S-shaped connected component of radial positive solutions for a prescribed mean curvature problem in an annular domain

Man Xu; Ruyun Ma

doi:10.1515/math-2019-0076

Article Open Access

S-shaped connected component of radial positive solutions for a prescribed mean curvature problem in an annular domain

Man Xu and Ruyun Ma

Published/Copyright: August 8, 2019

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$Open Mathematics$

From the journal Open Mathematics Volume 17 Issue 1

Abstract

In this paper, we show the existence of an S-shaped connected component in the set of radial positive solutions of boundary value problem

− div(ϕN(∇y))=λa(|x|)f(y)inA,∂y∂ν=0 on Γ1,y=0 onΓ2,

where R₂ ∈ (0, ∞) and R₁ ∈ (0, R₂) is a given constant, 𝓐 = {x ∈ ℝ^N : R₁ < ∣x∣ < R₂}, Γ₁ = {x ∈ ℝ^N : ∣x∣ = R₁}, Γ₂ = {x ∈ ℝ^N : ∣x∣ = R₂}, ϕN(s)=s1−|s|2, s ∈ ℝ^N, λ is a positive parameter, a ∈ C[R₁, R₂], f ∈ C[0, ∞), ∂y∂ν denotes the outward normal derivative of y and ∣⋅∣ denotes the Euclidean norm in ℝ^N. The proof of main result is based upon bifurcation techniques.

Keywords: S-shaped connected component; positive radial solutions; mean curvature operator; annular domain; Minkowski space; bifurcation

MSC 2010: 34B10; 34B18; 34C23; 35B40; 35J65

1 Introduction

Hypersurfaces of prescribed mean curvature in flat Minkowski space 𝕃^N+1 = {(x, t) : x ∈ ℝ^N, t ∈ ℝ}, with the Lorentzian metric ∑i=1Ndxi2−dt2, where (x, t) = (x₁, x₂, ⋯, x_N, t), are of interest in differential geometry and in general relativity. It is well-known that the study of spacelike submanifolds of codimension one in 𝕃^N+1 with prescribed mean extrinsic curvature leads to Dirichlet problems of the type

− div(∇u1−|∇u|2)=f(x,u)inΩ,u=0 on∂Ω, (1)

where Ω is a bounded domain in ℝ^N and the nonlinearity f : Ω × ℝ → ℝ is continuous, see [1, 2].

The existence, multiplicity and qualitative properties of solutions of (1) have been extensively studied by many authors in recent years, see Coelho et al. [3], Treibergs [4], Cano-Casanova et al. [5], Pan et al. [6], López [7], Corsato et al. [8, 9], Korman [10] as well as Ma et al. [11, 12], and the references therein. It is worth pointing out that the starting point of this type of problems is the seminal paper [13], and from Bartnik and Simon [2] as well as Bereanu and Mawhin [14], we know (1) has a solution whatever f is. This can be seen as a universal existence result for the above problem. However, in our study problem (1) generally admits the null solution, it may be interesting to investigate the existence of non-trivial solutions, especially the positive solutions. However, there are few works on positive solutions of (1), see Coelho et al. [15], Bereanu et al. [16, 17], Ma et al. [18] and Dai [19].

Specifically, depending on the behaviour of f = f(x, s) near s = 0, Coelho et al. [15] discussed the existence of either one, or two, or three, or infinitely many positive solutions of the quasilinear two-point boundary value problem

−(u′1−u′2)′=f(x,u),x∈(0,1),u(0)=u(1)=0,

where f is L^p-Carathéodory function, and the proof of main results are based upon the variational and topological methods. Bereanu et al. [16, 17] obtained some important existence, nonexistence and multiplicity results for the positive radial solutions of problem (1) in a ball by using Leray-Schauder degree argument and critical point theory. Recently, Ma et al. [18] are concerned with the global structure of radial positive solutions for the problem (1) in a ball by using global bifurcation techniques, and extended the results of [16, 17] to more general cases, all results, depending on the behavior of nonlinear term f near 0. Dai [19] investigated the intervals of the parameter λ in which the problem (1) has zero, one or two positive radial solutions corresponding to sublinear, linear, and superlinear nonlinearities f at zero, respectively. However, [18, 19] only give a full description of the set of radial positive solutions of (1) for certain classes of nonlinearities f, and give no any information about the directions of a bifurcation.

In 2015, Sim and Tanaka [20] proved the existence of S-shaped connected component in the set of positive solutions for the one-dimensional p-Laplacian problem with sign-changing weight

−(|u′|p−2u′)′=μm(x)f(u),x∈(0,1),u(0)=u(1)=0, (2)

where p > 1, m ∈ C[0, 1], f ∈ C[0, ∞) and μ is a positive parameter. They obtained the following result by bifurcation techniques.

Theorem A

([20, Thoerem 1.1]) Assume

there exist x₁, x₂ ∈ [0, 1] such that x₁ < x₂, m(x) > 0 on (x₁, x₂) and m(x) ≤ 0 on [0, 1]∖[x₁, x₂],
there exist α > 0, f₀ > 0 and f₁ > 0 such that lims→0+f(s)−f0sp−1sp−1+α=−f1,
f∞:=lims→∞f(s)sp−1=0,
there exists s₀ > 0 such that

mins∈[s0,2s0]f(s)sp−1≥f0(p−1)μ1m0(πpx2−x1)p,

where μ₁ > 0 is the first eigenvalue of the linear problem associated to (2), and

πp:=2πpsin⁡(πp),m0=minx∈[3x1+x24,x1+3x24]m(x).

Then there exist μ∗∈(0,μ1f0) and μ∗>μ1f0 such that
1. (2) has at least one positive solution if μ = μ_∗;
2. (2) has at least two positive solutions if μ_∗ < μ ≤ μ1f0 ;
3. (2) has at least three positive solutions if μ1f0 < μ < μ^∗;
4. (2) has at least two positive solutions if μ = μ^∗;
5. (2) has at least one positive solution if μ > μ^∗.

Of course, the natural question is whether or not the similar result can be established for the prescribed mean curvature problem (1)?

The purpose of this paper is to show the existence of the S-shaped connected component in the set of radial positive solutions for a prescribed mean curvature problem in an annular domain

− div(ϕN(∇y))=λa(|x|)f(y)in A,∂y∂ν=0on Γ1,y=0onΓ2, (3)

where R₂ ∈ (0, ∞) and R₁ ∈ (0, R₂) is a given constant, 𝓐 = {x ∈ ℝ^N : R₁ < ∣x∣ < R₂}, Γ₁ = {x ∈ ℝ^N : ∣x∣ = R₁}, Γ₂ = {x ∈ ℝ^N : ∣x∣ = R₂}, ϕN(s)=s1−|s|2, s ∈ ℝ^N, λ is a positive parameter, a ∈ C[R₁, R₂], f ∈ C[0, ∞), ∂y∂ν denotes the outward normal derivative of y and ∣⋅∣ denotes the Euclidean norm in ℝ^N. To the best of our knowledge, for problem (3), such bifurcation curve is completely new and has not been practically described before.

Setting, as usual ∣x∣ = r and y(x) = u(r), the problem (3) reduces to the mixed boundary value problem

−(rN−1ϕ1(u′))′=λrN−1a(r)f(u),r∈(R1,R2),u′(R1)=u(R2)=0, (4)

where ϕ₁ (s)=s1−s2, s ∈ ℝ, ϕ₁ : (−1, 1) → ℝ is an odd, increasing homeomorphism and ϕ₁(0) = 0. To find a positive radial solution of (3), it is enough to find a positive solution of (4). We say that a function u ∈ C¹[R₁, R₂] is a solution of (4) if maxr∈[R1,R2] ∣u′(r)∣ < 1, r^N−1ϕ₁(u′) ∈ C¹[R₁, R₂], and satisfies (4). Let X = C[R₁, R₂] with the norm ∥u∥_∞ := maxs∈[R1,R2] ∣u(s)∣. Let E = {u ∈ C¹[R₁, R₂] : u′(R₁) = u(R₂) = 0} with the norm ∥u∥ := ∥u′∥_∞.

Let λ_k(a, R₁) be the k-th eigenvalue of the eigenvalue problem

−(rN−1u′)′=λrN−1a(r)u,r∈(R1,R2),u′(R1)=u(R2)=0, (5)

and φ_k be the eigenfunction corresponding to λ_k(a, R₁). It is well-known that

0<λ1(a,R1)<λ2(a,R1)<⋯<λk(a,R1)<⋯→+∞ask→+∞,

and no other eigenvalues. Moreover, the algebraic multiplicity of λ_k(a, R₁) is 1, and the eigenfunction φ_k has exactly k − 1 zeros in (R₁, R₂), see [21].

The first eigenvalue λ₁(a, R₁) is the minimum of the Rayleigh quotient, namely,

λ1(a,R1)=inf{∫R1R2sN−1(u′(s))2ds∫R1R2sN−1a(s)(u(s))2ds:u∈H01(R1,R2),∫R1R2sN−1a(s)(u(s))2ds>0}.

Assume that:

a : [R₁, R₂] → [0, ∞) is a continuous function and a(s) > 0 for s ∈ (R₁, R₂);
f ∈ C([0, ∞), [0, ∞)) with f(s) > 0 for s > 0;
there exist f₀ ∈ (0, ∞), δ ∈ (0,R2−R132) and g ∈ C([0, ∞), [0, ∞)) with g(s) > 0 for s ∈ (0, δ] such that

f(s)=f0s−g(s)fors∈[0,δ],

where lims→0+g(s)s=0;
there exists s₀ : s₀ ∈ (R2−R132,3(R2−R1)32), such that

mins∈[s0,4s0]f(s)s≥64f077λ1(a,R1)a0⋅η1,

where

a0=mins∈[R1,R1+3R24]a(s),

η₁ is the first positive eigenvalue of the problem

(rN−1v′(r))′+η1rN−1v(r)=0,r∈(R1,R1+3R24),v′(R1)=v(R1+3R24)=0.

The main result of this paper is the following.

Theorem 1.1

Assume that (A1) and (F1)-(F3) hold. Then there exist λ∗∈(0,λ1(a,R1)f0) and λ∗>λ1(a,R1)f0 such that

(3) has at least one radial positive solution if λ = λ_∗;
(3) has at least two radial positive solutions if λ_∗ < λ ≤ λ1(a,R1)f0
(3) has at least three radial positive solutions if λ1(a,R1)f0 < λ < λ^∗;
(3) has at least two radial positive solutions if λ = λ^∗;
(3) has at least one radial positive solution if λ > λ^∗;
limλ→∞ ∥u∥_∞ = R₂ − R₁ and limλ→∞ ∥u∥ = 1.

Remark 1.1

Let (λ, u) be a solution of (4), then it follows from ∣u′(r)∣ < 1 that

||u||∞<R2−R1.

This leads to the bifurcation diagrams mainly depend on the behavior of f = f(s) near s = 0. This is a significant difference between the Minkowski-curvature problems and the p-Laplacian problems.

Remark 1.2

In the special case p = 2, (F1’) reduces to

f(s)=f0s−f1s1+αfors∈[0,χ], (6)

where χ > 0 is a sufficiently small constant. It is easy to see that condition (F2) is weaker than (6), in fact f₁ s^1+α is a special case of g(s).

The main result is obtained by reducing the problem (4) to an equivalent problem and use the Rabinowitz global bifurcation techniques [22]. Indeed, under (F1) and (F2) we get an unbounded connected component which is bifurcating from (λ1(a,R1)f0,0), and condition (F2) pushes the bifurcation to the right near u = 0. Condition (F3) leads the bifurcation curve to the left at some point, and finally to the right near λ = ∞.

For other results concerning the existence of an S-shaped connected component in the set of solutions for diverse boundary value problems, see [23, 24, 25, 26] for the semilinear boundary value problems, and [27] for the p-Laplacian boundary value problems.

The rest of the paper is organized as follows. In Section 2, we give an equivalent formulation of problem (4) and some preliminary results to show the change of direction of a bifurcation. Section 3 is devoted to proving the main result.

2 Some preliminary results

2.1 An equivalent formulation

Let us define a function f͠ : ℝ → ℝ by setting

f~(s)=f(s),0≤s≤R2−R1,linear,R2−R1<s<(R2−R1)+1,0,s≥(R2−R1)+1,−f~(−s),s<0. (7)

Notice that, within the context of positive solutions, problem (4) is equivalent to the same problem with f replaced by f͠. In the sequel, we shall replace f with f͠, however, for the sake of simplicity, the modified function f͠ will still be denoted by f. Next, let us define h as follows

h(s)=(1−s2)3/2,|s|≤1,0,|s|>1. (8)

Lemma 2.1

A function u ∈ C¹[R₁, R₂] is a positive solution of (4) if and only if it is a positive solution of the problem

−(rN−1u′)′=rN−1[λa(r)f(u)h(u′)−N−1ru′3],r∈(R1,R2),u′(R1)=u(R2)=0. (9)

Proof

It is clear that a positive solution u ∈ C¹[R₁, R₂] of (4) is a positive solution of (9). Conversely, assume that u ∈ C¹[R₁, R₂] is a positive solution of (9). We aim to show that ∥u′∥_∞ < 1. Assume on the contrary that this is not true. Then we can easily find an interval [c, d] ⊆ [R₁, R₂] such that, either u′(c) = 0, 0 < ∣u′(r)∣ < 1 in (c, d) and ∣u′(d)∣ = 1, or ∣u′(c)∣ = 1, 0 < ∣u′(r)∣ < 1 in (c, d) and u′(d) = 0. Assume the former case occurs. The function u satisfies the equation

(rN−1ϕ1(u′))′+λrN−1a(r)f(u)=0

in [c, d]. For each r ∈ (c, d), integrating over the interval [c, r], we obtain

|ϕ1(u′(r))|=|1rN−1∫crλtN−1a(t)f(u(t))dt|≤C1

for some constant C₁ > 0 and hence

|u′(r)|≤ϕ1−1(C1)

for every r ∈ [c, d]. Since ϕ1−1 (C₁) < 1, taking the limit as r → d⁻ we obtain the contradiction ∣u′(d)∣ < 1. Therefore ∥u′∥_∞ < 1 and accordingly, u is a positive solution of (4).□

Lemma 2.2

Assume that (A1) and (F1) hold. Let u be a nontrivial solution of (4). Then u > 0 on [R₁, R₂) and u is strictly decreasing.

Proof

From

ϕ1(u′)=−λrN−1∫R1rsN−1a(s)f(u(s))ds, (10)

it follows u′ ≤ 0 because (A1) and (F1), so u is decreasing. Since u(R₂) = 0, we have u ≥ 0 on [R₁, R₂]. As u is not identically zero, one has u(R₁) > 0 and, from (10) we deduce that u′ < 0 on (R₁, R₂], which ensures that actually u is strictly decreasing and u > 0 on [R₁, R₂).□

Lemma 2.3

Assume that (A1) and (F1) hold. Let u be a positive solution of (4). Then

14∥u∥∞≤u(r)≤∥u∥∞,r∈[R1,R1+3R24].

Proof

Since (r^N−1ϕ₁(u′))′ = −λ r^N−1a(r)f(u), condition (A1) and (F1) imply that u′(r) is decreasing on (R₁, R₂). Since u′(R₁) = u(R₂) = 0 and u(r) > 0 on (R₁, R₂), we have u′(R₂) < 0. Therefore, u is concave on (R₁, R₂).

Hence,

u(r)≥||u||∞R2−R1(R2−r),r∈[R1,R2].

Note that R2−rR2−R1≥14 is equivalent to r ≤ R1+3R24. Therefore, we get the conclusion.□

Next, we give some property of concave functions.

Lemma 2.4

Let ν ∈ (0, 1) and β₀ ∈ (0,(R2−R1)(1−ν)8) be given. Let Iν,β0:=[R1,R2−4β01−ν]. Then

|u′(s)|≤1−ν,∀u∈A,∀s∈Iν,β0,

where

A:={u∈E|uis concave and strictly decreasing on(R1,R2),u′(R2)>−1,||u||∞≤4β0}.

Proof

Assume on the contrary that for any [R1,R1+R2−R1n], there exist sequence u_n ∈ E with u_n is concave and strictly decreasing on (R₁, R₂),

un′(R2)>−1,||un||∞≤4β0,

and x_n ∈ [R1,R1+R2−R1n], such that

|un′(xn)|>1−ν,

that is to say

−un′(xn)>1−ν.

From the concavity and monotonicity of u_n on (R₁, R₂), we have

−un′(x)>1−ν,x∈[xn,R2].

Hence

un(xn)=∫xnR2−un′(s)ds>(1−ν)(R2−xn)>(R2−R1−R2−R1n)(1−ν)>R2−R12(1−ν).

This is a contradiction since ∥u∥_∞ ≤ 4β₀ < R2−R12 (1−ν).

And it is easy to see that (1−ν)(R₂ − θ) ≤ 4β₀, it follows that θ ≥ R₂ − 4β01−ν, hence we can take I_ν,β₀ = [R1,R2−4β01−ν]. □

If we let ν=14 and β0=3(R2−R1)64∈(0,3(R2−R1)32). Then we have the following

Corollary 2.1

For any concave function u ∈ E with

u′(R2)>−1,||u||∞≤3(R2−R1)16,

we have

|u′(r)|≤34,r∈[R1,R1+3R24].

2.2 The direction of bifurcation

Let h ∈ X be given. It is well-known that the solution u of problem

−(rN−1u′)′=rN−1h(r),r∈(R1,R2),u′(R1)=u(R2)=0 (11)

can be expressed by

u(r)=∫R1R2G(r,s)sN−1h(s)ds:=K(h),

where the Green’s function of (11) for N = 2 is explicitly given by

G(t,s)=ln⁡R2t,R1≤s≤t≤R2,ln⁡R2s,R1≤t≤s≤R2,

and the Green’s function of (11) for N ≥ 3 is explicitly given by

G(t,s)=12−N[R22−N−t2−N],R1≤s≤t≤R2,12−N[R22−N−s2−N],R1≤t≤s≤R2.

Let 𝓛 : X → E be defined by 𝓛(u) = 𝓚(au). Both 𝓚 and 𝓛 are completely continuous and (5) is equivalent to

u=λL(u),

so that the eigenvalues of (5) are the characteristic values of 𝓛.

If (F2) holds, then lims→0+f(s)s=f0. Moreover,

f(s)=(f0−g(s)s)s,

where g(s) is a continuous function and

lims→0g(s)s=0. (12)

Let us set, for convenience, k(y) = h(y) − 1 for y ∈ ℝ. We have

limy→0k(y)y=0. (13)

Define the operator 𝓗 : ℝ × E → E by

H(λ,u)=L(λ((f0−g(u)u)k(u′)−g(u)u)u−ν(⋅)u′3),

where ν(r)=N−1r. Clearly, 𝓗 is completely continuous and, by (12) and (13), we have

lim∥u∥→0∥H(λ,u)∥∥u∥=0

uniformly with respect to λ varying in bounded intervals. Observe that, for any λ, (λ, u) ∈ ℝ × E, with u > 0, is a solution of the equation

u=λf0L(u)+H(λ,u), (14)

if and only if u is a positive solution of (9). Denote by 𝓢 the closure in ℝ × E of the set of all non-trivial solutions (λ, u) of (14) with λ > 0. Let P = {u ∈ E : u(r) ≥ 0, r ∈ [R₁, R₂]}. Then P is a positive cone of E and intP ≠ ∅.

Notice that

||u′||∞<1,(λ,u)∈S,

which implies that

||u||∞<R2−R1,(λ,u)∈S.

Hence, by Theorem 1.3 in [22] or Theorem 1.1 in [18], we have the following result.

Lemma 2.5

Assume that (A1), (F1) and (F2) hold. Then there exists an unbounded connected component 𝓒 in 𝓢 which is bifurcating from (λ1(a,R1)f0,0) such that 𝓒 ⊆ (([0, +∞) × int P) ∪ {(λ1(a,R1)f0,0)}). Moreover, 𝓒 joins (λ1(a,R1)f0,0) with infinity in λ direction.

We next give a Sturm-type comparison theorem.

Lemma 2.6

([28, Lemma 2.1], [29, Lemma 4.1]) Let b₂(r) > b₁(r) > 0 for r ∈ (R₁, R₂) and b_i ∈ L^∞(R₁, R₂), i = 1, 2. Also let u₁, u₂ be solutions of

−(rN−1ui′)′=rN−1bi(r)ui,

respectively. If u₁ has k zeros in (R₁, R₂), then u₂ has at least k + 1 zeros in (R₁, R₂).

Lemma 2.7

Assume that (A1), (F1) and (F2) hold. Let {(λ_n, u_n)} be a sequence of positive solutions of (14) which satisfies ∥u_n∥ → 0 and λn→λ1(a,R1)f0. Let φ₁(r) be the first eigenfunction of (5) which satisfies ∥φ₁∥ = 1. Then there exists a subsequence of {u_n}, again denoted by {u_n}, such that un∥un∥ converges uniformly to φ₁ on [0, 1].

Proof

Let vn:=un∥un∥. Then ∥v_n∥ = ∥vn′∥∞=1, consequently, ∥v_n∥_∞ is bounded. By the Ascoli-Arzela theorem, there exists a subsequence of v_n which uniformly converges to v ∈ X. We again denote the subsequence by v_n. For any (λ_n, u_n), we have

un(r)=∫R1R2G(r,s)sN−1[λna(s)f(un(s))h(un′(s))−N−1s(un′(s))3]ds. (15)

Multiplying both sides of (15) by ∥u_n∥⁻¹, we have

vn(r)=∫R1R2G(r,s)sN−1[λna(s)f(un(s))un(s)h(un′(s))vn(s)−N−1s(un′(s))3∥un∥]ds.

Since ∥u_n∥ → 0 implies ∥u_n∥_∞ → 0. From (F2) and (8), we conclude that f(un(s))un(s)→f0 and h(un′(s))→1 as n → ∞ uniformly for s ∈ [0, 1]. By Lebesgue’s dominated convergence theorem we know that

v(r)=λ1(a,R1)f0∫R1R2G(r,s)sN−1f0a(s)v(s)ds,

which means that v is a nontrivial solution of (5) with λ = λ₁(a, R₁), and hence v ≡ φ₁.□

Lemma 2.8

Assume that (A1), (F1) and (F2) hold. Let 𝓒 be as in Lemma 2.5. Then there exists σ > 0 such that (λ, u) ∈ 𝓒 and |λ−λ1(a,R1)f0| + ∥u∥ ≤ σ imply λ>λ1(a,R1)f0.

Proof

By (F1) and (F2), there exists a sufficiently small δ > 0 such that

0≤f(s)≤f0sfors∈[0,δ].

This fact together with the definition of h, we have

0≤f(s)sh<f0in(0,δ). (16)

Fixed σ=δR2−R1, then for any (λ, u) ∈ 𝓒 which satisfies |λ−λ1(a,R1)f0| + ∥u∥ ≤ σ, we known u is a positive solution of the problem

−(rN−1u′)′=rN−1[λa(r)f(u)h(u′)−N−1ru′3],r∈(R1,R2),u′(R1)=u(R2)=0, (17)

and

0≤u≤δ.

Assume that there exists a sequence {(λ_n, u_n)} such that (λ_n, u_n) ∈ 𝓒, |λn−λ1(a,R1)f0| + ∥u_n∥ ≤ σ, λ_n → λ and ∥u_n∥ → 0. Let v_n = un||un||. By Lemma 2.7, there exists a subsequence of v_n, again denoted by v_n, such that v_n converges uniformly to v on [R₁, R₂], where v > 0 and satisfies ∥v∥ = 1. Dividing both sides of the equation of (17) with (λ, u) = (λ_n, u_n) by ∥u_n∥, we obtain

−(rN−1vn′)′=rN−1[λna(r)f(un)unh(un′)vn−N−1rvn′(un′)2],r∈(R1,R2),vn′(R1)=vn(R2)=0.

Since ∥u_n∥ → 0 implies ∥u_n∥_∞ → 0. Combining this fact with (16), we have

−(rN−1v′)′<λrN−1a(r)f0v,

and

v′(R1)=v(R2)=0.

On the other hand

−(rN−1φ1′)′=λ1(a,R1)rN−1a(r)φ1,r∈(R1,R2),φ1′(R1)=φ1(R2)=0,

therefore

λ1(a,R1)∫R1R2sN−1a(s)φ1(s)v(s)ds=∫R1R2sN−1v′(s)φ1′(s)ds<λf0∫R1R2sN−1a(s)φ1(s)v(s)ds,

that is λ₁(a, R₁) − λf₀ < 0, and accordingly, λ>λ1(a,R1)f0. □

Lemma 2.9

Assume that (A1), (F1) and (F2) hold. Let 𝓒 be as in Lemma 2.5. Then there exists λ_⋄ > 0 such that Proj_ℝ𝓒 = [λ_⋄, ∞) ⊂ (0, ∞).

Proof

Suppose on the contrary that λ_⋄ = 0. Then there exists a sequence {(μ_n, u_n)} ⊂ 𝓒 satisfying u_n > 0 such that

limn→∞(μn,un)=(0,u∗)inR×X

for some u^∗ ≥ 0. Then by the fact

−(rN−1ϕ1(un′))′=μnrN−1a(r)f(un),r∈(R1,R2),un′(R1)=un(R2)=0,

after taking a subsequence and relabeling, if necessary, we have u_n → 0.

From the fact un′ (R₁) = 0, it follows that

ϕ1(un′(r))=−μnrN−1∫R1rsN−1a(s)f(un(s))dsforr∈(R1,R2).

This fact together with f(0) = 0 yield that

limn→∞∥un′∥∞=0. (18)

On the other hand

−(rN−1un′)′=rN−1[μna(r)f(un)h(un′)−N−1run′3],r∈(R1,R2),un′(R1)=un(R2)=0.

Let, for all n, vn=un||un||. Then we have

−(rN−1vn′(r))′=rN−1[μna(r)f(un(r))un(r)h(un′(r))vn(r)−N−1r(un′(r))3||un||],r∈(R1,R2),vn′(R1)=vn(R2)=0. (19)

Similar to the proof of Lemma 2.7, by (18) and (19), it concludes that μn→λ1(a,R1)f0, which contradicts μ_n → 0.□

Lemma 2.10

Assume that (A1), (F1) and (F3) hold. Let (λ, u) ∈ 𝓒 with ∥u∥_∞ = 4s₀. Then λ < λ1(a,R1)f0 .

Proof

Let (λ, u) ∈ 𝓒. Then by Lemma 2.3, we obtain

s0≤u(r)≤4s0,r∈[R1,R1+3R24]. (20)

Fixed s0=3(R2−R1)64, then from Corollary 2.1, for any (λ, u) ∈ 𝓒 with ∥u∥_∞ = 4s₀, we have

0≤|u′(r)|≤34,r∈[R1,R1+3R24].

Now we assume on the contrary that λ ≥ λ1(a,R1)f0 . Then for r ∈ [R1,R1+3R24] , by (20) and (F3), we have

λa(r)f(u(r))h(u′(r))u(r)≥λ1(a,R1)f0⋅a0⋅64f0η177λ1(a,R1)a0⋅7764=η1,

where η₁ is the first positive eigenvalue of the problem

(rN−1v′(r))′+η1rN−1v(r)=0,r∈(R1,R1+3R24),v′(R1)=v(R1+3R24)=0.

Let v be the corresponding eigenfunction of η₁. Then

v(r)>0,r∈[R1,R1+3R24].

We notice that u is a solution of

−(rN−1u′(r))′=rN−1[λa(r)f(u(r))h(u′(r))u(r)u(r)−N−1ru′3(r)]

on [R1,R1+3R24] . Lemmas 2.2 and 2.6 imply that u has at least one zero on [R1,R1+3R24] . This contradicts the fact that u(r) > 0 on [R1,R1+3R24] .□

Lemma 2.11

Assume that (A1) and (F1) hold. Let 𝓒 be as in Lemma 2.5. Then lim(λ,u)∈C,λ→∞ ∥u∥ = 1 and lim(λ,u)∈C,λ→∞ ∥u∥_∞ = R₂ − R₁.

Proof

We divide the proof into four steps.

We claim that there exists a constant B₀ > 0 such that for every (λ, u) ∈ 𝓒, if λ ≥ B₀, then

||u||∞≥ρ∗

for some ρ_∗ > 0.

Suppose on the contrary that there exists a sequence {(μ_n, u_n)} ⊂ 𝓒 satisfying

(μn,un)→(∞,0)in(0,+∞)×X.

Then, similar to the proof of Lemma 2.9, we have un′ converges to 0 as n → ∞. From this fact and (19), after taking a subsequence and relabeling, if necessary, we have v_n → v_∗ in X for some v_∗ ∈ X, and,

−(rN−1v∗′(r))′=rN−1μna(r)f0v∗(r),r∈(R1,R2),v∗′(R1)=v∗(R2)=0,

that is μ_n → λ1(a,R1)f0 . This contradicts with the fact μ_n → ∞. Therefore, the claim is valid.
Fixed ε ∈ (0,R2−R14), we can show that there exists γ > 0 such that for every (λ, u) ∈ 𝓒 with λ > B₀,

minx∈[R1,R2−ε]u(x)≥γρ∗. (21)

In fact, it is an immediate consequence of the fact

u(r)=∫R1R2G(r,s)sN−1[λa(s)f(u(s))h(u′(s))−N−1su′3(s)]ds

and

G(r,s)≥γG(s,s),(r,s)∈[R1,R2−ε]×[R1,R2].
It follows from (21) and (F1), there exists some constant M₀ > 0 such that

f(u(r))≥M0>0forr∈[R1,R2−ε],

and accordingly,

limλ→∞λrN−1∫R1rsN−1a(s)f(u(s))ds=+∞uniformly inr∈[R1+ε1,R2−ε]

for arbitrary fixed ε₁ ∈ (0,R2−ε−R14). This together with (A1), (F1) and the relation

u′(r)=−ϕ1−1(λrN−1∫R1rsN−1a(s)f(u(s))ds)

imply that

u′→−1inC[R1+ε1,R2−ε]asλ→+∞. (22)

Therefore, by the arbitrariness of ε and ε₁, we get lim(λ,u)∈C,λ→∞ ∥u∥ = 1.
Since

−u′(r)≥0,r∈(R1,R2].

This fact together with (22) imply that for (λ, u) ∈ 𝓒,

limλ→∞||u||∞=limλ→∞u(R1)=limλ→∞∫R1R2−u′(s)ds≥limλ→∞∫R1+ε1R2−ε−u′(s)ds=R2−ε−R1−ε1. (23)

By the arbitrariness of ε and ε₁, we have

limλ→∞||u||∞≥R2−R1. (24)

On the other hand

u(R1)=∫R1R2−u′(s)ds≤R2−R1. (25)

Therefore, from (24) and (25), we have

limλ→∞||u||∞=R2−R1.

3 Proof of the main result

In this section, we shall prove Theorem 1.1. We divide the proof into three steps.

Rightward bifurcation.

By Lemmas 2.5 and 2.8, there exists an unbounded connected component 𝓒 in the set of radial positive solutions of (3), which is bifurcating from (λ1(a,R1)f0,0) , and for any (λ, u) ∈ 𝓒 which satisfies |λ−λ1(a,R1)f0| + ∥u∥ ≤ σ, where σ > 0 is a sufficiently small constant, 𝓒 goes rightward. Moreover, from Lemma 2.11, it follows that 𝓒 joins (λ1(a,R1)f0,0) with infinity in λ direction.
Direction turn of bifurcation.

By Lemma 2.11 again, we have lim(λ,u)∈C,λ→∞ ∥u∥ = 1 and lim(λ,u)∈C,λ→∞ ∥_∞ = R₂ − R₁. Then there exists (λ₀, u₀) ∈ 𝓒 such that ∥u₀∥_∞ = 4s₀. Lemma 2.10 implies that 𝓒 goes leftward.
Existence of λ_∗ and λ^∗.

By Lemma 2.9, if (λ, u) ∈ 𝓒, then there exists λ_⋄ > 0 such that λ ≥ λ_⋄. And by Lemmas 2.8, 2.10 and 2.11 again, 𝓒 passes through some points (λ1(a,R1)f0,v1) and (λ1(a,R1)f0,v2) with ∥v₁∥_∞ < 4s₀ < ∥v₂∥_∞, and there exists λ and λ which satisfy 0 < λ_<λ1(a,R1)f0<λ¯ and both (i) and (ii):
1. if λ ∈ (λ1(a,R1)f0,λ¯], then there exist u and v such that (λ, u), (λ, v) ∈ 𝓒 and ∥u∥_∞ < ∥v∥_∞ < 4s₀;
2. if λ ∈ [λ_,λ1(a,R1)f0], then there exist u and v such that (λ, u), (λ, v) ∈ 𝓒 and ∥u∥_∞ < 4s₀ < ∥v∥_∞.
  
  Define λ^∗ = sup{λ̄ : λ̄ satisfies (i)} and λ_∗ = inf{λ : λ satisfies (ii)}. Then by the standard argument, (3) has a radial positive solution at λ = λ_∗ and λ = λ^∗, respectively. This completes the proof.□

Acknowledgments

The authors are very grateful to the anonymous referees for their valuable suggestions.

This work was supported by the NSFC (No.11671322).

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Received: 2019-04-26

Accepted: 2019-05-29

Published Online: 2019-08-08

This work is licensed under the Creative Commons Attribution 4.0 Public License.

Articles in the same Issue

https://doi.org/10.1515/math-2019-0076

Keywords for this article

S-shaped connected component; positive radial solutions; mean curvature operator; annular domain; Minkowski space; bifurcation

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