and for t < t + h,

1 1 1

(x (t + h) ’ x (t)) + A(x(t + h) ’ x(t)) = (f (t + h) ’ f (t)).

h h h

For every h > 0 denote by y±,h , yh and gh the periodic functions:

1

(x± (t + h) ’ x± (t)), t ∈ R, ± > 0,

y±,h (t) =

h

1

yh (t) = (x(t + h) ’ x(t)), t ∈ R,

h

1

gh (t) = (f (t + h) ’ f (t)), t ∈ R,

h

and hence we have

t ∈ R,

±y±,h (t) + y±,h (t) + Ay±,h (t) = gh (t),

t ∈ R.

yh (t) + Ayh (t) = gh (t),

By the same computations we get

t

’±t

±e’±(t’s) yh (s) ds.

y±,h (t) ’ yh (t) ¤ e y±,h (0) ’ yh (0) +

0

Now by passing to the limit for h ’ 0 we deduce

t

’±t

±e’±(t’s) x (s) ds

x± (t) ’ x (t) ¤e x± (0) ’ x (0) +

0

’±t

x± (0) ’ x (0) + (1 ’ e’±t ) x

¤e t ∈ [0, T ].

L∞ ,

By the periodicity we obtain as before that

x± (t) ’ x (t) = x± (nT + t) ’ x (nT + t)

¤ e’±(nT +t) x± (0) ’ x (0) + (1 ’ e’±(nT +t) ) x L∞ ,

and hence by passing to the limit for n ’ ∞ we conclude that

x± ’ x ¤x L∞ , ± > 0.

L∞

Therefore, (x± )±>0 is also uniformly bounded in L∞

¤2 x

x± L∞ , ± > 0.

L∞

Conversely, the implication (ii) ’ (i) follows by using Arzela-Ascoli™s theorem

and by passing to the limit for ± ’ 0 in (39).

Let us continue the analysis of the previous example. The semigroup asso-

ciated to the equation (39) is given by

cos t, sin t

e’t(±+A) = e’±t e’tA = e’±t t ∈ R, ± > 0,

’ sin t, cos t

Mihai Bostan 25

and the periodic solution for equation (39) reads

T

’T (±+A) ’1

e’(T ’s)(±+A) f (s)ds

= (1 ’ e

x± (t) )

0

t

e’(t’s)(±+A) f (s)ds

+

0

T

1 ’ e’T (±’A)

e’(T ’s)(±+A) f (s)ds

=

(1 ’ e’±T cos T )2 + (e’±T sin T )2 0

t

e’(t’s)(±+A) f (s)ds,

+ t > 0, ± > 0.

0

As we have seen, proving the existence of periodic solutions reduces to ¬nding

uniform L∞ (]0, T [; H) estimates for (x± )±>0 and (x± )±>0 . Since A is linear

bounded operator ( A L(H;H) = 1) we have

f ’ ±x± ’ Ax± L∞ (]0,T [;H)

x± =

L∞ (]0,T [;H)

¤ f L∞ (]0,T [;H) + (± + A L(H;H) ) x± L∞ (]0,T [;H) , ± > 0,

and hence in this case it is su¬cient to ¬nd only uniform L∞ (]0, T [; H) estimates

for (x± )±>0 or uniform estimates for (x± (0))±>0 in H.

Case 1: T = 2nπ, n ≥ 0. We have

T

1

e’(T ’s)(±+A) f (s)ds.

lim x± (0) = lim

±’0 1 ’ e’±T

±’0 0

T

If 0 e’(T ’s)A f (s)ds = 0 , then (x± (0))±>0 is not bounded. In fact since

e’2nπA = 1 it is easy to check that equation (35) does not have any periodic

T

solution. If 0 e’(T ’s)A f (s)ds = 0 then every solution of (35) is T -periodic and

(x± (0))±>0 is convergent for ± ’ 0:

T ’±(T ’s)

’ 1)e’(T ’s)A f (s)ds

(e

0

lim x± (0) = lim

1 ’ e’±T

±’0 ±’0

T

T ’ s ’(T ’s)A

=’ e f (s)

T

0

T

1

se’(T ’s)A f (s).

=

T 0

Case 2: T = 2nπ for alln ≥ 0. In this case (1 ’ e’T A ) is invertible and

(x± (0))±>0 converges to x(0) where x is the unique T -periodic solution of (35):

T

’T (±+A) ’1

e’(T ’s)(±+A) f (s)ds

lim (1 ’ e

lim x± (0) = )

±’0 ±’0 0

T

(1 ’ e’T A )’1 e’(T ’s)A f (s)ds

=

0

T

1 T +π

e’( 2 ’s)A

= f (s)ds.

2 sin( T ) 0

2

26 Periodic solutions for evolution equations

We state now our main result of existence in the linear and symmetric case.

Theorem 3.8 Assume that A : D(A) ‚ H ’ H is a linear maximal monotone

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

the necessary and su¬cient condition for the existence of periodic solutions for

(33) is given by

1T

f (t)dt ∈ Range(A).

f :=

T0

In this case we have the estimates:

√

T T

¤ A’1 f

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

L∞ (]0,T [;H) L2 (]0,T [;H)

2 2

and

1

¤√ f

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

L∞ (]0,T [;H) L2 (]0,T [;H)

T

and the solution is unique up to a constant in A’1 (0).

Proof The condition is necessary (see Proposition 3.3). Let us show now that

it is also su¬cient. Consider the T -periodic solutions (x± )±>0 for

t ∈ [0, T ], ± > 0.

±x± (t) + x± (t) + Ax± (t) = f (t),

First we prove that (x± )±>0 is uniformly bounded in C 1 ([0, T ]; H). Let us

multiply by x± (t) and integrate on a period:

T T T

2

x± (t) dt + ±(x± (t), x± (t)) + (Ax± (t), x± (t))dt = (f (t), x± (t))dt.

0 0 0

Since A is symmetric and x± is T -periodic we have

T

±(x± (t), x± (t)) + (Ax± (t), x± (t))dt

0

T T

±d 1d

x± (t) 2 dt +

= (Ax± (t), x± (t))dt

2 dt 2 dt

0 0

1 2

+ (Ax± (t), x± (t)) |T = 0.

= ± x± (t) 0

2

Finally we get

2

¤ (f, x± )L2 (]0,T [;H) ¤ f · x±

x± L2 (]0,T [;H) ,

L2 (]0,T [;H)

L2 (]0,T [;H)

and hence

¤f

x± L2 (]0,T [;H) , ± > 0.

L2 (]0,T [;H)

Therefore we can write

1 1

x± (t) ¤ √ x± ¤√ f

min L2 (]0,T [;H) . (43)

L2 (]0,T [;H)

T T

t∈[0,T ]

Mihai Bostan 27

As seen before, since A is linear we can write

± 1

(x± (t + h) ’ x± (t)) + (x± (t + h) ’ x± (t))

h h

1 1

+ A(x± (t + h) ’ x± (t)) (f (t + h) ’ f (t)),

=

h h

and by standard calculations for s < t and h > 0, we get

1

x± (t + h) ’ x± (t)

h

t

1 1

¤ e’±(t’s) x± (s + h) ’ x± (s) + e’±(t’„ ) f („ + h) ’ f (t) d„ .

h h

s

Passing to the limit for h ’ 0 we deduce

t

’±(t’s)

e’±(t’„ ) f („ ) d„

¤e

x± (t) x± (s) +

s

t

¤ s ¤ t, ± > 0.

x± (s) + f („ ) d„, (44)

s

From (43) and (44) we conclude that the functions (x± )±>0 are uniformly

bounded in L∞ (]0, T [; H):

1

¤√ f

x± +f L1 (]0,T [;H) , ± > 0.

L∞ (]0,T [;H) L2 (]0,T [;H)

T

As shown before, since A is linear and x± is T -periodic we have also

± x± + A x± = f . (45)

By the hypothesis there is x0 ∈ D(A) such that f = Ax0 and hence

= (± + A)’1 f = (± + A)’1 Ax0 ¤ x0 ,

x± ± > 0.