Diffusion limits 217
14 Z
| u
3
|
2 p
+ |v
3
|
2 p
d x
12 p
≤ 4M2L
12 p
t. Finally, letting p → +∞, it is possible to show that, for any finite time, the solution of
Problem 5.3 is essentially bounded. In conclusion, considering our global problem, by linearity we have proved the following
theorem. T
HEOREM
5.1. Let u
ε
x, 0 = u x, v
ε
x, 0 = v x ∈ L
∞
and u
ε
− L , t =
ϕ
−
t , v
ε
+ L , t = ϕ
+
t ∈ W
1,∞
0, T , for all T 0. Then u
ε
x, t , v
ε
x, t ∈ L
∞
for
all t ∈ [0, T ], uniformly in ε.
3. The solution on the boundary
This section is devoted to the study of the limiting behaviour of u
ε
+ L , t and v
ε
− L , t on the
boundary. If 1 ≤ p ∞, equation 8 shows that
d dt
Z
u
2 p 1
+ v
2 p 1
d x + 1
ε h
u
2 p 1
L , t + v
2 p 1
− L , t
i ≤
and so, by Theorem 5.1 1
ε Z
T
h u
2 p 1
L , t + v
2 p 1
− L , t
i dt ≤ ku
k
2 p 2 p
+ kv k
2 p 2 p
= E ,
where E ∈ R
+
. Then we have lim
ε→
k u
1
L , t k
L
p
0,T
= 0,
lim
ε→
kv
1
− L , t k
L
p
0,T
= 0.
Furthermore, by Lemma 5.2, we have that lim
ε→
u
2
L , t = ϕ
+
t lim
ε→
v
2
− L , t = ϕ
−
t almost everywhere.
Finally, if we choose 1 ≤ p ∞, system 6 also implies that d
dt Z
| u
3
|
2 p
+ |v
3
|
2 p
d x + 1
ε h
u
2 p 3
L , t + v
2 p 3
− L , t
i ≤
2 pM Z
| u
3
|
2 p−1
+ |v
3
|
2 p−1
d x. By integration over 0, T we have
218 F. Salvarani
1 ε
Z
T
h u
2 p 3
L , t + v
2 p 3
− L , t
i dt ≤ 2 pM
Z
T
Z
| u
3
|
2 p−1
+ |v
3
|
2 p−1
d xdt + +
Z
| u
3
x, 0|
2 p
+ |v
3
x, 0|
2 p
− | u
3
x, T |
2 p
− |v
3
x, T |
2 p
d x. By inequality 14, we know that the L
p
-norms of u
3
and v
3
are bounded by a linear function of the time, and therefore all the integrals on the right-hand side of the above inequality are
bounded, provided that 1 ≤ p ∞, for all T 0. Also in this case, we then have lim
ε→
k u
3
L , t k
L
p
0,T
= 0,
lim
ε→
kv
3
− L , t k
L
p
0,T
= 0.
Therefore, since ρ
ε
=
3
X
i=1
u
i
+ v
i
, by using the properties of the norm we have proved the following theorem.
T
HEOREM
5.2. Let ρ
ε
= u
ε
+ v
ε
the macroscopic density of system 2 and ε 0. Then, on the boundary,
lim
ε→
ρ
ε
− L , t = 2ϕ
−
, lim
ε→
ρ
ε
+ L , t = 2ϕ
+
strongly in L
p
0, T , provided that 1 ≤ p ∞.
4. Behaviour of the flux
In this section we show that the flux is bounded in L
2
× 0, T .
We repeat the splitting of system 1 and study separately the behaviour of the three fluxes: j
1
, j
2
and j
3
respectively; by using the classical inequality 15
a + b + c
2
≤ 3a
2
+ b
2
+ c
2
, a, b, c, ∈ R
we will then derive a bound for the total flux j
ε
x, t . We can indeed multiply the two equations of system 4 by 2u
1
and 2v
1
respectively. Then we add and integrate on , obtaining:
d dt
Z
u
2 1
+ v
2 1
d x + 1
ε h
u
2 1
L , t + v
2 1
− L , t
i = −
2 Z
u
1
− v
1
ε
2
d x, and so
Z
| j
1
|
2
d x ≤ − 1
2 d
dt Z
u
2 1
+ v
2 1
d x
Diffusion limits 219
16 Z
T
Z
| j
1
|
2
d xdt ≤ 1
2 Z
[u
2
+ v
2
− u
2 1
x, T − v
2 1
x, T ]d x ≤ 1
2 k
u k
2 2
+ kv k
2 2
. Then, we must consider Problem 5.2: as usual, we use the explicit solution in Lemma 5.2 in
order to show that j
2
= 1
ε ϕ
−
t − ϕ
−
t − ϕ
+
t ε +
2L x + L − ϕ
+
t + ϕ
−
t − ϕ
+
t ε +
2L x − L
= 1
ε + 2L
[ϕ
−
t − ϕ
+
t ]. 17
Finally, we multiply the two equations of system 6 for 2u
3
and 2v
3
respectively. Then we add and integrate on , obtaining:
d dt
Z
u
2 3
+ v
2 3
d x + 1
ε h
u
2 3
L , t + v
2 3
− L , t
i =
= − 2
Z
u
3
− v
3
ε
2
d x + 2 Z
f
ε
u
3
+ g
ε
v
3
d x, that is, by the maximum principle for Problem 5.3 and using the properties of f
ε
and g
ε
: Z
| j
3
|
2
d x ≤ − 1
2 d
dt Z
u
2 3
+ v
2 3
d x + 2K t, where K is a positive constant. This means that
18 Z
T
Z
| j
3
|
2
d xdt ≤ K T
2
. Inequality 15, together with 16, 17 and 18, shows that j
ε
= j
1
+ j
2
+ j
3
is bounded in L
2
× 0, T .
T
HEOREM
5.3. Let u
ε
x, t , v
ε
x, t be the unique solution of the initial-boundary prob- lem 1. Then, for all T 0 there exists D ∈ R
+
such that: Z
T
Z
| j
ε
|
2
d xdt ≤ D, uniformly in ε.
5. The hydrodynamical limit
