Existence and Uniqueness of Weak Solution for Weighted p-bilaplacian (p-Biharmonic)
Bassam Al-Hamzah, Naji Yebari
University Abdelmalek Essaadi, Faculty of Sciences, Department of Mathematics, Tetouan, Morocco
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To cite this article:
Bassam Al-Hamzah, Naji Yebari. Existence and Uniqueness of Weak Solution for Weighted p-bilaplacian (p-Biharmonic). American Journal of Applied Mathematics. Vol. 3, No. 6, 2015, pp. 283-287. doi: 10.11648/j.ajam.20150306.17
Abstract: This paper deals with the equation in bounded domain Ω∈. Relying on Browder theorem, under conditions of the monotonous function f. We obtained the existence and uniqueness of weak solutions for the weighted p-bilaplacian boundary value of the form:
Keywords: Weak Solutions, p-biharmonic Operator, Browder Theorem
1. Introduction
In this paper, we are concerned with the existence and uniqueness of weak solution for a weighted p-bilaplacian boundary value of the form:
Let Ω be a bounded domain in and be a Caratheodry (CAR) Satisfies the following:
(H1) for a.e. ∈ Ω and , .
(H2) that there exists
Here and and
Satisfying the conditions:
We assume that , where
Our paper is organized as follow: Section 2 contains some basic definitions concerning the nonlinear operators that will be used throughout the paper. Also, we introduce the space setting of the problem and give some basic characteristics, as the equivalent norm and embedding results. In section 3 we state the main result on the existence and uniqueness of weak solutions of the problem (P).
The existence of solutions for the nonlinear eigenvalue problem with p-biharmonic see [3].
2. Preliminaries and Space Setting
First, we introduce some basic definitions concerning the nonlinear operators which we use extensively in this paper 3.
Definition 2.1 [4] Let A : V → V ‘ be an operator on a real Banach space V. We say that the operator A is :
(i) bounded iff it maps bounded sets into bounded i.e. for each r > 0 there exists M > 0 (M depending on r) such that
(ii) Coercive : iff
(iii) Monotone iff <A(u_{1}) − A(u_{2}), u_{1} − u_{2}> ≥ 0, ∀u_{1}, u_{2}∈ V .
(iv) Strictly monotone iff
<A(u_{1}) − A(u_{2} ), u1 − u_{2} > > 0, for all u1 , u_{2} ∈ V, u_{1} ≠ u_{2}.
(v) Strongly monotone iff there exists k > 0,
<A(u_{1} ) − A(u_{2}), u_{1} − u_{2}> ≥, for all u_{1}, u_{2} ∈ V, u_{1} ≠ u ;
(vi) Continuous iff un → u implies A(un ) → A(u), for all un , u∈V.
(vii) Strongly continuous iff unu implies A(un )→A(u), for all un ,u ∈V.
(viii) Demi continuous iff un → u implies A(un ) A(u), for all un , u ∈ V.
Theorem 2.1(Browder) [4]
Let A be a reflexive real Banach space. Moreover let A : V → V ‘ be an operator which is: bounded, demicontinuous, coercive, and monotone on the space V . Then, the equation A(u) = f has at least one solution u ∈ V for each f ∈ V ‘. If moreover, A is strictly monotone operator, then the equation (P) has precisely one solution u ∈ V for every f ∈ V ‘.
Proposition 2.1 [6] For any bounded domain Ω and 1 < p<∞, satisfies the following:
(i) is an hemicontinuous operator from (Ω) into .
(ii) is a bounded monotonous, and coercive operator.
(iii) : (Ω) is a bicontinuous operator.
Proof
See [6].
The norm is uniformly equivalent on to the usual norm of .
3. Existence and Uniqueness Results
In this section, using Browder theorem, we prove the existence and uniqueness of weak solution for equation (P).
Definition 3.1 We say that is a weak solution to equation (P) if
Our main results concerning problem (P) is the following theorem :
Theorem 3.1 Let p ≥ 2, λ > 0 and f (x, u) ∈ C AR(Ω×R) satisfy (H1) , (H2) and (H3). Then problem (P) has a unique weak solution.
Proof:
We define for λ >0 the operator A: , as A =J+ λG − F, where the
Operators J:, G: and F : Ω x are given by
And
for all u, φ ∈ . Thus, to find a weak solution of (P) is equivalent to finding which satisfies the operator equation A(u) = 0. Now, we have the following properties of the operators J, G, and F:
a) J, G and F are well defined. Using Holder’s inequality, we have
First case: and . Let u,φ By Hὅlder’s inequality, we have
Where and s is given by:
Therefore
The nit suffices to take so that G is well defined.
Second case: and . In this case embedded into the space for any , there is such that:
We obtain that . By Hὅlder’s inequality, we arrive at:
For any , and G is well defined.
Third case: and . In this case embedded into the space Then for any we have:
And G is well defined.
And
By Hὅlder’s inequality,
And hence J, G, F are well defined.
b) G is completely continuous. Let be as equence such that
un → u weakly in . We have to show that G (un) → G(u) strongly in , i .e.
For . Let s be as and and
where C is the constant of Sobolev. On the other hand, the Nemytskii’s operator
is continuous from Ls (Ω) into , and un → u weakly in . So, we deduce that un → u strongly in Ls (Ω) because s < p2. Hence as .
This completes the proof of the claim.
If then :
Where q is given . By Sobolev’s embedding there exist such that Thus
From the continuity of from into and from the compact embedding of in we have the desired result.
If , , then we obtain :
Where C is the constant given by embedding of in C (Ω) ∩ L∞(Ω). It is clear that,
as n→+∞.
Hence G is completely continuous, also in this case. J and F are bounded operators. Indeed, for every u such that
We have
.
Using Hὅlder’s inequality, we obtain
Also, we get
Where k is the constant of the embedding of into .
c) J and F are continuous operators. If in . Then, we have
Applying Dominated Convergence Theorem, we obtain
Hence
as.
Similarly, we have
as .
d) Let , we have the following inequality (see [8] )
(1)
Now,
Using (1), we get
,
for
So
(2)
Similarly, we have
Hence,
(3)
Also, we get
.
Since f is decreasing with respect to the second variable, we have
Consequently
(4)
Equations (2),(3), and (4) imply that
(5)
So A is strongly monotone.
Now, to apply Browder theorem, it remains to prove that A is a coercive operator. From (5) , we have
On the other hand
Then
So
.
This proves the coercivity condition and so, the existence of weak solution for (P). The uniqueness of weak solution of (P) is a direct consequence of (5). Suppose that u,φ be a weak solutions of (P) such that u≠φ. Now, from (5), we have
Therefore u=φ. This completes the proof.
References