ON SOME MIXED BOUNDARY VALUE PROBLEMS
WITH NONLOCAL DIFFUSION
N.-H. CHANG & M. CHIPOT
Abstract. We study here a class of nonlinear nonlocal problems. First we consider the
issue of existence and uniqueness for the parabolic setting. Then we study the asymptotic
behaviour of the solution for large time. This leads us to introduce and investigate in
details the associated stationary problem.
1. Introduction
Let Ω be a bounded open set of Rn , n ≥ 1 with a Lipschitz boundary Γ. We suppose
that Γ is split into two measurable subsets ΓD and ΓN = Γ \ ΓD . We denote by a = a(ζ)
a function such that
(
a is continous,
(1.1)
∃ m, M such that 0 < m ≤ a(ζ) ≤ M ∀ ζ ∈ R.
We consider then the problem of finding u = u(x, t) solution to


u − a(`(u(t)))∆u + u = f in Ω × R+ ,

 t
∂u
= 0 on ΓN × R+ ,
u = 0 on ΓD × R+ ,

∂ν


u(·, 0) = u0 in Ω.
(1.2)
In the above system ∆ is the usual Laplace operator, ν is the unit outward normal to Γ,
∂u
the outward normal derivative, ` is a linear form on L2 (Ω) so that
∂ν
Z
g(x)u(x, t) dx, g ∈ L2 (Ω),
(1.3)
`(u(t)) =
Ω
u0 and f are some functions such that
f ∈ L2 (Ω),
u0 ∈ L2 (Ω).
(1.4)
This kind of model problem arises for instance in diffusion of bacteria: u(x, t) is the
density of population located at x at the time t, f is the density of bacteria supplied
from outside, u0 is the initial density of population, a is the diffusion rate (depending
on `(u(t))), the lower order term u is the density of population eliminated by death at a
constant rate taken for the sake of simplicity equal to 1. The nonlocal dependence of a on
`(u(t)) includes for instance the case where the diffusion rate of the population depends
on the entire population – take g = 1 in (1.3). We could have also imposed a nonlocal
dependence on the death rate (see [2], [3]), however for simplicity we only consider here the
case (1.2). Among the interesting issues of (1.2) is the study of the asymptotic behaviour
of the density u(x, t) as t → +∞. This is one of the focus of our paper. Note that the
problem with no lower order term has been investigated in [6], [7], [8] – see also [15], but
the technique there could not apply here. For further aspects of nonlocal problems see
[9].
2. Existence and uniqueness of solution to (1.2)
First we consider the question of existence. We consider (1.2) in the slightly more general
setting where
f ∈ L2 (0, T ; L2 (Ω)),
(T is some arbitrary positive number). We define
Then we have:
u0 ∈ L2 (Ω),
V = { v ∈ H 1 (Ω) | v = 0 on ΓD }.
(2.1)
(2.2)
Theorem 2.1. Suppose (1.1), (1.3), (2.1). There exists a function u = u(x, t) solution
to:


u ∈ L2 (0, T ; V ) ∩ C([0, T ], L2 (Ω)), ut ∈ L2 (0, T ; V 0 ),




u(·, 0) = u0 ,
Z
(2.3)
d

(u, v) + a(`(u(t))) ∇u∇v dx + (u, v)


dt

Ω


= (f, v) in D0 (0, T ), ∀ v ∈ V.
Here (·, ·) is the usual scalar product on L2 (Ω), D0 (0, T ) is the space of distributions on
(0, T ), V 0 is the dual of V .
Proof. The proof is based on the Schauder fixed point theorem. We refer the reader
to [10], [1], [8], [4] for precise definitions of the functional spaces used here. For w ∈
L2 (0, T ; L2 (Ω)) the mapping
t 7→ a(`(w(·, t)))
is clearly measurable and thus belongs to L∞ (0, T ). From a well known result of J.L. Lions
(cf. [10], [4]) there exists a unique u solution to


u ∈ L2 (0, T ; V ) ∩ C([0, T ], L2 (Ω)), ut ∈ L2 (0, T ; V 0 ),




u(·, 0) = u0 ,
Z
(2.4)
d

(u, v) + a(`(w(t))) ∇u∇v dx + (u, v)


dt

Ω


= (f (t), v) in D0 (0, T ), ∀ v ∈ V.
(f (t) denotes the mapping f (·, t)). Then the idea is to show that the map
w 7→ u = R(w)
has a fixed point. For that taking v = u in (2.4) we get
Z
1d 2
|u| + a(`(w(t))) |∇u|2 dx + |u|22 = (f (t), u) ≤ |f (t)|2 |u|2
2 dt 2
Ω
(2.5)
(2.6)
(| · |2 denotes the usual L2 (Ω)-norm, | · | the usual euclidean norm).
From (1.1) and Young’s inequality we derive
¯
¯2
1d 2
1
1
|u|2 + m¯|∇u|¯2 + |u|22 ≤ |f (t)|22 + |u|22 .
2 dt
2
2
Hence
¯2
©¯
ª
d 2
|u|2 + m0 ¯|∇u|¯2 + |u|22 ≤ |f (t)|22
dt
where m0 = min(2m, 1). Integrating on (0, t) we obtain for t ∈ [0, T ]
|u(t)|22 + m0 |u|2L2 (0,t;V ) ≤ |f |2L2 (0,t;L2 (Ω)) + |u0 |22 .
(2.7)
(2.8)
(2.9)
(We denote by | · |L2 (0,T ;X) the usual norm in L2 (0, T ; X), cf. [10], [1], the norm in V is
the usual H 1 (Ω)-norm). Thus we have
|u|L2 (0,T ;V ) ≤ C
(2.10)
where C = {(|f |2L2 (0,T ;L2 (Ω)) + |u0 |22 )/m0 }1/2 is a constant independent of w. Writing the
last equation of (2.4) as
ut − a(`(w(t)))∆u + u = f
in V 0 ,
(2.11)
since f ∈ L2 (0, T ; L2 (Ω)), u ∈ L2 (0, T ; V ) then ut ∈ L2 (0, T ; V 0 ) and it holds that
e
|ut |L2 (0,T ;V 0 ) ≤ C
(2.12)
B = { v ∈ L2 (0, T ; L2 (Ω)) | |v|L2 (0,T ;L2 (Ω)) ≤ C }
(2.13)
e independent of w. We may then consider
for some constant C
where C is the constant in (2.10). It is then clear that the mapping w 7→ u = R(w) maps
B into itself. In order to apply the Schauder fixed point theorem we need to verify the
following:
(i) R(B) is relatively compact in B,
(ii) R is continuous in L2 (0, T ; L2 (Ω)).
(i) is obvious from (2.10), (2.12) (cf. [13]). The point (ii) follows the lines of [8] and is left
to the reader. This completes the proof of the theorem.
Regarding uniqueness, under the assumption of Theorem 2.1 we have:
Theorem 2.2. Suppose that a is locally Lipschitz continuous – i.e.
∀ z > 0 ∃ Az > 0 such that |a(ζ) − a(ζ 0 )| ≤ Az |ζ − ζ 0 |
∀ ζ, ζ 0 ∈ [−z, z].
(2.14)
Then, the problem (2.3) admits a unique solution.
Proof. Let u1 , u2 be two solutions of (2.3) with the same initial data u0 . We have
Z
d
(u1 , v) + a(`(u1 (t))) ∇u1 ∇v dx + (u1 , v)
dt
Ω
Z
(2.15)
d
0
= (u2 , v) + a(`(u2 (t))) ∇u2 ∇v dx + (u2 , v) in D (0, T ), ∀ v ∈ V.
dt
Ω
This leads to
d
(u1 − u2 , v) + a(`(u1 (t)))
dt
Z
∇(u1 − u2 )∇v dx + (u1 − u2 , v)
Z
= {a(`(u2 (t))) − a(`(u1 (t)))} ∇u2 ∇v dx ∀ v ∈ V.
Ω
(2.16)
Ω
2
Since u1 , u2 ∈ C([0, T ], L (Ω)) we have for some z > 0
`(ui (t)) ∈ [−z, z],
i = 1, 2.
Taking v = u1 − u2 in (2.16) we obtain easily
Z
1d
2
|u1 − u2 |2 + a(`(u1 (t))) |∇(u1 − u2 )|2 dx + |u1 − u2 |22
2 dt
Ω
¯
¯¯
¯
≤ Az |`(u2 (t)) − `(u1 (t))| ¯|∇u2 |¯2 ¯|∇(u1 − u2 |¯2
¯
¯¯
¯
≤ Az |g|2 ¯|∇u2 |¯ ¯|∇(u1 − u2 )|¯ |u1 − u2 |2
2
2
(we used Cauchy–Schwarz inequality). Using (1.1) and the Young inequality
1 2
m
b ∀ a, b ≥ 0
ab ≤ a2 +
2
2m
it comes
¯
¯2
1d
|u1 − u2 |22 + m¯|∇(u1 − u2 )|¯2 + |u1 − u2 |22
2 dt
¯2
¯2
1 2 2 ¯¯
m¯
Az |g|2 |∇u2 |¯2 |u1 − u2 |22
≤ ¯|∇(u1 − u2 )|¯2 +
2
2m
which leads to
d
|u1 − u2 |22 ≤ C(t)|u1 − u2 |22
dt
with
¯
¯2
1
C(t) = A2z |g|22 ¯|∇u2 |¯2 ∈ L1 (0, T ).
m
Rewriting (2.19) as
ª
d © − R t C(s) ds
e 0
|u1 − u2 |22 ≤ 0
dt
this shows that t 7→ e−
Rt
0
C(s) ds
(2.17)
(2.18)
(2.19)
(2.20)
|u1 − u2 |22 is nonincreasing. Since for t = 0
u1 (·, 0) = u2 (·, 0) = u0
this function vanishes at 0 and thus identically. This completes the proof of uniqueness.
Remark 2.1. For f = f (x) ∈ L2 (Ω) we have
f ∈ L2 (0, T ; L2 (Ω))
(2.21)
for any T > 0. Thus, in this case, under the assumptions of Theorem 2.2, for any
T > 0 there exists a unique solution to (2.3). This is from now on what we will call a
weak solution to (1.2) and this is the asymptotic behaviour of this solution that we will
investigate.
3. Stationary solutions
To study the asymptotic behaviour of the solution obtained in the preceding section we
start by investigating the corresponding stationary problem
(
−a(`(u))∆u + u = f in Ω,
(3.1)
∂u
u = 0 on ΓD ,
=
0
on
Γ
,
N
∂ν
with
`(u) =
Z
g(x)u(x) dx
and f, g ∈ L2 (Ω). Recalling the definition
V = { v ∈ H 1 (Ω) | v = 0 on ΓD }
the problem (3.1) can be written in a weak form as

u ∈ V, Z
Z
Z
a(`(u)) ∇u∇v dx + uv dx =
f v dx ∀ v ∈ V.
Ω
(3.2)
Ω
Ω
(3.3)
(3.4)
Ω
The idea to solve (3.1), (3.4) is to rely on an algebraic equation in R. For any a > 0 we
consider ϕ = ϕa the solution to

−a∆ϕa + ϕa = f in Ω,
(3.5)
∂ϕa
ϕa = 0 on ΓD ,
= 0 on ΓN ,
∂ν
or in a weak form

ϕaZ ∈ V,
Z
Z
(3.6)
a ∇ϕa ∇v dx + ϕa v dx =
f v dx ∀ v ∈ V.
Ω
Ω
Ω
(The existence of ϕa is a straightforward consequence of the Lax–Milgram theorem).
Then, we have
Theorem 3.1. The mapping
u 7→ `(u)
is a one-to-one mapping from the set of solutions of (3.1), (3.4) to the set of solutions of
the equation in R
Z
µ=
gϕa(µ) dx
(3.7)
Ω
– i.e. solving (3.7) will provide all the solutions to (3.4).
Proof. Suppose that u is solution to (3.4). Due to (3.6) we have
u = ϕa(`(u)) .
Multiplying this equality by g and integrating over Ω we get
Z
Z
gϕa(`(u)) dx
gu dx =
`(u) =
Ω
Ω
(3.8)
which implies that `(u) belongs to the set of solution of (3.7). Let now µ solve (3.7) and
set
u = ϕa(µ) .
(3.9)
Multiplying by g and integrating over Ω we obtain
Z
Z
`(u) =
gu dx =
gϕa(µ) dx = µ.
Ω
Ω
Moreover, by (3.9)
u = ϕa(`(u))
and u is solution to (3.4). This shows that the map ` is onto. If `(u1 ) = `(u2 ) where u1 ,
u2 are solutions to (3.1), (3.4) one has of course u1 = u2 and the injectivity of ` follows.
This completes the proof of the theorem.
After having established the correspondence between the solutions to (3.1), (3.4) and
(3.7) we are able to analyze more deeply the stationary system (3.1). For that – recall
(3.7) – we define for a > 0
Z
K(a) =
g(x)ϕa (x) dx.
(3.10)
Ω
The properties of K are given by the following lemma:
Lemma 3.1. We have:
K is continuous on (0, +∞),
Z
lim K(a) =
g(x)f (x) dx,
a→0
(3.11)
(3.12)
Ω
if |ΓD | 6= 0,
(3.13)
Z
Z
lim K(a) = |Ω|− g dx − f dx if |ΓD | = 0.
a→+∞
Ω
Ω
R
R
(|ΓRD | denotes the measure area of ΓD . − Ω h dx is the average of h on Ω given by − Ω h dx =
1
h(x) dx, |Ω| is the measure of Ω).
|Ω| Ω
lim K(a) = 0
a→+∞
Proof: (i) Proof of (3.11). By (3.10) we have for a1 , a2 ∈ (0, +∞)
¯
¯Z
¯
¯
|K(a1 ) − K(a2 )| = ¯¯ g(ϕa1 − ϕa2 ) dx¯¯ ≤ |g|2 |ϕa1 − ϕa2 |2 .
(3.14)
Ω
Let us then get a priori bounds for the solution ϕ = ϕa of (3.6). Choosing in (3.6)
v = ϕa
we obtain
¯2
¯
a¯|∇ϕa |¯2 + |ϕa |22 = (f, ϕa ) ≤ |f |2 |ϕa |2 .
(3.15)
Thus, we get, neglecting the first term above:
and it comes
|ϕa |22 ≤ |f |2 |ϕa |2
|ϕa |2 ≤ |f |2 .
(3.16)
Reporting in (3.15) this leads also to
¯
¯
√
¯|∇ϕa |¯ ≤ |f |2 / a.
2
(3.17)
Let us come back to the proof of the continuity of K. We have by (3.6)
Z
Z
Z
f v dx ∀ v ∈ V,
a1 ∇ϕa1 ∇v dx + ϕa1 v dx =
Ω
Ω
Ω
Z
Z
Z
a2 ∇ϕa2 ∇v dx + ϕa2 v dx =
f v dx ∀ v ∈ V,
Ω
and by subtraction
a1
Ω
Z
Ω
Ω
Z
∇(ϕa1 − ϕa2 )∇v dx + (ϕa1 − ϕa2 )v dx
Ω
Z
∀ v ∈ V.
= (a2 − a1 ) ∇ϕa2 ∇v dx
(3.18)
Ω
Taking v = ϕa1 − ϕa2 we derive
¯
¯
¯2
¯¯
¯
a1 ¯|∇(ϕa1 − ϕa2 )|¯2 + |ϕa1 − ϕa2 |22 ≤ |a2 − a1 | ¯|∇ϕa2 |¯2 ¯|∇(ϕa1 − ϕa2 |¯2 .
Using (3.17) we obtain
¯
¯2
¯
¯ √
a1 ¯|∇(ϕa1 − ϕa2 )|¯2 + |ϕa1 − ϕa2 |22 ≤ |a2 − a1 | |f |2 ¯|∇(ϕa1 − ϕa2 )|¯/ a2 .
The second term above being nonnegative this leads to
¯
¯
¯|∇(ϕa1 − ϕa2 )|¯ ≤ |a2 − a1 | |f |2 /a1 √a2
2
(3.19)
(3.20)
and thus going back to (3.19) we obtain
|ϕa1 − ϕa2 |22 ≤ (a2 − a1 )2 |f |22 /a1 a2 .
By (3.14) it comes
(3.21)
√
|K(a1 ) − K(a2 )| ≤ |g|2 |f |2 |a1 − a2 |/ a1 a2
and the continuity (local Lipschitz continuity) of K follows.
(ii) Proof of (3.12). Since by (3.16) ϕa is bounded in L2 (Ω) independently of a, we have
for some “subsequence” of a and some ϕ0
ϕa * ϕ0
in L2 (Ω) when a → 0.
(3.22)
Taking v ∈ D(Ω) – the space of C ∞ -functions with compact support in Ω – in (3.6) we
get after integration by parts
Letting a → 0 this leads to
a(ϕa , ∆v) + (ϕa , v) = (f, v).
(3.23)
(ϕ0 , v) = (f, v) ∀ v ∈ D(Ω),
(3.24)
i.e. to ϕ0 = f . Since this is true for every “subsequence” of a → 0 we have shown that
ϕa * f
in L2 (Ω) when a → 0.
(3.12) follows by passing to the limit in the definition of K.
(3.25)
(iii) Proof of (3.13). First if |ΓD | 6= 0 then
¯
¯
¯|∇v|¯
(3.26)
2
is a norm on V equivalent to the H 1 (Ω) = norm (see for instance [5]). Thus from (3.17)
we deduce that when a → +∞
ϕa → 0 in H 1 (Ω)
(3.27)
and the first part of (3.13) follows. Next if |ΓD | = 0 – i.e. when V = H 1 (Ω) – from (3.16)
we derive that for a “subsequence” of a → +∞ and some ϕ∞ it holds that
in L2 (Ω) when a → +∞.
ϕa * ϕ∞
This implies that for any i = 1, . . . , n we have
∂ϕ∞
∂ϕa
→
in D0 (Ω).
∂xi
∂xi
Now from (3.17) we have
∂ϕa
→ 0 in L2 (Ω)
∂xi
for every i and by uniqueness of the limit in D0 (Ω) we derive that
∂ϕ∞
= 0 ∀ i = 1, . . . , n
∂xi
that is to say ϕ∞ is a constant. Taking now v = 1 in (3.6) we get
Z
Z
ϕa dx =
f dx
Ω
(3.28)
(3.29)
(3.30)
Ω
and letting the “subsequence” a go to +∞ we obtain
Z
Z
ϕ∞ |Ω| =
f dx ⇔ ϕ∞ = − f dx.
Ω
Ω
Since the limit is independent of the subsequence considered we have shown that
Z
ϕa * − f dx in L2 (Ω) when a → +∞.
Ω
Passing to the limit in the definition of K(a), this completes the proof of the lemma.
As a consequence we can prove the following:
Theorem 3.2. Suppose that
a is a continuous function from R into (0, +∞).
(3.31)
Then if f, g ∈ L2 (Ω) there exists a solution to (3.1), (3.4).
Proof. It follows from Theorem 3.1 that the problem reduces to solve the equation
µ = K(a(µ)).
(3.32)
Due to Lemma 3.1 the function µ 7→ K(a(µ)) is continuous and bounded. Then it holds
that
lim µ − K(a(µ))) = −∞,
µ→−∞
lim µ − K(a(µ)) = +∞
µ→+∞
(3.33)
and clearly one can find µ such that (3.32) holds. This completes the proof of the theorem.
We now would like to investigate the number of solutions to (3.1), (3.4). For that we
will use the following lemma:
Lemma 3.2. Let f be a function such that
f ∈ H 1 (Ω),
f ≥ 0,
Z
∇f ∇v dx ≥ 0 ∀ v ∈ V,
(3.34)
v ≥ 0,
(3.35)
a.e. in Ω.
(3.36)
Ω
then the mapping a 7→ ϕa is nonincreasing, i.e.
a1 ≥ a2
If in addition
∆f 6≡ 0 in Ω
⇒
ϕ a1 ≤ ϕ a 2
f 6≡ 0 on ΓD
or
∂f
6≡ 0 on ΓN ,
∂ν
or
(3.37)
then the mapping is decreasing in the sense that
a1 > a2
⇒
ϕ a1 < ϕ ai
a.e. in Ω.
(3.38)
Proof. Let us first establish (3.36). We claim first that for any a ∈ (0, +∞) it holds that
(ϕa − f )+ = 0.
(3.39)
Indeed from (3.6) we obtain
Z
Z
Z
a ∇(ϕa − f )∇v dx + (ϕa − f )v dx = −a ∇f ∇v dx ∀ v ∈ V.
Ω
Ω
(3.40)
Ω
Since f ≥ 0 we have – see [12], [4]
(ϕa − f )+ ∈ V
and by (3.40), (3.35) we get taking v = (ϕa − f )+
Z
Z
Z
+ 2
+2
a |∇(ϕa − f ) | dx + (ϕa − f ) dx = −a ∇f ∇(ϕa − f )+ ≤ 0
Ω
Ω
Ω
which leads to (3.39). Next, using (3.6) again we have
Z
Z
a ∇ϕa ∇v dx = − (ϕa − f )v dx ∀ v ∈ V.
Ω
By (3.39) it follows that it holds that
Z
Z
a ∇ϕa ∇v dx = (ϕa − f )− v dx ≥ 0 ∀ v ∈ V, v ≥ 0.
Ω
(3.41)
Ω
(3.42)
Ω
We then go back to (3.18) we have for a1 > a2
Z
Z
a1 ∇(ϕa1 − ϕa2 )∇v dx + (ϕa1 − ϕa2 )v dx
Ω
Ω
Z
= (a2 − a1 ) ∇ϕa2 ∇v dx ∀ v ∈ V.
Ω
(3.43)
Taking v = (ϕa1 − ϕa2 )+ by (3.42) we derive
Z
Z
+ 2
a1 |∇(ϕa1 − ϕa2 ) | dx + (ϕa1 − ϕa2 )+2 dx ≤ 0
Ω
Ω
that is to say
(ϕa1 − ϕa2 )+ = 0
⇔
ϕ a1 ≤ ϕ a2 .
(3.44)
This completes the proof of the first part of the theorem. For the second part we notice
that by (3.43) it holds that – take v ∈ D(Ω):
−a1 ∆(ϕa1 − ϕa2 ) + ϕa1 − ϕa2 = (a1 − a2 )∆ϕa1
in D0 (Ω).
(3.45)
Now, from (3.6) we have also if we take a = a1
−a1 ∆ϕa1 = −(ϕa1 − f )
(3.46)
and by (3.39)
∆ϕa1 = −(ϕa1 − f )− /a1 ∈ H 1 (Ω)
and
∆ϕa1 ≤ 0.
If ∆ϕa1 6≡ 0 by (3.45) and the maximum principle we have
ϕ a1 < ϕ a2
a.e. in Ω
and we are done (see remark below). But if
∆ϕa1 ≡ 0
from (3.46) we derive
f = ϕ a1
and thus
∂f
= 0 on ΓN
∂ν
which contradicts (3.37). This completes the proof of the lemma.
∆f = 0 in Ω,
f = 0 on ΓD ,
Remark 3.1. The function u = ϕa1 − ϕa2 satisfies – see (3.45)
Then, we claim that if
it holds that
u ≤ 0,
−a1 ∆u + u = h ∈ L2 (Ω),
h ≤ 0,
h ≤ 0.
(3.47)
h 6≡ 0
(3.48)
u < 0 a.e. in Ω.
(3.49)
2,n
(This is what we used above). If u ∈ Wloc
(Ω) the result follows from Theorem 9.6 of [11].
2
If one assumes only h ∈ L (Ω) the result seems to be not available in the literature. To
convince the reader that it holds also in this case we can proceed as follows. If h ∈ L2 (Ω),
h ≤ 0 it is possible to find a sequence of simple functions hk such that
h ≤ hk ≤ 0,
hk & h a.e. x ∈ Ω.
(3.50)
Of course since h 6≡ 0 we can assume hk 6≡ 0. Thus if we introduce uk the solution to

uk Z∈ V,
Z
Z
(3.51)
a1 ∇uk ∇v dx + uk v dx =
hk v dx ∀ v ∈ V,
Ω
Ω
Ω
p
since hk ∈ L (Ω), p > n it holds that uk ∈
2,n
Wloc
(Ω)
and thus (cf. Theorem 9.6 of [11])
uk < 0 a.e. in Ω.
Since u is solution to

u ∈Z V,
a1
Ω
∇u∇v dx +
+
Z
uv dx =
Ω
Z
Ω
hv dx ∀ v ∈ V,
(3.52)
taking v = (u − uk ) in (3.51), (3.52) one derives easily that
u ≤ uk < 0 a.e. in Ω
which completes the proof.
As a consequence of Lemma 3.2 we have:
Lemma 3.3. Let f be a function satisfying (3.5)–(3.7). Suppose that
g ≥ 0,
g 6≡ 0
(3.53)
then it holds that
K is decreasing on (0, +∞).
(3.54)
Proof. This follows immediately from (3.38), (3.53) and the definition of K.
Remark 3.2. Somehow (3.54) cannot hold under the simple assumption that
f, g ≥ 0,
f, g 6≡ 0,
(3.55)
(cf. [14], [15]). Indeed, suppose for instance that (3.55) holds and that f , g have disjoint
supports. Then, by Lemma 3.1 we have
Z
lim K(a) =
g(x)f (x) dx = 0,
lim K(a) ≥ 0, K > 0
a→0
a→+∞
Ω
and thus K cannot be decreasing. By a perturbation argument since K depends continuously of f and g, K could fail also to be decreasing for f, g > 0.
Remark 3.3. For a ∈ (0, +∞) let us denote by ψa the weak solution to

−a∆ψa + ψa = g in Ω,
∂ψa
ψa = 0 on ΓD ,
= 0 on ΓN ,
∂ν
i.e.

ψaZ ∈ V,
Z
Z
a ∇ψa ∇v dx + ψa v dx =
gv dx ∀ v ∈ V.
Ω
Ω
Ω
(3.56)
(3.57)
It holds that
K(a) =
Z
gϕa dx =
Ω
Z
f ψa dx.
(3.58)
Ω
Indeed taking v = ψa in (3.6) and v = ϕa in (3.57) it comes
Z
Z
Z
Z
a ∇ψa ∇ϕa dx + ψa ϕa dx =
gϕa dx =
f ψa dx
Ω
Ω
Ω
(3.59)
Ω
which is (3.59). Then, f and g are playing a symmetric rôle and assuming that
f ≥ 0,
f ≡0
(3.60)
(3.34)–(3.36) holds for g in place of f ,
(3.61)
K is decreasing.
(3.62)
we have that
This is indeed a consequence of (3.58) since by (3.61), a 7→ ψa is decreasing. In particular
if |ΓD | 6= 0 then K is decreasing for
f = 1,
g ≥ 0,
g 6≡ 0,
or g = 1,
f ≥ 0,
f 6≡ 0.
(3.63)
(3.34), (3.35), (3.37) holds for f = 1 if |ΓD | 6= 0).
Let us see what can happen in the case where K is decreasing regarding the solutions
to (3.1). Suppose
for example that |ΓD | 6= 0 so that K is a one-to-one mapping from
R
(0, +∞) into ( Ω f 0 dx, g) – see Lemma 3.1. Denote by K −1 the inverse of K. Then the
equation (3.7) can be written – see (3.32) –
µ = K(a(µ))
⇔
a(µ) = K −1 (µ)
(3.64)
and (3.1) can have then a unique solution, several solutions, a continuum of solutions
depending on the number of solutions to (3.64) – see Theorem 3.1. The different situations
are explained in the figures below.
Z
f g dx
Ω
a
K −1
K
Z
µ
f g dx
Ω
Figure 3.1. A case with a single solution
µ2
µ1
Z
µ
f g dx
Ω
Figure 3.2. A case with two solutions
K −1
a
µ1 µ2
Z
µ
f g dx
Ω
Figure 3.3. A case with a continuum of solutions
4. Asymptotic behaviour
In this section we consider u = u(x, t) the weak solution to (1.2) – see Remark 2.1. We
would like to study the asymptotic behaviour of u(x, t) when t → +∞. Of course the
general situation is very complicated and we have to restrict ourselves to special cases.
4.1. A simple example. Let us consider u = u(x, t) solutions to

µZ
¶



u −a
u(x, t) dx ∆u + u = f in Ω × R+ ,

 t
Ω
∂u
= 0 on Γ × R+ ,



∂ν

u(·, 0) = u ,
0
(4.1)
i.e. u is solution to (1.2) with ΓD = ∅, Γ = ΓN , g ≡ 1. Assuming that a is locally Lipschitz
continuous it follows from Theorems 2.1, 2.2 that a unique weak solution to (4.1) does
exist. A stationary solution associated to (4.1) is a function u∞ such that
¶
 µZ

−a
u∞ (x) dx ∆u∞ + u∞ = f in Ω,
Ω
(4.2)

 ∂u∞ = 0 on Γ.
∂ν
Since ΓD = ∅ the condition (3.37) fails for f, g = 1. However, it is easy to show that (4.2)
admits a unique solution. Considering for instance (4.2) under its weak formulation

1

uµ
∞ ∈ V = H (Ω),

¶Z
Z



∇u∞ ∇v dx
u∞ (x) dx
a
(4.3)
ΩZ
Ω Z





+ u∞ v dx =
f v dx ∀ v ∈ H 1 (Ω),
Ω
we get by taking v = 1
Ω
Z
Ω
u∞ (x) dx =
Z
f dx
Ω
and u∞ is the unique solution to
¶
 µZ

−a
f (x) dx ∆u∞ + u∞ = f
in Ω,
(4.4)

 ∂u∞ = 0 on Γ.
∂ν
Similarly taking v = 1 in the weak formulation of (4.1) – or integrating the equation (4.1)
on Ω – we obtain:
¾ Z
½Z
Z
d
u(x, t) dx + u(x, t) dx =
f (x) dx
dt Ω
Ω
Ω
½ Z
¾
(4.5)
Z
d t
t
⇐⇒
u(x, t) dx = e
e
f (x) dx.
dt
Ω
Ω
Ω
Integrating in t between 0 and t it comes
Z
Z
Z
−t
−t
u(x, t) dx = e
u0 (x) dx + (1 − e ) f (x) dx.
Ω
Ω
Thus when t → +∞ it holds that
Z
Ω
u(x, t) dx →
(4.6)
Ω
Z
f (x) dx.
(4.7)
Ω
Since a is continuous it follows that
¶
¶
µZ
µZ
f (x) dx
u(x, t) dx → a
a
(4.8)
Ω
Ω
and it is then relatively easy to show that
u(·, t) → u∞
in L2 (Ω)
(4.9)
when t → +∞. We refer the reader to [8] or [4] for a proof.
4.2. A case with a single equilibrium. In this section we suppose that we are under
the conditions of Theorems 2.1, 2.2 and we denote by u = u(x, t) the unique weak solution
to (1.2) – see Remark 2.1. Moreover, let us assume that (3.34), (3.35), (3.37) holds in
such a way that K is a decreasing function. Finally, let us suppose that there exists a
unique µ∞ solution to
a(µ∞ ) = K −1 (µ∞ )
(4.10)
i.e. there exists a unique stationary solution to (1.2) given by u∞ where u∞ satisfies

−a(µ∞ )∆u∞ + u∞ = f in Ω,
(4.11)
∂u∞
u∞ = 0 on ΓD ,
= 0 on ΓN .
∂ν
We would like to show that in these conditions, when t → +∞ it holds that
u(·, t) → u∞
in L2 (Ω).
(4.12)
For that let us set
`0 = lim inf `(u(·, t)),
t→+∞
L0 = lim sup `(u(·, t)),
(4.13)
M0 = Sup a(ξ).
(4.14)
t→+∞
m0 = Inf a(ξ),
[`0 ,L0 ]
[`0 ,L0 ]
Then, we have the following lemma.
Lemma 4.1. Under the above conditions it holds that
`(ϕM0 ) ≤ `0 ≤ L0 ≤ `(ϕm0 ).
Proof. Recall that for a > 0, ϕa denotes the solution to (3.6) – i.e. to

ϕaZ ∈ V,
Z
Z
a ∇ϕa ∇v dx + ϕa v dx =
f v dx ∀ v ∈ V.
Ω
Ω
(4.15)
(4.16)
Ω
By the definition of `0 and L0 have
`0 = Sup inf `(u(·, t)),
t0
t≥t0
L0 = Inf Sup `(u(·, t)).
t0 t≥t0
(4.17)
Let ε > 0 be fixed. By the above definitions for t0 = t0 (ε) large enough we have
`0 − ε ≤ `(u(·, t)) ≤ L0 + ε ∀ t ≥ t0 .
(4.18)
This implies if we set u(·, t) = u(t)
m0 − δ(ε) ≤ a(`(u(t))) ≤ M0 + δ(ε) ∀ t ≥ t0
(4.19)
for some δ = δ(ε) such that
lim δ(ε) = 0.
ε→0
(4.20)
Let us prove the first inequality of (4.15). From the weak formulation of (1.2) and (4.16)
written for a = M0 + δ we derive
Z
d
(u − ϕM0 +δ , v) + a(`(u(t))) ∇(u − ϕM0 +δ )∇v dx + (u − ϕM0 +δ , v)
dt
ZΩ
(4.21)
= {M0 + δ − a(`(u(t)))} ∇ϕM0 +δ ∇v dx ∀ v ∈ V,
Ω
(this equality holding for instance in D0 (0, +∞).) Taking
v = −(u − ϕM0 +δ )−
(4.22)
it comes
¯
¯2
1d
|(u − ϕM0 +δ )− |22 + a(`(u(t)))¯|∇(u − ϕM0 +δ )− |¯2 + |(u − ϕM0 +δ )− |22
2 dt
Z
= {a(`(u(t))) − (M0 + δ)}
For t ≥ t0 we have by (4.19) and (3.42)
a(`(u(t))) − (M0 + δ) ≤ 0,
Ω
(4.23)
∇ϕM0 +δ ∇(u − ϕM0 +δ )− dx.
Z
Ω
∇ϕM0 +δ ∇(u − ϕM0 +δ )− dx ≥ 0.
It follows that for t ≥ t0
¯2
¯
1d
|(u − ϕM0 +δ )− |22 + a(`(u(t))¯|∇(u − ϕM0 +δ )− |¯2
2 dt
+ |(u − ϕM0 +δ )− |22 ≤ 0
(4.24)
d
|(u − ϕM0 +δ )− |22 + 2|(u − ϕM0 +δ )− |22 ≤ 0
dt
d 2t
{e |(u − ϕM0 +δ )− |22 } ≤ 0.
dt
(4.25)
and thus
⇐⇒
From this we derive that
=⇒
e2t |(u − ϕM0 +δ )− |22 ≤ e2t0 |(u(t0 ) − ϕM0 +δ )− |22
|(u(t) − ϕM0 +δ )− |2 ≤ e−(t−t0 )|(u(t0 ) − ϕM0 +δ )− |2 ,
Thus, we have when t → +∞,
From
t ≥ t0 .
(u(t) − ϕM0 +δ )− → 0 in L2 (Ω).
(4.26)
u − ϕM0 +δ = (u − ϕM0 +δ )+ − (u − ϕM0 +δ )− ≥ −(u − ϕM0 +δ )−
we derive since g ≥ 0, g ∈ L2 (Ω)
`(u − ϕM0 +δ ) ≥ −`((u − ϕM0 +δ )− ) ≥ −ε
(4.27)
`(u(t)) ≥ `(ϕM0 +δ ) − ε.
(4.28)
`0 = lim inf `(u(t)) ≥ `(ϕM0 +δ ) − ε.
(4.29)
for t large enough. Thus for t ≥ t1 = t1 (ε) we have
Passing to the limit this implies
t→+∞
Letting now ε → 0 the first inequality of (4.15) follows. The second inequality of (4.15)
is clear. The third can be obtained as above and for the sake of completeness we outline
briefly the proof. From the weak formulation of (1.2) and (4.16) written for a = m0 − δ
(we suppose ε small enough in such a way that m0 > δ) we derive as in (4.21):
Z
d
(u − ϕm0 −δ , v) + a(`(u(t))) ∇(u − ϕm0 −δ )∇v dx + (u − ϕm0 −δ , v)
dt
Ω
Z
(4.30)
0
= {(m0 − δ) − a(`(u(t)))} ∇ϕm0 −δ ∇v dx ∀ v ∈ V, in D (0, +∞).
Ω
Taking v = (u − ϕm0 −δ )+ and using (4.19), (3.42) we get for t ≥ t0
1d
|(u − ϕm0 −δ )+ |22 + |(u − ϕm0 −δ )+ |22 ≤ 0
2 dt
which implies as in (4.26) that
(u − ϕm0 −δ )+ → 0 in L2 (Ω)
(4.31)
as t → +∞. Thus arguing as in (4.27) we obtain
`(u − ϕm0 −δ ) ≤ `((u − ϕm0 −δ )+ ) ≤ ε
for t large enough and thus
L0 = lim sup `(u(t)) ≤ `(ϕm0 −δ ) + ε.
t→+∞
Letting ε → 0 the third inequality in (4.15) follows. Note that we have used here the
continuity of the mapping a 7→ ϕa which results from (3.21). This completes the proof of
the lemma.
In addition to the above assumptions we are going to assume that
a(µ) ≥ a(µ∞ ) ∀ µ ≤ µ∞
or
a(µ) ≤ a(µ∞ ) ∀ µ ≥ µ∞ ,
(4.32)
that is to say that we are in one of the cases of the figures below.
K −1
K −1
a
a
Figure 4
µ
µ∞
Figure 5
µ∞
µ
Remark 4.1. Note that (4.32) needs only to be satisfied on the domain of definition of
K −1 .
Then we can prove the following:
Theorem 4.1. Under the above assumptions we have when t → +∞
u(·, t) → u∞
in L2 (Ω)
(4.33)
where u∞ is the unique solution to (4.11).
Proof. If we can show that
lim `(u(t)) = µ∞
(4.34)
lim a(`(u(t))) = a(µ∞ )
(4.35)
t→+∞
we will have
t→+∞
and the result will follow – see [8], [4]. To prove (4.34) it is enough to show that
` 0 = L 0 = µ∞ .
(4.36)
` 0 = L0 .
(4.37)
` 0 < L0 .
(4.38)
Let us first show that
If not we have
Case 1. µ∞ ≤ `0 < L0
Denote by m1 a point of [`0 , L0 ] such that
m0 = Inf a(ξ) = a(m1 ).
(4.39)
L0 ≤ `(ϕm0 ) = `(ϕa(m1 ) ) = K(a(m1 )) = µ∞
(4.40)
[`0 ,L0 ]
If m1 = µ∞ , by (4.15) we have
which is impossible (recall that K(a) = `(ϕa )). If m1 > µ∞ , by (4.15) we have now
L0 ≤ `(ϕm0 ) = `(ϕa(m1 ) ) = K(a(m1 )) < m1
(4.41)
which contradicts the definition of m1 . In the above, the last inequality follows of the
uniqueness of the solution to
µ∞ = K(a(µ∞ )).
Indeed due to this uniqueness we have
µ > K(a(µ)) ∀ µ > µ∞ ,
Thus, the case 1 above is impossible.
µ < K(a(µ)) ∀ µ < µ∞ .
(4.42)
Case 2. ` < L0 ≤ µ∞
Let us denote by M1 a point of [`0 , L0 ] such that
M0 = Sup a(ξ) = a(M1 ).
(4.43)
[`0 ,L0 ]
If M1 = µ∞ , by (4.15) it holds that
`0 ≥ `(ϕM0 ) = `(ϕa(M1 ) ) = K(a(M1 )) = µ∞
(4.44)
`0 ≥ `(ϕM0 ) = `(ϕa(M1 ) ) = K(a(M1 )) > M1
(4.45)
which is impossible. If M1 < µ∞ then
(by (4.39)). This is also impossible and this case 2 cannot occur.
Case 3. `0 < µ∞ < L0
Define m1 and M1 as before. Suppose for instance that we are in the case of Figure 4 i.e.
a(µ) ≥ a(µ∞ ) ∀ µ ≤ µ∞ .
Then without loss of generality we can assume
m1 ∈ [µ∞ , L0 ].
(4.46)
If m1 = µ∞ we have (4.40) and a contradiction. If m1 > µ∞ we have (4.41) and another
impossibility. In the case of Figure 5 one argues similarly using M1 . Thus we have
established the impossibility of (4.38) and we have
` 0 = L0 .
(4.47)
`0 = L0 ≤ `(ϕa(`0 ) ) = K(a(`0 ))
(4.48)
In this case (4.15) becomes
and thus
` 0 = L 0 = µ∞ .
This completes the proof of the theorem.
Remark 4.2. If we drop the assumption (4.32) and consider an arbitrary a, the analysis
of the cases 1 and 2 together with (4.48) shows that we have in this case
` 0 < µ∞ < L 0
or
` 0 = L 0 = µ∞ .
(4.49)
4.3. The case of two equilibria. In this section we assume that we are in the case of
Figure 3.2. In particular we will assume
f satisfies (3.34), (3.35), (3.37),
(4.50)
g > 0 in Ω,
(4.51)
a(µi ) = K −1 (µi ) i = 1, 2,
a(µ) > K −1 (µ) ∀ µ ∈ (µ1 , µ2 ),
a(µ2 ) ≤ a(µ) ≤ a(µ1 ) ∀ µ ∈ (µ1 , µ2 ).
Let us denote by ui = ϕa(µi ) , i = 1, 2 the solution to

−a(`(ui ))∆ui + ui = f in Ω,
∂ui
ui = 0 on ΓD ,
= 0 on ΓN .
∂ν
Due to (3.38) we have
u1 < u2 .
(4.52)
(4.53)
(4.54)
(4.55)
We consider then u solution to (1.2) with u0 such that
We have:
u1 ≤ u 0 ≤ u 2 ,
u0 6= u2 .
(4.56)
Proposition 4.1. Let u be the weak solution to (1.2). Under the above assumptions it
holds that
u1 ≤ u(·, t) ≤ u2
∀ t.
(4.57)
Proof. Let us denote by E the set
E = { t | `(u(s)) ∈ [µ1 , µ2 ] ∀ s ≤ t }.
(4.58)
t∗ = Sup{ t | t ∈ E }.
(4.59)
By (4.56), E contains 0 (recall that g ≥ 0, `(ui ) = µi ). Set
By continuity of the mapping t 7→ u(t) in L2 (Ω), t 7→ `(u(t)) is continuous and
`(u(t∗ )) ∈ [µ1 , µ2 ],
t∗ ∈ E.
(4.60)
We claim next that
u1 ≤ u(t) ≤ u2
∀ t ∈ [0, t∗ ].
(4.61)
Suppose that we want to prove the left hand side inequality. Using the weak formulation
of (1.2) and (4.54) we have in D0 (0, t∗ )
Z
d
(u − u1 , v) + a(`(u(t))) ∇(u − u1 )∇v dx + (u − u1 , v)
dt
Ω
Z
(4.62)
= {a(µ1 ) − a(`(u(t)))} ∇u1 ∇v dx ∀ v ∈ V.
Ω
∗
Due to (4.53), (4.60), (3.42) we have on (0, t )
a(µ1 ) − a(`(u(t))) ≥ 0,
Z
∇u1 ∇v dx ≥ 0 ∀ v ∈ V, v ≥ 0.
Ω
−
Taking v = −(u − u1 ) in (4.62) we get easily
1d
|(u − u1 )− |22 + |(u − u1 )− |22 ≤ 0.
(4.63)
2 dt
Since (u − u1 )− (0) = (u0 − u1 )− = 0 it follows from the Gronwall inequality – see for
instance (4.25) – that (u−u1 )− = 0 and thus u ≥ u1 . The right hand side of the inequality
(4.61) can be proven in the same way. This established (4.61). Next, by definition of t∗ ,
if t∗ < +∞ we have
`(u(t∗ )) = `(u1 ) or `(u2 ).
By (4.51), (4.61) this implies
u(t∗ ) = u1
or u2
and by the uniqueness of the solution to (1.2) this equality remains valid for further time
which contradicts the definition of t∗ . We have thus t∗ = +∞ and (4.61) gives (4.57).
This completes the proof of the proposition.
Remark 4.3. Since we did not use the inequality in (4.52) Proposition 4.1 remains valid
in the case of Figure 3.3. The assumption (4.51) could be relaxed using the dynamical
system theory – see [8].
We can now show
Theorem 4.2. Assume that we are under the conditions of Proposition 4.1 with in addition
f = g.
(4.64)
If u is the weak solution to (1.2) with u0 satisfying (4.56), when t → +∞, it holds that
u(·, t) → u,
in L2 (Ω).
We first exhibit a Lyapunov function for the problem. Indeed we have:
Lemma 4.2. |u(t)|22 is a Lyapunov function – i.e. decreases with time.
(4.65)
Proof. The weak formulation of (1.2) reads in D0 (0, +∞)
hut , vi − a(`(u(t)))h∆u, vi + (u, v) = (f, v) ∀ v ∈ V
(4.66)
(h·i is the duality bracket between V 0 and V ). Taking v = ϕa(`(u(t)) = ϕa we get
⇐⇒
⇐⇒
hut , ϕa i − ah∆u, ϕa i + (u, ϕa ) = (f, ϕa ) = (g, ϕa )
hut , ϕa i + (u, −a∆ϕa + ϕa ) = (g, ϕa ) = K(a)
hut , ϕa(`(u(t))) i = K(a(`(u(t))) − `(u(t)).
Since `(u(t)) ∈ [µ1 , µ2 ], by (4.52) it holds that
a(`(u(t))) ≥ K −1 (`(u(t)))
and thus
⇔
K(a(`(u(t))) ≤ `(u(t))
hut , ϕa(`(u(t))) i ≤ 0.
Combining (3.6) written for a = a(`(u(t))) and (4.66) we obtain in D0 (0, +∞):
Z
hut , vi + a(`(u(t))) ∇(u − ϕa )∇v dx + (u − ϕa , v) = 0 ∀ v ∈ V.
(4.67)
(4.68)
Ω
Choosing in the above equality v = u − ϕa = u − ϕa(`(u(t))) we get by (4.67)
Z
hut , ui = hut , ϕa i − a(`(u(t))) |∇(u − ϕa )|2 dx − |u − ϕa |22
Ω
¯
¯2
¯
≤ −a(`(u(t))) |∇(u − ϕa )|¯2 − |u − ϕa |22
(4.69)
i.e.
hut , ui =
1d
|u(t)|22 ≤ 0.
2 dt
This completes the proof of the lemma.
Proof of Theorem 4.2. Note that (4.69) holds in the distributional sense in D0 (0, +∞) or
a.e. t. Then we claim that
limsupesshut , ui = Inf essuphut , ui(t) = 0.
t0
t→+∞
(4.70)
t≥t0
Indeed – due to (4.69) – if it is not true, for t0 large enough it holds that
essuphut , ui ≤ −α < 0
t≥t0
for some positive α. Thus we have
d
|u(t)|22 ≤ −2α a.e. t ≥ t0 .
dt
Integrating between t0 and t we obtain
|u(t)|22 ≤ −2α(t − t0 ) + |u0 |22
(4.71)
(recall that u is continuous). But then (4.71) is impossible for t large enough (the L2 (Ω)norm of u is bounded since |u(t)|2 is nonincreasing). This shows (4.70). Going back to
(4.69) it follows that
¯
¯2
0 = limsupesshut , ui ≤ limsupess{−a(`(u(t)))¯|∇(u − ϕa )|¯2 − |u − ϕa |22 } ≤ 0. (4.72)
t→+∞
t→+∞
It follows – recall that u is continuous – that we have
lim inf |u(t) − ϕa(`(u(t))) |22 = 0.
t→+∞
Thus, there exists a sequence tn → +∞ such that
(4.73)
u(tn ) − ϕa(`(u(tn ))) → 0
(4.74)
in L2 (Ω).
(4.75)
as tn → +∞. Since u(tn ) is bounded in L2 (Ω) we can extract from tn a subsequence, for
simplicity still labelled tn , such that for some u∞
u(tn ) * u∞
The set
C = { v ∈ L2 (Ω) | u1 (x) ≤ v(x) ≤ u2 (x) a.e. x ∈ Ω }
(4.76)
is closed and convex in L2 (Ω). It is also weakly closed and by Proposition 4.1 and (4.75)
we derive
Moreover, from (4.74) we get
u1 ≤ u∞ ≤ u 2
in Ω.
u∞ = ϕa(`(u∞ ))
(4.77)
(4.78)
that is to say u∞ is a stationary point and by (4.77)
u∞ = u1
or u2 .
Since |u(t)|22 is nonincreasing – recall that ui ≥ 0 – by (4.56) we can only have
u∞ = u1 .
Since the weak limit of any subsequence is unique we have shown that
u(tn ) * u1
in L2 (Ω).
(4.79)
Next consider another sequence t0n → +∞ such that
u(t0n ) * v∞
in L2 (Ω).
Since u(t0n ) ∈ C – cf. Proposition 4.1 – we also have v∞ ∈ C and in particular
From (4.74), (4.79) we deduce that
v∞ ≥ u1 .
u(tn ) → u1
in L2 (Ω).
Since |u(t)|22 is nonincreasing it admits a limit when t → +∞ and this limit can only be
|u1 |22 . Thus, passing to the limit in the inequality
we get
|u(t0n )|22 − (u(t0n ), u1 ) = (u(t0n ), u(t0n ) − u1 ) ≥ 0
|u1 |22 − (v∞ , u1 ) = (u1 − v∞ , u1 ) ≥ 0.
Since u1 > 0, v∞ − u1 ≥ 0 this clearly imposes
v ∞ = u1 .
Thus, every subsequence of u(t) converging weakly towards u1 , we have as t → +∞
u(t) * u1
in L2 (Ω).
The strong convergence follows from the fact that
|u(t)|2 → |u1 |2 .
This completes the proof of the theorem.
Remark 4.4. Some other cases can be treated. For instance in the case of Figure 3.3
and f = g one can show for u0 satisfying u1 ≤ u0 ≤ u2 that u(·, t) converges toward an
equilibrium when t → +∞. One can consider also cases where a is below K −1 . To keep
the paper short we postpone these issues to forthcoming works – cf. [3].
Acknowledgements: This work was done partly when the first author was visiting the
University of Zürich and the second the Academia Sinica. We thank both institutions for
their hospitality. The authors would also like to acknowledge the support of the Swiss
National Science Foundation under the contract # 20-67618.02.
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N.-H. Chang, Academia Sinica, Institute of Mathematics, Taipei 11529 Taiwan
E-mail address: [email protected]
M. Chipot, Institut für Mathematik, Abt. Angewandte Mathematik, Universität Zürich,
Winterthurerstrasse 190, CH–8057 Zürich, Switzerland
E-mail address: [email protected]