Periodic solutions for evolution equations

41 Pages·2002·0.334 MB·English

by  Sylvie Benzoni-Gavage, Denis Serre

Electronic Journal ofDifferential Equations,Monogrpah03,2002.
ISSN:1072-6691. URL:http://ejde.math.swt.eduorhttp://ejde.math.unt.edu
ftpejde.math.swt.edu(login: ftp)
∗
Periodic solutions for evolution equations
Mihai Bostan
Abstract
Westudytheexistenceanduniquenessofperiodicsolutionsforevolu-
tionequations. Firstweanalyzetheone-dimensionalcase. Thenforarbi-
trary 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
Manytheoreticalandnumericalstudiesinappliedmathematicsfocusonperma-
nentregimesforordinaryorpartialdifferentialequations. Themainpurposeof
this paper is to establish existence and uniqueness results for periodic solutions
in the general framework of evolution equations,
x(cid:48)(t)+Ax(t)=f(t), t∈R, (1)
∗Mathematics Subject Classifications: 34B05,34G10,34G20.
Key words: maximalmonotoneoperators,evolutionequations,Hille-Yosida’stheory.
(cid:13)c2002SouthwestTexasStateUniversity.
SubmittedMay14,2002. PublishedAugust23,2002.
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
1 T
(cid:104)f(cid:105):= (cid:90) f(t)dt∈Range(A). (2)
T
0
Unfortunately,thisconditionisnotalwayssufficientforexistence;seetheexam-
pleoftheorthogonalrotationofR2. Nevertheless,thecondition(2)issufficient
inthesymmetriccase. Thekeypointconsistsofconsideringfirsttheperturbed
equation
αx (t)+x(cid:48) (t)+Ax (t)=f(t), t∈R,
α α α
whereα>0. ByusingtheBanach’sfixedpointtheoremwededucetheexistence
and uniqueness of the periodic solutions x ,α > 0. Under the assumption (2),
α
in the linear symmetric case we show that (x ) is a Cauchy sequence in C1.
α α>0
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. Necessaryandsufficientconditionsfortheexistenceanduniquenessofpe-
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-differential 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 first the one dimensional case
x(cid:48)(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 ∈R there is an unique solution x∈C1(R;R) for (3). We
0
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 x ,x be two periodic solutions for (3). By taking the difference
1 2
between the two equations and multiplying by x (t)−x (t) we get
1 2
1 d
|x (t)−x (t)|2+[g(x (t))−g(x (t))][x (t)−x (t)]=0, t∈R. (4)
2dt 1 2 1 2 1 2
Since g is increasing we have (g(x )−g(x ))(x −x )≥0 for all x ,x ∈R and
1 2 1 2 1 2
thereforewededucethat|x (t)−x (t)|isdecreasing. Moreoverasx andx are
1 2 1 2
periodic it follows that |x (t)−x (t)| does not depend on t∈R and therefore,
1 2
from (4) we get
[g(x (t))−g(x (t))][x (t)−x (t)]=0, t∈R.
1 2 1 2
Finally, the strictly monotony of g implies that x =x .
1 2
Remark 2.2 If g is only increasing, it is possible that (3) has several periodic
solutions. Let us consider the function
x+ε x<−ε,
g(x)= 0 x∈[−ε,ε], (5)

x−ε x>ε,

and f(t) = εcost. We can easily check that x (t) = λ+ εsint are periodic
2 λ 2
solutions for (3) for λ∈[−ε,ε].
2 2
Generallywecanprovethateverytwoperiodicsolutionsdifferbyaconstant.
Proposition 2.3 Let g be an increasing function and x ,x two periodic solu-
1 2
tions of (3). Then there is a constant C ∈R such that
x (t)−x (t)=C, ∀t∈R.
1 2
Proof AsshownbeforethereisaconstantC ∈Rsuchthat|x (t)−x (t)|=C,
1 2
t ∈ R. Moreover x (t)−x (t) has constant sign, otherwise x (t ) = x (t ) for
1 2 1 0 2 0
some t ∈R and it follows that |x (t)−x (t)|=|x (t )−x (t )|=0, t∈R or
0 1 2 1 0 2 0
x =x . Finally we find that
1 2
x (t)−x (t)=sign(x (0)−x (0))C, t∈R.
1 2 1 2
Before analyzing in detail the uniqueness for increasing functions, let us define
the following sets.
x∈R:x+ t(f(s)−y)ds∈g−1(y)∀t∈R ⊂g−1(y), y ∈g(R),
O(y)=(cid:26) (cid:8)∅, (cid:82)0 (cid:9) y ∈/ g(R).
Proposition 2.4 Letgbeanincreasingfunctionandf periodic. Thenequation
(3) has different periodic solutions if and only if Int(O(cid:104)f(cid:105))(cid:54)=∅.
4 Periodic solutions for evolution equations
Proof Assume that (3) has two periodic solutions x (cid:54)= x . By the previous
1 2
proposition we have x −x =C >0. By integration on [0,T] one gets
2 1
T T T
(cid:90) g(x1(t))dt=(cid:90) f(t)dt=(cid:90) g(x2(t))dt. (6)
0 0 0
Since g is increasing we have g(x (t))≤g(x (t)), t∈R and therefore,
1 2
T T
(cid:90) g(x1(t))dt≤(cid:90) g(x2(t))dt. (7)
0 0
From (6) and (7) we deduce that g(x (t)) = g(x (t)), t ∈ R and thus g is
1 2
constant on each interval [x (t),x (t)] = [x (t),x (t)+C], t ∈ R. Finally it
1 2 1 1
implies that g is constant on Range(x )+[0,C] = {x (t)+y : t ∈ [0,T],y ∈
1 1
[0,C]} and this constant is exactly the time average of f:
g(x (t))=g(x (t))=(cid:104)f(cid:105), t∈[0,T].
1 2
Let x be an arbitrary real number in ]x (0),x (0)+C[. Then
1 1
t t
x+(cid:90) {f(s)−(cid:104)f(cid:105)}ds = x−x1(0)+x1(0)+(cid:90) {f(s)−g(x1(s))}ds
0 0
= x−x (0)+x (t)
1 1
> x (t), t∈R.
1
Similarly,
t t
x+(cid:90) {f(s)−(cid:104)f(cid:105)}ds = x−x2(0)+x2(0)+(cid:90) {f(s)−g(x2(s))}ds
0 0
= x−x (0)+x (t)
2 2
< x (t), t∈R.
2
Therefore,x+ t{f(s)−(cid:104)f(cid:105)}ds∈]x (t),x (t)[⊂g−1((cid:104)f(cid:105)), t∈Rwhichimplies
that x∈O(cid:104)f(cid:105)(cid:82)a0nd hence ]x (0),x (10)[⊂O2 (cid:104)f(cid:105).
1 2
Conversely,supposethatthereisxandC >0smallenoughsuchthatx,x+C ∈
O(cid:104)f(cid:105). It is easy to check that x ,x given below are different periodic solutions
1 2
for (3):
t
x1(t)=x+(cid:90) {f(s)−(cid:104)f(cid:105)}ds, t∈R,
0
t
x2(t)=x+C+(cid:90) {f(s)−(cid:104)f(cid:105)}ds=x1(t)+C, t∈R.
0
Remark 2.5 The condition Int(O(cid:104)f(cid:105))(cid:54)=∅ is equivalent to
diam(g−1(cid:104)f(cid:105))>diam(Range(cid:90) {f(t)−(cid:104)f(cid:105)}dt).
Mihai Bostan 5
Example: Consider the equation x(cid:48)(t)+g(x(t)) = ηcost,t ∈ R with g given
in Remark 2.2. We have <ηcost>=0∈g(R) and
t
O(0) = {x∈R|x+(cid:90) ηcossds∈g−1(0), t∈R} (8)
0
= {x∈R:x+ηsint∈g−1(0), t∈R}
= {x∈R:−ε≤x+ηsint≤ε,t∈R}
∅ |η|>ε,
=  {0} |η|=ε, (9)

[|η|−ε,ε−|η|] |η|<ε.

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.
Inthefollowingwesupposethatgisincreasingandweestablishanexistence
result.
2.2 Existence
Tostudytheexistence,notethatanecessaryconditionisgivenbythefollowing
proposition.
Proposition 2.6 Assume that equation (3) has T-periodic solutions. Then
there is x ∈R such that (cid:104)f(cid:105):= 1 T f(t)dt=g(x ).
0 T (cid:82)0 0
Proof Integrating on a period interval [0,T] we obtain
T T
x(T)−x(0)+(cid:90) g(x(t))dt=(cid:90) f(t)dt.
0 0
Since x is periodic and g◦x is continuous we get
T
Tg(x(τ))=(cid:90) f(t)dt, τ ∈]0,T[,
0
and hence
1 T
(cid:104)f(cid:105):= (cid:90) f(t)dt∈Range(g). (10)
T
0
♦
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 (t)+x(cid:48) (t)+g(x (t))=f(t), t∈R, α>0. (11)
α α α
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.
6 Periodic solutions for evolution equations
Remark 2.8 Before starting the proof let us observe that (11) reduces to an
equation of type (3) with g = α1 +g. Since g is increasing, is clear that g
α R α
is strictly increasing and by the Proposition 2.1 we deduce that the uniqueness
holds. Moreover since Range(g ) = R, the necessary condition (10) is trivially
α
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 ∈R such that the solution of the evolution problem
0
αx (t)+x(cid:48) (t)+g(x (t))=f(t), t∈[0,T],
α α α (12)
x(0)=x ,
0
verifies x(T ;0,x ) = x . Here we denote by x(·;0,x ) the solution of (12)
0 0 0
(existence and uniqueness assured by Picard’s theorem). We define the map
S :R→R given by
S(x )=x(T ;0,x ), x ∈R. (13)
0 0 0
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 x1 and x2. Using the
0 0
monotony of g we can write
1 d
α|x(t;0,x1)−x(t;0,x2)|2+ |x(t;0,x1)−x(t;0,x2)|2 ≤0,
0 0 2dt 0 0
which implies
1 d
{e2αt|x(t;0,x1)−x(t;0,x2)|2}≤0,
2dt 0 0
and therefore,
|S(x1)−S(x2)|=|x(T ;0,x1)−x(T ;0,x2)|≤e−αT|x1−x2|.
0 0 0 0 0 0
For α > 0 S is a contraction and the Banach’s fixed point theorem applies.
Therefore S(x ) = x for an unique x ∈ R and hence x(·;0,x ) is a periodic
0 0 0 0
solution of (3). ♦
Naturally, in the following proposition we inquire about the convergence of
(x ) to a periodic solution of (3) as α → 0. In view of the Proposition 2.6
α α>0
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 Ifequation(3)hasatleastoneperiodicsolution,then(x )
α α>0
is convergent in C0(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
1 d
α|x (t)−x(t)|2+ |x (t)−x(t)|2 ≤−αx(t)(x (t)−x(t)), t∈R, (17)
α 2dt α α
Mihai Bostan 7
which can be also written as
1 d
{e2αt|x (t)−x(t)|2}≤αeαt|x(t)|·eαt|x (t)−x(t)|, t∈R. (18)
2dt α α
Therefore, by integration on [0,t] we deduce
1 1 t
2{eαt|xα(t)−x(t)|}2 ≤ 2|xα(0)−x(0)|2+(cid:90) αeαs|x(s)|·eαs|xα(s)−x(s)|ds.
0
(19)
Using Bellman’s lemma, formula (19) gives
t
eαt|xα(t)−x(t)|≤|xα(0)−x(0)|+(cid:90) αeαs|x(s)|ds, t∈R. (20)
0
Letusconsiderα>0fixedforthemoment. Sincexisperiodicandcontinuous,
it is also bounded and therefore from (20) we get
|xα(t)−x(t)|≤e−αt|xα(0)−x(0)|+(1−e−αt)(cid:107)x(cid:107)L∞(R), t∈R. (21)
By periodicity we have
|x (t)−x(t)| = |x (nT +t)−x(nT +t)|
α α
≤ e−α(nT+t)|xα(0)−x(0)|+(1−e−α(nT+t))(cid:107)x(cid:107)L∞(R)
≤ e−α(nT+t)|xα(0)−x(0)|+(cid:107)x(cid:107)L∞(R), t∈R,n≥0.
Bypassingtothelimitasn→∞,wededucethat(x ) isuniformlybounded
α α>0
in L∞(R):
|xα(t)|≤|xα(t)−x(t)|+|x(t)|≤2(cid:107)x(cid:107)L∞(R), t∈R, α>0.
The derivatives x(cid:48) are also uniformly bounded in L∞(R) for α→0:
α
|x(cid:48) (t)|
α
= |f(t)−αx (t)−g(x (t))|
α α
≤ (cid:107)f(cid:107)L∞(R)+2α(cid:107)x(cid:107)L∞(R)+max{g(2(cid:107)x(cid:107)L∞(R)),−g(−2(cid:107)x(cid:107)L∞(R))}.
The uniform convergence of (x ) follows now from the Arzela-Ascoli’s the-
α α>0
orem. Denote by u the limit of (x ) as α→0. Obviously u is also periodic
α α>0
u(0)= lim x (0)= lim x (T)=u(T).
α α
α→0 α→0
To prove that u verifies (3), we write
t
xα(t)=xα(0)+(cid:90) {f(s)−g(xα(s))−αxα(s)}ds, t∈R.
0
Since the convergence is uniform, by passing to the limit for α→0 we obtain
t
u(t)=u(0)+(cid:90) {f(s)−g(u(s))}ds,
0
8 Periodic solutions for evolution equations
and hence u∈C1(R;R) and
u(cid:48)(t)+g(u(t))=f(t), t∈R.
From the previous proposition we conclude that the existence of periodic solu-
tions 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 ) is uniformly bounded in L∞(R). Moreover, in this
α α>0
case (x ) is convergent in C0(R;R) and the limit is a periodic solution for
α α>0
(3).
Note that generally we can not estimate (x ) uniformly in L∞(R). In-
α α>0
deed, by standard computations we obtain
1 d
α(x (t)−u)2+ (x (t)−u)2 ≤|f(t)−αu−g(u)|·|x (t)−u|, t,u∈R
α 2dt α α
and therefore
1 d
{e2αt(x (t)−u)2}≤eαt|f(t)−αu−g(u)|·eαt|x (t)−u|, t,u∈R.
2dt α α
Integration on [t,t+h], we get
1 t+h
2e2α(t+h)(xα(t+h)−u)2 ≤ (cid:90) e2αs|f(s)−αu−g(u)|·|xα(s)−u|ds
t
1
+ e2αt(x (t)−u)2, t<t+h, u∈R.
2 α
Now by using Bellman’s lemma we deduce
t+h
|xα(t+h)−u|≤e−αh|xα(t)−u|+(cid:90) e−α(t+h−s)|f(s)−αu−g(u)|ds, t<t+h.
t
Since x is T-periodic, by taking h=T we can write
α
1 T
|xα(t)−u|≤ 1−e−αT (cid:90) e−α(T−s)|f(s)−αu−g(u)|ds, t∈R,
0
and thus for u=0 we obtain
1 T 1
(cid:107)xα(cid:107)L∞(R) ≤ 1−e−αT (cid:90) |f(s)−g(0)|ds∼O(cid:18)α(cid:19), α>0.
0
Now we can state our main existence result.
Mihai Bostan 9
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 (cid:104)f(cid:105) := 1 T f(t)dt ∈ Range(g) (there is x ∈ R such that (cid:104)f(cid:105) = g(x )).
T (cid:82)0 0 0
Moreover in this case we have the estimate
T
(cid:107)x(cid:107)L∞(R) ≤|x0|+(cid:90) |f(t)−(cid:104)f(cid:105)|dt, ∀x0 ∈g−1(cid:104)f(cid:105),
0
and the solution is unique if and only if Int(O(cid:104)f(cid:105))=∅ or
diam(g−1(cid:104)f(cid:105))≤diam(Range(cid:90) {f(t)−(cid:104)f(cid:105)}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 ) of (11). Accordingly to the Proposition 2.10 we need to prove uniform
α α>0
estimates in L∞(R) for (x ) . Since x is T-periodic by integration on [0,T]
α α>0 α
we get
T
(cid:90) {αxα(t)+g(xα(t))}dt=T(cid:104)f(cid:105), α>0.
0
Using the average formula for continuous functions we have
T
(cid:90) {αxα(t)+g(xα(t))}dt=T{αxα(tα)+g(xα(tα))}, tα ∈]0,T[, α>0.
0
By the hypothesis there is x ∈R such that (cid:104)f(cid:105)=g(x ) and thus
0 0
αx (t )+g(x (t ))=g(x ), α>0. (22)
α α α α 0
Since g is increasing, we deduce
αx (t )[x −x (t )]=[g(x )−g(x (t ))][x −x (t )]≥0, α>0,
α α 0 α α 0 α α 0 α α
and thus
|x (t )|2 ≤x (t )x ≤|x (t )||x |.
α α α α 0 α α 0
Finally we deduce that x (t ) is uniformly bounded in R:
α α
|x (t )|≤|x |, ∀α>0.
α α 0
Now we can easily find uniform estimates in L∞(R) for (x ) . Let us take in
α α>0
the previous calculus u=x (t )and integrate on [t ,t]:
α α α
1 t
2e2αt(xα(t)−xα(tα))2 ≤(cid:90) e2αs|f(s)−αxα(tα)−g(xα(tα))|·|xα(s)−xα(tα)|ds.
tα
By using Bellman’s lemma we get
t
|xα(t)−xα(tα)|≤(cid:90) e−α(t−s)|f(s)−αxα(tα)−g(xα(tα))|ds, t>tα,
tα
10 Periodic solutions for evolution equations
and hence by (22) we deduce
T
|xα(t)| ≤ |x0|+(cid:90) |f(t)−αxα(tα)−g(xα(tα))|dt
0
T
= |x0|+(cid:90) |f(t)−(cid:104)f(cid:105)|dt, t∈R,α>0. (23)
0
Now by passing to the limit in (23) we get
T
|x(t)|≤|x0|+(cid:90) |f(t)−(cid:104)f(cid:105)|dt, t∈R, ∀x0 ∈g−1(cid:104)f(cid:105).
0
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 ∈ C1([0,T];R) is a sub-periodic solution for
(3) if
x(cid:48)(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 Ifequation(3)hassub-periodicsolutions, thenthereisx ∈
0
R such that g(x )≤ (cid:104)f(cid:105).
0
Proof Letxbeasub-periodicsolutionof(3). Byintegrationon[0,T]wefind
T
x(T)−x(0)+(cid:90) g(x(t))dt=T(cid:104)f(cid:105).
0
Since g◦x is continuous, there is τ ∈]0,T[ such that
1
g(x(τ))=(cid:104)f(cid:105)− (x(T)−x(0))≤ (cid:104)f(cid:105).
T
Similarly we define the notion of super-periodic solution.
Definition 2.14 We say that y ∈C1([0,T];R) is a super-periodic solution for
(3) if
(cid:48)
y (t)+g(y(t))=f(t), t[0,T],
and y(0)≥y(T).
The analogous necessary condition holds.