W01,p(x)Versus C1Local Minimizers for a Functional with Critical Growth

SAOUDI K.

College of arts and sciences at Nayriya,university of Dammam 31441 Dammam, Kingdom of Saudi Arabia.

W01,p(x)Versus C1Local Minimizers for a Functional with Critical Growth

SAOUDI K.?

College of arts and sciences at Nayriya,university of Dammam 31441 Dammam, Kingdom of Saudi Arabia.

Received 28 September 2012;Accepted 24 January 2014

AMS Subject Classifications:35J65,35J20,35J70

Chinese Library Classifications:O175.8,O175.25

p(x)-Laplacian equation;variational methods;local minimizer.

1 Introduction

The assumptions on the source terms f is as follows:

(f1)f:×R→R is a measurable function with respect to the first argument and continuous differentiable with respect to the second argument for a.e.x∈?.Moreover, f(x,0)=0 for(x,s)×R+.

The operator Δp(x)u:=divis called p(x)-Laplace where p is a continuous non-constant function.This differential operator is a natural generalization of the p-Laplace operator,where p＞1 is a real constant.However,the p(x)-Laplace operator possesses more complicated nonlinearity than p-Laplace operator,due to the fact that Δp(x)is not homogeneous.Our aim in this paper is to show the following

We remark that u0satisfies in the distributions sense the Euler-Lagrange equation associated to I,that is

with u=v=0 on??,where g,h∈L∞(?)are such that 0≤g＜h pointwise everywhere in ?.If

where n is the inward unit normal on??.Then,the following strong comparison principle holds:

Equation(P)appears in several models:electrorheological fluids,image processing, flow in porous media,calculus of variations,nonlinear elasticity theory,heterogeneous porous media models(see[8–10],Zhikov[11,12]).The study of such problems is a new and interesting topic some results can be found[8,9,12].

2 Proof of Theorem 1.1

We first deal with the subcritical case and then give the additional arguments to prove the result when q(x)=p?(x)-1.Case 1:r(x)＜p?(x)-1.We adapt the arguments in[16]. Let q(x)∈(r(x),p?(x)-1),define

and

We consider the following constraint minimization problem:

We now consider the following two cases:

1)Let K(v?)＜?.Then v?is also a local minimizer of I.Hence,v?is a solution of the problem(P).Therefore,due to the growth conditions on f and the smoothness of ? we have that(see[19,Theorem 1.2]for the detail)

for some α∈(0,1),and as ?→0+

which contradicts the fact that u0is a local minimizer in C1(?)∩C0(?).

Now,we deal with the second case:2)K(v?)=?:In this case,from the Lagrange multiplier rule we have

and then for t small we have

This contradicts the fact that v?is a global minimizer of I in S?.It follows thatμ?≤0.We deal now with two following cases:

case(i):μ?∈(-1,0).In this case,using

Now,sinceμ?≤0,there exists M,c＞0 independent of ? such that

It follows from[20,Theorem 4.1]that v?∈L∞(?)and|v?|L∞(?)≤c because||u?||W1,p(x)(?)is bounded uniformly for some ?∈(0,1),where c is a positive constant independen0t of ?. Hence,|v?|C1,α(?)≤c for some α∈(0,1)by using[19,Theorem 1.2].Then we conclude as above.

Let us consider the case(ii):μ?≤-1.

Furthermore,there exists a number M＞0,independent of ?,such that for

we have

Similarly,Z

Now,subtracting(2.6)with w=(v?-u)|v?-u0|β-1,where β≥1,we obtain

Using the bounds about v?,u0and the H¨older inequality we get

where C does not depend on β and ?.Passing to the limit β→+∞this leads to

So the right-hand side of(2.3)is uniformly bounded in L∞(?)from which as in the first case,we obtain that v?,(0＜?≤1)is bounded in C1,α(?)independentlyof ?.Finally,using Ascoli-Arzela Theorem we find a sequence ?n→0+such that

It follows that for ?＞0 sufficiently small,

which contradicts the fact that u0is a local minimizer of I for the C1(?)∩C0(?)topology. The proof of the Theorem 1.1 in the subcritical case is now complete.

and

We now consider the truncated functional

It follows that for each ?＞0,there is some j?(with j?→+∞as ?→0+)such that Ijε(v?)＜I(u0).On the other hand,since fj?has subcritical growth and since for some constants c,c1,c3independent of ?,

Then,it follows that for ?＞0 sufficiently small,

It remains to prove theClaim.For this purpose we write the Euler equation satisfied by u?:

whereμ?is a Lagrangemultiplier associated to theconstraint||u?-u0||Lp?(x)(?)≤?.Taking u0-u?as testing function in(2.12)and using the minimizing property of u?,one gets μ?≤0.

We again distinguish between the following two cases:

InCase(i)from the Euler equation:

satisfied by u?we get that

for some D＞0 independent of ?.Now using the Moser iterations as in Zhang-Liu[22], we get that{u?}are bounded in Lβp?(x)(?)for some β＞1 independently of ?.Then, using[20,Theorem 4.1]that u?∈L∞(?)and|u?|L∞(?)≤c,where c is a positive constant independent of ?.Hence,|u?|C1,α(?)≤c for some α∈(0,1)by using[19,Theorem 1.2].This proves the Claim in the case(i).

Let us consider now theCase(ii):

Again as in the subcritical case,there exists a number M＞0,independent of ?,such that for

Taking now(u?-M)+as a testing function in(2.12),one concludes by the weak comparison principle that u?(x)≤M in ?.So u?remains bounded in L∞(?)as ?→0.

Now,subtracting(2.12)with w=(u?-u)|u?-u0|β-1,where β≥1,we obtain

Using the bounds about u?,u0and the H¨older inequality we get

where C does not depend on β and ?.Passing to the limit β→+∞this leads to

So the right-hand side of(2.12)is uniformly bounded in L∞(?)from which as in the first case,from[19,Theorem 1.2]we obtain that u?,(0＜?≤1)is bounded in C1,α(?) independently of ?.This concludes the proof of the Claim in case(ii).

Acknowledgments

The author would like to thank the anonymous referees for their carefully reading this paper and their useful comments.

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10.4208/jpde.v27.n2.2 June 2014

?Corresponding author.Email address:kasaoudi@gmail.com(K.Saoudi)