Let \(\mathcal{C} = C([1,T],\mathbb{R})\) be the Banach space of all continuous functions from \([1,T]\) to \(\mathbb{R}\) endowed with the norm defined by \(\Vert x\Vert =\sup\{\vert x(t)\vert , t\in[1,T]\}\). Define an operator \(\mathcal{A}:\mathcal{C}\rightarrow\mathcal{C}\) by
$$\begin{aligned} \mathcal{A}x(t) =& I^{\alpha}\biggl( \frac{1}{p}I^{\beta}g_{x} \biggr) (t) - I^{\alpha}\biggl( \frac{r}{p} x \biggr) (t) \\ &{}+ \biggl( \frac{-\mu}{\mu+1} I^{\beta}(g_{x}) (T) + \frac{\xi}{\mu +1}x(T) \biggr) I^{\alpha}\biggl( \frac{1}{p} \biggr) (t) \\ & {}-\frac{1}{2} \biggl[I^{\alpha}\biggl( \frac{1}{p}I^{\beta}g_{x} \biggr) (T) - I^{\alpha}\biggl( \frac{r}{p} x \biggr) (T) \\ &{}+ \biggl(\frac{-\mu}{\mu+1} I^{\beta}(g_{x}) (T) + \frac{\xi}{\mu +1}x(T) \biggr) \eta \biggr]. \end{aligned}$$
(3.1)
Observe that the problem (1.2) has solutions if and only if the operator \(\mathcal{A}\) has fixed points.
Define the constants \(p_{*} = \inf_{t \in[1,T]} \vert p(t)\vert \) and \(r^{*}= \sup_{t \in[1,T]}\vert r(t)\vert \) and set
$$\begin{aligned}& \Omega_{1} = \frac{3(\log T)^{\alpha+\beta}}{2p_{*}\Gamma(\alpha +\beta+1)}+\frac{3\mu \vert \eta \vert (\log T)^{\beta}}{2(\mu+1)\Gamma (\beta+1)}, \end{aligned}$$
(3.2)
$$\begin{aligned}& \Omega_{2} = \frac{3r^{*}(\log T)^{\alpha}}{2p_{*}\Gamma(\alpha +1)}+\frac{3\vert \xi \vert \vert \eta \vert }{2(\mu+1)} , \end{aligned}$$
(3.3)
$$\begin{aligned}& \Omega_{3} = \Omega_{1}L+\Omega_{2}. \end{aligned}$$
(3.4)
3.1 Existence and uniqueness result
In this subsection we give one existence and uniqueness result, by using the Banach contraction mapping principle.
Theorem 3.1
Assume that:
-
\((H_{1})\)
:
-
there exists a constant
\(L>0\)
such that
\(\vert g(t,x)-g(t,y)\vert \leq L\vert x-y\vert \), for each
\(t \in[1,T]\)
and
\(x,y \in \mathbb{R}\).
If
$$\begin{aligned} \Omega_{3} < 1, \end{aligned}$$
(3.5)
where
\(\Omega_{3}\)
is defined by (3.3), then the problem (1.2) has a unique solution on
\([1,T]\).
Proof
To prove that the problem (1.2) has a unique solution, we consider a fixed point problem, \(x = \mathcal {A}x\), where the operator \(\mathcal{A}\) is defined as in (3.1). To accomplish our purpose, we apply the Banach contraction mapping principle to show that \(\mathcal{A}\) has a unique fixed point.
We define \(\sup_{t\in[1, T]}\vert g(t, 0)\vert =M<\infty\), and choose
$$ R\geq\frac{\Omega_{1} M}{1-\Omega_{3}}, $$
(3.6)
where \(\Omega_{1}\) and \(\Omega_{3}\) are defined by (3.2) and (3.4), respectively. Now, we show that \(\mathcal{A} B_{R}\subset B_{R}\), where \(B_{R}=\{x \in\mathcal{C} : \Vert x\Vert \leq R\}\). For any \(x\in B_{R}\), we have
$$\begin{aligned} \bigl\vert \mathcal{A}x(t)\bigr\vert \leq& I^{\alpha}\biggl( \frac{1}{\vert p\vert }I^{\beta} \bigl(\vert g_{x}\vert \bigr) \biggr) (t) + I^{\alpha}\biggl( \frac{\vert r\vert }{\vert p\vert } \vert x\vert \biggr) (t) \\ &{}+ \biggl( \frac{\mu}{(\mu+1)} I^{\beta} \bigl(\vert g_{x} \vert \bigr) (T) + \frac{\vert \xi \vert }{(\mu+1)}\bigl\vert x(T)\bigr\vert \biggr) I^{\alpha} \biggl( \frac{1}{\vert p\vert } \biggr) (t) \\ &{} +\frac{1}{2} \biggl[I^{\alpha}\biggl( \frac{1}{\vert p\vert }I^{\beta} \vert g_{x}\vert \biggr) (T) + I^{\alpha} \biggl( \frac{\vert r\vert }{\vert p\vert } \vert x\vert \biggr) (T) \\ &{}+ \biggl(\frac{\mu}{(\mu+1)} I^{\beta} \bigl(\vert g_{x} \vert \bigr) (T) + \frac{\vert \xi \vert }{(\mu+1)}\bigl\vert x(T)\bigr\vert \biggr) \vert \eta \vert \biggr] \\ \leq& I^{\alpha}\biggl( \frac{1}{\vert p\vert }I^{\beta}\bigl(\vert g_{x}-g_{0}\vert +\vert g_{0} \vert \bigr) \biggr) (T) + I^{\alpha}\biggl( \frac{\vert r\vert }{\vert p\vert }\vert x \vert \biggr) (T) \\ &{}+ \biggl( \frac{\mu}{(\mu+1)} I^{\beta}\bigl(\vert g_{x}-g_{0}\vert +\vert g_{0}\vert \bigr) (T) + \frac{\vert \xi \vert }{(\mu+1)}\bigl\vert x(T)\bigr\vert \biggr) I^{\alpha}\biggl( \frac {1}{\vert p\vert } \biggr) (T) \\ &{} +\frac{1}{2} I^{\alpha}\biggl( \frac{1}{\vert p\vert }I^{\beta}\bigl(\vert g_{x}-g_{0}\vert +\vert g_{0} \vert \bigr) \biggr) (T) + \frac{1}{2} I^{\alpha}\biggl( \frac{\vert r\vert }{\vert p\vert }\vert x\vert \biggr) (T) \\ &{}+ \frac{1}{2} \biggl( \frac{\mu}{(\mu+1)} I^{\beta}\bigl(\vert g_{x}-g_{0}\vert +\vert g_{0}\vert \bigr) (T) + \frac{\vert \xi \vert }{(\mu+1)}\bigl\vert x(T)\bigr\vert \biggr) \vert \eta \vert \\ \leq& \frac{3}{2} \biggl( \frac{RL+M}{p_{*}} \biggr) \frac{(\log T)^{\alpha+\beta}}{\Gamma(\alpha+ \beta+1)} + \frac{3}{2} \biggl( \frac{Rr^{*}}{p_{*}} \biggr) \frac{(\log T)^{\alpha}}{\Gamma(\alpha+ 1)} \\ &{}+ \frac{3}{2} \biggl( \frac{\mu \vert \eta \vert }{(\mu+1)} \biggr) (RL+M) \frac{(\log T)^{\beta}}{\Gamma(\beta+ 1)}+ \frac{3\vert \xi \vert \vert \eta \vert }{2(\mu+1)} R \\ = & \Omega_{1}M+\Omega_{3}R \leq R. \end{aligned}$$
This implies that \(\mathcal{A} B_{R}\subset B_{R}\).
By \((H_{1})\), for any \(x,y\in B_{R}\), we have
$$\begin{aligned}& \bigl\vert \mathcal{A}x(t)-\mathcal{A}y(t)\bigr\vert \\& \quad \leq \frac{1}{p_{*}} I^{\alpha+\beta}\bigl(\vert g_{x}-g_{y} \vert \bigr) (T)+\frac {r^{*}}{p_{*}}\Vert x-y\Vert I^{\alpha}(1) (T)+ \frac{\vert \eta \vert \mu}{(\mu+1)} I^{\beta }\bigl(\vert g_{x}-g_{y} \vert \bigr) (T) \\& \quad\quad{} + \vert \eta \vert \Vert x-y\Vert \frac{\vert \xi \vert }{(\mu+1)} + \frac{1}{2p_{*}}I^{\alpha +\beta}\bigl(\vert g_{x}-g_{y} \vert \bigr) (T)+ \frac{r^{*}}{2p_{*}}\Vert x-y\Vert I^{\alpha}(1) (T) \\& \quad\quad{} + \frac{\vert \eta \vert \mu}{2(\mu+1)} I^{\beta}\bigl(\vert g_{x}-g_{y} \vert \bigr) (T)+ \frac {\vert \eta \vert \vert \xi \vert }{2(\mu+1)} \Vert x-y\Vert \\& \quad \leq \frac{L\Vert x-y\Vert (\log T)^{\alpha+\beta}}{p_{*} \Gamma(\alpha +\beta+1)} + \frac{r^{*}\Vert x-y\Vert (\log T)^{\alpha}}{p_{*} \Gamma(\alpha +1)} + \frac{\vert \eta \vert \mu}{(\mu+1)} \frac{L\Vert x-y\Vert (\log T)^{\beta}}{ \Gamma(\beta+1)} \\& \quad\quad{} + \frac{\vert \eta \vert \vert \xi \vert }{(\mu+1)} \Vert x-y\Vert + \frac{L\Vert x-y\Vert (\log T)^{\alpha+\beta}}{2p_{*} \Gamma(\alpha+\beta+1)}+ \frac{r^{*}\Vert x-y\Vert (\log T)^{\alpha}}{2p_{*} \Gamma(\alpha+1)} \\& \quad\quad{} + \frac{\vert \eta \vert \mu}{2(\mu+1)} \frac{L\Vert x-y\Vert (\log T)^{\beta}}{\Gamma(\beta+1)}+\frac{\vert \eta \vert \vert \xi \vert }{2(\mu+1)} \Vert x-y\Vert \\& \quad = \Omega_{3}\Vert x-y\Vert . \end{aligned}$$
As \(\Omega_{3}<1\), \(\mathcal{A}\) is a contraction. Therefore, we see from the Banach contraction mapping principle that the operator \(\mathcal{A}\) has a fixed point which is the unique solution of the boundary value problem (1.2). This completes the proof. □
If \(r(t)\equiv0\) for \(t\in[1,T]\), then we have \(\xi=0\) and \(r^{*}=0\) and we also get the following result.
Corollary 3.1
Suppose that the condition
\((H_{1})\)
holds. If
\(\Omega _{1}L<1\), where
\(\Omega_{1}\)
is defined by (3.2), then the problem (1.3) has a unique solution on
\([1,T]\).
If \(p(t)\equiv1\) and \(r(t)\equiv\lambda\) for \(t\in[1,T]\) and \(\lambda\in\mathbb{R}\), then we obtain \(p_{*}=1\), \(r^{*}=\vert \lambda \vert \), \(\mu=1\), \(\xi=0\), \(\eta= \frac{(\log T)^{\alpha }}{\Gamma(\alpha+1)}\), and the following corollary.
Corollary 3.2
Assume that the condition
\((H_{1})\)
is satisfied. If
$$ \frac{3}{2} \biggl(\frac{1}{\Gamma(\alpha+\beta+1)}+\frac {1}{2\Gamma(\alpha+1)\Gamma(\beta+1)} \biggr)L(\log T)^{\alpha +\beta}+\frac{3}{2}\vert \lambda \vert \frac{(\log T)^{\alpha}}{\Gamma (\alpha+1)}< 1, $$
then the problem (1.4) has a unique solution on
\([1,T]\).
3.2 Existence results
Now we give an existence result via Leray-Schauder nonlinear alternative.
Theorem 3.2
Nonlinear alternative for single valued maps [31]
Let
E
be a Banach space, C
a closed, convex subset of
E, U
an open subset of
C
and
\(0\in U\). Suppose that
\(\mathcal{A}:\bar{U}\to C\)
is a continuous, compact (that is, \(\mathcal{A}(\bar{U})\)
is a relatively compact subset of
C) map. Then either
-
(i)
\(\mathcal{A}\)
has a fixed point in
Ū, or
-
(ii)
there is a
\(x\in\partial U\) (the boundary of
U
in
C) and
\(\lambda\in(0,1)\)
with
\(x = \lambda \mathcal{A}(x)\).
Theorem 3.3
Assume that:
-
\((H_{2})\)
:
-
there exist a continuous nondecreasing function
\(\psi: [0,\infty) \to(0,\infty)\)
and a function
\(\varphi\in C([1,T],\mathbb{R}^{+})\)
such that
$$ \bigl\vert g(t,x)\bigr\vert \le\varphi(t)\psi\bigl(\vert x\vert \bigr)\quad \textit{for each } (t,x) \in[1,T] \times\mathbb{R}; $$
(3.7)
-
\((H_{3})\)
:
-
there exists a constant
\(M>0\)
such that
$$ \frac{(1-\Omega_{2})M}{\Vert \varphi \Vert \psi(M)\Omega_{1} }> 1,\quad\Omega_{2}< 1, $$
(3.8)
where
\(\Omega_{1}\), \(\Omega_{2}\)
are defined by (3.2) and (3.3), respectively.
Then the boundary value problem (1.2) has at least one solution on
\([1,T]\).
Proof
Let the operator \(\mathcal{A}\) be defined as in (3.1). Now, we are going to prove that the operator \(\mathcal{A}\)
maps bounded sets (balls) into bounded sets in
\(C([1,T], \mathbb{R})\). For \(\rho>0\), we define a bounded ball \(B_{\rho}= \{x \in C([1,T], \mathbb{R}): \Vert x\Vert \le\rho \}\). Then, for \(t\in[1,T]\), we have
$$\begin{aligned} \bigl\vert (\mathcal{A}x) (t)\bigr\vert \leq& I^{\alpha}\biggl( \frac{1}{\vert p\vert }I^{\beta} \bigl(\vert g_{x}\vert \bigr) \biggr) (t) + I^{\alpha}\biggl( \frac{\vert r\vert }{\vert p\vert } \vert x\vert \biggr) (t)+ \biggl( \frac {\mu}{(\mu+1)} I^{\beta} \bigl(\vert g_{x} \vert \bigr) (T) \\ &{} + \frac{\vert \xi \vert }{(\mu+1)}\bigl\vert x(T)\bigr\vert \biggr) I^{\alpha} \biggl( \frac {1}{\vert p\vert } \biggr) (t)+\frac{1}{2} \biggl[I^{\alpha}\biggl( \frac {1}{\vert p\vert }I^{\beta} \vert g_{x}\vert \biggr) (T) \\ &{} + I^{\alpha} \biggl( \frac{\vert r\vert }{\vert p\vert } \vert x\vert \biggr) (T) + \biggl(\frac{\mu}{(\mu+1)} I^{\beta} \bigl(\vert g_{x}\vert \bigr) (T) + \frac{\vert \xi \vert }{(\mu +1)}\bigl\vert x(T)\bigr\vert \biggr) \vert \eta \vert \biggr] \\ \leq& \frac{\Vert \varphi \Vert \psi(\vert x\vert )}{p_{*}}I^{\alpha+\beta}(1) (T) + \frac{\rho r^{*}}{p_{*}}I^{\alpha}(1) (T) + \frac{\vert \eta \vert \mu \Vert \varphi \Vert \psi(\vert x\vert )}{(\mu+1)}I^{\beta}(1) (T) \\ &{}+\frac{\vert \eta \vert \vert \xi \vert \rho}{(\mu+1)} +\frac{\Vert \varphi \Vert \psi(\vert x\vert )}{2p_{*}}I^{\alpha+\beta}(1) (T)+ \frac {\rho r^{*}}{2p_{*}}I^{\alpha}(1) (T) \\ &{}+\frac{\vert \eta \vert \mu \Vert \varphi \Vert \psi(\vert x\vert )}{2(\mu+1)}I^{\beta}(1) (T) + \frac{\vert \eta \vert \vert \xi \vert \rho}{2(\mu+1)} \\ \leq& \Vert \varphi \Vert \psi(\rho)\Omega_{1}+\rho \Omega_{2}, \end{aligned}$$
which leads to
$$ \Vert \mathcal{A}x\Vert \le \Vert \varphi \Vert \psi(\rho) \Omega_{1}+\rho\Omega_{2}. $$
(3.9)
Next we will show that the operator \(\mathcal{A}\)
maps bounded sets into equicontinuous sets of
\(C([0,T], \mathbb{R})\)
. Let \(\tau_{1}, \tau_{2} \in[1,T]\) such that \(\tau_{1}< \tau_{2}\) and \(x \in B_{\rho}\). Then we have
$$\begin{aligned}& \bigl\vert (\mathcal{A}x) (\tau_{2})-(\mathcal{A}x) ( \tau_{1})\bigr\vert \\& \quad \le \frac{\Vert \varphi \Vert \psi(\rho)}{p_{*}\Gamma(\alpha +\beta)} \biggl[ \int_{1}^{\tau_{2}} \biggl(\log\frac{\tau_{2}}{s} \biggr)^{\alpha+\beta-1}\frac{ds}{s} - \int_{1}^{\tau_{1}} \biggl(\log\frac{\tau_{1}}{s} \biggr)^{\alpha+\beta -1}\frac{ds}{s} \biggr] \\& \quad\quad{} +\frac{\rho r^{*}}{p_{*}\Gamma(\alpha)} \biggl[ \int _{1}^{\tau_{2}} \biggl(\log\frac{\tau_{2}}{s} \biggr)^{\alpha-1}\frac{ds}{s} - \int_{1}^{\tau_{1}} \biggl(\log\frac{\tau_{1}}{s} \biggr)^{\alpha -1}\frac{ds}{s} \biggr] \\& \quad\quad{} + \biggl(\frac{\mu \Vert \varphi \Vert \psi(\rho)(\log T)^{\beta}}{(\mu+1)\Gamma(\beta+1)}+\frac{\rho \vert \xi \vert }{(\mu +1)} \biggr)\cdot\frac{1}{p_{*}\Gamma(\alpha)} \\& \quad\quad{} \times \biggl[ \int_{1}^{\tau_{2}} \biggl(\log\frac{\tau_{2}}{s} \biggr)^{\alpha-1}\frac{ds}{s} - \int_{1}^{\tau_{1}} \biggl(\log\frac{\tau_{1}}{s} \biggr)^{\alpha -1}\frac{ds}{s} \biggr] \\& \quad = \frac{\Vert \varphi \Vert \psi(\rho)}{p_{*}\Gamma(\alpha +\beta+1)}\bigl\vert (\log\tau_{2} )^{\alpha+\beta}- ( \log\tau_{1} )^{\alpha+\beta}\bigr\vert +\frac{\rho r^{*}}{p_{*}\Gamma(\alpha+1)}\bigl\vert (\log\tau_{2} )^{\alpha }- (\log\tau_{1} )^{\alpha}\bigr\vert \\& \quad\quad{} + \biggl(\frac{\mu \Vert \varphi \Vert \psi(\rho)(\log T)^{\beta}}{(\mu+1)\Gamma(\beta+1)}+\frac{\rho \vert \xi \vert }{(\mu +1)} \biggr)\cdot\frac{1}{p_{*}\Gamma(\alpha+1)} \cdot\bigl\vert (\log\tau_{2} )^{\alpha}- (\log \tau_{1} )^{\alpha }\bigr\vert . \end{aligned}$$
As \(\tau_{1}\rightarrow\tau_{2}\), the right-hand side of the above inequality tends to zero independently of \(x\in B_{\rho}\). Therefore by the Arzelá-Ascoli theorem the operator \(\mathcal{A}: C([0,T], \mathbb{R}) \to C([0,T], \mathbb{R})\) is completely continuous.
The result will follow from the Leray-Schauder nonlinear alternative (Theorem 3.2) once we have proved the boundedness of the set of the solutions to the equations \(x=\nu\mathcal{A} x\) for \(\nu\in(0,1)\).
Let x be a solution of the operator equation \(x=\mathcal{A}x\). Then, for \(t\in[1,T]\), by directly computation, we have
$$ \bigl\vert x(t)\bigr\vert \le \Vert \varphi \Vert \psi\bigl(\Vert x \Vert \bigr)\Omega_{1}+\Vert x\Vert \Omega_{2}, $$
(3.10)
which yields
$$ \frac{(1-\Omega_{2})\Vert x\Vert }{\Vert \varphi \Vert \psi(\Vert x\Vert )\Omega_{1}}\leq1, $$
where the constants \(\Omega_{1}\) and \(\Omega_{2}\) are defined by (3.2) and (3.3), respectively. From \((H_{3})\), there exists a positive constant M such that \(\Vert x\Vert \ne M\). Let us set
$$\begin{aligned} U=\bigl\{ x\in C\bigl([1, T], \mathbb{R}\bigr) : \Vert x\Vert < M\bigr\} . \end{aligned}$$
We observe that the operator \(\mathcal{A}:\overline{U}\rightarrow C([1, T], \mathbb{R})\) is continuous and completely continuous. From the choice of U, there is no \(x\in\partial U\) such that \(x=\nu \mathcal{A}x\) for some \(\nu\in(0,1)\). Consequently, by the nonlinear alternative of Leray-Schauder type (Theorem 3.2), we see that the operator \(\mathcal{A}\) has a fixed point \(x\in\overline{U}\) which is a solution of the problem (1.2). The proof is completed. □
Corollary 3.3
Suppose that the condition
\((H_{2})\)
is satisfied. If there exists a positive constant
M
such that
$$ \frac{M}{\Vert \varphi \Vert \psi(M)\Omega_{1}}> 1, $$
(3.11)
then the problem (1.3) has at least one solution on
\([1,T]\).
Corollary 3.4
Assume that the condition
\((H_{2})\)
is fulfilled. If there exists a positive constant
M
such that
$$ \frac{ (1-\frac{3}{2}\vert \lambda \vert \frac{(\log T)^{\alpha}}{\Gamma(\alpha+1)} )M}{ \frac{3}{2}\Vert \varphi \Vert \psi(M) (\frac{1}{\Gamma(\alpha+\beta+1)}+\frac {1}{2\Gamma(\alpha+1)\Gamma(\beta+1)} )(\log T)^{\alpha +\beta}}>1, $$
(3.12)
with
$$ \frac{3}{2}\vert \lambda \vert \frac{(\log T)^{\alpha}}{ \Gamma(\alpha+1)}< 1, $$
(3.13)
then the problem (1.4) has at least one solution on
\([1,T]\).
Our last existence result is based on Krasnoselskii’s fixed point theorem.
Theorem 3.4
Krasnoselskii’s fixed point theorem [32]
Let
M
be a closed, bounded, convex, and nonempty subset of a Banach space
X. Let
A, B
be operators such that
-
(a)
\(Ax+By \in M\)
where
\(x,y \in M\);
-
(b)
A
is compact and continuous;
-
(c)
B
is a contraction mapping.
Then there exists
\(z \in M\)
such that
\(z = Az+Bz\).
Theorem 3.5
Let
\(g : [1,T]\times\mathbb{R} \rightarrow\mathbb{R}\)
be a continuous function satisfying
\((H_{1})\)
in Theorem
3.1. In addition, assume that:
-
\((H_{4})\)
:
-
\(\vert g(t,x)\vert \leq\omega(t)\), \(\forall(t,x) \in [1,T]\times\mathbb{R}\), and
\(\omega\in C([1,T],\mathbb{R}^{+})\).
If
$$ L\Omega_{1} < 1 \quad\textit{and}\quad\Omega_{2}< 1, $$
(3.14)
where
\(\Omega_{1}\)
and
\(\Omega_{2}\)
are defined by (3.2) and (3.3), respectively, then the boundary value problem (1.2) has at least one solution on
\([1,T]\).
Proof
We decompose the operator \(\mathcal{A}\) defined in (3.1), into two operators \(\mathcal{A}_{1}\) and \(\mathcal {A}_{2}\) on \(B_{\varrho} = \{x \in\mathcal{C}:\Vert x\Vert \leq\varrho\}\) by
$$\begin{aligned}& \begin{aligned} \mathcal{A}_{1} x(t) &= - I^{\alpha}\biggl( \frac{r}{p} x \biggr) (t)+ \frac{\xi}{\mu+1}x(T)I^{\alpha}\biggl( \frac{1}{p} \biggr) (t)+\frac{1}{2}I^{\alpha}\biggl( \frac{r}{p} x \biggr) (T) \\ &\quad {}- \frac{\eta\xi}{2(\mu+1)}x(T), \end{aligned} \\ & \begin{aligned} \mathcal{A}_{2} x(t) &= I^{\alpha}\biggl( \frac{1}{p}I^{\beta}g_{x} \biggr) (t)+\frac{-\mu}{\mu+1} I^{\beta}(g_{x}) (T) I^{\alpha}\biggl( \frac {1}{p} \biggr) (t) -\frac{1}{2} I^{\alpha}\biggl( \frac{1}{p}I^{\beta}g_{x} \biggr) (T) \\ &\quad{} + \frac{\eta\mu}{2(\mu+1)} I^{\beta}(g_{x}) (T), \end{aligned} \end{aligned}$$
with ϱ satisfying
$$ \varrho\geq\frac{\Vert \omega \Vert \Omega_{1}}{1-\Omega_{2}}, $$
(3.15)
and \(\Vert \omega \Vert = \sup_{t\in[1,T]} \vert \omega(t)\vert \). Note that the ball \(B_{\varrho}\) is a closed, bounded, and convex subset of the Banach space \(\mathcal{C}\).
To prove that \(\mathcal{A}_{1} x + \mathcal{A}_{2} y \in B_{\varrho}\), we let \(x,y \in B_{\varrho}\). Then we get
$$\begin{aligned}& \bigl\vert \mathcal{A}_{1} x(t)+\mathcal{A}_{2} y(t) \bigr\vert \\ & \quad \leq I^{\alpha}\biggl( \frac{\vert r\vert }{\vert p\vert } \vert x\vert \biggr) (t)+ \frac{\vert \xi \vert }{(\mu+1)}\bigl\vert x(T)\bigr\vert I^{\alpha}\biggl( \frac{1}{\vert p\vert } \biggr) (t)+\frac {1}{2}I^{\alpha}\biggl( \frac{\vert r\vert }{\vert p\vert } \vert x\vert \biggr) (T) \\& \quad\quad{} + \frac{\vert \eta \vert \vert \xi \vert }{2(\mu+1)}\bigl\vert x(T)\bigr\vert +I^{\alpha}\biggl( \frac {1}{\vert p\vert }I^{\beta}\bigl(\vert g_{y}\vert \bigr) \biggr) (t)+\frac{\mu}{(\mu+1)} I^{\beta}\bigl(\vert g_{y} \vert \bigr) (T) I^{\alpha}\biggl( \frac{1}{\vert p\vert } \biggr) (t) \\& \quad\quad{} + \frac{1}{2} I^{\alpha}\biggl( \frac{1}{\vert p\vert }I^{\beta}\bigl(\vert g_{y}\vert \bigr) \biggr) (T)+\frac{\vert \eta \vert \mu}{2(\mu+1)} I^{\beta}\bigl(\vert g_{y}\vert \bigr) (T) \\& \quad \leq \varrho\Omega_{2}+\Vert \omega \Vert \Omega_{1} \leq \varrho. \end{aligned}$$
It follows that \(\mathcal{A}_{1} x + \mathcal{A}_{2} y \in B_{\varrho}\). Thus condition (a) of Theorem 3.4 is satisfied. To prove that \(\mathcal{A}_{2}\) is a contraction mapping, for \(x,y\in B_{\varrho}\), we have
$$\begin{aligned} \bigl\vert \mathcal{A}_{2} x(t) - \mathcal{A}_{2} y(t) \bigr\vert \leq& I^{\alpha}\biggl( \frac{1}{\vert p\vert }I^{\beta}\bigl(\vert g_{x}-g_{y}\vert \bigr) \biggr) (t)+ \frac{\mu}{(\mu+1)} I^{\beta}\bigl(\vert g_{x}-g_{y} \vert \bigr) (T) I^{\alpha}\biggl( \frac{1}{\vert p\vert } \biggr) (t) \\ &{}+ \frac{1}{2} I^{\alpha}\biggl( \frac{1}{\vert p\vert }I^{\beta}\bigl(\vert g_{x}-g_{y}\vert \bigr) \biggr) (T)+ \frac{\vert \eta \vert \mu}{2(\mu+1)} I^{\beta}\bigl(\vert g_{x}-g_{y} \vert \bigr) (T) \\ \leq& L\Omega_{1}\Vert x-y\Vert , \end{aligned}$$
which is a contraction, since \(L\Omega_{1}<1\). Therefore, the condition (c) of Theorem 3.4 is fulfilled.
By using the continuity of the function g, we deduce that the operator \(\mathcal{A}_{1}\) is continuous. For \(x \in B_{\varrho}\), it follows that
$$ \Vert \mathcal{A}_{1} x\Vert \leq\varrho\Omega_{2}, $$
which implies that the operator \(\mathcal{A}_{1}\) is uniformly bounded on \(B_{\varrho}\). Now we are going to prove that \(\mathcal{A}_{1}\) is equicontinuous. For \(\tau_{1}, \tau_{2}\in[1,T]\) such that \(\tau_{1} <\tau_{2}\) and for \(x\in B_{\varrho}\), we have
$$\begin{aligned}& \bigl\vert \mathcal{A}_{1} x(\tau_{2})- \mathcal{A}_{1} x(\tau_{1})\bigr\vert \\& \quad \leq \frac{\varrho r^{*}}{p_{*}\Gamma(\alpha+1)}\bigl\vert (\log \tau_{2} )^{\alpha}- (\log\tau_{1} )^{\alpha}\bigr\vert + \frac{\varrho \vert \xi \vert }{p_{*}(\mu+1)\Gamma(\alpha+1)}\bigl\vert (\log \tau_{2} )^{\alpha}- (\log\tau_{1} )^{\alpha}\bigr\vert , \end{aligned}$$
which is independent of x and tends to zero as \(\tau_{1} \rightarrow \tau_{2}\). Hence \(\mathcal{A}_{1}\) is equicontinuous. Therefore \(\mathcal {A}_{1}\) is relatively compact on \(B_{\varrho}\), and by Arzelá-Ascoli theorem, \(\mathcal{A}_{1}\) is compact on \(B_{\varrho}\). Thus the condition (b) of Theorem 3.4 is fulfilled. Therefore all conditions of Theorem 3.4 are satisfied, and consequently, the problem (1.2) has at least one solution on \([1,T]\). This completes the proof. □
Corollary 3.5
Suppose that the conditions
\((H_{1})\)
and
\((H_{4})\)
are satisfied. If
\(\Omega_{1}L<1\), where
\(\Omega_{1}\)
is defined by (3.2), then the problem (1.3) has at least one solution on
\([1,T]\).
Corollary 3.6
Assume that the conditions
\((H_{1})\)
and
\((H_{4})\)
are fulfilled. If
$$ \frac{3}{2} \biggl(\frac{1}{\Gamma(\alpha+\beta+1)}+\frac {1}{2\Gamma(\alpha+1)\Gamma(\beta+1)} \biggr)L(\log T)^{\alpha +\beta}< 1 \quad \textit{and} \quad \frac{3}{2}\vert \lambda \vert \frac{(\log T)^{\alpha}}{\Gamma(\alpha+1)}< 1, $$
then the problem (1.4) has at least one solution on
\([1,T]\).