∑-Shaped Connected Component of Positive Solutions for One- Dimensional Prescribed Mean Curvature Equation in Minkowski Space

Tiaoyan QI; Yanqiong LU

doi:10.1051/wujns/2025303241

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Volume 30 / No 3 (June 2025)

Wuhan Univ. J. Nat. Sci., 30 3 (2025) 241-252

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Issue		Wuhan Univ. J. Nat. Sci. Volume 30, Number 3, June 2025


Page(s)		241 - 252
DOI		https://doi.org/10.1051/wujns/2025303241
Published online		16 July 2025

Wuhan University Journal of Natural Sciences, 2025, Vol.30 No.3, 241-252

Mathematics

CLC number: O175

∑-Shaped Connected Component of Positive Solutions for One- Dimensional Prescribed Mean Curvature Equation in Minkowski Space

Minkowski 空间中一维给定平均曲率方程正解的∑-型连通分支

Tiaoyan QI (漆调艳) and Yanqiong LU (路艳琼)^†

College of Mathematics and Statistics, Northwest Normal University, Lanzhou 730070, Gansu, China

^† Corresponding author. E-mail: luyq8610@126.com

Received: 5 December 2024

Abstract

In this work, we demonstrate that the existence of an ∑-shaped connected component within the set of positive solutions for the one-dimensional prescribed mean curvature equation in Minkowski space ${\begin{array}{l} - {(\frac{u^{'} (t)}{\sqrt[]{1 - (u^{'} {(t))}^{2}}})}^{'} = λ h (t) f (u (t)), t \in (0,1), \\ u (0) = 0, u^{'} (1) + \sqrt[]{λ} u (1) = 0, \end{array}$ with boundary conditions having parameter in two cases $f (0) = 0$ and $f (0) > 0$ by using upper and lower solution method, where $λ > 0$ is a parameter, $f \in C^{2} ([0, \infty), R)$ is monotonically increasing and $\underset{u \to 1^{-}}{l i m} \frac{f (u)}{1 - u} = 0$ , $h \in C^{1} ([0,1], (0, \infty))$ is a nonincreasing function and $h (t) > 1$ .

摘要

运用上下解方法证明 Minkowski 空间中一维给定平均曲率方程 ${\begin{array}{l} - {(\frac{u^{'} (t)}{\sqrt[]{1 - (u^{'} {(t))}^{2}}})}^{'} = λ h (t) f (u (t)), t \in (0 1), \\ u (0) = 0, u^{'} (1) + \sqrt[]{λ} u (1) = 0, \end{array}$ $f (0) = 0$ 在和 $f (0) > 0$ 两种情形下正解集 $\sum$ 型连通分支的存在性，其中， $λ > 0$ 为参数， $f \in C^{2} ([0, \infty), R)$ 单调递增且满足 $\underset{u \to 1^{-}}{l i m} \frac{f (u)}{1 - u} = 0$ ， $h \in C^{1} ([0,1], (0, \infty))$ 是单调递减函数且 $h (t) > 1$ .

Key words: boundary conditions with parameters / positive solutions / the upper and lower solution method / asymptotic property

关键字 : 边界条件带参数 / 正解 / 上下解定理 / 渐进性质

Cite this article: QI Tiaoyan, LU Yanqiong. ∑-Shaped Connected Component of Positive Solutions for One- Dimensional Prescribed Mean Curvature Equation in Minkowski Space[J]. Wuhan Univ J of Nat Sci, 2025, 30(3): 241-252.

Biography: QI Tiaoyan, female, Master candidate, research direction: difference equations and their applications. E-mail: qity77@126.com

Foundation item: Supported by the National Natural Science Foundation of China (12361040)

© Wuhan University 2025

This 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

The boundary value problem in which the boundary conditions involve parameters is one of the most significant problems in the mathematical theory. In 1994, Binding was the first to propose the Sturm Liouville problems with boundary conditions dependent on eigenparameters^[1]. In 1999, Hai^[2] used the prior estimation method to prove the existence of solutions for boundary value problem

${\begin{array}{l} u^{″} (t) + a (t) u^{'} (t) + f (u (t)) = 0, t \in (0,1), \\ u (0) = 0, u (1) = λ \end{array}$

with boundray conditions having parameters. Since then, many different problems with parameters under boundary conditions have been studied respectively by Marinets et al^[3], and a great deal of fruitful results have been achieved, which can be found in Refs. [4-7]. Fonseka et al^[6-7] investigated the positive solution of the boundary value problem with two boundary conditions having parameters by employing the upper and lower solution method. They proved the existence and multiplicity of the positive solutions and obtained the bifurcation diagram of the above positive solutions by using the time mapping method. Particularly, Fonseka et al^[6] discussed the existence and multiplicity results of the steady-state reaction diffusion equation

${\begin{array}{l} - u^{″} (t) = λ h (t) f (u (t)), t \in (0,1), \\ - d u^{'} (0) + μ (λ) u (0) = 0, \\ u^{'} (1) + μ (λ) u (1) = 0 \end{array}$

via the upper and lower solution method in three cases of $f (0) = 0$ , $f (0) > 0$ and $f (0) < 0$ . Further, they established a unique result for $λ \approx 0$ and $λ ≫ 1$ . In the above equation, $λ > 0$ is a parameter, $f \in C^{2} ([0, \infty), R)$ is an increasing function which is sublinear at infinity, $h \in$ $C^{1} ([0,1], (0, \infty))$ is a nonin-creasing function with $h_{1} : = h (1) > 0$ and there exist constants $d_{0} > 0$ , $α \in [0,1]$ such that $h (t) \leq \frac{d_{0}}{t^{α}}$ for all $t \in (0,1]$ , and $μ \in C ([0, \infty), [0, \infty))$ is an increasing function such that $μ (0) \geq 0$ .

The prescribed mean curvature equation addressed in this paper is regarded as a significant problem in partial differential equations and possesses extensive applications in other domains like physics and biology, such as the shape of the human cornea^[8-9], drops of capillary droopiness^[10], micro electronic mechanical systems, and corresponding models of large spatial gradients^[11]. Specifically, the Dirichlet problem with the Minkowski-curvature operator

${\begin{array}{l} - d i v (\frac{\nabla u (t)}{\sqrt[]{1 - (1 - \nabla u {(t))}^{2}}}) = f (t, u (t)), t \in Ω, \\ u (0) = 0, t \in \partial Ω \end{array}$

has drawn the attention of numerous scholars. Especially, the existence and multiplicity of positive solutions for the Dirichlet problem with the one-dimensional Minkowski-curvature operator

${\begin{array}{l} - {(\frac{u^{'} (t)}{\sqrt[]{1 - (u^{'} {(t))}^{2}}})}^{'} = λ f (u (t)), t \in (0,1), \\ u (0) = u (1) = 0 \end{array}$

have been extensively studied by employing the variational method, the upper and lower solution method, and the time mapping method. See Refs. [12-20], where $λ > 0$ is parameter, $f \in C ([0, \infty), [0, \infty))$ and $f (u) > 0 (u > 0)$ . It is obvious that the problems in the above references merely study the positive solutions of the Dirichlet problem with the mean curvature operator in Minkowski space in the case where the nonlinear term has a parameter. However, the prescribed mean curvature boundary value problem where both the nonlinear term and the boundary conditions have parameters must be further considered to accurately describe the corresponding physical phenomena.

Inspired by the above-mentioned papers, we explore the existence and multiplicity of positive solutions for the one-dimensional prescribed mean curvature problem in Minkowski space

${\begin{array}{l} - {(\frac{u^{'} (t)}{\sqrt[]{1 - (u^{'} {(t))}^{2}}})}^{'} = λ h (t) f (u (t)), t \in (0,1), \\ u (0) = 0, u^{'} (1) + \sqrt[]{λ} u (1) = 0, \end{array}$ (1)

with boundary conditions having parameter in two cases $f (0) = 0$ and $f (0) > 0$ by using upper and lower solution method, where $λ > 0$ is a parameter, $f \in C^{2} ([0, \infty), R)$ is monotonically increasing and $\underset{u \to 1^{-}}{l i m} \frac{f (u)}{1 - u} = 0$ , $h \in$ $C^{1} ([0,1], (0, \infty))$ is a nonincreasing function and $h (t) > 1$ .

From the conclusion of Ref. [15], it follows that problem (1) has a positive solution $u$ if and only if the problem

${\begin{array}{l} - u^{″} (t) = λ (1 - (u^{'} {(t))}^{2})^{\frac{3}{2}} h (t) f (u (t)), t \in (0,1), \\ u (0) = 0, u^{'} (1) + \sqrt[]{λ} u (1) = 0 \end{array}$ (2)

has a positive solution, and $| u^{'} | < 1$ , ${‖ u ‖}_{\infty} < 1$ .

Let $ϕ : (- 1,1) \to R$ is monotone increasing homeomorphism defined by $ϕ (s) = \frac{s}{\sqrt[]{1 - s^{2}}}$ , then $ϕ (0) = 0$ and $ϕ^{- 1} (s) = \frac{s}{\sqrt[]{1 + s^{2}}}$ is also monotone increasing homeomorphism and bounded. Let us consider the eigenvalue problem

${\begin{array}{l} - u^{″} (t) = λ u (t), t \in (0,1), \\ u (0) = 0, u^{'} (1) + \sqrt[]{λ} u (1) = 0 . \end{array}$ (3)

Based on the results of Ref. [6], the problem (3) has principal eigenvalue $λ_{1} > 0$ , and the eigencurve $B_{1} (λ)$ is Lipschitz continuous, strictly decreasing and convex. Further $\underset{λ \to \infty}{l i m} λ_{1} = λ_{1 D}$ , where $λ_{1 D} > 0$ is the principal eigenvalue of

${\begin{array}{l} - u^{″} (t) = λ_{1 D} u (t), t \in (0,1), \\ u (0) = 0 = u (1) . \end{array}$

Throughout the paper we will assume that $f$ satisfies: $(C_{1})$ $f^{'} (0) = 1$ ; $(C_{2})$ $f^{″} (s) < 0, s \in [0, σ^{*})$ for some $σ^{*} > 0$ .

Let $a > 0, b > 0$ and $M^{*} > 0$ . Define

$L (a, b) = \frac{4 b / (τ (\frac{1}{4}) f (b))}{m i n {\frac{a}{{‖ v_{h} ‖}_{\infty} f (a)}, \frac{8 M^{*}}{f (b)}}},$

where $v_{h}$ is the solution of the problem

${\begin{array}{l} - v_{h}^{″} (t) = h (t), t \in (0,1), \\ v_{h} (0) = 0, v_{h}^{'} (1) + \sqrt[]{λ} v_{h}^{'} (1) = 0 . \end{array}$ (4)

First, we state the main results for the case: $f (0) = 0$ .

Theorem 1 Assume $(C_{1})$ and $f (0) = 0$ hold. Then problem (1) has no positive solution for $λ \approx 0$ and has at least one positive solution for $λ > λ_{1}$ . Further, if there exist $a > 0, b > 0, M^{*} > 0$ such that $a \in (0, b),$ $M^{*} > 2 b,$ $L (a, b) < 1$ and $\frac{4 b}{τ (\frac{1}{4}) f (b)} > λ_{1}$ . Then problem (1) has at least three positive solutions for $λ \in (\frac{4 b}{τ (\frac{1}{4}) f (b)}, m i n {\frac{a}{{‖ v_{h} ‖}_{\infty} f (a)}, \frac{8 M^{*}}{f (b)}})$ .

Theorem 2 Assume $(C_{1})$ , $(C_{2})$ and $f (0) = 0$ hold. Then problem (1) has least one positive solution $u_{λ}$ for $λ > λ_{1}$ such that ${‖ u_{λ} ‖}_{\infty} \to 0$ .

Next, we state the main results for case: $f (0) > 0$ .

Theorem 3 Assume $f (0) > 0$ . Then problem (1) as at least one positive solution for $λ > 0$ . Further, if there exist $a > 0$ , $b > 0$ and $M^{*} > 0$ such that $a \in (0, b)$ , $M^{*} > 2 b$ and $L (a, b) < 1$ . Then (1) has at least three positive solutions for $λ \in (\frac{4 b}{τ (\frac{1}{4}) f (b)}, m i n {\frac{a}{{‖ v_{h} ‖}_{\infty} f (a)}, \frac{8 M^{*}}{f (b)}}) .$

Note that the connected component of positive solutions of (1) is the $Σ$ shaped under the assumptions of Theorem 1 or Theorem 3. Figure 1(a) illustrated the main results of Theorem 1, Fig.1 (b) illustrated the main results of Theorem 3. The rest of the paper is organized as follows. In Section 1, we introduce some lemmas needed to prove the main theorems. In Section 2, we use the time-mapping method to prove that problem $(1)$ has at least one positive solution when $f (u) = u, h (t) = 1$ , which will help to construct the subsolution of $(1)$ . In Section 3, we give proofs of the main results. In Section 4, we show some examples.

Fig. 1 Bifurcation diagram for problem (1)

1 Preliminaries

Let $C_{0}^{1} [0,1] = {u \in C^{1} [0,1] | u (0) = 0}$ , it is not hard to verify that $C_{0}^{1} [0,1]$ endowed with the norm $‖ u ‖ = {‖ u ‖}_{\infty} + {‖ u' ‖}_{\infty}$ is Banach space.

We donote by $P, Q : C [0,1] \to C [0,1]$ , and the continuous projects defined by $P u (t) = u^{'} (0),$ $Q u (t) = \int_{0}^{1} u (t) d t$ , $t \in (0,1)$ and define the continuous linear operator $H : C [0,1] \to C_{0}^{1} [0,1]$ , $H u (t) = \int_{0}^{t} u (s) d s$ , $t \in (0,1)$ . Integration of both sides of the equation in problem (1) from $0$ to $t$ implies that

$\frac{u^{'} (t)}{\sqrt[]{1 - (u^{'} {(t))}^{2}}} = \frac{u^{'} (0)}{\sqrt[]{1 - (u^{'} {(0))}^{2}}} - λ \int_{0}^{1} h (s) f (u (s)) d s,$

i.e.

$ϕ (u^{'} (t)) = ϕ (u^{'} (0)) - λ \int_{0}^{t} h (s) f (u (s)) d s .$

Applying $ϕ^{- 1}$ to the above equation, we get that $u^{'} (t) = ϕ^{- 1} [ϕ (u^{'} (0)) - λ \int_{0}^{t} h (s) f (u (s)) d s]$ . Furthermore, we integrate the above equation from $0$ to $t$ , it follows that

$u (t) = \int_{0}^{t} ϕ^{- 1} [ϕ (u^{'} (0)) - λ \int_{0}^{t} h (s) f (u (τ)) d τ] d s .$ (5)

Combining $u^{'} (1) + \sqrt[]{λ} u (1) = 0$ , we obtain that $u^{'} (0)$ satisfies

$ϕ^{- 1} [ϕ (u^{'} (0)) - λ \int_{0}^{t} h (s) f (u (s)) d s] + \sqrt[]{λ} \int_{0}^{1} ϕ^{- 1} [ϕ (u^{'} (0)) - λ \int_{0}^{t} h (s) f (u (τ)) d τ] d s = 0$ (6)

which yields that

$\begin{array}{l} ϕ^{- 1} [ϕ (P u (t)) - λ Q (h (t) f (u (t)))] \\ + \sqrt[]{λ} Q [ϕ^{- 1} (ϕ (P u (t)) - λ H (h (t) f (u (t))))] = 0 \end{array}$ (7)

Lemma 1 For $h \in C [0,1]$ , there is a unique $γ : = u^{'} (0)$ such that $ϕ^{- 1} [ϕ (γ) - λ Q (h (t) f (u (t)))] + \sqrt[]{λ} Q [ϕ^{- 1} (ϕ (γ)$ $- λ H (h (t) f (u (t))))] = 0$ and $γ$ is continuous.

Proof For any given $λ > 0$ , define the function $g : (- 1,1) \to R$ as follows:

$\begin{array}{l} g (γ) = ϕ^{- 1} [ϕ (γ) - λ Q (h (t) f (u (t)))] \\ + \sqrt[]{λ} Q [ϕ^{- 1} (ϕ (γ) - λ H (h (t) f (u (t))))] \end{array}$

There are $γ_{1}, γ_{2} \in (- 1,1)$ such that

$\begin{array}{l} g (γ_{1}) = \underset{γ \in (0,1)}{m i n} g (γ) = - 1 - \sqrt[]{λ} < 0, \\ g (γ_{2}) = \underset{γ \in (0,1)}{m i n} g (γ) = 1 + \sqrt[]{λ} > 0, \end{array}$

then there exists $γ \in (γ_{1}, γ_{2})$ such that $g (γ) = 0$ .

Next we prove uniqueness of $γ$ . Let $γ_{*}, γ^{*} \in (- 1,1)$ such that $g (γ_{*}) = g (γ^{*}) = 0$ , hence there exists $t_{0} \in (0,1)$ such that

$\begin{array}{l} ϕ^{- 1} [ϕ (γ_{*}) - λ Q (h (t_{0}) f (u (t_{0})))] \\ + \sqrt[]{λ} Q [ϕ^{- 1} (ϕ (γ_{*}) - λ H (h (t_{0}) f (u (t_{0}))))] \\ = ϕ^{- 1} [ϕ (γ^{*}) - λ Q (h (t_{0}) f (u (t_{0})))] \\ + \sqrt[]{λ} Q [ϕ^{- 1} (ϕ (γ^{*}) - λ H (h (t_{0}) f (u (t_{0}))))] . \end{array}$

Since $ϕ^{- 1}$ and $Q$ is bijection, $γ_{*} = γ^{*}$ . Finally, from the continuity of $h$ , $ϕ$ and $ϕ^{- 1}$ , we can conclude that $γ$ is continuous.

Lemma 2 $u \in C_{0}^{1} [0,1]$ being a solution to problem (1) is equivalent to $u \in C_{0}^{1} [0,1]$ being a fixed point of

$A_{f} (u) : = H ϕ^{- 1} [ϕ (P u) - λ H (h f (u))]$

Proof It follows from Lemma 1 that there exists a unique $u^{'} (0)$ such that (7) holds, so the operator equation of problem (1) is $A_{f} (u) : = H ϕ^{- 1} [ϕ (P u) - λ H (h f (u))]$ . It is obvious that $ϕ : (- 1,1) \to R$ , hence its inverse mapping $ϕ^{- 1}$ is a bounded operator, and it follows from (5) that ${‖ u ‖}_{\infty} < 1$ .

Next, we introduce definitions of a (strict) subsolution and a (strict) supersolution of problem (1) and establish the upper and lower solution theorem that is used to prove existence and multiplicity results of (1).

By a supersolution of the problem (1) we define $α \in C^{2} (0,1) ⋂ C^{1} [0,1]$ that satisfies

${\begin{array}{l} - {(\frac{α^{'} (t)}{\sqrt[]{1 - (α^{'} {(t))}^{2}}})}^{'} \leq λ h (t) f (α (t)), t \in (0,1), \\ α (0) \leq 0, α^{'} (1) + \sqrt[]{λ} α (1) \leq 0 . \end{array}$ (8)

By a supersolution of the problem (1), we define $β \in$ $C^{2} (0,1) ⋂ C^{1} [0,1]$ that satisfies

${\begin{array}{l} - {(\frac{β^{'} (t)}{\sqrt[]{1 - (β^{'} {(t))}^{2}}})}^{'} \geq λ h (t) f (β (t)), t \in (0,1), \\ β (0) \geq 0, β^{'} (1) + \sqrt[]{λ} β (1) \geq 0 . \end{array}$ (9)

By a strict subsolution (supersolution) of (1) we mean a subsolution (supersolution) which is not a solution of (1).

Lemma 3 Let $M > 0$ and $v_{1}, v_{2} \in C^{1} [0,1]$ such that $ϕ (v_{i}^{'}) \in C^{1} [0,1] (i = 1,2)$ . If

$- (ϕ (v_{1}^{'} {(t)))}^{'} + M v_{1} (t) \leq - (ϕ (v_{2}^{'} {(t)))}^{'} + M v_{2} (t), \forall t \in (0,1),$ (10)

and $v_{1} (0) = 0$ , $v_{2} (0) = 0$ , then $v_{1} (t) \leq v_{2} (t)$ .

Proof Suppose on the contrary that there exists $t_{*} \in (0,1)$ such that $\underset{t \in (0,1)}{m a x} (v_{1} (t) - v_{2} (t)) = v_{1} (t_{*}) - v_{2} (t_{*}) > 0$ , then $v_{1}^{'} (t_{*}) = v_{2}^{'} (t_{*})$ , and there exists a sequence ${t_{k}} \to t_{*}$ on $(0, t_{*})$ such that $v_{1}^{'} (t_{k}) - v_{2}^{'} (t_{k}) \geq 0$ . The fact that $ϕ$ is monotone increasing homeomorphism implies that $ϕ (v_{1}^{'} (t_{k})) - ϕ (v_{1}^{'} (t_{*})) \geq ϕ (v_{2}^{'} (t_{k})) - ϕ (v_{2}^{'} (t_{*}))$ .

By the definition of the derivative we get that

$(ϕ (v_{1}^{'} (t_{k} {)))}_{t = t_{*}}^{'} \leq (ϕ (v_{1}^{'} (t_{*} {)))}_{t = t_{*}}^{'} .$ (11)

This together with (10) and (11) concludes that $0 < M (v_{1} (t_{*}) - v_{2} (t_{*})) \leq (ϕ (v_{1}^{'} {(t)))}_{t = t_{*}}^{'} - (ϕ (v_{2}^{'} {(t)))}_{t = t_{*}}^{'} \leq 0$ , which is contradictory. Therefore, $v_{1} (t) \leq v_{2} (t)$ for all $t \in (0,1)$ .

Corollary 1 Let $v_{1}, v_{2} \in C^{1} [0,1]$ such that $ϕ (v_{i}) \in$ $C^{1} [0,1] (i = 1,2)$ .If $- (ϕ (v_{1}^{'} {(t)))}^{'} < - (ϕ (v_{2}^{'} {(t)))}^{'}$ for any $t \in (0,1)$ , then $v_{1} (t) < v_{2} (t)$ .

Lemma 4 Let $α$ and $β$ be a subsolution and a supersolution of (1), respectively, such that $α \leq β$ , then (1) has a solution $u \in C^{2} (0,1) ⋂ C^{1} [0,1]$ such that $u \in [α, β]$ .

Proof Let $χ : R \to R$ be the continuous function which is defined by

$χ (u) = {\begin{array}{l} α (t), u (t) < α (t), \\ u (t), α (t) \leq u (t) \leq β (t), \\ β (t), u (t) > β (t), \end{array}$

and define $F : [0,1] \times R \to R$ by $F (t, u) = - λ h (t) f (χ (u))$ . Obviously, $F$ is continuous and bounded.

Next we consider the auxiliary problem

${\begin{array}{l} (ϕ (u^{'} {(t)))}^{'} = F (t, u (t)) + u (t) - χ (u (t)), t \in (0,1), \\ u (0) = 0, u^{'} (1) + \sqrt[]{λ} u (1) = 0 . \end{array}$ (12)

First, the problem (12) has at least one solution $u$ by the Schauder fixed theorem. Next, we only show that $α (t) \leq u (t) \leq β (t)$ , $t \in (0,1)$ , so $u$ is a solution of (1).

Suppose on the contrary that there exists $\bar{t} \in (0,1)$ such that $\underset{t \in (0,1)}{m a x} [α (t) - u (t)] = α (\bar{t}) - u (\bar{t})$ >0. Because $α^{'} (\bar{t}) - u^{'} (\bar{t}) = 0$ , there exist two sequence ${t_{k}} \in [\bar{t} - ε, \bar{t})$ and ${{\tilde{t}}_{k}} \in [\bar{t}, \bar{t} + ε)$ such that $α^{'} (t_{k}) - u^{'} (t_{k}) \geq 0$ and $α^{'} ({\tilde{t}}_{k}) - u^{'} ({\tilde{t}}_{k}) \leq 0$ . Without loss of generality, it follows from $α^{'} (t_{k}) - u^{'} (t_{k}) \geq 0$ that $α^{'} (t_{k}) \geq u^{'} (t_{k})$ . By $ϕ$ is an increasing homeomorphism, we get that $ϕ (α^{'} (t_{k})) \geq ϕ (u^{'} (t_{k}))$ .

$S i n c e ϕ (α^{'} (\bar{t})) = ϕ (u^{'} (\bar{t})), t h e n$

$\frac{ϕ (α^{'} (t_{k})) - ϕ (α^{'} (t))}{t_{k} - \bar{t}} \leq \frac{ϕ (u^{'} (t_{k})) - ϕ (u^{'} (\bar{t}))}{t_{k} - \bar{t}} (t_{k} < \bar{t}), w h i c h i m p l i e s (ϕ (α^{'} {(\bar{t})))}^{'} \leq (ϕ (u^{'} {(\bar{t})))}^{'} .$

Moreover, $α$ is a subsolution of (1), it yields that

$\begin{array}{l} (ϕ (α^{'} {(\bar{t})))}^{'} \leq (ϕ (u^{'} {(\bar{t})))}^{'} = F (\bar{t}, u (\bar{t})) + u (\bar{t}) + α (\bar{t}) \\ < F (\bar{t}, u (\bar{t})) = - λ h (t) f (χ (u (\bar{t}))) = - λ h (t) f (α (\bar{t})) \leq (ϕ (α^{'} {(\bar{t})))}^{'} . \end{array}$

This is a contradiction, thus $α (t) \leq u (t)$ . In addition, by the similar arguments, it follows that $u (t) \leq β (t)$ . Therefore $u \in [α, β]$ is a solution of problem (1).

Lemma 5 Let $α_{1}$ and $β_{1}$ be a subsolution and supersolution of (1) respectively such that $α_{1} \leq β_{1}$ . Let $α_{2}$ and $β_{2}$ be a strict subsoltion and a strict supersolution of (1) respectively satisfying $α_{2}, β_{2} \in [α_{1}, β_{1}]$ and $α_{2} \leq β_{2}$ . Then (1) has at least three solutions $u_{1}$ , $u_{2}$ and $u_{3}$ , where $u_{1} \in [α_{1}, β_{2}]$ , $u_{2} \in [α_{2}, β_{1}]$ and $u_{3} \in [α_{1}, β_{1}] \ ([α_{1}, β_{2}] ⋃ [α_{2}, β_{1}])$ .

Proof Let $e \in C^{2} [0,1]$ denote the the unique positive solution of

${\begin{array}{l} - {(\frac{e^{'} (t)}{\sqrt[]{1 - (e^{'} {(t))}^{2}}})}^{'} = 1, t \in (0,1), \\ e (0) = e (1) = 0 . \end{array}$

Let $I = [α_{1}, β_{1}]$ , $I_{1} = [α_{1}, β_{2}]$ , $I_{3} = [α_{2}, β_{1}]$ , $C_{e} [0,1] = {u \in C [0,1] | - t e \leq u \leq t e}$ . It is not difficult to verify that $C_{e} [0,1]$ endowed with the norm ${‖ u ‖}_{e} = i n f {t > 0 | - t e \leq u \leq t e}$ is Banach space, then $I$ , $I_{1}$ and $I_{2}$ are non-empty closed convex subsets of Banach space $C_{e} [0,1]$ . In the following, we will prove that $A_{f} (u) : C_{0}^{1} [0,1] \to C_{0}^{1} [0,1]$ is completely continuous and strongly increasing.

In fact, $I_{1}$ and $I_{2}$ are disjoint subsets of $I$ . The map $A_{f} (u)$ is restricted to $I$ . Since $A_{f} (u)$ is increasing and $α_{1}$ , $β_{2}$ are subsolution and supersolution of (1) respectively ( $α_{1} \leq β_{2}$ ), we have that $α_{1} \leq A_{f} (α_{1}) \leq A_{f} (β_{2}) \leq β_{2}$ , i.e. $A_{f} (u) \subset I$ . Similarly, $A_{f} (I_{k}) \subset I_{k}, k = 1,2$ . Since $β_{2}$ is a strict supersolution of (2), we have that $A_{f} (β_{2}) < β_{2}$ . By Ref. [21], Corollary 6.2, $A_{f} (u)$ has a maximal fixed point in $u_{1} \in I_{1}$ and $α_{1} \leq u_{1} < β_{2}$ . Similarly, $A_{f} (u)$ has a minimal fixed point in $u_{2} \in I_{2}$ and $α_{2} < u_{2} \leq β_{1}$ , because $α_{2}$ is a strict subsoluton of (1).

Notice that

$- (ϕ (u_{1}^{'} {(t)))}^{'} + M u_{1} (t) \leq - (ϕ (β_{2}^{'} {(t)))}^{'} + M β_{2} (t) (M > 0)$

it follows from Lemma 3 that $u_{1} \leq β_{2}$ , hence there exists a constant $c_{1} > 0$ such that $β_{2} - u_{1} \geq c_{1} e (t)$ . Similarly there exists a constant $c_{2} > 0$ such that $u_{2} - α_{2} \geq c_{2} e (t)$ . Define

$J_{k} = I ⋂ {z \in C_{e} [0,1] | {‖ z - u_{k} ‖}_{e} < t_{k}, k = 1,2}$

then $J_{k} \subset I_{k}$ is open set, $k = 1,2$ . Hence $I_{k}$ has non-empty interior. Let $J_{k}$ be the largest open set in $I_{k}$ , which contain $u_{k}$ such that $A_{f} (u)$ has no fixed point in $I_{k} \ J_{k}$ . Finally by Ref. [22], Lemma 3.8, $A_{f} (u)$ has a third fixed point $u_{3} \in I \ (I_{1} ⋃ I_{2})$ . Therefore, problem (1) has at least three solutions $u_{1} \in [α_{1}, β_{2}]$ , $u_{2} \in [α_{2}, β_{1}]$ and $u_{3} \in [α_{1}, β_{1}] \ ([α_{1}, β_{2}] ⋃ [α_{2}, β_{1}])$ .

2 The Autonomous Case of $f (u) = u$ , $h (t) = 1$

We consider the quasilinear problem

${\begin{array}{l} - {(\frac{u^{'} (t)}{\sqrt[]{1 - (u^{'} {(t))}^{2}}})}^{'} = λ u (t), t \in (0,1), \\ u (0) = 0, u^{'} (1) + \sqrt[]{λ} u (1) = 0, \end{array}$ (13)

where $λ > 0$ is a parameter.

Multiplying the differential equation in (13) by $u^{'} (t)$ and integrating from $0$ to $t$ , we get that $- (1 - (u^{'} {(t))}^{2})^{- \frac{1}{2}} = λ F (u (t)) + C$ , where $F (s) = \int_{0}^{s} z d z$ . From Ref. [8], Lemma 2.3, there exists $t_{0} \in (0,1)$ such that $u (t_{0}) = ρ = {‖ u ‖}_{\infty}$ . Now we choose $t = t_{0}$ in (13) and yield

$u^{'} (t) = \sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}, t \in (0, t_{0}),$ (14)

$u^{'} (t) = - \sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}, t \in (t_{0}, 1) .$ (15)

Let $u (1) = m > 0$ , further integration from $0$ to $t_{0}$ for (14) and integration from $t_{0}$ to $1$ for (15), it follows that

$\int_{0}^{ρ} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}} = t_{0},$ (16)

$\int_{ρ}^{m} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}} = t_{0} - 1 .$ (17)

We obtain by combining (16) and (17) that

$\int_{0}^{ρ} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}} + \int_{m}^{ρ} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}} = 1 .$ (18)

Finally, from boundary value $u^{'} (1) + \sqrt[]{λ} u (1) = 0$ and (15), we can get

$G_{λ} (m) : = (1 + λ (F (ρ) - F {(m)))}^{- 2} + λ m^{2} - 1 = 0 .$ (19)

Let $λ$ be a fixed value and $ρ \in (0,1)$ . By (19), it is easy to compute that

$\frac{\partial G_{λ}}{\partial m} = 2 λ m (1 + \frac{1}{{(1 + \frac{λ}{2} (ρ^{2} - m^{2}))}^{3}}) > 0, m \in (0, ρ),$ (20)

further $G_{λ} (0) = {(1 + \frac{λ}{2} ρ^{2})}^{- 2} - 1 < 0$ , $G_{λ} (ρ) = λ ρ^{2} > 0$ , so it follows from the intermediate value theorem that there exists $\tilde{m} = \tilde{m} (ρ) \in (0, ρ)$ such that (19) holds.

Lemma 6 For any fixed $ρ \in (0,1)$ , $\underset{ρ \to 0}{l i m} \frac{\tilde{m}}{ρ} = \frac{\sqrt[]{2}}{2}$ for $\tilde{m} \in (0, ρ)$ .

Proof By (19), we get that $G_{λ} (\tilde{m}) = 0$ . It is easy to compute that

$\frac{\partial G_{λ} (\tilde{m})}{\partial ρ} = \frac{- 2 λ ρ}{{(1 + \frac{λ}{2} (ρ^{2} - {\tilde{m}}^{2}))}^{3}}, \frac{\partial G_{λ} (\tilde{m})}{\partial \tilde{m}} = 2 λ \tilde{m} [1 + (1 + \frac{λ}{2} (ρ^{2} - {\tilde{m}}^{2} {))}^{- 3}] .$

Thus $\underset{ρ \to 0}{l i m} {(\frac{\tilde{m}}{ρ})}^{2} = \underset{ρ \to 0}{l i m} \tilde{m} (ρ) = \underset{ρ \to 0}{l i m} \frac{1}{[1 + (1 + \frac{λ}{2} (ρ^{2} - {\tilde{m}}^{2} {))]}^{- 3}} = \frac{1}{2}$ , i.e. $\underset{ρ \to 0}{l i m} \frac{\tilde{m}}{ρ} = \frac{\sqrt[]{2}}{2}$ .

Definition 1 For any given $ρ \in (0,1)$ , we define a map

$T_{λ, m} (ρ) = \int_{0}^{ρ} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}} + \int_{m}^{ρ} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}},$ (21)

called the time map of problem (13).

In fact, the existence of a solution to problem (13) is equivalent to the existence of a solution to $T_{λ, m} (ρ) = 1$ , see Refs. [19, 23].

Theorem 4 For any given $ρ \in (0,1)$ , there exists $λ_{1} = λ_{1} (ρ, \tilde{m})$ satisfying (18) and (19) such that problem (13) has at least one positive solution for any $λ > λ_{1}$ .

Proof Clearly, for any fixed $λ > 0$ , there exists $\tilde{m} = \tilde{m} (ρ)$ such that $G_{λ} (\tilde{m}) = 0$ . Now, let $m = \tilde{m}$ in (21), we show that

$T_{λ, \tilde{m}} (ρ) = \int_{0}^{ρ} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}} + \int_{\tilde{m}}^{ρ} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}} = 1 .$ (22)

It is easy to compute that $\frac{\partial T_{λ, \tilde{m}} (ρ)}{\partial ρ} = - \tilde{m^{'}} (ρ) \frac{1}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(\tilde{m})))}^{- 2}}} <$ $0, \tilde{m} \in (0, ρ)$ , and $T_{λ, \tilde{m}} (0) = 0 < 1,$

$T_{λ, \tilde{m}} (1) = \int_{0}^{1} \frac{d u}{\sqrt[]{1 - (1 + λ (F (1) - F {(u)))}^{- 2}}} + \int_{\tilde{m} (1)}^{1} \frac{d u}{\sqrt[]{1 - (1 + λ (F (1) - F {(u)))}^{- 2}}} > 1 .$

Therefore, there exist $ρ \in (0,1)$ and $\tilde{m} = \tilde{m} (ρ)$ , such that $T_{λ, \tilde{m}} (ρ) = 1$ holds. Then problem (13) has at least one positive solution.

Finally, we discuss the range of value of $λ$ when a positive solution exists for problem (13). The fact together with (18) concludes that

$\begin{array}{l} 1 = \underset{ρ \to 0^{+}}{l i m} (\int_{0}^{ρ} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}} + \int_{\tilde{m}}^{ρ} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}}) \\ = \underset{ρ \to 0^{+}}{l i m} (\int_{0}^{ρ} \frac{1 + λ (F (ρ) - F (u))}{\sqrt[]{(1 + λ (F (ρ) - F {(u)))}^{2} - 1}} d u + \int_{\tilde{m}}^{ρ} \frac{1 + λ (F (ρ) - F (u))}{\sqrt[]{(1 + λ (F (ρ) - F {(u)))}^{2} - 1}} d u) \\ = \underset{ρ \to 0^{+}}{l i m} (\int_{0}^{ρ} \frac{1 + \frac{λ}{2} (ρ^{2} - u^{2})}{\sqrt[]{\frac{λ^{2}}{4} (ρ^{2} - u^{2})^{2} + λ (ρ^{2} - u^{2})}} d u + \int_{\tilde{m}}^{ρ} \frac{1 + \frac{λ}{2} (ρ^{2} - u^{2})}{\sqrt[]{\frac{λ^{2}}{4} (ρ^{2} - u^{2})^{2} + λ (ρ^{2} - u^{2})}} d u) \end{array}$

$\begin{array}{l} = \underset{ρ \to 0^{+}}{l i m} (\int_{0}^{1} \frac{(1 + \frac{λ}{2} (ρ^{2} - {(ρ τ)}^{2})) ρ}{\sqrt[]{\frac{λ^{2}}{4} (ρ^{2} - {(ρ τ)}^{2})^{2} + λ (ρ^{2} - {(ρ τ)}^{2})}} d τ + \int_{\frac{\tilde{m}}{ρ}}^{1} \frac{(1 + \frac{λ}{2} (ρ^{2} - {(ρ τ)}^{2})) ρ}{\sqrt[]{\frac{λ^{2}}{4} (ρ^{2} - {(ρ τ)}^{2})^{2} + λ (ρ^{2} - {(ρ τ)}^{2})}} d τ) \\ = \underset{ρ \to 0^{+}}{l i m} (\int_{0}^{1} \frac{1 + \frac{λ}{2} (ρ^{2} - {(ρ τ)}^{2})}{\sqrt[]{\frac{λ^{2}}{4} (ρ - ρ τ^{2})^{2} + λ (1 - τ^{2})}} d τ \int_{\frac{\tilde{m}}{ρ}}^{1} \frac{1 + \frac{λ}{2} (ρ^{2} - {(ρ τ)}^{2})}{\sqrt[]{\frac{λ^{2}}{4} (ρ - ρ τ^{2})^{2} + λ (1 - τ^{2})}} d τ) \\ = \frac{1}{\sqrt[]{λ}} (\int_{0}^{1} \frac{1}{\sqrt[]{1 - τ^{2}}} d τ + \int_{\frac{\sqrt[]{2}}{2}}^{1} \frac{1}{\sqrt[]{1 - τ^{2}}} d τ) = \frac{1}{\sqrt[]{λ}} \frac{3 π}{4} . \end{array}$

Therefore, $\underset{ρ \to 0^{+}}{l i m} λ (ρ, \tilde{m}) = \frac{9}{16} π^{2} : = λ_{1}$ .

Next, we show that $\underset{ρ \to 1^{-}}{l i m} λ (ρ, \tilde{m}) = \infty$ . From the definition of $ϕ (u)$ , we get that $\underset{ρ \to 1^{-}}{l i m} \frac{u}{ϕ (u)} = 0$ . Hence, there exists $δ \in (0,1)$ for any $ε > 0$ such that $u \leq ε ϕ (u)$ , $\forall 1 - δ < u < 1$ and it follows that

$F (ρ) - F (u) = \int_{u}^{ρ} s d s < ε \int_{u}^{ρ} \frac{s}{\sqrt[]{1 - s^{2}}} d s = \frac{ε (ρ^{2} - u^{2})}{\sqrt[]{1 - u^{2}} + \sqrt[]{1 - ρ^{2}}} .$

This together with (18) implies that

$1 = \int_{0}^{ρ} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}} + \int_{\tilde{m}}^{ρ} \frac{d u}{\sqrt[]{1 - (1 + λ (F (ρ) - F {(u)))}^{- 2}}}$

$\begin{array}{l} \geq \int_{0}^{ρ} \frac{d u}{\sqrt[]{1 - [1 + λ (F (ρ) - F {(u))]}^{- 2}}} \\ \geq \int_{0}^{ρ} \frac{d u}{\sqrt[]{1 - [1 + λ ε (ρ^{2} - u^{2}) / {(\sqrt[]{1 - u^{2}} + \sqrt[]{1 - ρ^{2}})]}^{- 2}}} \\ = \int_{0}^{1} \frac{1 + λ ε ρ^{2} (1 - τ^{2}) / (\sqrt[]{1 - {(ρ τ)}^{2}} + \sqrt[]{1 - ρ^{2}})}{\sqrt[]{λ ε [\frac{2 (1 - τ^{2})}{\sqrt[]{1 - {(ρ τ)}^{2}} + \sqrt[]{1 - ρ^{2}}} + \frac{λ ε ρ^{2} (1 - τ^{2})^{2}}{{(\sqrt[]{1 - {(ρ τ)}^{2}} + \sqrt[]{1 - ρ^{2}})}^{2}}]}} d τ . \end{array}$

Therefore, we yield that for $ε > 0$ small enough as $ρ \to 1^{-}$ , $\underset{ρ \to 1^{-}}{l i m} \sqrt[]{λ (ρ, \tilde{m})} \geq \underset{ρ \to 1^{-}}{l i m} \frac{1}{\sqrt[]{ε}} \int_{0}^{1} \frac{1}{\sqrt[]{1 - τ^{2}}} d τ = \infty$ , i.e. $\underset{ρ \to 1^{-}}{l i m} λ (ρ, \tilde{m}) = \infty$ .

In consequence, the problem (13) has at least one positive solution for any $λ > λ_{1}$ .

Remark 1 In the case of $h (t) = 1$ , $f (u) = u$ , the principal eigenvalue of problem

${\begin{array}{l} - u^{″} (t) = λ u (t), t \in (0,1), \\ u (0) = 0, u^{'} (1) + \sqrt[]{λ} u (1) = 0 \end{array}$

has eigenvalue $λ_{1} = \frac{9}{16} π^{2}$ , and the corresponding eigenfunction is $B_{1} (λ) = s i n \frac{3 π}{4} t$ , $t \in (0,1)$ .

3 The Proof of Main Results

Proof of Theorem 1 We first show the nonexistence for $λ \approx 0$ . Let $σ_{λ}$ be the principal eigenvalue and $ω_{λ} > 0$ be the corresponding normalized eigenfunction of

${\begin{array}{l} - ω_{λ}^{″} (t) = (λ + σ) h (t) ω_{λ} (t), t \in (0,1), \\ ω_{λ} (0) = 0, ω_{λ}^{'} (1) + \sqrt[]{λ} ω_{λ} (1) = 0 . \end{array}$

We note that $σ_{λ} > 0$ for $λ < λ_{1}$ and $σ_{λ} < 0$ for $λ > λ_{1}$ , the detail see Ref. [24]. Supppose on the contrary that $u_{λ}$ be a positive solution of (1) for $λ \approx 0$ . Note that there exists $k_{0} > 0$ such that $f (s) \leq k_{0} s$ for $s \in [0, \infty)$ , and

$\begin{array}{l} 0 = \int_{0}^{1} (u_{λ}^{″} ω_{λ} - ω_{λ}^{″} u_{λ}) d s = \int_{0}^{1} h (s) [u_{λ} ω_{λ} (λ + σ_{λ}) - λ f (u_{λ}) ω_{λ} (1 - (u_{λ}^{'})^{2})^{\frac{3}{2}}] d s \\ \geq \int_{0}^{1} h (s) [(λ + σ_{λ}) - λ k_{0} (1 - (u_{λ}^{'})^{2})^{\frac{3}{2}}] u_{λ} ω_{λ} d s \geq \int_{0}^{1} h (s) [(λ + σ_{λ}) - λ k_{0}] u_{λ} ω_{λ} d s . \end{array}$ (23)

Clearly, it is easy to see that $\sqrt[]{λ} > λ$ for $λ \approx 0$ . Thus there exists constant $m > 0$ (independent of $λ$ ) such that $0 < m \sqrt[]{λ} - λ < σ_{λ}$ . Besides, by (23), it follows that $(λ + σ_{λ}) - λ k_{0} \leq 0$ , i.e. $\frac{σ_{λ}}{λ} \to \infty$ . This is a contradiction, hence (1) has no positive solution for $λ \approx 0$ .

Next, we show that the existence of positive solution of (1) for $λ > λ_{1}$ . Let $β_{1} = m_{λ} e_{h}$ , where $e_{h}$ is solution of

${\begin{array}{l} - {(\frac{u^{'} (t)}{\sqrt[]{1 - (u^{'} {(t))}^{2}}})}^{'} = h (t), t \in (0,1), \\ u (0) = 0, u^{'} (1) + \sqrt[]{λ} u (1) = 0 \end{array}$

and $1 < m_{λ} < \frac{1}{m i n {{‖ e_{h} ‖}_{\infty}, {‖ e_{h}^{'} ‖}_{\infty}}}$ . Then $1 - m_{λ} {‖ e_{h} ‖}_{\infty} > 0$ and $1 - m_{λ}^{2} e^{'}_{h}^{2} > 0$ .

Since $\underset{u \to 1^{-}}{l i m} \frac{f (u)}{1 - u} = 0$ , for all $ε > 0$ , there exists $0 < δ < 1$ such that $| \frac{f (u)}{1 - u} | < ε$ , $1 - δ < u < 1$ . Furthermore, choosing $ε = \frac{m_{λ}}{λ (1 - m_{λ} {‖ e_{h} ‖}_{\infty})}$ , we have that

$\frac{f (m_{λ} {‖ e_{h} ‖}_{\infty})}{1 - m_{λ} {‖ e_{h} ‖}_{\infty}} < \frac{m_{λ}}{λ (1 - m_{λ} {‖ e_{h} ‖}_{\infty})}, i . e . λ f (m_{λ} {‖ e_{h} ‖}_{\infty}) < m_{λ} .$ (24)

Next, we prove that $β_{1}$ is an upper solution to problem (1). By $m_{λ} > 1$ and (24), we conclude that

$- {(\frac{β_{1}^{'}}{\sqrt[]{1 - {β_{1}^{'}}^{2}}})}^{'} = - \frac{β_{1}^{″}}{(1 - {β_{1}^{'}}^{2})^{\frac{3}{2}}} = - \frac{m_{λ} e_{h}^{″}}{(1 - m_{λ}^{2} {e_{h}^{'}}^{2})^{\frac{3}{2}}} > - \frac{m_{λ} e_{h}^{″}}{(1 - {e_{h}^{'}}^{2})^{\frac{3}{2}}} = m_{λ} h (t) \geq λ f (m_{λ} e_{h}) h (t) = λ f (β_{1}) h (t) .$

In addition, we get that $β_{1} (0) = 0$ and ${β_{1}}^{'} (1) + \sqrt[]{λ} β_{1} (1) = 0$ , thus $β_{1}$ is a supersolution.

Let $α_{1} = ε v$ with $ε > 0$ , where $v$ is positive solution of (13) according to Theorem 4. Since $f^{'} (0) = 1$ , $\underset{u \to 0^{+}}{l i m} \frac{f (u)}{u} = 1$ . Therefore, for any $ε > 0$ , there exists $δ > 0$ such that $1 - ε \leq \frac{f (u)}{u} \leq 1 + ε$ . Hence, for $ε \approx 0$ , we have that

$- {(\frac{α_{1}^{'}}{\sqrt[]{1 - {α_{1}^{'}}^{2}}})}^{'} = - {(\frac{ε v^{'}}{\sqrt[]{1 - ε^{2} {v^{'}}^{2}}})}^{'} = \frac{λ ε (1 - {v^{'}}^{2})^{\frac{3}{2}} v}{(1 - ε^{2} {v^{'}}^{2})^{\frac{3}{2}}} \leq λ ε v h (t) \leq λ f (ε v) h (t) .$

Obviously $α_{1} (0) = 0$ and ${α_{1}}^{'} (1) + \sqrt[]{λ} α_{1} (1) = 0$ , thus $α_{1}$ is a subsolution of (2). Now choosing $ε \approx 0$ to ensure that $β_{1} \geq α_{1}$ . By Lemma 1 there exists a positive solution $u_{λ} \in [α_{1}, β_{1}]$ for $λ > λ_{1}$ .

Finally, we establish the multiplicity result of positive solution of (1). Let $m^{*}, M^{*} \in (0, η)$ such that $f$ is strictly increasing on $[m^{*}, M^{*}]$ . We first construct a strict subsolution of the Dirichlet problem

${\begin{array}{l} - {(\frac{u^{'} (t)}{\sqrt[]{1 - (u^{'} {(t))}^{2}}})}^{'} = λ h (t) \tilde{f} (u (t)), t \in (0,1), \\ u (0) = u (1) = 0 . \end{array}$ (25)

Define $\tilde{f} \in C^{2} ([0, \infty), R)$ satisfying

$\tilde{f} (u) = {\begin{array}{l} \hat{f} (u), u < m^{*}, \\ f (u), u \geq m^{*}, \end{array}$

where $\hat{f} (u)$ is defined such that the function $\tilde{f} (u)$ is strictly increasing on $[m^{*}, M^{*}]$ and $\tilde{f} (u) \leq f (u)$ . For $t \in [0, \frac{1}{2}]$ , let

$g (t) = {\begin{array}{l} 1 - (1 - {(\frac{t}{ξ})}^{μ})^{δ}, 0 \leq t < ξ, \\ 1, ξ \leq t \leq \frac{1}{2}, \end{array}$

and $g (t) = g (1 - t)$ , $t \in [\frac{1}{2}, 1]$ , where $0 < μ, δ < 1$ . By computing, we get that

$g^{'} (t) = {\begin{array}{l} \frac{δ μ}{ξ} {(\frac{t}{ξ})}^{μ - 1} {(1 - {(\frac{t}{ξ})}^{μ})}^{δ - 1}, 0 \leq t < ξ, \\ 0, ξ \leq t \leq \frac{1}{2} . \end{array}$

Assume that there exists a constant $b \in [m^{*}, M^{*})$ such that $M^{*} > 2 b$ , and define $z (t) = b g (t)$ . Since $| g^{'} (t) | < \frac{δ μ}{ξ}$ , $| z^{'} (t) | < \frac{b δ μ}{ξ}$ . Define $α_{*}$ on $[\frac{1}{2}, 1]$ to be the solution of

${\begin{array}{l} - {(\frac{α_{*}^{'} (t)}{\sqrt[]{1 - (α_{*}^{'} {(t))}^{2}}})}^{'} = λ \tilde{f} (z (t)), t \in (0,1), \\ α_{*} (0) = α_{*}^{'} (\frac{1}{2}) = 0, \end{array}$ (26)

and extend $α_{*}$ on $[0, \frac{1}{2}]$ such that $α_{*} (t) = α_{*} (1 - t)$ , where $α_{*} (t)$ has a maximum value at $t = \frac{1}{2}$ and is symmetric about $t = \frac{1}{2}$ , see Ref. [17].

Claim $α_{*} (t) \in (z (t), M^{*})$ , $t \in (0,1)$ .

Based on the claim, it follows that

$- α_{*}^{″} = λ (1 - (α_{*}^{'})^{2})^{\frac{3}{2}} \tilde{f} (z) < λ (1 - (α_{*}^{'})^{2})^{\frac{3}{2}} \tilde{f} (α_{*}) \leq λ (1 - (α_{*}^{'})^{2})^{\frac{3}{2}} h (t) \tilde{f} (α_{*} (t))$

Since $α_{*} (0) = α_{*} (1) = 0$ , $α_{*}^{'} (1) + \sqrt[]{λ} α_{*} (1) = α_{*}^{'} (1) \leq 0$ . Therefore, $α_{*}$ is a strict subsolution. Now, we prove the claim is true. First, we show that $α_{*} (t) > z (t)$ , $t \in [0,1]$ .

Define $τ = τ (ξ) : = {\int_{ξ}^{\frac{1}{2}} (1 - (α_{*}^{'} {(s))}^{2})}^{\frac{3}{2}} d s$ , then $0 < τ (ξ) < 1$ . Recall that (26). Integration from $t$ to $\frac{1}{2}$ with $t \in (0, \frac{1}{2})$ and noting that $α_{*}^{'} (\frac{1}{2}) = 0$ , we obtain that

$α_{*}^{'} (t) = λ \int_{t}^{\frac{1}{2}} (1 - (α_{*}^{'} {(s))}^{2})^{\frac{3}{2}} \tilde{f} (z (s)) d s \geq λ \int_{ξ}^{\frac{1}{2}} (1 - (α_{*}^{'} {(s))}^{2})^{\frac{3}{2}} \tilde{f} (b g (s)) d s = λ \tilde{f} (b) \int_{ξ}^{\frac{1}{2}} (1 - (α_{*}^{'} {(s))}^{2})^{\frac{3}{2}} d s = λ f (b) τ (ξ), t \in [0, ξ] .$ (27)

Now we choose $ξ = \frac{1}{4}$ . Since $λ > \frac{4 b}{τ (\frac{1}{4}) f (b)}$ , we can choose $0 < μ, δ < 1$ such that $λ > \frac{4 b (δ μ)}{τ (\frac{1}{4}) f (b)}$ . Hence for all $t \in [0, \frac{1}{4}]$ , $α_{*}^{'} (t) > λ f (b) τ (\frac{1}{4}) > 4 b δ μ = \frac{b δ μ}{ξ} > z^{'} (t)$ . Clearly, $z (t) = b$ , $z^{'} (t) = 0$ for all $t \in [\frac{1}{4}, \frac{1}{2})$ . We obtain that

$α_{*}^{'} (t) = λ \int_{t}^{\frac{1}{2}} (1 - (α_{*}^{'} {(s))}^{2})^{\frac{3}{2}} \tilde{f} (z (s)) d s > 0 = z^{'} (t) . i . e . α_{*}^{'} (t) > z^{'} (t) f o r a l l t \in [0, \frac{1}{2}) .$

Since $α_{*} (0) = z (0) = 0$ , we have $α_{*} (t) > z (t)$ for all $t \in [0, \frac{1}{2})$ . By symmetry of the solution of (26), $α_{*} (t) > z (t)$ for all $t \in [0,1]$ .

Next, we show that ${‖ α_{*} ‖}_{\infty} < M^{*} .$ Recall that $α_{*}^{'} (t) = λ \int_{t}^{\frac{1}{2}} (1 - (α_{*}^{'} {(s))}^{2})^{\frac{3}{2}} \tilde{f} (z (s)) d s$ , $t \in (0, \frac{1}{2})$ . Integrating from $0$ to $t$ , we have

$α_{*} (t) = \int_{0}^{t} (λ \int_{s}^{\frac{1}{2}} (1 - α_{*}^{'} {(τ))}^{\frac{3}{2}} \tilde{f} (z (τ)) d τ) d s, t \in (0, \frac{1}{2}) .$

Furthermore, ${‖ α_{*} ‖}_{\infty} = α_{*} (\frac{1}{2}) \leq λ \tilde{f} (b) \int_{0}^{\frac{1}{2}} \int_{s}^{\frac{1}{2}} (1 - (α_{*}^{'} {(τ))}^{2})^{\frac{3}{2}} d τ d s = \frac{λ \tilde{f} (b)}{8} = \frac{λ f (b)}{8}$ . This together with $λ < \frac{8 M^{*}}{f (b)}$ yields ${‖ α_{*} ‖}_{\infty} < M^{*}$ . Hence $v (t) < α_{*} (t) < M^{*}$ for $t \in (0,1)$ . Moreover $α_{*}$ is a strict subsolution of problem (1).

Let $α_{2}$ be the first interation of $α_{*}$ , then $α_{2}$ be the soluton to the problem

${\begin{array}{l} - {(\frac{α_{2}^{'} (t)}{\sqrt[]{1 - (α_{2}^{'} {(t))}^{2}}})}^{'} = λ h (t) f (α_{*} (t)), t \in (0,1), \\ α_{2} (0) = 0, α_{2}^{'} (1) + \sqrt[]{λ} α_{2} (1) = 0 . \end{array}$

Then $- {(\frac{α_{2}^{'}}{\sqrt[]{1 - (α_{2}^{'})^{2}}})}^{'} = λ h (t) f (α_{*}) > λ h (t) f (z) = - {(\frac{α_{*}^{'}}{\sqrt[]{1 - (α_{*}^{'})^{2}}})}^{'}$ . By Corollary 1, we have that $α_{2} > α_{*}$ . Hence, it is not difficult to verfy that $α_{2}$ is a strict subsolution of (1).

Finally, we construct a strict supersolution for $λ \in (\frac{4 b}{τ (\frac{1}{4}) f (b)}, \frac{a}{{‖ v_{h} ‖}_{\infty} f (a)})$ , where $a \in (0, b)$ is a constant. Let $β_{2} : = \frac{a v_{h}}{{‖ v_{h} ‖}_{\infty}}$ , where $v_{h}$ is defined by (4). Then

$- β_{2}^{″} = \frac{a h (t)}{{‖ v_{h} ‖}_{\infty}} > λ h (t) f (a) > λ (1 - (β_{2}^{'})^{2})^{\frac{3}{2}} h (t) f (a) > λ (1 - (β_{2}^{'})^{2})^{\frac{3}{2}} h (t) f (β_{2}) .$

On the other hand, $β_{2}$ satisifies $β_{2} (0) = 0$ and $β_{2}^{'} (1) + \sqrt[]{λ} β_{2} (1) = = \frac{a}{{‖ v_{h} ‖}_{\infty}} [v_{h}^{'} (1) + \sqrt[]{λ} v_{h} (1)] = 0$ . Therefore $β_{2}$ is a strict supersolution for $λ \in (\frac{4 b}{τ (\frac{1}{4}) f (b)}, \frac{a}{{‖ v_{h} ‖}_{\infty} f (a)})$ . Further, we can choose $ε \approx 0$ and $0 < a ≪ 1$ such that $α_{1} \leq α_{2} \leq β_{1}$ and $α_{1} \leq β_{2} \leq β_{1}$ . Since ${‖ α_{2} ‖}_{\infty} \geq b > a = {‖ β_{2} ‖}_{\infty}$ , $α_{2} \leq β_{2}$ . Therefore (1) has at least three positive solutions from Lemma 5.

Proof of Theorem 2 By $(C_{2})$ , there exists $A^{*} > 0$ such that $f^{″} (s) \leq - A^{*}$ for $s \approx 0$ . Let ${\hat{β}}_{1} : = δ_{λ} ω_{λ}$ , where $δ_{λ} = - \frac{2 σ_{λ}}{λ A^{*} \underset{t \in (0,1)}{m i n} ω_{λ}}$ . It follows that $δ_{λ} > 0$ and $δ_{λ} \to 0 (λ \to {λ_{1}}^{+})$ from $σ_{λ} < 0$ , $σ_{λ} \to 0 (λ \to {λ_{1}}^{+})$ and $\underset{t \in (0,1]}{m i n} ω_{λ} \to 0 (λ \to λ_{1}^{+})$ . By the Taylor's series $f ({\hat{β}}_{1}) = f (0) + f^{'} (0) {\hat{β}}_{1} + \frac{f^{″} (ζ)}{2} {\hat{β}}_{1}^{2} = {\hat{β}}_{1} + \frac{f^{″} (ζ)}{2} {\hat{β}}_{1}^{2}$ , $ζ \in [0, {\hat{β}}_{1}]$ , it concludes that

$\begin{array}{l} - {\hat{β}}_{1}^{″} - λ (1 - ({\hat{β}}_{1}^{'})^{2})^{\frac{3}{2}} h (t) f ({\hat{β}}_{1}) = δ_{λ} (λ + σ_{λ}) h (t) ω_{λ} - λ (1 - (β_{1}^{'})^{2})^{\frac{3}{2}} h (t) [δ_{λ} ω_{λ} + \frac{f^{″} (ζ)}{2} (δ_{λ} ω_{λ})^{2}] \\ \geq δ_{λ} (λ + σ_{λ}) (1 - (β_{1}^{'})^{2})^{\frac{3}{2}} h (t) ω_{λ} - λ (1 - (β_{1}^{'})^{2})^{\frac{3}{2}} h (t) [δ_{λ} ω_{λ} + \frac{f^{″} (ζ)}{2} (δ_{λ} ω_{λ})^{2}] \\ \geq δ_{λ} ω_{λ} (1 - (β_{1}^{'})^{2})^{\frac{3}{2}} h (t) [σ_{λ} + \frac{λ A^{*}}{2} δ_{λ} \underset{t \in (0,1]}{m i n} ω_{λ}] = 0 . \end{array}$

Therefore, ${\hat{β}}_{1}$ is a supersolution of (1) for $λ > λ_{1}$ and ${‖ {\hat{β}}_{1} ‖}_{\infty} \to 0 (λ \to λ_{1}^{+})$ . Let $α_{1} = ε v$ be as in the proof of Theorem 1. Choosing $ε \approx 0$ , we can note that $α_{1} \leq β_{1}$ . By Lemma 2, there exists a positive solution $u_{λ} \in [α_{1}, {\hat{β}}_{1}]$ for $λ > λ_{1}$ and $λ \approx λ_{1}$ such that ${‖ u_{λ} ‖}_{\infty} \to 0 (λ \to λ_{1}^{+})$ .

Proof of Theorem 3 First of all, we show the existence of positive solution for $λ > 0$ . Clearly, $α \equiv 0$ is a strict subsoluion of (1). Let $β =$ $m_{λ} e_{h}$ be as in the proof of Theorem 1, then is a supersolution of (1). Further, by Lemma 1, there exists $u_{λ} \in [α, β]$ . Next, the proof of multiplicity is similar to that of Theorem 1, we omit it.

4 Examples

Example 1 Let us consider the following problem

${\begin{array}{l} - {(\frac{u^{'} (t)}{\sqrt[]{1 - (u^{'} {(t))}^{2}}})}^{'} = λ 3 t^{- \frac{1}{2}} f (u (t)), t \in (0,1), \\ u (0) = 0, u^{'} (1) + \sqrt[]{λ} u (1) = 0, \end{array}$ (28)

where

$f (s) = {\begin{array}{l} (1 - s) c o s ((s - \frac{1}{2}) π), s \in [0, \frac{7}{20}), \\ (1 - s) s i n ((s - 1) π), s \in [\frac{7}{20}, 1] . \end{array}$

It is obvious that $f (0) = 0$ . Moreover, we note that $f (s)$ satisfies $(C_{1})$ , $(C_{2})$ and $\underset{s \to 1^{-}}{l i m} \frac{f (s)}{1 - s} = 0$ . Let $s = s_{0} = \frac{1}{10}$ , then $\frac{s_{0}}{f (s_{0})} \approx \frac{10}{27}$ . Let $a = s_{0}$ and $ι > 2 s_{0}$ , then we can choose $b_{ι} \in (s_{0}, \frac{ι}{2})$ such that $\frac{4 b}{ι (\frac{1}{4}) f (b_{ι})} < \frac{1}{6}$ . Let $v_{b_{ι}}$ be as in (4). It is easy to compute that $v_{b_{ι}} (t) = \frac{6 + 4 \sqrt[]{λ}}{3 (1 + \sqrt[]{λ})} - \frac{4}{3} t^{\frac{3}{2}}$ , $t \in (0,1)$ . Then we obtain that ${‖ v_{b_{ι}} ‖}_{\infty} \leq 2$ . Hence $(\frac{4 b}{τ (\frac{1}{4}) f (b_{ι})}) / (\frac{a}{{‖ v_{b} ‖}_{\infty} f (a)})$ <1. Next, let $M_{ι} = \frac{ι}{2}$ . Then $\frac{4 b_{ι}}{τ (\frac{1}{4}) f (b_{ι})} < m i n {\frac{a}{{‖ v_{b_{ι}} ‖}_{\infty} f (a)}, \frac{8 M_{ι}}{f (b_{ι})}}$ . Therefore, problem (28) has at least three positive solutions for $λ \in$ $(\frac{4 b_{ι}}{τ (\frac{1}{4}) f (b_{ι})}, m i n {\frac{a}{{‖ v_{b_{ι}} ‖}_{\infty} f (b_{ι})}, \frac{8 M^{*}}{f (b_{ι})}})$ by Theorem 1.

Example 2 Let us consider the following problem

${\begin{array}{l} - {(\frac{u^{'} (t)}{\sqrt[]{1 - (u^{'} {(t))}^{2}}})}^{'} = λ 3 t^{- \frac{1}{2}} f (u (t)), t \in (0,1), \\ u (0) = 0, u^{'} (1) + \sqrt[]{λ} u (1) = 0, \end{array}$ (29)

where

$f (s) = {\begin{array}{l} (\frac{1}{200} + s) (1 - s) l n (2 - s), s \in [0, \frac{7}{20}), \\ (\frac{1}{200} + s) (1 - s) l n (e^{1 - s} - 1), s \in [\frac{7}{20}, 1) . \end{array}$

It is obvious that $f (0) > 0$ . Moreover, we note that $f (s)$ satisfies $(C_{1})$ , $(C_{2})$ and $\underset{s \to 1^{-}}{l i m} \frac{f (s)}{1 - s} = 0$ . The following is similar to Example 1, so we omit it.

References

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Marinets V V, Pitovka O Y. On a problem with a parameter in the boundary conditions[J]. Nauk Visn Uzhgorod Univ Ser Mat Inform, 2005 (10-11): 70-76. [Google Scholar]
Li Z L. Second-order singular nonlinear boundary value problem with parameters in the boundary conditions[J]. Adv Math, 2008, 37(3): 353-360(Ch). [Google Scholar]
Aslanova N, Bayramoglu M, Aslanov K. On one class eigenvalue problem with eigenvalue parameter in the boundary condition at one end-point[J]. Filomat, 2018, 32(19): 6667-6674. [Google Scholar]
Fonseka N, Muthunayake A, Shivaji R, et al. Singular reaction diffusion equations where a parameter influences the reaction term and the boundary conditions[J]. Topological Methods in Nonlinear Analysis, 2021, 57(1): 221-242. [Google Scholar]
Acharya A, Fonseka N, Quiroa J, et al. σ-shaped bifurcation curves[J]. Adv in Nonl Anal, 2021, 10(1): 1255-1266. [Google Scholar]
Hai D D. Positive solutions to a class of elliptic boundary value problems[J]. J of Math Anal and Appl, 1998, 227(1): 195-199. [Google Scholar]
Njoku F I, Omari P, Zanolin F. Multiplicity of positive radial solutions of a quasilinear elliptic problem in a ball[J]. Adv in Diff Equ, 2000, 5(10-12): 1545-1570. [Google Scholar]
Bereanu C, Jebelean P, Mawhin J. Radial solutions for Neumann problems involving mean curvature operators in Euclidean and Minkowski spaces[J]. Mathematische Nachrichten, 2010, 283(3): 379-391. [Google Scholar]
Jean M. Radial solutions of Neumann problem for periodic perturbations of the mean extrinsic curvature operator[J]. Milan Journal of Mathematics, 2011, 79(1): 95-112. [Google Scholar]
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Azzollini A. On a prescribed mean curvature equation in Lorentz-Minkowski space[J]. Journal de Mathématiques Pures et Appliquées, 2016, 106(6): 1122-1140. [Google Scholar]
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Ma R Y, Lu Y Q. Multiplicity of positive solutions for second order nonlinear dirichlet problem with one-dimension Minkowski-curvature operator[J]. Adv Nonl Studies, 2015, 15(4): 789-803. [Google Scholar]
Dai G W. Bifurcation and positive solutions for problem with mean curvature operator in Minkowski space[J]. Cal of Vari and Part Diff Equ, 2016, 55(4): 72. [Google Scholar]
Xu M, Ma R Y. S-shaped connected component of radial positive solutions for a prescribed mean curvature problem in an annular domain[J]. Open Mathematics, 2008, 17: 929-941. [Google Scholar]
Lu Y Q, Li Z Q, Chen T L. Multiplicity of solutions for non-homogeneous dirichlet problem with one-dimension minkowski-curvature operator[J]. Qualitative Theory of Dynamical Systems, 2022, 21(4): 145. [Google Scholar]
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Goddard II J, Morris Q A, Robinson S B, et al. An exact bifurcation diagram for a reaction-diffusion equation arising in population dynamics[J]. Boundary Value Problems, 2018, 2018(1): 170. [Google Scholar]

All Figures

	Fig. 1 Bifurcation diagram for problem (1)
In the text

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[1] Binding P A, Browne P J, Seddighi K. Sturm-Liouville problems with eigenparameter dependent boundary conditions[J]. Proceedings of the Edinburgh Mathematical Society, 1994, 37(1): 57-72. [Google Scholar]

[2] Hai D D. Positive solutions for semilinear elliptic equations in annular domains[J]. Nonl Anal: Theo, Meth & Appl, 1999, 37(9): 1051-1058. [Google Scholar]

[3] Marinets V V, Pitovka O Y. On a problem with a parameter in the boundary conditions[J]. Nauk Visn Uzhgorod Univ Ser Mat Inform, 2005 (10-11): 70-76. [Google Scholar]

[4] Li Z L. Second-order singular nonlinear boundary value problem with parameters in the boundary conditions[J]. Adv Math, 2008, 37(3): 353-360(Ch). [Google Scholar]

[5] Aslanova N, Bayramoglu M, Aslanov K. On one class eigenvalue problem with eigenvalue parameter in the boundary condition at one end-point[J]. Filomat, 2018, 32(19): 6667-6674. [Google Scholar]

[6] Fonseka N, Muthunayake A, Shivaji R, et al. Singular reaction diffusion equations where a parameter influences the reaction term and the boundary conditions[J]. Topological Methods in Nonlinear Analysis, 2021, 57(1): 221-242. [Google Scholar]

[7] Acharya A, Fonseka N, Quiroa J, et al. σ-shaped bifurcation curves[J]. Adv in Nonl Anal, 2021, 10(1): 1255-1266. [Google Scholar]

[8] Hai D D. Positive solutions to a class of elliptic boundary value problems[J]. J of Math Anal and Appl, 1998, 227(1): 195-199. [Google Scholar]

[9] Njoku F I, Omari P, Zanolin F. Multiplicity of positive radial solutions of a quasilinear elliptic problem in a ball[J]. Adv in Diff Equ, 2000, 5(10-12): 1545-1570. [Google Scholar]

[10] Bereanu C, Jebelean P, Mawhin J. Radial solutions for Neumann problems involving mean curvature operators in Euclidean and Minkowski spaces[J]. Mathematische Nachrichten, 2010, 283(3): 379-391. [Google Scholar]

[11] Jean M. Radial solutions of Neumann problem for periodic perturbations of the mean extrinsic curvature operator[J]. Milan Journal of Mathematics, 2011, 79(1): 95-112. [Google Scholar]

[12] Bartnik R, Simon L. Spacelike hypersurfaces with prescribed boundary values and mean curvature[J]. Communications in Mathematical Physics, 1982, 87(1): 131-152. [Google Scholar]

[13] Azzollini A. On a prescribed mean curvature equation in Lorentz-Minkowski space[J]. Journal de Mathématiques Pures et Appliquées, 2016, 106(6): 1122-1140. [Google Scholar]

[14] Born M, Infeld L. Foundations of the new field theory[J]. Nature, 1933, 132(3348): 1004. [Google Scholar]

[15] Bereanu C, Mawhin J. Existence and multiplicity results for some nonlinear problems with singular -Laplacian[J]. J of Diff Equ, 2007, 243(2): 536-557. [Google Scholar]

[16] Anuradha V, Shivaji R. A quadrature method for classes of multi-parameter two point boundary value problems[J]. Applicable Analysis, 1994, 54(3-4): 263-281. [Google Scholar]

[17] Ma R Y, Lu Y Q. Multiplicity of positive solutions for second order nonlinear dirichlet problem with one-dimension Minkowski-curvature operator[J]. Adv Nonl Studies, 2015, 15(4): 789-803. [Google Scholar]

[18] Dai G W. Bifurcation and positive solutions for problem with mean curvature operator in Minkowski space[J]. Cal of Vari and Part Diff Equ, 2016, 55(4): 72. [Google Scholar]

[19] Xu M, Ma R Y. S-shaped connected component of radial positive solutions for a prescribed mean curvature problem in an annular domain[J]. Open Mathematics, 2008, 17: 929-941. [Google Scholar]

[20] Lu Y Q, Li Z Q, Chen T L. Multiplicity of solutions for non-homogeneous dirichlet problem with one-dimension minkowski-curvature operator[J]. Qualitative Theory of Dynamical Systems, 2022, 21(4): 145. [Google Scholar]

[21] Amann H. Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces[J]. SIAM Review, 1976, 18(4): 620-709. [Google Scholar]

[22] Dhanya R, Ko E, Shivaji R. A three solution theorem for singular nonlinear elliptic boundary value problems[J]. J of Math Anal and Appl, 2015, 424(1): 598-612. [Google Scholar]

[23] Gao H L, Xu J. Bifurcation curves and exact multiplicity of positive solutions for Dirichlet problems with the Minkowski-curvature equation[J]. Boundary Value Problems, 2021, 2021: 81. [Google Scholar]

[24] Goddard II J, Morris Q A, Robinson S B, et al. An exact bifurcation diagram for a reaction-diffusion equation arising in population dynamics[J]. Boundary Value Problems, 2018, 2018(1): 170. [Google Scholar]

∑-Shaped Connected Component of Positive Solutions for One- Dimensional Prescribed Mean Curvature Equation in Minkowski Space

Minkowski 空间中一维给定平均曲率方程正解的∑-型连通分支

0 Introduction

1 Preliminaries

2 The Autonomous Case of f(u)=u, h(t)=1

3 The Proof of Main Results

4 Examples

References

All Figures

2 The Autonomous Case of $f (u) = u$ , $h (t) = 1$