Now it is easy to check that (x± )±>0 is uniformly bounded in L∞ (]0, T [; H):

T

1

x± (t) ’ x± (x± (t) ’ x± (s))ds

=

T 0

1Tt

= x („ )d„ ds

T0 s±

√

T T

¤ f L2 (]0,T [;H) + f L1 (]0,T [;H) ,

2 2

and thus

√

T T

¤

x± x± + f L2 (]0,T [;H) + f L1 (]0,T [;H)

L∞ (]0,T [;H)

2 2

√

T T

¤ x0 + f L2 (]0,T [;H) + f L1 (]0,T [;H) .

2 2

28 Periodic solutions for evolution equations

Now we can prove that (x± )±>0 is convergent in C 1 ([0, T ]; H). Indeed, by

taking the di¬erence between the equations (39) written for ±, β > 0, after

multiplication by x± (t) ’ xβ (t) and integration on [0, T ] we get

T

{±(x± (t) ’ xβ (t), x± (t) ’ xβ (t))

0

2

+ x± (t) ’ xβ (t) + (A(x± (t) ’ xβ (t)), x± (t) ’ xβ (t))}dt

T

= ’(± ’ β) (xβ (t), x± (t) ’ xβ (t))dt.

0

Since A is symmetric, x± and xβ are T -periodic and uniformly bounded in

L∞ (]0, T [; H) we deduce that

x± ’ xβ ¤ |± ’ β| · sup xγ L2 (]0,T [;H) ,

L2 (]0,T [;H)

γ>0

or

|± ’ β|

√

x± ’ xβ ¤ · sup xγ L2 (]0,T [;H) + |± ’ β| · sup xγ L1 (]0,T [;H) ,

L∞ (]0,T [;H)

T γ>0 γ>0

and therefore (x± )±>0 converges in C([0, T ]; H).

We already know that ( x± )±>0 = ((± + A)’1 f )±>0 is bounded in H and

by the Proposition 3.6 it follows that ( x± )±>0 is convergent to the element of

minimal norm in A’1 f . We have

t

t ∈ R, ± > 0.

x± (t) = x± (0) + x± (s)ds,

0

t

By taking the average we deduce that x± (0) = x± ’ < 0 x± (s)ds > and

therefore, since (x± )±>0 is uniformly convergent, it follows that (x± (0))±>0 is

also convergent. Finally we conclude that (x± )±>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 mono-

tone and symmetric operator and f ∈ C 1 ([0, T ]; H) is a T -periodic function.

Then for every x0 ∈ D(A) we have

1

(x(t + T ; 0, x0 ) ’ x(t; 0, x0 )) = f ’ ProjR(A) f ,

lim (46)

t’∞ T

where x(·; 0, x0 ) represents the solution of (33) with the initial data x0 and R(A)

is the range of A.

Remark 3.10 A being maximal monotone, A’1 is also maximal monotone and

therefore D(A’1 ) = R(A) is convex.

Mihai Bostan 29

Proof of Proposition 3.9. Consider x0 ∈ D(A) and denote by x(·) the

corresponding solution. By integration on [t, t + T ] we get

t+T

1 1

(x(t + T ) ’ x(t)) + A x(s)ds = f . (47)

T T t

For each ± > 0 consider x± ∈ D(A) such that ±x± + Ax± = f . Denoting by

1 t+T

x(s)ds, t ≥ 0, equation (47) writes

y(·) the function y(t) = T t

t ≥ 0, ± > 0.

y (t) + Ay(t) = ±x± + Ax± ,

Let us search for y of the form y1 + y2 where

t ≥ 0,

y1 (t) + Ay1 (t) = ±x± ,

with the initial condition y1 (0) = 0 and

t ≥ 0,

y2 (t) + Ay2 (t) = Ax± , (48)

T

1

with the initial condition y2 (0) = y(0) = T 0 x(t)dt. We are interested on the

asymptotic behaviour of Ay(t) = Ay1 (t) + Ay2 (t) for large t. We have

t

’tA

e’(t’s)A ±x± ds

y1 (t) =e y1 (0) +

0

t

e’(t’s)A ±x± ds,

=

0

and therefore,

t t

Ae’(t’s)A ±x± ds = e’(t’s)A ±x± = (1 ’ e’tA )±x± .

Ay1 (t) =

0

0

By the other hand, after multiplication of (48) by y2 (t) = (y2 (t) ’ x± ) we get

2

+ (A(y2 (t) ’ x± ), (y2 (t) ’ x± ) ) = 0, t ≥ 0.

y2 (t)

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

t

1 1

y2 (s) 2 ds + (A(y2 (t) ’ x± ), y2 (t) ’ x± ) = (A(y2 (0) ’ x± ), y2 (0) ’ x± ),

2 2

0

and therefore, by the monotony of A it follows that

∞

1

y2 (t) 2 dt ¤ (A(y2 (0) ’ x± ), y2 (0) ’ x± ).

2

0

30 Periodic solutions for evolution equations

Thus limt’∞ y2 (t) = 0 and by passing to the limit in (48) we deduce that

limt’∞ Ay2 (t) = limt’∞ (Ax± ’ y2 (t)) = Ax± . Finally we ¬nd that

1

(x(t + T ) ’ x(t)) ’ e’tA ±x±

lim

t’∞ T

= lim {y (t) ’ e’tA ±x± }

t’∞

lim { f ’ Ay(t) ’ e’tA ±x± }

=

t’∞

lim { f ’ Ay1 (t) ’ Ay2 (t) ’ e’tA ±x± }

=

t’∞

f ’ ±x± ’ Ax± = 0,

= ± > 0. (49)

Now let us put y± = Ax± and observe that y± + ±A’1 y± = Ax± + ±x± =

f , ± > 0. Therefore,

lim (1 + ±A’1 )’1 f

lim y± =

± 0 ± 0

’1

A

= lim J± f

± 0

= ProjD(A’1 ) f

= ProjR(A) f ,

and it follows that

lim ±x± = lim ( f ’ Ax± ) = lim ( f ’ y± ) = f ’ ProjR(A) f .

± 0 ± 0 ± 0

Since Graph(A) is closed and [±x± , ±y± ] = [±x± , A(±x± )] ∈ A, ± > 0, by

0 we deduce that f ’ ProjR(A) f ∈ D(A) and

passing to the limit for ±

A( f ’ ProjR(A) f ) = 0. It is easy to see that we can pass to the limit for

0 in (49). Indeed, for µ > 0 let us consider ±µ > 0 such that lim± 0 ±x± ’

±

µ

±µ x±µ < 2 . We have

1

(x(t + T ) ’ x(t)) ’ e’tA lim ±x±

T ±0

1

(x(t + T ) ’ x(t)) ’ e’tA ±µ x±µ + e’tA ±µ x±µ ’ e’tA lim ±x±

¤

T ±0

1

(x(t + T ) ’ x(t)) ’ e’tA ±µ x±µ + ±µ x±µ ’ lim ±x±

¤

T ±0

µµ µ

¤ + = µ, t ≥ t(±µ , ) = t(µ),

22 2

and thus

1

lim { (x(t + T ) ’ x(t)) ’ e’tA ( f ’ ProjR(A) f )} = 0.

t’∞ T

But e’tA ( f ’ ProjR(A) f ) does not depend on t ≥ 0:

d ’tA

= ’Ae’tA ( f ’ ProjR(A) f )

( f ’ ProjR(A) f )

e

dt

= ’e’tA A( f ’ ProjR(A) f ) = 0,

Mihai Bostan 31

and thus the previous formula reads

1

(x(t + T ) ’ x(t)) = f ’ ProjR(A) f .

lim

t’∞ T

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

x(T ; 0, x0 ) ’ x0

f ’ ProjR(A) f

inf = = dist( f , R(A)).

T

x0 ∈D(A)

3.3 Periodic solutions for the heat equation

Let „¦ ‚ Rd , d ≥ 1, be an open bounded set with ‚„¦ ∈ C 2 . Consider the heat

equation

‚u

(t, x) ’ ∆u(t, x) = f (t, x), (t, x) ∈ R — „¦, (50)

‚t

with the Dirichlet boundary condition

(t, x) ∈ R — ‚„¦,

u(t, x) = g(t, x), (51)

or the Neumann boundary condition

‚u

(t, x) ∈ R — ‚„¦,

(t, x) = g(t, x), (52)

‚n

where we denote by n(x) the outward normal in x ∈ ‚„¦.

Theorem 3.12 Assume that f ∈ C 1 (R; L2 („¦)) is T -periodic and g(t, x) =

‚u0 1 2 2 2

‚n (t, x), (t, x) ∈ R—‚„¦ where u0 ∈ C (R; H („¦))©C (R; L („¦)) is T -periodic.

Then the heat problem (50), (52) has T -periodic solutions u ∈ C(R; H 2 („¦)) ©

C 1 (R; L2 („¦)) if and only if

T T

g(t, x)dtdσ + f (t, x)dtdx = 0.

‚„¦ 0 „¦ 0

In this case the periodic solutions satis¬es the estimates

1