Let \(X=C([0,1],R)\) be the Banach space of real-valued continuous functions on \([0,1]\) endowed with norm \(\| x \| =\max_{t \in [0,1]}| x(t) | \).
Throughout this paper, we make the following assumption on the singularity of nonlinear function \(f(t,x(t))\) in (1.1):
-
(H1)
\(f(t,x(t))\) has a singularity at \(t=0\) and \(t=1\), that is,
$$ \lim_{t\to 0^{+}} f(t,\cdot )=\infty ,\qquad \lim _{t \to 1^{-}} f(t,\cdot )=\infty . $$
Moreover, there exist constants \(0<\theta _{1}<1\) and \(0<\theta _{2}<1\) such that \(t^{\theta _{1}}(1-t)^{\theta _{2}}f(t,x(t))\) is continuous on \([0,1]\).
Based on condition (H1), we know that there is a positive constant \(M_{0}\) such that
$$ \bigl\vert t^{\theta _{1}}(1-t)^{\theta _{2}}f \bigl(t,x(t) \bigr) \bigr\vert \leq M_{0}, \quad x \in X,t\in [0,1]. $$
(3.1)
Let \(\lambda =\frac{3}{2+\gamma ^{3}}\). By Lemma 2.2 the operator \(A:X\rightarrow X\) can be represented as
$$\begin{aligned} (Ax) (t) =&\frac{1}{\Gamma (\alpha )} \int _{0}^{t}(t-\tau )^{\alpha -1}f \bigl( \tau ,x(\tau ) \bigr)\,d\tau +\frac{\lambda t^{2}}{\Gamma (\alpha +1)} \int _{0}^{1}(1- \tau )^{\alpha }f \bigl(\tau ,x( \tau ) \bigr)\,d\tau \\ &{}-\frac{\lambda t^{2}}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{ \alpha -1}f \bigl(\tau ,x( \tau ) \bigr)\,d\tau \\ &{}- \frac{\lambda t^{2}}{\Gamma (\alpha +1)} \int _{0}^{\gamma }(\gamma - \tau )^{\alpha }f \bigl( \tau ,x(\tau ) \bigr)\,d\tau . \end{aligned}$$
(3.2)
Then the solutions of problem (1.1) include the FPs of A.
Lemma 3.1
Suppose \(0<\theta _{1}<1\) and \(0<\theta _{2}<1\). Then the integral operator J defined as
$$ J(t)= \int _{0}^{t} (t-\tau )^{\alpha -1}\tau ^{-\theta _{1}}(1-\tau )^{- \theta _{2}}\,d\tau ,\quad t\in [0,1] $$
has the following specifications:
-
(1)
\(\lim_{t\to 0^{+}} J(t)=0\);
-
(2)
\(| J(t)-J(t_{0})| <(\alpha -1)B(1-\theta _{1},\alpha - \theta _{2}-1)| t-t_{0}|\) for all \(t,t_{0}\in [0,1]\),
where \(B(\cdot ,\cdot )\) denotes the beta function.
Proof
(1) By Lemma 2.6, for any \(p_{1}>1\), \(p_{2}>1\), \(p_{3}>1\) such that \(\frac{1}{p_{1}}+\frac{1}{p_{2}}+\frac{1}{p_{3}}=1\), \(0< p_{1}\theta _{1}<1\), and \(0< p_{2}\theta _{2}<1\), we have
$$\begin{aligned} J(t)\leq{}& \biggl[ \int _{0}^{t}\tau ^{-p_{1}\theta _{1}}\,d\tau \biggr]^{1/p_{1}} \biggl[ \int _{0}^{t}(1-\tau )^{-p_{2}\theta _{2}}\,d\tau \biggr]^{1/p_{2}} \bigg[ \biggl[ \int _{0}^{t}(t- \tau )^{p_{3}(\alpha -1)}\,d\tau \biggr]^{1/p_{3}} \\ ={}& \biggl[\frac{1}{1-p_{1}\theta _{1}}\tau ^{1-p_{1}\theta _{1}} \bigg| _{0}^{t} \biggr]^{1/p_{1}} \biggl[- \frac{1}{1-p_{2}\theta _{2}}(1-\tau )^{1-p_{2}\theta _{2}}\bigg| _{0}^{t} \biggr]^{1/p_{2}} \\ &{}\cdot \biggl[- \frac{1}{1+p_{3}(\alpha -1)}(t-\tau )^{1+p_{3}(\alpha -1)}\bigg| _{0}^{t} \biggr]^{1/p_{3}} \\ ={}&\frac{1}{\sqrt[p_{1}]{1-p_{1}\theta _{1}}} \frac{1}{\sqrt[p_{2}]{1-p_{2}\theta _{2}}} \frac{1}{\sqrt[p_{3}]{1+p_{3}(\alpha -1)}} \sqrt[p_{1}]{t^{1-p_{1}\theta _{1}}} \\ &{}\cdot \sqrt[p_{2}]{1-(1-t)^{1-p_{2}\theta _{2}}}\cdot \sqrt[p_{3}]{t^{1+p_{3}(\alpha -1)}}. \end{aligned}$$
Since \(J(t)\geq 0\), and \(\lim_{t\to 0^{+}}(\sqrt[p_{1}]{t^{1-p_{1}\theta _{1}}} \cdot \sqrt[p_{2}]{1-(1-t)^{1-p_{2}\theta _{2}}}\cdot \sqrt[p_{3}]{t^{1+p_{3}(\alpha -1)}})=0\), we get
$$ \lim_{t\to 0^{+}} J(t)=0. $$
(2) By the expression of \(J(t)\) we easily get
$$\begin{aligned} J'(t)& =(\alpha -1) \int _{0}^{t} (t-\tau )^{\alpha -2}\tau ^{-\theta _{1}}(1- \tau )^{-\theta _{2}}\,d\tau \\ & \leq (\alpha -1) \int _{0}^{1} (1-\tau )^{\alpha -2}\tau ^{-\theta _{1}}(1- \tau )^{-\theta _{2}}\,d\tau \\ &=(\alpha -1) \int _{0}^{1} (1-\tau )^{\alpha -\theta _{2}-2}\tau ^{- \theta _{1}}\,d\tau \\ & =(\alpha -1)B(1-\theta _{1},\alpha -\theta _{2}-1). \end{aligned}$$
Hence the mean value theorem gives us
$$ \bigl\vert J(t)-J(t_{0}) \bigr\vert =J'(\xi ) \vert t-t_{0} \vert < (\alpha -1)B(1- \theta _{1},\alpha - \theta _{2}-1) \vert t-t_{0} \vert , $$
where the number ξ is between t and \(t_{0}\). □
Lemma 3.2
Let \(2<\alpha \leq 3\), and let \(g:(0,1)\rightarrow R\) be a continuous function such that \(\lim_{t\to 0^{+}} g(t)=\infty \) and \(\lim_{t\to 1^{-}} g(t)=\infty \). Suppose that there exist two constants \(0<\theta _{1}<1\) and \(0<\theta _{2}<1\) such that \(t^{\theta _{1}}(1-t)^{\theta _{2}}g(t)\) is continuous in \([0,1]\). Then the function
$$\begin{aligned} G(t):={}&\frac{1}{\Gamma (\alpha )} \int _{0}^{t} (t-\tau )^{\alpha -1} g( \tau )\,d \tau +\frac{\lambda t^{2}}{\Gamma (\alpha +1)} \int _{0}^{1}(1- \tau )^{\alpha }g(\tau )\,d\tau \\ &{} -\frac{\lambda t^{2}}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{ \alpha -1}g(\tau )\,d\tau -\frac{\lambda t^{2}}{\Gamma (\alpha +1)} \int _{0}^{\gamma }(\gamma -\tau )^{\alpha }g(\tau ) \,d\tau \end{aligned}$$
is continuous in \([0,1]\).
Proof
Based on the expression of \(G(t)\), we easily find \(G(0)=0\). As \(t^{\theta _{1}}(1-t)^{\theta _{2}}g(t)\) is continuous in \([0,1]\), there is a positive constant \(M_{1}\) such that \(| t^{\theta _{1}}(1-t)^{\theta _{2}}g(t) | \leq M_{1}\) for all \(t\in [0,1]\). For all \(t_{0} \in [0,1]\), we will prove the continuity of \(G(t)\) in three cases.
(a) \(t_{0}=0\), \(t \in [0,1]\). We have
$$\begin{aligned}& \bigl\vert G(t)-G(0) \bigr\vert \\& \quad = \biggl\vert \frac{1}{\Gamma (\alpha )} \int _{0}^{t} (t- \tau )^{\alpha -1} \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}} \tau ^{ \theta _{1}}(1-\tau )^{\theta _{2}} g(\tau )\,d\tau \\& \qquad {}+\frac{\lambda t^{2}}{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{ \alpha } \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}} \tau ^{\theta _{1}}(1- \tau )^{\theta _{2}} g(\tau )\,d\tau \\& \qquad {}-\frac{\lambda t^{2}}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{ \alpha -1} \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}} \tau ^{ \theta _{1}}(1-\tau )^{\theta _{2}} g(\tau )\,d\tau \\& \qquad {}-\frac{\lambda t^{2}}{\Gamma (\alpha +1)} \int _{0}^{\gamma }(\gamma - \tau )^{\alpha } \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}} \tau ^{ \theta _{1}}(1-\tau )^{\theta _{2}} g(\tau )\,d\tau \biggr\vert \\& \quad \leq \frac{M_{1}}{\Gamma (\alpha )} \int _{0}^{t} (t-\tau )^{\alpha -1} \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau + \frac{\lambda M_{1}t^{2}}{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{ \alpha } \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau \\& \qquad {} +\frac{\lambda M_{1} t^{2}}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{ \alpha -1} \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau + \frac{\lambda M_{1} t^{2}}{\Gamma (\alpha +1)} \int _{0}^{\gamma }( \gamma -\tau )^{\alpha } \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau \\& \quad \leq \frac{M_{1}}{\Gamma (\alpha )}J(t)+ \frac{\lambda M_{1}t^{2}}{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{ \alpha -\theta _{2}} \tau ^{-\theta _{1}}\,d\tau + \frac{\lambda M_{1} t^{2}}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{ \alpha -\theta _{2}-1} \tau ^{-\theta _{1}}\,d\tau \\& \qquad {} +\frac{\lambda M_{1} t^{2}}{\Gamma (\alpha +1)} \int _{0}^{1}(1- \tau )^{\alpha -\theta _{2}} \tau ^{-\theta _{1}}\,d\tau \\& \quad = \frac{M_{1}}{\Gamma (\alpha )}J(t)+ \frac{2\lambda M_{1}t^{2}}{\Gamma (\alpha +1)}B(1-\theta _{1}, \alpha - \theta _{2}+1) +\frac{\lambda M_{1} t^{2}}{\Gamma (\alpha )}B(1- \theta _{1}, \alpha - \theta _{2} ) \\& \rightarrow{} 0\quad (t\rightarrow t_{0}=0). \end{aligned}$$
(b) \(t_{0}\in (0,1]\), \(t\in [0,t_{0})\). Then
$$\begin{aligned}& \bigl\vert G(t)-G(t_{0}) \bigr\vert \\& \quad = \biggl\vert \frac{1}{\Gamma (\alpha )} \int _{0}^{t_{0}} (t_{0}-\tau )^{\alpha -1} g(\tau )\,d\tau - \frac{1}{\Gamma (\alpha )} \int _{0}^{t} (t-\tau )^{\alpha -1} g(\tau )\,d \tau \\& \qquad {} +\frac{\lambda (t_{0}^{2}- t^{2})}{\Gamma (\alpha +1)} \int _{0}^{1}(1- \tau )^{\alpha } g(\tau )\,d \tau + \frac{\lambda (t_{0}^{2}- t^{2})}{\Gamma (\alpha )} \int _{0}^{1}(1- \tau )^{\alpha -1} g(\tau )\,d \tau \\& \qquad {} +\frac{\lambda (t_{0}^{2}- t^{2})}{\Gamma (\alpha +1)} \int _{0}^{\gamma }(\gamma -\tau )^{\alpha } g(\tau ) \,d\tau \biggr\vert \\& \quad \leq \biggl\vert \frac{1}{\Gamma (\alpha )} \int _{0}^{t} \bigl[(t_{0}-\tau )^{ \alpha -1}-(t-\tau )^{\alpha -1} \bigr] g(\tau )\,d\tau + \frac{1}{\Gamma (\alpha )} \int _{t}^{t_{0}} (t_{0}-\tau )^{\alpha -1} g( \tau )\,d\tau \biggr\vert \\& \qquad {} +\frac{\lambda M_{1}(t_{0}+ t)(t_{0}- t)}{\Gamma (\alpha +1)} \int _{0}^{1}(1- \tau )^{\alpha -\theta _{2}}\tau ^{-\theta _{1}}\,d\tau \\& \qquad {}+ \frac{\lambda M_{1}(t_{0}+ t)(t_{0}- t)}{\Gamma (\alpha )} \int _{0}^{1}(1- \tau )^{\alpha -\theta _{2}}\tau ^{-\theta _{1}}\,d\tau \\& \qquad {} +\frac{\lambda M_{1}(t_{0}+ t)(t_{0}- t)}{\Gamma (\alpha +1)} \int _{0}^{ \gamma }(\gamma -\tau )^{\alpha }\tau ^{-\theta _{1}}(1-\tau )^{- \theta _{2}}\,d\tau \\& \quad \leq \frac{M_{1}}{\Gamma (\alpha )} \int _{0}^{t} \bigl[(t_{0}-\tau )^{ \alpha -1}-(t-\tau )^{\alpha -1} \bigr] \tau ^{-\theta _{1}}(1-\tau )^{- \theta _{2}}\,d\tau \\& \qquad {} + \frac{M_{1}}{\Gamma (\alpha )} \int _{t}^{t_{0}} (t_{0}- \tau )^{\alpha -1} \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau \\& \qquad {} +\frac{2 \lambda M_{1}(t_{0}- t)}{\Gamma (\alpha +1)} \int _{0}^{1}(1- \tau )^{\alpha -\theta _{2}}\tau ^{-\theta _{1}}\,d\tau + \frac{2 \lambda M_{1}(t_{0}- t)}{\Gamma (\alpha )} \int _{0}^{1}(1- \tau )^{\alpha -\theta _{2}}\tau ^{-\theta _{1}}\,d\tau \\& \qquad {} +\frac{2\lambda M_{1}(t_{0}- t)}{\Gamma (\alpha +1)} \int _{0}^{1}(1- \tau )^{\alpha }\tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau \\& \quad = \frac{M_{1}}{\Gamma (\alpha )} \int _{0}^{t_{0}} (t_{0}-\tau )^{ \alpha -1} \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau - \frac{M_{1}}{\Gamma (\alpha )} \int _{0}^{t} (t-\tau )^{\alpha -1} \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau \\& \qquad {} +\frac{2 \lambda M_{1}(t_{0}- t)}{\Gamma (\alpha +1)} \int _{0}^{1}(1- \tau )^{\alpha -\theta _{2}}\tau ^{-\theta _{1}}\,d\tau + \frac{2 \lambda M_{1}(t_{0}- t)}{\Gamma (\alpha )} \int _{0}^{1}(1- \tau )^{\alpha -\theta _{2}}\tau ^{-\theta _{1}}\,d\tau \\& \qquad {} +\frac{2\lambda M_{1}(t_{0}- t)}{\Gamma (\alpha +1)} \int _{0}^{1}(1- \tau )^{\alpha -\theta _{2}}\tau ^{-\theta _{1}}\,d\tau \\& \quad = \frac{M_{1}}{\Gamma (\alpha )} \bigl[J(t_{0})-J(t) \bigr]+ \frac{4 \lambda M_{1}(t_{0}- t)}{\Gamma (\alpha +1)}B(1-\theta _{1}, \alpha -\theta _{2}+1) \\& \qquad {}+\frac{2\lambda M_{1}(t_{0}- t)}{\Gamma (\alpha )}B(1-\theta _{1}, \alpha -\theta _{2}+1). \end{aligned}$$
By the second result of Lemma 3.1 we have
$$ \bigl\vert G(t)-G(t_{0}) \bigr\vert \leq M_{1} \frac{\alpha (\alpha -1)+4\lambda +2\lambda \alpha }{\Gamma (\alpha +1)} \cdot B(1-\theta _{1},\alpha -\theta _{2}+1) (t_{0}-t)\rightarrow 0(t \rightarrow t_{0}). $$
(c) \(t_{0}\in (0,1)\), \(t\in (t_{0},1]\). Since the proof for this case is the same as that in case (b), we omit it. □
Lemma 3.3
Let \(2<\alpha \leq 3\), and let \(f:(0,1)\times R\rightarrow R\) be a continuous function satisfying the singularity condition (H1). Then the operator \(A:X\rightarrow X\) is completely continuous.
Proof
According to Lemma 3.2, \(A:X\rightarrow X\) is continuous. Let \(D\subset X=C([0,1],R)\) be a bounded set, that is, there is a positive constant \(L_{1}\) such that \(\| x\| \leq L_{1} \) for all \(x\in D\).
Relations (3.1) and (3.2) give
$$\begin{aligned} \vert Ax \vert \leq& \frac{1}{\Gamma (\alpha )} \biggl\vert \int _{0}^{t}(t- \tau )^{\alpha -1}f \bigl(\tau ,x(\tau ) \bigr)\,d\tau \biggr\vert + \frac{\lambda }{\Gamma (\alpha +1)} \biggl\vert \int _{0}^{1}(1-\tau )^{ \alpha }f \bigl(\tau ,x( \tau ) \bigr)\,d\tau \biggr\vert \\ &{}+\frac{\lambda }{\Gamma (\alpha )} \biggl\vert \int _{0}^{1}(1-\tau )^{ \alpha -1}f \bigl(\tau ,x( \tau ) \bigr)\,d\tau \biggr\vert + \frac{\lambda }{\Gamma (\alpha +1)} \biggl\vert \int _{0}^{\gamma }(\gamma - \tau )^{\alpha }f \bigl( \tau ,x(\tau ) \bigr)\,d\tau \biggr\vert \\ \leq& \frac{M_{0}}{\Gamma (\alpha )} \int _{0}^{t} (t-\tau )^{\alpha -1} \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau + \frac{\lambda M_{0}}{\Gamma (\alpha +1)} \int _{0}^{1} (1-\tau )^{ \alpha } \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau \\ &{} +\frac{\lambda M_{0}}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{ \alpha -1}\tau ^{-\theta _{1}} (1-\tau )^{-\theta _{2}}\,d\tau \\ &{} + \frac{ \lambda M_{0}}{\Gamma (\alpha +1)} \int _{0}^{\gamma }(\gamma - \tau )^{\alpha }\tau ^{-\theta _{1}} (1-\tau )^{-\theta _{2}}\,d\tau \\ \leq& \frac{M_{0}}{\Gamma (\alpha )} \int _{0}^{1} (1-\tau )^{\alpha - \theta _{2}-1} \tau ^{-\theta _{1}}\,d\tau + \frac{\lambda M_{0}}{\Gamma (\alpha +1)} \int _{0}^{1} (1-\tau )^{ \alpha -\theta _{2}} \tau ^{-\theta _{1}}\,d\tau \\ &{} +\frac{\lambda M_{0}}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{ \alpha -\theta _{2}-1}\tau ^{-\theta _{1}}\,d\tau + \frac{ \lambda M_{0}}{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{ \alpha -\theta _{2}}\tau ^{-\theta _{1}}\,d\tau \\ =& \frac{(1+\lambda )M_{0}}{\Gamma (\alpha )}B(1-\theta _{1},\alpha - \theta _{2})+ \frac{2\lambda M_{0}}{\Gamma (\alpha +1)}B(1-\theta _{1}, \alpha -\theta _{2}+1):=L_{2}, \end{aligned}$$
that is, \(\Vert Ax \Vert \leq L_{2}\), for all \(x\in D\). Thus the operator A is bounded on D. This yields the compactness of A. For every \(t\in [0,1]\), we have
$$\begin{aligned} \bigl\vert (Ax)'(t) \bigr\vert ={}& \biggl\vert \frac{1}{\Gamma (\alpha -1)} \int _{0}^{t}(t- \tau )^{\alpha -2}f \bigl(\tau ,x(\tau ) \bigr)\,d\tau + \frac{2\lambda t}{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{\alpha }f \bigl( \tau ,x( \tau ) \bigr)\,d\tau \\ &{}-\frac{2\lambda t}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{\alpha -1}f \bigl( \tau ,x(\tau ) \bigr)\,d\tau -\frac{2\lambda t}{\Gamma (\alpha +1)} \int _{0}^{\gamma }(\gamma -\tau )^{\alpha }f \bigl( \tau ,x(\tau ) \bigr)\,d\tau \biggr\vert \\ \leq{}& \frac{M_{0}}{\Gamma (\alpha -1)} \int _{0}^{t} (t-\tau )^{ \alpha -2} \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau \\ &{}+ \frac{2\lambda M_{0}}{\Gamma (\alpha +1)} \int _{0}^{1} (1-\tau )^{ \alpha } \tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau \\ & {}+\frac{2\lambda M_{0}}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{ \alpha -1}\tau ^{-\theta _{1}} (1-\tau )^{-\theta _{2}}\,d\tau \\ &{}+ \frac{ 2\lambda M_{0}}{\Gamma (\alpha +1)} \int _{0}^{\gamma }(\gamma - \tau )^{\alpha }\tau ^{-\theta _{1}} (1-\tau )^{-\theta _{2}}\,d\tau \\ \leq{}& \frac{M_{0}}{\Gamma (\alpha -1)} \int _{0}^{1} (1-\tau )^{ \alpha -\theta _{2}-2} \tau ^{-\theta _{1}}\,d\tau + \frac{2\lambda M_{0}}{\Gamma (\alpha +1)} \int _{0}^{1} (1-\tau )^{ \alpha -\theta _{2}} \tau ^{-\theta _{1}}\,d\tau \\ & {}+\frac{2\lambda M_{0}}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{ \alpha -\theta _{2}-1}\tau ^{-\theta _{1}}\,d\tau + \frac{ 2\lambda M_{0}}{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{ \alpha -\theta _{2}}\tau ^{-\theta _{1}}\,d\tau \\ ={}& \frac{M_{0}}{\Gamma (\alpha -1)}B(1-\theta _{1},\alpha -\theta _{2}-1)+ \frac{4\lambda M_{0}}{\Gamma (\alpha +1)}B(1-\theta _{1},\alpha - \theta _{2}+1) \\ & {}+ \frac{2\lambda M_{0}}{\Gamma (\alpha )}B(1-\theta _{1},\alpha - \theta _{2}):=L_{3}. \end{aligned}$$
Now the following inequality holds for \(t_{1},t_{2}\in [0,1]\) and \(t_{1}< t_{2}\):
$$ \bigl\vert (Ax) (t_{2})-(Ax) (t_{1}) \bigr\vert = \biggl\vert \int _{t_{1}}^{t_{2}} (Ax)'(s)\,ds \biggr\vert \leq L_{3}(t_{2}-t_{1}). $$
Therefore A is equicontinuous on D. Thus, by the Arzelà–Ascoli theorem the operator A is completely continuous on X. □
Now we present and demonstrate our fundamental results. The first result deals with the existence and uniqueness of the solution to problem (1.1).
Theorem 3.1
Let \(2<\alpha \leq 3 \) and \(0<\theta _{1}\), \(\theta _{2}<1\) be constants, and let \(f(t,x(t))\) satisfy condition (H1) and the following conditions:
-
(H2)
There is a function \(m(t)\in L^{p}([0,1],R^{+})\) (\(p>1\)) such that
$$ t^{\theta _{1}}(1-t)^{\theta _{2}} \bigl\vert f(t,x)-f(t,y) \bigr\vert \leq m(t) \vert x-y \vert . $$
-
(H3)
There exist three constants \(p_{1}\), \(p_{2}\), \(p_{3}\) satisfying \(p_{1}>1\), \(p_{2}>1\), \(p_{3}>1\), \(0< p_{1}\theta _{1}<1\), and \(\frac{1}{p_{1}}+\frac{1}{p_{2}}+\frac{1}{p_{3}}=1\). If
$$\begin{aligned}& \Vert m \Vert _{p_{3}} \frac{1}{\sqrt[p_{1}]{1-p_{1}\theta _{1}}}[ \frac{1+\lambda }{\Gamma (\alpha )} \frac{1}{\sqrt[p_{2}]{1+p_{2}(\alpha -\theta _{2}-1)}} \\& \quad {}+ \frac{2\lambda }{\Gamma (\alpha +1)} \frac{1}{\sqrt[p_{2}]{1+p_{2}(\alpha -\theta _{2})}} < 1, \end{aligned}$$
(3.3)
then the solution to problem (1.1) is unique.
Proof
For \(x,y \in X=C([0,1])\) and \(t\in [0,1]\), by (H2) we have
$$\begin{aligned} \bigl\vert (Ax) (t)-(Ay) (t) \bigr\vert \leq {}&\frac{1}{\Gamma (\alpha )} \int _{0}^{t}(t- \tau )^{\alpha -1} \bigl\vert f \bigl(\tau ,x(\tau ) \bigr)-f \bigl(\tau ,y(\tau ) \bigr) \bigr\vert \,d\tau \\ &{}+\frac{\lambda }{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{\alpha } \bigl\vert f \bigl(\tau ,x(\tau ) \bigr)-f \bigl(\tau ,y(\tau ) \bigr) \bigr\vert \,d\tau \\ &{} +\frac{\lambda }{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{\alpha -1} \bigl\vert f \bigl(\tau ,x(\tau ) \bigr)-f \bigl(\tau ,y(\tau ) \bigr) \bigr\vert \,d\tau \\ & {}+\frac{\lambda }{\Gamma (\alpha +1)}| \int _{0}^{\gamma }(\gamma - \tau )^{\alpha } \bigl\vert f \bigl(\tau ,x(\tau ) \bigr)-f \bigl(\tau ,y(\tau ) \bigr) \bigr\vert \,d \tau \\ \leq{}& \frac{1}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{\alpha -{ \theta _{2}}-1}\tau ^{-\theta _{1}}m(\tau ) \bigl\vert x(\tau )-y(\tau ) \bigr\vert \,d\tau \\ &{}+\frac{\lambda }{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{\alpha -{ \theta _{2}}}\tau ^{-\theta _{1}}m(\tau ) \bigl\vert x(\tau )-y(\tau ) \bigr\vert \,d \tau \\ &{}+\frac{\lambda }{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{\alpha -{ \theta _{2}}-1}\tau ^{-\theta _{1}}m(\tau ) \bigl\vert x(\tau )-y(\tau ) \bigr\vert \,d\tau \\ &{}+\frac{\lambda }{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{\alpha -{ \theta _{2}}}\tau ^{-\theta _{1}}m(\tau ) \bigl\vert x(\tau )-y(\tau ) \bigr\vert \,d \tau . \end{aligned}$$
By (H3) and the Hölder inequality we have
$$\begin{aligned}& \bigl\vert (Ax) (t)-(Ay) (t) \bigr\vert \\& \quad \leq \bigl\Vert x(\tau )-y(\tau ) \bigr\Vert \cdot \biggl\{ \frac{1}{\Gamma (\alpha )} \biggl[ \int _{0}^{1} \tau ^{- \theta _{1}p_{1}}\,d\tau \biggr]^{1/p_{1}} \biggl[ \int _{0}^{1}(1-\tau )^{(\alpha -{ \theta _{2}}-1)p_{2}}\,d\tau \biggr]^{1/p_{2}} \\& \qquad {}\times \biggl[ \int _{0}^{1} m^{p_{3}}(\tau )\,d\tau \biggr]^{1/p_{3}} \\& \qquad {} + \frac{\lambda }{\Gamma (\alpha +1)} \biggl[ \int _{0}^{1} \tau ^{-\theta _{1}p_{1}}\,d\tau \biggr]^{1/p_{1}} \biggl[ \int _{0}^{1}(1-\tau )^{(\alpha -{\theta _{2}})p_{2}}\,d\tau \biggr]^{1/p_{2}} \biggl[ \int _{0}^{1} m^{p_{3}}(\tau )\,d\tau \biggr]^{1/p_{3}} \\& \qquad {} + \frac{\lambda }{\Gamma (\alpha )} \biggl[ \int _{0}^{1} \tau ^{-\theta _{1}p_{1}}\,d\tau \biggr]^{1/p_{1}} \biggl[ \int _{0}^{1}(1-\tau )^{(\alpha -{\theta _{2}}-1)p_{2}}\,d\tau \biggr]^{1/p_{2}} \biggl[ \int _{0}^{1} m^{p_{3}}(\tau )\,d\tau \biggr]^{1/p_{3}} \\& \qquad {} + \frac{\lambda }{\Gamma (\alpha +1)} \biggl[ \int _{0}^{1} \tau ^{-\theta _{1}p_{1}}\,d\tau \biggr]^{1/p_{1}} \biggl[ \int _{0}^{1}(1-\tau )^{(\alpha -{\theta _{2}})p_{2}}\,d\tau \biggr]^{1/p_{2}} \biggl[ \int _{0}^{1} m^{p_{3}}(\tau )\,d\tau \biggr]^{1/p_{3}} \biggr\} \\& \quad = \Vert m \Vert _{p_{3}} \frac{1}{\sqrt[p_{1}]{1-p_{1}\theta _{1}}} \biggl[ \frac{1+\lambda }{\Gamma (\alpha )} \frac{1}{\sqrt[p_{2}]{1+p_{2}(\alpha -\theta _{2}-1)}} + \frac{2\lambda }{\Gamma (\alpha +1)} \frac{1}{\sqrt[p_{2}]{1+p_{2}(\alpha -\theta _{2})}} \biggr] \\& \qquad {}\times \bigl\Vert x( \tau )-y(\tau ) \bigr\Vert . \end{aligned}$$
Noticing (3.3), we conclude that A is a contraction mapping. Thus by Lemma 2.3 it has a unique FP, which is also the unique solution to problem (1.1). □
The second result states the existence of the solution to the BVP (1.1) derived from Lemma 2.4.
Theorem 3.2
Let \(2<\alpha \leq 3 \) and \(0<\theta _{1}\), \(\theta _{2}<1\) be constants, and let \(f(t,x(t))\) satisfy conditions (H1)–(H3) and the following condition:
$$\begin{aligned}& \Vert m \Vert _{p_{3}} \frac{1}{\sqrt[p_{1}]{1-p_{1}\theta _{1}}} \biggl[ \frac{\lambda }{\Gamma (\alpha )} \frac{1}{\sqrt[p_{2}]{1+p_{2}(\alpha -\theta _{2}-1)}} \\& \quad {}+ \frac{2\lambda }{\Gamma (\alpha +1)} \frac{1}{\sqrt[p_{2}]{1+p_{2}(\alpha -\theta _{2})}} \biggr]< 1. \end{aligned}$$
(3.4)
Then problem (1.1) has a solution.
Proof
We fix a constant
$$ r \geq M_{0} \biggl[\frac{1+\lambda }{\Gamma (\alpha )}B(1-\theta _{1}, \alpha -\theta _{2})+\frac{2\lambda }{\Gamma (\alpha +1)}B(1-\theta _{1}, \alpha -\theta _{2}+1) \biggr]. $$
Consider a ball \(B_{r} =\{x\in X=C([0,1],R):\Vert x \Vert \leq r\}\). Define two operators \(A_{1}\) and \(A_{1}\) on \(B_{r}\) as
$$\begin{aligned}& (A_{1}x) (t)=\frac{1}{\Gamma (\alpha )} \int _{0}^{t}(t-\tau )^{\alpha -1}f \bigl( \tau ,x(\tau ) \bigr)\,d\tau , \\& (A_{2}x) (t)=\frac{\lambda t^{2}}{\Gamma (\alpha +1)} \int _{0}^{1}(1- \tau )^{\alpha }f \bigl(\tau ,x( \tau ) \bigr)\,d\tau \\& \hphantom{(A_{1}x) (t)=}{}- \frac{\lambda t^{2}}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{\alpha -1}f \bigl( \tau ,x(\tau ) \bigr)\,d\tau \\& \hphantom{(A_{2}x) (t)=}{}-\frac{\lambda t^{2}}{\Gamma (\alpha +1)} \int _{0}^{\gamma }(\gamma -\tau )^{\alpha }f \bigl( \tau ,x(\tau ) \bigr)\,d\tau . \end{aligned}$$
For \(x,y\in B_{r} \), by (3.1) we can check that
$$\begin{aligned} \Vert A_{1}x+A_{2}y \Vert \leq{}& \frac{M_{0}}{\Gamma (\alpha )} \int _{0}^{1} (1-\tau )^{\alpha - \theta _{2}-1} \tau ^{-\theta _{1}}\,d\tau + \frac{\lambda M_{0}}{\Gamma (\alpha +1)} \int _{0}^{1} (1-\tau )^{ \alpha -\theta _{2}} \tau ^{-\theta _{1}}\,d\tau \\ & {}+\frac{\lambda M_{0}}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{ \alpha -\theta _{2}-1}\tau ^{-\theta _{1}}\,d\tau + \frac{ \lambda M_{0}}{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{ \alpha -\theta _{2}}\tau ^{-\theta _{1}}\,d\tau \\ ={}&M_{0} \biggl[ \frac{(1+\lambda )}{\Gamma (\alpha )}B(1-\theta _{1}, \alpha - \theta _{2})+ \frac{2\lambda }{\Gamma (\alpha +1)}B(1-\theta _{1}, \alpha - \theta _{2}+1) \biggr] \\ \leq{}& r. \end{aligned}$$
So \(A_{1}x+A_{2}y \in B_{r}\). Like in the proof of Theorem 3.1, from (H2), (H3), and (3.4) we can conclude that the operator \(A_{2}\) is also a contraction mapping. Lemma 3.2 and (H1) ensure the continuity of the operator \(A_{1}\). For any \(x \in B_{r}\), we have
$$\begin{aligned} \Vert A_{1}x \Vert & \leq \frac{M_{0}}{\Gamma (\alpha )} \int _{0}^{1} (1-\tau )^{\alpha -1} \tau ^{-\theta _{1}}(1-\tau )^{-}\theta _{2}\,d\tau \\ &\leq \frac{M_{0}}{\Gamma (\alpha )} \int _{0}^{1} (1-\tau )^{ \alpha -\theta _{2}-1} \tau ^{-\theta _{1}}\,d\tau \\ & = \frac{M_{0}}{\Gamma (\alpha )}B(1-\theta _{1},\alpha -\theta _{2}). \end{aligned}$$
Thus \(A_{1}\) is uniformly bounded on \(B_{r}\). For all \(t_{1},t_{2} \in [0,1]\) such that \(t_{1}< t_{2} \), we obtain
$$\begin{aligned}& \bigl\vert (A_{1}x) (t_{2})-(A_{1}x) (t_{1}) \bigr\vert \\& \quad = \frac{1}{\Gamma (\alpha )} \biggl\vert \int _{0}^{t_{2}}(t_{2}-\tau )^{ \alpha -1}f \bigl(\tau ,x(\tau ) \bigr)\,d\tau - \int _{0}^{t_{1}}(t_{1}-\tau )^{ \alpha -1}f \bigl(\tau ,x(\tau ) \bigr)\,d\tau \biggr\vert \\& \quad \leq \frac{1}{\Gamma (\alpha )} \biggl\vert \int _{0}^{t_{1}} \bigl[(t_{2}-\tau )^{ \alpha -1}-(t_{1}-\tau )^{\alpha -1} \bigr]f \bigl(\tau ,x( \tau ) \bigr)\,d\tau + \int _{t_{1}}^{t_{2}}(t_{2}- \tau )^{\alpha -1}f \bigl(\tau ,x(\tau ) \bigr)\,d\tau \biggr\vert \\& \quad \leq \frac{M_{0}}{\Gamma (\alpha )} \biggl[ \int _{0}^{t_{1}} \bigl[(t_{2}- \tau )^{\alpha -1}-(t_{1}-\tau )^{\alpha -1} \bigr]\tau ^{-\theta _{1}}(1- \tau )^{-\theta _{2}}\,d\tau \\& \qquad {} + \int _{t_{1}}^{t_{2}}(t_{2}-\tau )^{ \alpha -1}\tau ^{-\theta _{1}}(1-\tau )^{-\theta _{2}}\,d\tau \biggr] \\& \quad = \frac{M_{0}}{\Gamma (\alpha )} \bigl[J(t_{2})-J(t_{1}) \bigr]. \end{aligned}$$
By Lemma 3.1 we have
$$ \bigl\vert (A_{1}x) (t_{2})-(A_{1}x) (t_{1}) \bigr\vert = \frac{(\alpha -1)M_{0}}{\Gamma (\alpha )}B(1-\theta _{1}, \alpha - \theta _{2}-1) (t_{2}-t_{1}). $$
This means that \(A_{1} \) is equicontinuous and relatively compact on \(B_{r} \). Accordingly, by the Arzelà–Ascoli theorem \(A_{1} \) is compact on \(B_{r} \). Accordingly, Lemma 2.4 ensures the existence of a solution for problem (1.1) in \([0,1] \). □
The Schaefer fixed point theorem gives the last result.
Theorem 3.3
Let \(2<\alpha \leq 3\) and \(0<\theta _{1}\), \(\theta _{2}<1\) be constants, and let \(f(t,x(t))\) satisfy conditions (H1) and (3.1). Then problem (1.1) has a solution in \([0,1] \).
Proof
By Lemma 3.3 we know that the operator \(A:X\rightarrow X\) is completely continuous.
Next, we prove that the set \(V=\{x \in C([0,1],R):x=\mu Ax,0< \mu <1\}\) is bounded.
Let \(x \in V\). Then \(x=\mu (Ax)\). Thus, for each \(t\in [0,1]\), we have
$$\begin{aligned} \vert x \vert ={}& \mu \bigl\vert (Ax) (t) \bigr\vert \\ ={}&\mu \biggl\vert \frac{1}{\Gamma (\alpha )} \int _{0}^{t}(t-\tau )^{\alpha -1}f \bigl(\tau ,x( \tau ) \bigr)\,d\tau +\frac{\lambda t^{2}}{\Gamma (\alpha +1)} \int _{0}^{1}(1- \tau )^{\alpha }f \bigl(\tau ,x( \tau ) \bigr)\,d\tau \\ &{} -\frac{\lambda t^{2}}{\Gamma (\alpha )} \int _{0}^{t}(1-\tau )^{ \alpha -1}f \bigl(\tau ,x(\tau ) \bigr)\,d\tau - \frac{\lambda t^{2}}{\Gamma (\alpha +1)} \int _{0}^{\gamma }(\gamma - \tau )^{\alpha }f \bigl( \tau ,x(\tau ) \bigr)\,d\tau \biggr\vert \\ \leq{}& M_{0} \biggl[\frac{1}{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{ \alpha -\theta _{2}-1}\tau ^{-\theta _{1}}\,d\tau + \frac{\lambda }{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{\alpha - \theta _{2}}\tau ^{-\theta _{1}}\,d\tau \biggr] \\ &{}+\frac{\lambda }{\Gamma (\alpha )} \int _{0}^{1}(1-\tau )^{\alpha - \theta _{2}-1}\tau ^{-\theta _{1}}\,d\tau + \frac{\lambda }{\Gamma (\alpha +1)} \int _{0}^{1}(1-\tau )^{\alpha - \theta _{2}}\tau ^{-\theta _{1}}\,d\tau ] \\ ={} &M_{0} \biggl[\frac{1+\lambda }{\Gamma (\alpha )}B(1-\theta _{1},\alpha - \theta _{2})+\frac{2\lambda }{\Gamma (\alpha +1)}B(1-\theta _{1}, \alpha - \theta _{2}+1) \biggr]=L_{2}. \end{aligned}$$
Hence we have
$$ \Vert x \Vert \leq L_{2}. $$
This shows that the set V is bounded. Lemma 2.5 ensures the existence of fixed points of A. Accordingly, there is at least one solution to problem (1.1) in \([0,1]\). □