Existence and multiplicity of positive solutions for one-dimensional prescribed mean curvature equations

Yanqiong Lu; Ruyun Ma; Hongliang Gao

doi:10.1186/1687-2770-2014-120

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Existence and multiplicity of positive solutions for one-dimensional prescribed mean curvature equations

Yanqiong Lu¹,
Ruyun Ma¹Email author and
Hongliang Gao¹

Boundary Value Problems20142014:120

https://doi.org/10.1186/1687-2770-2014-120

Received: 3 January 2014

Accepted: 6 May 2014

Published: 16 May 2014

Abstract

In this work, we establish the existence and multiplicity results of positive solutions for one-dimensional prescribed mean curvature equations. Our approach is based on fixed point index theory for completely continuous operators which leave invariant a suitable cone in a Banach space of continuous functions.

MSC:34B10, 34B18.

Keywords

mean curvature equations positive solutions existence fixed point index

1 Introduction

The prescribed mean curvature problems like

{\begin{cases} - div (\frac{D u}{\sqrt{1 + κ {(D u)}^{2}}}) = f (x, u), x \in Ω, \\ u = 0, x \in \partial Ω \end{cases}

have attracted much attention in recent years, see [1–4] and the references therein. Since the problem is quasilinear non-uniformly elliptic, it is more difficult to study the existence of classical solutions. The greatest obstacle is the lack of gradient estimate, such kind of estimate does not hold in general and boundary gradient blow-up may occur. This leads to some new phenomena very different from those in semilinear problems. Many well-known results of semilinear problems have to be reconsidered for this quasilinear problem. Motivated by the search for solutions of the above problem, many authors (see [5–13]) studied the existence of (positive) solutions for one-dimensional prescribed mean curvature equations with Dirichlet boundary conditions

{\begin{cases} - {(\frac{u^{'}}{\sqrt{1 + κ {(u^{'})}^{2}}})}^{'} = λ f (u), x \in (0, 1), \\ u (0) = u (1) = 0, \end{cases}

(1.1)

where $κ > 0$ is a constant, $f \in C ([0, \infty), [0, \infty))$ and $f (u) > 0$ for $u > 0$ and $x \in [0, 1]$ .

Note that if

κ = 0

, problem (1.1) is degenerate to the second-order ordinary differential equation boundary value problems

{\begin{cases} - u^{″} = λ f (u), x \in (0, 1), \\ u (0) = u (1) = 0 . \end{cases}

(1.2)

The existence of (positive) solutions of (1.1) has been well known with various qualitative assumptions of the nonlinearity f, see [14, 15] and the references therein.

If $κ = 1$ , Bonheure et al. [6], Habets and Omari [9], Kusahara and Usami [10], Pan and Xing [12, 13] studied the existence of (positive) solutions of (1.2) by using the variational method, lower and upper solutions method and time mapping method, respectively.

However, to the best of our knowledge, the existence and multiplicity of positive solutions for (1.1) are relatively few by the fixed point index theory. In this paper, based on the fixed point index theory, we shall investigate the existence and multiplicity of positive solution of (1.1) when f is ϕ-superlinear and ϕ-sublinear at 0 and ∞, respectively, here $ϕ (s) = \frac{s}{\sqrt{1 + κ s^{2}}}$ .

Let

ϕ (s) = \frac{s}{\sqrt{1 + κ s^{2}}}

. Then (1.1) can be rewritten as

{\begin{cases} - {(ϕ (u^{'}))}^{'} = λ f (u (x)), x \in (0, 1), \\ u (0) = u (1) = 0 . \end{cases}

(1.3)

Obviously, $ϕ : R \to (- \frac{1}{\sqrt{κ}}, \frac{1}{\sqrt{κ}})$ is an odd, increasing homeomorphism with $ϕ (0) = 0$ .

For convenience, we introduce some notations

f_{0} = lim_{u \to 0} \frac{f (u)}{ϕ (u)} and f_{\infty} = lim_{u \to \infty} \frac{f (u)}{ϕ (u)} .

We will also need the function $f^{*} (u) = {max}_{0 \leq t \leq u} {f (t)}$ , and let $f_{0}^{*} = {lim}_{u \to 0} \frac{f^{*} (u)}{ϕ (u)}$ . By a similar method in [[14], Lemma 2.8], it is not difficult to verify that $f_{0}^{*} = f_{0}$ .

In the rest of this paper, we shall study the existence of positive solutions of (1.3) by using the fixed point index theory to give a brief and clear proof for the existence of positive solutions of (1.1). More concretely, we shall prove the following.

Theorem 1.1 Assume that

f \in C ([0, \infty), [0, \infty))

and

f (u) > 0

for

u > 0

.

(i)

If $f_{0} = 0$ , then there exists $0 < λ_{*} < λ^{*}$ such that (1.1) has a positive solution for $λ_{*} < λ < λ^{*}$ .
(ii)

If $f_{0} = \infty$ , then there exists $λ_{0} > 0$ such that (1.1) has a positive solution for $0 < λ < λ_{0}$ .
(iii)

If $f_{0} = f_{\infty} = 0$ , then there exists $0 < λ_{*} < λ^{*}$ such that (1.1) has at least two positive solutions for $λ_{*} < λ < λ^{*}$ .
(iv)

If $f_{0} = f_{\infty} = \infty$ , then there exists $0 < λ_{*} < λ^{*}$ such that (1.1) has at least two positive solutions for $λ_{*} < λ < λ^{*}$ .

Corollary 1.2 Assume that

f \in C ([0, \infty), [0, \infty))

and

f (u) > 0

for

u > 0

.

(a)

If $f_{0} = 0$ and $f_{\infty} = \infty$ , then there exists $0 < λ_{*} < λ^{*}$ such that (1.1) has a positive solution for $λ_{*} < λ < λ^{*}$ .
(b)

If $f_{0} = \infty$ and $f_{\infty} = 0$ , then there exists $λ_{0} > 0$ such that (1.1) has a positive solution for $0 < λ < λ_{0}$ .

Remark 1.1 The results of Theorem 1.1 and Corollary 1.2 are different with the case $κ = 0$ which is the classical Dirichlet boundary value problem (1.2). This phenomenon is a striking feature of problem (1.1), which is just the reason why we study the existence of positive solutions of problem (1.1). It is pointed out that in equation of (1.1), having replaced $f (u)$ with $f (x, u)$ , Theorem 1.1 and Corollary 1.2 also hold as well as all of the proofs with obvious changes.

2 Preliminaries

Throughout the paper

| \cdot |

will denote absolute value, and let

E = {u \in C [0, 1] ∣ u (0) = u (1) = 0} .

Then E is a Banach space endowed with the norm ${∥ u ∥}_{\infty} = {max}_{x \in [0, 1]} | u (x) |$ .

We first establish some preliminary results to prove our main result. An easy but useful property of ϕ and $ϕ^{- 1}$ is the following one.

Lemma 2.1 Let

ϕ (s) = \frac{s}{\sqrt{1 + κ s^{2}}}

. Then

ϕ : R \to (- \frac{1}{\sqrt{κ}}, \frac{1}{\sqrt{κ}})

is an odd, increasing homeomorphism with

ϕ (0) = 0

. Moreover, ϕ has the following properties:

(i)

ϕ is convex up on $[0, \infty)$ and $ϕ^{- 1}$ is concave up on $(0, \frac{1}{\sqrt{κ}})$ .
(ii)

For each $c \geq 1$ , there exists ${\hat{A}}_{c} \geq c$ such that $ϕ (c s) \leq {\hat{A}}_{c} ϕ (s)$ , $\forall s \geq 0$ with ${lim}_{c \to \infty} {\hat{A}}_{c} = \infty$ and for each $0 < c \leq 1$ , there exists $0 < {\hat{B}}_{c} \leq c$ such that $ϕ (c s) \geq {\hat{B}}_{c} ϕ (s)$ , $\forall s \geq 0$ with ${lim}_{c \to 0} {\hat{B}}_{c} = 0$ .
(iii)

For each $0 < c \leq 1$ , there exists $B_{c} \geq c$ such that $ϕ^{- 1} (c s) \leq B_{c} ϕ^{- 1} (s)$ , $\forall s \in (0, \frac{1}{\sqrt{κ}})$ . For each $c \geq 1$ with $- \frac{1}{\sqrt{κ}} < c s < \frac{1}{\sqrt{κ}}$ , there exists $A_{c} \leq c$ such that $ϕ^{- 1} (c s) \geq A_{c} ϕ^{- 1} (s)$ , $\forall s \in (0, \frac{1}{\sqrt{κ}})$ .

Proof By a simple computation, it follows that

ϕ (- s) = - ϕ (s)

and

ϕ^{'} (s) = \frac{1}{\sqrt{{(1 + κ s^{2})}^{3}}} > 0

. So ϕ is an odd, increasing homeomorphism with

ϕ (0) = 0

. Moreover, from

ϕ^{″} (s) = \frac{- 3 κ s}{\sqrt{{(1 + κ s^{2})}^{5}}}

, we get that ϕ is convex up on

[0, \infty)

. Notice that

ϕ^{- 1} (s) = \frac{s}{\sqrt{1 - κ s^{2}}}

is also an odd, increasing homeomorphism with

ϕ (0) = 0

. It is easy to verify that

ϕ^{- 1}

is concave up on

(0, \frac{1}{\sqrt{κ}})

.

(ii)

For each $0 < c \leq 1$ , there exists ${\hat{B}}_{c} \leq c$ such that
$ϕ (c s) = \frac{c s}{\sqrt{1 + κ c^{2} s^{2}}} \geq {\hat{B}}_{c} \frac{s}{\sqrt{1 + κ s^{2}}}, \forall s \geq 0;$

and for each

c \geq 1

, there exists

{\hat{A}}_{c} \geq c

such that

ϕ (c s) = \frac{c s}{\sqrt{1 + κ c^{2} s^{2}}} \leq {\hat{A}}_{c} \frac{s}{\sqrt{1 + κ s^{2}}}, \forall s \geq 0 .

(iii)

By a similar argument, it is not difficult to compute that for each $0 < c \leq 1$ , there exists $B_{c} \geq c$ such that $ϕ^{- 1} (c s) \leq B_{c} ϕ^{- 1} (s)$ , $\forall s \in (0, \frac{1}{\sqrt{κ}})$ . For each $c \geq 1$ with $- \frac{1}{\sqrt{κ}} < c s < \frac{1}{\sqrt{κ}}$ , there exists $A_{c} \leq c$ such that $ϕ^{- 1} (c s) \geq A_{c} ϕ^{- 1} (s)$ , $\forall s \in (0, \frac{1}{\sqrt{κ}})$ . □

Lemma 2.2 Let

h \in C ([0, 1], [0, \infty))

with

h ≢ 0

. Assume that w is the solution of

{\begin{cases} - {(ϕ (u^{'}))}^{'} = h (x), x \in (0, 1), \\ u (0) = u (1) = 0 . \end{cases}

(2.1)

Then $w > 0$ on $(0, 1)$ and ${∥ w^{'} ∥}_{\infty} \leq ϕ^{- 1} (M)$ , where $M = min {\frac{1}{\sqrt{κ}}, {sup}_{x \in [0, 1]} | h (x) |}$ .

Proof By integrating, it follows that (2.1) has the unique solution given by

w (x) = \int_{0}^{x} ϕ^{- 1} (C - \int_{0}^{s} h (t) d t) d s,

where C is such that $w (1) = 0$ . Hence we must have $0 < C < \int_{0}^{1} h (t) d t$ . Further, since $ϕ (u^{'}) \in (- \frac{1}{\sqrt{κ}}, \frac{1}{\sqrt{κ}})$ , by using $ϕ (w^{'} (x)) = C - \int_{0}^{x} h (t) d t$ , we obtain $- M \leq ϕ (w^{'} (x)) \leq M$ , $\forall x \in [0, 1]$ and ${∥ w^{'} ∥}_{\infty} \leq ϕ^{- 1} (M)$ follows, here $M = min {\frac{1}{\sqrt{κ}}, {sup}_{x \in [0, 1]} | h (x) |}$ .

Since $w (0) = w (1) = 0$ , there exists $x_{0} \in (0, 1)$ such that $w^{'} (x_{0}) = 0$ and it follows from $- {(ϕ (w^{'}))}^{'} \geq 0$ that $ϕ (w^{'})$ is decreasing on $(0, 1)$ . Then $w^{'} (x) > 0$ for $x < x_{0}$ and $w^{'} (x) < 0$ for $x > x_{0}$ . Hence, $w > 0$ on $(0, 1)$ . □

Note that from Lemma 2.2, there exists

τ_{i} \in (0, \frac{1}{2})

,

i = 1, 2

, such that

{min}_{x \in [τ_{1}, 1 - τ_{2}]} w (x) \geq σ {∥ w ∥}_{\infty}

with

0 < σ < 1

depending on

τ_{i}

. Define the cone P in E by

P = {u \in E | u (x) \geq 0 on [0, 1] and min_{x \in [τ_{1}, 1 - τ_{2}]} u (x) \geq σ {∥ u ∥}_{\infty}},

and for $r > 0$ , let $Ω_{r} = {u \in P ∣ {∥ u ∥}_{\infty} < r}$ .

Lemma 2.3 ([[11], Lemma 4.1 and Lemma 4.2])

For each

h \in C [0, 1]

, (2.1) has a unique solution given by

u (x) = \int_{0}^{x} ϕ^{- 1} (C - \int_{0}^{s} h (t) d t) d s = : T_{h} (u) (x),

where C is such that $u (1) = 0$ with $0 < C < \int_{0}^{1} h (t) d t$ . Moreover, the operator $T_{h} : E \to E$ is continuous and sends equicontinuous sets in $C [0, 1]$ into a relatively compact set in E.

We next state the fixed point index theorem which will be used to prove our results.

Lemma 2.4 ([[16], Chapter 6])

Let E be a Banach space and P be a cone in E. Assume that Ω is a bounded open subset of E with

0 \in Ω

, and let

T : P \cap \bar{Ω} \to P

be a completely continuous operator such that

T u \neq u

,

u \in \partial Ω \cap P

.

(i)

If $∥ T u ∥ \leq ∥ u ∥$ , $u \in \partial Ω \cap P$ , then $i (A, Ω \cap P, P) = 1$ .
(ii)

If $∥ T u ∥ \geq ∥ u ∥$ , $u \in \partial Ω \cap P$ , then $i (A, Ω \cap P, P) = 0$ .

From Lemma 2.2, problem (1.3) is equivalent to the fixed point problem

u (x) = \int_{0}^{x} ϕ^{- 1} (C_{0} - λ \int_{0}^{s} f (u (t)) d t) d s : = T_{λ} (u) (x)

in the space E, where

C_{0}

is such that

u (1) = 0

with

0 < C_{0} < λ \int_{0}^{1} f (u (t)) d t

, since otherwise,

u (1) = \int_{0}^{1} ϕ^{- 1} (C_{0} - λ \int_{0}^{s} f (u (t)) d t) d s \geq \int_{0}^{1} ϕ^{- 1} (λ \int_{s}^{1} f (u (t)) d t) d s > 0,

which is a contradiction. This together with Lemma 2.3 implies that

T_{λ} : E \to E

is a completely continuous mapping. Moreover, for any fixed

u \in P

, we have

T_{λ} (u) (x) = \int_{0}^{x} ϕ^{- 1} (C_{0} - λ \int_{0}^{s} f (u (t)) d t) d s \geq 0, x \in [0, 1]

(2.2)

and $T_{λ} (u) (0) = T_{λ} (u) (1) = 0$ . In addition, from Lemma 2.2, it follows that $T_{λ} (u) (x) > 0$ on $(0, 1)$ and there exist $τ_{i} \in (0, \frac{1}{2})$ , $i = 1, 2$ , such that ${min}_{x \in [τ_{1}, 1 - τ_{2}]} T_{λ} u \geq σ {∥ T_{λ} u ∥}_{\infty}$ . So $T_{λ} : P \to P$ is a completely continuous operator.

Lemma 2.5 Let

r > 0

be given. If there exists

ε > 0

small enough with

B_{λ ε} < 1

such that

f^{*} (r) \leq ε ϕ (r)

, then

{∥ T_{λ} u ∥}_{\infty} \leq B_{λ ε} {∥ u ∥}_{\infty} for u \in \partial Ω_{r},

where $B_{λ ε}$ is defined as in Lemma 2.1(iii).

Proof From the definition of

T_{λ}

, for any

u \in \partial Ω_{r}

, we have

\begin{aligned} {∥ T_{λ} u ∥}_{\infty} & = max_{x \in [0, 1]} | \int_{0}^{x} ϕ^{- 1} (C_{0} - λ \int_{0}^{s} f (u (t)) d t) d s | \\ \leq \int_{0}^{1} ϕ^{- 1} (λ \int_{s}^{1} f (u (t)) d t) d s \leq \int_{0}^{1} ϕ^{- 1} (λ \int_{s}^{1} f^{*} (r) d t) d s \\ \leq \int_{0}^{1} ϕ^{- 1} (λ ε ϕ (r)) d s \leq B_{λ ε} r . \end{aligned}

□

Lemma 2.6 Let

η > 0

be given. If

u \in P

and

f (u (x)) \geq η ϕ (u (x))

for

x \in [0, 1]

, then

{∥ T_{λ} u ∥}_{\infty} \geq σ x_{*} ϕ^{- 1} (λ (1 - σ) x_{*} η ϕ (σ {∥ u ∥}_{\infty})),

where $x_{*} = min {x_{0}, 1 - x_{0}}$ and $u (x_{0}) = {max}_{x \in [0, 1]} u (x) = {∥ u ∥}_{\infty}$ .

Proof From problem (1.3), since

u (0) = u (1) = 0

, it follows that there exists

x_{0} \in (0, 1)

such that

u^{'} (x_{0}) = 0

. Let

{∥ u ∥}_{\infty} = u (x_{0})

. Then u satisfies the following boundary value problem:

{\begin{cases} - {(ϕ (u^{'} (x)))}^{'} = λ f (u (x)), x \in (0, x_{0}), \\ u (0) = 0, u (x_{0}) = {∥ u ∥}_{\infty} . \end{cases}

(2.3)

Let v be the solution of the problem

{\begin{cases} - {(ϕ (v^{'} (x)))}^{'} = 0, x \in (0, x_{0}), \\ v (0) = 0, v (x_{0}) = {∥ u ∥}_{\infty} . \end{cases}

(2.4)

Then we have

{\begin{cases} - {(ϕ (u^{'} (x)) - ϕ (v^{'} (x)))}^{'} = λ f (u (x)), x \in (0, x_{0}), \\ (u - v) (0) = 0, (u - v) (x_{0}) = 0, \end{cases}

(2.5)

and by a comparison argument, we get that

u > v

on

(0, x_{0})

. In fact, from

(u - v) (0) = 0

,

(u - v) (x_{0}) = 0

, there exists

\hat{x} \in (0, x_{0})

such that

{(u - v)}^{'} (\hat{x}) = 0

, i.e.,

u^{'} (\hat{x}) = v^{'} (\hat{x})

. Thus, by a simple computation, we have that

ϕ (u^{'} (x)) = ϕ (v^{'} (x)) + \int_{x}^{\hat{x}} λ f (u (t)) d t \geq ϕ (v^{'} (x)) for x \in (0, \hat{x})

and

ϕ (v^{'} (x)) = ϕ (u^{'} (x)) + \int_{\hat{x}}^{x} λ f (u (t)) d t \geq ϕ (u^{'} (x)) for x \in (\hat{x}, x_{0}) .

This together with ϕ is an increasing homeomorphism implies that

u^{'} (x) \geq v^{'} (x) for x \in (0, \hat{x}),

(2.6)

v^{'} (x) \geq u^{'} (x) for x \in (\hat{x}, x_{0}) .

(2.7)

Integrating from 0 to x for (2.6) and integrating from x to $x_{0}$ for (2.7), respectively, we have that $u (x) \geq v (x)$ for $x \in [0, x_{0}]$ .

Note that

v (x) = {∥ u ∥}_{\infty} - \int_{x}^{x_{0}} ϕ^{- 1} (C_{1}) d s,

where

C_{1}

is such that

v (0) = 0

, and hence

{∥ u ∥}_{\infty} = \int_{0}^{x_{0}} ϕ^{- 1} (C_{1}) d s

. If

C_{1} > ϕ (\frac{{∥ u ∥}_{\infty}}{x_{0}})

, then it follows that

{∥ u ∥}_{\infty} = \int_{0}^{x_{0}} ϕ^{- 1} (C_{1}) d s > \int_{0}^{x_{0}} ϕ^{- 1} (ϕ (\frac{{∥ u ∥}_{\infty}}{x_{0}})) d s \geq {∥ u ∥}_{\infty},

which is a contradiction. Thus,

0 \leq C_{1} \leq ϕ (\frac{{∥ u ∥}_{\infty}}{x_{0}})

. Moreover, we have

\begin{aligned} v (x) & = {∥ u ∥}_{\infty} - \int_{x}^{x_{0}} ϕ^{- 1} (C_{1}) d s \geq {∥ u ∥}_{\infty} - (x_{0} - x) ϕ^{- 1} (ϕ (\frac{{∥ u ∥}_{\infty}}{x_{0}})) \\ = \frac{x}{x_{0}} {∥ u ∥}_{\infty}, x \in [x_{1}, x_{0}], \end{aligned}

where

x_{1} = σ x_{0}

. Consequently,

u (x) \geq v (x) \geq σ {∥ u ∥}_{\infty}

for

x \in [x_{1}, x_{0}]

. Obviously,

u (x) = \int_{0}^{x} ϕ^{- 1} (C_{2} - λ \int_{0}^{s} f (u (t)) d t) d s,

where

C_{2}

satisfies

u^{'} (x_{0}) = 0

. It follows from

ϕ^{- 1} (C_{2} - λ \int_{0}^{x_{0}} f (u (t)) d t) = 0

that

C_{2} = λ \int_{0}^{x_{0}} f (u (t)) d t

. Therefore,

u (x) = \int_{0}^{x} ϕ^{- 1} (λ \int_{s}^{x_{0}} f (u (t)) d t) d s = T_{λ} u (x), x \in [0, x_{0}] .

(2.8)

If

x_{0} \geq \frac{1}{2}

, then

\begin{aligned} {∥ T_{λ} u ∥}_{\infty} & \geq \int_{0}^{x_{1}} ϕ^{- 1} (λ \int_{s}^{x_{0}} f (u (t)) d t) d s \geq \int_{0}^{x_{1}} ϕ^{- 1} (λ \int_{x_{1}}^{x_{0}} η ϕ (u (t)) d t) d s \\ \geq \int_{0}^{x_{1}} ϕ^{- 1} (λ (x_{0} - x_{1}) η ϕ (σ {∥ u ∥}_{\infty})) d s \geq σ x_{0} ϕ^{- 1} (λ (1 - σ) x_{0} η ϕ (σ {∥ u ∥}_{\infty})) . \end{aligned}

If

x_{0} \leq \frac{1}{2}

, let w be the solution of

\begin{aligned} - {(ϕ (w^{'} (x)))}^{'} = 0, x \in (x_{0}, 1), \\ w (x_{0}) = {∥ u ∥}_{\infty}, w (1) = 0 . \end{aligned}

(2.9)

Then

w (x) = {∥ u ∥}_{\infty} + \int_{x_{0}}^{x} ϕ^{- 1} (C_{3}) d s,

where

C_{3} < 0

satisfies

w (1) = 0

, i.e.,

{∥ u ∥}_{\infty} = - \int_{x_{0}}^{1} ϕ^{- 1} (C_{3}) d s

. If

C_{3} < - ϕ (\frac{{∥ u ∥}_{\infty}}{1 - x_{0}})

, then

{∥ u ∥}_{\infty} = - \int_{x_{0}}^{1} ϕ^{- 1} (C_{3}) d s > - \int_{x_{0}}^{1} ϕ^{- 1} (- ϕ (\frac{{∥ u ∥}_{\infty}}{1 - x_{0}})) d s \geq {∥ u ∥}_{\infty},

which is contradiction. Hence $- ϕ (\frac{{∥ u ∥}_{\infty}}{1 - x_{0}}) \leq C_{3} \leq 0$ .

By a similar argument as before, it follows that

u \geq w

on

(x_{0}, 1)

. Moreover,

\begin{aligned} w (x) & = {∥ u ∥}_{\infty} + \int_{x_{0}}^{x} ϕ^{- 1} (C_{3}) d s \geq {∥ u ∥}_{\infty} + (x - x_{0}) ϕ^{- 1} (- ϕ (\frac{{∥ u ∥}_{\infty}}{1 - x_{0}})) \\ = \frac{1 - x}{1 - x_{0}} {∥ u ∥}_{\infty}, x \in [x_{0}, x_{2}], \end{aligned}

where

x_{2} = 1 - σ (1 - x_{0})

. So

u (x) \geq w (x) \geq σ {∥ u ∥}_{\infty}

,

x \in [x_{0}, x_{2}]

. Therefore, we have

u (x) = \int_{x}^{1} ϕ^{- 1} (λ \int_{x_{0}}^{s} f (u (t)) d t) d s = T_{λ} u (x), x \in [x_{0}, 1],

(2.10)

and subsequently,

\begin{aligned} {∥ T_{λ} u ∥}_{\infty} & \geq \int_{x_{2}}^{1} ϕ^{- 1} (λ \int_{x_{0}}^{s} f (u (t)) d t) d s \geq \int_{x_{2}}^{1} ϕ^{- 1} (λ \int_{x_{0}}^{x_{2}} η ϕ (u (t)) d t) d s \\ \geq \int_{x_{2}}^{1} ϕ^{- 1} (λ (x_{2} - x_{0}) η ϕ (σ {∥ u ∥}_{\infty})) d s \\ \geq σ (1 - x_{0}) ϕ^{- 1} (λ (1 - σ) (1 - x_{0}) η ϕ (σ {∥ u ∥}_{\infty})) . \end{aligned}

Let

x_{*} = min {x_{0}, 1 - x_{0}}

. Then

{∥ T_{λ} u ∥}_{\infty} \geq σ x_{*} ϕ^{- 1} (λ (1 - σ) x_{*} η ϕ (σ {∥ u ∥}_{\infty})) .

□

Lemma 2.7 Let

r > 0

be given. If

u \in \partial Ω_{r}

, then

{∥ T_{λ} u ∥}_{\infty} \leq ϕ^{- 1} (λ M_{r}),

where $M_{r} = 1 + {max}_{0 \leq u \leq r} {f (u)} > 0$ .

Proof Obviously, for any

u \in \partial Ω_{r}

, it follows that

f (u (x)) \leq M_{r}

for

x \in [0, 1]

. So we have

{∥ T_{λ} u ∥}_{\infty} \leq \int_{0}^{1} ϕ^{- 1} (λ \int_{s}^{1} f (u (t)) d t) d s \leq ϕ^{- 1} (λ M_{r}) .

□

Lemma 2.8 Let

r > 0

be given. If

u \in \partial Ω_{r}

, then

{∥ T_{λ} u ∥}_{\infty} \geq σ x_{*} ϕ^{- 1} (λ (1 - σ) x_{*} m_{r}),

where $m_{r} = {min}_{σ r \leq u \leq r} {f (u)} > 0$ and $x_{*} = min {x_{0}, 1 - x_{0}}$ .

Proof By using a similar argument of the proof of Lemma 2.6, we have that

u (x) \geq σ {∥ u ∥}_{\infty}

for

x \in [x_{1}, x_{2}]

. Meanwhile, (2.8) is true. If

x_{0} \geq \frac{1}{2}

, then we can get that

\begin{aligned} {∥ T_{λ} u ∥}_{\infty} & \geq \int_{0}^{x_{1}} ϕ^{- 1} (λ \int_{s}^{x_{0}} f (u (t)) d t) d s \geq \int_{0}^{x_{1}} ϕ^{- 1} (λ \int_{x_{1}}^{x_{0}} m_{r} d t) d s \\ \geq \int_{0}^{x_{1}} ϕ^{- 1} (λ (x_{0} - x_{1}) m_{r}) d s \geq x_{1} ϕ^{- 1} (λ (x_{0} - x_{1}) m_{r}) \\ = σ x_{0} ϕ^{- 1} (λ (1 - σ) x_{0} m_{r}) . \end{aligned}

If

x_{0} \leq \frac{1}{2}

, then (2.10) holds and it follows that

\begin{aligned} {∥ T_{λ} u ∥}_{\infty} & \geq \int_{x_{2}}^{1} ϕ^{- 1} (λ \int_{x_{0}}^{s} f (u (t)) d t) d s \geq \int_{x_{2}}^{1} ϕ^{- 1} (λ \int_{x_{0}}^{x_{2}} m_{r} d t) d s \\ \geq \int_{x_{2}}^{1} ϕ^{- 1} (λ (x_{2} - x_{0}) m_{r}) d s \geq (1 - x_{2}) ϕ^{- 1} (λ (x_{2} - x_{0}) m_{r}) \\ = σ (1 - x_{0}) ϕ^{- 1} (λ (1 - σ) (1 - x_{0}) m_{r}) . \end{aligned}

Let

x_{*} = min {x_{0}, (1 - x_{0})}

, then

{∥ T_{λ} u ∥}_{\infty} \geq σ x_{*} ϕ^{- 1} (λ (1 - σ) x_{*} m_{r}) .

□

3 Proof of the main results

Proof of Theorem 1.1 (i) Choose a suitable number

r_{1} > 0

. By Lemma 2.8, we have

{∥ T_{λ} u ∥}_{\infty} > {∥ u ∥}_{\infty} for u \in \partial Ω_{r_{1}} and λ_{*} < λ < λ^{*},

where

λ^{*} \leq \frac{1}{\sqrt{κ} (1 - σ) x_{*} m_{r_{1}}} and λ_{*} \geq \frac{1}{(1 - σ) x^{*} m_{r_{1}}} ϕ (\frac{r_{1}}{σ x^{*}}) .

If

f_{0} = 0

, then

f_{0}^{*} = 0

, and so we can choose

r_{2} \in (0, r_{1})

such that

f^{*} (r_{2}) \leq ε ϕ (r_{2})

, where

ε > 0

small enough satisfies

B_{λ ε} < 1 .

(3.1)

Then Lemma 2.5 implies that

{∥ T_{λ} u ∥}_{\infty} < {∥ u ∥}_{\infty} for u \in \partial Ω_{r_{2}} .

From Lemma 2.4, it follows that

i (T_{λ}, Ω_{r_{1}}, P) = 0

and

i (T_{λ}, Ω_{r_{2}}, P) = 1

. By using the additivity-excision property of the fixed point index [16], we have that

i (T_{λ}, Ω_{r_{1}} ∖ {\bar{Ω}}_{r_{2}}, P) = i (T_{λ}, Ω_{r_{1}}, P) - i (T_{λ}, Ω_{r_{2}}, P) = - 1 .

Therefore,

T_{λ}

has a fixed point in

{\bar{Ω}}_{r_{1}} ∖ Ω_{r_{2}}

. Consequently, (1.1) has a positive solution for

λ_{*} < λ < λ^{*}

.

(ii)

Choose a suitable number $r_{1} > 0$ . By Lemma 2.7, there exists
$0 < λ_{0} \leq min {\frac{1}{\sqrt{κ} M_{r_{1}}}, \frac{ϕ (r_{1})}{M_{r_{1}}}} = \frac{ϕ (r_{1})}{M_{r_{1}}}$

such that ${∥ T_{λ} u ∥}_{\infty} < {∥ u ∥}_{\infty}$ for $u \in \partial Ω_{r_{1}}$ and $0 < λ < λ_{0}$ . That is, $i (T_{λ}, Ω_{r_{1}}, P) = 1$ .

If

f_{0} = \infty

, then there exists

r_{2} \in (0, r_{1})

such that

f (u) \geq η ϕ (u)

for

0 \leq u \leq r_{2}

, where

η > 0

is chosen large enough so that

λ (1 - σ) x_{*} η ϕ (σ r_{2}) < \frac{1}{\sqrt{κ}} and σ^{2} λ (1 - σ) x_{*}^{2} η > 1 .

(3.2)

Clearly,

f (u) \geq η ϕ (u (x))

for

u \in \partial Ω_{r_{2}}

,

x \in [0, 1]

. From Lemma 2.6, we get that

\begin{aligned} {∥ T_{λ} u ∥}_{\infty} \geq σ x_{*} ϕ^{- 1} (λ (1 - σ) x_{*} η ϕ (σ {∥ u ∥}_{\infty})) \geq σ^{2} λ (1 - σ) x_{*}^{2} η {∥ u ∥}_{\infty} > {∥ u ∥}_{\infty} \\ for u \in \partial Ω_{r_{2}} . \end{aligned}

This together with Lemma 2.4 implies

i (T_{λ}, Ω_{r_{2}}, P) = 0

. By using the additivity-excision property of the fixed point index [16], we have

i (T_{λ}, Ω_{r_{1}} ∖ {\bar{Ω}}_{r_{2}}, P) = i (T_{λ}, Ω_{r_{1}}, P) - i (T_{λ}, Ω_{r_{2}}, P) = 1 .

Therefore,

T_{λ}

has a fixed point in

{\bar{Ω}}_{r_{1}} ∖ Ω_{r_{2}}

. Consequently, (1.1) has a positive solution for

0 < λ < λ_{0}

.

(iii)

Since ϕ is a bounded operator, multiplying (1.3) by $u^{'}$ and integrating from 0 to $x_{0}$ , we get that
$λ \int_{0}^{u (x_{0})} f (u) d u = \frac{1}{κ} [1 - \frac{1}{\sqrt{1 + κ {(u^{'} (0))}^{2}}}] \to \frac{1}{κ} as u^{'} (0) \to \infty .$

Let

λ^{*} > 0

be the solution of

λ \int_{0}^{r^{*}} f (u) d u = \frac{1}{κ}

. Then there exists

\bar{r} : = r^{*} - ϵ > 0

such that

λ \int_{0}^{\bar{r}} f (u) d u < \frac{1}{κ}

(3.3)

with $\bar{r} = u (x_{0}) = {∥ u ∥}_{\infty}$ .

Choose two numbers

0 < r_{3} < r_{4} < \bar{r}

satisfying

ϕ (\frac{r_{3}}{σ x_{*}}) < \frac{(1 - σ) x_{*} m_{r_{3}}}{M_{\bar{r}}} ϕ (\bar{r}) and ϕ (\frac{r_{4}}{σ x_{*}}) < \frac{(1 - σ) x_{*} m_{r_{4}}}{M_{\bar{r}}} ϕ (\bar{r}) .

By Lemma 2.8, there exist

λ_{*} = max {\frac{ϕ (\frac{r_{3}}{σ x^{*}})}{(1 - σ) x^{*} m_{r_{3}}}, \frac{ϕ (\frac{r_{4}}{σ x^{*}})}{(1 - σ) x^{*} m_{r_{4}}}}

and

λ^{*} = min {\frac{1}{\sqrt{κ} M_{\bar{r}}}, \frac{ϕ (\bar{r})}{M_{\bar{r}}}, \frac{1}{(1 - σ) x_{*} m_{r_{3}} \sqrt{κ}}, \frac{1}{(1 - σ) x_{*} m_{r_{4}} \sqrt{κ}}} = \frac{ϕ (\bar{r})}{M_{\bar{r}}}

such that for

λ_{*} < λ < λ^{*}

, we have

{∥ T_{λ} u ∥}_{\infty} > {∥ u ∥}_{\infty} for u \in \partial Ω_{r_{i}}, i = 3, 4 .

This together with Lemma 2.4 implies $i (T_{λ}, Ω_{r_{i}}, P) = 0$ , $i = 3, 4$ .

Since

f_{0} = 0

, from the proof of the case (i), it follows that we can choose

r_{1} \in (0, \frac{r_{3}}{2})

such that

i (T_{λ}, Ω_{r_{1}}, P) = 1

. Subsequently,

i (T_{λ}, Ω_{r_{3}} ∖ {\bar{Ω}}_{r_{1}}, P) = i (T_{λ}, Ω_{r_{3}}, P) - i (T_{λ}, Ω_{r_{1}}, P) = - 1 .

On the other hand,

f_{\infty} = 0

,

λ < λ^{*}

and (3.3) together with Lemma 2.7 implies

{∥ T_{λ} u ∥}_{\infty} \leq ϕ^{- 1} (λ M_{\bar{r}}) < {∥ u ∥}_{\infty} for any u \in \partial Ω_{\bar{r}} .

That is,

i (T_{λ}, Ω_{\bar{r}}, P) = 1

. Subsequently,

i (T_{λ}, Ω_{\bar{r}} ∖ {\bar{Ω}}_{r_{4}}, P) = i (T_{λ}, Ω_{\bar{r}}, P) - i (T_{λ}, Ω_{r_{4}}, P) = 1 .

Therefore,

T_{λ}

has two fixed points

u_{1}

and

u_{2}

such that

u_{1} \in {\bar{Ω}}_{r_{3}} ∖ Ω_{r_{1}}

and

u_{2} \in {\bar{Ω}}_{\bar{r}} ∖ Ω_{r_{4}}

. These are the desired distinct positive solutions of (1.1) for

λ_{*} < λ < λ^{*}

satisfying

r_{1} \leq {∥ u_{1} ∥}_{\infty} \leq r_{3} < r_{4} \leq {∥ u_{2} ∥}_{\infty} \leq r_{2} .

(3.4)

(iv)

Choose two numbers $0 < r_{3} < r_{4} < \bar{r}$ satisfying
$ϕ (r_{3}) > \frac{M_{r_{3}}}{(1 - σ) x_{*} m_{\bar{r}}} ϕ (\frac{\bar{r}}{σ x_{*}}) and ϕ (r_{4}) > \frac{M_{r_{4}}}{(1 - σ) x_{*} m_{\bar{r}}} ϕ (\frac{\bar{r}}{σ x_{*}}) .$

By Lemma 2.7, there exists

0 < λ_{0} \leq min {\frac{ϕ (r_{3})}{M_{r_{3}}}, \frac{ϕ (r_{4})}{M_{r_{4}}}}

such that for

0 < λ < λ_{0}

, we have

{∥ T_{λ} u ∥}_{\infty} < {∥ u ∥}_{\infty} for u \in \partial Ω_{r_{i}}, i = 3, 4 .

That is, $i (T_{λ}, Ω_{r_{3}}, P) = 1$ and $i (T_{λ}, Ω_{r_{4}}, P) = 1$ .

Since

f_{0} = \infty

, from the proof of the case (ii), choose

r_{1} \in (0, \frac{r_{3}}{2})

such that

i (T_{λ}, Ω_{r_{1}}, P) = 0

. Consequently,

i (T_{λ}, Ω_{r_{3}} ∖ {\bar{Ω}}_{r_{1}}, P) = i (T_{λ}, Ω_{r_{3}}, P) - i (T_{λ}, Ω_{r_{1}}, P) = 1 .

On the other hand,

f_{\infty} = \infty

and (3.3) together with Lemma 2.8 implies that

{∥ T_{λ} u ∥}_{\infty} \geq σ x_{*} ϕ^{- 1} (λ (1 - σ) x_{*} m_{\bar{r}}) .

Let

λ_{*} = \frac{ϕ (\frac{\bar{r}}{σ x_{*}})}{(1 - σ) x_{*} m_{\bar{r}}}

and

λ^{*} = min {λ_{0}, \frac{1}{\sqrt{κ} (1 - σ) x_{*} m_{\bar{r}}}}

. Then, for any

λ_{*} < λ < λ^{*}

, we have

{∥ T_{λ} u ∥}_{\infty} > {∥ u ∥}_{\infty} for any u \in u \in \partial Ω_{\bar{r}} .

That is,

i (T_{λ}, Ω_{\bar{r}}, P) = 0

. Subsequently,

i (T_{λ}, Ω_{\bar{r}} ∖ {\bar{Ω}}_{r_{4}}, P) = i (T_{λ}, Ω_{\bar{r}}, P) - i (T_{λ}, Ω_{r_{4}}, P) = - 1 .

Therefore, $T_{λ}$ has two fixed points $u_{1}$ and $u_{2}$ such that $u_{1} \in {\bar{Ω}}_{r_{3}} ∖ Ω_{r_{1}}$ and $u_{2} \in {\bar{Ω}}_{\bar{r}} ∖ Ω_{r_{4}}$ . These are the desired distinct positive solutions of (1.1) for $λ_{*} < λ < λ^{*}$ satisfying (3.4). □

Proof of Corollary 1.2 It is easy to show by the result of Theorem 1.1(i) and (ii). □

Example 3.1 Let us consider the following problem:

{\begin{cases} - {(\frac{u^{'}}{\sqrt{1 + κ {(u^{'})}^{2}}})}^{'} = λ e^{u}, x \in (0, 1), \\ u (0) = u (1) = 0 . \end{cases}

(3.5)

Obviously, $f_{0} = {lim}_{u \to 0} \frac{e^{u}}{ϕ (u)} = \infty$ and $f_{\infty} = {lim}_{u \to \infty} \frac{e^{u}}{ϕ (u)} = \infty$ . From Theorem 1.1(iv), there exists $0 < λ_{*} < λ^{*}$ such that (3.5) has at least two positive solutions for $λ_{*} < λ < λ^{*}$ .

Example 3.2 Let us consider the following problem:

{\begin{cases} - {(\frac{u^{'}}{\sqrt{1 + κ {(u^{'})}^{2}}})}^{'} = λ u^{p}, x \in (0, 1), \\ u (0) = u (1) = 0 . \end{cases}

(3.6)

Obviously, we divided the discussion into two cases as follows.

Case 1. $p > 1$

In this case, $f (u) = u^{p}$ is convex, and $f_{0} = {lim}_{u \to 0} \frac{u^{p}}{ϕ (u)} = 0$ and $f_{\infty} = {lim}_{u \to \infty} \frac{u^{p}}{ϕ (u)} = \infty$ . From Corollary 1.2(a), there exists $0 < λ_{*} < λ^{*}$ such that (3.6) has at least one positive solution for $λ_{*} < λ < λ^{*}$ .

Case 2. $p < 1$

In this case, $f (u) = u^{p}$ is concave, and $f_{0} = {lim}_{u \to 0} \frac{u^{p}}{ϕ (u)} = \infty$ and $f_{\infty} = {lim}_{u \to \infty} \frac{u^{p}}{ϕ (u)} = \infty$ . From Theorem 1.1(iv), there exists $0 < λ_{*} < λ^{*}$ such that (3.6) has at least two positive solutions for $λ_{*} < λ < λ^{*}$ .

Remark 3.1 It is worth to point out that our results only partly generalize the results of Habet and Omari [9] and Pan [12], since $f (u)$ is more general than $e^{u}$ , $u^{p}$ and due to the limitation of the fixed point index method.

Remark 3.2 Since $f (u)$ in (1.1) is autonomous, it is not difficult to show that the solutions of (1.1) are symmetric around $\frac{1}{2}$ , see [2, 12]. It follows that $x_{0} = \frac{1}{2}$ in the proof of Theorem 1.1, and we can set $τ_{i} = \frac{1}{4}$ , $i = 1, 2$ and $σ = \frac{1}{2}$ .

Declarations

Acknowledgements

The authors are very grateful to the anonymous referee for his valuable suggestions. This work was supported by the NSFC (No. 11361054, No. 11201378), SRFDP (No. 20126203110004), Gansu provincial National Science Foundation of China (No. 1208RJZA258).

Authors’ Affiliations

(1)

Department of Mathematics, Northwest Normal University

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