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 [1] and its extension obtained by Ge and Ren [2]; see [3–23] and the references cited therein. By using Leggett-Williams norm-type theorems due to O’Regan and Zima [24], the existence of positive solutions for the boundary value problems at resonance with a linear derivative operator has been investigated (see [25–28]). 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 [24].

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

[24]

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.

• (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;

• (C2) $Lx\neq\lambda Nx$ for all $x\in C\cap\partial\Omega _{2}\cap\operatorname{dom}L$ and $\lambda\in(0,1)$;

• (C3) γ maps subsets of $\overline{\Omega}_{2}$ into bounded subsets of C;

• (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;

• (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 $x\in C$;

• (C6) $(P+JQN)\gamma(\partial\Omega_{2})\subset C$;

• (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.