**Theorem 1** *Assume that the conditions* (A), (F), *and* (H) *hold and* $b=+\mathrm{\infty}$. *Then Eq*. (1.1) *has at least one positive solution*.

*Proof* Lemma 2 shows that $S:\mathcal{P}\to \mathcal{P}$. In addition, a standard argument involving the Arzela-Ascoli theorem implies that *S* is a completely continuous operator.

Now, we choose a positive constant

${r}_{1}$ such that

${r}_{1}\le {\int}_{0}^{a}{\varphi}^{-1}(\beta +{M}_{1}{\int}_{s}^{T}h(\tau )\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds$

and define

${\mathrm{\Omega}}_{1}:=\{u\in X:\parallel u\parallel <{r}_{1}\}$. For any

$u\in \mathcal{P}\cap \partial {\mathrm{\Omega}}_{1}$, we get from the condition (F) that

$\begin{array}{rl}\parallel Su\parallel & =(Su)(T)=\omega (u)+{\int}_{0}^{T}{\varphi}^{-1}(\beta +{\int}_{s}^{T}h(\tau )f(u(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\int}_{0}^{T}{\varphi}^{-1}(\beta +{\int}_{s}^{T}h(\tau )f(u(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\int}_{0}^{a}{\varphi}^{-1}(\beta +{M}_{1}{\int}_{s}^{T}h(\tau )\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds\\ \ge {r}_{1}=\parallel u\parallel .\end{array}$

Thus, for any

$u\in \mathcal{P}\cap \partial {\mathrm{\Omega}}_{1}$, we find that

$\parallel Su\parallel \ge \parallel u\parallel .$

(3.1)

From the hypothesis (A), we can let

${\sum}_{i\in \mathrm{\nabla}}{\alpha}_{i}=1-\epsilon $,

$\epsilon \in (0,1)$. Next, we choose a positive constant

${r}_{2}$ such that

${r}_{2}=max\{\frac{1}{\epsilon}{\int}_{0}^{T}{\varphi}^{-1}(\beta +{M}_{2}{\int}_{s}^{T}h(\tau )\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds,\frac{{r}_{1}}{\epsilon}\}$

and define

${\mathrm{\Omega}}_{2}:=\{u\in X:\parallel u\parallel <{r}_{2}\}$. Clearly, for any

$u\in \mathcal{P}\cap \partial {\mathrm{\Omega}}_{2}$, we obtain

$\begin{array}{rl}\parallel Su\parallel & =(Su)(T)=\omega (u)+{\int}_{0}^{T}{\varphi}^{-1}(\beta +{\int}_{s}^{T}h(\tau )f(u(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds\\ \le \sum _{i\in \mathrm{\nabla}}{\alpha}_{i}\parallel u\parallel +{\int}_{0}^{T}{\varphi}^{-1}(\beta +{M}_{2}{\int}_{s}^{T}h(\tau )\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds\\ \le \sum _{i\in \mathrm{\nabla}}{\alpha}_{i}\parallel u\parallel +\epsilon {r}_{2}\\ ={r}_{2}[\sum _{i\in \mathrm{\nabla}}{\alpha}_{i}+\epsilon ]={r}_{2}=\parallel u\parallel .\end{array}$

Then, for any

$u\in \mathcal{P}\cap \partial {\mathrm{\Omega}}_{2}$, it implies that

$\parallel Su\parallel \le \parallel u\parallel .$

(3.2)

Based on Lemma 4, we get from (3.1) and (3.2) that the operator *S* has at least one fixed point. Thus, it follows that Eq. (1.1) has at least one positive solution. □

**Remark 2** If the coefficients $\{{\alpha}_{i}:i\in I\}$ are nonnegative, then the condition (A) is replaced with

(A′) ${\sum}_{i=1}^{k}{\alpha}_{i}\in (0,1)$.

Applying the results in Remark 1 and Theorem 1, we get the following result.

**Corollary 1** *Assume that the conditions* (A′) *and* (F) *hold and* $b=+\mathrm{\infty}$. *Then Eq*. (1.1) *has at least one positive solution*.

*If*
$\varphi :\mathbb{R}\to (-b,b)$
*(*
$0<b<+\mathrm{\infty}$
*), then we have the following result.*

**Theorem 2** *Assume that the conditions* (A), (F), (B), *and* (H) *hold*. *Then Eq*. (1.1) *has at least one positive solution*.

*Proof* Using Lemma 3 and the proof of Theorem 1, we get that the conclusion holds. □

**Example 1** Consider the differential equation

${({|{u}^{\prime}(t)|}^{2}{u}^{\prime}(t))}^{\prime}=-t({e}^{-sinu(t)}+cos{u}^{2}(t)+5),\phantom{\rule{1em}{0ex}}t\in (0,4),$

(3.3)

subjected to the boundary conditions

$u(0)=\frac{1}{8}u(1)-\frac{1}{4}u(2)+\frac{1}{2}u(3),\phantom{\rule{2em}{0ex}}\varphi ({u}^{\prime}(4))=1.$

(3.4)

Clearly, we find

$\begin{array}{r}\varphi ={\varphi}_{4}\phantom{\rule{1em}{0ex}}\text{is one}p\text{-Laplacian operator,}p=4,\\ h(t)=t,\phantom{\rule{2em}{0ex}}f(u)={e}^{-sinu}+cos{u}^{2}+5,\\ \omega (u)=\frac{1}{8}u(1)-\frac{1}{4}u(2)+\frac{1}{2}u(3),\\ {\alpha}_{1}=\frac{1}{8},\phantom{\rule{2em}{0ex}}{\alpha}_{2}=-\frac{1}{4},\phantom{\rule{2em}{0ex}}{\alpha}_{3}=\frac{1}{2},\phantom{\rule{2em}{0ex}}\sum _{i=1}^{3}{\alpha}_{i}=\frac{3}{8}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\alpha}_{1}+{\alpha}_{3}=\frac{5}{8}.\end{array}$

Therefore, we conclude from Theorem 1 that Eq. (3.3)-(3.4) has at least one positive solution.

**Example 2** Consider the differential equation

${\left(\frac{{u}^{\prime}(t)}{\sqrt{1+{({u}^{\prime}(t))}^{2}}}\right)}^{\prime}=-{t}^{4}(sinu(t)+2),\phantom{\rule{1em}{0ex}}t\in (0,1),$

(3.5)

subjected to the boundary conditions

$u(0)=-\frac{1}{40}u(0.2)+\frac{1}{4}u(0.3)+\frac{1}{2}u(0.7),\phantom{\rule{2em}{0ex}}\varphi ({u}^{\prime}(1))=\frac{1}{4}.$

(3.6)

Obviously, we obtain

$\begin{array}{r}\varphi (u)=\frac{u}{\sqrt{1+{u}^{2}}},\\ h(t)={t}^{4},\phantom{\rule{2em}{0ex}}f(u)=sinu(t)+2,\\ \omega (u)=-\frac{1}{40}u(0.2)+\frac{1}{4}u(0.3)+\frac{1}{2}u(0.7),\\ {\alpha}_{1}=-\frac{1}{40},\phantom{\rule{2em}{0ex}}{\alpha}_{2}=\frac{1}{4},\phantom{\rule{2em}{0ex}}{\alpha}_{3}=\frac{1}{2},\phantom{\rule{2em}{0ex}}\sum _{i=1}^{3}{\alpha}_{i}=\frac{29}{40}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\alpha}_{2}+{\alpha}_{3}=\frac{3}{4}.\end{array}$

It is easy to verify that the conditions (B), (F), and (H) hold. Consequently, we get from Theorem 2 that the equation (3.5)-(3.6) has at least one positive solution.