FENG Tingfu(冯廷福)
(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)
whereΩ ⊂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 ∈∂Ω,whereν=ν(x) denotes the unit outward normal atx ∈∂Ω,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 Ω →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)(Ω)∩W1,p(x)0(Ω) be a solution to (1.3),1< p(x)< n,a(x)>0,Assume thatf(x,u):Ω×R→R is continuous and
satisfiesF(x,0)=0 forx ∈∂Ω.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),1
0,,a(x),b(x)∈C1(Ω),Fu(x,u,w) andFw(x,u,w):Ω×R×R→R are continuous and
satisfiesF(x,0,0)=0 forx ∈∂Ω.Then
where
Moreover,ifFu(x,u,w) andFw(x,u,w) satisfy
then (1.4) has no nontrivial solutions.
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) :Ω →(1,∞) is a Lipschitz continuous function.A variable exponent Lebesgue spaceLp(x)(Ω) is defined by
with the norm
A variable exponent Sobolev spaceWm,p(x)(Ω) is defined by
with the norm
Denote by(Ω) the closure ofC0∞(Ω) inWm,p(x)(Ω).It is known thatWm,p(x)(Ω)and(Ω) 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 overΩ.Then there holds
and
In virtue ofF(x,u)=F(x,0)=0 forx ∈∂Ωand 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).
SinceΩ ⊂Rnis star-shaped with respect to the origin,it showsx·ν ≥0 forx ∈∂Ω.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(Ω).Then (1.3) has no any positive solutions under the condition
In fact,due to
F(x,u)=F(x,0)=0 forx ∈∂Ω.Ifu>0 satisfies (1.3),then it concludes from (1.5),(2.9)and
13. A tiny little man: Rumpelstiltskin s size poses many questions unanswered by the text. Is he a dwarf? Most scholars do not think so, but some do. He may also be a demon, an elf or simply a small statured man. Jane Yolen proposes another interesting theory about the tale s meaning and Rumpelstiltskin s origin:
that
which is a contradiction.
Proof of Theorem 1.2Multiply the equation in (1.4) byuandrespectively,and integrate it overΩ.Then there holds
and
NotingF(x,u,w)=F(x,0,0)=0 forx ∈∂Ω,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 ∈∂Ω.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(Ω),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 ∈∂Ω.If (u,w)>(0,0) satisfies (1.4),then it is obtained from (1.7),(3.6) and
that
which is a contradiction.