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Now it is easy to check that (x± )±>0 is uniformly bounded in L∞ (]0, T [; H):
T
1
x± (t) ’ x± (x± (t) ’ x± (s))ds
=
T 0
1Tt
= x („ )d„ ds
T0 s±

T T
¤ f L2 (]0,T [;H) + f L1 (]0,T [;H) ,
2 2
and thus

T T
¤
x± x± + f L2 (]0,T [;H) + f L1 (]0,T [;H)
L∞ (]0,T [;H)
2 2

T T
¤ x0 + f L2 (]0,T [;H) + f L1 (]0,T [;H) .
2 2
28 Periodic solutions for evolution equations


Now we can prove that (x± )±>0 is convergent in C 1 ([0, T ]; H). Indeed, by
taking the di¬erence between the equations (39) written for ±, β > 0, after
multiplication by x± (t) ’ xβ (t) and integration on [0, T ] we get
T
{±(x± (t) ’ xβ (t), x± (t) ’ xβ (t))
0
2
+ x± (t) ’ xβ (t) + (A(x± (t) ’ xβ (t)), x± (t) ’ xβ (t))}dt
T
= ’(± ’ β) (xβ (t), x± (t) ’ xβ (t))dt.
0

Since A is symmetric, x± and xβ are T -periodic and uniformly bounded in
L∞ (]0, T [; H) we deduce that

x± ’ xβ ¤ |± ’ β| · sup xγ L2 (]0,T [;H) ,
L2 (]0,T [;H)
γ>0

or
|± ’ β|

x± ’ xβ ¤ · sup xγ L2 (]0,T [;H) + |± ’ β| · sup xγ L1 (]0,T [;H) ,
L∞ (]0,T [;H)
T γ>0 γ>0

and therefore (x± )±>0 converges in C([0, T ]; H).
We already know that ( x± )±>0 = ((± + A)’1 f )±>0 is bounded in H and
by the Proposition 3.6 it follows that ( x± )±>0 is convergent to the element of
minimal norm in A’1 f . We have
t
t ∈ R, ± > 0.
x± (t) = x± (0) + x± (s)ds,
0

t
By taking the average we deduce that x± (0) = x± ’ < 0 x± (s)ds > and
therefore, since (x± )±>0 is uniformly convergent, it follows that (x± (0))±>0 is
also convergent. Finally we conclude that (x± )±>0 is convergent in C 1 ([0, T ]; H)
to the periodic solution x for (33) such that < x > is the element of minimal
norm in A’1 f .
Before analyzing the periodic solution for the heat equation, following an
idea of [7], let us state the following proposition.

Proposition 3.9 Assume that A : D(A) ‚ H ’ H is a linear maximal mono-
tone and symmetric operator and f ∈ C 1 ([0, T ]; H) is a T -periodic function.
Then for every x0 ∈ D(A) we have
1
(x(t + T ; 0, x0 ) ’ x(t; 0, x0 )) = f ’ ProjR(A) f ,
lim (46)
t’∞ T

where x(·; 0, x0 ) represents the solution of (33) with the initial data x0 and R(A)
is the range of A.

Remark 3.10 A being maximal monotone, A’1 is also maximal monotone and
therefore D(A’1 ) = R(A) is convex.
Mihai Bostan 29


Proof of Proposition 3.9. Consider x0 ∈ D(A) and denote by x(·) the
corresponding solution. By integration on [t, t + T ] we get

t+T
1 1
(x(t + T ) ’ x(t)) + A x(s)ds = f . (47)
T T t

For each ± > 0 consider x± ∈ D(A) such that ±x± + Ax± = f . Denoting by
1 t+T
x(s)ds, t ≥ 0, equation (47) writes
y(·) the function y(t) = T t

t ≥ 0, ± > 0.
y (t) + Ay(t) = ±x± + Ax± ,

Let us search for y of the form y1 + y2 where

t ≥ 0,
y1 (t) + Ay1 (t) = ±x± ,

with the initial condition y1 (0) = 0 and

t ≥ 0,
y2 (t) + Ay2 (t) = Ax± , (48)

T
1
with the initial condition y2 (0) = y(0) = T 0 x(t)dt. We are interested on the
asymptotic behaviour of Ay(t) = Ay1 (t) + Ay2 (t) for large t. We have

t
’tA
e’(t’s)A ±x± ds
y1 (t) =e y1 (0) +
0
t
e’(t’s)A ±x± ds,
=
0

and therefore,
t t
Ae’(t’s)A ±x± ds = e’(t’s)A ±x± = (1 ’ e’tA )±x± .
Ay1 (t) =
0
0

By the other hand, after multiplication of (48) by y2 (t) = (y2 (t) ’ x± ) we get

2
+ (A(y2 (t) ’ x± ), (y2 (t) ’ x± ) ) = 0, t ≥ 0.
y2 (t)

Since A is symmetric, after integration on [0, t] we obtain

t
1 1
y2 (s) 2 ds + (A(y2 (t) ’ x± ), y2 (t) ’ x± ) = (A(y2 (0) ’ x± ), y2 (0) ’ x± ),
2 2
0

and therefore, by the monotony of A it follows that

1
y2 (t) 2 dt ¤ (A(y2 (0) ’ x± ), y2 (0) ’ x± ).
2
0
30 Periodic solutions for evolution equations


Thus limt’∞ y2 (t) = 0 and by passing to the limit in (48) we deduce that
limt’∞ Ay2 (t) = limt’∞ (Ax± ’ y2 (t)) = Ax± . Finally we ¬nd that
1
(x(t + T ) ’ x(t)) ’ e’tA ±x±
lim
t’∞ T
= lim {y (t) ’ e’tA ±x± }
t’∞
lim { f ’ Ay(t) ’ e’tA ±x± }
=
t’∞
lim { f ’ Ay1 (t) ’ Ay2 (t) ’ e’tA ±x± }
=
t’∞
f ’ ±x± ’ Ax± = 0,
= ± > 0. (49)
Now let us put y± = Ax± and observe that y± + ±A’1 y± = Ax± + ±x± =
f , ± > 0. Therefore,
lim (1 + ±A’1 )’1 f
lim y± =
± 0 ± 0
’1
A
= lim J± f
± 0
= ProjD(A’1 ) f
= ProjR(A) f ,
and it follows that
lim ±x± = lim ( f ’ Ax± ) = lim ( f ’ y± ) = f ’ ProjR(A) f .
± 0 ± 0 ± 0

Since Graph(A) is closed and [±x± , ±y± ] = [±x± , A(±x± )] ∈ A, ± > 0, by
0 we deduce that f ’ ProjR(A) f ∈ D(A) and
passing to the limit for ±
A( f ’ ProjR(A) f ) = 0. It is easy to see that we can pass to the limit for
0 in (49). Indeed, for µ > 0 let us consider ±µ > 0 such that lim± 0 ±x± ’
±
µ
±µ x±µ < 2 . We have
1
(x(t + T ) ’ x(t)) ’ e’tA lim ±x±
T ±0
1
(x(t + T ) ’ x(t)) ’ e’tA ±µ x±µ + e’tA ±µ x±µ ’ e’tA lim ±x±
¤
T ±0
1
(x(t + T ) ’ x(t)) ’ e’tA ±µ x±µ + ±µ x±µ ’ lim ±x±
¤
T ±0
µµ µ
¤ + = µ, t ≥ t(±µ , ) = t(µ),
22 2
and thus
1
lim { (x(t + T ) ’ x(t)) ’ e’tA ( f ’ ProjR(A) f )} = 0.
t’∞ T

But e’tA ( f ’ ProjR(A) f ) does not depend on t ≥ 0:
d ’tA
= ’Ae’tA ( f ’ ProjR(A) f )
( f ’ ProjR(A) f )
e
dt
= ’e’tA A( f ’ ProjR(A) f ) = 0,
Mihai Bostan 31


and thus the previous formula reads
1
(x(t + T ) ’ x(t)) = f ’ ProjR(A) f .
lim
t’∞ T

Remark 3.11 Under the same hypothesis as above we can easily check that
x(T ; 0, x0 ) ’ x0
f ’ ProjR(A) f
inf = = dist( f , R(A)).
T
x0 ∈D(A)


3.3 Periodic solutions for the heat equation
Let „¦ ‚ Rd , d ≥ 1, be an open bounded set with ‚„¦ ∈ C 2 . Consider the heat
equation
‚u
(t, x) ’ ∆u(t, x) = f (t, x), (t, x) ∈ R — „¦, (50)
‚t
with the Dirichlet boundary condition

(t, x) ∈ R — ‚„¦,
u(t, x) = g(t, x), (51)

or the Neumann boundary condition
‚u
(t, x) ∈ R — ‚„¦,
(t, x) = g(t, x), (52)
‚n
where we denote by n(x) the outward normal in x ∈ ‚„¦.

Theorem 3.12 Assume that f ∈ C 1 (R; L2 („¦)) is T -periodic and g(t, x) =
‚u0 1 2 2 2
‚n (t, x), (t, x) ∈ R—‚„¦ where u0 ∈ C (R; H („¦))©C (R; L („¦)) is T -periodic.
Then the heat problem (50), (52) has T -periodic solutions u ∈ C(R; H 2 („¦)) ©
C 1 (R; L2 („¦)) if and only if
T T
g(t, x)dtdσ + f (t, x)dtdx = 0.
‚„¦ 0 „¦ 0

In this case the periodic solutions satis¬es the estimates
1

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. 7
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