In this section, we will consider the existence of multiple positive solutions for the BVP (1.2)-(1.3).
Let \(E= \{u(t):u(t)\in C[-\tau_{1},1+\tau_{2}]\}\) denote a real Banach space with the norm \(\|\cdot\|\) defined by \(\|u\|= \max_{-\tau_{1}\leq t\leq1+\tau_{2}} |u(t) |\). Define the cone \(P\subset E\) by \(P = \{ u\in E:u(t)\geq0 \}\). Let
$$\begin{aligned}& K = \bigl\{ u\in P: u(t)\geq\sigma(t)\|u\|, \forall t\in[0,1] \bigr\} , \end{aligned}$$
(3.1)
$$\begin{aligned}& K_{r} =\bigl\{ u\in K:\|u\|< r\bigr\} , \qquad \partial K_{r}=\bigl\{ u\in K:\|u\| =r\bigr\} . \end{aligned}$$
(3.2)
Suppose that \(u(t)\) is a solution of (1.2)-(1.3); according to Lemma 2.3 and Remark 1.1, it can be written as
$$ u(t)=\left \{\begin{array}{l@{\quad}l} u(\tau_{1}, t), & -\tau_{1}\leq t\leq0, \\ \int_{0}^{1}G(t,s)f(s,u(s-\tau_{1}),u(s+\tau_{2}))\,ds, & 0\leq t\leq 1, \\ u(\tau_{2}, t), & 1\leq t\leq1+\tau_{2}, \end{array} \right . $$
where
$$ u(\tau_{1}, t)=\left \{\begin{array}{l@{\quad}l} e^{\frac{a}{b}t} (\frac{1}{b}\int_{t}^{0}e^{-\frac{a}{b}s}\eta (s)\,ds+u(0) ), & -\tau_{1}\leq t\leq0, b\neq0, \\ \frac{\eta(t)}{a}, & -\tau_{1}\leq t\leq0, b=0 \end{array} \right . $$
and
$$ u(\tau_{2}, t)=\left \{\begin{array}{l@{\quad}l} e^{-\frac{c}{d}t} (\frac{1}{d}\int_{1}^{t}e^{\frac{c}{d}s}\xi (s)\,ds+e^{\frac{c}{d}}u(1) ), & 1\leq t\leq1+\tau_{2}, d\neq0, \\ \frac{\xi(t)}{c}, & 1\leq t\leq1+\tau_{2}, d=0. \end{array} \right . $$
Throughout this paper, we assume that \(v_{0}(t)\) is the solution of (1.2)-(1.3) with \(f\equiv0\). Clearly, \(v_{0}(t)\) can be expressed as follows:
$$ v_{0}(t)=\left \{\begin{array}{l@{\quad}l} v_{0}(\tau_{1}, t), & -\tau_{1}\leq t\leq0, \\ 0, & 0\leq t\leq1, \\ v_{0}(\tau_{2}, t), & 1\leq t\leq1+\tau_{2}, \end{array} \right . $$
where
$$ v_{0}(\tau_{1}, t)=\left \{\begin{array}{l@{\quad}l} \frac{e^{\frac{a}{b}t}}{b}\int_{t}^{0}e^{-\frac{a}{b}s}\eta(s)\,ds, & -\tau_{1}\leq t\leq0, b\neq0, \\ \frac{\eta(t)}{a}, & -\tau_{1}\leq t\leq0, b=0 \end{array} \right . $$
and
$$ v_{0}(\tau_{2}, t)=\left \{\begin{array}{l@{\quad}l} \frac{e^{-\frac{c}{d}t}}{d}\int_{1}^{t}e^{\frac{c}{d}s}\xi(s)\,ds, & 1\leq t\leq1+\tau_{2}, d\neq0, \\ \frac{\xi(t)}{c}, & 1\leq t\leq1+\tau_{2}, d=0. \end{array} \right . $$
Obviously, \(v_{0}(t)\geq0\) for each \(t\in[-\tau_{1},1+\tau_{2}]\).
Let \(u(t)\) be a solution of (1.2)-(1.3) and \(v(t)=u(t)-v_{0}(t)\). Noting that \(v(t)\equiv u(t)\) for \(0\leq t\leq1\), we have
$$ v(t)=\left \{\begin{array}{l@{\quad}l} v(\tau_{1}, t), & -\tau_{1}\leq t\leq0, \\ \int_{0}^{1}G(t,s)f(s,(v+v_{0})(s-\tau_{1}),(v+v_{0})(s+\tau_{2}))\,ds, & 0\leq t\leq1, \\ v(\tau_{2}, t), & 1\leq t\leq1+\tau_{2}, \end{array} \right . $$
where
$$\begin{aligned}& (v+v_{0}) (s-\tau_{1})=v(s-\tau_{1})+v_{0}(s- \tau_{1}), \\& (v+v_{0}) (s+\tau_{2})=v(s+\tau_{2})+v_{0}(s+ \tau_{2}), \\& v(\tau_{1}, t)=\left \{\begin{array}{l@{\quad}l} e^{\frac{a}{b}t}v(0), & -\tau_{1}\leq t\leq0, b\neq0, \\ 0, & -\tau_{1}\leq t\leq0, b=0 \end{array} \right . \end{aligned}$$
and
$$ v(\tau_{2}, t)=\left \{\begin{array}{l@{\quad}l} e^{-\frac{c}{d}(t-1)}v(1), & 1\leq t\leq1+\tau_{2}, d\neq0, \\ 0, & 1\leq t\leq1+\tau_{2}, d=0. \end{array} \right . $$
Define an operator \(A:E\rightarrow E\) as follows:
$$ (Av) (t)=\left \{\begin{array}{l@{\quad}l} (B_{1}v)(t), & -\tau_{1}\leq t\leq0, \\ \int_{0}^{1}G(t,s)f(s,(v+v_{0})(s-\tau_{1}),(v+v_{0})(s+\tau_{2}))\,ds, & 0\leq t\leq1, \\ (B_{2}v)(t), & 1\leq t\leq1+\tau_{2}, \end{array} \right . $$
(3.3)
where
$$ (B_{1}v) (t)=\left \{\begin{array}{l@{\quad}l} e^{\frac{a}{b}t}\int_{0}^{1}G(0,s)f(s,(v+v_{0})(s-\tau_{1}),(v+v_{0})(s+\tau _{2}))\,ds, & -\tau_{1}\leq t\leq0, b\neq0, \\ 0, & -\tau_{1}\leq t\leq0, b=0 \end{array} \right . $$
and
$$ (B_{2}v) (t)=\left \{\begin{array}{l@{\quad}l} e^{-\frac{c}{d}}\int_{0}^{1}f(s,(v+v_{0})(s-\tau_{1}),(v+v_{0})(s+\tau_{2})) \\ \quad {}\times G(1,s)\,ds, & 1\leq t\leq1+\tau_{2}, d\neq0, \\ 0, & 1\leq t\leq1+\tau_{2}, d=0. \end{array} \right . $$
It is easy to derive that u is a positive solution of BVP (1.2)-(1.3) if \(v=u-v_{0}\) is a nontrivial fixed point of \(A: K \rightarrow K\), where \(v_{0}\) is defined as before.
Lemma 3.1
\(A:K \rightarrow K\)
defined by (3.3) is completely continuous.
Proof
For \(v\in K\), we find from Lemma 2.4 and the definition of A that \(0\leq(Av)(t)\leq(Av)(0)\), for \(t\in[-\tau_{1},0]\), and \(0\leq(Av)(t)\leq(Av)(1)\) for \(t\in[1,1+\tau_{2}]\). Thus, \(\|Av\|=\|Av\|_{[0,1]}=\max_{0\leq t\leq1}|(Av)(t)|\). It follows from Lemma 2.4 that
$$\begin{aligned} (Av) (t)&\geq\sigma(t)\int_{0}^{1}G(s)f \bigl(s,(v+v_{0}) (s-\tau _{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &\geq\sigma(t)\|Av\|_{[0,1]}=\sigma(t)\|Av\|. \end{aligned}$$
Thus, \(A(K)\subset K\). In addition, since f is continuous, it follows that A is continuous.
Let \(Q_{1} \in K\) be bounded, that is, there exists a positive constant \(M_{1}>0\) such that \(\|v\|_{[-\tau_{1},1+\tau_{2}]}\leq M_{1}\) for all \(v \in Q_{1}\). Then \(\|v+v_{0}\|_{[-\tau_{1},1+\tau_{2}]}\leq M_{1}+M_{0}\triangleq M_{2}\) for \(v\in Q_{1}\), where \(v_{0}\) is defined as before. Define a set \(Q_{2}\subset E\) as follows:
$$ Q_{2}=\bigl\{ \psi\in E:\Vert \psi \Vert _{[-\tau_{1},1+\tau_{2}]}\leq M_{2}\bigr\} . $$
Hence, \(\max_{-\tau_{1}\leq t\leq1+\tau_{2}}|(v+v_{0})(t)|\leq \|v+v_{0}\|_{[-\tau_{1},1+\tau_{2}]}\leq \|v\|_{[-\tau_{1},1+\tau_{2}]}+\|v_{0}\|_{[-\tau_{1},1+\tau_{2}]}\leq M_{2}\). Noting that f is continuous on \([0,1]\times[0,M_{2}]\times[0,M_{2}]\), there exists a constant \(M_{3}>0\) such that on \([0,1]\times[0,M_{2}]\times[0,M_{2}]\),
$$ \bigl\vert f\bigl(t,(v+v_{0}) (t-\tau_{1}),(v+v_{0}) (t+\tau_{2})\bigr)\bigr\vert \leq M_{3}. $$
Therefore,
$$ \|Av\|_{[-\tau_{1},1+\tau_{2}]}=\|Av\|_{[0,1]}\leq M_{3}\int _{0}^{1}G(s)\,ds. $$
Hence, \(A (Q_{1})\) is bounded.
Finally, we show the operator A is equicontinuous. For \(v\in Q_{1}\), we have
$$\begin{aligned}& (Av)'(t)=\int_{0}^{1} \frac{\partial}{\partial t}G(t,s)f\bigl(s,(v+v_{0}) (s-\tau_{1}),(v+v_{0}) (s+\tau_{2})\bigr)\,ds, \quad t\in [0,1], \\& (Av)'(t)=\frac{a}{b}e^{\frac{a}{b}t}\int _{0}^{1}G(0,s)f\bigl(s,(v+v_{0}) (s- \tau_{1}),(v+v_{0}) (s+\tau_{2})\bigr)\,ds,\quad t \in[-\tau_{1},0] \end{aligned}$$
and
$$ (Av)'(t)=-\frac{c}{d}e^{-\frac{c}{d}(t-1)}\int _{0}^{1}G(1,s)f\bigl(s,(v+v_{0}) (s- \tau_{1}),(v+v_{0}) (s+\tau_{2})\bigr)\,ds,\quad t \in[1,1+\tau_{2}]. $$
In the light of \(f\leq M_{3}\) and
$$\begin{aligned} \frac{\partial}{\partial t}G(t,s)&\leq\frac{(q-1)t^{q-2}(1-s)^{q-1}+(q-1)(t-s)^{q-2}}{\Gamma (q)}+\sum _{i=1}^{m-2}\frac{\alpha_{i}(q-1)t^{q-2}}{\Omega \Gamma(p_{i}+q)}g_{i}( \zeta_{i},s) \\ &\leq\frac{1}{\Gamma(q-1)}+\sum_{i=1}^{m-2} \frac{\alpha _{i}(q-1)}{\Omega\Gamma(p_{i}+q)}g_{i}(\zeta_{i},s)\triangleq M', \end{aligned}$$
we have \(\|(Av)'\|\leq M\) for some positive constant M. Thus, for \(-\tau_{1}\leq t_{1}< t_{2}\leq1+\tau_{2}\), we have
$$ \bigl\Vert (Av) (t_{2})-(Av) (t_{1})\bigr\Vert \leq\int _{t_{1}}^{t_{2}}\bigl\vert (Av)'(s)\bigr\vert \,ds\leq M(t_{2}-t_{1})\rightarrow 0, \quad \mbox{as }t_{1}\rightarrow t_{2}. $$
Therefore, for any \(\epsilon>0\), there exists \(\delta=\delta(\epsilon)>0\) which is independent of \(t_{1}\), \(t_{2}\), and v such that \(\|(Av)(t_{2})-(Av)(t_{1})\|\leq\epsilon\), whenever \(|t_{2}-t_{1}|\leq\delta\). Thus, \(A(Q_{1})\) is equicontinuous. In view of the Ascoli-Arzela theorem, we can easily see that \(A:K \rightarrow K\) is a completely continuous operator. The proof is complete. □
Further we make the following assumptions for \(f(t,x,y)\):
- (H1):
-
There exists a constant \(r_{1}>0\) such that \(0\leq x\leq r_{1}+\|v_{0}\|_{[-\tau_{1},0]}\), \(0\leq y\leq r_{1}+\|v_{0}\|_{[1,1+\tau_{2}]}\), and \(0\leq t\leq1\) implies \(f(t,x,y)<\rho_{1} r_{1}\), where \(\rho_{1}=\frac{1}{\int_{0}^{1}G(s)\,ds}\).
- (H2):
-
There exists a constant \(r_{2}>0\) such that \(\sigma _{1}r_{2}\leq x\leq r_{2}\), \(\sigma_{2}r_{2}\leq y\leq r_{2}\), and \(\theta\leq t\leq1-\theta\) implies \(f(t,x,y)>\rho_{2} r_{2}\), where \(\rho_{2}=\frac{1}{\sigma(\theta)\int_{\theta}^{1-\theta}G(s)\,ds}\), \(\sigma_{1}=\min_{\theta-\tau_{1}\leq t\leq1-\theta-\tau_{1} }|\sigma(t)|\), \(\sigma_{2}=\min_{\theta+\tau_{2}\leq t\leq1-\theta+\tau_{2} }|\sigma(t)|\).
- (H3):
-
\(f_{\infty}=\liminf_{x+y\rightarrow +\infty}\min_{t\in[0,1]}\frac{f(t,x,y)}{x+y}=\infty\).
- (H4):
-
\(f^{\infty}=\limsup_{x+y \rightarrow +\infty}\max_{t\in[0,1]}\frac{f(t,x,y)}{x+y}=0\).
- (H5):
-
\(f_{0}=\liminf_{x+y\rightarrow 0}\min_{t\in[0,1]}\frac{f(t,x,y)}{x+y}=\infty\).
- (H6):
-
\(f^{0}=\limsup_{x+y \rightarrow 0}\max_{t\in[0,1]}\frac{f(t,x,y)}{x+y}=0\).
Theorem 3.1
Assume that (H1), (H2), and (H3) are satisfied. If
\(r_{1}>r_{2}>0\), then BVP (1.2)-(1.3) has at least two positive solutions
\(u_{1}\)
and
\(u_{2}\)
such that
$$ 0< r_{2}<\|u_{1}\|_{[0,1]}<r_{1}< \|u_{2}\|_{[0,1]}. $$
Proof
Let \(A:K\rightarrow K\) be the cone preserving completely continuous that is defined by (3.3). Let \(\Phi_{r_{1}}=\{v\in E:\|v\|< r_{1}\}\), then for any \(v\in K\cap\partial\Phi_{r_{1}}\), we get
$$\begin{aligned}& 0\leq(v+v_{0}) (t-\tau_{1})\leq \|v\|+\|v_{0} \|_{[-\tau_{1},1]}=r_{1}+\|v_{0}\|_{[-\tau_{1},0]},\quad 0\leq t \leq1, \\& 0\leq(v+v_{0}) (t+\tau_{2})\leq \|v\|+\|v_{0} \|_{[0,1+\tau_{2}]}=r_{1}+\|v_{0}\|_{[1,1+\tau_{2}]},\quad 0\leq t \leq1. \end{aligned}$$
Thus, from (A3) and (H1), we have
$$\begin{aligned} \|Av\| &=\|Av\|_{[0,1]}=\max_{t\in[0,1]}(Av) (t) \\ & =\max_{t\in[0,1]}\int_{0}^{1}G(t,s)f \bigl(s,(v+v_{0}) (s-\tau _{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &\leq\int_{0}^{1}G(s)f\bigl(s,(v+v_{0}) (s-\tau_{1}),(v+v_{0}) (s+\tau_{2})\bigr)\,ds \\ &< \rho_{1}r_{1}\int_{0}^{1}G(s) \,ds=r_{1}=\|v\|. \end{aligned}$$
Therefore,
$$ \|Av\|< \|v\|, \quad v\in K\cap\partial\Phi_{r_{1}}. $$
(3.4)
Let \(\Phi_{r_{2}}=\{v\in E:\|v\|< r_{2}\}\), then for any \(v\in K\cap\partial\Phi_{r_{2}}\), we have
$$\begin{aligned}& r_{2}=\|v\|\geq v(t-\tau_{1})\geq \sigma(t- \tau_{1})\|v\|\geq\sigma_{1}r_{2},\quad \theta\leq t \leq 1-\theta, \\& r_{2}=\|v\|\geq v(t+\tau_{2})\geq \sigma(t+ \tau_{2})\|v\|\geq\sigma_{2}r_{2},\quad \theta\leq t \leq 1-\theta. \end{aligned}$$
Thus, from (H2) and (A3) of Lemma 2.4, we get
$$\begin{aligned} \|Av\| &=\|Av\|_{[0,1]}=\max_{t\in[0,1]}(Av) (t) \\ & =\max_{t\in[0,1]}\int_{0}^{1}G(t,s)f \bigl(s,(v+v_{0}) (s-\tau _{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ & \geq\min_{t\in[\theta,1-\theta]}\int_{0}^{1}G(t,s)f \bigl(s,(v+v_{0}) (s-\tau_{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &\geq\sigma(\theta)\int_{0}^{1}G(s)f \bigl(s,(v+v_{0}) (s-\tau _{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &\geq\sigma(\theta)\int_{\theta}^{1-\theta }G(s)f \bigl(s,(v+v_{0}) (s-\tau_{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &=\sigma(\theta)\int_{\theta}^{1-\theta}G(s)f\bigl(s,v(s- \tau _{1}),v(s+\tau_{2})\bigr)\,ds \\ &> \rho_{2}r_{2}\sigma(\theta)\int_{\theta}^{1-\theta}G(s) \,ds=r_{2}=\|v\|. \end{aligned}$$
So
$$ \|Av\|>\|v\|, \quad v\in K\cap\partial\Phi_{r_{2}}. $$
(3.5)
Choose \(L>0\) such that
$$ (\sigma_{1}+\sigma_{2})L\sigma(\theta)\int _{\theta}^{1-\theta}G(s)\,ds>1. $$
(3.6)
From (H3), there exists \(R_{1}>0\) such that
$$ f(t,x,y)\geq L(x+y), \quad x, y\geq0, x+y\geq R_{1}, 0 \leq t\leq1. $$
(3.7)
Choose
$$ R_{0}>\max \biggl\{ r_{1},\frac{R_{1}}{2\min\{\sigma_{1},\sigma_{2}\}} \biggr\} . $$
Let \(\Phi_{R}=\{v\in E:\|v\|< R,R\geq R_{0}\}\), then for any \(v\in K\cap\partial\Phi_{R}\), we have
$$\begin{aligned}& v(t-\tau_{1})\geq\sigma(t-\tau_{1})\|v\|\geq \sigma_{1}R\geq\min\{ \sigma_{1},\sigma_{2} \}R_{0}>\frac{1}{2}R_{1},\quad \theta\leq t\leq1- \theta, \\& v(t+\tau_{2})\geq\sigma(t+\tau_{2})\|v\|\geq \sigma_{2}R\geq\min\{ \sigma_{1},\sigma_{2} \}R_{0}>\frac{1}{2}R_{1},\quad \theta\leq t\leq1- \theta. \end{aligned}$$
Then, from (3.6) and (3.7) we have
$$\begin{aligned} \|Av\| &=\|Av\|_{[0,1]}=\max_{t\in[0,1]}(Av) (t) \\ & =\max_{t\in[0,1]}\int_{0}^{1}G(t,s)f \bigl(s,(v+v_{0}) (s-\tau _{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ & \geq\min_{t\in[\theta,1-\theta]}\int_{0}^{1}G(t,s)f \bigl(s,(v+v_{0}) (s-\tau_{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &\geq\sigma(\theta)\int_{0}^{1}G(s)f \bigl(s,(v+v_{0}) (s-\tau _{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &\geq\sigma(\theta)\int_{\theta}^{1-\theta }G(s)f \bigl(s,(v+v_{0}) (s-\tau_{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &=\sigma(\theta)\int_{\theta}^{1-\theta}G(s)f\bigl(s,v(s- \tau _{1}),v(s+\tau_{2})\bigr)\,ds \\ &\geq L\sigma(\theta)\int_{\theta}^{1-\theta}G(s)\bigl[v(s- \tau _{1})+v(s+\tau_{2})\bigr]\,ds \\ &\geq(\sigma_{1}+\sigma_{2})RL\sigma(\theta)\int _{\theta }^{1-\theta}G(s)\,ds>R=\|v\|. \end{aligned}$$
Therefore,
$$ \|Av\|> \|v\|, \quad v\in K\cap\partial\Phi_{R}. $$
(3.8)
Applying Lemma 2.2 to (3.4) and (3.5) yields the result that A has a fixed point \(v_{1}\in K\cap(\overline{\Phi}_{r_{1}}\setminus\Phi_{r_{2}})\) with \(v_{1}(t)\geq \sigma_{1} \|u\|>0\), \(t\in[0,1]\). Similarly, Lemma 2.2 associated with (3.4) and (3.8) shows that A has another fixed point \(v_{2}\in K\cap(\overline{\Phi}_{R}\setminus \Phi_{r_{1}})\) with \(v_{2}(t)\geq\sigma_{2} \|u\|>0\), \(t\in[0,1]\), which means that \(u_{1}(t)=v_{1}(t)+v_{0}(t)\) and \(u_{2}(t)=v_{2}(t)+v_{0}(t)\) are two positive solutions of BVP (1.2)-(1.3). Since
$$ \|u_{i}\|_{[0,1]}=\|v_{i}+v_{0} \|_{[0,1]}=\|v_{i}\|_{[0,1]}=\|Av_{i} \|_{[0,1]}=\| Av_{i}\|=\|v_{i}\|,\quad i=1,2, $$
it follows that \(u_{1}(t)\) and \(u_{2}(t)\) satisfy
$$ 0< r_{2}<\|u_{1}\|_{[0,1]}=\|v_{1} \|<r_{1}<\|v_{2}\|=\|u_{2}\|_{[0,1]}. $$
The proof is complete. □
Theorem 3.2
Assume that (H1), (H3), and (H5) are satisfied. There exist constants
\(R>r_{1}>r>0\), then BVP (1.2)-(1.3) has at least two positive solutions
\(u_{1}\)
and
\(u_{2}\)
such that
$$ 0< r<\|u_{1}\|_{[0,1]}<r_{1}<\|u_{2} \|_{[0,1]}. $$
Proof
Choose \(0< r<r_{1}<R\), let \(\Phi_{r}=\{v\in E:\|v\|< r,r<r_{1}\}\). For the same \(L>0\) satisfying (3.6), (H5) implies that
$$ f(t,x,y)\geq L(x+y), \quad x, y\geq0, x+y\leq2r, 0\leq t\leq1. $$
(3.9)
Then for any \(v\in K\cap\partial\Phi_{r}\), we have
$$\begin{aligned}& r=\|v\|\geq v(t-\tau_{1})\geq\sigma(t-\tau_{1})\|v\|\geq \sigma _{1}r,\quad \theta\leq t\leq1-\theta, \\& r=\|v\|\geq v(t+\tau_{2})\geq\sigma(t+\tau_{2})\|v\|\geq \sigma _{2}r,\quad \theta\leq t\leq1-\theta. \end{aligned}$$
Then, from (3.6) and (3.9) we have
$$\begin{aligned} \|Av\| &=\|Av\|_{[0,1]}=\max_{t\in[0,1]}(Av) (t) \\ & =\max_{t\in[0,1]}\int_{0}^{1}G(t,s)f \bigl(s,(v+v_{0}) (s-\tau _{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &\geq\sigma(\theta)\int_{0}^{1}G(s)f \bigl(s,(v+v_{0}) (s-\tau _{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &\geq\sigma(\theta)\int_{\theta}^{1-\theta }G(s)f \bigl(s,(v+v_{0}) (s-\tau_{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &= \sigma(\theta)\int_{\theta}^{1-\theta}G(s)f\bigl(s,v(s- \tau _{1}),v(s+\tau_{2})\bigr)\,ds \\ &\geq L\sigma(\theta)\int_{\theta}^{1-\theta}G(s)\bigl[v(s- \tau _{1})+v(s+\tau_{2})\bigr]\,ds \\ &\geq(\sigma_{1}+\sigma_{2})rL\sigma(\theta)\int _{\theta }^{1-\theta}G(s)\,ds>r=\|v\|. \end{aligned}$$
Therefore,
$$ \|Av\|> \|v\|, \quad v\in K\cap\partial\Phi_{r}. $$
(3.10)
Applying Lemma 2.2 to (3.4) and (3.10) yields that A has a fixed point \(v_{1}\in K\cap(\overline{\Phi}_{r_{1}}\setminus\Phi_{r})\) with \(v_{1}(t)\geq \sigma_{1} \|u\|>0\), \(t\in[0,1]\). Similarly, Lemma 2.2 associated with (3.4) and (3.8) yields the result that A has another fixed point \(v_{2}\in K\cap(\overline{\Phi}_{R}\setminus \Phi_{r_{1}})\) with \(v_{2}(t)\geq\sigma_{2} \|u\|>0\), \(t\in[0,1]\). This means that \(u_{1}(t)=v_{1}(t)+v_{0}(t)\) and \(u_{2}(t)=v_{2}(t)+v_{0}(t)\) are two positive solutions of BVP (1.2)-(1.3). Since
$$ \|u_{i}\|_{[0,1]}=\|v_{i}+v_{0} \|_{[0,1]}=\|v_{i}\|_{[0,1]}=\|Av_{i} \|_{[0,1]}=\| Av_{i}\|=\|v_{i}\|,\quad i=1,2, $$
it follows that \(u_{1}(t)\) and \(u_{2}(t)\) satisfy
$$ 0< r<\|u_{1}\|_{[0,1]}=\|v_{1}\|<r_{1}< \|v_{2}\|=\|u_{2}\|_{[0,1]}. $$
The proof is complete. □
Theorem 3.3
Assume that (H2), (H4), and (H6) are satisfied. There exist constants
\(R>r_{2}>r>0\), then BVP (1.2)-(1.3) has at least two positive solutions
\(u_{1}\)
and
\(u_{2}\)
such that
$$ 0< r<\|u_{1}\|_{[0,1]}<r_{2}<\|u_{2} \|_{[0,1]}. $$
Proof
Choose \(0< r<r_{2}<R\). By (H4), for any \(0<\varepsilon<\frac{1}{2\int_{0}^{1}G(s)\,ds}\), there exists \(R'>0\) such that
$$ f(t,x,y)\leq\varepsilon(x+y),\quad x, y\geq0, x+y\geq R', 0\leq t \leq1. $$
Putting
$$ C\triangleq\max_{0\leq t\leq1}\max_{0\leq x,y,x+y\leq R'}\bigl\vert f(t,x,y)\bigr\vert +1, $$
then
$$ f(t,x,y)\leq\varepsilon(x+y)+C,\quad x, y\geq0, 0\leq t\leq1. $$
Choose
$$ R_{0}>\bigl(C+2\varepsilon\|v_{0}\|\bigr)\frac{\int_{0}^{1}G(s)\,ds}{1-2\varepsilon\int_{0}^{1}G(s)\,ds}. $$
Let \(\Phi_{R}=\{v\in E: \|v\|< R, R\geq\max\{r_{2},R_{0}\}\}\). Then, for any \(v\in K\cap\partial\Phi_{R}\), we have
$$\begin{aligned} \|Av\| &=\|Av\|_{[0,1]}=\max_{t\in[0,1]}(Av) (t) \\ & =\max_{t\in[0,1]}\int_{0}^{1}G(t,s)f \bigl(s,(v+v_{0}) (s-\tau _{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &\leq\int_{0}^{1}G(s)f\bigl(s,(v+v_{0}) (s-\tau_{1}),(v+v_{0}) (s+\tau_{2})\bigr)\,ds \\ &\leq\int_{0}^{1}G(s) \bigl[\varepsilon \bigl((v+v_{0}) (s-\tau _{1})+(v+v_{0}) (s+ \tau_{2}) \bigr)+C \bigr]\,ds \\ &\leq\int_{0}^{1}G(s) \bigl[2\varepsilon \bigl(\|v\|+ \|v_{0}\| \bigr)+C \bigr]\,ds \\ &=2\varepsilon R\int_{0}^{1}G(s)\,ds+\bigl(C+2 \varepsilon\|v_{0}\|\bigr)\int_{0}^{1}G(s)\,ds < R=\|v\|. \end{aligned}$$
Therefore,
$$ \|Av\|< \|v\|, \quad v\in K\cap\partial\Phi_{R}. $$
(3.11)
Let \(\Phi_{r}=\{v\in E:\|v\|< r,r<r_{2}\}\). By (H6), for any \(0<\varepsilon<\frac{1}{2\int_{0}^{1}G(s)\,ds}\), there exists \(0< r'<r\) such that
$$ f(t,x,y)\leq\varepsilon(x+y),\quad x, y\geq0, x+y< r', 0\leq t\leq1. $$
Putting
$$ C'\triangleq\max_{0\leq t\leq1}\max_{0\leq x,y,r'\leq x+y\leq 2r} \bigl\vert f(t,x,y)\bigr\vert +1, $$
then
$$ f(t,x,y)\leq\varepsilon(x+y)+C', \quad x, y\in \partial \Phi_{r}, 0\leq t\leq1. $$
Choose
$$ r>\bigl(C'+2\varepsilon \Vert v_{0}\Vert \bigr) \frac{\int_{0}^{1}G(s)\,ds}{1-2\varepsilon\int_{0}^{1}G(s)\,ds}. $$
Then, for any \(v\in K\cap\partial\Phi_{r}\), we have
$$\begin{aligned} \|Av\| &=\|Av\|_{[0,1]}=\max_{t\in[0,1]}(Av) (t) \\ & =\max_{t\in[0,1]}\int_{0}^{1}G(t,s)f \bigl(s,(v+v_{0}) (s-\tau _{1}),(v+v_{0}) (s+ \tau_{2})\bigr)\,ds \\ &\leq\int_{0}^{1}G(s)f\bigl(s,(v+v_{0}) (s-\tau_{1}),(v+v_{0}) (s+\tau_{2})\bigr)\,ds \\ &\leq\int_{0}^{1}G(s) \bigl[\varepsilon \bigl((v+v_{0}) (s-\tau _{1})+(v+v_{0}) (s+ \tau_{2}) \bigr)+C' \bigr]\,ds \\ &\leq\int_{0}^{1}G(s) \bigl[2\varepsilon \bigl(\|v\|+ \|v_{0}\| \bigr)+C' \bigr]\,ds \\ &=2\varepsilon r\int_{0}^{1}G(s)\,ds+ \bigl(C'+2\varepsilon\|v_{0}\|\bigr)\int _{0}^{1}G(s)\,ds < r=\|v\|. \end{aligned}$$
So,
$$ \|Av\|< \|v\|, \quad v\in K\cap\partial\Phi_{r}. $$
(3.12)
Applying Lemma 2.2 to (3.5) and (3.12) yields the result that A has a fixed point \(v_{1}\in K\cap(\overline{\Phi}_{r_{1}}\setminus\Phi_{r})\) with \(v_{1}(t)\geq \sigma_{1} \|u\|>0\), \(t\in[0,1]\). Similarly, from Lemma 2.2 associated with (3.5) and (3.11) one derives that A has another fixed point \(v_{2}\in K\cap(\overline{\Phi}_{R}\setminus \Phi_{r_{1}})\) with \(v_{2}(t)\geq\sigma_{2} \|u\|>0\), \(t\in[0,1]\). This means that \(u_{1}(t)=v_{1}(t)+v_{0}(t)\) and \(u_{2}(t)=v_{2}(t)+v_{0}(t)\) are two positive solutions of BVP (1.2)-(1.3). Since
$$ \|u_{i}\|_{[0,1]}=\|v_{i}+v_{0} \|_{[0,1]}=\|v_{i}\|_{[0,1]}=\|Av_{i} \|_{[0,1]}=\| Av_{i}\|=\|v_{i}\|,\quad i=1,2, $$
it follows that \(u_{1}(t)\) and \(u_{2}(t)\) satisfy
$$ 0< r<\|u_{1}\|_{[0,1]}=\|v_{1}\|<r_{2}< \|v_{2}\|=\|u_{2}\|_{[0,1]}. $$
This completes the proof. □
We account for the control functions
$$\begin{aligned}& \varphi(r)=\max\bigl\{ f(t,x,y):(t,x,y)\in[0,1]\times[0,r+r_{0}]\times \bigl[0,r+r'_{0}\bigr]\bigr\} , \\& \psi(r)=\min\bigl\{ f(t,x,y):(t,x,y)\in[\theta,1-\theta]\times[\sigma _{1}r,r]\times[\sigma_{2}r,r]\bigr\} , \end{aligned}$$
where \(r_{0}=\|v_{0}\|_{[-\tau_{1},0]}\), \(r'_{0}=\|v_{0}\|_{[1,1+\tau_{2}]}\).
Theorem 3.4
Suppose that there exist two positive numbers
\(\xi_{2}<\xi_{1}\)
such that one of the following conditions is satisfied:
- (B1):
-
\(\varphi(\xi_{1})< \rho_{1}\xi_{1}\), \(\psi(\xi _{2})>\rho_{2}\xi_{2}\).
- (B2):
-
\(\psi(\xi_{1})>\rho_{2}\xi_{1}\), \(\varphi(\xi_{2})<\rho _{1}\xi_{2}\).
Then BVP (1.2)-(1.3) has at least one positive solution
\(u\in K\)
such that
$$ \xi_{2}<\|u\|_{[0,1]}<\xi_{1}. $$
Proof
Because of the similarity of the proof, we prove only this theorem under condition (B1). By assumption (B1), we have
$$ f(t,x,y)\leq\varphi(\xi_{1})< \rho_{1}\xi_{1}, \qquad f(t,x,y)\geq\psi(\xi_{2})>\rho_{2}\xi_{2}, $$
which are the assumptions (H1) and (H2). By Theorem 3.1, we find that A has a fixed point \(v\in K\cap(\overline{\Phi}_{\xi_{1}}\setminus\Phi_{\xi_{2}})\), which means that (1.2)-(1.3) has at least one positive solution u and \(\xi_{2}<\|u\|_{[0,1]}<\xi_{1}\). This completes the proof. □
Similarly, we can obtain the existence of multiple positive solutions for BVP (1.2)-(1.3).
Theorem 3.5
Suppose that there exist three positive numbers
\(\xi_{3}<\xi_{2}<\xi_{1}\)
such that one of the following conditions is satisfied:
- (B3):
-
\(\varphi(\xi_{1})<\rho_{1}\xi_{1}\), \(\psi(\xi_{2})>\rho _{2}\xi_{2}\), \(\varphi(\xi_{3})<\rho_{1}\xi_{3} \).
- (B4):
-
\(\psi(\xi_{1})>\rho_{2}\xi_{1}\), \(\varphi(\xi_{2})<\rho _{1}\xi_{2}\), \(\psi(\xi_{3})>\rho_{2}\xi_{3} \).
Then BVP (1.2)-(1.3) has at least two positive solutions
\(u_{1}, u_{2}\in K\)
such that
$$ \xi_{3}<\|u_{1}\|_{[0,1]}<\xi_{2}< \|u_{2}\|_{[0,1]}<\xi_{1}. $$
Theorem 3.6
Suppose that there exist four positive numbers
\(\xi_{4}<\xi_{3}<\xi_{2}<\xi_{1}\)
such that one of the following conditions is satisfied:
- (B5):
-
\(\varphi(\xi_{1})<\rho_{1}\xi_{1}\), \(\psi(\xi_{2})>\rho _{2}\xi_{2}\), \(\varphi(\xi_{3})<\rho_{1}\xi_{3}\), \(\psi(\xi_{4})>\rho_{2}\xi _{4} \).
- (B6):
-
\(\psi(\xi_{1})>\rho_{2}\xi_{1}\), \(\varphi(\xi_{2})<\rho _{1}\xi_{2}\), \(\psi(\xi_{3})>\rho_{2}\xi_{3}\), \(\varphi(\xi_{4})<\rho _{1}\xi_{4}\).
Then BVP (1.2)-(1.3) has at least three positive solutions
\(u_{1}, u_{2}, u_{3}\in K\)
such that
$$ \xi_{4}<\|u_{1}\|_{[0,1]}<\xi_{3}< \|u_{2}\|_{[0,1]}<\xi_{2}<\|u_{3}\| _{[0,1]}<\xi_{1}. $$
Theorem 3.7
Suppose that there exist
\(n+1\)
positive numbers
\(\xi_{n+1}<\xi_{n}<\cdots<\xi_{2}<\xi_{1}\)
such that one of the following conditions is satisfied:
- (B7):
-
\(\varphi(\xi_{2k-1})< \rho_{1}\xi_{2k-1}\), \(\psi (\xi_{2k})>\rho_{2}\xi_{2k}\), \(k=1,2,\ldots, [\frac {n+2}{2} ]\).
- (B8):
-
\(\psi(\xi_{2k-1})>\rho_{2}\xi_{2k-1}\), \(\varphi (\xi_{2k})<\rho_{1}\xi_{2k}\), \(k=1,2,\ldots, [\frac {n+2}{2} ] \).
Then BVP (1.2)-(1.3) has at least
n
positive solutions
\(u_{i}\in K\) (\(i=1,2,\ldots,n\)) such that
$$ \xi_{i+1}<\|u_{i}\|_{[0,1]}<\xi_{i}. $$