We now provide sufficient conditions for the existence of positive solutions for PBVP (1). For convenience and ease of exposition, we make use of the following notation:

$e(t)=exp(-{\int}_{0}^{t}h(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ),\phantom{\rule{2em}{0ex}}\phi (t)={\int}_{0}^{t}e(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,\phantom{\rule{2em}{0ex}}\mathrm{\Phi}(t)={\int}_{0}^{t}\phi (\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,$

(4)

and

$\psi (t)=\frac{1}{e(t)}(\frac{1}{1-e(T)}-\frac{\phi (t)}{\phi (T)}),\phantom{\rule{1em}{0ex}}t\in [0,T].$

(5)

We observe that

$0<\psi (t)<\frac{1}{e(T)(1-e(T))}$ on

$[0,T]$. Moreover, we put

$k(t,s)=\frac{1}{Te(s)}\{\begin{array}{c}\frac{\phi (s)}{\phi (T)}[\phi (T)s-T\phi (t)+\mathrm{\Phi}(T)]-\mathrm{\Phi}(s),\phantom{\rule{1em}{0ex}}0\le s\le t\le T,\hfill \\ \frac{\phi (s)}{\phi (T)}[\phi (T)(s-T)-T\phi (t)+\mathrm{\Phi}(T)]+T\phi (t)-\mathrm{\Phi}(s),\hfill \\ \phantom{\rule{1em}{0ex}}0\le t\le s\le T,\hfill \end{array}$

(6)

and

$K(t,s)=k(t,s)+\frac{M-{\int}_{0}^{T}k(t,\tau )\phantom{\rule{0.2em}{0ex}}d\tau}{{\int}_{0}^{T}\psi (\tau )\phantom{\rule{0.2em}{0ex}}d\tau}\psi (s),\phantom{\rule{1em}{0ex}}t,s\in [0,T],$

(7)

where *M* is a positive constant.

We assume that

(H1) $f:[0,T]\times [0,\mathrm{\infty})\times \mathbb{R}\to \mathbb{R}$ and $h:[0,T]\to (0,\mathrm{\infty})$ are continuous functions.

We also assume that there exist $R>0$, $0<\alpha \le \beta $, $0<M\le \frac{e(T)(1-e(T)){\int}_{0}^{T}\psi (\tau )\phantom{\rule{0.2em}{0ex}}d\tau}{\alpha T}$, $r\in (0,R)$, $m\in (0,1)$, $\eta \in [0,T]$ and a continuous function $g:[0,T]\to [0,\mathrm{\infty})$ such that

(H2) $f(t,x,y)>-\alpha x+\beta |y|$ for $(t,x,y)\in [0,T]\times [0,R]\times [-R,R]$,

(H3) $f(t,R,0)<0$ for $t\in [0,T]$,

(H4) $f(0,x,R)=f(T,x,R)$ and $f(0,x,-R)=f(T,x,-R)$ for $x\in [0,R]$,

(H5) $f(t,x,-R)\le h(t)R$ for $t\in [0,T]$ and $x\in [0,R)$,

(H6) $f(t,x,y)\ge g(t)(x+|y|)$ for $(t,x,y)\in [0,T]\times (0,r]\times [-r,r]$,

(H7) $\frac{1}{\alpha T}\ge K(t,s)\ge 0$ for $t,s\in [0,T]$ and $m{\int}_{0}^{T}K(\eta ,s)g(s)\phantom{\rule{0.2em}{0ex}}ds\ge 1$.

**Theorem 2** *Under the assumptions* (H1)-(H7), *PBVP* (1) *has a positive solution on* $[0,T]$.

*Proof* Let ${\parallel \cdot \parallel}_{\mathrm{\infty}}$ denote the supremum norm in the space $C[0,T]$, that is, ${\parallel x\parallel}_{\mathrm{\infty}}={sup}_{t\in [0,T]}|x(t)|$. Consider the Banach spaces $X={C}^{1}[0,T]$ with the norm $\parallel x\parallel =max\{{\parallel x\parallel}_{\mathrm{\infty}},{\parallel {x}^{\prime}\parallel}_{\mathrm{\infty}}\}$, and $Y=C[0,T]$ with the norm ${\parallel \cdot \parallel}_{\mathrm{\infty}}$.

We write problem (1) as a coincidence equation

where

$Lx(t)=-{x}^{\u2033}(t)-h(t){x}^{\prime}(t),\phantom{\rule{1em}{0ex}}t\in [0,T],$

and

$Nx(t)=f(t,x(t),{x}^{\prime}(t)),\phantom{\rule{1em}{0ex}}t\in [0,T],$

with

$domL=\{x\in X:{x}^{\u2033}\in C[0,T],x(0)=x(T),{x}^{\prime}(0)={x}^{\prime}(T)\}$. Then

$KerL=\{x\in X:x(t)=c,t\in [0,T],c\in \mathbb{R}\}$

and

$ImL=\{y\in Y:{\int}_{0}^{T}\psi (s)y(s)\phantom{\rule{0.2em}{0ex}}ds=0\},$

where *ψ* is given by (5).

Clearly, Im

*L* is closed and

$Y={Y}_{1}+ImL$ with

${Y}_{1}=\{{y}_{1}\in Y:{y}_{1}=\frac{1}{{\int}_{0}^{T}\psi (s)\phantom{\rule{0.2em}{0ex}}ds}{\int}_{0}^{T}\psi (s)y(s)\phantom{\rule{0.2em}{0ex}}ds,y\in Y\}.$

Since ${Y}_{1}\cap ImL=\{0\}$, we have $Y={Y}_{1}\oplus ImL$. Moreover, $dim{Y}_{1}=1$, which gives $codimImL=1$. Consequently, *L* is Fredholm of index zero, and the assumption 1^{∘} is satisfied.

Define the projections

$P:X\to X$ by

$Px(t)=\frac{1}{T}{\int}_{0}^{T}x(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\in [0,T],$

and

$Q:Y\to Y$ by

$Qy(t)=\frac{1}{{\int}_{0}^{T}\psi (s)\phantom{\rule{0.2em}{0ex}}ds}{\int}_{0}^{T}\psi (s)y(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\in [0,T].$

It is a routine matter to show that for

$y\in ImL$, the inverse

${K}_{P}$ of

${L}_{P}$ is given by

$({K}_{P}y)(t)={\int}_{0}^{T}k(t,s)y(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\in [0,T],$

with the kernel

*k* defined by (6). Clearly, the assumption 2

^{∘} is satisfied. For

$y\in ImQ$, define

Then

*J* is an isomorphism from Im

*Q* to Ker

*L*. Next, consider a cone

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

For

${u}_{0}(t)\equiv 1$, we have

$\sigma ({u}_{0})=1$ and

$C({u}_{0})=\{x\in C:x(t)>0\text{on}[0,T]\}.$

Let

${\mathrm{\Omega}}_{1}=\{x\in X:\parallel x\parallel <r,|x(t)|>m{\parallel x\parallel}_{\mathrm{\infty}}\text{and}|{x}^{\prime}(t)|m{\parallel {x}^{\prime}\parallel}_{\mathrm{\infty}}\text{on}[0,T]\},$

and

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

Obviously, ${\mathrm{\Omega}}_{1}$ and ${\mathrm{\Omega}}_{2}$ are open bounded subsets of *X*, and ${\overline{\mathrm{\Omega}}}_{1}\subset {\mathrm{\Omega}}_{2}$.

To verify 3

^{∘}, suppose that there exist

${x}_{0}\in C\cap \partial {\mathrm{\Omega}}_{2}\cap domL$ and

${\lambda}_{0}\in (0,1)$ such that

$L{x}_{0}={\lambda}_{0}N{x}_{0}$. Then

$x(t)\ge 0$ on

$[0,T]$,

$\parallel {x}_{0}\parallel =R$,

$-{x}_{0}^{\u2033}(t)-h(t){x}_{0}^{\prime}(t)={\lambda}_{0}f(t,{x}_{0}(t),{x}_{0}^{\prime}(t)),\phantom{\rule{1em}{0ex}}t\in [0,T],$

(8)

and

${x}_{0}(0)={x}_{0}(T),\phantom{\rule{2em}{0ex}}{x}_{0}^{\prime}(0)={x}_{0}^{\prime}(T).$

(9)

There are two cases to consider.

- 1.
If $\parallel {x}_{0}\parallel ={\parallel {x}_{0}\parallel}_{\mathrm{\infty}}$, then there exists ${t}_{0}\in [0,T]$ such that $x({t}_{0})=R$. For ${t}_{0}\in (0,T)$, we get $0\le -{x}^{\u2033}({t}_{0})={\lambda}_{0}f({t}_{0},R,0)$, contrary to the assumption (H3). Similarly, if ${t}_{0}=0$ or ${t}_{0}=T$, BCs (9) imply ${x}^{\prime}(0)={x}^{\prime}(T)=0$. Hence, $0\le -{x}^{\u2033}({t}_{0})={\lambda}_{0}f({t}_{0},R,0)$ which contradicts (H3) again.

- 2.
If

$\parallel {x}_{0}\parallel ={\parallel {x}_{0}^{\prime}\parallel}_{\mathrm{\infty}}>{\parallel {x}_{0}\parallel}_{\mathrm{\infty}}$, then there exists

${t}_{0}\in [0,T]$ such that

$|{x}^{\prime}({t}_{0})|=R$. Observe that (H2) implies

$f(t,x,\pm R)>0$ for

$t\in [0,T]$ and

$x\in [0,R]$. Suppose that

${t}_{0}\in (0,T)$. If

${x}^{\prime}({t}_{0})=R$, we get from (8)

$-h({t}_{0})R={\lambda}_{0}f({t}_{0},{x}_{0}({t}_{0}),R),$

(10)

a contradiction. For

${x}^{\prime}({t}_{0})=-R$, we have

$h({t}_{0})R={\lambda}_{0}f({t}_{0},{x}_{0}({t}_{0}),-R)<f({t}_{0},{x}_{0}({t}_{0}),-R),$

(11)

contrary to (H5). By similar arguments, if ${t}_{0}=0$ or ${t}_{0}=T$, BCs (9) and (H4) imply either (10) or (11). Thus, 3^{∘} is fulfilled.

Next, for

$x\in X$, define (see [

15])

$\rho x(t)=\{\begin{array}{cc}x(t)\hfill & \text{if}x(t)\ge 0\text{on}[0,T],\hfill \\ \frac{1}{2}(x(t)-min\{x(t):t\in [0,T]\})\hfill & \text{if}x(\tilde{t})0\text{for some}\tilde{t}\in [0,T].\hfill \end{array}$

Clearly, *ρ* is a retraction and maps subsets of ${\overline{\mathrm{\Omega}}}_{2}$ into bounded subsets of *C*, so 4^{∘} holds.

To verify 5

^{∘}, it is enough to consider, for

$x\in KerL\cap {\mathrm{\Omega}}_{2}$ and

$\lambda \in [0,1]$, the mapping

$H(x,\lambda )(t)=x(t)-\lambda (\frac{1}{T}{\int}_{0}^{T}(\rho x)(s)\phantom{\rule{0.2em}{0ex}}ds+\frac{M}{{\int}_{0}^{T}\psi (s)\phantom{\rule{0.2em}{0ex}}ds}{\int}_{0}^{T}\psi (s)f(s,(\rho x)(s),{(\rho x)}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds).$

Observe that if

$x\in KerL\cap {\mathrm{\Omega}}_{2}$, then

$x(t)=c$ on

$[0,T]$ and

$\parallel x\parallel <R$. Suppose

$H(x,\lambda )=0$ for

$x\in \partial {\mathrm{\Omega}}_{2}$. Then

$c=\pm R$. For

$c=R$, we have

$(\rho x)(t)=x(t)$ and in view of (H3), we get

$0\le R(1-\lambda )=\lambda \frac{M}{{\int}_{0}^{T}\psi (s)\phantom{\rule{0.2em}{0ex}}ds}{\int}_{0}^{T}\psi (s)f(s,R,0)\phantom{\rule{0.2em}{0ex}}ds<0,$

which is a contradiction. If

$c=-R$, then

$(\rho x)(t)=0$, hence

$-R=\lambda \frac{M}{{\int}_{0}^{T}\psi (s)\phantom{\rule{0.2em}{0ex}}ds}{\int}_{0}^{T}\psi (s)f(s,0,0)\phantom{\rule{0.2em}{0ex}}ds,$

which contradicts (H2). Thus,

$H(x,\lambda )\ne 0$ for

$x\in \partial {\mathrm{\Omega}}_{2}$ and

$\lambda \in [0,1]$. This implies

${d}_{B}(H(x,0),KerL\cap {\mathrm{\Omega}}_{2},0)={d}_{B}(H(x,1),KerL\cap {\mathrm{\Omega}}_{2},0),$

and

${d}_{B}([I-(P+JQN)\rho ]{|}_{KerL},KerL\cap {\mathrm{\Omega}}_{2},0)={d}_{B}(H(c,1),KerL\cap {\mathrm{\Omega}}_{2},0)\ne 0.$

We next show that 6

^{∘} holds. Let

$x\in C({u}_{0})\cap \partial {\mathrm{\Omega}}_{1}$. Then for

$t\in [0,T]$, we have

$r\ge x(t)\ge m{\parallel x\parallel}_{\mathrm{\infty}}>0$,

$r\ge |{x}^{\prime}(t)|\ge {\parallel {x}^{\prime}\parallel}_{\mathrm{\infty}}$, and by (H6) and (H7), we obtain

$\begin{array}{rcl}\mathrm{\Psi}x(\eta )& =& \frac{1}{T}{\int}_{0}^{T}x(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{T}K(\eta ,s)f(s,x(s),{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\int}_{0}^{T}K(\eta ,s)g(s)[x(s)+|{x}^{\prime}(s)|]\phantom{\rule{0.2em}{0ex}}ds\ge m{\int}_{0}^{T}K(\eta ,s)g(s)[{\parallel x\parallel}_{\mathrm{\infty}}+{\parallel {x}^{\prime}\parallel}_{\mathrm{\infty}}]\phantom{\rule{0.2em}{0ex}}ds\\ \ge & m\parallel x\parallel {\int}_{0}^{T}K(\eta ,s)g(s)\phantom{\rule{0.2em}{0ex}}ds\ge \parallel x\parallel .\end{array}$

This implies $\parallel x\parallel \le \parallel \mathrm{\Psi}x\parallel $ for $x\in C({u}_{0})\cap \partial {\mathrm{\Omega}}_{1}$, so 6^{∘} is satisfied.

Finally, we must check if 7

^{∘} holds. If

$x\in \partial {\mathrm{\Omega}}_{2}$, then in view of (H2), we get

$\begin{array}{rcl}(P+JQN)(\rho x)(t)& =& \frac{1}{T}{\int}_{0}^{T}(\rho x)(s)\phantom{\rule{0.2em}{0ex}}ds+\frac{M}{{\int}_{0}^{T}\psi (s)\phantom{\rule{0.2em}{0ex}}ds}{\int}_{0}^{T}\psi (s)f(s,(\rho x)(s),{(\rho x)}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \frac{1}{T}{\int}_{0}^{T}(\rho x)(s)\phantom{\rule{0.2em}{0ex}}ds+\frac{M}{{\int}_{0}^{T}\psi (s)\phantom{\rule{0.2em}{0ex}}ds}{\int}_{0}^{T}\psi (s)[-\alpha (\rho x)(s)+\beta |{(\rho x)}^{\prime}(s)|]\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\int}_{0}^{T}[\frac{1}{T}-\frac{\alpha M\psi (s)}{{\int}_{0}^{T}\psi (\tau )\phantom{\rule{0.2em}{0ex}}d\tau}](\rho x)(s)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\int}_{0}^{T}[\frac{1}{T}-\frac{\alpha M}{e(T)(1-e(T)){\int}_{0}^{T}\psi (\tau )\phantom{\rule{0.2em}{0ex}}d\tau}](\rho x)(s)\phantom{\rule{0.2em}{0ex}}ds\ge 0.\end{array}$

Moreover, for

$x\in {\overline{\mathrm{\Omega}}}_{2}\setminus {\mathrm{\Omega}}_{1}$, we have from (H2) and (H7)

$\begin{array}{rcl}{\mathrm{\Psi}}_{\rho}x(t)& =& \frac{1}{T}{\int}_{0}^{T}(\rho x)(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{T}K(t,s)f(s,(\rho x)(s),{(\rho x)}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \frac{1}{T}{\int}_{0}^{T}(\rho x)(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{T}K(t,s)[-\alpha (\rho x)(s)+\beta |{(\rho x)}^{\prime}(s)|]\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\int}_{0}^{T}[\frac{1}{T}-\alpha K(t,s)](\rho x)(s)\phantom{\rule{0.2em}{0ex}}ds\ge 0.\end{array}$

Thus, 7^{∘} is fulfilled and the assertion follows. □

We now give two examples illustrating Theorem 2. Some calculations have been made with *Mathematica*. In the first example, the function *h* is constant, while in the second $h(t)=1/(1+t)$ and *f* is independent of *t*.

**Example 1**

Consider the following PBVP:

$\{\begin{array}{c}{x}^{\u2033}(t)+{x}^{\prime}(t)+(t(1-t)+1)(-\frac{2}{9}x(t)+\frac{3}{4}|{x}^{\prime}(t)|+1)=0,\phantom{\rule{1em}{0ex}}t\in [0,1],\hfill \\ x(0)=x(1),\phantom{\rule{2em}{0ex}}{x}^{\prime}(0)={x}^{\prime}(1).\hfill \end{array}$

(12)

Then

$e(t)={e}^{-t}$,

$\phi (t)=1-{e}^{-t}$,

$\mathrm{\Phi}(t)=t+{e}^{-t}-1$,

$\psi (t)=\frac{e}{e-1}$, and

$k(t,s)=\{\begin{array}{cc}-s+\frac{{e}^{s-t+1}-{e}^{1-t}}{e-1},\hfill & 0\le s\le t\le 1,\hfill \\ -s+1+\frac{{e}^{s-t}-{e}^{1-t}}{e-1},\hfill & 0\le t\le s\le 1.\hfill \end{array}$

Moreover, (7) with

$M=\frac{3}{2}$ reads

$K(t,s)=\{\begin{array}{cc}t-s+\frac{{e}^{s-t+1}}{e-1},\hfill & 0\le s\le t\le 1,\hfill \\ t-s+1+\frac{{e}^{s-t}}{e-1},\hfill & 0\le t\le s\le 1,\hfill \end{array}$

and the assumptions (H2)-(H7) are met with $R=20$, $\alpha =\frac{2}{9}$, $\beta =\frac{3}{4}$, $r=\frac{36}{53}$, $m\in [\frac{12(e-1)}{17+7e},1)$, $\eta =0$ and $g(t)=t(1-t)+1$. By Theorem 2, problem (12) has a positive solution.

**Example 2**

Consider the PBVP

$\{\begin{array}{c}{x}^{\u2033}(t)+\frac{1}{1+t}{x}^{\prime}(t)+\frac{1}{10}-\frac{1}{9}x(t)+{({x}^{\prime}(t))}^{4/5}=0,\phantom{\rule{1em}{0ex}}t\in [0,\frac{1}{2}],\hfill \\ x(0)=x(\frac{1}{2}),\phantom{\rule{2em}{0ex}}{x}^{\prime}(0)={x}^{\prime}(\frac{1}{2}).\hfill \end{array}$

(13)

In this case, we have

$e(t)=\frac{1}{1+t}$,

$\phi (t)=ln(1+t)$,

$\mathrm{\Phi}(t)=-t+ln(1+t)+tln(1+t)$ and

$\psi (t)=(1+t)(3-\frac{ln(1+t)}{ln(\frac{3}{2})}).$

The assumptions of Theorem 2 are fulfilled with $M=1$, $R=10$, $\alpha =\frac{1}{3}$, $\beta =\frac{1}{2}$, $r=\frac{1}{100}$, $m=0.9$, $\eta =\frac{1}{4}$ and $g(t)=3$.