In this section, for the convenience of reader, we present some notation and lemmas that will be used in the proof of our main results. They can be found in the literature; see [5, 6, 12, 13].
Definition 2.1
The Riemann-Liouville fractional integral of order \(\alpha>0\) of a function \(y : (0, \infty) \rightarrow\mathbb{R} \) is given by
$$I_{0^{+}}^{\alpha}y(t)=\frac{1}{\Gamma(\alpha)} \int _{0}^{t}(t-s)^{\alpha-1}y(s)\,ds, $$
provided that the right-hand side is pointwise defined on \((0, \infty )\).
Definition 2.2
The Riemann-Liouville fractional derivative of order \(\alpha>0\) of a continuous function \(y : (0, \infty) \rightarrow\mathbb{R}\) is given by
$$D_{0^{+}}^{\alpha}y(t)=\frac{1}{\Gamma(n-\alpha)} \biggl( \frac{d}{dt} \biggr) ^{n} \int_{0}^{t}\frac{y(s)}{(t-s)^{\alpha-n+1}}\,ds, $$
where \(n = [\alpha]+1\) with \([\alpha]\) denoting the integer part of a number α, provided that the right-hand side is pointwise defined on \((0, \infty)\).
Lemma 2.1
[6]
Let
\(\alpha>0\). If
\(u \in C(0, 1) \cap L^{1}(0, 1)\), then the fractional differential equation
$$D_{0^{+}}^{\alpha}u(t)=0 $$
has
$$u(t) = C_{1}t^{\alpha-1}+C_{2}t^{\alpha-2}+ \cdots+C_{N}t^{\alpha -N}, \quad C_{i} \in\mathbb{R}, i =1, 2, \dots, N, $$
as the unique solution, where
\(N = [\alpha]+1\).
From the definition of the Riemann-Liouville derivative we can obtain the following statement.
Lemma 2.2
[6]
Assume that
\(u \in C(0, 1) \cap L^{1}(0, 1)\)
with a fractional derivative of order
\(\alpha>0\)
that belongs to
\(C(0, 1) \cap L^{1}(0, 1)\). Then
$$I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}u(t)=u(t)+ C_{1}t^{\alpha -1}+C_{2}t^{\alpha-2}+ \cdots+C_{N}t^{\alpha-N} $$
for some
\(C_{i}\in\mathbb{R} \) (\(i =1, 2, \dots, N\)), where
\(N = [\alpha]+1\).
In the following, we present the Green function of the fractional differential equation boundary value problem.
Lemma 2.3
Given
\(y\in C(0, 1) \cap L^{1}(0, 1)\), \(n-1< \alpha\leq n\), the problem
$$ \left \{ \textstyle\begin{array}{@{}l} D_{0^{+}}^{\alpha}u(t)+y(t)=0, \quad 0< t< 1, \\ u(0)=u'(0)=\cdots=u^{(n-2)}(0)=0, \\ u(1)=\int_{0}^{1}h(s)u(s)\,dA(s) \end{array}\displaystyle \right . $$
(2.1)
is equivalent to
$$u(t)= \int_{0}^{1}G(t, s)y(s)\,ds, $$
where
$$ G(t, s)=G_{0}(t, s)+\frac{t^{\alpha-1}}{1-\Gamma}g_{A}(s) $$
(2.2)
with
$$\begin{aligned} &G_{0}(t, s)=\frac{1}{\Gamma(\alpha)} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} {[}t(1-s)]^{\alpha-1}-(t-s)^{\alpha-1}, & 0\leq s\leq t\leq1 , \\ {[}t(1-s)]^{\alpha-1}, & 0\leq t\leq s\leq1 , \end{array}\displaystyle \right .\\ &\Gamma= \int_{0}^{1}t^{\alpha-1}h(t)\,dA(t),\qquad g_{A}(s)= \int_{0}^{1}G_{0}(t, s)h(t)\,dA(t). \end{aligned}$$
Proof
We may apply Lemma 2.2 to reduce (2.1) to the equivalent integral equation
$$u(t)=-I_{0^{+}}^{\alpha}y(t)+ C_{1}t^{\alpha-1}+C_{2}t^{\alpha -2}+ \cdots+C_{n}t^{\alpha-n} $$
for some \(C_{i}\in\mathbb{R} \) (\(i =1, 2, \ldots, n\)). Consequently, the general solution of (2.1) is
$$ u(t)=-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}y(s)\,ds+C_{1}t^{\alpha-1}+C_{2}t^{\alpha-2}+ \cdots+C_{n}t^{\alpha -n}. $$
(2.3)
Since \(u(0)=u'(0)=\cdots=u^{(n-2)}(0)=0\), we get that \(C_{2}=C_{3}=\cdots=C_{n}=0\) by (2.3). Then, we obtain
$$ u(t)=-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}y(s)\,ds+C_{1}t^{\alpha-1}. $$
(2.4)
On the other hand, we can combine (2.1) with
$$u(1)=-\frac{1}{\Gamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-1}y(s)\,ds+C_{1}, $$
yielding
$$ C_{1}=\frac{1}{\Gamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-1}y(s)\,ds+ \int _{0}^{1}h(s)u(s)\,dA(s). $$
(2.5)
So, by (2.4) and (2.5) we have
$$\begin{aligned} u(t) ={}&{-}\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}y(s)\,ds+ \frac {t^{\alpha-1}}{\Gamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-1}y(s)\,ds \\ &{}+ t^{\alpha-1} \int_{0}^{1}h(s)u(s)\,dA(s) \\ ={}&\frac{1}{\Gamma(\alpha)} \int_{0}^{t}\bigl[t^{\alpha-1}(1-s)^{\alpha -1}-(t-s)^{\alpha-1} \bigr]y(s)\,ds \\ &{}+\frac{1}{\Gamma(\alpha)} \int_{t}^{1}t^{\alpha-1}(1-s)^{\alpha -1}y(s)\,ds+t^{\alpha-1} \int_{0}^{1}h(s)u(s)\,dA(s). \end{aligned}$$
Let
$$G_{0}(t, s)=\frac{1}{\Gamma(\alpha)} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} {[}t(1-s)]^{\alpha-1}-(t-s)^{\alpha-1}, & 0\leq s\leq t\leq1 , \\ {[}t(1-s)]^{\alpha-1}, & 0\leq t\leq s\leq1 . \end{array}\displaystyle \right . $$
Then
$$ u(t)= \int_{0}^{1}G_{0}(t, s)y(s)\,ds+t^{\alpha-1} \int _{0}^{1}h(s)u(s)\,dA(s). $$
(2.6)
Multiplying both sides of (2.6) by \(h(t)\) and then integrating from 0 to 1, we have
$$\begin{aligned} \int_{0}^{1}h(t)u(t)\,dA(t)={}& \int_{0}^{1}h(t) \int_{0}^{1}G_{0}(t, s)y(s)\,ds\,dA(t) \\ &{}+ \int_{0}^{1}t^{\alpha-1}h(t) \int _{0}^{1}h(s)u(s)\,dA(s)\,dA(t). \end{aligned}$$
(2.7)
Then, by (2.7) we obtain
$$ \biggl(1- \int_{0}^{1}t^{\alpha-1}h(t)\,dA(t) \biggr) \int _{0}^{1}h(t)u(t)\,dA(t)= \int_{0}^{1} \int_{0}^{1}h(t)G_{0}(t, s)\,dA(t)y(s)\,ds. $$
(2.8)
Let
$$ \Gamma= \int_{0}^{1}t^{\alpha-1}h(t)\,dA(t), \qquad g_{A}(s)= \int _{0}^{1}G_{0}(t, s)h(t)\,dA(t). $$
(2.9)
From (2.8) and (2.9) we deduce that
$$ \int_{0}^{1}h(t)u(t)\,dA(t) =\frac{1}{1-\Gamma} \int _{0}^{1}g_{A}(s)y(s)\,ds. $$
(2.10)
Substituting (2.10) into (2.6), we obtain
$$u(t)= \int_{0}^{1}G_{0}(t, s)y(s)\,ds+ \frac{t^{\alpha-1}}{1-\Gamma} \int _{0}^{1}g_{A}(s)y(s)\,ds = \int_{0}^{1}G(t, s)y(s)\,ds . $$
The proof is complete. □
Lemma 2.4
Let
\(0\leq\Gamma<1\)
and
\(g_{A}(s)\geq 0\)
for
\(s\in[0, 1]\). Then the Green function
\(G(t, s)\)
defined by (2.2) satisfies
-
(1)
\(G:[0, 1]\times[0, 1]\rightarrow[0, \infty)\)
is continuous.
-
(2)
For any
\(t, s \in[0, 1]\), we have
\(c_{0}(t)\phi(s) \leq G(t, s)\leq\phi(s)\), where
$$\begin{aligned}& \phi(s)=\phi_{0}(s)+\frac{g_{A}(s)}{1-\Gamma},\qquad \phi_{0}(s)= \frac {\tau(s)^{\alpha-2}s(1-s)^{\alpha-1}}{\Gamma(\alpha-1)}, \\& \tau(s)=\frac{s}{1-(1-s)^{\frac{\alpha-1}{\alpha-2}}},\qquad c_{0}(t)=\min \biggl\{ \frac{(\alpha-1)^{\alpha-1}t^{\alpha -2}(1-t)}{(\alpha-2)^{\alpha-2}}, t^{\alpha-1} \biggr\} . \end{aligned}$$
-
(3)
Then, taking
\(\theta\in(0, \frac{1}{2})\), for any
\(t\in [\theta, 1-\theta]\)
and
\(s\in[0, 1]\), we have
\(G(t, s)\geq\gamma _{0}\phi(s)\), where
\(\gamma_{0}=\min_{\theta\leq t\leq1-\theta }c_{0}(t)\in(0,1)\).
The proof of Lemma 2.4 is similar to that in [14], so we omit it.
Remark 2.1
By Lemma 2.4 we can obtain that \(\frac{\alpha -2}{\alpha-1}\leq\tau(s)\leq1\), \(\phi_{0}(s)\leq\frac{1}{\alpha-1}\). That is, \(\phi(s)\) is bounded, and there exists a constant \(\overline {g}>0\) such that \(\phi(s)\leq\overline{g}\). It is easy to verify that \(\max_{0\leq t\leq1}c_{0}(t)\leq1\).
Let \(E=C[0, 1]\) and \(\|u\|=\sup_{0\leq t\leq1}\mid u(t)\mid\). Then \((E, \|\cdot\|)\) is a Banach space. Let
$$P=\bigl\{ u\in E:u(t)\geq0, t\in[0, 1]\bigr\} $$
and
$$K=\bigl\{ u\in P:u(t)\geq c_{0}(t)\|u\|, t\in[0, 1]; u(t)\geq \gamma_{0}\| u\|, t\in[\theta, 1-\theta]\bigr\} , $$
and let \(K_{r}=\{u\in K: \|u\|< r \}\). It is easy to see that K is a cone in E and \(\overline{K_{R}}\setminus K_{r}\subset K\subset P\).
Throughout the paper, we need the following conditions:
- (H1):
-
\(A: [0, 1]\rightarrow\mathbb{R}\) is a function of bounded variation, and \(g_{A}(s)\geq0\) for all \(s\in[0, 1]\);
- (H2):
-
\(h\in C(0, 1)\cap L^{1}(0, 1)\), and \(0\leq\Gamma =\int_{0}^{1}t^{\alpha-1}h(s)\,dA(s)<1\);
- (H3):
-
\(p, q:(0, 1)\rightarrow[0, \infty)\) are continuous, \(p(t)\not\equiv0\), \(q(t)\not\equiv0\), \(t\in[0, 1]\), and
$$\int_{0}^{1}\phi(s)p(s)\,ds< +\infty,\qquad \int_{0}^{1}\phi(s)q(s)\,ds< +\infty; $$
- (H4):
-
\(f, g:[0, 1]\times(0, \infty)\rightarrow[0, \infty)\) are continuous, and for any \(0< r< R<+\infty\), \(\lim_{m\rightarrow \infty}\sup_{u\in\overline{K_{R}} \setminus K_{r}}\int_{H(m)}(p(s)f(s, u(s))+q(s)g(s, u(s)))\,ds=0\), where \(H(m)=[0, \frac{1}{m}]\cup [\frac{m-1}{m}, 1]\).
In what follows, let us define the nonlinear operator \(T:\overline {K_{R}} \setminus K_{r}\rightarrow K\) by
$$ (Tu) (t)= \int_{0}^{1}G(t, s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds,\quad t\in[0, 1]. $$
(2.11)
Lemma 2.5
Suppose that (H1)-(H4) hold. Then
\(T:\overline{K_{R}} \setminus K_{r}\rightarrow K\)
is a completely continuous operator and the fixed point of
T
in
\(\overline{K_{R}} \setminus K_{r}\)
is the positive solution to BVP (1.1).
Proof
It follows from (H4) that there exists a natural number \(m\geq2\) such that
$$\sup_{u\in\overline{K_{R}} \setminus K_{r}} \int_{H(m)}p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\,ds< 1. $$
It is easy to see that for any \(u\in\overline{K_{R}} \setminus K_{r}\) and \(t\in[\frac{1}{m}, \frac{m-1}{m}]\), we have \(\lambda_{0}r\leq u(t)\leq R\), where \(\lambda_{0}=\min\{c_{0}(t): \frac{1}{m}\leq t\leq \frac{m-1}{m}\}>0\). Let
$$\begin{aligned}& M_{1}=\max \biggl\{ f(t, x):\frac{1}{m}\leq t\leq \frac{m-1}{m} , \lambda _{0}r\leq x\leq R \biggr\} , \\& M_{2}=\max \biggl\{ g(t, x):\frac{1}{m}\leq t\leq \frac{m-1}{m} , \lambda _{0}r\leq x\leq R \biggr\} . \end{aligned}$$
So, by (H3), (H4), and Remark 2.1 we have
$$\begin{aligned} & \sup_{u\in\overline{K_{R}} \setminus K_{r}} \int_{0}^{1}G(t, s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \\ &\quad\leq\sup_{u\in\overline{K_{R}} \setminus K_{r}} \int_{0}^{1}\phi (s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \\ &\quad\leq\overline{g}\sup_{u\in\overline{K_{R}} \setminus K_{r}} \int _{H(m)}p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\,ds \\ &\qquad{} +\sup_{u\in\overline{K_{R}}\setminus K_{r}} \int_{\frac{1}{m}}^{\frac {m-1}{m}}\phi(s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \\ &\quad\leq\overline{g}+(M_{1}+M_{2}) \int_{\frac{1}{m}}^{\frac{m-1}{m}}\phi (s) \bigl(p(s)+q(s)\bigr)\,ds \\ &\quad\leq\overline{g}+(M_{1}+M_{2}) \int_{0}^{1}\phi(s) \bigl(p(s)+q(s)\bigr)\,ds< +\infty. \end{aligned}$$
This implies that the operator T defined by (2.11) is well defined.
Now, we show that \(T:\overline{K_{R}}\setminus K_{r}\subset K\). For any \(u\in\overline{K_{R}}\setminus K_{r}\), \(t\in[0, 1]\), we have
$$\begin{aligned} (Tu) (t) &= \int_{0}^{1}G(t, s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \\ &\leq \int_{0}^{1}\phi(s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds. \end{aligned}$$
Hence,
$$\|Tu\|\leq \int_{0}^{1}\phi(s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds. $$
On the other hand, by Lemma 2.4 we have
$$(Tu) (t)\geq c_{0}(t) \int_{0}^{1}\phi(s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \geq c_{0}(t)\|Tu\|, \quad t \in[0, 1], $$
and at the same time, we can get
$$(Tu) (t)\geq\gamma_{0}\|Tu\|, \quad t\in[\theta, 1-\theta]. $$
Thus, \(Tu\in K\). Consequently, \(T:\overline{K_{R}}\setminus K_{r}\subset K\). Finally, we prove that T, which maps \(\overline{K_{R}}\setminus K_{r}\) into K, is a completely continuous map. Let \(D\subset\overline {K_{R}}\setminus K_{r}\) be an arbitrary bounded set. Then from the previous proof we know that \(T(D)\) is uniformly bounded.
Next, we show \(T(D)\) is equicontinuous. In fact, for any \(\epsilon>0\), there exists a natural number \(m\geq2\) such that
$$ \sup_{u\in\overline{K_{R}} \setminus K_{r}} \int_{H(m)}p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\,ds< \frac{\epsilon}{4\overline{g}}. $$
(2.12)
Since \(G(t, s)\) is uniformly continuous on \([0, 1]\times[0, 1]\), for the above \(\epsilon>0\), there exists \(\delta>0\) such that for any \(t_{1}, t_{2}\in[0, 1]\), \(|t_{1}-t_{2}|<\delta\), \(s\in[\frac{1}{m}, \frac{m-1}{m}]\),
$$\bigl|G(t_{1}, s)-G(t_{2}, s)\bigr|< \frac{\epsilon}{2(\overline{p}M_{1}+\overline {q}M_{2})}. $$
Consequently, for any \(u\in D\), \(t_{1}, t_{2}\in[0, 1]\), \(|t_{1}-t_{2}|<\delta\), we have
$$\begin{aligned} &\bigl|(Tu) (t_{1})-(Tu) (t_{2})\bigr| \\ &\quad=\biggl| \int_{0}^{1}\bigl(G(t_{1}, s)-G(t_{2}, s)\bigr) \bigl(p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s) \bigr)\bigr)\,ds\biggr| \\ &\quad\leq 2 \int_{H(m)}\phi(s)\bigl[p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \\ &\qquad{}+ \sup_{u\in D} \int_{\frac{1}{m}}^{\frac{m-1}{m}}\bigl|\bigl(G(t_{1}, s)-G(t_{2}, s)\bigr)\bigr|\bigl(p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s) \bigr)\bigr)\,ds \\ &\quad< 2\overline{g}\frac{\epsilon}{4\overline{g}}+\frac{\epsilon }{2(\overline{p}M_{1}+\overline{q}M_{2})}(\overline{p}M_{1}+ \overline {q}M_{2}) \\ &\quad=\epsilon, \end{aligned}$$
where
$$\overline{p}=\max \biggl\{ p(t):\frac{1}{m}\leq t\leq\frac {m-1}{m} \biggr\} , \qquad \overline{q}=\max \biggl\{ q(t):\frac{1}{m}\leq t\leq \frac {m-1}{m} \biggr\} . $$
This shows that \(T(D)\) is equicontinuous. By the Arzela-Ascoli theorem, \(T:D\rightarrow K\) is compact.
Finally, we prove that \(T:\overline{K_{R}}\setminus K_{r}\rightarrow K\) is continuous. Assume that \(u_{0}, u_{n}\in\overline{K_{R}}\setminus K_{r}\) (\(n=0,1,2,\dots\)) and \(\|u_{n}-u_{0}\|\rightarrow0\) (\(n\rightarrow \infty\)). Then \(r\leq\|u_{n}\|\leq R\) and \(r\leq\|u_{0}\|\leq R\). For any \(\epsilon>0\), by (H4) there exists a natural number \(m\geq 2\) such that
$$\sup_{u\in\overline{K_{R}} \setminus K_{r}} \int_{H(m)}p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\,ds< \frac{\epsilon}{\overline{g}}. $$
On the other hand, for any \(t\in[\frac{1}{m}, \frac{m-1}{m}]\), we have \(\gamma_{0}r\leq u_{n}(t)\leq R \) (\(n=0, 1, 2,\dots\)). Since \(f(t, u)\), \(g(t, u)\) are uniformly continuous in \([\frac{1}{m}, \frac{m-1}{m}]\times[\gamma_{0}r, R]\), we have that
$$\begin{aligned}& \lim_{n\rightarrow\infty}\bigl|f\bigl(s, u_{n}(s)\bigr)-f\bigl(s, u_{0}(s)\bigr)\bigr|=0, \\& \lim_{n\rightarrow\infty}\bigl|g\bigl(s, u_{n}(s)\bigr)-g\bigl(s, u_{0}(s)\bigr)\bigr|=0 \end{aligned}$$
uniformly on \([\frac{1}{m}, \frac{m-1}{m}]\). Then the Lebesgue dominated convergence theorem yields that
$$\begin{aligned} &\int_{\frac{1}{m}}^{\frac{m-1}{m}}\phi(s)\bigl[p(s) \bigl(f\bigl(s, u_{n}(s)\bigr)-f\bigl(s, u_{0}(s)\bigr)\bigr)+q(s) \bigl(g \bigl(s, u_{n}(s)\bigr)-g\bigl(s, u_{0}(s)\bigr)\bigr) \bigr]\,ds\rightarrow0, \\ &\quad n\rightarrow\infty. \end{aligned}$$
Thus, for the above \(\epsilon>0\), there exists a natural number N such that for \(n>N\), we have
$$ \int_{\frac{1}{m}}^{\frac{m-1}{m}}\phi(s)\bigl[p(s) \bigl(f\bigl(s, u_{n}(s)\bigr)-f\bigl(s, u_{0}(s)\bigr)\bigr)+q(s) \bigl(g \bigl(s, u_{n}(s)\bigr)-g\bigl(s, u_{0}(s)\bigr)\bigr) \bigr]\,ds< \frac{\epsilon}{2} . $$
(2.13)
It follows from (2.12), (2.13) that for \(n>N\),
$$\begin{aligned} \|Tu_{n}-Tu_{0}\| \leq{}&\sup_{u\in\overline{K_{R}}\setminus K_{r} } \int_{H(m)}\phi (s)\bigl[p(s) \bigl(f\bigl(s, u_{n}(s) \bigr)-f\bigl(s, u_{0}(s)\bigr)\bigr) \\ &{}+q(s) \bigl(g\bigl(s, u_{n}(s)\bigr)-g\bigl(s, u_{0}(s) \bigr)\bigr)\bigr]\,ds \\ &{}+\sup_{u\in\overline {K_{R}}\setminus K_{r}} \int_{\frac{1}{m}}^{\frac{m-1}{m}}\phi (s)\bigl[p(s) \bigl(f\bigl(s, u_{n}(s)\bigr) -f\bigl(s, u_{0}(s)\bigr)\bigr) \\ &{}+q(s) \bigl(g\bigl(s, u_{n}(s)\bigr)-g\bigl(s, u_{0}(s) \bigr)\bigr)\bigr]\,ds \\ < {}& \overline{g}\frac{\epsilon}{2\overline{g}}+\frac{\epsilon}{2} =\epsilon. \end{aligned}$$
This implies that \(T:\overline{K_{R}}\setminus K_{r}\rightarrow K\) is continuous. Thus, \(T:\overline{K_{R}}\setminus K_{r}\rightarrow K\) is completely continuous. It is clear that if u is a fixed point of T in \(\overline{K_{R}}\setminus K_{r}\), then u satisfies (1.1) and is a positive solution of BVP (1.1). □
To prove the main results, we need the following well-known fixed point theorem.
Lemma 2.6
[15]
Let
P
be a positive cone in a Banach space
E, \(\Omega_{1}\)
and
\(\Omega_{2}\)
be two bounded open sets in
E
such that
\(\theta\in\Omega_{1}\)
and
\(\overline{\Omega _{1}}\subset\Omega_{2}\), \(A:P\cap(\overline{\Omega_{2}}\setminus\Omega _{1})\rightarrow P\)
be a completely continuous operator, where
θ
denotes the zero element of
E. Suppose that one of the following two conditions holds:
-
(i)
\(\|Au\|\leq\|u\|\), \(\forall u\in P\cap\partial\Omega _{1}\); \(\|Au\|\geq\|u\|\), \(\forall u\in P\cap\partial\Omega_{2}\);
-
(ii)
\(\|Au\|\geq\|u\|\), \(\forall u\in P\cap\partial\Omega _{1}\); \(\|Au\|\leq\|u\|\), \(\forall u\in P\cap\partial\Omega_{2} \).
Then
A
has a fixed point in
\(P\cap(\overline{\Omega_{2}}\setminus \Omega_{1})\).