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.

t±

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±

t

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

t±

¤ 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.