Almost periodic equations and conditions of
Ambrosetti-Prodi type ∗
Rafael Ortega1 and Massimo Tarallo2
1
Departamento de Matemática Aplicada, Facultad de Ciencias
Universidad de Granada, 18071 Granada, Spain
[email protected]
2
Dipartimento di Matematica dell’Università
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
ẍ + cẋ + 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 = ẍ + cẋ
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 Fréchet 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
Fréchet 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
ÿ + cẏ + 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 ϕ̇ 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 ÿ + cẏ = 0. In general, given
ϕ ∈ BC 2 (R), ϕ >> 0, we can construct α in P from the formula
α(t) = −
ϕ̈(t) + cϕ̇(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
ÿ + cẏ + 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
ÿ − cẏ + 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
ẏ(t)ψ(t) + y(t)(cψ(t) − ψ̇(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
ẏ + a(t)y = b(t)
has a unique bounded solution.
This remark can be applied to
ẏ + (c −
ψ̇(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
ÿ + cẏ + 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)
z̈ + (c +
2ϕ̇(t)
)ż = (α(t) − β(t))z.
ϕ(t)
6
If we assume that α >> β then ξ = ż satisfies the differential inequality
2ϕ̇(t)
)ξ > 0.
ξ˙ + (c +
ϕ(t)
From here we deduce that, for each τ ∈ R,
ż(t) < ż(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 ż(τ ) < 0.
This would imply that ż(t) is unbounded and this leads us to the searched
contradiction because y(t) and ẏ(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, ϕ̇(τ ) > 0. The function y(t) = e 2 t ϕ(t) satisfies
y(τ ) = 0,
ẏ(τ ) > 0,
7
(10)
and
lim y(t) = lim ẏ(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
ÿ + [α(t) −
ÿ(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 ϕ̂(t) = 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 ϕ̂(t) 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 ẋ 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 ẋ(t) is uniformly continuous and this is
sufficient to guarantee that ẋ ∈ 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 Fréchet 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
ÿ + cẏ + g 0 (x(t))y = 0.
(12)
We shall consider the family of linear equations
ÿ + cẏ + 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
ẍ + cẋ + 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
ẋ2 (hn ) − ẋ1 (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
ẍ + cẋ + 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
ẍ + cẋ + 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
ÿ + cẏ + 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
ẍ + cẋ + 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
ẍ + cẋ = 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{||ẋ||∞ + c||x||∞ }.
By integrating the equation between t and t + ` one obtains
Z t+`
F (s, x(s))ds = ẋ(t + `) − ẋ(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 ẍ + cẋ ≤ −Γ implies
ect1 ẋ(t1 ) ≤ ect2 ẋ(t2 ) −
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 ẋ(τ ) ≤ 0 (≥ 0). As a consequence of Lemma 11, x(t) would
be unbounded.
2)
If x(τ ) = σ and ẋ(τ ) ≥ 0 then ẋ(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 ẋ(τ ) = 0 as a consequence of ẍ(τ ) + cẋ(τ ) ≤ −Γ.
The differential inequality ẍ + cẋ ≤ Cσ then implies:
ẋ(t) ≥ ec(τ −t) [ẋ(τ ) −
Cσ
Cσ
]+
.
c
c
If, by contradiction, ẋ(τ ) ≥ Cσ /c then the same holds for every t ∈ (α, τ ).
Thus α would be finite, x(α) = −σ and ẋ(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 τ̂ be an instant where x(t) reaches a local
12
maximum with x(τ̂ ) > σ. Then we can find τ < τ̂ such that x(τ ) = σ and
x(t) > σ if t ∈ (τ, τ̂ ]. From ẍ + cẋ ≤ −Γ one obtains
ẋ(τ̂ ) − ẋ(τ ) + c(x(τ̂ ) − x(τ )) ≤ −Γ(τ̂ − τ ) < 0,
and Step 2 leads to
cx(τ̂ ) ≤ ẋ(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
α̈ + cα̇ >> F (·, α),
β̈ + cβ̇ << F (·, β)
(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
ÿ + cẏ − 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
α̈n+1 + cα̇n+1 − λαn+1 = F (t, αn (t)) − λαn (t),
β̈n+1 + cβ̇n+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) := ẍ0 (t) + cẋ0 (t) + λx0 (t)
2
where λ > 0. This is possible whenever λ > c4 . All the solutions of ẍ +
cẋ + 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
ÿ + cẏ +
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 Fréchet 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 ẍ(t) in terms
of x(t) and ẋ(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
ẍ(t) + cẋ(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
β̈ + cβ̇ + 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
α̈ + cα̇ + 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 Fréchet
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 Fréchet 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? ).
16
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
Fréchet 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
ÿ + cẏ + 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
ÿ + cẏ + α(t)y = 0
(29)
has a solution ϕ satisfying
ϕ ∈ BC 2 (R),
ϕ(t) > 0 ∀t ∈ R,
inf {ϕ(t)2 + ϕ̇(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 Ṫδ (t) 2 1 d Ṫδ (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 +
ϕ̇(tn )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+ .
19
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) := ẍ(t) + cẋ(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).
References
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set of a scalar convex differential equation. To appear.
[2] A. Ambrosetti and G. Prodi. On the inversion of some differentiable
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Appl. 93 (1972), 231-246.
[3] W.A. Coppel. Dichotomies in Stability Theory, LNM 629, SpringerVerlag, 1978.
[4] J. Favard. Sur les équations différentielles à coefficients presquepériodiques. Acta Math. 51 (1928), 31-81.
[5] A.M. Fink. Almost periodic differential equations, LNM 377, SpringerVerlag, 1974.
[6] M.A. Krasnosel’skii, V.S. Burd and Y.S. Kolesov. Nonlinear
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20