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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.

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