Almost periodic equations and conditions of Ambrosetti-Prodi type ∗ Rafael Ortega1 and Massimo Tarallo2 1
Departamento de Matem´atica Aplicada, Facultad de Ciencias Universidad de Granada, 18071 Granada, Spain
[email protected] 2 Dipartimento di Matematica dell’Universit`a Via Saldini 50, 20133 Milano, Italy
[email protected]
Abstract We discuss the exact number of almost periodic solutions of certain ordinary differential equations of the second order. The class of equations under consideration is inspired by a well known result in the area of elliptic boundary value problems.
1
Introduction
We consider the differential equation x ¨ + cx˙ + g(x) = p(t) + s
(1)
where c > 0 is a fixed constant, g is in C 1 (R), p is an almost periodic function and s ∈ R is a parameter. In addition we impose two conditions on g, g 0 is strictly increasing (2) and
c2 . (3) 4 We will refer to (2) as the convexity condition and to (3) as the jumping condition. The main result of the paper is − ∞ ≤ g 0 (−∞) < 0 < g 0 (+∞) ≤
∗
work supported by Azione Integrata Italia-Spagna HI2000-0112.
1
Theorem 1 There exists s? = s? (g, p) ∈ R such that the equation (1) has: a) exactly two almost periodic solutions if s > s? ; b) at most one almost periodic solution if s = s? ; c) only unbounded solutions if s < s? . Moreover, mod(ϕ) ⊂ mod(p) holds for any almost periodic solution to (1). Here mod(ϕ) is the module of ϕ; this is the smallest additive subgroup of R containing all the exponents in the Fourier series of ϕ. The conditions imposed on g as well as the result are inspired by [2]. In that paper Ambrosetti and Prodi considered the Dirichlet problem ∆u + g(u) = p(x), x ∈ Ω ⊂⊂ RN (4) u = 0 on ∂Ω and assumed that g satisfies a convexity condition slightly more restrictive than (2) and the jumping condition 0 < g 0 (−∞) < λ1 < g 0 (+∞) < λ2 , where λ1 and λ2 are the first eigenvalues of −∆. The conclusion was an exact count on the number of solutions of (4) depending on p (two, one or zero solutions). In our almost periodic equation we have replaced the laplacian by the linear operator Lx = x ¨ + cx˙ and the spectrum of −L is interpreted in the sense of Sacker and Sell [14]. More precisely, Σ = {λ ∈ C / Lx + λx = 0 has not an exponential dichotomy}. Since the operator L is of constant coefficients the set Σ can also be described as Σ = {λ ∈ C / Lx + λx = 0 has nontrivial almost periodic solutions}. This is perhaps more natural from the point of view of the theory of boundary value problems. A computation shows that Σ is the parabola {σ 2 + 2 iσc / σ ∈ R} with vertex at the origin and focus at ( c4 , 0). For this reason 2 the number 0 plays the role of λ1 in our result. The connection between c4 and Σ is more intricate but nevertheless this number is optimal. This can 2
be deduced from the results in [11]. Assuming that p(t) is T -periodic for some T > 0, keeping the convexity condition and replacing (3) by −∞ ≤ g 0 (−∞) < 0 < g 0 (+∞) ≤
c2 2π + ( )2 , 4 T
one can prove that (1) has at most two T -periodic solutions. There are cases when (1) has T -periodic solutions ϕ1 and ϕ2 and also another periodic solution with minimal period 2T , say ϕ3 . In such a situation we have at least four almost periodic solutions, ϕ1 (t), ϕ2 (t), ϕ3 (t) and ϕ3 (t + T ), and the conclusion of the Theorem does not remain valid. Going back to the comparison with the result in [2] for the Dirichlet problem (4) we notice that our Theorem 1 is less precise. For if s = s? we do not know if it is always possible to find an almost periodic solution. The proof of the Theorem will show that there is a bounded solution in this case; from here it is possible to derive some consequences about existence of recurrent solutions. We refer to [13] for the case of a quasi-periodic forcing p(t). The existence of almost periodic solutions for equations of the type (1) has been discussed by several authors. We refer to the monographs [5] and [6] for a description of the most classical results. In many of these works the existence of an almost periodic solution is obtained after proving that every equation in the hull of (1) has a unique bounded solution. This strategy does not seem adequate in our case since we can have a continuum of bounded solutions. The main idea will be to prove that the set of bounded solutions is ordered and has maximal and minimal solutions. These two extreme solutions will be almost periodic. This is reminiscent of the arguments employed in the theory of almost periodic solutions of first order equations of Riccati type (see [6], page 118, and [1] for recent results). Since our equation is of the second order our proofs need more machinery. The main tools employed in this paper will be the theory of exponential dichotomies for linear equations and the method of upper and lower solutions for bounded solutions. The rest of the paper is organized in seven sections. The proof of Theorem 1 is in §7. From §2 to §6 we present preliminary results to prepare this proof. The last section §8 contains some observations about the critical case s = s? . To conclude this introduction we present some terminology. The space of bounded and continuous functions ϕ : R → R will be denoted by BC(R). There are two natural topologies on BC(R). We can consider the structure of Banach space induced by the norm of the uniform convergence on the
3
whole line, ||ϕ||∞ = sup |ϕ(t)|, t∈R
and we can also look at BC(R) as a Fr´echet space with the topology induced by the uniform convergence on bounded subintervals of R. The space BC(R) will be ordered in the natural way but it will be convenient to specify a notion of strong ordering. Given two functions ϕ and ψ in BC(R), the notation ϕ >> ψ will indicate that inf t∈R {ϕ(t) − ψ(t)} > 0. The space of almost periodic functions AP (R) will be viewed as a subspace of BC(R). It is closed with respect to the Banach topology. Other spaces which will also be employed are BC p (R) = {ϕ ∈ C p (R) / ϕ, ϕ0 . . . ϕ(p) ∈ BC(R)},
p≥1
and BP (R) = {continuous functions with bounded primitive}. The space BC p (R) has also two natural topologies. The norm ||ϕ||BC p (R) =
p X
||ϕ(i) ||∞
i=0
induces a structure of Banach space. The uniform convergence on bounded intervals of the function and its derivatives (up to the order p) induces the Fr´echet topology. An almost periodic solution of (1) will be a solution x(t) such that x and 0 x belong to AP (R).
2
A class of linear equations
Consider the equation y¨ + cy˙ + α(t)y = 0
(5)
where α ∈ BC(R). We shall say that α is in the class P if there is a solution ϕ(t) satisfying ϕ ∈ BC(R) and ϕ >> 0. From this definition one can deduce that ϕ is indeed in BC 2 (R). This is a consequence of a well known result that goes back to Escanglon (see [5], page 82 and [6], page 9). Once we know that ϕ and ϕ˙ are bounded we can apply the Liouville formula to deduce that the solutions which are linearly independent with ϕ are unbounded. The simplest function in the class P 4
is α = 0 because ϕ = 1 is a solution of y¨ + cy˙ = 0. In general, given ϕ ∈ BC 2 (R), ϕ >> 0, we can construct α in P from the formula α(t) = −
ϕ(t) ¨ + cϕ(t) ˙ . ϕ(t)
This is a way to describe all the functions in P. Next we want to discuss the existence of bounded solutions of the nonhomogeneous equation y¨ + cy˙ + α(t)y = p(t)
(6)
when α ∈ P and p ∈ BC(R). Repeating some previous arguments one can prove that it is equivalent to search for solutions in BC(R) or in BC 2 (R). We are interested in obtaining a sort of Fredholm alternative and so it will be convenient to consider the adjoint equation y¨ − cy˙ + α(t)y = 0.
(7)
This equation may be obtained from (5) via the change of variables x = y e−ct . Thus ϕ(t)ect is a particular solution of (7) and the method of reduction of the order implies that Z ∞ 1 −cs ψ(t) = ect ϕ(t) e ds 2 ϕ(s) t is also a solution of (7). A direct computation shows that it is bounded and satisfies ψ >> 0. Proposition 2 Assume that α ∈ P and p ∈ BC(R). Then (6) has a bounded solution if and only if pψ ∈ BP (R).
(8)
Remarks. 1. For α = 0 this result is related to Lemma 2.4 in [12]. 2. When α and p are almost periodic one can apply Favard theory [4] to conclude that (8) is also necessary and sufficient for the existence of a solution in AP (R). Proof. Assume first that (6) has a bounded solution y(t). We multiply the equation by ψ and integrate by parts to obtain Z t ˙ y(t)ψ(t) ˙ + y(t)(cψ(t) − ψ(t)) = k + p(s)ψ(s)ds 0
5
where k is an appropriate constant. This implies that the function Z t P (t) = p(s)ψ(s)ds 0
is bounded and so pψ belongs to BP (R). To prove the converse we recall a well known fact about first order equations (see for instance [3]): assume that a, b ∈ BC(R) and Z t a(s)ds = at + O(1) as |t| → ∞ 0
for some a 6= 0, then the equation y˙ + a(t)y = b(t) has a unique bounded solution. This remark can be applied to y˙ + (c −
˙ ψ(t) P (t) )y = ψ(t) ψ(t)
and the bounded solution of this equation is also a solution of (6). Now we derive a consequence of this Fredholm alternative which will be used later. Proposition 3 Consider the equation y¨ + cy˙ + β(t)y = 0
(9)
with β ∈ BC(R) and assume that there is α ∈ P such that α >> β or β >> α. Then every bounded solution of (9) must vanish somewhere. Proof. By a contradiction argument we assume that y(t) is a bounded and positive solution of (9). We can see this function as a solution of (6) for p = (α − β)y. Now we apply the Fredholm alternative to deduce that pψ is in BP (R). Since ψ and |α − β| are bounded and uniformly positive we conclude that also y has a bounded primitive. This implies that y(t) → 0 as |t| → ∞ because y is positive and uniformly continuous. The function y(t) is a solution of z(t) = ϕ(t) z¨ + (c +
2ϕ(t) ˙ )z˙ = (α(t) − β(t))z. ϕ(t) 6
If we assume that α >> β then ξ = z˙ satisfies the differential inequality 2ϕ(t) ˙ )ξ > 0. ξ˙ + (c + ϕ(t) From here we deduce that, for each τ ∈ R, z(t) ˙ < z(τ ˙ )e−c(t−τ )
ϕ(τ )2 ∀t ∈ (−∞, τ ). ϕ(t)2
Since z(t) decays to zero as |t| → ∞ we can pick τ such that z(τ ˙ ) < 0. This would imply that z(t) ˙ is unbounded and this leads us to the searched contradiction because y(t) and y(t) ˙ are bounded. The case α << β is treated in a similar way.
3
Some properties of the bounded solutions
In this section we go back to the nonlinear equation (1) and assume that (2) and (3) hold. A bounded solution will be a solution in BC(R) or, equivalently, in BC 2 (R). A crucial fact about these solutions is that they cannot cross. Proposition 4 Let x1 and x2 be bounded solutions of (1) with x1 (t0 ) < x2 (t0 ) for some t0 ∈ R. Then x1 (t) < x2 (t) ∀t ∈ R. This result is a consequence of the following property of linear equations. Lemma 5 Assume that α ∈ BC(R) satisfies α(t) ≤
c2 , ∀t ∈ R 4
and let ϕ(t) be a bounded and nontrivial solution of (5). Then ϕ(t) 6= 0 ∀t ∈ R. Proof. This result was stated and proved in [13] and a related result can be seen in [5], page 248. Next we present a sketch of the proof. By a contradiction argument assume that ϕ(t) vanishes at some instant τ , say c ϕ(τ ) = 0, ϕ(τ ˙ ) > 0. The function y(t) = e 2 t ϕ(t) satisfies y(τ ) = 0,
y(τ ˙ ) > 0, 7
(10)
and lim y(t) = lim y(t) ˙ = 0,
t→−∞
t→−∞
(11)
because ϕ is bounded. On the other hand y(t) is a solution of c2 ]y = 0 4 and the Sturm comparison theory implies that y(t) only vanishes at τ . Now one uses the equation to deduce that y¨ + [α(t) −
y¨(t) ≤ 0 ∀t ∈ (−∞, τ ) and this is not compatible with (10) and (11). Proof of Proposition 4. From (2) and (3) we deduce that c2 ∀x ∈ R. 4 The difference ϕ(t) = x2 (t) − x1 (t) satisfies (5) with Z 1 α(t) = g 0 (λx2 (t) + (1 − λ)x1 (t))dλ. g 0 (x) <
0
This function is in the conditions of the previous Lemma and so ϕ(t) must be positive everywhere. Our next result implies that it is not possible to find three bounded solutions which are strongly ordered. Proposition 6 Assume that x1 , x2 and x3 are bounded solutions of (1) satisfying x1 < x2 < x3 . Then inf {x3 (t) − x2 (t)} = inf {x2 (t) − x1 (t)} = 0.
t∈R
t∈R
Proof. Once again we proceed by contradiction. Let us assume for instance that x3 >> x2 . The function ϕ(t) = x3 (t) − x2 (t) is a solution of (5) with Z 1 α(t) = g 0 (λx3 (t) + (1 − λ)x2 (t))dλ 0
and so α belongs to P. On the other hand ϕ(t) ˆ = x2 (t) − x1 (t) is a bounded solution of (9) with Z 1 β(t) = g 0 (λx2 (t) + (1 − λ)x1 (t))dλ. 0
For each λ ∈ (0, 1], λx3 + (1 − λ)x2 >> λx2 + (1 − λ)x1 and this implies that α >> β. We can apply Proposition 3 and conclude that ϕ(t) ˆ must vanish. This is not possible since x2 > x1 . 8
4
Almost periodic solutions and exponential dichotomies
In the introduction of the paper we defined an almost periodic solution of (1) as a solution x(t) such that x and x˙ belong to AP (R). Indeed it would be sufficient that x ∈ AP (R). In that case one deduces from previous sections that x ∈ BC 2 (R). In particular x(t) ˙ is uniformly continuous and this is sufficient to guarantee that x˙ ∈ AP (R). See for instance [5], page 10. Given f ∈ AP (R) and h ∈ R, the translate Th f is defined as Th f (t) = f (t + h),
t ∈ R.
The hull of f , H(f ), is the closure in AP (R) of {Th f / h ∈ R}. Here it is indifferent whether we refer to the Banach or the Fr´echet topology. A generic element of H(f ) will be denoted by f ? . Assume now that x(t) is an almost periodic solution of (1) with linearized equation y¨ + cy˙ + g 0 (x(t))y = 0.
(12)
We shall consider the family of linear equations y¨ + cy˙ + g 0 (x? (t))y = 0, x? ∈ H(x).
(13)
The solution x(t) will be called hyperbolic if y = 0 is the only bounded solution of (13) for arbitrary x? . To justify this terminology we notice that in the periodic case (p(t) and x(t) periodic of the same period) hyperbolicity just means that the Floquet multipliers of (12) are not on the unit circle. In the general situation hyperbolicity is equivalent to saying that (12) has an exponential dichotomy. This is a consequence of a result of Sacker and Sell which can be seen in [3], page 78. From this equivalence it is easy to deduce that hyperbolic solutions are persistent under small perturbations of the equation. In particular the equation x ¨ + cx˙ + g(x) = q(t) + σ will have an almost periodic solution close to x(t) if q ∈ AP (R), σ ∈ R and ||p − q||∞ + |s − σ| is small enough. See for instance [5], page 149. The main result of this Section is Proposition 7 Assume that x1 (t) and x2 (t) are two different almost periodic solutions of (1). Then both of them are hyperbolic.
9
We know from Proposition 4 that the solutions x1 and x2 must be ordered. To prepare the proof of Proposition 7 we first show that they are strongly ordered. Lemma 8 Assume that x1 (t) and x2 (t) are as in the previous Proposition. Then either x1 >> x2 or x1 << x2 . Proof. We assume that x1 < x2 and inf t∈R {x2 (t) − x1 (t)} = 0 and try to reach a contradiction. Since x2 − x1 is positive, it is always possible to select a sequence (hn ) ⊂ R in such a way that the sequences x2 (hn ) − x1 (hn ) and x˙ 2 (hn ) − x˙ 1 (hn ) converge simultaneously to zero. The almost periodicity of x1 (t), x2 (t) and p(t) allows us to extract a common subsequence (hk ) ⊂ (hn ) such that Thk xi → x?i , i = 1, 2, Thk p → p? . The convergence is uniform in R and x?1 (t), x?2 (t) are almost periodic solutions of x ¨ + cx˙ + g(x) = p? (t) + s. These solutions satisfy the same initial conditions at t = 0 and so x?1 = x?2 . Thus ||Thk x1 − Thk x2 ||∞ → 0 and we have arrived at the contradiction because ||Thk x1 − Thk x2 ||∞ = ||x1 − x2 ||∞ > 0. Proof of Proposition 7. Assume x2 >> x1 . We prove that one of them, say x1 (t), is hyperbolic. Given x?1 ∈ H(x1 ) we find a sequence (hn ) ⊂ R for which Thn x1 → x?1 . After extracting a subsequence we can assume that Thn x2 and Thn p also converge to functions x?2 and p? . Then x?2 >> x?1 and both x?1 (t) and x?2 (t) are solutions of x ¨ + cx˙ + g(x) = p? (t) + s. The function Z α(t) =
1
g 0 (λx?2 (t) + (1 − λ)x?1 (t))dλ
0
belongs to P because ϕ(t) = x?2 (t) − x?1 (t) is a solution of (5). Since g 0 ◦ x?2 >> α >> g 0 ◦ x?1 , we combine Proposition 3 with Lemma 5 to deduce that the linear equations y¨ + cy˙ + g 0 (x?i (t))y = 0, i = 1, 2 do not have bounded solutions different from y = 0. This implies that x1 (t) is hyperbolic. We conclude this Section with a preliminary result about the number of almost periodic solutions. 10
Proposition 9 The equation (1) has at most two almost periodic solutions. Moreover, when (1) has exactly two almost periodic solutions then the same holds for x ¨ + cx˙ + g(x) = p(t) + σ with σ close to s. Proof. The first part of the Proposition follows from Lemma 8 and Proposition 6. To prove the second part one applies Proposition 7 and uses the persistence under small perturbations of hyperbolic solutions.
5
An estimate for bounded solutions
In this Section we will consider the differential equation x ¨ + cx˙ = F (t, x)
(14)
where F : R×R → R is continuous and satisfies the two following conditions: there exist Γ > 0 and σ > 0 such that F (t, x) ≤ −Γ if |x| ≥ σ, t ∈ R
(15)
and for each R > 0 there exists CR > 0 such that |F (t, x)| ≤ CR if |x| ≤ R, t ∈ R.
(16)
Later we shall check that our original equation (1) is in this setting. The techniques employed in this Section become more transparent when applied to (14). Proposition 10 Assume that (15) and (16) hold and let x(t) be a solution of (14) in BC(R). Then x is also in BC 2 (R) and there exists M (depending only upon σ, Γ and Cσ ) such that ||x||BC 2 (R) ≤ M. First of all we notice that it is sufficient to obtain a uniform bound of x in BC(R). The same arguments as in Section 2 show that x ∈ BC 2 (R). They also lead to the bound in BC 2 . The proof will follow after a preliminary result. Lemma 11 Let x(t) be a bounded solution of (14). Then |x(t)| < σ holds for a relatively dense set of t’s. 11
Proof. We pick ` > 0 with Γ` > 2{||x|| ˙ ∞ + c||x||∞ }. By integrating the equation between t and t + ` one obtains Z t+` F (s, x(s))ds = x(t ˙ + `) − x(t) ˙ + c(x(t + `) − x(t)). t
From the way ` was chosen we deduce that Z t+` 1 | F (s, x(s))ds| < Γ ` t and so |x(s)| < σ for some s ∈ [t, t + `]. The proof is complete because t is arbitrary. Proof of Proposition 10. It is structured in several steps. Throughout the proof, x(t) is an arbitrary bounded solution of (14). 1) x(t) ≥ −σ, ∀t ∈ R. To prove this we notice that the differential inequality x ¨ + cx˙ ≤ −Γ implies ect1 x(t ˙ 1 ) ≤ ect2 x(t ˙ 2) −
Γ ct1 (e − ect2 ), t1 > t2 . c
Given τ ∈ R with x(τ ) < −σ it is easy to show that the same holds for t ≥ τ (≤ τ ) if x(τ ˙ ) ≤ 0 (≥ 0). As a consequence of Lemma 11, x(t) would be unbounded. 2)
If x(τ ) = σ and x(τ ˙ ) ≥ 0 then x(τ ˙ ) ≤ Cσ /c.
Let (α, τ ) be the maximal interval to the left where x(t) satisfies |x(t)| < σ. Note that α < τ also when x(τ ˙ ) = 0 as a consequence of x ¨(τ ) + cx(τ ˙ ) ≤ −Γ. The differential inequality x ¨ + cx˙ ≤ Cσ then implies: x(t) ˙ ≥ ec(τ −t) [x(τ ˙ )−
Cσ Cσ ]+ . c c
If, by contradiction, x(τ ˙ ) ≥ Cσ /c then the same holds for every t ∈ (α, τ ). Thus α would be finite, x(α) = −σ and x(α) ˙ > 0. This is not compatible with Step 1. 3)
x(t) ≤ σ + Cσ /c2
∀t ∈ R.
It is sufficient to obtain a bound on the local maxima of x(t). This is a consequence of Lemma 11. Let τˆ be an instant where x(t) reaches a local 12
maximum with x(ˆ τ ) > σ. Then we can find τ < τˆ such that x(τ ) = σ and x(t) > σ if t ∈ (τ, τˆ]. From x ¨ + cx˙ ≤ −Γ one obtains x(ˆ ˙ τ ) − x(τ ˙ ) + c(x(ˆ τ ) − x(τ )) ≤ −Γ(ˆ τ − τ ) < 0, and Step 2 leads to cx(ˆ τ ) ≤ x(τ ˙ ) + cx(τ ) ≤
6
Cσ + cσ. c
The method of upper and lower solutions
We shall present a variant of this method for bounded solutions. It goes back to the work by Opial [10] and we refer to [7] and [8] for recent applications. In this section we also deal with the equation (14) but the assumptions on F are not as in the previous section. We assume now that F is a continuous function satisfying (16) and such that for each R > 0 there exists L with |F (t, x1 ) − F (t, x2 )| ≤ L|x1 − x2 |
if |x1 | , |x2 | ≤ R, t ∈ R.
(17)
Proposition 12 Assume that α, β ∈ BC 2 (R) satisfy β¨ + cβ˙ << F (·, β)
α ¨ + cα˙ >> F (·, α),
(18)
and α(t) ≤ β(t)
∀t ∈ R.
(19)
Then in fact α << β and (14) has a solution ϕ(t) defined in (−∞, +∞) and such that α << ϕ << β. Proof. First we recall that given p ∈ BC(R), the unique solution in BC 2 (R) of y¨ + cy˙ − λy = p(t), λ > 0, is given by the formula Z
+∞
y(t) =
G(t, s)p(s)ds
(20)
−∞
where the Green function G(t, s) is defined by 1 c G(t, s) = − e 2 (s−t) e−ν|t−s| , 2ν 13
r ν =+ λ+
c2 . 4
In particular p >> 0 implies y << 0. More information on this type of positivity principles can be found in [6]. Now we construct sequences {αn } and {βn } in BC 2 (R) as α ¨ n+1 + cα˙ n+1 − λαn+1 = F (t, αn (t)) − λαn (t), β¨n+1 + cβ˙ n+1 − λβn+1 = F (t, βn (t)) − λβn (t), where λ > 0 is such that the function H(t, x) = F (t, x) − λx satisfies, for some positive , H(t, x1 ) − H(t, x2 ) ≤ −(x1 − x2 ) if α(t) ≤ x2 ≤ x1 ≤ β(t), t ∈ R. It follows from a standard argument that α0 = α << α1 << . . . << αn << . . . << βn << . . . << β1 << β0 = β. Then we can define ϕ(t) as the limit of αn (t). When the order is reversed in Proposition 12 and, instead of (19), it is assumed that β(t) ≤ α(t) ∀t ∈ R, (21) it is not always possible to find a solution of (14) between α and β. To show this consider a function x0 (t) which is periodic (for some period T > 0) and such that it changes sign but the function p(t) defined below is positive, p(t) := x ¨0 (t) + cx˙ 0 (t) + λx0 (t) 2
where λ > 0. This is possible whenever λ > c4 . All the solutions of x ¨+ cx˙ + λx = p(t) converge to x0 (t) as t → +∞ and so they must change sign. On the other hand the constants α > ||p||λ ∞ and β = 0 satisfy (18) for F (t, x) = −λx + p(t) and it is clear that there is no solution between them. After this example the following result seems natural. Proposition 13 Assume that α, β ∈ BC 2 (R) satisfy (18) and (21). In addition, F (t, x1 ) − F (t, x2 ) c2 ≥− x1 − x2 4 for each t ∈ R and β(t) ≤ x2 < x1 ≤ α(t). Then in fact β << α and (14) has a solution ϕ(t) defined in (−∞, +∞) and such that β << ϕ << α. 14
The strategy of the proof is the same as in the previous case. The function H(t, x) = F (t, x) + c2 x/4 satisfies H(t, x1 ) ≤ H(t, x2 ) if β(t) ≤ x1 ≤ x2 ≤ α(t), t ∈ R , and the unique bounded solution of y¨ + cy˙ +
c2 y = p(t), 4
p ∈ BC(R)
is such that x >> 0 as soon as p >> 0. This is a consequence of formula (20) where the Green function is now − 2c (t−s) s≤t, G(t, s) = (t − s)e 0 s≥t. We notice that the previous proposition can be seen as a very special case of Theorem 3 in [9].
7
Proof of the main Theorem
First of all we notice that g satisfies lim g(x) = +∞.
|x|→∞
(22)
This is a consequence of (3). In particular (1) is always in the conditions of Proposition 10. Let B = B(s, p) be the (possibly empty) set of bounded solutions of (1). This is a bounded subset of the Banach space BC 2 (R) and it is compact with respect to the Fr´echet topology. The C 1 compactness follows from Ascoli Theorem and the C 2 compactness is a consequence of the equation itself. From (1) one can express the second derivative x ¨(t) in terms of x(t) and x(t). ˙ In addition we know that B is totally ordered (Proposition 4) and so, if it is non-empty, the maximal and minimal bounded solutions can be defined by means of x# (t) = x# (t; s, p) = sup{x(t) : x ∈ B(s, p)} , x# (t) = x# (t; s, p) = inf{x(t) : x ∈ B(s, p)} . Every solution x(t) in B \ {x# , x# } will satisfy x# > x > x# and, from Proposition 6, inf {x# (t) − x(t)} = inf {x(t) − x# (t)} = 0.
t∈R
t∈R
15
(23)
Let us now consider the dependence of B with respect to s. This set is empty for s large and negative. For s ≤ −1 − ||p||∞ + minR g, every solution x(t) of (1) will satisfy x ¨(t) + cx(t) ˙ ≤ −1, t ∈ R, and so it cannot be bounded. On the other side, for s large and positive, the set B(s, p) will be non-empty. This is an easy application of the method of upper and lower solutions. In fact the function β = 0 satisfies β¨ + cβ˙ + g(β) << p + s
(24)
for s > g(0) + ||p||∞ . Using the coercivity condition (22) we notice that, for a fixed s, every large constant α (positive or negative) is a solution of the reversed inequality α ¨ + cα˙ + g(α) >> p + s. (25) Let us now fix s1 ∈ R such that B(s1 , p) 6= ∅ and take s > s1 . Then x# (t; s1 , p) and x# (t; s1 , p) are solutions of (24). Using again large constants as solutions of (25) we deduce that B(s, p) is non-empty and x# (·; s, p) << x# (·; s1 , p) ≤ x# (·; s1 , p) << x# (·; s, p).
(26)
Define s? = inf{s ∈ R / B(s, p) 6= ∅}. We have just proved that B(s, p) is non-empty for all s > s? . We prove now that the same holds for s = s? . Proposition 10 implies that, for s ∈ (s? , s? + 1], the sets B(s, p) are uniformly bounded in the Banach space BC 2 (R). Let sn ↓ s? and xn ∈ B(sn , p) be sequences with supn ||xn ||BC 2 (R) < ∞. After extracting a subsequence one can assume that xn converges (in the Fr´echet topology) to some x ∈ BC 2 (R) which is a solution of (1) for s = s? . Thus B(s? , p) is non-empty. The number s? depends upon p, s? = s? (p), but it is interesting to notice that it is invariant on the hull of p. More precisely, s? (p) = s? (p? )
∀p? ∈ H(p).
(27)
To justify this property we take a sequence (hn ) ⊂ R with Thn p → p? . Given x ∈ B(s? (p), p) we consider the sequence (Thn x) ⊂ BC 2 (R) and extract a subsequence converging in the Fr´echet topology. The limit function clearly belongs to B(s? (p), p? ) and so s? (p? ) ≤ s? (p). This suffices to prove (27) because H(p) = H(p? ).
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The definition of s? says that all the solutions of (1) are unbounded for s < s? . We shall prove that for s > s? the functions x# (t) and x# (t) are in AP (R). In view of Proposition 9 this will be sufficient to complete the proof of the first part of Theorem 1. The following result will be useful. Lemma 14 Let F be a function in AP (R) with hull H(F ) and assume that there exists a mapping Φ : H(F ) → BC(R), F ? 7→ Φ(F ? ) satisfying i) Th ◦ Φ = Φ ◦ Th for each h ∈ R ii) Given a sequence (Fn? ) ⊂ H(F ) and F ? ∈ H(F ) with ||Fn? − F ? ||∞ → 0, the functions Φ(Fn? ) converge pointwise to Φ(F ? ). Then f = Φ(F ) belongs to AP (R) and mod(f ) ⊂ mod(F ). This Lemma can be proved using standard techniques in the theory of almost periodic functions. In particular it can be obtained from Theorems 1.17 and 4.5 in [5]. Our candidate for the application of the Lemma will be the maximal solution, Φ : H(p) → BC(R), p? 7→ x# (·; s, p? ). In view of (27) we know that Φ is well defined for s ≥ s? (p). The condition i) of Lemma 14 is clearly satisfied and we shall prove that ii) holds when s > s? . Equivalently, we shall prove that from every sequence (p?n ) ⊂ H(p) converging to some p? , one can extract (p?k ) ⊂ (p?n ) with x# (t; s, p?k ) → x# (t; s, p? ) ∀t ∈ R. To do this we start by observing that the sequences {x# (·; s, p?n )} and {x# (·; s? , p?n )} are bounded in the Banach space BC 2 (R). This is again a consequence of Proposition 10. From these bounded sequences we extract common subsequences x# (·; s, p?k ) and x# (·; s? , p?k ) which converge in the Fr´echet topology to some y ∈ B(s, p? ) and z ∈ B(s? , p? ), respectively. It remains to prove that y = x# (·; s, p? ). To this end we notice that, by (26), x# (·; s, p?k ) >> x# (·; s? , p?k ) and x# (·; s, p? ) << x# (·; s? , p? ). From the first inequality we deduce that y ≥ z ≥ x# (·; s? , p? ). We combine this with the second inequality to deduce that y >> x# (·; s, p? ). The 17
property (23) implies that y = x# (·; s, p? ) and Lemma 14 implies that x# is almost periodic for s > s? . In the same way one proves that x# is almost periodic. To complete the proof of the Theorem, we have to show that any almost periodic solution to (1) has the module containment property. This follows from Lemma 14 only when s > s? . In fact, we will show below that it is a general consequence of the ordering property stated in Proposition 4. Assume, by contradiction, that ϕ is an almost periodic solution to (1) without module containment. This is equivalent to say that it is possible to select an ε > 0 and a sequence (hn ) ⊂ R in such a way: T hn p → p
||Thn ϕ − ϕ||∞ ≥ ε
and
(see [5] page 61). The almost periodicity of ϕ allows us to extract a subsequence (hk ) ⊂ (hn ) such that Thk ϕ → ψ uniformly in R. The function ψ is again an almost periodic solution to (1) and satisfies ||ψ − ϕ||∞ ≥ ε; in particular, it is different from ϕ. Proposition 4 then says that ψ cannot cross ϕ. Assume for instance that ϕ > ψ, and consider a sequence (tn ) ⊂ R such that ϕ(tn ) → inf ϕ. Since inf ϕ = inf ψ ≤ ψ(tn ) < ϕ(tn ), it follows that inf(ϕ − ψ) = 0. This contradicts Lemma 8.
8
Comments on the critical case
We do not know if there is a function p(t) in AP (R) for which the equation (1) has bounded solutions but no almost periodic solutions. It is a consequence of Theorem 1 that this could only happen at s = s? . We shall show how to construct a less ambitious example: an equation (1) having a unique almost periodic solution x(t) and such that the linearized equation y¨ + cy˙ + g 0 (x(t))y = 0
(28)
is somehow pathological. 1. Given a function α ∈ BC(R), we say that α is in the class W P if the linear equation y¨ + cy˙ + α(t)y = 0 (29) has a solution ϕ satisfying ϕ ∈ BC 2 (R),
ϕ(t) > 0 ∀t ∈ R,
inf {ϕ(t)2 + ϕ(t) ˙ 2 } = 0.
t∈R
It is clear that the solutions which are linearly independent with ϕ must be unbounded. This implies that W P is disjoint with the class P introduced in 18
Section 2. When α ∈ W P the equation (29) is non-reducible. We use this term in the sense of Lyapunov: our equation is not kinematically similar to an autonomous system (see [3] for the terminology). The reason for the nonreducibility is that ϕ becomes arbitrarily small and this cannot happen for bounded solutions in autonomous or periodic systems. In this direction it is also worth noticing that for autonomous (or periodic) systems all bounded solutions are almost periodic. In contrast ϕ will never be almost periodic if α ∈ W P ∩ AP (R). The previous observations imply that there are no periodic functions in W P. The next result shows that W P contains almost periodic functions. Lemma 15 Given > 0 there exists α ∈ W P ∩ C ω (R) with ||α||∞ < and α(k) ∈ AP (R) for each k = 0, 1, 2, . . . Proof. It is inspired by the last section of Favard’s paper [4]. We start with a function φ ∈ AP (R) ∩ C ω (R) with the properties listed below, i) φ has zero average Rt ii) The primitive of φ, Φ(t) = 0 φ(s)ds, is unbounded and non-negative iii) φ(k) ∈ AP (R), k = 0, 1, 2, . . . A concrete example is ∞ X 1 t φ(t) = sin( k ). 2k 3 k=0
Define
c − δφ(t). 2 For small δ > 0 this function is uniformly positive and a computation shows that the family Tδ (t) =
y(t) = Tδ (t)−1/2 [k1 e−δΦ(t) + k2 e−ct+δΦ(t) ],
k1 , k2 ∈ R
is the general solution of (29) for α = αδ , αδ (t) =
c2 1 T˙δ (t) 2 1 d T˙δ (t) − Tδ (t)2 − ( ) + ( ). 4 4 Tδ (t) 2 dt Tδ (t)
The solution ϕ(t) = Tδ (t)−1/2 e−δΦ(t) is positive and bounded and ϕ(tn )2 + ϕ(t ˙ n )2 → 0 for any sequence (tn ) such that Φ(tn ) → ∞. Thus αδ ∈ W P ∩ AP (R) and the proof is complete because ||αδ ||∞ = O(δ) as δ → 0+ .
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2. Now we shall find p ∈ AP (R) such that (1) has an almost periodic solution x(t) with g 0 ◦ x ∈ W P. To do this we need stronger assumptions than those in Theorem 1. Let us assume that g ∈ C 3 (R),
g 00 (x) > 0 ∀x ∈ R
and (3) holds. We apply Lemma 15 with < min{g 0 (+∞), −g 0 (−∞)} and find a corresponding α. The equation g 0 (x(t)) = α(t) defines x(t) uniquely. It is clear that x is of class C 2 and belongs to AP (R) as well as the first and second derivative. Finally we define p(t) and s so that x(t) becomes a solution. For example, p(t) := x ¨(t) + cx(t) ˙ + g(x(t)), s = 0. We notice that the linearized equation (28) has not an exponential dichotomy and so we are at the critical case (s? = 0).
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