Theorem 3.1
Assume that
\((\mathbf{H}_{0})\)-\((\mathbf {H}_{2})\)
hold, then
\(A_{n} : K\rightarrow K \)
is a completely continuous operator for any fixed
\(n \in\mathbb{N}\).
Proof
Let \(\lambda> 0\) and \(n \in\mathbb{N}\) be fixed. For any \(x\in K\) and \(t\in[0, T]\), by Lemma 2.2, we have
$$\begin{aligned} \lambda \int_{0}^{T} G(s, s)f_{n} \bigl(s, x(s) \bigr)\,ds \leq&(A_{n} x) (t)=\lambda \int_{0}^{T} G(t, s)f_{n} \bigl(s, x(s) \bigr)\,ds \\ \leq& \frac{\lambda }{\xi} \int_{0}^{T} G(t, s)f_{n} \bigl(s, x(s) \bigr)\,ds. \end{aligned}$$
This implies that \((A_{n} x)(t)\geq\xi \Vert A_{n} x \Vert \), therefore \(A_{n}(K)\subset K \). By a standard argument, under assumptions \((\mathbf {H}_{0})\)-\((\mathbf{H}_{2})\), we know that \(A_{n} : K\rightarrow K \) is well defined.
Next, for any positive integers \(n, m\in\mathbb{N}\), we define an operator \(A_{n, m}: K\rightarrow X\) by
$$\begin{aligned}& (A_{n, m} x) (t)=\lambda \int_{\frac {1}{m}}^{T-\frac{1}{m}} G(t, s)f_{n} \bigl(s, x(s) \bigr)\,ds,\quad t\in[0, T]. \end{aligned}$$
(11)
In a similar discussion, \(A_{n, m}: K\rightarrow X\) is well defined and \(A_{n,m}(K)\subseteq K\). In what follows, we will prove that \(A_{n, m}:K\to K\) is completely continuous for each \(m\geq1\). Firstly, we show that \(A_{n, m}:K\to K\) is continuous. Let \(x_{\upsilon}, x\in K\) satisfy \(\Vert x_{\upsilon}-x \Vert \to0\) as \(\upsilon\to+\infty\). Notice that \(t\in [\frac{1}{m}, {T-\frac{1}{m}} ]\), \(\vert f_{n}(t, x_{\upsilon}(t))-f_{n}(t, x(t)) \vert \to0\) as \(\upsilon\to+\infty\). Using the Lebesgue dominated convergence theorem, we have
$$\begin{aligned}& \biggl\vert \lambda \int_{\frac {1}{m}}^{T-\frac {1}{m}} G(t, s)f_{n} \bigl(s, x_{\upsilon}(s) \bigr)\,ds-\lambda \int_{\frac {1}{m}}^{T-\frac{1}{m}} G(t, s)f_{n} \bigl(s, x(s) \bigr)\,ds \biggr\vert \\& \quad \leq \frac{\lambda}{\xi} \int_{\frac{1}{m}}^{T-\frac{1}{m}} G(s, s) \bigl\vert f_{n} \bigl(s, x_{\upsilon}(s) \bigr)-f_{n} \bigl(s, x(s) \bigr) \bigr\vert \,ds\to 0,\quad\upsilon\to+\infty. \end{aligned}$$
Therefore
$$\begin{aligned} \Vert A_{n, m}x_{\upsilon}-A_{n, m}x \Vert \leq \frac{\lambda}{\xi} \int_{\frac{1}{m}}^{T-\frac{1}{m}} G(s, s) \bigl\vert f_{n} \bigl(s, x_{\upsilon}(s) \bigr)-f_{n} \bigl(s, x(s) \bigr) \bigr\vert \,ds\to 0,\quad\upsilon\to+\infty. \end{aligned}$$
So, \(A_{n, m}:K \to C[0, T]\) is continuous for any natural numbers n, m. Then \(A_{n, m}:K \to K\) is continuous for any natural numbers n, m.
Let \(D \subset K\) be any bounded set, then for any \(x\in D\), we have \(\Vert x \Vert \leq r\), and then \(0<\xi r\leq x(t)\leq r\) for any \(t\in[0, T]\). By \((\mathbf{H}_{1})\)-\((\mathbf{H}_{2})\), for any \(x\in D\), we have
$$\begin{aligned}& \biggl\vert \lambda \int_{\frac {1}{m}}^{T-\frac {1}{m}} G(t, s)f_{n} \bigl(s, x(s) \bigr)\,ds \biggr\vert \\& \quad \leq \frac{\lambda}{\xi} \int_{\frac{1}{m}}^{T-\frac{1}{m}} G(s, s)\phi(s) \biggl(g \biggl(x(s)+ \frac{1}{n} \biggr)+h \biggl(x(s)+\frac{1}{n} \biggr) \biggr)\,ds \\& \quad \leq \frac{\lambda}{\xi} \int_{\frac{1}{m}}^{T-\frac{1}{m}} G(s, s)\phi(s) \biggl(g \biggl(\xi r+ \frac{1}{n} \biggr)+h \biggl(x(s)+\frac{1}{n} \biggr) \biggr)\,ds \\& \quad \leq \frac{\lambda}{\xi} \int_{\frac{1}{m}}^{T-\frac{1}{m}} G(s, s)\phi(s) \Bigl(g (\xi r )+\max _{y\in[\xi r, r+1]}h(y) \Bigr)\,ds \\& \quad < +\infty. \end{aligned}$$
(12)
So, \(A_{n,m}D\) is bounded in K.
In order to show that \(A_{n,m}\) is a compact operator, we only need to show that \(A_{n,m}D\) is equicontinuous. For any \(\varepsilon> 0\), by the continuity of \(G(t, s)\) on \([0, T]\times[0, T]\), there exists \(\delta > 0\) such that for any \(t_{1}, t_{2} \in[0, T]\), \(s\in [\frac {1}{m}, {T-\frac{1}{m}} ]\), and \(\vert t_{1} - t_{2} \vert < \delta\), we have
$$\bigl\vert G(t_{1}, s)-G(t_{2}, s) \bigr\vert < \varepsilon \biggl(\lambda \int_{\frac{1}{m}}^{T-\frac{1}{m}} \phi(s) \Bigl(g (\xi r )+\max _{y\in[\xi r, r+1]}h(y) \Bigr)\,ds \biggr)^{-1}. $$
Then, for any \(x\in D\), for any \(t_{1}, t_{2} \in[0, T]\), \(s\in [\frac{1}{m}, {T-\frac{1}{m}} ]\), and \(\vert t_{1} - t_{2} \vert < \delta\), we have
$$\begin{aligned}& \bigl\vert (A_{n, m}) (t_{1})-(A_{n, m}) (t_{2}) \bigr\vert \\& \quad \leq \lambda \int_{\frac{1}{m}}^{T-\frac{1}{m}} \bigl\vert G(t_{1}, s)-G(t_{2}, s) \bigr\vert f_{n} \bigl(s, x(s) \bigr)\,ds \\& \quad \leq \lambda \int_{\frac{1}{m}}^{T-\frac{1}{m}} \bigl\vert G(t_{1}, s)-G(t_{2}, s) \bigr\vert \phi(s) \biggl(g \biggl(x(s)+ \frac {1}{n} \biggr)+h \biggl(x(s)+\frac{1}{n} \biggr) \biggr)\,ds \\& \quad \leq \lambda \int_{\frac{1}{m}}^{T-\frac{1}{m}} \bigl\vert G(t_{1}, s)-G(t_{2}, s) \bigr\vert \phi(s) \Bigl(g (\xi r )+\max _{y\in[\xi r, r+1]}h(y) \Bigr)\,ds \\ & \quad < \varepsilon, \end{aligned}$$
which means that \(A_{n, m}D\) is equicontinuous. By the Arzela-Ascoli theorem, \(A_{n, m}D\) is a relatively compact set and so \(A_{n, m}: K\to K\) is a completely continuous operator.
Finally, we show that \(A_{n}: K \to K\) is a completely continuous operator. For any \(t\in[0, T]\) and \(x\in S=\{x\in K, \Vert x \Vert \leq1\}\), by (10), (11), we have
$$\begin{aligned}& \lambda \int_{0}^{\frac{1}{m}} G(t, s)f_{n} \bigl(s, x(s) \bigr)\,ds+\lambda \int_{T-\frac{1}{m}}^{T} G(t, s)f_{n} \bigl(s, x(s) \bigr)\,ds \\ & \quad \leq\frac{\lambda}{\xi} \biggl( \int_{0}^{\frac{1}{m}} + \int_{T-\frac{1}{m}}^{T} \biggr) G(s, s)\phi(s) \biggl(g \biggl(x(s)+\frac{1}{n} \biggr)+h \biggl(x(s)+\frac{1}{n} \biggr) \biggr)\,ds \\ & \quad \leq\frac{\lambda}{\xi} \biggl( \int_{0}^{\frac{1}{m}} + \int_{T-\frac{1}{m}}^{T} \biggr) G(s, s)\phi(s) \biggl(g \biggl( \frac {1}{n} \biggr)+\max_{y\in [\frac{1}{n}, 2 ]}h(y) \biggr)\,ds \\ & \quad \to0,\quad m\to+\infty. \end{aligned}$$
Hence
$$\Vert A_{n}-A_{n, m} \Vert =\sup_{x\in S } \Vert A_{n}x-A_{n, m}x \Vert \to0,\quad m\to+\infty. $$
Therefore, by \(A_{n, m}: K \to K \) is a completely continuous operator, we get that \(A_{n}: K \to K \) is a completely continuous operator. □
Theorem 3.2
Assume that
\((\mathbf {H}_{0})\)-\((\mathbf{H}_{2})\)
hold and
f
satisfies the following condition:
-
\((\mathbf{H}_{3})\)
:
-
There exists
\([a, b]\subset(0, T)\)
such that
$$\lim_{u\to+\infty}\min_{t\in[a, b]}\frac{f(t, u)}{u}=+ \infty. $$
Then there exists
\(\overline{\lambda} > 0\)
such that PBVP (5) has at least one positive solution for any
\(\lambda\in(0, \overline{\lambda})\).
Proof
Choose \(r_{1}>0\), let
$$\overline{\lambda}=\min \biggl\{ 1, \frac{\xi r_{1}}{\int_{0}^{T } G(s, s)\phi(s) (g (\xi r )+\max_{y\in[\xi r_{1}, r_{1}+1]}h (y ) )\,ds} \biggr\} . $$
Let \(K_{r_{1}}=\{x\in K: \Vert x \Vert < r_{1} \}\). For any \(x\in \partial K_{r_{1}}\), \(t\in[0, T]\), by the definition of \(\Vert \cdot \Vert \), we have
$$x(t)\leq \Vert x \Vert \leq r_{1}, \quad\quad x(t)\geq\xi \Vert x \Vert \geq\xi r_{1}. $$
For any \(\lambda\in(0, \overline{\lambda})\), we have
$$\begin{aligned} \bigl\vert (A_{n}x) (t) \bigr\vert =& \biggl\vert \lambda \int_{0}^{T } G(t, s)f_{n} \bigl(s, x(s) \bigr)\,ds \biggr\vert \\ \leq& \frac{\lambda}{\xi} \int_{0}^{T} G(s, s)\phi(s) \biggl(g \biggl(x(s)+ \frac{1}{n} \biggr)+h \biggl(x(s)+\frac{1}{n} \biggr) \biggr)\,ds \\ \leq& \frac{\lambda}{\xi} \int_{0}^{T } G(s, s)\phi(s) \biggl(g \biggl(\xi r+ \frac{1}{n} \biggr)+h \biggl(x(s)+\frac{1}{n} \biggr) \biggr)\,ds \\ \leq& \frac{\lambda}{\xi} \int_{0}^{T } G(s, s)\phi(s) \Bigl(g (\xi r )+\max _{y\in[\xi r_{1}, r_{1}+1]}h (y ) \Bigr)\,ds \\ < &r_{1}. \end{aligned}$$
Thus,
$$\begin{aligned} \Vert A_{n}x \Vert \leq\Vert x\Vert \quad \mbox{for any } x\in \partial K_{r_{1}}. \end{aligned}$$
(13)
On the other hand, by the inequality in \((\mathbf{H}_{3})\), choose \(l>0\) such that \(\lambda l\xi r_{2}\int_{a}^{b} G(s, s)\,ds>1\), then there exists \(N^{*}>0\) such that
$$f(t,u)\geq lu, \quad u\geq N^{*}, \quad t\in[a, b]. $$
Let \(r_{2}>\max \{r_{1}, \frac{N^{*}}{\xi} \}\), \(K_{r_{2}}=\{x\in K: \Vert x \Vert < r_{2}\}\). Take \(q_{1}\equiv 1\in\partial K_{1}=\{x\in K: \Vert x \Vert =1\}\). For any \(x\in \partial K_{r_{2}} \), \(\mu>0\), \(n\in\mathbb{N}\), we will show
$$\begin{aligned} x\neq A_{n}x+\mu q_{1}. \end{aligned}$$
(14)
Otherwise, there exist \(x_{0}\in\partial K_{r_{2}}\) and \(\mu_{0}>0\) such that \(x_{0}= A_{n}x_{0}+\mu_{0} q_{1}\). From \(x_{0}\in\partial K_{r_{2}}\), we know that \(\Vert x_{0} \Vert =r_{2}\). Then, for \(t\in[a,b]\), we have
$$x(t)\geq\xi \Vert x \Vert \geq\xi r_{2}\geq N^{*}. $$
Hence, we conclude that
$$\begin{aligned} x_{0}(t) =&\lambda \int_{0}^{T } G(t, s)f_{n} \bigl(s, x_{0}(s) \bigr)\,ds+\mu_{0} \\ \geq& \lambda \int_{a}^{b} G(s, s)f_{n} \bigl(s, x_{0}(s) \bigr)\,ds+\mu_{0} \\ \geq& \lambda \int_{a}^{b} G(s, s)l\xi r_{2} \,ds+ \mu_{0} \\ \geq& r_{2}+\mu_{0} >r_{2}. \end{aligned}$$
This implies that \(r_{2}>r_{2}\), which is a contradiction. This yields that (14) holds.
It follows from the above discussion, (13), (14), Lemma 2.3 and Theorem 3.1 that, for any \(n\in\mathbb{N}\), \(\lambda\in(0, \overline{\lambda})\), \(A_{n}\) has a fixed point \(x_{n}\in\overline{K}_{r_{2}}\setminus K_{r_{1}}\).
Let \(\{x_{n}\}_{n=1}^{\infty}\) be the sequence of solutions of PBVP (5). It is easy to see that they are uniformly bounded. From \(x_{n}\in\overline {K}_{r_{2}}\setminus K_{r_{1}}\), we know that
$$r_{2}\geq \Vert x_{n} \Vert \geq x_{n}(t) \geq\xi \Vert x_{n} \Vert \geq\xi r_{1}, \quad t\in[0, T]. $$
For any \(\varepsilon> 0\), by the continuity of \(G(t, s)\) on \([0, T]\times[0, T]\), there exists \(\delta _{1} > 0\) such that for any \(t_{1}, t_{2}, s \in[0, T]\), \(\vert t_{1} - t_{2} \vert < \delta_{1}\), we have
$$\bigl\vert G(t_{1}, s)-G(t_{2}, s) \bigr\vert < \varepsilon \biggl(\lambda \int_{0}^{T} \phi(s) \Bigl(g (\xi r_{1} )+\max_{y\in[\xi r_{1}, r_{2}+1]}h(y) \Bigr)\,ds \biggr)^{-1}. $$
Then, for any \(t_{1}, t_{2}, s \in[0, T]\), \(\vert t_{1} - t_{2} \vert < \delta_{1}\), we obtain
$$\begin{aligned}& \bigl\vert x_{n}(t_{1})-x_{n}(t_{2}) \bigr\vert \\& \quad \leq\lambda \int_{0}^{T} \bigl\vert G(t_{1}, s)-G(t_{2}, s) \bigr\vert f_{n} \bigl(s, x_{n}(s) \bigr)\,ds \\& \quad \leq\lambda \int_{0}^{T} \bigl\vert G(t_{1}, s)-G(t_{2}, s) \bigr\vert \phi(s) \biggl(g \biggl(x_{n}(s)+ \frac{1}{n} \biggr)+h \biggl(x_{n}(s)+\frac{1}{n} \biggr) \biggr)\,ds \\& \quad \leq\lambda \int_{0}^{T} \bigl\vert G(t_{1}, s)-G(t_{2}, s) \bigr\vert \phi(s) \Bigl(g (\xi r_{1} )+\max _{y\in[\xi r_{1}, r_{2}+1]}h(y) \Bigr)\,ds \\& \quad < \varepsilon. \end{aligned}$$
(15)
Similarly to (12), together with (15), by the Ascoli-Arzela theorem, the sequence \(\{x_{n}\}_{n=1}^{\infty}\) has a subsequence being uniformly convergent on \([0, T]\). Without loss of generality, we still assume that \(\{x_{n}\}_{n=1}^{\infty}\) itself uniformly converges to x on \([0, T]\). Since \(\{x_{n}\}_{n=1}^{\infty}\in\overline{K}_{r_{2}}\setminus K_{r_{1}}\subset K \), we have \(x_{n}\geq0\). Besides, we have
$$\begin{aligned} x_{n}(t) =&x_{n} \biggl(\frac{1}{2} \biggr)+x'_{n} \biggl(\frac{1}{2} \biggr) \biggl(t- \frac{1}{2} \biggr)- a \int_{\frac{1}{2}}^{t} \biggl(x_{n}(s)-x_{n} \biggl(\frac{1}{2} \biggr) \biggr)\,ds \\ &{}-k^{2} \int_{\frac{1}{2}}^{t} \int_{\frac{1}{2}}^{s} (x_{n}(\varsigma )-\lambda f_{n} \bigl(\varsigma, x_{n}(\varsigma) \bigr) \,d\varsigma \,ds, \quad t\in(0,T). \end{aligned}$$
(16)
Since \(\{x'_{n}(\frac{1}{2})\}_{n=1}^{\infty}\) is bounded, without loss of generality, we may assume \(x'_{n}(\frac{1}{2})\rightarrow c_{0}\) as \(n\rightarrow+\infty\). Then, by (16) and the Lebesgue dominated convergence theorem, we have
$$\begin{aligned} x(t) =&x \biggl(\frac{1}{2} \biggr)+c_{0} \biggl(t- \frac {1}{2} \biggr)- a \int_{\frac{1}{2}}^{t} \biggl(x(s)-x \biggl(\frac{1}{2} \biggr) \biggr)\,ds \\ &{}-k^{2} \int_{\frac{1}{2}}^{t} \int_{\frac{1}{2}}^{s} (x(\varsigma )-\lambda f \bigl( \varsigma, x(\varsigma) \bigr) \,d\varsigma \,ds, \quad t\in(0,T). \end{aligned}$$
(17)
By (17), the direct computation shows that
$$x''(t)+ax'(t)+k^{2}x(t)=\lambda f \bigl(t, x(t) \bigr), \quad t\in(0,T). $$
On the other hand, let \(n\rightarrow+\infty\) in the following boundary conditions:
$$x_{n}(0)=x_{n}(T), \quad\quad x'_{n}(0)=x'_{n}(T). $$
Therefore, we deduce that x is a solution of PBVP (5). The proof is completed. □
Theorem 3.3
Assume that
\((\mathbf {H}_{0})\)-\((\mathbf{H}_{2})\)
hold and
f
satisfies the following condition:
-
\((\mathbf{H}_{4})\)
:
-
There exists
\([c, d]\subset(0, T)\)
such that
$$\liminf_{u\to+\infty}\min_{t\in[c,d]} f(t, u)> \frac{1}{\int_{c}^{d} G(s, s)\,ds},\quad\quad \lim_{u\rightarrow+\infty}\frac{h(u)}{u}=0. $$
Then there exists
\(\overline{\lambda}>0\)
such that PBVP (5) has at least one positive solution for any
\(\lambda\in(\overline{\lambda}, +\infty)\).
Proof
By the first inequality of \((\mathbf{H}_{4})\), we have that there exists \(N_{*}>0\) such that for any \(t\in[c, d]\), \(u>N_{*}\), we have
$$\begin{aligned} f(t, u)>\frac{1}{\int_{c}^{d} G(s, s)\,ds}. \end{aligned}$$
(18)
Select \(\overline{\lambda}=\max \{1, \frac{N_{*}}{\xi} \}\). In the following proof, we suppose \(\lambda>\overline{\lambda}\), choose \(R_{1}=\lambda\), \(K_{R_{1}}=\{x\in K: \Vert x \Vert < R_{1} \}\). For any \(x\in \partial K_{R_{1}}\), \(t\in[c, d]\), we have
$$x(t)\geq\xi \Vert x \Vert \geq\xi R_{1}>N_{*}. $$
Then, by (18), we have
$$\begin{aligned} \bigl\vert (Tx_{n}) (t) \bigr\vert =&\lambda \int_{0}^{T } G(t, s)f_{n} \bigl(s, x(s) \bigr)\,ds \geq\lambda \int_{c}^{d} G(s, s)f_{n} \bigl(s, x(s) \bigr)\,ds \\ \geq& \lambda \int_{c}^{d} G(s, s)\frac{1}{\int_{c}^{d} G(s, s)} \,ds \geq R_{1}. \end{aligned}$$
Therefore, we have
$$\begin{aligned}& \Vert A_{n}x \Vert \geq \Vert x \Vert \quad \mbox{for any } x\in\partial K_{R_{1}}. \end{aligned}$$
(19)
Based on the second inequality in \((\mathbf{H}_{4})\) and the continuity of \(h(u)\) on \([0, +\infty)\), for
$$\overline{c}=\max \biggl\{ 1, \biggl(\frac{4\lambda}{\xi} \int _{0}^{T } G(s, s)\phi(s)\,ds \biggr)^{-1} \biggr\} , $$
there exists \(N^{*}>0\) such that when \(x\geq N^{*}\), for any \(0\leq z\leq x\), we have \(h(z)\leq\overline{c} x\). Select
$$R_{2}\geq \biggl\{ 2, R_{1}, N^{*}, \frac{2\lambda}{\xi} \int_{0}^{T } G(s, s)\phi(s)g (\xi R_{2} ) \,ds \biggr\} . $$
Then, for any \(x\in\partial K_{R_{2}}\), \(t\in[0, +\infty)\), we have
$$x(t)\leq \Vert x \Vert \leq R_{2}, \quad\quad x(t)\geq\xi \Vert x \Vert \geq\xi R_{2}. $$
Hence, we gain
$$\begin{aligned} \bigl\vert (A_{n}x) (t) \bigr\vert =& \biggl\vert \lambda \int_{0}^{T } G(t, s)f_{n} \bigl(s, x(s) \bigr)\,ds \biggr\vert \\ \leq& \frac{\lambda}{\xi} \int_{0}^{T} G(s, s)\phi(s) \biggl(g \biggl(x(s)+ \frac{1}{n} \biggr)+h \biggl(x(s)+\frac{1}{n} \biggr) \biggr)\,ds \\ \leq& \frac{\lambda}{\xi} \int_{0}^{T } G(s, s)\phi(s) \biggl(g \biggl(\xi R_{2}+ \frac{1}{n} \biggr)+h \biggl(x(s)+\frac{1}{n} \biggr) \biggr)\,ds \\ \leq& \frac{\lambda}{\xi} \int_{0}^{T } G(s, s)\phi(s) \bigl(g (\xi R_{2} )+\overline{c}(R_{2}+1) \bigr)\,ds \leq R_{2}. \end{aligned}$$
Thus,
$$\begin{aligned} \Vert A_{n}x \Vert \leq \Vert x \Vert \quad \mbox{for any } x\in\partial K_{R_{2}}. \end{aligned}$$
(20)
It follows from the above discussion, (19), (20), Lemma 2.4 and Theorem 3.1 that, for \(n\in\mathbb{N}\), \(\lambda\in(\overline {\lambda}, +\infty)\), \(A_{n}\) has a fixed point \(x_{n}\in \overline{K}_{R_{2}}\setminus K_{R_{1}}\) satisfying \(R_{1}\leq \Vert x_{n} \Vert \leq R_{2}\). The rest of the proof is similar to Theorem 3.2. That is the proof of Theorem 3.3. □
Corollary 3.1
The conclusion of Theorem
3.3
is valid if
\((\mathbf{H}_{4})\)
is replaced by the following:
-
\((\mathbf{H}^{*}_{4})\)
:
-
There exists
\([c, d]\subset(0, T)\)
such that
$$\liminf_{u\to+\infty}\min_{t\in[c,d]} f(t, u)=+\infty, \quad\quad \lim_{u\rightarrow+\infty}\frac{h(u)}{u}=0. $$
Remark 3.1
From the proof of Theorems 3.2 and 3.3, we can obtain the main results under the condition that the function \(f(t, u) \) not only has singularity on t but also has singularity on u, and we use the approximation method to overcome the difficulty caused by singularity.
Remark 3.2
In this paper, we can get the positive solution of PBVP (5) when the parameter λ is sufficiently large and small; concretely, we can choose \(\lambda\in (0, 1)\) and \(\lambda\in(1, +\infty)\). What is more, the solution x in PBVP (5) satisfies \(x(t)>0\) for any \(t\in[0, T]\).