### Lemma 3.1

*Assume that* (H_{1})-(H_{2}) *holds*. *Then there exist a solution*
\(x=A_{1}\in(0,\pi)\)
*such that*

$$ g_{1}(x):=\frac{\alpha[1-\cos(\eta x)]}{x\sin x}=1 $$

(3.1)

*and a solution*
\(x=A_{2}\in(0,\pi)\)
*such that*

$$ g_{2}(x):=\frac{\alpha\eta\sin(\eta x)}{2\sin x}=1. $$

(3.2)

### Proof

It is not difficult to show that

$$\lim_{x\rightarrow0^{+}}g_{1}(x)=\frac{\alpha\eta^{2} }{2}< 1, \qquad \lim _{x\rightarrow\pi^{-}}g_{1}(x)=\infty>1. $$

Since the function \(g_{1}(x)\) is continuous on \((0,\pi)\), there must exist a constant \(A_{1}\in(0,\pi)\) such that \(g_{1}(A_{1})=1\).

Similarly,

$$\lim_{x\rightarrow0^{+}}g_{2}(x)=\frac{\alpha\eta^{2}}{2}< 1, \qquad \lim _{x\rightarrow\pi^{-}}g_{2}(x)=\infty>1. $$

Thus, there exists a positive constant \(A_{2}\in(0,\pi)\) such that \(g_{2}(A_{2})=1\). □

### Theorem 3.1

*Assume that* (H_{1})-(H_{2}) *holds*. *Suppose one of the following conditions holds*:

$$(\mathrm{i}) \quad 0\leq \bar{f}_{0}< \frac{\underline {A}^{2}}{a^{L}} ,\qquad \underline{f}_{\infty}> \frac {\bar{A}^{2}}{a^{l}}; \qquad (\mathrm{ii})\quad 0\leq \bar{f}_{\infty}< \frac{\underline{A}^{2}}{a^{L}}, \qquad\underline {f}_{0}> \frac{\bar{A}^{2}}{a^{l}}. $$

*Then problem* (1.1)-(1.2) *has at least one positive solution*, *where*

$$\underline{A}=\min\{A_{1},A_{2}\},\qquad \bar{A}=\max \{A_{1},A_{2}\}, $$

*and*
\(A_{1}\), \(A_{2}\)
*is defined in* (3.1) *and* (3.2), *respectively*.

### Proof

(i) Since \(0\leq \bar{f}_{0}< \frac{\underline{A}^{2}}{a^{L}}\), there exists a positive number *r* such that

$$ \frac{f(u)}{u}< \frac{{\underline{A}}^{2}}{a^{L}}\leq\frac{{A_{1}}^{2}}{a^{L}}, \quad 0< u\leq r. $$

(3.3)

Let \(0< m_{1}^{*}< r\), then from the Sturm comparison theorem and the concavity of \(u(t,m_{1}^{*})\), it follows that \(0\leq u(t,m_{1}^{*})\leq m_{1}^{*}t\leq m_{1}^{*}< r\) for \(t\in[0,1]\). Thus

$$ 0\leq a(t)f \bigl(u \bigl(t,m_{1}^{*} \bigr) \bigr)< a^{L} \frac{A_{1}^{2}}{a^{L}}u \bigl(t,m_{1}^{*} \bigr)=A_{1}^{2}u\bigl(t,m_{1}^{*}\bigr)< \pi ^{2}u \bigl(t,m_{1}^{*} \bigr),\quad t\in(0,1]. $$

By Lemma 2.3, it gives \(u(t,m_{1}^{*})>0\) for \(t\in(0,1]\).

Let \(Z(t)=(m_{1}^{*}/A_{1})\sin(A_{1} t)\) for \(t\in[0,1]\), then

$$ Z^{\prime\prime}(t)+A_{1}^{2}Z(t)=0,\qquad Z(0)=0,\qquad Z'(0)=m_{1}^{*}. $$

(3.4)

From Lemma 2.2 and Lemma 3.1, we have

$$ k \bigl(m_{1}^{*} \bigr)= \frac{\alpha\int_{0}^{\eta}u(s,m_{1}^{*})\,ds}{u(1,m_{1}^{*})} < \frac{\alpha\int_{0}^{\eta}m_{1}^{*}\sin (A_{1}s)\,ds}{m_{1}^{*}\sin A_{1}}= \frac{\alpha[1-\cos(\eta A_{1})]}{A_{1}\sin A_{1}}=1, $$

(3.5)

that is, \(\varphi(m_{1}^{*})\leq0\).

On the other hand, the second inequality in (i) implies that there exists a number *L* large enough such that

$$ \frac{f(u)}{u}>\frac{{\bar{A}}^{2}}{a^{l}}\geq\frac{A_{2}^{2}}{a^{l}}, \quad u\geq L, $$

(3.6)

and there exists a positive number \(\epsilon< A_{2}(1-\eta)/\eta\) small enough that

$$ \frac{f(u)}{u}\geq\frac{(A_{2}+\epsilon)^{2}}{a^{l}},\quad u\geq L. $$

(3.7)

Next, we will find a positive number \(m_{2}^{*}\) such that \(\varphi (m_{2}^{*})\geq0\).

*Claim*. There exist a slope \(m_{2}^{*}\) and two positive numbers *ρ* and *σ* such that

$$0< \rho\leq\eta\leq\frac{A_{2}}{A_{2}+\epsilon} \leq\sigma\leq1 \quad \mbox{and}\quad u \bigl(t,m_{2}^{*} \bigr)\geq L \quad\mbox{for } t\in[\rho, \sigma]. $$

Since the solution \(u(t,m)\) is concave, it hits the line \(u=L\) at most two times for the constant *L* defined in (3.6) and \(t\in(0,1]\). We denote the left intersecting time by \(\underline{\delta}_{m}\) and the right one by \(\overline{\delta}_{m}\) provided they exist. Henceforth, denote \(I_{m}=[\underline{\delta}_{m},\overline{\delta}_{m}]\subseteq (0,1]\). If \(u(1,m)\geq L\), then \(\overline{\delta}_{m}=1\).

The discussion is divided into three steps.

*Step* 1. We claim that there exists a slope \(m_{0}\) large enough such that \(0\leq u(t,m_{0})\leq L\) for \(t\in[0,\underline{\delta }_{m_{0}}]\) and \(u(t,m_{0})\geq L\) for \(t\in I_{m_{0}}\).

Otherwise, provided \(u(t,m)\leq L\) for all \(t\in[0,1]\) as \(m\rightarrow\infty\), then by integrating both sides of (1.1) from 0 to *t*, we have

$$ u(t,m)= mt-\int_{0}^{t} (t-s)a(s)f \bigl(u(s,m) \bigr)\,ds. $$

(3.8)

Hence, from (3.3) and the continuity of \(f(u)\), we have

$$ m= u(1,m)+\int_{0}^{1}(1-s) a(s)f \bigl(u(s,m) \bigr)\,ds\leq L+L_{f} a^{L}, $$

(3.9)

where \(L_{f}=\max_{u\in[0,L]}{f(u)}\). If we choose \(m>L+L_{f} a^{L}\), (3.9) will lead to a contradiction.

Since \(u(t,m)\) is continuous and concave, there exists a number \(m_{0}\) large enough such that \(u(t,m_{0})\geq L\) for \(t\in I_{m_{0}}\).

*Step* 2. There exists a monotonically increasing sequence \(\{m_{k}\} \) such that the sequence \({\underline{\delta}_{m_{k}}}\) is decreasing on \(m_{k}\) and \({\overline{\delta}_{m_{k}}}\) is increasing on \(m_{k}\). That is,

$$I_{m_{0}}\subset I_{m_{1}}\subset\cdots\subset I_{m_{k}} \subset\cdots \subseteq(0,1] $$

and \(u(t,m_{k})\geq L\) for \(t\in I_{m_{k}}\).

First, we prove that

$$ \underline{\delta}_{m_{k}}< \underline{\delta}_{m_{k-1}},\quad k=1,2 \ldots \mbox{ for } m_{k}>m_{k-1}. $$

(3.10)

When \(k=1\), we have

$$u(\underline{\delta}_{m_{0}},m_{1})>u(\underline{ \delta}_{m_{0}},m_{0}) $$

in the case

$$ m_{1}>m_{0}+2a^{L}L_{f}\underline{ \delta}_{m_{0}}. $$

(3.11)

Otherwise, provided

$$ u(\underline{\delta}_{m_{0}},m_{1})\leq u(\underline{\delta }_{m_{0}},m_{0})=L, $$

(3.12)

then from (3.8) and (3.11), we have

$$\begin{aligned} & u(\underline{\delta}_{m_{0}},m_{1})-u(\underline{\delta }_{m_{0}},m_{0}) \\ &\quad= (m_{1}-m_{0})\underline{\delta}_{m_{0}}-\int _{0}^{\underline {\delta}_{m_{0}}} (\underline{\delta }_{m_{0}}-s)a(s) \bigl[f \bigl(u(s,m_{1}) \bigr)-f \bigl(u(s,m_{0}) \bigr) \bigr] \,ds \\ &\quad> (m_{1}-m_{0})\underline{\delta}_{m_{0}}-2a^{L}L_{f} \underline {\delta}_{m_{0}}^{2} \\ &\quad= \underline{\delta}_{m_{0}} \bigl[(m_{1}-m_{0})-2a^{L}L_{f} \underline {\delta}_{m_{0}} \bigr]>0, \end{aligned}$$

which contradicts (3.12).

Hence, for a slope \(m_{1}>m_{0}+2a^{L}L_{f}\underline{\delta}_{m_{0}}\), there exists a number \(0<\underline{\delta}_{m_{1}}<\underline{\delta}_{m_{0}}\) such that

$$u(\underline{\delta}_{m_{1}},m_{1})=L, \quad \mbox{and} \quad u(t,m_{1})\leq L \quad \mbox{for } t\in (0,\underline{ \delta}_{m_{1}}]. $$

See Figure 1.

By mathematical induction, it is not difficult to show that \(\underline{\delta}_{m_{k}}<\underline{\delta}_{m_{k-1}}\), \(k=1,2,\ldots\) .

Further, we turn to the right hand of the interval \(I_{m_{k}}\). Since *f* guarantees that \(u(t, m)\) is uniquely defined, the solutions \(u(t,m_{k-1})\) and \(u(t,m_{k})\) have no intersection in the interval \([\underline{\delta}_{m_{k-1}},1)\). It follows from

$$u(\underline{\delta}_{m_{k-1}},m_{k})>u(\underline{ \delta}_{m_{k-1}},m_{k-1}) $$

that

$$u(\overline{\delta}_{m_{k-1}},m_{k})>u(\overline{ \delta}_{m_{k-1}},m_{k-1}). $$

Thus we have

$$ \overline{\delta}_{m_{k}}>\overline{\delta}_{m_{k-1}},\quad k=1,2, \ldots\mbox{ for } m_{k}>m_{k-1}. $$

(3.13)

When \(k=1\), also see Figure 1.

*Step* 3. Seek out a slope \(m_{2}^{*}\) and two positive numbers *ρ* and *σ* such that \(0<\rho\leq\eta\leq\frac{A_{2}}{A_{2}+\epsilon} \leq\sigma\leq1\) and \(u(t,m_{2}^{*})\geq L\) for \(t\in[\rho,\sigma]\).

*Subcase* 1. \(\eta\in[\underline{\delta}_{m_{0}},\overline{\delta }_{m_{0}}]\) and \(u(1,m_{0})\geq L\). In this case, we take \(m_{2}^{*}=m_{0}\) and \(\rho=\underline{\delta}_{m_{0}}\), \(\sigma=\overline{\delta}_{m_{0}}=1\).

*Subcase* 2. \(\eta\nsubseteq[\underline{\delta}_{m_{0}},\overline {\delta}_{m_{0}}]\) or \(u(1,m_{0})< L\). Following the step 1, step 2, and the extension principle of solutions, there exists a positive integer *n* large enough such that

$$ \underline{\delta}_{m_{n}}< \eta, \qquad \overline{\delta}_{m_{n}} \geq\frac {A_{2}}{A_{2}+\epsilon}. $$

(3.14)

If we take \(m_{2}^{*}=m_{n}\) and \(\rho=\underline{\delta}_{m_{n}}\), \(\sigma =\overline{\delta}_{m_{n}}\), then

$$ \sigma(A_{2}+\epsilon)\geq A_{2}. $$

(3.15)

Two of the possible cases of \(I_{m_{0}}\) can be seen in Figure 2.

In the following, we prove that \(k(m_{2}^{*})\geq1\) or \(\varphi(m_{2}^{*})>0\) for the selected \(m_{2}^{*}\) and *ρ*, *σ*.

Set \(z(t)= (m_{2}^{*}/\sigma(A_{2}+\epsilon) )\sin (\sigma (A_{2}+\epsilon)t )\), then

$$ z''(t)+\sigma^{2}(A_{2}+ \epsilon)^{2}z(t)=0, \qquad z(0)=0, \qquad z'(0)=m_{2}^{*}, \quad t\in[\rho,\sigma], $$

(3.16)

where \(\rho\leq\eta<\sigma\leq1\). From (3.7), we have

$$\frac{f(u)}{u}\geq\frac{\sigma^{2}(A_{2}+\epsilon)^{2}}{a^{l}},\quad u\geq L. $$

Further, noting that \(u(1,m_{2}^{*})>L\) (this time \(\sigma=1\)) or \(u(1,m_{2}^{*})\leq u(\sigma,m_{2}^{*})=L\) and the function

$$S(x)=\frac{\sin\eta x}{\sin x} $$

is increasing for \(x\in(0,\pi)\), then by Lemma 2.2, Lemma 3.1, and inequality (3.15), we have

$$\begin{aligned} k \bigl(m_{2}^{*} \bigr)&= \frac{\alpha\int_{0}^{\eta}u(s,m_{2}^{*})\,ds}{u(1,m_{2}^{*})} \geq \frac{\alpha\eta u(\eta,m_{2}^{*})}{2u(1,m_{2}^{*})}\geq \frac{\alpha\eta u(\eta,m_{2}^{*})}{2u(\sigma,m_{2}^{*})} \\ &\geq \frac{\alpha\eta\sin\eta\sigma(A_{2}+\epsilon)}{2\sin \sigma(A_{2}+\epsilon)}\geq \frac{\alpha\eta\sin(\eta A_{2})}{2\sin A_{2}}=1, \end{aligned}$$

(3.17)

which implies \(\varphi(m_{2}^{*})\geq0\).

From (3.5) and (3.17), we can find a \(m^{*}\) between \(m_{1}^{*}\) and \(m_{2}^{*}\) such that \(u(t,m^{*})\) is the solution of (1.1)-(1.2). The theorem is complete.

The proof for (ii) is similar, so we omit it. □

Now, we present the result for BVP (1.1) with (1.7), which is also the correction of Theorem 3.1 and Theorem 3.2 in [9].

### Theorem 3.2

*Assume that* (H_{1})-(H_{2}) *hold*. *Suppose one of the following conditions holds*:

$$(\mathrm{i}) \quad 0\leq \bar{f}_{0}< \frac{\underline {A}^{2}}{a^{L}} , \qquad \underline{f}_{\infty}> \frac {\bar{A}^{2}}{a^{l}}; \qquad (\mathrm{ii}) \quad 0\leq \bar{f}_{\infty}< \frac{\underline{A}^{2}}{a^{L}}, \qquad \underline {f}_{0}> \frac{\bar{A}^{2}}{a^{l}}. $$

*Then problem* (1.1) *with* (1.7) *has at least one positive solution*, *where*

$$\underline{A}=\min\{A_{1},A_{2}\},\qquad \bar{A}=\max \{A_{1},A_{2}\} $$

*and*
\(A_{1}\), \(A_{2}\)
*is defined by*

$$ \frac{\sum_{i=1}^{n}\alpha_{i}[1-\cos(A_{1}\eta_{i} )]}{A_{1}\sin A_{1}}=1 $$

(3.18)

*and*

$$ \frac{\sum_{i=1}^{n}\alpha_{i}\eta_{i}\sin(A_{2}\eta_{i} )}{2\sin A_{2}}=1. $$

(3.19)

### Proof

Similar to (3.5) and (3.17), it follows from (1.7) and (3.18)-(3.19) that

$$\begin{aligned} k \bigl(m_{1}^{*} \bigr)&= \frac{\sum_{i=1}^{n}\alpha_{i}\int_{0}^{\eta _{i}}u(s,m_{1}^{*})\,ds}{u(1,m_{1}^{*})} < \frac{\sum_{i=1}^{n}\alpha_{i}\int_{0}^{\eta _{i}}m_{1}^{*}\sin(A_{1}s)\,ds}{m_{1}^{*}\sin A_{1}} \\ &= \frac{\sum_{i=1}^{n}\alpha_{i}[1-\cos(A_{1}\eta_{i} )]}{A_{1}\sin A_{1}}=1 \end{aligned}$$

(3.20)

and

$$\begin{aligned} k \bigl(m_{2}^{*} \bigr)&= \frac{\sum_{i=1}^{n}\alpha_{i}\int_{0}^{\eta _{i}}u(s,m_{2}^{*})\,ds}{u(1,m_{2}^{*})} \geq \frac{\sum_{i=1}^{n}\alpha_{i}\eta_{i} u(\eta _{i},m_{2}^{*})}{2u(1,m_{2}^{*})}\geq \frac{\sum_{i=1}^{n}\alpha_{i}\eta_{i} u(\eta_{i},m_{2}^{*})}{2u(\sigma,m_{2}^{*})} \\ &\geq \frac{\sum_{i=1}^{n}\alpha_{i}\eta_{i}\sin(\eta _{i}\sigma(A_{2}+\epsilon))}{2\sin\sigma(A_{2}+\epsilon)}\geq \frac{\sum_{i=1}^{n}\alpha_{i}\eta_{i}\sin(A_{2}\eta_{i} )}{2\sin A_{2}}=1, \end{aligned}$$

(3.21)

where \(\eta_{n}<\sigma\leq1\) and (3.15) holds.

The remainder of the proof is similar, so we omit it. □