Boundary Value Problems

, 2016:175 | Cite as

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

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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.

Keywords

positive solutions p-Laplacian operator boundary value problem resonance Fredholm operator 

MSC

34B15 

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, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 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, 26, 27, 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.
  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.

Notes

Acknowledgements

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

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© Jiang et al. 2016

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Authors and Affiliations

  1. 1.College of SciencesHebei University of Science and TechnologyShijiazhuangP.R. China

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