Electronic Journal of Differential Equations, Monogrpah 03, 2002. ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu (login: ftp)

Periodic solutions for evolution equations ^{∗}

Mihai Bostan

Abstract

We study the existence and uniqueness of periodic solutions for evolution equations. First we analyze the one-dimensional case. Then for arbitrary dimensions (finite or not), we consider linear symmetric operators. We also prove the same results for non-linear sub-differential operators A = ∂φ where φ is convex.

Contents

1 Introduction 1

2 Periodic solutions for one dimensional evolution equations 2 2.1 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2.2 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2.3 Sub(super)-periodic solutions . . . . . . . . . . . . . . . . . . . . 10

3 Periodic solutions for evolution equations on Hilbert spaces 16 3.1 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 3.2 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 3.3 Periodic solutions for the heat equation . . . . . . . . . . . . . . 31 3.4 Non-linear case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

1 Introduction

Many theoretical and numerical studies in applied mathematics focus on permanent regimes for ordinary or partial differential equations. The main purpose of this paper is to establish existence and uniqueness results for periodic solutions in the general framework of evolution equations,

x′(t) + Ax(t) = f(t), t ∈ ℝ,  (1)

∗Mathematics Subject Classifications: 34B05, 34G10, 34G20.

Key words: maximal monotone operators, evolution equations, Hille-Yosida’s theory.

\begin{matrix}c \\ ◯ \end{matrix}2002 Southwest Texas State University.

Submitted May 14, 2002. Published August 23, 2002.

by using the penalization method. Note that in the linear case a necessary condition for the existence is

⟨f⟩ := \frac{1}{T} ∫_{0}^{T} f(t)dt ∈ Range(A).  (2)

Unfortunately, this condition is not always sufficient for existence; see the example of the orthogonal rotation of ℝ^{2}. Nevertheless, the condition (2) is sufficient in the symmetric case. The key point consists of considering first the perturbed equation

αx_{α}(t) + x′_{α}(t) + Ax_{α}(t) = f(t), t ∈ ℝ,

where α > 0. By using the Banach’s fixed point theorem we deduce the existence and uniqueness of the periodic solutions x_{α}, α > 0. Under the assumption (2), in the linear symmetric case we show that (x_{α})_{α>0} is a Cauchy sequence in C^{1}. Then by passing to the limit for α → 0 it follows that the limit function is a periodic solution for (1).

These results have been announced in [2]. The same approach applies for the study of almost periodic solutions (see [3]). Results concerning this topic have been obtained previously by other authors using different methods. A similar condition (2) has been investigated in [5] when studying the range of sums of monotone operators. A different method consists of applying fixed point techniques, see for example [4, 7].

This article is organized as follows. First we analyze the one dimensional case. Necessary and sufficient conditions for the existence and uniqueness of periodic solutions are shown. Results for sub(super)-periodic solutions are proved as well in this case. In the next section we show that the same existence result holds for linear symmetric maximal monotone operators on Hilbert spaces. In the last section the case of non-linear sub-differential operators is considered.

2 Periodic solutions for one dimensional evolution equations

To study the periodic solutions for evolution equations it is convenient to consider first the one dimensional case

x′(t) + g(x(t)) = f(t), t ∈ ℝ,  (3)

where g : ℝ → ℝ is increasing Lipschitz continuous in x and f : ℝ → ℝ is T -periodic and continuous in t. By Picard’s theorem it follows that for each initial data x(0) = x_{0} ∈ ℝ there is an unique solution x ∈ C^{1}(ℝ; ℝ) for (3). We are looking for T -periodic solutions. Let us start by the uniqueness study.

2.1 Uniqueness

Proposition 2.1 Assume that g is strictly increasing and f is periodic. Then there is at most one periodic solution for (3).

Proof Let x_{1}, x_{2} be two periodic solutions for (3). By taking the difference between the two equations and multiplying by x_{1}(t) − x_{2}(t) we get

\frac{1}{2} \frac{d}{dt}|x_{1}(t) − x_{2}(t)|^{2} + [g(x_{1}(t)) − g(x_{2}(t))][x_{1}(t) − x_{2}(t)] = 0, t ∈ ℝ. (4)

Since g is increasing we have (g(x_{1}) − g(x_{2}))(x_{1} − x_{2}) ≥ 0 for all x_{1}, x_{2} ∈ ℝ and therefore we deduce that |x_{1}(t)−x_{2}(t)| is decreasing. Moreover as x_{1} and x_{2} are periodic it follows that |x_{1}(t) − x_{2}(t)| does not depend on t ∈ ℝ and therefore, from (4) we get

[g(x_{1}(t)) − g(x_{2}(t))][x_{1}(t) − x_{2}(t)] = 0, t ∈ ℝ.

Finally, the strictly monotony of g implies that x_{1} = x_{2}.

Remark 2.2 If g is only increasing, it is possible that (3) has several periodic solutions. Let us consider the function

g(x) = \{ \begin{matrix}x + ε & x < −ε, \\ 0 & x ∈ [−ε, ε], \\ x − ε & x > ε, \end{matrix}  (5)

and f(t) = \frac{ε}{2} cos t. We can easily check that x_{λ}(t) = λ + \frac{ε}{2} sin t are periodic solutions for (3) for λ ∈ [−\frac{ε}{2}, \frac{ε}{2}].

Generally we can prove that every two periodic solutions differ by a constant.

Proposition 2.3 Let g be an increasing function and x_{1}, x_{2} two periodic solutions of (3). Then there is a constant C ∈ ℝ such that

x_{1}(t) − x_{2}(t) = C, ∀t ∈ ℝ.

Proof As shown before there is a constant C ∈ ℝ such that |x_{1}(t)−x_{2}(t)| = C, t ∈ ℝ. Moreover x_{1}(t) − x_{2}(t) has constant sign, otherwise x_{1}(t_{0}) = x_{2}(t_{0}) for some t_{0} ∈ ℝ and it follows that |x_{1}(t) − x_{2}(t)| = |x_{1}(t_{0}) − x_{2}(t_{0})| = 0, t ∈ ℝ or x_{1} = x_{2}. Finally we find that

x_{1}(t) − x_{2}(t) = sign(x_{1}(0) − x_{2}(0))C, t ∈ ℝ.

Before analyzing in detail the uniqueness for increasing functions, let us define the following sets.

O(y) = \{ \begin{matrix}\{x ∈ ℝ : x + ∫_{0}^{t}(f(s) − y)ds ∈ g^{−1}(y) ∀t ∈ ℝ\} ⊂ g^{−1}(y), & y ∈ g(ℝ), \\ ∅, & y ∉ g(ℝ). \end{matrix}

Proposition 2.4 Let g be an increasing function and f periodic. Then equation (3) has different periodic solutions if and only if Int(O⟨f⟩) ≠ ∅.

Proof Assume that (3) has two periodic solutions x_{1} ≠ x_{2}. By the previous proposition we have x_{2} − x_{1} = C > 0. By integration on [0, T] one gets

∫_{0}^{T} g(x_{1}(t))dt = ∫_{0}^{T} f(t)dt = ∫_{0}^{T} g(x_{2}(t))dt.  (6)

Since g is increasing we have g(x_{1}(t)) ≤ g(x_{2}(t)), t ∈ ℝ and therefore,

∫_{0}^{T} g(x_{1}(t))dt ≤ ∫_{0}^{T} g(x_{2}(t))dt.  (7)

From (6) and (7) we deduce that g(x_{1}(t)) = g(x_{2}(t)), t ∈ ℝ and thus g is constant on each interval [x_{1}(t), x_{2}(t)] = [x_{1}(t), x_{1}(t) + C], t ∈ ℝ. Finally it implies that g is constant on Range(x_{1}) + [0, C] = {x_{1}(t) + y : t ∈ [0, T], y ∈ [0, C]} and this constant is exactly the time average of f:

g(x_{1}(t)) = g(x_{2}(t)) = ⟨f⟩, t ∈ [0, T].

Let x be an arbitrary real number in ]x_{1}(0), x_{1}(0) + C[. Then

x + ∫_{0}^{t}{f(s) − ⟨f⟩}ds = x − x_{1}(0) + x_{1}(0) + ∫_{0}^{t}{f(s) − g(x_{1}(s))}ds
= x − x_{1}(0) + x_{1}(t)
> x_{1}(t), t ∈ ℝ.

Similarly,

x + ∫_{0}^{t}{f(s) − ⟨f⟩}ds = x − x_{2}(0) + x_{2}(0) + ∫_{0}^{t}{f(s) − g(x_{2}(s))}ds
= x − x_{2}(0) + x_{2}(t)
< x_{2}(t), t ∈ ℝ.

Therefore, x+∫_{0}^{t}{f(s)−⟨f⟩}ds ∈]x_{1}(t), x_{2}(t)[⊂ g^{−1}(⟨f⟩), t ∈ ℝ which implies that x ∈ O⟨f⟩ and hence ]x_{1}(0), x_{2}(0)[⊂ O⟨f⟩. Conversely, suppose that there is x and C > 0 small enough such that x, x+C ∈ O⟨f⟩. It is easy to check that x_{1}, x_{2} given below are different periodic solutions for (3):

x_{1}(t) = x + ∫_{0}^{t}{f(s) − ⟨f⟩}ds, t ∈ ℝ,
x_{2}(t) = x + C + ∫_{0}^{t}{f(s) − ⟨f⟩}ds = x_{1}(t) + C, t ∈ ℝ.

Remark 2.5 The condition Int(O⟨f⟩) ≠ ∅ is equivalent to

diam(g^{−1}⟨f⟩) > diam(Range ∫ {f(t) − ⟨f⟩}dt).

Example: Consider the equation x′(t) + g(x(t)) = η cos t, t ∈ ℝ with g given in Remark 2.2. We have < η cos t >= 0 ∈ g(ℝ) and

O(0) = {x ∈ ℝ | x + ∫_{0}^{t} η cos s ds ∈ g^{−1}(0), t ∈ ℝ}  (8)
= {x ∈ ℝ : x + η sin t ∈ g^{−1}(0), t ∈ ℝ}
= {x ∈ ℝ : −ε ≤ x + η sin t ≤ ε, t ∈ ℝ}
= \{ \begin{matrix}∅ & |η| > ε, \\ \{0\} & |η| = ε, \\ [|η| − ε, ε − |η|] & |η| < ε. \end{matrix}  (9)

Therefore, uniqueness does not occur if |η| < ε, for example if η = ε/2, as seen before in Remark 2.2. If |η| ≥ ε there is an unique periodic solution.

In the following we suppose that g is increasing and we establish an existence result.

2.2 Existence

To study the existence, note that a necessary condition is given by the following proposition.

Proposition 2.6 Assume that equation (3) has T -periodic solutions. Then there is x_{0} ∈ ℝ such that ⟨f⟩ := \frac{1}{T} ∫_{0}^{T} f(t)dt = g(x_{0}).

Proof Integrating on a period interval [0, T] we obtain

x(T) − x(0) + ∫_{0}^{T} g(x(t))dt = ∫_{0}^{T} f(t)dt.

Since x is periodic and g ∘ x is continuous we get

T g(x(τ)) = ∫_{0}^{T} f(t)dt, τ ∈]0, T [,

and hence

⟨f⟩ := \frac{1}{T} ∫_{0}^{T} f(t)dt ∈ Range(g).  (10)
♢

In the following we will show that this condition is also sufficient for the existence of periodic solutions. We will prove this result in several steps. First we establish the existence for the equation

αx_{α}(t) + x′_{α}(t) + g(x_{α}(t)) = f(t), t ∈ ℝ, α > 0.  (11)

Proposition 2.7 Suppose that g is increasing Lipschitz continuous and f is T -periodic and continuous. Then for every α > 0 the equation (11) has exactly one periodic solution.

Remark 2.8 Before starting the proof let us observe that (11) reduces to an equation of type (3) with g_{α} = α1_{ℝ} + g. Since g is increasing, is clear that g_{α} is strictly increasing and by the Proposition 2.1 we deduce that the uniqueness holds. Moreover since Range(g_{α}) = ℝ, the necessary condition (10) is trivially verified and therefore, in this case we can expect to prove existence.

Proof First of all remark that the existence of periodic solutions reduces to finding x_{0} ∈ ℝ such that the solution of the evolution problem

\begin{matrix}αx_{α}(t) + x′_{α}(t) + g(x_{α}(t)) = f(t), t ∈ [0, T], \\ x(0) = x_{0}, \end{matrix}  (12)

verifies x(T ; 0, x_{0}) = x_{0}. Here we denote by x(·; 0, x_{0}) the solution of (12) (existence and uniqueness assured by Picard’s theorem). We define the map S : ℝ → ℝ given by

S(x_{0}) = x(T ; 0, x_{0}), x_{0} ∈ ℝ.  (13)

We demonstrate the existence and uniqueness of the periodic solution of (12) by showing that the Banach’s fixed point theorem applies. Let us consider two solutions of (12) corresponding to the initial datas x_{0}^{1} and x_{0}^{2}. Using the monotony of g we can write

α|x(t; 0, x_{0}^{1}) − x(t; 0, x_{0}^{2})|^{2} + \frac{1}{2} \frac{d}{dt}|x(t; 0, x_{0}^{1}) − x(t; 0, x_{0}^{2})|^{2} ≤ 0,

which implies

\frac{1}{2} \frac{d}{dt}\{e^{2αt}|x(t; 0, x_{0}^{1}) − x(t; 0, x_{0}^{2})|^{2}\} ≤ 0,

and therefore,

|S(x_{0}^{1}) − S(x_{0}^{2})| = |x(T ; 0, x_{0}^{1}) − x(T ; 0, x_{0}^{2})| ≤ e^{−αT} |x_{0}^{1} − x_{0}^{2}|.

For α > 0 S is a contraction and the Banach’s fixed point theorem applies. Therefore S(x_{0}) = x_{0} for an unique x_{0} ∈ ℝ and hence x(·; 0, x_{0}) is a periodic solution of (3). ♢

Naturally, in the following proposition we inquire about the convergence of (x_{α})_{α>0} to a periodic solution of (3) as α → 0. In view of the Proposition 2.6 this convergence does not hold if (10) is not verified. Assume for the moment that (3) has at least one periodic solution. In this case convergence holds.

Proposition 2.9 If equation (3) has at least one periodic solution, then (x_{α})_{α>0} is convergent in C^{0}(ℝ; ℝ) and the limit is also a periodic solution of (3).

Proof Denote by x a periodic solution of (3). By elementary calculations we find

α|x_{α}(t) − x(t)|^{2} + \frac{1}{2} \frac{d}{dt}|x_{α}(t) − x(t)|^{2} ≤ −αx(t)(x_{α}(t) − x(t)), t ∈ ℝ, (17)

which can be also written as

\frac{1}{2} \frac{d}{dt}\{e^{2αt}|x_{α}(t) − x(t)|^{2}\} ≤ αe^{αt}|x(t)| · e^{αt}|x_{α}(t) − x(t)|, t ∈ ℝ.  (18)

Therefore, by integration on [0, t] we deduce

\frac{1}{2}\{e^{αt}|x_{α}(t) − x(t)|\}^{2} ≤ \frac{1}{2}|x_{α}(0) − x(0)|^{2} + ∫_{0}^{t} αe^{αs}|x(s)| · e^{αs}|x_{α}(s) − x(s)|ds.

(19)

Using Bellman’s lemma, formula (19) gives

e^{αt}|x_{α}(t) − x(t)| ≤ |x_{α}(0) − x(0)| + ∫_{0}^{t} αe^{αs}|x(s)|ds, t ∈ ℝ.  (20)

Let us consider α > 0 fixed for the moment. Since x is periodic and continuous, it is also bounded and therefore from (20) we get

|x_{α}(t) − x(t)| ≤ e^{−αt}|x_{α}(0) − x(0)| + (1 − e^{−αt})∥x∥_{L^{∞}(ℝ)}, t ∈ ℝ. (21)

By periodicity we have

|x_{α}(t) − x(t)| = |x_{α}(nT + t) − x(nT + t)|
≤ e^{−α(nT +t)}|x_{α}(0) − x(0)| + (1 − e^{−α(nT +t)})∥x∥_{L^{∞}(ℝ)}
≤ e^{−α(nT +t)}|x_{α}(0) − x(0)| + ∥x∥_{L^{∞}(ℝ)}, t ∈ ℝ, n ≥ 0.

By passing to the limit as n → ∞, we deduce that (x_{α})_{α>0} is uniformly bounded in L^{∞}(ℝ):

|x_{α}(t)| ≤ |x_{α}(t) − x(t)| + |x(t)| ≤ 2∥x∥_{L^{∞}(ℝ)}, t ∈ ℝ, α > 0.

The derivatives x′_{α} are also uniformly bounded in L^{∞}(ℝ) for α → 0:

|x′_{α}(t)|
= |f(t) − αx_{α}(t) − g(x_{α}(t))|
≤ ∥f∥_{L^{∞}(ℝ)} + 2α∥x∥_{L^{∞}(ℝ)} + max{g(2∥x∥_{L^{∞}(ℝ)}), −g(−2∥x∥_{L^{∞}(ℝ)})}.

The uniform convergence of (x_{α})_{α>0} follows now from the Arzela-Ascoli’s theorem. Denote by u the limit of (x_{α})_{α>0} as α → 0. Obviously u is also periodic

u(0) = lim_{α→0} x_{α}(0) = lim_{α→0} x_{α}(T) = u(T).

To prove that u verifies (3), we write

x_{α}(t) = x_{α}(0) + ∫_{0}^{t}{f(s) − g(x_{α}(s)) − αx_{α}(s)}ds, t ∈ ℝ.

Since the convergence is uniform, by passing to the limit for α → 0 we obtain

u(t) = u(0) + ∫_{0}^{t}{f(s) − g(u(s))}ds,

and hence u ∈ C^{1}(ℝ; ℝ) and

u′(t) + g(u(t)) = f(t), t ∈ ℝ.

From the previous proposition we conclude that the existence of periodic solutions for (3) reduces to uniform estimates in L^{∞}(ℝ) for (x_{α})_{α>0}.

Proposition 2.10 Assume that g is increasing Lipschitz continuous and f is T -periodic and continuous. Then the following statements are equivalent:

(i) equation (3) has periodic solutions; (ii) the sequence (x_{α})_{α>0} is uniformly bounded in L^{∞}(ℝ). Moreover, in this case (x_{α})_{α>0} is convergent in C^{0}(ℝ; ℝ) and the limit is a periodic solution for (3).

Note that generally we can not estimate (x_{α})_{α>0} uniformly in L^{∞}(ℝ). Indeed, by standard computations we obtain

α(x_{α}(t) − u)^{2} + \frac{1}{2} \frac{d}{dt}(x_{α}(t) − u)^{2} ≤ |f(t) − αu − g(u)| · |x_{α}(t) − u|, t, u ∈ ℝ

and therefore

\frac{1}{2} \frac{d}{dt}\{e^{2αt}(x_{α}(t) − u)^{2}\} ≤ e^{αt}|f(t) − αu − g(u)| · e^{αt}|x_{α}(t) − u|, t, u ∈ ℝ.

Integration on [t, t + h], we get

\frac{1}{2}e^{2α(t+h)}(x_{α}(t + h) − u)^{2} ≤ ∫_{t}^{t+h} e^{2αs}|f(s) − αu − g(u)| · |x_{α}(s) − u|ds

+\frac{1}{2}e^{2αt}(x_{α}(t) − u)^{2}, t < t + h, u ∈ ℝ.

Now by using Bellman’s lemma we deduce

|x_{α}(t+h)−u| ≤ e^{−αh}|x_{α}(t)−u|+∫_{t}^{t+h}e^{−α(t+h−s)}|f(s)−αu−g(u)|ds, t < t+h.

Since x_{α} is T -periodic, by taking h = T we can write

|x_{α}(t) − u| ≤ \frac{1}{1 − e^{−αT}} ∫_{0}^{T} e^{−α(T −s)}|f(s) − αu − g(u)|ds, t ∈ ℝ,

and thus for u = 0 we obtain

∥x_{α}∥_{L^{∞}(ℝ)} ≤ \frac{1}{1 − e^{−αT}} ∫_{0}^{T} |f(s) − g(0)|ds ∼ O (\frac{1}{α}) , α > 0.

Now we can state our main existence result.

Theorem 2.11 Assume that g is increasing Lipschitz continuous, and f is T - periodic and continuous. Then equation (3) has periodic solutions if and only if ⟨f⟩ := \frac{1}{T} ∫_{0}^{T} f(t)dt ∈ Range(g) (there is x_{0} ∈ ℝ such that ⟨f⟩ = g(x_{0})). Moreover in this case we have the estimate

∥x∥_{L^{∞}(ℝ)} ≤ |x_{0}| + ∫_{0}^{T} |f(t) − ⟨f⟩|dt, ∀ x_{0} ∈ g^{−1}⟨f⟩,

and the solution is unique if and only if Int(O⟨f⟩) = ∅ or

diam(g^{−1}⟨f⟩) ≤ diam(Range ∫ {f(t) − ⟨f⟩}dt).

Proof The condition is necessary (see Proposition 2.6). We will prove now that it is also sufficient. Let us consider the sequence of periodic solutions (x_{α})_{α>0} of (11). Accordingly to the Proposition 2.10 we need to prove uniform estimates in L^{∞}(ℝ) for (x_{α})_{α>0}. Since x_{α} is T -periodic by integration on [0, T] we get

∫_{0}^{T} {αx_{α}(t) + g(x_{α}(t))}dt = T ⟨f⟩, α > 0.

Using the average formula for continuous functions we have

∫_{0}^{T} {αx_{α}(t) + g(x_{α}(t))}dt = T {αx_{α}(t_{α}) + g(x_{α}(t_{α}))}, t_{α} ∈]0, T [, α > 0.

By the hypothesis there is x_{0} ∈ ℝ such that ⟨f⟩ = g(x_{0}) and thus

αx_{α}(t_{α}) + g(x_{α}(t_{α})) = g(x_{0}), α > 0.  (22)

Since g is increasing, we deduce

αx_{α}(t_{α})[x_{0} − x_{α}(t_{α})] = [g(x_{0}) − g(x_{α}(t_{α}))][x_{0} − x_{α}(t_{α})] ≥ 0, α > 0,

and thus

|x_{α}(t_{α})|^{2} ≤ x_{α}(t_{α})x_{0} ≤ |x_{α}(t_{α})||x_{0}|.

Finally we deduce that x_{α}(t_{α}) is uniformly bounded in ℝ:

|x_{α}(t_{α})| ≤ |x_{0}|, ∀ α > 0.

Now we can easily find uniform estimates in L^{∞}(ℝ) for (x_{α})_{α>0}. Let us take in the previous calculus u = x_{α}(t_{α})and integrate on [t_{α}, t]:

\frac{1}{2}e^{2αt}(x_{α}(t)−x_{α}(t_{α}))^{2} ≤ ∫_{t_{α}}^{t} e^{2αs}|f(s)−αx_{α}(t_{α})−g(x_{α}(t_{α}))|·|x_{α}(s)−x_{α}(t_{α})|ds.

By using Bellman’s lemma we get

|x_{α}(t) − x_{α}(t_{α})| ≤ ∫_{t_{α}}^{t} e^{−α(t−s)}|f(s) − αx_{α}(t_{α}) − g(x_{α}(t_{α}))|ds, t > t_{α},

and hence by (22) we deduce

|x_{α}(t)| ≤ |x_{0}| + ∫_{0}^{T} |f(t) − αx_{α}(t_{α}) − g(x_{α}(t_{α}))|dt
= |x_{0}| + ∫_{0}^{T} |f(t) − ⟨f⟩|dt, t ∈ ℝ, α > 0.  (23)

Now by passing to the limit in (23) we get

|x(t)| ≤ |x_{0}| + ∫_{0}^{T} |f(t) − ⟨f⟩|dt, t ∈ ℝ, ∀ x_{0} ∈ g^{−1}⟨f⟩.

2.3 Sub(super)-periodic solutions

In this part we generalize the previous existence results for sub(super)-periodic solutions. We will see that similar results hold. Let us introduce the concept of sub(super)-periodic solutions.

Definition 2.12 We say that x ∈ C^{1}([0, T]; ℝ) is a sub-periodic solution for (3) if

x′(t) + g(x(t)) = f(t), t ∈ [0, T],

and x(0) ≤ x(T).

Note that a necessary condition for the existence is given next.

Proposition 2.13 If equation (3) has sub-periodic solutions, then there is x_{0} ∈ ℝ such that g(x_{0}) ≤ ⟨f⟩.

Proof Let x be a sub-periodic solution of (3). By integration on [0, T] we find

x(T) − x(0) + ∫_{0}^{T} g(x(t))dt = T ⟨f⟩.

Since g ∘ x is continuous, there is τ ∈]0, T [ such that

g(x(τ)) = ⟨f⟩ − \frac{1}{T} (x(T) − x(0)) ≤ ⟨f⟩.

Similarly we define the notion of super-periodic solution.

Definition 2.14 We say that y ∈ C^{1}([0, T]; ℝ) is a super-periodic solution for (3) if

y′(t) + g(y(t)) = f(t), t[0, T],

and y(0) ≥ y(T).

The analogous necessary condition holds.

Proposition 2.15 If equation (3) has super-periodic solutions, then there is y_{0} ∈ ℝ such that g(y_{0}) ≥ ⟨f⟩.

Remark 2.16 It is clear that x is periodic solution for (3) if and only if is in the same time sub-periodic and super-periodic solution. Therefore there are x_{0}, y_{0} ∈ ℝ such that

g(x_{0}) ≤ ⟨f⟩ ≤ g(y_{0}).

Since g is continuous, we deduce that ⟨f⟩ ∈ Range(g) which is exactly the necessary condition given by the Proposition 2.6.

As before we will prove that the necessary condition of Proposition 2.13 is also sufficient for the existence of sub-periodic solutions.

Theorem 2.17 Assume that g is increasing Lipschitz continuous and f is T - periodic continuous. Then equation (3) has sub-periodic solutions if and only if there is x_{0} ∈ ℝ such that g(x_{0}) ≤ ⟨f⟩.

Proof The condition is necessary (see Proposition 2.13). Let us prove now that it is also sufficient. Consider z_{0} an arbitrary initial data and denote by x : [0, ∞[→ ℝ the solution for (3) with the initial condition x(0) = z_{0}. If there is t_{0} ≥ 0 such that x(t_{0}) ≤ x(t_{0} + T), thus x_{t_{0}}(t) := x(t_{0} + t), t ∈ [0, T] is a sub-periodic solution. Suppose now that x(t) > x(t+T), ∀t ∈ ℝ. By integration on [nT, (n + 1)T], n ≥ 0 we get

x((n + 1)T) − x(nT) + ∫_{0}^{T} g(x(nT + t))dt = T ⟨f⟩, n ≥ 0.

Using the hypothesis and the average formula we have

g(x(nT + τ_{n})) = ⟨f⟩ + \frac{1}{T} \{x(nT) − x((n + 1)T)\} > g(x_{0}),

for τ_{n} ∈]0, T [ and n ≥ 0. Since g is increasing we deduce that x(nT + τ_{n}) > x_{0}, n ≥ 0. We have also x(nT + τ_{n}) ≤ x((n − 1)T + τ_{n}) ≤ ⋯ ≤ x(τ_{n}) ≤ sup_{t∈[0,T]} |x(t)| and thus we deduce that (x(nT + τ_{n}))_{n≥0} is bounded:

|x(nT + τ_{n})| ≤ K, n ≥ 0.

Consider now the functions x_{n} : [0, T] → ℝ given by

x_{n}(t) = x(nT + t), t ∈ [0, T].

By a standard computation we get

\frac{1}{2} \frac{d}{dt}|x_{n}(t)|^{2} + [g(x_{n}(t)) − g(0)]x_{n}(t) = [f(t) − g(0)]x_{n}(t), t ∈ [0, T].

Using the monotony of g we obtain

|x_{n}(t)| ≤ |x_{n}(s)| + ∫_{s}^{t} |f(u) − g(0)|du, 0 ≤ s ≤ t ≤ T.

Taking s = τ_{n} ∈]0, T [ we can write

|x_{n}(t)| ≤ |x_{n}(τ_{n})| + ∫_{τ_{n}}^{t} |f(u) − g(0)|du ≤ K + ∫_{0}^{T} |f(u) − g(0)|du, t ∈ [τ_{n}, T].

For t ∈ [0, τ_{n}], n ≥ 1 we have

|x_{n}(t)| = |x(nT + t)| ≤ |x((n − 1)T + τ_{n−1})| + ∫_{(n−1)T +τ_{n−1}}^{nT +t} |f(u) − g(0)|du

≤ K + ∫_{(n−1)T}^{(n+1)T} |f(u) − g(0)|du

≤ K + 2 ∫_{0}^{T} |f(u) − g(0)|du.

Therefore, the sequence (x_{n})_{n≥0} is uniformly bounded in L^{∞}(ℝ) and

∥x_{n}∥_{L^{∞}(ℝ)} ≤ K + 2 ∫_{0}^{T} |f(t) − g(0)|dt := M.

Moreover, (x′_{n})_{n≥0} is also uniformly bounded in L^{∞}(ℝ). Indeed we have

|x′_{n}(t)| = |f(t) − g(x_{n}(t))| ≤ ∥f∥_{L^{∞}(ℝ)} + max{g(M), −g(−M)},

and hence, by Arzela-Ascoli’s theorem we deduce that (x_{n})_{n≥0} converges in C^{0}([0, T], ℝ):

lim_{n→∞} x_{n}(t) = u(t), uniformly for t ∈ [0, T].

As usual, by passing to the limit for n → ∞ we find that u is also solution for (3). Moreover since (x(nT))_{n≥0} is decreasing and bounded, it is convergent and we can prove that u is periodic:

u(0) = lim_{n→∞} x_{n}(0) = lim_{n→∞} x(nT) = lim_{n→∞} x((n + 1)T) = lim_{n→∞} x_{n}(T) = u(T).

Therefore, u is a sub-periodic solution for (3). An analogous result holds for super-periodic solutions.

Proposition 2.18 Under the same assumptions as in Theorem 2.17 the equation (3) has super-periodic solutions if and only if there is y_{0} ∈ ℝ such that g(y_{0}) ≥ ⟨f⟩.

We state now a comparison result between sub-periodic and super-periodic solutions.

Proposition 2.19 If g is increasing, x is a sub-periodic solution and y is a super-periodic solution we have

x(t) ≤ y(t), ∀t ∈ [0, T],

provided that x and y are not both periodic.

Proof Both x and y verify (3), thus

(x − y)′(t) + g(x(t)) − g(y(t)) = 0, t ∈ [0, T].

With the notation

r(t) = \{ \begin{matrix}\frac{g(x(t))−g(y(t))}{x(t)−y(t)} & t ∈ [0, T], x(t) ≠ y(t) \\ 0 & t ∈ [0, T], x(t) = y(t), \end{matrix}  (24)

we can write g(x(t)) − g(y(t)) = r(t)(x(t) − y(t)), t ∈ [0, T] and therefore,

(x − y)′(t) + r(t)(x(t) − y(t)) = 0, t ∈ [0, T]

which implies

x(t) − y(t) = (x(0) − y(0))e^{− ∫_{0}^{t} r(s)ds}.  (25)

Now it is clear that if x(0) ≤ y(0) we also have x(t) ≤ y(t), t ∈ [0, T]. Suppose now that x(0) > y(0). Taking t = T in (25) we obtain

x(T) − y(T) = (x(0) − y(0))e^{− ∫_{0}^{T} r(t)dt}.  (26)

Since g is increasing, by the definition of the function r we have r ≥ 0. Two cases are possible: (i) either ∫_{0}^{T} r(t)dt > 0, (ii) either ∫_{0}^{T} r(t)dt = 0 in which case r(t) = 0, t ∈ [0, T] (r vanishes in all points of continuity t such that x(t) ≠ y(t) and also in all points t with x(t) = y(t) by the definition). Let us analyse the first case (i). By (26) we deduce that x(T) − y(T) < x(0) − y(0) or x(T) − x(0) < y(T) − y(0). Since x is sub-periodic we have x(0) ≤ x(T) which implies that y(T) > y(0) which is in contradiction with the super-periodicity of y ( y(T) ≤ y(0)). In the second case (ii) we have g(x(t)) = g(y(t)), t ∈ [0, T] so (x − y)′ = 0 and therefore there is a constant C ∈ ℝ such that x(t) = y(t) + C, t ∈ [0, T]. Taking t = 0 and t = T we obtain

0 ≥ x(0) − x(T) = y(0) − y(T) ≥ 0,

and thus x and y are both periodic which is in contradiction with the hypothesis. In the following we will see how it is possible to retrieve the existence result for periodic solutions by using the method of sub(super)-periodic solutions. Suppose that ⟨f⟩ ∈ Range(g). Obviously both sufficient conditions for existence of sub(super)-periodic solutions are satisfied and thus there are x_{0}(y_{0}) sub(super)- periodic solutions. If y_{0} is even periodic the proof is complete. Assume that y_{0} is not periodic (y_{0}(0) > y_{0}(T)). Denote by M the set of sub-periodic solutions for (3):

M = {x : [0, T] → ℝ : x sub-periodic solution , x_{0}(t) ≤ x(t), t ∈ [0, T]}.

Since x_{0} ∈ M we have M ≠ ∅. Moreover, from the comparison result since y_{0} is super-periodic but not periodic we have x ≤ y_{0}, ∀x ∈ M. We prove that M contains a maximal element in respect to the order:

x_{1} ≺ x_{2} (if and only if) x_{1}(t) ≤ x_{2}(t), t ∈ [0, T].

Finally we show that this maximal element is even a periodic solution for (3) since otherwise it would be possible to construct a sub-periodic solution greater than the maximal element. We state now the following generalization.

Theorem 2.20 Assume that g : ℝ × ℝ → ℝ is increasing Lipschitz continuous function in x, T -periodic and continuous in t and f : ℝ → ℝ is T -periodic and continuous in t. Then the equation

x′(t) + g(t, x(t)) = f(t), t ∈ ℝ,  (27)

has periodic solutions if and only if there is x_{0} ∈ ℝ such that

⟨f⟩ := \frac{1}{T} ∫_{0}^{T} f(t)dt = \frac{1}{T} ∫_{0}^{T} g(t, x_{0})dt = G(x_{0}).  (28)

Moreover, in this case we have the estimate

∥x∥_{L^{∞}(ℝ)} ≤ |x_{0}| + ∫_{0}^{T} |f(t) − g(t, x_{0})|dt, ∀ x_{0} ∈ G^{−1}⟨f⟩.

Proof Consider the average function G : ℝ → ℝ given by

G(x) = \frac{1}{T} ∫_{0}^{T} g(t, x)dt, x ∈ ℝ.

It is easy to check that G is also increasing and Lipschitz continuous with the same constant. Let us prove that the condition (28) is necessary. Suppose that x is a periodic solution for (27). By integration on [0, T] we get

\frac{1}{T} ∫_{0}^{T} g(t, x(t))dt = ⟨f⟩.  (29)

We can write

m ≤ x(t) ≤ M, t ∈ [0, T],

and thus

g(t, m) ≤ g(t, x(t)) ≤ g(t, M), t ∈ [0, T],

which implies

G(m) = \frac{1}{T} ∫_{0}^{T} g(t, m)dt ≤ \frac{1}{T} ∫_{0}^{T} g(t, x(t))dt ≤ \frac{1}{T} ∫_{0}^{T} g(t, M)dt = G(M).

Since G is continuous it follows that there is x_{0} ∈ [m, M] such that G(x_{0}) = \frac{1}{T} ∫_{0}^{T} g(t, x(t))dt and from (29) we deduce that ⟨f⟩ = G(x_{0}). Let us show that the condition (28) is also sufficient. As before let us consider the unique periodic solution for

αx_{α}(t) + x′_{α}(t) + g(t, x_{α}(t)) = f(t), t ∈ [0, T], α > 0,

(existence and uniqueness follow by the Banach’s fixed point theorem exactly as before). All we need to prove is that (x_{α})_{α>0} is uniformly bounded in L^{∞}(ℝ) (then (x′_{α})_{α>0} is also uniformly bounded in L^{∞}(ℝ) and by Arzela-Ascoli’s theorem we deduce that x_{α} converges to a periodic solution for (27)). Taking the average on [0, T] we get

\frac{1}{T} ∫_{0}^{T} \{αx_{α}(t) + g(t, x_{α}(t))\}dt = ⟨f⟩ = G(x_{0}), α > 0.

As before we can write

αm_{α} + g(t, m_{α}) ≤ αx_{α}(t) + g(t, x_{α}(t)) ≤ αM_{α} + g(t, M_{α}), t ∈ [0, T], α > 0,

where

m_{α} ≤ x_{α}(t) ≤ M_{α}, t ∈ [0, T], α > 0,

and hence

αm_{α} + G(m_{α}) ≤ \frac{1}{T} ∫_{0}^{T} \{αx_{α}(t) + g(t, x_{α}(t))\}dt ≤ αM_{α} + G(M_{α}), α > 0.

Finally we get

G(x_{0}) = \frac{1}{T} ∫_{0}^{T} \{αx_{α}(t)+g(t, x_{α}(t))\}dt = αu_{α}+G(u_{α}), u_{α} ∈]m_{α}, M_{α}[, α > 0.

(30)

Multiplying by u_{α} − x_{0} we obtain

αu_{α}(u_{α} − x_{0}) = −(G(x_{0}) − G(u_{α}))(x_{0} − u_{α}), α > 0.

Since G is increasing we deduce that |u_{α}|^{2} ≤ u_{α}x_{0} ≤ |u_{α}|·|x_{0}|, α > 0 and hence (u_{α})_{α>0} is bounded:

|u_{α}| ≤ |x_{0}|, α > 0.

Now using (30) it follows

\frac{1}{T} ∫_{0}^{T} \{αx_{α}(t) + g(t, x_{α}(t))\}dt = \frac{1}{T} ∫_{0}^{T} \{αu_{α} + g(t, u_{α})\}dt,

and thus there is t_{α} ∈]0, T [ such that

αx_{α}(t_{α}) + g(t_{α}, x_{α}(t_{α})) = αu_{α} + g(t_{α}, u_{α}), α > 0.

Since α(x_{α}(t_{α}) − u_{α})^{2} = −[g(t_{α}, x_{α}(t_{α})) − g(t_{α}, u_{α})][x_{α}(t_{α}) − u_{α}] ≤ 0 we find that x_{α}(t_{α}) = u_{α}, α > 0 and thus (x_{α}(t_{α}))_{α>0} is also bounded

|x_{α}(t_{α})| ≤ |x_{0}|, α > 0.

Now by standard calculations we can write

\frac{1}{2} \frac{d}{dt}|x_{α}(t) − x_{α}(t_{α})|^{2} + [g(t, x_{α}(t)) − g(t, x_{α}(t_{α}))][x_{α}(t) − x_{α}(t_{α})]

≤ [f(t) − αx_{α}(t_{α}) − g(t, x_{α}(t_{α}))][x_{α}(t) − x_{α}(t_{α})], t ∈ ℝ,

and thus

|x_{α}(t) − x_{α}(t_{α})| ≤ ∫_{t_{α}}^{t} |f(s) − αx_{α}(t_{α}) − g(s, x_{α}(t_{α}))|ds, t > t_{α}, α > 0,

which implies

|x_{α}(t)| ≤ |x_{0}| + ∫_{0}^{T} |f(t) − αx_{α}(t_{α}) − g(t, x_{α}(t_{α}))|dt, t ∈ [0, T], α > 0. (31)

Since (x_{α}(t_{α}))_{α>0} is bounded we have

u_{α} = x_{α}(t_{α}) → x_{1},

such that

G(x_{0}) = lim_{α→0}{αu_{α} + G(u_{α})} = G(x_{1}).

Moreover, if x_{0} ≤ x_{1} we have

0 ≤ \frac{1}{T} ∫_{0}^{T} [g(t, x_{1}) − g(t, x_{0})]dt = G(x_{1}) − G(x_{0}) = 0,

and hence g(t, x_{1}) = g(t, x_{0}) for all t ∈ [0, T]. Obviously the same equalities hold if x_{0} > x_{1}. Now by passing to the limit in (31) we find

|x(t)| ≤ |x_{0}| + ∫_{0}^{T} |f(t) − g(t, x_{1})|dt  (32)
= |x_{0}| + ∫_{0}^{T} |f(t) − g(t, x_{0})|dt, t ∈ [0, T], ∀ x_{0} ∈ G^{−1}⟨f⟩,

and therefore (x_{α})_{α>0} is uniformly bounded in L^{∞}(ℝ).

3 Periodic solutions for evolution equations on Hilbert spaces

In this section we analyze the existence and uniqueness of periodic solutions for general evolution equations on Hilbert spaces

x′(t) + Ax(t) = f(t), t > 0,  (33)

where A : D(A) ⊂ H → H is a maximal monotone operator on a Hilbert space H and f ∈ C^{1}(ℝ; H) is a T -periodic function. As known by the theory of Hille-Yosida, for every initial data x_{0} ∈ D(A) there is an unique solution x ∈ C^{1}([0, +∞[; H) ∩ C([0, +∞[ ; D(A)) for (33), see [6, p. 101]. Obviously, the periodic problem reduces to find x_{0} ∈ D(A) such that x(T) = x_{0}. As in the one dimensional case we demonstrate uniqueness for strictly monotone operators. We state also necessary and sufficient condition for the existence in the linear symmetric case. Finally the case of non-linear sub-differential operators is considered. Let us start with the definition of periodic solutions for Definition 3.1 Let A : D(A) ⊂ H → H be a maximal monotone operator on a Hilbert space H and f ∈ C^{1}(ℝ; H) a T -periodic function. We say that x ∈ C^{1}([0, T]; H) ∩ C([0, T]; D(A)) is a periodic solution for (33) if and only if

x′(t) + Ax(t) = f(t), t ∈ [0, T],

and x(0) = x(T).

3.1 Uniqueness

Generally the uniqueness does not hold (see the example in the following paragraph). However it occurs under the hypothesis of strictly monotony.

Proposition 3.2 Assume that A is strictly monotone ((Ax_{1}−Ax_{2}, x_{1}−x_{2}) = 0 implies x_{1} = x_{2}). Then (33) has at most one periodic solution.

Proof Let x_{1}, x_{2} be two different periodic solutions. By taking the difference of (33) and multiplying both sides by x_{1}(t) − x_{2}(t) we find

\frac{1}{2} \frac{d}{dt}∥x_{1}(t) − x_{2}(t)∥^{2} + (Ax_{1}(t) − Ax_{2}(t), x_{1}(t) − x_{2}(t)) = 0, t ∈ [0, T].

By the monotony of A we deduce that ∥x_{1} − x_{2}∥^{2} is decreasing and therefore we have

∥x_{1}(0) − x_{2}(0)∥ ≥ ∥x_{1}(t) − x_{2}(t)∥ ≥ ∥x_{1}(T) − x_{2}(T)∥, t ∈ [0, T].

Since x_{1} and x_{2} are T -periodic we have

∥x_{1}(0) − x_{2}(0)∥ = ∥x_{1}(T) − x_{2}(T)∥,

which implies that ∥x_{1}(t) − x_{2}(t)∥ is constant for t ∈ [0, T] and thus

(Ax_{1}(t) − Ax_{2}(t), x_{1}(t) − x_{2}(t)) = 0, t ∈ [0, T].

Now uniqueness follows by the strictly monotony of A.

3.2 Existence

In this section we establish existence results. In the linear case we state the following necessary condition.

Proposition 3.3 Let A : D(A) ⊂ H → H be a linear maximal monotone operator and f ∈ L^{1}(]0, T [; H) a T -periodic function. If (33) has T -periodic solutions, then the following necessary condition holds.

⟨f⟩ := \frac{1}{T} ∫_{0}^{T} f(t)dt ∈ Range(A),

(there is x_{0} ∈ D(A) such that ⟨f⟩ = Ax_{0}).

Proof Suppose that x ∈ C^{1}([0, T]; H)∩C([0, T]; D(A)) is a T -periodic solution for (33). Let us consider the divisions Δ_{n} : 0 = t_{0}^{n} < t_{1}^{n} < ⋯ < t_{n}^{n} = T such that

lim_{n→∞} max_{1≤i≤n} |t_{i}^{n} − t_{i−1}^{n}| = 0.  (34)

We can write

(t_{i}^{n} − t_{i−1}^{n})x′(t_{i−1}^{n}) + (t_{i}^{n} − t_{i−1}^{n})Ax(t_{i−1}^{n}) = (t_{i}^{n} − t_{i−1}^{n})f(t_{i−1}^{n}), 1 ≤ i ≤ n.

Since A is linear we deduce

\frac{1}{T} ∑_{i=1}^{n}(t_{i}^{n} −t_{i−1}^{n})x′(t_{i−1}^{n})+A(\frac{1}{T} ∑_{i=1}^{n}(t_{i}^{n} −t_{i−1}^{n})x(t_{i−1}^{n})) = \frac{1}{T} ∑_{i=1}^{n}(t_{i}^{n} −t_{i−1}^{n})f(t_{i−1}^{n}),

and hence

[\frac{1}{T} ∑_{i=1}^{n}(t_{i}^{n} − t_{i−1}^{n})x(t_{i−1}^{n})), \frac{1}{T} ∑_{i=1}^{n}(t_{i}^{n} − t_{i−1}^{n})[f(t_{i−1}^{n}) − x′(t_{i−1}^{n})]] ∈ A.

By (34) we deduce that

\frac{1}{T} ∑_{i=1}^{n}(t_{i}^{n} − t_{i−1}^{n})x(t_{i−1}^{n})) → \frac{1}{T} ∫_{0}^{T} x(t)dt,

and

\frac{1}{T} ∑_{i=1}^{n}(t_{i}^{n} − t_{i−1}^{n})[f(t_{i−1}^{n}) − x′(t_{i−1}^{n})] → \frac{1}{T} ∫_{0}^{T} [f(t) − x′(t)]dt

= \frac{1}{T} ∫_{0}^{T} f(t)dt − \frac{1}{T} x(t)|_{0}^{T}
= \frac{1}{T} ∫_{0}^{T} f(t)dt.

Since A is maximal monotone Graph(A) is closed and therefore

[\frac{1}{T} ∫_{0}^{T} x(t)dt, \frac{1}{T} ∫_{0}^{T} f(t)dt] ∈ A.

Thus \frac{1}{T} ∫_{0}^{T} x(t)dt ∈ D(A) and ⟨f⟩ = A(\frac{1}{T} ∫_{0}^{T} x(t)dt). Generally the previous condition is not sufficient for the existence of periodic solutions. For example let us analyse the periodic solutions x = (x_{1}, x_{2}) ∈ C^{1}([0, T]; ℝ^{2}) for

x′(t) + Ax(t) = f(t), t ∈ [0, T],  (35)

where A : ℝ^{2} → ℝ^{2} is the orthogonal rotation:

A(x_{1}, x_{2}) = (−x_{2}, x_{1}), (x_{1}, x_{2}) ∈ ℝ^{2},

and f = (f_{1}, f_{2}) ∈ L^{1}(]0, T [; ℝ^{2}) is T -periodic. For a given initial data x(0) = x_{0} ∈ ℝ^{2} the solution writes

x(t) = e^{−tA}x_{0} + ∫_{0}^{t} e^{−(t−s)A}f(s)ds, t > 0,  (36)

where the semigroup e^{−tA} is given by

e^{−tA} = (\begin{matrix}cos t & sin t \\ − sin t & cos t \end{matrix}) .  (37)

Since e^{−2πA} = 1 we deduce that the equation (35) has 2π-periodic solutions if and only if

∫_{0}^{2π} e^{tA}f(t)dt = 0.  (38)

Thus if ∫_{0}^{2π}{f_{1}(t) cos t − f_{2}(t) sin t}dt ≠ 0 or ∫_{0}^{2π}{f_{1}(t) sin t + f_{2}(t) cos t}dt ≠ 0 equation (35) does not have any 2π-periodic solution and the necessary condition still holds because Range(A) = ℝ^{2}. Moreover if (38) is satisfied then every solution of (35) is periodic and therefore uniqueness does not occur (the operator A is not strictly monotone). Let us analyse now the existence. As in the one dimensional case we have

Proposition 3.4 Suppose that A : D(A) ⊂ H → H is maximal monotone and f ∈ C^{1}(ℝ; H) is T -periodic. Then for every α > 0 the equation

αx(t) + x′(t) + Ax(t) = f(t), t ∈ ℝ,  (39)

has an unique T -periodic solution in C^{1}(ℝ; H) ∩ C(ℝ; D(A)).

Proof Since α+A is strictly monotone the uniqueness follows from Proposition 3.2. Indeed,

α∥x − y∥^{2} + (Ax − Ay, x − y) = 0, x, y ∈ D(A),

implies α∥x − y∥^{2} = 0 and hence x = y. Consider now an arbitrary initial data x_{0} ∈ D(A). By the Hille-Yosida’s theorem, there is x ∈ C^{1}([0, +∞[; H) ∩ C([0, +∞[; D(A)) solution for (39). Denote by (x_{n})_{n≥0} the functions

x_{n}(t) = x(nT + t), t ∈ [0, T], n ≥ 0.

We have

αx_{n+1}(t) + x′_{n+1}(t) + Ax_{n+1}(t) = f((n + 1)T + t), t ∈ [0, T],

and

αx_{n}(t) + x′_{n}(t) + Ax_{n}(t) = f(nT + t), t ∈ [0, T].

Since f is T -periodic, after usual computations we get

α∥x_{n+1}(t) − x_{n}(t)∥^{2} + \frac{1}{2} \frac{d}{dt}∥x_{n+1}(t) − x_{n}(t)∥^{2}

+(Ax_{n+1}(t) − Ax_{n}(t), x_{n+1}(t) − x_{n}(t)) = 0, t ∈ [0, T].

Taking into account that A is monotone we deduce

∥x_{n+1}(t) − x_{n}(t)∥ ≤ e^{−αt}∥x_{n+1}(0) − x_{n}(0)∥, t ∈ [0, T],

and hence

∥x_{n+1}(0) − x_{n}(0)∥ = ∥x_{n}(T) − x_{n−1}(T)∥
≤ e^{−αT} ∥x_{n}(0) − x_{n−1}(0)∥
≤ e^{−2αT} ∥x_{n−1}(0) − x_{n−2}(0)∥
≤ ...
≤ e^{−nαT} ∥x_{1}(0) − x_{0}(0)∥, n ≥ 0.  (40)

Finally we get the estimate

∥x_{n+1}(t) − x_{n}(t)∥ ≤ e^{−α(nT +t)}∥S_{α}(T; 0, x_{0}) − x_{0}∥, t ∈ [0, T], n ≥ 0.

Here S_{α}(t; 0, x_{0}) represents the solution of (39) for the initial data x_{0}. From the previous estimate it is clear that (x_{n})_{n≥0} is convergent in C^{0}([0, T]; H):

x_{n}(t) = x_{0}(t) + ∑_{k=0}^{n−1}(x_{k+1}(t) − x_{k}(t)), t ∈ [0, T],

where

∥ ∑_{k=0}^{n−1}(x_{k+1}(t) − x_{k}(t))∥ ≤ ∑_{k=0}^{n−1} ∥x_{k+1}(t) − x_{k}(t)∥

≤ ∑_{k=0}^{n−1} e^{−α(kT +t)}∥S_{α}(T; 0, x_{0}) − x_{0}∥

≤ \frac{e^{−αt}}{1 − e^{−αT}} ∥S_{α}(T; 0, x_{0}) − x_{0}∥.

Moreover ∥x_{n}(t)∥ ≤ ∥S_{α}(t; 0, x_{0})∥ + \frac{1}{1−e^{−αT}} ∥S_{α}(T; 0, x_{0}) − x_{0}∥. Denote by x_{α} the limit of (x_{n})_{n≥0} as n → ∞. We should note that without any other hypothesis (x_{α})_{α>0} is not uniformly bounded in L^{∞}(]0, T [; H). We have only estimate in O(1 + \frac{1}{α}),

∥x_{α}∥_{L^{∞}([0,T];H)} ≤ C(1 + \frac{1}{1 − e^{−αT}} ) ∼ O(1 + \frac{1}{α}).

The above estimate leads immediately to the following statement.

Remark 3.5 The sequence (αx_{α})_{α>0} is uniformly bounded in L^{∞}([0, T]; H).

Let us demonstrate that x_{α} is T -periodic and solution for (39). Indeed,

x_{α}(0) = lim_{n→∞} x_{n}(0) = lim_{n→∞} x_{n−1}(T) = x_{α}(T).

Now let us show that (x′_{n})_{n≥0} is also uniformly bounded in L^{∞}(]0, T [; H). By taking the difference between the equations (39) at the moments t and t + h we have

α(x(t+h)−x(t))+x′(t+h)−x′(t)+Ax(t+h)−Ax(t) = f(t+h)−f(t), t < t+h.

Multiplying by x(t + h) − x(t) we obtain

α∥x(t+h)−x(t)∥^{2} + \frac{1}{2} \frac{d}{dt}∥x(t+h)−x(t)∥^{2} ≤ ∥f(t+h)−f(t)∥·∥x(t+h)−x(t)∥,

which can be also rewritten as

\frac{1}{2}e^{2αt}∥x(t + h) − x(t)∥^{2} ≤ ∫_{0}^{t} e^{αs}∥f(s + h) − f(s)∥ · e^{αs}∥x(s + h) − x(s)∥ds

+\frac{1}{2}∥x(h) − x(0)∥^{2}, t < t + h.

By using Bellman’s lemma we conclude that

\frac{1}{h}∥x(t + h) − x(t)∥ ≤ ∫_{0}^{t} e^{−α(t−s)} \frac{1}{h}∥f(s + h) − f(s)∥ds
+e^{−αt} \frac{1}{h}∥x(h) − x(0)∥, 0 ≤ t < t + h. (41)

By passing to the limit for h → 0 the previous formula yields

∥x′(t)∥ ≤ e^{−αt}∥x′(0)∥ + ∫_{0}^{t} e^{−α(t−s)}∥f′(s)∥ds
≤ e^{−αt}∥f(0) − αx_{0} − Ax_{0}∥ + \frac{1}{α}(1 − e^{−αt})∥f′∥_{L^{∞}(]0,T [;H)}
≤ ∥f(0) − αx_{0} − Ax_{0}∥ + \frac{1}{α}∥f′∥_{L^{∞}(]0,T [;H)} < +∞.

Therefore (x′_{n})_{n≥0} is uniformly bounded in L^{∞}(]0, T [; H) since

∥x′_{n}∥_{L^{∞}(]0,T [;H)} = ∥x′(nT + (·))∥_{L^{∞}(]0,T [;H)} ≤ ∥x′∥_{L^{∞}([0,+∞[;H)},

and thus we have x′_{n}(t) ⇀ y_{α}(t), t ∈ [0, T]. We can write

(x_{n}(t), z) = (x_{n}(0), z) + ∫_{0}^{t}(x′_{n}(s), z)ds, z ∈ H, t ∈ [0, T], n ≥ 0,

and by passing to the limit for n → ∞ we deduce

(x_{α}(t), z) = (x_{α}(0), z) + ∫_{0}^{t}(y_{α}(s), z)ds, z ∈ H, t ∈ [0, T],

which is equivalent to

x_{α}(t) = x_{α}(0) + ∫_{0}^{t} y_{α}(s)ds, t ∈ [0, T].

Therefore x_{α} is differentiable and x′_{α} = y_{α}. Finally we can write x′_{n}(t) ⇀ x′_{α}(t), t ∈ [0, T]. Let us show that x_{α} is also solution for (39). We have

[x_{n}(t), f(t) − αx_{n}(t) − x′_{n}(t)] ∈ A, n ≥ 0, t ∈ [0, T].

Since x_{n}(t) → x_{α}(t), x′_{n}(t) ⇀ x′_{α}(t) and A is maximal monotone we conclude that

[x_{α}(t), f(t) − x′_{α}(t)] ∈ A, t ∈ [0, T], α > 0,

which means that x_{α}(t) ∈ D(A) and Ax_{α}(t) = f(t) − x′_{α}(t), t ∈ [0, T].

Now we establish for the linear case the similar result stated in Proposition 2.10. Before let us recall a standard result concerning maximal monotone operators on Hilbert spaces

Proposition 3.6 Assume that A is a maximal monotone operator (linear or not) and αu_{α} + Au_{α} = f, u_{α} ∈ D(A), f ∈ H, α > 0. Then the following statements are equivalent:

(i) f ∈ Range(A); (ii) (u_{α})_{α>0} is bounded in H. Moreover, in this case (u_{α})_{α>0} is convergent in H to the element of minimal norm in A^{−1}f.

Proof it (i) → (ii) By the hypothesis there is u ∈ D(A) such that f = Au. After multiplication by u_{α} − u we get

α(u_{α}, u_{α} − u) + (Au_{α} − Au, u_{α} − u) = 0, α > 0.

Taking into account that A is monotone we deduce

∥u_{α}∥^{2} ≤ (u_{α}, u) ≤ ∥u_{α}∥ · ∥u∥, α > 0,

and hence ∥u_{α}∥ ≤ ∥u∥, α > 0, u ∈ A^{−1}f which implies that u_{α} ⇀ u_{0}. We have [u_{α}, f − αu_{α}] ∈ A, α > 0 and since A is maximal monotone, by passing to the limit for α → 0 we deduce that [u_{0}, f] ∈ A, or u_{0} ∈ A^{−1}f. Moreover

∥u_{0}∥ = ∥w − lim_{α→0} u_{α}∥ ≤ lim inf_{α→0} ∥u_{α}∥ ≤ lim sup_{α→0} ∥u_{α}∥ ≤ ∥u∥, ∀u ∈ A^{−1}f.

In particulat taking u = u_{0} ∈ A^{−1}f we get

∥w − lim_{α→0} u_{α}∥ = lim_{α→0} ∥u_{α}∥,

and hence, since any Hilbert space is strictly convex, by Mazur’s theorem we deduce that the convergence is strong

u_{α} → u_{0} ∈ A^{−1}f, α → 0,

where ∥u_{0}∥ = inf_{u∈A^{−1}f} ∥u∥ = min_{u∈A^{−1}f} ∥u∥. (ii) → (i) Conversely, suppose that (u_{α})_{α>0} is bounded in H. Therefore u_{α} ⇀ u in H. We have [u_{α}, f − αu_{α}] ∈ A, α > 0 and since A is maximal monotone by passing to the limit for α → 0 we deduce that [u, f] ∈ A or u ∈ D(A) and f = Au.

Theorem 3.7 Assume that A : D(A) ⊂ H → H is a linear maximal monotone operator on a compact Hilbert space H and f ∈ C^{1}(ℝ; H) is a T -periodic function. Then the following statements are equivalent:

(i) equation (33) has periodic solutions; (ii) the sequence of periodic solutions for (39) is bounded in C^{1}(ℝ; H). Moreover in this case (x_{α})_{α>0} is convergent in C^{0}(ℝ; H) and the limit is also a T -periodic solution for (33).

Proof (i) → (ii) Denote by x, x_{α} the periodic solutions for (33) and (39). By taking the difference and after multiplication by x_{α}(t) − x(t) we get:

α∥x_{α}(t)−x(t)∥^{2} + \frac{1}{2} \frac{d}{dt}∥x_{α}(t)−x(t)∥^{2} ≤ α∥x(t)∥·∥x_{α}(t)−x(t)∥, t ∈ ℝ. (42)

Finally, after integration and by using Bellman’s lemma, formula (42) yields

∥x_{α}(t) − x(t)∥ ≤ e^{−αt}∥x_{α}(0) − x(0)∥ + ∫_{0}^{t} αe^{−α(t−s)}∥x(s)∥ds
≤ e^{−αt}∥x_{α}(0) − x(0)∥ + (1 − e^{−αt})∥x∥_{L^{∞}}, t ∈ ℝ.

Since x_{α} and x are T -periodic we can also write

∥x_{α}(t) − x(t)∥ = ∥x_{α}(nT + t) − x(nT + t)∥
≤ e^{−α(nT +t)}∥x_{α}(0) − x(0)∥ + (1 − e^{−α(nT +t)})∥x∥_{L^{∞}}.

By passing to the limit for n → ∞ we obtain

∥x_{α} − x∥_{L^{∞}} ≤ ∥x∥_{L^{∞}}, α > 0,

and hence

∥x_{α}∥_{L^{∞}} ≤ 2∥x∥_{L^{∞}}, α > 0.

Since A is linear we can write

\frac{α}{h}(x_{α}(t + h) − x_{α}(t)) + \frac{1}{h}(x′_{α}(t + h) − x′_{α}(t)) + \frac{1}{h}A(x_{α}(t + h) − x_{α}(t))

= \frac{1}{h}(f(t + h) − f(t)), t < t + h, α > 0,

and for t < t + h,

\frac{1}{h}(x′(t + h) − x′(t)) + \frac{1}{h}A(x(t + h) − x(t)) = \frac{1}{h}(f(t + h) − f(t)).

For every h > 0 denote by y_{α,h}, y_{h} and g_{h} the periodic functions:

y_{α,h}(t) = \frac{1}{h}(x_{α}(t + h) − x_{α}(t)), t ∈ ℝ, α > 0,

y_{h}(t) = \frac{1}{h}(x(t + h) − x(t)), t ∈ ℝ,

g_{h}(t) = \frac{1}{h}(f(t + h) − f(t)), t ∈ ℝ,

and hence we have

αy_{α,h}(t) + y′_{α,h}(t) + Ay_{α,h}(t) = g_{h}(t), t ∈ ℝ,
y′_{h}(t) + Ay_{h}(t) = g_{h}(t), t ∈ ℝ.

By the same computations we get

∥y_{α,h}(t) − y_{h}(t)∥ ≤ e^{−αt}∥y_{α,h}(0) − y_{h}(0)∥ + ∫_{0}^{t} αe^{−α(t−s)}∥y_{h}(s)∥ds.

Now by passing to the limit for h → 0 we deduce

∥x′_{α}(t) − x′(t)∥ ≤ e^{−αt}∥x′_{α}(0) − x′(0)∥ + ∫_{0}^{t} αe^{−α(t−s)}∥x′(s)∥ds
≤ e^{−αt}∥x′_{α}(0) − x′(0)∥ + (1 − e^{−αt})∥x′∥_{L^{∞}}, t ∈ [0, T].

By the periodicity we obtain as before that

∥x′_{α}(t) − x′(t)∥ = ∥x′_{α}(nT + t) − x′(nT + t)∥
≤ e^{−α(nT +t)}∥x′_{α}(0) − x′(0)∥ + (1 − e^{−α(nT +t)})∥x′∥_{L^{∞}},

and hence by passing to the limit for n → ∞ we conclude that

∥x′_{α} − x′∥_{L^{∞}} ≤ ∥x′∥_{L^{∞}}, α > 0.

Therefore, (x′_{α})_{α>0} is also uniformly bounded in L^{∞}

∥x′_{α}∥_{L^{∞}} ≤ 2∥x′∥_{L^{∞}}, α > 0.

Conversely, the implication (ii) → (i) follows by using Arzela-Ascoli’s theorem and by passing to the limit for α → 0 in (39).

Let us continue the analysis of the previous example. The semigroup associated to the equation (39) is given by

e^{−t(α+A)} = e^{−αt}e^{−tA} = e^{−αt} (\begin{matrix}cos t, & sin t \\ − sin t, & cos t \end{matrix}) t ∈ ℝ, α > 0,

and the periodic solution for equation (39) reads

x_{α}(t) = (1 − e^{−T (α+A)})^{−1} ∫_{0}^{T} e^{−(T −s)(α+A)}f(s)ds
+ ∫_{0}^{t} e^{−(t−s)(α+A)}f(s)ds
= \frac{1 − e^{−T (α−A)}}{(1 − e^{−αT} cos T)^{2} + (e^{−αT} sin T)^{2}} ∫_{0}^{T} e^{−(T −s)(α+A)}f(s)ds
+ ∫_{0}^{t} e^{−(t−s)(α+A)}f(s)ds, t > 0, α > 0.

As we have seen, proving the existence of periodic solutions reduces to finding uniform L^{∞}(]0, T [; H) estimates for (x_{α})_{α>0} and (x′_{α})_{α>0} . Since A is linear bounded operator (∥A∥_{L(H;H)} = 1) we have

∥x′_{α}∥_{L^{∞}(]0,T [;H)} = ∥f − αx_{α} − Ax_{α}∥_{L^{∞}(]0,T [;H)}
≤ ∥f∥_{L^{∞}(]0,T [;H)} + (α + ∥A∥_{L(H;H)})∥x_{α}∥_{L^{∞}(]0,T [;H)}, α > 0,

and hence in this case it is sufficient to find only uniform L^{∞}(]0, T [; H) estimates for (x_{α})_{α>0} or uniform estimates for (x_{α}(0))_{α>0} in H. Case 1: T = 2nπ, n ≥ 0. We have

lim_{α→0} x_{α}(0) = lim_{α→0} \frac{1}{1 − e^{−αT}} ∫_{0}^{T} e^{−(T −s)(α+A)}f(s)ds.

If ∫_{0}^{T} e^{−(T −s)A}f(s)ds ≠ 0 , then (x_{α}(0))_{α>0} is not bounded. In fact since e^{−2nπA} = 1 it is easy to check that equation (35) does not have any periodic solution. If ∫_{0}^{T} e^{−(T −s)A}f(s)ds = 0 then every solution of (35) is T -periodic and (x_{α}(0))_{α>0} is convergent for α → 0:

lim_{α→0} x_{α}(0) = lim_{α→0} \frac{∫_{0}^{T} (e^{−α(T −s)} − 1)e^{−(T −s)A}f(s)ds}{1 − e^{−αT}}
= − ∫_{0}^{T} \frac{T − s}{T}e^{−(T −s)A}f(s)
= \frac{1}{T} ∫_{0}^{T} se^{−(T −s)A}f(s).

Case 2: T ≠ 2nπ for alln ≥ 0. In this case (1 − e^{−T A}) is invertible and (x_{α}(0))_{α>0} converges to x(0) where x is the unique T -periodic solution of (35):

lim_{α→0} x_{α}(0) = lim_{α→0}(1 − e^{−T (α+A)})^{−1} ∫_{0}^{T} e^{−(T −s)(α+A)}f(s)ds
= (1 − e^{−T A})^{−1} ∫_{0}^{T} e^{−(T −s)A}f(s)ds
= \frac{1}{2 sin(\frac{T}{2} )} ∫_{0}^{T} e^{−( \frac{T +π}{2} −s)A}f(s)ds.

We state now our main result of existence in the linear and symmetric case.

Theorem 3.8 Assume that A : D(A) ⊂ H → H is a linear maximal monotone and symmetric operator and f ∈ C^{1}([0, T]; H) is a T -periodic function. Then the necessary and sufficient condition for the existence of periodic solutions for (33) is given by

⟨f⟩ := \frac{1}{T} ∫_{0}^{T} f(t)dt ∈ Range(A).

In this case we have the estimates:

∥x∥_{L^{∞}(]0,T [;H)} ≤ ∥A^{−1}⟨f⟩∥ + \frac{\sqrt{T}}{2} ∥f∥_{L^{2}(]0,T [;H)} + \frac{T}{2} ∥f′∥_{L^{1}(]0,T [;H)},

and

∥x′∥_{L^{∞}(]0,T [;H)} ≤ \frac{1}{\sqrt{T}} ∥f∥_{L^{2}(]0,T [;H)} + ∥f′∥_{L^{1}(]0,T [;H)},

and the solution is unique up to a constant in A^{−1}(0).

Proof The condition is necessary (see Proposition 3.3). Let us show now that it is also sufficient. Consider the T -periodic solutions (x_{α})_{α>0} for

αx_{α}(t) + x′_{α}(t) + Ax_{α}(t) = f(t), t ∈ [0, T], α > 0.

First we prove that (x_{α})_{α>0} is uniformly bounded in C^{1}([0, T]; H). Let us multiply by x′_{α}(t) and integrate on a period:

∫_{0}^{T} ∥x′_{α}(t)∥^{2}dt + ∫_{0}^{T} α(x_{α}(t), x′_{α}(t)) + (Ax_{α}(t), x′_{α}(t))dt = ∫_{0}^{T} (f(t), x′_{α}(t))dt.

Since A is symmetric and x_{α} is T -periodic we have

∫_{0}^{T} α(x_{α}(t), x′_{α}(t)) + (Ax_{α}(t), x′_{α}(t))dt
= ∫_{0}^{T} \frac{α}{2} \frac{d}{dt}∥x_{α}(t)∥^{2}dt + ∫_{0}^{T} \frac{1}{2} \frac{d}{dt}(Ax_{α}(t), x_{α}(t))dt
= \frac{1}{2} \{α∥x_{α}(t)∥^{2} + (Ax_{α}(t), x_{α}(t))\} |_{0}^{T} = 0.

Finally we get

∥x′_{α}∥_{L^{2}(]0,T [;H)}^{2} ≤ (f, x′_{α})_{L^{2}(]0,T [;H)} ≤ ∥f∥_{L^{2}(]0,T [;H)} · ∥x′_{α}∥_{L^{2}(]0,T [;H)},

and hence

∥x′_{α}∥_{L^{2}(]0,T [;H)} ≤ ∥f∥_{L^{2}(]0,T [;H)}, α > 0.

Therefore we can write

min_{t∈[0,T]} ∥x′_{α}(t)∥ ≤ \frac{1}{\sqrt{T}} ∥x′_{α}∥_{L^{2}(]0,T [;H)} ≤ \frac{1}{\sqrt{T}} ∥f∥_{L^{2}(]0,T [;H)}.  (43)

Mihai Bostan 27 As seen before, since A is linear we can write

\frac{α}{h}(x_{α}(t + h) − x_{α}(t)) + \frac{1}{h}(x′_{α}(t + h) − x′_{α}(t))

+\frac{1}{h}A(x_{α}(t + h) − x_{α}(t)) = \frac{1}{h}(f(t + h) − f(t)),

and by standard calculations for s < t and h > 0, we get

\frac{1}{h}∥x_{α}(t + h) − x_{α}(t)∥

≤ e^{−α(t−s)} \frac{1}{h}∥x_{α}(s + h) − x_{α}(s)∥ + ∫_{s}^{t} e^{−α(t−τ)} \frac{1}{h}∥f(τ + h) − f(t)∥dτ .

Passing to the limit for h → 0 we deduce

∥x′_{α}(t)∥ ≤ e^{−α(t−s)}∥x′_{α}(s)∥ + ∫_{s}^{t} e^{−α(t−τ)}∥f′(τ)∥dτ
≤ ∥x′_{α}(s)∥ + ∫_{s}^{t} ∥f′(τ)∥dτ, s ≤ t, α > 0.  (44)

From (43) and (44) we conclude that the functions (x′_{α})_{α>0} are uniformly bounded in L^{∞}(]0, T [; H):

∥x′_{α}∥_{L^{∞}(]0,T [;H)} ≤ \frac{1}{\sqrt{T}} ∥f∥_{L^{2}(]0,T [;H)} + ∥f′∥_{L^{1}(]0,T [;H)}, α > 0.

As shown before, since A is linear and x_{α} is T -periodic we have also

α⟨x_{α}⟩ + A⟨x_{α}⟩ = ⟨f⟩.  (45)

By the hypothesis there is x_{0} ∈ D(A) such that ⟨f⟩ = Ax_{0} and hence

∥⟨x_{α}⟩∥ = ∥(α + A)^{−1}⟨f⟩∥ = ∥(α + A)^{−1}Ax_{0}∥ ≤ ∥x_{0}∥, α > 0. Now it is easy to check that (x_{α})_{α>0} is uniformly bounded in L^{∞}(]0, T [; H):

∥x_{α}(t) − ⟨x_{α}⟩∥ = ∥\frac{1}{T} ∫_{0}^{T} (x_{α}(t) − x_{α}(s))ds∥
= ∥\frac{1}{T} ∫_{0}^{T} ∫_{s}^{t} x′_{α}(τ)dτds∥
≤ \frac{\sqrt{T}}{2} ∥f∥_{L^{2}(]0,T [;H)} + \frac{T}{2} ∥f′∥_{L^{1}(]0,T [;H)},

and thus

∥x_{α}∥_{L^{∞}(]0,T [;H)} ≤ ∥⟨x_{α}⟩∥ + \frac{\sqrt{T}}{2} ∥f∥_{L^{2}(]0,T [;H)} + \frac{T}{2} ∥f′∥_{L^{1}(]0,T [;H)}
≤ ∥x_{0}∥ + \frac{\sqrt{T}}{2} ∥f∥_{L^{2}(]0,T [;H)} + \frac{T}{2} ∥f′∥_{L^{1}(]0,T [;H)}.

Now we can prove that (x_{α})_{α>0} is convergent in C^{1}([0, T]; H). Indeed, by taking the difference between the equations (39) written for α, β > 0, after multiplication by x′_{α}(t) − x′_{β}(t) and integration on [0, T] we get

∫_{0}^{T} {α(x_{α}(t) − x_{β}(t), x′_{α}(t) − x′_{β}(t))
+∥x′_{α}(t) − x′_{β}(t)∥^{2} + (A(x_{α}(t) − x_{β}(t)), x′_{α}(t) − x′_{β}(t))}dt
= −(α − β) ∫_{0}^{T} (x_{β}(t), x′_{α}(t) − x′_{β}(t))dt.

Since A is symmetric, x_{α} and x_{β} are T -periodic and uniformly bounded in L^{∞}(]0, T [; H) we deduce that

∥x′_{α} − x′_{β}∥_{L^{2}(]0,T [;H)} ≤ |α − β| · sup_{γ>0} ∥x_{γ}∥_{L^{2}(]0,T [;H)},

or

∥x′_{α} −x′_{β}∥_{L^{∞}(]0,T [;H)} ≤ \frac{|α − β|}{\sqrt{T}} ·sup_{γ>0} ∥x_{γ}∥_{L^{2}(]0,T [;H)} +|α−β|·sup_{γ>0} ∥x′_{γ}∥_{L^{1}(]0,T [;H)},

and therefore (x′_{α})_{α>0} converges in C([0, T]; H).

We already know that (⟨x_{α}⟩)_{α>0} = ((α + A)^{−1}⟨f⟩)_{α>0} is bounded in H and by the Proposition 3.6 it follows that (⟨x_{α}⟩)_{α>0} is convergent to the element of minimal norm in A^{−1}⟨f⟩. We have

x_{α}(t) = x_{α}(0) + ∫_{0}^{t} x′_{α}(s)ds, t ∈ ℝ, α > 0.

By taking the average we deduce that x_{α}(0) = ⟨x_{α}⟩− < ∫_{0}^{t} x′_{α}(s)ds > and therefore, since (x′_{α})_{α>0} is uniformly convergent, it follows that (x_{α}(0))_{α>0} is also convergent. Finally we conclude that (x_{α})_{α>0} is convergent in C^{1}([0, T]; H) to the periodic solution x for (33) such that < x > is the element of minimal norm in A^{−1}⟨f⟩.

Before analyzing the periodic solution for the heat equation, following an idea of [7], let us state the following proposition.

Proposition 3.9 Assume that A : D(A) ⊂ H → H is a linear maximal monotone and symmetric operator and f ∈ C^{1}([0, T]; H) is a T -periodic function. Then for every x_{0} ∈ D(A) we have

lim_{t→∞} \frac{1}{T} (x(t + T; 0, x_{0}) − x(t; 0, x_{0})) = ⟨f⟩ − Proj\overline{R(A)}⟨f⟩,  (46)

where x(·; 0, x_{0}) represents the solution of (33) with the initial data x_{0} and R(A) is the range of A.

Remark 3.10 A being maximal monotone, A^{−1} is also maximal monotone and therefore \overline{D(A^{−1})} = \overline{R(A)} is convex.

Proof of Proposition 3.9. Consider x_{0} ∈ D(A) and denote by x(·) the corresponding solution. By integration on [t, t + T] we get

\frac{1}{T} (x(t + T) − x(t)) + A(\frac{1}{T} ∫_{t}^{t+T} x(s)ds) = ⟨f⟩.  (47)

For each α > 0 consider x_{α} ∈ D(A) such that αx_{α} + Ax_{α} = ⟨f⟩. Denoting by y(·) the function y(t) = \frac{1}{T} ∫_{t}^{t+T} x(s)ds, t ≥ 0, equation (47) writes

y′(t) + Ay(t) = αx_{α} + Ax_{α}, t ≥ 0, α > 0.

Let us search for y of the form y_{1} + y_{2} where

y′_{1}(t) + Ay_{1}(t) = αx_{α}, t ≥ 0,

with the initial condition y_{1}(0) = 0 and

y′_{2}(t) + Ay_{2}(t) = Ax_{α}, t ≥ 0,  (48)

with the initial condition y_{2}(0) = y(0) = \frac{1}{T} ∫_{0}^{T} x(t)dt. We are interested on the asymptotic behaviour of Ay(t) = Ay_{1}(t) + Ay_{2}(t) for large t. We have

y_{1}(t) = e^{−tA}y_{1}(0) + ∫_{0}^{t} e^{−(t−s)A}αx_{α}ds
= ∫_{0}^{t} e^{−(t−s)A}αx_{α}ds,

and therefore,

Ay_{1}(t) = ∫_{0}^{t} Ae^{−(t−s)A}αx_{α}ds = e^{−(t−s)A}αx_{α}|_{0}^{t} = (1 − e^{−tA})αx_{α}.

By the other hand, after multiplication of (48) by y′_{2}(t) = (y_{2}(t) − x_{α})′ we get

∥y′_{2}(t)∥^{2} + (A(y_{2}(t) − x_{α}), (y_{2}(t) − x_{α})′) = 0, t ≥ 0.

Since A is symmetric, after integration on [0, t] we obtain

∫_{0}^{t} ∥y′_{2}(s)∥^{2}ds + \frac{1}{2}(A(y_{2}(t) − x_{α}), y_{2}(t) − x_{α}) = \frac{1}{2}(A(y_{2}(0) − x_{α}), y_{2}(0) − x_{α}),

and therefore, by the monotony of A it follows that

∫_{0}^{∞} ∥y′_{2}(t)∥^{2}dt ≤ \frac{1}{2}(A(y_{2}(0) − x_{α}), y_{2}(0) − x_{α}).

Thus lim_{t→∞} y′_{2}(t) = 0 and by passing to the limit in (48) we deduce that lim_{t→∞} Ay_{2}(t) = lim_{t→∞}(Ax_{α} − y′_{2}(t)) = Ax_{α}. Finally we find that

lim_{t→∞} \{\frac{1}{T} (x(t + T) − x(t)) − e^{−tA}αx_{α}\}

= lim_{t→∞}{y′(t) − e^{−tA}αx_{α}}
= lim_{t→∞}{⟨f⟩ − Ay(t) − e^{−tA}αx_{α}}
= lim_{t→∞}{⟨f⟩ − Ay_{1}(t) − Ay_{2}(t) − e^{−tA}αx_{α}}
= ⟨f⟩ − αx_{α} − Ax_{α} = 0, α > 0.  (49)

Now let us put y_{α} = Ax_{α} and observe that y_{α} + αA^{−1}y_{α} = Ax_{α} + αx_{α} = ⟨f⟩, α > 0. Therefore,

lim_{α↘0} y_{α} = lim_{α↘0}(1 + αA^{−1})^{−1}⟨f⟩
= lim_{α↘0} J_{α}^{A^{−1}}⟨f⟩
= Proj\overline{D(A^{−1})}⟨f⟩
= Proj\overline{R(A)}⟨f⟩,

and it follows that

lim_{α↘0} αx_{α} = lim_{α↘0}(⟨f⟩ − Ax_{α}) = lim_{α↘0}(⟨f⟩ − y_{α}) = ⟨f⟩ − Proj\overline{R(A)}⟨f⟩.

Since Graph(A) is closed and [αx_{α}, αy_{α}] = [αx_{α}, A(αx_{α})] ∈ A, α > 0, by passing to the limit for α ↘ 0 we deduce that ⟨f⟩ − Proj\overline{R(A)}⟨f⟩ ∈ D(A) and A(⟨f⟩ − Proj\overline{R(A)}⟨f⟩) = 0. It is easy to see that we can pass to the limit for α ↘ 0 in (49). Indeed, for ε > 0 let us consider α_{ε} > 0 such that ∥ lim_{α↘0} αx_{α}− α_{ε}x_{α_{ε}}∥ < \frac{ε}{2}. We have

∥\frac{1}{T} (x(t + T) − x(t)) − e^{−tA} lim_{α↘0} αx_{α}∥

≤ ∥\frac{1}{T} (x(t + T) − x(t)) − e^{−tA}α_{ε}x_{α_{ε}} ∥ + ∥e^{−tA}α_{ε}x_{α_{ε}} − e^{−tA} lim_{α↘0} αx_{α}∥
≤ ∥\frac{1}{T} (x(t + T) − x(t)) − e^{−tA}α_{ε}x_{α_{ε}} ∥ + ∥α_{ε}x_{α_{ε}} − lim_{α↘0} αx_{α}∥
≤ \frac{ε}{2} + \frac{ε}{2} = ε, t ≥ t(α_{ε}, \frac{ε}{2}) = t(ε),

and thus

lim_{t→∞}\{\frac{1}{T} (x(t + T) − x(t)) − e^{−tA}(⟨f⟩ − Proj\overline{R(A)}⟨f⟩)\} = 0.

But e^{−tA}(⟨f⟩ − Proj\overline{R(A)}⟨f⟩) does not depend on t ≥ 0:

\frac{d}{dt}e^{−tA}(⟨f⟩ − Proj\overline{R(A)}⟨f⟩) = −Ae^{−tA}(⟨f⟩ − Proj\overline{R(A)}⟨f⟩)
= −e^{−tA}A(⟨f⟩ − Proj\overline{R(A)}⟨f⟩) = 0,

and thus the previous formula reads

lim_{t→∞} \frac{1}{T} (x(t + T) − x(t)) = ⟨f⟩ − Proj\overline{R(A)}⟨f⟩.

Remark 3.11 Under the same hypothesis as above we can easily check that

inf_{x_{0}∈D(A)} \frac{∥x(T; 0, x_{0}) − x_{0}∥}{T} = ∥⟨f⟩ − Proj\overline{R(A)}⟨f⟩∥ = dist(⟨f⟩, \overline{R(A)}).

3.3 Periodic solutions for the heat equation

Let Ω ⊂ ℝ^{d}, d ≥ 1, be an open bounded set with ∂Ω ∈ C^{2}. Consider the heat equation

\frac{∂u}{∂t}(t, x) − Δu(t, x) = f(t, x), (t, x) ∈ ℝ × Ω,  (50)

with the Dirichlet boundary condition

u(t, x) = g(t, x), (t, x) ∈ ℝ × ∂Ω,  (51)

or the Neumann boundary condition

\frac{∂u}{∂n}(t, x) = g(t, x), (t, x) ∈ ℝ × ∂Ω,  (52)

where we denote by n(x) the outward normal in x ∈ ∂Ω.

Theorem 3.12 Assume that f ∈ C^{1}(ℝ; L^{2}(Ω)) is T -periodic and g(t, x) = \frac{∂u_{0}}{∂n} (t, x), (t, x) ∈ ℝ×∂Ω where u_{0} ∈ C^{1}(ℝ; H^{2}(Ω))∩C^{2}(ℝ; L^{2}(Ω)) is T -periodic. Then the heat problem (50), (52) has T -periodic solutions u ∈ C(ℝ; H^{2}(Ω)) ∩ C^{1}(ℝ; L^{2}(Ω)) if and only if

∫_{∂Ω} ∫_{0}^{T} g(t, x)dtdσ + ∫_{Ω} ∫_{0}^{T} f(t, x)dtdx = 0.

In this case the periodic solutions satisfies the estimates

∥u′ − u′_{0}∥_{L^{∞}([0,T];L^{2}(Ω))} ≤ \frac{1}{\sqrt{T}} ∥f − u′_{0} + Δu_{0}∥_{L^{2}(]0,T [;L^{2}(Ω))}
+ ∥f′ − u′′_{0} + Δu′_{0}∥_{L^{1}(]0,T [;L^{2}(Ω))},  (53)

and the solution is unique up to a constant.

Proof Let us search for solutions u = u_{0} + v where

\frac{∂v}{∂t} (t, x) − Δv(t, x) = f(t, x) − \frac{∂u_{0}}{∂t} (t, x) + Δu_{0}(t, x), (t, x) ∈ ℝ × Ω, (54)

and

\frac{∂v}{∂n}(t, x) = g(t, x) − \frac{∂u_{0}}{∂n} (t, x) = 0, (t, x) ∈ ℝ × ∂Ω.  (55)

Consider the operator A_{N} : D(A_{N}) ⊂ L^{2}(Ω) → L^{2}(Ω) given as

A_{N}v = −Δv

with domain

D(A_{N}) = \{v ∈ H^{2}(Ω) : \frac{∂v}{∂n}(x) = 0, ∀ x ∈ ∂Ω\}.

The operator A_{N} is linear monotone:

(A_{N}v, v) = − ∫_{Ω} Δv(x)v(x)dx
= − ∫_{∂Ω} \frac{∂v}{∂n}(x)v(x)dσ + ∫_{Ω} ∥∇v(x)∥^{2}dx
= ∫_{Ω} ∥∇v(x)∥^{2}dx ≥ 0, ∀ v ∈ D(A_{N}).  (56)

Since the equation λv − Δv = f has unique solution in D(A_{N}) for every f ∈ L^{2}(Ω), λ > 0 it follows that A_{N} is maximal (see [6]). Moreover, it is symmetric

(A_{N}v_{1}, v_{2}) = ∫_{Ω} ∇v_{1}(x) · ∇v_{2}(x)dx = (v_{1}, A_{N}v_{2}), ∀ v_{1}, v_{2} ∈ D(A_{N}).

Note that by the hypothesis the second member in (54) f − u′_{0} + Δu_{0} belongs to C^{1}(ℝ; L^{2}(Ω)). Therefore the Theorem 3.8 applies and hence the problem (54),

(55) has periodic solutions if and only if there is w ∈ D(A_{N}) such that

−Δw = \frac{1}{T} ∫_{0}^{T} \{f(t) − \frac{du_{0}}{dt} (t) + Δu_{0}(t)\}dt.

Since u_{0} is T -periodic we have ∫_{0}^{T} \frac{du_{0}}{dt} (t)dt = 0 and thus w + \frac{1}{T} ∫_{0}^{T} u_{0}(t)dt is solution for the elliptic problem

−Δ(w + \frac{1}{T} ∫_{0}^{T} u_{0}(t)dt) = \frac{1}{T} ∫_{0}^{T} f(t)dt = F,

with the boundary condition

\frac{∂}{∂n}(w + \frac{1}{T} ∫_{0}^{T} u_{0}(t)dt) = \frac{∂w}{∂n} + \frac{1}{T} ∫_{0}^{T} \frac{∂u_{0}}{∂n} (t)dt
= \frac{1}{T} ∫_{0}^{T} g(t)dt = G.

As known from the general theory of partial differential equations (see [6]) this problem has solution if and only if ∫_{∂Ω} G(x)dσ + ∫_{Ω} F (x)dx = 0 or

∫_{∂Ω} ∫_{0}^{T} g(t, x)dtdσ + ∫_{Ω} ∫_{0}^{T} f(t, x)dtdx = 0.

The estimate (53) follows from Theorem 3.8.

For the heat equation with Dirichlet boundary condition we have the following existence result.

Theorem 3.13 Assume that f ∈ C^{1}(ℝ; L^{2}(Ω)) is T -periodic and g(t, x) = u_{0}(t, x), (t, x) ∈ ℝ × ∂Ω where u_{0} ∈ C^{1}(ℝ; H^{2}(Ω)) ∩ C^{2}(ℝ; L^{2}(Ω)) is T -periodic. Then the heat problem (50), (51) has an unique T -periodic solution u in C(ℝ; H^{2}(Ω)) ∩ C^{1}(ℝ; L^{2}(Ω)) and there is a constant C(Ω) such that

∥u − u_{0}∥_{L^{∞}([0,T];L^{2}(Ω))} ≤ C(Ω)∥f + Δu_{0}∥_{L^{∞}([0,T];L^{2}(Ω))}

+\frac{\sqrt{T}}{2} ∥f − u′_{0} + Δu_{0}∥_{L^{2}(]0,T [;L^{2}(Ω))}

+\frac{T}{2} ∥f′ − u′′_{0} + Δu′_{0}∥_{L^{1}(]0,T [;L^{2}(Ω))}, (57)

and

∥u′ − u′_{0}∥_{L^{∞}([0,T];L^{2}(Ω))} ≤ \frac{1}{\sqrt{T}} ∥f − u′_{0} + Δu_{0}∥_{L^{2}(]0,T [;L^{2}(Ω))}
+∥f′ − u′′_{0} + Δu′_{0}∥_{L^{1}(]0,T [;L^{2}(Ω))}.  (58)

Proof This time we consider the operator A_{D} : D(A_{D}) ⊂ L^{2}(Ω) → L^{2}(Ω) given as

A_{D}v = −Δv

with domain

D(A_{D}) = {v ∈ H^{2}(Ω) : v(x) = 0, ∀x ∈ ∂Ω},

As before A_{D} is linear, monotone and symmetric and thus our problem reduces to the existence for an elliptic equation:

−Δw = \frac{1}{T} ∫_{0}^{T} \{f(t) + Δu_{0}(t)\}dt,

with homogenous Dirichlet boundary condition w = 0 on ∂Ω. Since the previous problem has a unique solution verifying

∥w∥_{L^{2}(Ω)} ≤ C(Ω)∥\frac{1}{T} ∫_{0}^{T} \{f(t) + Δu_{0}(t)\}dt∥_{L^{2}(Ω)}
≤ C(Ω)∥f + Δu_{0}∥_{L^{∞}([0,T];L^{2}(Ω))},  (59)

we prove the existence for (50), (51). Here we denote by C(Ω) the Poincaré’s constant,

( ∫_{Ω} |w(x)|^{2}dx)^{1/2} ≤ C(Ω)( ∫_{Ω} ∥∇w(x)∥^{2}dx)^{1/2}, ∀w ∈ H_{0}^{1}(Ω).

Moreover in this case the operator A_{D} is strictly monotone. Indeed, by using the Poincaré’s inequality, for each v ∈ D(A_{D}), we have have

( ∫_{Ω} |v(x)|^{2}dx)^{1/2} ≤ C(Ω)( ∫_{Ω} ∥∇v(x)∥^{2}dx)^{1/2} = C(Ω)(A_{D}v, v)^{1/2}.

Hence if (A_{D}v, v) = 0 we deduce that v = 0. Therefore, by Proposition 3.2 we deduce the uniqueness of the periodic solution for (50), (51). The estimates of the solution follow immediately from (59) and Theorem 3.8.

3.4 Non-linear case

Throughout this section we will consider evolution equations associated to subdifferential operators. Let φ : H →]−∞, +∞] be a lower-semicontinuous proper convex function on a real Hilbert space H. Denote by ∂φ ⊂ H × H the subdifferential of φ,

∂φ(x) = {y ∈ H; φ(x) − φ(u) ≤ (y, x − u), ∀u ∈ H},  (60)

and denote by D(φ) the effective domain of φ:

D(φ) = {x ∈ H; φ(x) < +∞}.

Under the previous assumptions on φ we recall that A = ∂φ is maximal monotone in H × H and \overline{D(A)} = \overline{D(φ)}. Consider the equation

x′(t) + ∂φx(t) ∋ f(t), 0 < t < T.  (61)

We say that x is solution for (61) if x ∈ C([0, T]; H), x is absolutely continuous on every compact of ]0, T [ (and therefore a.e. differentiable on ]0, T [) and satisfies x(t) ∈ D(∂φ) a.e. on ]0, T [ and x′(t) + ∂φx(t) ∋ f(t) a.e. on ]0, T [. We have the following main result [1]

Theorem 3.14 Let f be given in L^{2}(]0, T [; H) and x_{0} ∈ \overline{D(∂φ)}. Then the Cauchy problem (61) with the initial condition x(0) = x_{0} has a unique solution x ∈ C([0, T]; H) which satisfies:

x ∈ W^{1,2}(]δ, T [; H) ∀ 0 < δ < T, \sqrt{t} · x′ ∈ L^{2}(]0, T [; H), φ ∘ x ∈ L^{1}(0, T).

Moreover, if x_{0} ∈ D(φ) then

x′ ∈ L^{2}(]0, T [; H), φ ∘ x ∈ L^{∞}(0, T).

We are interested in finding sufficient conditions on A = ∂φ and f such that equation (61) has unique T -periodic solution, i.e. x(0) = x(T). Obviously, if such a solution exists, by periodicity we deduce that it is absolutely continuous on [0, T] and belongs to W^{1,2}(]0, T [; H). It is well known that if φ is strictly convex then ∂φ is strictly monotone and therefore the uniqueness holds

Proposition 3.15 Assume that φ : H →] − ∞, +∞] is a lower-semicontinuous proper, strictly convex function. Then equation (61) has at most one periodic solution.

Proof By using Proposition 3.2 it is sufficient to prove that ∂φ is strictly monotone. Suppose that there are u_{1}, u_{2} ∈ D(∂φ), u_{1} ≠ u_{2} such that

(∂φ(u_{1}) − ∂φ(u_{2}), u_{1} − u_{2}) = 0.

We have

φ(u_{2}) − φ(u_{1}) ≥ (∂φ(u_{1}), u_{2} − u_{1})
= −(∂φ(u_{2}), u_{1} − u_{2})
≥ φ(u_{2}) − φ(u_{1}),

and hence

φ(u_{2}) − φ(u_{1}) = (∂φ(u_{1}), u_{2} − u_{1}).

We can also write for λ ∈]0, 1[

φ((1 − λ)u_{1} + λu_{2}) = φ(u_{1} + λ(u_{2} − u_{1}))
≥ φ(u_{1}) + (∂φ(u_{1}), λ(u_{2} − u_{1}))
= φ(u_{1}) + λ(∂φ(u_{1}), u_{2} − u_{1})
= φ(u_{1}) + λ(φ(u_{2}) − φ(u_{1}))
= (1 − λ)φ(u_{1}) + λφ(u_{2}).

Since φ is strictly convex we have also

φ((1 − λ)u_{1} + λu_{2}) < (1 − λ)φ(u_{1}) + λφ(u_{2}),

which is in contradiction with the previous inequality. Thus u_{1} = u_{2} and hence ∂φ is strictly monotone. We state now the result concerning the existence of periodic solutions.

Theorem 3.16 Suppose that φ : H →] − ∞, +∞] is a lower-semicontinuous proper convex function and f ∈ L^{2}(]0, T [; H) such that

lim_{∥x∥→∞}{φ(x) − (x, ⟨f⟩)} = +∞,  (62)

and every level subset {x ∈ H; φ(x) + ∥x∥^{2} ≤ M} is compact. Then equation (61) has T -periodic solutions x ∈ C([0, T]; H) ∩ W^{1,2}(]0, T [; H) which satisfy

∥x′∥_{L^{2}(]0,T [;H)} ≤ ∥f∥_{L^{2}(]0,T [;H)}, x(t) ∈ D(φ) ∀ t ∈ [0, T], φ ∘ x ∈ L^{∞}(0, T).

Before showing this result, notice that the condition (62) implies that the lower-semicontinuous proper convex function ψ : H →] − ∞, +∞] given by ψ(x) = φ(x) − (x, ⟨f⟩) has a minimum point x_{0} ∈ H and therefore ⟨f⟩ ∈ Range(∂φ) since 0 = ∂ψ(x_{0}) = ∂φ(x_{0}) − ⟨f⟩.

Proof As previous for every α > 0 we consider the unique periodic solution x_{α} for

αx_{α}(t) + x′_{α}(t) + ∂φx_{α}(t) = f(t), 0 < t < T.  (63)

(In order to prove the existence and uniqueness of the periodic solution for (63) consider the application S_{α} : \overline{D(∂φ)} → \overline{D(∂φ)} defined by S_{α}(x_{0}) = x(T; 0, x_{0}), where x(·; 0, x_{0}) denote the unique solution of (63) with the initial condition x_{0} and apply the Banach’s fixed point theorem. By the previous theorem it follows that the periodic solution x_{α} is absolutely continuous on [0, T] and belongs to C([0, T]; H)∩W^{1,2}(]0, T [; H)). First of all we will show that (x′_{α})_{α>0} is uniformly bounded in L^{2}(]0, T [; H). Indeed, after multiplication by x′_{α}(t) we obtain

∫_{0}^{T} ∥x′_{α}(t)∥^{2}dt+∫_{0}^{T} {α(x_{α}(t), x′_{α}(t))+(∂φx_{α}(t), x′_{α}(t))}dt = ∫_{0}^{T} (f(t), x′_{α}(t))dt.

Since x_{α} is T -periodic we deduce that

∫_{0}^{T} {α(x_{α}(t), x′_{α}(t)) + (∂φx_{α}(t), x′_{α}(t))}dt
= ∫_{0}^{T} \frac{d}{dt}\{\frac{α}{2}∥x_{α}(t)∥^{2} + φ(x_{α}(t))\}dt
= \frac{α}{2}∥x_{α}(t)∥^{2} + φ(x_{α}(t))|_{0}^{T} = 0.  (64)

Therefore, ∥x′_{α}∥_{L^{2}(]0,T [;H)}^{2} ≤ (f, x′_{α})_{L^{2}(]0,T [;H)} and thus

∥x′_{α}∥_{L^{2}(]0,T [;H)} ≤ ∥f∥_{L^{2}(]0,T [;H)}, α > 0.

Before estimate (x_{α})_{α>0}, let us check that (αx_{α})_{α>0} is bounded. By taking x_{0} ∈ D(∂φ), after standard calculation we find that

∥x_{α}(t) − x_{0}∥ ≤ e^{−αt}∥x_{α}(0) − x_{0}∥ + ∫_{0}^{t} e^{−α(t−s)}∥f(s) − αx_{0} − ∂φ(x_{0})∥ ds. (65)

Since x_{α} is T -periodic we can write

∥x_{α}(t) − x_{0}∥ = lim_{n→∞} ∥x_{α}(nT + t) − x_{0}∥
≤ lim_{n→∞} {e^{−α(nT +t)}∥x_{α}(0) − x_{0}∥

+ ∫_{0}^{nT +t} e^{−α(nT +t−s)}∥f(s) − αx_{0} − ∂φ(x_{0})∥ ds}

≤ \frac{1}{α}∥αx_{0} + ∂φ(x_{0})∥ + lim_{n→∞} ∫_{0}^{nT +t} e^{−α(nT +t−s)}∥f(s)∥ ds
≤ \frac{1}{α}∥αx_{0} + ∂φ(x_{0})∥

+ lim_{n→∞} {[1 + e^{−αt}(e^{−α(n−1)T} + ⋯ + e^{−αT} + 1)] · ∥f∥_{L^{1}}}

= \frac{1}{α}∥αx_{0} + ∂φ(x_{0})∥ + (1 + \frac{e^{−αt}}{1 − e^{−αT}} ) · ∥f∥_{L^{1}(]0,T [;H)}
≤ C_{1}(x_{0}, T, ∥f∥_{L^{2}(]0,T [;H)})(1 + \frac{1}{α}), 0 ≤ t ≤ T, α > 0.

It follows that α∥x_{α}(t)∥ ≤ C_{2}(x_{0}, T, ∥f∥_{L^{2}(]0,T [;H)}), 0 ≤ t ≤ T, 0 < α < 1. Now we can estimate x_{α}, α > 0. After multiplication by x_{α}(t) and integration on [0, T] we obtain

∫_{0}^{T} α∥x_{α}(t)∥^{2}dt + ∫_{0}^{T} (∂φ(x_{α}(t)), x_{α}(t))dt = ∫_{0}^{T} (f(t), x_{α}(t))dt.  (66)

We have

φ(x_{0}) ≥ φ(x_{α}(t)) + (∂φ(x_{α}(t)), x_{0} − x_{α}(t)), t ∈ [0, T], α > 0.

Thus we deduce that for α > 0,

∫_{0}^{T} (∂φ(x_{α}(t)), x_{α}(t))dt ≥ ∫_{0}^{T} φ(x_{α}(t))dt + ∫_{0}^{T} {(∂φ(x_{α}(t)), x_{0}) − φ(x_{0})}dt.

On the other hand for 0 < α < 1,

∫_{0}^{T} (∂φ(x_{α}(t)), x_{0}) dt = ∫_{0}^{T} (f(t) − αx_{α}(t) − x′_{α}(t), x_{0}) dt
= ( ∫_{0}^{T} f(t) dt, x_{0}) − ∫_{0}^{T} (αx_{α}(t), x_{0}) dt
≥ −C_{3}(x_{0}, T, ∥f∥_{L^{2}(]0,T [;H)}).

Therefore,

∫_{0}^{T} (∂φ(x_{α}(t)), x_{α}(t))dt ≥ ∫_{0}^{T} φ(x_{α}(t))dt − C_{4}(x_{0}, T, ∥f∥_{L^{2}(]0,T [;H)}). (67)

Combining (66) and (67) we deduce that

∫_{0}^{T} φ(x_{α}(t))dt ≤ C_{4} + ∫_{0}^{T} (∂φ(x_{α}(t)), x_{α}(t))dt
= C_{4} + ∫_{0}^{T} (f(t), x_{α}(t))dt − ∫_{0}^{T} α∥x_{α}(t)∥^{2}dt
≤ C_{4} + ∫_{0}^{T} (f(t), x_{α}(t))dt, 0 < α < 1.  (68)

On the other hand we have

∫_{0}^{T} (f(t), x_{α}(t))dt
= ∫_{0}^{T} (f(t) − ⟨f⟩, x_{α}(t))dt + ( ∫_{0}^{T} x_{α}(t)dt, ⟨f⟩)
= ∫_{0}^{T} (f(t) − ⟨f⟩, x_{α}(0) + ∫_{0}^{t} x′_{α}(s) ds)dt + T (⟨x_{α}⟩, ⟨f⟩)
= ∫_{0}^{T} (f(t) − ⟨f⟩, ∫_{0}^{t} x′_{α}(s) ds)dt + T (⟨x_{α}⟩, ⟨f⟩)
≤ ∫_{0}^{T} ∥f(t) − ⟨f⟩∥ · ( ∫_{0}^{t} ∥x′_{α}(s)∥^{2} ds)^{1/2} · t^{1/2} dt + T (⟨x_{α}⟩, ⟨f⟩)
≤ ∥f − ⟨f⟩∥_{L^{2}(]0,T [;H)} · ∥f∥_{L^{2}(]0,T [;H)} · \frac{T}{\sqrt{2}} + T (⟨x_{α}⟩, ⟨f⟩).

Finally we deduce that

∫_{0}^{T} {φ(x_{α}(t)) − (x_{α}(t), ⟨f⟩)} dt ≤ C_{5}(x_{0}, T, ∥f∥_{L^{2}(]0,T [;H)}), 0 < α < 1, (69)

and thus there is t_{α} ∈ [0, T] such that

φ(x_{α}(t)) − (x_{α}(t), ⟨f⟩) ≤ \frac{C_{5}}{T} , 0 < α < 1.  (70)

From Hypothesis (62) we get that (x_{α}(t_{α}))_{0<α<1} is bounded and therefore from (65), for t ∈ [t_{α}, t_{α} + T],

∥x_{α}(t) − x_{0}∥ ≤ e^{−α(t−t_{α})}∥x_{α}(t_{α}) − x_{0}∥ + ∫_{t_{α}}^{t} e^{−α(t−s)}∥f(s) − αx_{0} − ∂φ(x_{0})∥ ds.

we deduce that (x_{α})_{0<α<1} is bounded in L^{∞}(]0, T [; H) and that there is x ∈ L^{∞}(]0, T [; H) such that x_{α}(t) ⇀ x(t) when α goes to 0 for t ∈ [0, T]. Moreover, from (70) it follows that (φ(x_{α}(t_{α})))_{0<α<1} is bounded from above and we deduce that

φ(x_{α}(t)) = φ(x_{α}(t_{α})) + ∫_{t_{α}}^{t}(∂φ(x_{α}(s)), x′_{α}(s)) ds
≤ φ(x_{α}(t_{α})) + ∫_{t_{α}}^{t}(f(s) − αx_{α}(s) − x′_{α}(s), x′_{α}(s)) ds
≤ C_{6}(x_{0}, T, ∥f∥_{L^{2}(]0,T [;H)}), 0 < α < 1.

On the other hand, by writing φ(x_{α}(t)) ≥ φ(x_{0}) + (∂φ(x_{0}), x_{α}(t) − x_{0}), 0 ≤ t ≤ T, α > 0 we deduce that φ(x_{α}(t)) is also bounded from below so that finally (φ ∘ x_{α})_{0<α<1} is bounded in L^{∞}(]0, T [; H).

Now, using the second hypothesis of the theorem (every level subset is compact) we deduce that x_{α}(0) → x(0) when α goes to 0 (at least for a subsequence α_{n} ↘ 0). In fact we can easily check that x_{α} converges uniformly to x on [0, T] since

∥x_{α}(t) − x_{β}(t)∥ ≤ ∥x_{α}(0) − x_{β}(0)∥ + |α − β| · T · sup_{0<γ<1} ∥x_{γ}∥_{L^{∞}(]0,T [;H)},

for 0 ≤ t ≤ T, 0 < α, β < 1. Now, since lim_{α↘0} dx_{α}/dt = dx/dt in the sense of H-valued vectorial distribution on ]0, T [ and (x′_{α})_{α>0} is bounded in L^{2}(]0, T [; H) it follows that x′ belongs to L^{2}(]0, T [; H) and in particular x is absolutely continuous on every compact of ]0, T [ and therefore a.e. differentiable on ]0, T [.

To complete the proof we need to show that x(t) ∈ D(φ) a.e. on ]0, T [ and x′(t) + ∂φx(t) ∋ f(t) a.e. on ]0, T [. For arbitrarily [u, v] ∈ ∂φ we have

\frac{1}{2}e^{2αt}∥x_{α}(t) − u∥^{2} ≤ \frac{1}{2}e^{2αs}∥x_{α}(s) − u∥^{2} + ∫_{s}^{t} e^{2ατ} (f(τ) − αu − v, x_{α}(τ) − u) dτ,

with 0 ≤ s ≤ t ≤ T, α > 0. Passing to the limit for α ↘ 0 we get

\frac{1}{2}∥x(t) − u∥^{2} ≤ \frac{1}{2}∥x(s) − u∥^{2} + ∫_{s}^{t}(f(τ) − v, x(τ) − u) dτ, 0 ≤ s ≤ t ≤ T.

Thus

(x(t)−x(s), x(s)−u) ≤ \frac{1}{2}∥x(t)−u∥^{2} − \frac{1}{2}∥x(s)−u∥^{2} ≤ ∫_{s}^{t}(f(τ)−v, x(τ)−u) dτ,

for 0 ≤ s ≤ t ≤ T. Since x is a.e. differentiable on ]0, T [ we find that

(x′(t), x(t) − u) = lim_{s↗t} \frac{1}{t − s}(x(t) − x(s), x(s) − u)
≤ lim_{s↗t} \frac{1}{t − s} ∫_{s}^{t}(f(τ) − v, x(τ) − u) dτ
= (f(t) − v, x(t) − u), a.e. t ∈]0, T [, ∀ [u, v] ∈ ∂φ.

Finally, since ∂φ is maximal monotone and (f(t) − x′(t) − v, x(t) − u) ≥ 0 for all [u, v] ∈ ∂φ we deduce that x(t) ∈ D(∂φ) a.e. on ]0, T [ and x′(t) + ∂φx(t) ∋ f(t) a.e. on ]0, T [. Since φ is lower-semicontinuous we also have

φ(x(t)) ≤ lim_{α↘0} inf φ(x_{α}(t)) ≤ lim_{α↘0} inf ∥φ ∘ x_{α}∥_{L^{∞}} ≤ sup_{0<γ<1} ∥φ ∘ x_{γ}∥_{L^{∞}}.

As previous, by writing

φ(x(t)) ≥ φ(x_{0}) + (∂φ(x_{0}), x(t) − x_{0})
≥ φ(x_{0}) − ∥∂φ(x_{0})∥ · (∥x_{0}∥ + lim_{α↘0} inf ∥x_{α}(t)∥)
≥ φ(x_{0}) − ∥∂φ(x_{0})∥ · (∥x_{0}∥ + sup_{0<γ<1} ∥x_{γ}∥_{L^{∞}}), 0 ≤ t ≤ T,

we deduce finally that φ ∘ x ∈ L^{∞}(0, T).

Remark 3.17 If dim H < +∞ then the level subsets {x ∈ H ; φ(x) + ∥x∥^{2} ≤ M} are compact as bounded sets.

Remark 3.18 Assume that φ : H →] − ∞, +∞] is a lower-semicontinuous proper convex function such that Range(∂φ) = H which is equivalent to

lim_{∥x∥→∞}{φ(x) − (x, y)} = +∞, ∀y ∈ H,

see [4], pp.41. In particular, by taking y = ⟨f⟩ we deduce that the hypothesis (62) is verified.

Remark 3.19 Assume that φ is coercive

lim_{∥x∥→∞} \frac{(∂φ(x), x − x_{0})}{∥x∥} = +∞, ∀x_{0} ∈ D(φ),

which is equivalent to lim_{∥x∥→∞} \frac{φ(x)}{∥x∥} = +∞ (see [4], pp.42). Then Range(φ) = H because the previous condition is satisfied: lim_{∥x∥→∞}{φ(x) − (x, y)} = +∞, for all y ∈ H and therefore (62) is verified.

Theorem 3.20 Suppose that φ : H →] − ∞, +∞] is a lower-semicontinuous proper convex function and f ∈ W^{1,1}(]0, T [; H) such that

lim_{∥x∥→∞}{φ(x) − (x, ⟨f⟩)} = +∞,  (71)

and every level subset {x ∈ H; φ(x) + ∥x∥^{2} ≤ M} is compact. Then equation (61) has T -periodic solutions x ∈ C([0, T]; H) ∩ W^{1,∞}(]0, T [; H) which satisfy

x(t) ∈ D(∂φ), ∀ t ∈ [0, T], \frac{d^{+}}{dt} x(t) + (∂φx(t) − f(t))^{∘} = 0, ∀ t ∈ [0, T],

where (∂φ − f)^{∘} denote the minimal section of ∂φ − f.

Proof Since W^{1,1}(]0, T [; H) ⊂ L^{2}(]0, T [; H) the previous theorem applies. Consider x ∈ C([0, T]; H) ∩ W^{1,2}(]0, T [; H) a T -periodic solution for (61). Since ∥x′∥_{L^{2}(]0,T [;H)} ≤ ∥f∥_{L^{2}(]0,T [;H)} it follows that there is t^{⋆} ∈]0, T [ such that x is differentiable in t^{⋆} and ∥x′(t^{⋆})∥ ≤ \frac{1}{\sqrt{T}} ∥f∥_{L^{2}(]0,T [;H)}. By standard calculation we find that:

∥\frac{1}{h}(x(t + h) − x(t))∥ ≤ ∥\frac{1}{h}(x(t^{⋆} + h) − x(t^{⋆}))∥ + ∫_{t^{⋆}}^{t} ∥\frac{1}{h}(f(τ + h) − f(τ)∥ dτ,

and therefore sup_{0≤t≤T, h>0} ∥\frac{1}{h}(x(t + h) − x(t))∥ ≤ C which implies that x ∈ W^{1,∞}(]0, T [; H). Making use of the inequality

\frac{1}{h}(x(t+h)−x(t), x(t)−u) ≤ \frac{1}{h} ∫_{t}^{t+h}(f(τ)−v, x(τ)−u) dτ, 0 ≤ t < t+h ≤ T,

which holds for every [u, v] ∈ ∂φ we deduce that x(t) ∈ D(∂φ) for all t ∈ [0, T] and the weak closure of the set \{\frac{1}{h}(x(t + h) − x(t)), h > 0\} belongs to f(t) − ∂φx(t), ∀t ∈ [0, T]. On the other hand by writing

∥x(t + h) − u∥ ≤ ∥x(t) − u∥ + ∫_{t}^{t+h} ∥f(τ) − v∥ dτ, 0 ≤ t < t + h ≤ T,

for u = x(t) and v ∈ ∂φx(t) we find that

∥(∂φx(t) − f(t))^{∘}∥ ≤ ∥w − lim_{h↘0} \frac{1}{h}(x(t + h) − x(t))∥
≤ lim sup_{h↘0} ∥\frac{1}{h}(x(t + h) − x(t))∥
≤ ∥(∂φx(t) − f(t))^{∘}∥.

This shows that lim_{h↘0} \frac{1}{h}(x(t + h) − x(t)) = \frac{d^{+}}{dt} x(t) exists for every t ∈ [0, T] and coincides with −(∂φx(t) − f(t))^{∘}.

References

[1] V. Barbu, Nonlinear Semigroups and Differential Equations in Banach Spaces, Noordhoff (1976).

[2] M. Bostan, Solutions périodiques des équations d’évolution, C. R. Acad. Sci. Paris, Sér. I Math. t.332, pp. 1-4, Équations dérivées partielles, (2001).

[3] M. Bostan, Almost periodic solutions for evolution equations, article in preparation.

[4] H. Brezis, Opérateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert, Noth-Holland, Lecture Notes no. 5 (1972).

[5] H. Brezis, A Haraux, Image d’une somme d’opérateurs monotones et applications, Israel J. Math. 23 (1976), 2, pp. 165-186.

[6] H. Brezis, Analyse fonctionnelle, Masson, (1998).

[7] A. Haraux, Équations d’évolution non linéaires: solutions bornées périodiques, Ann. Inst. Fourier 28 (1978), 2, pp. 202-220.

Mihai Bostan Université de Franche-Comté 16 route de Gray F-25030 Besançon Cedex, France mbostan@math.univ-fcomte.fr