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EXISTENCE OF SOLUTION FOR A SINGULAR ELLIPTIC EQUATION WITH CRITICAL SOBOLEV-HARDY EXPONENTS

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EXISTENCE OF SOLUTION FOR A SINGULAR ELLIPTIC EQUATION WITH CRITICAL SOBOLEV-HARDY EXPONENTS

JUAN LI

Received 9 June 2005 and in revised form 21 September 2005

Via the variational methods, we prove the existence of a nontrivial solution to a singular semilinear elliptic equation with critical Sobolev-Hardy exponent under certain condi- tions.

1. Introduction

In this paper, we consider the following elliptic problem:

uµ u

|x|2 =|u|2(s)2

|x|s u+a(x)|u|r2u+λu, xRN, (1.1) where N3, 0µ <µ¯=. ((N2)/2)2, 0s <2, λ0, and 2(s)=. 2(Ns)/(N 2) is the critical Sobolev-Hardy exponent; note that 2(0)=2=. 2N/(N2) is the critical Sobolev exponent. The spaceH .

=H(RN) is the completion ofC0(RN) in the norm

u=.

RN

u2µ u2

|x|2

dx 1/2

. (1.2)

By the Hardy inequality [8,9], this norm is equivalent to the usual norm (RN|∇u|2dx)1/2. The scalar product inHis

(u,v)=.

RN

uvµ uv

|x|2

dx u,vH. (1.3)

We defineHrHwith

Hr=. uH,u(x)=u |x|

. (1.4)

Copyright©2005 Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences 2005:20 (2005) 3213–3223 DOI:10.1155/IJMMS.2005.3213

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The hypothesis fora(x) is as follows:

(A) a(x) is nonnegative and locally bounded in RN\{0}, a(x)=O(|x|s) in the bounded neighborhoodGof the origin,a(x)=O(|x|t) as|x| → ∞, 0s < t <2, 2(t)<

r <2(s), where 2(t) .

=2(Nt)/(N2) for 0t <2.

The singular elliptic problems have received some attention in recent years. For exam- ple, Janneli [10] and Ferrero and Gazzolo [7] studied the semilinear elliptic equation

∆uµ u

|x|2= |u|22u+λu, xΩ,

u(x)=0, x∂Ω, (1.5)

whereΩRN(N3) is a smooth bounded domain containing the origin 0. They proved that (1.5) has a nontrivial solution under certain conditions forλandµ. Moreover, Cao in [4,5] and Chen in [6] also studied the semilinear elliptic equation (1.5). They show that (1.5) has nontrivial solutions and a sign-changing solution under some conditions forµ,λ. Ghoussoub and Yuan in [9] considered the quasilinear problem

pu=µ|u|q2u

|x|s +λ|u|r2u, xΩ, u(x)=0, x.

(1.6)

They get that (1.6) has a positive solution and a sign-changing solution under some con- ditions forλ,µ,r,q.

In the case whenΩis an unbounded domain inRN, the corresponding problem be- comes more complicated since the Sobolev embeddingW1,p(Ω)Lq(Ω)(p2) is not compact for allq[p,p]. However, by the Strauss lemma (see [13]), the embedding Hr(RN)Lq(RN) is compact for allq[2, 2). Therefore, we can discuss the nontriv- ial solutions of (1.1) inHrby variational methods. But there are also some difficulties for (1.1), because the embeddingHrL2(s)(RN,|x|s) is still not compact. In [11], asλ=0, the existence of a nontrivial solution is given for (1.1) withs=0, so it will be meaningful to study the existence of nontrivial solutions for (1.1) ass[0, 2) andλ=0. In this paper, we obtain the following existence results.

Theorem1.1. Suppose (A) and0s <2,0µ <µ,¯ λ0. Assume that one of the following conditions holds:

(i)λ=0and max

Ns

µ¯+µ¯µ,Ns2µ¯µ µ¯ , 2(t)

< r <2(s), (1.7) (ii) 0< λ < λ1(µ)and0µµ¯1, whereλ1(µ) .

=infuH\{0}(u2/RNu2dx).

Then problem (1.1) has at least a nontrivial solution inHr.

Throughout this paper, we will use the letterCto denote the natural various constants independent ofu, and·dxinstead ofRN·dx.

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2. Proof of the main result

We first give some definitions and lemmas.

Definition 2.1. Let{um}be a sequence inHr, if there exists a constantcR1such that J um

−→c, J um

−→0 inHr1 (2.1)

asm→ ∞, then{um}is called a (PS)csequence inHr.

Lemma2.2 (Hardy inequality [8,9]). Assume that1< p < NanduW1,p(RN). Then |u|p

|x|pdx p

Np p

updx. (2.2)

Lemma 2.3 (Sobolev-Hardy inequality [9]). Assume that 1< p < N and that p(s) .

= ((Ns)/(Np))p,0sp. Then there exists a constantC >0 such that for anyu W1,p(RN),

|u|p(s)

|x|s dx p/ p(s)

C updx. (2.3)

Lemma2.4 [11]. Assume that hypothesis (A) holds. Then the embeddingHLr(RN,a(x)) is compact.

Consider the energy functional

J(u)=1

2u 1 2(s)

|u|2(s)

|x|s dx1 r

a(x)|u|rdxλ 2

|u|2dx, (2.4) byLemma 2.4,J(u)is well defined andJC1(H,R); the critical points of the functionalJ correspond to weak solutions of problem (1.1).

For0µ <µ, define the best Sobolev-Hardy constant:¯ As=. As(µ)= inf

uH{0}

|∇u|2µu2/|x|2 dx

|u|2(s)/|x|sdx2/2(s). (2.5) In[12], the author found thatAsis attained by the functions

yε(x)=µ¯µ(Ns)/µ¯

µ/(2¯ s)

|x|µ¯µ¯µ ε+|x|(2s)µ¯µ/µ¯(N2)/(2s) (2.6) for allε >0. Moreover, the functionsyε(x)solve the equation

uµ u

|x|2=|u|2(s)2

|x|s u inRN\{0}, (2.7)

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and satisfy

yε2µyε2

|x|2

dx= yε2(s)

|x|s dx=A(Ns s)/(2s). (2.8)

In the following, we first give some estimates for the extremal functions.

Let

Cε=

2ε( ¯µµ)(Ns) µ¯

µ/(2¯ s)

, Uε(x)= yε(x)

Cε , (2.9)

B2l= {xRN,|x|<2l} ⊂Gwithl >0 andGis the domain in hypothesis (A), let 0φ 1 be a cutting-offfunction inC0(RN)Hr, such thatφ(x)=1 inBlandφ(x)=0 inRN\ B2l. Setuε(x)=φ(x)yε(x) andvε=uε(x)/(|uε|2(s)/|x|s)1/2(s), so that(|vε|2(s)/|x|s)= 1. In [12], the author proved that the following estimates are true:

vε2=As+O ε(N2)/(2s), (2.10)

vεqdx=

Oεµq/(2¯ s), 1q < N

µ¯+µ¯µ, Oεµq/(2¯ s)|lnε|

, q= N

µ¯+µ¯µ, Oεµ(N¯ qµ)/((2¯ s)µ¯µ), N

µ¯+µ¯µ< q <2.

(2.11)

Moreover, we also need the following results.

Lemma2.5. Suppose thatγ=

¯

µ+µ¯µ,γ´=

¯ µ

¯

µµ,0µ <µ, and¯ 0s <2, then, vε(x)satisfies the following estimates:

vεq

|x|s dx

c1εµq/(2¯ s), 1q <Ns γ , c2εµq/(2¯ s)|lnε|, q=Ns

γ , c3ε(µ(N¯ s)µq)/(2¯ s)µ¯µ, Ns

γ < q <2(s),

(2.12)

whereci(i=1, 2, 3)are positive constants.

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Proof. LetωNdenote the surface area of the (N1) sphereSN1inRN. For 1q <2(s), we have

vεq

|x|s dx= uε(x)q

|x|s dx· uε2(s)

|x|s dx q/2(s)

=B φ(x)CεUεq

|x|s dx

=BCqε

O(1) +ωN

l

0

ε+r(2s)µ¯µ/µ¯q(N2)/(2s)rNs1qγ´dr

=BCqε

O(1) +ωNεq((N2)/(2s))+(µ(N¯ sγq)/(2´ s)µ¯µ)

×

µ/((s¯ 2)µ¯µ)

0

1 +r(2s)µ¯µ/µ¯q(N2)/(2s)rNs1qγ´dr

, (2.13) whereB=(|uε|2(s)/|x|sdx)q/2(s).

If2qµ¯µ+Nsγq´ =0, that is,q=(Ns)/γ, vεq

|x|s dx=BCqε

O(1) +ωN

µ/((s¯ 2)µ¯µ)

1

1 rdr

Bc´1εµq/(2¯ s)|lnε|, (2.14) where ´c1>0 is a constant.

If2qµ¯µ+Nsγq <´ 0, that is,q >(Ns)/γ, vεq

|x|s dx=BCqε

O(1) +O

εq((N2)/(2s))+(µ(N¯ sγq)/(2´ s)µ¯µ)

Bc´2ε(µ(N¯ s)µq)/(2¯ s)µ¯µ,

(2.15)

wherec2>0 is a constant.

If2qµ¯µ+Nsγq >´ 0, that is,q <(Ns)/γ, vεq

|x|s dx=BCεq

O(1) +ωN l

0

ε+r(2s)µ¯µ/µ¯q(N2)/(2s)rNs1qγ´dx

=BCεq·O(1)Bc´3εµq/(2¯ s),

(2.16)

where ´c3>0 is a constant.

By

B= uε2(s)

|x|s dx q/2(s)

= φ(x)yε2(s)

|x|s dx

q/2(s)

yε2(s)

|x|s dx q/2(s)

=A(2s N)q/2(2s),

(2.17)

we have finished the proof ofLemma 2.5.

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Lemma2.6. Suppose (A) and0s <2,0µ <µ,¯ λ0. Assume that one of the following conditions holds:

(i)λ=0and max

Ns

µ¯+µ¯µ,Ns2µ¯µ µ¯ , 2(t)

< r <2(s), (2.18) (ii) 0< λ < λ1(µ)and0µµ¯1.

Then, there existsu0Hr,u0=0, such that the following inequality holds:

0<sup

t0

J tu0

< 2s

2(Ns)A(Ns s)/(2s). (2.19)

Proof. Fort0, we consider the functions g(t) .

=J tvε

=t2

2vε2 t2(s) 2(s)

tr r

a(x)vεrdxλt2

2 vε2dx,

¯ g(t)=t2

2vε2 t2(s) 2(s).

(2.20)

Note that limt→∞g(t)= −∞,g(0)=0, andg(t)>0 ast0+, therefore, supt0g(t)>0 must be attained by some 0< tε<+andg(tε)=0. So we have

g tε

=tεvε2tε2(s)1tεr1

a(x)vεrdxλtε vε2dx=0. (2.21) Then

vε2=t2ε(s)2+tεr2

a(x)vεrdx+λ vε2dxt2ε(s)2, tεvε2/(2(s)2). (2.22) Moreover, by hypothesis (A), we have

vε2tε2(s)2+Cvε2(r2)/(2(s)2)

B2l

vεr

|x|s +λ vε2dx. (2.23) From (2.23) and (2.10)–(2.12), asεsmall enough, we get

t2ε(s)2As

2 . (2.24)

By the simple computation, we know that the function ¯g(t) attains its maximum at t0= vε2/(2(s)2)and is increasing in the interval [0,t0]. So, by (2.10), (2.22), and (2.24),

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we have g tε

g¯ t0

1 r

As 2

r/(2(s)2) vεr

|x|s dxλ 2

As 2

2/(2(s)2)

vε2dx

2s

2(Ns)vε2(Ns)/(2s)C vεr

|x|s C vε2dx

= 2s

2(Ns)A(Ns s)/(2s)+Oε(N2)/(2s)C vεr

|x|s C vε2dx.

(2.25)

In case (i), since

r >max Ns

γ ,Ns2µ¯µ µ¯ , 2(t)

, (2.26)

by (2.12), we have

vεr

|x|s c3εµ(N¯ sµr)/(2¯ s)µ¯µ, µ¯ Ns

¯ µr

(2s)µ¯µ <N2 2s.

(2.27)

Letu0=vε, choosingεsmall enough, from (2.25), we can deduce that sup

t0

J tu0

=g tε< 2s

2(Ns)A(Ns s)/(2s). (2.28) In case (ii), 0< λ < λ1(µ). By (2.11), asµ=µ¯1,

vε2=Oε(N2)/(2s)|lnε|

, (2.29)

as 0µ <µ¯1,

vε2=Oε(N2)/((2s)µ¯µ). (2.30) Choosingεsmall enough, we also get (2.28). The proof ofLemma 2.6is completed.

Lemma2.7. Suppose thatc(0, (2s)/(2(Ns))A(Ns s)/(2s)). ThenJ(u)satisfies(PS)c

condition.

Proof. Let{um} ∈Hrbe a (PS)csequence. Then we have J um

=1

2um2 1 2(s)

um2(s)

|x|s dx1 r

a(x)umrdxλ

2 um2dx=c+o(1), (2.31) J um

,um

=um2 um2(s)

|x|s dx

a(x)umrdxλ um2dx=o(1)um. (2.32)

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Let (2.31)×2(2.32), we have 2c+o(1) +o(1)um

1 2

2(s)

um2(s)

|x|s dx+

12 r

a(x)umrdx. (2.33) From

um2=2J um

+ 2 2(s)

um2(s)

|x|s dx+2 r

a(x)umrdx+λ um2dx, (2.34) we get

1 λ

λ1(µ)

um22J um

+ 2 2(s)

um2(s)

|x|s dx+2 r

a(x)umrdx

o(1) +o(1)um+C.

(2.35)

So, we conclude that{um}is bounded inHr. Passing to a subsequence (still denoted by {um}), asm→ ∞, we get that

umuweakly inHr, um−→ustrongly inLq RN

, q[2, 2), um−→ua.e. inRN,

um−→ustrongly inLr RN,a(x).

(2.36)

It follows from the Sobolev-Hardy inequality (see [9]) that|um|2(s)2umis bounded in L2(s)/(2(s)1)(RN,|x|s), thus we have that

um2(s)2um|u|2(s)2uweakly inL2(s)/(2(s)1) RN,|x|s

. (2.37)

SinceJ(um)0, from (2.36) and (2.37), we obtain J(u),u= u2

|u|2(s)

|x|s dx

a(x)|u|rdxλ

|u|2dx= lim

m→∞

J um ,u=0.

(2.38) Setvmumu, by Brezis-Lieb lemma [2], we have

um2=vm2+u2+o(1), (2.39) um2(s)

|x|s dx=

|u|2(s)

|x|s dx+ vm2(s)

|x|s dx+o(1). (2.40)

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It follows directly from (2.31)–(2.40) that o(1)um=

J um,um=um2 um2(s)

|x|s dx

a(x)umrdxλ um2dx

=

J(u),u+vm2 vm2(s)

|x|s dx+o(1)=vm2 vm2(s)

|x|s dx+o(1), J(u)=J um

1

2vm2+ 1 2(s)

vm2(s)

|x|s dx+o(1)

=c1

2vm2+ 1 2(s)

vm2(s)

|x|s dx+o(1).

(2.41) Since{vm}is bounded, without loss of generality, we may assume that

mlim→∞vm2=k. (2.42)

Then we get that

mlim→∞

vm2(s)

|x|s dx=k. (2.43)

By the Sobolev-Hardy inequality, vm2(s)

|x|s dxAs2(s)/2vm2(s) (2.44)

for allmN. Then by takingm+, we obtain

kAs2(s)/2k2(s)/2. (2.45)

Ifk >0, we have thatkA2s(s)/(2(s)2). By (2.41) we deduce that J(u)=c1

2 1 2(s)

kc2(s)2

22(s) A2s(s)/(2(s)2)=c 2s

2(Ns)A(Ns s)(2s)<0, (2.46) but from (2.38), we get

J(u)=J(u)1 2

J(u),u= 1

2 1 2(s)

|u|2(s)

|x|s dx+ 1

2 1 r

a(x)|u|rdx0, (2.47) this contradiction impliesk=0. By the definition ofvm, we conclude thatJ(u) satisfies (PS)ccondition. We have completed the proof ofLemma 2.7.

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