Lemma 3.1
Let
\(h(t)\in L^{1}[0,\infty)\)
be a nonnegative continuous function. Then the boundary value problem of fractional differential equation
$$\begin{aligned}& ^{c}D^{\alpha}_{0^{+}}u(t)=h(t),\quad 0< t< + \infty, \end{aligned}$$
(3.1)
$$\begin{aligned}& u(0)=0,\quad\quad u^{(q)}(0)=0,\quad\quad{}^{c}D^{\alpha-1}_{0^{+}}u(+ \infty)=\sum^{m-2}_{i=1}\beta_{i}u( \xi_{i}), \end{aligned}$$
(3.2)
has a unique solution
$$u(t)= \int^{\infty}_{0}G(t,s)h(s)\,ds, $$
where
$$G(t,s)= \textstyle\begin{cases} \frac{(t-s)^{\alpha-1}}{\Gamma({\alpha})}+\frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}-\frac{t\sum^{m-2}_{i=1}\beta_{i}(\xi _{i}-s)^{\alpha-1}}{\Gamma(\alpha)\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}, \\ \quad0\leq s\leq\min(t,\xi_{1})< \infty,\\ \frac{(t-s)^{\alpha-1}}{\Gamma({\alpha})}+\frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}},\\ \quad0\leq\xi_{m-2}\leq s\leq t< \infty,\\ \frac{(t-s)^{\alpha-1}}{\Gamma({\alpha})}+\frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}-\frac{t\sum^{m-2}_{i=\kappa}\beta_{i}(\xi _{i}-s)^{\alpha-1}}{\Gamma(\alpha)\sum^{m-2}_{i=\kappa}\beta_{i}\xi _{i}}, \\ \quad0\leq\xi_{\kappa-1}< s\leq\xi_{\kappa}\leq t< \infty,\kappa =2,3,\ldots,m-2,\\ \frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}-\frac{t\sum^{m-2}_{i=1}\beta_{i}(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha)\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}, \\ \quad0\leq t\leq s\leq\xi_{1}< \infty,\\ \frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}-\frac{t\sum^{m-2}_{i=\kappa}\beta_{i}(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha) \sum^{m-2}_{i=\kappa}\beta_{i}\xi_{i}}, \\ \quad0\leq t\leq\xi_{\kappa-1}< s\leq \xi_{\kappa}< \infty,\kappa=2,3,\ldots,m-2,\\ \frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}},\\ \quad0\leq\max(t,\xi _{m-2})\leq s< \infty. \end{cases} $$
Proof
In view of Lemma 2.4, it is clear that equation (3.1) is equivalent to the integral form
$$u(t)=I^{\alpha}_{0^{+}}h(t)+c_{1}+c_{2}t+c_{3}t^{2}+ \cdots+c_{n}t^{n-1}, $$
for some \(c_{i}\in \mathbb{R}\), \(i=1,2,\ldots,n\), \(n=\lceil\alpha\rceil\).
By the boundary value conditions \(u(0)=0\), \(u^{(q)}(0)=0\), we imply that
$$c_{1}=0,\quad\quad c_{3}=c_{4}=c_{5}= \cdots=c_{n}=0 $$
and
$$u(t)=I^{\alpha}_{0^{+}}h(t)+c_{2}t. $$
Applying Lemma 2.1 and the boundary condition \({}^{c}D^{\alpha -1}_{0^{+}}u(+\infty)=\sum^{m-2}_{i=1}\beta_{i}u(\xi_{i})\), we obtain
$$\begin{aligned}& {}^{c}D^{\alpha-1}_{0^{+}}u(+\infty)=\lim_{t\rightarrow+\infty } \bigl(I_{0^{+}}h(t)+{}^{c}D^{\alpha-1}_{0^{+}}(c_{2}t) \bigr) \\& \hphantom{{}^{c}D^{\alpha-1}_{0^{+}}u(+\infty)}=\lim_{t\rightarrow +\infty}I_{0^{+}}h(t)= \int^{\infty}_{0}h(s)\,ds, \\& u(\xi_{i})=I^{\alpha}_{0^{+}}h(\xi_{i})+c_{2} \xi_{i}, \end{aligned}$$
consequently
$$c_{2}=\frac{\int^{\infty}_{0}h(s)\,ds}{\sum^{m-2}_{i=1}\beta_{i}\xi _{i}}-\frac{I^{\alpha}_{0^{+}}\sum^{m-2}_{i=1}\beta_{i}h(\xi _{i})}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}. $$
Substituting \(c_{2}\) by its value, it yields
$$\begin{aligned} u(t) =&\frac{1}{\Gamma(\alpha)} \int^{t}_{0}(t-s)^{\alpha-1}h(s)\,ds+ \frac {t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \int^{\infty}_{0}h(s)\,ds\\ &{}-\frac {t\sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}(\xi_{i}-s)^{\alpha -1}h(s)\,ds}{\Gamma(\alpha)\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \\ =& \int^{\infty}_{0}G(t,s)h(s)\,ds. \end{aligned}$$
The proof is completed. □
Define \(C_{\infty}=\{u\in C([0,+\infty), \mathbb{R}):\sup_{t\in [0,+\infty)}\frac{|u(t)|}{1+t^{\alpha-1}}<+\infty\}\) endowed with the norm
$$\|u\|_{C_{\infty}}=\sup_{t\in [0,+\infty)}\frac{|u(t)|}{1+t^{\alpha-1}}. $$
Lemma 3.2
([27])
\(C_{\infty}\)
is a Banach space.
Define a cone \(K\subset C_{\infty}\) by
$$K=\bigl\{ u\in C_{\infty}:u(t)\geq0, t\in[0,+\infty)\bigr\} . $$
Define an operator \(T:K\rightarrow C_{\infty}\) as follows:
$$Tu(t)= \int_{0}^{\infty}G(t,s)a(s)f\bigl(s,u(s)\bigr)\,ds. $$
Set \(h(t)=a(t)f(t,u(t))\) in Lemma 3.1. We deduce that u is a solution of the boundary value problem (1.1)-(1.2) if and only if it is a fixed point of the operator T.
Lemma 3.3
The function
\(G(t, s)\)
in Lemma
3.1
satisfies the following properties:
-
(i)
\(G(t, s)\)
is continuous on
\([0,+\infty)\times[0,+\infty)\);
-
(ii)
\(G(t, s)> 0\), for any
\(t,s\in(0,+\infty)\);
-
(iii)
\(0< \frac{G(t, s)}{1+t^{\alpha-1}} \leq\frac{2}{\sum^{m-2}_{i=1}\beta_{i}}\), for any
\(t,s\in(0, +\infty)\).
Proof
It is easy to see that (i) holds. So we prove that the rest are true. Let
$$\begin{aligned}& g_{1}(t, s)=\frac{(t-s)^{\alpha-1}}{\Gamma({\alpha})}+\frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}-\frac{t\sum^{m-2}_{i=1} \beta_{i}(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha)\sum^{m-2}_{i=1}\beta _{i}\xi_{i}},\quad 0\leq s\leq\min(t,\xi_{1})< \infty, \\& g_{2}(t, s)=\frac{(t-s)^{\alpha-1}}{\Gamma({\alpha})}+\frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}},\quad 0\leq \xi_{m-2}\leq s\leq t< \infty, \\& g_{3}(t, s)=\frac{(t-s)^{\alpha-1}}{\Gamma({\alpha})}+\frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}- \frac{t\sum^{m-2}_{i=\kappa} \beta_{i}(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha)\sum^{m-2}_{i=\kappa }\beta_{i}\xi_{i}},\\& \quad 0\leq\xi_{\kappa-1}< s\leq\xi_{\kappa} \leq t< \infty, \kappa=2,3,\ldots,m-2, \\& g_{4}(t, s)= \frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}-\frac{t\sum^{m-2}_{i=1}\beta_{i}(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha) \sum^{m-2}_{i=1}\beta_{i}\xi_{i}},\quad 0\leq t\leq s\leq \xi_{1}< \infty, \\& g_{5}(t, s)= \frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}-\frac{t\sum^{m-2}_{i=\kappa}\beta_{i}(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha) \sum^{m-2}_{i=\kappa}\beta_{i}\xi_{i}},\\& \quad 0\leq t\leq \xi_{\kappa -1}< s\leq\xi_{\kappa}< \infty,\kappa=2,3,\ldots,m-2, \\& g_{6}(t,s)= \frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}},\quad 0\leq\max(t,\xi _{m-2})\leq s< \infty. \end{aligned}$$
Let \(g_{k}(t, s)\) (\(k=1, 2, 3, 4,5,6\)) be defined by the above formulas. We will show that
$$g_{4}(t, s)\geq0,\quad 0\leq t\leq s\leq\xi_{1}< \infty. $$
Since
$$\begin{aligned} g_{4}(t, s) =&\frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}-\frac{t\sum^{m-2}_{i=1}\beta_{i}(\xi_{i}-s)^{\alpha-1}}{\Gamma({\alpha})\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \\ \geq& \frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}-\frac{t\sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha-1}}{\Gamma({\alpha})\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \\ =&t\biggl(\frac{\Gamma({\alpha})-\sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha -1}}{\Gamma({\alpha})\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}\biggr) \\ \geq& 0, \end{aligned}$$
we deduce
$$g_{4}(t, s)\geq0,\quad 0\leq t\leq s\leq\xi_{1}< \infty. $$
By using an analogous argument, we can conclude that
$$\begin{aligned}& g_{1}(t, s)\geq0,\quad 0\leq s\leq\min(t,\xi_{1})< \infty, \\& g_{2}(t, s)\geq 0,\quad0\leq\xi_{m-2}\leq s\leq t< \infty, \\& g_{3}(t, s)\geq0,\quad0\leq\xi_{\kappa-1}< s\leq\xi_{\kappa}\leq t< \infty , \\& g_{5}(t, s)\geq0,\quad0\leq t\leq\xi_{\kappa-1}< s\leq \xi_{\kappa}< \infty ,\kappa=2,3,\ldots,m-2, \end{aligned}$$
and
$$g_{6}(t, s)\geq0,\quad 0\leq\max(t,\xi_{m-2})\leq s< \infty. $$
Therefore, we get \(G(t, s)>0\), for any \(t,s\in(0,+\infty)\).
Next, we will prove (iii) is true. We will show that \(\frac {g_{2}(t,s)}{1+t^{\alpha-1}}\leq\frac{2}{\sum^{m-2}_{i=1}\beta _{i}}\), \(0\leq\xi_{m-2}\leq s\leq t<\infty\).
$$\begin{aligned} \frac{g_{2}(t,s)}{1+t^{\alpha-1}} =&\frac{(t-s)^{\alpha-1}}{(1+t^{\alpha -1})\Gamma({\alpha})}+\frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi _{i}(1+t^{\alpha-1})} \\ =&\frac{(t-s)^{\alpha-1}\sum^{m-2}_{i=1}\beta _{i}\xi_{i}+t\Gamma({\alpha})}{\sum^{m-2}_{i=1}\beta_{i}\xi _{i}(1+t^{\alpha-1})\Gamma({\alpha})} \\ \leq& \frac{t^{\alpha-1}\sum^{m-2}_{i=1}\beta_{i}\xi_{i}+t\Gamma ({\alpha})}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}(1+t^{\alpha-1})\Gamma ({\alpha})}\leq\frac{t^{\alpha-1}\sum^{m-2}_{i=1}\beta_{i}\xi _{i}+t\Gamma({\alpha})}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}\Gamma({\alpha })} \\ \leq&\frac{t^{\alpha-1}}{\Gamma({\alpha})}+\frac{t}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}\leq\frac{t^{\alpha-1}}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha-1}}+ \frac{t}{\sum^{m-2}_{i=1}\beta _{i}\xi_{i}} \\ \leq&\frac{1}{\sum^{m-2}_{i=1}\beta_{i}}(1+1)=\frac{2}{\sum^{m-2}_{i=1}\beta_{i}}, \end{aligned}$$
so we have \(0\leq\frac{g_{2}(t,s)}{1+t^{\alpha-1}} \leq\frac{2}{\sum^{m-2}_{i=1}\beta_{i}}\), \(0\leq\xi_{m-2}\leq s\leq t<\infty\).
By using an analogous argument, we can conclude that
$$\begin{aligned}& 0\leq\frac{g_{1}(t,s)}{1+t^{\alpha-1}}\leq\frac{2}{\sum^{m-2}_{i=1}\beta_{i}},\quad 0\leq s\leq\min(t, \xi_{1})< \infty, \\& 0\leq\frac {g_{3}(t,s)}{1+t^{\alpha-1}}\leq\frac{2}{\sum^{m-2}_{i=1}\beta _{i}},\quad0\leq \xi_{\kappa-1}< s\leq\xi_{\kappa}\leq t< \infty, \\& 0\leq\frac{g_{4}(t,s)}{1+t^{\alpha-1}}\leq\frac{2}{\sum^{m-2}_{i=1}\beta_{i}},\quad0\leq t\leq s\leq \xi_{1}< \infty, \\& 0\leq\frac {g_{5}(t,s)}{1+t^{\alpha-1}}\leq\frac{2}{\sum^{m-2}_{i=1}\beta _{i}},\quad0\leq t \leq\xi_{\kappa-1}< s\leq\xi_{\kappa}< \infty, \end{aligned}$$
\(\kappa=2,3,\ldots,m-2\), and
$$0\leq\frac{g_{6}(t,s)}{1+t^{\alpha-1}}\leq\frac{2}{\sum^{m-2}_{i=1}\beta_{i}},\quad0\leq\max(t,\xi_{m-2})\leq s< \infty. $$
Therefore, we get \(0<\frac{G(t, s)}{1+t^{\alpha-1}}\leq\frac{2}{\sum^{m-2}_{i=1}\beta_{i}}\), for any \(s,t\in(0,+\infty)\). The proof is completed. □
Lemma 3.4
If (H1) and (H2) hold, the operator
\(T:K\rightarrow K\)
is completely continuous.
Proof
We divide the proof into the following five steps.
Step 1: We show that \(T: K\rightarrow K\).
In view of the continuous and nonnegative of \(G(t,s)\), \(f\in C([0,+\infty)\times[0,+\infty),[0,+\infty))\), and \(a(t)\in L^{1}[0,\infty)\) is nonnegative, it is easy to see that \(Tu(t)\geq0\) for \(t\in[0,+\infty)\).
By condition (H1) and Lemma 3.3, for any fixed \(u\in K\), we have
$$\frac{u(t)}{1+t^{\alpha-1}}\leq\|u\|_{C_{\infty}},\quad t\in[0,+\infty) $$
and there exists \(\Upsilon_{u}\) such that
$$\begin{aligned} \sup_{t\in[0,+\infty)}\frac{|(Tu)(t)|}{1+t^{\alpha-1}} =&\sup_{t\in [0,+\infty)} \int_{0}^{\infty}\frac{G(t,s)}{1+t^{\alpha -1}}a(s)f\bigl(s,u(s) \bigr)\,ds \\ \leq&\sup_{t\in[0,+\infty)}\frac{2}{\sum^{m-2}_{i=1}\beta_{i}} \int ^{\infty}_{0}a(s)f \biggl(s,\bigl(1+s^{\alpha-1} \bigr)\frac{u(s)}{1+s^{\alpha -1}} \biggr)\,ds \\ \leq&\frac{2\Upsilon_{u}}{\sum^{m-2}_{i=1}\beta_{i}} \int^{\infty }_{0}a(s)\,ds< \infty, \end{aligned}$$
so \(T(K)\subset K\).
Step 2: We show that \(T: K\rightarrow K\) is continuous.
Let \(u_{n}\rightarrow u\) as \(n\rightarrow+\infty\) in K. Then \(\frac{u_{n}(s)}{1+s^{\alpha-1}}\rightrightarrows \frac{u(s)}{1+s^{\alpha-1}}\) as \(n\rightarrow+\infty\) on \([0,\infty)\). Hence
$$\begin{aligned}& \biggl\vert \frac{Tu_{n}(t)}{1+t^{\alpha-1}}-\frac{Tu(t)}{1+t^{\alpha -1}}\biggr\vert \\& \quad = \biggl\vert \int_{0}^{\infty}\frac{G(t,s)}{1+t^{\alpha-1}} a(s)f \bigl(s,u_{n}(s)\bigr)\,ds- \int_{0}^{\infty}\frac{G(t,s)}{1+t^{\alpha -1}}a(s)f\bigl(s,u(s) \bigr)\,ds\biggr\vert \\& \quad\leq \int_{0}^{\infty}\frac{G(t,s)}{1+t^{\alpha-1}}a(s)\bigl\vert f \bigl(s,u_{n}(s)\bigr)-f\bigl(s,u(s)\bigr)\bigr\vert \,ds \\& \quad\leq \frac{2}{\sum^{m-2}_{i=1}\beta_{i}} \int_{0}^{\infty}a(s) \biggl\vert f \biggl(s, \bigl(1+s^{\alpha-1}\bigr)\frac{u_{n}(s)}{1+s^{\alpha-1}} \biggr)-f \biggl(s, \bigl(1+s^{\alpha-1}\bigr)\frac{u(s)}{1+s^{\alpha-1}} \biggr)\biggr\vert \,ds. \end{aligned}$$
With the help of Lebesgue’s dominated convergence theorem and the continuity of f, we have
$$\|Tu_{n}-Tu\|_{C_{\infty}}=\sup_{t\in[0,+\infty)}\biggl\vert \frac {(Tu_{n})(t)}{1+t^{\alpha-1}}-\frac{(Tu)(t)}{1+t^{\alpha-1}}\biggr\vert \rightarrow0, \quad\mbox{as }n\rightarrow\infty, $$
that is, T is continuous.
Now take \(\Omega\subset K\) be bounded, i.e., there exists a positive constant l such that \(\|u\|_{C_{\infty}}\leq l\) for all \(u\in\Omega\).
Step 3: \(T(\Omega)\) is uniformly bounded.
By condition (H1), let
$$\Upsilon_{l}=\sup\bigl\{ f\bigl(t,\bigl(1+t^{\alpha-1}\bigr)u \bigr),(t,u)\in[0,+\infty)\times [0,l]\bigr\} . $$
For any \(u\in\Omega\), by Lemma 3.3, we have
$$\begin{aligned} \|Tu\|_{C_{\infty}} =&\sup_{t\in[0,+\infty)}\frac {|(Tu)(t)|}{1+t^{\alpha-1}} \\ =&\sup _{t\in[0,+\infty)} \int_{0}^{\infty }\frac{G(t,s)}{1+t^{\alpha-1}}a(s)f\bigl(s,u(s) \bigr)\,ds \\ \leq&\sup_{t\in[0,+\infty)}\frac{2}{\sum^{m-2}_{i=1}\beta_{i}} \int ^{\infty}_{0}a(s)f \biggl(s,\bigl(1+s^{\alpha-1} \bigr)\frac{u(s)}{1+s^{\alpha -1}} \biggr)\,ds \\ \leq&\frac{2\Upsilon_{l}}{\sum^{m-2}_{i=1}\beta_{i}} \int^{\infty }_{0}a(s)\,ds< \infty, \end{aligned}$$
therefore \(T(\Omega)\) is uniformly bounded.
Step 4: We show that \(T(\Omega)\) is locally equicontinuous on any finite subinterval of \([0,+\infty)\).
For any \(\theta>0\), \(t_{1},t_{2}\in[0,\theta]\) and \(u\in\Omega\), without loss of generality, we assume that \(t_{2}>t_{1}\). Then
$$\begin{aligned}& \biggl\vert \frac{(Tu)(t_{2})}{1+t_{2}^{\alpha-1}}-\frac {(Tu)(t_{1})}{1+t_{1}^{\alpha-1}}\biggr\vert \\& \quad= \biggl\vert \int_{0}^{\infty}\frac {G(t_{2},s)}{1+t_{2}^{\alpha-1}}a(s)f\bigl(s,u(s) \bigr)\,ds- \int_{0}^{\infty}\frac {G(t_{1},s)}{1+t_{1}^{\alpha-1}}a(s)f\bigl(s,u(s) \bigr)\,ds\biggr\vert \\& \quad= \biggl\vert \int_{0}^{\infty} \biggl(\frac{G(t_{2},s)}{1+t_{2}^{\alpha -1}}- \frac{G(t_{1},s)}{1+t_{1}^{\alpha-1}} \biggr)a(s)f\bigl(s,u(s)\bigr)\,ds\biggr\vert \\ & \quad= \int_{0}^{\infty}\biggl\vert \frac{G(t_{2},s)}{1+t_{2}^{\alpha-1}}- \frac {G(t_{1},s)}{1+t_{2}^{\alpha-1}}+\frac{G(t_{1},s)}{1+t_{2}^{\alpha -1}}-\frac{G(t_{1},s)}{1+t_{1}^{\alpha-1}}\biggr\vert a(s)f \bigl(s,u(s)\bigr)\,ds \\ & \quad\leq \int_{0}^{\infty} \biggl(\biggl\vert \frac{G(t_{2},s)}{1+t_{2}^{\alpha -1}}-\frac{G(t_{1},s)}{1+t_{2}^{\alpha-1}}\biggr\vert +\biggl\vert \frac {G(t_{1},s)}{1+t_{2}^{\alpha-1}}-\frac{G(t_{1},s)}{1+t_{1}^{\alpha -1}}\biggr\vert \biggr)a(s)f\bigl(s,u(s)\bigr) \,ds. \end{aligned}$$
Furthermore, we deduce that
$$\begin{aligned}& \int_{0}^{\infty}\biggl\vert \frac{G(t_{2},s)}{1+t_{2}^{\alpha-1}}- \frac {G(t_{1},s)}{1+t_{2}^{\alpha-1}}\biggr\vert a(s)f\bigl(s,u(s)\bigr)\,ds \\ & \quad=\frac{1}{1+t_{2}^{\alpha-1}} \biggl|\frac{1}{\Gamma(\alpha)} \int ^{t_{2}}_{0}(t_{2}-s)^{\alpha-1}a(s)f \bigl(s,u(s)\bigr)\,ds +\frac{t_{2}}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \int^{\infty }_{0}a(s)f\bigl(s,u(s)\bigr)\,ds \\ & \quad\quad{}-\frac{t_{2}\sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}(\xi _{i}-s)^{\alpha-1} a(s)f(s,u(s))\,ds}{\Gamma(\alpha)\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} -\frac{1}{\Gamma(\alpha)} \int^{t_{1}}_{0}(t_{1}-s)^{\alpha -1}a(s)f \bigl(s,u(s)\bigr)\,ds \\ & \quad\quad{}-\frac{t_{1}}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \int^{\infty }_{0}a(s)f\bigl(s,u(s)\bigr)\,ds+ \frac{t_{1} \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}(\xi_{i}-s)^{\alpha -1}a(s)f(s,u(s))\,ds}{\Gamma(\alpha)\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \biggr| \\ & \quad\leq\frac{1}{1+t_{2}^{\alpha-1}} \biggl(\biggl\vert \frac{1}{\Gamma(\alpha)} \int ^{t_{2}}_{0} (t_{2}-s)^{\alpha-1}a(s)f \bigl(s,u(s)\bigr)\,ds \\ & \quad\quad{}-\frac{1}{\Gamma(\alpha)} \int ^{t_{1}}_{0}(t_{1}-s)^{\alpha-1}a(s)f \bigl(s,u(s)\bigr)\,ds\biggr\vert \\ & \quad\quad{}+\biggl\vert \frac{t_{2}}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \int^{\infty}_{0} a(s)f\bigl(s,u(s)\bigr)\,ds- \frac{t_{1}}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \int ^{\infty}_{0}a(s)f\bigl(s,u(s)\bigr)\,ds\biggr\vert \\ & \quad\quad{}+\biggl\vert \frac{t_{1}\sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}(\xi _{i}-s)^{\alpha-1}a(s)f(s,u(s))\,ds}{ \Gamma(\alpha)\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \\ & \quad\quad{}-\frac{t_{2}\sum^{m-2}_{i=1}\beta_{i} \int^{\xi_{i}}_{0}(\xi_{i}-s)^{\alpha-1}a(s)f(s,u(s))\,ds}{\Gamma(\alpha )\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}\biggr\vert \biggr) \\ & \quad\leq\frac{1}{1+t_{2}^{\alpha-1}} \biggl(\biggl\vert \frac{1}{\Gamma(\alpha)} \int ^{t_{1}}_{0} \bigl((t_{2}-s)^{\alpha-1}-(t_{1}-s)^{\alpha-1} \bigr)a(s)f \biggl(s,\bigl(1+s^{\alpha -1}\bigr)\frac{u(s)}{1+s^{\alpha-1}} \biggr)\,ds \biggr\vert \\ & \quad\quad{} +\biggl\vert \frac{1}{\Gamma(\alpha)} \int^{t_{2}}_{t_{1}}(t_{2}-s)^{\alpha-1}a(s) f \biggl(s,\bigl(1+s^{\alpha-1}\bigr)\frac{u(s)}{1+s^{\alpha-1}} \biggr)\,ds\biggr\vert \\ & \quad\quad{}+\biggl\vert \frac{(t_{2}-t_{1})}{ \sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \int^{\infty}_{0}a(s)f \biggl(s,\bigl(1+s^{\alpha-1} \bigr)\frac{u(s)}{1+s^{\alpha-1}} \biggr)\,ds\biggr\vert \\ & \quad\quad{}+\biggl\vert \frac{(t_{1}-t_{2})\sum^{m-2}_{i=1}\beta_{i} \int^{\xi_{i}}_{0}(\xi_{i}-s)^{\alpha-1}a(s)f (s,(1+s^{\alpha -1})\frac{u(s)}{1+s^{\alpha-1}} )\,ds}{ \Gamma(\alpha)\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}\biggr\vert \biggr) \\ & \quad\leq\frac{\Upsilon_{l}}{1+t_{2}^{\alpha-1}} \biggl(\frac {t_{2}-t_{1}}{\Gamma(\alpha)} \int^{t_{1}}_{0}a(s)\,ds +\biggl\vert \frac{1}{\Gamma(\alpha)} \int^{t_{2}}_{t_{1}}(t_{2}-s)^{\alpha -1}a(s) \,ds\biggr\vert \\ & \quad\quad{}+\frac{(t_{2}-t_{1})}{\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \int^{\infty}_{0}a(s)\,ds +\frac{(t_{2}-t_{1})\sum^{m-2}_{i=1}\beta_{i}\int^{\xi _{i}}_{0}(\xi_{i}-s)^{\alpha-1}a(s)\,ds}{ \Gamma(\alpha)\sum^{m-2}_{i=1}\beta_{i}\xi_{i}} \biggr). \end{aligned}$$
(3.3)
Since \(0<\int^{+\infty}_{0}a(s)\,ds<\infty\), by the integration of Cauchy’s test for convergence, we can get
$$\int_{0}^{\infty}\biggl\vert \frac{G(t_{2},s)}{1+t_{2}^{\alpha-1}}- \frac {G(t_{1},s)}{1+t_{2}^{\alpha-1}}\biggr\vert a(s)f\bigl(s,u(s)\bigr)\,ds\rightarrow0, $$
uniformly as \(t_{1}\rightarrow t_{2}\).
Similar to (3.3), we can deduce that
$$\int_{0}^{\infty}\biggl\vert \frac{G(t_{1},s)}{1+t_{2}^{\alpha-1}}- \frac {G(t_{1},s)}{1+t_{1}^{\alpha-1}}\biggr\vert a(s)f\bigl(s,u(s)\bigr)\,ds\rightarrow0, $$
uniformly as \(t_{1}\rightarrow t_{2}\).
Thus, we conclude that
$$\biggl\vert \frac{(Tu)(t_{2})}{1+t_{2}^{\alpha-1}}-\frac {(Tu)(t_{1})}{1+t_{1}^{\alpha-1}}\biggr\vert \rightarrow0 $$
uniformly as \(t_{1}\rightarrow t_{2}\), and hence \(T(\Omega)\) is locally equicontinuous on any finite subinterval of \([0,+\infty)\).
Step 5: We show that \(T:K\rightarrow K\) is equiconvergent at ∞.
For convenience, we denote \(\Delta=\frac{\sum^{m-2}_{i=1}\beta_{i}(\xi_{i}-s)^{\alpha-1}}{\Gamma (\alpha)\sum^{m-2}_{i=1}\beta_{i}\xi_{i}}\) and \(\Lambda=\sum^{m-2}_{i=1}\beta_{i}\xi_{i}\).
Since \(\lim_{t\rightarrow\infty}\frac{t}{1+t^{\alpha-1}}=0\), there exists \(N_{1}>0\) such that
$$ \biggl\vert \frac{t_{2}}{1+t_{2}^{\alpha-1}}-\frac{t_{1}}{1+t_{1}^{\alpha -1}}\biggr\vert \leq\biggl\vert 1-\frac{t_{2}}{1+t_{2}^{\alpha-1}}\biggr\vert +\biggl\vert 1- \frac{t_{1}}{1+t_{1}^{\alpha-1}}\biggr\vert < \varepsilon_{1} $$
(3.4)
for any \(t_{2}>t_{1}>N_{1}\), \(\varepsilon_{1}>0\).
Similarly, there exist \(M>0\), \(N_{2}>0\) such that \(\lim_{t\rightarrow \infty}\frac{(t-M)^{\alpha-1}}{1+t^{\alpha-1}}=1\) and
$$\begin{aligned} \biggl\vert \frac{(t_{2}-s)^{\alpha-1}}{1+t_{2}^{\alpha-1}}-\frac {(t_{1}-s)^{\alpha-1}}{1+t_{1}^{\alpha-1}}\biggr\vert \leq&\biggl\vert 1-\frac {(t_{2}-s)^{\alpha-1}}{1+t_{2}^{\alpha-1}}\biggr\vert +\biggl\vert 1-\frac {(t_{1}-s)^{\alpha-1}}{1+t_{1}^{\alpha-1}} \biggr\vert \\ \leq&\biggl\vert 1-\frac{(t_{2}-M)^{\alpha-1}}{1+t_{2}^{\alpha-1}}\biggr\vert +\biggl\vert 1- \frac{(t_{1}-M)^{\alpha-1}}{1+t_{1}^{\alpha-1}}\biggr\vert < \varepsilon_{2} \end{aligned}$$
(3.5)
for any \(t_{2}>t_{1}>N_{2}\), \(\varepsilon_{2}>0\) and \(0\leq s\leq M\).
Let \(N>\max\{N_{1},N_{2}\}\). For any \(u\in\Omega\), by (3.4) and (3.5), we have
$$\begin{aligned}& \biggl\vert \frac{(Tu)(t_{2})}{1+t_{2}^{\alpha-1}}-\frac {(Tu)(t_{1})}{1+t_{1}^{\alpha-1}}\biggr\vert \\& \quad= \biggl\vert \int_{0}^{\infty}\frac {G(t_{2},s)}{1+t_{2}^{\alpha-1}}a(s)f\bigl(s,u(s) \bigr)\,ds- \int_{0}^{\infty}\frac {G(t_{1},s)}{1+t_{1}^{\alpha-1}}a(s)f\bigl(s,u(s) \bigr)\,ds\biggr\vert \\& \quad= \biggl\vert \int_{0}^{\infty} \biggl(\frac{G(t_{2},s)}{1+t_{2}^{\alpha -1}}- \frac{G(t_{1},s)}{1+t_{1}^{\alpha-1}} \biggr)a(s)f\bigl(s,u(s)\bigr)\,ds\biggr\vert \\& \quad\leq \int_{0}^{t_{1}} \biggl(\biggl\vert \frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)(1+t_{2}^{\alpha-1})} -\frac{(t_{1}-s)^{\alpha-1}}{\Gamma(\alpha)(1+t_{1}^{\alpha-1})}\biggr\vert +\biggl\vert \frac{t_{2}}{\Lambda(1+t_{2}^{\alpha-1})}-\frac{t_{1}}{\Lambda (1+t_{1}^{\alpha-1})}\biggr\vert \\& \quad\quad{}+\biggl\vert \frac{t_{1}\Delta}{(1+t_{1}^{\alpha-1})}-\frac{t_{2}\Delta }{(1+t_{2}^{\alpha-1})}\biggr\vert \biggr)a(s)f\bigl(s,u(s)\bigr)\,ds \\& \quad\quad{}+ \int_{t_{1}}^{t_{2}} \biggl(\biggl\vert \frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)(1+t_{2}^{\alpha-1})}\biggr\vert +\biggl\vert \frac{t_{2}}{\Lambda (1+t_{2}^{\alpha-1})}- \frac{t_{1}}{\Lambda(1+t_{1}^{\alpha-1})}\biggr\vert \\& \quad\quad{} +\biggl\vert \frac{t_{1}\Delta}{(1+t_{1}^{\alpha-1})}- \frac{t_{2}\Delta }{(1+t_{2}^{\alpha-1})}\biggr\vert \biggr)a(s)f\bigl(s,u(s)\bigr)\,ds \\& \quad\quad{}+ \int_{t_{2}}^{+\infty} \biggl(\biggl\vert \frac{t_{2}}{\Lambda (1+t_{2}^{\alpha-1})}-\frac{t_{1}}{\Lambda(1+t_{1}^{\alpha-1})}\biggr\vert +\biggl\vert \frac{t_{1}\Delta}{(1+t_{1}^{\alpha-1})}-\frac{t_{2}\Delta }{(1+t_{2}^{\alpha-1})}\biggr\vert \biggr)a(s)f\bigl(s,u(s)\bigr) \,ds \\& \quad\leq \int_{0}^{t_{1}} \biggl(\frac{\varepsilon_{2}}{\Gamma(\alpha)}+ \frac {\varepsilon_{1}}{\Lambda}+\varepsilon_{1}\Delta \biggr)a(s)f \biggl(s, \bigl(1+s^{\alpha-1}\bigr)\frac{u(s)}{1+s^{\alpha-1}} \biggr)\,ds \\& \quad\quad{}+ \int_{t_{1}}^{t_{2}} \biggl(1+\frac{\varepsilon_{1}}{\Lambda }+ \varepsilon_{1}\Delta \biggr)a(s)f \biggl(s,\bigl(1+s^{\alpha-1} \bigr)\frac {u(s)}{1+s^{\alpha-1}} \biggr)\,ds \\& \quad\quad{}+ \int_{t_{2}}^{+\infty} \biggl(\frac{\varepsilon_{1}}{\Lambda }+ \varepsilon_{1}\Delta \biggr)a(s)f \biggl(s,\bigl(1+s^{\alpha-1} \bigr)\frac {u(s)}{1+s^{\alpha-1}} \biggr)\,ds \\& \quad\leq \Upsilon_{l} \biggl( \biggl(\frac{\varepsilon_{2}}{\Gamma(\alpha )}+ \frac{\varepsilon_{1}}{\Lambda}+\varepsilon_{1}\Delta \biggr) \int _{0}^{t_{1}}a(s)\,ds+ \biggl(1+ \frac{\varepsilon_{1}}{\Lambda}+\varepsilon _{1}\Delta \biggr) \int_{t_{1}}^{t_{2}}a(s)\,ds \\& \quad\quad{}+ \biggl(\frac{\varepsilon_{1}}{\Lambda}+\varepsilon_{1}\Delta \biggr) \int_{t_{2}}^{+\infty}a(s)\,ds \biggr) \\& \quad\rightarrow0 \end{aligned}$$
uniformly as \(t_{1}\rightarrow t_{2}\).
In conclusion, for any \(\varepsilon>0\), there exists a sufficiently large \(N>0\) such that for any \(u\in\Omega\),
$$\biggl\vert \frac{u(t_{1})}{1+t^{\alpha-1}_{1}}-\frac{u(t_{2})}{1+t^{\alpha -1}_{2}}\biggr\vert < \varepsilon, $$
\(\forall t_{1},t_{2}>N\).
This implies that \(T:K\rightarrow K\) is equiconvergent at ∞.
By the Arzela-Ascoli theorem, we see that \(T:K\rightarrow K\) is completely continuous. The proof is completed. □
Now we will list the following condition in this section:
- (H3):
-
\(f(t,\cdot)\) is nondecreasing for any \(t\in[0,+\infty)\), and there exists a constant \(b>0\), such that \(f(t,(1+t^{\alpha-1})u)\leq \frac{b\sum^{m-2}_{i=1}\beta_{i}}{2\int^{+\infty}_{0}a(s)\,ds}\) for \((t,u)\in[0,+\infty)\times[0,b]\).
Theorem 3.1
Assume (H1), (H2), and (H3) hold. Then the multi-point boundary value problem (1.1)-(1.2) has the minimal and maximal positive solutions
\(v^{\ast}\), \(u^{\ast}\)
in
\((0,b]\), which can be obtained by the following two explicit monotone iterative sequences:
$$v_{n+1}= \int^{+\infty}_{0}G(t,s)a(s)f\bigl(s,v_{n}(s) \bigr)\,ds $$
with initial value
\(v_{0}(t)=0\),
$$u_{n+1}= \int^{+\infty}_{0}G(t,s)a(s)f\bigl(s,u_{n}(s) \bigr)\,ds $$
with initial value
\(u_{0}(t)=b\). Moreover,
$$\begin{aligned} v_{0} \leq& v_{1}\leq\cdots\leq v_{n}\leq\cdots \leq v^{\ast}\leq\cdots \\ \leq& u^{\ast}\leq\cdots\leq u_{n}\leq\cdots\leq u_{1}\leq u_{0}. \end{aligned}$$
Proof
Denote \(\Phi=\{u\in K,\|u\|_{C_{\infty}}\leq b\}\). Then we have \(T(\Phi)\subset\Phi\). In fact, let \(u\in\Phi\). Then by (H1), (H3), and Lemma 3.3, we get
$$\begin{aligned} \|Tu\|_{C_{\infty}} =&\sup_{t\in[0,+\infty)}\biggl\vert \int^{+\infty }_{0}\frac{G(t,s)}{1+t^{\alpha-1}}a(s)f\bigl(s,u(s) \bigr)\biggr\vert \,ds \\ \leq& \frac{2}{\sum^{m-2}_{i=1}\beta_{i}} \int^{+\infty}_{0}a(s)f \biggl(s,\bigl(1+s^{\alpha-1} \bigr)\frac{u(s)}{1+s^{\alpha-1}} \biggr)\,ds \\ \leq&\frac{2}{\sum^{m-2}_{i=1}\beta_{i}} \int^{+\infty}_{0}a(s)\,ds\frac {b\sum^{m-2}_{i=1}\beta_{i}}{2\int^{+\infty}_{0}a(s)\,ds} \\ =&b. \end{aligned}$$
So \(T(\Phi)\subset\Phi\).
Denote that \(v_{0}(t)=0\), \(v_{1}=Tv_{0}\), and \(v_{2}=T^{2}v_{0}=Tv_{1}\), for all \(t\in[0,+\infty)\). Since \(v_{0}(t)=0\in\Phi\) and \(T:\Phi\rightarrow\Phi\), \(v_{1}\in T(\Phi)\subset\Phi\) and \(v_{2}\in T(\Phi)\subset\Phi\). We have
$$v_{1}(t)=(Tv_{0}) (t)\geq0=v_{0}(t), $$
for all \(t\in[0,+\infty)\).
By (H3), for \(u,v\in\Phi\) and \(u\geq v\), we deduce
$$\begin{aligned} Tu(t) =& \int^{+\infty}_{0}G(t,s)a(s)f\bigl(s,u(s)\bigr)\,ds \\ \geq& \int^{+\infty}_{0}G(t,s)a(s)f\bigl(s,v(s)\bigr)\,ds \\ =&Tv(t). \end{aligned}$$
We know that T is a nondecreasing operator.
So we have
$$v_{2}(t)=(Tv_{1}) (t)\geq(Tv_{0}) (t)=v_{1}(t), $$
for all \(t\in[0,+\infty)\).
By the induction, define \(v_{n+1}=Tv_{n}\), \(n=0,1,2,\ldots \) . The sequence \(\{v_{n}\}^{\infty}_{n=1}\subset T(\Phi)\subset\Phi\) and satisfies the following relation:
$$v_{n+1}(t)\geq v_{n}(t), $$
for all \(t\in[0,+\infty)\), \(n=0,1,2,\ldots\) .
In view of T is completely continuous and \(v_{n+1}=Tv_{n}\), \(\{v_{n}\}^{\infty}_{n=1}\) is relative compact. That is to say, \(\{v_{n}\}^{\infty}_{n=1}\) has a convergent subsequence \(\{v_{n_{k}}\}^{\infty}_{k=1}\) and there exists a \(v^{\ast}\in\Phi\) such that \(v_{n_{k}}\rightarrow v^{\ast}\) as \(k\rightarrow\infty\).
By the above part and \(v_{n+1}(t)\geq v_{n}(t)\), for all \(t\in [0,+\infty)\), \(n=0,1,2,\ldots \) , we can get \(\lim_{n\rightarrow\infty}v_{n}=v^{\ast}\).
Since T is continuous and \(v_{n+1}=Tv_{n}\), we have \(Tv^{\ast}=v^{\ast }\). That is to say, \(v^{\ast}\) is a fixed point of the operator T.
Denote \(u_{0}(t)=b\), \(u_{1}=Tu_{0}\), and \(u_{2}=T^{2}u_{0}=Tu_{1}\), for all \(t\in[0,+\infty)\). Since \(u_{0}(t)\in\Phi\) and \(T:\Phi\rightarrow\Phi\), \(u_{1}\in T(\Phi)\subset\Phi\), and \(u_{2}\in T(\Phi)\subset\Phi\).
By (H3), we deduce
$$\begin{aligned} u_{1}(t) =& \int^{+\infty}_{0}G(t,s)a(s)f\bigl(s,u_{0}(s) \bigr)\,ds \\ \leq&\frac{2}{\sum^{m-2}_{i=1}\beta_{i}} \int^{+\infty }_{0}a(s)f\bigl(s,\bigl(1+s^{\alpha-1} \bigr)u_{0}(s)\bigr)\,ds \\ \leq&\frac{2}{\sum^{m-2}_{i=1}\beta_{i}} \int^{+\infty}_{0}a(s)\,ds\frac {b\sum^{m-2}_{i=1}\beta_{i}}{2\int^{+\infty}_{0}a(s)\,ds} \\ =&b=u_{0}(t), \end{aligned}$$
for all \(t\in[0,+\infty)\).
Since T is a nondecreasing operator, we have
$$u_{2}(t)=(Tu_{1}) (t)\leq(Tu_{0}) (t)=u_{1}(t), $$
for all \(t\in[0,+\infty)\).
By the induction, define \(u_{n+1}=Tu_{n}\), \(n=0,1,2,\ldots \) . The sequence \(\{u_{n}\}^{\infty}_{n=1}\subset T(\Phi)\subset\Phi\) and satisfies the following relation:
$$u_{n+1}(t)\leq u_{n}(t), $$
for all \(t\in[0,+\infty)\), \(n=0,1,2,\ldots \) .
With an analysis exactly parallel to the proving process of \(\lim_{n\rightarrow\infty}v_{n}=v^{\ast}\), we see that there exists a \(u^{\ast}\in\Phi\) such that \(\lim_{n\rightarrow\infty}u_{n}=u^{\ast}\).
Since T is completely continuous and \(u_{n+1}=Tu_{n}\), we have \(Tu^{\ast}=u^{\ast}\). That is to say, \(u^{\ast}\) is a fixed point of the operator T.
Now, we will show that \(u^{\ast}\) and \(v^{\ast}\) are the maximal and minimal positive solutions of the boundary value problem (1.1)-(1.2) in \((0,b]\).
Let \(\phi\in[0,b]\) be any solution of the boundary value problem (1.1)-(1.2). That is, \(T\phi=\phi\). Noting that T is nondecreasing and \(v_{0}(t)=0\leq\phi(t)\leq b=u_{0}(t)\), we have \(v_{1}(t)=Tv_{0}(t)\leq\phi(t)\leq Tu_{0}(t)=u_{1}(t)\), for all \(t\in[0,+\infty)\).
Similarly, we can obtain
$$v_{n}(t)\leq\phi(t)\leq u_{n}(t), $$
for all \(t\in[0,+\infty)\), \(n=0,1,2,\ldots\) .
Since \(\lim_{n\rightarrow\infty}u_{n}=u^{\ast}\) and \(\lim_{n\rightarrow\infty}v_{n}=v^{\ast}\), by the above formulas we obtain
$$\begin{aligned} v_{0} \leq& v_{1}\leq\cdots\leq v_{n}\leq\cdots \leq v^{\ast}\leq\cdots \\ \leq& u^{\ast}\leq\cdots\leq u_{n}\leq\cdots\leq u_{1}\leq u_{0}. \end{aligned}$$
Since \(f(t,0)\not\equiv0\), for all \(t\in[0,+\infty)\), 0 is not a solution of the boundary value problem (1.1)-(1.2). We know that \(u^{\ast}\) and \(v^{\ast}\) are the maximal and minimal positive solutions of the boundary value problem (1.1)-(1.2) in \((0,b]\), which can be obtained by the corresponding iterative sequences in
$$v_{n+1}= \int^{+\infty}_{0}G(t,s)a(s)f\bigl(s,v_{n}(s) \bigr)\,ds $$
with initial value \(v_{0}(t)=0\),
$$u_{n+1}= \int^{+\infty}_{0}G(t,s)a(s)f\bigl(s,u_{n}(s) \bigr)\,ds $$
with initial value \(u_{0}(t)=b\). The proof is completed. □