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±m± + g(t, m± ) ¤ ±x± (t) + g(t, x± (t)) ¤ ±M± + g(t, M± ), t ∈ [0, T ], ± > 0,
where
m± ¤ x± (t) ¤ M± , t ∈ [0, T ], ± > 0,
and hence
T
1
±m± + G(m± ) ¤ {±x± (t) + g(t, x± (t))}dt ¤ ±M± + G(M± ), ± > 0.
T 0

Finally we get
T
1
{±x± (t)+g(t, x± (t))}dt = ±u± +G(u± ), u± ∈]m± , M± [, ± > 0.
G(x0 ) =
T 0
(30)
Multiplying by u± ’ x0 we obtain
±u± (u± ’ x0 ) = ’(G(x0 ) ’ G(u± ))(x0 ’ u± ), ± > 0.
Since G is increasing we deduce that |u± |2 ¤ u± x0 ¤ |u± | · |x0 |, ± > 0 and hence
(u± )±>0 is bounded:
|u± | ¤ |x0 |, ± > 0.
Now using (30) it follows
T T
1 1
{±x± (t) + g(t, x± (t))}dt = {±u± + g(t, u± )}dt,
T T
0 0

and thus there is t± ∈]0, T [ such that
±x± (t± ) + g(t± , x± (t± )) = ±u± + g(t± , u± ), ± > 0.
Since ±(x± (t± ) ’ u± )2 = ’[g(t± , x± (t± )) ’ g(t± , u± )][x± (t± ) ’ u± ] ¤ 0 we ¬nd
that x± (t± ) = u± , ± > 0 and thus (x± (t± ))±>0 is also bounded
|x± (t± )| ¤ |x0 |, ± > 0.
Now by standard calculations we can write
1d
|x± (t) ’ x± (t± )|2 + [g(t, x± (t)) ’ g(t, x± (t± ))][x± (t) ’ x± (t± )]
2 dt
¤ [f (t) ’ ±x± (t± ) ’ g(t, x± (t± ))][x± (t) ’ x± (t± )], t ∈ R,
16 Periodic solutions for evolution equations


and thus
t
|x± (t) ’ x± (t± )| ¤ |f (s) ’ ±x± (t± ) ’ g(s, x± (t± ))|ds, t > t± , ± > 0,


which implies
T
|x± (t)| ¤ |x0 | + |f (t) ’ ±x± (t± ) ’ g(t, x± (t± ))|dt, t ∈ [0, T ], ± > 0. (31)
0

Since (x± (t± ))±>0 is bounded we have
u± = x± (t± ) ’ x1 ,
such that
G(x0 ) = lim {±u± + G(u± )} = G(x1 ).
±’0
Moreover, if x0 ¤ x1 we have
T
1
0¤ [g(t, x1 ) ’ g(t, x0 )]dt = G(x1 ) ’ G(x0 ) = 0,
T 0

and hence g(t, x1 ) = g(t, x0 ) for all t ∈ [0, T ]. Obviously the same equalities
hold if x0 > x1 . Now by passing to the limit in (31) we ¬nd
T
|x(t)| ¤ |x0 | + |f (t) ’ g(t, x1 )|dt (32)
0
T
t ∈ [0, T ], ∀ x0 ∈ G’1 f ,
= |x0 | + |f (t) ’ g(t, x0 )|dt,
0

and therefore (x± )±>0 is uniformly bounded in L∞ (R).


3 Periodic solutions for evolution equations on
Hilbert spaces
In this section we analyze the existence and uniqueness of periodic solutions for
general evolution equations on Hilbert spaces
x (t) + Ax(t) = f (t), t > 0, (33)
where A : D(A) ‚ H ’ H is a maximal monotone operator on a Hilbert
space H and f ∈ C 1 (R; H) is a T -periodic function. As known by the theory
of Hille-Yosida, for every initial data x0 ∈ D(A) there is an unique solution
x ∈ C 1 ([0, +∞[; H) © C([0, +∞[ ; D(A)) for (33), see [6, p. 101]. Obviously,
the periodic problem reduces to ¬nd x0 ∈ D(A) such that x(T ) = x0 . As
in the one dimensional case we demonstrate uniqueness for strictly monotone
operators. We state also necessary and su¬cient condition for the existence
in the linear symmetric case. Finally the case of non-linear sub-di¬erential
operators is considered. Let us start with the de¬nition of periodic solutions for
(33).
Mihai Bostan 17


De¬nition 3.1 Let A : D(A) ‚ H ’ H be a maximal monotone operator
on a Hilbert space H and f ∈ C 1 (R; H) a T -periodic function. We say that
x ∈ C 1 ([0, T ]; H) © C([0, T ]; D(A)) is a periodic solution for (33) if and only if

t ∈ [0, T ],
x (t) + Ax(t) = f (t),

and x(0) = x(T ).

3.1 Uniqueness
Generally the uniqueness does not hold (see the example in the following para-
graph). However it occurs under the hypothesis of strictly monotony.
Proposition 3.2 Assume that A is strictly monotone ((Ax1 ’Ax2 , x1 ’x2 ) = 0
implies x1 = x2 ). Then (33) has at most one periodic solution.

Proof Let x1 , x2 be two di¬erent periodic solutions. By taking the di¬erence
of (33) and multiplying both sides by x1 (t) ’ x2 (t) we ¬nd
1d 2
x1 (t) ’ x2 (t) + (Ax1 (t) ’ Ax2 (t), x1 (t) ’ x2 (t)) = 0, t ∈ [0, T ].
2 dt
2
By the monotony of A we deduce that x1 ’ x2 is decreasing and therefore
we have

x1 (0) ’ x2 (0) ≥ x1 (t) ’ x2 (t) ≥ x1 (T ) ’ x2 (T ) , t ∈ [0, T ].

Since x1 and x2 are T -periodic we have

x1 (0) ’ x2 (0) = x1 (T ) ’ x2 (T ) ,

which implies that x1 (t) ’ x2 (t) is constant for t ∈ [0, T ] and thus

(Ax1 (t) ’ Ax2 (t), x1 (t) ’ x2 (t)) = 0, t ∈ [0, T ].

Now uniqueness follows by the strictly monotony of A.

3.2 Existence
In this section we establish existence results. In the linear case we state the
following necessary condition.
Proposition 3.3 Let A : D(A) ‚ H ’ H be a linear maximal monotone
operator and f ∈ L1 (]0, T [; H) a T -periodic function. If (33) has T -periodic
solutions, then the following necessary condition holds.
T
1
f (t)dt ∈ Range(A),
f :=
T 0

(there is x0 ∈ D(A) such that f = Ax0 ).
18 Periodic solutions for evolution equations


Proof Suppose that x ∈ C 1 ([0, T ]; H)©C([0, T ]; D(A)) is a T -periodic solution
for (33). Let us consider the divisions ∆n : 0 = tn < tn < · · · < tn = T such
n
0 1
that
lim max |tn ’ tn | = 0. (34)
i i’1
n’∞ 1¤i¤n

We can write

(tn ’ tn )x (tn ) + (tn ’ tn )Ax(tn ) = (tn ’ tn )f (tn ), 1 ¤ i ¤ n.
i i’1 i’1 i i’1 i’1 i i’1 i’1

Since A is linear we deduce
n n n
1 1 1
(tn ’tn )x (tn )+A (tn ’tn )x(tn ) (tn ’tn )f (tn ),
=
i i’1 i’1 i i’1 i’1 i i’1 i’1
T T T
i=1 i=1 i=1

and hence
n n
1 1
(tn tn )x(tn )), (tn ’ tn )[f (tn ) ’ x (tn )] ∈ A.

i i’1 i’1 i i’1 i’1 i’1
T T
i=1 i=1

By (34) we deduce that
n T
1 1
(tn ’ tn )x(tn )) ’ x(t)dt,
i i’1 i’1
T T 0
i=1

and
n T
1 1
(tn tn )[f (tn ) (tn )]
’ ’x ’ [f (t) ’ x (t)]dt
i i’1 i’1 i’1
T T 0
i=1
T
1 1
x(t)|T
f (t)dt ’
= 0
T T
0
T
1
= f (t)dt.
T 0

Since A is maximal monotone Graph(A) is closed and therefore
T T
1 1
f (t)dt ∈ A.
x(t)dt,
T T
0 0

T T
1 1
Thus T 0 x(t)dt ∈ D(A) and f = A( T 0 x(t)dt). Generally the previous
condition is not su¬cient for the existence of periodic solutions. For example
let us analyse the periodic solutions x = (x1 , x2 ) ∈ C 1 ([0, T ]; R2 ) for

x (t) + Ax(t) = f (t), t ∈ [0, T ], (35)

where A : R2 ’ R2 is the orthogonal rotation:

A(x1 , x2 ) = (’x2 , x1 ), (x1 , x2 ) ∈ R2 ,
Mihai Bostan 19


and f = (f1 , f2 ) ∈ L1 (]0, T [; R2 ) is T -periodic. For a given initial data x(0) =
x0 ∈ R2 the solution writes
t
’tA
e’(t’s)A f (s)ds,
x(t) = e x0 + t > 0, (36)
0

where the semigroup e’tA is given by

cos t sin t
e’tA = . (37)
’ sin t cos t

Since e’2πA = 1 we deduce that the equation (35) has 2π-periodic solutions if
and only if

etA f (t)dt = 0. (38)
0
2π 2π
Thus if 0 {f1 (t) cos t ’ f2 (t) sin t}dt = 0 or 0 {f1 (t) sin t + f2 (t) cos t}dt = 0
equation (35) does not have any 2π-periodic solution and the necessary condition
still holds because Range(A) = R2 . Moreover if (38) is satis¬ed then every
solution of (35) is periodic and therefore uniqueness does not occur (the operator
A is not strictly monotone). Let us analyse now the existence. As in the one
dimensional case we have

Proposition 3.4 Suppose that A : D(A) ‚ H ’ H is maximal monotone and
f ∈ C 1 (R; H) is T -periodic. Then for every ± > 0 the equation

t ∈ R,
±x(t) + x (t) + Ax(t) = f (t), (39)

has an unique T -periodic solution in C 1 (R; H) © C(R; D(A)).

Proof Since ±+A is strictly monotone the uniqueness follows from Proposition
3.2. Indeed,
2
± x’y + (Ax ’ Ay, x ’ y) = 0, x, y ∈ D(A),

implies ± x ’ y 2 = 0 and hence x = y.
Consider now an arbitrary initial data x0 ∈ D(A). By the Hille-Yosida™s theo-
rem, there is x ∈ C 1 ([0, +∞[; H) © C([0, +∞[; D(A)) solution for (39). Denote

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