In this section, we give a theorem on the existence of solutions for BVP (1.1). Let
$$\begin{aligned} &{X = \biggl\{ {x\Big\vert {x, D_{0 + }^{\alpha}x} \in C[0, + \infty), \sup_{t \in[0, + \infty)} \frac{{\vert x(t)\vert }}{{1 + t^{\alpha}}} < + \infty, }} \\ &{\hphantom{X =} { \sup_{t \in[0, + \infty)} \frac{{\vert D_{0 + }^{\alpha-1} x(t)\vert }}{ {1 + t^{\alpha}}} < + \infty, \sup_{t \in[0, + \infty)} \bigl\vert D_{0 + }^{\alpha}x(t) \bigr\vert < + \infty} \biggr\} ,} \\ &{Y = L^{1} [0, + \infty),} \end{aligned}$$
with norms
$$\Vert x \Vert _{X} = \max \bigl\{ {\Vert x \Vert _{0} ,\bigl\Vert {D_{0 + }^{\alpha - 1} x} \bigr\Vert _{0} , \bigl\Vert {D_{0 + }^{\alpha}x} \bigr\Vert _{\infty}} \bigr\} ,\qquad \Vert y \Vert _{Y} = \Vert y\Vert _{1}, $$
where \(\Vert y \Vert _{1} = \int_{0}^{ + \infty} {\vert {y(t)} \vert \,dt} \), \(\Vert {D_{0 + }^{\alpha}x} \Vert _{\infty}= \sup_{t \in[0, + \infty)} \vert {D_{0 + }^{\alpha}x(t)} \vert \), \(\Vert x \Vert _{0} = \Vert {\frac{{x(t)}}{ {1 + t^{\alpha}}}} \Vert _{\infty}\). Clearly, \((X,\Vert \cdot \Vert _{X} ) \) and \((Y,\Vert \cdot \Vert _{Y} )\) are Banach spaces.
Define the operators \(M:\operatorname{dom}M \subset X \to Y\) and \(N_{\lambda}:X \to Y\) as follows:
$$Mx = \bigl(\phi_{p} \bigl(D_{0 + }^{\alpha}x \bigr) \bigr)',\qquad N_{\lambda}x(t) = - \lambda f \bigl(t,x(t),D_{0 + }^{\alpha-1} x(t),D_{0 + }^{\alpha}x(t) \bigr),\quad \lambda\in[0,1], x \in X, $$
where
$$\begin{aligned} \operatorname{dom}M =& \Biggl\{ x \in X \Big\vert {\phi_{p} \bigl(D_{0 + }^{\alpha}x \bigr)} \in AC[0, + \infty), x(0) = x'(0) = 0, \\ & \phi_{p} \bigl(D_{0 +}^{\alpha}x( + \infty) \bigr) = \sum_{i = 1}^{n} {\alpha _{i} \phi_{p} \bigl(D_{0 +}^{\alpha}x(\xi _{i}) \bigr)} \Biggr\} . \end{aligned}$$
Then the BVP (1.1) is equivalent to \(Mx=Nx\), \(x \in\operatorname{dom}M\).
Lemma 3.1
For
M
defined as before, we have
$$\begin{aligned} &{\operatorname{Ker}M = \bigl\{ {x \in\operatorname{dom}M \vert x(t) = ct^{\alpha}, \forall t \in[0, + \infty),c \in\mathbb{R}} \bigr\} ,} \end{aligned}$$
(3.1)
$$\begin{aligned} &{\operatorname{Im}M = \Biggl\{ {y \in Y\Big\vert \sum _{i = 1}^{n} {\alpha_{i} \int_{\xi_{i} }^{ + \infty} {y(s)\,ds = 0} } } \Biggr\} ,} \end{aligned}$$
(3.2)
and
M
is a quasi-linear operator.
Proof
For \(x \in\operatorname{Ker}M\), \(Mx= 0\), that is, \((\phi_{p} (D_{0 + }^{\alpha}x))' = 0\), by \(\phi_{p} (D_{0 +}^{\alpha}x( + \infty)) = \sum_{i = 1}^{n} \alpha_{i} \phi_{p}(D_{0 +}^{\alpha}x(\xi_{i}))\) we easily get
$$\phi_{p} \bigl(D_{0 + }^{\alpha}x(t) \bigr) = \phi _{p} \bigl(D_{0 +}^{\alpha}x( + \infty) \bigr) = \sum _{i = 1}^{n} {\alpha_{i} \phi _{p} \bigl(D_{0 +}^{\alpha}x(\xi_{i}) \bigr)}. $$
Based on Lemma 2.1, since \(x(0) = x'(0) = 0\), we have
$$\begin{aligned} x(t) =& c_{1} t^{\alpha - 1} + c_{2} t^{\alpha - 2} + \frac{{\phi_{q} (\sum_{i = 1}^{n} {\alpha_{i} \phi_{p} (D_{0 +}^{\alpha}x(\xi_{i}))} )}}{ {\Gamma(\alpha)}} \int_{0}^{t} {(t - s)^{\alpha - 1}\,ds} \\ =& \frac{{\phi_{q} (\sum_{i = 1}^{n} {\alpha_{i} \phi_{p} (D_{0 +}^{\alpha}x(\xi_{i}))} )}}{ {\Gamma(\alpha+ 1)}}t^{\alpha}. \end{aligned}$$
Conversely, if \(x = ct^{\alpha}\), then \(Mx = 0\) by (3.1). If \(y \in\operatorname{Im} M\), then there exists a function \(x \in \operatorname{dom}M\) such that \(y(t) = (\phi_{p} (D_{0 + }^{\alpha}x(t)))'\). Then
$$\begin{aligned} \phi_{p} \bigl(D_{0 +}^{\alpha}x(t) \bigr) =& \phi _{p} \bigl(D_{0 +}^{\alpha}x( + \infty) \bigr) - \int_{t}^{ + \infty} {y(s)\,ds} \\ =& \sum_{i = 1}^{n} {\alpha_{i} \phi_{p} \bigl(D_{0 +}^{\alpha}x(\xi_{i}) \bigr)} - \int_{t}^{ + \infty} {y(s)\,ds} \\ =&\sum_{i = 1}^{n} {\alpha_{i} } \biggl[ {\phi_{p} \bigl(D_{0 +}^{\alpha}x( + \infty) \bigr) - \int_{\xi_{i} }^{ + \infty} {y(s)\,ds} } \biggr] - \int_{t}^{ + \infty} {y(s)\,ds}, \end{aligned}$$
that is,
$$ \sum_{i = 1}^{n} \alpha_{i} \int_{\xi_{i} }^{ + \infty} y(s)\,ds = 0. $$
(3.3)
On the other hand, if \(y \in Y\) satisfies (3.3), then take
$$x(t) = I_{0 + }^{\alpha}\phi_{q} \biggl({- \int_{t}^{ + \infty} {y(s)\,ds} } \biggr). $$
Then \(x \in\operatorname{dom} M\) and \(Mx = y\). Hence, (3.2) holds. Clearly, \(\operatorname{dim}\operatorname{Ker}M = 1<+\infty\) and \(\operatorname{Im}M:=M(\operatorname{dom}M \cap X)\) is a closed subset of Y. Therefore, we get that M is a quasi-linear operator. □
Lemma 3.2
(See [1])
Let
\(V \subset C_{\infty}= \{u \in C[0,\infty), \lim_{t \to + \infty} u(t)\textit{ exists} \}\). Then
V
is relatively compact if the following conditions hold:
- (\(b_{1}\)):
-
all functions from
V
are uniformly bounded,
- (\(b_{2}\)):
-
all functions from
V
are equicontinuous on any compact interval of
\([0,+\infty)\),
- (\(b_{3}\)):
-
all functions from
V
are equiconvergent at infinity.
Remark 3.1
By Lemma 3.2, any set \(V \subset X\) (defined as before) is relatively compact, we only need to show that the sets
$$V_{1} = \biggl\{ {\frac{{x(t)}}{ {1 + t^{\alpha}}}\Big\vert {x \in V} } \biggr\} ,\qquad V_{2} = \biggl\{ {\frac{{D_{0 + }^{\alpha - 1} x(t)}}{ {1 + t^{\alpha}}}\Big\vert {x \in V} } \biggr\} , \qquad V_{3} = \bigl\{ {D_{0 + }^{\alpha}x(t)\vert {x \in V} } \bigr\} $$
are uniformly bounded in X, equicontinuous on any compact intervals of \([0, + \infty)\), and equiconvergent at infinity.
Lemma 3.3
Let
\(\Omega \subset X\)
be nonempty, open, and bounded. Then
\(N_{\lambda}\)
is
M-compact on Ω̄.
Proof
Define two projectors \(P:X \to X_{1}\) and \(Q:Y \to Y_{1}\) by
$$Px(t) = \frac{{D_{0 + }^{\alpha}x( + \infty)}}{ {\Gamma(\alpha + 1)}}t^{\alpha},\qquad Qy(t) = \frac{{\sum_{i = 1}^{n} {\alpha _{i} \int_{\xi_{i} }^{ + \infty} {y(s)\,ds} } }}{ {\sum_{i = 1}^{n} {\alpha_{i} e^{ - \xi_{i} } } }}e^{ - t}, \quad \forall t \in[0, + \infty), $$
where \(X_{1} =\operatorname{Ker}M\) and \(Y_{1} =\operatorname {Im} Q\).
First, we show that (\(a_{1}\)) and (\(a_{2}\)) in Definition 2.4 hold. In fact, by the definition of P we get \(\operatorname{Im} P=\operatorname{Ker}M\) and \(P^{2} x(t) = Px(t)\). For \(x \in X\), since \(x = (x - Px) + Px\) and \(\operatorname{Im} P=\operatorname{Ker}M\), we have \((x - Px) \in\operatorname{Ker}P\), \(Px \in\operatorname{Ker}M\). We easily see that \(\operatorname{Ker}M \cap\operatorname{Ker}P= \{ 0 \}\). So, \(X=\operatorname{Ker}M \oplus\operatorname{Ker}P=X_{1} \oplus X_{2}\). Similarly, by the definition of Q, we can obtain
$$Q^{2} y = Q(Qy) = \frac{{\sum_{i = 1}^{n} {\alpha_{i} \int_{\xi_{i} }^{ + \infty} {Qy\,ds} } }}{ {\sum_{i = 1}^{n} {\alpha_{i} e^{ - \xi_{i} } } }}e^{ - t} = Qy, $$
and \(\operatorname{Ker}Q = \operatorname{Im} M\). For \(y \in Y\), since \(y = (y - Qy) + Qy\) and \(\operatorname{Ker}Q = \operatorname{Im} M\), we have \((y - Qy) \in\operatorname{Ker}Q\), \(Qy\in\operatorname{Im} M\). Clearly, \(\operatorname{Im} Q \cap\operatorname{Im} M = \{ 0 \}\). So, we have \(Y = \operatorname{Im} Q \oplus\operatorname{Im} M = Y_{1} \oplus Y_{2}\) and \(\operatorname{dim} X_{1} = \operatorname{dim}\operatorname{Ker}M = \operatorname{dim} \operatorname{Im} Q = \operatorname{dim} Y_{1}\), where \(X_{2} =\operatorname{Ker}P\), \(Y_{2} = \operatorname{Im} M\). Let \(\Omega \subset X\) be open bounded, and let \(\theta \in\Omega\). On the one hand, for \(x \in\bar{\Omega}\), since \(Q(I - Q)\) is a zero operator, we have \(Q[(I - Q)N_{\lambda}x] = 0\); thus, \((I - Q)N_{\lambda}x \in\operatorname{Ker}Q=\operatorname{Im} M\), that is, \((I - Q)N_{\lambda}(\bar{\Omega}) \subset\operatorname{Im} M\). On the other hand, for \(y \in\operatorname{Im} M\), since \(y = (y - Qy) + Qy\) and \(\operatorname{Ker}Q=\operatorname{Im} M\), we have \(y \in(I - Q)Y\), that is, \(\operatorname{Im} M \subset(I - Q)Y\). Clearly, \(QN_{\lambda}x = 0\), \(\lambda \in(0,1) \Leftrightarrow QNx = 0\). So, conditions (\(a_{1}\)) and (\(a_{2}\)) of Definition 2.4 hold.
Second, we give the definition of operator R and aim to show that R is compact. For notational convenience, let
$$l(t,x,\lambda) = \int_{t}^{ + \infty} {(Q - I)N_{\lambda}x(s)}\,ds + \phi_{p} \bigl(D_{0 + }^{\alpha}x( + \infty) \bigr), h(t) = \phi_{q} \bigl(l(t,x,\lambda) \bigr)-D_{0 + }^{\alpha}x( + \infty). $$
Define the operator \(R:\bar{\Omega}\times[0,1] \to X_{2}\) by
$$R(x,\lambda) (t) = I_{0 + }^{\alpha}h(t). $$
By (\(A_{1}\)) it is easy to know that \(R(x,\lambda)(t)\) is continuous on \(\bar{\Omega}\times[0,1]\).
Step 1. We prove that \(R(x,\lambda)(\bar{\Omega})\) is both uniformly bounded in X and equicontinuous on any compact interval of \([0, + \infty)\). In fact, since \(\Omega \subset X\) is nonempty, open, and bounded, by (\(A_{1}\)) there exist a constant \(r > 0\) and a nonnegative function \(g_{r} (t) \in L^{1} [0, + \infty)\) such that
$$\Vert x \Vert _{X} \leqslant r,\qquad \bigl\vert {f \bigl(t,x(t),D_{0 + }^{\alpha-1} x(t),D_{0 + }^{\alpha}x(t) \bigr)} \bigr\vert \leq g_{r} (t),\quad \mbox{a.e. } t \in[ {0, + \infty} ),x \in\bar{\Omega}. $$
Since
$$\begin{aligned} & \biggl\vert \int_{s}^{ + \infty} {{(Q - I)N_{\lambda}x}\,d\tau} \biggr\vert \\ &\quad \leq \int_{s}^{ + \infty} {\vert {QN_{\lambda}x - N_{\lambda}x} \vert \,d\tau} \\ &\quad \leq \int_{0}^{ + \infty} \biggl\vert f \bigl(\tau,x( \tau),D_{0 + }^{\alpha-1} x(\tau),D_{0 + }^{\alpha}x( \tau) \bigr) \\ &\qquad {} - \frac{{\sum_{i = 1}^{n} {\alpha_{i} \int_{\xi_{i} }^{ + \infty} {f(s,x(s),D_{0 + }^{\alpha-1} x(s),D_{0 + }^{\alpha}x(s))\,ds} } }}{ {\sum_{i = 1}^{n} {\alpha_{i} e^{ - \xi_{i} } } }}e^{ - \tau} \biggr\vert \,d\tau \\ &\quad \leq \int_{0}^{ + \infty} \bigl\vert f \bigl(\tau,x( \tau),D_{0 + }^{\alpha-1} x(\tau),D_{0 + } ^{\alpha}x( \tau) \bigr) \bigr\vert \,d\tau \\ &\qquad {}+ \biggl\vert \frac{\sum_{i = 1}^{n} \alpha_{i} \int_{\xi_{i} }^{ + \infty} {f(s,x(s),D_{0 + }^{\alpha-1} x(s),D_{0 + }^{\alpha}x(s))\,ds} }{ {\sum_{i = 1}^{n} {\alpha_{i} e^{ - \xi_{i} } } }} \biggr\vert \\ &\quad \leq \Vert {g_{r} } \Vert _{1} \biggl( 1 + \frac{1}{ \sum_{i = 1}^{n} {\alpha_{i} e^{ - \xi_{i} } }} \biggr) =:\tilde{r}, \end{aligned}$$
(3.4)
we have
$$\begin{aligned} \bigl\vert {h(s)} \bigr\vert =& \biggl\vert {{\phi_{q} \biggl[ { \int_{s}^{ + \infty} {(Q - I)N_{\lambda}x(\tau)}\,d \tau+ \phi_{p} \bigl(D_{0 + }^{\alpha}x( + \infty) \bigr)} \biggr]}- D_{0 + }^{\alpha}x( + \infty)} \biggr\vert \\ \leq& \phi_{q} \bigl[ {\tilde{r} + \phi_{p} (r)} \bigr] + r =: m. \end{aligned}$$
Therefore, for any \(x \in\bar{\Omega}\), we have
$$\begin{aligned}& \begin{aligned} \bigl\Vert R(x,\lambda) (t) \bigr\Vert _{0} &= \sup_{t \in[0, + \infty)} \biggl\vert {\frac{{R(x,\lambda)(t)}}{ {1 +t^{\alpha}}}} \biggr\vert \\ &= \sup_{t \in[0, + \infty)} \frac{1}{ {\Gamma(\alpha)}} { \biggl\vert { \int_{0}^{t } {\frac{{(t - s)^{\alpha - 1} }}{ {1 +t^{\alpha}}}h(s)\,ds} } \biggr\vert } \\ &\leq \frac{m}{{\Gamma(\alpha)}} \sup_{t \in[0, + \infty)} { \int_{0}^{t } {\frac{{(t - s)^{\alpha - 1} }}{ {1 +t^{\alpha}}}\,ds} }\\ & \leq \frac{m}{{\Gamma(\alpha+1)}} \leq m, \end{aligned} \\& \begin{aligned} \bigl\Vert {D_{0 + }^{\alpha-1} R(x,\lambda) (t)} \bigr\Vert _{0} &= \sup_{t \in[0, + \infty)} \biggl\vert { \frac{{D_{0 + }^{\alpha-1}R(x,\lambda)(t)}}{ {1 +t^{\alpha}}}} \biggr\vert \\ &= \sup_{t \in[0, + \infty)} \biggl\vert { \int_{0}^{t } \frac{1}{ {1 +t^{\alpha}}}h(s)\,ds} \biggr\vert \\ &\leq m \sup_{t \in[0, + \infty)} { \int_{0}^{t } {\frac{1}{ {1 +t^{\alpha}}}\,ds} } \leq m, \end{aligned} \end{aligned}$$
and
$$\begin{gathered} \bigl\Vert {D_{0 + }^{\alpha}R(x, \lambda) (t)} \bigr\Vert _{\infty}= \sup_{t \in[0, + \infty)} \bigl\vert {D_{0 + }^{\alpha}R(x,\lambda) (t)} \bigr\vert = \sup _{t \in[0, + \infty)} \bigl\vert h(t) \bigr\vert \leq m. \end{gathered} $$
That is, \(R(x,\lambda)(\bar{\Omega})\) is uniformly bounded in X. Next, we show that \(R(x,\lambda)(\bar{\Omega})\) is equicontinuous on any compact interval of \([0, + \infty)\). In fact, for any \(K > 0\), \(t_{1},t_{2} \in[0,K]\), \(x \in\bar{\Omega}\), \(\lambda \in [0,1]\), we have
$$\begin{aligned} & \biggl\vert {\frac{{R(x,\lambda)(t_{1})}}{ {1 +t_{1}^{\alpha}}} - \frac{{R(x,\lambda)(t_{2})}}{ {1 +t_{2}^{\alpha}}}} \biggr\vert \\ &\quad =\frac{1}{ {\Gamma(\alpha)}} \biggl[ { \biggl\vert { \int_{0}^{t_{1} } {\frac{{(t_{1} - s)^{\alpha - 1} }}{ {1 +t_{1}^{\alpha}}}h(s)\,ds} } } - \int_{0}^{t_{2} } {\frac{{(t_{2} - s)^{\alpha - 1} }}{ {1 +t_{2}^{\alpha}}}h(s)\,ds} \biggr\vert \biggr] \\ &\quad \leq\frac{1}{ {\Gamma(\alpha)}} \biggl[ { \biggl\vert { \int_{0}^{t_{1} } { \biggl( {\frac{{(t_{1} - s)^{\alpha - 1} }}{ {1 +t_{1}^{\alpha}}} - \frac{{(t_{2} - s)^{\alpha - 1} }}{ {1 +t_{2}^{\alpha}}}} \biggr)h(s)\,ds} } \biggr\vert } + { \biggl\vert { \int_{t_{1} }^{t_{2} } {\frac{{(t_{2} - s)^{\alpha - 1} }}{ {1 +t_{2}^{\alpha}}}h(s)\,ds} } \biggr\vert } \biggr] \\ &\quad \leq\frac{m}{ {\Gamma(\alpha)}} \biggl[ { \int_{0}^{t_{1} } { \biggl\vert {\frac{{(t_{1} - s)^{\alpha - 1} }}{ {1 +t_{1}^{\alpha}}} - \frac{{(t_{2} - s)^{\alpha - 1} }}{ {1 +t_{2}^{\alpha}}}} \biggr\vert \,ds} + { \int_{t_{1} }^{t_{2} } {\frac{{(t_{2} - s)^{\alpha - 1} }}{ {1 +t_{2}^{\alpha}}}\,ds} }} \biggr] \\ &\quad \leq\frac{m}{ {\Gamma(\alpha)}} \biggl[ { \int_{0}^{t_{1} } { \biggl\vert {\frac{{(t_{1} - s)^{\alpha - 1} }}{ {1 +t_{1}^{\alpha}}} - \frac{{(t_{2} - s)^{\alpha - 1} }}{ {1 +t_{2}^{\alpha}}}} \biggr\vert \,ds} + \frac{1}{ \alpha}\frac{{(t_{2} - t_{1})^{\alpha}}}{ {1 +t_{2}^{\alpha}}}} \biggr] \to0\quad \mbox{as } t_{1} \to t_{2}. \end{aligned} $$
So, \(\{ {\frac{{R(x,\lambda)(t)}}{ {1 +t^{\alpha}}},x \in\bar{\Omega}} \}\) is equicontinuous on \([0,K]\). Similarly, we obtain that \(\{ {\frac{{D_{0 + }^{\alpha - 1} R(x,\lambda)(t)}}{ {1 +t^{\alpha}}},x \in\bar{\Omega}} \}\) is equicontinuous on \([0,K]\). In addition, since
$$\begin{aligned} \bigl\vert {D_{0 + }^{\alpha}R(x,\lambda) (t_{1}) - D_{0 + }^{\alpha}R(x,\lambda) (t_{2})} \bigr\vert =& \bigl\vert {h(t_{1}) - h(t_{2})} \bigr\vert \\ =& \bigl\vert \phi_{q} \bigl(l(t_{1},x,\lambda) \bigr)- \phi_{q} \bigl(l(t_{2},x,\lambda) \bigr) \bigr\vert \end{aligned}$$
and
$$\begin{aligned} \bigl\vert {l(t,x,\lambda)} \bigr\vert =& \biggl\vert { \int_{t}^{ + \infty} {(Q - I)N_{\lambda}x(s)}\,ds + \phi_{p} \bigl(D_{0 + }^{\alpha}x( + \infty) \bigr)} \biggr\vert \\ \leq& \tilde{r}+ \phi_{p} (r)\quad \bigl(t \in[0,K], x \in\bar{\Omega}\bigr), \end{aligned}$$
we have
$$\bigl\vert {l(t_{1},x,\lambda) - l(t_{2},x,\lambda)} \bigr\vert = \biggl\vert { \int_{t_{1} }^{t_{2} } {(Q - I)N_{\lambda}x(s)\,ds} } \biggr\vert \leq \int_{t_{1} }^{t_{2} } { \biggl( {g_{r} (s) + \frac{{\Vert {g_{r} } \Vert _{L^{1} } }}{ {\sum_{i = 1}^{n} {\alpha_{i} e^{ - \xi_{i} } } }}e^{ - s} } \biggr)\,ds} . $$
By the absolute continuity of the integral, \(\{ {l(t,x,\lambda),x \in \bar{\Omega}} \}\) is equicontinuous on \([0,K]\), which combined with the uniform continuity of \(\phi_{q} (x)\) on \([ - \tilde{r} - \phi_{p} (r),\tilde{r} + \phi_{p} (r)]\), gives that \(\{ {D_{0 + }^{\alpha}R(x,\lambda)(t),x \in\bar{\Omega}} \}\) is equicontinuous on \([0,K]\).
Step 2. We establish the fact that \(R(x,\lambda)(\bar{\Omega})\) is equiconvergent at infinity. In fact, for any \(x \in\bar{\Omega}\), by (3.4) we have
$$\begin{gathered} \lim_{s \to + \infty}{{ \int_{s}^{ + \infty} {(Q - I)N_{\lambda}x(\tau)\,d \tau } } } = 0. \end{gathered} $$
Since \(\phi_{q} (x)\) is uniformly continuous on \([ - \tilde{r} - \phi_{p} (r),\tilde{r} + \phi_{p} (r)]\), for any \(\varepsilon > 0\), there exists a constant \(L_{1} > 0\) such that, for \(s \geq L_{1} \), we have
$$\begin{aligned} \bigl\vert {h(s)} \bigr\vert =& \biggl\vert {{\phi _{q} \biggl[ { \int_{s}^{ + \infty} {(Q - I)N_{\lambda}x(\tau)}\,d \tau+ \phi_{p} \bigl(D_{0 + }^{\alpha}x( + \infty) \bigr)} \biggr]} - \phi_{q} \bigl[ {\phi_{p} \bigl(D_{0 + }^{\alpha}x( + \infty) \bigr)} \bigr]} \biggr\vert \\ < & \varepsilon,\quad \forall x \in\bar{\Omega}. \end{aligned}$$
Therefore, \(\vert {h(s)} \vert < \varepsilon\) for \(s \geq L_{1}\) and \(\vert {h(s)} \vert \leq m\) for \(s < L_{1}\). On the other hand, since \(\lim_{t \to + \infty} \frac {{t^{\alpha - 1} }}{ {1 + t^{\alpha}}} = 0\) and \(\lim_{t \to + \infty} \frac{1}{ {1 + t^{\alpha}}} = 0\) for the above \(\varepsilon > 0\), there exists a constant \(L > L_{1} > 0\) such that, for any \(t_{1},t_{2} \geq L\) and \(0 \leq s\leq L_{1}\), we have
$$\biggl\vert {\frac{{(t_{1} - s)^{\alpha - 1} }}{ {1 + t_{1} ^{\alpha}}} - \frac{{(t_{2} - s)^{\alpha - 1} }}{ {1 + t_{2} ^{\alpha}}}} \biggr\vert \leq \frac{{t_{1}^{\alpha - 1} }}{ {1 + t_{1} ^{\alpha}}} + \frac{{t_{2}^{\alpha - 1} }}{ {1 + t_{2} ^{\alpha}}} < \varepsilon $$
and
$$\biggl\vert {\frac{1}{ {1 + t_{1} ^{\alpha}}} - \frac{1}{ {1 + t_{2} ^{\alpha}}}} \biggr\vert \leq \frac{1}{ {1 + t_{1} ^{\alpha}}} + \frac{1}{ {1 + t_{2} ^{\alpha}}} < \varepsilon. $$
Then, for \(t_{1},t_{2} \geq L\), from the above we obtain
$$\begin{aligned} & \biggl\vert {\frac{{R(x,\lambda)(t_{1})}}{ {1 + t_{1} ^{\alpha}}} - \frac{{R(x,\lambda)(t_{2})}}{ {1 + t_{2} ^{\alpha}}}} \biggr\vert \\ &\quad = \frac{1}{ {\Gamma(\alpha)}} \biggl\vert { \int_{0}^{t_{1} } {\frac{{(t_{1} - s)^{\alpha - 1} }}{ {1 + t_{1} ^{\alpha}}}h(s)\,ds - \int_{0}^{t_{2} } {\frac{{(t_{2} - s)^{\alpha - 1} }}{ {1 + t_{2} ^{\alpha}}}h(s)\,ds} } } \biggr\vert \\ &\quad \leq\frac{1}{ {\Gamma(\alpha)}} \int_{0}^{L_{1} } { \biggl\vert {\frac{{(t_{1} - s)^{\alpha - 1} }}{ {1 + t_{1} ^{\alpha}}} - \frac{{(t_{2} - s)^{\alpha - 1} }}{ {1 + t_{2} ^{\alpha}}}} \biggr\vert \bigl\vert {h(s)} \bigr\vert \,ds} + \frac{1}{ {\Gamma(\alpha)}} \int_{L_{1} }^{t_{1} } {\frac{{(t_{1} - s)^{\alpha - 1} }}{ {1 + t_{1} ^{\alpha}}} \bigl\vert {h(s)} \bigr\vert \,ds} \\ &\qquad {}+ \frac{1}{ {\Gamma(\alpha)}} \int_{L_{1} }^{t_{2} } {\frac{{(t_{2} - s)^{\alpha - 1} }}{ {1 + t_{2} ^{\alpha}}} \bigl\vert {h(s)} \bigr\vert \,ds} \\ &\quad \leq\frac{m}{ {\Gamma(\alpha)}}L_{1} \varepsilon+ \frac{2m}{ {\Gamma(\alpha + 1)}} \varepsilon. \end{aligned} $$
Similarly, we get
$$\biggl\vert {\frac{{D_{0 + }^{\alpha - 1} R(x,\lambda)(t_{1})}}{ {1 + t_{1} ^{\alpha}}} - \frac{{D_{0 + }^{\alpha - 1} R(x,\lambda)(t_{2})}}{ {1 + t_{2} ^{\alpha}}}} \biggr\vert \leq mL_{1} \varepsilon+ 2\varepsilon $$
and, in addition,
$$\begin{aligned} \bigl\vert {D_{0 + }^{\alpha}R(x,\lambda) (t_{1}) - D_{0 + }^{\alpha}R(x,\lambda) (t_{2})} \bigr\vert =& \bigl\vert {h(t_{1}) - h(t_{2})} \bigr\vert \\ \leq& \bigl\vert h(t_{1}) \bigr\vert + \bigl\vert h(t_{2}) \bigr\vert \\ < & 2\varepsilon. \end{aligned}$$
So, \(R(x,\lambda)(\bar{\Omega})\) is equiconvergent at infinity. By Lemma 3.2, \(R:\bar{\Omega}\times[0,1] \to X_{2}\) is completely continuous.
Finally, we prove that the (\(a_{3}\)) and (\(a_{4}\)) in Definition 2.4 hold. Let \(x \in\sum_{\lambda}= \{ x \in\bar{\Omega}\vert Mx = N_{\lambda}x \}\). Then \((\phi_{p} (D_{0 +}^{\alpha}x(t)))'=N_{\lambda}x(t) \in\operatorname{Im} M = \operatorname{Ker}Q\) and
$$\begin{aligned} R(x,\lambda) (t) =& I_{0 + }^{\alpha}h(t) \\ =& \frac{1}{ {\Gamma(\alpha)}} \int_{0}^{t} (t - s)^{\alpha - 1} \biggl[\phi _{q} \biggl( { \int_{s}^{ + \infty} {(Q - I)N_{\lambda}x(\tau)\,d \tau + \phi_{p} \bigl(D_{0 + }^{\alpha}x( + \infty) \bigr)} } \biggr) \\ &{}-D_{0 + }^{\alpha}x( + \infty) \biggr]\,ds \\ =& \frac{1}{ {\Gamma(\alpha)}} \int_{0}^{t} {(t - s)^{\alpha - 1} \phi _{q} \biggl( { \int_{s}^{ + \infty} { - N_{\lambda}x(\tau)\,d\tau+ \phi_{p} \bigl(D_{0 + }^{\alpha}x( + \infty) \bigr)} } \biggr)\,ds} - (Px) (t) \\ =&\frac{1}{ {\Gamma(\alpha)}} \int_{0}^{t} {(t - s)^{\alpha - 1} \phi _{q} \biggl( { \int_{s}^{ + \infty} { - \bigl(\phi_{p} \bigl(D_{0 +}^{\alpha}x(\tau) \bigr) \bigr)'\,d\tau+ \phi_{p} \bigl(D_{0 + }^{\alpha}x( + \infty) \bigr)} } \biggr)\,ds} \\ &{}- (Px) (t) \\ =& \frac{1}{ {\Gamma(\alpha)}} \int_{0}^{t} {(t - s)^{\alpha - 1} D_{0 +}^{\alpha}x(s)\,ds} - (Px) (t), \end{aligned}$$
which, combined with boundary conditions, yields that
$$R(x,\lambda) (t)=x(t) - (Px) (t) = \bigl[(I - P)x \bigr](t). $$
It is clear that \(R(x,0)(t)\) is a zero operator, and for any \(x \in\bar{\Omega}\), we have
$$\begin{aligned}& M \bigl[Px + R(x,\lambda) \bigr](t) \\& \quad = M \biggl[ {\frac{1}{ {\Gamma(\alpha)}} \int_{0}^{t} {(t - s)^{\alpha - 1} \phi _{q} \biggl( { \int_{s}^{ + \infty} {(Q - I)N_{\lambda}x(\tau)\,d \tau + \phi_{p} \bigl(D_{0 + }^{\alpha}x( + \infty) \bigr)} } \biggr)\,ds} } \biggr] \\& \quad = \biggl[ { \int_{t}^{ + \infty} {{(Q - I)N_{\lambda}x(s)\,ds + \phi _{p} \bigl(D_{0 + }^{\alpha}x( + \infty) \bigr)}} } \biggr]^{\prime}\\& \quad = (I - Q)N_{\lambda}x(t). \end{aligned}$$
By the above, \(N_{\lambda}\) is M-compact on Ω̄. □
Theorem 3.1
Suppose that (\(A_{1}\)) and the following conditions hold:
- (\(A_{2}\)):
-
there exist nonnegative functions
\(a(t),b(t),c(t),d(t) \in Y\)
such that
$$\begin{aligned} \bigl\vert {f(t,u,v,w)} \bigr\vert \leq& a(t) + b(t)\frac{\vert u \vert ^{p - 1}}{{(1+t^{\alpha})^{p - 1}}} + c(t) \frac{\vert v \vert ^{p - 1}}{{(1+t^{\alpha})^{p - 1}}} \\ &{} + d(t)\vert w \vert ^{p - 1} ,\quad \forall t \in[ {0, + \infty} ),(u,v,w) \in\mathbb{R}^{3}; \end{aligned}$$
- (\(A_{3}\)):
-
there exists a positive constant
B
such that one of the following inequalities hold:
$$\begin{aligned} &{ wf(t,u,v,w) > 0,\quad \forall t \in[ {0, + \infty} ),u,v \in \mathbb{R},\vert w \vert > {B},} \end{aligned}$$
(3.5)
$$\begin{aligned} &{ wf(t,u,v,w) < 0,\quad \forall t \in[ {0, + \infty} ),u,v \in \mathbb{R},\vert w \vert > {B}.} \end{aligned}$$
(3.6)
Then BVP (1.1) has at least one solution in
X, provided that
\(\Vert {b} \Vert _{1}+\Vert {c} \Vert _{1}+ \Vert {d} \Vert _{1} < 1\).
Before we prove Theorem 3.1, we show two lemmas.
Lemma 3.4
Let
\(\Omega_{1} = \{ {x \in\operatorname{dom}M\setminus\operatorname{Ker}M \vert Mx = N_{\lambda}x,\lambda \in(0,1)} \}\). Suppose that (\(A_{2}\)) and (\(A_{3}\)) hold. Then
\(\Omega_{1}\)
is bounded in
X.
Proof
Let \(x \in\Omega_{1}\). Then \(Mx = N_{\lambda}x\) and thus \(QN_{\lambda}x = 0\), that is,
$$\sum_{i = 1}^{n} {\alpha_{i} } \int_{\xi_{i} }^{ + \infty} {f \bigl(s,x(s),{D_{0 + }^{\alpha-1}} x(s),D_{0 + }^{\alpha}x(s) \bigr)\,ds = 0}. $$
By Lemma 2.1 and the boundary conditions we have
$$x(t) = c_{1} t^{\alpha - 1} + c_{2} t^{\alpha - 2} + I_{0 + }^{\alpha}D_{0 + }^{\alpha}x(t) = I_{0 + }^{\alpha}D_{0 + }^{\alpha}x(t) = \frac{1}{ {\Gamma(\alpha)}} \int_{0}^{t} {(t - s)^{\alpha - 1} D_{0 + }^{\alpha}x(s)\,ds}. $$
Thus,
$$\Vert x \Vert _{0} \leq \bigl\Vert {D_{0 + }^{\alpha}x} \bigr\Vert _{\infty},\qquad \bigl\Vert {D_{0 + }^{\alpha - 1} x} \bigr\Vert _{0} \leq \bigl\Vert {D_{0 + }^{\alpha}x} \bigr\Vert _{\infty}. $$
By (\(A_{3}\)) there exists a constant \(s_{0} \in[0, + \infty)\) such that \(\vert {D_{0 + }^{\alpha}x(s_{0} )} \vert \leqslant{B}\), which, combined with \(Mx = N_{\lambda}x\) and (\(A_{2}\)), gives
$$\begin{aligned}& \bigl\vert {\phi_{p} \bigl(D_{0 +}^{\alpha}x(t) \bigr)} \bigr\vert \\& \quad = \biggl\vert {\phi_{p} \bigl(D_{0 +}^{\alpha}x(s_{0} ) \bigr) - \int_{s_{0} }^{t} {\lambda f \bigl(s,x(s),D_{0 +}^{\alpha-1} x(s),D_{0 +}^{\alpha}x(s) \bigr)\,ds} } \biggr\vert \\& \quad \leq \bigl\vert {\phi_{p} \bigl(D_{0 +}^{\alpha}x(s_{0} ) \bigr)} \bigr\vert + \biggl\vert { \int_{s_{0} }^{t} {\lambda f \bigl(s,x(s),D_{0 +}^{\alpha-1} x(s),D_{0 +}^{\alpha}x(s) \bigr)\,ds} } \biggr\vert \\& \quad \leq \phi_{p} (B) + \int_{0}^{ + \infty} { \bigl\vert {f \bigl(s,x(s),D_{0 +}^{\alpha-1} x(s),D_{0 +}^{\alpha}x(s) \bigr)} \bigr\vert \,ds } \\& \quad \leq \phi_{p} (B) + \int_{0}^{ + \infty} { \biggl[ {a(s) + b(s) \frac{\vert x \vert ^{p - 1}}{{(1+s^{\alpha})^{p - 1}}}+c(s)\frac {\vert {D_{0 +}^{\alpha-1} x}\vert ^{p - 1}}{{(1+s^{\alpha})^{p - 1}}} + d(s) \bigl\vert {D_{0 +}^{\alpha}x} \bigr\vert ^{p - 1} } \biggr]\,ds} \\& \quad \leq \phi_{p} (B) + \Vert a \Vert _{1} + \Vert {b} \Vert _{1} \phi_{p} \bigl( {\Vert x\Vert _{0} } \bigr) +\Vert {c} \Vert _{1} \phi_{p} \bigl( { \bigl\Vert {D_{0 +}^{\alpha-1} x} \bigr\Vert _{0} } \bigr)+ \Vert d \Vert _{1} \phi_{p} \bigl( { \bigl\Vert {D_{0 +}^{\alpha}x} \bigr\Vert _{\infty}} \bigr) \\& \quad \leq\phi_{p} (B) + \Vert a \Vert _{1} + \bigl[ { \Vert {b} \Vert _{1} +\Vert {c} \Vert _{1} + \Vert d \Vert _{1} } \bigr]\phi_{p} \bigl( { \bigl\Vert {D_{0 +}^{\alpha}x} \bigr\Vert _{\infty}} \bigr). \end{aligned}$$
Then
$$\begin{aligned} \bigl\Vert {\phi_{p} \bigl(D_{0 +}^{\alpha}x \bigr)} \bigr\Vert _{\infty} =& \phi_{p} \bigl( { \bigl\Vert {D_{0 +}^{\alpha}x} \bigr\Vert _{\infty}} \bigr) \\ \leq& \phi_{p} (B) + \Vert a \Vert _{1} + \bigl[ { \Vert {b} \Vert _{1} +\Vert {c} \Vert _{1} + \Vert d \Vert _{1} } \bigr]\phi_{p} \bigl( { \bigl\Vert {D_{0 +}^{\alpha}x} \bigr\Vert _{\infty}} \bigr). \end{aligned}$$
Thus,
$$\phi_{p} \bigl( { \bigl\Vert {D_{0 +}^{\alpha}x} \bigr\Vert _{\infty}} \bigr) \leq\frac{{\phi_{p} (B) + \Vert a \Vert _{1} }}{ {1 - [ {\Vert {b} \Vert _{1} + \Vert {c} \Vert _{1} + \Vert d \Vert _{1} } ]}}: = A. $$
That is,
$$\bigl\Vert {D_{0 +}^{\alpha}x} \bigr\Vert _{\infty}\leq \phi_{q} ( A ). $$
Therefore,
$$\Vert x \Vert _{X} = \max \bigl\{ {\Vert x \Vert _{0} , \bigl\Vert {D_{0 + }^{\alpha - 1} x} \bigr\Vert _{0} , \bigl\Vert {D_{0 + }^{\alpha}x} \bigr\Vert _{\infty}} \bigr\} \leq\phi_{q} ( A ). $$
So, \(\Omega_{1}\) is bounded in X. □
Lemma 3.5
Let
\(\Omega_{2} = \{ {x \in\operatorname{Ker}M\vert Nx \in\operatorname{Im} M} \}\). Suppose that (\(A_{3}\)) holds. Then
\(\Omega_{2}\)
is bounded in
X.
Proof
Let \(x \in\Omega_{2}\), that is, \(x = ct^{\alpha}\), \(c \in\mathbb{R}\), \(QNx = 0\), so that
$$\sum_{i = 1}^{n} {\alpha_{i} \int_{\xi_{i} }^{ + \infty} {f \bigl(t,ct^{\alpha},c\Gamma( \alpha+1)t,c\Gamma(\alpha) \bigr)}\,dt}= 0. $$
By (\(A_{3}\)) we have \(\vert c \Gamma(\alpha) \vert \leq{B}\), that is, \(\vert c \vert \leq \frac{B}{ {\Gamma(\alpha)}}\). Therefore,
$$\begin{aligned} \Vert x \Vert _{X} =& \max \bigl\{ \Vert x \Vert _{0} , \bigl\Vert {D_{0 + }^{\alpha - 1} x} \bigr\Vert _{0} , \bigl\Vert {D_{0 + }^{\alpha}x} \bigr\Vert _{\infty}\bigr\} \\ =& \max \biggl\{ \biggl\Vert \frac{ct^{\alpha}}{ 1 + t^{\alpha}} \biggr\Vert _{\infty}, \biggl\Vert \frac{c\Gamma(\alpha + 1)t}{ 1 + t^{\alpha}} \biggr\Vert _{\infty}, \bigl\vert c \Gamma(\alpha) \bigr\vert \biggr\} \\ \leq& \biggl\Vert {\frac{{ct^{\alpha}}}{ {1 + t^{\alpha}}}} \biggr\Vert _{\infty}+ \biggl\Vert {\frac{{c\Gamma(\alpha + 1)t}}{ {1 + t^{\alpha}}}} \biggr\Vert _{\infty}+ \bigl\vert { c \Gamma( \alpha)} \bigr\vert \leq \bigl( {1 + \Gamma (\alpha) + \Gamma(\alpha+ 1)} \bigr) \vert c \vert : = C. \end{aligned}$$
So, \(\Omega_{2}\) is bounded in X. □
Proof of Theorem 3.1
Set \(\Omega = \{ {x \in X\vert {\Vert x \Vert } _{X} < \max \{ {B,\phi_{q} ( A ),C} \} + 1} \}\). By Lemma 3.1 and Lemma 3.3 we know that M is quasi-linear and \(N_{\lambda}\) is M-compact on Ω̄. From Lemma 3.4 and Lemma 3.5 we obtain:
- (\(B_{1}\)):
-
\(Mx \ne N_{\lambda}x\), \(\forall(x,\lambda) \in\partial\Omega \times(0,1)\),
- (\(B_{2}\)):
-
\(QNx \ne0\), \(\forall x \in\operatorname{Ker}M \cap\partial\Omega\).
Now we show (\(B_{3}\)) holds. Let \(J:\operatorname{Im} Q \to\operatorname{Ker}M\) be the homeomorphism defined by
$$J \bigl(ce^{ - t} \bigr) = ct^{\alpha},\quad c\in\mathbb{R}, t \in[0,+ \infty). $$
Without loss of generality, we suppose that (3.6) holds. Define the homotopic mapping
$$H(x,\lambda) = \lambda x - (1 - \lambda)JQNx,\quad \forall x \in\bar{\Omega}\cap \operatorname{Ker}M, \lambda\in[0,1]. $$
Then \(H(x,\lambda) \ne0\), \(x \in\partial\Omega \cap\operatorname{Ker}M\), \(\lambda \in[0,1]\). Indeed, for \(x \in\partial\Omega \cap \operatorname{Ker}M\), we have \(x = ct^{\alpha}\) and thus
$$H(x,\lambda) = \lambda ct^{\alpha}- (1 - \lambda)\frac{{\sum_{i = 1}^{n} {\alpha_{i} \int_{\xi_{i} }^{ + \infty} {f(s,cs^{\alpha},c\Gamma(\alpha +1)s,c\Gamma(\alpha))\,ds} } }}{ {\sum_{i = 1}^{n} {\alpha_{i} e^{ - \xi_{i} } } }}t^{\alpha}. $$
Clearly, \(H(x,1) \ne0\), \(x \in\partial\Omega \cap\operatorname{Ker}M\). For \(\lambda \in[0,1)\) and \(x = ct^{\alpha}\in\partial\Omega \cap \operatorname{Ker}M\), if \(H(x,\lambda) = 0\), then
$$\frac{{\sum_{i = 1}^{n} {\alpha_{i} \int_{\xi_{i} }^{ + \infty} {c\Gamma (\alpha)f(s,cs^{\alpha},c\Gamma(\alpha+1)s,c\Gamma(\alpha))\,ds} } }}{ {\sum_{i = 1}^{n} {\alpha_{i} e^{ - \xi_{i} } } }} = \frac{\lambda}{ {1 - \lambda}}c^{2} \Gamma(\alpha) \geq0, $$
which contradicts (3.6). If (3.5) holds, then defining the homotopic mapping
$$H(x,\lambda) = \lambda x + (1 - \lambda)JQNx,\quad x \in\bar{\Omega}\cap \operatorname{Ker}M, \lambda\in[0,1], $$
we also get contradiction in a similar way. Therefore, via the homotopy property of degree, we obtain
$$\begin{aligned} \deg(JQN,\Omega\cap\operatorname{Ker}M,0) =& \deg \bigl( {H( \cdot,0),\Omega \cap \operatorname{Ker}M,0} \bigr) \\ =& \deg \bigl( {H( \cdot,1),\Omega\cap\operatorname{Ker}M,0} \bigr) \\ =& \deg( {I,\Omega\cap\operatorname{Ker}M,0} ) \ne0. \end{aligned}$$
Applying Lemma 2.2, we conclude that (1.1) has at least one solution in Ω̄. □