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Volume 2009, Article ID 147308, 19 pagesdoi:10.1155/2009/147308 Research Article On Multiple Solutions of Concave and Convex Kuan-Ju Chen Department of Applied Science, Naval Academy, 90

Volume 2009, Article ID 147308, 19 pages

doi:10.1155/2009/147308

Research Article

On Multiple Solutions of Concave and Convex

Kuan-Ju Chen

Department of Applied Science, Naval Academy, 90175 Zuoying, Taiwan

Correspondence should be addressed to Kuan-Ju Chen,kuanju@mail.cna.edu.tw

Received 18 February 2009; Accepted 28 May 2009

Recommended by Martin Schechter

We consider the existence of multiple solutions of the elliptic equation onRNwith concave and convex nonlinearities

Copyrightq 2009 Kuan-Ju Chen 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

First, we look for positive solutions of the following problem:

u > 0, inRN ,

u ∈ H1

RN

,

1.1

where λ > 0 is a real parameter, 1 < p < 2 < q < 2∗ 2N/N − 2, N ≥ 3 We will impose some assumptions on ax and bx Assume

a1 ax ≥ 0, ax ∈ L α/α−1RN  ∩ L∞RN , where 1 < α < 2∗/p,

Such problems occur in various branches of mathematical physics and population

In the present paper, we discuss the Nehari manifold and examine carefully the connection between the Nehari manifold and the fibrering maps, then using arguments

In5, Ambrosetti et al showed that for λ > 0 small with respect to μ > 0 there exist

−Δu  λ|u| p−2 u  μ |u| q−2 u, inΩ,

with negative energy:

2



Ω|∇u|2−λ

p



Ω|u| p−μ

q



Ω|u| q

proved by few people

−Δu  u  μax|u| p−2 u  λb x|u| q−2 u, inRN ,

u ∈ H1

RN

,

1.4

by taking advantage of the oddness of the nonlinearity

Our main results state the following

Theorem 1.1 Under the assumptions (a1) and (b1), there exists λ∗> 0, such that for all λ ∈ 0, λ∗,

problem1.1 has at least two positive solutions u0 and u1, u0is a local minimizer of I λ and I λ u0 < 0,

where I λ is the energy functional of problem1.1.

Theorem 1.2 Under the assumptions (a1) and (b1), for every λ > 0 and μ ∈ R, the problem 1.4

has infinitely many solutions with positive energy and for every μ > 0 and λ ∈ R, infinitely many solutions with negative energy.

2 The Existence of Two Positive Solutions

I λ u  1

2



|∇u|2 u2

p



a x|u| p−λ

q



indication

Through this paper, we denote the universal positive constant by C unless some

Λλu ∈ H1

RN :

||u||2−



a x|u| p − λ



I λ u 

1

p ||u||2− λ

1

q−1



1

q ||u||2−

1

p− 1

2.4

The Nehari manifold is closely linked to the behavior of the functions of the form

we have

φ u t  t2

2||u||2−t p

p



a x|u| p − λ t q

q



b x|u| q ,

φu t  t||u||2− t p−1



a x|u| p − λt q−1



b x|u| q

,

φ u t  ||u||2−p − 1

t p−2



a x|u| p − λq − 1

t q−2



b x|u| q

2.5

minima, local maxima, and points of inflection, and so we define

Λ

λ u ∈ Λ λ : φ u 1 > 0,

Λ−

λ u ∈ Λ λ : φ u 1 < 0,

Λ0

λ u ∈ Λ λ : φ u1  0,

2.6

and note that if u ∈ Λ λ , that is, φu1  0, then

φ u 1 2− p||u||2− λq − p 

b x|u| q

2− q||u||2−p − q 

a x|u| p

2.7

This section will be devoted to proveTheorem 1.2 To proveTheorem 1.2, several preliminary results are in order

Lemma 2.1 Under the assumptions (a1), (b1), there exists λ∗ > 0 such that when 0 < λ < λ∗, for every u ∈ H1RN , u /≡ 0, there exist unique t  tu > 0, t−  t−u > 0 such that tu ∈ Λ−λ ,

t−u ∈ Λλ In particular, one has

t >

2− qu2

p − q 

a x|u| p

1/p−2

I λ t−u  min t∈0,t I λ tu < 0 and I λ tu  max t≥t−I λ tu.

Proof Given u ∈ H1RN  \ {0}, set ϕ u t  t2−q||u||2− t p−q

ax|u| p Clearly, for t > 0, tu ∈ Λ λ

if and only if t is a solution of

ϕ u t  λ



Moreover,

ϕu t 2− qt1−q||u||2−p − q

t p−q−1



a x|u| p

tmax



2− q||u||2

p − q 

a x|u| p

1/p−2

If λ > 0 is sufficiently large, 2.9 has no solution, and so φ u t  I λ tu has no critical

If, on the other hand, λ > 0 is sufficiently small, then there exist exactly two solutions

tu > t−u > 0 of 2.9, where t tu, t−  t−u, ϕ

u t− > 0, and ϕ

u t < 0.

It follows from 2.7 and 2.10 that φ

tu 1  t q1 ϕu t, and so tu ∈ Λ−λ , t−u ∈ Λλ;

moreover φ uis decreasing in0, t−, increasing in t−, t, and decreasing in t, ∞.

exist exactly two solutions of problem2.9 for all u ∈ H1RN \ {0}, that is,

λ



b x|u| q

< ϕ u tmax 

2− q

p − q

2−q/p−2

p − 2

p − q

||u|| 2p−2q/p−2





a x|u| p ≤ ||a|| L α/α−1 ||u|| p

L αp ≤ ||a|| L α/α−1 S p αp ||u|| p

ϕ u tmax ≥

p − q

2−q/p−2

p − 2

p − q

||u|| 2p−2q/p−2



||a|| L α/α−1 S p αp ||u|| p2−q/p−2



p − q

2−q/p−2

p − 2

p − q

||u|| q



||a|| L α/α−1 S p αp

2−q/p−2 ,

2.14

and then



b x|u| q ≤ M||u|| q

L q ≤ MS q

q ||u|| q

q

p − q

2− q

2−q/p−2

p − q

p − 2 ||a|| L α/α−1 S p αp

2−q/p−2

ϕ u tmax

 cϕ u tmax,

2.15

independent of u, hence

ϕ u tmax − λ



bx|u| q < ϕ u tmax for all u ∈ H1R N  \ {0} provided λ < 1/2c  λ∗

t−u and a local maximum at t tu; moreover I λ t−u  min t∈0,t I λ tu < 0 and I λ tu 

maxt≥t−I λ tu.

In particular, we have the following result

Corollary 2.2 Under the assumptions (a1), (b1), when 0 < λ < λ∗, for every u ∈ Λ λ , u / ≡ 0, one has

2− q||u||2−p − q 

a x|u| p

/

(i.e.,Λ0

λ  ∅).

Proof Let us argue by contradiction and assume that there exists u ∈ Λ λ\ {0} such that

2 − q||u||2− p − qax|u| p  0, this implies

λ



b x|u| q  ||u||2−



a x|u| p



p − 2



p − 2

p−q/p−2 

a x|u| p q−2/p−2



p − 2

2− q

1

p − q

p−q/p−2

p − q 

a x|u| p p−q/p−2 

a x|u| p q−2/p−2



p − 2

p − q

2− q

p − q

2−q/p−2

||u||2p−q/p−2



a x|u| p q−2/p−2

 ϕ u tmax

2.18 which contradicts2.12 for 0 < λ < λ∗

Lemma 2.3 Under the assumptions (a1), (b1), if 0 < λ < λ∗, for every u ∈ Λ λ , u / ≡ 0, then there exist

a  > 0 and a C1-map t  tw > 0, w ∈ H1RN , ||w|| <  satisfying that



t0, w 2



uw − λq

b x|u| q−2

uw

2− q||u||2−p − q 

Proof We define F : R × H1RN → R by

F t, w  t||u − w||2− t p−1



a x|u − w| p − λt q−1



Since F1, 0  0 and F t 1, 0  ||u||2− p − 1ax|u| p − λq − 1bx|u| q  2 − q||u||2− p −

q

ax|u| p /≡ 0 byCorollary 2.2, we can apply the implicit function theorem at the point

1, 0 and get the result.

can establish the existence of the first positive solution

Proposition 2.4 If 0 < λ < λ∗, then the minimization problem:

c0 inf I λ

Λλ

 inf I λ

Λ 

λ

2.21

is achieved at a point u0∈ Λ

λ which is a critical point for I λ with u0> 0 and I λ u0 < 0 Furthermore,

u0is a local minimizer of I λ

Proof First, we show that I λis bounded from below inΛλ Indeed, for u ∈ Λ λ, from2.13,

we have

I λ u  1

2||u||2− 1

p



a x|u| p−λ

q



b x|u| q



1

q ||u||2−

1

p− 1

≥

1

q ||u||2−

1

p− 1

q ||a|| L α/α−1 S p αp ||u|| p

2.22

0 < t−v < tv such that t−vv ∈ Λ λ Thus,

c0≤ I λ

t∈0,tv I λ tv < 0. 2.23

c0≤ I λ u n  < c0 1

I λ v ≥ I λ u n − 1

I λ u n 

1

q ||u n||2−

1

p− 1

< c0 1

n < 0, 2.26

from which we deduce that for n large



a x|u n|p≥ pq

p − q c0, ||u n||2≤ 2

q − p

p

q − 2



which yields

for suitable b1, b2 > 0.

Now we will show that

I

Since u n∈ Λλ, byLemma 2.3, we can find a  n > 0 and a C1-map t n  t n w > 0, w ∈ H1RN,

||w|| <  nsatisfying that

that 1/2 ≤ t n w ≤ 3/2 for ||w|| <  n

It follows from2.25 that

I λ t n wu n − w − I λ u n ≥ −1

that is,



I λu n , t n wu n − w − u n



 o||t n wu n − w − u n||

Consequently,

t n wI λu n , w 1 − t n wI λu n , u n



By the choice of  n, we obtain



I λu n , w≤ C

nt

n 0, w  o||w||  C

n ||w||

 ot

n 0, w||u n ||  ||w||.

2.34

ByLemma 2.3,Corollary 2.2, and the estimate2.28, we have



tn 0, w 2



∇u n ∇w  u n w  − pa x|u n|p−2 u n w − λq

b x|u n|q−2 u n w

2− q||u n||2−p − q 

a x|u n|p

≤ C||w||,

2.35



I λu n , w≤ C

Hence, for any  ∈ 0,  n, we have

I

λ u n  1

 ||w||sup



I λu n , w≤ C

n 1

Let u0 ∈ H1RN  be the weak limit in H1RN  of u n From2.29,



I λu0, w 0, ∀w ∈ H1

RN

c0≤ I λ u0 ≤ lim

n → ∞ I λ u n   c0; 2.39

that is,

c0 I λ u0  inf

Λλ

λ In fact, if u0 ∈ Λ−

λ, byLemma 2.1, there exists only one t > 0

such that tu0∈ Λ−

λ , we have t tu0  1, t−  t−u0 < 1 Since

dI λ t−u0

dt  0, d2I λ t−u0

there exists t≥ t > t−such that I λ tu0 > I λ t−u0 ByLemma 2.1,

I λ

t−u0

< I λ



tu0



this is a contradiction

I λ su ≥ I λ

t−u

∀0 < s <



2− q||u||2

p − q 

a x|u| p

1/p−2

In particular, for u  u0∈ Λ

t−u0  1 <



2− q||u0||2

p − q 

a x|u0| p

1/p−2

Let  > 0 sufficiently small to have

1 <



2− q||u0 − w||2

p − q 

a x|u0 − w| p

1/p−2

FromLemma 2.3, let tw > 0 satisfy twu0 − w ∈ Λ λfor every||w|| <  By the continuity

of tw and t0  1, we can always assume that

t w <



2− q||u0 − w||2

p − qax |u0 − w| p

1/p−2

0 < s <



2− q||u0 − w||2

p − q 

a x|u0 − w| p

1/p−2

we have

Taking s  1, we conclude

Furthermore, taking t−|u0| > 0 with t−|u0||u0| ∈ Λ

λ, therefore,

I λ u0 ≤ I λ

can show that u0 > 0 in R N

λ and c0  infΛλ I λ  infΛ

λ I λ , thus, in the search of our second positive

solution, it is natural to consider the second minimization problem:

c1 inf

Λ −

λ

u > 0, inRN ,

u ∈ H1

RN

.

2.52

We state here some known results for problem2.52 First of all, we recall that Lions 14

λ  inf{I∞

λ u : u ∈ H1RN , u / 0 , I∞

λ u  0} > 0, where I∞

λ u  1/2||u||2− 1/qλb∞

|u| q

For future reference, note also that a minimum exists and is realized by a ground state ω > 0 in

RN such that S∞λ  I∞

λ ω  sup s≥0 I λ∞sω Gidas et al 15 showed that there exist a1,a2 > 0

such that for all x ∈ R N,

a1|x|  1−N−1/2 e −|x| ≤ ωx ≤ a2|x|  1 −N−1/2 e −|x| 2.53

Lemma 2.5 Let ax ∈ L α/α−1RN  ∩ L∞RN , where 1 < α < 2∗/p and 1 < p < 2 If u n weakly in H1RN , then a subsequence of {u n }, still denoted by {u n }, satisfies

lim

n → ∞



Proof Since ax ∈ L α/α−1RN , then for every  > 0, there exists R0 > 0 such that



|x|>R0

|ax| α/α−1

dx

α−1/α

locRN , 1 ≤ s < 2N/N − 2, then we

have



|x|≤R0

|u n − u| αp

dx

1/αp

Observe that by H ¨older inequality we have



a x|u n − u| p

dx 



|x|≤R0

a x|u n − u| p

dx 



|x|>R0

a x|u n − u| p

dx ≤ C, 2.57

hence limn → ∞

ax|u n − u| p 0

Our first task is to locate the levels free from this noncompactness effect

Proposition 2.6 Every sequence {u n } ⊂ H1RN , satisfying

a I λ u n   c  o1 with c < c0  S∞

λ ,

b I

λ u n   o1 strongly in H−1RN ,

has a convergent subsequence.

Proof It is easy to see that {u n } is bounded in H1RN , so we can find a u ∈ H1RN such

locRN,



I λu, w 0, ∀ w ∈ H1

RN

that is, u is a weak solution of problem 1.1 and u ∈ Λ λ Set v n  u n − u to get v n

locRN , 1 ≤ s < 2N/N − 2, we

such that||v n || ≥ β > 0 Apply the Brezis-Lieb theorem see 16 andLemma 2.5,

I λ u n  1

2||u n||2−1

p



a x|u n|p−λ

q



b x|u n|q

 I λ u  1

2||v n||2−1

p



a x|v n|p−λ

q



b x|v n|q  o1

 I λ u  1

2||v n||2−λb∞

q



|v n|q−λ

q



bx − b∞|v n|q  o1.

2.59

o1 I λu n , u n



 ||u n||2−



a x|u n|p − λ



b x|u n|q

I λu, u ||v n||2−



a x|v n|p − λ



b x|v n|q  o1

 ||v n||2− λb∞

|v n|q − λ



bx − b∞|v n|q  o1.

2.60

Byb1, for any  > 0, there exist: R0 > 0 such that |bx − b∞| <  for |x| ≥ R0 Since v n → 0

strongly in L s locRN  for 1 ≤ s < 2N/N − 2, {v n } is a bounded sequence in H1RN, therefore,



bx − b∞|v n|q ≤ CB



||v n||2− λb∞

|v n|q  o1, I λ u n  ≥ c01

2||v n||2− λ

q b

∞

Since||v n || ≥ β > 0, we can find a sequence {s n }, s n > 0, s n → 1 as n → ∞, such that t n  s n v n

satisfying||t n||2− λb∞

|t n|q  0 Hence

I λ u n  ≥ c01

2||t n||2−λ

q b

∞

|t n|q  o1 ≥ c0  S∞

that is, c  lim n → ∞ I λ u n  ≥ c0  S∞

strongly

Lete  1, 0, , 0 be a fixed unit vector in R N and ω be a ground state of problem

Proposition 2.7 Under the assumptions (a1) and (b1), Then

I λ u0  tω < c0  I∞

Proof.

I λ u0  tω  12||u0  tω||2− 1

p



a x|u0  tω| p−λ

q



b x|u0  tω| q

< I λ u0 1

2||tω||2−λ

q t q



b x|ω| q

≤ I λ u0 1

2||tω||2−λ

q b

∞

|tω| q

 I λ u0  I∞

λ tω

≤ c0  I∞

λ ω.

2.65

Proposition 2.8 If 0 < λ < λ∗, for c1  infΛ−

λ I λ , one can find a minimizing sequence {u n} ⊂ Λ−

λ such that

a I λ u n   c1  o1,

b I

λ u n   o1 strongly in H−1RN ,

c c1 < c0 S∞

λ Proof Set Σ  {u ∈ H1RN  : ||u||  1} and define the map Ψ : Σ → Λ−

λ

U1



u  0 or u : ||u|| < t

u

U2



u : ||u|| > t

u

2.66

λ ⊂ U1

We will prove that there exists t1 such that u0 t1 ω ∈ U2 Denote t0 tu0tω/||u0

tω|| Since tu0 tω/||u0  tω||u0  tω/||u0  tω|| ∈ Λ−

t20−t

q

0λ

b x|u0  tω| q

p

0



a x|u0  tω| p

Thus

t0≤



||u0  tω||

λ

bx |u0  tω| q1/q

q/q−2





||u0 /t  ω||

λ

bx |u0 /t  ω|q1/q

q/q−2

≤



||u0 /t  ω||

λ

b∞|u0 /t  ω|q1/q

q/q−2

−→ ||ω|| < ∞ as t −→ ∞.

2.68

Therefore, there exists t2 > 0 such that t0  tu0  tω/||u0  tω|| < 2||ω||, for t ≥ t2 Set

t1> t2 2, then

||u0  t1 ω||2  ||u0||2 t2

1||ω||2 2t1



 ||u0||2 t2

1||ω||2 2t1 λb∞



|ω| q−1 u0

> t21||ω||2> 4 ||ω||2> t20,

2.69

0, 1 such that u0  st1 ω ∈ Λ−λ Therefore, c1 ≤ I λ u0  st1 ω < c0 S∞

Proposition 2.7

principle8 gives a sequence {u n} ⊂ Λ−

Proposition 2.9 If 0 < λ < λ∗, then the minimization problem c1  infΛ−

λ I λ is achieved at a point

u1∈ Λ−

λ which is a critical point for I λ and u1> 0.

Proof Applying Propositions 2.6 and 2.8, we have u n → u1 strongly in H1RN

Consequently, u1 is a critical point for I λ , u1∈ Λ−

λsince Λ−

λis closed and Iλ u1  c1.

Let t|u1| > 0 satisfy t|u1||u1| ∈ Λ−

λ Since u1 ∈ Λ−

conclude that

t≥tmaxu1 I λ tu1 ≥ I λ t|u1|u1 ≥ I λ t|u1||u1| ≥ c1 2.71

Hence, It|u1||u1|  c1, So we can always take u1≥ 0 By standard regularity method and

Proof of Theorem 1.1 Applying Propositions 2.4 and 2.9, we can obtain the conclusion of

Theorem 1.1

X

j : spane1, , e j

, X k: ⊕j≥k X

j

, X k: ⊕j≤k X

j

and C, C1 , C2, , denote possibly different positive constants.

I u  1

2



|∇u|2 u2

p



a x|u| p−λ

q



b x|u| q

Proposition 3.1 Under the assumptions (a1) and (b1), for every λ > 0 and μ ∈ R, the problem 1.4

has infinitely many solutions with positive energy Iu.

Proof We will show that the energy functional Iu satisfies the assumptions of Fountain

convergent subsequence

A3 infρ>0supu∈X k ,||u||≥ρ Iu ≤ 0, for every k ∈ N.

A4 supr>0infu∈X k ,||u||r Iu → ∞ as k → ∞.

We define

u∈X k−{0}



b x|u| q 1/q

then

Indeed, clearly we have

λ  ∅).

Proof Let us argue by contradiction and assume that there exists u ∈ Λ λ\... class="text_page_counter">Trang 4

This section will be devoted to proveTheorem 1.2 To proveTheorem 1.2, several preliminary results are in order... t−, increasing in t−, t, and decreasing in t, ∞.

exist exactly two solutions of problem2.9 for