Let

$\mathrm{\Omega}=C[0,1]$,

$u\in \mathrm{\Omega}$ equipped the norm

$\parallel u\parallel =\underset{0\le t\le 1}{sup}|u(t)|,$

(3.1)

then Ω is a Banach space.

We first give some lemmas as follows:

**Lemma 3.1** *Problem* (1.1)-(1.2) *is equivalent to the following integral equation*

$u(t)={\int}_{0}^{1}G(t,s)f(s,u(s))\phantom{\rule{0.2em}{0ex}}ds+u(1){t}^{\alpha -1},$

(3.2)

*where*
$G(t,s)=\{\begin{array}{c}\frac{{t}^{\alpha -1}{(1-s)}^{\alpha -1}-{t}^{\alpha -1}{(\eta -s)}^{\alpha -1}-(1-{\eta}^{\alpha -1}){(t-s)}^{\alpha -1}}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})},\hfill \\ \phantom{\rule{1em}{0ex}}0\le s\le min\{t,\eta \}\le 1;\hfill \\ \frac{{t}^{\alpha -1}{(1-s)}^{\alpha -1}-{t}^{\alpha -1}{(\eta -s)}^{\alpha -1}}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})},\phantom{\rule{1em}{0ex}}0\le t\le s\le \eta \le 1;\hfill \\ \frac{{t}^{\alpha -1}{(1-s)}^{\alpha -1}-(1-{\eta}^{\alpha -1}){(t-s)}^{\alpha -1}}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})},\phantom{\rule{1em}{0ex}}0\le \eta \le s\le t\le 1;\hfill \\ \frac{{t}^{\alpha -1}{(1-s)}^{\alpha -1}}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})},\phantom{\rule{1em}{0ex}}0\le max\{t,\eta \}\le s\le 1.\hfill \end{array}$

(3.3)

*Proof* The sufficiency is obvious, we only need to prove the necessity.

Suppose that

$u(t)$ is a solution of the problem (1.1)-(1.2). Integrating both sides of (1.1) of

*α* order from 0 to

*t* with respect to

*t*, it follows that

$u(t)=-\frac{1}{\mathrm{\Gamma}(\alpha )}{\int}_{0}^{t}{(t-s)}^{\alpha -1}f(s,u(s))\phantom{\rule{0.2em}{0ex}}ds+{c}_{1}{t}^{\alpha -1}+{c}_{2}{t}^{\alpha -2}.$

(3.4)

According to (1.2) and (3.4), we have

Combining (3.4) and (3.5), we obtain

$\begin{array}{rcl}u(t)& =& -\frac{1}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})}{\int}_{0}^{t}(1-{\eta}^{\alpha -1}){(t-s)}^{\alpha -1}f(s,u(s))\phantom{\rule{0.2em}{0ex}}ds\\ +\frac{1}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})}\{{\int}_{0}^{1}{t}^{\alpha -1}{(1-s)}^{\alpha -1}f(s,u(s))\phantom{\rule{0.2em}{0ex}}ds-{\int}_{0}^{\eta}{t}^{\alpha -1}{(\eta -s)}^{\alpha -1}f(s,u(s))\phantom{\rule{0.2em}{0ex}}ds\}\\ +u(1){t}^{\alpha -1}.\end{array}$

According to (3.3), it is easy to show that (3.2) holds. The proof is completed. □

**Lemma 3.2** *For any*$(t,s)\in [0,1]\times [0,1]$, $G(t,s)$*is continuous*, *and*$G(t,s)>0$*for any*$(t,s)\in (0,1)\times (0,1)$.

*Proof* The continuity of $G(t,s)$ for $(t,s)\in [0,1]\times [0,1]$ is obvious.

Let

${g}_{1}(t,s)={t}^{\alpha -1}{(1-s)}^{\alpha -1}-{t}^{\alpha -1}{(\eta -s)}^{\alpha -1}-(1-{\eta}^{\alpha -1}){(t-s)}^{\alpha -1},$

we only need to show that

${g}_{1}(t,s)>0$ for

$0\le s\le min\{t,\eta \}\le 1$, the rest of the proof is similar or obvious. From the definition of

${g}_{1}(t,s)$, we have

$\begin{array}{rcl}{g}_{1}(t,s)& =& {t}^{\alpha -1}\{{(1-s)}^{\alpha -1}-{(\eta -s)}^{\alpha -1}-(1-{\eta}^{\alpha -1}){(1-\frac{s}{t})}^{\alpha -1}\}\\ \ge & {t}^{\alpha -1}\{{(1-s)}^{\alpha -1}-{(\eta -s)}^{\alpha -1}-(1-{\eta}^{\alpha -1}){(1-s)}^{\alpha -1}\}\\ \ge & {t}^{\alpha -1}\{{\eta}^{\alpha -1}{(1-s)}^{\alpha -1}-{(\eta -s)}^{\alpha -1}\}\\ \ge & {t}^{\alpha -1}\{{(\eta -\eta s)}^{\alpha -1}-{(\eta -s)}^{\alpha -1}\}>0\end{array}$

for $0\le s\le min\{t,\eta \}\le 1$. The proof is completed. □

Let

$G(t,s)={t}^{\alpha -1}{G}^{\ast}(t,s),$

then

${G}^{\ast}(t,s)=\{\begin{array}{c}\frac{{(1-s)}^{\alpha -1}-{(\eta -s)}^{\alpha -1}-(1-{\eta}^{\alpha -1}){(1-\frac{s}{t})}^{\alpha -1}}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})},\hfill \\ \phantom{\rule{1em}{0ex}}0\le s\le min\{t,\eta \}\le 1;\hfill \\ \frac{{(1-s)}^{\alpha -1}-{(\eta -s)}^{\alpha -1}}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})},\phantom{\rule{1em}{0ex}}0\le t\le s\le \eta \le 1;\hfill \\ \frac{{(1-s)}^{\alpha -1}-(1-{\eta}^{\alpha -1}){(1-\frac{s}{t})}^{\alpha -1}}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})}\phantom{\rule{1em}{0ex}}0\le \eta \le s\le t\le 1;\hfill \\ \frac{{(1-s)}^{\alpha -1}}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})},\phantom{\rule{1em}{0ex}}0\le max\{t,\eta \}\le s\le 1.\hfill \end{array}$

(3.6)

The new Green’s function ${G}^{\ast}(t,s)$ has the following properties:

**Lemma 3.3**${G}^{\ast}(t,s)$*is continuous for*$(t,s)\in (0,1)\times (0,1)$,

*and*$\underset{t\to 0}{lim}{G}^{\ast}(t,s):={G}^{\ast}(0,s)=\{\begin{array}{c}\frac{1}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})}\{{(1-s)}^{\alpha -1}-{(\eta -s)}^{\alpha -1}\},\phantom{\rule{1em}{0ex}}0\le s\le \eta ,\hfill \\ \frac{1}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})}{(1-s)}^{\alpha -1},\phantom{\rule{1em}{0ex}}\eta \le s\le 1.\hfill \end{array}$

*Furthermore*, ${G}^{\ast}(t,s)>0$*for*$(t,s)\in (0,1)\times (0,1)$.

**Lemma 3.4** *For any*$s\in (0,1)$, ${G}^{\ast}(t,s)$*is nonincreasing with respect to*$t\in [0,1]$. *Especially*, *for any*$s\in [0,1]$, $\frac{\partial {G}^{\ast}}{\partial t}\le 0$*for*$t\in [s,1]$, *and*$\frac{\partial {G}^{\ast}}{\partial t}=0$*for*$t\in [0,s]$. *That is*${G}^{\ast}(1,s)\le {G}^{\ast}(t,s)\le {G}^{\ast}(s,s)$, *where*

${G}^{\ast}(1,s)=\frac{1}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})}\{\begin{array}{c}{\eta}^{\alpha -1}{(1-s)}^{\alpha -1}-{(\eta -s)}^{\alpha -1},\phantom{\rule{1em}{0ex}}0\le s\le \eta ,\hfill \\ {\eta}^{\alpha -1}{(1-s)}^{\alpha -1},\phantom{\rule{1em}{0ex}}\eta \le s\le 1,\hfill \end{array}$

(3.7)

*and*
${G}^{\ast}(s,s)=\frac{1}{\mathrm{\Gamma}(\alpha )(1-{\eta}^{\alpha -1})}\{\begin{array}{c}{(1-s)}^{\alpha -1}-{(\eta -s)}^{\alpha -1},\phantom{\rule{1em}{0ex}}0\le s\le \eta ,\hfill \\ {(1-s)}^{\alpha -1},\phantom{\rule{1em}{0ex}}\eta \le s\le 1.\hfill \end{array}$

(3.8)

Let

$u(t)={t}^{\alpha -1}x(t).$

(3.9)

Then,

$u(1)=x(1)$, we have from Lemma 3.1, (3.6) and (3.9) that the integral Equation (

3.2) can be rewritten as follows:

$x(t)={\int}_{0}^{1}{G}^{\ast}(t,s)f(s,{s}^{\alpha -1}x(s))\phantom{\rule{0.2em}{0ex}}ds+x(1).$

(3.10)

Then,

$y(1)=0$ and (3.10) is equivalent to the following

$y(t)={\int}_{0}^{1}{G}^{\ast}(t,s)f(s,{s}^{\alpha -1}(y(s)+x(1)))\phantom{\rule{0.2em}{0ex}}ds.$

(3.12)

We can divide our proof into the following two steps:

First, we replace

$x(1)$ by any real number

*μ*, then (3.12) can be rewritten as

$y(t)={\int}_{0}^{1}{G}^{\ast}(t,s)f(s,{s}^{\alpha -1}(y(s)+\mu ))\phantom{\rule{0.2em}{0ex}}ds.$

(3.13)

It suffices to show that for any given real number *μ*, (3.13) has a solution $y(t)$, which implies that Equation (1.1) has a solution $u(t)$ which satisfies the first boundary value condition $u(0)=0$.

Second, we show that there exists a *μ* such that the solution $y(t)$ of (3.13) satisfies $y(1)=0$, which implies that the solution $u(t)$ of (1.1) also satisfies the boundary value condition $u(1)=\frac{1}{{\eta}^{\alpha -1}}u(\eta )$.

In this section, we will prove the first step. For convenience sake, we define an operator

*T* on the set Ω as follows:

$Ty(t)={\int}_{0}^{1}{G}^{\ast}(t,s)f(s,{s}^{\alpha -1}(y(s)+\mu ))\phantom{\rule{0.2em}{0ex}}ds.$

(3.14)

**Lemma 3.5** *Suppose that*$f\in C([0,1]\times R,R)$, *and* (2.4) *hold*, *then the operator* *T* *is completely continuous in* Ω.

*Proof* It is easy to show that the operator *T* maps Ω into itself. We divide the proof into the following three steps.

Step 1. $Ty(t)$ is continuous with respect to $y(t)\in \mathrm{\Omega}$.

In fact, suppose that

$\{{y}_{n}(t)\}$ is a sequence in Ω, and

$\{{y}_{n}(t)\}$ converges to

$y(t)\in \mathrm{\Omega}$. Since

$f(t,{t}^{\alpha -1}y)$ is continuous with respect to

$y\in R$, and it is obvious that

${G}^{\ast}(t,s)$ is uniformly continuous with respect to

$(t,s)\in [0,1]\times [0,1]$ from Lemma 3.3, then for any

$\epsilon >0$, there exists an integer

*N*, when

$n>N$,

$\parallel f(t,{t}^{\alpha -1}({y}_{n}(t)+\mu ))-f(t,{t}^{\alpha -1}(y(t)+\mu ))\parallel \le \frac{\epsilon}{{\int}_{0}^{1}{G}^{\ast}(t,s)\phantom{\rule{0.2em}{0ex}}ds},$

(3.15)

which follows from (3.14)-(3.15) that

$\begin{array}{rcl}\parallel (T{y}_{n})(t)-(Ty)(t)\parallel & =& \parallel {\int}_{0}^{1}{G}^{\ast}(t,s)\{f(s,{s}^{\alpha -1}({y}_{n}(s)+\mu ))-f(s,{s}^{\alpha -1}(y(s)+\mu ))\}\phantom{\rule{0.2em}{0ex}}ds\parallel \\ \le & {\int}_{0}^{1}{G}^{\ast}(t,s)\phantom{\rule{0.2em}{0ex}}ds\parallel f(t,{t}^{\alpha -1}({y}_{n}(t)+\mu ))-f(t,{t}^{\alpha -1}(y(t)+\mu ))\parallel \\ \le & \epsilon .\end{array}$

Thus, the operator *T* is continuous in Ω.

Step 2. *T* maps bounded set in Ω into bounded set.

Suppose that

$B\in \mathrm{\Omega}$ is a bounded set with

$\parallel y(t)\parallel \le r$ for any

$y\in B$. Then, we have from (2.4) and (3.14) that

$\begin{array}{rcl}\parallel (Ty)(t)\parallel & =& \parallel {\int}_{0}^{1}{G}^{\ast}(t,s)f(s,{s}^{\alpha -1}(y(s)+\mu ))\phantom{\rule{0.2em}{0ex}}ds\parallel \\ \le & {\int}_{0}^{1}{G}^{\ast}(t,s)a(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{1}{G}^{\ast}(t,s)b(s){|y(s)+\mu |}^{p}\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int}_{0}^{1}{G}^{\ast}(t,s)a(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{1}{G}^{\ast}(t,s)b(s)\phantom{\rule{0.2em}{0ex}}ds{(\parallel y(t)\parallel +\parallel \mu \parallel )}^{p}\\ \le & {\int}_{0}^{1}{G}^{\ast}(t,s)a(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{1}{G}^{\ast}(t,s)b(s)\phantom{\rule{0.2em}{0ex}}ds{(r+\parallel \mu \parallel )}^{p}:=l.\end{array}$

This gives that the operator *T* maps bounded set into bounded set in Ω.

Step 3. *T* is equicontinuous in Ω.

It suffices to show that for any

$y(t)\in B$ and any

$0<{t}_{1}<{t}_{2}<1$,

$Ty({t}_{1})\to Ty({t}_{2})$ as

${t}_{1}\to {t}_{2}$. We consider the following three cases:

- (i)
$0<{t}_{1}<{t}_{2}<\eta $;

- (ii)
$0<{t}_{1}<\eta <{t}_{2}$;

- (iii)
$0<\eta <{t}_{1}<{t}_{2}$.

We only prove the case (i), the rest two cases are similar. Since

*B* is bounded, then there exists a

$M>0$ such that

$f\le M$. According to (3.14), we have

According to Step 1-Step 3, the operator *T* is completely continuous in Ω. The proof is completed. □

Further, we have

**Lemma 3.6** *Suppose that*$f\in C([0,1]\times R,R)$, *and* (2.4) *and* (2.6) *holds*, *then*, *for any real number* *μ*, *the integral Equation* (3.13) *has at least one solution*.

*Proof* We only need to show that the operator

*T* is priori bounded. Let

$r=max\{1,\frac{{\int}_{0}^{1}{G}^{\ast}(s,s)a(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{1}{G}^{\ast}(s,s)b(s)\phantom{\rule{0.2em}{0ex}}ds{|\mu |}^{p}}{1-{\int}_{0}^{1}{G}^{\ast}(s,s)b(s)\phantom{\rule{0.2em}{0ex}}ds}\}.$

(3.16)

Define a set

$K\in \mathrm{\Omega}$ as follows

$K=\{y\in \mathrm{\Omega}|\parallel y(t)\parallel \le r\}.$

(3.17)

To show the existence of a fixed point of *T* by Lemma 2.3, we need to verify that the second possibility in Lemma 2.3 cannot happen.

In fact, assume that there exists

$y\in \partial K$ with

$\parallel y(t)\parallel =r$ and

$\gamma \in (0,1)$ such that

$y=\gamma Ty$. It follows that

$y(t)=\gamma |(Ty)(t)|=\gamma {\int}_{0}^{1}{G}^{\ast}(t,s)f(s,{s}^{\alpha -1}(y(s)+\mu ))\phantom{\rule{0.2em}{0ex}}ds,$

and

$\begin{array}{rl}\parallel y(t)\parallel & =\parallel \gamma {\int}_{0}^{1}{G}^{\ast}(t,s)f(s,{s}^{\alpha -1}(y(s)+\mu ))\phantom{\rule{0.2em}{0ex}}ds\parallel \\ \le \gamma {\int}_{0}^{1}{G}^{\ast}(s,s)\left|f(s,{s}^{\alpha -1}(y(s)+\mu ))\right|\phantom{\rule{0.2em}{0ex}}ds\\ <{\int}_{0}^{1}{G}^{\ast}(s,s)a(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{1}{G}^{\ast}(s,s)b(s)\phantom{\rule{0.2em}{0ex}}ds{\parallel y(t)+\mu \parallel}^{p}\\ \le {\int}_{0}^{1}{G}^{\ast}(s,s)a(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{1}{G}^{\ast}(s,s)b(s)\phantom{\rule{0.2em}{0ex}}ds{\parallel \mu \parallel}^{p}+{\int}_{0}^{1}{G}^{\ast}(s,s)b(s)\phantom{\rule{0.2em}{0ex}}ds{\parallel r\parallel}^{p}\\ \le {\int}_{0}^{1}{G}^{\ast}(s,s)a(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{1}{G}^{\ast}(s,s)b(s)\phantom{\rule{0.2em}{0ex}}ds{\parallel \mu \parallel}^{p}+{\int}_{0}^{1}{G}^{\ast}(s,s)b(s)\phantom{\rule{0.2em}{0ex}}ds\parallel r\parallel \\ \le \parallel r\parallel .\end{array}$

(3.18)

Here we have the use of the inequality

${(a+b)}^{p}\le {a}^{p}+{b}^{p}\phantom{\rule{1em}{0ex}}\text{for}a,b\ge 0,0\le p\le 1.$

It is obvious that (3.18) contradicts our assumption that $\parallel y(t)\parallel =r$. Therefore, by Lemma 2.3, it follows that *T* has a fixed point $y\in \overline{K}$. Hence, the integral Equation (3.14) has at least a solution $y(t)$. The proof is completed. □