Existence of solutions for a nonlinear elliptic Dirichlet boundary value problem with an inverse square potential

DIRICHLET BOUNDARY VALUE PROBLEM WITH

AN INVERSE SQUARE POTENTIAL

SHENGHUA WENG AND YONGQING LI

Received 11 January 2006; Accepted 24 March 2006

Via the linking theorem, the existence of nontrivial solutions for a nonlinear elliptic Dir-ichlet boundary value problem with an inverse square potential is proved.

Copyright © 2006 S. Weng and Y. Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

This paper is concerned with the existence of nontrivial solutions to the following prob-lem:

−u− μ

|x|2u= |u|

p−2_{u+λu inΩ\ {0},}

u(x)=0 on∂Ω,

(1.1)

where 0∈Ω⊂RN _{(N≥3) is a bounded domain with smooth boundary, 0≤μ < μ=}

((N−2)/2)2_{, andμis the best constant in the Hardy inequality:}

C

RN u2 |x|2dx≤

RN|∇u|

2_{dx (1.2)}

(cf. [3, Lemma 2.1]), 2< p <2∗, where 2∗=2N/(N−2) is the so-called critical Sobolev exponent andλ >0 is a parameter.

Finally, inTheorem 1.3we prove, for smallλ >0, the existence of a solution to

−u− μ |x|2u=u

p−1_{+λu inΩ\ {0},}

u(x)>0 inΩ\ {0}, u(x)=0 on∂Ω.

(1.3)

Hindawi Publishing Corporation Boundary Value Problems

In the caseμ=0, problem (1.1) has been studied extensively. For example, whenp= 2∗, Capozzi et al. [1] have shown that (1.1) has at least one positive solution forN≥ 5. When 2< p <2∗, the existence of positive solutions of (1.1) has been shown in [5, Chapter 1].

Our results are the following.

Theorem1.1. Let0∈Ω⊂RN_{(N≥3)be an open bounded domain. If0≤μ < μ, then for}

anyλ >0, problem (1.1) possesses a nontrivial solution.

Remark 1.2. We mention that whenp=2∗, the existence of nontrivial solutions of (1.1) has been proved in [2, Theorem 1.3].

Theorem1.3. Let0∈Ω⊂RN _{(N≥3)be an open bounded domain. If0≤μ <μ, prob-¯}

lem (1.3) has a positive solution for0< λ < λ1, whereλ1 denotes the first eigenvalue of the operator− −μ/|x|2_.

This paper is organized as follows. InSection 2, we give some preliminaries.Section 3 is devoted to the proof ofTheorem 1.1. The proof ofTheorem 1.3is contained inSection 4.

2. Notations and preliminaries

Throughout this paper,c,ciwill denote various positive constants whose exact values are

not important.H1

0(Ω) will be denoted as the standard Sobolev space, whose norm · is deduced by the standard inner product. By| · |p, we denote the norm ofLp(Ω). All

integrals are taken overΩunless stated otherwise. OnH1

0(Ω), we use the norm

u 2

μ= |∇u|2−

μ |x|2u

2

dx. (2.1)

It follows from the Hardy inequality that the norm · μis equivalent to the usual norm

· ofH1

0(Ω).H01(Ω) with the norm · μis simply denoted byH.

By using the critical point theory, we define the action function onH:

Jμ(u)=1₂ |∇u|2−_|x|μ2u 2

dx−1

p

|u|p_dx−λ

2

|u|2_{dx. (2.2)}

It is well known that a weak solutionu∈H1

0(Ω) of (1.1) is precisely a critical point ofJμ.

That is,

Jμ(u),ϕ

= ∇u∇ϕ−_|x|μ₂uϕ

dx−

|u|p−2

uϕdx−λ

uϕdx=0 (2.3)

Definition 2.1(see [6, Definition 1.16]). Letc∈R, letEbe a Banach space, and letI∈ C1_{(E,R). Say thatIsatisfies (PS)}

ccondition if any sequence{un}inEsuch thatI(un)→c

and I(un) E−1→0 has a convergent subsequence. If this holds for everyc∈R,Isatisfies (PS) condition.

Now we will prove thatJμsatisfies (PS) condition, which is contained in the following

two lemmas.

Lemma2.2. If0≤μ < μ=((N−2)/2)2_{, then any sequence{u}

n} ⊂H01(Ω)satisfying Jμ

un

−→c, J_μun

−→0, n−→ ∞, (2.4)

is bounded inH01(Ω). Proof. Since Jμ un =1 2 ∇un 2 − μ

|x|2u 2

n

dx−1

p

un pdx−λ₂

un 2 dx, Jμ un

,ϕ= ∇un∇ϕ−_|x|μ2unϕ

dx− un p−2unϕdx−λ

unϕdx.

(2.5)

Choose 2< q < p, and letϕ=unin (2.5). Fornlarge enough,

c+ 1 +o(1)unμ

≥Jμ

un −1 q Jμ un ,un

= 1 2− 1 q

un2_μ+

1 q− 1 p

un pdx+

1 q− 1 2

λ un 2dx

≥1 2− 1 q un 2 μ+ ₁ q− 1 p

un pdx+

₁

q− 1 2

λCun

2

μ.

(2.6)

It follows fromp >2 that{un}is bounded inH01(Ω).

Lemma2.3. Under the assumption ofLemma 2.2,{un}possesses a convergent subsequence

inH.

Proof. ByLemma 2.2, going if necessary to a subsequence, we can assume that

unu inH,

un−→u inLr(Ω) for 1≤r <2∗. (2.7)

Let f(u)= |u|p−2_{u, [5, Theorem A.2] implies that f(u}

n)→ f(u) inLs, wheres=r/(r−

1). Observe that

_un−u2

μ=

J_μun−J

μ(u),un−u

+ fun−f(u)un−u+λun−u2

dx.

It is clear that

J_μun

−J_μ(u),un−u

−→0, n−→ ∞. (2.9)

It follows from the H¨older inequality that

fun−f(u)un−udx≤ fun−f(u) r/(r−1) un−u r−→0, n−→ ∞.

(2.10)

Thus we have proved that un−u μ→0,n→ ∞.

3. Proof ofTheorem 1.1

In this section, we will proveTheorem 1.1via the following linking theorem from Rabi-nowitz [5, Theorem 5.3] (see also [6]).

Proposition3.1. LetEbe a Banach space withE=Y⊕X, where dimY <∞. Suppose that I∈C1_{(E,R)and satisfies that}

(i)there existρ,α >0such thatI|∂BρX≥α;

(ii)there existe∈∂B1XandR > ρsuch that ifQ≡(BρY)⊕ {re; 0< r < R}, then

I|∂Q≤0.

IfIsatisfies (PS)ccondition with

c=inf

h∈Γmaxu∈QI

h(u), (3.1)

where

Γ=h∈C(Q,E);h|∂Q=id

, (3.2)

thencis a critical value ofIandc≥α.

Remark 3.2(see [5, Remark 5.5(iii)]). SupposeI|Y≤0 and there are ane∈∂B1

Xand

T > ρsuch that I(u)≤0 foru∈Y⊕span{e}and u ≥T, then for any largeT,Q= (Bρ

Y)⊕ {te; 0< t < T}satisfiesI|∂Q≤0.

To continue our discussion, we may assume that there isk such thatλk≤λ < λk+1,

whereλk is thekth eigenvalue of the operator (− −μ/|x|2) with Dirichlet boundary

condition (see [2,4]). Let

Y:=Yk=span

φ1,φ2,...,φk

, (3.3)

hereφidenotes the eigenfunction corresponding toλi. DecomposeH01(Ω)=Y⊕X(where Xis the topological complement ofYinH01(Ω)). For anyy∈Y, we have that

∇y 2− μ |x|2y

2 dx≤λk

y2dx, (3.4)

∇u 2− μ |x|2u

2

dx≥λk+1

u2_{dx for anyu∈ X. (3.5)}

Proposition3.3. There existρ,α >0such thatJμ|∂BρX≥α.

Proof. For anyu∈X,λk≤λ < λk+1, we obtain from (3.5) and Sobolev inequality that

Jμ(u)=1₂ |∇u|2−_|x|μ2u 2

dx−1

p

|u|p_dx−λ

2

|u|2_dx

≥1 2

λk+1−λ

λk+1 |∇u|

2₋ μ |x|2u

2

dx−1 p

|u|p_dx

≥1 2

λk+1−λ λk+1 u

2

μ−c u μp.

(3.6)

Then we can choose u μ=ρsufficiently small andα >0 such thatJμ|∂BρX≥α.

Proposition3.4. Jμverifies (ii) ofProposition 3.1.

Proof. First, for anyy∈Y, we obtain from (3.4) that

Jμ(y)=1₂ |∇y|2−_|x|μ2y 2

dx−1

p

|y|p_dx−λ

2

|y|2_dx

≤1 2

λk−λ

λk |∇y|

2₋ μ |x|2y

2_dx−1 p

|y|p_dx

=1 2

λk−λ

λk y

2

μ−

1 p|y|

p p.

(3.7)

ThusJμ(y)≤0 since all norms are equivalent onY. Lete:=φk+1be the (k+ 1)th eigen-function of (− −μ/|x|2_{), since for anyy∈Y,}

Jμ

y+tφk+1

−→ −∞ ast−→ ∞. (3.8)

It follows fromRemark 3.2that we can takeTsufficiently large and defineQ=(BTY)⊕

{re; 0< t < T}such thatProposition 3.4holds.

The proof in the case ofc≥αis the same as in the proof of [5, Theorem 5.3], by now we have completed the proof ofTheorem 1.1.

4. Proof ofTheorem 1.3

In this section, we will proveTheorem 1.3. Here we define the following Euler-Lagrange functional of (1.3) onH:

Jμ(u)=1₂ |∇u|2−_|x|μ2u 2

dx−1

p u

+p dx−λ

2 u

+2

dx, (4.1)

whereu+_{=max{u, 0}, and for anyϕ∈C}∞ 0(Ω),

Jμ(u),ϕ

= ∇u∇ϕ− μ |x|2uϕ

dx− u+p−1ϕdx−λ u+ϕdx. (4.2)

By using the same method in the proof ofTheorem 1.1, we obtain thatJμ satisfies (PS)

It is easy to check thatJμ(u)∈C1(H01(Ω),R), we will verify the assumptions of the mountain pass theorem. By the Sobolev theorem, there existsc1>0, such that foru∈ H, u Lp_(Ω)≤c₁ u _μ. Hence we have

Jμ(u)=1₂ |∇u|2−_|x|μ2u 2

dx−1

p u

+p_dx−λ

2 u

+2_dx

≥1 2 u

2

μ−

c1 p u

p μ− λ

2λ1 u 2

μ

=1 2

1− λ λ1

u 2

μ−c

1 p u

p μ.

(4.3)

So there isr >0 such that

b:= inf

u μ=r

Jμ(u)>0=Jμ(0). (4.4)

Letu∈Hwithu >0 onΩ, we have, fort≥0,

Jμ(tu)=t

2

2 |∇u|

2₋ μ |x|2u

2

dx−tp p

(u+₎p_dx−λt2

2 u

+2_{dx. (4.5)}

Since p >2, there existse:=tu, such that e μ> r andJμ(e)≤0. By the mountain pass

theorem,Jμhas a positive critical value, and problem

−u− μ |x|2u=

u+p−1_+λu+ _{inΩ\ {0},}

u∈H01(Ω)

(4.6)

has a nontrivial solutionu. Multiplying the equation byu−and integrating overΩ, we find

0= ∇u− 2− μ |x|2

u−2

dx= u− 2

μ. (4.7)

Henceu−=0, that is, u≥0. A standard elliptic regularity argument implies thatu∈ C2_{(Ω\ {0}), in which case, by the strong maximum principle,uis positive, thus is the} solution of problem (1.3).

Acknowledgments

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Shenghua Weng: Department of Mathematics, Fujian Normal University, Fuzhou 350007, China E-mail address:[email protected]