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Three solutions for equations involving nonhomogeneous operators of p-Laplace type in R N

Abstract

In this paper, we are concerned with the following elliptic equation

div ( φ ( x , u ) ) =λf(x,u)in  R N ,

where the function φ(x,v) is of type | v | p 2 v and f: R N ×RR satisfies a Carathéodory condition. We establish the existence of at least three weak solutions for the problem above which is based on an abstract three critical points theory due to Ricceri. Moreover, we determine precisely the intervals of λ’s for which the given problem possesses either only the trivial solution or at least two nontrivial solutions.

MSC:35D30, 35D50, 35J15, 35J60, 35J62.

1 Introduction

In this paper, we establish the existence of at least three solutions for equations of the p-Laplace type

(P λ ) div ( φ ( x , u ) ) =λf(x,u)  in R N ,

where the function φ(x,v) is of type |v | p 2 v and f: R N ×RR satisfies a Carathéodory condition. A Ricceri-type three critical points theorem has been extensively studied by many researchers (see [15] and the references therein), but the results on the localization of the interval for the existence of three solutions are rare. The authors in [3, 4] investigated the existence of multiple solutions for quasilinear nonhomogeneous problems with Dirichlet boundary conditions by applying an abstract three critical points theorem which is the extension of the famous result of Ricceri [6, 7].

Ricceri’s theorems in [68] gave no further information on the size and location of an interval of values λR for the existence of at least three critical points. However, further information concerning these points was given in [9]. Also the authors in [3] investigated the localization of the interval for the existence of three solutions for the Dirichlet problem involving the p-Laplace type operators which was motivated by the work of Arcoya and Carmona [2]. It is well known that the first eigenvalue of the p-Laplacian plays a decisive role in obtaining these results in [3, 9]. Hence, by using the positivity of the principal eigenvalue of the p-Laplacian in R N , which was given in [1012], we localize a three critical points interval for the problem above as in [3, 9]. Especially, the main aim of this paper is to determine precisely the intervals of λ’s for which problem (P λ ) admits only the trivial solution and for which problem (P λ ) has at least two nontrivial solutions, following the basic idea in [3]. To do this, we consider some of the basic properties for the integral operator corresponding to problem (P λ ) in the setting of weighted Sobolev spaces.

To this end, we recall in what follows some definitions of the basic function space which will be treated in the next sections. For a deeper treatment on these spaces, we refer to [12, 13].

Let , be the Euclidean scalar product on R N or the usual pairing of X and X, where X denotes the dual space of X. Let 1<p<N and set p :=Np/(Np). Let ω be a weight function defined by

ω(x)= 1 ( 1 + | x | ) p for x R N .

Assume that

  1. (A)

    a belongs to L ( R N ) and there is a positive constant a 0 such that

    a(x) a 0 for almost all x R N .

Let X be the completion of C 0 ( R N ) with respect to the norm

u X = ( R N a ( x ) | u | p d x + R N ω ( x ) | u | p d x ) 1 p .

From Hardy’s inequality and assumption (A), it follows that

R N ω(x) | u | p dx 1 a 0 ( p N p ) p R N a(x) | u | p dx,

which implies that on X, the norm X is equivalent to the other norm a given by

u a = ( R N a ( x ) | u | p d x ) 1 p .

Note that there exist positive constants c and c such that

c u X u a c u X
(1.1)

for all uX. The following Sobolev inequality will be used in the sequel:

( R N | u | p d x ) 1 p c 0 ( R N a ( x ) | u | p d x ) 1 p

for some positive constant c 0 (see [12]).

This paper is organized as follows. We first present some properties of the corresponding integral operators. Then we give and prove our main results in Theorem 2.12 and Theorem 2.14.

2 Main results

Definition 2.1 We say that uX is a weak solution of problem (P λ ) if

R N φ(x,u)vdx=λ R N f(x,u)vdx

for all vX.

We assume that φ(x,v): R N × R N R N is a continuous derivative with respect to v of the mapping Φ 0 : R N × R N R, Φ 0 = Φ 0 (x,v), that is, φ(x,v)= d d v Φ 0 (x,v). Suppose that φ and Φ 0 satisfy the following assumptions:

(J1) The following equalities

Φ 0 (x,0)=0and Φ 0 (x,v)= Φ 0 (x,v)

hold for all x R N and for all v R N .

(J2) φ: R N × R N R N satisfies the following conditions: φ(,v) is measurable for all v R N and φ(x,) is continuous for almost all x R N .

(J3) There are a function σ 0 L p ( R N ) and a positive constant d such that

| φ ( x , v ) | σ 0 (x)+d | v | p 1

for almost all x R N and for all v R N .

(J4) Φ 0 (x,) is strictly convex in R N for all x R N .

(J5) The following relations

c 1 a(x) | v | p φ(x,v)vand c 1 a(x) | v | p p Φ 0 (x,v)

hold for all x R N and v R N , where c 1 is a positive constant.

Let us define the functional Φ:XR by

Φ(u)= R N Φ 0 (x,u)dx

for any uX. Under assumptions (J1)-(J3) and (J5), it follows from [[14], Lemma 3.2] that the functional Φ is well defined on X, Φ C 1 (X,R) and its Fréchet derivative is given by

Φ ( u ) , v = R N φ(x,u)vdx
(2.1)

for any uX.

Next, taking inspiration from the argument given in [3], we will show that the operator Φ is a mapping of type ( S + ) which plays an important role in obtaining our main results.

Lemma 2.2 Assume that (A) and (J1)-(J5) hold. Then the functional Φ:XR is convex and weakly lower semicontinuous on X. Moreover, the operator Φ is a mapping of type ( S + ), i.e., if u n u in X as n and lim sup n Φ ( u n ) Φ (u), u n u0, then u n u in X as n.

Proof From assumption (J4), the operator Φ is strictly convex and thus Φ is strictly monotone (see [[15], Proposition 25.10]), namely

Φ ( u ) Φ ( v ) , u v >0
(2.2)

for uv. The convexity of Φ 0 also implies that Φ is weakly lower semicontinuous in X, that is, u n u implies

Φ(u) lim inf n Φ( u n ).
(2.3)

Now we claim that the operator Φ is a mapping of type ( S + ). Let { u n } be a sequence in X such that u n u in X as n and

lim sup n Φ ( u n ) Φ ( u ) , u n u 0.
(2.4)

From relations (2.2) and (2.4), we have

lim n R N ( φ ( x , u n ) φ ( x , u ) ) ( u n u)dx= lim n Φ ( u n ) Φ ( u ) , u n u =0,

that is, the sequence {(φ(x, u n )φ(x,u))( u n u)} converges to 0 in L 1 ( R N ) as n. Hence the sequence { u n } has a subsequence { u n k } such that

lim k ( φ ( x , u n k ( x ) ) φ ( x , u ( x ) ) ) ( u n k ( x ) u ( x ) ) =0
(2.5)

for almost all x R N . Thus there exists M>0 such that

φ ( x , u n k ( x ) ) u n k ( x ) M + | φ ( x , u n k ( x ) ) | | u ( x ) | + | φ ( x , u ( x ) ) | | u n k ( x ) | + | φ ( x , u ( x ) ) | | u ( x ) |

for almost all x R N . It follows from conditions (A), (J3) and (J5) that

c 1 a 0 | u n k ( x ) | p c 1 a ( x ) | u n k ( x ) | p φ ( x , u n k ( x ) ) u n k ( x ) M + | φ ( x , u n k ( x ) ) | | u ( x ) | + | φ ( x , u ( x ) ) | | u n k ( x ) | + | φ ( x , u ( x ) ) | | u ( x ) | M + ( σ 0 ( x ) + d | u n k ( x ) | p 1 ) | u ( x ) | + | φ ( x , u ( x ) ) | | u n k ( x ) | + | φ ( x , u ( x ) ) | | u ( x ) |
(2.6)

for almost all x R N . By using Young’s inequality, we deduce that

d| u n k (x) | p 1 |u(x)| c 1 a 0 3 | u n k (x) | p + ( 3 d p c 1 a 0 ) p 1 |u(x) | p ,

and

|φ ( x , u ( x ) ) || u n k (x)| ( 3 c 1 a 0 ) 1 p 1 |φ ( x , u ( x ) ) | p + c 1 a 0 3 | u n k (x) | p

for almost all x R N . These together with relation (2.6) imply that

c 1 a 0 3 | u n k ( x ) | p M + σ 0 ( x ) | u ( x ) | + ( 3 d p c 1 a 0 ) p 1 | u ( x ) | p + ( 3 c 1 a 0 ) 1 p 1 | φ ( x , u ( x ) ) | p + | φ ( x , u ( x ) ) | | u ( x ) |

for almost all x R N . Since c 1 and a 0 are positive constants, the above inequality implies that the sequence {| u n k (x)|} is bounded, and so { u n k (x)} is bounded in R N for almost all x R N . By passing to a subsequence, we can suppose that u n k (x)ξ as k for some ξ R N and for almost all x R N . Then we have φ(x, u n k (x))φ(x,ξ) as k for almost all x R N . It follows from (2.5) that

0 = lim k ( φ ( x , u n k ( x ) ) φ ( x , u ( x ) ) ) ( u n k ( x ) u ( x ) ) = ( φ ( x , ξ ) φ ( x , u ( x ) ) ) ( ξ u ( x ) )

for almost all x R N . Since φ is strictly monotone by (J4), this means ξ=u(x), that is, u n k (x)u(x) as k for almost all x R N . The arguments above hold for any subsequence { u n k } of the sequence { u n }. Hence we obtain u n (x)u(x) as n for almost all x R N . Then it implies that

lim n R N φ(x, u n )( u n u)dx=0.
(2.7)

Since the functional Φ is convex, it is obvious that

Φ(u)+ R N φ(x, u n )( u n u)dxΦ( u n ),

and so we get Φ(u) lim sup n Φ( u n ). Therefore, it is derived from (2.3) that

Φ(u)= lim n Φ( u n ).
(2.8)

Consider the sequence { g n } in L 1 ( R N ) defined pointwise by

g n (x)= 1 2 ( Φ 0 ( x , u n ) + Φ 0 ( x , u ) ) Φ 0 ( x , 1 2 ( u n u ) ) .

Then g n 0 for all nN by (J1) and (J4). Since Φ 0 (x,) is continuous for almost all x R N , we obtain that g n Φ 0 (x,u) as n for almost all x R N . Therefore, by the Fatou lemma and relation (2.8), we have

Φ(u) lim inf n R N g n (x)dx=Φ(u) lim sup n R N Φ 0 ( x , 1 2 ( u n u ) ) dx.

Hence

lim sup n R N Φ 0 ( x , 1 2 ( u n u ) ) dx0,

that is,

lim n R N Φ 0 ( x , 1 2 ( u n u ) ) dx=0,

in other words, lim n u n u a =0 by (J5). Since u n u X 1 c u n u a by (1.1), in conclusion, lim n u n u X =0, as claimed. □

Corollary 2.3 Assume that (A) and (J1)-(J5) hold. Then the operator Φ :X X is bounded homeomorphism onto X .

Proof It is immediate that the operator Φ is strictly monotone, coercive, and hemicontinuous. Hence the Browder-Minty theorem implies that the inverse operator ( Φ ) 1 : X X exists and is bounded; see Theorem 26.A in [15]. Since the operator Φ is a mapping of type ( S + ) by Lemma 2.2, it is easy to prove that the inverse operator ( Φ ) 1 is continuous and is omitted here. □

Before dealing with our main results in this section, we need the following assumptions for f. Let us put F(x,t)= 0 t f(x,s)ds.

(F1) f: R N ×RR satisfies the Carathéodory condition in the sense that f(,t) is measurable for all tR and f(x,) is continuous for almost all x R N .

(F2) f satisfies the following growth condition: for all (x,t) R N ×R,

| f ( x , t ) | σ(x)+ρ(x)|t | γ 1 ,

where σ L ( p ) ( R N ) L ( R N ), γR such that γ<p, ρ L s ( R N ) L ( R N ) with (1/s)+(γ/ p )=1.

(F3) There exist a real number s 0 and a positive constant r 0 so small that

B N ( x 0 , r 0 ) F(x, s 0 )dx>0,

and F(x,t)0 for almost all x B N ( x 0 , r 0 ) B N ( x 0 ,σ r 0 ) with σ(0,1) and for all 0t| s 0 |, where B N ( x 0 , r 0 )={x R N :|x x 0 | r 0 } R N .

Then we define the functionals Ψ, I λ :XR by

Ψ(u)= R N F(x,u)dxand I λ (u)=Φ(u)+λΨ(u)

for any uX. It is easy to check that Ψ C 1 (X,R) and its Fréchet derivative is

Ψ ( u ) , v = R N f(x,u)vdx
(2.9)

for any u,vX.

Lemma 2.4 Assume that (A), and (F1)-(F2) hold. Then Ψ and Ψ are weakly-strongly continuous on X.

Proof The analogous arguments as in Lemma 4.4 of [12] imply that functionals Ψ and Ψ are weakly-strongly continuous on X. □

Lemma 2.5 Assume that (A), (J1)-(J3), (J5), and (F1)-(F2) hold. Then we have

lim u X ( Φ ( u ) + λ Ψ ( u ) ) =+

for all λR.

Proof If u X is large enough and λR, then it follows from (J5), (F2) and Hölder’s inequality that

Φ ( u ) + λ Ψ ( u ) = R N Φ 0 ( x , u ) d x λ R N F ( x , u ) d x c 1 p R N a ( x ) | u | p d x | λ | R N | σ ( x ) | | u | d x | λ | R N 1 γ | ρ ( x ) | | u | γ d x c 1 p u a p | λ | σ L ( p ) ( R N ) u L p ( R N ) | λ | γ ρ L s ( R N ) u L p ( R N ) γ c 1 c p p u X p | λ | C 1 u X | λ | C 2 γ u X γ

for some positive constants C 1 and C 2 . Since p>γ, we get that

lim u X ( Φ ( u ) + λ Ψ ( u ) ) =+

for all λR. □

Now we will localize the interval for which problem (P λ ) has at least three solutions as the application of three critical points theorems given in [9] and [2], respectively. To do this, we consider the following eigenvalue problem:

  1. (E)

    div ( a ( x ) | u | p 2 u ) =λm(x)|u | p 2 u  in R N .

Proposition 2.6 ([11, 12])

Assume that (A) and (J1)-(J5) hold. Moreover, suppose that

  1. (M)

    m(x)>0 for all x R N such that m L ( R N ) L N / p ( R N ), and m L κ 1 ( R N ), where

    κ 1 = p p κ with p<κ< p .

Denote the quantity

λ 1 = inf u X { 0 } ( R N a ( x ) | u | p d x R N m ( x ) | u | p d x ) .

Then the eigenvalue problem (E) has a pair ( λ 1 , u 1 ) of a principal eigenvalue λ 1 and an eigenfunction u 1 with λ 1 >0 and 0< u 1 X L ( R N ). Moreover, λ 1 is simple and u 1 (x) decays uniformly as |x|.

Definition 2.7 Let X be a real Banach space. We call that W X is the class of all functionals Φ:XR satisfying the following property: if { u n } is a sequence such that u n u in X as n and lim inf n Φ( u n )Φ( u n ), then { u n } has a subsequence { u n k } and u n k u in X as k.

The following lemma is three critical points theory which was introduced by Ricceri [9].

Lemma 2.8 ([9])

Let X be a separable and reflexive real Banach space; let Φ:XR be a coercive, sequentially weakly lower semicontinuous C 1 -functional, belonging to W X , bounded on each bounded subset of X and whose derivative admits a continuous inverse on X . Let Ψ:XR be a C 1 -functional with compact derivative. Assume that Φ has a strict local minimum u 0 with Φ( u 0 )=Ψ( u 0 )=0. Finally, set

α=max { 0 , lim sup u X ( Ψ ( u ) Φ ( u ) ) , lim sup u u 0 ( Ψ ( u ) Φ ( u ) ) } ,β= sup u Φ 1 ( ( 0 , + ) ) ( Ψ ( u ) Φ ( u ) ) .

Assume that α<β. Then, for each compact interval [a,b]( 1 β , 1 α ) (with the conventions 1 0 =+, 1 + =0), there exists R>0 with the following property: for every λ[a,b], the equation Φ (u)+λ Ψ (u)=0 has at least three solutions whose norms are less than R.

In order to apply the above lemma to (P λ ), we have to show that the functional Φ belongs to W X . To do this, we need the following additional assumption:

(J6) The following relation holds for all u,v R N :

1 2 ( Φ 0 ( x , u ) + Φ 0 ( x , v ) ) Φ 0 ( x , u + v 2 ) + Φ 0 ( x , u v 2 ) .

To consider some examples that satisfy hypothesis (J6), we observe the following argument which is given in [16].

Remark 2.9 If ϕ(t) is a continuous, strictly increasing function for t0 with ϕ(0)=0 and

tϕ( t )is convex for all t[0,),
(2.10)

then the following estimate

1 2 ( ϕ ( | u | ) + ϕ ( | v | ) ) ϕ ( | u + v 2 | ) +ϕ ( | u v 2 | )

holds for all u,v R N .

Example 2.10 Let us consider

φ(x,v)=|v | p 2 vand Φ 0 (x,v)= 1 p |v | p

for all v R N . If p2, then we obtain a Clarkson-type inequality for the function Φ 0 , i.e.,

1 2 ( | u | p + | v | p ) | u + v 2 | p +| u v 2 | p

for all u,v R N . Therefore assumption (J6) holds.

Example 2.11 Let p2. Suppose that w L 2 p ( R N ) and there exists a positive constant w 0 such that w(x) w 0 for almost all x R N . Let us consider

φ(x,v)= ( w ( x ) + | v | 2 ) p 2 1 vand Φ 0 (x,v)= 1 p [ ( w ( x ) + | v | 2 ) p 2 1 ]

for all v R N . Set ϕ(t)=(1/p)[ ( w ( x ) + t 2 ) p / 2 1] for t0. Then it is easy to calculate that ϕ satisfies all the assumptions of Remark 2.9 and therefore condition (J6) is verified.

Combining with Proposition 2.6 and Lemma 2.8, we derive the following consequence.

Theorem 2.12 Assume that conditions (A), (J1)-(J6), (F1)-(F3) and (M) hold. Moreover, suppose that

(F4) lim sup | s | F ( x , s ) m ( x ) | s | p 0 for x R N uniformly.

(F5) lim sup s 0 F ( x , s ) m ( x ) | s | p 0 for x R N uniformly.

(F6) For all compact KR, there exists a function ψ K L 1 ( R N ) such that

F(x,s) ψ K (x)

for almost all x R N and for all sK.

Assume also that the condition γ<p is removed and replaced by the more general condition γ< p in assumption (F2). Set ξ= sup u X { 0 } ( Ψ ( u ) Φ ( u ) ). Then, for each compact interval [a,b]( 1 ξ ,+), there exists R>0 with the following property: for every λ[a,b], problem (P λ ) has at least three solutions whose norms are less than R.

Proof It is obvious that the functional Φ is coercive, sequentially weakly lower semicontinuous of class C 1 , bounded on each subset of X, and whose derivative is a homeomorphism by Corollary 2.3. Moreover, the functional Ψ C 1 (X,R) has a compact derivative due to Lemma 2.4.

First of all, let us claim that the functional Φ belongs to W X . It follows from the same argument as in the proof of Theorem 3.1 in [17]. For the sake of convenience, we give the proof. Let { u n } be a sequence in X that converges weakly to u in X as n and lim inf n Φ( u n )Φ(u). By Lemma 2.2, Φ is sequentially weakly lower semicontinuous, namely Φ(u) lim inf n Φ( u n ). Thus there exists a subsequence of { u n }, still denoted by { u n }, such that lim n Φ( u n )=Φ(u). Since u n u as n, the sequence {( u n +u)/2} also converges weakly to u in X as n, and we get

Φ(u) lim inf n Φ ( u n + u 2 ) .
(2.11)

If { u n } does not converge to u as n approaches infinity, the sequence {( u n u)/2} also does not converge to 0 as n. So we can choose ε 0 >0 and a subsequence { u n k } of { u n } such that ( u n k u ) / 2 X ε 0 for all kN. By assumption (J5) and (1.1), we deduce that

Φ ( u n k u 2 ) = R N Φ 0 ( x , u n k u 2 ) d x c 1 p R N a ( x ) | u n k u 2 | p d x = c 1 p u n k u 2 a p c 1 c p p u n k u 2 X p c 1 c p p ε 0 p

for all kN. From (J6), we know

1 2 ( R N Φ 0 ( x , u n k ) d x + R N Φ 0 ( x , u ) d x ) R N Φ 0 ( x , u n k + u 2 ) d x + R N Φ 0 ( x , u n k u 2 ) d x .

Thus we deduce that the following relation

1 2 ( Φ ( u n k ) + Φ ( u ) ) Φ ( u n k + u 2 ) +Φ ( u n k u 2 ) Φ ( u n k + u 2 ) + c 1 c p p ε 0 p
(2.12)

holds for all kN. From (2.11) and (2.12), we have Φ(u)Φ(u)+( c 1 c p /p) ε 0 p as k, a contradiction. Therefore, we conclude that u n u as n and so Φ W X .

Observe now that Φ(u)>0 for every uX{0}. Then 0 is a strict local (even global) minimum with Φ(0)=Ψ(0)=0. By assumptions (F4) and (F6), for every ε>0, we get

F(x,s)εm(x)|s | p + ψ ε (x)

for almost all x R N and for all sR, where ψ ε L 1 ( R N ). It implies that

R N F(x,u)dxε R N m(x)|u | p dx+ R N ψ ε (x)dx.
(2.13)

Notice that

λ 1 = inf u X { 0 } ( R N a ( x ) | u | p d x R N m ( x ) | u | p d x ) >0
(2.14)

by Proposition 2.6. Then it follows from (2.13), (2.14) and (J5) that

R N F ( x , u ) d x ε λ 1 R N a ( x ) | u | p d x + R N ψ ε ( x ) d x ε p λ 1 c 1 R N Φ 0 ( x , u ) d x + R N ψ ε ( x ) d x ε p λ 1 c 1 Φ ( u ) + R N ψ ε ( x ) d x .

Hence we have

lim sup u X R N F ( x , u ) d x Φ ( u ) ε ( p λ 1 c 1 ) .

Since ε is arbitrary, the following inequality holds:

lim sup u X ( Ψ ( u ) Φ ( u ) ) = lim sup u X R N F ( x , u ) d x Φ ( u ) 0.
(2.15)

On the other hand, by conditions (F4) and (F5), we have that for every ε>0, there exists C ε >0 verifying that

F(x,s)εm(x)|s | p + C ε m(x)|s | κ
(2.16)

for almost all x R N and for all sR. From (2.14), (2.16) and (J5), we deduce

R N F ( x , u ) d x ε R N m ( x ) | u | p d x + C ε R N m ( x ) | u | κ d x ε λ 1 R N a ( x ) | u | p d x + C ε m L κ 1 ( R N ) ( R N | u | p d x ) κ p ε p λ 1 c 1 R N Φ 0 ( x , u ) d x + C ε m L κ 1 ( R N ) u L p ( R N ) κ ε p λ 1 c 1 Φ ( u ) + C ε C 3 u X κ

for some positive constant C 3 . Then it follows that

R N F ( x , u ) d x Φ ( u ) ε ( p λ 1 c 1 ) + C ε C 3 u X κ Φ ( u ) .

Hence we obtain

lim sup u X 0 ( Ψ ( u ) Φ ( u ) ) = lim sup u X 0 R N F ( x , u ) d x Φ ( u ) ε ( p λ 1 c 1 )

for all ε>0, which leads to

lim sup u X 0 ( Ψ ( u ) Φ ( u ) ) 0.
(2.17)

Taking now assumption (F3) into account, it follows from (2.15) and (2.17) that

max { 0 , lim sup u X ( Ψ ( u ) Φ ( u ) ) , lim sup u 0 ( Ψ ( u ) Φ ( u ) ) } =0< sup u Φ 1 ( ( 0 , + ) ) ( Ψ ( u ) Φ ( u ) ) .

Therefore, all the conditions of Lemma 2.8 are fulfilled and thus the proof is completed. □

In the rest of this section, we determine precisely the intervals of λ’s for which problem (P λ ) possesses either only the trivial solution or at least two nontrivial solutions. To do this, we assume that

(F7) lim sup s 0 | f ( x , s ) | m ( x ) | s | κ 1 <+ uniformly for almost all x R N .

Then we get that lim sup s 0 | F ( x , s ) | m ( x ) | s | κ <+ uniformly for almost all x R N by the L’Hôspital’s rule. Let us consider that two functions

χ 1 (r)= inf u Ψ 1 ( ( , r ) ) inf v Ψ 1 ( r ) Φ ( v ) Φ ( u ) Ψ ( u ) r ,
(2.18)
χ 2 (r)= sup u Ψ 1 ( ( r , + ) ) inf v Ψ 1 ( r ) Φ ( v ) Φ ( u ) Ψ ( u ) r
(2.19)

for every r( inf u X Ψ(u), sup u X Ψ(u)). Also we consider the following crucial value:

C f =ess sup s 0 , x R N | f ( x , s ) | m ( x ) | s | p 1 .

Then the same arguments in [3] imply that C f is a positive constant. From this fact, we obtain

ess sup s 0 , x R N | F ( x , s ) | m ( x ) | s | p = C f p .
(2.20)

The next lemma represents the differentiable form of the Arcoya and Carmona Theorem 3.4 in [2].

Lemma 2.13 Let Φ and Ψ be two functionals on X such that Φ and Ψ are weakly lower semicontinuous and continuously Gâteaux differentiable in X, and Ψ is nonconstant. Let also Φ :X X have the ( S + ) property, and that Ψ is a compact operator. Assume that there exists an interval IR such that the one parameter family of functionals I λ =Φ+λΨ is coercive in X for all λI. Let us assume that there exists

r ( inf u X Ψ ( u ) , sup u X Ψ ( u ) ) such that χ 1 (r)< χ 2 (r),
(2.21)

then the following properties hold.

  1. (i)

    The functional I λ admits at least one critical point for every λI.

  2. (ii)

    If furthermore ( χ 1 (r), χ 2 (r))I0, then

  3. (a)

    I λ has at least three critical points for every λ( χ 1 (r), χ 2 (r))I.

  4. (b)

    I χ 1 ( r ) has at least two critical points provided that χ 1 (r)I.

  5. (c)

    I χ 2 ( r ) has at least two critical points provided that χ 2 (r)I.

Theorem 2.14 Assume that (A), (J1)-(J5), (F1)-(F3) and (M) hold. Then we have

  1. (i)

    If λ[0, ), where = c 1 λ 1 / C f , then problem (P λ ) has only the trivial solution, where λ 1 is the principal eigenvalue of problem (E), c 1 is a positive constant in (J5), and both of c and c are positive constants from (1.1).

  2. (ii)

    If furthermore f satisfies condition (F7), then there exists a positive constant with such that problem (P λ ) has at least two nontrivial solutions for all λ( ,+).

Proof By Lemma 2.2, the functional Φ:XR is a sequentially weakly lower semicontinuous C 1 -functional and the operator Φ is a mapping of type ( S + ). It follows from Lemma 2.4 that the functional Ψ is also sequentially weakly lower semicontinuous C 1 -functional and the operator Ψ :X X is compact. Due to Lemma 2.5, we have

lim u X ( Φ ( u ) + λ Ψ ( u ) ) =+

for all uX and for all λR.

First we claim the assertion (i). Let uX be a nontrivial weak solution of problem (P λ ), that is,

R N φ(x,u)vdx=λ R N f(x,u)vdx

for all vX. If we put v=u, then it follows from (J5) that

c 1 λ 1 u a p λ 1 R N φ ( x , u ) u d x = λ 1 λ R N f ( x , u ) u d x = λ 1 λ R N f ( x , u ) m ( x ) | u | p 1 m ( x ) | u | p d x λ 1 λ C f R N m ( x ) | u | p d x λ C f R N a ( x ) | u | p d x = λ C f u a p .

Thus if u is a nontrivial weak solution of problem (P λ ), then necessarily λ = c 1 λ 1 / C f , as required.

Next let us prove assertion (ii). Let s 0 0 be from (F3). For σ(0,1), define

u σ (x)={ 0 if  x R N B N ( x 0 , r 0 ) , | s 0 | if  x B N ( x 0 , σ r 0 ) , | s 0 | r 0 ( 1 σ ) ( r 0 | x x 0 | ) if  x B N ( x 0 , r 0 ) B N ( x 0 , σ r 0 ) .
(2.22)

Then it is obvious that 0 u σ (x)| s 0 | for all x R N and u σ X. From condition (F3),

Ψ ( u σ ) = B N ( x 0 , σ r 0 ) F ( x , | s 0 | ) d x + B N ( x 0 , r 0 ) B N ( x 0 , σ r 0 ) F ( x , | s 0 | r 0 ( 1 σ ) ( r 0 | x x 0 | ) ) d x > 0 .

It follows that the crucial number

= χ 1 (0)= inf u Ψ 1 ( ( , 0 ) ) Φ ( u ) Ψ ( u )

is well defined. Let u be in X with u0. Using (J5) and (2.20), we have

Φ ( u ) | Ψ ( u ) | = R N Φ 0 ( x , u ) d x R N F ( x , u ) d x c 1 p R N a ( x ) | u | p d x R N | F ( x , u ) | m ( x ) | u | p m ( x ) | u | p d x c 1 p R N a ( x ) | u | p d x C f p R N m ( x ) | u | p d x c 1 λ 1 C f = .

Hence we get . To employ Lemma 2.13, we have to verify assumption (2.21). For all u Ψ 1 ((,0)), we have that

χ 1 (r)= inf u Ψ 1 ( ( , r ) ) inf v Ψ 1 ( r ) Φ ( v ) Φ ( u ) Ψ ( u ) r inf v Ψ 1 ( r ) Φ ( v ) Φ ( u ) Ψ ( u ) r Φ ( u ) r Ψ ( u )

for all r(Ψ(u),0), and hence

lim sup r 0 χ 1 (r) Φ ( u ) Ψ ( u )

for all u Ψ 1 ((,0)). Then it implies that

lim sup r 0 χ 1 (r) χ 1 (0)= .

By assumption (F7), there exists a positive real number M >0 such that

|F(x,s)| M m(x)|s | κ
(2.23)

for almost all x R N and for all sR. Indeed, denote

M 0 = lim sup s 0 | F ( x , s ) | m ( x ) | s | κ .

Then there exists δ>0 such that |F(x,s)|( M 0 +1)m(x)|s | κ for almost all x R N and for all sR with |s|<δ. Let s be fixed with |s|δ. According to (2.20),

|F(x,s)| C f p |s | p κ m(x)|s | κ C f δ p κ p m(x)|s | κ

for almost all x R N . Put M =max{ M 0 +1, C f δ p κ /p}. Then relation (2.23) holds.

Hence we deduce that

|Ψ(u)| R N M m(x)|u | κ dx C 4 m L κ 1 ( R N ) u X κ

for some positive constant C 4 . If r<0 and v Ψ 1 (r), then we obtain by (J5) that

r=Ψ(v) C 4 m L κ 1 ( R N ) v X κ C 4 m L κ 1 ( R N ) ( p c 1 c p Φ ( v ) ) κ p .

Since u=0 Ψ 1 ((r,)), by using (2.19), we have

χ 2 (r) 1 | r | inf v Ψ 1 ( r ) Φ(v) | r | p κ 1 C 4 p κ m L κ 1 ( R N ) p κ c 1 c p p ,

and so lim r 0 χ 2 (r)= since κ>p. Therefore, we conclude

lim sup r 0 χ 1 (r) χ 1 (0)= < lim r 0 χ 2 (r)=+.

It means that there exists a negative sequence { r n } such that r n 0 as n, so that χ 1 ( r n )< +1/n<n< χ 2 ( r n ) for all integers n with n n =2+[ ]. By Lemma 2.5, we put I=R. Since u0 is a critical point of I λ , according to the part (a) of (ii) in Lemma 2.13, problem (P λ ) admits at least two nontrivial solutions for all

λ ( , + ) = n = n [ + 1 n , n ] n = n ( χ 1 ( r n ) , χ 2 ( r n ) ) ,

as claimed. □

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The authors would like to thank the referees for useful comments and remarks.

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Choi, E.B., Kim, Y. Three solutions for equations involving nonhomogeneous operators of p-Laplace type in R N . J Inequal Appl 2014, 427 (2014). https://doi.org/10.1186/1029-242X-2014-427

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Keywords

  • p-Laplacian
  • weighted Lebesgue-Sobolev spaces
  • three critical points theorem
  • multiple solutions