In this section, we discuss the existence of positive solutions for boundary value problem (1.1). For convenience, we give some conditions, which will play roles in this paper.
- (H1):
-
There exists a nonnegative function \(\rho\in C(0,1)\cap L(0,1)\) such that
$$\varphi_{2}(t)h_{2}(x)\leq f\bigl(t,v(t)x\bigr)+\rho(t) \leq\varphi_{1}(t) \bigl(g(x)+h_{1}(x)\bigr) $$
for all \((t,x)\in(0,1)\times\mathbb{R}^{+}\), where \(\varphi_{1},\varphi _{2}\in L(0,1)\) are nonnegative for \(t\in(0,1)\), \(h_{1},h_{2}\in C(\mathbb{R}_{0}^{+},\mathbb {R}_{0}^{+})\) are nondecreasing, \(g\in C(\mathbb{R}^{+},\mathbb{R}_{0}^{+})\) is nonincreasing, \(\mathbb{R}_{0}^{+}=[0,+\infty)\), and
$$v(t)= \left \{ \textstyle\begin{array}{l@{\quad}l} 1, & t\in(0,\tau], \\ (t-\tau)^{\alpha-5},& t\in(\tau,1). \end{array}\displaystyle \right . $$
- (H2):
-
$$0< \int_{0}^{\tau}s(1-s)^{\alpha-2} \varphi_{1}(s)g\bigl(\eta(s-\tau)\bigr)\,ds< +\infty, $$
and there exists a constant \(k>0\) such that
$$\int_{\tau}^{1}s(1-s)^{\alpha-2} \varphi_{1}(s)g\biggl(\frac{k}{2}(s-\tau )^{2}\biggr) \,ds< +\infty. $$
- (H3):
-
There exists a subinterval \([a,b]\subset(\tau,1)\) such that \(\int_{a}^{b}s(1-s)^{\alpha-1}\varphi_{2}(s)\,ds>0\).
Let \(X:=\{x(t):x\in C([-\tau,1],\mathbb{R}),x(t)=0 \text{ for } t\in [-\tau,0],x'(1)=x'(0)=0\}\) be a Banach space with the maximum norm \(\|x\|_{[-\tau,1]}=\max_{-\tau\leq t\leq1}|x(t)|=\max_{0\leq t\leq1}|x(t)|\) for \(x\in X\). Let K be a cone in X defined by
$$K=\bigl\{ x\in X; x(t)\geq0 \text{ for } t\in[-\tau,1]\bigr\} . $$
Define
$$\begin{aligned}& \bar{\eta}(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \eta(t), &t\in[-\tau,0], \\ 0, &t\in(0,1], \end{array}\displaystyle \right . \\& \omega(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} 0, &t\in[-\tau,0], \\ \int_{0}^{1}\lambda G(t,s)\rho(s)\,ds, &t\in(0,1], \end{array}\displaystyle \right . \end{aligned}$$
and
$$\begin{aligned} x^{*}(t)&=\max\bigl\{ x(t)+\bar{\eta}(t)-\omega(t),0\bigr\} \\ &= \left \{ \textstyle\begin{array}{l@{\quad}l} \eta(t), & t\in[-\tau,0], \\ \max\{x(t)-\omega(t),0\}, & t\in(0,1] \end{array}\displaystyle \right . \end{aligned}$$
for any \(x\in K\).
It is easy to know that the restriction \(\omega|_{[0,1]}\) of ω on \([0,1]\) is exactly the solution of
$$\left \{ \textstyle\begin{array}{l@{\quad}l} D^{\alpha}x(t)+\lambda\rho(t)=0, & t\in(0,1), \alpha\in (2,3], \\ x'(1)=x'(0)=x(0)=0. \end{array}\displaystyle \right . $$
Since \(f:[0,1]\times C[-\tau,1]\rightarrow\mathbb{R}\) is a continuous function, setting \(f(t,x(t-\tau)):=h(t)\) in Lemma 2.1, we see by Lemma 2.1 that a function x is a solution of boundary value problem (1.1) if and only if it satisfies
$$x(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \int_{0}^{1}\lambda G(t,s)f(s,x(s-\tau))\,ds, &t\in(0,1), \\ \eta(t), &t\in[-\tau,0]. \end{array}\displaystyle \right . $$
Consider the following operator:
$$ (Tx) (t)= \left \{ \textstyle\begin{array}{l@{\quad}l} \int_{0}^{1}\lambda G(t,s) (f(s,x^{*}(s-\tau))+\rho(s) )\,ds, &t\in(0,1], \\ 0, &t\in[-\tau,0]. \end{array}\displaystyle \right . $$
(3.1)
Set
$$y(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} t^{5-\alpha}x(t), &t\in(0,1), \\ 0, &t\in[-\tau,0] \end{array}\displaystyle \right . $$
and
$$y^{*}(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \max\{t^{\alpha-5}y(t)-\omega(t),0\}, &t\in(0,1], \\ \eta(t), &t\in[-\tau,0]. \end{array}\displaystyle \right . $$
Then (3.1) is equivalent to
$$ (Ty) (t)= \left \{ \textstyle\begin{array}{l@{\quad}l} \int_{0}^{1}\lambda G^{*}(t,s) (f(s,y^{*}(s-\tau))+\rho(s) )\,ds, &t\in(0,1], \\ 0, &t\in[-\tau,0]. \end{array}\displaystyle \right . $$
(3.2)
Obviously, if ỹ is a fixed point of operator T in (3.2), then
$$\tilde{x}(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} t^{\alpha-5}\tilde{y}(t), &t\in(0,1], \\ 0, &t\in[-\tau,0], \end{array}\displaystyle \right . $$
is a fixed point of operator T defined by (3.1). Lemma 2.1 implies that
$$\left \{ \textstyle\begin{array}{l@{\quad}l} D^{\alpha}\tilde{x}(t)+\lambda f ((t,\tilde{x}^{*}(t-\tau))+\rho (t) )=0, &t\in(0,1)\backslash\{\tau\}, \\ \tilde{x}(t)=0, &t\in[-\tau,0], \\ \tilde{x}'(1)=\tilde{x}'(0)=0. \end{array}\displaystyle \right . $$
Thus if
$$ \tilde{x}(t-\tau)+\bar{\eta}(t-\tau)-\omega(t-\tau)\geq0\quad \text{for } t\in[0,1], $$
(3.3)
then
$$\tilde{x}^{*}(t-\tau)=\tilde{x}(t-\tau)+\bar{\eta}(t-\tau)-\omega(t-\tau). $$
Let
$$x(t)=\tilde{x}(t)+\bar{\eta}(t)-\omega(t). $$
Then one can verify easily that the function x satisfied boundary value problem (1.1). As a result, in the following we will concentrate our study on finding the fixed points of operator T defined by (3.2).
Define the cone
$$K_{1}=\bigl\{ y\in K: y(t)\geq t^{2}\|y\| \text{ for } t \in[0,1]\bigr\} $$
and
$$\begin{aligned}& \Omega_{1}=\bigl\{ y\in K_{1}:\Vert y\Vert < r_{1}\bigr\} , \\& \Omega_{2}=\bigl\{ y\in K_{1}:\Vert y\Vert < r_{2}\bigr\} , \\& \Omega_{3}=\bigl\{ y\in K_{1}:\Vert y\Vert < R_{2}\bigr\} , \\& \Omega_{4}=\bigl\{ y\in K_{1}:\Vert y\Vert < R_{1}\bigr\} \end{aligned}$$
for any \(r_{2}>r_{1}\geq\max\{k,2c\}\), \(R_{2}>R_{1}\geq\max\{k,2c\}\), where
$$ c:=\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{1}s(1-s)^{\alpha-2}\rho (s)\,ds< + \infty $$
(3.4)
and k is the constant in (H2).
Lemma 3.1
Let (H1) and (H2) hold. Then the operator
\(T:K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\rightarrow K_{1}\)
is completely continuous.
Proof
First we show that operator T is well defined on \(K_{1}\cap(\bar{\Omega}_{2} \backslash\Omega_{1})\). For any \(y\in K_{1}\cap(\bar{\Omega}_{2} \backslash\Omega_{1})\), we know that
$$r_{1}\leq\|y\|\leq r_{2} $$
and
$$y(t)\geq t^{2}\|y\|\geq t^{2} r_{1} \quad \text{for } t\in[0,1]. $$
Then, for \(t\in[0,1]\), we get
$$\begin{aligned} t^{5-\alpha}\omega(t)&=t^{5-\alpha} \int_{0}^{1}\lambda G(t,s)\rho(s)\,ds \\ &\leq\frac{t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{1}\lambda s(1-s)^{\alpha -2}\rho(s)\,ds \\ &\leq\frac{t^{2}}{\Gamma(\alpha)} \int_{0}^{1}\lambda s(1-s)^{\alpha-2}\rho (s) \,ds \\ &\leq t^{2}c, \end{aligned}$$
(3.5)
where c is defined as (3.4). Thus, for \(t\in[0,1]\),
$$\begin{aligned} \begin{aligned}[b] y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(r_{1}-c) \\ &\geq\frac{r_{1}}{2}t^{2}. \end{aligned} \end{aligned}$$
(3.6)
In view of (H1), (H2), and Lemma 2.3, we show
$$\begin{aligned} (Ty) (t) =& \int_{0}^{\tau}\lambda G^{*}(t,s) \bigl(f\bigl(s,\eta(s- \tau)\bigr)+\rho(s) \bigr)\,ds \\ &{} + \int_{\tau}^{1}\lambda G^{*}(t,s) \bigl(f\bigl(s,(s- \tau)^{\alpha-5}y(s-\tau )-\omega(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \\ &{}\times\biggl(g \biggl(\frac{r_{1}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau)-(s-\tau )^{5-\alpha}\omega(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{r_{1}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+h_{1}(r_{2}) \biggr)\,ds \\ < &+\infty. \end{aligned}$$
Hence, T is uniformly bounded and T is well defined.
In fact, for \(y\in K_{1}\cap(\bar{\Omega}_{2} \backslash\Omega_{1})\), \(t\in[0,1]\), in view of Lemma 2.3, we have
$$\begin{aligned} \|Ty\|&\leq\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int _{0}^{1}s(1-s)^{\alpha-2} \bigl(f \bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ &\leq\frac{\lambda t^{2}}{\Gamma(\alpha)} \int_{0}^{1}s(1-s)^{\alpha-2} \bigl(f \bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \end{aligned}$$
and
$$\begin{aligned} (Ty) (t)&\geq\frac{\lambda t^{4}}{\Gamma(\alpha)} \int_{0}^{1}s(1-s)^{\alpha -2} \bigl(f \bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ &\geq t^{2}\|Ty\|. \end{aligned}$$
Hence, \(T: K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\rightarrow K_{1}\).
Next we show \(T:K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\rightarrow K_{1}\) is continuous and compact. For any \(y_{n}, y\in K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\), \(n=1,2,\ldots\) with \(\|y_{n}-y\|_{[-\tau,1]}\rightarrow0\) as \(n\rightarrow\infty\). Since \(r_{1}\leq\|y_{n}\|\leq r_{2}\) and \(r_{1}\leq\| y\|\leq r_{2}\), for \(t\in(0,1)\), we know
$$y_{n}(t)-t^{5-\alpha}\omega(t)\geq\frac{r_{1}}{2}t^{2} $$
and
$$y(t)-t^{5-\alpha}\omega(t)\geq\frac{r_{1}}{2}t^{2}. $$
Then, for \(t\in[0,1]\), we have
$$\begin{aligned} \bigl\vert (Ty_{n}) (t)-(Ty) (t)\bigr\vert =&\biggl\vert \int_{\tau}^{1}\lambda G^{*}(t,s) \bigl(f\bigl(s,(s- \tau)^{\alpha-5}y_{n}(s-\tau )-\omega(s-\tau)\bigr)+\rho(s) \\ &{}-f\bigl(s,(s-\tau)^{\alpha-5}y(s-\tau)-\omega(s-\tau)\bigr)-\rho(s) \bigr)\,ds\biggr\vert \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau }^{1}s(1-s)^{\alpha-2}\bigl\vert f \bigl(s,(s-\tau)^{\alpha-5}y_{n}(s-\tau)-\omega(s-\tau)\bigr) \\ &{}-f\bigl(s,(s-\tau)^{\alpha-5}y(s-\tau)-\omega(s-\tau)\bigr)\bigr\vert \,ds \\ \leq&\frac{2\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau }^{1}s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g\biggl(\frac{k}{2}(s-\tau )^{2} \biggr)+h_{1}(r_{2}) \biggr)\,ds \\ < &+\infty. \end{aligned}$$
This implies that \(\|Ty_{n}-Ty\|_{[-\tau,1]}\rightarrow0\) as \(n\rightarrow\infty\). Hence T is continuous.
Next we prove T is equicontinuous.
Since \(G^{*}\) in uniformly continuous for \(t\in(0,1)\), that is, for any \(\epsilon>0\), there exists \(\xi_{0}>0\), when \(t_{1},t_{2}\in[0,1]\) and \(|t_{1}-t_{2}|<\xi_{0}\), we have
$$\begin{aligned} \bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s)\bigr\vert =& \frac{\epsilon}{2} \biggl( \int_{0}^{\tau}\lambda\varphi_{1}(s) \bigl(g \bigl(\eta (s-\tau)\bigr)+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr)\,ds \\ &{}+ \biggl.\biggl.\int_{\tau}^{1}\lambda\varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau )^{2}\biggr)+h_{1}(r_{2}) \biggr)\,ds \biggr)\biggr.^{-1}. \end{aligned}$$
Thus, for any \(y\in K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\), we get
$$\begin{aligned} \bigl\vert (Ty) (t_{1})-(Ty) (t_{2})\bigr\vert \leq& \int_{0}^{\tau}\lambda\bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s)\bigr\vert \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr) \,ds \\ &{} + \int_{\tau}^{1}\lambda\bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s) \bigr\vert \varphi_{1}(s) \biggl(g\biggl(\frac{k}{2}(s- \tau)^{2}\biggr)+h_{1}(r_{2}) \biggr)\,ds \\ < &\frac{\epsilon}{2}+\frac{\epsilon}{2} \\ =&\epsilon. \end{aligned}$$
Thus T is equicontinuous. Accordingly to the Ascoli-Arzelà theorem, T is completely continuous. The proof is completed. □
Now we prove the existence of positive solutions for boundary value problem (1.1) by using the Guo-Krasnoselskii fixed point theorem.
For convenience, we denote
$$\begin{aligned} \xi_{1} :=&\frac{1}{\Gamma(\alpha)} \biggl( \int_{0}^{\tau}s(1-s)^{\alpha -2} \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr) \\ &{}+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr)\,ds+ \int_{\tau}^{1}s(1-s)^{\alpha-2}\varphi _{1}(s) \biggl( g\biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+\epsilon r_{2} \biggr)\,ds \biggr) \\ >&0, \end{aligned}$$
and there exists a subinterval \([\beta,\gamma]\subset(\tau,1)\),
$$\begin{aligned}& \zeta_{1}:=\min_{t\in[\beta,\gamma]}(t-\tau)^{2}, \qquad \zeta _{2}:=\min_{t\in[\beta,\gamma]}t^{4}, \\& \xi_{2}(t):=\frac{\zeta_{2}}{\Gamma(\alpha)}h_{2}\biggl( \frac{r_{1}\zeta_{1}}{2}\biggr) \int _{\beta}^{\gamma} s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds>0. \end{aligned}$$
Theorem 3.1
Let (H1), (H2), and
\(\xi_{2}^{-1}r_{1}<\xi_{1}^{-1}r_{2}\)
hold. Then the boundary value problem (1.1) has at least one positive solution if
$$ \lim_{y\rightarrow+\infty}\frac{h_{1}(y)}{y}=0 $$
(3.7)
for each
$$\lambda\in\bigl(\xi_{2}^{-1}r_{1}, \xi_{1}^{-1}r_{2}\bigr). $$
Proof
Let \(\epsilon>0\). Then in view of (3.7), there exists a \(M>0\) such that
$$ h_{1}(y)\leq\epsilon y \quad \text{for } y>M. $$
(3.8)
Choose
$$r_{2}\geq\max\{M+1,r_{1}+1\}, $$
then for \(y\in\partial\Omega_{2}\), like for (3.6), for \(t\in [0,1]\), we obtain
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(r_{2}-c) \\ &\geq\frac{r_{2}}{2}t^{2}. \end{aligned}$$
(3.9)
Then from (H1), (3.8), (3.9), and Lemma 2.3, we get
$$\begin{aligned} (Ty) (t) \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \\ &{}\times\biggl(g \biggl(\frac{r_{2}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau)-(s-\tau )^{5-\alpha}\omega(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{r_{2}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau}s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+h_{1}(r_{2}) \biggr)\,ds \\ \leq&\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau}s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+\epsilon r_{2} \biggr)\,ds \\ \leq&\lambda\xi_{1}< r_{2}. \end{aligned}$$
Therefore, for \(y\in\partial\Omega_{2}\), we have \(\|Ty\|\leq\|y\|\).
On the other hand, for \(y\in\partial\Omega_{1}\), like for (3.6), for \(t\in[0,1]\), we obtain
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(r_{1}-c) \\ &\geq\frac{r_{1}}{2}t^{2}. \end{aligned}$$
(3.10)
Thus from (H1), (3.10), and Lemma 2.3, we get
$$\begin{aligned} \|Ty\| \geq& \int_{\beta}^{\gamma}\lambda\min_{t\in[\beta,\gamma ]}G^{*}(t,s) \bigl(f\bigl(s,(s-\tau)^{\alpha-5}y(s-\tau)-\omega(s-\tau)\bigr)+\rho (s) \bigr)\,ds \\ \geq& \int_{\beta}^{\gamma}\lambda\min_{t\in[\beta,\gamma ]}G^{*}(t,s) \varphi_{2}(s)h_{2} \biggl(\frac{r_{1}}{2}(s- \tau)^{2} \biggr)\,ds \\ \geq&\frac{\lambda\zeta_{2}}{\Gamma(\alpha)}h_{2}\biggl(\frac{r_{1}\zeta _{1}}{2}\biggr) \int_{\beta}^{\gamma}s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds \\ \geq&\lambda\xi_{2}>r_{1}. \end{aligned}$$
Therefore, for \(y\in\partial\Omega_{1}\), we have \(\|Ty\|\geq\|y\|\). Then T defined by (3.2) has a fixed point \(\tilde{y}\in K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega _{1})\). In view of (3.10), we have
$$\begin{aligned} t^{\alpha-5}\tilde{y}(t)-\omega(t)&=t^{\alpha-5} \bigl(\tilde {y}(t)-t^{5-\alpha}\omega(t) \bigr) \\ &\geq\frac{r_{1}}{2}t^{\alpha-3} \\ &>0. \end{aligned}$$
It is easy to know (3.3) is satisfied. The proof is completed. □
Denote
$$\begin{aligned} \xi_{3} :=&\frac{1}{\Gamma(\alpha)} \biggl( \int_{0}^{\tau}s(1-s)^{\alpha -2} \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr)\,ds \\ &{}+ \int_{\tau}^{1}s(1-s)^{\alpha-2}\varphi _{1}(s) \biggl(g\biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+h_{1}(R_{1}) \biggr)\,ds \biggr) \\ >&0. \end{aligned}$$
Theorem 3.2
Let (H1)-(H3) and
$$\frac{2\Gamma(\alpha)}{M^{*}A\zeta_{2}} \biggl( \int_{a}^{b}s(1-s)^{\alpha-2}\varphi _{2}(s)\,ds \biggr)^{-1}< \xi_{3}^{-1}R_{1} $$
hold. Then boundary value problem (1.1) has at least one positive solution if
$$ \lim_{y\rightarrow+\infty}\frac{h_{2}(y)}{y}=+\infty $$
(3.11)
for each
$$ \lambda\in \biggl(\frac{2\Gamma(\alpha)}{M^{*}A\zeta_{2}} \biggl( \int _{a}^{b}s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds \biggr)^{-1},\xi_{3}^{-1}R_{1} \biggr), $$
(3.12)
where
\(A:=\min_{t\in[a,b]}(t-\tau)^{2}\), \(M^{*}\)
is a positive constant.
Proof
It follows from (3.11) that there exists a \(M^{*}>0\) such that
$$ h_{2}(y)\geq M^{*}y \quad \text{for } y>M^{*}. $$
(3.13)
Choose
$$R_{2}\geq\max\biggl\{ R_{1}+1,\frac{2M^{*}}{A}\biggr\} . $$
Then for \(y\in\partial\Omega_{3}\), like for (3.6), for \(t\in [0,1]\), we obtain
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(R_{2}-c) \\ &\geq\frac{R_{2}}{2}t^{2}. \end{aligned}$$
(3.14)
Thus from (H1), (3.13), (3.14), and Lemma 2.3, we get
$$\begin{aligned} \|Ty\| \geq& \int_{a}^{b}\lambda\min_{t\in[a,b]}G^{*}(t,s) \bigl(f\bigl(s,(s-\tau)^{\alpha-5}y(s-\tau)-\omega(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ \geq& \int_{a}^{b}\lambda\min_{t\in[a,b]}G^{*}(t,s) \varphi_{2}(s)h_{2} \biggl(\frac{R_{2}}{2}(s- \tau)^{2} \biggr)\,ds \\ \geq& \int_{a}^{b}\lambda\min_{t\in[a,b]}G^{*}(t,s) \varphi _{2}(s)h_{2}\biggl(\frac{R_{2}A}{2}\biggr)\,ds \\ \geq&\frac{M^{*}A\lambda R_{2}}{2} \int_{a}^{b}\min_{t\in [a,b]}G^{*}(t,s) \varphi_{2}(s)\,ds \\ \geq&\frac{\lambda M^{*}AR_{2}\zeta_{2}}{2\Gamma(\alpha)} \int_{a}^{b} s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds \\ \geq& R_{2}. \end{aligned}$$
Therefore, for \(y\in\partial\Omega_{3}\), we have \(\|Ty\|\geq\|y\|\).
On the other hand, for \(y\in\partial\Omega_{4}\), like for (3.6), for \(t\in[0,1]\), we obtain
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(R_{1}-c) \\ &\geq\frac{R_{1}}{2}t^{2}. \end{aligned}$$
(3.15)
Thus from (H1), (3.12), and (3.15), we have
$$\begin{aligned} (Ty) (t) \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \\ &{}\times\biggl(g \biggl(\frac{R_{1}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau)-(s-\tau )^{5-\alpha}\omega(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau}s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+h_{1}(R_{1}) \biggr)\,ds \\ \leq&\lambda\xi_{3}< R_{1}. \end{aligned}$$
Therefore, for \(y\in\partial\Omega_{4}\), we have \(\|Ty\|\leq\|y\|\). Arguments similar to those at the end of the proof of Theorem 3.1 show that boundary value problem (1.1) has a positive solution. The proof is completed. □
Theorem 3.3
Let (H1) and (H2) hold. Furthermore assume that
- (H4):
-
there exists a subinterval
\([\beta,\gamma]\subset(\tau,1)\)
and a positive constant
r
such that
$$\begin{aligned} r >&\max\biggl\{ k,2c,\frac{\lambda}{\Gamma(\alpha)} \biggl( \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl( \eta(s-\tau)\bigr) \bigr)\,ds \\ &{} + \int_{\tau}^{1}s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g\biggl(\frac {k}{2}(s-\tau)^{2} \biggr)+h_{1}(r)\,ds \biggr) \biggr)\biggr\} , \end{aligned}$$
where
k
is defined in (H2), c
is defined as (3.4) and
\(\lambda\in(0,+\infty)\).
Then boundary value problem (1.1) has at least one positive solution
y
with
\(0<\|y\|<r\).
Proof
In view of (H4), we choose \(n_{0}\in\{1,2,\ldots\}\) such that
$$\begin{aligned} r >&\frac{\lambda}{\Gamma(\alpha)} \biggl( \int_{0}^{\tau}s(1-s)^{\alpha -2} \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl( \eta(s-\tau)\bigr)\bigr)\,ds \\ &{} + \int_{\tau}^{1}s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g\biggl(\frac{k}{2}(s-\tau )^{2} \biggr)+h_{1}(r)\biggr)\,ds \biggr)+\frac{1}{n_{0}}. \end{aligned}$$
Let \(N_{0}=\{n_{0},n_{0}+1,\ldots\}\). Fix \(n\in N_{0}\) and consider the family of integral equations
$$ y(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \kappa\int_{0}^{1}\lambda G^{*}(t,s) (f_{n}(s,y^{*}(s-\tau))+\rho(s) )\,ds+\frac{1}{n}, & t\in(0,1), \\ \frac{1}{n}, & t\in[-\tau,0], \end{array}\displaystyle \right . $$
(3.16)
where \(\kappa\in(0,1)\),
$$f_{n}\bigl(t,y^{*}(t-\tau)\bigr)+\rho(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} f(t,y^{*}(t-\tau))+\rho(t), & y^{*}(t-\tau)\geq\frac{1}{n}, \\ f(t,\frac{1}{n})+\rho(t), & y^{*}(t-\tau)< \frac{1}{n}. \end{array}\displaystyle \right . $$
We claim that any solution y of (3.16) for any \(\kappa\in (0,1)\) must satisfy \(\|y\|\neq r\). Otherwise, assume that y is a solution of (3.16) for some \(\kappa\in(0,1)\) such that \(\|y\|=r\). Then \(y^{*}(t-\tau)\geq\frac{1}{n}\) for \(t\in(0,1)\). In view of Lemma 2.3, we have
$$ \|y\|\leq\frac{\kappa\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int _{0}^{1}s(1-s)^{\alpha-2} \bigl(f_{n}\bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds+ \frac{1}{n}. $$
(3.17)
Thus, for \(t\in(0,1)\), we have
$$\begin{aligned} y(t) \geq&\frac{1}{n}+\frac{\kappa\lambda t^{4}}{\Gamma(\alpha)} \int _{0}^{1}s(1-s)^{\alpha-2} \bigl(f_{n}\bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ \geq&\frac{1}{n}+t^{\alpha-1}\biggl(\|y\|-\frac{1}{n}\biggr) \\ \geq&\bigl(1-t^{\alpha-1}\bigr)\frac{1}{n}+t^{\alpha-1}\|y\| \\ \geq& t^{\alpha-1}\|y\| \geq t^{2}r. \end{aligned}$$
Then like for (3.6), for \(t\in(0,1)\), we have
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(r-c) \\ &\geq\frac{r}{2}t^{2}. \end{aligned}$$
Then from (H1), for \(t\in(0,1)\), \(\kappa\in(0,1)\), we have
$$\begin{aligned} y(t) =&\frac{1}{n}+\kappa\lambda \int_{0}^{1}G^{*}(t,s) \bigl(f_{n} \bigl(s,y^{*}(s-\tau )\bigr)+\rho(s) \bigr)\,ds \\ =&\frac{1}{n}+\kappa\lambda \int_{0}^{1}G^{*}(t,s) \bigl(f\bigl(s,y^{*}(s-\tau) \bigr)+\rho (s) \bigr)\,ds \\ \leq&\frac{1}{n}+\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int _{0}^{1}s(1-s)^{\alpha-2} \bigl(f \bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ \leq&\frac{1}{n_{0}}+\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \bigl( \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl( \eta(s-\tau)\bigr) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1}s(1-s)^{\alpha-2} \\ &{}\times \biggl( \varphi_{1}(s) \biggl(g\biggl(\frac{r}{2}(s-\tau)^{2} \biggr) +h_{1}\bigl(y(s-\tau)-(s-\tau)^{5-\alpha}\omega(s-\tau) \bigr) \biggr) \biggr)\,ds. \end{aligned}$$
Hence we obtain
$$\begin{aligned} r =&\bigl\Vert y(t)\bigr\Vert \\ \leq&\frac{1}{n_{0}}+\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \bigl( \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl( \eta(s-\tau)\bigr) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1}s(1-s)^{\alpha-2} \\ &{}\times \biggl( \varphi_{1}(s) \biggl(g\biggl(\frac{r}{2}(s-\tau)^{2} \biggr) +h_{1}\bigl(y(s-\tau)-(s-\tau)^{5-\alpha}\omega(s-\tau) \bigr) \biggr) \biggr)\,ds. \end{aligned}$$
This is a contradiction and the claim is proved.
Now the Leray-Schauder nonlinear alternative theorem guarantees that the equation
$$y(t)= \int_{0}^{1}\lambda G^{*}(t,s) \bigl(f_{n} \bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds+\frac{1}{n} $$
has a solution \(y_{n}\), in \(\bar{\Omega}_{r}=\{y\in C[0,1]:\|y\|\leq r\}\), for \(t\in(0,1)\).
Next we claim that \(y_{n}(t)\) has a uniform sharper lower bound. In view of (H1) and \(\|y_{n}(t)\|\leq r\), we obtain
$$\begin{aligned} y_{n}(t) =&\frac{1}{n}+\lambda \int_{0}^{1}G^{*}(t,s) \bigl(f_{n} \bigl(s,y^{*}_{n}(s-\tau )\bigr)+\rho(s) \bigr)\,ds \\ \geq&\frac{1}{n}+\lambda \int_{\beta}^{\gamma}G^{*}(t,s) \bigl(f\bigl(s,y^{*}_{n}(s- \tau)\bigr)+\rho(s) \bigr)\,ds \\ \geq&\frac{\lambda t^{4}}{\Gamma(\alpha)} \int_{\beta}^{\gamma} s(1-s)^{\alpha-2} \bigl(f \bigl(s,y^{*}_{n}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ \geq&\frac{\lambda t^{4}}{\Gamma(\alpha)} \int_{\beta}^{\gamma }s(1-s)^{\alpha-2} \biggl( \varphi_{2}(s)h_{2}\biggl(\frac{r}{2}(s- \tau)^{2}\biggr) \biggr)\,ds \\ \geq&\frac{\lambda t^{4}}{\Gamma(\alpha)}h_{2}\biggl(\frac{r\zeta_{1}}{2}\biggr) \int _{\beta}^{\gamma}s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds. \end{aligned}$$
Choosing \(\delta(t)=\frac{\lambda t^{4}}{\Gamma(\alpha)}h_{2}(\frac{r\zeta _{1}}{2})\int_{\beta}^{\gamma}s(1-s)^{\alpha-2}\varphi_{2}(s)\,ds\). Then we conclude that there exists a function \(\delta\in C(0,1)\) that is unrelated to n such that \(\delta(t)>0\) for a.e. \(t\in(0,1)\) and for any \(n\in N_{0}\),
$$y_{n}(t)\geq\delta(t). $$
Then we prove \(\{y_{n}\}_{n\in N_{0}}\) is an equicontinuous family on \((0,1)\). Since \(G^{*}\) in uniformly continuous for \(t\in(0,1)\), that is, for any \(\epsilon>0\), there exists \(\zeta _{0}>0\), when \(t_{1},t_{2}\in[0,1]\) and \(|t_{1}-t_{2}|<\zeta_{0}\), we have
$$\begin{aligned} \bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s)\bigr\vert =& \frac{\epsilon}{2\lambda} \biggl( \int_{0}^{\tau}\varphi_{1}(s) \bigl(g\bigl( \eta (s-\tau)\bigr)+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr)\,ds \\ &{} +\biggl.\biggl. \int_{\tau}^{1}\varphi_{1}(s) \biggl(g\biggl( \frac{k}{2}(s-\tau)^{2}\biggr)+h_{1}(r) \biggr)\,ds \biggr)\biggr.^{-1}. \end{aligned}$$
Thus
$$\begin{aligned} \bigl\vert (y_{n}) (t_{1})-(y_{n}) (t_{2})\bigr\vert \leq&\lambda \int_{0}^{\tau}\bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s) \bigr\vert \varphi_{1}(s) \bigl(g\bigl(\eta (s-\tau) \bigr)+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr)\,ds \\ &{} +\lambda \int_{\tau}^{1}\bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s) \bigr\vert \varphi_{1}(s) \biggl(g\biggl(\frac{k}{2}(s- \tau)^{2}\biggr)+h_{1}(r) \biggr)\,ds \\ < &\frac{\epsilon}{2}+\frac{\epsilon}{2} \\ =&\epsilon. \end{aligned}$$
Therefore, \(\{y_{n}\}_{n\in N_{0}}\) is an equicontinuous family on \((0,1)\). By the Arzelà-Ascoli theorem, there exist a subsequence \(N_{1}\) of \(N_{0}\) and \(y\in C(0,1)\) such that \(\{y_{n}\}_{n\in N_{1}}\) is uniformly convergent to y and y satisfies \(\delta(t)\leq y(t)\leq r\) for any \(t\in(0,1)\). By the Lebesgue dominated convergence theorem, in view of
$$y_{n}(t)= \int_{0}^{1}\lambda G^{*}(t,s) \bigl(f_{n} \bigl(s,y^{*}_{n}(s-\tau)\bigr)+\rho(s) \bigr)\,ds, $$
we have
$$y(t)= \int_{0}^{1}\lambda G^{*}(t,s) \bigl(f\bigl(s,y^{*}(s- \tau)\bigr)+\rho(s) \bigr)\,ds. $$
Then boundary value problem (1.1) has one positive solution with \(0<\|y\|<r\). The proof is completed. □
Theorem 3.4
Let (H1) and (H2) hold. Assume that there exists a subinterval
\([\beta,\gamma]\subset(\tau,1)\)
satisfying
$$\begin{aligned}& \max g(\cdot)+\max h_{1}(\cdot)\leq\frac{r_{2}}{\frac{\lambda}{\Gamma (\alpha)}\int_{0}^{1}s(1-s)^{\alpha-2}\varphi_{1}(s)\,ds}, \\ & \min_{0\leq y\leq r_{1}} h_{2}(y)\geq\frac{r_{1}}{\frac{\lambda\zeta _{2}}{\Gamma(\alpha)}\int_{\beta}^{\gamma}s(1-s)^{\alpha-2}\varphi_{2}(s)\,ds}, \end{aligned}$$
where
\(\lambda\in(0,+\infty)\). Then boundary value problem (1.1) has at least one positive solution.
Proof
In view of Theorem 3.1, for \(y\in\partial\Omega _{2}\), \(t\in[0,1]\), we obtain
$$ y(t)-t^{5-\alpha}\omega(t)\geq\frac{r_{2}}{2}t^{2}. $$
(3.18)
Then from (H1), (3.18), and Lemma 2.3, we get
$$\begin{aligned} (Ty) (t) \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \\ &{}\times\biggl(g \biggl(\frac{r_{2}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau)-(s-\tau )^{5-\alpha}\omega(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{r_{2}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau}s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+h_{1}(r_{2}) \biggr)\,ds \\ \leq&\frac{\lambda}{\Gamma(\alpha)}\frac{r_{2}}{\frac{\lambda}{\Gamma (\alpha)}\int_{0}^{1}s(1-s)^{\alpha-2}\varphi_{1}(s)\,ds} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)}\frac{r_{2}}{\frac{\lambda}{\Gamma (\alpha)}\int_{0}^{1}s(1-s)^{\alpha-2}\varphi_{1}(s)\,ds} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s)\,ds \leq r_{2}. \end{aligned}$$
Therefore, for \(y\in\partial\Omega_{2}\), we have \(\|Ty\|\leq\|y\|\).
On the other hand, for \(y\in\partial\Omega_{1}\), like for (3.6), for \(t\in[0,1]\), we obtain
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(r_{1}-c) \\ &\geq\frac{r_{1}}{2}t^{2}. \end{aligned}$$
(3.19)
Thus from (H1), (3.19), and Lemma 2.3, we get
$$\begin{aligned} \|Ty\| \geq& \int_{\beta}^{\gamma}\lambda\min_{t\in[\beta,\gamma ]}G^{*}(t,s) \bigl(f\bigl(s,(s-\tau)^{\alpha-5}y(s-\tau)-\omega(s-\tau)\bigr)+\rho (s) \bigr)\,ds \\ \geq& \int_{\beta}^{\gamma}\lambda\min_{t\in[\beta,\gamma ]}G^{*}(t,s) \varphi_{2}(s)h_{2} \biggl(\frac{r_{1}}{2}(s- \tau)^{2} \biggr)\,ds \\ \geq&\frac{\lambda\zeta_{2}}{\Gamma(\alpha)}\frac{r_{1}}{\frac{\lambda \zeta_{2}}{\Gamma(\alpha)}\int_{\beta}^{\gamma}s(1-s)^{\alpha-2}\varphi_{2}(s)\,ds} \int_{\beta}^{\gamma}s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds \\ \geq& r_{1}. \end{aligned}$$
Therefore, for \(y\in\partial\Omega_{1}\), we have \(\|Ty\|\geq\|y\|\). Arguments similar to those at the end of the proof of Theorem 3.1 show that boundary value problem (1.1) has a positive solution. The proof is completed. □