# Nodal Solutions for Problems with Mean Curvature Operator in Minkowski Space with Nonlinearity Jumping Only at the Origin

Authors: Wenguo Shen
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we establish a unilateral global bifurcation result for half-linear perturbation problems with mean curvature operator in Minkowski space. As applications of the abovementioned result, we shall prove the existence of nodal solutions for the following problem−div∇v/1−∇v2=αxv++βxv−+λaxfv,inBR0,vx=0,on∂BR0, where λ ≠ 0 is a parameter, R is a positive constant, and BR0=x∈ℝN:x<R is the standard open ball in the Euclidean space ℝNN≥1 which is centered at the origin and has radius R. a(|x|) ∈ C[0, R] is positive, v+ = max{v, 0}, v− = −min{v, 0}, α(|x|), β(|x|) ∈ C[0, R]; f∈Cℝ,ℝ, s f (s) > 0 for s ≠ 0, and f0 ∈ [0, ∞], where f0 = lim|s|⟶0 f(s)/s. We use unilateral global bifurcation techniques and the approximation of connected components to prove our main results.


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## Body

## 1. Introduction

We first consider the following problem with mean curvature operator in Minkowski space:(1)−div∇v1−∇v2=λaxv+Fx,v,λ,inBR0,vx=0,on∂BR0,where λ ≠ 0 is a parameter, R is a positive constant, and BR0=x∈ℝN:x<R is the standard open ball in the Euclidean space ℝNN≥1, which is centered at the origin and has radius R. Here, the nonlinear function F∈CBR0¯×ℝ2,ℝ and a(|x|) is a weighted function. Dirichlet problem (1) is associated to the mean curvature operator in the flat Minkowski space LN+1 with (x1, …, xN, t) and the Lorentzian metric ∑i=1Ndxi2−dt2.Some important and interesting results [1–3] for this type of problems have been obtained. Some specialists have studied problem (1); for example, Cheng and Yau [4] and Treibergs [5] studied problem (1) with λav ≡ F ≡ 0 and λav + F ≡ C, respectively. Bidaut-Véron and Ratto [6] and López [7] studied problem (1) with a ≡ 0, F = f(u) and a ≡ 0, F = kv + λ, respectively.Recently, Bereanu et al. [8, 9] have proved existence of classical positive radial solutions for problem (1) by Leray–Schauder degree argument and critical point theory.In 2016, Ma et al. [10] and Dai and Wang [11, 12] studied the existence of radial positive solutions and radial nodal solutions for problem (1) (where λav + g = λf(|x|, v)) by bifurcation techniques, respectively.On the contrary, among the abovementioned papers, the nonlinearities are differentiable at the origin. In [13], Berestycki established an important global bifurcation theorem from intervals for a class of second-order problems involving nondifferentiable nonlinearity.Recently, Dai and Ma [14, 15] considered interval bifurcation problem for second-order and high-dimensional p-Laplacian problems involving nondifferentiable nonlinearity, respectively.In 2016, whena(x) ≡ 1, R ≡ 1, Dai and Yang [16] have established a global bifurcation result from interval for the following problem with nondifferentiable nonlinearity:(2)−Δu=λaxu+Fx,u,λ,inBR0,ux=0,on∂BR0.It is clear that the radial solutions of (2) is equivalent to the solutions of the following problem:(3)−rN−1u′′=λrN−1aru+rN−1Fr,u,λ,r∈0,R,u′0=uR=0,where λ is a parameter, r = |x| and x ∈ BR(0), the nonlinear term F has the form F = f + g, where f,g∈C0,R×ℝ2 are radially symmetric with respect to r, and a, f, and g satisfy the following conditions:(H1) a ∈ C([0, R], (0, ∞)) is radially symmetricH2fr,s,λ/s≤M1, for all r ∈ (0, R), 0 < |s| ≤ R and all λ∈ℝ, where M1 is a positive constant(H3) g(r, s, λ) = o(|s|) near s = 0 uniformly in r ∈ (0, R) and λ on bounded setsLetC[0, R] is a real Banach space with the norm ‖u‖∞ = maxr∈[0,R]|u(r)|. Let E = {u(r) ∈ C1[0, R]: u′(0) = u(R) = 0} with the usual norm ‖u‖ = max{‖u‖∞, ‖u′‖∞}. Let Sk+ denote the set of functions in E which have exactly k − 1 interior nodal (i.e., nondegenerate) zeros in (0, R) and are positive at r = 0, and set Sk−=−Sk+ and Sk=Sk+∪Sk−. Let Φk±=ℝ×Sk± under the product topology. We use S to denote the closure of the nontrivial solutions set of problem (3) in ℝ×E, and Sk± to denote the subset of S with u∈Sk± and Sk=Sk+∪Sk−.Using the same method to prove ([16, Theorem 1]) with obvious changes, we may get the following global bifurcation result about Lemma 1.Lemma 1 (see [16, Theorem 1]).
Let (H1), (H2), and (H3) hold. Let Ik = [λk − M1, λk + M1]. The component Ck+ of Sk+∪Ik×0, containing Ik ×{0}, is unbounded and lies in Sk+∪Ik×0 and the component Ck− of Sk−∪Ik×0, containing Ik × {0}, is unbounded and lies in Sk−∪Ik×0.Letλk be the kth eigenvalue of the following eigenvalue problem:(4)−rN−1u′′=λrN−1aru,r∈0,R,u′0=uR=0.It is well known thatλk is positive and simple, and the eigenfunction corresponding to it has exactly k − 1 simple zeros in (0, R) (see [17, Theorem 2.1]).Furthermore, Dai and Yang [16] (a(x) ≡ 1, R ≡ 1) established spectrum of the following problem:(5)−rN−1u′′=λarrN−1u+αrrN−1u++βrrN−1u−,r∈0,R,u′0=uR=0,where a satisfies (H1), u+ = max{u, 0}, u− = − min{u, 0}, α(r) and β(r) satisfy:(H4) α(r), β(r) ∈ C[0, R] are radially symmetricProblem (5) is called half linear because it is positively homogeneous and linear in the cones u > 0 and u < 0. Similar to [13], we say that λ is a half-eigenvalue of problem (5), if there exists a nontrivial solution (λ, uλ). The eigenvalue λ is said to be simple if v = cuλ, c > 0, for all solutions (λ, v) of problem (5).Using the same method to prove ([16], Theorem 2) with obvious changes, we may get the following result about Lemma 2.Lemma 2 (see [16, Theorem 2]).
Let (H1) and (H4) hold. There exist two sequences of simple halfeigenvalues for problem (5), λ1+<λ2+…<λk+<… and λ1−<λ2−…<λk−<…. The corresponding half-linear solutions are in λk+×Sk+ and λk−×Sk−. Furthermore, aside from these solutions and the trivial one, there are no other solutions of problem (5).Motivated by the abovementioned papers, in this paper, we shall firstly establish a Dancer-type unilateral global bifurcation result (see Theorem2) for the following problem:(6)−div∇v1−∇v2=λaxv+αxv++βxv−+gx,v,λ,inBR0,vx=0,on∂BR0.It is clear that the radial solutions of (6) is equivalent to the solutions of the following problem:(7)−rN−1u′1−u′2′=rN−1λaru+αu++βu−+gr,u,λ,r∈0,R,u′0=uR=0,where λ ≠ 0 is a parameter, r = |x| and u(r) = v(|x|), a satisfies (H1), α, β satisfy (H4). Assume g∈C0,R×−R,R×ℝ satisfies (H3) and the following condition:(H5) There exists a function ht,u∈C0,R×−R,R,ℝ with h(t, s)s > 0 for any r ∈ [0, R] and s ≠ 0, such that g(t, u, λ) = λh(t, u)By a solution to problem (7), we mean a function u = u(r) ∈ C1[0, R] with ‖u′‖∞ < 1, such that rN−1u′/1−u′2 is differentiable and (7) is satisfied. Here, ‖⋅‖∞ denotes the usual sup-norm.Furthermore, we shall investigate the existence of solutions for the following problems:(8)−div∇v1−∇v2=αxv++βxv−+λaxfv,inBR0,vx=0,on∂BR0.It is clear that the radial nodal solutions of (8) is equivalent to the solutions of the following problem:(9)−rN−1u′1−u′2′=αrN−1u++βrN−1u−+λrN−1arfu,r∈0,R,u′0=uR=0,where λ ≠ 0 is a parameter, a satisfies (H1) and α and β satisfy (H4). Clearly, the nonlinear term of (9) is not necessarily differentiable at the origin because of the influence of the term αu+ + βu−. So, the bifurcation theory of [11, 12] cannot be applied directly to obtain our results. Fortunately, using the global interval bifurcation, we can obtain some results of the existence of radial solutions which extend the corresponding results of [12].The rest of this paper is arranged as follows. In Section2, we establish a unilateral global bifurcation result for half-linear perturbation problems (7). In Section 3, on the basis of the unilateral global bifurcation result (see Theorem 2), we shall investigate the existence of nodal solutions for a class of the half-linear perturbation problems (9).


## 2. Unilateral Global Bifurcation for Problem (7)

From Lemma1 (or see Theorem 1 and its proof of [16]), we can easily get the following result.Theorem 1.
There exist two unbounded subcontinuaDk+ and Dk− of solutions of (3) in ℝ×E, bifurcating from Ik × {0}, for ν ∈ {+, −}, such that Dkν⊂Φkν∪Ik×0.Similar to the process of ([11, (2.1)-p.60]) or ([12, (2.1)-p.469]) obtained, problem (7) is equivalent to(10)−rN−1u′1−u′2′=rN−1H1r,u,λ,r∈0,R,u′0=uR=0,where H1(r, u, λ) = λa(r)u + αu+ + βu− + g(r, u, λ). If u is a solution of problem (7), then for any r ∈ (0, R), one has that(11)rN−1H1r,u,λ=−rN−1u′′11−u′2−rN−1u′2u″11−u′21−u′2=−N−1rN−2u′11−u′2−rN−1u″11−u′2⋅11−u′2.By (11), one obtains that(12)−u″=H1r,u,λ1−u′23/2+N−1ru′1−u′2.Substituting (12) into (11), it follows that(13)−rN−1u′′=rN−1H1r,u,λ1−u′23/2+N−1rN−2u′3.Thus, problem (7) is equivalent to(14)−rN−1u′′=λrN−1aru+rN−1αu++βu−+rN−1K1r,u,λ,r∈0,R,u′0=uR=0,where(15)K1r,u,λ=λaru+αu++βu−1−u′23/2−1+gr,u,λ1−u′23/2−N−1ru′3.In Theorem2, we shall prove that K1(r, u, λ) satisfies (H3). Similar to ([13, (3.2)-p.383]), problem (14) is the perturbation of problem (5). In other words, problem (7) is the perturbation of problem (5).Next, it is clear that problem (14) can be equivalently written as(16)u=λLNu+LN∘K2=Fλ,u,where(17)K2r,u,λ=αu++βu−+K1r,u,λ,LNu=∫0RGt,ssN−1asusds,LN∘K2=∫0RGt,ssN−1K2s,us,λds,where G(r, s) be Green’s function associated with the operator Lu:=−rN−1u′′ with the same boundary condition as in problem (14) (see [18]). L: E ⟶ E is linear completely continuous (see ([18, (2.5)-p.502 to line 4-p.503])). L∘K2:ℝ×E⟶E is completely continuous (see [18]). Moreover, F is completely continuous from ℝ×E⟶E and Fλ,0=0,∀λ∈ℝ.Let(18)g¯r,u,λ=max0≤|s|≤ugr,s,λ,forr∈0,Randλon bounded sets,then g¯ is nondecreasing and(19)limu⟶0+g¯r,u,λu=0,uniformly for r ∈ (0, R) and λ on bounded sets.Similar to Lemma2.2 in [12], with obvious changes, we may get the following lemma.Lemma 3.
For fixedλ > 0, if {uk} is a sequence of solutions of problem (7) satisfying limk⟶+∞uk = 0 uniformly in r ∈ [0, R], then limk⟶+∞uk′ = 0 and limk⟶+∞uk″ = 0 uniformly in r ∈ [0, R].Proof.
Integrating the first equation of problem (7) from 0 to r for any r ∈ [0, R], we get that(20)rN−1uk′1−uk′2=−∫0rαrN−1uk++βrN−1uk−+rN−1gr,uk,λdr.
By (H3), we have that limk⟶+∞g(r, uk, λ) = 0 uniformly r ∈ [0, R] and λ on bounded sets. It follows that limk⟶+∞uk″ = 0 uniformly in r ∈ [0, R].
By (12), for any r ∈ (0, R), one has that(21)−uk″=λaruk+αuk++βuk−+gr,uk,λ1−uk′23/2+N−1ruk′1−uk′2.
By limk⟶+∞uk = 0 and limk⟶+∞uk′ = 0 uniformly in r ∈ [0, R] together with (21), it follows that limk⟶+∞uk′ = 0 uniformly in r ∈ (0, R).
Taking the limitr ⟶ 0+ on both sides of equation (21), together with limr⟶0+ukr=0 and limr⟶0+uk′r=0, by L’Hospital’s rule, we have that(22)−limr⟶0+uk″r=limr⟶0+N−1uk′r1−uk′2rr=limr⟶0+N−1uk″r1−uk′2r+N−1uk′−2uk′uk″1=limr⟶0+N−1uk″r.
We obtain thatlimr⟶0+uk″r=0.
Bylimr⟶R−ukr=limr⟶R−uk′r=0, taking the limit r ⟶ R− on both sides of equation (21), we may get that limr⟶R−uk″r=0.
Moreover, it follows that limk⟶+∞uk″ = 0 uniformly in r ∈ [0, R].Lemma 4.
For fixedλ > 0, if u is a solutions of problem (7) satisfying u ⟶ 0 uniformly in r ∈ [0, R], then(23)limu⟶0+1−u′23/2−1u=0,limu⟶0+N−1u′3u=0.Proof.
From Lemma3, we have that limu⟶0u′ = 0 and limu⟶0u″ = 0 uniformly in r ∈ [0, R].
By L’Hospital’s rule, we have(24)limu⟶0+1−u′23/2−1u=limu⟶0+3/21−u′21/2⋅−2u′u″u′=limu⟶0+321−u′21/2⋅−2u″=0,limu⟶0+N−1u′3u=limu⟶0+N−13u′2u″u′=limu⟶0+3N−1u′u″=0.
Using a similar method to prove ([14, Theorem 3.3]) or ([15, Theorem 3.1]) with obvious changes, we may obtain the following result.Theorem 2.
Let (H1) and (H3)–(H6) hold. For v ∈ {+, −}, λkν,0 is a bifurcation point for problem (7). Moreover, there exists a subcontinuum Dkν of solutions of problem (7), for v ∈ {+, −}, such that(i)
Dkν⊂Φkν∪λkν,0;(ii)
Dkν is unbounded;(iii)
λkν,0 is the unique bifurcation point for problem (7);(iv)
limλ⟶∞‖u‖ = max{1, R} for λ,u∈Dkν\λkν,0;(v)
limk⟶∞‖u‖ = 0 for λ,u∈Dkν\λkν,0.Proof.
By (19), one obtains that(25)gr,u,λu≤g¯r,u,λu≤g¯r,u,λu⟶0,asu⟶0,uniformly for r ∈ (0, R) and λ on bounded sets. It follows that(26)limu⟶0gr,u,λ1−u′23/2u=0.
By Lemma4, we can get that(27)λaru+αu++βu−1−u′23/2−1u⟶0,N−1u′3ru⟶0,asu⟶0.
By (26) and (27), we have that(28)K1r,u,λu⟶0,asu⟶0,uniformly for r ∈ (0, R) and λ on bounded sets. Let α0 ≔ maxr∈[0,R]|α(r)| and β0 ≔ maxr∈[0,R]|β(r)|. For 0 < |u| ≤ R, one get that |(αu+ + βu−)/u| ≤ α0 + β0. Furthermore, let(29)Ik0=λk−α0+β0a0,λk+α0+β0a0.
Theorem1 shows that there exist two unbounded subcontinua Dk+ and Dk− of solutions of (14) in ℝ×E, bifurcating from Ik0×0, and Dkν⊂Φkν∪Ik0×0 for v = + and v = −, in other words, (i) and (ii) hold.
We divide the rest of proofs into the following several steps:
(iii) Let us show thatDkν∩ℝ×0=λkν,0, i.e., λkν,0 is the unique bifurcation point for problem (7). Indeed, if there exists (λn, un) be a sequence of solutions of problem (7) converging to (λ, 0). By ‖un‖ for the two side of (14), let wn=un/un, then wn should be a solution of problem:(30)wn=LNλrN−1arwn+αrrN−1wn++βrrN−1wn−+rN−1K1r,un,λnun.
By (19), it follows that(31)gr,un,λnun≤g¯r,un,λnun≤g¯r,un∞,λnun≤g¯r,un,λnun⟶0,asun⟶0,uniformly for r ∈ (0, R) and λ on bounded sets.
By proof of Theorem 1.1 of [11] or proof of Theorem 1.1 of [12], we have that(32)N−1un′3run⟶0,λnarun+αrun++βrun−1−un′23/2−1un⟶0asun⟶0.
By (31) and (32), we have that(33)K1r,un,λnun⟶0asun⟶0,uniformly for r ∈ (0, R) and λn on bounded sets.
By (33) and the compactness of L, we obtain that, for some convenient subsequence wn⟶w0 as n ⟶ +∞. Now, w0 verifies the following equation:(34)−rN−1w0′′=λarrN−1w0+αrrN−1w0++βrrN−1w0−,and ‖w0‖ = 1. This implies that λ=λkν for v ∈ {+, −}.
(iv) Now, we shall prove that limλ⟶∞‖u‖ = max{1, R}. The idea is similar to the proof of Theorem 1.1 of [12], but we give a rough sketch of the proof for readers’ convenience. We only prove the case of v = + since the case of v = − is similar. For any λn,un∈Dkν\λkν,0 with λn ⟶ +∞ as n ⟶ +∞. Let t0n=0<t1n<t2n<⋅⋅⋅<tk−1n<tkn=R denote the zeros of un and r0n∈tjn,tj+1n (j ∈ {0, …, k − 1}) such that un′r0n=un′∞. Let r1n=minr∈tjn,tj+1n:un′r=0 (see [12, lines 1–4 in p.472]). Letting ρn∈r1n,r0n (or ρn∈r0n,r1n), and ρ∗=liminfn⟶+∞ρn,r1∞=liminfn⟶+∞r1n, by [12, line 23: p.472–line 25: p.473], one may obtain that unr1∞+ρ≥σ0 for n large enough and constant σ0 > 0, where ρ∈0,ρ∗−r1∞. By (H5), letting h1=minr1∞+ρ/4,r1∞+ρ×σ0,Rht,un. It follows that g(r, un, λn) ≥ λnh1. Now, integrating the first equation of problem (7) from r1∞+ρ/4 to r for any r∈r1∞+ρ/2,r1∞+ρ and n large enough, we get that(35)rN−1un′1−un′2=−∫r1∞+ρ/4rtN−1λnarun+αun++βun−+gr,un,λndt.
Seta0 = mint∈[0,R]a(t), α0 ≔ maxt∈[0,R]|α(t)| and β0 ≔ maxt∈[0,R]|β(t)|. By simple computation, we can show that(36)λnarun+αun++βun−+gr,un,λn≥λnσ0a0+h1−Rα0−β0for allr∈0,R.
Moreover, we have that(37)11−un′∞2≥11−un′2≥un′1−un′2=1rN−1∫r1∞+ρ/4rrN−1λnaru+αun++βun−+gr,un,λndr≥λnσ0a0+h1−Rα0−β0NrN−1∫r1∞+ρ/4rtN−1dt=λnσ0a0+h1−Rα0−β0rN−1rN−r1∞+ρ/4N≥λnσ0a0+h1−Rα0−β0Nr1∞+ρN−1r1∞+ρ/2N−r1∞+ρ/4N.
By limn⟶+∞λn = + ∞, it follows that limn⟶+∞un′∞=1. Noting that(38)unr=∫Rruntdt≤∫Rrunt′dt≤un′∞R.
Thus, one may obtain that limλ⟶∞‖u‖ = max{1, R}.
(v) Finally, we show that(39)limk⟶∞uk=0.
By Lemma4, we only show that limk⟶∞uk∞=0. On the contrary, one supposes limk⟶∞uk∞=ρ>0.
Lettk0=0<tk1<tk2<⋅⋅⋅<tkk−1<tkk=R denote the zeros of uk. By [12, p. 475], there exists rk∈tkj,tkj+1 (j ∈ {0, …, k − 1}) such that ukrk=uk∞=ρk, where ρk ≥ ρ/2 for any k ≥ k0, k0∈ℕ. Let r0k∈rk,tkj+1 satisfy ukr0k=θρk, where θ ∈ (0, 1). Set a0 = maxt∈[0,R]a(t), where h1 ≔ max(t,u)∈[0,R]×[−R,R]|h(t, uk)|. By [12, p. 475], one may obtain that(40)uk′≥uk′1−uk′2=1rN−1∫rkrrN−1λaruk+αuk++βuk−+gr,uk,λdr≥−λRa0+h1+Rα0+β0NrN−1∫rkrtN−1dt≥−λRa0+h1+Rα0+β0Nr.
Furthermore, we have that(41)−1−θρk=ukr0k−ρk≥−λRa0+h1+Rα0+β0N⋅12r0k2−rk2.
Therefore, we may obtain that(42)tkj+1−tkj>r0k−rk≥N1−θρ2λRa0+h1+Rα0+β0R>0.
However, by an argument similar to that of [19, Proposition 3.7], one obtains that tkj+1−tkj≤C/k⟶0, as k ⟶ +0, and we get a contradiction.
Thus, we have thatlimsupk⟶+∞uk∞=ρ=0.Remark 1.
Theorem2 indicates that the bifurcation interval Ik0=λk+,λk−, i.e., for problem (7), the bifurcation interval Ik0 is a finite point set. What conditions can ensure that the component indeed bifurcating from an interval is still an open problem for the problems with the mean curvature operator in Minkowski space.
To prove Theorem4, we use Whyburn type superior limit theorems. From [20], if the collection of the infinite sequence of sets is unbounded, Whyburn’s limit theorem ([21, Theorem 9.1]) cannot be used directly because the collection may not be relatively compact (where the definitions of superior limit and inferior limit, see [20, line 11 to line 16]). Dai [20] overcomed this difficulty and established the following results.Lemma 5 (see [20, Lemma 2.5]).
LetX be a normal space and let {Cn|n = 1, 2, …} be a sequence of unbounded connected subsets of X. Assume that(i)
There existsz∗ ∈ lim infn⟶+∞Cn with ‖z∗‖ < + ∞.(ii)
For everyR > 0, ∪n=1+∞Cn∩B¯R is a relatively compact set of X, where(43)BR=x∈Xx≤R.Then,D:=limsupn⟶∞Cn is unbounded, closed, and connected.In order to treat the problems with nonasymptotic nonlinearity at∞, we shall need the following lemmas.Lemma 6 (see [20, Corollary 2.1]).
LetX be a normal vector space and let {Cn|n = 1, 2, …} be a sequence of unbounded connected subsets of X. Assume that(i)
There existsz∗ ∈ lim infn⟶+∞Cn with ‖z∗‖ = +∞.(ii)
There exists a homeomorphismT: X ⟶ X such that ‖T(z∗)‖ < + ∞ and {T(Cn)} be a sequence of unbounded connected subsets in X.(iii)
For everyR > 0, ∪n=1+∞Cn∩B¯R is a relatively compact set of X, where(44)BR=x∈Xx≤R.
Then,D:=limsupn⟶∞Cn is unbounded, closed, and connected.Lemma 7 (see [20, Lemma 2.6]).
Let (X, ρ) be a metric space. If Cii∈ℕ is a sequence of sets whose limit superior is L and there exists a homeomorphism T: X ⟶ X such that, for every R > 0, ∪i=1+∞TCi∩B¯R is a relatively compact set, then for each ϵ > 0 there exists an m such that, for every n > m, Cn ⊂ Vϵ(L), where VϵL denotes the set of all points p with ρ(p, x) < ϵ for any x ∈ L.


## 3. Nodal Solutions for Problem (9)

Following Theorem2, we shall investigate the existence of nodal solutions for problem (9), where a(t), α(t), and β(t) satisfy conditions (H1) and (H4), respectively. We assume that f satisfies the following assumptions:(H6) sf(s) > 0 for s ≠ 0.(H7) f0 ∈ (0, ∞).(H8) f0 = ∞.(H9) f0 = 0, where(45)f0=lims⟶0fss.Applying Theorem2 to problem (1.10), we have the following results.Theorem 3.
Let (H1), (H4), (H6), and (H7) hold. For ν ∈ {+, −}, λkν/f0,0 is a bifurcation point for problem (9). Moreover, there exists an unbounded continuum Dν of solutions of problem (9), for ν ∈ {+, −} such that Dkν joins λkν/f0,0 to (∞, max{1, R}), Dkν⊂ℝ×Skν∪λkν/f0,0, and Dkν∩ℝ×0=λkν/f0,0. Furthermore, one obtains that limk⟶∞‖u‖ = 0 for λ,uλ∈Dkν\λkν/f0,0.Theorem 4.
Let (H1), (H4), (H6), and (H8) hold. For ν ∈ { + , −}, (0, 0) is a bifurcation point for problem (9). Moreover, there exists an unbounded continuum Dkν of solutions of problem (9), for ν ∈ {+, −} such that Dkν joins (0, 0) to (∞, max{1, R}), Dkν⊂ℝ×Skν∪0,0, and Dkν∩ℝ×0=0,0. Furthermore, one obtains that limk⟶∞‖u‖ = 0 for λ,u∈Dkν\0,0.Theorem 5.
Let (H1), (H4), (H6), and (H9) hold. (∞, 0) is a bifurcation point for problem (9). Moreover, there exists an unbounded continuum Dkν of solutions of problem (9), for ν ∈ {+, −} and such that Dkν joins (∞, 0) to (∞, max{1, R}), Dkν⊂ℝ×Skν, and ProjℝDkν≠Ø. Furthermore, one obtain that limk⟶∞‖u‖ = 0 for λ,u∈Dkν.From Theorems3–5, we can easily derive the following corollary, which give the ranges of parameter guaranteeing problem (9) has zero, two, or four radial nodal solutions.Corollary 1.
Assume that (H1), (H2), (H6), and (H7) hold. We may get the following results:(i)
Ifλkν>0 for ν ∈ {+, −}, then there exists μ1∈0,λkν/f0 such that problem (9) has no radial solution for all λ ∈ (0, μ1) and has at least two radial nodal solutions for all λ ∈ (μ1, +∞)(ii)
Ifνλkν>0 for ν ∈ {+ , −}, then there exist μ1+∈0,λk+/f0 and μ1−∈λk−/f0,0 such that problem (9) has no radial solution for all λ∈μ1−,0∪0,μ1+ and has at least two radial nodal solutions for all λ∈−∞,μ1−∪μ1+,+∞(iii)
Ifνλkν<0 for ν ∈ {+, −}, then there exist μ1−∈0,λk+/f0 and μ1+∈λk−/f0,0 such that problem (9) has no radial solution for all λ∈μ1+,0∪0,μ1− and has at least two radial nodal solutions for all λ∈−∞,μ1+∪μ1−,+∞(iv)
Ifλkν<0 for ν ∈ {+, −}, then there exists μ1∈λkν/f0,0 such that problem (9) has no radial solution for all λ ∈ (μ1, 0) and has at least two radial nodal solutions for all λ ∈ (−∞, μ1)Corollary 2.
Assume that (H1), (H2), (H6), and (H8) hold. We may get the following results:(i)
Ifλkν>0 for ν ∈ {+, −}, then problem (9) has at least two radial nodal solutions for all λ ∈ (0, +∞)(ii)
Ifνλkν≠0 for ν ∈ {+, −}, then problem (9) has at least two radial nodal solutions for all λ ∈ (−∞, 0) ∪ (0, +∞)(iii)
Ifλkν<0 for ν ∈ {+, −}, then problem (9) has at least two radial nodal solutions for all λ ∈ (−∞, 0)Corollary 3.
Assume that (H1), (H2), (H6), and (H9) hold. We may get the following results:(i)
Ifλkν>0 for ν ∈ {+, −}, then there exists 0 < μ2 ≤ μ3 such that problem (9) has no radial solution for all λ ∈ (0, μ2) and has at least four radial nodal solutions for all λ ∈ (μ3, +∞)(ii)
Ifνλν≠0 for ν ∈ { + , −}, then there exists 0≠νμ2ν≤νμ3ν such that problem (9) has no radial solution for all λ∈μ2−ν,0∪0,μ2ν and has at least four radial nodal solutions for all λ∈−∞,μ3−ν∪μ3ν,+∞(iii)
Ifλkν<0 for ν ∈ {+, −}, then there exists μ3 ≤ μ2 < 0 such that problem (9) has no radial solution for all λ ∈ (μ2, 0)and has at least four radial nodal solutions for all λ ∈ (−∞, μ3)Proof of Theorem 3.
Letζ∈Cℝ,ℝ be such that(46)fu=f0u+ζu,with(47)limu⟶0ζuu=0.
By [11, (2.1)-p. 60] or [12, (2.1)-p. 469], problem (9) is equivalent to(48)−rN−1u′′=λrN−1arf0u+rN−1αu++βu−+rN−1K3r,u,λ,r∈0,R,u′0=uR=0,where(49)K3r,u,λ=λarf0u+αu++βrN−1u−1−u′23/2−1+λarζu1−u′23/2−N−1ru′3.
Using a similar method to prove (28) with obvious changes, we have that(50)K3r,u,λu⟶0,asu⟶0,uniformly for r ∈ (0, R) and λ on bounded sets.
Let us consider problem (48) as a bifurcation problem from the trivial solution axis.
Applying Theorem2 to problem (48), we have the following result.
Forν ∈ {+, −}, λkν/f0,0 is a bifurcation point for problem (48). Moreover, there exists an unbounded continuum Dkν of solutions of problem (48), such that Dkν joins λkν/f0,0 to (∞, max{1, R}), Dkν⊂ℝ×Skν∪λkν/f0,0, Dkν∩ℝ×0=λkν/f0,0, and limk⟶∞‖u‖ = 0 for λ,uλ∈Dkν\λkν/f0,0.Proof of Theorem4.
Inspired by the idea of [22], we define the cut-off function of f as the following:(51)fns:=ns,s∈−1n,1n,f2n−1ns−2+f2n,s∈1n,2n,−f−2n+1ns+2+f−2n,s∈−2n,−1n,fs,s∈−∞,−2n∪2n,+∞.
We consider the following problem:(52)−rN−1u′1−u′2′=αrN−1u++βrN−1u−+λrN−1arfnu,r∈0,R,u′0=uR=0.
Clearly, we can see that limn⟶+∞f[n](s) = f(s), fn0=n.
Similar to the proof of Theorem3, there exists an unbounded continuum Dkνn of solutions of problem (52) emanating from λkν/n,0 and joining to (∞, max{1, R}) such that Dkνn⊂ℝ×Skν∪λkν/n,0.
Takingz∗ = (0, 0), we easily obtain that z∗∈liminfn⟶+∞Dkνn. So, condition (i) in Lemma 5 is satisfied with z∗ = (0, 0).
SinceF is completely continuous from ℝ×E⟶E, we have that ∪n=1∞Dkνn∩B¯R is precompact, and accordingly (ii) in Lemma 5 holds.
Therefore, by Lemma5, Dkν=limsupn⟶∞Dkνn is unbounded closed connected such that z∗∈Dkν, and ∞,max1,R∈Dkν. Clearly, u is the solution of problem (9) for any λ,u∈Dkν. From the definition of superior limit (see [21, P. 7]), we can easily see that Dkν⊆∪n=1∞Dkνn. So, one has that Dkν⊂ℝ×Skν∪0,0.
We may claim thatDkν∩ℝ×0=0,0, i.e., z∗ = (0, 0), is the unique bifurcation point of Dkν. Suppose on the contrary that there exists a sequence λn,un∈Dkν\0,0=limsupn⟶∞Dkνn\0,0 such that limn⟶∞λn ≠ 0 and limn⟶∞un = 0. Hence, for any N0∈ℕ, there exists n0 ≥ N0 such that λn,un∈Dkνn0. By (52), it follows that λn0=λkν/n0 for n0 ≥ N0. From the arbitrary of N0, it implies that n0 ⟶ ∞, i.e., μ = 0, which contradicts the assumption of μ ≠ 0.
ByDkν⊆∪n=1∞Dkνn and Theorem 3, one obtains that limk⟶∞‖u‖ = 0 for λ,u∈Dkν\0,0.Proof of Theorem 5.
Inspired by the idea of [22], we define the cut-off function of f as the following:(53)fns:=1ns,s∈−1n,1n,f2n−1n2ns−2+f2n,s∈1n,2n,−f−2n+1n2ns+2+f−2n,s∈−2n,−1n,fs,s∈−∞,−2n∪2n,+∞.
We consider the following problem:(54)−rN−1u′1−u′2′=αrN−1u++βrN−1u−+λrN−1arfnu,r∈0,R,u′0=uR=0.
Clearly, we can see that limn⟶+∞f[n](s) = f(s), fn0=1/n.
Similar to the proof of Theorem3, there exists an unbounded continuum Dkνn of solutions of the problem (54) emanating from λkνn,0, such that Dkνn⊂ℝ×Skν∪λνn,0 and ∞,max1,R∈Dkνn.
Takingz∗ = (∞, 0), we easily obtain that z∗∈liminfn⟶+∞Dνn with z∗ℝ×E=+∞ So, condition (ii) in Lemma 6 is satisfied with z∗ = (+∞, 0).
Define a mappingT: R × X ⟶ R × X such that(55)Tλ,u=1λ,u,λ∈−∞,0∪0,+∞,0,u,λ=∞,∞,u,λ=0.
It is easy to verify thatT is a homeomorphism and Tz∗ℝ×X=0. Obviously, TDkνn be a sequence of unbounded connected subsets in X. So, (ii) in Lemma 6 holds. Since F is completely continuous from ℝ×E⟶E, we have that ∪n=1+∞TDνn∩B¯R is precompact, and accordingly (iii) in Lemma 6 holds.
Therefore, by Lemma6, Dkν=limsupn⟶∞Dkνn is unbounded closed connected such that z∗∈Dkν and ∞,max1,R∈Dkν. Obviously, u is the solution of problem (9) for any λ,u∈Dkν. From the definition of superior limit (see [21, P. 7]), we can easily see that Dkν⊆∪n=1∞Dkνn. So, one has that Dkν⊂ℝ×Skν∪0. Next, we show that Dkν∩ℝ×0=Ø. Suppose on the contrary that there exists a sequence λn,un⊆Dkν such that limn⟶+∞λn = μ and limn⟶+∞‖un‖ = 0. From (9), we can easily get that(56)un=∫0RGr,ssN−2sαun++sβun−+λnarsfun1−un′23/2−N−1un′3dr.
Lettingwn = un/‖un‖, we have that(57)wn=∫0RGr,ssN−2sαwn++sβwn−+λnarsfunun1−un′23/2−N−1un′3undr.
Similar to (31), we can show that(58)limn⟶+∞funun=0.
By (32), (57), and (58), we have that(59)wn≤M∫0RGr,ssN−1wndr,where M = max{α0, β0}. By the Gronwall–Bellman inequality ([23, Lemma 2.1]), we obtain that ‖w‖ = 0. This contradicts the fact of ‖w‖ = 1. Hence, we have that Dν⊂ℝ×Pν.
Finally, we show thatProjℝDkν≠Ø. From the argument of Theorem 3, we have known that Dkνn has unbounded projection on ℝ for any fixed n∈ℕ. By Lemma 7, for each fixed ϵ > 0 there exists an m such that, for every n>m,Dkνn⊂VϵDkν. This implies that(60)λνn,+∞⊆ProjℝDkνn⊆ProjℝVεDkν,where ProjℝDkν denotes the projection of Dkν on ℝ. It follows that the projection of Dkν is nonempty on ℝ.
By an argument similar to that of Theorem4, we can get the asymptotic behavior of limk⟶∞‖u‖ = 0 for λ,u∈Dkν.



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