Let

*X* be the Banach space

$C([0,T])$ with the maximum norm

$\parallel x\parallel ={max}_{t\in [0,T]}|x(t)|$. Define a cone by

$P=\{x\in C([0,T]):x(t)\ge 0\}.$

For the sake of convenience, we give the following conditions:

- (i)
The nonlinearity

$f:[0,T]\times [A-aT,A+aT]\times (-a,a)\to {\mathbb{R}}^{+}$ satisfies

for any

$(t,u,v)\in [0,T]\times [A-aT,A+aT]\times (-a,a)$;

- (ii)
$\mathrm{\Delta}=A{\sum}_{i=1}^{k+1}{\tau}_{i}>0$;

- (iii)
There exists a positive constant

*r* such that

$\mathrm{\Delta}+\lambda {\varphi}^{-1}(r)+MT\le r<b,$

(3.1)

where

$\lambda ={\sum}_{i\in \mathrm{\Gamma}}{\tau}_{i}{\zeta}_{i}$,

$\mathrm{\Gamma}=\{i:{\tau}_{i}\ge 0\}$;

- (iv)
${min}_{0\le v(t)<b}{\sum}_{i=1}^{k+1}{\tau}_{i}{\int}_{0}^{{\zeta}_{i}}{\varphi}^{-1}(v(s))\phantom{\rule{0.2em}{0ex}}ds\ge 0$.

**Remark 3.1**
- (1)
If ${\tau}_{i}\ge 0$, for all $i=1,2,\dots ,k+1$, then the condition (iv) clearly holds.

- (2)
If

$A=0$, instead of conditions (i) and (ii), we assume that for any

$(t,u,v)\in [0,T]\times [A-aT,A+aT]\times (-a,a)$, the following inequalities hold:

${M}_{1}\le f(t,u,v)\le {M}_{2},$

where

$0<{M}_{1}<{M}_{2}$ are two constants. Also, the condition (3.1) is replaced by

$\lambda {\varphi}^{-1}(r)+{M}_{2}T\le r<b.$

*Case* I. Singular *ϕ*-Laplacian operator: $\varphi :(-a,a)\to (-\mathrm{\infty},+\mathrm{\infty})$ ($0<a<+\mathrm{\infty}$).

**Theorem 3.1** *If* *f* *is continuous*, *then the problem* (1.1) *has at least one solution*.

*Proof* Define a set by

$\mathrm{\Omega}:=\{u,{u}^{\prime}\in X:\parallel u\parallel <A+aT,\parallel {u}^{\prime}\parallel <a\}.$

From the definition of

${\varphi}^{-1}:(-\mathrm{\infty},+\mathrm{\infty})\to (-a,a)$, (2.2), (2.7), and (2.8), we get, for any

$u\in \mathrm{\Omega}$,

$\begin{array}{rl}\parallel Su\parallel & =\underset{t\in [0,T]}{max}|(Su)(t)|\\ =\underset{t\in [0,T]}{max}|A+{\int}_{0}^{t}{\varphi}^{-1}((Bx)(s))\phantom{\rule{0.2em}{0ex}}ds|\le A+aT\end{array}$

and

$\parallel {(Su)}^{\prime}\parallel =\underset{t\in [0,T]}{max}|{(Su)}^{\prime}(t)|=\underset{t\in [0,T]}{max}\left|{\varphi}^{-1}((Bx)(t))\right|\le a.$

Thus $S:\overline{\mathrm{\Omega}}\to \overline{\mathrm{\Omega}}$. Utilizing (2.9) and Arzela-Ascoli theorem, it is easy to verify that the nonlinear operator *S* is a completely continuous operator. Therefore, the nonlinear operator *S* has at least one fixed point by Schauder fixed point theorem.

From the definition of

*S* in (2.8), we have

$\begin{array}{c}(Su)(0)=A,\hfill \\ {(Su)}^{\prime}(t)={\varphi}^{-1}((Bu)(t)),\hfill \end{array}$

*i.e.*,

$\varphi ({(Su)}^{\prime}(t))=(Bu)(t).$

Then we obtain

$\begin{array}{rl}\varphi ({(Su)}^{\prime}(T))& =(Bu)(T)\\ =A\sum _{i=1}^{k+1}{\tau}_{i}+\sum _{i=1}^{k+1}{\tau}_{i}{\int}_{0}^{{\zeta}_{i}}{u}^{\prime}(s)\phantom{\rule{0.2em}{0ex}}ds\\ =\tau u(T)+\sum _{i=1}^{k}{\tau}_{i}u({\zeta}_{i})+\sum _{i=1}^{k+1}{\tau}_{i}[u(0)-A]\end{array}$

and

${\left(\varphi ({(Su)}^{\prime}(t))\right)}^{\prime}=-f(t,u(t),{u}^{\prime}(t)).$

Consequently, we conclude that the fixed point of *S* is a solution of the problem (1.1). □

**Remark 3.2** Theorem 3.1 shows that if *ϕ* is singular ($a<\mathrm{\infty}$) and *f* is continuous on $[0,T]\times {\mathbb{R}}^{2}$, then the problem (1.1) is always solvable.

*Case* II. Bounded *ϕ*-Laplacian operator: $\varphi :(-a,a)\to (-b,b)$, $0<b<+\mathrm{\infty}$.

**Theorem 3.2** *If the conditions* (i)-(iv) *hold*, *then the nonlinear operator* *B* *defined by* (2.7) *has at least one fixed point*. *Further*, *the problem* (1.1) *has at least one positive solution*.

*Proof* Define a set by

${\mathrm{\Omega}}_{b}:=\{x\in X:\parallel x\parallel <b\}.$

From the conditions (ii) and (iv), we obtain, for any

$x\in P\cap {\mathrm{\Omega}}_{b}$,

$\begin{array}{rcl}(Bx)(t)& =& \mathrm{\Delta}+\sum _{i=1}^{k+1}{\tau}_{i}{\int}_{0}^{{\zeta}_{i}}{\varphi}^{-1}(x(s))\phantom{\rule{0.2em}{0ex}}ds\\ +{\int}_{t}^{T}f(s,A+{\int}_{0}^{s}{\varphi}^{-1}(x(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau ,{\varphi}^{-1}(x(s)))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & 0.\end{array}$

Clearly, the nonlinear operator $B:P\cap {\mathrm{\Omega}}_{b}\to P$ is well defined.

The condition (iii) implies that we find a constant

${r}_{2}$ such that

$\mathrm{\Delta}+\lambda {\varphi}^{-1}({r}_{2})+MT\le {r}_{2}<b.$

Define a set by

${\mathrm{\Omega}}_{2}:=\{x\in X:\parallel x\parallel <{r}_{2}\}.$

In virtue of the increasing property of

${\varphi}^{-1}$, we get, for any

$x\in P\cap \partial {\mathrm{\Omega}}_{2}$,

$\begin{array}{rcl}\parallel Bx\parallel & =& (Bx)(0)\\ =& \mathrm{\Delta}+\sum _{i=1}^{k+1}{\tau}_{i}{\int}_{0}^{{\zeta}_{i}}{\varphi}^{-1}(x(s))\phantom{\rule{0.2em}{0ex}}ds\\ +{\int}_{0}^{T}f(s,A+{\int}_{0}^{s}{\varphi}^{-1}(x(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau ,{\varphi}^{-1}(x(s)))\phantom{\rule{0.2em}{0ex}}ds\\ \le & \mathrm{\Delta}+\lambda {\varphi}^{-1}({r}_{2})+MT\\ \le & {r}_{2}=\parallel x\parallel .\end{array}$

Thus, for any

$x\in P\cap \partial {\mathrm{\Omega}}_{2}$, we have

$\parallel Bx\parallel \le \parallel x\parallel .$

We choose a small positive constant

${r}_{1}$ such that

${r}_{1}\le min\{\mathrm{\Delta},\frac{{r}_{2}}{2}\}$ and define

${\mathrm{\Omega}}_{1}:=\{x\in X:\parallel x\parallel <{r}_{1}\}$. Then for any

$x\in P\cap \partial {\mathrm{\Omega}}_{1}$, we find

$\begin{array}{rcl}\parallel Bx\parallel & =& (Bx)(0)\\ =& \mathrm{\Delta}+\sum _{i=1}^{k+1}{\tau}_{i}{\int}_{0}^{{\zeta}_{i}}{\varphi}^{-1}(x(s))\phantom{\rule{0.2em}{0ex}}ds\\ +{\int}_{0}^{T}f(s,A+{\int}_{0}^{s}{\varphi}^{-1}(x(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau ,{\varphi}^{-1}(x(s)))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \mathrm{\Delta}+\sum _{i=1}^{k+1}{\tau}_{i}{\int}_{0}^{{\zeta}_{i}}{\varphi}^{-1}(x(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {r}_{1}=\parallel x\parallel .\end{array}$

Thus, for any

$x\in P\cap \partial {\mathrm{\Omega}}_{1}$, we get

$\parallel Bx\parallel \ge \parallel x\parallel .$

In addition, a standard argument involving the Arzela-Ascoli theorem implies that

$B:P\cap ({\overline{\mathrm{\Omega}}}_{2}\mathrm{\setminus}{\mathrm{\Omega}}_{1})\to P$ is a completely continuous operator. Therefore, the nonlinear operator

*B* has at least one fixed point by the use of Lemma 2.1. Let

$x\in P\cap ({\overline{\mathrm{\Omega}}}_{2}\mathrm{\setminus}{\mathrm{\Omega}}_{1})$ be a fixed point of

*B*, then, from (2.6), we obtain

$u(t)=A+{\int}_{0}^{t}{\varphi}^{-1}(x(s))\phantom{\rule{0.2em}{0ex}}ds.$

Consequently, we get from Remark 2.1 that the problem (1.1) has a positive solution $u(t)$. □

**Remark 3.3** In order to prove the existence of a positive solution of the problem (1.1), we make a change of variable and introduce a first-order differential equation, and investigate the existence of a fixed point of the corresponding nonlinear operator *B*. The technique can be used for the different domains and ranges of ${\varphi}^{-1}$ and give an *a priori* estimate of the solution.

**Theorem 3.3**
*If*
${\tau}_{1}={\tau}_{2}=\cdots ={\tau}_{k}=\tau =0$
*and*
*f*
*satisfies the condition*
*for any* $(t,u,v)\in [0,T]\times [A-aT,A+aT]\times (-a,a)$, *then the problem* (1.1) *has at least one solution*.

*Proof* Adopting a similar technique to (2.2)-(2.6), we define a nonlinear operator

$\mathfrak{B}$ by

$(\mathfrak{B}x)(t):={\int}_{t}^{T}f(s,A+{\int}_{0}^{s}{\varphi}^{-1}(x(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau ,{\varphi}^{-1}(x(s)))\phantom{\rule{0.2em}{0ex}}ds.$

Then we get, for any

$x\in {\mathrm{\Omega}}_{b}$,

$\begin{array}{rl}\parallel \mathfrak{B}x\parallel & =\underset{t\in [0,T]}{max}|(\mathfrak{B}x)(t)|\\ \le {\int}_{0}^{T}\left|f(s,A+{\int}_{0}^{s}{\varphi}^{-1}(x(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau ,{\varphi}^{-1}(x(s)))\right|\phantom{\rule{0.2em}{0ex}}ds\\ <b.\end{array}$

Then the operator $\mathfrak{B}:{\mathrm{\Omega}}_{b}\to {\mathrm{\Omega}}_{b}$ has at least one fixed point by the use of Schauder fixed point theorem. Applying expression (2.6), we conclude that the problem (1.1) has at least one solution. □

**Remark 3.4** Observe that the solution provided by Theorem 3.3 could be trivial or negative.

**Theorem 3.4** *If the conditions* (ii)

*and* (iv)

*hold and* *f* *satisfies the condition* $f(t,u,v)\ge \frac{b}{T}$

(3.2)

*for any* $(t,u,v)\in [0,T]\times [A-aT,A+aT]\times (-a,a)$, *then problem* (1.1) *has no solution*.

*Proof* Taking arbitrarily

$x\in {\mathrm{\Omega}}_{b}$, we get from (3.2)

$\begin{array}{rcl}(Bx)(0)& =& \mathrm{\Delta}+\sum _{i=1}^{k+1}{\tau}_{i}{\int}_{0}^{{\zeta}_{i}}{\varphi}^{-1}(x(s))\phantom{\rule{0.2em}{0ex}}ds\\ +{\int}_{0}^{T}f(s,A+{\int}_{0}^{s}{\varphi}^{-1}(x(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau ,{\varphi}^{-1}(x(s)))\phantom{\rule{0.2em}{0ex}}ds\\ >& {\int}_{0}^{T}f(s,A+{\int}_{0}^{s}{\varphi}^{-1}(x(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau ,{\varphi}^{-1}(x(s)))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & b.\end{array}$

Thus, it implies that there exists a neighborhood ${N}_{\delta}=[0,\delta )$ such that $(Bx)(t)>b$ for any $t\in [0,\delta )$. This implies that the nonlinear operator $(Su)(t)=A+{\int}_{0}^{t}{\varphi}^{-1}((Bx)(s))\phantom{\rule{0.2em}{0ex}}ds$ is not well defined, since the domain of ${\varphi}^{-1}$ is the interval $(-b,b)$ and thus a solution of (1.1) cannot exist. □

**Example 3.1** We consider the following nonhomogeneous boundary value problem with

*ϕ*-Laplacian operator:

$\{\begin{array}{c}{(\varphi ({u}^{\prime}(t)))}^{\prime}=-f(t,u(t),{u}^{\prime}(t)),\phantom{\rule{1em}{0ex}}t\in (0,1),\hfill \\ u(0)=1,\phantom{\rule{2em}{0ex}}\varphi ({u}^{\prime}(1))=\frac{1}{8}u(1)+\frac{1}{2}u(\frac{3}{20})-\frac{1}{8}u(\frac{1}{5}),\hfill \end{array}$

(3.3)

where

$\begin{array}{r}\varphi (x)=\frac{5x}{\sqrt{25+{x}^{2}}},\\ f(t,u,v)=\frac{t}{2}|sinu|+t|cosv|.\end{array}$

It is easy to see that the nonlinearity

*f* satisfies the condition (i), that is,

$0\le f(t,u,v)\le \frac{3}{2}$

for any

$(t,u,v)\in [0,1]\times (-\mathrm{\infty},+\mathrm{\infty})\times (-\mathrm{\infty},+\mathrm{\infty})$. Computation yields

$\mathrm{\Delta}=A\sum _{i=1}^{k+1}{\tau}_{i}=\frac{1}{2},$

and there exists a positive constant

$3\le r\le 4.683$ such that the following inequalities hold:

$\frac{1}{2}+\frac{1}{5}{\varphi}^{-1}(r)+\frac{3}{2}\le r<5$

and

$\underset{0\le v(t)<5}{min}\sum _{i=1}^{3}{\tau}_{i}{\int}_{0}^{{\zeta}_{i}}{\varphi}^{-1}(v(s))\phantom{\rule{0.2em}{0ex}}ds\ge 0.$

Then conditions (ii)-(iv) also hold. Therefore, we find from Theorem 3.2 that the differential equation (3.3) has at least one positive solution.