In this section, we introduce some notations and definitions of fractional calculus [1] and present preliminary results needed in our proofs later.
Definition 2.1
[1]
The Hadamard fractional integral of order q for a function g is defined as
$$I^{q} g(t)=\frac{1}{\Gamma(q)} \int_{1}^{t} \biggl(\log\frac{t}{s} \biggr)^{q-1}\frac{g(s)}{s}\,ds,\quad q>0, $$
provided the integral exists.
Definition 2.2
[1]
The Hadamard derivative of fractional order q for a function \(g: [1, \infty)\to\mathbb{R}\) is defined as
$$D^{q} g(t)=\frac{1}{\Gamma(n-q)} \biggl(t\frac{d}{dt} \biggr)^{n} \int _{1}^{t} \biggl(\log\frac{t}{s} \biggr)^{n-q-1}\frac{g(s)}{s}\,ds, \quad n-1 < q < n, n=[q]+1, $$
where \([q]\) denotes the integer part of the real number q and \(\log(\cdot) =\log_{e} (\cdot)\).
Lemma 2.3
([1], Property 2.24)
If
\(a,\alpha, \beta>0\)
then
$$\biggl(D^{\alpha}_{a} \biggl(\log\frac{t}{a} \biggr)^{\beta-1} \biggr) (x)=\frac{\Gamma(\beta)}{\Gamma(\beta-\alpha)} \biggl(\log\frac {x}{a} \biggr)^{\beta-\alpha-1}. $$
Lemma 2.4
[1, 2]
Let
\(q > 0\)
and
\(x\in C[1,\infty)\cap L^{1}[1,\infty)\). Then the Hadamard fractional differential equation
\(D^{q}x(t)=0\)
has the solutions
$$ x(t) = \sum_{i=1}^{n}c_{i} ( \log t )^{q-i}, $$
and the following formula holds:
$$ I^{q}D^{q}x(t)=x(t)+\sum_{i=1}^{n}c_{i} (\log t )^{q-i}, $$
where
\(c_{i} \in\mathbb{R}\), \(i = 1,2,\ldots,n\), and
\(n-1< q< n\).
Lemma 2.5
Let
\(h\in C[1,\infty)\)
with
\(0<\int_{1}^{\infty}h(s)\frac{ds}{s}<\infty\), and
$$ \Omega=\Gamma(\alpha)-\sum_{i=1}^{m} \frac{\lambda_{i}\Gamma(\alpha )}{\Gamma(\alpha+\beta_{i})}(\log\eta)^{\alpha+\beta_{i}-1}>0. $$
(2.1)
Then the unique solution of the following fractional differential equation:
$$ D^{\alpha}u(t)+h(t)=0,\quad t\in(1,\infty), \alpha\in(1,2), $$
(2.2)
subject to the boundary conditions
$$ u(1)=0, \qquad D^{\alpha-1}u(\infty)=\sum _{i=1}^{m}\lambda_{i}I^{\beta _{i}}u( \eta), $$
(2.3)
is given by the integral equation
$$ u(t) = \int_{1}^{\infty}G(t,s)h(s)\frac{ds}{s}, $$
(2.4)
where
$$ G(t,s)=g(t,s)+\sum_{i=1}^{m} \frac{\lambda_{i}(\log t)^{\alpha -1}}{\Omega\Gamma(\alpha+\beta_{i})}g_{i}(\eta,s) $$
(2.5)
and
$$\begin{aligned}& g(t,s) = \frac{1}{\Gamma(\alpha)} \textstyle\begin{cases} (\log t)^{\alpha-1}- (\log\frac{t}{s} )^{\alpha-1}, &1\leq s\leq t< \infty, \\ (\log t)^{\alpha-1},& 1\leq t\leq s< \infty, \end{cases}\displaystyle \end{aligned}$$
(2.6)
$$\begin{aligned}& g_{i}(\eta,s) = \textstyle\begin{cases} (\log\eta)^{\alpha+\beta_{i}-1}- (\log\frac{\eta }{s} )^{\alpha+\beta_{i}-1},& 1\leq s\leq\eta< \infty, \\ (\log\eta)^{\alpha+\beta_{i}-1},& 1\leq\eta\leq s< \infty. \end{cases}\displaystyle \end{aligned}$$
(2.7)
Proof
Applying the Hadamard fractional integral of order α to both sides of (2.2), we have
$$ u(t)=c_{1}(\log t)^{\alpha-1}+c_{2}(\log t)^{\alpha-2}-\frac{1}{\Gamma (\alpha)} \int_{1}^{t} \biggl(\log\frac{t}{s} \biggr)^{\alpha -1}h(s)\frac{ds}{s}, $$
(2.8)
where \(c_{1}, c_{2}\in\mathbb{R}\).
The first condition of (2.3) implies \(c_{2}=0\). Therefore,
$$ u(t)=c_{1}(\log t)^{\alpha-1}-\frac{1}{\Gamma(\alpha)} \int_{1}^{t} \biggl(\log\frac{t}{s} \biggr)^{\alpha-1}h(s)\frac{ds}{s}. $$
(2.9)
In accordance with Lemma 2.3, we have
$$D^{\alpha-1}u(t)=c_{1}\Gamma(\alpha)- \int_{1}^{t} h(s)\frac{ds}{s}. $$
The second condition of (2.3) leads to
$$ c_{1}=\frac{1}{\Omega} \Biggl( \int_{1}^{\infty}h(s)\frac{ds}{s}-\sum _{i=1}^{m}\frac{\lambda_{i}}{\Gamma(\alpha+\beta_{i})} \int_{1}^{\eta}\biggl(\log\frac{\eta}{s} \biggr)^{\alpha+\beta_{i}-1}h(s)\frac {ds}{s} \Biggr), $$
(2.10)
where Ω is defined by (2.1). Therefore, the unique solution of fractional boundary value problem (2.2)-(2.3) is
$$\begin{aligned} u(t) =& \frac{(\log t)^{\alpha-1}}{\Omega} \int _{1}^{\infty}h(s)\frac{ds}{s}-\sum _{i=1}^{m}\frac{\lambda_{i}(\log t)^{\alpha-1}}{\Omega\Gamma(\alpha+\beta_{i})} \int_{1}^{\eta}\biggl(\log\frac{\eta}{s} \biggr)^{\alpha+\beta_{i}-1}h(s)\frac{ds}{s} \\ &{}- \frac{1}{\Gamma(\alpha)} \int_{1}^{t} \biggl(\log\frac{t}{s} \biggr)^{\alpha-1}h(s)\frac{ds}{s} \\ =& \frac{ (\log t)^{\alpha-1}\Gamma (\alpha)}{ \Gamma(\alpha)-\sum_{i=1}^{m}\frac{\lambda _{i}\Gamma(\alpha)}{\Gamma(\alpha+\beta_{i})}(\log\eta)^{\alpha +\beta_{i}-1}} \int_{1}^{\infty}\frac{h(s)}{\Gamma(\alpha)}\frac {ds}{s} \\ &{}- \sum_{i=1}^{m}\frac{\lambda_{i}(\log t)^{\alpha-1}}{\Omega \Gamma(\alpha+\beta_{i})} \int_{1}^{\eta}\biggl(\log\frac{\eta }{s} \biggr)^{\alpha+\beta_{i}-1}h(s)\frac{ds}{s} - \frac{1}{\Gamma(\alpha)} \int_{1}^{t} \biggl(\log\frac{t}{s} \biggr)^{\alpha-1}h(s)\frac{ds}{s} \\ =&\frac{ (\log t)^{\alpha-1} (\Gamma(\alpha )-\sum_{i=1}^{m}\frac{\lambda_{i}\Gamma(\alpha)}{\Gamma(\alpha +\beta_{i})}(\log\eta)^{\alpha+\beta_{i}-1}+\sum_{i=1}^{m}\frac {\lambda_{i}\Gamma(\alpha)}{\Gamma(\alpha+\beta_{i})}(\log\eta )^{\alpha+\beta_{i}-1} )}{ \Gamma(\alpha)-\sum_{i=1}^{m}\frac{\lambda_{i}\Gamma(\alpha)}{\Gamma(\alpha+\beta _{i})}(\log\eta)^{\alpha+\beta_{i}-1}} \\ &{}\times \int_{1}^{\infty}\frac{h(s)}{\Gamma(\alpha)}\frac{ds}{s}- \sum_{i=1}^{m}\frac{\lambda_{i}(\log t)^{\alpha-1}}{\Omega\Gamma (\alpha+\beta_{i})} \int_{1}^{\eta}\biggl(\log\frac{\eta}{s} \biggr)^{\alpha+\beta_{i}-1}h(s)\frac{ds}{s} \\ &{}- \frac{1}{\Gamma(\alpha)} \int_{1}^{t} \biggl(\log\frac{t}{s} \biggr)^{\alpha-1}h(s)\frac{ds}{s} \\ =&(\log t)^{\alpha-1} \int_{1}^{\infty}\frac{h(s)}{\Gamma(\alpha )}\frac{ds}{s}+ \sum_{i=1}^{m}\frac{\lambda_{i}(\log t)^{\alpha -1}}{\Omega\Gamma(\alpha+\beta_{i})}(\log \eta)^{\alpha+\beta _{i}-1} \int_{1}^{\infty}h(s)\frac{ds}{s} \\ &{}- \sum_{i=1}^{m}\frac{\lambda_{i}(\log t)^{\alpha-1}}{\Omega \Gamma(\alpha+\beta_{i})} \int_{1}^{\eta}\biggl(\log\frac{\eta }{s} \biggr)^{\alpha+\beta_{i}-1}h(s)\frac{ds}{s} \\ &{}- \frac{1}{\Gamma(\alpha)} \int_{1}^{t} \biggl(\log\frac{t}{s} \biggr)^{\alpha-1}h(s)\frac{ds}{s} \\ =&\frac{1}{\Gamma(\alpha)} \int_{1}^{t} \biggl[(\log t)^{\alpha -1}- \biggl( \log\frac{t}{s} \biggr)^{\alpha-1} \biggr]h(s)\frac {ds}{s}+ \frac{1}{\Gamma(\alpha)} \int_{t}^{\infty}(\log t)^{\alpha -1}h(s) \frac{ds}{s} \\ &{}+ \sum_{i=1}^{m}\frac{\lambda_{i}(\log t)^{\alpha-1}}{\Omega \Gamma(\alpha+\beta_{i})} \int_{1}^{\eta}\biggl[(\log\eta)^{\alpha +\beta_{i}-1}- \biggl(\log\frac{\eta}{s} \biggr)^{\alpha+\beta _{i}-1} \biggr]h(s)\frac{ds}{s} \\ &{}+ \sum_{i=1}^{m}\frac{\lambda_{i}(\log t)^{\alpha-1}}{\Omega \Gamma(\alpha+\beta_{i})} \int_{\eta}^{\infty}(\log\eta)^{\alpha +\beta_{i}-1}h(s) \frac{ds}{s} \\ =& \int_{1}^{\infty}g(t,s)h(s)\frac{ds}{s}+\sum _{i=1}^{m}\frac{\lambda _{i}(\log t)^{\alpha-1}}{\Omega\Gamma(\alpha+\beta_{i})} \int_{1}^{\infty}g_{i}(\eta,s)h(s) \frac{ds}{s} \\ =& \int_{1}^{\infty}G(t,s)h(s)\frac{ds}{s}. \end{aligned}$$
The proof is completed. □
Lemma 2.6
The Green’s function
\(G(t,s)\)
defined by (2.5) satisfies the following conditions:
- (C1):
-
\(G(t,s)\)
is a continuous function for
\((t,s)\in [1,\infty)\times[1,\infty)\);
- (C2):
-
\(G(t,s) \geq0\)
for all
\(s,t \in[1,\infty)\);
- (C3):
-
\(\frac{G(t,s)}{1+(\log t)^{\alpha -1}}\leq\frac{1}{\Gamma(\alpha)}+\sum_{i=1}^{m} \frac{\lambda_{i} g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta_{i})}\)
for all
\(s,t \in [1,\infty)\);
- (C4):
-
\(\min_{\eta\leq t\leq k\eta} \frac {G(t,s)}{1+(\log t)^{\alpha-1}}\geq\sum_{i=1}^{m} \frac{\lambda _{i}(\log\eta)^{\alpha-1}g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta _{i}) (1+(\log\eta)^{\alpha-1} )}\)
for
\(k>1\)
and
\(s\in [1,\infty)\).
Proof
It is easy to check that (C1) and (C2) hold.
To prove (C3), we have, for \(s,t\in[1,\infty)\),
$$\begin{aligned} \frac{G(t,s)}{1+(\log t)^{\alpha-1}} =&\frac{g(t,s)}{1+(\log t)^{\alpha-1}}+\sum_{i=1}^{m} \frac{\lambda_{i}(\log t)^{\alpha -1}g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta_{i}) (1+(\log t)^{\alpha-1} )} \\ \leq&\frac{1}{\Gamma(\alpha)}\cdot\frac{(\log t)^{\alpha -1}}{1+(\log t)^{\alpha-1}}+\sum _{i=1}^{m} \frac{\lambda_{i}(\log t)^{\alpha-1}g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta_{i}) (1+(\log t)^{\alpha-1} )} \\ \leq&\frac{1}{\Gamma(\alpha)}+\sum_{i=1}^{m} \frac{\lambda_{i} g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta_{i})}. \end{aligned}$$
To prove (C4), from \(g(t,s)\geq0\) and \(g_{i}(\eta,s)\geq0\), \(i=1,2,\ldots,m\), for all \(s,t\in[1,\infty)\), we have, for \(k>1\),
$$\begin{aligned} \min_{\eta\leq t\leq k\eta}\frac{G(t,s)}{1+(\log t)^{\alpha -1}} =&\min_{\eta\leq t\leq k\eta} \Biggl[\frac{g(t,s)}{1+(\log t)^{\alpha-1}}+\sum_{i=1}^{m} \frac{\lambda_{i}(\log t)^{\alpha -1}g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta_{i}) (1+(\log t)^{\alpha-1} )} \Biggr] \\ \geq&\min_{\eta\leq t\leq k\eta}\frac{g(t,s)}{1+(\log t)^{\alpha -1}}+\min_{\eta\leq t\leq k\eta} \sum_{i=1}^{m} \frac{\lambda_{i}(\log t)^{\alpha-1}g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta_{i}) (1+(\log t)^{\alpha-1} )} \\ \geq&\min_{\eta\leq t\leq k\eta}\sum_{i=1}^{m} \frac{\lambda _{i}(\log t)^{\alpha-1}g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta _{i}) (1+(\log t)^{\alpha-1} )} \\ \geq&\sum_{i=1}^{m} \frac{\lambda_{i}(\log\eta)^{\alpha-1}g_{i}(\eta ,s)}{\Omega\Gamma(\alpha+\beta_{i}) (1+(\log\eta)^{\alpha -1} )} \end{aligned}$$
for \(s\in[1,\infty)\). The proof is completed. □
In this paper, we will use the Banach space E, which is defined by
$$ E= \biggl\{ u\in C\bigl( [1,\infty ),\mathbb{R}\bigr) : \sup_{t\in[1,\infty )} \frac{|u(t)|}{1+(\log t)^{\alpha-1}}< \infty \biggr\} $$
and is equipped with the norm
$$ \|u\|_{E}=\sup_{t\in[1,\infty)}\frac{|u(t)|}{1+(\log t)^{\alpha-1}}. $$
Lemma 2.7
\((E,\|\cdot\|_{E})\)
is Banach space.
Proof
Let \(\{u_{n}\}_{n=1}^{\infty}\) be any Cauchy sequence in the space \((E,\|\cdot\|_{E})\). Then \(\forall\varepsilon >0\), \(\exists N>0\) such that
$$ \|u_{n}-u_{m}\|_{E}=\sup_{t\in[1,\infty)} \frac{|u_{n}(t)-u_{m}(t)|}{1+(\log t)^{\alpha-1}}< \varepsilon $$
for \(n,m>N\). Therefore, any fixed \(t_{0}\in[1,\infty)\), we have \(\{ u_{n}(t_{0})\}_{n=1}^{\infty}\) is a Cauchy sequence in \(\mathbb{R}\). In this way, we can associate to each \(t\in[1,\infty)\) for a unique \(u(t)\). Letting \(n\to\infty\), we obtain \(|u(t)-u_{m}(t)|\leq \varepsilon\) for all \(m>N\) and \(t\in[1,\infty)\). It is easy to show that \(u_{m}\to u\) in E as \(m\to\infty\). Therefore, we see that \((E,\| \cdot\|_{E})\) is Banach space. □
Lemma 2.8
Let
\(U\subset E\)
be a bounded set. Then
U
is relatively compact in
E
if the following conditions hold:
-
(i)
for any
\(u(t)\in U\), \(\frac{u(t)}{1+(\log t)^{\alpha-1}}\)
is equicontinuous on any compact interval of
\([1,\infty)\);
-
(ii)
for any
\(\varepsilon>0\), there exists a constant
\(T=T(\varepsilon)>0\)
such that
$$ \biggl\vert \frac{u(t_{1})}{1+(\log t_{1})^{\alpha-1}}-\frac{u(t_{2})}{1+(\log t_{2})^{\alpha-1}}\biggr\vert < \varepsilon $$
(2.11)
for any
\(t_{1},t_{2}\geq T\)
and
\(u\in U\).
Proof
Evidently, it is sufficient to prove that U is totally bounded. In the following we divide the proof into two steps.
Step 1. Let us consider the case \(t\in[1,T]\).
Define
$$ U_{[1,T]}=\bigl\{ u(t):u(t)\in U, t\in[1,T]\bigr\} . $$
Then clearly, \(U_{[1,T]}\), with the norm \(\|u\|_{\infty}=\sup_{t\in[1,T]}\frac{|u(t)|}{1+(\log t)^{\alpha-1}}\) is a Banach space. The condition (i) combined with the Arzelá-Ascoli theorem indicates that \(U_{[1,T]}\) is relatively compact, hence \(U_{[1,T]}\) is totally bounded, namely, for any \(\varepsilon>0\), there exist finitely many balls \(B_{\varepsilon}(u_{i})\) such that
$$ U_{[1,T]}\subset\bigcup_{i=1}^{n} B_{\varepsilon}(u_{i}), $$
where
$$ B_{\varepsilon}(u_{i})= \biggl\{ u(t)\in U_{[1,T]}: \|u-u_{i}\|_{\infty}=\sup_{t\in[1,T]}\biggl\vert \frac{u(t)}{1+(\log t)^{\alpha-1}}-\frac {u_{i}(t)}{1+(\log t)^{\alpha-1}}\biggr\vert < \varepsilon \biggr\} . $$
Step 2. Define
$$ U_{i}= \bigl\{ u(t)\in U:u_{[1,T]}\in B_{\varepsilon}(u_{i}) \bigr\} . $$
It is clear that \(U_{[1,T]}\subset\bigcup_{1\leq i\leq n}U_{i[1,T]}\). Now let us take \(u_{i}\in U_{i}\); then U can be covered by the balls \(B_{3\varepsilon}(u_{i})\), \(i=1,2,\ldots,n\) where
$$ B_{3\varepsilon}(u_{i})= \bigl\{ u(t)\in U : \|u-u_{i} \|_{E}< 3\varepsilon \bigr\} . $$
In fact, for \(u(t)\in U\), the arguments in Step 1 imply that there exist i such that \(u_{[1,T]}\in B_{\varepsilon}(u_{i})\). Hence, for \(t\in[1,T]\), we have
$$ \biggl\vert \frac{u(t)}{1+(\log t)^{\alpha-1}}-\frac{u_{i}(t)}{1+(\log t)^{\alpha-1}}\biggr\vert < \varepsilon. $$
(2.12)
For \(t\in[T,+\infty)\), (2.11), and (2.12) yield
$$\begin{aligned}& \biggl\vert \frac{u(t)}{1+(\log t)^{\alpha-1}}-\frac{u_{i}(t)}{1+(\log t)^{\alpha-1}}\biggr\vert \\& \quad \leq \biggl\vert \frac{u(t)}{1+(\log t)^{\alpha-1}}-\frac{u(T)}{1+(\log T)^{\alpha-1}}\biggr\vert + \biggl\vert \frac{u(T)}{1+(\log T)^{\alpha -1}}-\frac{u_{i}(T)}{1+(\log T)^{\alpha-1}}\biggr\vert \\& \qquad {}+\biggl\vert \frac{u_{i}(T)}{1+(\log T)^{\alpha-1}}-\frac {u_{i}(t)}{1+(\log t)^{\alpha-1}}\biggr\vert \\& \quad < \varepsilon+\varepsilon+\varepsilon=3\varepsilon. \end{aligned}$$
(2.13)
Equations (2.12) and (2.13) show that \(\|u(t)-u_{i}(t)\| _{E}<3\varepsilon\). Therefore, U is totally bounded and Lemma 2.8 is proved. □
We define the cone \(P\subset E\) by
$$ P=\bigl\{ u\in E : u(t)\geq0 \text{ on } [1,\infty )\bigr\} , $$
and the operator \(T:P\rightarrow E\) by
$$ Tu(t)= \int_{1}^{\infty}G(t,s)a(s)f\bigl(u(s)\bigr) \frac{ds}{s},\quad t\in [1,\infty), $$
(2.14)
where \(G(t,s)\) defined by (2.5).
Throughout this paper, we assume that the following conditions hold:
- (A1):
-
\(f\in C([0,\infty),[0,\infty))\), \(f(u)\neq0\) on any subinterval of \((0,\infty)\) and \(f ((1+(\log t)^{\alpha -1})u )\) is bounded on \([0,\infty)\);
- (A2):
-
\(a : [1,\infty)\rightarrow[0,\infty)\) is not identical zero on any closed subinterval of \([1,\infty)\) and
$$ 0< \int_{1}^{\infty}a(s)\frac{ds}{s}< \infty. $$
Lemma 2.9
Let (A1) and (A2) hold. Then
\(T:P\rightarrow P\)
is completely continuous.
Proof
We divide the proof into four steps.
Step 1: We show that
T
is uniformly bounded on
P.
From the definition of E, we can choose \(r_{0}\) such that \(\sup_{n\in{\mathbb{N}}}\|u_{n}\|_{E}< r_{0}\). Let \(B_{r_{0}}=\sup \{f((1+(\log t)^{\alpha-1})u), u\in[1,r_{0}] \}\) and Φ be any bounded subset of P, then there exists \(r>0\) such that \(\|u\|_{E}\leq r\) for all \(u\in\Phi\). From (C3), it follows that
$$\begin{aligned} \|Tu\|_{E} =&\sup_{t\in[1,\infty)}\frac{1}{1+(\log t)^{\alpha -1}}\biggl\vert \int_{1}^{\infty}G(t,s)a(s)f\bigl(u(s)\bigr) \frac{ds}{s}\biggr\vert \\ \leq& \int_{1}^{\infty}\Biggl(\frac{1}{\Gamma(\alpha)}+\sum _{i=1}^{m} \frac{\lambda_{i} g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta_{i})} \Biggr)a(s)f\bigl(u(s) \bigr)\frac{ds}{s} \\ \leq& \Biggl(\frac{1}{\Gamma(\alpha)}+\sum_{i=1}^{m} \frac{\lambda _{i} (\log\eta)^{\alpha+\beta_{i}-1}}{\Omega\Gamma(\alpha+\beta _{i})} \Biggr)B_{r} \int_{1}^{\infty}a(s)\frac{ds}{s}< \infty \end{aligned}$$
for \(u\in\Phi\). Therefore TΦ is uniformly bounded.
Step 2: We show that
T
is equicontinuous on any compact interval of
\([1,\infty)\).
For any \(S>1\), \(t_{1},t_{2}\in[1,S]\), and \(u\in\Phi\), without loss of generality, we assume that \(t_{1}< t_{2}\). In fact,
$$\begin{aligned}& \biggl\vert \frac{Tu(t_{2})}{1+(\log t_{2})^{\alpha-1}}-\frac {Tu(t_{1})}{1+(\log t_{1})^{\alpha-1}}\biggr\vert \\& \quad = \biggl\vert \int_{1}^{\infty}\frac{G(t_{2},s)}{1+(\log t_{2})^{\alpha -1}}a(s)f\bigl(u(s)\bigr) \frac{ds}{s}- \int_{1}^{\infty}\frac{G(t_{1},s)}{1+(\log t_{1})^{\alpha-1}}a(s)f\bigl(u(s)\bigr) \frac{ds}{s}\biggr\vert \\& \quad \leq \biggl\vert \int_{1}^{\infty}\biggl(\frac{g(t_{2},s)}{1+(\log t_{2})^{\alpha-1}}- \frac{g(t_{1},s)}{1+(\log t_{1})^{\alpha-1}} \biggr)a(s)f\bigl(u(s)\bigr)\frac{ds}{s}\biggr\vert \\& \qquad {} +\Biggl\vert \int_{1}^{\infty}\biggl(\frac{(\log t_{2})^{\alpha -1}}{1+(\log t_{2})^{\alpha-1}}- \frac{(\log t_{1})^{\alpha-1}}{1+(\log t_{1})^{\alpha-1}} \biggr) \sum_{i=1}^{m} \frac{\lambda_{i} g_{i}(\eta ,s)}{\Omega\Gamma(\alpha+\beta_{i})} a(s)f\bigl(u(s)\bigr)\frac{ds}{s} \Biggr\vert \\& \quad \leq \int_{1}^{\infty}\biggl\vert \frac{g(t_{2},s)}{1+(\log t_{2})^{\alpha -1}}- \frac{g(t_{1},s)}{1+(\log t_{2})^{\alpha-1}}\biggr\vert a(s)f\bigl(u(s)\bigr)\frac {ds}{s} \\& \qquad {} + \int_{1}^{\infty}\biggl\vert \frac{g(t_{1},s)}{1+(\log t_{2})^{\alpha -1}}- \frac{g(t_{1},s)}{1+(\log t_{1})^{\alpha-1}}\biggr\vert a(s)f\bigl(u(s)\bigr)\frac {ds}{s} \\& \qquad {} + \int_{1}^{\infty}\biggl\vert \frac{(\log t_{2})^{\alpha-1}}{1+(\log t_{2})^{\alpha-1}}- \frac{(\log t_{1})^{\alpha-1}}{1+(\log t_{1})^{\alpha -1}}\biggr\vert \sum_{i=1}^{m} \frac{\lambda_{i} g_{i}(\eta,s)}{\Omega \Gamma(\alpha+\beta_{i})} a(s)f\bigl(u(s)\bigr)\frac{ds}{s} \\& \quad = \int_{1}^{\infty}\biggl\vert \frac{g(t_{2},s)}{1+(\log t_{2})^{\alpha -1}}- \frac{g(t_{1},s)}{1+(\log t_{2})^{\alpha-1}}\biggr\vert a(s)f\bigl(u(s)\bigr)\frac {ds}{s} \\& \qquad {} + \int_{1}^{\infty}\frac{(\log t_{2})^{\alpha-1}-(\log t_{1})^{\alpha -1}}{(1+(\log t_{2})^{\alpha-1})(1+(\log t_{1})^{\alpha -1})}g(t_{1},s)a(s)f \bigl(u(s)\bigr)\frac{ds}{s} \\& \qquad {} + \int_{1}^{\infty}\frac{(\log t_{2})^{\alpha-1}-(\log t_{1})^{\alpha -1}}{(1+(\log t_{2})^{\alpha-1})(1+(\log t_{1})^{\alpha-1})} \sum _{i=1}^{m} \frac{\lambda_{i} g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta _{i})} a(s)f\bigl(u(s)\bigr) \frac{ds}{s}. \end{aligned}$$
We now consider
$$\begin{aligned}& \int_{1}^{\infty}\biggl\vert \frac{g(t_{2},s)}{1+(\log t_{2})^{\alpha -1}}- \frac{g(t_{1},s)}{1+(\log t_{2})^{\alpha-1}}\biggr\vert a(s)f\bigl(u(s)\bigr)\frac {ds}{s} \\& \quad = \int_{1}^{t_{1}} \biggl\vert \frac{g(t_{2},s)}{1+(\log t_{2})^{\alpha -1}}- \frac{g(t_{1},s)}{1+(\log t_{2})^{\alpha-1}}\biggr\vert a(s)f\bigl(u(s)\bigr)\frac {ds}{s} \\& \qquad {} + \int_{t_{1}}^{t_{2}} \biggl\vert \frac{g(t_{2},s)}{1+(\log t_{2})^{\alpha -1}}- \frac{g(t_{1},s)}{1+(\log t_{2})^{\alpha-1}}\biggr\vert a(s)f\bigl(u(s)\bigr)\frac {ds}{s} \\& \qquad {} + \int_{t_{2}}^{\infty}\biggl\vert \frac{g(t_{2},s)}{1+(\log t_{2})^{\alpha -1}}- \frac{g(t_{1},s)}{1+(\log t_{2})^{\alpha-1}}\biggr\vert a(s)f\bigl(u(s)\bigr)\frac {ds}{s} \\& \quad \leq \frac{1}{\Gamma(\alpha)} \int_{1}^{t_{1}} \frac{(\log t_{2})^{\alpha-1}-(\log t_{1})^{\alpha-1}+(\log\frac{t_{2}}{s})^{\alpha -1}-(\log\frac{t_{1}}{s})^{\alpha-1}}{1+(\log t_{2})^{\alpha -1}}a(s)f\bigl(u(s)\bigr) \frac{ds}{s} \\& \qquad {}+\frac{1}{\Gamma(\alpha)} \int_{t_{1}}^{t_{2}} \frac{(\log t_{2})^{\alpha-1}-(\log t_{1})^{\alpha-1}+(\log\frac{t_{2}}{s})^{\alpha -1}}{1+(\log t_{2})^{\alpha-1}}a(s)f\bigl(u(s) \bigr)\frac{ds}{s} \\& \qquad {}+\frac{1}{\Gamma(\alpha)} \int_{t_{2}}^{\infty} \frac{(\log t_{2})^{\alpha-1}-(\log t_{1})^{\alpha-1}}{1+(\log t_{2})^{\alpha -1}}a(s)f\bigl(u(s) \bigr)\frac{ds}{s} \\& \quad \rightarrow 0 \quad \text{uniformly as } t_{1}\rightarrow t_{2}. \end{aligned}$$
(2.15)
Similarly, we have
$$ \int_{1}^{\infty}\frac{(\log t_{2})^{\alpha-1}-(\log t_{1})^{\alpha -1}}{(1+(\log t_{2})^{\alpha-1})(1+(\log t_{1})^{\alpha -1})}g(t_{1},s)a(s)f \bigl(u(s)\bigr)\frac{ds}{s}\rightarrow0 $$
(2.16)
uniformly as \(t_{1}\rightarrow t_{2}\), and
$$ \int_{1}^{\infty}\frac{(\log t_{2})^{\alpha-1}-(\log t_{1})^{\alpha -1}}{(1+(\log t_{2})^{\alpha-1})(1+(\log t_{1})^{\alpha-1})} \sum _{i=1}^{m} \frac{\lambda_{i} g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta _{i})} a(s)f\bigl(u(s)\bigr) \frac{ds}{s}\rightarrow0 $$
(2.17)
uniformly as \(t_{1}\rightarrow t_{2}\).
Hence, from (2.15), (2.16), and (2.17), we get
$$ \biggl\vert \frac{Tu(t_{2})}{1+(\log t_{2})^{\alpha-1}}-\frac {Tu(t_{1})}{1+(\log t_{1})^{\alpha-1}}\biggr\vert \rightarrow0 \quad \text{uniformly as } t_{1}\rightarrow t_{2}. $$
Thus TΦ is equicontinuous on \([1,\infty)\).
Step 3: We show that
T
is equiconvergent at ∞.
For any \(u\in\Phi\), we have
$$\int_{1}^{\infty}a(s)f\bigl(u(s)\bigr)\frac{ds}{s} \le B_{r} \int_{1}^{\infty}a(s)\frac {ds}{s}< \infty $$
and
$$\begin{aligned}& \lim_{t\rightarrow\infty}\biggl\vert \frac{(Tu)(t)}{1+(\log t)^{\alpha -1}}\biggr\vert \\& \quad = \lim_{t\rightarrow\infty}\biggl\vert \frac{1}{1+(\log t)^{\alpha -1}} \int_{1}^{\infty}G(t,s)a(s)f\bigl(u(s)\bigr) \frac{ds}{s}\biggr\vert \\& \quad \leq \lim_{t\rightarrow\infty} \int_{1}^{\infty}\Biggl(\frac {1}{\Gamma(\alpha)}\cdot \frac{(\log t)^{\alpha-1}}{1+(\log t)^{\alpha-1}}+\frac{(\log t)^{\alpha-1}}{1+(\log t)^{\alpha -1}}\cdot\sum_{i=1}^{m} \frac{\lambda_{i} g_{i}(\eta,s)}{\Omega\Gamma (\alpha+\beta_{i})} \Biggr)a(s)f\bigl(u(s)\bigr)\frac{ds}{s} \\& \quad \leq \int_{1}^{\infty}\Biggl(\frac{1}{\Gamma(\alpha)}+\sum _{i=1}^{m} \frac{\lambda_{i} g_{i}(\eta,s)}{\Omega\Gamma(\alpha+\beta_{i})} \Biggr)a(s)f\bigl(u(s) \bigr)\frac{ds}{s} \\& \quad \leq \Biggl(\frac{1}{\Gamma(\alpha)}+\sum_{i=1}^{m} \frac{\lambda _{i} (\log\eta)^{\alpha+\beta_{i}-1}}{\Omega\Gamma(\alpha+\beta _{i})} \Biggr)B_{r} \int_{1}^{\infty}a(s)\frac{ds}{s} \\& \quad < \infty. \end{aligned}$$
Hence, TΦ is equiconvergent at infinity.
Step 4: We show that
T
is continuous.
Let \(u_{n}\rightarrow u\) as →∞ in P. We have
$$ \int_{1}^{\infty}a(s)f\bigl(u(s)\bigr) \frac{ds}{s}< \infty. $$
Hence the Lebesgue dominated convergence and the continuity of f guarantee that
$$ \int_{1}^{\infty}a(s)f\bigl(u_{n}(s)\bigr) \frac{ds}{s}\rightarrow \int_{1}^{\infty}a(s)f\bigl(u(s)\bigr)\frac{ds}{s} \quad \text{as } n\rightarrow\infty. $$
Therefore, we get
$$\begin{aligned}& \|Tu_{n}-Tu\|_{E} \\& \quad = \sup_{t\in[1,\infty)}\frac{1}{1+(\log t)^{\alpha-1}} \vert Tu_{n}-Tu\vert \\& \quad = \sup_{t\in[1,\infty)}\biggl\vert \int_{1}^{\infty}\frac{G(t,s)}{1+(\log t)^{\alpha-1}}a(s) \bigl[f \bigl(u_{n}(s)\bigr)-f\bigl(u(s)\bigr) \bigr]\frac{ds}{s} \biggr\vert \\& \quad \leq \Biggl(\frac{1}{\Gamma(\alpha)}+\sum_{i=1}^{m} \frac{\lambda _{i} (\log\eta)^{\alpha+\beta_{i}-1}}{\Omega\Gamma(\alpha+\beta _{i})} \Biggr)\biggl\vert \int_{1}^{\infty}a(s)f\bigl(u_{n}(s)\bigr) \frac{ds}{s}- \int _{1}^{\infty}a(s)f\bigl(u(s)\bigr) \frac{ds}{s}\biggr\vert \\& \quad \rightarrow 0\quad \text{as } n\rightarrow\infty. \end{aligned}$$
So, T is continuous.
Using Lemma 2.8, we see that \(T:P\rightarrow P\) is completely continuous. The proof is completed. □