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 [AaT,A+aT]\times (a,a)\to {\mathbb{R}}^{+} satisfies
for any (t,u,v)\in [0,T]\times [AaT,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 [AaT,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 ϕLaplacian 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 ArzelaAscoli 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 ϕLaplacian 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 ArzelaAscoli 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 firstorder 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 [AaT,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}\leftf(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 [AaT,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 ϕLaplacian 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+tcosv.\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.