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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:

−∆u−µ u

|x|2 =|u|2∗(s)−2

|x|s u+a(x)|u|r−2u+λu, x∈RN, (1.1) where N≥3, 0≤µ <µ¯=. ((N−2)/2)2, 0≤s <2, λ≥0, and 2∗(s)=. 2(N−s)/(N− 2) is the critical Sobolev-Hardy exponent; note that 2∗(0)=2∗=. 2N/(N−2) is the critical Sobolev exponent. The spaceH .

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

u=.

RN

∇u2−µ 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

∇u∇v−µ uv

|x|2

dx ∀u,v∈H. (1.3)

We defineHr⊂Hwith

Hr=. u∈H,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| → ∞, 0≤s < t <2, 2∗(t)<

r <2∗(s), where 2∗(t) .

=2(N−t)/(N−2) for 0≤t <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

−∆u−µ u

|x|2= |u|2∗−2u+λu, x∈Ω,

u(x)=0, x∈∂Ω, (1.5)

whereΩ⊂RN(N≥3) 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|q−2u

|x|s +λ|u|r−2u, x∈Ω, u(x)=0, x∈∂Ω.

(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)(p≥2) 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) and0≤s <2,0≤µ <µ,¯ λ≥0. Assume that one of the following conditions holds:

(i)λ=0and max

N−s

µ¯+µ¯−µ,N−s−2µ¯−µ µ¯ , 2∗(t)

< r <2∗(s), (1.7) (ii) 0< λ < λ1(µ)and0≤µ≤µ¯−1, whereλ1(µ) .

=infu∈H\{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 constantc∈R1such that J um

−→c, J um

−→0 inHr−1 (2.1)

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

Lemma2.2 (Hardy inequality [8,9]). Assume that1< p < Nandu∈W1,p(RN). Then |u|p

|x|pdx≤ p

N−p p

∇updx. (2.2)

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

= ((N−s)/(N−p))p,0≤s≤p. 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 dx−1 r

a(x)|u|rdx−λ 2

|u|2dx, (2.4) byLemma 2.4,J(u)is well defined andJ∈C1(H,R); the critical points of the functionalJ correspond to weak solutions of problem (1.1).

For0≤µ <µ, define the best Sobolev-Hardy constant:¯ As=. As(µ)= inf

u∈H{0}

|∇u|2−µu2/|x|2 dx

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

yε(x)= 2ε µ¯−µ(N−s)/µ¯

√µ/(2¯ −s)

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

−∆u−µ u

|x|2=|u|2∗(s)−2

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

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and satisfy

∇yε2−µyε2

|x|2

dx= yε2∗(s)

|x|s dx=A(Ns −s)/(2−s). (2.8)

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

Let

Cε=

2ε( ¯µ−µ)(N−s) µ¯

√µ/(2¯ −s)

, Uε(x)= yε(x)

Cε , (2.9)

B2l= {x∈RN,|x|<2l} ⊂Gwithl >0 andGis the domain in hypothesis (A), let 0≤φ≤ 1 be a cutting-offfunction inC∞0(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(N−2)/(2−s), (2.10)

vεqdx=

Oε√µq/(2¯ −s), 1≤q < N

µ¯+µ¯−µ, Oε√µq/(2¯ −s)|lnε|

, q= N

µ¯+µ¯−µ, Oε√µ(N¯ −q√µ)/((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−

¯

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

vεq

|x|s dx≥

c1ε√µq/(2¯ −s), 1≤q <N−s γ , c2ε√µq/(2¯ −s)|lnε|, q=N−s

γ , c3ε(√µ(N¯ −s)−µq)/(2¯ −s)√µ¯−µ, N−s

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 (N−1) sphereSN−1inRN. For 1≤q <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(2−s)√µ¯−µ/√µ¯−q(N−2)/(2−s)rN−s−1−qγ´dr

=BCqε

O(1) +ωNε−q((N−2)/(2−s))+(√µ(N¯ −s−γq)/(2´ −s)√µ¯−µ)

×

lε√µ/((s¯ −2)√µ¯−µ)

0

1 +r(2−s)√µ¯−µ/√µ¯−q(N−2)/(2−s)rN−s−1−qγ´dr

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

If−2qµ¯−µ+N−s−γq´ =0, that is,q=(N−s)/γ, vεq

|x|s dx=BCqε

O(1) +ωN

lε√µ/((s¯ −2)√µ¯−µ)

1

1 rdr

≥Bc´1ε√µq/(2¯ −s)|lnε|, (2.14) where ´c1>0 is a constant.

If−2qµ¯−µ+N−s−γq <´ 0, that is,q >(N−s)/γ, vεq

|x|s dx=BCqε

O(1) +O

ε−q((N−2)/(2−s))+(√µ(N¯ −s−γq)/(2´ −s)√µ¯−µ)

≥Bc´2ε(√µ(N¯ −s)−µq)/(2¯ −s)√µ¯−µ,

(2.15)

wherec2>0 is a constant.

If−2qµ¯−µ+N−s−γq >´ 0, that is,q <(N−s)/γ, vεq

|x|s dx=BCεq

O(1) +ωN l

0

ε+r(2−s)√µ¯−µ/√µ¯−q(N−2)/(2−s)rN−s−1−qγ´dx

=BCεq·O(1)≥Bc´3ε√µq/(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(2−s),

(2.17)

we have finished the proof ofLemma 2.5.

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Lemma2.6. Suppose (A) and0≤s <2,0≤µ <µ,¯ λ≥0. Assume that one of the following conditions holds:

(i)λ=0and max

N−s

µ¯+µ¯−µ,N−s−2µ¯−µ µ¯ , 2∗(t)

< r <2∗(s), (2.18) (ii) 0< λ < λ1(µ)and0≤µ≤µ¯−1.

Then, there existsu0∈Hr,u0=0, such that the following inequality holds:

0<sup

t≥0

J tu0

< 2−s

2(N−s)A(Ns −s)/(2−s). (2.19)

Proof. Fort≥0, we consider the functions g(t) .

=J tvε

=t2

2vε2− t2∗(s) 2∗(s)−

tr r

a(x)vεrdx−λt2

2 vε2dx,

¯ g(t)=t2

2vε2− t2∗(s) 2∗(s).

(2.20)

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

g tε

=tεvε2−tε2∗(s)−1−tεr−1

a(x)vεrdx−λtε vε2dx=0. (2.21) Then

vε2=t2ε∗(s)−2+tεr−2

a(x)vεrdx+λ vε2dx≥t2ε∗(s)−2, tε≤vε2/(2∗(s)−2). (2.22) Moreover, by hypothesis (A), we have

vε2≤tε2∗(s)−2+Cvε2(r−2)/(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)−2≥As

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 dx−λ 2

As 2

2/(2∗(s)−2)

vε2dx

≤ 2−s

2(N−s)vε2(N−s)/(2−s)−C vεr

|x|s −C vε2dx

= 2−s

2(N−s)A(Ns −s)/(2−s)+Oε(N−2)/(2−s)−C vεr

|x|s −C vε2dx.

(2.25)

In case (i), since

r >max N−s

γ ,N−s−2µ¯−µ µ¯ , 2∗(t)

, (2.26)

by (2.12), we have

vεr

|x|s ≥c3ε√µ(N¯ −s−√µr)/(2¯ −s)√µ¯−µ, µ¯ N−s−

¯ µr

(2−s)µ¯−µ <N−2 2−s.

(2.27)

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

t≥0

J tu0

=g tε< 2−s

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

vε2=Oε(N−2)/(2−s)|lnε|

, (2.29)

as 0≤µ <µ¯−1,

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

Lemma2.7. Suppose thatc∈(0, (2−s)/(2(N−s))A(Ns −s)/(2−s)). 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 dx−1 r

a(x)umrdx−λ

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

,um

=um2− um2∗(s)

|x|s dx−

a(x)umrdx−λ 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+

1−2 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)

um2≤2J 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|rdx−λ

|u|2dx= lim

m→∞

J um ,u=0.

(2.38) Setvm≡um−u, 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)umrdx−λ 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)

=c−1

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 dx≤A−s2∗(s)/2vm2∗(s) (2.44)

for allm∈N. Then by takingm→+∞, we obtain

k≤A−s2∗(s)/2k2∗(s)/2. (2.45)

Ifk >0, we have thatk≥A2s∗(s)/(2∗(s)−2). By (2.41) we deduce that J(u)=c−1

2− 1 2∗(s)

k≤c−2∗(s)−2

22∗(s) A2s∗(s)/(2∗(s)−2)=c− 2−s

2(N−s)A(Ns −s)(2−s)<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|rdx≥0, (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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