Based on the lemmas mentioned in the previous section, we define an operator \(S:\mathcal{C}\rightarrow\mathcal{C}\) as follows:
$$ \begin{aligned}[b] (Sx) (t)&= \int_{0}^{1}G(t,s)f\bigl(s,x(s)\bigr)\,ds+ \frac{t^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \beta_{i} \int_{0}^{1}H(\eta_{i},s)f\bigl(s,x(s) \bigr)\,ds \\ &\quad{}+\frac{t^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \gamma_{i} \int _{0}^{1}G(\eta_{i},s)f\bigl(s,x(s) \bigr)\,ds, \end{aligned} $$
(3.1)
where \(\mathcal{C}=C([0,1],\textbf{R})\) denotes the Banach space of all continuous functions defined on \([0,1]\) that are mapped into R with the norm defined as \(\Vert x \Vert = \sup_{t\in[0,1]} \vert x(t) \vert \). And G and H are provided in Lemma 2.6.
Solutions to the problem exist if and only if the operator S has fixed points. In order to make the analysis clear, we introduce some fixed point theorems that play a role in our proof.
Lemma 3.1
(Krasnoselskii [19])
Let
Q
be a closed, convex, bounded and nonempty subset of a Banach space
Y. Let
\(\psi_{1}\), \(\psi_{2}\)
be operators such that
-
(i)
\(\psi_{1}v_{1}+\psi_{2}v_{2}\in Q\)
whenever
\(v_{1}, v_{2}\in Q\);
-
(ii)
\(\psi_{1}\)
is compact and continuous;
-
(iii)
\(\psi_{2}\)
is a contraction mapping.
Then there exists
\(v\in Q\)
such that
\(v=\psi_{1}v_{1}+\psi_{2}v_{2}\).
Lemma 3.2
([19])
Let
X
be a Banach space. Assume that
\(T: X\rightarrow X\)
is a completely continuous operator and the set
\(V=\{u\in X: u=\epsilon Tu, 0<\epsilon<1\}\)
is bounded. Then
T
has a fixed point in
X.
Lemma 3.3
([20])
Let
E
be a Banach space, \(E_{1}\)
be a closed, convex subset of
E, V
be an open subset of
\(E_{1}\)
and
\(0\in V\). Suppose that
\(U:\overline{V} \rightarrow E_{1}\)
is a continuous, compact (that is, \(U(\overline{V})\)
is a relatively compact subset of
\(E_{1}\)) map. Then either
-
(i)
U
has a fixed point in
V̅, or
-
(ii)
there are
\(x\in\partial V\) (the boundary of
V
in
\(E_{1}\)) and
\(k\in(0,1)\)
with
\(x=kU(x)\).
For convenience, we set some notations:
$$ \begin{aligned} &\alpha_{1}= \frac{1}{\Gamma(\sigma)}+\frac{1}{\xi}\sum_{i=1}^{m-2} \beta _{i}\frac{(\sigma+1)\eta_{i}^{\sigma}-\sigma\eta_{i}^{\sigma+1}}{\sigma (\sigma+1)\Gamma(\sigma+1)}+\frac{1}{\xi}\sum _{i=1}^{m-2}\gamma_{i}\frac {\eta_{i}^{\sigma-1}-\eta_{i}^{\sigma}}{\Gamma(\sigma+1)}, \\ &\alpha_{2}=\alpha_{1}-\frac{1}{\Gamma(\sigma)}. \end{aligned} $$
(3.2)
Theorem 3.4
Let
\(f:[0,1]\times R\rightarrow R\)
be a continuous function that satisfies the conditions:
- (\(H_{1}\)):
-
\(\vert f(t,x)-f(t,y) \vert \leq l \vert x-y \vert \)
for all
\(t\in[0,1]\)
and
\(x,y\in\textbf{R}\);
- (\(H_{2}\)):
-
\(\vert f(t,x) \vert \leq\omega(t)\)
for all
\((t,x)\in[0,1]\times\textbf{R}\)
and
\(\omega\in C([0,1],\textbf {R}^{+})\),
then BVP (1.3) has at least one solution on
\([0,1]\)
when
\(l\alpha_{2}\leq1\)
with
\(\alpha_{2}\)
given in (3.2).
Proof
Define a set \(B_{\rho}=\{x\in C: \Vert x \Vert \leq\rho\}\), where \(\rho\geq \Vert \omega \Vert \alpha_{1}\) with \(\alpha_{1}\) defined in (3.2) and \(\Vert \omega \Vert =\sup_{t\in[0,1]} \vert \omega (t) \vert \). Let the operators \(S_{1}\) and \(S_{2}\) on \(B_{\rho}\) be defined as
$$\begin{aligned} &(S_{1}x) (t)= \int_{0}^{1}G(t,s)f\bigl(s,x(s)\bigr)\,ds, \\ &(S_{2}x) (t)=\frac{t^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \beta_{i} \int _{0}^{1}H(\eta_{i},s)f\bigl(s,x(s) \bigr)\,ds+\frac{t^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \gamma_{i} \int_{0}^{1}G(\eta_{i},s)f\bigl(s,x(s) \bigr)\,ds. \end{aligned}$$
It is easy to understand \(\Vert S_{1}x+S_{2}y \Vert \leq \Vert \omega \Vert \alpha_{1}\) for any \(x,y\in B_{\rho}\) that means \(S_{1}x+S_{2}y\in B_{\rho}\).
By assumption(\(H_{1}\)),
$$\begin{aligned} \Vert S_{2}x-S_{2}y \Vert &=\sup_{t\in[0,1]} \vert S_{2}x-S_{2}y \vert \\ &\leq \sup_{t\in[0,1]}\frac{t^{\sigma-1}}{\xi}\sum _{i=1}^{m-2}\beta _{i} \int_{0}^{1}H(\eta_{i},s) \bigl\vert f(s,x)-f(s,y) \bigr\vert \,ds \\ &\quad{}+\sup_{t\in[0,1]} \frac{t^{\sigma-1}}{\xi}\sum _{i=1}^{m-2}\gamma_{i} \int_{0}^{1}G(\eta_{i},s) \bigl\vert f(s,x)-f(s,y)\bigr\vert \,ds \\ &\leq \frac{l}{\xi} \Biggl\{ \sum_{i=1}^{m-2} \beta_{i}\frac{(\sigma +1)\eta_{i}^{\sigma}-\sigma\eta_{i}^{\sigma+1}}{\sigma(\sigma+1)\Gamma (\sigma+1)}+\sum_{i=1}^{m-2} \gamma_{i}\frac{\eta_{i}^{\sigma-1}-\eta _{i}^{\sigma}}{\Gamma(\sigma+1)} \Biggr\} \Vert x-y \Vert \\ &\leq l\alpha_{2} \Vert x-y \Vert . \end{aligned}$$
The operator \(S_{2}\) is a contraction because of \(l\alpha_{2}<1\). As we all know, \(S_{1}\) is a continuous result from the continuity of f. Moreover,
$$\Vert S_{1}x \Vert =\sup_{t\in[0,1]} \biggl\{ \int _{0}^{1}G(t,s)f\bigl(s,x(s)\bigr)\,ds \biggr\} \leq\sup_{t\in[0,1]} \biggl\{ \int _{0}^{1}\frac{ \vert f(s,x(s)) \vert }{\Gamma(\sigma)}\,ds \biggr\} \leq \frac{ \Vert \omega \Vert }{\Gamma(\sigma)}, $$
which implies that \(S_{1}\) is uniformly bounded on \(B_{\rho}\). Apart from that, the following inequalities hold with \(\sup_{(t,x)\in[0,1]\times B_{\rho}} \vert f(t,x) \vert =f_{m}<+\infty\) and \(0< t_{1}< t_{2}<1\):
$$\begin{aligned} \bigl\vert (S_{1}x) (t_{2})-(S_{1}x) (t_{1}) \bigr\vert &= \biggl\vert \int _{0}^{1}G(t_{2},s)f\bigl(s,x(s) \bigr)\,ds- \int_{0}^{1}G(t_{1},s)f\bigl(s,x(s) \bigr)\,ds \biggr\vert \\ &\leq \int_{0}^{1}\bigl[G(t_{2},s)-G(t_{1},s) \bigr] \bigl\vert f\bigl(s,x(s)\bigr) \bigr\vert \,ds \\ &\leq\frac{f_{m}}{\Gamma(\sigma+1)}\bigl[\bigl(t_{2}^{\sigma-1}-t_{1}^{\sigma -1} \bigr)+\bigl(t_{2}^{\sigma}-t_{1}^{\sigma}\bigr) \bigr]. \end{aligned}$$
The operator \(S_{1}\) is compact due to the Arzela-Ascoli theorem. Since three conditions are satisfied, BVP (1.3) has at least one solution on \([0,1]\) by application of Krasnoselskii’s fixed point theorem. □
Theorem 3.5
Assume that there exists a constant
L
such that
\(\vert f(t,x) \vert \leq L\)
for any
\(t\in[0,1]\)
and
\(x\in C[0,1]\). Then there exists at least one solution to BVP (1.3).
Proof
Firstly, we set out to verify that the operator S given in (3.1) is completely continuous. Define a bounded set \(U\subset C([0,1],\textbf{R}^{+})\), then \(\vert (Sx)(t) \vert \leq L\alpha_{1}\) holds when we take \(x\in U\). On the top of that,
$$\begin{aligned} \bigl\vert (Sx) (t_{2})-(Sx) (t_{1}) \bigr\vert &= \Biggl\vert \int _{0}^{1}\bigl[G(t_{2},s)-G(t_{1},s) \bigr]f\bigl(s,x(s)\bigr)\,ds \\ &\quad{}+\frac{t_{2}^{\sigma-1}-t_{1}^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \beta _{i} \int_{0}^{1}H(\eta_{i},s)f\bigl(s,x(s) \bigr)\,ds \\ &\quad{}+\frac{t_{2}^{\sigma-1}-t_{1}^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \gamma_{i} \int_{0}^{1}G(\eta_{i},s)f\bigl(s,x(s) \bigr)\,ds \Biggr\vert \\ &\leq L \biggl\{ \frac{(t_{2}^{\sigma-1}-t_{1}^{\sigma-1})+(t_{2}^{\sigma }-t_{1}^{\sigma})}{\Gamma(\sigma+1)}+\alpha_{2}\bigl(t_{2}^{\sigma-1}-t_{1}^{\sigma -1} \bigr) \biggr\} . \end{aligned}$$
Hence, S is equicontinuous on \([0,1]\) in view of \(t^{\sigma}\) and \(t^{\sigma-1}\) is equicontinuous on \([0,1]\). The operator S is deduced to be completely continuous by the Arzela-Ascoli theorem along with the continuity of S decided by f.
Secondly, we consider the set \(V=\{x\in C:x=\lambda Sx,0<\lambda<1\}\) and prove that V is bounded. In fact, for each \(x\in V\) and \(t\in[0,1]\),
$$\begin{aligned} \bigl\Vert x(t) \bigr\Vert &= \bigl\Vert \lambda(Sx) (t) \bigr\Vert \\ &\leq L \Biggl\{ \frac{1}{\Gamma(\sigma)}+\frac{1}{\xi}\sum _{i=1}^{m-2}\beta _{i}\frac{(\sigma+1)\eta_{i}^{\sigma}-\sigma\eta_{i}^{\sigma+1}}{\sigma (\sigma+1)\Gamma(\sigma+1)}+ \frac{1}{\xi}\sum_{i=1}^{m-2} \gamma_{i}\frac {\eta_{i}^{\sigma-1}-\eta_{i}^{\sigma}}{\Gamma(\sigma+1)} \Biggr\} \\ &=L\alpha_{1}. \end{aligned}$$
Consequently, the set V is bounded by definition.
Finally, we conclude that BVP (1.3) has at least one solution according to Lemma 3.5 and the proof is completed. □
Theorem 3.6
Assume that
\(f:[0,1]\times\textbf{R}\rightarrow\textbf{R}\)
is a continuous function and satisfies condition(\(H_{1}\)) with
\(l\alpha_{1}<1\), where
\(\alpha_{1}\)
is defined in (3.2). Then the BVP has a unique solution on
\([0,1]\).
Proof
Let \(P_{r}=\{x\in C: \Vert x \Vert \leq r\}\) be a bounded set. To show \(\mathcal{S}P_{r}\subset P_{r}\) with the operator S defined in (3.1), \(\vert f(s,x(s)) \vert \leq lr+\delta\) holds when we take \(x\in P_{r}\) for \(t\in[0,1]\) and with the condition provided by \(\sup_{t\in[0,1]} \vert f(t,0) \vert =\delta\) and \(r\geq\frac{\delta\alpha_{1}}{1-l\alpha_{1}}\). Beyond that,
$$\begin{aligned} \bigl\Vert (Sx) \bigr\Vert &\leq \sup_{t\in[0,1]} \Biggl\{ \int_{0}^{1}G(t,s) \bigl\vert f\bigl(s,x(s)\bigr) \bigr\vert \,ds+\frac{t^{\sigma -1}}{\xi}\sum_{i=1}^{m-2} \beta_{i} \int_{0}^{1}H(\eta_{i},s) \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert \,ds \\ &\quad{}+\frac{t^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \gamma_{i} \int _{0}^{1}G(\eta_{i},s) \bigl\vert f\bigl(s,x(s)\bigr) \bigr\vert \,ds \Biggr\} \\ &\leq(lr+\delta)\alpha_{1}\leq r. \end{aligned} $$
\(SP_{r}\subset P_{r}\) is strictly proved. Next, choosing \(x,y\in C\) with \(t\in[0,1]\), we get
$$\begin{aligned} \Vert Sx-Sy \Vert &\leq \sup_{t\in[0,1]} \Biggl\{ \int _{0}^{1}G(t,s)\bigl\vert f\bigl(s,x(s)\bigr)-f \bigl(s,y(s)\bigr)\bigr\vert \,ds \\ &\quad{}+\frac{t^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \beta_{i} \int _{0}^{1}H(\eta_{i},s)\bigl\vert f \bigl(s,x(s)\bigr)-f\bigl(s,y(s)\bigr)\bigr\vert \,ds \\ &\quad{}+\frac{t^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \gamma_{i} \int _{0}^{1}G(\eta_{i},s)\bigl\vert f \bigl(s,x(s)\bigr)-f\bigl(s,y(s)\bigr)\bigr\vert \,ds \Biggr\} \\ &\leq l\alpha_{1} \Vert x-y \Vert . \end{aligned}$$
The operator S is a contraction with the assumption \(l\alpha_{1}<1\). Therefore, BVP (1.3) has a unique solution by Banach’s contraction mapping principle. □
Theorem 3.7
Let
\(f:[0,1]\times R\rightarrow R\)
be a continuous function, and assume that
- (\(H_{3}\)):
-
there exist a function
\(p\in C([0,1],\textbf{R}^{+})\)
and a nondecreasing function
\(q:\textbf{R}^{+}\rightarrow\textbf{R}^{+}\)
such that
$$\begin{aligned} \bigl\vert f(t,x) \bigr\vert \leq p(t)q\bigl( \Vert x \Vert \bigr)\quad \textit{for all } (t,x)\in[0,1]\times \textbf{R}; \end{aligned}$$
- (\(H_{4}\)):
-
there exists a constant
\(N>0\)
such that
$$\begin{aligned} N \Biggl\{ q(v)\Vert p\Vert \Biggl\{ \frac{1}{\Gamma(\sigma)}+\frac{1}{\xi}\sum _{i=1}^{m-2}\beta_{i} \frac{(\sigma+1)\eta_{i}^{\sigma}-\sigma\eta _{i}^{\sigma+1}}{\sigma(\sigma+1)\Gamma(\sigma+1)}+\frac{1}{\xi}\sum_{i=1}^{m-2} \gamma_{i}\frac{\eta_{i}^{\sigma-1}-\eta_{i}^{\sigma }}{\Gamma(\sigma+1)} \Biggr\} \Biggr\} ^{-1}\leq 1. \end{aligned}$$
Then BVP (1.3) has at least one solution on
\([0,1]\).
Proof
The first step is to show that the operator S given in (3.1) maps bounded sets into bounded sets in C. Let v be a positive number and \(B_{v}=\{x\in C: \Vert x \Vert \leq v\}\) be a bounded set in C. For each \(x\in B_{v}\) and by(\(H_{3}\)), the following equalities are obtained:
$$\begin{aligned} \bigl\vert (Sx) (t) \bigr\vert &\leq \int_{0}^{1}G(t,s)p(s)q\bigl( \Vert x \Vert \bigr)\,ds +\frac{t^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \beta_{i} \int_{0}^{1}H(\eta _{i},s)p(s)q\bigl( \Vert x \Vert \bigr)\,ds \\ &\quad{}+\frac{t^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \gamma_{i} \int _{0}^{1}G(\eta_{i},s)p(s)q\bigl( \Vert x \Vert \bigr)\,ds \\ &\leq q(v) \Vert p \Vert \Biggl\{ \frac{1}{\Gamma(\sigma)}+\frac {1}{\xi}\sum _{i=1}^{m-2}\beta_{i} \frac{(\sigma+1)\eta_{i}^{\sigma }-\sigma\eta_{i}^{\sigma+1}}{\sigma(\sigma+1)\Gamma(\sigma+1)}+\frac {1}{\xi}\sum_{i=1}^{m-2} \gamma_{i}\frac{\eta_{i}^{\sigma-1}-\eta _{i}^{\sigma}}{\Gamma(\sigma+1)} \Biggr\} \\ &=q(v) \Vert p \Vert \alpha_{1}. \end{aligned}$$
The next step is to verify that the operator S maps bounded sets into equicontinuous sets of C.
Choose \(t_{1}, t_{2}\in[0,1]\) with \(t_{1}< t_{2}\) and take \(x\in B_{v}\),
$$\begin{aligned} \bigl\vert (Sx)t_{2}-(Sx)t_{1} \bigr\vert &= \Biggl\vert \int _{0}^{1}\bigl[G(t_{2},s)-G(t_{1},s) \bigr]f\bigl(s,x(s)\bigr)\,ds \\ &\quad{}+\frac{t_{2}^{\sigma-1}-t_{1}^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \beta _{i} \int_{0}^{1}H(\eta_{i},s)f\bigl(s,x(s) \bigr)\,ds \\ &\quad{}+\frac{t_{2}^{\sigma-1}-t_{1}^{\sigma-1}}{\xi}\sum_{i=1}^{m-2} \gamma_{i} \int_{0}^{1}G(\eta_{i},s)f\bigl(s,x(s) \bigr)\,ds \Biggr\vert \\ &\leq q(v)\Vert p\Vert \biggl\{ \frac{(t_{2}^{\sigma-1}-t_{1}^{\sigma-1})+(t_{2}^{\sigma }-t_{1}^{\sigma})}{\Gamma(\sigma+1)}+\alpha_{2} \bigl(t_{2}^{\sigma-1}-t_{1}^{\sigma -1}\bigr) \biggr\} . \end{aligned} $$
By the Arzela-Ascoli theorem, the operator S is completely continuous since the right-hand side tends to zero independent of \(x\in B_{v}\) as \(t_{2}\rightarrow t_{1}\). Let x be a solution of BVP (1.3), then for \(\mu\in(0,1)\) and by the same method applied to show the boundedness of S, the results hold:
$$\begin{aligned} \bigl\vert x(t) \bigr\vert &= \bigl\vert \mu(Sx) (t) \bigr\vert \\ &\leq q\bigl( \Vert x \Vert \bigr)\Vert p\Vert \Biggl\{ \frac{1}{\Gamma(\sigma)}+ \frac{1}{\xi}\sum_{i=1}^{m-2} \beta_{i}\frac{(\sigma+1)\eta_{i}^{\sigma}-\sigma\eta _{i}^{\sigma+1}}{\sigma(\sigma+1)\Gamma(\sigma)}+\frac{1}{\xi}\sum _{i=1}^{m-2}\gamma_{i}\frac{\eta_{i}^{\sigma-1}-\eta_{i}^{\sigma }}{\Gamma(\sigma+1)} \Biggr\} , \end{aligned}$$
i.e.,
$$\begin{aligned} & \Vert x \Vert \Biggl\{ q(v)\Vert p\Vert \Biggl\{ \frac{1}{\Gamma(\sigma)}+ \frac{1}{\xi}\sum_{i=1}^{m-2} \beta_{i}\frac{(\sigma+1)\eta_{i}^{\sigma}-\sigma\eta _{i}^{\sigma+1}}{\sigma(\sigma+1)\Gamma(\sigma+1)}+\frac{1}{\xi}\sum _{i=1}^{m-2}\gamma_{i}\frac{\eta_{i}^{\sigma-1}-\eta_{i}^{\sigma }}{\Gamma(\sigma+1)} \Biggr\} \Biggr\} ^{-1}\leq 1. \end{aligned}$$
The final step is to select a set \(Q=\{x\in C: \Vert x \Vert < N+1\}\) in order to take(\(H_{4}\)) into consideration where there exists N such that \(\Vert x \Vert \neq N\). Even though \(S:\overline {Q}\rightarrow C\) is completely continuous can be verified, there is no \(x\in\partial N\) that can satisfy \(x=\mu S(x)\) for \(\mu\in(0,1)\) decided by the selection of N.
Above all, we complete the proof that the operator S has a fixed point in Q̅ which is a solution to BVP (1.3). □