American Journal of Applied Mathematics
Volume 3, Issue 6, December 2015, Pages: 283-287

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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(B. Al-Hamzah)
(N. Yebari)
(N. Yebari)

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(u1) A(u2), u1 u2> ≥ 0, u1, u2 V .

(iv)  Strictly monotone iff

<A(u1) A(u2 ), u1 u2 > > 0, for all u1 , u2 V, u1 u2.

(v)  Strongly monotone iff there exists k > 0,

<A(u1 ) A(u2), u1 u2> ≥, for all u1, u2 V, u1 ≠ u ;

(vi)  Continuous iff un u implies A(un ) A(u), for all un , uV.

(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.


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