Pohoaev Identities of Elliptic Equations and Systems with Variable Exponents and Some Applications

FENG Tingfu(馮廷福)

(Department of Mathematics,Kunming University,Kunming 650214,China)

Abstract: In this paper,we obtain Pohoaev identities for the elliptic equation and system with variable exponent,respectively.Moreover,nonexistence of nontrivial solutions is derived under some suitable conditions,which extends the results of Dinca and Isaia(2010)and Lopez(2014).

Key words: Pohoaev identity; Nonexistence; Variable exponent; Elliptic equation(system)

1.Introduction and Main Results

where? ?Rn(n ≥3) is a star-shaped bounded smooth domain with respect to the origin,namely have (x ?0)·ν=x·ν ≥0 for allx ∈??,whereν=ν(x) denotes the unit outward normal atx ∈??,and then applied it to prove nonexistence of nontrivial solutions.This identity was later called the Pohoaev identity and extended to general elliptic equations and systems[2?11].

Elliptic equations with variable exponents as appropriate models in image processing,nonlinear electrorheological fluids and elastic mechanics etc.were extensively studied[12?14].The study of existence and multiplicity of positive solutions for the elliptic equations with variable exponents were given in [15–26] and references therein .

Dinca and Isaia[27]considered the Dirichlet problem

for all k ∈R.

Recently,Lopez[28]considered the elliptic problem with variable exponent

where p(x) and q(x) are ? →R Lipschitz continuous functions and 2

where

In this paper,based on the previous works of Dinca and Isaia[27]and Lopez[28],for the weighted functions and more general nonlinear terms in elliptic equation (system) with variable exponent,we consider the following elliptic equation and system with variable exponent,respectively,

and

Main results are stated in the following.

Theorem 1.1Letu ∈W2,p(x)(?)∩W1,p(x)0(?) be a solution to (1.3),1< p(x)< n,a(x)>0,Assume thatf(x,u):?×R→R is continuous and

satisfiesF(x,0)=0 forx ∈??.Then

then (1.3) has no nontrivial solutions.

Remark 1.1In the special casep(x)=2 anda(x)=1,we have by (1.5) that

Theorem 1.2Let (u,w)be a pair of solutions to(1.4),10,,a(x),b(x)∈C1(?),Fu(x,u,w) andFw(x,u,w):?×R×R→R are continuous and

satisfiesF(x,0,0)=0 forx ∈??.Then

where

Moreover,ifFu(x,u,w) andFw(x,u,w) satisfy

then (1.4) has no nontrivial solutions.

2.Proof of Theorem 1.1

Proof of Theorem 1.1Before giving proofs,we need some theory of variable exponent Lebesgue spaces and Sobolev spaces,see [18,29-30] and references therein.For convenience,we give a simple description here.

Assume thatp(x) :? →(1,∞) is a Lipschitz continuous function.A variable exponent Lebesgue spaceLp(x)(?) is defined by

with the norm

A variable exponent Sobolev spaceWm,p(x)(?) is defined by

with the norm

Denote by(?) the closure ofC0∞(?) inWm,p(x)(?).It is known thatWm,p(x)(?)and(?) are both separable and reflexive Banach spaces.

We now prove Theorem 1.1 and give an example.

Proof of Theorem 1.1Multiply the equation in (1.3) byuandx·respectively,and integrate it over?.Then there holds

and

In virtue ofF(x,u)=F(x,0)=0 forx ∈??and the divergence theorem,we obtain

which implies

Using the left hand side of (2.2) and the divergence theorem,we have

In (2.4),we note that

Combining (2.4),(2.5) and (2.6),we obtain

Substituting (2.3) and (2.7) into (2.2),we obtain (1.5).

Since? ?Rnis star-shaped with respect to the origin,it showsx·ν ≥0 forx ∈??.If0 satisfies (1.3),then from (1.5) and (1.6),

which is impossible.Then (1.3) has no any nontrivial solutions.

Example 2.1Letf(x,u)=λh(x)uq?1in (1.3),whereλ,q ∈R,0< h(x)∈C1(?).Then (1.3) has no any positive solutions under the condition

In fact,due to

F(x,u)=F(x,0)=0 forx ∈??.Ifu>0 satisfies (1.3),then it concludes from (1.5),(2.9)and

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that

which is a contradiction.

3.Proof of Theorem 1.2

Proof of Theorem 1.2Multiply the equation in (1.4) byuandrespectively,and integrate it over?.Then there holds

and

NotingF(x,u,w)=F(x,0,0)=0 forx ∈??,by the divergence theorem,we obtain

and then

With the help of (3.1),(3.3) and the divergence theorem,we have

We prove (1.7) by substituting (3.4) and (3.5) into (3.3).

Let us notex·ν ≥0 forx ∈??.If (u,w)(0,0) satisfies (1.4),then it attains from(1.7) and (1.8) that

which is impossible.

Example 3.1Letin (1.4),whereλ ∈R,g(x)∈C1(?),then (1.4) has no any positive solutions under the condition

Actually,noting that

we getF(x,u,w)=F(x,0,0)=0 forx ∈??.If (u,w)>(0,0) satisfies (1.4),then it is obtained from (1.7),(3.6) and

that

which is a contradiction.