Periodic solutions for evolution equations

ftp ejde.math.swt.edu login: ftpMihai Bostan Abstract We study the existence and uniqueness of periodic solutions for evolu-tion equaevolu-tions.. The main purpose of this paper is to es

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Mihai Bostan

Abstract

We study the existence and uniqueness of periodic solutions for evolu-tion equaevolu-tions First we analyze the one-dimensional case Then for arbi-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

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 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,

x0(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 c

Submitted May 14, 2002 Published August 23, 2002.

1

by using the penalization method Note that in the linear case a necessarycondition for the existence is

hfi := 1T

Z T 0

in the linear symmetric case we show that (xα)α>0 is a Cauchy sequence in C1.Then by passing to the limit for α→ 0 it follows that the limit function is aperiodic solution for (1)

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

This article is organized as follows First we analyze the one dimensionalcase Necessary and sufficient 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 resultholds for linear symmetric maximal monotone operators on Hilbert spaces Inthe last section the case of non-linear sub-differential operators is considered

2 Periodic solutions for one dimensional tion equations

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

con-x0(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 eachinitial data x(0) = x0∈ R there is an unique solution x ∈ C1(R; R) for (3) Weare looking for T -periodic solutions Let us start by the uniqueness study

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

Proof Let x1, x2 be two periodic solutions for (3) By taking the differencebetween the two equations and multiplying by x1(t)− x2(t) we get

1

2

d

dt|x1(t)− x2(t)|2+ [g(x1(t))− g(x2(t))][x1(t)− x2(t)] = 0, t∈ R (4)Since g is increasing we have (g(x1)− g(x2))(x1− x2)≥ 0 for all x1, x2∈ R andtherefore we deduce that|x1(t)−x2(t)| is decreasing Moreover as x1and x2areperiodic 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 periodicsolutions Let us consider the function

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) forsome t0∈ R and it follows that |x1(t)− x2(t)| = |x1(t0)− x2(t0)| = 0, t ∈ R or

x1= x2 Finally we find that

Proof Assume that (3) has two periodic solutions x1 6= x2 By the previousproposition we have x2− x1= C > 0 By integration on [0, T ] one gets

Z T 0

g(x1(t))dt =

Z T 0

f (t)dt =

Z T 0

g(x2(t))dt (6)

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

Z T 0

g(x1(t))dt≤

Z T 0

From (6) and (7) we deduce that g(x1(t)) = g(x2(t)), t ∈ R and thus g isconstant on each interval [x1(t), x2(t)] = [x1(t), x1(t) + C], t ∈ R Finally itimplies 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:

x1(t) = x +

Z t 0

{f(s) − hfi}ds, t∈ R,

x2(t) = x + C +

Z t 0

{f(s) − hfi}ds = x1(t) + C, t∈ R

Remark 2.5 The condition Int(Ohfi) 6= ∅ is equivalent to

diam(g−1hfi) > diam(Range

Z{f(t) − hfi}dt)

Example: Consider the equation x0(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 +

Z t 0

g(x(t))dt =

Z T 0

f (t)dt

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

T g(x(τ )) =

Z T 0

f (t)dt, τ ∈]0, T [,and hence

hfi := 1T

Z T 0

T -periodic and continuous Then for every α > 0 the equation (11) has exactlyone periodic solution

Remark 2.8 Before starting the proof let us observe that (11) reduces to anequation of type (3) with gα= α1R+ g Since g is increasing, is clear that gα

is strictly increasing and by the Proposition 2.1 we deduce that the uniquenessholds Moreover since Range(gα) = R, the necessary condition (10) is triviallyverified and therefore, in this case we can expect to prove existence

Proof First of all remark that the existence of periodic solutions reduces tofinding x0∈ R such that the solution of the evolution problem

We demonstrate the existence and uniqueness of the periodic solution of (12)

by showing that the Banach’s fixed point theorem applies Let us considertwo solutions of (12) corresponding to the initial datas x1 and x2 Using themonotony of g we can write

α|x(t ; 0, x1

0)− x(t ; 0, x2

0)|2+12

12

d

dt{e2αt|x(t ; 0, x10)− x(t ; 0, x20)|2} ≤ 0,and therefore,

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 wefind

which can be also written as

αeαs|x(s)| · eαs

|xα(s)− x(s)|ds

(19)Using Bellman’s lemma, formula (19) gives

eαt|xα(t)− x(t)| ≤ |xα(0)− x(0)| +

Z t 0

αeαs|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

and hence u∈ C1(R; R) and

u0(t) + g(u(t)) = f (t), t∈ R

From the previous proposition we conclude that the existence of periodic tions for (3) reduces to uniform estimates in L∞(R) for (xα)α>0

solu-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 (xα)α>0 is uniformly bounded in L∞(R) Moreover, in thiscase (xα)α>0 is convergent in C0(R; R) and the limit is a periodic solution for(3)

Note that generally we can not estimate (xα)α>0 uniformly in L∞(R) deed, by standard computations we obtain

e2αs|f(s) − αu − g(u)| · |xα(s)− u|ds

|f(s) − g(0)|ds ∼ O 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

|f(t) − hfi|dt, ∀ x0∈ g−1hfi,

and the solution is unique if and only if Int(Ohfi) = ∅ or

diam(g−1hfi) ≤ diam(Range

Z{f(t) − hfi}dt)

Proof The condition is necessary (see Proposition 2.6) We will prove nowthat it is also sufficient Let us consider the sequence of periodic solutions(xα)α>0 of (11) Accordingly to the Proposition 2.10 we need to prove uniformestimates in L∞(R) for (xα)α>0 Since xαis T -periodic by integration on [0, T ]

we get

Z T 0

αxα(tα)[x0− xα(tα)] = [g(x0)− g(xα(tα))][x0− xα(tα)]≥ 0, α > 0,and thus

and hence by (22) we deduce

|xα(t)| ≤ |x0| +

Z T 0

|f(t) − αxα(tα)− g(xα(tα))|dt

= |x0| +

Z T 0

|f(t) − hfi|dt, t∈ R, α > 0 (23)Now by passing to the limit in (23) we get

|x(t)| ≤ |x0| +

Z T 0

|f(t) − hfi|dt, t∈ R, ∀ x0∈ g−1hfi

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

Definition 2.12 We say that x ∈ C1([0, T ]; R) is a sub-periodic solution for(3) if

x0(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 x0∈

Rsuch that g(x0)≤ hfi

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

x(T )− x(0) +

Z T 0

g(x(t))dt = Thfi

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

g(x(τ )) =hfi − 1

T(x(T )− x(0)) ≤ hfi

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

y0(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

y0∈ R such that g(y0)≥ hfi

Remark 2.16 It is clear that x is periodic solution for (3) if and only if is inthe same time sub-periodic and super-periodic solution Therefore there are

-Proof The condition is necessary (see Proposition 2.13) Let us prove nowthat it is also sufficient Consider z0 an arbitrary initial data and denote by

x : [0,∞[→ R the solution for (3) with the initial condition x(0) = z0 If there

is t0 ≥ 0 such that x(t0)≤ x(t0+ T ), thus xt0(t) := x(t0+ t), t ∈ [0, T ] is asub-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 ) +

Z T 0

g(x(nT + t))dt = Thfi, n ≥ 0

Using the hypothesis and the average formula we have

g(x(nT + τn)) =hfi + 1

T{x(nT ) − x((n + 1)T )} > g(x0),for τn ∈]0, T [ and n ≥ 0 Since g is increasing we deduce that x(nT + τn) >

x0, n ≥ 0 We have also x(nT + τn) ≤ x((n − 1)T + τn) ≤ · · · ≤ x(τn) ≤supt∈[0,T ]|x(t)| and thus we deduce that (x(nT + τn))n ≥0 is bounded:

|xn(t)| ≤ |xn(s)| +

Z t

|f(u) − g(0)|du, 0≤ s ≤ t ≤ T

Taking s = τn ∈]0, T [ we can write

|f(u) − g(0)|du, t ∈ [τn, T ]

For t∈ [0, τn], n≥ 1 we have

|xn(t)| = |x(nT + t)| ≤ |x((n − 1)T + τn −1)| +

Z nT +t (n −1)T +τ n−1

|f(u) − g(0)|du

≤ K +

Z (n+1)T (n −1)T

|f(u) − g(0)|du

≤ K + 2

Z T 0

|f(u) − g(0)|du

Therefore, the sequence (xn)n≥0 is uniformly bounded in L∞(R) and

kxnkL ∞ (R)≤ K + 2

Z T 0

n →∞xn(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 ≥0is decreasing and bounded, it is convergent and

we can prove that u is periodic:

Proposition 2.18 Under the same assumptions as in Theorem 2.17 the tion (3) has super-periodic solutions if and only if there is y0 ∈ R such thatg(y0)≥ hfi

equa-We state now a comparison result between sub-periodic and super-periodicsolutions

Proposition 2.19 If g is increasing, x is a sub-periodic solution and y is asuper-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)0(t) + g(x(t))− g(y(t)) = 0, t∈ [0, T ]

With the notation

r(t) =

( g(x(t)) −g(y(t)) x(t) −y(t) t∈ [0, T ], x(t) 6= y(t)

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

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

x(t)− y(t) = (x(0) − y(0))e−R0tr(s)ds (25)Now it is clear that if x(0)≤ y(0) we also have x(t) ≤ y(t), t ∈ [0, T ] Supposenow that x(0) > y(0) Taking t = T in (25) we obtain

x(T )− y(T ) = (x(0) − y(0))e−R0Tr(t)dt (26)Since g is increasing, by the definition of the function r we have r ≥ 0 Twocases are possible: (i) either R0Tr(t)dt > 0, (ii) either R0Tr(t)dt = 0 in whichcase r(t) = 0, t ∈ [0, T ] (r vanishes in all points of continuity t such thatx(t)6= y(t) and also in all points t with x(t) = y(t) by the definition) Let usanalyse the first case (i) By (26) we deduce that x(T )− y(T ) < x(0) − y(0) orx(T )− x(0) < y(T ) − y(0) Since x is sub-periodic we have x(0) ≤ x(T ) whichimplies that y(T ) > y(0) which is in contradiction with the super-periodicity of

In the following we will see how it is possible to retrieve the existence result forperiodic solutions by using the method of sub(super)-periodic solutions Sup-pose that hfi ∈ Range(g) Obviously both sufficient conditions for existence ofsub(super)-periodic solutions are satisfied and thus there are x0(y0) sub(super)-periodic solutions If y0 is even periodic the proof is complete Assume that y0

is not periodic (y0(0) > y0(T )) Denote byM the set of sub-periodic solutionsfor (3):

M = {x : [0, T ] → R : x sub-periodic solution , x0(t)≤ x(t), t ∈ [0, T ]}.Since x0∈ M we have M 6= ∅ Moreover, from the comparison result since y0

is super-periodic but not periodic we have x≤ y0,∀x ∈ M We prove that Mcontains a maximal element in respect to the order:

x ≺ x (if and only if) x (t)≤ x (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 greaterthan the maximal element We state now the following generalization.

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

x0(t) + g(t, x(t)) = f (t), t∈ R, (27)has periodic solutions if and only if there is x0∈ R such that

hfi := 1T

Z T 0

f (t)dt = 1

T

Z T 0

g(t, x0)dt = G(x0) (28)

Moreover, in this case we have the estimate

kxkL ∞ (R)≤ |x0| +

Z T 0

Z T 0

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(t, x(t))dt≤ 1

T

Z T 0

0 g(t, x(t))dt and from (29) we deduce thathfi = G(x0)

Let us show that the condition (28) is also sufficient As before let us considerthe unique periodic solution for

αxα(t) + x0 (t) + g(t, xα(t)) = f (t), t∈ [0, T ], α > 0,

(existence and uniqueness follow by the Banach’s fixed point theorem exactly asbefore) All we need to prove is that (xα)α>0 is uniformly bounded in L∞(R)(then (x0α)α>0 is also uniformly bounded in L∞(R) and by Arzela-Ascoli’s the-orem we deduce that xα converges to a periodic solution for (27)) Taking theaverage on [0, T ] we get

mα≤ xα(t)≤ Mα, t∈ [0, T ], α > 0,and hence

αmα+ G(mα)≤ 1

T

Z T 0

αuα(uα− x0) =−(G(x0)− G(uα))(x0− uα), α > 0

Since G is increasing we deduce that|uα|2

≤ uαx0≤ |uα| · |x0|, α > 0 and hence(uα)α>0 is bounded:

{αuα+ g(t, uα)}dt,

and thus there is tα∈]0, T [ such that

αxα(tα) + g(tα, xα(tα)) = αuα+ g(tα, uα), α > 0

Since α(xα(tα)− uα)2 =−[g(tα, xα(tα))− g(tα, uα)][xα(tα)− uα] ≤ 0 we findthat xα(tα) = uα, α > 0 and thus (xα(tα))α>0is also bounded

|f(t) − αxα(tα)− g(t, xα(tα))|dt, t∈ [0, T ], α > 0 (31)

Since (xα(tα))α>0 is bounded we have

uα= xα(tα)→ x1,such that

= |x0| +

Z T 0

|f(t) − g(t, x0)|dt, t∈ [0, T ], ∀ x0∈ G−1hfi,

and therefore (xα)α>0is 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 forgeneral evolution equations on Hilbert spaces

x0(t) + Ax(t) = f (t), t > 0, (33)where A : D(A) ⊂ H → H is a maximal monotone operator on a Hilbertspace H and f ∈ C1(R; H) is a T -periodic function As known by the theory

of Hille-Yosida, for every initial data x0 ∈ D(A) there is an unique solution

x ∈ C1([0, +∞[; H) ∩ C([0, +∞[ ; D(A)) for (33), see [6, p 101] Obviously,the periodic problem reduces to find x0 ∈ D(A) such that x(T ) = x0 As

in the one dimensional case we demonstrate uniqueness for strictly monotoneoperators We state also necessary and sufficient condition for the existence

in the linear symmetric case Finally the case of non-linear sub-differentialoperators is considered Let us start with the definition of periodic solutions for(33)

Definition 3.1 Let A : D(A) ⊂ H → H be a maximal monotone operator

on a Hilbert space H and f ∈ C1(R; H) a T -periodic function We say that

x∈ C1([0, T ]; H)∩ C([0, T ]; D(A)) is a periodic solution for (33) if and only if

x0(t) + Ax(t) = f (t), t∈ [0, T ],and x(0) = x(T )

Proof Let x1, x2 be two different periodic solutions By taking the difference

of (33) and multiplying both sides by x1(t)− x2(t) we find

kx1(0)− x2(0)k = kx1(T )− x2(T )k,which implies thatkx1(t)− x2(t)k is constant for t ∈ [0, T ] and thus

hfi := 1T

Z T 0

f (t)dt∈ Range(A),(there is x ∈ D(A) such that hfi = Ax )

Proof Suppose that x∈ C1([0, T ]; H)∩C([0, T ]; D(A)) is a T -periodic solutionfor (33) Let us consider the divisions ∆n : 0 = tn0 < tn1 <· · · < tn

n = T suchthat

[f (t)− x0(t)]dt

T

Z T 0

f (t)dt− 1

Tx(t)|T 0

T

Z T 0

f (t)dt

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

h1T

Z T 0

x(t)dt, 1T

Z T 0

x0(t) + Ax(t) = f (t), t∈ [0, T ], (35)

where A : R2→ R2 is the orthogonal rotation:

A(x , x ) = (−x , x ), (x , x )∈ R2,

and f = (f1, f2)∈ L1(]0, T [; R2) is T -periodic For a given initial data x(0) =

x0∈ R2the solution writes

x(t) = e−tAx0+

Z t 0

e−(t−s)Af (s)ds, t > 0, (36)

where the semigroup e−tAis given by

e−tA= cos t sin t

Thus ifR2π

0 {f1(t) cos t− f2(t) sin t}dt 6= 0 or R2π

0 {f1(t) sin t + f2(t) cos t}dt 6= 0equation (35) does not have any 2π-periodic solution and the necessary conditionstill holds because Range(A) = R2 Moreover if (38) is satisfied then everysolution 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 onedimensional case we have

Proposition 3.4 Suppose that A : D(A)⊂ H → H is maximal monotone and

f ∈ C1(R; H) is T -periodic Then for every α > 0 the equation

αx(t) + x0(t) + Ax(t) = f (t), t∈ R, (39)

has an unique T -periodic solution in C1(R; H)∩ C(R; D(A))

Proof Since α+A is strictly monotone the uniqueness follows from Proposition3.2 Indeed,

αkx − yk2+ (Ax− Ay, x − y) = 0, x, y∈ D(A),

implies αkx − yk2= 0 and hence x = y

Consider now an arbitrary initial data x0∈ D(A) By the Hille-Yosida’s rem, there is x∈ C1([0, +∞[; H) ∩ C([0, +∞[; D(A)) solution for (39) Denote

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