1. No category

arXiv:1410.4042v1 [math.AP] 15 Oct 2014
EXISTENCE AND UNIQUENESS FOR A QUASILINEAR
ELLIPTIC PROBLEM WITH NONLINEAR ROBIN CONDITIONS
DJAMEL AITAKLI
Abstract. This paper deals with an existence and uniqueness result of the
weak solution for a quasilinear elliptic PDE with nonlinear Robin boundary
conditions.This problem is defined on a domain whose boundary is the union of
two disjoint surfaces.where a Dirichelet condition is prescribed on the exterior
boundary and a nonlinear Robin condition on the interior one.To achieve that,
we use the theory of monotone operators.
1. Introduction
In this paper we deal with the question of existence and uniqueness of the following
quasilinear elliptic problem in an open set Ω of Rn
(
−div(a(x, u, ∇u)) = f
Ω
(1)
a(x, u, ∇u) · n + h(x, u) = 0 Γ1
u=0
Γ0
Where Ω is an open bounded subset of RN of the form Ω = Ω0 \Ω1 . Ω0 ,Ω1 are two open
sets such that Ω1 represents a hole.
As it is seen, there is two boundary conditions.One is a Dirichelet condition on
Γ0 = ∂Ω0 , and the other is a nonlinear Robin condition on Γ1 = ∂Ω1 .n is the exterior normal to Ω.
The case considered in this paper is when the vector field a depends nonlinearly of ∇u.
Note that the existence in the particular case where the vector field a is linear with respect
to ∇u was done in [2] .
we prove the existence of a weak solution for the problem via the theory of monotone
operators which, in order to apply, we make coreciveness, boundedness and monotonicity
assumptions on the quasilinear term a.
Of course, the main difficulties come from the nonlinear charachter of both the quasilinear term and the second boundary condition on which we have to make adequate assumptions, we use a result proven in the above mentionned article which is based on the
study of the nemytski’i operator associated to h. Regarding uniqueness, we adapt to our
problem the same idea as in [2]. An interesting and classical example is the p-laplacean:
a(x, s, ζ) = A(x, s)|ζ|p−2 ζ
where p > 1 and A is a matrix field.
2. Assumptions and statement of the main result
We make the following assumptions. The source term
′
f ∈ Lp (Ω)
(2)
p > 1 and p′ :=
p
p−1
denote the conjugate of p.
Date: October 16, 2014.
2010 Mathematics Subject Classification. 35j.
Key words and phrases. Existence and uniqueness, quasilinear elliptic, Robin conditions.
1
2
DJAMEL AITAKLI
a : Ω × R × Rn −→ Rn
(x, s, ζ) −→ a(x, s, ζ)
is a caratheodory function i.e:
1.1) a(., s, ζ) is measurable, ∀(s, ζ) ∈ Rn+1
1.2) a(x, ., .) is continuous a.e x ∈ Ω
There exists a constant c > 0, an exponent p > 1 and a positive measurable function
α ∈ L∞ (Ω) such that:
|a(x, s, ζ)| ≤ cα(x)(1 + |s|p−1 + |ζ|p−1 )
(3)
a.e x ∈ Ω ,∀(s, ζ) ∈ Rn+1
There exists a function β : (x, s) −→ β(x, s) such that β(x, s) ≥ β0 > 0
and such that
a(x, s, ζ)ζ ≥ c−1 β(x, s)|ζ|p
(4)
a.e x ∈ Ω,∀(s, ζ) ∈ Rn+1
(5)
∀ζ1 , ζ2 ∈ Rn ζ1 6= ζ2 =⇒ (a(x, s, ζ1 ) − a(x, s, ζ2 ))(ζ1 − ζ2 ) >0
∀(x, s) ∈ Rn+1
∃θ > 0, ∃r ∈ ]0, 1], ∀s1 , s2 ∈ R a.e x ∈ Ω ∀ζ ∈ Rn
(6)
|a(x, s1 , ζ) − a(x, s2 , ζ)| ≤ θ|s1 − s2 |r (1 + |s1 |p−1−r + |s2 |p−1−r + |ζ|p−1−r )
The function h : Ω × R −→ R is such that
i) h(x, .) is increasing a.e x ∈ Ω,
ii) h(x, 0) = 0,a.e x ∈ Ω
iii) h(x, .) ∈ C 1 (R) a.e x ∈ Ω,
iv)
|h′ (x, s)| ≤ c(1 + |s|p∗−1 )
a.e x ∈ Ω with
1 ≤ p∗ ≤
N
if N > 2 and 1 ≤ p∗ < ∞ if N = 2
N −2
Set V := WΓ1,2
(Ω) ∩ W 1,p (Ω), which contains functions of the sobolev space H 1 (Ω)
0
with trace vanishing on Γ0
The set V is endowed with the norm kvkV := k∇vkLp (Ω) .Recall the statement of
poincare inequality:
kvkW 1,p (Ω) ≤ C(p, Ω)k∇vkLp (Ω)
∀v ∈ W 1,p (Ω),with v = 0 on Γ0 , we easily see that the two norms are equivalent
In order to use the trace, sobolev and compact imbbedding theorems, we assume that the
boundary Γ1 is Lipschitz. The variational formulation of problem (1) is:
F
R
R
R ind u ∈ V, ∀v ∈ V
(7)
f · v dx
h(x, u) · v dσ =
a(x, u, ∇u) · ∇v dx +
Ω
Γ1
Ω
The main result of this paper is :
Theorem 1. under the assumptions above, problem (7) admits a unique solution
EXISTENCE AND UNIQUENESS FOR A QUASILINEAR ELLIPTIC PROBLEM WITH NONLINEAR ROBIN CONDITIONS
3
(8)
This problem is equivalent to:
Where
< A(u), v >:=
Z
F ind u ∈ V, ∀v ∈ V
< A(u), v >= F (v).
a(x, u, ∇u) · ∇v dx +
Ω
Z
F (v) :=
f · v dx
Z
h(x, u) · v dσ
Γ1
Ω
3. Auxilliary results
First of all we need to show that the operators A and F are well defined:
Let u, v ∈ V .Using Holder inequality we obtain:
Z
| a(x, u, ∇u) · ∇v dx| ≤ ckαkL∞ ka(x, u, ∇u)kLp′ (Ω) k∇vkLp (Ω)
Ω
But, from hypothesis (3) we have:
p
p
p
p−1
|a(x, u, ∇u)| p−1 ≤ ckαkLp−1
+ |∇u|p−1 ) p−1
∞ (Ω) (1 + |u|
Z
p
p−1
p
≤ ckαkL∞ (Ω) ( (1 + |u|p−1 + |∇u|p−1 ) p−1 ) p dx
ka(x, u, ∇u)k p−1
L
(Ω)
Ω
Then, using the triangular inequality we obtain:
ka(x, u, ∇u)k
p
L p−1 (Ω)
p−1
≤ ckαkL∞ (C + kukp−1
Lp (Ω) + k∇ukLp (Ω) )
Since Ω is bounded,
ka(x, u, ∇u)k
p
L p−1 (Ω)
< +∞
On the other hand we know that:
W 1,2 (Ω) −→ Lq (Γ1 )
. We know from (proposition (3.1) in [2]) that,
(continuous embedding) ∀1 ≤ q ≤ (N−1)p
N−p
if h satisfies the above hypothesis, then the application
′
Φ : Lr (Γ1 ) −→ Lr (Γ1 )
u −→ h(u)
whith r :=
2(N−1)
N−2
is bounded and continuous.Write now, using holder inequality:
Z
h(x, u)v dσ ≤ kh(u)kLr′ (Γ1 ) kvkLr (Γ1 )
Γ1
We conclude using the trace theorem that:
Z
|
h(u)v dσ| ≤ ckvkV
Γ1
Again by the Holder inequality:
Z
| f v dx| ≤ kf kLp′ (Ω) kvkLp (Ω) ≤ ckvkV
Ω
It follows that the operators A : V −→ V ′ and F : V −→ R are well defined.We recall the
result that we will use to prove the existence of a solution of problem (1):
Theorem 2. Let A be an operator from a reflexive and separable Banach space V into
its dual such that, A is bounded, Type M, and satisfy the following property
(9)
∃ρ > 0, ∀v ∈ V , kvkV > ρ, < A(u), u >> F (u)
Then F ∈ Rg(A)
Before starting the proof, we state some lemmas
4
DJAMEL AITAKLI
Lemma 1. Let 1 < p < 2 then
∀ρ0 > 0, ∃ρ > 0, ∀v ∈ L2 (Ω), kvkL2 (Ω) > ρ =⇒ kvkLp (Ω) > ρ0
Proof. Suppose the converse .i.e.
∃ρ0 > 0, ∀ρ > 0, ∃vρ ∈ L2 (Ω), kvρ kL2 (Ω) > ρ and kvρ kLp (Ω) ≤ ρ0
it means that the sequence vρ is divergent in L2 (Ω) and bounded in Lp (Ω) which implies
that :
∃Ω′ ⊂ Ω, meas(Ω′ ) > 0, ∀g ∈ L2 (Ω),∃ρ > 0; g(x) ≤ vρ (x) a.e x ∈ Ω′ .
Since Ω is bounded, choose g(x) = ρ0 + 1 for x ∈ Ω′ , then, kρ0 + 1kLp (Ω) = ρ0 + 1 ≤
ρ0 .Contradiction. The lemma is proved
Lemma 2. The operator A′ : V −→ V ′
u −→ A′ (u) =
Z
a(x, u, ∇u) · ∇v dx
Ω
is type M
Proof. Let un ⇀ u, A(un ) ⇀ F , limsupA(un )un ≤ F u then ([4] lemma 6.3) we have
Gn −→ 0 in L1 (Ω) where
Z
Z
Gn =
(a(x, un , ∇un ) − a(x, un , ∇u)) · (∇un − ∇u) dx
Ω
Ω
Then passing to a subsequence Gn (x) −→ 0 a.e x ∈ Ω. Choose x ∈ Ω such that
un (x) −→ u(x) and Gn (x) −→ G(x) From hypothesis (3).we have
Gn (x) ≥ a(x, un (x), ∇un (x)) · ∇un (x) − Cx (1 + |∇un (x)|p−1 )
By hypothesis (4).|∇un (x)| is bounded. if ζ is an acumulation point of |∇un (x)| passing
to the limit
(a(x, u, ∇u) − a(x, u, ζ)) · (∇u − ζ) = 0
by hypothesis (5) we deduce that ζ = ∇u,repeating the same reasoning to any subsequence,
′
we conclude that ∇un (x) −→ ∇u(x) a.e x ∈ Ω Since a(x, un , ∇un ) is bounded in Lp (Ω)
′
then we infer by proposition 3.4 in [4] that a(x, un , ∇un ) ⇀ a(x, u, ∇u) weakly in Lp (Ω)
by hypothesis Aun ⇀ F and by the uniqueness of the weak limit we conclude that Au =
F .This ends the proof of the lemma
We need the following lemma in the proof of uniqueness
Lemma 3. Let uζ , g be measurable functions such that uζ ∈ Lp (Ω) and g ∈ L∞ (Ω) then
if
kuζ gkL1 (Ω) ≤ C∗
we have
kuζ kLp (Ω) ≤ C
The constants C and C∗ are independant of ζ.
Proof. Suppose the converse i.e
∀δ > 0 ∃ζδ ∈ N kuζδ kLp (Ω) > δ
This implies the existence of a divergent subsequence uζδ then,
∃Ω′ meas(Ω′ ) ∀L ∈ Lp (Ω′ ) > 0 ∃ζδ uζδ (x) > L(x)
a.e in Ω′ .Choose L =
2C∗
kgkL1 (Ω)
then
∃nδ 2C∗ ≤ kuζδ gkL1 (Ω) ≤ C∗
The contradiction proves the lemma
We will use the following lemma
Lemma 4. Let T be a distribution on D(Ω) such that ∇T ∈ Lp (Ω) then T ∈ Lp (Ω)
EXISTENCE AND UNIQUENESS FOR A QUASILINEAR ELLIPTIC PROBLEM WITH NONLINEAR ROBIN CONDITIONS
5
Rx
Proof. Indeed, Set fi := ∂xi T and gi (x) := 0 i fi (t, x′ ) dt then ∂xi gi = fi We have by
using Holder inequality
Z
kgi kpLp (Ω) ≤ c
|fi |p dx
Ω
We conclude that T = gi ∈ Lp (Ω)
4. Proof of the main result
Proof. First, we prove that A is bounded.Let S be a bounded subset of V .i.e. ∃a > 0,
such that,
kukV ≤ a
∀v ∈ S
Let u ∈ S, we know from above that:
ka(x, u, ∇u)k
p
L p−1 (Ω)
and
≤ a′
kh(u)kLr′ (Γ1 ) ≤ a′′
The norm of A is defined as:
kA(u)kV ′ = supv∈V,kvkV =1 |< A(u), v >|
From the preceding, using the Poincare inequality and the trace theorem, we have
|< A(u), v >V ′ ×V | ≤ ckvkV
so that
kA(u)kV ′ ≤ c
c is independent of u.We conclude that the image of the set S under the operator A is
bounded.
Now, we prove that the operator A satisfies the estimate (9)
p
1
Set ρ0 := (C(p, Ω)) p−1 (cβ0−1 kf kLp′ (Ω) ) p−1 then, by lemma, (1)
∃ρ > 0 kvkV > ρ =⇒ kvkLp (Ω) > ρ0
Using hypothesis (4) and since h is increasing:
Z
Z
a(x, u, ∇u) · ∇u dx +
h(x, u)u dσ > c−1 β0 k∇ukpLp (Ω)
Ω
Γ1
−1
< A(u), u >> c
β0 C(p, Ω)−p kukW 1,p (Ω) kukp−1
W 1,p (Ω)
< A(u), u >> c−1 β0 C(p, Ω)−p kukLp (Ω) ρp−1
0
< A(u), u >> kf kLp′ (Ω) kukLp (Ω
< A(u), u >> kukLp (Ω) kf kLp′ (Ω)
Z
< A(u), u >>
f · u dx
Ω
Let us now prove that the operator A is Type M. We know from [4] that the sum of
a type M operator and a monotone, weakly continuous one is type M. Using the result
of(theorem 3.2 in [2]) we deduce that the application
′
Φ′ : W 1,2 (Ω) −→ Lr (Γ1 )
u −→ h(x, u)
is weakly continuous and since h is increasing it defines a monotone operator from V into
V′
Thanks to lemma (2), A′ is Type M.We deduce that A is Type M too.By the theorem
(2), we conclude the existence of a solution to problem (8).i.e the existence part of Theorem
(1)
Before proving uniqueness, we give a lemma
6
DJAMEL AITAKLI
Let u, v be two distinct solutions of problem (7)
Lemma 5. Set Ω0 := {x ∈ Ω ∇(u − v) 6= 0} There exist a constant α > 0 and a nonzero
measure subset Ω′ ⊂ Ω0 such that:
α|∇(u − v)(x)| ≤ (a(x, u(x), ∇u(x)) − a(x, u(x), ∇v(x))) · ∇(u − v)(x)
′
a.e x ∈ Ω
Proof. suppose the converse i.e
∀α > 0, ∀Ω′ ⊂ Ω0 , meas(Ω′ ) > 0, α|∇(u − v)(x)| > (a(x, u(x), ∇u(x))−a(x, u(x), ∇v(x)))·∇(u − v)(x) > 0
a.e in Ω0 which implies by (5) that ∇u(x) = ∇v(x) a.e in Ω0 , this proves the lemma
Proof. (Uniqueness) u, v verify:
Z
Z
Z
Z
(10)
a(x, u, ∇u) · ∇w dx +
h(x, u)w dσ =
a(x, v, ∇v) · ∇w dx +
Ω
Γ1
Ω
∀w ∈ V ,
Define
Fζ (x) =
if x > ζ
Z
x
ζ
h(x, v)w dσ
Γ1
dt
tγ
Fζ (x) = 0
otherwise, with γ such that 1 < γ < 1 + r. Set for a positive real ζ > 0:
E = {x ∈ Ω; (u − v)(x) > ζ}
By the definition of the function F , we know from [3], Fζ (u − v) ∈ V
∇Fζ (u − v) = ∇(u − v)G′ζ (u − v) ∈ Lp (Ω) ∩ L2 (Ω)
∇Fζ (u − v) = 0 outside of E. Let w be the distribution such that:
∇w =
since
1
∇Fζ (u − v)
(1 + |u|p−1−r + |v|p−1−r + |∇v|p−1−r )
1
∈ L∞ (Ω)
(1 + |u|p−1−r + |v|p−1−r + |∇v|p−1−r )
Applying lemma (4), w ∈ Lp (Ω) ∩ L2 (Ω) so that w ∈ VR .
Test equation (10) with the function w and substract Ω a(x, u, ∇v) · ∇w from both sides
we obtain:
Z
Z
(a(x, u, ∇u)−a(x, u, ∇v))·∇Fζ (u − v) dx =
(a(x, v, ∇v)−a(x, u, ∇v))·∇Fζ (u − v) dx
Ω
Ω
−
Z
(h(x, u) − h(x, v)) · Fζ (u − v) dσ
Γ1
We have also (see [2]) that (h(u) − h(v))Fζ (u − v) ≥ 0, a.e x ∈ Γ1
Note that lemma (5) is valid only on a subset of Ω so that we have to prove uniqueness of
the solution on every nonzero measure subsets and then conclude the uniqueness on the
whole set. Then we can write using the precedent lemma, with E0 = E ∩ Ω′ and Ω′ is as
defined in this lemma, using (6)
Z
|∇(u − v)|
1
dx
γ (1 + |u|p−1−r + |v|p−1−r + |∇v|p−1−r )
|u
−
v|
E0
Z
(a(x, u, ∇u) − a(x, u, ∇v))∇(u − v)
dx
≤
γ (1 + |u|p−1−r + |v|p−1−r + |∇v|p−1−r )
|u
−
v|
E0
Z
|∇(u − v)|(1 + |u|p−1−r + |v|p−1−r + |∇v|p−1−r )
≤θ
dx
γ−r (1 + |u|p−1−r + |v|p−1−r + |∇v|p−1−r )
E0 |u − v|
EXISTENCE AND UNIQUENESS FOR A QUASILINEAR ELLIPTIC PROBLEM WITH NONLINEAR ROBIN CONDITIONS
7
Then
Z
E0
1
|∇(u − v)|
dx ≤ θ
|u − v|γ (1 + |u|p−1−r + |v|p−1−r + |∇v|p−1−r )
i.e
Z
Z
E0
|∇(u − v)|
dx
|u − v|γ−r
Z
|∇(u − v)|
1
dx
≤
θ
|∇Gζ (u − v)(x)| dx
γ (1 + |u|p−1−r + |v|p−1−r + |∇v|p−1−r )
E0 |u − v|
E0
R x ds
Since γ − r < 1
Where Gζ (x) = ζ sγ−r
Z
|∇(u − v)|
dx ≤ C
γ
p−1−r + |v|p−1−r + |∇v|p−1−r )
E0 |u − v| (1 + |u|
and g :=
C is independent of ζ Set uζ := 1E0 |∇(u−v)|
|u−v|γ
Applying lemma(3) to these functions we deduce that:
Z
|∇(u − v)|
dx ≤ C ′
γ
E0 |u − v|
1
(1+|u|p−1−r +|v|p−1−r +|∇v|p−1−r )
∀ζ > 0
i.e
(11)
Z
E0
|∇(u − v)|
|∇(u − v)|
dx = k
k 1
≤ C′
|u − v|γ
|u − v|γ L (E0)
The inequality (11) is written, by the defintion of F :
kFζ (u − v)kW 1,1 (Ω) = k∇Fζ (u − v)kL1 (Ω) ≤ C ′
∀ζ > 0
Then there exist a subsequence ζk and a function F ∈ L1 (Ω); Fζk ⇀ F weakly in W 1,1 (Ω)
and strongly in L1 (Ω) Consequently: limFζk (u − v)(x) < +∞ a.e in Ω. On the other
hand, by the definition of Fζ , we have
Z (u−v)(x)
ds
limζk −→0 Fζk (u − v)(x) = limζk −→0
= +∞
γ
s
ζk
a.e in E0 , we infer that meas(E0 ) = meas([u − v > 0]) = 0, interchanging the roles of u
and v, we get u = v a.e in Ω′ . Applying the same reasoning to any subset Ω′ ⊂ Ω gives
the desired result i.e the uniqueness on the whole set Ω of the solution for (7).
References
[1] Adams R A Sobolev spaces.Pure and applied Mathematics Series
[2] Cabarrubias B., Donato P Existence and uniqueness for a quasilinear elliptic problem with Robin conditions.Carpathian J.Math.,(2)2011
[3] Chipot M.,Elliptic equations: An introductory course: Birkhauser advanced text,
Basler Lehrbcher, Birkhauser verlag, Basel, 2009
[4] Showalter., R.EMonotone operators in Banach space and nonlinear partial differential equations, Mathematical suveys and monographs, volume 49, American
Mathematical Society.
E-mail address: [email protected]
Houari Boumediene sciences and technologies university (USTHB), Algiers, Algeria