Weak formulation 71
P
ROBLEM
1. Find u ∈ C [0, T ]; V ∩ C
1
[0, T ]; H and w ∈ L
∞
0, T ; H satisfying u
t t
∈ L
1
0, T ; V
′
, 12
hu
t t
t , vi + hwt , vi + hLut , vi = hS[u]t , vi 13
for a.a. t ∈]0, T [, ∀v ∈ V ,
w t ∈ Bu
t
t for a.a. t ∈]0, T [ ,
14 u0 = u
, dudt 0 = v
. 15
T
HEOREM
1
EXISTENCE AND UNIQUENESS
. Assume that Hypotheses a, b, c hold and let u
, v , s
20
be given such that u
∈ V, v
∈ H, s
20
∈ V
′
. 16
Then there exists one and one solution of Problem 1. T
HEOREM
2
CONTINUOUS DEPENDENCE
. Assume that Hypotheses a, b, c hold. Let { ˆu
, ˆ v
, ˆ s
20
}, { ˜u , ˜
v , ˜
s
20
} be two sets of data satisfying 16, and let ˆu, ˆ w
, ˜ u, ˜
w be the cor-
responding solutions of Problem 1. Then there is a constant N , depending only on ℓ, c, C, C
S
and T , such that k ˆu − ˜uk
C [0,T ];V ∩C
1
[0,T ];H
≤ Nk ˆu − ˜u
k + | ˆ v
− ˜v | + kˆs
20
− ˜s
20
k
∗
. 17
The above theorems are shown in Section 5, after proving an auxiliary lemma in Section 3 and looking at the explicit problem in Section 4. The last section is devoted to the mentioned
application.
3. Auxiliary lemma
L
EMMA
1. Let V , H, V
′
be a Hilbert triplet and B denote a maximal monotone operator from DB = H to H . If the condition
∃3 0 :
∀u ∈ H, ∀ω ∈ Bu,
|ω| ≤ 31 + |u| , 18
is fulfilled, then the restriction A of B to V is maximal monotone from V to V
′
. Proof. Without loss of generality we may assume that 0 ∈ B0: this can be achieved by shifting
the range of B. The monotonicity of A is obvious. We check its maximal monotonicity. Given f in V
′
, we try to solve in V the equation f ∈ u + Au by approximating it with the equation
f ∈ u
ε
+ B
ε
u
ε
ε 0, where is the Riesz operator from V to V
′
, that is, h u, vi = u, v
∀u, v ∈ V , and B
ε
the Yosida approximation of B in H . Being I the identity operator of H , we recall that B
ε
= I − J
ε
ε ,
where J
ε
= I + ε B
− 1
denotes the resolvent of B , B
ε
u =
u − u
ε
ε ,
where u
ε
is defined by I + ε Bu
ε
= u . It is important to distinguish between the single-valued operator B
ε
of H and the multivalued operator B J
ε
. We have B
ε
u ∈ B J
ε
u for all u ∈ H . In fact, B
ε
u ∈ B J
ε
u means I − J
ε
u ∈
72 S. Durando
ε B J
ε
u, that is, u ∈ J
ε
u + ε B J
ε
u, and this is true owing to the definition of J
ε
. As B
ε
is maximal monotone and Lipschitz continuous of constant 1ε, B
ε V
is also monotone and Lipschitz continuous from V to V
′
. As : V → V
′
is obviously coercive, [1, Corollary 1.3, p. 48] ensures the existence of
u
ε
∈ V satisfying u
ε
+ B
ε
u
ε
= f . Multiply this equation by u
ε
. Note that h
u
ε
, u
ε
i = k u
ε
k
2
, B
ε
u
ε
, u
ε
≥ 0 because B
ε
0 = 0, h f, u
ε
i ≤ k f k
∗
k u
ε
k. Then we easily get
k u
ε
k
2
≤ k f k
∗
k u
ε
k , whence {
u
ε
}
ε
is bounded in V . Setting w
ε
:= B
ε
u
ε
, from [1, Proposition 1.1. iii, p. 42], 4, and 18 we recover the estimate
| w
ε
| = |B
ε
u
ε
| ≤ inf
w ∈
B u
ε
| w
| ≤ 31 + | u
ε
| ≤ 31 + Ck u
ε
k ≤ c
, for some constant
c independent of ε. Therefore, there are a subsequence {
u
ε
n
} weakly converg- ing to
u in V and a subsequence {
w
ε
n
} weakly converging to w
in H . As n goes to ∞, ε
n
↓ 0 and the equality
h u
ε
n
, vi + B
ε
n
u
ε
n
, v = h f, vi,
v ∈ V ,
tends to h
u , vi +
w , v = h
f, vi, v ∈
V , thanks to the continuity of
. Now we show that lim sup
n↑∞
w
ε
n
, u
ε
n
≤ w
, u
. 19
In order to simplify the notation, we replace ε
n
with n. On account of the relation h
u
n
, u
n
i + w
n
, u
n
= h f,
u
n
i , we deduce that
k u
k
2
≤ lim inf k
u
n
k
2
= − lim sup−h
u
n
, u
n
i =
− lim sup w
n
, u
n
− h f,
u
n
i =
h f, u
i − lim sup w
n
, u
n
and consequently lim sup
w
n
, u
n
≤ h f,
u i − k
u k
2
= h f, u
i − h u
, u
i = w
, u
. Hence, 19 is true and [1, Proposition 1.1 iv, p. 42] allows us to conclude that
w ∈ B
u . Thus
u ∈ V solves
u + B
u ∋ f and the lemma is completely proved.
R
EMARK
1. In fact, the assumption 18 has been used only to deduce the boundedness of {
w
ε
} from that of { u
ε
}. Therefore the same proof hold if 18 is replaced by the more general condition
B is bounded on bounded sets .
Weak formulation 73
4. The explicit problem
