In this section, we apply Lemma 2.4 and Lemma 2.5 to establish the existence of triple positive solutions for problem (1.1). We consider the following three cases for \(\omega\in L^{p}[0,1]\): \(p> 1\), \(p=1\), and \(p=\infty\). Case \(p>1\) is treated in the following theorem.
For convenience, we introduce the following notation:
$$\begin{aligned}& D=\gamma\gamma_{1}\|e\|_{q}\|\omega\|_{p}, \qquad D_{1}=\frac{m}{1-\nu},\\& \delta_{1}=\min _{t\in J_{\delta},s\in(0,1)}H_{1s}'(t,s) (1-\nu), \qquad \delta^{*}=\min\biggl\{ \frac{\delta}{\gamma_{1}},\delta_{1}\biggr\} , \\& f^{\infty}=\limsup_{x \rightarrow\infty}\max_{t\in J} \frac{f(t,x)}{x},\qquad I^{\infty}(k)=\limsup_{x \rightarrow\infty}\max _{t\in J} \frac{I_{k}(t,x)}{x},\quad k=1,2,\ldots,m. \end{aligned}$$
Theorem 3.1
Assume that (H1)-(H3) hold. Furthermore, suppose that there exist constants
\(0< d<a<\frac{a}{\delta^{*}}\leq c\)
such that
- (H4):
-
\(f^{\infty}<\frac{1}{2D}\), \(I^{\infty}(k)<\frac {1}{2D_{1}}\), \(k=1,2,\ldots,m\);
- (H5):
-
\(f(t,x)> \frac{3a}{\delta^{2}(1-2\delta)n}\)
for
\((t,x)\in J_{\delta}\times[a,\frac{a}{\delta^{*}}]\);
- (H6):
-
\(f(t,x)<\frac{d}{2D}\), \(I_{k}(t,x)<\frac{d}{2D_{1}}\)
for
\((t,x)\in J\times[0,d]\), \(k=1,2,\ldots,m\).
Then problem (1.1) has at least three positive solutions
\(x_{1}\), \(x_{2}\), and
\(x_{3}\)
such that
$$\|x_{1}\|< d, \qquad a<\beta(x_{2}), \qquad \textit{and} \quad x_{3}>d \quad\textit{with }\beta(x_{3})<a. $$
Proof
By the definition of operator T and its properties, it suffices to show that the conditions of Lemma 2.4 hold with respect to T.
Let \(\beta(x)=\min_{t\in J_{\delta}}x(t)\). Then \(\beta(x)\) is a nonnegative continuous concave functional on the cone K satisfying \(\beta(x)\leq\|x\|\) for all \(x\in K\).
For convenience, we denote \(b=\frac{a}{\delta^{*}}\).
Considering (H4), there exist \(0<\sigma<\frac{1}{2D}\), \(0<\sigma _{1}<\frac{1}{2D_{1}}\), and \(l>0\) such that
$$f(t,x)\leq\sigma x,\qquad I_{k}(t,x)\leq\sigma_{1} x,\quad k=1,2,\ldots,m, \forall t\in J, x\geq l. $$
Let
$$\eta=\max_{0\leq x\leq l, t\in J}f(t,x),\qquad \eta_{1}=\max _{0\leq x\leq l, t\in J}I_{k}(t,x),\quad k=1,2,\ldots,m. $$
Then
$$ f(t,x)\leq\sigma x+\eta, \qquad I_{k}(t,x)\leq\sigma_{1} x+ \eta_{1}, \quad \forall t\in J, 0\leq x\leq+\infty. $$
(3.1)
Set
$$c>\max \biggl\{ \frac{2D\eta}{1-2D\sigma},\frac{2D_{1}\eta _{1}}{1-2D_{1}\sigma_{1}},\frac{a}{\delta^{*}} \biggr\} . $$
Then, for \(x\in\bar{K}_{c}\), it follows from (2.19), (2.22), and (3.1) that
$$\begin{aligned} (Tx) (t)&=\int_{0}^{1}\int_{0}^{1}H_{1}(t,s)H(s, \tau)\omega(\tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds+\sum _{k=1}^{m}H'_{1s}(t,t_{k})I_{k} \bigl(t_{k},x(t_{k})\bigr) \\ &\leq\int_{0}^{1}\int_{0}^{1} \gamma_{1}\gamma e(\tau)\omega(\tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds+ \frac{1}{1-\nu}\sum_{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k})\bigr) \\ &\leq\int_{0}^{1}\int_{0}^{1} \gamma_{1}\gamma e(\tau)\omega(\tau) (\sigma x+\eta)\,d\tau \,ds+ \frac{1}{1-\nu}\sum_{k=1}^{m}( \sigma_{1} x+\eta_{1}) \\ &\leq\int_{0}^{1}\int_{0}^{1} \gamma_{1}\gamma e(\tau)\omega(\tau) \bigl(\sigma\|x\|+\eta\bigr)\,d\tau \,ds+ \frac{1}{1-\nu}\sum_{k=1}^{m}\bigl( \sigma_{1} \|x\|+\eta_{1}\bigr) \\ &\leq\gamma_{1}\gamma(\sigma c+\eta)\int_{0}^{1} e(\tau)\omega(\tau)\,d\tau+\frac{m}{1-\nu}(\sigma_{1} c+ \eta_{1}) \\ &\leq\gamma_{1}\gamma(\sigma c+\eta)\|e\|_{q}\| \omega \|_{p} +\frac{m}{1-\nu}(\sigma_{1} c+\eta_{1}) \\ &<\frac{c}{2}+\frac{c}{2} = c, \end{aligned}$$
which shows that \(Tx\in K_{c}\).
Hence, we have shown that if (H4) holds, then T maps \(\bar{K}_{c}\) into \(K_{c}\).
Next, we verify that \(\{x\in K(\beta,a,b):\beta(x)>a\}\neq\emptyset\) and \(\beta(Tu)>a\) for all \(x\in K(\beta,a,b)\).
Take \(\varphi_{0}(t)\equiv\frac{\delta^{*}+1}{2\delta^{*}}a\), for \(t\in J\). Then
$$\varphi_{0}\in \biggl\{ x\Bigm| x\in K\biggl(\beta, a, \frac{a}{\delta^{*}}\biggr), \beta(x)>a \biggr\} . $$
This shows that
$$\bigl\{ x\in K(\beta,a,b):\beta(x)>a\bigr\} \neq\emptyset. $$
Therefore, it follows from (H5) that
$$\begin{aligned} \beta(Tx)&=\min_{t\in J_{\delta}}(Tx) (t) \\ &=\min_{t\in J_{\delta}}\int_{0}^{1}\int _{0}^{1}H_{1}(t,s)H(s, \tau)\omega(\tau)f \bigl(\tau,x(\tau)\bigr)\,d\tau \,ds+\sum_{k=1}^{m}H'_{1s}(t,t_{k})I_{k} \bigl(t_{k},x(t_{k})\bigr) \\ &\geq\min_{t\in J_{\delta}}\int_{0}^{1}\int _{0}^{1}H_{1}(t,s)H(s, \tau)\omega(\tau)f \bigl(\tau,x(\tau)\bigr)\,d\tau \,ds \\ &\geq\delta\int_{0}^{1}\int_{0}^{1}e(s)H(s, \tau)\omega(\tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds \\ &\geq\delta\int_{0}^{1}\int_{\delta}^{1-\delta}e(s)H(s, \tau)\omega(\tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds \\ &\geq\delta^{2}\int_{0}^{1}s^{2}\,ds \int_{\delta}^{1-\delta }\omega(\tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \\ &>\frac{1}{3}\delta^{2}n(1-2\delta)\frac{3a}{\delta ^{2}(1-2\delta)n} \\ &=a. \end{aligned}$$
If \(x\in\bar{K}_{d}\), then it follows from (H6) that
$$\begin{aligned} (Tx) (t)&=\int_{0}^{1}\int_{0}^{1}H_{1}(t,s)H(s, \tau)\omega(\tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds+\sum _{k=1}^{m}H'_{1s}(t,t_{k})I_{k} \bigl(t_{k},x(t_{k})\bigr) \\ &\leq\int_{0}^{1}\int_{0}^{1} \gamma_{1}\gamma e(\tau)\omega(\tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds+ \frac{1}{1-\nu}\sum_{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k})\bigr) \\ &< \int_{0}^{1}\int_{0}^{1} \gamma_{1}\gamma e(\tau )\omega(\tau)\frac{d}{2D}\,d\tau \,ds+ \frac{1}{1-\nu}\sum_{k=1}^{m} \frac {d}{2D_{1}} \\ &= d. \end{aligned}$$
Finally, we assert that if \(x\in K(\beta,a,c)\) and \(\|Tx\|>b\), then \(\beta(Tx)>a\).
Suppose \(x\in K(\beta,a,c)\) and \(\|Tx\|>b\), then it follows from (2.18), (2.20), and (2.23) that
$$\begin{aligned} \beta(Tx)={}&\min_{t\in J_{\delta}}(Tx) (t) \\ ={}&\min_{t\in J_{\delta}} \Biggl[\int_{0}^{1} \int_{0}^{1}H_{1}(t,s)H(s, \tau)\omega( \tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds+\sum_{k=1}^{m}H'_{1s}(t,t_{k})I_{k} \bigl(t_{k},x(t_{k})\bigr) \Biggr] \\ \geq{}&\delta\int_{0}^{1}\int _{0}^{1}e(s)H(s, \tau)\omega(\tau)f\bigl(\tau,x( \tau)\bigr)\,d\tau \,ds \\ &{}+\min_{t\in J_{\delta}}\sum_{k=1}^{m}H'_{1s}(t,t_{k}) (1-\nu)\frac{1}{1-\nu}I_{k}\bigl(t_{k},x(t_{k}) \bigr) \\ \geq&{}\frac{\delta}{\gamma_{1}}\int_{0}^{1}\int _{0}^{1}\gamma_{1}e(s)H(s, \tau)\omega( \tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds+\delta_{1}\sum _{k=1}^{m}\frac {1}{1-\nu}I_{k} \bigl(t_{k},x(t_{k})\bigr) \\ \geq{}&\min\biggl\{ \frac{\delta}{\gamma_{1}},\delta_{1}\biggr\} \Biggl[\int _{0}^{1}\int_{0}^{1} \gamma_{1}e(s)H(s, \tau)\omega(\tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds+\! \sum_{k=1}^{m}\frac{1}{1-\nu }I_{k} \bigl(t_{k},x(t_{k})\bigr)\! \Biggr] \\ \geq&{}\delta^{*}\|Tx\| \\ >{}&a. \end{aligned}$$
To sum up, the hypotheses of Lemma 2.5 hold. Therefore, an application of Lemma 2.5 implies problem (1.1) has at least three positive solutions \(x_{1}\), \(x_{2}\), and \(x_{3}\) such that
$$\|x_{1}\|< d,\qquad a<\beta(x_{2}), \quad \mbox{and} \quad x_{3}>d \quad\mbox{with }\beta(x_{3})<a. $$
□
The following theorem deals with the case \(p=\infty\).
Corollary 3.1
Assume that (H1)-(H6) hold. Then problem (1.1) has at least three positive solutions
\(x_{1}\), \(x_{2}\), and
\(x_{3}\)
such that
$$\|x_{1}\|< d, \qquad a<\beta(x_{2}), \quad \textit{and}\quad x_{3}>d \quad\textit{with }\beta(x_{3})<a. $$
Proof
Let \(\|e\|_{1}\|\omega\|_{\infty}\) replace \(\|e\|_{p}\| \omega\|_{q}\) and repeat the argument above. □
Finally we consider the case of \(p=1\). Let
-
\((\mathrm{H}_{4})'\)
:
-
\(f^{\infty}<\frac{1}{D'}\), \(I^{\infty}(k)<\frac {1}{D_{1}}\), \(k=1,2,\ldots,m\);
-
\((\mathrm{H}_{6})'\)
:
-
\(f(t,x)\leq\frac{d}{2D'}\), \(I_{k}(t,x)\leq\frac {d}{2D_{1}}\) (\(k=1,2,\ldots,m\)) for \((t,x)\in J\times[0,d]\),
where
$$D'=\gamma\gamma_{1}\|\omega\|_{1}. $$
Corollary 3.2
Assume that (H1)-(H3), \((\mathrm{H}_{4})'\), (H5), and
\((\mathrm{H}_{6})'\)
hold. Then problem (1.1) has at least three positive solutions
\(x_{1}\), \(x_{2}\), and
\(x_{3}\)
such that
$$\|x_{1}\|< d, \qquad a<\beta(x_{2}), \quad\textit{and}\quad x_{3}>d \quad\textit{with }\beta(x_{3})<a. $$
Proof
Set
$$c'>\max \biggl\{ \frac{2D'\eta}{1-2D'\sigma'},\frac{2D_{1}\eta _{1}}{1-2D_{1}\sigma_{1}}, \frac{a}{\delta^{*}} \biggr\} , $$
where \(0<\sigma'<\frac{1}{2D'}\). Then, for \(x\in\bar{K}_{c'}\), it follows from (2.19), (2.22), and (3.1) that
$$\begin{aligned} (Tx) (t)&=\int_{0}^{1}\int_{0}^{1}H_{1}(t,s)H(s, \tau)\omega(\tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds+\sum _{k=1}^{m}H'_{1s}(t,t_{k})I_{k} \bigl(t_{k},x(t_{k})\bigr) \\ &\leq\int_{0}^{1}\int_{0}^{1} \gamma_{1}\gamma e(\tau)\omega(\tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds+ \frac{1}{1-\nu}\sum_{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k})\bigr) \\ &\leq\int_{0}^{1}\int_{0}^{1} \gamma_{1}\gamma e(\tau)\omega(\tau) (\sigma x+\eta)\,d\tau \,ds+ \frac{1}{1-\nu}\sum_{k=1}^{m}( \sigma_{1} x+\eta_{1}) \\ &\leq\int_{0}^{1}\int_{0}^{1} \gamma_{1}\gamma e(\tau)\omega(\tau) \bigl(\sigma\|x\|+\eta\bigr)\,d\tau \,ds+ \frac{1}{1-\nu}\sum_{k=1}^{m}\bigl( \sigma_{1} \|x\|+\eta_{1}\bigr) \\ &\leq\gamma_{1}\gamma\bigl(\sigma c'+\eta\bigr)\int _{0}^{1}\omega(\tau)\,d\tau+\frac{m}{1-\nu}\bigl( \sigma_{1} c'+\eta_{1}\bigr) \\ &\leq\gamma_{1}\gamma\bigl(\sigma c'+\eta\bigr)\|\omega\| _{1} +\frac{m}{1-\nu}\bigl(\sigma_{1} c'+ \eta_{1}\bigr) \\ &<\frac{c'}{2}+\frac{c'}{2} =c', \end{aligned}$$
which shows that \(Tx\in K_{c'}\).
Hence, we have shown that if \((\mathrm{H}_{4})'\) holds, then T maps \(\bar{K}_{c'}\) into \(K_{c'}\).
If \(x\in\bar{K}_{d}\), then it follows from \((\mathrm{H}_{6})'\) that
$$\begin{aligned} (Tx) (t)&=\int_{0}^{1}\int_{0}^{1}H_{1}(t,s)H(s, \tau)\omega(\tau)f\bigl(\tau,x(\tau)\bigr)\,d\tau \,ds+\sum _{k=1}^{m}H'_{1s}(t,t_{k})I_{k} \bigl(t_{k},x(t_{k})\bigr) \\ &\leq\int_{0}^{1}\int_{0}^{1} \gamma_{1}\gamma e(\tau)\omega(\tau)\frac{d}{2D'}\,d\tau \,ds+ \frac{1}{1-\nu}\sum_{k=1}^{m} \frac{d}{2D_{1}} \\ &\leq\gamma_{1}\gamma\frac{d}{2D'}\int_{0}^{1} \omega(\tau)\,d\tau+\frac{1}{1-\nu}\sum_{k=1}^{m} \frac {d}{2D_{1}} \\ &= d. \end{aligned}$$
Similar to the proof of Theorem 3.1, one can find the results of Corollary 3.2. □
We remark that the condition (H6) in Theorem 3.1 can be replaced by the following condition:
-
\((\mathrm{H}_{6})''\)
:
-
\(f_{0}^{d}\leq\frac{1}{2D}\), \(I_{0}^{d}(k)\leq\frac {1}{2D_{1}}\), \(k=1,2,\ldots,m\), where
$$f_{0}^{d}=\max \biggl\{ \max_{t\in J} \frac{f(t,x)}{d}:x\in [0,d] \biggr\} ,\qquad I_{0}^{d}(k)= \max \biggl\{ \max_{t\in J}\frac {I_{k}(t,x)}{d}:x\in [0,d] \biggr\} . $$
-
\((\mathrm{H}_{6})'''\)
:
-
\(f^{0}\leq\frac{1}{2D}\), \(I^{0}(k)\leq\frac {1}{2D_{1}}\), \(k=1,2,\ldots,m\).
Corollary 3.3
If the condition (H6) in Theorem
3.1
is replaced by
\((\mathrm{H}_{6})''\)
or
\((\mathrm{H}_{6})'''\), respectively, then the conclusion of Theorem
3.1
also holds.
Proof
It follows from the proof of Theorem 3.1 that Corollary 3.3 holds. □
Remark 3.1
Comparing with Zhang and Ge [45], the main features of this paper are as follows.
-
(i)
Triple positive solutions are available.
-
(ii)
\(I_{k}\neq0\) (\(k=1, 2, \ldots, m\)) is considered.
-
(iii)
\(\omega(t)\) is \(L^{p}\)-integrable, not only \(\omega(t)\in C(0,1)\) for \(t\in J\).