# The existence of positive solutions for p-Laplacian boundary value problems at resonance

## Abstract

By using the Leggett-Williams norm-type theorem due to O’Regan and Zima and constructing suitable Banach spaces and operators, we investigate the existence of positive solutions for fractional p-Laplacian boundary value problems at resonance. An example is given to illustrate the main results.

## 1 Introduction

Boundary value problems at resonance have attracted more and more attention. Many authors studied the existence of solutions for these problems by using Mawhin’s continuous theorem  and its extension obtained by Ge and Ren ; see  and the references cited therein. By using Leggett-Williams norm-type theorems due to O’Regan and Zima , the existence of positive solutions for the boundary value problems at resonance with a linear derivative operator has been investigated (see ). To the best of our knowledge, there is no paper to show the existence of a positive solution for boundary value problems with a nonlinear derivative operator (for instance, p-Laplacian operator) at resonance by using Leggett-Williams norm-type theorems. Motivated by the excellent results mentioned above, we will discuss the existence of positive solutions for the p-Laplacian boundary value problem

$$\left \{ \textstyle\begin{array}{l} {}^{\mathrm{C}}D_{0^{+}}^{\beta}[\varphi_{p}({}^{\mathrm{C}}D_{0^{+}}^{\alpha }x)](t)=f(t,({}^{\mathrm{C}}D_{0^{+}}^{\alpha}x)(t)),\quad t\in(0,1), \\ ({}^{\mathrm{C}}D_{0^{+}}^{\alpha}x)(0)=({}^{\mathrm{C}}D_{0^{+}}^{\alpha }x)(1),\qquad x^{(i)}(0)=0, \quad i=0,1,2,\ldots,n-1, \end{array}\displaystyle \right .$$
(1.1)

where $$n-1<\alpha\leq n$$, $$0<\beta<1$$, $$\varphi _{p}(s)=|s|^{p-2}s$$, $$p>1$$, $${}^{\mathrm{C}}D_{0^{+}}^{\beta}$$ is the Caputo fractional derivative (see [29, 30]).

## 2 Preliminaries

For convenience, we introduce some notations and a theorem. For more details see .

Assume that X, Y are real Banach spaces. A linear mapping $$L:\operatorname{dom}L\subset X\rightarrow Y$$ is a Fredholm operator of index zero (i.e. $$\operatorname{dim}\operatorname{Ker}L= \operatorname {codim}\operatorname{Im}L<+\infty$$ and ImL is closed in Y) and an operator $$N: X\rightarrow Y$$ is nonlinear. $$P:X\rightarrow X$$ and $$Q:Y\rightarrow Y$$ are projectors with $$\operatorname{Im}P= \operatorname{Ker}L$$ and $$\operatorname {Ker}Q= \operatorname{Im}L$$. $$J:\operatorname{Im}Q\rightarrow \operatorname{Ker}L$$ is a isomorphism since $$\operatorname{dim} \operatorname {Im}Q=\operatorname{dim} \operatorname{Ker}L$$. Denote by $$L_{P}$$ the restriction of L to $$\operatorname{Ker}P \cap\operatorname {dom}L\rightarrow\operatorname{Im}L$$ and its inverse by $$K_{P}$$. So, x is a solution of $$Lx=Nx$$ if and only if it satisfies $$x=(P+JQN)x+K_{P}(I-Q)Nx$$.

Let $$C\subset X$$ be a cone, $$\gamma:X\rightarrow C$$ be a retraction, $$\Psi:=P+JQN+K_{P}(I-Q)N$$ and $$\Psi_{\gamma}:=\Psi \circ\gamma$$.

### Theorem 2.1



Let $$\Omega_{1}$$, $$\Omega_{2}$$ be open bounded subsets of X with $$\overline{\Omega}_{1}\subset \Omega_{2}$$ and $$C\cap(\overline{\Omega}_{2}\setminus \Omega_{1})\neq\emptyset$$. Assume that $$L:\operatorname {dom}L\subset X\rightarrow Y$$ is a Fredholm operator of index zero and the following conditions are satisfied.

1. (C1)

$$QN:X\rightarrow Y$$ is continuous and bounded and $$K_{P}(I-Q)N:X\rightarrow X$$ is compact on every bounded subset of X;

2. (C2)

$$Lx\neq\lambda Nx$$ for all $$x\in C\cap\partial\Omega _{2}\cap\operatorname{dom}L$$ and $$\lambda\in(0,1)$$;

3. (C3)

γ maps subsets of $$\overline{\Omega}_{2}$$ into bounded subsets of C;

4. (C4)

$$d_{B}([I-(P+JQN)\gamma]|_{\operatorname{Ker}L},\operatorname {Ker}L\cap\Omega _{2},0)\neq0$$, where $$d_{B}$$ stands for the Brouwer degree;

5. (C5)

there exists $$u_{0}\in C\setminus\{0\}$$ such that $$\|x\| \leq\sigma(u_{0})\|\Psi x\|$$ for $$x \in C(u_{0})\cap\partial \Omega_{1}$$, where $$C(u_{0})=\{ x\in C:\mu u_{0}\preceq x\textit{ for some }\mu> 0\}$$ and $$\sigma(u_{0})$$ is such that $$\|x+u_{0}\|\geq\sigma(u_{0})\|x\|$$ for every $$x\in C$$;

6. (C6)

$$(P+JQN)\gamma(\partial\Omega_{2})\subset C$$;

7. (C7)

$$\Psi_{\gamma}(\overline{\Omega}_{2}\setminus\Omega _{1})\subset C$$.

Then the equation $$Lx=Nx$$ has at least one solution in the set $$C\cap(\overline{\Omega}_{2}\setminus\Omega_{1})$$.

Now, we present some fundamental facts on the fractional calculus theory which can be found in [29, 30].

### Definition 2.1

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

$$I_{0^{+}}^{\alpha}y(t)=\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}y(s)\,ds,$$

provided the right-hand side is pointwise defined on $$(0,\infty)$$.

### Definition 2.2

The Caputo fractional derivative of order $$\delta>0$$ of a function $$y:(0,\infty)\rightarrow\mathbb{R}$$ is given by

$${}^{\mathrm{C}}D^{\delta}_{0^{+}}y(t)=\frac{1}{\Gamma(n-\delta)} \int _{0}^{t}(t-s)^{n-\delta -1}y^{(n)}(s) \,ds,$$

provided that the right-hand side is pointwise defined on $$(0,\infty)$$, where $$n=[\delta]+1$$.

### Lemma 2.1

[29, 30]

Assume $$f\in L[0,1]$$, $$q> p\geq0$$, $$q>1$$, then

$${}^{\mathrm{C}}D_{0^{+}}^{p}I_{0^{+}}^{q}f(t)=I_{0^{+}}^{q-p}f(t), \qquad {}^{\mathrm{C}}D_{0^{+}}^{p}I_{0^{+}}^{p}f(t)=f(t).$$

### Lemma 2.2

[29, 30]

Assume $$p>0$$, then

$$I_{0^{+}}^{p} {{}^{\mathrm{C}}D_{0^{+}}^{p}}f(t)=f(t)+c_{0}+c_{1}t+ \cdots+c_{n-1}t^{n-1},$$

where n is an integer and $$n-1< p\leq n$$.

Since $${}^{\mathrm{C}}D_{0^{+}}^{\beta}[\varphi_{p}({}^{\mathrm{C}}D_{0^{+}}^{\alpha}\cdot)]$$ is a nonlinear operator, we cannot solve the problem (1.1) by Theorem 2.1. Based on this, we prove the following lemma.

### Lemma 2.3

$$u(t)$$ is a solution of the following problem:

$$\left \{ \textstyle\begin{array}{l} ({}^{\mathrm{C}}D_{0^{+}}^{\beta}u)(t)=f(t,\varphi_{q}(u(t))),\quad t\in[0,1], \\ u(0)=u(1), \end{array}\displaystyle \right .$$
(2.1)

if and only if $$x(t)$$ is a solution of (1.1), where $$x(t)=I_{0^{+}}^{\alpha}\varphi_{q}(u(t))$$, $$\frac{1}{p}+\frac{1}{q}=1$$.

### Proof

Assume that $$u(t)$$ is a solution of the problem (2.1) and $$x(t)=I_{0^{+}}^{\alpha}\varphi_{q}(u(t))$$. Then $$u(t)=[\varphi_{p}({}^{\mathrm{C}}D_{0^{+}}^{\alpha}x)](t)$$ and $$x^{(i)}(0)=0$$, $$i=0,1,2,\ldots,n-1$$. Replaces $$u(t)$$ with $$[\varphi_{p}({}^{\mathrm{C}}D_{0^{+}}^{\alpha}x)](t)$$ in (2.1), we can see that $$x(t)$$ is a solution of (1.1).

On the other hand, if $$x(t)$$ is a solution of (1.1) and $$u(t)=[\varphi _{p}({}^{\mathrm{C}}D_{0^{+}}^{\alpha}x)](t)$$, substituting $$u(t)$$ for $$[\varphi _{p}({}^{\mathrm{C}}D_{0^{+}}^{\alpha}x)](t)$$ in (1.1), we can see that $$u(t)$$ satisfies (2.1). □

In this paper, we will always suppose that $$f \in[0,1]\times \mathbb{R}\rightarrow\mathbb{R}$$ is continuous, $$p>1$$, $$\varphi_{p}(s)=s\cdot|s|^{p-2}$$, $$\frac{1}{p}+\frac {1}{q}=1$$, $$\alpha>0$$, $$0<\beta<1$$.

## 3 Main result

Let $$X=Y=C[0,1]$$ with the norm $$\|u\|=\max_{t\in[0,1]}|u(t)|$$. Take a cone

$$C=\bigl\{ u(t)\in X\mid u(t)\geq0, t\in[0,1]\bigr\} .$$

Define operator $$L:\operatorname{dom}L\subset X\rightarrow Y$$ and $$N:X\rightarrow Y$$ as follows:

$$(Lu) (t)= \bigl({}^{\mathrm{C}}D_{0^{+}}^{\beta}u \bigr) (t), \qquad (Nu) (t)=f \bigl(t,\varphi _{q}\bigl(u(t)\bigr) \bigr),$$

where

$$\operatorname{dom}L= \bigl\{ u(t) \mid u(t),{}^{\mathrm{C}}D_{0^{+}}^{\beta}u(t) \in X,u(0)=u(1) \bigr\} .$$

Then the problem (2.1) can be written by

$$Lu=Nu, \quad u\in\operatorname{dom}L.$$

### Lemma 3.1

L is a Fredholm operator of index zero. $$K_{P}$$ is the inverse of $$L|_{\operatorname{dom}L\cap \operatorname{Ker}P}$$, where $$K_{P}:\operatorname{Im}L\rightarrow \operatorname{dom}L\cap\operatorname{Ker}P$$ is given by

$$K_{P}y(t)=\frac{1}{\Gamma(\beta)} \biggl[ \int_{0}^{t}(t-s)^{\beta -1}y(s)\,ds- \frac{1}{\beta} \int_{0}^{1}(1-s)^{\beta}y(s)\,ds \biggr].$$

### Proof

It is easy to see that

$$\operatorname{Ker}L=\{c \mid c\in\mathbb{R}\},\qquad \operatorname {Im}L= \biggl\{ y\in Y \Bigm| \int_{0}^{1}(1-s)^{\beta-1}y(s)\,ds =0 \biggr\} ,$$

and $$\operatorname{Im}L\subset Y$$ is closed.

Define $$P:X\rightarrow X$$, $$Q:Y\rightarrow Y$$ as

$$Pu= \int_{0}^{1}u(t)\,dt,\qquad Qy=\beta \int _{0}^{1}(1-s)^{\beta-1}y(s)\,ds.$$

Obviously, $$P:X\rightarrow X$$, $$Q:Y\rightarrow Y$$ are projectors and $$\operatorname{Im}P=\operatorname{Ker}L$$, $$X=\operatorname{Ker}P\oplus \operatorname{Ker}L$$.

It is easy to see that $$\operatorname{Im}L\subset\operatorname {Ker}Q$$. Conversely, if $$y(t)\in\operatorname{Ker}Q$$, take $$u(t)= \frac{1}{\Gamma(\beta)}\int_{0}^{t}(t-s)^{\beta-1}y(s)\,ds$$. Then $$u(t)\in\operatorname{dom}L$$ and $$Lu={}^{\mathrm{C}}D_{0^{+}}^{\beta }u(t)=y(t)$$. These imply $$\operatorname{Ker}Q\subset\operatorname{Im}L$$. Therefore $$\operatorname{Im}L=\operatorname{Ker}Q$$. For $$y\in Y$$, $$y=(y-Qy)+Qy\in\operatorname{Im}L+\operatorname{Im}Q$$. If $$y\in\operatorname{Im}L \cap\operatorname{Im}Q$$, then $$y=Qy$$ and $$y\in\operatorname{Im}L=\operatorname{Ker}Q$$. This means that $$y=0$$, i.e. $$Y=\operatorname{Im}L \oplus\operatorname{Im}Q$$. So, $$\operatorname {dim}\operatorname{Ker}L=\operatorname{codim}\operatorname {Im}L=1<+\infty$$. L is a Fredholm operator of index zero.

For $$y\in\operatorname{Im}L$$, it is clear that $$K_{P}y\in\operatorname {dom}L\cap\operatorname{Ker}P$$ and $$LK_{P}y=y$$. On the other hand, if $$u\in\operatorname{dom}L\cap \operatorname{Ker}P$$, by Lemma 2.2, we get

\begin{aligned} K_{P}Lu(t) =& \frac{1}{\Gamma(\beta)} \biggl[ \int _{0}^{t}(t-s)^{\beta-1}Lu(s)\,ds- \frac{1}{\beta} \int _{0}^{1}(1-s)^{\beta }Lu(s)\,ds \biggr] \\ =&I_{0^{+}}^{\beta} {{}^{\mathrm{C}}D_{0^{+}}^{\beta}}u(t)-I_{0^{+}}^{\beta+1} {{}^{\mathrm{C}}D_{0^{+}}^{\beta}}u(1) \\ =&u(t)+c-I_{0^{+}}^{\beta+1} {{}^{\mathrm{C}}D_{0^{+}}^{\beta}}u(1). \end{aligned}

Thus, $$\int_{0}^{1}K_{P}Lu(t)\,dt=\int_{0}^{1}u(t)\,dt+c-I_{0^{+}}^{\beta+1} {{}^{\mathrm{C}}D_{0^{+}}^{\beta}}u(1)$$. It follows from $$u\in\operatorname{Ker}P$$ and $$K_{P}Lu\in \operatorname{Ker}P$$ that $$c-I_{0^{+}}^{\beta+1} {{}^{\mathrm{C}}D_{0^{+}}^{\beta}}u(1)=0$$. So, we have $$K_{P}Lu=u$$, $$u\in\operatorname{dom}L\cap\operatorname {Ker}P$$. □

Define $$J:\operatorname{Im}Q\rightarrow\operatorname{Ker}L$$ as $$J(c)=c$$, $$c\in\mathbb{R}$$.

Thus, $$JQN+K_{P}(I-Q)N:X\rightarrow X$$ is given by

$$\bigl[JQN+K_{P}(I-Q)N\bigr]u(t)= \int_{0}^{1}G(t,s)f\bigl(s,\varphi _{q} \bigl(u(s)\bigr)\bigr)\,ds,$$
(3.1)

where

$$G(t,s)=\left \{ \textstyle\begin{array}{l@{\quad}l} \beta(1-s)^{\beta-1} (1-\frac{t^{\beta}}{\Gamma(\beta +1)}+\frac{1}{\Gamma(\beta+2)} )-\frac{(1-s)^{\beta}}{\Gamma (\beta +1)}+\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}, &0\leq s< t\leq1, \\ \beta(1-s)^{\beta-1} (1-\frac{t^{\beta}}{\Gamma(\beta +1)}+\frac{1}{\Gamma(\beta+2)} )-\frac{(1-s)^{\beta}}{\Gamma (\beta +1)},& 0\leq t\leq s< 1. \end{array}\displaystyle \right .$$

### Lemma 3.2

$$QN:X\rightarrow Y$$ is continuous and bounded and $$K_{P}(I-Q)N:\overline{\Omega}\rightarrow X$$ is compact, where $$\Omega\subset X$$ is bounded.

### Proof

Assume that $$\Omega\subset X$$ is bounded. There exists a constant $$M>0$$, such that $$|Nu|=|f(t,\varphi_{q}(u(t)))|\leq M$$, $$t\in[0,1]$$, $$u\in\overline{\Omega}$$. So, $$|QNu|\leq M$$, $$u\in\overline{\Omega}$$, i.e. $$QN(\overline{\Omega})$$ is bounded. Based on the definition of Q and the continuity of f we know that $$QN:X\rightarrow Y$$ is continuous.

For $$u\in\overline{\Omega}$$, we have

\begin{aligned}& \bigl\vert K_{P}(I-Q)Nu(t)\bigr\vert \\& \quad = \biggl\vert \frac{1}{\Gamma(\beta)}\biggl[ \int _{0}^{t}(t-s)^{\beta-1}(I-Q)Nu(s)\,ds- \frac{1}{\beta} \int _{0}^{1}(1-s)^{\beta}(I-Q)Nu(s)\,ds \biggr]\biggr\vert \\& \quad \leq \frac{1}{\Gamma(\beta)} \int _{0}^{t}(t-s)^{\beta-1}\bigl\vert Nu(s) \bigr\vert \,ds+\frac{1}{\Gamma(\beta)} \int _{0}^{t}(t-s)^{\beta-1}\bigl\vert QNu(s)\bigr\vert \,ds \\& \qquad {}+\frac{1}{\beta\Gamma(\beta)} \int _{0}^{1}(1-s)^{\beta}\bigl\vert Nu(s) \bigr\vert \,ds+\frac{1}{\beta\Gamma(\beta)} \int _{0}^{1}(1-s)^{\beta}\bigl\vert QNu(s) \bigr\vert \,ds \\& \quad \leq \frac{4M}{\Gamma(\beta +1)}< +\infty . \end{aligned}

Thus, $$|K_{P}(I-Q)N(\overline{\Omega})$$ is bounded.

For $$u\in\overline{\Omega}$$, $$0\leq t_{1}< t_{2}\leq1$$, we get

\begin{aligned}& \bigl\vert K_{P}(I-Q)Nu(t_{2})-K_{P}(I-Q)Nu(t_{1}) \bigr\vert \\& \quad = \frac{1}{\Gamma(\beta)}\biggl\vert \int _{0}^{t_{2}}(t_{2}-s)^{\beta-1}(I-Q)Nu(s) \,ds- \int _{0}^{t_{1}}(t_{1}-s)^{\beta-1}(I-Q)Nu(s) \,ds\biggr\vert \\& \quad = \frac{1}{\Gamma(\beta)}\biggl\vert \int _{0}^{t_{1}}\bigl[(t_{2}-s)^{\beta-1}-(t_{1}-s)^{\beta -1} \bigr](I-Q)Nu(s)\,ds+ \int _{t_{1}}^{t_{2}}(t_{2}-s)^{\beta-1}(I-Q)Nu(s) \,ds\biggr\vert \\& \quad \leq\frac{2M}{\Gamma(\beta)}\biggl[ \int _{0}^{t_{1}}\bigl[(t_{1}-s)^{\beta-1}-(t_{2}-s)^{\beta-1} \bigr]\,ds+ \int _{t_{1}}^{t_{2}}(t_{2}-s)^{\beta-1} \,ds\biggr] \\& \quad = \frac{2M}{\Gamma(\beta+1)}\bigl[t_{1}^{\beta}-t_{2}^{\beta }+2(t_{2}-t_{1})^{\beta} \bigr] . \end{aligned}

It follows from the uniform continuity of $$t^{\beta}$$ and t on $$[0,1]$$ that $$K_{P}(I-Q)N(\overline{\Omega})$$ are equicontinuous on $$[0,1]$$. By the Arzela-Ascoli theorem, we see that $$K_{P}(I-Q)N(\overline{\Omega})$$ is compact. □

In order to prove our main result, we need the following conditions.

(H1):

There exists a constant $$R_{0}>0$$, such that $$f(t,u)<0$$, $$t\in [0,1]$$, $$u> R_{0}$$.

(H2):

There exist nonnegative functions $$a(t)$$, $$b(t)$$ with $$\max_{t\in[0,1]}\frac{1}{\Gamma(\beta)}\int _{0}^{t}(t-s)^{(\beta-1)}a(s)\,ds:=A<+\infty$$, $$\max_{t\in[0,1]}\frac{1}{\Gamma(\beta)}\int _{0}^{t}(t-s)^{(\beta-1)}b(s)\,ds:=B<1/2$$, such that

$$\bigl\vert f(t,u)\bigr\vert \leq a(t)+b(t)\varphi_{p}\bigl( \vert u\vert \bigr).$$
(H3):

$$f(t,u)\geq-(1-t)^{1-\beta}\varphi_{p}(u)/\beta$$, $$t\in [0,1]$$, $$u>0$$.

(H4):

There exist $$r>0$$, $$t_{0}\in[0,1]$$, and $$M_{0}\in(0,1)$$ such that

$$G(t_{0},s)f(s,u)\geq\frac{1-M_{0}}{M_{0}}\varphi_{p}(u),\quad s\in [0,1), M_{0}r\leq u\leq r.$$
(H5):

$$G(t,s)f(s,u)\geq-\varphi_{p}(u)$$, $$s\in[0,1)$$, $$t\in[0,1]$$, $$u\geq 0$$.

### Lemma 3.3

If the conditions (H1) and (H2) hold, the set

$$\Omega_{0}= \bigl\{ u(t) \mid (Lu) (t)=\lambda Nu(t), u(t)\in C\cap \operatorname{dom}L, \lambda\in(0,1) \bigr\}$$

is bounded.

### Proof

For $$u(t)\in\Omega_{0}$$, we get $$QNu(t)=0$$ and $$u(t)=\lambda I_{0^{+}}^{\beta} Nu(t)+u(0)$$. By (H1) and $$QNu(t)=0$$, there exists $$t_{0}\in[0,1]$$ such that $$\varphi_{q}(u(t_{0}))\leq R_{0}$$. This, together with $$u(t)=\lambda I_{0^{+}}^{\beta} Nu(t)+u(0)$$, means

$$u(0)\leq u(t_{0})+\bigl\vert \lambda I_{0^{+}}^{\beta} Nu(t_{0})\bigr\vert \leq \varphi_{p}(R_{0})+ \bigl\vert I_{0^{+}}^{\beta} Nu(t_{0})\bigr\vert .$$

Thus, we have

$$u(t) \leq u(0)+ \bigl\vert \lambda I_{0^{+}}^{\beta} Nu(t)\bigr\vert \leq\varphi _{p}(R_{0})+\bigl\vert I_{0^{+}}^{\beta} Nu(t_{0})\bigr\vert + \bigl\vert I_{0^{+}}^{\beta} Nu(t)\bigr\vert .$$
(3.2)

It follows from (H2) that

\begin{aligned} u(t) \leq& \varphi_{p}(R_{0})+ \frac{1}{\Gamma(\beta)} \int _{0}^{t_{0}}(t_{0}-s)^{\beta-1} \bigl\vert f\bigl(s,\varphi_{q}\bigl(u(s)\bigr)\bigr)\bigr\vert \,ds \\ &{}+\frac{1}{\Gamma (\beta )} \int_{0}^{t}(t-s)^{\beta-1}\bigl\vert f \bigl(s,\varphi_{q}\bigl(u(s)\bigr)\bigr)\bigr\vert \,ds \\ \leq& \varphi_{p}(R_{0})+ \frac{1}{\Gamma(\beta)} \int _{0}^{t_{0}}(t_{0}-s)^{\beta-1} \bigl[a(s)+b(s)u(s)\bigr]\,ds \\ &{}+\frac{1}{\Gamma(\beta )} \int _{0}^{t}(t-s)^{\beta-1}\bigl[a(s)+b(s)u(s) \bigr]\,ds \\ \leq&\varphi_{p}(R_{0})+2\bigl(A+B\Vert u\Vert \bigr). \end{aligned}

Therefore,

$$\|u\|\leq\frac{\varphi_{p}(R_{0})+2A }{1-2B}< +\infty.$$

This means that $$\Omega_{0}$$ is bounded. □

### Theorem 3.1

Assume that the conditions (H1)-(H5) hold. Then the boundary value problem (1.1) has at least one positive solution.

### Proof

Set

$$\Omega_{1}= \bigl\{ u\in X\mid M_{0}\|u\|< \bigl|u(t)\bigr|< r< R, t \in[0,1] \bigr\} ,\qquad \Omega_{2}=\bigl\{ u\in X\mid\|u\|< R\bigr\} ,$$

where $$R>\max\{\varphi_{p}(R_{0}), \Gamma(\beta+1)A\}$$ is large enough such that $$\Omega_{2}\supset\Omega_{0}$$. Clearly, $$\Omega_{1}$$ and $$\Omega_{2}$$ are open bounded sets of X, $$\overline{\Omega}_{1}\subset\Omega_{2}$$ and $$C\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\neq\emptyset$$.

In view of Lemmas 3.1, 3.2, and 3.3, L is a Fredholm operator of index zero and the conditions (C1), (C2) of Theorem 2.1 are fulfilled.

Define $$\gamma:X\rightarrow C$$ as $$(\gamma u)(t)=|u(t)|$$, $$u(t)\in X$$. Then $$\gamma:X\rightarrow C$$ is a retraction and (C3) holds.

Let $$u(t)\in\operatorname{Ker}L\cap\partial\Omega_{2}$$, then $$u(t)\equiv c=\pm R$$, $$t\in[0,1]$$. Define

$$H(c,\lambda)= c-\lambda \vert c\vert -\lambda\beta \int _{0}^{1}(1-s)^{\beta-1}f\bigl(s, \varphi_{q}\bigl(\vert c\vert \bigr)\bigr)\,ds.$$

If $$c=R$$, $$\lambda\in[0,1]$$, by (H1), we get

$$H(R,\lambda)= R-\lambda R-\lambda\beta \int _{0}^{1}(1-s)^{\beta-1}f\bigl(s, \varphi_{q}(R)\bigr)\,ds>0.$$

If $$c=-R$$, $$\lambda\in[0,1]$$, by (H3), we obtain

$$H(-R,\lambda)= -R-\lambda R-\lambda\beta \int _{0}^{1}(1-s)^{\beta-1}f\bigl(s, \varphi_{q}(R)\bigr)\,ds< -(1+\lambda )R+\lambda R=-R.$$

So, we have $$H(u,\lambda)\neq0$$, $$u\in\operatorname{Ker}L\cap \partial\Omega _{2}$$, $$\lambda\in[0,1]$$.

Therefore,

\begin{aligned}& d_{B}\bigl(\bigl[I-(P+JQN)\gamma\bigr]|_{\operatorname{Ker}L}, \operatorname {Ker}L\cap\Omega_{2}, 0\bigr) \\& \quad =d_{B}\bigl(H(\cdot,1)|_{\operatorname{Ker}L}, \operatorname{Ker}L\cap \Omega_{2}, 0\bigr) =d_{B}\bigl(H(\cdot,0)|_{\operatorname{Ker}L}, \operatorname{Ker}L\cap \Omega_{2}, 0\bigr) \\& \quad =d_{B}(I| _{\operatorname{Ker}L}, \operatorname{Ker}L\cap \Omega_{2}, 0)=1\neq 0. \end{aligned}

Thus, (C4) holds.

Set $$u_{0}(t)=1$$, $$t \in[0,1]$$, then $$u_{0}\in C\setminus \{0\}$$, $$C(u_{0})=\{u\in C \mid u(t)>0, t \in[0,1]\}$$. Take $$\sigma(u_{0})=1$$ and $$u \in C(u_{0})\cap \partial\Omega_{1}$$. Then $$M_{0}r\leq u(t)\leq r$$, $$t \in[0,1]$$. By (H4), we get

\begin{aligned} \Psi u(t_{0}) =& \int_{0}^{1}u(s)\,ds+ \int_{0}^{1}G(t_{0},s)f\bigl(s,\varphi _{q}\bigl(u(s)\bigr)\bigr)\,ds \\ \geq& \int_{0}^{1}u(s)\,ds+ \int_{0}^{1}\frac {1-M_{0}}{M_{0}}u(s)\,ds \\ \geq& M_{0}r+(1-M_{0})r=r=\|u\|. \end{aligned}

Thus, $$\|u\|\leq\sigma(u_{0})\|\Psi u\|$$, for $$u\in C(u_{0})\cap \partial \Omega_{1}$$. So, (C5) holds.

For $$u(t)\in\partial\Omega_{2}$$, $$t\in[0,1]$$, by the condition (H3), we have

\begin{aligned} (P+JQN)\gamma(u) =& \int _{0}^{1}\bigl\vert u(s)\bigr\vert \,ds+ \beta \int_{0}^{1}(1-s)^{\beta-1}f\bigl(s,\varphi _{q}\bigl(\bigl\vert u(s)\bigr\vert \bigr)\bigr)\,ds \\ \geq& \int_{0}^{1}\bigl\vert u(s)\bigr\vert \,ds- \int_{0}^{1}\bigl\vert u(s)\bigr\vert \,ds=0. \end{aligned}

So, $$(P+JQN)\gamma(\partial\Omega_{2})\subset C$$. Hence, (C6) holds.

For $$u(t)\in\overline{\Omega}_{2}\setminus\Omega_{1}$$, $$t\in[0,1]$$, it follows from (H5) that

$$(\Psi_{\gamma}u) (t)= \int_{0}^{1}\bigl\vert u(s)\bigr\vert \,ds+ \int _{0}^{1}G(t,s)f\bigl(s,\varphi_{q} \bigl(\bigl\vert u(s)\bigr\vert \bigr)\bigr)\,ds \geq \int_{0}^{1}\bigl\vert u(s)\bigr\vert \,ds- \int_{0}^{1}\bigl\vert u(s)\bigr\vert \,ds = 0.$$

So, (C7) is satisfied.

By Theorem 2.1, we confirm that the equation $$Lu=Nu$$ has a positive solution u. Based on Lemma 2.3, the problem (1.1) has at least one positive solution. □

## 4 Examples

To illustrate our main result, we present an example.

Let us consider the following boundary value problem:

$$\left \{ \textstyle\begin{array}{l} {}^{\mathrm{C}}D_{0^{+}}^{\frac{3}{4}} [\varphi_{\frac{5}{4}} ({}^{\mathrm{C}}D_{0^{+}}^{\frac{1}{2}}x ) ](t)= \frac {1}{4}(1-t)^{\frac{1}{4}}- \frac{1}{20}(1-t)^{\frac{1}{4}}\vert {}^{\mathrm{C}}D_{0^{+}}^{\frac {1}{2}}x(t)\vert ^{\frac{1}{4}},\quad t\in(0,1), \\ x(0)=0,\qquad ({}^{\mathrm{C}}D_{0^{+}}^{\frac{1}{2}}x)(0)=({}^{\mathrm{C}}D_{0^{+}}^{\frac{1}{2}}x)(1). \end{array}\displaystyle \right .$$
(4.1)

On the basis of Lemma 2.3, it is sufficient to examine the issue

$$\left \{ \textstyle\begin{array}{l} {}^{\mathrm{C}}D_{0^{+}}^{\frac{3}{4}}u(t)= \frac{1}{4}(1-t)^{\frac {1}{4}}-\frac{1}{20}(1-t)^{\frac{1}{4}}|u(t)|, \quad t\in[0,1], \\ u(0)=u(1). \end{array}\displaystyle \right .$$
(4.2)

Corresponding to the problem (2.1), we have $$f(t,u)= \frac{1}{4}(1-t)^{\frac{1}{4}}-\frac {1}{20}(1-t)^{\frac{1}{4}}|u|^{\frac{1}{4}}$$, $$p=\frac{5}{4}$$, $$q=5$$, $$\alpha =\frac{1}{2}$$, $$\beta=\frac{3}{4}$$. So,

$$G(t,s)=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{3}{4}(1-s)^{-\frac{1}{4}} (1-\frac{t^{\frac {3}{4}}}{\Gamma(\frac{7}{4})}+\frac{1}{\Gamma(\frac{11}{4})} )-\frac {(1-s)^{\frac{3}{4}}}{\Gamma(\frac{7}{4})}+\frac{(t-s)^{-\frac {1}{4}}}{\Gamma(\frac{3}{4})},&0\leq s< t\leq1, \\ \frac{3}{4}(1-s)^{-\frac{1}{4}} (1-\frac{t^{\frac {3}{4}}}{\Gamma(\frac{7}{4})}+\frac{1}{\Gamma(\frac{11}{4})} )-\frac {(1-s)^{\frac{3}{4}}}{\Gamma(\frac{7}{4})},& 0\leq t\leq s< 1. \end{array}\displaystyle \right .$$

Take $$R_{0}=625$$, $$a(t)=1$$, $$b(t)=\frac{1}{4}$$, $$r=0.006$$, $$t_{0}=0$$, and $$M_{0}=0.95$$.

Clearly, (H1) holds. By simple calculations, we can see that

\begin{aligned}& \bigl\vert f(t,u)\bigr\vert \leq a(t)+b(t)\varphi_{p}\bigl( \vert u\vert \bigr), \\& A=\max_{t\in[0,1]}\frac{1}{\Gamma(\frac{3}{4})} \int _{0}^{t}(t-s)^{-\frac{1}{4}}\,ds= \frac{4}{3.6762}< +\infty, \\& B=\max_{t\in[0,1]}\frac{1}{\Gamma(\frac{3}{4})} \int _{0}^{t}(t-s)^{-\frac{1}{4}}\cdot \frac{1}{4}\,ds=\frac {1}{3.6762}< \frac{1}{2}, \\& f(t,u)\geq-\frac{4}{3}(1-t)^{\frac{1}{4}}u^{\frac{1}{4}},\quad u>0, \\& G(t_{0},s)f(s,u)\geq\frac{0.12828103}{4}-\frac{1.21630192}{20}u^{\frac {1}{4}} \\& \hphantom{G(t_{0},s)f(s,u)}\geq\frac{0.05}{0.95}u^{\frac{1}{4}}, \quad 0.0057\leq u\leq 0.006,s\in[0,1), \\& G(t,s)f(s,u)\geq-u^{\frac{1}{4}},\quad u\geq0, s\in[0,1), t\in[0,1]. \end{aligned}

So, the conditions (H1)- (H5) hold. By Theorem 3.1, we can conclude that the problem (4.1) has at least one positive solution.

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## Acknowledgements

This work is supported by the Natural Science Foundation of Hebei Province (A2017208101).

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Correspondence to Weihua Jiang.

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All results belong to WJ, JQ, and CY. All authors read and approved the final manuscript.

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