Positive solutions for a class of fractional 3-point boundary value problems at resonance

Abstract

In this paper, we study the nonlocal fractional differential equation:

$$\left \{ \textstyle\begin{array}{@{}l} D^{\alpha}_{0+}u(t)+f(t,u(t))=0 ,\quad 0< t< 1,\\ u(0)=0,\qquad u(1)=\eta u(\xi), \end{array}\displaystyle \right . $$

where $1 < \alpha< 2$, $0 < \xi< 1$, $\eta\xi^{\alpha-1}= 1$, $D^{\alpha}_{0+}$ is the standard Riemann-Liouville derivative, $f:[0,1]\times[0,+\infty)\rightarrow\mathbb{R}$ is continuous. The existence and uniqueness of positive solutions are obtained by means of the fixed point index theory and iterative technique.

1 Introduction

In this paper, we consider the following fractional differential equation:

$$ \left \{ \textstyle\begin{array}{@{}l} D^{\alpha}_{0+}u(t)+f(t,u(t))=0 ,\quad 0< t< 1,\\ u(0)=0,\qquad u(1)=\eta u(\xi), \end{array}\displaystyle \right . $$

(1.1)

where $1 < \alpha< 2$, $0 < \xi< 1$, $\eta\xi^{\alpha-1}= 1$, $D^{\alpha}_{0+}$ is the standard Riemann-Liouville derivative, $f:[0,1]\times[0,+\infty)\rightarrow\mathbb{R}$ is continuous. Problem (1.1) happens to be at resonance, since $\lambda=0$ is an eigenvalue of the linear problem

$$ \left \{ \textstyle\begin{array}{@{}l} -D^{\alpha}_{0+}u=\lambda u , \quad 0< t< 1,\\ u(0)=0,\qquad u(1)=\eta u(\xi), \end{array}\displaystyle \right . $$

(1.2)

and $ct^{\alpha-1},c\in\mathbb{R,}$ is the corresponding eigenfunction.

Fractional differential equations occur frequently in various fields such as physics, chemistry, engineering and control of dynamical systems, etc. During the last few decades, many papers and books on fractional calculus and fractional differential equations have appeared (see [1–22] and the references therein).

When $0<\eta\xi^{\alpha-1}< 1$, problem (1.1) is non-resonant. In [9], the author studied the existence of positive solutions for the non-resonant case by means of the fixed point index theory under sublinear conditions.

In [18], the authors investigated the existence and multiplicity results of positive solutions by using of the fixed point theorem for the fractional differential equation given by

$$ \left \{ \textstyle\begin{array}{@{}l} D^{\alpha}_{0+}u(t)=f(t,u(t)), \quad 0< t< 1,\\ u(0)=0,\qquad D^{\beta}_{0+}u(1)=a D^{\beta}_{0+}u(\xi), \end{array}\displaystyle \right . $$

(1.3)

where $1 <\alpha\leq2$, $0 \leq\beta\leq1$, $0 < \xi< 1$, $0 \leq a\leq1$ with $a\xi^{\alpha-\beta-2}< 1-\beta$, $0\leq \alpha-\beta-1$.

Recently, there are some papers dealing with the existence of solutions of fractional boundary value problem at resonance by using the coincidence degree theory due to Mawhin (see [19–22]). In [22], the authors investigated the following fractional three-point boundary value problem (BVP for short) at resonance:

$$ \left \{ \textstyle\begin{array}{@{}l} D^{\alpha}_{0+}u(t)+f(t,u(t),D^{\alpha -1}_{0+}u(t))=0 ,\quad 0< t< 1,\\ u(0)=0,\qquad u(1)=\eta u(\xi), \end{array}\displaystyle \right . $$

(1.4)

where $1 < \alpha\leq2$, $0 < \xi< 1$, $\eta\xi^{\alpha-1}= 1$, $D^{\alpha}_{0+}$ is the standard Riemann-Liouville derivative, $f:[0,1]\times\mathbb{R}^{2}\rightarrow\mathbb{R}$ is continuous. By using the coincidence degree theory, the existence of solutions for BVP (1.4) are obtained under certain growth conditions.

To the best of our knowledge, there are only very few papers dealing with the existence of positive solutions for resonant boundary value problems since the corresponding linear operator is non-reversible. For the case that α is an integer, some work has been done dealing with the existence of positive solutions for resonant boundary value problems by using Leggett-Williams norm-type theorem for coincidence (see [23–25]). Webb [26] established existence of positive solutions for second order boundary value problems at resonance by considering equivalent non-resonant perturbed problems with the same boundary conditions.

Inspired by the work mentioned above, in this paper we aim to establish the existence of positive solutions for resonant problem (1.1). The paper is organized as follows. Firstly, we reduce non-perturbed boundary value problems at resonance to equivalent non-resonant perturbed problems with the same boundary conditions. Then we derive the corresponding Green’s function and argue its properties. Finally, the existence and uniqueness results of positive solutions are obtained by using of the fixed point index and iterative technique.

2 Basic definitions and preliminaries

In this section, we present some preliminaries and lemmas. The definitions and properties of fractional derivative can be found in the literature [1–22].

Definition 2.1

The fractional integral of order $\alpha> 0$ of a function $u:(0,+\infty)\rightarrow R$ is given by

$$I^{\alpha}_{0+}u(t)=\frac{1}{\Gamma(\alpha)} \int^{t}_{0}(t-s)^{\alpha-1}u(s)\,ds $$

provided that the right-hand side is point-wise defined on $(0,+\infty)$.

Definition 2.2

The Riemann-Liouville fractional derivative of order $\alpha> 0$ of a function $u:(0,+\infty)\rightarrow R$ is given by

$$D^{\alpha}_{0+}u(t)=\frac{1}{\Gamma(n-\alpha)} \biggl(\frac{d}{dt} \biggr)^{n} \int^{t}_{0}(t-s)^{n-\alpha-1}u(s)\,ds, $$

where $n=[\alpha]+1$, $[\alpha]$denotes the integer part of number α, provided that the right-hand side is point-wise defined on $(0,+\infty)$.

Denote

$$ g(t)=\frac{\alpha-2}{\Gamma(\alpha-1)}+\sum^{+\infty}_{k=1} \frac {t^{k}}{\Gamma((k+1)\alpha-2)}. $$

(2.1)

It is easy to check that $g'(t)>0$ on $(0,+\infty)$, and

$$g(0)=\frac{\alpha-2}{\Gamma(\alpha-1)}< 0, \qquad\lim_{t\rightarrow +\infty}g(t)=+\infty. $$

Therefore, there exists a unique $b^{\ast}>0$ such that

$$ g\bigl(b^{\ast}\bigr)=0. $$

(2.2)

For the convenience in presentation, we here list the assumptions to be used throughout the paper.

(H₁):

$b\in(0, b^{\ast}]$ is a constant.

(H₂):

$f:[0,1]\times[0,+\infty)\rightarrow\mathbb{R}$ is continuous and

$$ f(t,x)+bx\geq0. $$

(2.3)

Set

$$ G_{b}(t)=t^{\alpha-1}E_{\alpha,\alpha}\bigl(bt^{\alpha} \bigr), $$

(2.4)

where

$$ E_{\alpha,\alpha}(x)=\sum^{+\infty}_{k=0} \frac{x^{k}}{\Gamma((k+1)\alpha )} $$

(2.5)

is the Mittag-Leffler function (see [1, 2]).

Next we consider the following boundary value problem:

$$ \left \{ \textstyle\begin{array}{@{}l} -D^{\alpha}_{0+}u(t)+bu(t)=f(t,u(t))+bu(t) , \quad 0< t< 1,\\ u(0)=0,\qquad u(1)=\eta u(\xi). \end{array}\displaystyle \right . $$

(2.6)

It is clear that (1.1) is equivalent to (2.6).

Set

$$\begin{aligned}& K_{0}(t,s)=\frac{1}{G_{b}(1)}\left \{ \textstyle\begin{array}{@{}l@{\quad}l} G_{b}(t)G_{b}(1-s), & 0\leq t\leq s\leq1 ,\\ G_{b}(t)G_{b}(1-s)-G_{b}(t-s)G_{b}(1), & 0\leq s\leq t\leq1; \end{array}\displaystyle \right . \end{aligned}$$

(2.7)

$$\begin{aligned}& q(s)=\frac{\eta K_{0}(\xi,s)}{ G_{b}(1)-\eta G_{b}(\xi)}; \end{aligned}$$

(2.8)

$$\begin{aligned}& K(t,s)=K_{0}(t,s)+G_{b}(t)q(s). \end{aligned}$$

(2.9)

Lemma 2.1

Suppose that (H₁) holds, and $y\in L[0,1]$. Then the unique solution of the problem

$$ \left \{ \textstyle\begin{array}{@{}l} -D^{\alpha}_{0+}u(t)+bu(t)=y(t) , \quad 0< t< 1,\\ u(0)=0,\qquad u(1)=\eta u(\xi), \end{array}\displaystyle \right . $$

(2.10)

is

$$u(t)= \int_{0}^{1}K(t,s)y(s)\,ds. $$

Proof

By [1, 2], we know that the solution of (2.10) can be expressed by

$$u(t)=- \int^{t}_{0}G_{b}(t-s)y(s)\,ds+c_{1}G_{b}(t)+c_{2}G'_{b}(t). $$

By $u(0)=0$, we have $c_{2}=0$.

On the other hand, we have

$$\begin{aligned}& u(1)=- \int^{1}_{0}G_{b}(1-s)y(s)\,ds+c_{1}G_{b}(1), \end{aligned}$$

(2.11)

$$\begin{aligned}& u(\xi)=- \int^{\xi}_{0}G_{b}(\xi-s)y(s)\,ds+c_{1}G_{b}( \xi). \end{aligned}$$

(2.12)

Noting that $\eta\xi^{\alpha-1}= 1$, and $0 < \xi< 1$, we have

$$ G_{b}(1)-\eta G_{b}(\xi)=\sum ^{+\infty}_{k=0}\frac{b^{k}[1-\eta \xi^{(k+1)\alpha-1}]}{\Gamma((k+1)\alpha)}=\sum ^{+\infty}_{k=0}\frac{b^{k}(1- \xi^{k\alpha})}{\Gamma((k+1)\alpha)}>0. $$

(2.13)

Equations (2.11) and (2.12) yield

$$c_{1}=\frac{\int^{1}_{0}G_{b}(1-s)y(s)\,ds-\eta\int^{\xi}_{0}G_{b}(\xi -s)y(s)\,ds}{ G_{b}(1)-\eta G_{b}(\xi)}. $$

Therefore, the solution of (2.10) is

$$\begin{aligned} u(t)={}&{-} \int^{t}_{0}G_{b}(t-s)y(s)\,ds+ \frac{\int ^{1}_{0}G_{b}(1-s)y(s)\,ds-\eta\int^{\xi}_{0}G_{b}(\xi-s)y(s)\,ds}{ G_{b}(1)-\eta G_{b}(\xi)}G_{b}(t) \\ ={}&\frac{\int^{1}_{0}G_{b}(t)G_{b}(1-s)y(s)\,ds-\int^{t}_{0} G_{b}(1)G_{b}(t-s)y(s)\,ds}{ G_{b}(1)} \\ &{}-\frac{\int^{1}_{0}G_{b}(1-s)y(s)\,ds}{ G_{b}(1)}G_{b}(t)+\frac{\int^{1}_{0}G_{b}(1-s)y(s)\,ds}{ G_{b}(1)-\eta G_{b}(\xi)}G_{b}(t) \\ &{}-\frac{\eta \int^{\xi}_{0}G_{b}(\xi-s)y(s)\,ds}{ G_{b}(1)-\eta G_{b}(\xi)}G_{b}(t) \\ ={}& \int_{0}^{1}K_{0}(t,s)y(s)\,ds + \frac{\eta \int^{1}_{0}G_{b}(\xi)G_{b}(1-s)y(s)\,ds}{ G_{b}(1)[G_{b}(1)-\eta G_{b}(\xi)]}G_{b}(t) \\ &{}-\frac{\eta \int^{\xi}_{0}G_{b}(1)G_{b}(\xi-s)y(s)\,ds}{G_{b}(1)[ G_{b}(1)-\eta G_{b}(\xi)]}G_{b}(t) \\ ={}& \int_{0}^{1}K(t,s)y(s)\,ds. \end{aligned}$$

This completes the proof. □

Lemma 2.2

Suppose that (H₁) holds. The function $K(t,s)$ has the following properties:

• (1) $K(t,s) > 0, \forall t,s\in(0,1)$;

• (2) $\omega_{2}(s)t^{\alpha-1}\leq K(t,s) \leq\omega_{1}(s)t^{\alpha-1}, \forall t,s\in[0,1]$, where

  $$ \omega_{1}(s)=G_{b}(1-s)+G_{b}(1)q(s), \qquad \omega_{2}(s)=\frac {q(s)}{\Gamma(\alpha)}. $$

  (2.14)

Proof

It is clear that we just need to prove that (2) holds.

By (2.4), we can get

$$\begin{aligned}& \frac{t^{\alpha-1}}{\Gamma(\alpha)}\leq G_{b}(t)=t^{\alpha-1} \sum ^{+\infty}_{k=0}\frac{b^{k}t^{\alpha k}}{\Gamma((k+1)\alpha)} \leq t^{\alpha-1}G_{b}(1), \quad t \in [0,1], \end{aligned}$$

(2.15)

$$\begin{aligned}& G'_{b}(t)=\sum^{+\infty}_{k=0} \frac{b^{k}t^{(k+1)\alpha-2}}{\Gamma ((k+1)\alpha-1)}>0,\quad t \in (0,1], \end{aligned}$$

(2.16)

and

$$\begin{aligned} G''_{b}(t)&= t^{\alpha-3} \Biggl[\frac{\alpha-2}{\Gamma(\alpha-1)}+\sum^{+\infty}_{k=1} \frac{b^{k}t^{k\alpha}}{\Gamma((k+1)\alpha-2)} \Biggr] \\ &= t^{\alpha-3}g\bigl(bt^{\alpha}\bigr)< t^{\alpha-3}g(b)\leq t^{\alpha-3}g\bigl(b^{\ast}\bigr)=0, \quad t \in (0,1), \end{aligned}$$

(2.17)

which implies $G_{b}(t)$ is strictly increasing on $[0,1]$, and $G'_{b}(t)$ is strictly decreasing on $(0,1]$.

By (2.15), we have

$$\begin{aligned} K(t,s)&= K_{0}(t,s)+G_{b}(t)q(s)\leq \frac{G_{b}(t)G_{b}(1-s)}{G_{b}(1)}+G_{b}(t)q(s) \\ &\leq G_{b}(1-s)t^{\alpha-1}+t^{\alpha-1}G_{b}(1)q(s)= \omega_{1}(s)t^{\alpha-1}. \end{aligned}$$

(2.18)

On the other hand, when $0< t\leq s< 1$, noticing $G_{b}(0)=0$, and the monotonicity of $G_{b}(t)$, it is clear that

$$ G_{b}(t)G_{b}(1-s)>0. $$

(2.19)

When $0< s< t< 1$, we have

$$\begin{aligned} &\frac{\partial }{\partial s}\bigl[G_{b}(t)G_{b}(1-s)-G_{b}(t-s)G_{b}(1) \bigr] \\ &\quad=G'_{b}(t-s)G_{b}(1)-G_{b}(t)G'_{b}(1-s) \\ &\quad \geq \bigl[G_{b}(1)-G_{b}(t)\bigr]G'_{b}(1-s). \end{aligned}$$

(2.20)

Integrating (2.20) with respect to s, we obtain

$$\begin{aligned} & G_{b}(t)G_{b}(1-s)-G_{b}(t-s)G_{b}(1) \\ &\quad\geq \int^{s}_{0}\bigl[G_{b}(1)-G_{b}(t) \bigr]G'_{b}(1-\tau)\,d\tau \\ &\quad=\bigl[G_{b}(1)-G_{b}(t)\bigr] \bigl[G_{b}(1)-G_{b}(1-s) \bigr]>0. \end{aligned}$$

(2.21)

By (2.7), (2.19), and (2.21), we get $K_{0}(t,s) > 0, \forall t,s\in(0,1)$. Then

$$\begin{aligned} K(t,s)&=K_{0}(t,s)+G_{b}(t)q(s)\geq G_{b}(t)q(s) \\ &\geq \frac{t^{\alpha-1}}{\Gamma(\alpha)}q(s)=\omega_{2}(s)t^{\alpha-1}> 0, \quad \forall t,s\in(0,1). \end{aligned}$$

This completes the proof. □

Let $E=C[0,1]$ be endowed with the maximum norm $\| u\| = \max_{0\leq t\leq1}| u(t)|$, θ is the zero element of E, $B_{r}=\{u\in E : \| u\|< r\}$. Define a cone P by

$$P= \bigl\{ u\in E : u(t)\geq0, t\in[0,1] \bigr\} . $$

Let

$$\begin{aligned}& Au(t)= \int_{0}^{1}K(t,s) \bigl[f\bigl(s,u(s)\bigr)+bu(s) \bigr]\,ds. \end{aligned}$$

(2.22)

$$\begin{aligned}& T u(t)= \int_{0}^{1}K(t,s) u(s)\,ds. \end{aligned}$$

(2.23)

By means of Lemma 2.2 and the Arzela-Ascoli theorem, we can get $A: P\rightarrow P$ is completely continuous, $T: P\rightarrow P$ is completely continuous linear operator. By virtue of the Krein-Rutmann theorem and Lemma 2.2, we have the spectral radius $r(T)> 0$ and T has a positive eigenfunction corresponding to its first eigenvalue $\lambda_{1}=(r(T))^{-1}$. Since $\lambda=0$ is the eigenvalue of the linear problems (1.2), and $t^{\alpha-1}$ is the corresponding eigenfunction, we have the following lemma.

Lemma 2.3

Suppose that (H₁) holds, then the first eigenvalue of T is $\lambda_{1}=b$, and $\varphi_{1}(t)=t^{\alpha-1}$ is the positive eigenfunction corresponding to $\lambda_{1}$, that is, $\varphi_{1}=bT\varphi_{1}$.

Lemma 2.4

[27]

Let P be a cone in a Banach space E, and Ω be a bounded open set in E. Suppose that A: $\overline{\Omega}\cap P\rightarrow P$ is a completely continuous operator. If there exists $u_{0}\in P$ with $u_{0}\neq\theta$ such that

$$u-Au\neq\lambda u_{0},\quad\forall\lambda\geq0,x\in\partial\Omega\cap P, $$

then $i(A,\Omega\cap P,P)=0$.

Lemma 2.5

[27]

Let P be a cone in a Banach space E, and Ω be a bounded open set in E. Suppose that A: $\overline{\Omega}\cap P\rightarrow P$ is a completely continuous operator. If

$$Au\neq\lambda u,\quad\forall\lambda\geq1,u\in\partial\Omega\cap P, $$

then $i(A,\Omega\cap P,P)=1$.

3 The uniqueness result

Theorem 3.1

Assume that there exists $\lambda\in(0,b)$ such that

$$\bigl|f(t,u)+bu-f(t,v)-bv\bigr|\leq \lambda|u-v|, \quad \textit{for } t \in[0,1], u, v \in[0,\infty), $$

then (1.1) has a unique nonnegative solution.

Proof

Firstly, we will prove A has fixed point in P.

Set

$$ Q=\bigl\{ u\in P: \exists l_{1}, l_{2}>0 \mbox{ such that } l_{1}t^{\alpha-1} \leq u(t)\leq l_{2}t^{\alpha-1}\bigr\} . $$

(3.1)

For any $u \in P\setminus \{\theta\}$, let

$$ l_{i}(u)= \int_{0}^{1}\omega_{i}(s) u(s)\,ds,\quad i=1,2. $$

(3.2)

By Lemma 2.2, it is obvious that $l_{1}(u), l_{2}(u)> 0$, and

$$ l_{1}(u)t^{\alpha-1} \leq(Tu) (t)\leq l_{2}(u)t^{\alpha-1}, $$

(3.3)

that is,

$$T:P\setminus\{\theta\} \rightarrow Q. $$

For any $u_{0}\in P\setminus\{\theta\}$, let

$$ u_{n}=A(u_{n-1}), \quad n=1,2,\ldots. $$

(3.4)

We may suppose that $u_{1}-u_{0} \neq\theta$ (otherwise, the proof is finished). Then there exists $l_{2}(|u_{1}-u_{0}|)> 0$, such that

$$ T\bigl(|u_{1}-u_{0}|\bigr)\leq l_{2}\bigl(|u_{1}-u_{0}|\bigr)t^{\alpha -1}=l_{2}\bigl(|u_{1}-u_{0}|\bigr) \varphi_{1}. $$

(3.5)

Thus

$$\begin{aligned} |u_{2}-u_{1}|&=\biggl\vert \int_{0}^{1}K(t,s) \bigl[f\bigl(s,u_{1}(s) \bigr)+bu_{1}(s)-f\bigl(s,u_{0}(s)\bigr)-bu_{0}(s) \bigr]\,ds\biggr\vert \\ &\leq \int_{0}^{1}K(t,s) \bigl|f\bigl(s,u_{1}(s) \bigr)+bu_{1}(s)-f\bigl(s,u_{0}(s)\bigr)-bu_{0}(s)\bigr|\,ds \\ &\leq \lambda \int_{0}^{1}K(t,s) \bigl|u_{1}(s)-u_{0}(s)\bigr|\,ds \\ & =\lambda T\bigl(|u_{1}-u_{0}|\bigr)\leq\lambda l_{2}\bigl(|u_{1}-u_{0}|\bigr) \varphi_{1}, \end{aligned}$$

(3.6)

$$\begin{aligned} |u_{3}-u_{2}|&=\biggl\vert \int_{0}^{1}K(t,s) \bigl[f\bigl(s,u_{2}(s) \bigr)+bu_{2}(s)-f\bigl(s,u_{1}(s)\bigr)-bu_{1}(s) \bigr]\,ds\biggr\vert \\ &\leq \int_{0}^{1}K(t,s) \bigl|f\bigl(s,u_{2}(s) \bigr)+bu_{2}(s)-f\bigl(s,u_{1}(s)\bigr)-bu_{1}(s)\bigr|\,ds \\ &\leq \lambda \int_{0}^{1}K(t,s) \bigl|u_{2}(s)-u_{1}(s)\bigr|\,ds \\ &\leq \lambda^{2} l_{2}\bigl(|u_{1}-u_{0}|\bigr)T \varphi_{1}=\frac{\lambda^{2} }{b}l_{2}\bigl(|u_{1}-u_{0}|\bigr) \varphi_{1},\\ &\cdots. \end{aligned}$$

(3.7)

By induction, we can get

$$ |u_{n+1}-u_{n}|\leq \biggl(\frac{\lambda}{b} \biggr)^{n-1}\lambda l_{2}\bigl(|u_{1}-u_{0}|\bigr) \varphi_{1}. $$

(3.8)

Then, for any $n, m \in\mathbb{N}$, we have

$$\begin{aligned} |u_{n+m}-u_{m}|&\leq |u_{n+m}-u_{n+m-1}|+ \cdots+|u_{n+1}-u_{n}| \\ &\leq \biggl[ \biggl(\frac{\lambda}{b} \biggr)^{n+m-1}+\cdots+ \biggl( \frac{\lambda}{b} \biggr)^{n} \biggr]\lambda l_{2}\bigl(|u_{1}-u_{0}|\bigr) \varphi_{1} \\ & \leq\frac{ (\frac{\lambda }{b} )^{n}}{1-\frac{\lambda}{b}}\lambda l_{2}\bigl(|u_{1}-u_{0}|\bigr) \varphi_{1}=\frac{\lambda^{n+1}l_{2}(|u_{1}-u_{0}|)}{b ^{n-1}(b-\lambda)} \varphi_{1}. \end{aligned}$$

(3.9)

So,

$$ \|u_{n+m}-u_{m}\|\leq \frac{\lambda^{n+1}l_{2}(|u_{1}-u_{0}|)}{b ^{n-1}(b-\lambda)}\rightarrow0,\quad n \rightarrow \infty, $$

(3.10)

which implies $\{u_{n}\}$ is a Cauchy sequence. Therefore, there exists a $u^{\ast}\in P$, such that $\{u_{n}\}$ converges to $u^{\ast}$. Clearly, $u^{\ast}$ is a fixed point of A.

In the following, we will prove the fixed point of A is unique.

Suppose $v\neq u^{\ast}$ is a fixed point of A. Then there exists $l_{2}(|u^{\ast}-v|)> 0$, such that

$$ T\bigl(\bigl|u^{\ast}-v\bigr|\bigr)\leq l_{2}\bigl(\bigl\vert u^{\ast}-v\bigr\vert \bigr)\varphi_{1}. $$

(3.11)

Then

$$\begin{aligned} \bigl|Au^{\ast}-Av\bigr|&=\biggl\vert \int_{0}^{1}K(t,s)\bigl[f\bigl(s,u^{\ast }(s) \bigr)+bu^{\ast}(s)-f\bigl(s,v(s)\bigr)-bv(s)\bigr]\,ds\biggr\vert \\ &\leq \int_{0}^{1}K(t,s) \bigl|f\bigl(s,u^{\ast}(s) \bigr)+bu^{\ast}(s)-f\bigl(s,v(s)\bigr)-bv(s)\bigr|\,ds \\ & \leq\lambda \int_{0}^{1}K(t,s) \bigl|u^{\ast}(s)-v(s)\bigr|\,ds \\ &=\lambda T\bigl(\bigl|u^{\ast}-v\bigr|\bigr) \leq\lambda l_{2} \bigl(\bigl|u^{\ast}-v\bigr|\bigr) \varphi_{1}. \end{aligned}$$

(3.12)

By induction, we can get

$$ \bigl\vert A^{n}u^{\ast}-A^{n}v\bigr\vert \leq \biggl(\frac{\lambda}{b} \biggr)^{n-1}\lambda l_{2}\bigl(\bigl\vert u^{\ast}-v\bigr\vert \bigr) \varphi_{1}. $$

(3.13)

So,

$$ \bigl\Vert u^{\ast}-v\bigr\Vert =\bigl\| A^{n}u^{\ast}-A^{n}v \bigr\| \leq \biggl(\frac{\lambda}{b} \biggr)^{n-1}\lambda l_{2} \bigl(\bigl|u^{\ast}-v\bigr|\bigr)\rightarrow0, \quad n\rightarrow \infty. $$

(3.14)

Consequently, the fixed point of A is unique.

This completes the proof. □

Remark 3.1

The unique nonnegative solution $u^{\ast}$ of (1.1) can be approximated by the iterative schemes: for any $u_{0}\in P\setminus\{\theta\}$, let

$$u_{n}=A(u_{n-1}), \quad n=1,2,\ldots, $$

then $u_{n}\rightarrow u^{\ast}$.

Remark 3.2

If $f(t,0)\equiv0$ on $[0,1]$, then θ is the unique solution of (1.1) in P; If $f(t,0)\not\equiv0$ on $[0,1]$, then the unique solution $u^{\ast }$ is a positive solution.

4 Existence of positive solutions

Theorem 4.1

Assume that (H₁), (H₂), and the following assumptions hold:

$$\begin{aligned}& \liminf_{x\rightarrow 0+}\min_{t\in[0,1]}\frac{f(t,x)}{x} >0, \end{aligned}$$

(4.1)

$$\begin{aligned}& \limsup_{x\rightarrow +\infty}\max_{t\in[0,1]}\frac{f(t,x)}{x} < 0. \end{aligned}$$

(4.2)

Then (1.1) has at least one positive solution.

Proof

It follows from (4.1) that there exists $r_{1} > 0$ such that

$$ f(t,x)\geq0,\quad \forall(t,x)\in[0,1]\times[0,r_{1}]. $$

(4.3)

Thus, for any $u\in\partial B_{r_{1}}\cap P$, we have

$$ Au(t)= \int_{0}^{1}K(t,s)\bigl[f\bigl(s,u(s)\bigr)+bu(s) \bigr]\,ds \geq bTu(t). $$

(4.4)

We may suppose that A has no fixed points on $\partial B_{r_{1}}\cap P$ (otherwise, the proof is finished). Now we show that

$$ u-Au \neq\mu\varphi_{1},\quad\forall u \in\partial B_{r_{1}} \cap P,\mu > 0. $$

(4.5)

If otherwise, there exist $u_{1}\in\partial B_{r_{1}}\cap P$ and $\mu_{0}> 0$ such that

$$u_{1}-Au_{1}= \mu_{0} \varphi_{1}. $$

Then

$$u_{1}=Au_{1}+\mu_{0} \varphi_{1}\geq \mu_{0} \varphi_{1}. $$

Denote

$$ \mu^{\ast}=\sup\{\mu:u_{1}\geq\mu \varphi_{1} \}. $$

(4.6)

It is clear that $\mu^{\ast}\geq \mu_{0}$ and $u_{1}\geq\mu^{\ast}\varphi_{1}$. Since $T(P)\subset P$, we have $b Tu_{1}\geq\mu^{\ast}b T\varphi_{1}=\mu^{\ast}\varphi_{1}$. Then

$$u_{1}=Au_{1}+\mu_{0}\varphi_{1}\geq bTu_{1}+\mu_{0}\varphi_{1}\geq \bigl( \mu^{\ast}+\mu_{0}\bigr)\varphi_{1}, $$

contradicts the definition of $\mu^{\ast}$. Hence (4.5) holds and we see from Lemma 2.4 that

$$ i(A,B_{r_{1}}\cap P,P)=0. $$

(4.7)

On the other hand, it follows from (4.2) that there exist $0 < \sigma< 1$ and $r_{2} > r_{1}$ such that

$$ f(t,x)\leq (\sigma-1)bx, \quad\forall t\in[0,1], x\geq r_{2}. $$

(4.8)

Let $T_{1}u=\sigma b Tu$. Then $T_{1}$ is a bounded linear operator and $T_{1}(P)\subset P$. Set

$$ W= \{u\in P\mid u=\mu Au,0 \leq\mu\leq1 \}. $$

(4.9)

In the following, we will prove that W is bounded.

For any $u \in W$, set $\tilde{u}(t)=\min\{u(t),r_{2}\}$. Then

$$f\bigl(t,u(t)\bigr)+bu(t)\leq \sigma b u(t)+f\bigl(t,\tilde{u}(t)\bigr)+b \tilde{u}(t). $$

Therefore,

$$u(t)=\mu Au(t)\leq Au(t) \leq\sigma b Tu(t)+A\tilde{u}(t) \leq T_{1}u(t)+M, $$

where

$$ M=\max_{(t,x)\in[0,1]\times[0, r_{2}]}\bigl\{ f(t,x)+bx\bigr\} \int_{0}^{1}\omega_{1}(s)\,ds. $$

(4.10)

Thus $(I-T_{1})u(t)\leq M,t\in[0,1]$. Noticing b is the first eigenvalue of T and $0 < \sigma< 1$, we have $(r(T_{1}))^{-1}=\sigma^{-1}>1$. So the inverse operator of $I-T_{1}$ exists, and

$$(I-T_{1})^{-1}=I+T_{1}+T_{1}^{2}+ \cdots+T_{1}^{n}+\cdots. $$

It follows from $T_{1}(P)\subset P$ that $(I-T_{1})^{-1}(P)\subset P$. We have

$$u(t)\leq(I-T_{1})^{-1}M\leq M\bigl\| (I-T_{1})^{-1} \bigr\| , \quad t\in[0,1], $$

which implies W is bounded.

Select $r_{3}> \max\{r_{2},M\| (I-T_{1})^{-1} \|\}$. Then by Lemma 2.5, we have

$$ i(A,B_{r_{3}}\cap P,P)=1. $$

(4.11)

By (4.7) and (4.11), we have

$$i\bigl(A,(B_{r_{3}}\backslash\bar{B}_{r_{1}})\cap P,P \bigr)=i(A,B_{r_{3}}\cap P,P)-i(A,B_{r_{1}}\cap P,P)=1, $$

which implies that A has at least one fixed point on $(B_{r_{3}}\backslash\bar{B}_{r_{1}})\cap P$. This means that BVP (1.1) has at least one positive solution.

This completes the proof. □

Theorem 4.2

Assume that (H₁), (H₂), and the following assumptions hold:

$$ \limsup_{x\rightarrow 0+}\max_{t\in[0,1]}\frac{f(t,x)}{x} < 0, $$

(4.12)

and $f(t,0)\not\equiv0$ on $[0,1]$. Then (1.1) has at least one positive solution.

Proof

It follows from (4.12) that there exists $r_{1} > 0$ such that

$$ f(t,x)\leq0,\quad \forall(t,x)\in[0,1]\times[0,r_{1}]. $$

(4.13)

Denote $T_{2}u=bTu$. Obviously, $r(T_{2})=1$.

We may suppose that A has no fixed points on $\partial B_{r_{1}}\cap P$ (otherwise, the proof is finished). In the following, we prove that

$$ Au\neq\mu u, \quad\forall u \in\partial B_{r_{1}}\cap P,\mu> 1. $$

(4.14)

If otherwise, there exist $u_{1}\in \partial B_{r_{1}}\cap P,\mu_{0}> 1$, such that $Au_{1}= \mu_{0} u_{1}$. It is clear that $\mu_{0} u_{1}=Au_{1}\leq T_{2}u_{1}$, and

$$\mu_{0}^{n} u_{1}\leq T_{2}^{n}u_{1}, \quad n=1,2,\ldots. $$

Therefore,

$$\mu_{0}^{n} u_{1}\leq T_{2}^{n}u_{1} \leq\bigl\| T_{2}^{n}\bigr\| \| u_{1}\|. $$

Thus $r(T_{2})=\lim_{n\rightarrow+\infty}\sqrt[n]{\| T_{2}^{n}\|}\geq \mu_{0} > 1$, which contradicts $r(T_{2})=1$. We have from Lemma 2.5

$$ i(A,B_{r_{1}}\cap P,P)=1. $$

(4.15)

Since $f(t,0)\not\equiv0$ on $[0,1]$, clearly we have $A\theta\neq \theta$, here θ is the zero element of E. So (4.15) implies that the problem (1.1) has at least one positive solution. □

Remark 4.1

Suppose u is a positive solution of (1.1), then there exist $l_{1}, l_{2}>0$, such that

$$l_{1}t^{\alpha-1} \leq u(t)\leq l_{2}t^{\alpha-1}. $$

Example 4.1

A 3-point boundary value problem at resonance

Consider the following problem:

$$ \left \{ \textstyle\begin{array}{@{}l} D^{\frac{3}{2}}_{0+}u(t)+f(t,u(t))=0 , \quad 0< t< 1,\\ u(0)=0,\qquad u(1)=2 u(\frac{1}{4}). \end{array}\displaystyle \right . $$

(4.16)

Since $\Gamma(\cdot)$ is strictly increasing on $[2,+\infty)$, for any $t\in[0,+\infty)$, we have

$$\begin{aligned} g(t)&=-\frac{1}{2\sqrt{\pi}}+\sum^{+\infty}_{k=1} \frac {t^{k}}{\Gamma(\frac{3}{2}k-\frac{1}{2})} =-\frac{1}{2\sqrt{\pi}}+t+\sum^{+\infty}_{k=2} \frac{t^{k}}{\Gamma(\frac {3}{2}k-\frac{1}{2})} \\ &\leq-\frac{1}{2\sqrt{\pi}}+t+\sum^{+\infty}_{k=2} \frac{t^{k}}{\Gamma(k)} =-\frac{1}{2\sqrt{\pi}}+t \Biggl[1+\sum ^{+\infty}_{k=1}\frac {t^{k}}{k!} \Biggr] \\ &=-\frac{1}{2\sqrt{\pi}}+te^{t}. \end{aligned}$$

Noticing $\frac{1}{2\sqrt{\pi}}\thickapprox0.282$, $\frac{1}{5}e^{\frac {1}{5}}\thickapprox0.243$, we have $g(\frac{1}{5})< 0$. Therefore $b^{\ast}> b_{1}:=\frac{1}{5}$.

Let

$$f(t,x)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{1}{4}(1+t)(1-2b)x,& (t,x)\in [0,1]\times[0,4),\\ \frac{1}{2}(1+t)(\sqrt{x}-bx),& (t,x)\in[0,1]\times[4,+\infty), \end{array}\displaystyle \right . $$

where $b\in(0,b_{1}]$. It is clear that (H₁) and (H₂) hold. Moreover,

$$\begin{aligned}& \liminf_{u\rightarrow 0+}\min_{t\in[0,1]}\frac{f(t,u)}{u}= \frac{1-2b}{4} >0, \\& \limsup_{x\rightarrow +\infty}\max_{t\in[0,1]}\frac{f(t,x)}{x}=- \frac{b}{2} < 0. \end{aligned}$$

Therefore the assumptions of Theorem 4.1 are satisfied. Thus Theorem 4.1 ensures that (4.16) has at least one positive solution.