74 S. Durando
5 yield w, u
t
≥ 0 a.e. in ]0, T [, the procedure followed in [4, Lemma 3.2, p. 87] enables us
to infer k ˆu
ε
− ˜u
ε
k
2 t
≤C
1
k ˆu
ε
− ˜u
ε
k
2
+ | ˆv
ε
− ˜v
ε
|
2
+ kˆs
ε 20
− ˜s
ε 20
k
2 ∗
+ kˆs
ε 1
− ˜s
ε 1
k
2 L
1
0,t ;H
+ kˆs
ε 2
− ˜s
ε 2
t
k
2 L
1
0,t ;V
′
25 with obvious notation, and note that 0 solves Problem 2 with null data
ku
ε
k
2 t
≤ C
1
ku
ε
k
2
+ |v
ε
|
2
+ ks
ε 20
k
2 ∗
+ ks
ε 1
k
2 L
1
0,t ;H
+ ks
ε 2
t
k
2 L
1
0,t ;V
′
26 for all t ∈]0, T ] and for some constant C
1
independent of ε. Consequently, also kw
ε
k
L
∞
0,T ;H
is uniformly bounded. Due to well-known compactness results, we can find subsequences converging weakly star
∗
⇀ . Let u, w, and ε
n
↓ 0 fulfill u
n ∗
⇀ u
in L
∞
0, T ; V ∩ W
1,∞
0, T ; H , w
n ∗
⇀ w in
L
∞
0, T ; H , where u
n
and w
n
stand for u
ε
n
and w
ε
n
, respectively. Now, one can show that the pair u, w solves the equations of Problem 2. In fact, it turns out that
u
n
→ u strongly in C
[0, T ]; V ∩ C
1
[0, T ]; H , 27
w ∈ Bu
t
a.e. in ]0, T [ . 28
The proof of 27 consists in a direct check of the Cauchy condition in C [0, T ]; V ∩C
1
[0, T ]; H for u
n
, by applying again [4, estimate 3.5, p. 87]. Further, accounting for the strong convergence of {u
n t
} in L
2
0, T ; H and arguing as in the proof of 19, thanks to [1, Lemma 1.3, p. 42] one verifies the second condition 28.
At this point, we can first take the limit in 25 on some subsequence ε
n
↓ 0 and recover 24. Then, the uniqueness of u and w follows from 24 and a comparison in 23.
5. Existence, uniqueness and continuous dependence
. The next step for proving Theorem 1 consists in showing that Problem 1 has one and only one solution in some time interval [0, τ ], τ ∈]0, T ]. The main tool is the classical Contraction
Mapping Principle. Introduce the metric spaces
X :=
C [0, τ ]; V ∩ C
1
[0, τ ]; H , X
:= {ξ ∈ X : ξ0 = u
, ξ
t
0 = v , kξ k
2 τ
≤ C
2
} , where the constants C
2
and τ are specified later. Fix an element z of X and let u ∈ X and
w ∈ L
∞
0, τ ; V
′
satisfy 14-15 and hd
2
udt
2
t , vi + hLut , vi + hwt , vi = hS[z]t , vi 29
for a.a. t ∈]0, τ [ and any v ∈ V . Due to the assumptions on S and to Theorem 3, the operator D
mapping z into the unique solution u of the above explicit problem is well defined. Since
Weak formulation 75
u0 = u and u
t
0 = v , the claim is that
D X
⊆ X
provided C
2
is suitably chosen and τ is small enough. Indeed, 24 and 11 enable us to deduce that
kuk
2 τ
≤ C
1
ku k
2
+ |v |
2
+ ks
20
k
2 ∗
+ C
1
C
′ S
1 + kzk
2 L
1
0,τ ;V
+ kz
t
k
2 L
1
0,τ ;H
, whence, taking C
2
= 2C
1
C
′ S
+ ku k
2
+ |v |
2
+ ks
20
k
2 ∗
and C
3
= C
1
C
′ S
, it results that kuk
2 τ
≤ C
2
2 + C
3
τ
2
kzk
2 τ
, 30
on account of 20. Therefore, if τ
2
≤
1 2C
3
it is ensured that D
: X → X
. On the other hand, X
is a complete metric space with respect to the distance dˆz, ˜z := kˆz − ˜zk
τ
, ˆz, ˜z ∈ X . Thus,
it becomes important to find conditions on τ in order that D
yields a contraction mapping. Let ˆz, ˜z ∈ X
and ˆ u =
D ˆ
z, ˜ u =
D ˜
z. Owing to 24 and 10, one can infer that k ˆu − ˜uk
2 τ
≤ C
4
τ
2
kˆz − ˜zk
2 τ
, 31
with C
4
= C
1
C
S
. Then, choosing τ = min T,
1 2C
4 1
2
, we can conclude that the operator D
has a unique fixed point u ∈ X . Clearly, such u provides, along with the related w, the solution
to Problem 1 in the preliminary time interval [0, τ ]. To complete the proof of Theorem 1 it remains to verify the possibility of extending the
solution to the whole interval [0, T ]. Here we perform that by deriving a global estimate. Owing to the assumptions 10-d2.9, we can argue as in the deduction of 30 recalling 13
in place of 29 and, in view of Theorem 3 and the H¨older inequality, conclude that kuk
2 τ
≤ C
2
+ C
3
T Z
τ
kuk
2 s
ds , 32
where τ is now an arbitrary value in [0, T ]. As the function t 7→ kuk
t
is continuous, the Gronwall lemma implies that kuk
2 τ
≤ C
2
e
C
3
T τ
. Then, integrating from 0 to t ∈ [0, T ], by 4 we find a constant C
5
, independent of t , such that kuk
C [0,t ];V ∩C
1
[0,t ];H
≤ C
5
33 for any t ∈]0, T ]. Moreover, 14 and 6 provide us a bound for kwk
L
∞
0,T ;H
. To prove Theorem 2, consider two sets of initial data ˆ
u , ˆ
v , ˆs
20
, ˜ u
, ˜ v
, ˜s
20
and denote by ˆ
u, ˆ w
, ˜ u, ˜
w the corresponding solutions of Problem 1. Thanks to 24 and 10, we have
that k ˆu − ˜uk
2 t
≤ C
1
k ˆu − ˜u
k
2
+ | ˆv − ˜v
|
2
+ kˆs
20
− ˜s
20
k
2 ∗
+C
S
k ˆu − ˜uk
2 L
1
0,t ;V
+ C
S
k ˆu
t
− ˜u
t
k
2 L
1
0,t ;H
, whence, estimating the norms in L
1
by the norms in L
2
, we infer k ˆu − ˜uk
2 t
≤ C
1
k ˆu − ˜u
k
2
+ | ˆv − ˜v
|
2
+ kˆs
20
− ˜s
20
k
2 ∗
+ C
4
T Z
t
k ˆu − ˜uk
2 s
ds 34
for any t ∈]0, T ]. Finally, 17 comes out from applying the Gronwall lemma to 34.
76 S. Durando
6. The Stefan problem
