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JOURNAL OF PARTIAL DIFFERENTIAL EQUATIONSJ. Part. Diff. Eq., Vol.26, No. 1, pp. 25-38

doi: 10.4208/jpde.v26.n1.3March 2013

Existence of Nontrivial Weak Solutions to Quasi-Linear

Elliptic Equations with Exponential Growth

WANG Chong

Department of Mathematics, the George Washington University,Washington DC 20052, USA.

Received 13 June 2012; Accepted 1 December 2012

Abstract. In this paper, we study the existence of nontrivial weak solutions to thefollowing quasi-linear elliptic equations

nu+V(x)|u|n2u =

f(x,u)

|x| , x

n (n2),

wherenu=div(|u|n2u), 0

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26 C. Wang / J. Partial Diff. Eq.,26(2013), pp. 25-38

In this paper we consider quasi-linear elliptic equations in the whole Euclidean space

nu+V(x)|u|

n2

u =

f(x,u)

|x| , x

n

(n2), (1.2)

where nu =div(|u|n2u), 00 such that for all(x,s)RnR+,

|f(x,s)|b1sn1 +b2

e0|s|

nn1

n2

k=0

k0|s| knn1

k!

;

(H2)There exists>n, such that for allxRn

ands>0,

00, such that for all xRn ands>R0,

F(x,s)M0f(x,s);

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2 Compactness analysis

We will give some preliminary results before proving Theorem 1.1. Define a function:NRR by

(n,s) = esn2

k=0

sk

k!=

k=n1

sk

k!. (2.1)

Lets0,p1 be real numbers andn2 be an integer, then there holds (see [17])

(n,s)

p(n,ps). (2.2)

Problem (1.2) is closely related to a singular Trudinger-Moser type inequality [15].That is, for all>0, 0

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Existence of Nontrivial Weak Solutions to Quasi-Linear Elliptic Equations 29

Lemma 2.1. Assume V(x)V0in Rn,(H1), (H2)and (H3)hold. Then for any nonnegative,compactly supported function uW1,n(Rn)\{0}, there holds J(tu) as t+ .

Proof. We follow the line of [15]. (H2)and (H3) imply that there existsR0> 0 such thatfor alls>R0,

s

s F(x,s)

F(x,s) =

s

lnF(x,s)

,

therefore, s

R0

F(x,s)

F(x,R0).

It follows thatF(x,s)F(x,R0)R

0 s

.

Letc1= F(x,R0)R0 , then we have for all(x,s)[0,+), F(x,s) c1s

c2, which is

under the assumption thatu is supported in a bounded domain and c2 is a positiveconstant. This implies that

J(tu)tn

nun

c1tu

|x| dx = tn

un

n c1t

n

u

|x|dx

.

Since>n, this impliesJ(tu) ast+.

Lemma 2.2. Assume that V(x) V0 in Rn, (H1) and (H4) are satisfied. Then there exist>0and r>0such that J(u)for all u= r.

Proof. According to(H4), there exist,>0 such that if|s|,

n|F(x,s)||s|n

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whereC = C. Here we also use the inequality

Rn

|u|n+1R(0,u)

|x| dx

Cu

n+1

, (2.9)

which is taken from Lemma 4.2 in [15]. According to the definition of, we get

Rn

|u|n

|x|dx

un

. (2.10)

Thanks to (2.8), (2.9), and (2.10), we obtain

J(u)1

nun

Rn

n

|u|n

|x| dx

Rn

C|u|n+1R(0,u)

|x| dx

1

nun

n un

Cun+1

=u

nun1Cun

. (2.11)

For sufficiently smallr>0, we have

nrn1Crn

2nrn1, (2.12)

which is due to>0. Therefore, according to (2.11) and (2.12), for all u = r,

J(u) r

2n

rn1 =

2n

rn.

Finally, let= 2n rn, we have J(u)for all u= r.

Lemma 2.3. Critical points of J are weak solutions of(1.2).

Proof. Though the proof is standard, we write it for completeness. Define a functiong(t) =J(u+t), namely

g(t) =1

n

Rn

(|(u+t)|n +V(x)|u+t|n)dxRn

F(x,u+t)

|x| dx.

By a simple calculation,

g(t)

t=0

=J(u+t)

t=0

=J(u).

Let f1(t) = |(u+t)|n, f2(t) = |u+t|n, and

f3(t) =Rn

F(x,u+t)

|x| dx.

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Clearly we have

f

1(t)

t=0 =

n

22|u|

n2

u= n|u|

n2

u,

f

2(t)

t=0

= n

2|u|n22u= n|u|n2u,

f

3(t)

t=0

=Rn

f(x,u)

|x| dx.

Combining the above, we have for allE,

J(u)=

Rn

(|u|n2 u+V(x)|u|n2u)dxRn

f(x,u)

|x| dx. (2.13)

Therefore,J(u)= 0 is equivalent toRn

(|u|n2 u+V(x)|u|n2u)dxRn

f(x,u)

|x| dx = 0.

Hence we get the desired result.

Lemma 2.4. Assume(H5)is satisfied, then there exists a function upE which satisfies up=Sp, and for t [0,+), we define

J(tup)tn

nup

nRn

F(x,tup)

|x| dx.

There holds

maxt0

J(tup)0, there exists a constantKsuch that whenk>K,

Rn

|uk|p|up|p

|x| dx

< .Therefore,

Rn

|uk|p

|x| dx

Rn

|up|p

|x| dx = 1. (2.15)

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Next we will proveup lim

kinfuk= Sp. (2.16)

Since uk up weakly in E, we know ukup weakly in Ln(Rn). According to thedefinition of weak convergence and the Holder inequality, we get

Rn|up|

n dx limk

infRn|uk|

n dx. (2.17)

Similarly to the proof of (2.15), we knowRn

V(x)|uk|n dx

Rn

V(x)|up|n dx. (2.18)

Thanks to (2.17) and (2.18), (2.16) holds. Meanwhile, by the definition ofSp , we knowSpup . Therefore, we know up = Sp. According to(H5), we have

Rn

F(x,tup)|x|

dxCptp

p. (2.19)

Due to the definition ofJ(tup)and (2.19), we have

J(tup)tn

nS np Cp

tp

p.

Let

f(t) =tn

nS np Cp

tp

p,

and by calculation we know for any real number t,

f(t)f

S np

Cp

1pn

.

This means

tn

n S np Cp

tp

p

pn

np

Snp

pnp

Cn

pnp

.

If we set

Cp> (pn

p )

pnn

n0(n)n

(n1)(pn)n

S p

p,

then we have

pn

np

Snp

pnp

Cn

pnp

0, there existsM> C such that

|u|M

f(x,u)

|x| dx< , (2.28)

whereCis the constant in (2.25). According to (2.25), we know

|uk|M

f(x,uk)

|x| dx

1

M

|uk|M

f(x,uk)uk|x|

dx

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according to the generalized Lebesgues dominated convergence theorem, we know

limk|uk|0, J(uk)0,

inE, where

c = min

maxu

J(u) and =C([0,1],E) :(0) = 0,(1) = e

.

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According to Lemma 2.5, we know that up to a subsequence

uku weakly inE,

uku strongly inLq(Rn), for anyq1,

limk

Rn

F(x,uk)

|x| dx =

Rn

F(x,u)

|x| dx,

u is a weak solution of(1.2).

Next we will prove that the solutionu which we get in the above is nontrivial. Supposeu0. Due toF(x,u)0 for allxRn, we get

limk

Rn

F(x,uk)

|x| dx =

Rn

F(x,u)

|x| dx = 0. (3.1)

According to (2.20), we have

limk

1

nuk

nRn

F(x,uk)

|x| dx

= c>0. (3.2)

Combining (3.1) and (3.2), we can obtain that

limk

ukn= nc>0. (3.3)

By Lemma 2.4, we get

00, such that

ukn

n

n

n00

n1, (3.5)

for allk>K. According to (3.5), we can chooseq>1 sufficiently close to 1 such that

q0 uk nn1

1

n

n

002

, (3.6)

for allk>K. By(H1

)and (2.1), we have

|f(x,uk)uk|b1|uk|n +b2|uk| (n,0|uk|

nn1 ).

It follows that

Rn

|f(x,uk)uk|

|x| dxb1

Rn

|uk|n

|x| dx+b2

Rn

|uk| (n,0|uk| nn1 )

|x| dx. (3.7)

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Letting 1/q+1/q=1, and according to(3.6),(3.7), the Holder Inequality,(2.2), and(2.4),

we have

Rn

|f(x,uk)uk|

|x| dxb1

Rn

|uk|n

|x| dx+b2

Rn

|uk|q

|x| dx

1q

Rn

(n,q0|uk| nn1 )

|x| dx

1q

b1

Rn

|uk|n

|x| dx+C

Rn

|uk|q

|x| dx

1q

0, as k. (3.8)

According to (2.22) and (3.8), we have

uk0, as k.

This is in contradiction with (3.3). Thus the solutionuof (1.2) is nontrivial.

Testing Eq. (1.2) with u

, the negative part ofu, we conclude that u

0. Henceu0.

Acknowledgments

The author is supported by the China Scholarship Council.

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