unit density in Sa is a2
r = 0,
- Co, 2
g(r, 0) +
u(x) =
00
EC,, (a)
n
0 < r < a,
cos n0,
n=0
a2# l og a
2
r
+
a2 7r
00
E
( cn
+ dn )
(a)"rcos n 0 ,
r > - a,
n=0
where (if 0 < r < a, z = rei0 and Re denotes the real part) si n 2/i z 2 log a + cos 2# Re z2 4 r2 , g(r, 0) := si 2# Re z 2 [log a + i n , fl < 0 < 2 ir
-
n
g(r, 0)
=
s in 2/3 41T
2
a
r log r cos 20 +
r2 it
40
E do cos nO
for all 0 E Il8,
n=0
and where /f
co = 2 log
1
a
/3
+4,
1
1
c2 = 4 sin 2/3 - 4 /f cos 2/3,
sin n/i
cn = - n2(n - 2)
if n = 1 or n >- 3, do = - 9 , 4
d2 = -
3
sin 2/3 +
1
/3 cos 2/3,
6 if n=1 or4 n>3.
do =
n
2 (n2n
4)
-
A.6 Exercises
219
n/2. (ii) Given that in Theorem A.2 the hypotheses v E C2 (RN \ G) and Av = 0 in RN \ G are implied by the conditions v E C1(RN) and (i) Show that laiu(x)l --> oo as x -> 0 in 1182\aSS, provided that
f
for all (pEC,°(RN\G), N
verify that the foregoing formulae for u do indeed represent the Newtonian potential of unit density in Sp.
Exercise A.31 In order to discuss the Newtonian potential of a density function f on an unbounded open set G c RN, we demand that
where W(x) :=
W f E L1(G),
l+lxl
if N=1,
log(e + 21x1)
if N = 2, if N>_3.
1
(A.78)
As preparation for extending certain theorems, prove that, if (A.78) holds, if a is a multi-index and if 6 E (0, 1], then
f
(aK)(xo-x)f(x)dx\f(xo,d)
if N =1, loci = 0,
(1 + Ixol) Il w.f ll
{log1+62Ixol+11
IIWfll
1
Cab-N+2-iai
11111
J
if N=2, Ial=0, if N or N
(A.79)
3, loci = 0 1,
loci
1,
L,(G)11, the constant Ca is as in (A.31), and G may be . 11 = II . unbounded. Prove also that, if f E Lp(G) for some p > 1, then (A.78) ensures that f c- LS(G) for l < s < p. [Observe that, with the notation (A.22b), we have log(1/R) < log(116) where 11
I
if 6 < R < 1; also, log R < log 2ro if R > 1 and r < ro ;
finally,
log R < log 2r if R >_ 1 and ro < r.]
Exercise A.32 Use the results of Exercise A.31, and inspection of proofs for a bounded set G, to establish the following results for an unbounded open set G c RN. (i) Theorem A.5 remains valid if the hypothesis f E L1(G) is replaced by (A.78).
Appendix A. On the Newtonian Potential
220
(ii) Theorems A.6 and A.8, Lemma A.9 and Theorem A.11 remain valid if the hypothesis (A.78) is added. (iii) Theorem A.16 remains valid if the symbol CO,,"(.) is replaced by Cb'µ(. ), and the hypothesis (A.78) is added.
Exercise A.33 This exercise concerns an extension of Theorem A.16 to very smooth density functions f. (i) Let G be a bounded open subset of RN, N >- 2; assume that either aG is of class C1 or G is listed in Remark D.4. Let u be the Newtonian potential of a density function f E C°° (G) and let x° E G. Prove that
(aiu)(x°) _
JO ni(x)K(x-x°) f(x) dS(x)+J K(x-x°) (aif)(x) dx, c
c
(A.80)
and that, for each multi-index a, in terms of integers i = i(a, j) and multi-indices p = p(a, j) and a = ar(a, j) that need not be specified precisely, I"I
(a"u)(x°) =
L(-1)1«l+.i-1
f n(x) (aPK)(x-x°) (f)(x) dS(x)
+ J K(x - x°) (a"f) (x) dx,
(A.81)
G
where IPI + Ioi = lal -1 and the sum over j is zero when Joel = 0. (If a" is written as ai,, a;Za;,, where m = Joel, then a precise but labyrinthine form of (A.81) can be found). (ii) Now let f E Ck (G) and let Of E C°'` (G) if Jf I = k; here k E No, µ E (0,1) and y is the same for all Of of order k. Prove that a7u E Co," (S2) if IyI = k + 2 and SZ c G, and that u E Ck+2 (G).
Appendix B. Rudimentary Facts about Harmonic Functions and the Poisson Equation
B.1 Real-analytic functions
The formula W(t) :_
expl\ 11
t2
-1
I,
Itl
0,
-
1,
defines a function W E C,°(R); it is the case N = 1 of the function w : RN -> R considered in Exercise 1.15. Such functions have many uses (as test functions and, after scaling, as kernels in the smoothing operation), but we may feel that they are imperfect in some way, that they are man-made, that they are nylon rather than silk. Our instinct is sound, for the flaw of W is this: if we form the Taylor expansion about t = 1, namely, k-1
n
W(1 +x) = EW(")(1)
HI
+&(x),
n=o
where Rk(W ) = fox ( (k
y)
1)!
W(')(1 + y) dy = W("(1 + Ox) k
0 < 0 < 1,
then there is no number p > 0 such that Rk(x) - 0 for all x E (-p, p) as k -> oo. If there were such a number p E (0, 1), say, then we would have 0"
Zp)
n=0
j
(n)(1)(-2p)n n!
which is a contradiction because W(1 - Zp) > 0, while W(n)(1) = 0 for each n.
Real-analytic functions are real-valued functions that lack this flaw; 221
Appendix B. Rudimentary Facts
222
for a real-analytic function f and for every point p in its domain, the Taylor series of f about p has a positive radius of convergence. Here we make this statement precise for functions defined on open subsets of RN; then we establish a test for real-analyticity that is often more convenient than the definition. Notation As always, S is an open, non-empty subset of RN; unless the
contrary is stated, N is any positive integer. We use multi-indices a (Definition A.3) with a! := a1! a2! ... aN!. Theorem B.1 (Taylor's formula). If f E Ck(fl) and if the line segment { (1 - t)p + tx 10 < t < 1 } is a subset of 52, then
f(x) _
(aaf)(P) (x
a!p)a+E(0-f)((1-0)p+Ox) (x
lal
a!p)a,
(B.1)
Ial=k
for some number 0 = 0(p, x) E [0, 1].
Proof The proof is very similar to several calculations in Appendix A, in particular, to the proof of Theorem A.12, but there the mean-value form of the remainder was not used.
Let h := x - p and g(t) := f (p + th), 0 < t < 1, so that d/dt = h1a1 +
+ hNBN; the multinomial theorem (Exercise A.22) states that
(-)mg(t) =
E
y1« (a"f)(p+th),
Ial=m
a1
(B.2)
and repeated integration by parts gives +g(k-1)(0)
f(p + h) = g(1) = g(0) +
Si(0) + ...
(k - 1)!
(1 - t)k-1
+ I01 (k -1)! k-1
_
(k)
(m)(0)
M=O
g
(k)
(t) dt
(0)
by the first mean-value theorem for integrals. (The refinement 0 < 0 < 1 is not needed here.) Inserting (B.2) into (B.3), we obtain k-1 hot
f(p+h) M=O IaI=m
(3 f)(p)+V h" (00'f)(p+0h), to l=k
which is (B.1) in a slightly different notation.
B.1 Real-analytic functions
223
Definition B.2 A function f E C00(0) is real-analytic in ) if, for each p E S2, there is a radius p = p(p) > 0 such that (8"f)(p) (x
f(x) _ M
a
a!
whenever x E M(p,p)f1S2.
(B.4)
Here Ea means summation over all multi-indices of length N, and convergence to f (x) of the series is the main part of the statement. Lemma B.3 For each n E N, e
(n )n e
e2 (n+ I)
(B.5)
Proof For n = 1, there is equality on both sides of (B.5). For n + log n >
log(n!) = log 2 + log 3 +
2,
flogtdt,
because log 2 > log t for 1 < t < 2, ..., log n > log t for n - 1 < t <
n.
Similarly, for n >- 2,
log(n!) = log2+log3+ +logn <
jn+I
logt dt.
Evalu ating the integrals and taking exponentials, one obtains (B.5).
Theorem B.4 A function f E C00(0) is real-analytic in S2 if and only if, for each compact subset E c 92, there are positive constants A = A(E) and B = B(E) such that «'
I8"f (x)l < A(BIaI) l
for all x E E
(B.6)
and for all multi-indices a of length N. (It is to be understood that 10CIIaI = 1
for dal = 0.)
Proof of sufficiency Let (B.6) hold and let p E S be given. Referring to (B.1), we write h := x - p,
(8"f)(p+6h)
Rk(h) lal=k
a
(0 < 0 < 1),
(B.7)
and wish to find p = p(p) > 0 such that Rk(h) -+ 0 as k - oo whenever IhI < p. First, let p Zdp, where dp := dist(p,Of)) if fl * j N, or dp := 2
Appendix B. Rudimentary Facts
224
if 0 = RN. Then E := M(p, Zdp) is suitable for application of (B.6); if A and B are the constants corresponding to this set, we have
<
Rk(h)I
A(Bk)k
EI
Ihil°fj...IhNIaN
I«I=k
(Ih1I + ... + IhNI )k A(Bk)k
k!
by the multinomial theorem (Exercise A.22). In view of (B.5) and the Cauchy-Schwarz inequality,
<
1
k! - e
(e)
k
}1/2
N
and
Ihl <
1/2
h
12
j
j=1
so that IRk(h)I <
= Nl/2lhl,
i
(BeNI nlhl)k -+ 0
as k -> oo
if IhI < p and if we chooseep to be the smaller of Zdp and
1/BeN1/2.
The proof of necessity [that real-analyticity in S2 implies (B.6) for every
compact set E c Q] is somewhat longer and harder; hints are provided in Exercises B.34 and B.35. These exercises are for enthusiasts; we shall use the sufficiency of (B.6), but we need the necessity only to know that the condition is a good one. B.2 Smoothness and mean-value properties of harmonic functions
The word harmonic has many meanings in mathematics and science, but the phrase harmonic function usually means a function that satisfies the Laplace equation in some sense. In this book, the precise meaning must be the following, for reasons to be given presently.
Definition B.5 A function v : 0 --+ R is harmonic in !n if it is locally integrable in S2 (integrable on each compact subset of fl) and
f(Lq)v = 0
whenever
E C,°(fl)
and p >_ 0.
(B.8)
Henceforth Ll°°(!n) will denote the set of functions that are locally integrable in S2.
This definition coincides with a part of Definition 2.20 and states that, according to Definitions 2.10 and A.7, a harmonic function v is
B.2 Smoothness and mean-value properties of harmonic functions 225
a distributional solution of the Laplace equation Av = 0 in f2. In Definition A.7, the condition cp > 0 is omitted; that this has no effect is explained in Remark 8 of §A.2. The need to adopt Definition B.5 in this book comes from our need to have maximum principles for subharmonic functions that lack second derivatives here and there. That such maximum principles are required in applications is explained before Definition 2.10. Once the word subharmonic has been defined in a distributional sense, the corresponding meanings must be given to superharmonic and harmonic, otherwise our confusion would be total. Terminology Here as elsewhere, almost everywhere or a.e. means: except on a set of measure zero. The measure will be Lebesgue measure on RN (volume measure) until the contrary is stated. Theorem B.6 A function v is harmonic in 92 if and only if it is equal almost everywhere in f2 to a function u satisfying
and Au = 0 in
u E C°°())
(B.9)
92.
Proof (i) 'If. Let (B.9) hold; then [by (A.47) and (A.48)]
0=
Jn
coLu =
Jn
(A4p)u
whenever (p E Cc(f2)
and p >_ 0.
Since v = u a.e. in n, it follows that v E L110C(92) and that v satisfies (B.8).
(ii) `Only if'. Let v E L11' °(f2) and let it satisfy (B.8). In this proof a set G will be called admissible if G c S2 and G is bounded, open and not empty. It suffices to prove that, for each admissible set G, there is a function u(. ; G) E C00(G) such that u(x; G) = v(x) a.e. in G. This is sufficient for the following reasons.
(a) Compatibility: if G and G' are admissible sets that intersect, then u(x; G) = u(x; G') for x E G n G'; first, this holds almost everywhere in G n G', by equality a.e. to v(x), then it holds everywhere in G n G', by continuity of u(. ; G) and u(. ; G').
(b) Because of (a), we can define the desired function u by setting u(x) := u(x; G) whenever x E G. [For each point x E 12, there is no shortage of admissible sets G containing x.] Then u E C00(0) because each function u(. ; G) E C40(G). That Du = 0 in f2 follows from (B.8),
(D(y), N>_ 3
Fig. B.1.
which now implies that (PAu = 0
whenever cp E CC°(I) and cp >- 0,
and from Exercise 1.16.
(c) The boundedness of an admissible set G will be needed, when 0 is unbounded but f # RN, to ensure that dist (G, 8Q > 0 [because then is compact, 8S2 is closed and the two are disjoint]. (iii) We prepare to form a special test function cp. First, let µ E C°° [0, oo)
be non-increasing and such that µ(t) = 1 for 0 < t < 1/2, while µ(t) = 0 for t >- 1. [To construct such a function, apply the smoothing operation of Exercise 1.23, with p = 1/5, say, to the function that equals 1 on (-oo, 3/4) and vanishes elsewhere on R] Let K be the Newtonian kernel, introduced by (A.18), and define, for m E N and X E RN \ {0}, K(x; N) µ(mJxJ)
if N
2n (log mlxl) µ(mIxl)
if N = 2.
K(x)
2,
The function Km, illustrated in Figure B.1, has two useful offshoots, as follows.
(a) Let
h(x)
L Km(x) 0
if x
0,
if x = 0.
We shall see presently that hm is a smoothing kernel, of smoothing radius 1/m, that has every property of the smoothing kernel in Exercise
B.2 Smoothness and mean-value properties of harmonic functions 227
1.23 except (at least in some cases) non-negativity. In other words, hm(x) = mNhi(mx) and hm E Cc°(RN),
supphm c -4(0,1/m), f hm = 1.
(B.10)
N
(b) For a given admissible set G, let b := dist(G, 00) if 12 * RN, or let 8 = 1 if 0 = RN. Define (Figure B.1) (D(y) := Km(Y - x) - Kn(Y - x) y E RN \ {x}, X E G and n > m > 2/8 ;
for
(B.
il)
here x, m and n are parameters. Define D(x) by continuity, so that I(x) := 0 if N 2 and I(x) := (1/2n) log(n/m) if N = 2. The definition implies that D < 0 if N = 1 and that (D > 0 if N >- 2; we shall see that b E C,°(0). Since (B.8) extends to functions W E C,°(S2) such that p < 0 [choose cp = -ip], we may use 1 as a test function in (B.8).
Regarding the kernel hm. An easy calculation shows that hm(x) _ mNh1(mx). The first two claims in (B.10) are implied by the definitions of Km and hm; indeed, hm(x) = 0 also for IxI < 1/2m, because L K(x; N) = 0
if x 0. That f hm = 1 follows from the divergence theorem [or from integration of (d/dr)(rN-1 dKm /dr) with respect to r] :
dS - f dKm dS = 0 + 1 =1/2m dr dr by the basic property (A.17) of the Newtonian kernel. Regarding the function (D. It is clear that t E CI(RN) with supp 1 -4(x, l/m). Moreover, 4(x,1/m) c 12 because x E G, so that dist(x, 812) > S > 2/m when 12 * RN; consequently, D E C,°(12).
J
AKm(x) dx =
dKm
-1/m
(0,1/m)
(iv) For a given admissible set G, let S be the number defined before (B.11), and let B
U -I(x,16) , xEG
VG(Y) =
v(y) 0
if y E B, if y E RN \ B,
vG,m(x) := JIRN hm(x - Y) VG(Y) dY
for m E N and X E RN. Then VG,m E C00(RN) [by Exercise 1.23, (i), the proof of which does not need the condition k,, >- 0]. Also VG E L1(RN), so that IIvG - vG,m LI(RN)II --> 0 as m --+ oo [by Exercise 1.25 with I
p = 1; the hypotheses there are amply satisfied]. This last implies that a subsequence of (VG,m) converges pointwise to VG almost everywhere in RN (Rudin 1970, p.73; Weir 1973, p.171), hence to v a.e. in G.
Appendix B. Rudimentary Facts
228
But, if x E G and n > m > 2/6, then hn(x-.) and hm(x-.) have supports within -4(x, that VG,m(x) - VG,n(x)
=
28),
which is in n and on which vG(y) = v(y), so
jj(AK.)(x-y)-(AK.)(x-y)jv(y)dy
= f(&1)(y)v(Y) dy = 0
(B.12)
by (B.8) with the test function D in (B.11). Thus all functions vG,m are equal on G for m > 2/8; since a 'subsequence converges pointwise' to v a.e. in G, we have VG,m = v a.e. in G for m > 2/6. Let u(x; G) := VG,m(X) for x E G and m > 2/8; then u(. ; G) E C00(G) and u(x; G) = v(x) a.e. in G, as desired. Definition B.7 We shall say that u is smoothly harmonic in n if u E C°°(f2)
and Au = 0 in fl. Note that, if v is harmonic and continuous in fl, then v is smoothly harmonic in Q. For, in Theorem B.6, equality of v and u almost everywhere in 0 becomes equality everywhere in f) if v is continuous.
Theorem B.8 (the mean-value property of harmonic functions). If u is smoothly harmonic in a ball B := M(c, p), and u E C(B), then the mean values of u over 8B and over B are both equal to the value of u at the centre of B. That is, LB u = u(c),
BI
I aBI
,
f u = u(c).
(B.13a,b)
Proof It will be helpful to write integrals over 89(c, r) as integrals over the unit sphere Y := 8-4(0,1); we shall do this by means of the transformation x = c + ry, in which r := Ix - cl and y E Y. First, let B0 := M(c, ro) with 0 < ro < p, and apply the divergence theorem :
0=0Au=JoB0
8u
= Jr
ry) I
fB13
ro -1 d dro
f
Y
ro -' dS(y) r=ro
u(c + roy) dS(y).
a
(8n
:=
n V)
B.2 Smoothness and mean-value properties of harmonic functions 229
an
an Fig. B.2.
Now discard the factor ro -1 and integrate the rest over (0, rl ), where
0 < rl
JY
u(c + r1 y) dS(y) -
JY
u(c) dS(y) = N 1 J rl -
u - I Y l u(c), (B.14)
881
where B1 :_ a (c, r1). Since I OBI I = ri -1 Y 1, the result (B.14) is (B.13a)
for the sphere aB1. To prove (B.13a) for aB, let rl -> p and use the continuity of u on B. To prove (B.13b), multiply (B.14) by ri _i and integrate over (0, p) with respect to r1. O Theorem B.9 If u is smoothly harmonic in n and S2 * jaau(x)j <_
Nlal
{ dist(x, 00)
RN, then
lMI
supyen lu(y)I
(B.15)
for all x e S2 and all multi-indices a of length N. (It is to be understood that lall'I =1 for I«I = 0). Proof Since u E C°°(S1), each derivative a"u E C°°(c2); since also a" and
A commute, a"u is smoothly harmonic in 0. The proof of (B.15) is by induction and uses the mean-value property of each derivative over balls. We may suppose that sup lu(y)I < oo (otherwise, there is nothing to prove); let M := supyEn l u(y)I,
dx := dist(x, On),
dx,E := dx(1 - E)
for some s E (0, 1) that is fixed until we reach step (iv).
(i) Given p E S2, we estimate (aju)(p) as follows for each j. Let B1 := -V(p, dp,E) (Figure B.2), apply the mean-value result (B.13b) for
Appendix B. Rudimentary Facts
230
balls to aju, and use the divergence theorem: l (aju)(p)l =
If B1I
< I B iI M = pEM. (B.16)
nJul
aJul
I
I
I
(ii) Given q E 0, we estimate (aj ju)(q) as follows for each i and j. Let B2 := M(q,
Zdq,E) (Figure B.2), and begin as before:
N
1
l (a,a,u)(q)l =
niaju
1
LB,
IB21
2
Now, for x c aB2 we have dx >- dq N
l aju(x) l<
M<
ax
dq,E
Zdq, so that (B.16) yields
Zdq,E >
N 1
dq,E
SUPxEaB2 1 a,u(x)I
M (x E aB2),
2
whence
(2Nl z
(B.17)
dq,E
(iii) We now have l a"u(x)l < 1Jl
1 I«
IaIN
Il M
for all x E 0
(B.18)
dx,E
if I a I = 0,1 or 2. Assume that (B.18) holds for I a I = k -1; to extend it to Ia1 = k at a given point c E S2, choose Bk := . (c, (1/k)d,,E) and proceed as in step (ii). Then N 1(a;aau)(c)l
supxEaBk
l a"u(x)l
(j al = k - 1);
for x E aBk, dx > do
-
1
dC,E >
k- 1 k
(kN
d,
l aau(x) l < t
k-1
d1
M,
C,E
whence
kN l k
M. { dC,E 1 Thus (B.18) is extended by induction to every multi-index a of length N. l (araau)(c)l
(iv) Since (B.18) holds for every e e (0, 1), it holds also for e = 0. [Otherwise, for some point y E S2 and some multi-index /J, J I aI N 1
J
I (01u)(Y)I> l d
Y
lal
M,
whence
(a&u)(y)l =
j I-IN l M l dy,a f
for some number 6 > 0, and this contradicts (B.18) if e = 6/2.]
B.2 Smoothness and mean-value properties of harmonic functions 231
Theorem B.9 is remarkable in that, outside the class of holomorphic (or complex analytic) functions, it is rare for the supremum of the modulus of a function to control the magnitudes of all derivatives. For example, the real-analytic function defined by u(x) = cos AX,
x E (-b, b) c R,
A = const. > 0,
satisfies (B.15), for lal = 2m, m E N, and x = 0, only if A < 2m/b. Perhaps better evidence of the strength of Theorem B.9 is that it yields rather easily Theorems B.10 and B.11, both of which are reminiscent of properties of holomorphic functions. Theorem B.10 If u is smoothly harmonic in 92, then it is real-analytic in 0.
Proof By Theorem B.4 it is sufficient to prove that, for each compact set E c Q, there are positive constants A = A(E) and B = B(E) such that 17"u(x)l < A(BlaI)I'l
for all x E E
(B.19)
and for all multi-indices a of length N. Let E be given; observing that dist(E, 752) > 0 when Q # RN [because E is compact, On is closed and the two are disjoint], choose a bounded open set G such that E c G and G c 0, and define A := supYEG u(Y)I = maxyec1 u(Y)
By Theorem B.9, applied to the set G rather than to 52, I7u(x) <_ A {
l°`
Nlal
dist(x, G) }
}II Nlal
for all x E E,
where dist(E, 7G) > 0 for the same reason that dist(E, 7Q) > 0 when 52 * RN. Therefore (B.19) holds if we choose B := N/dist(E, 7G). Theorem B.11 (a Liouville theorem). If u is smoothly harmonic in RN and u(x) = o(r) as r := IxI -> oo, then u is a constant.
Proof We shall prove that (7iu) (xo) = 0 for each xo E RN and each
j E {1,...,N} by applying Theorem B.9 with lal = 1 and with S = _V(O,Rm). Here (Rm) is an increasing sequence such that Rl > 21xol for given x0 E RN and such that Rm oo as m -> oo. Then (7Ju)(xo)j <_
Nlxol
Iu(xI < R
lu(ym) I
,
(B.20)
Appendix B. Rudimentary Facts
232
where Iu(ym) I is the maximum of (u(x)i for jxi < R,,,. If Iu(ym+1) I > I u(ym) I, then lym+1I > lyml; if Iu(ym+1) I = Iu(ym) I, then we either have, or may take, ym+1 = ym. Thus the sequence (I ym I) is non-decreasing. The theorem now follows from (B.20) because I u(ym) I/ Rm , 0 as m --> oo. [If I ym I -* oo, then u (ym) = o (I.ym I) and I ym l < Rm ; if I ym I is bounded,
then so is u(ym).]
B.3 The Kelvin transformation
This section concerns a conformal map of RN \ {0}, and a related transformation of functions, under which smoothly harmonic functions remain smoothly harmonic. Definition B.12 Given a sphere 8-4(0, a) in RN, we call 2
Sx := 72x a
(x E RN \ {0}, r := lxi)
(B.21)
the reflection or inverse point of x relative to 0-4(0,a). The Kelvin transform T f relative to 8-4(0, a), of a function f : S2 -> R, is then defined by
a N-2
(Tf)(x) := (p)
f (Sx)
(Sx E S2),
(B.22)
provided that 92 c RN \ {0}.
To calculate with S and T, we use points l; E RN with p = Sx, then pr = a2 and
r2 (a2 I Y2 x S = p2 = a2 a2
If
x,
so that S is its own inverse operator: S-1 = S. If
= Sx and qp = T f,
then (ar
Ox) =
N-2 f(Sx)
N-2
f( )_ \p)
so that T is also its own inverse operator: T-1 = T. As regards the geometry of the transformation = Sx, it is obvious that 0-4(0,a), the sphere of inversion, is mapped onto itself, and that the sets -4(0, a) \ {0} and RN \ (0, a) are mapped onto each other. The Frechet or total or linear derivative of the transformation is 2
S'(x) _
(ax
(x)) =
r2
M(x),
2 where Mij(x) := Sid - r2 j.
(13.23)
B.3 The Kelvin transformation
233
This shows that the matrix M(x) is orthogonal as well as symmetric, because one checks easily that M(x)M(x) = I, the identity matrix. The orthogonality of M(x) implies that, for all h and k in RN \ {0}, the angle between S'(x)h and S'(x)k equals that between h and k. Consequently, the map S is conformal in the sense that, if two smooth arcs intersect
at a point, then the angle between them at that point is conserved by the mapping. [For, let x = f (t) and x = g(t) be descriptions of two smooth arcs that intersect at xo = f (to) = g(to). The orthogonality of M(xo) ensures that the angle between S'(xo)f'(to) and S'(xo)g'(to) equals that between f(to) and g'(to); the latter angle is that between tangents at xo to the original arcs x = f (t) and x = g(t), while the former angle is that between tangents at Sx0 to the transformed arcs = S(f(t)) and
= S require a small definition.
Definition B.13 By a half-space in RN we mean a set { x E RN where k E RN \ {0} and p E R.
I
>,u
Exercise B.14 Let S be the reflection operator in (B.21), and let G be a ball or a half-space in RN. (i) Prove that S(G) is a ball if 0 G, that S(G) is a half-space if 0 E 8G, and that S(G \ {0}) is the complement of a closed ball if 0 E G.
(ii) Prove that, for N > 2, S(G) = G if and only if 8G intersects 0-4(0, a) orthogonally.
[Orthogonality means here that normals nG to 7G and nB to 8,4(0, a) satisfy nG(y) nB(y) = 0 at a point y of intersection.]
The next theorem shows that, if u is smoothly harmonic in 1, then its Kelvin transform Tu is smoothly harmonic in S(fl); Definition 2.20 and Exercise B.37 deal with an extension to merely subharmonic functions. It is this property of the Kelvin transform that makes it useful. Indeed, almost everything that can be done with holomorphic (or complex analytic) functions in the complex plane C by means of a Mobius (or `bilinear') transformation, can be done with smoothly harmonic functions in R' by a composition of translation (u H u(. +c), where c E RN), dilation (u'--* u(2. ), where A E R \ {0}) and the Kelvin transformation. The proof of Theorem B.15 is not the shortest route to the formula (B.24), but includes details that are useful in applications of the transformation.
Appendix B. Rudimentary Facts
234
Fig. B.3.
Theorem B.15 Let f E C2(ST ), where fl c RN \ {0}, and let T f be its Kelvin transform, as in (B.22). Then, for x E S(c2),
A(Tf)(x) =
(ar)N+2
(L.f)(Sx).
(B.24)
Proof (i) Let = Sx, x * 0, so that x = 0. Consider two copies of RN \ {01: the first has Cartesian co-ordinates c,. .. , N ; the second (Figure B.3) is the image of the first by S and has Cartesian co-ordinates x1, ... , XN, while 1, ... , N are curvilinear co-ordinates there.
Let ax/ak := (akS)(). In the second copy of RN \ {0}, the vector ax/ask at any given point is tangential to a curve along which k increases while the other j are fixed. The vectors ax/al,...,ax/aN at the given point are mutually orthogonal because they are images, under the conformal map S, of vectors along the co-ordinate lines in the first copy of RN \ {0}. To verify this orthogonality of ax/a 1, ... , ax/a N, we infer from (B.23) that ax
ax -
aj ak
a4 N
a4
= P4 E Mpj( )Mpk( ) = P4 Sjk.
(B.25)
Thus 1,...1;N are orthogonal curvilinear co-ordinates in the second copy of RN \ {0}.
(ii) We use arc-length functions h j defined by az
axJ
2'
B.4 On the Dirichlet and Neumann problems
235
and basis vectors bj := for j = 1,...,N. The rules (Kellogg 1929, pp. 181 and 183; Spiegel 1959, pp. 148 and 151) a
...
axl a2
ax1
+ ... +
a
_b' a
axN
h1
a2
1
axN
h1... hN a
bN a
(B.26a)
hN
(h2... hN a
a h,
... hN_1
+ aCN
hN
+...
T
h1
)
a
(B.26b)
NN-
simplify here because all the h j are equal.
(iii) Let g(x) := f(Sx) and recall that Or-N+2 = 0 in RN \ {0} for all N E N; then, by the definition (B.22) of T f and the Leibniz rule for repeated differentiation of products,
A(T f)(x) = 2
v
(ar
(ar)N-2
N-2
Og(x),
Vg(x) +
where air = p/a and g(x) = f (Sx) = f
Application of (B.26a, b) now
gives
O(T f)(x) N
= 2(N - 2)a-aN-2pN Y j
af( )
j=1=1
+a-N-2
P
3N-2
1` j(-2N + 4)p-2N+2 j ON) j=1
p N+2
=
+ p_2N+4
J(Z >
j
(a
and this is the desired formula (B.24).
B.4 On the Dirichlet and Neumann problems In Appendix A we encountered the Poisson equation, -Au = f in G, as an equation satisfied by the Newtonian potential u of a smooth density
function f on G. Here we regard the Poisson equation as one to be solved when a boundary condition for u is specified.
Let f : S2 -> R and g : au -> R be given functions. The Dirichlet problem for -A in S2 is to find u E C(S2) n C2(SZ) such that
- Au = f in u,
ul an = g.
(B.27)
236
Appendix B. Rudimentary Facts
The Neumann problem for -,L in 0 is to find u e CI(S2) n C2(0) such that
-Du=f in
au 12,
an an
= g,
(B.28)
where n denotes the unit normal outward from S2 and a/an := n V. These statements omit a great deal. Only pointwise solutions (C2solutions) have been mentioned. We have not said what smoothness
the data f, g and aS must have in order that the solution u have the continuity properties that we have demanded. When 0 is bounded, we can hope to solve the Neumann problem only if the compatibility condition
-J nf =Jang
(B.29)
holds, because both sides of (B.29) equal fan au/an for a smooth solution of (B.28) with smooth data. In fact, if 52 is not connected, (B.29) must hold for each component (each maximal connected subset) of fl. When S2 is unbounded, growth conditions (or decay conditions), specifying how large Iu(x)I and perhaps IVu(x)I are allowed to become as IxI -+ co, must
be added to the statements of the Dirichlet and Neumann problems. Some of these gaps will be filled as we proceed.
Reduction to the Laplace equation or to zero boundary data (i) Often one can remove the term f from the Poisson equation in (B.27) and (B.28) by the substitution u = Uf+v, where Uf is the Newtonian potential off. For example, if 0 is bounded and f E C°'µ(S2) for some p E (0, 1) [Definition A.10, with the suffix b omitted from Cy°" because S2 is bounded], then
Theorems A.11 and A.16 ensure that Uf E CI(S2) n C2(f) and that -,L Uf = f pointwise in S2. The Dirichlet problem is then to find v E C(S2) n C2(S2) such that
Ov = 0 in n,
vI
an = g - U f Ion;
(B.30)
the Neumann problem is to find v E CI(S2) n C2(S2) such that
Ov = 0 in
S2,
av
=g-
an an
a Uf an
.
(B.31)
an
In both cases, v (if it exists) will be not merely in C2 (K2) but in C°°(S2) (Theorem B.6), indeed, it will be real-analytic (Theorem B.10). Note that, if the compatibility condition (B.29) holds for f and g in the Neumann
case, then it holds also for 0 and g - 0Uf/an.
B.4 On the Dirichlet and Neumann problems
237
(ii) Alternatively, if one can find a sufficiently smooth function h : S2
R such that h = g on OS2 in the Dirichlet case, or Oh/an = g on OS2 in the Neumann case, then the substitution u = h + w yields a problem for w with boundary condition w = 0 on OS2 or aw/an = 0 on OS2. However, for a Poisson equation with forcing function in C°4'(S2), it is necessary that Ah E C°°µ(S2); to find such a function h may not be easy if g and On are less than beautiful. We begin work on the Dirichlet and Neumann problems by proving uniqueness of solutions in some cases. Then any method of solution becomes acceptable in those cases, however contrived or squalid the method may seem, because what it produces is the solution. Theorem B.16 The Dirichlet problem for -L in n has at most one solution if either (a) S2 is bounded, or (b) S2 is a half-space (Definition B.13) and we add the growth condition:
u(x)=o(r)for xES2and r:=xj-+oc. Proof Let v := u1 - u2 be the difference of two solutions; then v E C(S2) n C2(S2), we have Av = 0 in n, and v = 0 on On. The additional smoothness implied by Theorems B.6 and B.10 is not needed here. (a) If 92 is bounded, then Theorem 2.5 (our first and simplest maximum
principle) states that v < 0 on fl, because v is a C2-subsolution relative to A and n, and that v >- 0 on n, because -v is also a C2-subsolution. Thus v = 0 on S2, as desired. (b) If 92 is a half-space, we so choose co-ordinates that n = { x E ][8N
XN > 0 }, and note that v(x) = o(r) as r -+ oc. For N = 1, the equation Lv = 0 implies that v(x) = co + clx, where co and cl are constants; then
co = 0 because v(0) = 0, and cl = 0 because v(x) = o(x) as x
oo.
For N >- 2, we apply Theorem 2.30 (a relatively advanced maximum principle). The hypotheses of that theorem are satisfied by both v and -v (with room to spare in both smoothness and growth condition); hence supra v = 0 and info v = 0, so that v = 0 on S once again. For the Neumann problem stated in (B.28), uniqueness in the strictest sense is impossible: if we add to a solution u any function k that
is constant on each component [on each maximal connected subset] of S2, then u + k is again a solution. However, in many applications only uniqueness of Vu is required, and we have already encountered an example of such uniqueness in Theorem A.1 and the remark following it.
Appendix B. Rudimentary Facts
238
Theorem B.17 The Neumann problem for -A in S2 admits at most one function Vu for solutions u if (a) S2 is bounded and as2 is of class C', or (b) 92 is bounded and is listed in Remark D.4, or (c) 0 is a half-space and we add the growth condition: u(x) = o(r) for
xES2and r:=lxl-+oo. Proof The difference v := u1-u2 of two solutions is now in C'(S2)r1C2(S2)
and satisfies Av = 0 in 0, av/an = 0 on M. (a),(b) If 5 is bounded, and either as2 is of class C' or S2 is listed in Remark D.4, then we have both a divergence theorem and approximations S2to u as in Theorem D.9. Applying the divergence theorem to the vector field vVv and one of these sets [which is legitimate because V E C2(1 m)], and observing that V (vVv) = IVv12 because Av = 0 in S2 [a fortiori in 52,,,], we obtain IvvI2 = fa.
van anm an
Let m -> oo; by Theorem D.9 and because v E C'(S2), av/an = 0 on aft, IVv12 = 0,
and this implies that IDvI = 0 on S2, because v E C'(S2).
(c) If 0 is a half-space, we choose co-ordinates again that make 0 = { x E RN
XN > 0 }, and notice again that v(x) = o(r) as r -+ oo. Let nn := { x E RN XN > 1 /m } for m E N. Integration by parts along co-ordinate lines, very much as in Remark 7 of §A.2, shows that
f(Lp)v = -.f m
a.m
{ (N(p)v
NV} + f Cty if (p E C(RN); m
the last integral vanishes because Av = 0 in 0. Let m -+ oo and recall that v E C'(S2,) and aNV = 0 on 00; then
(A(P)v = -
f
(aN(p)v
if (p E CC°(RN).
(B.32)
asp
Now extend v to RN as an even function of XN by setting v(x', -XN) :_ v(x), where x' := (x1,...,XN_1) and XN > 0. Let G := {x E R"' XN < 0 }; repeating forfG G the steps that led to (B.32), we obtain (
= Jac
if
E C(RN).
(B.33)
B.4 On the Dirichlet and Neumann problems
239
Also, V E C(RN). (In fact, v E C'(RN), but we do not need this.) Add (B.32) and (B.33) for the same function q ; the boundary terms cancel and the result shows, in view of Definition B.5, that v is harmonic in RN. Since also v is continuous, it is smoothly harmonic in RN, by the remark following Definition B.7, and the extended function v is still o(r) as r co. Therefore Theorem B.11 states that v is a constant. Regarding existence of solutions, we shall consider only some aspects of the method of Green functions, which is perhaps the most classical and constructive of the various approaches that are now available. Our limited treatment will yield results only for balls and half-spaces, with
a slender hint of how the method proceeds for other sets. The main ingredients of the method are the notion of a fundamental solution, and the representation formula (B.35) below. Definition B.18 Let K be the Newtonian kernel introduced as K(. ; N) in (A.18). A function F defined by F(xo, x) := K(xo - x) + q(xo, x)
for xo E S2, x E f2
and xo * x
is a fundamental solution of -A in 0 if, for each fixed x0 E 12,
q(xo,.) E C'(n) n C2(f), Aq(xo, x) = 0
for all x E S2,
(B.34)
where A is with respect to x. The function q will be called the non-singular
part of F. Theorems B.6 and B.10 imply once again that q(xo,.) is real-analytic in n; we have written C2 (fl) as a demand for a pointwise solution, not as a result. Obviously q is not determined uniquely by (B.34); presently we shall add one of several boundary conditions. Exercise B.19 In both the following situations F is a fundamental solution of -A in S2, and u E C1(S2) n C2(S) with Du E L1(S2). (i) Assume that 12 is bounded and that either 812 is of class C' or 92 is listed in Remark D.4. Prove that u(xo)
= - in F(xo, x) (Lu)(x) dx + I {F(xo, x) au(x)
7F(xn, x)
n
where xo E S2 and a/an is with respect to x.
u(x)} dS(x),
(B.35)
Appendix B. Rudimentary Facts
240
(ii) Assume that fl is a half-space and that (in addition to the foregoing hypotheses about F and u) one or other of the following growth conditions holds. For x E Sl and r := lxl -+ oo, for multi-indices a (Definition A.3) of order dal = 0 or 1, for each fixed xo E 0 and for some constant 6 > 0, either a
a (_)F(xox) = 0 (rN+1)
aU(x) = 0 (rl)
,
,
(B.36a,b)
or
a Ox
s
F(xo, x) =
0 (r
N+2-lal)
if N 2, if N = 2,
l 0 (r lal log r)
aaU(x) = O (r-b-l al
.
(B.37a,b)
Prove that (B.35) still holds. [For both (i) and (ii), adapt the proof of Theorem A.2, which contains all essential steps.]
Now we specify boundary conditions for q(xo,.) and hence for F(xo,. ).
In order to have a fundamental solution that is useful for the Dirichlet problem, we demand that
q(xo, x) = -K(xo - x)
if xo E Sl
and X E Of).
(B.38)
In other words, F(xo,.) is to vanish on afl; then on the right-hand side of (B.35) only the terms involving L u and ul an remain, and these functions are prescribed in the Dirichlet problem. By Theorem B.16, there is at most one solution of (B.34) and (B.38) if 1 is a bounded set or a half-space, provided that for a half-space we add the growth condition: q(xo, x) = o(r) as r := lxi --> oc with x0 fixed in Q. For the Neumann problem, if fl is bounded and either at) is of class C1 or f is listed in Remark D.4, a suitable boundary condition is aq(xo, x) _ aK(x - xo) 1 if xo E S and x E Of); (B.39) an
an
laslI
here a/an is with respect to x and Ia
l
is the surface area of M. We
must not demand that aF(xo, x)/an = 0, because (A.20) and (B.34) imply that
P aK(x - xo) Jan
an
- -1
and
f
n
aq(xo, x) = 0
an
for fixed x0 E Q. For the Neumann problem with Sl a half-space, we do demand that aq(xo, x) an
_ aK(x - xo) an
if x o E Sl
and x E asl;
(B.40)
B.4 On the Dirichlet and Neumann problems
241
in this case, whatever emerges from the point source at x0 can escape to infinity.
Exercise B.20 Let 0 be bounded and either have 8Q of class C1 or be listed in Remark D.4. (a) Assume that the solution q(xo,.) of (B.34) and (B.38) exists, and
denote the corresponding fundamental solution by G. Show that, if the solution u of the Dirichlet problem (B.27) exists and belongs to C1(52) n C2(52) with Au E L1(52), then it is given by
u(xo) = f G(xo, x) f (x) dx -
Jon
7G(xo, x) g(x) dS(x),
(B.41)
On
where x0 E SZ and 8/8n is with respect to x.
(b) Assume that a solution q(xo,.) of (B.34) and (B.39) exists, and denote the corresponding fundamental solution by H. Show that, if a solution u of the Neumann problem (B.28) exists and belongs to C1(S2) n C2(52) with Au E L1O), then it is given by u(xo) =
Jn
H(xo, x) f (x) dx +
Jan
H(xo, x) g(x) dS(x) + k,
(B.42)
where xo E 0 and the constant k is the mean value of u over 852.
Remarks 1. A fundamental solution that satisfies a useful boundary condition, and leads to a representation formula like (B.41) or (B.42), is called a Green function. This phrase can be lengthened; for example, the function G in Exercise B.20, (a), is (if it exists) the Green function of the Dirichlet problem for -A in 52. Green functions exist for differential operators other than A and for boundary conditions other than those of the Dirichlet and Neumann problems. 2. At this stage of the theory, the formula (B.41) rests on two assumptions :
that the non-singular part q of the Green function G, and the
solution u of the Dirichlet problem, both exist. This state of affairs can be improved.
(a) For a few simple sets 52, such as a ball in R1, there is an explicit formula for the Green function G. (In the case of the half-space, this formula is an obvious one, but at the moment we are considering bounded sets 52.) For these particular sets 52, the first assumption is not needed, and we can dispose of the second by a change of direction. Instead
of proceeding from the Dirichlet problem (B.27) to the representation formula (B.41) by means of assumptions about the solution u, we can
Appendix B. Rudimentary Facts
242
show (when G is known explicitly) that the function u defined by (B.41) satisfies equations (B.27) for suitable data f and g. This will be done for a ball in §B.5. (b) For an arbitrary set 0, one cannot expect an explicit formula for the Green function (however smooth the boundary 8S2 may be). Rather, a more abstract argument is needed to prove existence of solutions of the Dirichlet problem. This material is outside the range of this book, but we remark that one can cast the Dirichlet problem into a form to which the Fredholm alternatives apply. This means, in effect, that uniqueness of solutions implies their existence, so that Theorem B.16 is again a corner stone. For a pleasant boundary On, the existence theory for the Dirichlet problem implies existence of q(xo,.) and hence of the Green function; then the formula (B.41) has the advantage of representing the solution u for all admissible data f and g. Similar remarks apply to the Neumann problem. 3. For the Neumann problem, we observed before Theorem B.17 that a constant can always be added, on each component of S2, to a solution u. For the function q(xo,.) this arbitrary constant becomes an arbitrary function of x0 that can be added to q(xo,.) and hence to
F(xo,. ). To remedy this partly, we demand that, for a Green function F of the Neumann problem for -,L in a bounded set S2 (with On of class C1 or with f listed in Remark D.4),
I F(xo, x) dS(x) = c
Ion
for all xo E 0,
(B.43)
where c is independent of x0 but is otherwise arbitrary.
Given a fundamental solution F* suitable for the present Neumann problem in that its non-singular part satisfies (B.39), and such that fan F* (xo,.) = y* (xo) for all xo E 0, we must now define c - Y* (xo)
F(xo x) = F (xo x) .4 and here only c is arbitrary.
4. In our discussion up to now of fundamental solutions F, the field point xo has been fixed, although an arbitrary point of 0. However, when we come to use the representation formulae (B.41) and (B.42), the dependence on xo of the Green functions there (that is, the behaviour of G(xo, x) and H(xo, x) as xo varies) will be of importance. This dependence on the first variable is given to us cheaply by the next theorem, which
B.4 On the Dirichlet and Neumann problems
243
ensures that results postulated or established for F(p,. ), with p fixed in S2, apply also to F(. , p). Theorem B.21 Let 52 be bounded and either have as2 of class C1 or be listed
in Remark D.4. Assume the existence of Green functions of the Dirichlet and Neumann problems for -,L in 52. In the Dirichlet case, the non-singular part is to satisfy (B.34) and (B.38); in the Neumann case, conditions (B.34), (B.39) and (B.43) are to hold. Then, after extension by continuity, both Green functions have the symmetry property
F(y,z)=F(z,y)
if (y,z)E(52xS2)U(52x52) and y*z.
Proof (i) We need prove only that F(y, z) = F(z, y) for (y, z) E 52 x 5 and y * z. For, suppose that this has been shown and that y E 852, z E 0. There is a sequence (y") in 52 such that, for each n, y" * z and such that y" -+ y as n -> oo. Then F(y", z) = F(z, y") for each n, and F(z, y") -* F(z, y) by the continuity of F(z,.) on 52 \ {z}. Extending F(. , z) by continuity, we have F(y, z) := limy ,y F(y", z) = F(z, y). (ii) To prove symmetry for (y, z) E 52 x n and y z, it will be sufficient to consider the Neumann case; the proof for the Dirichlet case is similar but a little easier. We apply the Green identity (A.26),
fA
OW {vLw-w/ v}= fOA{va
n
Ov
-w8n
to v = F(y,. ), w = F(z,.) and A = 52,"
4(y, s) U (z, s) }, where is the mth approximation to n in Theorem D.9; once again this
approximation is required in order that v E C2(A) and W E C2(A). Given distinct points y and z in 52, we can choose m E N so large, and e > 0 so small, that 4-(Y18) and ,,t7(z, e) are within and are disjoint. The integral over A vanishes because Lv = L w = 0 in A. Dealing with the integrals over a9(y, s) and 02(z, s) as we dealt with the integral over a. (xo, e) in the proof of Theorem A.2, we find that
0 = F'(y, z) - F(z, y) + fan.
aF(yan, x)
aF(8zn, x)
F(y, x)
- F(z' x)
dS(x),
where a/an is with respect to x. Suppose that our choice of m was m >- M. Since F(y,.) and F(z,.) are in C 1(S2 \ 52,M ), it follows from Theorem D.9 that the integral over 3 m tends to the corresponding integral over a52
Appendix B. Rudimentary Facts
244
as m -> oo. By (B.39),
F(y, z) - F(z, y) = lal
J
n{ fly, x) - F(z, x) } dS(x),
(B.44)
and this integral over Of) vanishes for all y and z in SZ if and only if (B.43) holds. Thus F(y, z) = F(z, y).
We turn to particular cases: the Green functions of the Dirichlet and Neumann problems for half-space and ball. The ordering of this material is: notation, formulae for the half-space (presented first because of their simplicity), formulae for the ball, explanations and comments. Notation Let D := { x E RN XN > 0 } and B := M(0, a) in RN, for all N E N in both cases. The reflection in 8D of any point x E RN will be written x* := (x', -xN), where x' := (X1, ... , XN_1). The reflection in 8B, or inverse point, of any point x E 1RN\101 will be written Sx := (a2/r2)x, where r := IxI, as in Definition B.12. The Newtonian kernel, introduced as K(. ; N) in (A.18), continues to be denoted by K. Field points will often be denoted by y instead of xo. I
Statement B.22 The following formulae define Green functions on S2 x S2 \ { (y, x)
I
x = y }, where either ) = D or 12 = B. The non-singular
parts are defined on
S2x52\{(y,x) I this domain is bigger than that demanded in Definition B.18, and bigger than that established by extension in Theorem B.21. The reader should
verify that, if x E D, y E D and x = y*, then x = y E OD; also that, if
xEB,yEBandx=Sy,then x=yEBB. (i) The Green function of the Dirichlet problem for -,L in D is
GD(y, X) = K(y - x) - K(y* - x),
(B.45)
where
IY* - XI = { Iv' - x'IZ + (YN + xN)2
}112
= IY - x* I.
(B.46)
(ii) A Green function of the Neumann problem for -A in D is HD(y, x) = K(y - x) + K(y* - x).
(B.47)
(iii) The Green function of the Dirichlet problem for -A in B is
GB(Y,x)=K(y-x)-K (J-t(sy - x) I,
(B.48)
B.4 On the Dirichlet and Neumann problems
245
where IYI
a ISy - xI =
a
2 2 f a4 - 2a yz x +IYI IxI
}
1/2
IY - SxI. = IxI a
(B.49)
It is to be understood that limiting values are taken for y = 0 or x = 0, both here and, later, in formulae for derivatives of GB.
(iv) A Green function of the Neumann problem for -A in B is
HB(Y,x)=
K(y - x)
if N = 1,
K(y - x) + K (1X-t(sy - x) I
if N = 2,
K(y-x)+K (!i(Sy_x)) + E(y, 2 if N> 3, (B.50a,b,c)
where UN := Ia.N(0,1)I and E(y,x) is defined as follows. Let r := IxI, s := IyI and rs cos l :=x-y; then
fly, x)
f {PN_3(P2_2Prcos2+r2)_( N-2)
-1
dp,
(B.51a)
P
a2/s
00
J
tN-3 (2
-
2t rs cos A a2
+ r 2s2 1 -z(N-2) a4 J
-_
1 t
dt, (B.51 b)
00
f {PN_3(p2_2PscOsl+s2)_N_2)_ 1 } dp.
J
(B.51c)
P
a2/r
Again it is to be understood that limiting values are taken for y = 0 or
x=0. Remarks 5. Construction of these Green functions. (a) Definition B.18
obliges us always to begin with the potential K(y-.) [equally, K(. -y)] of a unit source at y. The artifice of making GD(Y, x) an odd function of xN (hence vanishing for xN = 0) by adding a negative unit source at the reflected point y*, and of making HD(y,x) an even function of xN (hence satisfying 8HD(y,x)/8xN = 0 for xN = 0) by adding a positive unit source at y*, is an application of a standard device in mathematics and physics. A shadow of this method can be found in virtually all formulae for Green functions.
(b) In the case of GB, there is an obvious resemblance of Sy to y*, but the factor IyI/a, accompanying Sy - x in the formula (B.48), requires
246
Appendix B. Rudimentary Facts
Sy
aB
Fig. B.4.
explanation. The situation is this: with y a fixed parameter and y E B, we seek the function -q(y,.) that is smoothly harmonic in B and equals K(y-.) on B. These conditions suggest the Kelvin transform, relative to 7B, ofK(y-.)IRN\B. [Here it is the dot that is restricted to RN\B, and with respect to which the transform is taken] By Definition B.12 and Theorem B.15, this Kelvin transform is defined on B \ {0}, is smoothly harmonic in B \ {0} and equals K(y-.) on B. If N # 2, we extend it to R by continuity to obtain -q(y,.). If N = 2, this Kelvin transform tends to minus infinity at the origin and is not symmetrical in . and y, but addition of the function x H (1 /27r) log(a/r) corrects both faults. (c) For the Green function HB of the Neumann problem, addition of the point singularity K ((IyI/a)(Sy-. )) is completely successful only for N = 2. When N = 1, this term is not needed for the boundary condition 8HB/8n = -1/IaBI; when N >- 3, it is not enough. Equation (B.51a) shows that the field -VE(y,.) of the extra term (for N >- 3 and y 0) is that of sources distributed on the radial line outward from Sy (Figure B.4); their strength (charge per length or mass per length) is proportional to pN-3 at the point distant p from the origin. The term -1/p in (B.51a) is required for convergence of the integral defining the potential E(y,. ), but makes no contribution when we differentiate with respect to x. 6.
Verification of basic properties. To check that the Green functions
in Statement B.22 have non-singular parts q(y,.) that are smoothly harmonic in S2, and satisfy appropriate auxiliary conditions, is a matter of inspection and direct calculation that is left mainly to the reader. Here are a few details that may be helpful.
B.4 On the Dirichlet and Neumann problems
247
(a) That the non-singular part of GB(y,.) is smoothly harmonic in B, for fixed parameter y E B (hence for fixed Sy B), is shown by the formulae yl
C
N+2
a
if N
K(x - Sy)
KNa N+2
K
(ii(SY_x)) _
if
K(x-Sy)+ I
I
2, y
0,
N*2, y=0,
logy if N=2, y*0, IYI
if N=2, y = 0,
og 1
a
271
(B.52)
where the values for y = 0 are limits of those for y # 0 [because JSyj = a2/lyl when y
0].
(b) That the function E(y,.), occurring in HB(y,.) for N > 3,
is
smoothly harmonic in B, for fixed parameter y E B, is established as follows. For y * 0, we use (B.51a); differentiation with respect to x under the integral sign is legitimate and is justified very much as in the proof of Theorem A.S. For y = 0, we observe from (B.51b) that E(0,x) = 0 for
allxEB. (c) To verify that HB, for N > 3, satisfies the boundary condition (B.39), we use the notation introduced before (B.51), write R
r2
- 2rs cos A + s2 }
1/2,
r2s2 l 1/2 A
{ a2 - 2rs cos A +
}
a2
JJJ
and use (B.51c) for the function E. Then (for r # 0) BHB( y, x)
N J R-N(r - s cos A) + A-N I/ a2
2
- s cos 2.
- a-N+2) } . - I (A-N+2 l setting r = a and observing a pleasing cancellation, we obtain 8HB(y, x) Or
1
r=a
aN aN-I '
(d) The function HB satisfies the normalization condition (B.43) because it has the symmetry property HB(y,z) = HB(z, y); the equivalence of these two conditions was established by (B.44).
Appendix B. Rudimentary Facts
248
(e) For the half-space D, the Green function GD satisfies the growth condition (B.36a), while HD satisfies (B.37a). This shows that the hypotheses leading to the representation formula (B.35) for the half-space were not artificial.
7. On uniqueness of the Green functions. (a) As was remarked after (B.38), Theorem B.16 implies uniqueness of GD and GB, provided that, in the case of D, the non-singular part gD(xo, x) = o(r) as r := Ixl -+ oo. This
growth condition is not satisfied for N = 1. However, if the condition is changed to GD(xo, x) = o(r) as r - oo, then this is satisfied even for N = 1, and the proof of Theorem B.16 still applies to the difference of two Green functions, because this difference has no singularity at
x=xoE0. (b) The function HD is unique apart from an additive constant, provided that for N = 1 we demand symmetry: HD(x,y) = HD(y,x), and provided that for N >- 2 we impose the growth condition: HD(xo, x) = o(r)
or gD(xo, x) = o(r) as r -* oo. For N = 1, this uniqueness follows from elementary consideration of what may be added to the function HD in (B.47); for N > 2, it follows from Theorem B.17. (c) One can concoct a partial uniqueness result for the function HB, but this is hardly worthwhile, because there are alternatives to the conditions (B.39) and (B.43), and because it is the uniqueness statement in Theorem B.17 that is the important one.
B.5 The solution of the Dirichlet problem for a ball
According to the representation formula (B.41), the solution of the Dirichlet problem for the ball B := -4(0, a) is u(xo) =
,fB
0G(xo, x)
G(xo, x) f (x) dx JOB OB
On
g(x) dS(x),
x0 E SZ,
(B.53)
where G = GB is now given explicitly by (B.48). In this section we discard
the strong assumptions on u under which (B.41) was derived. Rather, we impose various conditions on the data f and g, and show that (B.53) yields various kinds of solution of the Dirichlet problem.
Notation Throughout this §B.5 we write r := Ixl, ro := 1X01, UN I09N(0,1)1 and [as in §B.3] Sx := (a2/r2)x, Sx0 := (a2/ro)xo.
It is helpful to decompose the right-hand member of (B.53) into three
B.5 The solution of the Dirichlet problem for a ball
249
parts by the definitions
uf(xo) :_ f G(xo, x) f (x) dx = v(xo) - w(xo),
(B.54)
where
v(xo) := w(xo) :=
JaB
f
JB
fK(xo - x) f (x) dx,
K (r (xo - Sx)) f (x) dx,
(B.55)
xo E
(B.56)
if xo E B,
(B.57a)
if xo E 3B,
(B.57b)
a
P(xo, x) g(x) dS(x)
g(xo)
-aGxn, x)
P(xo x)
xo E RN,
.-a
Ixo z-z r°xN UN a
(xo E B, x E aB). (B.58)
Here of may be called the Green volume potential of the ball B and the density function f ; the function ug is the Poisson integral for B of the boundary-value function g; and P is the Poisson kernel of B. The two parts of the volume potential are the Newtonian potential
v of f and a function -w that has two useful representations. The first form of w results from changing the variable of integration to := Sx = (a2/r2)x; the Jacobian determinant of the transformation is det(ax;/al;j) = by (B.23). Then, with the notation a2
a
N+2
f*O :=f (2,) I -
p
(B.59)
and apart from an added constant when N = 2, w(xo) is the Newtonian potential at xo E B of the density function f*, which has support outside B. That is,
forN#2, w(xo) =
Jp>a
K(xo - )
d
(xo E B),
(B.60a)
Appendix B. Rudimentary Facts
250
for N = 2, w(xo) =
f
C=
where
a
K(xo - )f) d +C (xo E W),, f
1J 27r
p>a
log P
do
a
(B.60b)
.
The second form of w results from the identity (B.49), now written as (r/a)jxo - Sxj = (ro/a)ISxo - xl. This implies that, apart from an added (a/ro)N-2 times the Newtonian harmonic function when N = 2, w(xo) is potential, at the field point o := Sxo outside B, of the original density function f. That is, w(xo) _ B
K ( PO a PO
a )
x)) f (x) dx
(Po := I'o j ? a)
N-2
vo
v(o) + I log ao f f (x) dx
if N = 2.
(B.61b)
B
The case N = 1 of these formulae is discussed in Remark 1.5 and Exercise B.32; it is to be understood in the remainder of this §B.5 that the dimension N >- 2. Theorem B.23 Let u denote the Green volume potential called u f in (B.54).
(i) If f E Lp(B) for some p > N/2, then u E C(W), ulaB = 0 and u is a distributional solution (Definition A.7) of -Au = f in B; it is the only function with these three properties.
(ii) If f E LP(B) for some p > N, then u E C' (K), ul aB = 0 and u is a generalized solution (Definition A.7) of -Au = f in B; again it is the only function with these three properties. Proof We use the decomposition u = v - w described above, and results in Appendix A. (i) (a) First, v c C(W) when f e Lp(B) with p > N/2 because v E C(RN) by Theorem A.6. The same is true for w, by the extension of Theorem A.6 in Exercise A.32, if the function f * introduced in (B.59) has two
properties: f* E Lp(RN \ B) with p > N/2, and Wf* E L1(RN \ W), where W is the weight function in (A.78). For the first of these we use
B.5 The solution of the Dirichlet problem for a ball
251
the same p as in f E Lp(B); then P
a
J
>a
P(n't2)
d
P
()2N
P
Q= lB
P
If (x)Ip dx,
because p(N + 2) > z N2 + N > 2N. For the L1 property of W f*, first let N = 2. Then f * (l;) = f (x)(r/a)4 and d = (a/r)4 dx, so that
log (e+ p>a
If(x)I dx
Ixl
B
<-
z 2a)
III Lq(B)II I
If
I
Lp(B)II,
where t(x) := log(e + 2a2/IxI), 1/p + 1/q = 1 and p > 1 implies that q < oo, hence that e E Lq(B). For N > 3, the L1 property of W f* = f* is established again by means of the Holder inequality. Thus w E C(B). (b) That UI aB = 0 follows from (B.61), which shows that wI aB =
vI
aB
(c) Theorem A.8 establishes v as a distributional solution of -Av = f in B, while Aw = 0 in B pointwise by (B.60), which shows B to be outside the support of f*, and by the extension of Theorem A.5 in Exercise A.32.
(d) Let uo be the difference of two functions u having the three properties stated in (i) of the theorem. Then uo E C(B), uoI aB = 0 and both uo and -uo are distributional subsolutions (Definition 2.10) relative to A and B. It follows from Theorem 2.11 that uo = 0 on B.
(ii) The proof for f E Lp(B) with p > N is entirely analogous to the foregoing proof for p > N/2. In place of Theorems A.6 and A.8 we now use Theorem A. 11.
It is to be expected that, if f E C°'µ(B) for some p E (0, 1), then of is the
pointwise solution of the Dirichlet problem for -A in B with boundary condition u = 0 on OB. To prove this, we need a long (but straightforward) calculation like that in Lemma A.14. Reverting to the symbol q for the non-singular part of a fundamental solution, we now write q(xo, x)
q `j(xo,x)
where again
_
a2q(xo, x)
:= Sx and p = I
-K (a (xo - Sx)) , (a)_N+2
-P I.
(aiaj K)
(B.62a)
(xo - ),
(B.62b)
Appendix B. Rudimentary Facts
252
Lemma B.24 Let qij be as in (B.62) and let H :_ -4(xo,21h1) with xo E B and 0 < Ihl < !a. Then 2
E IB\H
x) dx
< const.,
where the constant depends only on N.
Proof To a large extent the proof follows that of Lemma A. 14. Although
gij(xo, x) has no singularity when xo E B and x E B, it is unbounded because Ixo - I can be arbitrarily small when xo and x are near 8B and near each other.
(i) In contrast to the function Kij(xo-.) in the earlier proof, gij(xo,.) is integrable on B and on HnB, so that the integrals over these two sets can be calculated separately. Moreover, gij(xo,.) E C00(B) and is smoothly harmonic in B [because q(xo,.) has these properties by (B.49) and (B.52), and because the operators 82/7xoi8xoj and A commute]. Therefore we can use the mean-value property of harmonic functions, Theorem B.8, to integrate gij(xo,.) over any ball contained in B. In particular, for all i, j E { 1,...,N},
fqiJ(xox) dx = IBlgij(xo,0) = 0,
(B.63)
because (B.62b) shows that qi.j (xo, x) = O(p2) as x -+ 0 and p - oo. (ii) Under a rotation of the co-ordinate frame, again written x = T(xo)x and xo = T (xo)zo, where T(xo) is an orthogonal N x N matrix depending on xo, we also have = T(xo)Z, where Z = Sz = (Sx), and qij transforms exactly as Kid did in the proof of Lemma A.14. Again we so choose T(xo) that zo = (Ixol, 0, ... , 0), and use these special co-ordinates zj in the rest of this proof, but omit the tilde for ease of writing. For the same reasons as before,
L\H
f
\H
if i * j,
gij(xo, x) dx = 0
gJJ(xo, x) dx,
ql l (xo, x) dx = j=2
B\H
so that we need bound only the integrals of qjj for j = 2,..., N, and these integrals are all equal.
(iii) For H c B, we evaluate the integral of qjj(xo,. ), j# 1, over H as
B.5 The solution of the Dirichlet problem for a ball
253
0
Fig. B.5.
follows. Set Ixol =: ta, 0 < t < 1, and apply the mean-value property of harmonic functions once more; there results
= IHI Igjj(xo,xo)I
.f gjj(xo,x) dx H
=
t2)-N
6N(21hl)N a-Nt2(1 -
N
(J
6N
1)
Since H c B, we have 21hl < a - Ixol = all - t) and 0< t < 1, so that gjj(xo,x) dx <
.fH
t2(l -{- t)-N
N
I
(H c B, j * 1).
(B.64)
(iv) It remains to bound the integral of qjj(xo,.) over H f1 B when H intersects 8B and j 1. Changing the variable of integration from x to := Sx, we find from the solution of Exercise B.14 that
where z :=
S(H) = R(z,Y),
a r02
-4lhl2xo,
Y
ro
a _4IhI22IhI (B.65)
also, ro - 2IhI >- 3a because H intersects aB and Ihl < !a. The Jacobian
determinant of the transformation is -(a/p)2N. Let J S(H); the configuration is shown in Figure B.5. Accordingly, Aj(xo,h)
:
LflB gjj(xo,x) dx =
-f
(\
(K)(-xo) 1\ P /
f d\s
3
1: AJ,k(xo, h) k=1
(B.66)
Appendix B. Rudimentary Facts
254
if we define
Aj,2(xo, h) _ Aj,3(xo,h)
N+2
a
dSO,
(ajK)( -xo) ( a
Aj,l(xo,h)
:=
-f
snJ
I
ni(b) (OjK) ( - xo) dS( ), p1l
8
xo)
-(OjK)(
\B
N+2
d.
a/
Here is the unit normal outward from J \ B; use of the divergence theorem is legitimate because J \ B is a set listed in Remark D.4.
For the contribution Aj,l of 7J \ B, we set
1=z1-ycos(p, N)1/2=ysin(p
(
EBJ, 0
(
Then, very much as in the proof Lemma A.14, step (iv),
I-xo12=Y2{(C-1)2+4Csin2
d(
C:-IZI-Ixol>0,
where
J,
Y
2 and, for j * 1 and some angle µ E [0, ir/2) that depends on Ixol and Ihl, IAi,I(xo,h)I =
- 1)aN
N/2
(a) N+2
C - 12 +4sin2 i c/2
6N-12N+1
(N -1)6N
1
4cos2 (p sin2
Jµir
6N-1
(sin
(P)N
d(p
(j
1),
(B.67)
JO
where we have used first that a/p < 1 and then the step from (A.69a) to (A.69b); this step remains valid if C > 1. In the contribution Aj,2 of 8B n J, the integrand, apart from sign, is precisely that considered on 8B \ H in the earlier proof; therefore, a coarse bound is IAj,2(xo, h)I <
6N-12N+1
(N - l)6N
rr/2
f (sin 8)N d8
1).
(J
(B.68)
For the contribution A j,3 of J \ B, we use once more the cylindrical symmetry about the xl-axis: 2)a +2'1 Sj IAj,3(xo, h)l = (N + +4 (f 1) p fj\ii 6NI - xoIN
d
KK
(N + 2)aN+2
(N - 1)6N
p
JJ\B
2 +,
1 - xoIN
N
1
pN+4 dS
B.5 The solution of the Dirichlet problem for a ball
<
(N + 2)a-2 1 d (N -1)aN Jr\B I - xoIN-2
255
.
One verifies from (B.65) that J \ B c .(xo, 7a/3) for all xo E B and
Ihi<6a,sothat IA1,3(xo, h)I
<
(N + 2)a-2 (N -1)QN
I
49N+2
7a/3
R-N+26NRN-I dR
U* 0.
18N-1
(B.69)
In view of remarks in (ii), the Lemma now follows from taking the bigger
of two constants: that implied by (B.63) and (B.64) if B c B, and that implied by (B.63) and (B.66) to (B.69) if H intersects aB.
We observe in passing that, with q as in (B.62a), 2
lB
K(xo - x) dx
i
L
-- 2N o + CN(a)
(B.70)
a2
q(xo, x) dx G(xo, x) dx
(xo E B),
2N
_
- cN(a)
a2 - r2
2N °
(xo E B),
(xo E B);
(B.71)
(B.72)
here (B.70) is a part of (A.23a) and cN(a) is as in (A.23b). The identity (B.71) can be derived by means of the mean-value property of harmonic functions, Theorem B.8; then addition of (B.70) and (B.71) yields (B.72). Alternatively, we may observe that (B.72) is a particular case of the representation formula (B.41), because u(x) = (a2-r2)/2N is the (unique, pointwise) solution of -Au = 1 in B, u = 0 on aB; in the present case, the formula extends to B by continuity. Note that (B.71) implies (B.63). Evidently the Green function G and the Green volume potential of have much in common with the Newtonian kernel K and the Newtonian
potential v of f. Two small lemmas now precede a theorem in this direction.
Lemma B.25 (i) If xo E B and X E B \ {0}, then
a Ixo - Sxj > ISxI Ixo - xI.
(B.73)
(ii) If xo (=- B, x E B \ {xo} and limiting values are taken for xo = 0 or x = 0, then
Ca
axo
a
q(xo, x) < C,j Ixo -
xI-N+2-1#1
(N > 2),
(B.74)
Appendix B. Rudimentary Facts
256
where q is the function in (B.62a), Cp is the constant in (A.31), and IQI >- 1
ifN=2. Proof (i) For xo = 0, the inequality (B.73) reduces to a > Ixl, which is true. Therefore assume that xo # 0 and let A be the angle between the vectors xo and x. Continuing to write := Sx, p and hence r = a2/p, we wish to prove that a2 (r20 - trop cos A + p2)
> p2 (r_2rocosA+ a4
(
P
P2
equivalently, that p2 (a2 - ro) > a2 (a2 - ro), which is true. (ii) First, a
- (/_) q(xo,x)
\)1
(
=
Ca
((xO_)) (K) N+2 (01K) (xo - )
a
because OK is algebraically homogeneous of degree -N + 2 - Ial, as is displayed in the proof of Lemma A.4. Next, (A.31) and then (B.73) imply that
I
( a) -N+2 C xo 1\
< Ci Ixo <
Cfl Ixo
XI-N+2 Ixo
- xl-
IN+2-Irsl
- j I-IflI
N+2-IQI
because (B.73), in which a < ISxI, certainly implies that
Ixo-xl
Lemma B.26 In the Green volume potential uf, defined by (B.54), let f E C°,µ(B) for some p c- (0, 1). Then
(0iajuf)(xo) = I 8xoiaxN{f(x) -f(xo)}
dx-bijfN),
(B.75)
whenever i, j E { 1, ... , N} and xo E B.
Proof Let (. )ij denote 02(. )/8xo,8xoj, except in the Kronecker delta bij. We use the decomposition u f = v - w and the formulae (to be explained
B.5 The solution of the Dirichlet problem for a ball
f f 0 = -f
257
Kij(xo - x){ f (x) - f (xo) } dx -Sid f N )
B
=
qi (xo, x) f (x) dx,
B
gij(xo, x) f (xo) dx,
B
where xo E B. The first is the result of Lemma A.13. The second comes from the form (B.60) of w and from the extension of Theorem A.5 in Exercise A.32. The third is (B.63) multiplied by -f(xo); alternatively, it is an implication of (B.71). The sum of the three formulae gives (B.75).
Theorem B.27 Let u denote the Green volume potential called of in (B.54).
If f E C°'µ(B) for some p E (0,1), then u E C2(B), UI aB = 0 and u is a C2-solution of -,Lu = f in B. Indeed, each second derivative 8;8fu1 B has an extension to B that is in C°-µ(B).
Proof Apart from the result ul aB = 0, which follows from (B.61), the proof is an exact parallel of the proof of Theorem A.15. The properties enjoyed there by the Newtonian kernel K and Newtonian potential v of f are now shared by the Green function G and Green volume potential u; this is the content of Lemmas B.24 to B.26. In other words, to establish the uniform Holder continuity of second derivatives of u, and to prove that -Au = f in B pointwise, we merely replace K(xo-.) by G(xo,.) in the proof of Theorem A.15, and replace the condition 0 < Jhi < 4a there
by0
to solve the problem Au = 0 in B, u = g on B. The Poisson kernel P, introduced in (B.58), has two essential properties: first, as (B.58) shows, P(xo, x) --- 0 as xo tends to any point of 8B other than x [so that ro -+ a
with ixo - xi > 0]; second, P satisfies the following analogue of the identity (B.72) satisfied by G. Lemma B.28
P (xo,x) dS(x) = 1
for all xE B.
(B.76)
LB
First proof Recall from (B.58) that P(xo, x) = -8G(xo, x)/8n, where
Appendix B. Rudimentary Facts
258
x E OB. Therefore (B.76) is a particular case of the representation formula (B.41), because u(x) = 1 is the (unique, pointwise) solution of
Lu=O inB,u=1 on OB. Second proof By the basic property (A.20) of the Newtonian kernel,
-
/' OK(x - xo)
dS(x) = 1.
an
aB
Since q(xo,.) E C°°(B) and (L q)(xo,.) = 0 in B [by (B.52) and the accompanying remark],
-f
aq(xo, x) dS(x)
(,Lq)(xo, x) dx = 0. fE
On
aB
Finally, P(xo, x) = -a{ K(x - xo) + q(xo, x) }/an evaluated at x E 3B. To define the space LJaB) of essentially bounded functions on 3B, we merely replace the open set S2 by the sphere aB in the definition of L,(S2) in Chapter 0, (xii), it being understood that sets of measure zero are now those having zero surface area on 3B. Theorem B.29 Let u denote the function called ug in (B.57); let A be a closed (possibly empty) subset of aB. If g E L,,(aB) f1 C(aB \ A), with the understanding that Ig(x)I <_ IIg L,,(OB)II at every x E aB, then I
(a) u E C(B \ A) f1 C°°(B), (b) lu(x)I < IIg L,,,(aB)II for all x E B, I
(c) Lu = 0 in B pointwise. Proof (i) Note first that, for fixed x E aB, the function P (. , x) is smoothly harmonic in B, because this is true for G(. , x) and because a'/ax'; and -a/an commute for x0 * x. The the operators LO
proof that u c C '(B) and that Au = 0 in B is now very like that of Theorem A.5. We omit the details because, for given xo E B, we have b0 := dist(xo, aB) > 0.
(ii) The inequality lu(x)I <_ IIglI, in which . denotes the norm of LoJaB), holds at every x E aB by hypothesis. To prove it for xo E B, we use the Holder inequality with exponents 1 and oo, the fact that II
II
P(xo, x) > 0 for x0 E B and X E aB, and the result (B.76); accordingly,
fs P(xo,x)g(x) dS(x) < Ilgll JOB
,
f
P(xo,x) dS(x) = Ilgll
aB
(iii) It remains to prove that u E C(B \ A). Since u E C0°(B) and
B.5 The solution of the Dirichlet problem for a ball
259
u E C (OB \ A), we need prove only this: for given p E OB \ A ands > 0, there is a number S = 6(p, e) > 0 such that Iu(xo) - u(p) I < e
whenever x0 E B and Ixo - pI < 6.
In the following argument, (B.76) and the positivity of P are used repeatedly; we also use an abbreviated (but self-explanatory) notation for subsets of aB. Observe that, for every p > 0,
- u(P)I
Iu(xo)
= -<
dS(x)
f
LB
P(xo, x) I g(x) - g(p)I dS(x) + J
P(xo, x) 211%11 dS(x)
Ix-PI?P
x-PI
< suplx-PI
(B.77)
Since g is continuous at p, and OB \A is open relative to the metric space OB, we can choose p = p(p, e) to be positive and so small that suplx-PI
With p now fixed, we prepare to choose 6. If Ix - pl >- p and Ixo - pl < S < 1 p, then Ix - xol > Zp and a - ro < 6, so that P
(xo,
x) =
a2 - r02
UNa
< 26 Ix - xolUN 2 p) (1
-N
N
Therefore we can choose S to be in (0, 1p] and to be so small that the last term of (B.77) is less than Zs. El
If in Theorem B.29 the set A is empty, then g E C(OB), mention of L,,(OB) is not necessary, and the theorem shows that ug e C(B) n C°°(B). In that case, Theorem B.29 establishes existence of the solution of the
particular Dirichlet problem Au = 0 in B, u = g on OB, and Theorem B.16 establishes uniqueness. (In addition, the solution is real-analytic in B, by Theorem B.10.) If A is not empty and g is discontinuous on A, then Theorem B.29 may still be useful (there is an application of the theorem in Appendix C), but the question of uniqueness depends on details of A and g.
The next theorem states that uniform Holder continuity of g on OB implies uniform Holder continuity of ug on B with the same exponent. This requires a proof longer than that of Theorem B.29 because Holder
continuity with exponent 2 at every point of a compact set does not imply uniform Holder continuity with exponent 2 (Exercise B.38).
Appendix B. Rudimentary Facts
260
we mean the normed By Cb'A(7B), usually abbreviated to linear space that results from Definition A.10 if both C1 and S2 are replaced by 8B there.
Theorem B.30 Again let u denote the function called ug in (B.57). If g E for some A E (0, 1), then u E C°'2(B).
Proof (i) Theorem B.29 ensures that u E C(B), the set A being empty; therefore, it remains to establish uniform Holder continuity in B, with exponent A. This is immediate on any closed ball -4(0, p) with p < a because u E C00(B): if x,y E -4(0, p), we bound Iu(x)-u(y)I by integrating Vu along the line segment from x to y. In the remainder of this proof we consider field points
xo and xo + h in B \ M (0,2 a), with 0 < Ihl < 6a;
(B.78)
the point p :_ (a/ro)xo E aB will be the centre of a small ball 4(p, 31h1); and the norm III := IIg C°''(OB)II is to be so defined that I
Ig(x) - g(p)I < lg II Ix -
pIA
for all x E B.
(B.79)
The identity (B.76) implies that
u(y) =
P(y, x){g(x) - g(p) } dS(x) + g(p) s whence, with the abbreviation dS := dS(x),
u(xo + h) - u(xo) =
for all y E B,
f{P(xo+hx)_P(xox)}{g(x)_g(p)} dS B
= h(xo, h) + I2(xo, h)
(B.80)
if we define S1 := 7B n 1(p, 3Ih1), S2 := 8B \ Sl and Ij(xo, h) := j{P(xo + h, x) - P(xo, x) } { g(x) - g(p) } dS
(j = 1, 2). (B.81)
(ii) It is easy to bound I,: using first (B.79), then the positivity of P and (B.76), we obtain IIi(xo,h)I
< IISII(31h1)' f IP(xo+h,x)-P(xo,x)I dS ,
<
I1g1l(31hl)2 f B{P(xo+h,x)+P(xo,x)} dS
=
211g1l (31h1)2.
(B.82)
B.5 The solution of the Dirichlet problem for a ball
261
Fig. B.6.
(iii) In order to bound I2, we let 0 denote the angle between the radiusvectors to x E 7B and p E 8B (see Figure B.6);then
= 4a2 sing 0
Ix -pI2 R2
Ix
(x E 7B, 0 < 0 < 7t),
- xo I2 = (a - ro)2 + 4aro sine
8 .
Since ro >- (2/3)a by (B.78), we have Ix - pI2/R2 < a/ro < 3/2; then, for
0
>- R- 1
(B.83)
ZR.
The inequality (B.83) allows us to estimate P(xo + h, x) - P(xo, x), for x outside 2(p, 3IhI), by the method used repeatedly in Appendix A, from Theorem A.5 onwards, for difference kernels with x outside a small ball.
Let r denote constants independent of xo, x, h, g and a; then, for Ix - pl >- 3IhI P (xo + h, x) - P (xo, x) l 1
GNa
(x012-Ixo+h12+(a2_1x012) 1x0 + h -xIN
( Ixo + hI - xIN -
I
Ixo -xIN
)
(B.84)
< r IhI R-N,
since a - ro < R. Accordingly, 1I2(xo,h)I
<-
rf
IhI R-N IIgI1 Ix-pI2 dS
sZ
r IIg II
a'+N-1
I hI
f
"
(sin(0/2))A (sin 0)N-2
v(h) { (a - ro)2 + 4aro sin2(0/2) }N/2
d0,
Appendix B. Rudimentary Facts
262
where y(h) := 2 sin-' (31hi/2a) with sin-1 taking values in [-7r/2,7r/2]. Discarding the (a-ro)2 in the denominator of the integrand, and recalling that ro >- 2a/3, we obtain 0
IF IIgiI Ihi a-1 f
112(xo,h)I
2
dO < I' III IhI.
(B.85)
1h1/a
The uniform Holder continuity of u in B, with exponent A, now follows from (B.80), (B.82) and (B.85).
B.6 Exercises Exercise B.31 Use the Taylor formula (1.6) to solve the Neumann problem
- u" = f in
(a, b),
u'(a) = a,
u(b)
(B.86)
where f E C [a, b] and - f b f = fi - a; this last is the compatibility condition (B.29) for the present case. Verify that your solution satisfies (B.86), and adapt it to the problem
-v" = f in
v(a) = a,
(a, oo),
where f c C [a, oo).
Exercise B.32 Consider the Dirichlet problem for a ball in one dimension (N = 1):
- u" = f
in (-a, a),
u(-a) = c1,
u(a) = c2.
(B.87)
(i) Deferring conditions on f, derive the formula u = of + u, where
uf(x) := a2ax f x(a+t) f(t) dt+a ax
ja
(a- t) f(t) dt (-a < x < a),
a
u, (x) := 2a { cl(a - x) + c2(a + x) }
(-a < x < a),
from the proof of Remark 1.5, or else from (B.41) and (B.48), or, ideally, from both. (ii) Verify that, if f E C [-a, a], then u f E C2 [-a, a] and u f + uc satisfies (B.87).
(iii) Prove that, if f E L1(-a, a), then of E C1 [-a, a], of satisfies the
B.6 Exercises
263
differential equation almost everywhere in (-a, a), and of + u, satisfies the boundary conditions. In what respects is this result better than the result in Theorem B.23, (ii), for N >- 2?
[If g E Ll (a, fl), then the function x H f x g is continuous on [a, and (d/ dx) fx g = g(x) almost everywhere in (a, /3). But for the proof in (iii) that of e C1 [-a, a], almost everywhere is not enough; one method is to show that uf(x) =
1
a
2a
fa
x
where F(x) := J " f .
F - F(x),
Exercise B.33 Consider the Dirichlet problem for a half-line:
- u" = f in
u(x) = o(x) as x
u(0) = c,
(0, oc),
co.
(B.88)
(i) Again deferring conditions on f, derive the formula u = of + c, where 00
tf (t) dt +x
uf(x) ox
Jx
f (t) dt
(x > 0),
from Exercise B.19, (ii) [that is, from (B.35)] and from (B.45).
(ii) Verify that, if f E C [0, oo) and f (t) = O (t-1-S) as t - oc, for some constant 6 > 0, then of E C2 [0, oo) and of + c satisfies (B.88).
(iii) Prove that, if f E L1(0, oo), then of E C1 [0, co), u f satisfies the differential equation almost everywhere in (0, oo), and of + c satisfies the second and third conditions in (B.88).
[The hint about g in Exercise B.32 extends from (a,#) to (a, co) and here
uf(x) =
F,
J
where F(x) :=
0X
Jx
f.]
Exercise B.34 This exercise and the next concern the necessity of condition (B.6) in Theorem B.4. Let Q(p, a) := (p - a, p + a)' [an open cube in Il8'v
with centre p and edges of length 2a], and let aE := all - E). (i) Given that
f (x) = E c"(x - p)"
if x E Q(p, a),
a
where c" := (a"f)(p)/a! and summation is over all multi-indices a of length N, prove that for every e E (0, 1) there is a number ME such that c"
MEaE
for all a,
Appendix B. Rudimentary Facts
264
of length N,
and hence that, for each multi-index al f (x)
a t.
6E "I
ME
if s E (0,1
2E
and x E Q(p,a2E)
Here a > /3 means that aj /3j for each j. (ii) Noting that, for -1 < t < 1, 00
n!
t= (d)ktnk!(lt)_1_k,
prove that Q-IaI
a!
6 1a-/31
=a
1 -8
l
N
(E6)-ICI
(0
and hence that, for each multi-index /3 of length N, 4
IPI
aflf(x)I < M1143N$! (_)
if x E Q (p, ZQ).
Exercise B.35 (i) Prove that /3! < 1$1! for all multi-indices /3. [Induction on N is possible]
(ii) Use the result of Exercise B.34 to prove that, if f : S2 -* ![8 is real-analytic, then for each compact subset E c 92 there are constants A = A(E) and B = B(E) such that
la'f(x)I
for all xeE
and for all multi-indices /3 of length N.
Exercise B.36 Under a co-ordinate transformation (r, P11.... , 1N-1) H x,
in which r = IxI and j :_ (rii,...,riN_1) labels points of the unit sphere 8-IN(0,1), N >_ 2, we have
A = rN+1
-(8r rN-1
1
-
rO
n,
where On is independent of r. (Exercises 1.18, 1.20 and D.18 give examples of such transformations.) Use this form of D to prove Theorem
B.15forN>2. Exercise B.37 Let the operators S and T of the Kelvin transformation be as in Definition B.12, and let 0 c RN \ {0}. Show that, if u is subharmonic in S2 (Definition 2.20), then Tu is subharmonic in S(I).
B.6 Exercises
265
Exercise B.38 Holder continuity on a compact set need not be uniform. Define u(x) :
- x sin(1/x) 0
if 0 < x < 1, if x = 0,
v(x) :
= x sin(el/x) if 0 < x < 1, if x = 0.
Recall Definition A.10 and the remarks preceding it.
(i) Show that u and v are Lipschitz continuous (Holder continuous with exponent 2 = 1) at each point of [0, 1]. (ii) Prove that there are constants A and B, independent of x and h, such that Iu(x + h) - u(x)I < Ah112,
Iv(x + h) - v(x)I < B/ log h
whenever x E [0,1), x + h E (0,1] and 0 < h< 2 [For all such x and h we have both lu(x + h) - u(x)I < 2x + h and
Iu(x + h) - u(x)l < h/x + h if x > 0;
the former is better for x < (h/2) 1/2 and the latter for x > (h/2)'/2. In the case of v, the value of x at which two such bounds are equal must be estimated.]
(iii) Prove that in (ii) the bound Ah1/2 cannot be replaced by one that is o (h1/2) as h 10, and that B/log(1/h) cannot be replaced by a bound that is o(1/log(1/h)) as h 10. [For u, contemplate points x, := (2nm + it/2)-1 and x + h := with nEN.] (2nir)-1
Exercise B.39 Show that, for the Green function HB of the Neumann problem for a ball, the formula given in (B.50c) for N >- 3 is valid for all N E N. [For N = 1, this function differs by a constant from that in (B.50a).]
Exercise B.40 Let B :=.N(0, a), let r := jxj and write 0/8r := (x/r) 0 if x * 0. Consider for N > 2 the Neumann problem of finding u such that u E C(B) n C2(B),
Au = 0 in B,
8u/ar E C(B \ {0}), 8u Or as
= g,
(B.89)
where g E C(8B) and fB g = 0 (the compatibility condition). The solution of Exercise 2.42 shows (even though less smoothness is demanded here) that any two solutions of (B.89) differ only by a constant.
Appendix B. Rudimentary Facts
266
Let HB be the Green function in (B.50), and define
g() dS()
u(xo) :=
for xo E
(B.90)
LB
Given the result that this function u e C(aB), prove that it is a solution of (B.89).
[One can verify the boundary condition, and prove that au/ar E C(B \ {0}), by means of the identity ro
allB
(xo,
) = a PB (x0, ) -
for x0 E B \ {0},
6N aN_2
COB,
Or oo
(B.91)
and by use of Theorem B.29; here PB denotes the Poisson kernel called P in (B.58). The proof that u e C(aB) is not difficult but not short.]
Exercise B.41 Show by an example that a function u satisfying (B.89) need not belong to C1(B). [One possibility is to choose N = 2, to write (x, y) for points of R2 and z = x + iy for points of C, and to contemplate the real part of z+a
w(z) := 3ial; log(logwhere
3a
and a convenient branch of arg(log ) is chosen.]
Exercise B.42 (i) Show that the function u in (B.90) may be written
u=v+wif,forN-2 andxoEB, K (xo - )
v(xo) := 2
(B.92)
LB
w(xo)
__
a
aN
rN-3
-
f>a Ixo - xIN-2 where
1
r
}
g()
rN-1
dx
:= ax. r
r := IxI,
}
(B.93)
Note that w = 0 (the zero function) if N = 2. (ii) By a lateral derivative of u at x we mean any directional derivative
m(x) Vu(x)
with x E B \ {0},
m(x)
0,
x m(x) = 0.
Show that every lateral derivative of u at x is a linear combination of N - 1 lateral derivatives of form u;;(x) := x j (aju) (x) - xi (a ju) (x) if xi
0, i is fixed and j E { 1, 2, ... , N} \ {i}.
B.6 Exercises
267
Caution This (. )ij must not be confused with that in (A.65) and (B.62b).
(iii) Prove that lateral derivatives of the v in (B.92) and w in (B.93) are given by (B.94)
g(oo) }
vii(xo) = 2 fs Kii (xo, ) { a
wij(xo) = a(N - 2)
/'
Jr>a
Kij (xo, x)
gO -g(oo) r2
dx,
(B.95)
where xo E B \ {0}, o := (a/ro)xo, i:# j and, for x E RN \ {xo}, Kid (xo, x) := xoi (aiK) (xo - x) - xoi (ajK) (xo - x).
(B.96)
Exercise B.43 This exercise concerns the following result. If in the Neumann problem (B.89) we have g E C°°µ(aB) for some u E (0,1), and LB g = 0, then not only does the solution u in (B.90) belong to C' (B) [cf. Exercise B.41], but also Vu E C°^µ(B,RN).
Give the main parts of the proof by showing that the functions x H rau(x)/ar and vii, all defined to be zero at the origin, belong to C°'µ(B). Here vij(xo) is as in (B.94) for X0 E B \ {0}.
[The uniform Holder continuity of rau/ar follows from (B.91) and For vii, the proof of Theorem B.30 is relevant, but perhaps the key step is a decomposition of vi j (xo + h) - vii (xo) that
Theorem B.30.
resembles (A.71), despite the different meaning here of (.)ij. The proof for wig, which is not demanded here, is similar to that for vii but easier.]
Exercise B.44 Consider for N > 2 the Green volume potential U f (xo) := in GD (xo, x) f (x) dx,
xo E D,
(B.97)
D
of the half-space D := { x c RN xN > 0 1; here, GD is the Green function in (B.45). Assume that f is measurable and that, for some I
6 E (0,1), IfI
I
I := ess supXED(l + r)'+a I f (x)I < oo
(r := Ix1).
(B.98)
Prove that, for all xo E D, I (a"u1) (xo) I -< 1111f I I (1 + ro) I
if (a1 = 0 or
1,
(B.99)
where the constants f depend only on N, 6 and a. Note that (B.98) fails to imply that f E L1(D) [since N > 2], hence
Appendix B. Rudimentary Facts
268
fails to imply the condition (A.78) assumed for f in Exercise A.31, but that (B.98) is enough to imply the growth conditions assumed for u in Theorem B.16 and in (B.36b) of Exercise B.19.
[It may be helpful to separate the cases ro < 1 and rp >- 1; and, for ro > 1, to integrate separately over the subsets of D in which (a) R := Ix - x0 l < 12 ro, (b) R > 12 ro and r < 2ro, (c) r > 2ro.]
Exercise B.45 For the half-space D and for N > 2, the Poisson kernel -0GD/7n is 2XON
-2(aNK)
PD (xp,
with xo E D,
YY
6N Ix0 - b IN
E OD,
and we shall also need the Marcel Riesz kernels
M'M1(xp,
-2 ajK)(xo
)
-
2(xo.i- 0) ) = 6N
Ixo - SIN
with j E {1,...,N - 1}, xoED \
E 10D.
The corresponding Poisson and Riesz integrals are, respectively, g
f
)
PD (xo, )
if xo E 8D,
and
Mj (xo,
if xoED,
aJD Mi(xo,
ID
if xo E 3D,
where m is a 'mollifier': m E C°° [0, oo), m(t) = 1 for 0 < t < 1, m(t) = 0 for t > 2, and m is non-increasing.
Still for N > 2, prove (or merely accept, in order to do Exercises B.46 and B.47) that the map g H ug is a bounded linear operator from C °'µ(OD) into C°'µ (D), and that g --> vg,j is a bounded linear operator from C°'µ(7D) n Lp(0D) into C°'µ(D) ; here µ E (0,1), p E [l, oo) and 7D may be identified with R' [For ug, the proof resembles that of Theorem B.30. Let o := (x', 0) ; 1
then vg,.i
(xo)
ID
for all xo E D.
B.6 Exercises
269
The uniform Holder continuity of vg,j follows once again from a decomposition like (A.71).]
Exercise B.46 Use the results of Exercise B.45 to prove an analogue, for the Neumann problem in the half-space D with N > 3, of the theorem in Exercise B.43.
u(xo) :=
More precisely, defining
JaD
for xo E D, N> 3,
HD(xo, )
where HD is the Green function in (B.47) and g c Cb'µ(aD) n Lp(aD) for some p E (0, 1) and p c- [1, N - 1), prove that u E C' (D) n C00(D), that Du E Ce''` (D, RN) and that
Du = 0 in
D,
-aNUI aD = g'
Exercise B.47 Here we are concerned with the smoothness on D of the Poisson integral u,y, defined in Exercise B.45, when ip E Cc°(aD). The compact support of ip could be the result of applying a partition of unity to a more general function in C0°(3D). Derive the (formulae (aN uw) (xo)
(aN+tuw)(xo)
= JfD PD (xo =
)
2JD(0'K)(xo-(0
+ aN_, and O' :_ (al, ... , ON-I)For xo E D, write any derivative (a"uy,) (xo) as an integral with kernel PD or kernel 2V'K. Use the results of Exercise B.45 to show that u. E C`° (D).
in which k E No, xo E D, A' := a, +
Appendix C. Construction of the Primary Function of Siegel Type
In this appendix we use the Poisson integral for a ball, which is the subject of Theorems B.29 and B.30, to prove Theorem 2.27, which describes the
primary function of Siegel type. The dimension N > 2. There is a clash of notation between §B.5 and §2.5; the primary function of Siegel type, denoted by g in Theorem 2.27, will be called v in this appendix. We recall from (B.57) and (B.58) that the Poisson integral for the ball B :_ R(0, a) may be written
fP(x, )
if x E B,
(Cl.a)
if x E aB,
(Cl.b)
B
g(x)
the Poisson kernel P being defined by P(x,
a
alx 2
)
r2IN
(r := lxi, x E B,
E aB).
(C.2)
QN
The symbol , used consistently in Appendix B for the inverse point (a2/r2)x of x, is now merely a variable of integration. Theorem B.29 shows that, if g E C(aB), then u E C(B)nC°°(B) and the formula (C.1) describes the unique pointwise solution of the particular
Dirichlet problem Au = 0 in B, u = g on
B.
The theorem also
establishes properties of the function u in (C.1) when g is discontinuous but bounded. In the present application, however, we shall have
g(x) = xN on aB \ E,
where E := aB n { x E I[8N I xN = 0 } , (C.3)
so that g is neither continuous nor bounded on aB; something beyond the results of Theorem B.29 will be needed. Our first task is to examine 270
Appendic C. The Primary Function of Siegel Type
271
the particular form of Poisson's integral for B when g is an odd function of its Nth argument.
Notation For upper and lower hemispheres we write
5+{xE18N Ixl=a, xN>0}, Ixl=a, xN<0}.
S- := {xERN
As elsewhere, x' = (x1,...,XN_1) and x* := (x',-XN) for any point x E RN, so that
Ix* -
I2
= Ix - 1I2 + (XN + N)2 = Ix -* I2
The dimension N >- 2.
If g is an odd function of its Nth argument, so that g(x*) = -g(x), then the substitution
= rl* yields
.f P(x, )
fs+ P(x, i*) g(n) ds(n);
s
) s+
u(x) :=
Q(x,
)
g(x)
if x E B,
(C.4a)
if x E aB,
(C.4b)
where
Q(x,) := P(x,) - P(x, * ) a 2-r2
1
1
UN a
Ix - SIN
Ix - *IN
(C.5) '
Lemma C.1 The kernel Q(., ) is an odd function of its Nth argument. If XN >- 0 (and x e B, D E S+), then
0
2N(a2 -
(C.6)
and
My
()Y
Q(x,
)
Myalyl
Ix SN Ix -
I-N-3
if IYI =1,
(C.7a)
I-N-21y1
if IYI ? 2;
(C.7b)
Appendix C. The Primary Function of Siegel Type
272
the constant My depends only on the multi-index y (the length of which specifies N).
Proof (i) Replacing xN by -xN changes Ix - I to Ix - * I and changes Ix - * I to Ix - I; then Q(x, ) changes sign, by (C.5).
(ii) To derive (C.6) we write R := Ix - I and R* := Ix - * I. In the remainder of this proof we assume that xN >- 0; then 0 < R < R* and R* - R2 = 4xNSN
R* - R = R* +R
2XNSN
R*+R -
R
Accordingly, for m E N,
R* - Rm
0 < R-m - R-m _ *
(R *
- R)
2
(R* R)-
R Rm + R2 Rm-1
+ ... +
1
N
(C . 8)
Rm+2
The definition (C.5) of Q, and the inequality (C.8) with m = N, imply (C.6).
(iii) To derive (C.7), we begin with an application of the Leibniz rule (Exercise A.23) to obtain
l
/
z
_ UNa
/
Y
(Y-N)!N!
{()Y
a )' (R-N-RN Y
(ax)
(a2
- r2)
<
2a2-IYI+tRI.
(a 2
- r) 2
x
(C.9)
(C.10)
This last is an overestimate if /3 = y or ly - $I > 2, but a harmless one. (iv) For derivatives of R-N - R N, we use a slight variant of (A.32) : aflr-N = r-N-21RIpp(x)
(x * 0),
Appendic C. The Primary Function of Siegel Type
273
where pp is a homogeneous polynomial in x1, ... , xN of degree I# I ; say
pp(x) = E
C#'-X11.
1"1=1f1
It follows that le
a
ax C-/
(R_N
- R-N)
*J
=
(R_N_21P1
-R -N-21flI ) pp(x -
+R*N 211{pp(x-
pp(x- *)}.(C.11)
In this proof, const. will denote positive numbers independent of x, and a. By (C.8), and because Ipp(x - )I < const.RlB', (R_N_2Ifh
-
R*N-2IQ1)
lpfl(x - )I < const.
Next, with the notation a' RN-21,1I pp(x
aN_1),
- ) - pp(x - * )I
= RN-21#1
Cp a(x - 0", 1 (XN - N)"N - (XN + N)"N I I"I=1p1
< const. a1B1-1 NR-N-21BI ; RI" I because we have ignored the favourable inequality I (x' -')" there are terms with Cp," 0, Ia'I = 0 and aN = IQI Insertion of the last two estimates into (C.11) yields
C aX
I
(R-N - R*N) < const.
RI#I-2 + aI#I-2) R-N-21 1
(C.12)
where RIPI-2 + a1#1-2 < const. a' 2 if Ia I > 2, because R < 2a; the term R1#1-2 will be retained only for I#I < 1. '
(v) Finally, we return to (C.9) and use (C.10), (C.12) and the remark following (C.12); there results
(a)Y
Q(x,
) ICI
const. N
a2-IYIR-N-2
+
21QI
a IPI_
C2
IYIR-N
B-Y
(-)
Since 1 < 2a/R < oo, the terms with I#I = 1 dominate the first sum, and
274
Appendix C. The Primary Function of Siegel Type
that with $ = y dominates the second; therefore Q(x,
)
< const. N { a3-IvI R-N-3 + a1Y1 R-N-21Y1 } .
(C.13)
For 1vI = 1, the first term on the right of (C.13) dominates [because 1 < 2a/R < oo] ; for ly I >- 2, the second term does.
The primary function of Siegel type is now defined by
J
v(x) = v(x; a) :=
Q(x,
dS(g)
)
if x E B,
(C.14a)
if x E aB \ E,
(C.14b)
N
a/xN
where E denotes the equator of B, as in (C.3). Recall that the signum
function sgn is defined by sgn t = -1, 0,1 for t < 0, t = 0, t > 0 respectively.
Theorem C.2 The function v = v(. ;a) defined by (C.14) belongs to C(B \ E) n C00(B). Also,
Lv=0 for x E
E,
in B;
v(x', -xN) = -v(x', XN)
and sgn v(x) = sgn XN;
v(x; a) depends only on x/a; and
Iv(x)I <- const. IxNI/a
if IxI <- a/2,
where the constant depends only on N.
Proof (i) To prove that v E C00(B) and Av = 0 in B, we note first that E C°°(B), by Lemma C.1, and that 0 in B, because this is true for In addition, for x E B and E S+ the integrand Q(x, )/N is bounded in terms of dist(x, aB), because of the factor N in the bound (C.6) for Q(x, ). Similarly, (C.7) shows that (a/ax)vQ(x, )/N is bounded in terms of dist(x, 3B) for each multi-index y. Then an argument like the proof of Theorem A.5 shows that v E C°°(B) and
,Lv=0inB. (ii) On its domain B \ E, v is an odd function of its Nth argument, and sgn v(x) = sgn xN. This is true in B by the definition (C.14a) and because
Q(., ) has these two properties; it is true on aB \ E by the definition (C.14b).
Appendic C. The Primary Function of Siegel Type
275
31pN)
-4 (p,
Fig. C.1.
(iii) Now we prove that v E C(B \ E). Since v E C(aB \ E) by definition,
and v E C°°(B) by (i) above, it remains only to prove that v(x) -* v(p) whenever x E B, p E aB \ E and x p. Let p E aB \ E by given; in view of (ii), we may assume that PN > 0. We shall prove that v
is continuous on 4 (p, 3 pN) n B,
(C.15)
which is more than enough. Partition aB and v by defining (Figure C.1)
S1 := aB n { x IIxNI > 3 pN 1,
S2 := aB \ S1,
A:=aBn{x I IXNI=3pN}, J a/xN on Si, 0
on S2,
g2(x)
0
on Si U E,
a/xN
on S2 \ E,
and, for j = 1, 2, ve(x) :=
Then v=v1+v2 on
Q(x,) g1O dSO if x E B, s gj(x) if x E aB. E.
Since g1 E L,,,(aB)nC(aB\A), Theorem B.29 ensures that v1 E C(B\A); certainly v1 satisfies (C.15). Now consider v2 on B \ S2, observing that
S2 is closed. Adapting remarks in (i), we see that (a/ax)YQ(x, )$2()
276
Appendix C. The Primary Function of Siegel Type
is bounded in terms of dist(x, S2) for each y, and that v2 E C°°(B \ S2).
(Lemma C.1 extends to points x E OB, provided that Ix - I > 0 .) In particular, (C.7a) shows that for points x and x + h in B, such that I xN I > 3 PN and I xN + hNI > 3 PN, we have
Iv2(x + h) - V2(x)I < const. aN+2 (3 pN)-N-3 IhI,
with a constant that depends only on N. Thus v2 satisfies (C.15) with room to spare. (iv) To prove that v(x; a) depends only on x/a, we write Q(x, ; a) for the function defined by (C.5) and set x =: az, a in the definition (C.14a); there results
v(x; a) = fQ(;1) (1/iN)
if x E B,
I = 1, N > 0 }. This proves the claim for x E B; the result is immediate from (C.14b) if the integral being over the unit hemisphere {
I
I
xEBB\E. (v) It remains to bound Iv(x)I for xl < a/2. Now, Ix - I >- a/2 when IxI < a/2 and c E S+; then the definition of v and the bound (C.6) for Q imply that 1V(x)1 <
a
a
i+ QNa(a/2)N+2 'N
Remark C.3 For N = 2, let (x, y) E R2 and z = x + iy E C. Then, on
B\EcR2, v(x,y;a) = Im
a
(a-z
a - a+z
ay
ay
_ (a-x)2+y2 + (a+x)2+y2 (C.16)
Proof The proof is an exercise in contour integration for which we choose a = 1 (as we may, because v(x, y; a) depends only on x/a and y/a) and use (C.1) rather than its special form (C.4) for anti-symmetric functions g. The Poisson kernel may be written 1
-
P(z, O := 2 Iz r 12'
where
Izi
= r < 1, (=e it ,
Appendic C. The Primary Function of Siegel Type
277
CE
eiE
e-iE
Fig. C.2.
and, if we define
IE(z) := j,
z- 7 1
JE :_ (-n + e, -s) U (e, 7t - s),
sdtt'
(C.17)
then
Ox, y;1) =
1
- r2 27r
liiIE(z)
(C.18)
Elo
for Izl = Ix + iyl < 1. The result for IzI = 1, z # ±1, is known a priori and need not be calculated.
Now, since Iz - (I2 = (( - z)(1g - z), where z := x - iy, and sin t = -(i/2)(t' - 1/4), it follows from (C.17) that IE(z) = f F(z,C) dC,
where CE :_ { eit
I
t E JE }
E
and
-24
F(z Ti
1).
,
(! - 1)(( - z)((2 -
Let AE and BE be circular arcs in the closed unit disc, centred at t' = 1 and -1, respectively; AE proceeds from a iE to eic and BE from -e 'E to -e" (Figure C.2). Chooses to be positive and so small that z belongs to the indented open disc bounded by AE U BE U CE. The only singularity of F(z,.) in this indented disc is the pole at = z; consequently,
IE(z) + f F(z, ) dC +.f
F(z,
e
C) d( = (zz
- )(z2 z - 1)
278
Appendix C. The Primary Function of Siegel Type
Calculation of -iti times the residues of F(z,.) at t' = 1 and yields
I F(z, t') dt
Z)
E
J
F(z,
) d= (1 +
)7Ti
+ 0(E),
(1 + z) + 0(E).
In view of (C.18), the rest is manipulation.
Appendix D. On the Divergence Theorem and Related Matters
D.1 A first divergence theorem
Our task in this appendix is to extend the fundamental theorem of the calculus to functions defined on subsets of IRN. If the result is to have a modest generality and to be of some use, then this task cannot be short and easy, for several reasons. First, it is not obvious how to pass from the merely local description of 812 in Chapter 0, (viii), to the evaluation of integrals over 8S2 as a whole; Definition D.1 and Lemma D.2 are preparations for this step. Second, we must attend to the smoothness both of the function being integrated and of the boundary 812. Third, conditions that allow a straightforward proof of the divergence theorem (such as those in Theorem D.3) are often too restrictive for applications. Although we take only two primitive steps towards relaxing the conditions in Theorem D.3, those steps require a certain length if they are to be elementary and transparent. Notation Throughout this appendix, RN is to have dimension N >_ 2. The notation of Cartesian products will be taken beyond the definition of A x B in Chapter 0, (i); for example, (-IJ, /3)N-1 denotes the cube Q'(0,/3) described rather fully in Definition D.1, and {a} x denotes a face of the cube (O, a)', more precisely, the intersection of the hyperplane {x E RN I x1 = a} and the closed cube [0,a]N. [O,a]N-1
Definition D.1 Let 812 be [at least] of class C and let p E 812. Referring to Chapter 0, (viii) and Figure 0.3, we note that 12 is defined globally in terms of co-ordinates xj, while yj are `local' co-ordinates such that 812n U(p) (where U(p) is open in RN and contains p) has a representation YN = h(Y'),
where h = h(., p), 279
y' = (Y1, ... , YN-1).
(D.1)
280 Appendix D. On the Divergence Theorem and Related Matters
Q'(0, l3)
. xl Fig. D.1.
(a) The co-ordinate transformation is y = Yp(x) := A(p)(x - p),
(D.2)
where A(p) is an orthogonal N x N matrix depending on p. (b) Denote a cube about the origin in l N_i, with edges parallel to the co-ordinate axes and of length 2/i > 0, by
Q'(0,/3):={y'ERN-1
-/i
Define (Figure D.1)
'_(p) = { y E RN I Y E Q'(0, /l),
-l3 < YN - h(Y) < fl }
and its x-image Gll(p) := Yp 1('K(p)), where /3 = 3(p) and /3 is so small that V(p) c U(p). The sets 't(p) and W(p) will be called quasi-cubical neighbourhoods of the points 0 and p, respectively (because they are easily transformed to cubes in RN).
The yN-axis will point into S throughout this appendix; then the definition implies that
x E X n'W(p) x E n n 61l(p)
if y = Yp(x) E Y,-(p) and YN = h(y'), if y = Yp(x) E Y"(p) and YN > h(y').
(D3.a) (D3.b)
The functions Wm in the following lemma are an example of a partition of unity on S2. They are a key to our proof of the divergence theorem because we use them to reduce the proof to one for functions of small support in convenient neighbourhoods. [The support of a function
D. 1 A first divergence theorem
281
is defined in Chapter 0, (iv); essentially, it is the smallest closed set outside which the function equals zero.] The hatched set in Figure D.1 represents the support of a function W. when p = pm in the figure.
Lemma D.2 Let n be bounded and 3) of class C. Then for some integer M there exist functions P1,1P2,. .. ,'PM as follows. (a) im 1 ipm(x) = 1 whenever x E 52. (b) W. E C°°(S2) and 0 < lPm(x) < 1 for each m E { 1, ... , M} and all
xE52. (c) The functions 1m are of two kinds: there is an integer K < M such that, if m E {1,...,K - 11, then Pm E Cc(S2), while, if m e {K, ... , M}, then supp'WVm intersects 852. In the second case supp Wm c S2nW(pm), where
W(pm) is a quasi-cubical neighbourhood of a point pm E 852.
Proof (i) Assign to every point c E S2 a ball .1(c, p) as follows. If c E 52, choose p = p(c) so small that 4(c, p) c 52; if c E 852, choose p = p(c) so small that _4(c, p) c W(c), where W(c) is a quasi-cubical neighbourhood of c. The family {M(c, p) c E S2} is then an open cover of S2 [in other words, S2 is a subset of the union of all these open balls]. Since S2 is compact, there is a finite subfamily, say {M(cm, pm) I m = 1, ... , M}, that covers S2, and we so label this subfamily that cm E 52 if m < K, while cm EOn if m K. I
-
(ii) Define for each m E {1, ... , M} (see Figure D.2) exp -
0-(x)
1
if s := ifs>-1,
1 - S2
0
iVmx)
Pk(x)
Pm(X)
Ix - cmI
< 1,
Pm
if x E S2 ,
(D.4)
(D.5)
k=1
and let pm := cm when cm E On (when m > K). We observe that
>M 1 (pk(x) > 0 when x E S2, because then x E *c", Pk) for at least one k, and that (pm E Cc (RN) for each m, by Exercise 1.15. The properties (a) to (c) of the functions W, now follow directly from the definitions (D.4) and (D.5).
Notation We shall use the abbreviations
j g :=
Jn
g(x) dx
g := j g(x) dS(x),
and sZ
sp
282 Appendix D. On the Divergence Theorem and Related Matters
an Fig. D.2.
where dx := dxl dx2... dxN and dS(x) is as in Chapter 0, (viii), unless a transformation of co-ordinates or detailed evaluation of an integral should demand a fuller display. Theorem D.3 (a first divergence theorem). Let 91 be bounded, with On of class C1, and let n = (n1i...,nN) denote the outward unit normal on On. Then
f alf = fn n.if
(D.6)
whenever f E C 1(Sl) and j E{ I,-, N}.
Proof (i) Let p',. .. , W m be as in Lemma D.2, and define fn := f Wm for each m E { 1, ... , M}. By (a) of Lemma D.2, M
M
f .(x) = f (x) E Wm(x) = f (x) m=1
for x E S2,
(D.7)
m=1
and each fm E C1(S2) by (b) of the lemma. (If f should happen to be smoother than C', then so is fm.) Since
f ajfm = m=1
I fm) = f 8jf and inn al (\m=1 ,
m=1
.f nlf, Ja4 n.ifm = 84
it is sufficient to prove (D.6) for each fm; the result for f then follows by summation over m. Let Bm := e(cm, pm), where cm and pm are as in (D.4). There are two cases: if cm E S (if m < K), then supp fm c Bm c f ; if cm E 00 (if m >_ K), then pm := cm and supp fm C S2 n Bm, where Bm c WW(pm).
D.1 A first divergence theorem
283
(ii) Let cm E 52 (let m < K). We shall prove /that
in
ajfm = 0 = f
ajfm
(cm E n).
(D.8)
The right-hand equality is immediate because fm = 0 on 852. To prove the left-hand equality, we define fm to be zero not merely on S2 \ Bm but on RN \ Bm; then in ajfm = j ajfm m
for any cube Qm (in RN) that contains Bm. Suppose that j = 1 and that Q. =: (a,b) x Q", where Q" is a cube in RN-1, and let x" := (x2,...,XN). By Fubini's theorem,
JQ
m
alfm = f
l
Qf(f)(")
l dx1 } dx" JJJ
f {fm(b,x")-fm(a,x")} dx" = 0 because fm = 0 on BQm. The argument is similar for j E 12,..., NJ: the Fubini theorem allows us to replace the volume integral by a repeated integral, and we integrate 8 j f m first with respect to x j.
(iii) Let pm = cm E 852 (let m Z K). With m fixed, we abbreviate pm to p, adopt the notation in Definition D.1, and let fm(y) := fm(x) under the transformation y = Yp(x). Define
V+ := { y E K(p) I YN > h(y') 1,
F := { y E 'K(p) I YN = h(y') } ,
and recall that fm vanishes in S2 \ B. Then the desired result,
fJfm =
whenever j E {1,...,N},
(D.9)
is equivalent to the statement
f+1m)dy = f vk(y') fm(y) d(y') whenever k E {1,...,N}; r
(D.10)
here v and da denote [as in Chapter 0, (viii)] the outward unit normal and element of surface area, respectively, in terms of the local co-ordinates. [To pass from (D.9) to (D.10), multiply (D.9) by Akj(p) and sum over j; to pass from (D.10) to (D.9), multiply (D.10) by Akj(p) and sum over k.] Upon use of the formulae (0.2) and (0.3) for v and da, equation (D.10)
284 Appendix D. On the Divergence Theorem and Related Matters becomes
J(akh)(Y')Jm(Y'h(Y')) dy' if kE{1,...,N-1},
J v+
Q'(o,P)
(akjm)(Y) dy =
if k = N.
ffm(Y'h(Y')) dy'
(D.11)
(iv) In order to prove (D.11), we make the further co-ordinate transformation z
'=Y
zN = YN - h(Y)
,
(Y E "i(p)),
which maps Y" (p) onto the cube Q'(0, /3) x (-/3, /3) shown in Figure D.3. Then
if k E {1,...,N- 1},
aYk = aak - (kh)(Y )aN
aN aY
= a
,
N
and the Jacobian determinant is 1
0
0
1
0
-(aih)(Y')
-(a2h)(Y')
1
z;= detGyi) I
...
0
=1. 1
Let W+ := Q'(0, /3) x (0, /3) and let fm(z) := Jm(y); by the rule for subjecting integrals to co-ordinate transformations (Apostol 1974, p.421; Weir 1973, p.158), the desired result (D{.11) becomes
JW' { (akim)z) - (akh)(z) (aNfm)(z) } dz
J
(akh)(z') fm(z',0) dz'
if k E {1,...,N-1},
(D.12a)
,(o,a)
fm(z', 0) dz'
if k = N.
(D.12b)
Now in (D.12a) we have (ak.fm)(z) dz = 0
if k E {1,...,N - 1},
(D.13)
fw+
because, very much as in step (ii), Fubini's theorem allows us to replace the integral over W+ by a repeated integral and to integrate akfm first
D.2 Extension to some sets with edges and vertices
285
ZN
aW+
ZI
Fig. D.3.
with respect to xk; moreover, fm(z) = 0 on 8W+ \ { z (D.13) established, (D.12) is seen to be true because 0J'
I
ZN = 0 }. With
(aNfm) (Z', ZN) dZN = -fmW, 0),
and because we may use repeated integrals again, integrating aNfm first with respect to ZN.
D.2 Extension to some sets with edges and vertices
In many applications one needs a divergence theorem for sets f) with boundaries M that are not of class C' because they have edges and vertices. A general treatment of such boundaries would lead us into deep water, but results for some particular cases are useful and not difficult.
Remark D.4 We shall derive a divergence theorem for the following bounded open sets in RN (N (i) An open interval, say I
2). As elsewhere, x" := (x2, ... , xN). x (aN, bN). (al, b1) x (a2, b2) x
(ii) A circular cylinder, say C := { x E RN I a < x1 < b, Ix"I < p }. (iii) A segment of a ball, by which we mean the (non-empty) intersection
of a ball and a half-space, say D := M(0, b) n { x E RN I x1 > b cos a },
where a E (0, 7C).
(iv) A finite sector in R2, or circular cone in RN with N >_ 3, possibly
truncated near its vertex: in terms of r := IxI and 01 such that x1 =
286 Appendix D. On the Divergence Theorem and Related Matters r cos 01 and Ix"I = r sin 01, say
K :={xERN I a
E :_ °.A(0,R)\R(c,p)
with R-p < cl
.(c, p) \ .(0, R)
with R - p < cy < R + p,
or
E'
where 0 < p < R in both cases. (vi) Sets resulting from those in (i) to (v) when the co-ordinate frame is translated and rotated.
(vii) Complements of the foregoing in a larger admissible set: fl = S22 \ 521, where S21 c 522 and 521, f22 are listed in (i) to (vi), or else one of
them is listed while the other is a ball.
The reader may wish to lengthen the list in Remark D.4; the method that will be used to prove Theorem D.5 can be applied to other configurations. Alternatively, the reader may prefer to proceed to more advanced forms of the divergence theorem (Evans & Gariepy 1992, p.209; Maz'ja 1985, p.304; Ziemer 1969, p.248) which allow sets SZ and functions f that are far less restricted than those considered here. A first question is: how should one write the normal n(x) and element dS(x) of surface area on boundaries with edges and vertices? But this is hardly a difficulty for the catalogue in Remark D.4; for these sets, a direct application of Cartesian, cylindrical or spherical co-ordinates is possible; alternatively, one can apply the definitions in Chapter 0, (viii), to each smooth piece of the boundary. We shall exhibit both methods. Consider as an example the segment D of a ball in Remark D.4, (iii). Partition 8D into a flat part {x x1 = b cos a, Ix"I < b sin a}, a spherical part {x IxI = b, x1 > b cos a} and an edge {x x1 = b cos a, Ix"I = b sin a}. The edge does not contribute to the boundary integral because it has zero surface area (has (N - 1)-dimensional measure zero). On the flat part of 8D, I
I
n(x) = (-1, 0,..., 0),
dS(x) = dx" := dx2 dx3 ... dxN .
(D.14)
D.2 Extension to some sets with edges and vertices
287
For the spherical part of 3D we use spherical co-ordinates as in Exercise D.18; then on the spherical part of 8D,
dS(x) = b'-'A(O) d0,
n(x) = x/JxJ, where
A(O) := (sin
01)N-2
(sin 02)N-3
... (sin ON-2),
(D. 15)
d0 := d01 d02 ... dON-1
This formula for dS(x) is not invalidated at x = (b, 0, ... , 0) by the fact that the Jacobian determinant det8(x;)/7(r,0j) vanishes there. [Consider
the contribution to an integral over 8D of any piece of surface that contains the point x = (b, 0, ... , 0) and has diameter e.] In order to illustrate how the formulae in Chapter 0, (viii), extend to a boundary with an edge, we now represent 8D, near an edge point p, as the graph of a function that fails to be C' only at the projection of the edge onto the domain of the function. Let 2 E
(0, 2),
p := (b cos 2#, 0,..., 0, -b sin 2#),
and let the transformation y = Yp(x) to local co-ordinates (Figure D.4) be Y1
Y2
YN-1
YN
= (cos/i)(x1-p1)-(sinf)(xN-PN), = x2, = xN-1, = (sin/3)(x1 -p1) +(cosf3)(xN -PN)
The flat part of 8D has radius R := b sin 2/1 = -PN, and its projection onto the hyperplane {y YN = 0} is the elliptic set I
E :=
y' E RN_1
Yi + R sin Q R sin
2
2
+ 11 R2
<1 ,
where Y2+...+Y2v-1
112
Our local representation of 3D is restricted to the cylinder
V := {y E RN I
1y'I
having radius p E (0, b - b cos /3) and cross-section
B:={
y'ERN-1
I
Y1
+n2
Appendix D. On the Divergence Theorem and Related Matters
288
1
Fig. D.4.
If 1 - cos /3 < sin 2/3 (as is the case in Figure D.4), then p < R, so that a part of E is outside B; however, if /3 is so near it/2 that 1-cos /3 > sin 2/3,
then it can happen that E c B. A calculation now yields the following representation of YP(aD) n V :
-ylcot# yN
= h(Y) :=
bsin/3-
if y'
B
E,
(bsin/3)2-2bylcos/3-y2-rl2 ify'EB\E; (D.16)
it is easy to check that these two expressions are equal on B n .E. The formulae (0.2) and (0.3) for the outward unit normal v(y) and element do(y') of surface area are valid in B \ OE for this function h. Theorem D.5 If G is one of the (open, bounded) sets listed in Remark D.4, then
(D.17)
fGjGn;f
whenever f E C'(6) and j E { 1, ... , N}. The boundary integral is over those parts of aG on which the outward unit normal n is defined and continuous.
Proof (i) We shall introduce a sequence (Gm) of open subsets of G, each having closure Gm c G and boundary aGm of class C', such that
,
fGjf
-f
G
ajf
and f
Gm
IG
n1f
as m --oo, (D.18)
D.2 Extension to some sets with edges and vertices
289
Fig. D.S.
whenever f E C' (U) and j E {1,..., N}. Since fGm if = faGs njf for each m, by Theorem D.3, this will prove the present theorem. (ii) Consider the particular set G := Q \ D (Figure D.5), where Q is the cube (-a, a)N and D is the segment of a ball in Remark D.4, (iii), with b < a. It will suffice to display Gm for this configuration, because a similar construction is possible for all cases of our list. (The symmetry of Q and the alignment of Q and D save words, but are not necessary for the construction. Exercise D.12 deals with another set in the list.) Our method is simply to round off all edges and vertices; in order to make the resulting set a subset of G, we must first enlarge D slightly. At the same time we shall shrink Q slightly, in order to have Gm c G; the virtue of this condition will be seen in §D.3. The approximating sets Gm := Q,,, \ Dm are formed as follows. Choose and fix bl E (0, (a - b)) and set am := 61 /m for m E N. The rounded set Dm (Figure D.6)3is D,,, := U M(c, 6m)
x E RN I dist (x, D) < bm I,
cED
and its boundary 8Dm is partitioned into a flat part Fm, a spherical part Sm and a toroidal part Tm by the definitions
Fm :=8Dm(1{x I x1 =bcosoc-bm}, Sm := 8Dm n 8.4(0, b +'5m)
and Tm := 8Dm \ (Fm U Sm).
290 Appendix D. On the Divergence Theorem and Related Matters aQm\Em
aQ\rm
F.
Fig. D.6.
The toroidal part has surface area I Tm I < const.8m, where the constant is independent of m (and may be taken to be Or - x) 10-4N-1(0, 01 x (b sin a + bl )N-2). Since aDm is a surface of revolution, one checks without difficulty that it is of class C'. Now consider the approximation Qm to Q. First form the smaller cube Mm := (-a + 28m, a then define `Lm
U Ac, Sm). cEM,,
Defining an intermediate cube by Pm :_ (-a + Sm, a - 8m)N, we set Em := aQm fl aPm and so partition aQm into a flat part Em and a curved
part aQm \ Em. Note that Em consists of 2N squares (that is, (N - 1)dimensional cubes) like
{a-6m) x (-a + 2am, a -
28m)N-1
Correspondingly, the original boundary aQ is partitioned into a part Fm 26)N-1 and a part consisting of 2N squares like {a} x (-a + 28m, a aQ \ rm containing the vertices of Q. The curved part of aQm and the corner part of aQ have surface areas I aQm \ Eml <- const.Sm,
I3Q \ rmI < const.Sm,
D.2 Extension to some sets with edges and vertices
291
where the constants are independent of m. That OQ,,, is of class C' is easily seen for N = 2 or 3 with the help of figures, and is the subject of Exercise D.13 for arbitrary N >- 2.
(iii) To establish the first limit in (D.18), we fix j and write h := aJf, hm := xmh, where Xm is the characteristic function of Gm [equal to 1 on Gm, and to 0 on RN \ Gm]. Then hm(x) h(x) at each fixed x E G as m -> co; also Ihm(x)I < Ih(x)I and Ihl E C(G). By the Lebesgue dominated convergence theorem,
f
I h=J
as m - co.
h
(iv) Consider the second limit in (D.18). Let If II
:= supxEG{ f(x)I + IVf(x)1 };
we have If 11 < oo because f E C'(G). In view of remarks made in (ii) about surface area,
f
T.
njf +JaQm\Em nif +
OQ\r,
nJf
-< 11f1111T.1+JaQmEm1+IOQ\rm1 }->O as m-cc.(D.19)
Let F and S denote, respectively, the flat and spherical parts of aD. It remains to prove that (as m -> oo)
JFfF nJf O, jnJf_fnJfO fnjf_fnjfO. ,
(D.20)
We shall prove the second of these; the first and third are similar but easier, because only Cartesian co-ordinates are required. In order to have a convenient pairing of points of Sm and points of S, we use spherical co-ordinates r, 01, ...,ON-, as in Exercise D.18 ; we also
use the notation in (D.15) and write 0 := (01,...,ON-1), nJ(0) := nJ(x) and 1(r, 0) := f (x). The labelling 01 < a of the domain of integration
means-a<01
I.
nJf
- f nJ
is
_
h j(0){ 1(b + 5,,,, 0) - 1(b, 0) } (b + B,
+
bm)N-lA(O)
dO
<«
fnJ(O) 1(b, 0) { (b +
< const. If Ilam,
$m)N-1
- bN-1 }A(O) dd (D.21)
292 Appendix D. On the Divergence Theorem and Related Matters
where the constant is independent of f and m because I1(b+8m,0)- 1(b, 0)1 =
(m)N-1 < (b
(b +
+
fb+Sm a f I
Sm)N-1 -b N-I
(b +
81)N-1,
ar(r,0) dr
Jb
<- If Ilbm,
< (N - 1)(b +
8i)N-26m
The inequality (D.21) establishes the second limit in (D.20), and we have noted that the first and third are proved similarly but more easily.
Remark D.6 The result (D.18) can be improved. We note again that Gm c G and that aGm is of class C' for each m E N, and now claim that, asm ->oo,
fG, u-> JG u
LG
,
nj v -* j nv
whenever u E LI(G),
whenever v E C() and j
(D.22a)
1,...,N}.
G
(D.22b)
[Recall that the condition u E Li(G) means merely that u is integrable on G] As a description of how well Gm approximates G, the statement (D.22) is more satisfactory than (D.18), because the condition f E C'(G) is not natural for such a description. The proof of (D.22a) is virtually the same as step (iii) of the foregoing proof: Xm(x)u(x) - u(x) at each fixed x E G, and the dominant function is now Jul E L,(G).
is compact, v is bounded To prove (D.22b), we note that, since and uniformly continuous on G: there are positive numbers K and y, (depending on v) such that lv(x)l < K for all x E G and such that, for every e>0, lv(x) -
e
whenever x,
E G and Ix - I < yE.
Accordingly, K replaces If II in (D.19) when v replaces f ; thereafter, we use the fact that, for every e > 0 and for the relevant (x,, x") and (r, 0),
Iv(xl - 8m, x") - v(xi, x")l < e and if m is so large that 6 m < yE.
v(r + 5m, 0) - v(r, 0)l < s (D.23)
D.3 Interior approximations to the boundary aS2
293
D.3 Interior approximations to the boundary ac Not infrequently the condition f E C1(I2) in Theorem D.3 is an embarrassment; here is an example. If in (D.6) or in (D.17) we set f = uaj u and sum over j, there results the identity
in
{ Ioul2 + uLu } = Jn u
(an :- n
an
V),
(D.24)
for admissible Q, provided that u E C2(S2). In the uniqueness theorems
for which this identity is commonly used, it is natural to assume that u E C1(S2) n C2(S2) and that Au = 0 in 0, but artificial to assume that U E C2(52). As far as we know, an individual second derivative such as ai u(x) might misbehave as x approaches some boundary point. To overcome this difficulty, we shall develop not a third divergence theorem but, rather, approximations S2m to fl. These sets S2m satisfy S2m c 12, have boundaries acim of class C' and have the approximation properties for integrals that are displayed in (D.22) for the sets Gm considered there. With such approximations S2m in hand, we can apply (D.24) to S2m
rather than to n, noting that u E C2(S2m) when u E C2(0). In fact, if U E C1(S2) n C2(S2) and Du = 0 in S2, we shall have
fIVuI2 =
llmm
llmmy
Jm
f
IQUI2
anm
u-au = an
-
au
,
which vanishes, as desired in uniqueness proofs, if u = 0 on au or au/an = 0 on aft. If aft were of class C2, then the approximations urn could be formed,
for sufficiently small numbers Sm > 0, by defining Mn to be the set consisting of all points q(p, 6m)
P - 8m n(p),
p E au.
(D.25)
It can be shown that aum is then a boundary of class C2. This is the conventional method in geometry for the construction of what are called tubular neighbourhoods of a surface (the word `tubular' actually referring to layers on both sides of the surface). When a52 is merely of class C1, however, points of S2 arbitrarily close to au may be on more than one line normal to aKI (Exercise D.17) and the set of points q(p, bm) need not form a boundary of class C', however small bm may be. Therefore we use a different method and proceed with due caution.
294 Appendix D. On the Divergence Theorem and Related Matters
Notation We continue to write a/an := n V, where n(x) denotes the outward unit normal at x E an. Lemma D.7 If 91 is bounded and an is of class C1, there exists a function w E C' (S2) such that
_ w1az = 0,
aw an
>1 an
andw>0in5:. Proof We shall use the local representations YN = h(y', p) of 00 in Definition D.1; the transformations (D.2) to local co-ordinates will now be written y = Y(x,p),
where Y = (Y', YN)
and Y' :_ (Y1,..., YN_1),
and the yN-axis points into 0. We shall also use the partition functions , WM of the second kind in Lemma D.2. Since the functions WK, , WK-1 of the first kind all vanish on On, those of the second kind W1, form a partition of unity on an: M
EWk(x)=1
if xE00;
k=K
also, Wk E C%2), suppWk S2 n °u(pk) and 0 < Wk (X) < 1 for each k E {K_., M} and all x E S2. We define M
xE
w(x) E Wk(x)11k(x),
(D.26a)
k=K
where
kx
( YN(x, pk) - h(Y'(x, pk), pk) 1I
0
if x E S2 n &(pk), elsewhere in
S2.
D.26b (
)
[The definition of nk(x) for x 0 suppWk is unimportant, because Wk(x)Y1k(x) = 0 outside suppWk under any definition of Ilk Note that, if x E S2 n oh(pk) and y = Y(x, pk), then t1k(x) = YN - h(y, pk )
R.]
D.3 Interior approximations to the boundary 00
295
therefore, at points x E as2 n 1(pk) and with the abbreviation h(y) h(y', pk ),
=
N
- an (x)
aye
>vj(Y'){h(Y')-YN}
N-1
ah = vJ(Y')a
=
.(Y)-vN(Y')
y'
I=1
+... + (ON-1h)(Y )2 + 1 }1/2
{ (01h) (Y')2
(D.27)
upon application of the formula (0.2) for v(y'). The function w E C I (S2) because each 1pk E C'(0), each h(. , pk) E C1(Qk), where Qk := Q1(0,/3(pk)), and each Y(.,pk) is an affine function (that is, a linear function plus a constant). Moreover, wlan = 0 because nk an = 0 for all k ; and the definition (D.26) also shows that w > 0 in 0. Finally, for each k and all x E an n °h(pk) we have -a,)k(x)/an >_ 1 by (D.27); since also 11k I an = 0, this yields I
Ow an an
M Wk
k=K
M
_
E Wk
On
k=K
00
an
1'
as desired.
To construct nn, we shall choose a suitably small number al > 0, set a,n := al/m for m E N, and define to be the part of the level set {x E 12 w(x) = a,n} that lies close to 00. It will save trouble later to I
record here some implications of Lemma D.7.
Remark D.8 (i) Since Vw is continuous on the compact set S2, it
is
uniformly continuous: for every a > 0 there is a number SE > 0 such that x,
E SZ
and Ix - I < 8E
I(Vw)(x) -
E.
(D.28)
(ii) On as2 the vector field Vw is normal to as2 because w I an = 0, and it
points inwards to n because -Ow/On > 0. Therefore IVwl = -Ow/On > 1 on an, whence x E f2, p E as2
and Ix - pI < 8E
I(Vw)(x)l > 1 - E.
(D.29)
(iii) Define
d(x) := dist(x, an), if x E SZ and d(x) < 61/2 e(x) := Vw(x)/IVw(x)I
(D.30)
296 Appendix D. On the Divergence Theorem and Related Matters
Fig. D.7.
[This last ensures, in view of (D.29), that IVw(x)I > 2.] The direction field (or unit-vector field) e is a continuous extension of the inward unit
normal field -n on 852; that is, le(x)I = 1 and eI au = -n. Condition (D.28) now implies the existence of a number 6* > 0 as follows: if x E 52,
p E % and Ix - pI < 6*, then
e(p) e(x) >
2 = cos 6
and
e(p) Ow(x) > 5.
(D.31)
(A calculation, outlined in Exercise D.14, shows that 51 4 can serve as 6*
(iv) We use (D.31) to construct a useful cylinder D(p) about any boundary point p E 852. Take local co-ordinates zj such that z := Zv(x) := C(p)(x - p), (CNI(h))N
where
OxZN =
J=1 = e(p) = -n(p),
(D.32)
and where C(p) is an orthogonal N x N matrix depending on p. Define (Figure D.7) 6*
fl:_
,
E:_{zERN I IZ'I
observing that E is independent of p, so that a translation and rotation of the same cylinder E gives D(p) for all p E 852. The inward normal -n(x) at any point x E as2 n D(p) makes an angle with e(p) that is less than n/6, by (D.31); hence z E ZP(a.Q) n E
IZNI < (tan
6
) Iz'I <
-,
(D.34)
D.3 Interior approximations to the boundary Al
297
with equality only if z' = 0. Thus Zp(af2) \ {0} lies between the two conical surfaces shown by broken lines in Figure D.7.
(v) Let w(z) := w(x) under the transformation z = Zp(x), so that aw(z)/azN = e(p) Vw(x). Integrating aw/3ZN along a vertical line z' = const.} in E from a point of Zp(a92) to the point above it on the ceiling of E, and applying (D.31) and (D.34), we obtain {z
I
3
w(z" Q) > 5
(#_
)
> 4Q
(Iz'I < /3).
(D.35)
It follows from (D.31) and (D.35) that every vertical line segment z' = const.} in E n Zp(S2) contains exactly one point of Zp(af2), at which w(z) = 0, and exactly one point of Zp(af2,n), at which '(z) = a,,,, if 0 < al < 4 l3 and a,n := al /m form E N. We label these points (z', g(z', p)) in the case of Zp(aQ), and (z', g,n(z', p)) in the case of Zp(aS2,n). {z
I
Theorem D.9 Suppose that S2 is bounded and that either a ) is of class C1 or S2 is one of the sets listed in Remark D.4. Then there exists a sequence (S2m) of open sets such that (a) S2m c S2 for each m,
(b) each boundary 0I2,n is of class C', (c) fnm u --> fn u as m
(d) f a n.
oo whenever u E L, (SL),
njv - fa n njv as m -+ oo whenever v E C(S2) and j E {1, ... , N}.
Proof (i) If n is one of the sets listed in Remark D.4, then the present theorem merely repeats the second sentence of Remark D.6. The truth of that sentence rests on the construction of G n in the proof of Theorem D.5, step (ii); on analogous constructions, like that in Exercise D.12, for
other sets in the list; on the proof in Remark 6 of (D.22a, b); and on slight variants of (D.23) for sets other than Q \ D. (ii) For the rest of this proof, let E3 be of class C' and let w be as in Lemma D.7. We retain the notation in Remark D.8 [in particular, d(x) := dist(x, aS2)] and shall use the estimates there. The graph of w on the inward normal line from any point p e Of) is qualitatively as in Figure D.8.
We define Stn, by setting al = 5an, := al/m for m E N and
I w(x)>o
or d(x)>/3},
equivalently,
S2\52,,,_{xES2 I w(x)
298
Appendix D. On the Divergence Theorem and Related Matters
Fig. D.B.
In the definition of 52,,,, the alternative condition d(x) > /3 merely ensures that 52,,, contains those points x relatively far from 892 at which w(x) < a,,. We now show that, in fact, d(x) < / at all x r= 92 \ 92,, ; therefore 892 3 < /3 } and is determined there by lies well inside the set { x E 92 d(x) the condition w(x) = a,,,. Let xo E 92 \ 92m and let po E 092 with Ixo - poi = d(xo). There is at least one such nearest boundary point po for xo. The line segment from I
po to xo has direction e(po); let dt'(x) denote the element of length at a point x of this line segment. Then, by (D.31), xo
e(po) Vw(x) d((x) > s Ixo - pol = s d(xo),
am >- w(xo) _ JP0
whence
d(xo) < 3 am < 3 al =
a
(xo e 92 \ 92m).
(D.36)
3
(iii) We must prove that 892m is of class C'. Let a point q E 092 be given; again there is at least one point p E 092 such that Iq - pl = d(q), and we take local co-ordinates zj at p as in (D.32). The final observation of Remark D.8 states that the set Zp(092m) n E has a representation ZN = 9.(Z', P) [and ZP(q) must satisfy this equation because it is the only point on the line {z z' = 0} in E at which '(z) = am]. The implicitfunction theorem ensures that gm(., p) is a C' function on some set open in RN-1 and containing 0 E RN-1; we may take this set to be -4N_1(0, /3), again by the final observation of Remark D.B. Thus g,,(.,p) E C1(G) whenever G c -IN_1(0, /3). The implicit-function theorem also ensures existence of a set U(q, m), open in RN and containing q, such that I
x E U(q, m)
and
w(x) = am
ZN = gm(z', p),
(D.37)
D.3 Interior approximations to the boundary 5Q
299
under the co-ordinate transformation z = ZP(x). The uniqueness part (=>) of this result deserves emphasis here. Suppose that q has another nearest boundary point: there exists a point p* E 3Q \ {p} such that Iq - p* I = d(q). Taking local co-ordinates cj at p*, again of the type (D.32), we obtain another local representation (N = T n((', p*) of 852,,, near q. Suppose that under the affine transformation H z this second representation becomes ZN = yn,(z', p* ). Then p*) = g,,,(. , p) wherever both are defined, otherwise (D.37) is contradicted.
f
(iv) The proof that fn. U fl u whenever u E L1(S2) is like step (iii) in the proof of Theorem D.5 and the extension of that step in Remark D.6.
(v) The proof of condition (d) of the theorem involves a partition of 521 very reminiscent of that in Lemma D.2. The family { R(p, 3 fl) p E 8S2 } is an open cover of the compact set 52 \ 521, because of (D.36). Hence there is a finite subcover, say { _q(pk, 3/3) k = unity on SZ
1, ... , K } with each pk E 8S2 , and we define, for each k E { 1, ... , K },
exp C- 1 -1s2 //
I
9k(x)
if s :=
3 I x- pk I< 1
2/3
if s > 1,
0
0n(x)
Wk(x) := (Pk(x)
if x E 52 \521.
n=1
Each partition function 1k has support well within the cylinder D(pk) defined by (D.33).
(vi) It remains to prove condition (d). Given V E C(O), we define Vk := VWk. It is sufficient to prove that
jfliVk-4jfliVk
as m -+ oo,
(D.38)
for j E { 1, ... , N} and k E {1,...,K); then summation over k yields the result for v because v = > vk on 52 \ 521.
We fix k, abbreviate pk to p, make the transformation z = Zp(x) to local co-ordinates as in (D.32) and write 6(z) := vk(x), with the understanding that v is continuous on Zp(S2) and that supp v c E. The local representations of 80 and of 852, are abbreviated to ZN = g(z') and to ZN = g(z'), respectively, and B := MN_1(0, /3) denotes the crosssection of E.
300
Appendix D. On the Divergence Theorem and Related Matters
In view of the formulae (0.2) and (0.3) for the unit normal and element of surface area, we must prove that, if j E { 1, ... , N - 11,
f(Jgm)(z'gm) dz'
, j(ig)(z'g) dz',
(D.39)
where gm = gm(z') and g = g(z'), and that, if j = N, f(z',gm) dz' -*
dz' .
(D.40)
For (D.40) it is enough that u is uniformly continuous and that 0 < gm(z') - g(z') < 35 am
3m
for all z' E B,
(D.41)
which is proved as (D.36) was. For (D.39) we must also compare O'gm and V'g, where 0' := (01,.. - aN_1); these gradients are calculated by differentiation of the equations w (z, gm(z')) = am
and w (z', g(z')) = 0.
Accordingly, for every s > 0,
lv,gm(z') - o'g(z,)l
_
-
(
`V'x') `(z', g(z')) (ONO) (z', g(z'))
< const. s if
3m
gm(z')) (ONO) (z', 9,, W))
< 6, and z' E B,
(D.42)
where we have used (D.41) and the basic inequality (D.28). A calculation based on (D.31) shows that the constant may be taken to be (5/3) (1 +
3_1/2).
D.4 Exercises Exercise D.Provve the divergence theorem,
a,f = fG
njf
if f E C'() and j E {1,2},
(D.43)
aG
for an interval in R2, say G := (al, bl) x (a2, b2), by a method that is more direct for this case than the proof of Theorem D.S.
Exercise D.11 Prove the divergence theorem (D.43) for the truncated sector G := {(r cos O, r sin O) E R2 1
a < r < b, ) < 0 < µ},
D.4 Exercises
301
where a > 0 and y - 2 < 2ir, by writing f (XI, x2) =: I (r, 0), observing that
II. al f dx1 dx2 = JIG
a2 f dxl dx2
JJ
{ r cos 0 or - sin 0 Of } dr dO, JJJ
IJa,b)x(A,p) { rsin0+cos0} dr dO, 111
111
and integrating by parts along lines of constant 0 and lines of constant r.
Exercise D.12 This exercise concerns verification of Theorem D.5 for the set
W := 9(0, R) \ M(c, p),
where c := (R - p, 0,..., 0) and 0 < p < R,
which perhaps is the worst of those in Remark D.4 because OW is not of class C. Choose b1 E (0, 1(R - p)), set am := 6 1/m for m E N, and define Vm := M (0, R - 2bm) \ °.l (c, p + 26m),
Wm
U .4 (z, b zEV,
Prove that Wm c W, that 3Wm is of class C1, that ffm u -> fW u as few n1 v as m -+ oo m -+ oo whenever u E L,(W) and that few n1 v whenever v E C(S2) and j E { 1, ... , N}.
Exercise D.13 The following result implies that, in the proof of Theorem D.5, the boundary aQm is of class C1 for each dimension N >- 2. Let A denote the closed `negative octant' in RN, that is, A :_ (-c0,0]N, and let B :_ UaEA 4(a, p) for some p > 0. Prove that aB is of class C1. [The boundary aB is the union of 2N - 1 pieces as follows. Call
a boundary piece of type k if, for some subset J := { jl, j2, ... , jk } of {1,2,...,N}, the piece is
{xERN I
=p2, x1 > -0 if jEJ, xm<0 if
The outward unit normal n on this piece has components n1(x) = x1/p if j e J, nm(x) = 0 if m J. There are N(N - 1) (N - k + 1)/k! pieces of type k; the N pieces of type 1 are flat; the single piece of type N is spherical.
Since (0.2) states that ash = -v1/vN in the notation used there, it suffices here to prove that n is continuous on S U T whenever S and T are boundary pieces that intersect.]
302 Appendix D. On the Divergence Theorem and Related Matters
Exercise D.14 Prove for the vector field e := Ow/Iowl in (D.30) that, if
xEQ,pEOJ, x-pI<SEand0<6<1,then I e(x) - e(p)I < 26,
e(p) - Ow(x) > (1 - 262) (1
e(p) - e(x) > 1- 262,
6).
Infer that 514 can serve as S* in (D.31). [If a :_ (Vw)(x) and b := (Vw)(p), then
e(x) - e(p) _
(IbI - IaD)a + IaI(a - b)
and IbI > 1.]
Exercise D.15 Let G(S) := { x E 1 dist(x, 8s2) > S } for a given (open) set 52 and given number S > 0. Prove that I
(a)G(S)={xEl[R' 14(xS)cS2}; (b) if 52 is convex [Apostol 1974, p.66; Rudin 1976, p.31], then G(S) is also convex.
Exercise D.16 Let x = (x, y) and
n) denote points of R2. Given
g E C 1(R), define
52:={xER2 I xER, y>g(x)}, (-91(t),1)
(t, S) := (t, g(t)) + S
g(t + 1
t E R,
S = const. > 0,
and let P(S) t E ][R }. Note that P(S) is the set of points g(t, 6) called q(.,S) in (D.25); also that P(S), as the range of the continuous function g(., S) : R R2, may be called a path (or curve or arc). (i) Show that P(S) is parallel to 852 in the sense that I
dy
dx
PS)
_
11r(t, S)
-S(t)
at any point t such that g"(t) happens to exist and jt, S) (ii) Prove that, if G(S) is as in Exercise D.15, then 8G(6)
0.
P(S).
Exercise D.17 Consider the particular case of Exercise D.16 in which g(x) = 3 IxI3/2 for all x E JR. Show that (a) each point (0, y) E 52 is the point of intersection of three straight lines normal to 852; (b) the path P(S) is qualitatively as in Figure D.9, the numbers to = to(o) and
tl = t1(S) being uniquely determined by (to,S) = 0, to > 0 and by l;Jtl, S) = 0, t1 > 0; (c) neither P(S) nor 8G(S) is a boundary of class C1.
D.4 Exercises
303
Y
Fig. D.9.
Exercise D.18 Spherical co-ordinates r,01,.--,ON-1 for RN, N >_ 2, are defined as follows. If N = 2, x1 = r cos 01, x2 = r sin 01i
if N = 3, x1 = r cos 01, x2 = r sin 01. cos 02,
x3 = r sin 01. sin 02i
if N = 4, x, = rcos01, x2 = r sin 01. cos 02, x3 = r sin 01. sin 02. cos 03,
x4 = rsin01.sin02.sin03i if N >- 5, x1 = r cos 01, x2 = r sin 01. cos 02,
XN_1 = r sin 01 ... sin 0N_2 cos ON-1,
XN = r sin 01 ... sin 0N_2 sin ON-1
Write this transformation as x = f (v), v E E, where v := (r, 01,..., ON-1) and E := (0, oo) x (0, 7C)N-2 x (-n, it), with the understanding that E (0, oo) x (-7G, 7C) if N = 2.
304 Appendix D. On the Divergence Theorem and Related Matters Prove the following.
(i) xi + + xN = r2 for all v E E. [Begin with xN + 4_i.] (ii) xN + xN_1 >0 for N > 3 if and only if r > 0 and 0 < Oj < n for
j=1,...,N-2. (iii) If f is restricted to
(0, 00) x (0, n)N-2 x (-n, n], then its range
is
f (E) = { x E RN I x,
0 },
and f It is injective (one-to-one). (iv) Spherical co-ordinates are orthogonal co-ordinates: for v E Ox
Ox
avi
avj
j,
i
fax/avjl are given by
and the arc-length functions hj hi(v) = 1,
if
=0
h2(V)=r, h3(v) = r sin O1, ... , hN(v) = r sin 91 ... sin ON-2.
[This can be proved by induction, since y E RN+1 can be represented as
y = (xi, .... XN-i, XN COS ON, XN sin ON),
where x E RN and xj = f j(r, 01, ... , ON-1) for j = 1, ... , N.]
(v) The Jacobian determinant of the transformation is J(v)
:= det (avxij) =
h1h2...hN
rN-1 (sin 01)N-2 (sin 02)
N-3
... (sin ON-2).
(vi) The Laplace operator becomes, for v E E, hih3 ... hN a _ 1 a N-i a 1 a r
rN-i ar
a +aeN i
ar
(
+ J(v)
a01
hih2 ... hN_i
a
hN
aeN i
h2
aai
+
[See, for example, Kellogg 1929, p.183; or Spiegel 1959, p.151]
Appendix E. The Edge-Point Lemma
E.1 Preliminaries In this appendix we shall try to maintain a certain similarity between the boundary-point lemma (Lemma 2.12 and Theorem 2.15) and the edgepoint lemma (Theorems E.8 and E.9 below), but a greater complexity of the present situation cannot be avoided. Example E.5 will illustrate this, after we have established an appropriate terminology. The edge-point lemma will be described loosely after Example E.5. Definition E.1 By a neighbourhood of a point p E RN we mean a set that contains p and is open in RN.
A point p E 811 will be called an edge point of S2 if, for some neighbourhood U of p, (a) 52 n U = (521 n 5211) n U, where 521 and 011 are open subsets of RN having boundaries of class C2 and outward unit normals n' and n11, respectively; (b)
p E 8521 n 85211
and
-1 < n1(p) n11(p) < 1,
(E.1)
as is shown in Figure E.1.
A connected set of edge points of 52 will be called an edge of Q.
If, for example, 52 := R3 \ [0, oo) x [0, oo) x R (the set obtained by removing two octants from R3), then the origin is not an edge point, even though there is a corner there. This set 52 is a union of sets with smooth boundaries, rather than an intersection. Also, this set .0 has the interior-ball property at the origin (Definition 2.14), whereas there is never an interior ball at an edge point. 305
Appendix E. The Edge-Point Lemma
306
as21
Fig. E.1.
Definitions E.1 to E.3 would still have meaning if 8f21 and 89211 were
only of class C' and if cpi and cpj, were only in C'(U). However, C2 smoothness of all these objects will be required in Theorems E.8 and E.9, and it seems simpler to adopt this smoothness ab initio.
Definition E.2 We shall say that cpi is an admissible function describing 921 near p if, for some neighbourhood U of p, (a) (pi E C2(U); (b) IVgpi(x)I > 0 for all x E U; (c) for x E U, the value (pi(x) is negative, zero or positive according
asxE01,xE8921 or Admissible functions describing 9211 near p are defined similarly.
Definition E.3 Let p be an edge point of 92 (with 92 = 921 n 9211 near p); let 9, and cpii be admissible functions describing 921 and 9211, respectively,
in a neighbourhood U of p; and let a = (a, ), now defined on S2, be the matrix of leading coefficients of an elliptic operator L of order two. Then the function B defined by B(x)
:= V pi(x) . a(x) . V pj1(x) N
a1W1(x) a11(x) 8icpii(x),
x E s2 n U,
(E.2)
i,j=1
will be called a bluntness function for the edge 8921 n 89211 relative to the operator L.
We use the word bluntness because at an edge point B(x) is a weighted form of ni(x) n1I(x), and this latter varies from -1 at a perfectly sharp
E.1 Preliminaries
307
edge to 1 at a perfectly blunt edge. Of course, both extremes are excluded by (E.1).
In §E.3 we shall encounter the following alternative bluntness conditions at an edge point p E 8521 n On,,. B(p) > 0.
(E.3)
B(p) = 0 and (aB/OT)(p) = 0 for every differential operator O/aT tangential to On, n as111 at p.
(E.4)
The coefficients aid are to be sufficiently smooth for continuity of B at p in the case of (E.3), and for continuity of VB in ?in U in the case of (E.4). The derivatives (aB/OT)(p) may be evaluated either as (a) lim,- (T VB)(x), where x E n n U and T is a non-zero vector satisfying T n'(p) = 0 and T n11(p) = 0 [equivalently, (T V p1)(p) = 0 and (T - V pji)(p) = 0] ; or as
0 < t < 1 } is an arc in '(t)I > 0 on [0,1] and (to) = p for
(b) dB (fi(t)) / dt at t = to, where { (t) as21 n aQii for which some to E [0, 1].
E C' [0, 1],
1
Remark E.4 Each of the bluntness conditions (E.3) and (E.4) has a meaning independent of the choice of cp1 and T jj, provided that these are admissible in the sense of Definition E.2. To demonstrate this, we define
/3(x) := n1(x) a(x)
n11(x)
and
g(x) := Iow1(x)I
for x E (as21 n af)11) n U, and adopt the evaluation (b) of (aB/aT)(p). Then B(x) = f3(x)g(x),
with /3 independent of the choice of cp1 and p11, while g(x) > 0. Thus (E.3)
is equivalent to /3(p) > 0, and the pair of conditions (E.4) is equivalent to the pair /3(p) = 0 and (8/i/3T)(p) = 0. Example E.5 Consider the sector
S2 :_ {(rcosO,rsin0)ER2
I 0
aE(0,n).
The points (0, 0), (R cos 2, -R sin Z) and (R cos 2, R sin 2) are all edge points, but only the origin will be considered. Defining half-planes S21 :_ { x E R2
cpi(x) := -xi sin
<01
- X2 cos 2
1
2
Appendix E. The Edge-Point Lemma
308
S21I = { x E Il82
I
cpII(x) := -xl sin 2 + x2 cos
<0},
2 we have S2 n 9(0, p) = (01 n 1211) n -4(0, p) whenever p < R. Relative
to the Laplace operator L [for which aij(x) = Sij], a suitable bluntness function is B(x) = V(p1(x) V p11(x) = -cosac
(IxI < R).
We restrict attention to harmonic functions u having a maximum at the origin :
Lu = 0 in
u E CI(S2) n C2(S2),
S2,
u(0) = 0 = maxjj u.
In fact, at this stage it is sufficient to contemplate [cf. Exercise 2.41] u(x;
-rte/p cost ,
E[a,iv],
-2<0<2. 06
Throughout this appendix we pursue mainly first and second outward derivatives at an edge point; in the present case we merely calculate limits, as r 10 with 0 fixed in (-a/2, a/2), of -Ur(x;
Iro
7r
COS
(x E 92),
Or - F')r(rz-2#)l# (x E 0). cos Q r2 It is instructive to list four cases. (i) $ = it, u(x; 7t) = -xi. For each a E (0, 7t) and all 0 E (-a/2, a/2), urr(x;N) _
as r 10 with 0 fixed.
-ur(x; 7t) --a cos 0 > 0
The bluntness function is irrelevant.
(ii) a < fi < 7t and $ > it/2. For each a E (0, it) and all 0 E (-a/2, a/2), -Ur(0; $) = 0,
urr(x; $) -+ -oo
as r J, 0 with 0 fixed.
The bluntness function has a little significance: if a = fl, then B(0) > 0.
(iii) a < $ = n/2, so that B(0) < 0; u(x; 7t/2) = -xi + x22. For each a E (0, n/2] and all 0 E(-a/2, a/2), -ur(0; it/2) = 0, urr(x; 7t/2) -* -2 cos 20 < 0
as r 10 with 0 fixed.
E.2 Bluntness and ellipticity under co-ordinate transformations 309
(iv) a < /I < n/2, so that B(0) < 0. For each a E (0,7r/2) and all 0 E (-Lx/2,a/2),
-ur(O;Q) = 0,
urr(0;Q) = 0.
In §E.3, Theorems E.8 and E.9 will exclude the analogue of (iv), in Example E.5, by means of the bluntness condition (E.3) in Theorem E.8 and (E.4) in Theorem E.9. Both theorems establish that, under these bluntness conditions, first and second outward derivatives of a subsolution u, at an edge point where supra u is attained, must be essentially as in (i) or (ii) or (iii) of Example E.5. Second outward derivatives at an edge point, of a sufficiently smooth
function u
:
S2 -> R, result from setting v = 8u/8m in the following
definition.
Definition E.6 Let p be an edge point as in Definition E.1. A constant vector m is outward from 52 at p if it is outward from 521 and from 5211
at p : that is, m ni(p) > 0 and m n"(p) > 0. In that case, 8/8m := m - V at points x E 52 near p, while 8v
8m
(p) := limo
v(p) - v(p - tm) t
whenever this limit exists [cf. (2.21)].
E.2 Bluntness and ellipticity under co-ordinate transformations
The proofs of Theorems E.8 and E.9 depend on the conservation of various conditions under co-ordinate transformations more general than the translations and rotations of axes considered previously. By a Cc co-ordinate transformation or C' diffeomorphism, where t' E {1,2,3 .2,3,.. .},
we mean a bijection f of an open set U in RN onto an open set f (U) in RN such that f E C"(U, ][8N) and f-' E Ce(f (U), RN). In the present context, U will always be bounded. It is often convenient to use the archaic notation y = f(x),
axr := (0ifr)(x),
ay' := (asf; 1) (y),
(E.5)
where x E U, y E f (U) and f 1 := (f -1)j. We specify the following transformation rules. (a) Scalar-valued functions transform in the obvious way: SP(Y) := P(f-1(Y)) = cp(x).
(E.6)
Appendix E. The Edge-Point Lemma
310
(b) Any vector field k used to define a directional derivative a/ak :=
(for example, the i tangential to rill n afii in (E.4), or the m outward from f) in Definition E.6) transforms contravariantly; so does the matrix (a;j) of leading coefficients of an elliptic operator L. This means that, for r,s E {1,..., N}, N
aYr
kr(Y) i=1
N
ars(Y) : =
aYr
i,j=1
x=f-1(Y),
ki(x),
axi
ax, aij(x)
ay"
= f-1(Y).
axJ,
(E.7a)
E.7b)
(c) Test vectors l; for the ellipticity condition [Ei j a;j(x)l;ii;j >_ transform covariantly. This means that, for r e { 1, ... , N}, N
br :=
axj
T, aYr i.
(E.8)
All these transformations are invertible. For example, multiplying (E.7a) by axj/ayr and summing over r, we obtain N
kJ(x ) _ -
axI r-1 ayr kr(Y)
The reason for transforming (a;j) as in (E.7b) is essentially the same as in Exercise 2.2: contravariance implies that a; j(x)
a
a
ax; axj
= E ars(Y) r ,s
a
a
aYr ays
+
a lower-order operator.
Remark E.7 Definitions (E.5) to (E.8) and the chain rule imply the following invariance or conservation results. It is to be understood that y = f (x) and x = f -1(y) throughout the list.
(i) Directional derivatives are invariant: with a/ak := k V, and
a/ak := k V, 00 ak (Y)
A
(x).
(E.9)
(ii) m is outward from f (S21) at f (p) if and only if m is outward from S21 at p. [In other words, if cpI is an admissible function describing S21
near p, then am
which is immediate from (E.9).]
a
a
M (P) > 0,
am
E.3 Two stages of the edge-point lemma
311
(iii) Let p be an edge point; say p E E := 00, n afIll. Then i is tangential to f (E) at f (p) if and only if r is tangential to E at p. [In other words, (8ipj/ai)(f(p)) = 0 for J = 1,11 if and only if (acpj/&r)(p) = 0 for J = I, II, which is immediate from (E.9).] (iv) Bluntness functions are invariant: B(y) := (V 1)(y) - a(y) - (Doll) (y) = (V(p1)(x) - a(x) - (V pu)(x) =: B(x). (E.10)
(v) Uniform ellipticity is conserved when U is bounded, although moduli of ellipticity are changed in general. Let o be the uniform modulus of ellipticity of the operator having leading coefficients aij(x), x E U. Then, for all Z E ][8N,
rs(y)rs =
affi(x)
20
M :=
20
IA(y)jj
if we define N
N
A(y)e :_ ((i3rf)(Y)i) i=1
r=1
for
E.3 Two stages of the edge-point lemma In the following theorems, the operators Lo, L and L1 are as in Definition 2.3 except for the greater smoothness of the coefficients aid that is stated in the theorems. The words C2-subsolution and generalized subsolution
retain the meanings in Definitions 2.4 and 2.10, respectively. The set C1 (SZ U {p}) consists of those functions f E C (f2 U {p}) n C' (Q) for which all first derivatives ajf have extensions continuous on S2 U W.
Theorem E.8 Suppose that (a) p is an edge point of Q (the relevant edge being BS, n 8f2n);
(b) u is a C2-subsolution relative to Lo or L and Q, or a generalized subsolution relative to L, and f2;
(c) u E C'(Q U {p}) and u(x) < u(p) for all x E f2, with u(p) > 0 when the coefficient c is not the zero function;
Appendix E. The Edge-Point Lemma
312
DL21I
Fig. E.2.
(d) a bluntness function B, for the edge 8521n85211 relative to the operator
considered in (b), satisfies B(p) > 0; (e) all coefficients a1 are continuous at p.
Let m be a unit vector outward from 52 at p. Then either (a) am(p) > 0; or (fi)
un
(p)
0 and lim supt10 tam (p - tm) > 0,
which implies that (82u/8m2)(p) < 0 whenever this derivative exists.
Proof (i) Assume that we can find a set H and a function v as follows. The set H (Figure E.2) is a bounded open subset of fl such that H c 52 U {p}
and p is an edge point of H; moreover, the representation HI n HII of H near p is such that the outward unit normals at p to 8H1 and 8H11 coincide with those to 8521 and 85211, respectively.
The function v E C2(H) and v(p) = 0;
8v
8m
(I)
2
( p)
_0
and
8m2 ( p )
>0
(Ila ,b)
for every vector m outward from H (equally, from 52) at p;
u+v <M on
H,
(III)
where M := u(p) = supn u.
With such H and v in hand, we can prove the theorem as follows. Let w := u + v and let a vector m as in (II) be given. By (I) and (III),
E.3 Two stages of the edge-point lemma
313
w(p) = M = SUPH w; hence (aw/am)(p) > 0 [otherwise, w(x) > M at certain x in H]. Consequently, au
av am(P)?-am(P)=0.
If (au/am)(p) > 0, then the theorem is true. Suppose therefore that (au/am)(p) = 0, and consider the difference quotient Q(t)
t
am (P)
- am (p - tm)))) _ -tam (P - tm)
(t > 0).
There must exist a sequence (tn) such that to 10 and Q(tn) < 0, because
0 < w(p) - w(p - tm) = J c dsw(p - sm) ds =
f,
am (p - sm) ds,
and this is contradicted if there is a number e > 0 such that Q(s) > 0 [hence (aw/am)(p - sm) < 0] for all s E (0, e). Now consider
P(t)
t {
am(P) - am(p - tm)
t
am(p - tm)
(t > 0).
Since lim,1o P(t) > 0, by (IIb), we have P(tn) >- c for some constant c > 0 whenever to is sufficiently small. For all such tn,
I au am(P - tnm) = NO - Q(tn) > c,
n
and this proves the theorem for case (Q). (ii) The function v will be of the form 6V, where S is a small, positive constant to be chosen presently, while V E C2(H) and satisfies (Ila,b); in addition,
V =O on aH\F,
(I')
where IF is a compact subset of OH that is disjoint from {p}, and
LV > 0 in H,
(III')
where L denotes whichever of LO, L and Ll is considered in hypothesis (b).
Consider the values of u+bV on OH. On aH\F we have u< M, V = 0 and hence u + b V < M, with equality at p. On F, which is a subset of 0, we have u < M by hypothesis (c); since F is compact and u continuous, u < M - a on F for some constant a > 0. Choose b = Za/ maxr V ; then
u+6V<Mon F, so that maxaH (u + 6 V) = M.
Appendix E. The Edge-Point Lemma
314
Condition (III) now follows from (III') and one of our three versions of the weak maximum principle, in precisely the way that was spelled out in the proof of Lemma 2.12. (iii) In order to construct the set H and the function V, we simplify the geometry by means of co-ordinate transformations. First, let Xj be local co-ordinates of the usual kind (resulting from a translation and rotation of axes as in Exercise 5.5) such that (a) X = 0 at x = p; n'(p) + n"(p) (b) oxXN = InI(p) + nII(p)I
(c) the (two-dimensional) plane spanned by VxX1 and VxXN is that spanned by n'(p) and n"(p). Conditions (E.1) and (b) imply that n'(p) oxXN > 0 for J = I, II; then, by Exercise 5.5, 0921 and 75211 have local representations XN = gi(X')
(for J = I, 11) such that XN < gj(X') when x is near p and x E f1j. Moreover, conditions (a) to (c) and the C2 nature of the boundaries ffij (Definition E.1) allow us to write gi(X') = yXi + r1(X'), rI E CZ(S), gII(X') _ -yX1 +rII(X'), rII E C2(S),
(E.12)
where the constant y > 0; the set S is a convex neighbourhood of 0 in RN-1; and, as X' -+ 0, rj(X') = O(IX'12) and Vrj(X') = O(IX'I) for
J=I,II. The next co-ordinate transformation is YI
= XI + I {rI(X') - rII(X')},
yA
= XA
Y
for each A E {2,._, N -
11,
(E.13)
YN = XN - 2 { r1(X') + rII(X') }. If we write this as y = f(X), then the Freshet or total or linear derivative
f'(0) = I, the identity operator RN
RN.
Thus f'(0) is certainly
invertible (is certainly a linear homeomorphism) and the inverse-function theorem ensures that f is a C2 diffeomorphism on some neighbourhood of 0 in RN. [For a form of the inverse-function theorem that yields the C2 property of f -1, see Dieudonne 1969, p.273.] Define open half-spaces 61
{yERN I YN
921I
{ y E RN I YN < -YY1 }
(E.14) ,
E.3 Two stages of the edge-point lemma
315
aG11
aG1
Fig. E.3.
and set C2 := S21 n 0211. Then, for sufficiently small Ix - pI and IyI, we have
x E Q if and only if y E C2, because (E.13) transforms XN - $1(X') to YN - yyi and transforms XN - 811(X') to yN + yy1.
(iv) Now choose at pleasure a constant K > 0 (which will be a curvature), let GI G11
y E RN I YN < yYi -
K2IY'I2
y E RN I yN < -yY1 -
K2IY1I2
set G := GI n G11 (Figure E.3) and define
H :=GnR(0,p)
(E.15)
for some small p > 0 that remains to be chosen. The inverse image of k under the C2 diffeomorphism x -+ y is to be the set H specified in step (i).
(v) To construct the comparison function V, we first define (p(Y)
YN - yYl +
w(Y) := YN + yY1 +
K2IY,I2,
P(y)
8 KW(Y) - 1,
(E.16)
K2IY I
,
Q(v) := e-KW(Y) -
1,
for all y E RN and for some large K > 0 that also remains to be chosen. Observe that cp and -P are admissible functions describing GI (Definition E.2) and that ip and -Q are admissible functions describing GII; in particular, P(y) > 0 and Q(y) > 0 if y E G, and P, Q vanish on 8G1i aG11 respectively.
Appendix E. The Edge-Point Lemma
316
Now define V E C00(RN) by V := PQ, and let y = Y(x) denote the C2 diffeomorphism that we have constructed on a neighbourhood of p. Denoting this neighbourhood by #(p), we set V(x) := V (Y (x)),
x E ./#(P).
(E.17)
Then V E C2(H) if H c .gy(p), and we proceed to show that V satisfies (I'), (Ila,b) and (III') if p and K are chosen suitably. This will complete the proof. (vi) Condition (I') holds if we define IF to be the inverse image of G n 89(0, p) under the map Y [because PQ vanishes on 0G]. For (Ila,b), let a unit vector m, outward from H at p, be given; define, for r E {1,..., N}, N
mr(Y) i=1
a
a Y,
axi (x) mi,
am
N
= E mr(Y) r=1
a ay'.
Then (aV/am)(p) = (aV/ain)(0) and m(0) is outward from if at 0, by Remark E.7, (i) and (ii). Now VV = (VP)Q+P(VQ) and P(0) = Q(0) = 0, so that certainly (aV/am)(p) = (aV/am)(0) = 0. Next, using once more the condition T(O) W(0) = P(0) = Q(0) = 0, we obtain am
(p) =
a z01V (0)
8P
ainn
am
=2
(0)
8Q amp aye (0) = 2K2 (0) (0) > 0, am am am
since m is outward from GI and from GII at 0.
(vii) It remains to prove that V satisfies (III') if p and K are chosen suitably; equivalently, that LV > 0 in fl, where L is the operator to which L is transformed (§E.2) by the map y = Y(x). Let air(y), ba(y) and c(y) denote the coefficients of L. Then
LV = (LP)Q + 2VP a VQ + P(LQ) - cPQ.
(E.18)
Since P > 0 and Q > 0 in ft, the essentially new term relative to Lemma 2.12 is
2VP where
BG, say, is a bluntness function for the edge 0GInaGII
relative to L. By hypothesis (d) and the two kinds of invariance of the condition B(p) > 0 [Remark E.4 and (E.10)], we have BG(0) > 0, and BG is continuous at 0 by hypothesis (e) and the definitions of (p and W. Therefore, if p is chosen sufficiently small in the definition (E.15) of ft, then
in ft.
E.3 Two stages of the edge-point lemma
317
Consider the remaining terms in (E.18). We have -cPQ >- 0 in H, and
make LP > 0 and LQ > 0 in fl by choosing K sufficiently large, very much as in the proof of Lemma 2.12. In fact, LP
=
e-Kw
K2 E aid (a,cp)(ai(,)
-K
aii ajaiq ii
i,i
-K
- c,
bi 8icp +
(E.19)
where -c >- 0 and
=, = (-Y + KYI, KY2, ..., KYN-1, 1),
(aj(p(Y))N
a,aj(o(Y)
K
if i=jE{1,...,N-1},
0
otherwise.
(E.20a)
(E.20b)
Since uniform ellipticity, say with modulus L > 0 [Remark E.7, (v)], implies that a,j (0,(p)(ai(P) >- oIV(P12 > 1o
in fl,
(E.21)
we can certainly choose K so large that LP > 0 in ft, and similarly for Q. Theorem E.9 Suppose that hypotheses (a) to (c) of Theorem E.8 stand,
while (d) and (e) are replaced by (d') a bluntness function B, for the edge 8521 n a011 relative to the op-
erator considered in (b), satisfies B(p) = 0 and (8B/ar)(p) = 0 for every differential operator 0/0,r tangential to 8521 n 001, at p; (e') all coefficients aij E C2 (S2 n 1(p, a)) for some a > 0. Then the conclusion of Theorem E.8 remains true : we have either (a) or (fl) there.
Proof Steps (i) to (iii) in the proof of Theorem E.8 are not changed. However, in order to have LV >- 0 in H, we introduce here: (a) a further co-ordinate transformation, which will simplify the matrix a(y) of leading coefficients; (b) a further parameter, in the definitions of H and V, such that LV >- 0 in H when the new parameter is sufficiently large. (iv') Proceeding from the co-ordinates y defined by (E. 13) and the set C2 defined by (E. 14) and S2 := C21nC211, we transform linearly to co-ordinates
Appendix E. The Edge-Point Lemma
318
z, obtaining a set A and coefficients ai j(z), b j(z), c(z) such that 15 = C2 and
aAN(0) = O for A E {2,...,N -1}.
Notation In this proof, Greek subscripts take only the values 2,...,N-1; `for each 2' will mean for each A E {2,...,N - 1}; and E. := EN 2
Define, for constants CA that we are about to determine, Z1 = yt,
ZA = YA + CA YN for each A,
then aZA aYyi
= h + CA 6M,
a
ZN = YN;
(E.22)
aZN
= SNj,
Yj
so that 8ZA
aAN(Z) _ i,j
ayi
aij(Y)
-?N
ayj
= aAN(Y) + CA aNN(Y)
Here aNN(O) >- .10 > 0 by the uniform ellipticity of L with modulus .1o and
with test vector (0,...,0,1); therefore we choose CA = -aAN(0)/aNN(0). The transformation (E.22) is invertible [since yA = zA - CAZN for each A]; its only effect is that aAN(0) = 0
for each 2;
(E.23)
by (E.14) and (E.22), z E S2 := C2 if and only if y E C2. Now define a new set W := WI n W11 (Figure E.4) by WI
:=
{zEI''
ZN < YZl - 2 Zl - 2 E ZA j A
W11
:=
{ZERN
ZN < -yZ1 - 2Z1 - 2
} ZA
, >
1
where K > 0 and A > K/y. Any K > 0 will serve, as before, but the main choice of the large parameter A has yet to be made. Next, let
k := W n 9(0, p)
(E.24)
for some small p > 0 that also remains to be chosen. The inverse image of H under the C2 diffeomorphism x --* z is to be the set H specified in step (i).
E.3 Two stages of the edge-point lemma ZN
319
ZN
edge of S2
SID
edge F of W
Fig. E.4.
(v') To construct the comparison function V, we first define [cf. (E.16)]
f(z) := ZN - Yz1 + 2zI + 2
z22,
g(z) = ZN + YZ1 + 2 Zi + 2 E zA,
(F(z) := e-/{f(z) - 1
T(z)
e-Kg(z)
- 1, (E.25)
for all z E RN and for some large K > 0 that will be chosen after A has been fixed. We note that f and -4) are admissible functions describing WI, while g and -'F are admissible functions describing W11. Now define V E C°°(RN) by V := (F'F, and let z = Z(x) denote the C2 diffeomorphism that we have constructed on a neighbourhood, say ,*'(p), of p. We set V(x) := V (Z(x)),
X E .N(p);
(E.26)
then V E C2(H) if H c .K(p).
(vi') The argument in step (vi) of the earlier proof shows that the function V in (E.26) satisfies conditions (I') and (Ita,b) [stated in steps (ii) and (i), respectively]. Condition (III'), or equivalently that LV > 0 in H, requires further labour. Here L denotes whichever of Lo, L and L1 is considered in hypothesis (b); L is the operator to which L is transformed by the map z = Z(x); and the coefficients of L are aij(z), bi(z) and c(z). Accordingly,
LP = (L(D)'' + 2v(D . a . V'Y + 4)(L') - i^(F'P,
(E.27)
and the troublesome term is 2e -Kf-Kg K
where
Bw := V f a Vg on H.
1
(E.28)
Appendix E. The Edge-Point Lemma
320
The bluntness hypothesis (d') concerns not the edge of W and the function Bw, but [in view of Remark E.4 and the transformation law (E.10)] the edge of 6 and the function
(-y,0,...,0,1)'a(z)'(Y,O,...,0,1)
BA(z)
_
-Y2a11(Z) + aNN(Z)
(z E S2,
IzI
small),
in which the alN-terms have cancelled each other. Thus hypothesis (d') implies that and
-Y2all(0) + aNN(O) = 0, -y2 (&Aal1) (0) + (BAaNN) (0) = 0
for each A.
(vii') We begin with the restriction of Bw to the edge F := a Wl n a Wit of W ; more precisely, to F n.4(0, p* ), the radius p* begin so small that (E.30)
A0, p*) - Z (gy(p) n °M(p, a)),
where X(p) is the domain of Z and a is as in hypothesis (e'). The edge F of W has a representation 2A IC 12),
Z = ZF(O := (0,C,-
RN-2
C :_ (b2,...,CN-1) E
Let aij(C) := aiJ (ZF(c)) and let
BF(C) := Bw (zF(O) for 1C1<_
A
< 2 p*;
(E.31)
this last condition ensures that IzF(0I < p*. We shall see that hypothesis (d') and uniform ellipticity make BF(C) > 0 if A is sufficiently large. First, because
ali-terms cancel for j
1,
BF(I) = A2 E aAµ(0 CACµ + 2A A,µ
aANG) CA - y2al l(C) + aNN(). A
Second, BF(I) is of the second order in C: by (E.23) and (E.29),
BF(b) = A2EaAµ(C)CA(µ+2AE{cANG)-aAN(0)}CA A,µ
-Y2
A
{l1_l1o_ DaAall)(0) A
+ { aNN(b) - aNN(0) -
l CA
1 (aAaNN)(0) CA
1
E.3 Two stages
of.
the edge-point lemma
321
By the uniform ellipticity of L, say with modulus 2o, and the C2 smoothness of the a,j, we have 2
1
BF(t')> (A2.o - const.A - const.) 112
ICI-<
where the constants are independent of t; and A. We choose and fix A so large that
A>
KI
<
p*
and
BF(t;) > 0 for ICI <- A.
(E.32)
A
(viii') In order to estimate Bw effectively in fl, define
dF(z) := dist(z,F),
K 2A
2
2 {f (z) + 9(z)} _ -zN - 2zi - 2 E ZA,
h(z)
A
so that h(z) > 0 in W and h(z) = 0 on F; we establish an inequality
0 < ci <
dF(Z)
< Ci
for z c= W n 4(0,1/A),
(E.33)
in which cl and Cl are independent of z. First, consider points in WO := W n { z zl = 0 }. Both dF and h are continuous and positive in WO; so, therefore, is dF/h. Moreover, if I
z E Woandz -+zoEF,then h(z) = IVh(z)I dF(z) + O(dF(z)2), whence
h(zl)
1 +A 2
zOA A
Thus dF/h, extended to Wo by these limiting values, is continuous and positive on WO, hence is bounded between positive constants on any compact subset of Wo. In particular, there are constants co and Co such that 0 < co < dF(0, z") < Co
h(0 z") -
for (0, z") E W n -4(0,1/A);
here z" := (z2, ... , zN ), so that (0, z") E WO. In addition, we have A
Y Izi I < -zN - 2 / ZA 2 = h(0, z")
for all z E W,
(E.34)
Appendix E. The Edge-Point Lemma
322
from which it follows that, for z E W n -4(0,1/A), h
1 < h(z))
-
(I_ 2Ay) 1
1<
d
z) ) <
C
(E.35)
) 1/2 (E.36)
1 + 22 Y CO
where K/2Ay < Z by (E.32). The inequalities (E.34), (E.35) and (E.36) imply (E.33).
(ix') We now estimate Bw in H := W n.(0, p). If p is sufficiently small,
then to each point z e H there corresponds a point zo E F such that Iz - zol = dF(z) and zo E R(O,1/A). In fact, the line segment from z to zo is then contained in _4(0,1/A). A calculation shows that p < 1/2A is quite small enough for this condition; we adopt this bound for definiteness. Then p < p* /.,15-, where p* is as in (E.30), so that Bw E C2 (x). Since Bw(zo) > 0 when zo E FnR(0,1/A), by (E.32), there is a constant C2 > 0 such that
Bw(z) > Bw(z) - Bw(zo) > -C2IZ - Zol = -C2dF(z) for all z E ff. By the inequality (E.33),
Bw(z) >- -C1C2h(z) = 1CiC2{f(z)+g(z)}
for all z E ft,
(E.37)
provided that p< 1/2A. (x') It remains to choose K, and to make a final choice of p, in order that LP > 0 in f7. First we observe [cf. (E.19) to (E.21)] that
L) > e-Kf
ail alai f - K
K 2 .o I Vf 12 - K i,i
>-
ZeKf K2 ,o
bi ai f + c
i
in fi
if we make a first, sufficiently large choice of K and if p < 1/2A. The function LT is treated similarly. The formula (E.27) for LV now yields
LP > Ze
Kg-1}+2e Kf-KgK2Bw
+1e-KQK2 o { eKf - 1 }
= ZeKf-KgK2 o l 2 - eKf -eKg + (4/),o)Bw 1 > ZeKf-KgK2 o { 2 -eKf -eKg + C3f + C3g } , (E.38)
E.3 Two stages of the edge-point lemma
323
where C3 := (2/.10) Ci C2 and we have used (E.37). The result (E.38) holds in ft if p < 1/2A. Now define
q(t):=1-eKt-C3t
for
0
whence q(O) = 0
and
q(t) = KeKt - C3 >-
Ke-i - C3.
If necessary, increase K beyond our choice for L' and L'Y so that K >- C3e; then q(t) > 0 for 0 < t < 1/K. If necessary, decrease p below
1/2A so that -f(z) < 1/K and -g(z) < 1/K when z E ff. Then (E.38) implies that, in fl,
LV > 2i e Kr-Kg K2 o {q(-f) + q(-g)} > 0, as desired.
Notes on Sources
I apologize for the inadequacy of these notes. They describe material that I used, rather than all the material that I should have used. The notes also point to further theorems that I happen to have encountered; they do not amount to a sketch of the present state of the subject. Chapter 1 It is worth mention that Aharonov, Schiffer & Zalcman (1981) obtained the conclusion of Theorem 1.1 from quite different hypotheses.
Their first result concerns a compact set P in RN which, inter alia, is connected and such that ][8N \ P is connected. They proved that, if in a (non-empty) open set outside P the Newtonian potential of constant density in P is equal to the potential of a point mass, then P is a ball. In using a flower to illustrate a symmetry group, I have followed Weyl (1952, p.66), who used vinca herbacea for the cyclic group C5 and used a geranium for the dihedral group D5. Weyl's book is full of charming examples; every page reflects the erudition and humanity of a master from an age less barbarous than ours. For problems that allow symmetry of solutions and that can be formulated as exercises in the calculus of variations, there is an approach to proofs of symmetry that is older than, and quite different from, the methods set out in this book. This approach is based on various geometrical operations (devised by J. Steiner and H.A. Schwarz in the nineteenth century) called symmetrization with respect to a plane, a line or a point. Typically, such an operation assigns to a given set A c R3 (which may be the set below the graph of a function of two variables) a set A* that has a specified symmetry and has the same cross-sectional area as A for each of a family of cross-sections. In particular, symmetrization with respect to a hyperplane, also called Steiner symmetrization, yields sets of the form called Steiner symmetric in Definition 3.2. The books of Polya 324
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325
and Szego (1951), of Bandle (1980) and of Kawohl (1985) explain these procedures fully.
Chapter 2 In §§2.2 and 2.3 the weak maximum principle, the boundarypoint lemma and the strong maximum principle for C2-subsolutions are taken from Gilbarg & Trudinger's book (1983, Chapter 3); of course, with expanded proofs and some differences of detail. The boundary-point lemma in this form is due to E. Hopf (1952b) and to Oleinik (1952); the others are somewhat older. Theorem 2.11 (the weak maximum principle
for distributional subsolutions relative to the operator L1, which has constant coefficients) is a very particular case of maximum principles due to Littman (1959, 1963). The elementary proof of Theorem 2.11 seems (to me) to be new, but may well occur somewhere in the literature of the 1950s or 1960s; my colleague Buffoni suggested the use of (2.16) after I had used (2.17).
The maximum principle for thin sets (Theorem 2.19) is a very restricted version, although not quite a particular case, of theorems due to Berestycki & Nirenberg (1991, Proposition 1.1), to Dancer (1992, Lemma 1) and to Healey, Kielhofer & Stuart (1994, Theorem 3.2). It seems that, for theorems of this kind, the basic observation appeared first in work of Bakelman preparatory to publication of his book (1994) and in unpublished remarks of S.R.S. Varadhan. Theorem 2.19 is more naive than its counterparts elsewhere because the constant coefficients of the operator L10 make it easy to avoid derivatives of the subsolution (and thus to avoid Sobolev spaces). The Phragmen-Lindelof theory in §§2.4 and 2.6 is a mixture based on my reading of Protter & Weinberger (1984, §2.9) and of Stein & Weiss (1971, §11.5.2), but with different comparison functions. As is mentioned in §2.6, the primary comparison function g, described by Theorem 2.27, is an extension to all dimensions N >_ 2 of a function introduced by D. Siegel (1988) for N = 2. Recently, Siegel has abandoned this function in favour of a more elaborate one, defined on polytopes for all N >_ 2 (Siegel & Talvila 1996, Theorem 4.1), that allows larger rates of growth of a subharmonic function when lxi --* oo and xN/Ixi -> 0 in the halfspace { x E RN I XN > 0 }. I have not seen previously the comparison functions ge and $2, constructed from g in §2.6 by means of the Kelvin
transformation; the trio g, ge and $2 is intended to be a compromise between the simplest comparison functions and difficult ones that yield results close to best possible. Chapter 2 provides only an introduction to the subject of maximum
326
Notes on Sources
principles for elliptic equations. For example, information about positive solutions of Lu + f (u) = 0 in a bounded set S2 can be gleaned by means of P-functions; these are of form IVuI2g(u)+h(u) and, for suitable choices of g and h, attain their supremum either on On or at a stationary point of u. The construction and use of such functions are the main topics of the book by Sperb (1981). In another direction, Bakelman (1994) considers fully non-linear equations, such as those of Monge-Ampere type, and derives for such equations maximum principles (Chapter 8 of his book) that make those in our Chapter 2 look like toys.
Chapter 3 For boundaries 00 of class C2, monotonicity and symmetry results like Theorem 3.3 and its corollaries, but for more general elliptic equations, were proved by Gidas, Ni & Nirenberg (1979), building on work of Alexandrov (see H. Hopf 1956, p.147) and of Serrin (1971). A maximum principle for narrow sets ), and use of the compact set called F in the proof of Theorem 3.3, were introduced into problems of this kind by Berestycki & Nirenberg (1991); thereby the need for smoothness of On was removed and proofs were shortened. Parallel steps were taken by Dancer (1992), who made more use of Sobolev spaces and proved monotonicity results like those in Theorem 3.3 under hypotheses on u that are weaker than those of Berestycki & Nirenberg (and considerably weaker than those in Theorem 3.3). Theorem 3.6 and Corollary 3.9 have their roots in the paper of Gidas, Ni & Nirenberg (1979), in the work of Berestycki & Nirenberg (1991) and in an extension of two results of the 1979 paper to the case of discontinuous f (Amick & Fraenkel 1986). The combination of discontinuous f and a set ) that is merely Steiner symmetric seems to be new. Exercises 3.18 and 3.19 (which show that one cannot relax significantly
the conditions on f used in most of the book) extend a construction of Gidas, Ni & Nirenberg (1979, p.220) in a way demonstrated for ordinary differential equations by work of C.A. Stuart (1976) and by unpublished examples of J.F. Toland. Gidas, Ni & Nirenberg (1979, p.220) raised the question of whether results like those of Theorem 3.3 remain true if the hypothesis u > 0 in
n is changed to u >- 0 in n with u > 0 somewhere in n. They observed that for N = 1 (for Q c ll8) the answer is No because u(x) = 1 + cos xx
u" + m2(u - 1) = 0 in (-3,3),
u(±3) = 0,
and u' has the wrong sign in (1,2); they proved that the answer is Yes for all N E N if f(0) >_ 0. Castro & Shivaji (1989) have shown that for
Notes on Sources
327
f (0) < 0 the wider hypothesis is allowable in Theorem 3.3 if c2 is a ball and N > 2. Note that extensions of the functions u in Exercises 3.10 and 3.11 are consistent with all this. Allegretto & Siegel (1995) have considered the problem Au + A p(x) g(u) = 0
in K2,
u=0,
where f2 := (-1, 1) x n' is a cylinder, S2' being open and bounded in RN-1, and p(-x1, x") = p(x1, x"). Using Sobolev-space methods, a family of non-negative eigenfunctions defined on caps of fl, and corresponding hypotheses on p and g, they avoid pointwise estimates near 092 and prove symmetry under hypotheses on u slightly weaker than those of Berestycki & Nirenberg (1991).
Chapter 4 Regarding the behaviour of u(x) as r := IxI --f oo: by now there are many different sets of conditions that lead to symmetry for solutions u of suitable equations in RN. Gidas, Ni & Nirenberg (1979, Theorem 4) considered positive solutions of rather general equations and introduced hypothesis (C) of Definition 4.1, but with three explicit terms before a remainder. Caffarelli & Friedman (1980, Theorem 7.1) observed for the equation Au + f (u) = 0 in R2 that positivity of u could be abandoned, that logarithmic growth at infinity could be tolerated and that the asymptotic approximation could be shortened. Definition 4.1, and the sufficiency of its conditions for Theorem 4.2, evolved from these
two sources in the course of my lectures at Bath, given in the years 1989-1992.
The approach of Gidas, Ni & Nirenberg in their 1981 paper is quite different: for positive C2-solutions of Au + f (u) = 0 in IRN, with N z 2 in some cases and N >- 3 in others, f is specified rather precisely but very little is demanded of u as r -+ oo; for example, merely that u(x) = O(r-') with m > 0. More detailed asymptotic behaviour is then deduced. Li & Ni (1993) have taken this further, even for fully non-linear equations, by demanding only that u(x) -> 0 as r -> oo; no rate of decay is prescribed.
Of course, the function f or its fully non-linear counterpart must be specified carefully. The reader who is bewildered but attracted by remarks in the text about
Sobolev spaces and variational principles (for example, after Theorem 4.13 and in Exercises 4.22, 5.29 and 5.30) will find lucid and satisfying explanations in the books by Berger (1977), by Brezis (1983) and by Friedman (1982).
Notes on Sources
328
Regarding steady vortex motions. Turkington (1983), considering various steady vortex flows in two dimensions, used reflection in hyperplanes
(in this case reflection in lines) to prove that intense vortices of small diameter are approximately circular and, suitably scaled, become exactly circular in a certain limit. Concerning Hill's vortex and Hicks's vortex: the solutions of Exercises 4.22 and 4.23 can be found, to a large extent, in papers by Amick & Fraenkel (1986) and by Fraenkel (1992). Uniqueness of the vortex pair in R2 that corresponds to Hill's vortex has been proved by Burton (1996).
Two conspicuous omissions from this book are (a) overdetermined problems, by which we mean here that two boundary conditions ul an
= constant
and
8u
an
= constant an
accompany an elliptic equation of order two; (b) problems on an exterior domain S2, by which we mean a connected set in RN the complement
of which is compact. The first of these omissions is unfortunate in that overdetermined problems in a bounded set 0 were the subject of Serrin's early and fruitful use of reflection in hyperplanes (Serrin 1971); the conclusion was that 92 must be a ball and the solution u spherically symmetric and monotonic. Reichel (1996, 1997) has considered overdetermined problems on ex-
terior domains for the equation Au + f (u, IVul) = 0 and for certain fully non-linear equations, recovering Serrin's conclusions under suitable
hypotheses on f (u, JVuj) or on its fully non-linear counterpart. These hypotheses, however, exclude the case f (u, JVul) = fo(u) with fo positive and increasing on (0, oo). This defect has been made good by Aftalion & Busca (1998), who combined skilful use of the Kelvin transformation with a maximum principle for thin sets.
Chapter 5 Most authors seem to regard results like Theorem 5.7 (two pleasant properties of boundaries of class C1) and Proposition 5.20 (resolution of a topological question) as obvious; however, colleagues attending my lectures at Bath insisted on proof of such matters. Theorem 5.7 is taken from Amick & Fraenkel (1986, Lemma A.1) and from Keady
& Kloeden (1987, Lemma 2.1); Proposition 5.20 appears to be new. Rhomboid neighbourhoods (Definition 5.4) have a long history in the theory of Sobolev spaces, although they are seldom given a name; their treatment in Chapter 5 follows that of Fraenkel (1979). References to their earlier use are given in that paper.
Notes on Sources
329
The monotonicity results in §5.3 are due to Gidas, Ni & Nirenberg (1979). In our version, somewhat more attention is given to the components of a cap, and less smoothness is required of 852, f and u. On the other hand, our equations (5.1) and (5.2) are far less general than those considered by Gidas, Ni & Nirenberg. The result in Exercise 5.30, that symmetry results for a ball do not extend to an annulus or spherical shell A, is not new; examples of positive solutions in A that lack spherical symmetry have been given by Brezis & Nirenberg (1983, p.453) and by Coffman (1984). Possibly the example in Exercise 5.30 is simpler than previous ones; it is a modified form of an example suggested to me by G. Keady. The annulus or spherical shell A has also been used by Dancer (1997)
to prove that positive solutions u E C2(S2), of Au + f(u) = 0 in 52 with u = 0 on 852, can have stationary points arbitrarily close to 052. Dancer used 52 = A and a family of functions f = f (., e) to construct spherically symmetric, positive solutions u(. , e) having a stationary point
that approaches the inner part of aA as a --* 0. Thereby he answered a question of Gidas, Ni & Nirenberg (1979, p.223), who had gathered evidence for the opposite conclusion.
Keady & Kloeden (1987) applied monotonicity results like those in Theorem 5.10 to discuss solutions of
Au + 2fH(u - k) = 0 in a bounded set 52 c ><82,
ulan = 0,
where 2 and k are positive constants and fH denotes the Heaviside function. They showed that, for convex 52 with smooth boundary, the solutions resemble those when n is a ball, provided that the set C := u(x) > k }, which they called the core, is connected. In { x E 52 particular, there is a branch of solutions along which diam C -> 0 and I
oo. (When Q is a ball, all solutions are known explicitly; see Exercises 3.15 and 3.16, in which the same equation occurs and 52 is a ball in I[8 or 2
in 1185.)
Two important techniques, beyond those in Chapters 3 to 5, were introduced by Berestycki & Nirenberg (1988, 1990) and used to prove monotonicity, symmetry or anti-symmetry in a variety of problems with boundary conditions more general than ul an = 0. These techniques were also simplified and extended in their 1991 paper. First, equations more general than Du + f (u) = 0 may lead to Lw + b(x, u) Vw + c(x, µ)w - 2 /3(x, µ) aW ->O for x E Z (µ), µ
(j')
Notes on Sources
330
in place of our (3.6); here l3(x,µ) > 0 and the identity 8x1
u(2µ - xI, x") = 1 2
a
aµ
{ u(x) - u(2µ - xi, x")} =
1 8w 2 aµ
(x µ)
has been used. Now, if µ is regarded as a time-like variable, then (t) is an inequality for a parabolic partial differential operator (although a degenerate one wherever /3(x,µ) = 0); somewhat surprisingly, boundary conditions that suit such an inequality also arise in the (basically elliptic) problems considered by Berestycki & Nirenberg. Thus maximum principles for parabolic operators play an important part. Second, the method of sliding domains was introduced and used to good effect. In this method, one compares the graph of a function not with a reflection of that graph, but with a translation of it, defining, for example, 92T:={xE11RN
w(x, T) := u(x) - u(x - Tel)
I x-Te'ES2}, for T > 0 and x E 12 n f'T.
When the boundary conditions suggest it, monotonicity in the whole of S2 is often established by an application of maximum principles that is not unlike that used for reflection and symmetry. In the same direction and exhibiting further developments, the work of Craig & Sternberg (1988, 1992) established the symmetry of certain
gravity waves in hydrodynamics; in particular, the symmetry of the classical solitary wave. This is of interest because, when surface tension is included, there is no shortage of asymmetric solitary waves. Appendix A Little need be said about this material except to acknowledge that I first learned the main results, some thirty years ago, from the book of Gunter (1967). However, the resemblance of Appendix A to Ginter's
book now seems rather faint; for example, in the context of Theorems A.15 and A.16, Gunter (1967, p.308) was willing to accept hIµlog(1/1hl) in place of the jhV L for which we work so hard. Exercise A.25 is a fragment of generalized axially symmetric potential theory (Weinstein 1953).
Appendix B In the proof of Theorem B.6, the choice of test function is that in Sobolev's monograph (1963, p.92). Theorem B.9 is taken from Gilbarg & Trudinger (1983, p.23). Theorems B.29 and B.30, on the Poisson integral for a ball, should really be accompanied by a description
Notes on Sources
331
of how boundary values are attained when the boundary-value function g is merely in Lp(8B); Folland (1976, p.124) demonstrates this beautifully. Of course, my hope is that Appendices A and B may prepare beginners
for the more concise and extensive material in, say, Dautray & Lions (1990), Folland (1976) and Gilbarg & Trudinger (1983).
Appendix E As was remarked in the Preface, the edge-point lemma (or
boundary-point lemma at a corner) is due to Serrin (1971) and was extended by Gidas, Ni & Nirenberg (1979, pp.214 and 237-243), who called it Lemma S. The version in Appendix E is essentially Lemma S, with slight changes intended to increase the resemblance to our boundarypoint lemma, and with a vastly expanded proof.
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Index
admissible asymptotic behaviour, 107, 122 admissible for direction k, 142 admissible function describing Q1, 306 affine, 28
affinely independent, 28 aligned with k, 111 almost everywhere, 11, 225 alpha cap, 142, 153 annulus, 163, 329 averaging functions in L,(IRN), 38 averaging kernel, 37 ball, 3 Bernoulli function, 134 Bessel function, 35, 36, 101, 137 beta cap, 151 bluntness conditions, 307 bluntness function, 306 boundary, 4 boundary of class Ck, 7 162 boundary of class
boundary-point lemma for balls, 50 boundary-point theorem for 52, 54 bounded continuous functions, 196 Buniakowsky's inequality, 13 C2-solution, 44, 190 C2-subsolution, 41 C2-supersolution, 43 cap, 88, 142 carissa grandii lora, 28 Cartesian product, 2 centre of charge, 202 centre of mass, 202 characteristic function, 4, 122 closure, 4 co-domain, 4 co-ordinate transformation, 309 compact, 7 compact support, 7
comparison function of first kind, 64 of second kind, 68 of Siegel type, 72-77 compatibility condition, 236 complex potential, 213 complex velocity, 213 component, 4 component of alpha cap, 159 conformal map, 79, 81, 233 connected, 4 conservation under co-ordinate transformation of bluntness, 311 of uniform ellipticity, 311 constants 1', 187 continuity of force field, 194 continuity of translation, 38 continuously differentiable functions, 6 contravariant, 310 convex, 8 Coulomb, 168 covariant, 310 cyclic group, 30 cylindrical co-ordinates, 173 cylindrical symmetry, 199 cylindrically symmetric, 173
decreasing, 2 diameter, 4 diffeomorphism, 309 dihedral group, 30 dipole, 172, 199, 202 Dirichlet problem for L, 44
for -0, 235-248
for -0 in a ball, 248-262 for -A in a half-space, 263, 267, 269 distance function, 3 distributional sense, 57 337
338
distributional solution, 46, 191 distributional subsolution, 46 distributional supersolution, 46 divergence of a vector field, 10 divergence theorem, 9, 279-292 domain, 4 downward jumps, 94 dyadic cube, 91, 99, 102 dyadic square, 99 edge, 286 edge point, 55, 287, 305 edge-point lemma, 305-323 eigenfunctions, 19, 45 electric field, 169 electric potential, 176 electrostatics, 168 elliptic disk, 33, 136, 162 elliptic operators, 39 LO, Land L1, 41 equator, 64, 72 essential supremum, 11 Euclidean space, 3 extended Minkowski inequality, 14 extended real-number system, 2 extension, 5 exterior, 3 exterior domain, 328 exterior function of Siegel type, 73 exterior-ball property, 54
factorial function, 33, 223 field lines, 199, 214 field point, 176 first divergence theorem, 282 function, 4 fundamental solution, 239 gamma function, 33 generalized sense, 94 generalized solution, 191 generalized subsolution, 45 generalized supersolution, 46 geometry of caps, 142 geometry of reflected caps, 142 gradient operator, 6 gravitation, 169 gravitational potential, 176 Green function, 241-248 for ball, 245 for half-space, 244-245 of Dirichlet problem, 241-244 of Neumann problem, 241-244, 265 symmetry property, 243 Green identity, 180 Green volume potential, 249, 255, 267 group isomorphism, 30
Index
growth condition, 65, 236-238
Holder conjugate, 12 Holder constant, 196 Holder continuity, 196, 203, 265 Holder continuous, 94 Holder exponent, 196 Holder's inequality, 13 half-space, 61 half-space in RN, 233 half-strip, 80 harmonic, 62, 224 harmonic polynomials, 62 heat equation, 40 Heaviside function, 6, 122 Hicks equation, 134 Hicks's vortex, 137 Hill's vortex, 137 Hill's vortex in a ball, 103 Hooke, 169 hydrodynamics, 133, 170, 216 hyperplane, 24 image, 4 implicit-function theorem, 146 increasing, 2 inequality for parabolic operator, 330 integrable, 8 integrals, 8 interior, 3 interior approximations to boundary 852, 293
interior-ball property, 54, 84 interpolation between LP spaces, 15 interval in R2, 300 invariance, 310 invariance group, 31 invariance transformation, 30 inverse image, 4 inverse-square law, 169 isometry, 28 Kelvin transform, 232 Kelvin transformation, 72, 232, 264 Kronecker delta, 5
Landau symbols O(.) and o(.), 5 Laplace operator, 6 lateral derivative, 266 Lebesgue spaces LP(S2), 10 Lebesgue spine, 85 Legendre function, 36 Leibniz rule, 214 level set, 4, 173 limit point, 4 Liouville theorem, 231 Lipschitz constant, 87
Index
Lipschitz continuous, 87, 196 locally, 87 uniformly, 87 local co-ordinates, 8 local maximum, 20 locally integrable, 224 logarithmic growth at infinity, 327
339
of density function with unbounded support, 219 of density r' in spherical shell, 217 of unit density in a ball, 177 of unit density in sector, 218 second characterization, 179 solves Poisson equation, 193, 197, 210, 211
magnetohydrostatics, 138 map, 4 mapping, 4 matrix multiplication, 5 maximum, 20 maximum principles, 19-24, 39-84 boundary-point for balls, 50 boundary-point for Q, 54 edge point, 311, 317 for thin sets n, 60 Phragmen-Lindelof for bad boundary point, 69, 81, 82, 86 Phragmen-Lindelof for unbounded S2,
67,77-80,86 strong, 53 weak for Lo, 41 weak for L, 43 weak for Lt, 46 mean function, 37 mean-value property of harmonic functions, 228 measurable, 9 measure zero, 10 metric space, 3 Minkowski's inequality, 13 modified Newtonian kernel, 56 modified Newtonian potential, 57 moduli of ellipticity, 39 monotonicity of positive solutions, 141 multi-index, 183 multinomial theorem, 214 multipole, 199, 214
neighbourhood, 305 Neumann problem for L, 85
for -A, 235-248, 236 for -A in a ball, 262, 265, 267 for -A in a half-space, 269 Newton, 169 Newtonian kernel, 122, 174 Newtonian potential, 174-220 behaviour outside support of density function, 185 continuity of first derivatives, 197 continuity of potential, 188 continuity of second derivatives, 210 far field, 200
symmetry of potential, 17, 122-133, 137-139 non-increasing, 2 non-negative part, 5 non-positive part, 5 non-singular part of F, 239 normed linear spaces, 10 one-step solution, 128 open cover, 149 open interval in RN, 10 orthogonal curvilinear co-ordinates, 234 overdetermined problems, 328
partial derivatives, 6 partition of unity, 280 pathwise connected, 53 Phragmen-Lindelof theory, 61 for bad boundary point, 69, 81, 82, 86 for unbounded 4, 67, 77-80, 86 point charge, 168 point mass, 169 point source, 167 pointwise modulus of ellipticity, 40 Poisson equation, 176 Poisson integral, 249, 257, 270 Poisson kernel, 63, 249, 268 positive, 2 potential of unit density in a ball, 177 potential of unit source, 167 primary function of Siegel type, 72, 270 proper subset, 2 protuberance, 151
quadrupole, 199 quasi-cubical neighbourhood, 280 range, 4 Rankine solid, 171, 216 real-analytic, 62, 223 real-analytic functions, 221 reflected cap, 88, 142 reflection in hyperplane of a function, 25 of a point, 25 reflection in hyperplanes (method), 24-28 reflection operator, 32 reflection relative to 8&(0, a), 232 region, 3
340
regularization, 37 repeated integrals, 9 representation formula, 239, 241 restriction, 5 rhomboid neighbourhood, 145, 146 Riesz kernel, 268 right-hand boundary of cap, 142 Schwarz inequality, 13 sector, 79 semi-norm, 10 separated, 4 sets with edges and vertices, 285 sign of potential functions, 169 signum function, 72 sliding domain, 330 small-ball technique, 187 smoothing kernel, 37 smoothing operation, 37 smoothing radius, 37 smoothly harmonic, 228 solution, 21 source of unit strength, 167 source point, 176 space Cb(S2), 196 space Cb°' (52), 197 spherical co-ordinates, 172, 303 spherical shell, 163, 329 spherically symmetric, 19, 25 Steiner symmetric, 88, 101 Stokes stream function, 134, 173, 213, 214 stream function, 134, 213 streamlines, 135 strict meaning, 2 subharmonic, 62, 225 subset, 2 subsolution, 21 superharmonic, 62, 225 supersolution, 21 support, 4, 185
Index
supremum, 20 surface area, 8 surface area of sphere, 33 symmetrization with respect to a line, 324 with respect to a plane, 324 with respect to a point, 324 symmetry group, 29 symmetry of potential, 18 symmetry transformation, 28 Taylor's theorem, 222 test functions, 190 test particle, 168, 169 thin set, 56 Tricomi operator, 40 truncated sector, 300
two-ball function of Siegel type, 76 two-step solution, 128, 138, 139 uniform Holder continuity, 196 uniformly elliptic, 39, 41 uniqueness of solutions Dirichlet problem, 44, 237, 250 idealized flow past body, 170, 213 Neumann problem, 85, 238 unit normal, 8 upward jumps, 94 vanishes, 4 variables aligned with k, 111 velocity potential, 170 vinca herbacea, 324 volume of ball, 33 vorticity, 134
wave equation, 40 Weyl, H., 324 Young's inequality, 12