Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 29, Number 4, August 2024
Page(s) 365 - 373
DOI https://doi.org/10.1051/wujns/2024294365
Published online 04 September 2024

© Wuhan University 2024

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

Consider the following semi-linear elliptic problem

{ - Δ u = λ a ( x ) f ( u ) ,    x R N , u = 0 , a s | x | + , (1)

where λ is a real parameter, N3, and aClocα(RN,R) for some α(0,1) is a weighted function which can be sign-changing and fC(R,R), and f(s)s>0 for any s0. Edelson et al[1,2] studied the existence of positive solution and the existence of global branches of minimal solutions of the problem (1) by the Schauder-Tychonoff fixed point theorem and Dancer global bifurcation theorems[3], respectively. By using Rabinowitz global bifurcation method[4], Rumbos et al[5] showed the existence of positive minimal solution of the problem (1). In 2017, Dai et al[6] established a global bifurcation result for problem (1).

On the other hand, Lions[7] studied the following Kirchhoff type problem

{ - M ( Ω | u | 2 d x ) Δ u = λ a ( x ) u + g ( x , u , λ ) ,   i n   Ω , u = 0 ,    o n   Ω , (2)

where Ω is a bounded domain in RN with a smooth boundary Ω. The problem (2) is nonlocal as the appearance of the term Ω|u|2dx which implies that it is not a pointwise identity. By applying the bifurcation techniques, Liang et al[8] and Figueiredo et al[9] also studied equation (2). Dai et al[10] studied the problem (2) by Rabinowitz[4].

Motivated by the above papers, we shall study the following Kirchhoff type problem

{ - M ( R N | u | 2 d x ) Δ u = λ a ( x ) [ u + g ( u ) ] ,    x R N , u = 0 , a s   | x | + , (3)

where λ is a real parameter. By Ref. [6], set I(Ω):={aClocα(Ω,R):{xΩ: a(x)>0}}.

For any uCc(Ω) with ΩRN, we define u1=(Ω|u|2dx)1/2. Denote by D1,2(Ω) the completion of Cc(Ω) with respect to the norm u1. Denote by S(RN) the set of all measurable real functions defined on RN. Two functions in S(RN) are considered as the same element of S(RN) when they are equal almost everywhere. Let L2(RN;|a|):={uS(RN):RN|a|u2dx<+}.

We assume that a, g() and M() satisfy the following conditions:

(A1) Let aI(RN). If there exist two continuous positive radially symmetric functions p and P, where PL2q'r(RN) (where q and r are given in (A3) and Section 1) such that 0<pa(x)P(|x|), xRN and RN|x|2-NP(|x|)dx<+.

Furthermore, if P satisfies the following stronger condition (with r=|x|)

0 + r N - 1 P ( r ) d r < + . (4)

(A2) gC(R,R) is a Holder continuous function with exponent α and limsg(s)/s=0.

(A3) There exist c>0 and q(1,2*] such that |g(s)|c(1+|s|q-1), where

2 * = { 2 N N - 2 ,   N > 2 , + , N 2 .

(A4) M(t)C(R+) is increasing, M(0)>0, R+=[0,+).

(A5) There exists m1>0, such that limt+M(t)=m1.

Furthermore, we shall investigate the existence of one-sign solutions for the following Kirchhoff type problems

{ - M ( R N | u | 2 d x ) Δ u = λ a ( x ) f ( u ) ,   x R N , u = 0 ,   a s   | x | + . (5)

We assume that a satisfies (A1), and f satisfies the following assumptions:

(H1) fC(R,R) is a Holder continuous function with exponent α such that sf(s)>0 for any s0.

(H2) f0, f(0,).

(H3) f0(0,), f=.

(H4) f0(0,), f=0.

(H5) f0=, f(0,).

(H6) f0=0, f(0,).

(H7) f0=, f=0.

(H8) f0=0, f=.

(H9) f0=0, f=0.

(H10) f0=, f=.

where f0=lim|s|0f(s)s, f=lim|s|f(s)s.

Finally, we shall study the exact multiplicity of one-sign solutions for (5) by Implicit Function Theorem, the stability properties and condition (A6).

(A6) fCα(R,R) such that f(s)/s is decreasing in (0,+) and is increasing in (-,0).

The rest of this paper is arranged as follows. In Section 1, we give some preliminaries and establish the unilateral global bifurcation result for the problem (3). In Section 2, on the above unilateral global bifurcation result, we prove the existence of one-sign solutions for the Kirchhoff type problem (5). In Section 3, we study the exact multiplicity of one-sign solutions for (5).

1 Preliminaries

Let E:=H1(RN) with the norm u=(RN|u|2dx)1/2. Let P+={uE|u(x)>0, xRN} and set P-=-P+ and P=P+P-.

Now, from Theorem 1.1 in Ref. [6], we know that the following eigenvalue problem

{ - Δ u = λ a ( x ) u , i n   R N , u ( x ) 0 ,   a s   | x | + , (6)

possesses a unique principal eigenvalue λ1, and λ1 is simple and isolated.

To prove Theorem 1, by Section 4 of Ref. [6], we first consider the following problem

{ - Δ u = h ( u ) ,   i n   R N , u ( x ) 0 , a s   | x | + . (7)

Let us define the operator T: EE by

u ( x ) = T [ h ] ( x ) = R N Γ N ( x - y ) h ( y ) d y , (8)

where ΓN(x-y)=1N(N-2)ωN|x-y|2-N,ωN being the volume of the unit ball in RN.

Then by an argument similar to that of Ref. [1], we can show that u is a one-sign C2+α solution of problem (7) if and only if u is a solution of the operator equation u(x)=T(h). Similar to proposition 1 in Ref. [5], we also can show that T: EE is linear completely continuous and (8) is equivalent to (7).

The first main result for (3) is the following unilateral global bifurcation theorem.

Theorem 1   Assume that (A1)-(A5) hold. The pair (λ1M(0),0) is a bifurcation point of the problem (3) and there are two distinct unbounded continua D+ and D- in R×H1(RN) of solutions of the problem (3) emanating from (λ1M(0),0). Moreover, we have Dν((R×Pν){(λ1M(0),0)}), where μ{+,-}.

Proof   By p.5960-5961 in Ref. [6], it is clear that the problem (3) can be equivalently written as u=G(λ,u)=λT(au)M(0)+H(λ,u), where

H ( λ , u ) = λ ( M ( 0 ) - M ( u 2 ) ) M ( 0 ) M ( u 2 ) T ( a u ) + T [ λ a ( x ) g ( u ) ] M ( u 2 ) .

From conditions (A1)-(A5) and noting 2<2*, we can see that H: R×EE is completely continuous. Furthermore, it follows that G: R×EE is completely continuous and G(λ,0)=0, λR.

Next, we show limu0H(λ,u)/u=0 at u=0 uniformly on bounded λ sets. Without loss of generality, we may assume that q>2. Otherwise, we can consider q˜=cq, c>1 such that q˜(2,2*). From q<2*, we can see q'(q-2)2*<2-q'2*. So we can choose a real number r>1 such that q'(q-2)2*1r2-q'2*.

It follows

q ' r ( q - 2 ) 2 * ,   q ' r ' 2 * . (9)

By (A2) and (A3), for any ε>0, we can choose positive numbers δ=δ(ε) and M=M(δ) such that the following relations hold: |g(s)/s|ε, for |s|δ.|g(s)/s|M|s|q-2, for|s|>δ.

Then we can obtain

R N | a ( x ) g ( u ) u | q ' r d x ε R N ( a ( x ) ) q ' r d x + M q ' r R N ( a ( x ) ) q ' r | u | q ' r ( q - 2 ) d x ε R N ( P ( | x | ) ) q ' r d x    + M q ' r ( R N ( P ( | x | ) ) 2 q ' r d x ) 1 2 ( R N | u | 2 q ' r ( q - 2 ) d x ) 1 2 .

By PL2q'r(RN) and the continuous embedding of L2q'r(RN)Lq'r(RN), we have

R N ( P ( | x | ) ) 2 q ' r d x < + ,   R N ( P ( | x | ) ) q ' r d x < +

Moreover, as u+, we obtain that

| a ( x ) g ( u ) u | q ' 0   i n L r ( R N )

Let v=u/u, by the boundedness of vE, q'r'2* and the continuous embedding of EL2*(RN), we have RN|v|q'r'dx<c, furthermore, we can get

R N | a ( x ) g ( u ) u | q ' d x = R N | a ( x ) g ( u ) | u | | q ' | v | q ' d x ( R N | a ( x ) g ( u ) | u | | q ' r d x ) 1 r ( R N | v | q ' r ' d x ) 1 r ' 0 .

We obtain

l i m u 0 a ( x ) g ( u ) / u = 0 ,   i n   L q ' ( R N ) (10)

uniformly xRN.

By (10), we have limu0H(λ,u)/u=0 uniformly for xRN and λ on bounded sets, i.e. H(λ,u)=o(u) at u=0 uniformly on bounded λ sets.

Furthermore, applying the similar proof method of Theorem 1.3 in Ref. [6] and the Rabinowitz global bifurcation theorem[4], one can obtain that (λ1M(0),0) is a bifurcation point of the problem (3) and there exists one unbounded continua D of solutions of the problem (3) emanating from (λ1M(0),0).

Moreover, by the Dancer unilateral global bifurcation theorem[11], we have that there are two distinct unbounded bifurcation continua D+ and D- in R×H1(RN) of solutions of the problem (3) emanating from (λ1M(0),0). Moreover, we have

D ν ( ( R × P ν ) { ( λ 1 M ( 0 ) , 0 ) } ) , where ν{+,-}.

2 One-Sign Solutions for Kirchhoff Type Problem

We first have the following results.

Remark 1   From (H1) and (H2), we can see that there exist two positive constants 0<ρ<σ such that ρf(s)sσ for all s0.

By an argument similar to that of Lemma 4.1, 4.2 in Ref. [6], we can obtain Lemma 1 and 2.

Lemma 1   Let (H1) and (H2) hold. By Remark 1, the problem (5) has no one-sign solution for any λ(λ1m1/ρ,+).

Lemma 2   Let (H1) and (H2) hold. By Remark 1, the problem (5) has no positive solution for any λ(0,λ1M(0)/σ). The main results of this section are the following theorem.

Theorem 2   Let (A1), (A4), (A5), (H1) and (H2) hold. For any λ(min {λ1fm1,λ1f0M(0)}, max {λ1fm1,λ1f0M(0)}), the problem (5) possesses two solutions u1+, u1- such that u1+>0 and u1-<0 in RN. Therefore, we have

l i m | x | + | x | N - 2 u 1 + ( x ) = c 1 > 0 ,   l i m | x | + | x | N - 2 u 1 - ( x ) = c 2 < 0

for some constants c1 and c2.

Proof   Let ζ: R+R+ be a continuous function such that f(u)=f0u+ζ(u), with lim|s|0ζ(s)s=0,lim|s|+ζ(s)s=f-f0, uniformly a.e. in RN. Equation (5) can be divided in the form

{ - Δ u = λ f 0 a ( x ) u ( x ) M ( 0 ) + H 1 ( λ , u ) , i n   R N , u ( x ) 0 ,     a s   | x | + , (11)

where

H 1 ( λ , u ) = λ ( M ( 0 ) - M ( u 2 ) ) M ( 0 ) M ( u 2 ) ( f 0 a ( x ) u ) + λ a ( x ) ζ ( u ) M ( u 2 ) .

Using the same method to prove (10) with obvious changes, it follows that

l i m u 0 ζ ( u ) / u = 0 ,   i n   L q ' ( R N ) .

Moreover, we have

l i m u 0 H 1 ( λ , u ) / u = 0 ,   i n   L q ' ( R N )

uniformly on bounded λ sets.

By Theorem 1, there are two distinct unbounded continua D+ and D- in R×H1(RN) of solutions of the problem (5) emanating from (λ1f0M(0),0), such that

D ν ( ( R × P ν ) { ( λ 1 f 0 M ( 0 ) , 0 ) } ) , where μ{+,-}.

We will show that Dν joins (λ1f0M(0), 0) to (λ1fm1, +). Let

( μ n , u n ) D ν \ { ( λ 1 f 0 M ( 0 ) , 0 ) } satisfy |μn|+|un|+.

By Remark 1 and Lemma 2, one can obtain μn>0 for all nN. It follows from Lemma 1 that there exists a constant M such that μn(0,M] for any nN. Therefore, we get un+.

One can get that Dν joins (λ1f0M(0),0) to (λ1fm1,+). Let ξ: R+R+ be a continuous function such that f(u)=fu+ξ(u), with

l i m s + ξ ( s ) s = 0 ,   l i m s 0 + ξ ( s ) s = f 0 - f (12)

uniformly a.e. in RN. We divide the equation

{ - Δ u = λ f a ( x ) u ( x ) m 1 + H 2 ( λ , u ) ,   i n   R N , u ( x ) 0 , a s   | x | + , (13)

where

H 2 ( λ , u ) = λ ( m 1 - M ( u 2 ) ) m 1 M ( u 2 ) ( f a ( x ) u ) + λ a ( x ) ξ ( u ) M ( u 2 ) .

By (12), for any ε>0, we can choose positive numbers δ=δ(ε) and M=M(ε) such that for a.e. xRN,the following relations hold:

| ξ ( s ) / s | ε ,   f o r | s | > δ ; | ξ ( s ) / s | M ,   f o r | s | δ .

Then we can obtain

R N | a ( x ) ξ ( u ) u | q ' r d x ε R N ( a ( x ) ) q ' r d x + M q ' r R N | a ( x ) u | q ' r d x ε R N ( P ( | x | ) ) q ' r d x    + M q ' r ( R N ( P ( | x | ) ) 2 q ' r d x ) 1 2 ( R N | u | 2 q ' r ( q - 2 ) d x ) 1 2 .

By PL2q'r(RN), we have

R N ( P ( | x | ) ) 2 q ' r d x < + ,   R N ( P ( | x | ) ) q ' r d x < + .

Moreover, as u+, we obtain

| a ( x ) ξ ( u ) u | q ' 0   i n L r ( R N ) .

Let v=u/u, by the boundedness of vE, (9) and the continuous embedding of EL2*(RN), we have

R N | v | q ' r ' d x < c .

Furthermore, we can get

R N | a ( x ) ξ ( u ) u | q ' d x = R N | a ( x ) ξ ( u ) | u | | q ' | v | q ' d x ( R N | a ( x ) ξ ( u ) | u | | q ' r d x ) 1 r ( R N | v | q ' r ' d x ) 1 r ' 0 .

It follows from that

l i m u a ( x ) ξ ( u ) / u = 0 ,   i n   L q ' ( R N ) (14)

uniformly xRN. Furthermore, one obtain

l i m u H 2 ( λ , u ) / u = 0 ,   i n L q ' ( R N )

uniformly for λ on bounded sets.

By the compactness of T-1, we obtain

{ - Δ u = μ f a ( x ) u ( x ) m 1 , i n   R N , u ( x ) 0 , a s   | x | + , (15)

where μ=limnμn, again choosing a subsequence and relabeling if necessary. Thus it is clear that uD˙νDν since Dν is closed in R×E. Moreover, by (15), μf=λ1m1, so that μ=λ1fm1. Thus Dν joins (λ1f0M(0),0) to (λ1fm1,+). Now the existence of u1+ and u1- is clear.

Similar to the proof of the Theorem 1.3 in Ref. [6], we have

l i m | x | + | x | N - 2 u 1 + ( x ) = c 1 > 0 ,   l i m | x | + | x | N - 2 u 1 - ( x ) = c 2 < 0

for some constants c1 and c2.

Theorem 3   Let (A1), (A4), (A5), (H1) and (H3) hold. If λ1(0,λ1f0M(0)), the problem (5) possesses two solutions u1+, u1- such that u1+>0 and u1-<0 in RN. Therefore, we have

l i m | x | + | x | N - 2 u 1 + ( x ) = c 1 > 0 ,   l i m | x | + | x | N - 2 u 1 - ( x ) = c 2 < 0

for some constants c1 and c2.

Proof   Inspired by the idea of Ref. [12], we define the cut-off function of f as the following

f [ n ] ( s ) = { n s ,   s [ - , - 2 n ] [ 2 n , + ] ,                        2 n 2 + f ( - n ) n ( s + n ) + f ( - n ) ,   s ( - 2 n , - n ) , 2 n 2 - f ( n ) n ( s - n ) + f ( n ) ,   s ( n , 2 n ) ,             f ( s ) ,   s [ - n , n ] .                                                

We consider the following problem

{ - M ( R N | u | 2 ) Δ u = λ a ( x ) f [ n ] ( u ) ,   x R N , u = 0 , a s   | x | + . (16)

Clearly, we can see that limn+f[n](s)=f(s), and (f[n])=n.

The Proposition 4.1 in Ref.[13] implies that there exist two sequence unbounded continua Dν[n] of solution set of problem (16) emanating from (λ1nfm1,),such that Dν[n]((R×Pν){(λ1nfm1,)}), where ν{+,-}.

Taking z*=(0,), we easily obtain that z*limn+inf Dν[n] with z*R×E=+. So condition (i) of Theorem 1.2 in Ref.[13] is satisfied with z*=(0,).

Define a mapping T: R×XR×X such that

T ( λ , u ) = { ( λ , u u 2 ) ,    i f   0 u < + , ( λ , 0 ) ,            i f   u = + , ( λ , ) ,           i f   u = 0 .

It is easy to verify that T is a homeomorphism and T(z*)R×E=0. Obviously, {T(Dν[n])} is a sequence of unbounded connected subsets in E, so (ii) of the Theorem 1.2 in Ref. [13] holds. Since F(λ,0) is completely continuous from R×EE, we have (n=1+T(Dν[n]))B¯R is pre-compact, and accordingly (iii) of the Theorem 1.2 in Ref.[13] holds. Therefore, by the Theorem 1.2 in Ref.[13], Dν=limn+sup Dν[n] is unbounded closed connected of solutions of the problem (5) emanating from (0,), and Dν((R×Pν){(0,)}) by the Proposition 5.1 in Ref.[13], such that either Dν is unbounded in the direction of λ or meets some point on {(λ*,0),λ*R}.

From (H1) and (H3), we obtain that there exists a positive constant τ such that f(s)/sτ for any s>0. So, Lemma 1 implies Dν is bounded in the direction of λ.Hence, Dν meets (λ*,0) for some λ*0. From Theorem 2, we can obtain λ*=λ1f0M(0) and (λ1f0M(0),0)Dν, where ν={+,-}. Now the desired conclusion is obvious.

Theorem 4   Let (A1), (A4), (A5), (H1) and (H4) hold. If λ(λ1f0M(0),+), then the problem (5) possesses two solutions u1+, u1- such that u1+>0 and u1-<0 in RN. Therefore, we have

l i m | x | + | x | N - 2 u 1 + ( x ) = c 1 > 0 ,   l i m | x | + | x | N - 2 u 1 - ( x ) = c 2 < 0

for some constants c1 and c2.

Proof   In view of Theorem 2, there are two distinct unbounded continua D+ and D- in R×H1(RN) of solutions of the problem (5) emanating from (λ1f0M(0),0), such that Dν((R×Pν){(λ1f0M(0),0)}), where ν{+,-}.

We only need to show that Dν joins (λ1f0M(0),0) to (,). We shall only prove the case ν=+ since the proof for the other case is completely analogous.

Suppose on the contrary that there exists λM be a blow-up point and λM<+. Then there exists a sequence (λn,un) such that limnλn=λM and limnun=+. Let vn=un/un. Then vn should be the solutions of problem

{ - Δ v n = λ a ( x ) M ( u n 2 ) f ( u n ) u n ,   i n   R N , v n ( x ) 0 , a s   | x | + . (17)

Similar to the proof of (14), we can show

l i m n f ( u n ) u n = 0 ,   i n   L q ' ( R N ) .

By the compactness of G(λ,) and (17), we obtain that for some convenient subsequence vnv00. This contradicts v0=1.

Similar to the proof of Theorem 2, we have

l i m | x | + | x | N - 2 u 1 + ( x ) = c 1 > 0 ,   l i m | x | + | x | N - 2 u 1 - ( x ) = c 2 < 0

for some constants c1 and c2.

Theorem 5   Let (A1), (A4), (A5), (H1) and (H5) hold. If λ(0,λ1fm1), then the problem (5) possesses two solutions u1+, u1- such that u1+>0 and u1-<0 in RN. Therefore, we have

l i m | x | + | x | N - 2 u 1 + ( x ) = c 1 > 0 ,   l i m | x | + | x | N - 2 u 1 - ( x ) = c 2 < 0

for some constants c1 and c2.

Proof   If (λ,u) is any nontrivial solution of problem (5), dividing problem (5) by u2 and setting v=uu2 yields

{ - M ( u 2 ) Δ v = λ a ( x ) f ( u ) u 2 ,   i n   R N , v ( x ) 0 ,   a s   | x | + . (18)

Define

f ˜ ( v ) = { v 2 f ( v v 2 ) ,   i f   v 0 , 0 ,                   i f   v = 0 ,

and

M ˜ ( v 2 ) : = { M ( 1 v 2 ) ,   i f   v 0 , m 1 ,        i f   v = 0 .

Evidently, problem (18) is equivalent to

{ - M ˜ ( v 2 ) Δ v = λ a ( x ) f ˜ ( v ) ,   i n   R N , v ( x ) 0 , a s   | x | + . (19)

It is obvious that (λM˜(0),0) is always the solution of problem (19). By simple computation, we can show that f˜0=f(0,) and f˜=f0=.

Now, applying Theorem 3 and the inversion vvv=u, we achieve the conclusion.

Theorem 6   Let (A1), (A4), (A5), (H1) and (H6) hold. If λ(λ1fm1,+), then the problem (5) possesses two solutions u1+, u1- such that u1+>0 and u1-<0 in RN. Therefore, we have

l i m | x | + | x | N - 2 u 1 + ( x ) = c 1 > 0 ,   l i m | x | + | x | N - 2 u 1 - ( x ) = c 2 < 0

for some constants c1 and c2.

Proof   Applying a similar method as the proof of Theorem 5 and the conclusion of Theorem 4, we can easily get the desired conclusion.

Theorem 7   Let (A1), (A4), (A5), (H1) and (H7) hold. If λ(0,+), then the problem (5) possesses two solutions u1+, u1- such that u1+>0 and u1-<0 in RN. Therefore, we have

l i m | x | + | x | N - 2 u 1 + ( x ) = c 1 > 0 ,   l i m | x | + | x | N - 2 u 1 - ( x ) = c 2 < 0

for some constants c1 and c2.

Proof   Define

f [ n ] ( s ) = { n s ,                                                     s [ - 1 n , 1 n ] , [ f ( 2 n ) - 1 ] ( n s - 2 ) + f ( 2 n ) ,               s ( 1 n , 2 n ) , - [ f ( - 2 n ) + 1 ] ( n s + 2 ) + f ( - 2 n ) ,   s ( - 2 n , 1 n ) , f ( s ) ,                            s ( - , - 2 n ] [ 2 n , + ) .

Clearly, we can see that limn+f[n](s)=f(s), and (f[n])0=n.

Theorem 4   implies that there exists a sequence of unbounded components Dν[n]of solutions to problem (20) emanating from (λ1f0nM(0),0) and joins to (,).

The Lemma 2.5 in Ref. [13] implies that there exists an unbounded component Dνof limn+sup Dν[n] such that (0,0)Dν and (,)Dν where ν=+,-.

Theorem 8   Let (A1), (A4), (A5), (H1) and (H8) hold. If λ(0,+), then the problem (5) possesses two solutions u1+, u1- such that u1+>0 and u1-<0 in RN. Therefore, we have

l i m | x | + | x | N - 2 u 1 + ( x ) = c 1 > 0 ,   l i m | x | + | x | N - 2 u 1 - ( x ) = c 2 < 0

for some constants c1 and c2.

Proof   Define

f [ n ] ( s ) = { 1 n s ,                                                       s [ - 1 n , 1 n ] , [ f ( 2 n ) - 1 n 2 ] ( n s - 2 ) + f ( 2 n ) ,              s ( 1 n , 2 n ) , - [ f ( - 2 n ) + 1 n 2 ] ( n s + 2 ) + f ( - 2 n ) ,   s ( - 2 n , 1 n ) , f ( s ) ,                                 s ( - , - 2 n ] [ 2 n , + ) .

Clearly, we can see limn+f[n](s)=f(s), and (f[n])0=1n(f[n])=f.

Theorem 3   implies that there exists a sequence of unbounded components Dν[n]of solutions to problem (21) emanating from (λ1f0M(0)n,0) and joins to (0,).

The Corollary 2.1 in Ref. [13] implies that there exists an unbounded component Dν[n]of limn+sup Dν[n] such that (,0)Dν and (0,)Dν, where ν=+,-.

Theorem 9   Let (A1), (A4), (A5), (H1) and (H9) hold. There exists a λ+>0, such that λ(λ+,+), then the problem (5) possesses two solutions u1+, u1- such that u1+>0 and u1-<0 in RN. Therefore, we have

l i m | x | + | x | N - 2 u 1 + ( x ) = c 1 > 0 ,   l i m | x | + | x | N - 2 u 1 - ( x ) = c 2 < 0

for some constants c1 and c2.

Proof   In view of Theorem 8, there are two distinct unbounded continua Dν of solutions of the problem (5) emanating from (,0). Similar to the proof of Theorem 4, we can obtain that Dν joins (,0) to (,).

Theorem 10   Let (A1), (A4), (A5), (H1) and (H10) hold. There exists a λ+>0, such that λ(0,λ+), then the problem (5) possesses two solutions u1+, u1- such that u1+>0 and u1-<0 in RN. Therefore, we have

l i m | x | + | x | N - 2 u 1 + ( x ) = c 1 > 0 ,   l i m | x | + | x | N - 2 u 1 - ( x ) = c 2 < 0

for some constants c1 and c2.

Proof   In view of Theorem 7, there are two distinct unbounded continua Dν of solutions of the problem (5) emanating from (0,0). Similar to the proof of Theorem 3, we can obtain that Dν joins (0,0) to (0,).

3 Exact Multiplicity of One-Sign Solutions for Problem (5)

Refs. [14,15] studied exact multiplicity of solutions for a semi-linear elliptic equation, respectively.

In this section, we study exact multiplicity of one-sign solutions for problem (5). We first study the local structure of the bifurcation branch Dν (ν=+,-) near (λ1M(0),0), which is obtained in Theorem 1. Let Φ(λ,u)=u-G(λ,u) and

S = { ( λ , u ) R × E :   Φ ( λ , u ) = 0 , u 0 } ¯ R × E

For λR and 0<s<+, define an open neighborhood of (λ1M(0),0) in R×E as follows.

B s ( λ 1 M ( 0 ) , 0 ) = { ( λ , u ) R × E :   u + | λ - λ 1 M ( 0 ) | < s } .

Let E0 be a closed subset of E satisfying E=span{ψ1}E0, where ψ1 is an eigenfunction corresponding to λ1M(0) with ψ1=1. According to the Hahn-Banach theorem, we have lE* satisfying l(ψ1)=1 and E0={uE: l(u)=0}, where E* denotes the dual space of E. For any 0<ε<+ and 0<η<1, define

K ε , η + = { ( λ , u ) R × E : | λ - λ 1 M ( 0 ) | < ε ,   | l ( u ) | > η u } .

Obviously, Kε,η+ is an open subset of E, Kε,η=Kε,η+Kε,η-, with Kε,η-=-Kε,η+ which are disjoint and open in E.

Similar to the Lemma 6.4.1 in Ref. [16], we can show the following lemma.

Lemma 3   Let η(0,1), there is δ0>0 such that for each δ: 0<δ<δ0, it holds that

( ( S \ { ( λ 1 M ( 0 ) , 0 ) } ) B δ ( λ 1 M ( 0 ) , 0 ) ) K ε , η .

And there exist sR and a unique yE0 such that v=sψ1+y and |s|>ηv, for each λ=λ1M(0)+o(1).

Further,((S\{(λ1M(0),0)})Bδ(λ1M(0),0))Kε,η. λ=λ1M(0)+o(1) and y=o(s) as s0 for these solutions (λ,v).

Remark 2   From (H2) and (A6), we can see that f0f(s)/sf>0 for any s0, f(0)=0 and f0>f.

Remark 3   From Lemma 3, we can see that D=D+D- near (λ1M(0),0) is given by a curve (λ(s),u(s))=(λ1M(0)+o(1),sψ1+o(1)) for s near 0. Moreover, we can distinguish between two portions of this curve by s0 and s0.

Now, when a, M() and f satisfy the conditions (A1), (A4), (A5), (A6), by Dai et al[17] and Afrouzi et al[18], we give the definition of linearly stable solution for the problem (5) first.

For any φE and positive solution u of problem (5), we can calculate that the linearized eigenvalue problem of (5) at the direction φ is

{ - Δ φ - λ M ( u 2 ) a ( x ) f ' ( u ) φ = μ M ( u 2 ) φ ,    i n   R N , φ ( x ) 0 ,                    a s   | x | + . (20)

Definition 1   Suppose u is a solution of problem (5). The linear stability of u can be determined by the linearized eigenvalue problem (20). If all eigenvalues of problem (20) are positive, then we call u is stable, otherwise we call it unstable.

The Morse index M(u) of u is defined as the number of negative eigenvalues of problem (20). Call u is degenerate if 0 is an eigenvalue of problem (20), otherwise it is non-degenerate.

The main results of this paper are the following:

Theorem 11   Let (A1), (A4), (A5), (A6) and (H2) hold. If λ(λ1f0M(0),λ1fm1), then the problem (5) has exactly two solutions u1+(λ,) and u1-(λ,) such that u1+>0 and u1-<0 in RN, and has only the trivial solution for any λ(0,λ1f0M(0)][λ1fm1,+).

The following lemma is stability result for the positive solution.

Lemma 4   Under the assumptions of Theorem 11, then any solution u of problem (5) is stable and non-degenerate, and their Morse index are M(u)=0.

Proof   Let u be a solution of problem (5), and let (μ1,φ1) be the corresponding principal eigenpair of problem (20) with φ1>0 in RN. Notice that u and φ1 satisfy

{ - Δ u = λ M ( u 2 ) a ( x ) f ( u ) ,    i n   R N , u ( x ) 0 , a s   | x | + , (21)

and

{ - Δ φ 1 - λ M ( u 2 ) a ( x ) f ' ( u ) φ 1 = λ M ( u 2 ) φ 1 ,   i n   R N , φ 1 ( x ) 0 ,                    a s   | x | + . (22)

Multiplying the first equation of problem (22) by u and the first equation of problem (21) by φ1, subtracting and integrating, we obtain

μ 1 R N φ 1 u d x = λ M ( u 2 ) R N a ( x ) ( f ( u ) - f ' ( u ) u ) d x .

By some simple computations, we can show that it follows from (A6) that f(s)-f'(s)s0 for any s0. Since u0 and φ1>0 in RN, we have μ1>0 and the positive solution u must be stable. Similarly, we also have:

Lemma 5   Under the assumptions of Lemma 4, any negative solution u of problem (5) is stable, hence, non-degenerate and Morse index M(u)=0.

Proof of Theorem 11   Define F: R×ER by

F ( λ , u ) = - Δ u - λ M ( u 2 ) a ( x ) f ( u ) .

From Lemma 4 and Lemma 5, we know that any one sign solution (λ,u) of problem (5) is stable. Therefore, at any one-sign solution (λ*,u*) for the problem (5), we can apply the Implicit Function Theorem to F(λ,u)=0, and all the solutions of F(λ,u)=0 near (λ*,u*) are on a curve (λ,u(λ)) with |λ-λ*|ε for some small ε>0. Furthermore, by virtue of Remark 3, the unbounded continua D+ and D- are all curves.

To complete the proof, it suffices to show that u1+(λ,) (u1-(λ,)) is increasing (decreasing) with respect to λ. We only prove the case of u1+(λ,). The proof of u1-(λ,) can be given similarly. Since u1+(λ,) is differentiable with respect to λ (as a consequence of Implicit Function Theorem), taking the derivative of the first equation of problem (21) by λ, one can obtain that

- Δ ( d u 1 + d λ ) = - λ M ( u 1 + 2 ) a ( x ) f ' ( u 1 + ) d u 1 + d λ                      + 1 M ( u 1 + 2 ) a ( x ) f ' ( u 1 + ) . (23)

Multiplying the first equation of problem (23) by u and the first equation of problem (21) by du1+dλ, subtracting and integrating, we obtain 1M(u1+2)vRN[λa(x)(f'(u1+)u1+

- f ( u 1 + ) ) d u 1 + d λ + f ( u 1 + ) u 1 + ] d x = 0 .

Remark 2   implies f(s)s>0 for any sR. So we get (f'(u1+)u1+-f(u1+))du1+dλ0 by (A1). While (A6) shows that f'(u1+)u1+-f(u1+)0. Therefore, we have du1+dλ0.

Next we only prove the case of the uniqueness of positive solution of problem (5) since the proof of the uniqueness of negative solution of problem (5) is similar.

Suppose on the contrary that there exist two solutions u11+ and u12+ corresponding to λ with u11+D+ of the problem (5) for λ(λ1/f0,+). For ε>0, take (λ-ε,uλ-ε+), (λ+ε,uλ+ε+)D+, then uλ±ε+u11+ as ε0. By the monotonicity of u12+ with respect to λ, we get uλ-ε+u12+uλ+ε+. Then u11+=u12+.

By an argument as the above, we can show that problem (5) with λ=λ1f0M(0) has only the trivial solution. We can show that problem (5) has no one-sign solution for any λ(0,λ1f0M(0)). Suppose on the contrary that there exists a positive solution u for the problem (5), we multiply the first equation of problem (21) by ϕ1, and obtain after integrations by

λ 1 R N a ( x ) u ϕ 1 d x = 1 M ( u 2 ) R N f ( u ) u a ( x ) u ϕ 1 d x < λ f 0 M ( 0 ) R N a ( x ) u ϕ 1 d x

where ϕ1 is a positive eigenfunction associated to λ1. It follows that λ>λ1f0M(0), which contradicts λ(0,λ1f0M(0)). Similar to the above proof, we can obtain that the problem (5) has no positive solution for any λ(λ1fm1,+). Furthermore, we can obtain that the problem (5) has only the trivial solution for any λ(0,λ1f0M(0)][λ1fm1,+).

Theorem 12   Let (A1), (A4), (A5), (A6) and (H4) hold. If λ(λ1f0M(0),+), then the problem (5) has exactly two solutions u1+(λ,) and u1-(λ,) for such that u1+>0 and u1-<0 in RN, and has only the trivial solution for any λ(0,λ1f0M(0)].

Proof   By Theorem 4 and an argument similar to that of Theorem 11, we can prove it.

Theorem 13   Let (A1), (A4), (A5), (A6) and (H5) hold. If λ(0, λ1fm1), then problem (5) has exactly two solutions u1+(λ,) and u1-(λ,) for such that u1+>0 and u1-<0 in RN, and has only the trivial solution for any λ[λ1fm1,+).

Proof   By Theorem 5 and an argument similar to that of Theorem 11, we can obtain it.

Theorem 14   Let (A1), (A4), (A5), (A6) and (H7) hold. If λ(0,+), then the problem (5) has exactly two solutions u1+(λ,) and u1-(λ,) for such that u1+>0 and u1-<0 in RN.

Proof   By Theorem 7 and an argument similar to that of Theorem 11, we can prove it.

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