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—

Periodic solutions for evolution equations

Mihai Bostan

Abstract

We study the existence and uniqueness of periodic solutions for evolu-

tion equations. First we analyze the one-dimensional case. Then for arbi-

trary dimensions (¬nite or not), we consider linear symmetric operators.

We also prove the same results for non-linear sub-di¬erential operators

A = ‚• where • 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 perma-

nent regimes for ordinary or partial di¬erential equations. The main purpose of

this paper is to establish existence and uniqueness results for periodic solutions

in the general framework of evolution equations,

t ∈ R,

x (t) + Ax(t) = f (t), (1)

— Mathematics Subject Classi¬cations: 34B05, 34G10, 34G20.

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

c 2002 Southwest Texas State University.

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

1

2 Periodic solutions for evolution equations

by using the penalization method. Note that in the linear case a necessary

condition for the existence is

T

1

f (t)dt ∈ Range(A).

f := (2)

T 0

Unfortunately, this condition is not always su¬cient for existence; see the exam-

ple of the orthogonal rotation of R2 . Nevertheless, the condition (2) is su¬cient

in the symmetric case. The key point consists of considering ¬rst the perturbed

equation

±x± (t) + x± (t) + Ax± (t) = f (t), t ∈ R,

where ± > 0. By using the Banach™s ¬xed point theorem we deduce the existence

and uniqueness of the periodic solutions x± , ± > 0. Under the assumption (2),

in the linear symmetric case we show that (x± )±>0 is a Cauchy sequence in C 1 .

Then by passing to the limit for ± ’ 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 di¬erent methods. A

similar condition (2) has been investigated in [5] when studying the range of

sums of monotone operators. A di¬erent method consists of applying ¬xed

point techniques, see for example [4, 7].

This article is organized as follows. First we analyze the one dimensional

case. Necessary and su¬cient conditions for the existence and uniqueness of pe-

riodic 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-di¬erential operators is considered.

2 Periodic solutions for one dimensional evolu-

tion equations

To study the periodic solutions for evolution equations it is convenient to con-

sider ¬rst the one dimensional case

t ∈ R,

x (t) + g(x(t)) = f (t), (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) = x0 ∈ 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).

Mihai Bostan 3

Proof Let x1 , x2 be two periodic solutions for (3). By taking the di¬erence

between the two equations and multiplying by x1 (t) ’ x2 (t) we get

1d

|x1 (t) ’ x2 (t)|2 + [g(x1 (t)) ’ g(x2 (t))][x1 (t) ’ x2 (t)] = 0, t ∈ R. (4)

2 dt

Since g is increasing we have (g(x1 ) ’ g(x2 ))(x1 ’ x2 ) ≥ 0 for all x1 , x2 ∈ R and

therefore we deduce that |x1 (t)’x2 (t)| is decreasing. Moreover as x1 and x2 are

periodic it follows that |x1 (t) ’ x2 (t)| does not depend on t ∈ R and therefore,

from (4) we get

[g(x1 (t)) ’ g(x2 (t))][x1 (t) ’ x2 (t)] = 0, t ∈ R.

Finally, the strictly monotony of g implies that x1 = x2 .

Remark 2.2 If g is only increasing, it is possible that (3) has several periodic

solutions. Let us consider the function

±

x < ’µ,

x+µ

x ∈ [’µ, µ],

0

g(x) = (5)

x’µ x > µ,

µ µ

and f (t) = 2 cos t. We can easily check that x» (t) = » + sin t are periodic

2

µµ

solutions for (3) for » ∈ [’ 2 , 2 ].

Generally we can prove that every two periodic solutions di¬er by a constant.

Proposition 2.3 Let g be an increasing function and x1 , x2 two periodic solu-

tions of (3). Then there is a constant C ∈ R such that

x1 (t) ’ x2 (t) = C, ∀t ∈ R.

Proof As shown before there is a constant C ∈ R such that |x1 (t)’x2 (t)| = C,

t ∈ R. Moreover x1 (t) ’ x2 (t) has constant sign, otherwise x1 (t0 ) = x2 (t0 ) for

some t0 ∈ R and it follows that |x1 (t) ’ x2 (t)| = |x1 (t0 ) ’ x2 (t0 )| = 0, t ∈ R or

x1 = x2 . Finally we ¬nd that

x1 (t) ’ x2 (t) = sign(x1 (0) ’ x2 (0))C, t ∈ R.

Before analyzing in detail the uniqueness for increasing functions, let us de¬ne

the following sets.

t

’ y)ds ∈ g ’1 (y) ∀t ∈ R ‚ g ’1 (y),

x∈R:x+ y ∈ g(R),

(f (s)

O(y) = 0

…, y ∈ g(R).

/

Proposition 2.4 Let g be an increasing function and f periodic. Then equation

(3) has di¬erent periodic solutions if and only if Int(O f ) = ….

4 Periodic solutions for evolution equations

Proof Assume that (3) has two periodic solutions x1 = x2 . By the previous

proposition we have x2 ’ x1 = C > 0. By integration on [0, T ] one gets

T T T

g(x1 (t))dt = f (t)dt = g(x2 (t))dt. (6)

0 0 0

Since g is increasing we have g(x1 (t)) ¤ g(x2 (t)), t ∈ R and therefore,

T T

g(x1 (t))dt ¤ g(x2 (t))dt. (7)

0 0

From (6) and (7) we deduce that g(x1 (t)) = g(x2 (t)), t ∈ R and thus g is

constant on each interval [x1 (t), x2 (t)] = [x1 (t), x1 (t) + C], t ∈ R. Finally it

implies that g is constant on Range(x1 ) + [0, C] = {x1 (t) + y : t ∈ [0, T ], y ∈

[0, C]} and this constant is exactly the time average of f :

t ∈ [0, T ].

g(x1 (t)) = g(x2 (t)) = f ,

Let x be an arbitrary real number in ]x1 (0), x1 (0) + C[. Then

t t

{f (s) ’ f }ds = x ’ x1 (0) + x1 (0) + {f (s) ’ g(x1 (s))}ds

x+

0 0

= x ’ x1 (0) + x1 (t)

> x1 (t), t ∈ R.

Similarly,

t t

{f (s) ’ f }ds = x ’ x2 (0) + x2 (0) + {f (s) ’ g(x2 (s))}ds

x+

0 0

= x ’ x2 (0) + x2 (t)

< x2 (t), t ∈ R.

t

Therefore, x + 0 {f (s) ’ f }ds ∈]x1 (t), x2 (t)[‚ g ’1 ( f ), t ∈ R which implies

that x ∈ O f and hence ]x1 (0), x2 (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 x1 , x2 given below are di¬erent periodic solutions

for (3):

t

{f (s) ’ f }ds, t ∈ R,

x1 (t) = x +

0

t

{f (s) ’ f }ds = x1 (t) + C, t ∈ R.

x2 (t) = x + C +

0

Remark 2.5 The condition Int(O f ) = … is equivalent to

diam(g ’1 f ) > diam(Range {f (t) ’ f }dt).

Mihai Bostan 5

Example: Consider the equation x (t) + g(x(t)) = · cos t, t ∈ R with g given

in Remark 2.2. We have < · cos t >= 0 ∈ g(R) and

t

· cos s ds ∈ g ’1 (0),

O(0) = {x ∈ R | x + t ∈ R} (8)

0

= {x ∈ R : x + · sin t ∈ g ’1 (0), t ∈ R}

= {x ∈ R : ’µ ¤ x + · sin t ¤ µ, t ∈ R}

±

… |·| > µ,

{0} |·| = µ,

= (9)

[|·| ’ µ, µ ’ |·|] |·| < µ.

Therefore, uniqueness does not occur if |·| < µ, for example if · = µ/2, as seen

before in Remark 2.2. If |·| ≥ µ 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

1T

there is x0 ∈ R such that f := T 0 f (t)dt = g(x0 ).

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

T T

x(T ) ’ x(0) + g(x(t))dt = f (t)dt.

0 0

Since x is periodic and g —¦ x is continuous we get

T

„ ∈]0, T [,

T g(x(„ )) = f (t)dt,

0

and hence

T

1

f (t)dt ∈ Range(g).

f := (10)

T 0

™¦

In the following we will show that this condition is also su¬cient for the