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by (xn )n≥0 the functions

t ∈ [0, T ], n ≥ 0.
xn (t) = x(nT + t),

We have

t ∈ [0, T ],
±xn+1 (t) + xn+1 (t) + Axn+1 (t) = f ((n + 1)T + t),

and
t ∈ [0, T ].
±xn (t) + xn (t) + Axn (t) = f (nT + t),
20 Periodic solutions for evolution equations


Since f is T -periodic, after usual computations we get
1d2
xn+1 (t) ’ xn (t) 2
± xn+1 (t) ’ xn (t) +
2 dt
+(Axn+1 (t) ’ Axn (t), xn+1 (t) ’ xn (t)) t ∈ [0, T ].
= 0,

Taking into account that A is monotone we deduce

xn+1 (t) ’ xn (t) ¤ e’±t xn+1 (0) ’ xn (0) , t ∈ [0, T ],

and hence

xn+1 (0) ’ xn (0) ’ xn’1 (T )
= xn (T )
’±T
¤ xn (0) ’ xn’1 (0)
e
e’2±T xn’1 (0) ’ xn’2 (0)
¤
¤ ...
e’n±T x1 (0) ’ x0 (0) , n ≥ 0.
¤ (40)

Finally we get the estimate

xn+1 (t) ’ xn (t) ¤ e’±(nT +t) S± (T ; 0, x0 ) ’ x0 , t ∈ [0, T ], n ≥ 0.

Here S± (t; 0, x0 ) represents the solution of (39) for the initial data x0 . From the
previous estimate it is clear that (xn )n≥0 is convergent in C 0 ([0, T ]; H):
n’1
(xk+1 (t) ’ xk (t)), t ∈ [0, T ],
xn (t) = x0 (t) +
k=0

where
n’1 n’1
(xk+1 (t) ’ xk (t)) ¤ xk+1 (t) ’ xk (t)
k=0 k=0
n’1
e’±(kT +t) S± (T ; 0, x0 ) ’ x0
¤
k=0
e’±t
¤ S± (T ; 0, x0 ) ’ x0 .
1 ’ e’±T
Moreover xn (t) ¤ S± (t; 0, x0 ) + 1’e1 S± (T ; 0, x0 ) ’ x0 . Denote by
’±T

x± the limit of (xn )n≥0 as n ’ ∞. We should note that without any other
hypothesis (x± )±>0 is not uniformly bounded in L∞ (]0, T [; H). We have only
1
estimate in O(1 + ± ),

1 1
¤C 1+ ∼O 1+
x± .
L∞ ([0,T ];H)
1 ’ e’±T ±
The above estimate leads immediately to the following statement.
Mihai Bostan 21


Remark 3.5 The sequence (±x± )±>0 is uniformly bounded in L∞ ([0, T ]; H).

Let us demonstrate that x± is T -periodic and solution for (39). Indeed,

x± (0) = lim xn (0) = lim xn’1 (T ) = x± (T ).
n’∞ n’∞

Now let us show that (xn )n≥0 is also uniformly bounded in L∞ (]0, T [; H). By
taking the di¬erence between the equations (39) at the moments t and t + h we
have

±(x(t+h)’x(t))+x (t+h)’x (t)+Ax(t+h)’Ax(t) = f (t+h)’f (t), t < t+h.

Multiplying by x(t + h) ’ x(t) we obtain

1d
2 2
± x(t + h) ’ x(t) x(t + h) ’ x(t) ¤ f (t + h) ’ f (t) · x(t + h) ’ x(t) ,
+
2 dt
which can be also rewritten as
t
1 2±t 2
e±s f (s + h) ’ f (s) · e±s x(s + h) ’ x(s) ds
x(t + h) ’ x(t) ¤
e
2 0
1
x(h) ’ x(0) 2 ,
+ t < t + h.
2
By using Bellman™s lemma we conclude that
t
1 1
e’±(t’s)
x(t + h) ’ x(t) ¤ f (s + h) ’ f (s) ds
h h
0
1
+e’±t x(h) ’ x(0) , 0 ¤ t < t + h. (41)
h
By passing to the limit for h ’ 0 the previous formula yields
t
’±t
e’±(t’s) f (s) ds
¤e
x (t) x (0) +
0
1
¤ e’±t f (0) ’ ±x0 ’ Ax0 + (1 ’ e’±t ) f L∞ (]0,T [;H)
±
1
¤ f (0) ’ ±x0 ’ Ax0 + f < +∞.
L∞ (]0,T [;H)
±
Therefore (xn )n≥0 is uniformly bounded in L∞ (]0, T [; H) since

¤x
xn = x (nT + (·)) L∞ ([0,+∞[;H) ,
L∞ (]0,T [;H) L∞ (]0,T [;H)


y± (t), t ∈ [0, T ]. We can write
and thus we have xn (t)
t
z ∈ H, t ∈ [0, T ], n ≥ 0,
(xn (t), z) = (xn (0), z) + (xn (s), z)ds,
0
22 Periodic solutions for evolution equations


and by passing to the limit for n ’ ∞ we deduce
t
z ∈ H, t ∈ [0, T ],
(x± (t), z) = (x± (0), z) + (y± (s), z)ds,
0

which is equivalent to
t
t ∈ [0, T ].
x± (t) = x± (0) + y± (s)ds,
0

Therefore x± is di¬erentiable and x± = y± . Finally we can write xn (t) x± (t),
t ∈ [0, T ]. Let us show that x± is also solution for (39). We have

[xn (t), f (t) ’ ±xn (t) ’ xn (t)] ∈ A, n ≥ 0, t ∈ [0, T ].

Since xn (t) ’ x± (t), xn (t) x± (t) and A is maximal monotone we conclude
that
[x± (t), f (t) ’ x± (t)] ∈ A, t ∈ [0, T ], ± > 0,
which means that x± (t) ∈ D(A) and Ax± (t) = f (t) ’ x± (t), t ∈ [0, T ].

Now we establish for the linear case the similar result stated in Proposition
2.10. Before let us recall a standard result concerning maximal monotone oper-
ators on Hilbert spaces

Proposition 3.6 Assume that A is a maximal monotone operator (linear or
not) and ±u± + Au± = f , u± ∈ D(A), f ∈ H, ± > 0. Then the following
statements are equivalent:
(i) f ∈ Range(A);
(ii) (u± )±>0 is bounded in H. Moreover, in this case (u± )±>0 is convergent in
H to the element of minimal norm in A’1 f .

Proof it (i) ’ (ii) By the hypothesis there is u ∈ D(A) such that f = Au.
After multiplication by u± ’ u we get

±(u± , u± ’ u) + (Au± ’ Au, u± ’ u) = 0, ± > 0.

Taking into account that A is monotone we deduce
2
¤ (u± , u) ¤ u± · u ,
u± ± > 0,

and hence u± ¤ u , ± > 0, u ∈ A’1 f which implies that u± u0 . We have
[u± , f ’ ±u± ] ∈ A, ± > 0 and since A is maximal monotone, by passing to the
limit for ± ’ 0 we deduce that [u0 , f ] ∈ A, or u0 ∈ A’1 f . Moreover

∀u ∈ A’1 f.
u0 = w ’ lim u± ¤ lim inf u± ¤ lim sup u± ¤ u ,
±’0 ±’0 ±’0

In particulat taking u = u0 ∈ A’1 f we get

w ’ lim u± = lim u± ,
±’0 ±’0
Mihai Bostan 23


and hence, since any Hilbert space is strictly convex, by Mazur™s theorem we
deduce that the convergence is strong

u± ’ u0 ∈ A’1 f, ± ’ 0,

where u0 = inf u∈A’1 f u = minu∈A’1 f u .
(ii) ’ (i) Conversely, suppose that (u± )±>0 is bounded in H. Therefore u± u
in H. We have [u± , f ’ ±u± ] ∈ A, ± > 0 and since A is maximal monotone
by passing to the limit for ± ’ 0 we deduce that [u, f ] ∈ A or u ∈ D(A) and
f = Au.
Theorem 3.7 Assume that A : D(A) ‚ H ’ H is a linear maximal monotone
operator on a compact Hilbert space H and f ∈ C 1 (R; H) is a T -periodic func-
tion. Then the following statements are equivalent:
(i) equation (33) has periodic solutions;
(ii) the sequence of periodic solutions for (39) is bounded in C 1 (R; H). Moreover
in this case (x± )±>0 is convergent in C 0 (R; H) and the limit is also a T -periodic
solution for (33).

Proof (i) ’ (ii) Denote by x, x± the periodic solutions for (33) and (39). By
taking the di¬erence and after multiplication by x± (t) ’ x(t) we get:

1d
2 2
± x± (t) ’ x(t) x± (t) ’ x(t) ¤ ± x(t) · x± (t) ’ x(t) , t ∈ R. (42)
+
2 dt
Finally, after integration and by using Bellman™s lemma, formula (42) yields
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 ∈ R.
L∞ ,

Since x± and x are T -periodic we can also write

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

By passing to the limit for n ’ ∞ we obtain

x± ’ x ¤x L∞ , ± > 0,
L∞

and hence
¤2 x
x± L∞ , ± > 0.
L∞

Since A is linear we can write
± 1 1
(x± (t + h) ’ x± (t)) + (x± (t + h) ’ x± (t)) + A(x± (t + h) ’ x± (t))
h h h
1
(f (t + h) ’ f (t)), t < t + h, ± > 0,
=
h
24 Periodic solutions for evolution equations

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