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{±(x± (t), x± (t))+(‚•x± (t), x± (t))}dt =
x± (t) dt+ (f (t), x± (t))dt.
0 0 0

Since x± is T -periodic we deduce that
T
{±(x± (t), x± (t)) + (‚•x± (t), x± (t))}dt
0
T
d± 2
{ x± (t)
= + •(x± (t))}dt
dt 2
0
± 2
+ •(x± (t))|T = 0.
= x± (t) (64)
0
2
2
¤ (f, x± )L2 (]0,T [;H) and thus
Therefore, x± L2 (]0,T [;H)


¤f
x± L2 (]0,T [;H) , ± > 0.
L2 (]0,T [;H)


Before estimate (x± )±>0 , let us check that (±x± )±>0 is bounded. By taking
x0 ∈ D(‚•), after standard calculation we ¬nd that
t
’±t
e’±(t’s) f (s) ’ ±x0 ’ ‚•(x0 ) ds. (65)
x± (t) ’ x0 ¤ e x± (0) ’ x0 +
0

Since x± is T -periodic we can write

x± (t) ’ x0 x± (nT + t) ’ x0
= lim
n’∞

e’±(nT +t) x± (0) ’ x0
¤ lim
n’∞
nT +t
e’±(nT +t’s) f (s) ’ ±x0 ’ ‚•(x0 ) ds
+
0
nT +t
1
e’±(nT +t’s) f (s) ds
¤ ±x0 + ‚•(x0 ) + lim
± n’∞ 0
1
¤ ±x0 + ‚•(x0 )
±
1 + e’±t (e’±(n’1)T + · · · + e’±T + 1) · f
+ lim L1
n’∞
e’±t
1
· f L1 (]0,T [;H)
= ±x0 + ‚•(x0 ) + 1 +
1 ’ e’±T
±
1
¤ C1 (x0 , T, f L2 (]0,T [;H) ) 1 + , 0 ¤ t ¤ T, ± > 0.
±
Mihai Bostan 37


It follows that ± x± (t) ¤ C2 (x0 , T, f L2 (]0,T [;H) ), 0 ¤ t ¤ T , 0 < ± < 1. Now
we can estimate x± , ± > 0. After multiplication by x± (t) and integration on
[0, T ] we obtain

T T T
2
± x± (t) dt + (‚•(x± (t)), x± (t))dt = (f (t), x± (t))dt. (66)
0 0 0



We have


•(x0 ) ≥ •(x± (t)) + (‚•(x± (t)), x0 ’ x± (t)), t ∈ [0, T ], ± > 0.


Thus we deduce that for ± > 0,

T T T
(‚•(x± (t)), x± (t))dt ≥ {(‚•(x± (t)), x0 ) ’ •(x0 )}dt.
•(x± (t))dt +
0 0 0



On the other hand for 0 < ± < 1,

T T
(f (t) ’ ±x± (t) ’ x± (t), x0 ) dt
(‚•(x± (t)), x0 ) dt =
0 0
T T
f (t) dt, x0 ’
= (±x± (t), x0 ) dt
0 0
≥ ’C3 (x0 , T, f L2 (]0,T [;H) ).



Therefore,

T T
(‚•(x± (t)), x± (t))dt ≥ •(x± (t))dt ’ C4 (x0 , T, f L2 (]0,T [;H) ). (67)
0 0



Combining (66) and (67) we deduce that

T T
¤ C4 +
•(x± (t))dt (‚•(x± (t)), x± (t))dt
0 0
T T
± x± (t) 2 dt
(f (t), x± (t))dt ’
= C4 +
0 0
T
¤ C4 + (f (t), x± (t))dt, 0 < ± < 1. (68)
0
38 Periodic solutions for evolution equations


On the other hand we have
T
(f (t), x± (t))dt
0
T T
(f (t) ’ f , x± (t))dt +
= x± (t)dt, f
0 0
T t
(f (t) ’ f , x± (0) +
= x± (s) ds)dt + T ( x± , f )
0 0
T t
(f (t) ’ f ,
= x± (s) ds)dt + T ( x± , f )
0 0
T t 1/2
2
· t1/2 dt + T ( x± , f )
¤ f (t) ’ f · x± (s) ds
0 0
T
· √ + T ( x± , f ).
¤ f’ f ·f
L2 (]0,T [;H) L2 (]0,T [;H)
2
Finally we deduce that
T
{•(x± (t)) ’ (x± (t), f )} dt ¤ C5 (x0 , T, f L2 (]0,T [;H) ), 0 < ± < 1, (69)
0

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

C5
•(x± (t)) ’ (x± (t), f ) ¤ , 0 < ± < 1. (70)
T
From Hypothesis (62) we get that (x± (t± ))0<±<1 is bounded and therefore from
(65), for t ∈ [t± , t± + T ],
t
x± (t) ’ x0 ¤ e’±(t’t± ) x± (t± ) ’ x0 + e’±(t’s) f (s) ’ ±x0 ’ ‚•(x0 ) ds.


we deduce that (x± )0<±<1 is bounded in L∞ (]0, T [; H) and that there is x ∈
L∞ (]0, T [; H) such that x± (t) x(t) when ± goes to 0 for t ∈ [0, T ]. Moreover,
from (70) it follows that (•(x± (t± )))0<±<1 is bounded from above and we deduce
that
t
•(x± (t)) = •(x± (t± )) + (‚•(x± (s)), x± (s)) ds

t
¤ •(x± (t± )) + (f (s) ’ ±x± (s) ’ x± (s), x± (s)) ds

¤ C6 (x0 , T, f L2 (]0,T [;H) ), 0 < ± < 1.

On the other hand, by writing •(x± (t)) ≥ •(x0 ) + (‚•(x0 ), x± (t) ’ x0 ), 0 ¤ t ¤
T , ± > 0 we deduce that •(x± (t)) is also bounded from below so that ¬nally
(• —¦ x± )0<±<1 is bounded in L∞ (]0, T [; H).
Mihai Bostan 39


Now, using the second hypothesis of the theorem (every level subset is com-
pact) we deduce that x± (0) ’ x(0) when ± goes to 0 (at least for a subsequence
±n 0). In fact we can easily check that x± converges uniformly to x on [0, T ]
since
x± (t) ’ xβ (t) ¤ x± (0) ’ xβ (0) + |± ’ β| · T · sup xγ L∞ (]0,T [;H) ,
0<γ<1

for 0 ¤ t ¤ T , 0 < ±, β < 1. Now, since lim± 0 dx± /dt = dx/dt in the
sense of H-valued vectorial distribution on ]0, T [ and (x± )±>0 is bounded in
L2 (]0, T [; H) it follows that x belongs to L2 (]0, T [; H) and in particular x is
absolutely continuous on every compact of ]0, T [ and therefore a.e. di¬erentiable
on ]0, T [.
To complete the proof we need to show that x(t) ∈ D(•) a.e. on ]0, T [ and
x (t) + ‚•x(t) f (t) a.e. on ]0, T [. For arbitrarily [u, v] ∈ ‚• we have
t
1 2±t 1
2
¤ e2±s x± (s) ’ u 2
e2±„ (f („ ) ’ ±u ’ v, x± („ ) ’ u) d„,
x± (t) ’ u
e +
2 2 s

with 0 ¤ s ¤ t ¤ T , ± > 0. Passing to the limit for ± 0 we get
t
1 1
2 2
x(t) ’ u ¤ x(s) ’ u (f („ ) ’ v, x(„ ) ’ u) d„, 0 ¤ s ¤ t ¤ T.
+
2 2 s

Thus
t
1 1
2 2
(x(t)’x(s), x(s)’u) ¤ ’ x(s)’u ¤
x(t)’u (f („ )’v, x(„ )’u) d„,
2 2 s

for 0 ¤ s ¤ t ¤ T . Since x is a.e. di¬erentiable on ]0, T [ we ¬nd that
1
(x (t), x(t) ’ u) (x(t) ’ x(s), x(s) ’ u)
= lim
s tt’s
t
1
¤ lim (f („ ) ’ v, x(„ ) ’ u) d„
s tt’s s

= (f (t) ’ v, x(t) ’ u), a.e. t ∈]0, T [, ∀ [u, v] ∈ ‚•.
Finally, since ‚• is maximal monotone and (f (t) ’ x (t) ’ v, x(t) ’ u) ≥ 0 for all
[u, v] ∈ ‚• we deduce that x(t) ∈ D(‚•) a.e. on ]0, T [ and x (t) + ‚•x(t) f (t)
a.e. on ]0, T [. Since • is lower-semicontinuous we also have
•(x(t)) ¤ lim inf •(x± (t)) ¤ lim inf • —¦ x± ¤ sup • —¦ xγ L∞ .
L∞
± 0 ± 0 0<γ<1

As previous, by writing
•(x(t)) ≥ •(x0 ) + (‚•(x0 ), x(t) ’ x0 )
≥ •(x0 ) ’ ‚•(x0 ) · ( x0 + lim inf x± (t) )
± 0
≥ •(x0 ) ’ ‚•(x0 ) · ( x0 + sup 0 ¤ t ¤ T,
xγ L∞ ),
0<γ<1

we deduce ¬nally that • —¦ x ∈ L∞ (0, T ).
40 Periodic solutions for evolution equations


2
Remark 3.17 If dim H < +∞ then the level subsets {x ∈ H ; •(x) + x ¤
M } are compact as bounded sets.

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