. 1
( 10 .)



>>

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 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

. 1
( 10 .)



>>