In this section, we shall use the fixed point theorems stated in Section 2 to obtain the existence of at least three positive solutions of the complementary Lidstone boundary value problem (1.1). By a *positive solution* *y* of (1.1), we mean a nontrivial $y\in C[0,1]$ satisfying (1.1) and $y(t)\ge 0$ for $t\in [0,1]$.

To tackle (1.1), we first consider the initial value problem

$\begin{array}{r}{y}^{\prime}(t)=x(t),\phantom{\rule{1em}{0ex}}t\in [0,1],\\ y(0)=0\end{array}$

(3.1)

whose solution is simply

$y(t)={\int}_{0}^{t}x(s)\phantom{\rule{0.2em}{0ex}}ds.$

(3.2)

Taking into account (3.1) and (3.2), the

*complementary Lidstone* boundary value problem (1.1) reduces to the

*Lidstone* boundary value problem

$\begin{array}{r}{(-1)}^{m}{x}^{(2m)}(t)=F(t,{\int}_{0}^{t}x(s)\phantom{\rule{0.2em}{0ex}}ds,x(t)),\phantom{\rule{1em}{0ex}}t\in [0,1],\\ {x}^{(2k-2)}(0)={x}^{(2k-2)}(1)=0,\phantom{\rule{1em}{0ex}}1\le k\le m.\end{array}$

(3.3)

If (3.3) has a solution

${x}^{\ast}$, then by virtue of (3.2), the boundary value problem (1.1) has a solution given by

${y}^{\ast}(t)={\int}_{0}^{t}{x}^{\ast}(s)\phantom{\rule{0.2em}{0ex}}ds.$

(3.4)

So the existence of a solution of the *complementary Lidstone* boundary value problem (1.1) follows from the existence of a solution of the *Lidstone* boundary value problem (3.3). It is clear from (3.4) that $\parallel {y}^{\ast}\parallel \le \parallel {x}^{\ast}\parallel $; moreover if ${x}^{\ast}$ is positive, so is ${y}^{\ast}$. With the tools in Section 2 and a technique to handle the nonlinear term *F*, we shall study the boundary value problem (1.1) via (3.3).

Let the Banach space

$B=C[0,1]$ be equipped with the norm

$\parallel x\parallel ={sup}_{t\in [0,1]}|x(t)|$ for

$x\in B$. Define the operator

$S:C[0,1]\to C[0,1]$ by

$\begin{array}{rcl}Sx(t)& =& {\int}_{0}^{1}{(-1)}^{m}{g}_{m}(t,s)F(s,{\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))\phantom{\rule{0.2em}{0ex}}ds\\ =& {\int}_{0}^{1}|{g}_{m}(t,s)|F(s,{\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\in [0,1],\end{array}$

(3.5)

where ${g}_{m}(t,s)$ is the Green’s function given in (2.4). A fixed point ${x}^{\ast}$ of the operator *S* is clearly a solution of the boundary value problem (3.3), and as seen earlier ${y}^{\ast}(t)={\int}_{0}^{t}{x}^{\ast}(s)\phantom{\rule{0.2em}{0ex}}ds$ is a solution of (1.1).

For easy reference, we shall list the conditions that are needed later. In these conditions the sets

*K* and

$\tilde{K}$ are defined by

$\begin{array}{r}\tilde{K}=\{x\in B\mid x(t)\ge 0,t\in [0,1]\},\\ K=\{x\in \tilde{K}\mid x(t)>0\text{on some subset of}[0,1]\text{of positive measure}\}.\end{array}$

(3.6)

(C1) $F:[0,1]\times {\mathbb{R}}^{2}\to \mathbb{R}$ is an ${L}^{1}$-Carathéodory function.

(C2) We have

$\begin{array}{c}F(t,u,v)\ge 0,\phantom{\rule{1em}{0ex}}u,v\in \tilde{K},\text{a.e.}t\in (0,1)\phantom{\rule{1em}{0ex}}\text{and}\hfill \\ F(t,u,v)0,\phantom{\rule{1em}{0ex}}u,v\in K,\text{a.e.}t\in (0,1).\hfill \end{array}$

(C3) There exist continuous functions

*f*,

*ν*,

*μ* with

$f:[0,\mathrm{\infty})\times [0,\mathrm{\infty})\to [0,\mathrm{\infty})$ and

$\nu ,\mu :(0,1)\to [0,\mathrm{\infty})$ such that

$\mu (t)f(u,v)\le F(t,u,v)\le \nu (t)f(u,v),\phantom{\rule{1em}{0ex}}u,v\in \tilde{K},\text{a.e.}t\in (0,1).$

(C4) There exists a number

$0<\theta \le 1$ such that

$\mu (t)\ge \theta \nu (t),\phantom{\rule{1em}{0ex}}\text{a.e.}t\in (0,1).$

If (C2) and (C3) hold, then it follows from (3.5) that for

$x\in \tilde{K}$ and

$t\in [0,1]$,

$\begin{array}{rcl}0& \le & {\int}_{0}^{1}|{g}_{m}(t,s)|\mu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))\phantom{\rule{0.2em}{0ex}}ds\le Sx(t)\\ \le & {\int}_{0}^{1}|{g}_{m}(t,s)|\nu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

(3.7)

Let

$\delta \in (0,\frac{1}{2})$ be fixed. We define a cone

*C* in

*B* as

$C=\{x\in B|x(t)\ge 0\text{for}t\in [0,1],\text{and}\underset{t\in [\delta ,1-\delta ]}{min}x(t)\ge \frac{2\delta \theta}{\pi}\parallel x\parallel \},$

(3.8)

where *θ* is given in (C4). Clearly, we have $C\subseteq \tilde{K}$.

**Lemma 3.1** *Let* (C1)-(C4) *hold*. *Then the operator* *S* *defined in* (3.5) *is continuous and completely continuous*, *and* *S* *maps* *C* *into* *C*.

*Proof* From (2.4) we have ${g}_{m}(t,s)\in C[0,1]\subseteq {L}^{\mathrm{\infty}}[0,1]$, $t\in [0,1]$ and the map $t\to {g}_{m}(t,s)$ is continuous from $[0,1]$ to $C[0,1]$. This together with $F:[0,1]\times {\mathbb{R}}^{2}\to \mathbb{R}$ is an ${L}^{1}$-Carathéodory function ensures (as in [[42], Theorem 4.2.2]) that *S* is continuous and completely continuous.

Let

$x\in C$. From (3.7) we have

$Sx(t)\ge 0$ for

$t\in [0,1]$. Next, using (3.7) and Lemma 2.1 gives for

$t\in [0,1]$,

$\begin{array}{rl}Sx(t)& \le {\int}_{0}^{1}|{g}_{m}(t,s)|\nu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))\phantom{\rule{0.2em}{0ex}}ds\\ \le \frac{1}{{\pi}^{2m-1}}{\int}_{0}^{1}\nu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))sin\pi s\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

(3.9)

Hence, we have

$\parallel Sx\parallel \le \frac{1}{{\pi}^{2m-1}}{\int}_{0}^{1}\nu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))sin\pi s\phantom{\rule{0.2em}{0ex}}ds.$

(3.10)

Now, employing (3.7), Lemma 2.2, (C4) and (3.10), we find for

$t\in [\delta ,1-\delta ]$,

$\begin{array}{rcl}Sx(t)& \ge & {\int}_{0}^{1}|{g}_{m}(t,s)|\mu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \frac{2\delta}{{\pi}^{2m}}{\int}_{0}^{1}\mu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))sin\pi s\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \frac{2\delta}{{\pi}^{2m}}{\int}_{0}^{1}\theta \nu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))sin\pi s\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \frac{2\delta \theta}{\pi}\parallel Sx\parallel .\end{array}$

This leads to

$\underset{t\in [\delta ,1-\delta ]}{min}Sx(t)\ge \frac{2\delta \theta}{\pi}\parallel Sx\parallel .$

We have shown that $Sx\in C$. □

For subsequent results, we define the following constants for fixed

$\delta \in (0,\frac{1}{2})$ and

${\tau}_{1},{\tau}_{2},{\tau}_{3},{\tau}_{4}\in [0,1]$:

$\{\begin{array}{l}q=\frac{1}{{\pi}^{2m-1}}{\int}_{0}^{1}\nu (s)sin\pi s\phantom{\rule{0.2em}{0ex}}ds,\\ r={min}_{t\in [\delta ,1-\delta ]}{\int}_{\frac{1}{2}}^{1-\delta}|{g}_{m}(t,s)|\mu (s)\phantom{\rule{0.2em}{0ex}}ds,\\ {p}_{1}={min}_{t\in [{\tau}_{2},{\tau}_{3}]}{\int}_{\frac{1}{2}}^{{\tau}_{3}}|{g}_{m}(t,s)|\mu (s)\phantom{\rule{0.2em}{0ex}}ds,\\ {p}_{2}=\frac{1}{{\pi}^{2m-1}}{\int}_{{\tau}_{1}}^{{\tau}_{4}}\nu (s)sin\pi s\phantom{\rule{0.2em}{0ex}}ds,\\ {p}_{3}=\frac{1}{{\pi}^{2m-1}}{\int}_{0}^{{\tau}_{1}}\nu (s)sin\pi s\phantom{\rule{0.2em}{0ex}}ds+\frac{1}{{\pi}^{2m-1}}{\int}_{{\tau}_{4}}^{1}\nu (s)sin\pi s\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

(3.11)

**Lemma 3.2** *Let* (C1)-(C4) *hold*, *and assume*

(C5) *the function* $\nu (s)sin\pi s>0$ *on a subset of* $[0,1]$ *of positive measure*.

*Suppose that there exists a number* $d>0$ *such that for* $u,v\in [0,d]$,

$f(u,v)<\frac{d}{q}.$

(3.12)

*Then*
$S(\overline{C}(d))\subseteq C(d)\subset \overline{C}(d).$

(3.13)

*Proof* Let

$x\in \overline{C}(d)$. So

$\parallel x\parallel \le d$, which implies immediately that

${\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau \le d\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x(s)\le d,\phantom{\rule{1em}{0ex}}s\in [0,1].$

Then, using (3.9), (C5) and (3.12), we find for

$t\in [0,1]$,

$\begin{array}{rcl}|Sx(t)|& \le & \frac{1}{{\pi}^{2m-1}}{\int}_{0}^{1}\nu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))sin\pi s\phantom{\rule{0.2em}{0ex}}ds\\ <& \frac{1}{{\pi}^{2m-1}}{\int}_{0}^{1}\nu (s)sin\pi s\phantom{\rule{0.2em}{0ex}}ds\cdot \frac{d}{q}\\ =& q\cdot \frac{d}{q}=d.\end{array}$

This implies $\parallel Sx\parallel <d$. Together with the fact that $Sx\in C$ (Lemma 3.1), we have shown that $Sx\in C(d)$. Conclusion (3.13) is now immediate. □

Using a similar argument as Lemma 3.2, we have the following lemma.

**Lemma 3.3** *Let* (C1)-(C4)

*hold*.

*Suppose that there exists a number* $d>0$ *such that for* $u,v\in [0,d]$,

*Then*
$S(\overline{C}(d))\subseteq \overline{C}(d).$

We are now ready to establish the existence of three positive solutions for the complementary Lidstone boundary value problem (1.1). The first result below uses Leggett-Williams’ fixed point theorem (Theorem 2.1).

**Theorem 3.1** *Let* $\delta \in (0,\frac{1}{2})$ *be fixed*. *Let* (C1)-(C5) *hold*, *and assume*

(C6) *for each* $t\in [\delta ,1-\delta ]$, *the function* $|{g}_{m}(t,s)|\mu (s)>0$ *on a subset of* $[\frac{1}{2},1-\delta ]$ *of positive measure*.

*Suppose that there exist numbers* ${w}_{1}$,

${w}_{2}$,

${w}_{3}$ *with* $0<{w}_{1}<{w}_{2}(\frac{1}{2}-\delta )<{w}_{2}<\frac{\pi {w}_{2}}{2\delta \theta}\le {w}_{3}$

*such that the following hold*:

- (P)
$f(u,v)<\frac{{w}_{1}}{q}$ *for* $u,v\in [0,{w}_{1}]$;

- (Q)
*one of the following holds*:

- (Q1)
${lim\hspace{0.17em}sup}_{u\to \mathrm{\infty},v\to \mathrm{\infty}}\frac{f(u,v)}{u}<\frac{1}{q}$ *or* ${lim\hspace{0.17em}sup}_{u\to \mathrm{\infty},v\to \mathrm{\infty}}\frac{f(u,v)}{v}<\frac{1}{q}$;

- (Q2)
*there exists a number* *d* ($\ge {w}_{3}$) *such that* $f(u,v)\le \frac{d}{q}$ *for* $u,v\in [0,d]$;

- (R)
$f(u,v)>\frac{{w}_{2}}{r}$ *for* $u\in [{w}_{2}(\frac{1}{2}-\delta ),{w}_{3}(\frac{1}{2}-\delta )]$ *and* $v\in [{w}_{2},{w}_{3}]$.

*Then we have the following conclusions*:

- (a)
*The Lidstone boundary value problem* (3.3)

*has* (

*at least*)

*three positive solutions* ${x}_{1},{x}_{2},{x}_{3}\in C$ (

*where* *C* *is defined in* (3.8))

*such that* $\{\begin{array}{l}\parallel {x}_{1}\parallel <{w}_{1};\\ {x}_{2}(t)>{w}_{2},\phantom{\rule{1em}{0ex}}t\in [\delta ,1-\delta ];\\ \parallel {x}_{3}\parallel >{w}_{1}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{min}_{t\in [\delta ,1-\delta ]}{x}_{3}(t)<{w}_{2}.\end{array}$

(3.14)

- (b)
*The complementary Lidstone boundary value problem* (1.1)

*has* (

*at least*)

*three positive solutions* ${y}_{1}$,

${y}_{2}$,

${y}_{3}$ *such that for* $i=1,2,3$,

$\{\begin{array}{l}{y}_{i}(t)={\int}_{0}^{t}{x}_{i}(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\in [0,1];\\ \parallel {y}_{i}\parallel \le \parallel {x}_{i}\parallel ;\phantom{\rule{2em}{0ex}}{y}_{i}(t)\ge \frac{2\delta \theta}{\pi}\parallel {x}_{i}\parallel (t-\delta ),\phantom{\rule{1em}{0ex}}t\in [\delta ,1-\delta ]\end{array}$

(3.15)

(

*where* ${x}_{i}$’

*s are those in conclusion* (a)).

*We further have* $\{\begin{array}{l}\parallel {y}_{1}\parallel <{w}_{1};\\ {y}_{2}(t)>{w}_{2}(t-\delta ),\phantom{\rule{1em}{0ex}}t\in [\delta ,1-\delta ];\\ {y}_{3}(t)>\frac{2\delta \theta}{\pi}{w}_{1}(t-\delta ),\phantom{\rule{1em}{0ex}}t\in [\delta ,1-\delta ].\end{array}$

(3.16)

*Proof* We shall employ Theorem 2.1 with the cone

*C* defined in (3.8). First, we shall prove that condition (Q) implies the existence of a number

${w}_{4}$, where

${w}_{4}\ge {w}_{3}$, such that

$S(\overline{C}({w}_{4}))\subseteq \overline{C}({w}_{4}).$

(3.17)

Suppose that (Q2) holds. Then by Lemma 3.3 we immediately have (3.17) where we pick

${w}_{4}=d$. Suppose now that

${lim\hspace{0.17em}sup}_{u\to \mathrm{\infty},v\to \mathrm{\infty}}\frac{f(u,v)}{u}<\frac{1}{q}$ of (Q1) is satisfied. Then there exist

$N>0$ and

$\u03f5<\frac{1}{q}$ such that

$\frac{f(u,v)}{u}<\u03f5,\phantom{\rule{1em}{0ex}}u,v>N.$

(3.18)

Let

$L=\underset{u,v\in [0,N]}{max}f(u,v).$

Noting (3.18), it is then clear that for

$u,v\ge 0$,

$f(u,v)\le L+\u03f5u.$

(3.19)

Now, pick the number

${w}_{4}$ so that

${w}_{4}>max\{{w}_{3},L{(\frac{1}{q}-\u03f5)}^{-1}\}.$

(3.20)

Let

$x\in \overline{C}({w}_{4})$. Using (3.9), (3.19) and (3.20) yields for

$t\in [0,1]$,

$\begin{array}{rl}|Sx(t)|& \le \frac{1}{{\pi}^{2m-1}}{\int}_{0}^{1}\nu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))sin\pi s\phantom{\rule{0.2em}{0ex}}ds\\ \le \frac{1}{{\pi}^{2m-1}}{\int}_{0}^{1}\nu (s)(L+\u03f5{\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau )sin\pi s\phantom{\rule{0.2em}{0ex}}ds\\ \le \frac{1}{{\pi}^{2m-1}}{\int}_{0}^{1}\nu (s)(L+\u03f5{w}_{4})sin\pi s\phantom{\rule{0.2em}{0ex}}ds\\ =q(L+\u03f5{w}_{4})\\ <q[{w}_{4}(\frac{1}{q}-\u03f5)+\u03f5{w}_{4}]={w}_{4}.\end{array}$

Hence, $\parallel Sx\parallel <{w}_{4}$ and so $Sx\in C({w}_{4})\subset \overline{C}({w}_{4})$. Thus, (3.17) follows immediately. Note that the argument is similar if we assume that ${lim\hspace{0.17em}sup}_{u\to \mathrm{\infty},v\to \mathrm{\infty}}\frac{f(u,v)}{v}<\frac{1}{q}$ of (Q1) is satisfied.

Let

$\psi :C\to [0,\mathrm{\infty})$ be defined by

$\psi (x)=\underset{t\in [\delta ,1-\delta ]}{min}x(t).$

Clearly, *ψ* is a nonnegative continuous concave functional on *C* and $\psi (x)\le \parallel x\parallel $ for all $x\in C$.

We shall verify that condition (a) of Theorem 2.1 is satisfied. It is obvious that

$x(t)=\frac{{w}_{2}+{w}_{3}}{2}\in \{x\in C(\psi ,{w}_{2},{w}_{3})\mid \psi (x)>{w}_{2}\}$

and so

$\{x\in C(\psi ,{w}_{2},{w}_{3})\mid \psi (x)>{w}_{2}\}\ne \mathrm{\varnothing}$. Next, let

$x\in C(\psi ,{w}_{2},{w}_{3})$. Then

$\psi (x)\ge {w}_{2}$ and

$\parallel x\parallel \le {w}_{3}$ which imply

$x(s)\in [{w}_{2},{w}_{3}],\phantom{\rule{1em}{0ex}}s\in [\delta ,1-\delta ]\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\int}_{\delta}^{\frac{1}{2}}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau \in [{w}_{2}(\frac{1}{2}-\delta ),{w}_{3}(\frac{1}{2}-\delta )].$

(3.21)

Using (3.7), (3.21), (C6) and (R), it follows that

$\begin{array}{rcl}\psi (Sx)& =& \underset{t\in [\delta ,1-\delta ]}{min}Sx(t)\\ \ge & \underset{t\in [\delta ,1-\delta ]}{min}{\int}_{0}^{1}|{g}_{m}(t,s)|\mu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \underset{t\in [\delta ,1-\delta ]}{min}{\int}_{\frac{1}{2}}^{1-\delta}|{g}_{m}(t,s)|\mu (s)f({\int}_{\delta}^{\frac{1}{2}}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))\phantom{\rule{0.2em}{0ex}}ds\\ >& \underset{t\in [\delta ,1-\delta ]}{min}{\int}_{\frac{1}{2}}^{1-\delta}|{g}_{m}(t,s)|\mu (s)\phantom{\rule{0.2em}{0ex}}ds\cdot \frac{{w}_{2}}{r}\\ =& r\cdot \frac{{w}_{2}}{r}={w}_{2}.\end{array}$

Therefore, we have shown that $\psi (Sx)>{w}_{2}$ for all $x\in C(\psi ,{w}_{2},{w}_{3})$.

Next, by condition (P) and Lemma 3.2 (with $d={w}_{1}$), we have $S(\overline{C}({w}_{1}))\subseteq C({w}_{1})$. Hence, condition (b) of Theorem 2.1 is fulfilled.

Finally, we shall show that condition (c) of Theorem 2.1 holds. Let

$x\in C(\psi ,{w}_{2},{w}_{4})$ with

$\parallel Sx\parallel >{w}_{3}$. Using (3.7), Lemma 2.2, (C4), (3.10) and the inequality

$\frac{\pi {w}_{2}}{2\delta \theta}\le {w}_{3}$, we find

$\begin{array}{rcl}\psi (Sx)& \ge & \underset{t\in [\delta ,1-\delta ]}{min}{\int}_{0}^{1}|{g}_{m}(t,s)|\mu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \frac{2\delta}{{\pi}^{2m}}{\int}_{0}^{1}\mu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))sin\pi s\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \frac{2\delta}{{\pi}^{2m}}{\int}_{0}^{1}\theta \nu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))sin\pi s\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \frac{2\delta \theta}{\pi}\parallel Sx\parallel \\ >& \frac{2\delta \theta}{\pi}{w}_{3}\ge {w}_{2}.\end{array}$

Hence, we have proved that $\psi (Sx)>{w}_{2}$ for all $x\in C(\psi ,{w}_{2},{w}_{4})$ with $\parallel Sx\parallel >{w}_{3}$.

It now follows from Theorem 2.1 that the Lidstone boundary value problem (3.3) has (at least) three positive solutions ${x}_{1},{x}_{2},{x}_{3}\in \overline{C}({w}_{4})$ satisfying (2.1). It is easy to see that here (2.1) reduces to (3.14). This completes the proof of conclusion (a).

Finally, it is observed from (3.4) that the complementary Lidstone boundary value problem (1.1) has (at least) three positive solutions

${y}_{1}$,

${y}_{2}$,

${y}_{3}$ such that for

$i=1,2,3$,

${y}_{i}(t)={\int}_{0}^{t}{x}_{i}(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\in [0,1]\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\parallel {y}_{i}\parallel \le \parallel {x}_{i}\parallel .$

(3.22)

Moreover, since

${x}_{i}\in C$, we get for

$t\in [\delta ,1-\delta ]$,

${y}_{i}(t)={\int}_{0}^{t}{x}_{i}(s)\phantom{\rule{0.2em}{0ex}}ds\ge {\int}_{\delta}^{t}{x}_{i}(s)\phantom{\rule{0.2em}{0ex}}ds\ge {\int}_{\delta}^{t}\frac{2\delta \theta}{\pi}\parallel {x}_{i}\parallel \phantom{\rule{0.2em}{0ex}}ds=\frac{2\delta \theta}{\pi}\parallel {x}_{i}\parallel (t-\delta ).$

(3.23)

Combining (3.22) and (3.23) gives (3.15) immediately.

Further, since

${x}_{2}(t)>{w}_{2}$ for

$t\in [\delta ,1-\delta ]$, we have for

$t\in [\delta ,1-\delta ]$,

${y}_{2}(t)={\int}_{0}^{t}{x}_{2}(s)\phantom{\rule{0.2em}{0ex}}ds>{\int}_{\delta}^{t}{x}_{2}(s)\phantom{\rule{0.2em}{0ex}}ds>{\int}_{\delta}^{t}{w}_{2}\phantom{\rule{0.2em}{0ex}}ds={w}_{2}(t-\delta ).$

(3.24)

Hence, noting (3.14), (3.15) and (3.24), we get (3.16). This completes the proof of conclusion (b). □

We shall now employ the five-functional fixed point theorem (Theorem 2.2) to give other existence criteria. In applying Theorem 2.2 it is possible to choose the functionals and constants in different ways, indeed we shall do so and derive two results. Our first result below turns out to be a generalization of Theorem 3.1.

**Theorem 3.2** *Let* $\delta \in (0,\frac{1}{2})$ *be fixed*.

*Let* (C1)-(C4)

*hold*.

*Assume that there exist numbers* ${\tau}_{j}$,

$1\le j\le 4$,

*with* $0\le {\tau}_{1}\le \delta \le {\tau}_{2}<\frac{1}{2}<{\tau}_{3}\le 1-\delta \le {\tau}_{4}\le 1$

*such that*

(C7) *for each* $t\in [{\tau}_{2},{\tau}_{3}]$, *the function* $|{g}_{m}(t,s)|\mu (s)>0$ *on a subset of* $[\frac{1}{2},{\tau}_{3}]$ *of positive measure*;

(C8) *the function* $\nu (s)sin\pi s>0$ *on a subset of* $[{\tau}_{1},{\tau}_{4}]$ *of positive measure*.

*Suppose that there exist numbers* ${w}_{i}$,

$2\le i\le 5$,

*with* $\{\begin{array}{l}0<{w}_{2}<{w}_{3}<\frac{\pi {w}_{3}}{2\delta \theta}\le {w}_{4}\le {w}_{5}<\frac{q{w}_{2}}{{p}_{3}},\\ {\tau}_{1}{w}_{5}+({\tau}_{4}-{\tau}_{1}){w}_{2}<{w}_{3}(\frac{1}{2}-{\tau}_{2})\end{array}$

*such that the following hold*:

- (P)
$f(u,v)<\frac{1}{{p}_{2}}({w}_{2}-\frac{{w}_{5}{p}_{3}}{q})$ *for* $u\in [0,{\tau}_{1}{w}_{5}+({\tau}_{4}-{\tau}_{1}){w}_{2}]$ *and* $v\in [0,{w}_{2}]$;

- (Q)
$f(u,v)\le \frac{{w}_{5}}{q}$ *for* $u,v\in [0,{w}_{5}]$;

- (R)
$f(u,v)>\frac{{w}_{3}}{{p}_{1}}$ *for* $u\in [{w}_{3}(\frac{1}{2}-{\tau}_{2}),{w}_{4}(\frac{1}{2}-{\tau}_{2})]$ *and* $v\in [{w}_{3},{w}_{4}]$.

*Then we have the following conclusions*:

- (a)
*The Lidstone boundary value problem* (3.3)

*has* (

*at least*)

*three positive solutions* ${x}_{1},{x}_{2},{x}_{3}\in \overline{C}({w}_{5})$ (

*where* *C* *is defined in* (3.8))

*such that* $\{\begin{array}{l}{x}_{1}(t)<{w}_{2},\phantom{\rule{1em}{0ex}}t\in [{\tau}_{1},{\tau}_{4}];\\ {x}_{2}(t)>{w}_{3},\phantom{\rule{1em}{0ex}}t\in [{\tau}_{2},{\tau}_{3}];\\ {max}_{t\in [{\tau}_{1},{\tau}_{4}]}{x}_{3}(t)>{w}_{2}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{min}_{t\in [{\tau}_{2},{\tau}_{3}]}{x}_{3}(t)<{w}_{3}.\end{array}$

(3.25)

- (b)
*The complementary Lidstone boundary value problem* (1.1)

*has* (

*at least*)

*three positive solutions* ${y}_{1}$,

${y}_{2}$,

${y}_{3}$ *such that* (3.15)

*holds for* $i=1,2,3$.

*We further have* $\{\begin{array}{l}{y}_{1}(t)<{\tau}_{1}{max}_{s\in [0,{\tau}_{1}]}{x}_{1}(s)+({\tau}_{4}-{\tau}_{1}){w}_{2},\phantom{\rule{1em}{0ex}}t\in [{\tau}_{1},{\tau}_{4}];\\ {y}_{2}(t)>{w}_{3}(t-{\tau}_{2}),\phantom{\rule{1em}{0ex}}t\in [{\tau}_{2},{\tau}_{3}];\\ {y}_{3}(t)>\frac{2\delta \theta}{\pi}{w}_{2}(t-\delta ),\phantom{\rule{1em}{0ex}}t\in [\delta ,1-\delta ].\end{array}$

(3.26)

*Proof* We shall apply Theorem 2.2 with the cone

*C* defined in (3.8). We define the following five functionals on the cone

*C*:

$\{\begin{array}{l}\gamma (x)=\parallel x\parallel ,\\ \psi (x)={min}_{t\in [\delta ,1-\delta ]}x(t),\\ \beta (x)=\mathrm{\Theta}(x)={max}_{t\in [{\tau}_{1},{\tau}_{4}]}x(t),\\ \alpha (x)={min}_{t\in [{\tau}_{2},{\tau}_{3}]}x(t).\end{array}$

(3.27)

First, we shall show that the operator *S* maps $\overline{P}(\gamma ,{w}_{5})$ into $\overline{P}(\gamma ,{w}_{5})$. Note that $\overline{P}(\gamma ,{w}_{5})=\overline{C}({w}_{5})$. By (Q) and Lemma 3.3 (with $d={w}_{5}$), we immediately have $S(\overline{C}({w}_{5}))\subseteq \overline{C}({w}_{5})$.

Next, to see that condition (a) of Theorem 2.2 is fulfilled, we note that

$\{x\in P(\gamma ,\mathrm{\Theta},\alpha ,{w}_{3},{w}_{4},{w}_{5})\mid \alpha (x)>{w}_{3}\}\ne \mathrm{\varnothing}$

since it has an element

$x(t)=\frac{1}{2}({w}_{3}+{w}_{4})$. Let

$x\in P(\gamma ,\mathrm{\Theta},\alpha ,{w}_{3},{w}_{4},{w}_{5})$. Then by definition we have

$\alpha (x)\ge {w}_{3}$ and

$\mathrm{\Theta}(x)\le {w}_{4}$, which imply

$\begin{array}{r}x(s)\in [{w}_{3},{w}_{4}],\phantom{\rule{1em}{0ex}}s\in [{\tau}_{2},{\tau}_{3}]\phantom{\rule{1em}{0ex}}\text{and}\\ {\int}_{{\tau}_{2}}^{\frac{1}{2}}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau \in [{w}_{3}(\frac{1}{2}-{\tau}_{2}),{w}_{4}(\frac{1}{2}-{\tau}_{2})].\end{array}$

(3.28)

Noting (3.7), (3.28), (C7) and (R), we find

$\begin{array}{rcl}\alpha (Sx)& =& \underset{t\in [{\tau}_{2},{\tau}_{3}]}{min}Sx(t)\\ \ge & \underset{t\in [{\tau}_{2},{\tau}_{3}]}{min}{\int}_{0}^{1}|{g}_{m}(t,s)|\mu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \underset{t\in [{\tau}_{2},{\tau}_{3}]}{min}{\int}_{\frac{1}{2}}^{{\tau}_{3}}|{g}_{m}(t,s)|\mu (s)f({\int}_{{\tau}_{2}}^{\frac{1}{2}}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))\phantom{\rule{0.2em}{0ex}}ds\\ >& \underset{t\in [{\tau}_{2},{\tau}_{3}]}{min}{\int}_{\frac{1}{2}}^{{\tau}_{3}}|{g}_{m}(t,s)|\mu (s)\phantom{\rule{0.2em}{0ex}}ds\cdot \frac{{w}_{3}}{{p}_{1}}\\ =& {p}_{1}\cdot \frac{{w}_{3}}{{p}_{1}}={w}_{3}.\end{array}$

Hence, $\alpha (Sx)>{w}_{3}$ for all $x\in P(\gamma ,\mathrm{\Theta},\alpha ,{w}_{3},{w}_{4},{w}_{5})$.

We shall now verify that condition (b) of Theorem 2.2 is satisfied. Let

${w}_{1}$ be such that

$0<{w}_{1}<{w}_{2}$. Note that

$\{x\in Q(\gamma ,\beta ,\psi ,{w}_{1},{w}_{2},{w}_{5})\mid \beta (u)<{w}_{2}\}\ne \mathrm{\varnothing}$

because it has an element

$x(t)=\frac{1}{2}({w}_{1}+{w}_{2})$. Let

$x\in Q(\gamma ,\beta ,\psi ,{w}_{1},{w}_{2},{w}_{5})$. Then we have

$\beta (x)\le {w}_{2}$ and

$\gamma (x)\le {w}_{5}$,

*i.e.*,

$x(s)\in [0,{w}_{2}],\phantom{\rule{1em}{0ex}}s\in [{\tau}_{1},{\tau}_{4}]\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x(s)\in [0,{w}_{5}],\phantom{\rule{1em}{0ex}}s\in [0,1],$

(3.29)

which lead to the following:

$\begin{array}{c}{\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau \le {w}_{5},\phantom{\rule{1em}{0ex}}s\in [0,{\tau}_{1}]\cup [{\tau}_{4},1];\hfill \\ {\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau \le {\int}_{0}^{{\tau}_{4}}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ={\int}_{0}^{{\tau}_{1}}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau +{\int}_{{\tau}_{1}}^{{\tau}_{4}}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau \hfill \\ \phantom{{\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau}\le {\tau}_{1}{w}_{5}+({\tau}_{4}-{\tau}_{1}){w}_{2},\phantom{\rule{1em}{0ex}}s\in [{\tau}_{1},{\tau}_{4}].\hfill \end{array}$

(3.30)

Using (3.9), (3.29), (3.30), (C8), (P) and (Q) successively, we find

$\begin{array}{rcl}\beta (Sx)& =& \underset{t\in [{\tau}_{1},{\tau}_{4}]}{max}Sx(t)\\ \le & \frac{1}{{\pi}^{2m-1}}{\int}_{0}^{1}\nu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))sin\pi s\phantom{\rule{0.2em}{0ex}}ds\\ =& \frac{1}{{\pi}^{2m-1}}({\int}_{0}^{{\tau}_{1}}+{\int}_{{\tau}_{1}}^{{\tau}_{4}}+{\int}_{{\tau}_{4}}^{1})\nu (s)f({\int}_{0}^{s}x(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,x(s))sin\pi s\phantom{\rule{0.2em}{0ex}}ds\\ <& [\frac{1}{{\pi}^{2m-1}}{\int}_{0}^{{\tau}_{1}}\nu (s)sin\pi s\phantom{\rule{0.2em}{0ex}}ds+\frac{1}{{\pi}^{2m-1}}{\int}_{{\tau}_{4}}^{1}\nu (s)sin\pi s\phantom{\rule{0.2em}{0ex}}ds]\frac{{w}_{5}}{q}\\ +[\frac{1}{{\pi}^{2m-1}}{\int}_{{\tau}_{1}}^{{\tau}_{4}}\nu (s)sin\pi s\phantom{\rule{0.2em}{0ex}}ds]\frac{1}{{p}_{2}}({w}_{2}-\frac{{w}_{5}{p}_{3}}{q})\\ =& {p}_{3}\cdot \frac{{w}_{5}}{q}+{p}_{2}\cdot \frac{1}{{p}_{2}}({w}_{2}-\frac{{w}_{5}{p}_{3}}{q})={w}_{2}.\end{array}$

Therefore, $\beta (Sx)<{w}_{2}$ for all $x\in Q(\gamma ,\beta ,\psi ,{w}_{1},{w}_{2},{w}_{5})$.

Next, we shall show that condition (c) of Theorem 2.2 is met. Let

$x\in C$. Clearly, we have

$\mathrm{\Theta}(Sx)=\underset{t\in [{\tau}_{1},{\tau}_{4}]}{max}Sx(t)\le \parallel Sx\parallel .$

(3.31)

Moreover, using the fact that

*S* maps

*C* into

*C*, we find

$\alpha (Sx)=\underset{t\in [{\tau}_{2},{\tau}_{3}]}{min}Sx(t)\ge \underset{t\in [\delta ,1-\delta ]}{min}Sx(t)\ge \frac{2\delta \theta}{\pi}\parallel Sx\parallel .$

(3.32)

Combining (3.31) and (3.32) yields

$\alpha (Sx)\ge \frac{2\delta \theta}{\pi}\mathrm{\Theta}(Sx),\phantom{\rule{1em}{0ex}}x\in C.$

(3.33)

Now, let

$x\in P(\gamma ,\alpha ,{w}_{3},{w}_{5})$ with

$\mathrm{\Theta}(Sx)>{w}_{4}$. Then it follows from (3.33) and the inequality

$\frac{\pi {w}_{3}}{2\delta \theta}\le {w}_{4}$ that

$\alpha (Sx)\ge \frac{2\delta \theta}{\pi}\mathrm{\Theta}(Sx)>\frac{2\delta \theta}{\pi}{w}_{4}\ge \frac{2\delta \theta}{\pi}\frac{\pi {w}_{3}}{2\delta \theta}={w}_{3}.$

(3.34)

Thus, $\alpha (Sx)>{w}_{3}$ for all $x\in P(\gamma ,\alpha ,{w}_{3},{w}_{5})$ with $\mathrm{\Theta}(Sx)>{w}_{4}$.

Finally, we shall prove that condition (d) of Theorem 2.2 is fulfilled. Let $x\in Q(\gamma ,\beta ,{w}_{2},{w}_{5})$ with $\psi (Sx)<{w}_{1}$. Then we have $\beta (x)\le {w}_{2}$ and $\gamma (x)\le {w}_{5}$ which give (3.29) and (3.30). As in proving condition (b), we get $\beta (Sx)<{w}_{2}$. Hence, condition (d) of Theorem 2.2 is satisfied.

It now follows from Theorem 2.2 that the Lidstone boundary value problem (3.3) has (at least) three positive solutions ${x}_{1},{x}_{2},{x}_{3}\in \overline{P}(\gamma ,{w}_{5})=\overline{C}({w}_{5})$ satisfying (2.2). Furthermore, (2.2) reduces to (3.25) immediately. This completes the proof of conclusion (a).

Finally, as in the proof of Theorem 3.1, we see that (3.15) holds for the positive solutions

${y}_{i}$,

$i=1,2,3$, of the complementary Lidstone boundary value problem (1.1). Moreover, noting that

${x}_{1}(t)<{w}_{2}$ for

$t\in [{\tau}_{1},{\tau}_{4}]$, we find for

$t\in [{\tau}_{1},{\tau}_{4}]$,

$\begin{array}{rcl}{y}_{1}(t)& =& {\int}_{0}^{t}{x}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds\le {\int}_{0}^{{\tau}_{4}}{x}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds={\int}_{0}^{{\tau}_{1}}{x}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{{\tau}_{1}}^{{\tau}_{4}}{x}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds\\ <& {\tau}_{1}\underset{s\in [0,{\tau}_{1}]}{max}{x}_{1}(s)+({\tau}_{4}-{\tau}_{1}){w}_{2}.\end{array}$

Next, noting

${x}_{2}(t)>{w}_{3}$ for

$t\in [{\tau}_{2},{\tau}_{3}]$, we get for

$t\in [{\tau}_{2},{\tau}_{3}]$,

${y}_{2}(t)={\int}_{0}^{t}{x}_{2}(s)\phantom{\rule{0.2em}{0ex}}ds>{\int}_{{\tau}_{2}}^{t}{x}_{2}(s)\phantom{\rule{0.2em}{0ex}}ds>{w}_{3}(t-{\tau}_{2}).$

Lastly, using (3.15) and

$\parallel {x}_{3}\parallel \ge {max}_{t\in [{\tau}_{1},{\tau}_{4}]}{x}_{3}(t)>{w}_{2}$, we find for

$t\in [\delta ,1-\delta ]$,

${y}_{3}(t)\ge \frac{2\delta \theta}{\pi}\parallel {x}_{3}\parallel (t-\delta )>\frac{2\delta \theta}{\pi}{w}_{2}(t-\delta ).$

The proof of conclusion (b) is complete. □

We shall now consider the special case of Theorem 3.2 when

${\tau}_{1}=0,\phantom{\rule{2em}{0ex}}{\tau}_{2}=\delta ,\phantom{\rule{2em}{0ex}}{\tau}_{3}=1-\delta \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\tau}_{4}=1.$

Then, from definitions (3.11), we see that

${p}_{1}=r,\phantom{\rule{2em}{0ex}}{p}_{2}=q\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{p}_{3}=0.$

In this case Theorem 3.2 yields the following corollary.

**Corollary 3.1** *Let* $\delta \in (0,\frac{1}{2})$ *be fixed*. *Let* (C1)-(C4) *hold*, *and assume*

(C7)′ *for each* $t\in [\delta ,1-\delta ]$, *the function* $|{g}_{m}(t,s)|\mu (s)>0$ *on a subset of* $[\frac{1}{2},1-\delta ]$ *of positive measure*;

(C8)′ *the function* $\nu (s)sin\pi s>0$ *on a subset of* $[0,1]$ *of positive measure*.

*Suppose that there exist numbers* ${w}_{i}$,

$2\le i\le 5$,

*with* $0<{w}_{2}<{w}_{3}(\frac{1}{2}-\delta )<{w}_{3}<\frac{\pi {w}_{3}}{2\delta \theta}\le {w}_{4}\le {w}_{5}$

*such that the following hold*:

- (P)
$f(u,v)<\frac{{w}_{2}}{q}$ *for* $u,v\in [0,{w}_{2}]$;

- (Q)
$f(u,v)\le \frac{{w}_{5}}{q}$ *for* $u,v\in [0,{w}_{5}]$;

- (R)
$f(u,v)>\frac{{w}_{3}}{r}$ *for* $u\in [{w}_{3}(\frac{1}{2}-\delta ),{w}_{4}(\frac{1}{2}-\delta )]$ *and* $v\in [{w}_{3},{w}_{4}]$.

*Then we have the following conclusions*:

- (a)
*The Lidstone boundary value problem* (3.3)

*has* (

*at least*)

*three positive solutions* ${x}_{1},{x}_{2},{x}_{3}\in \overline{C}({w}_{5})$ (

*where* *C* *is defined in* (3.8))

*such that* $\{\begin{array}{l}\parallel {x}_{1}\parallel <{w}_{2};\\ {x}_{2}(t)>{w}_{3},\phantom{\rule{1em}{0ex}}t\in [\delta ,1-\delta ];\\ \parallel {x}_{3}\parallel >{w}_{2}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{min}_{t\in [\delta ,1-\delta ]}{x}_{3}(t)<{w}_{3}.\end{array}$

(3.35)

- (b)
*The complementary Lidstone boundary value problem* (1.1)

*has* (

*at least*)

*three positive solutions* ${y}_{1}$,

${y}_{2}$,

${y}_{3}$ *such that* (3.15)

*holds for* $i=1,2,3$.

*We further have* $\{\begin{array}{l}\parallel {y}_{1}\parallel <{w}_{2};\\ {y}_{2}(t)>{w}_{3}(t-\delta ),\phantom{\rule{1em}{0ex}}t\in [\delta ,1-\delta ];\\ {y}_{3}(t)>\frac{2\delta \theta}{\pi}{w}_{2}(t-\delta ),\phantom{\rule{1em}{0ex}}t\in [\delta ,1-\delta ].\end{array}$

(3.36)

**Remark 3.1** Corollary 3.1 is actually Theorem 3.1. Since Corollary 3.1 is a special case of Theorem 3.2, this shows that Theorem 3.2 is *more general* than Theorem 3.1.

The next theorem illustrates another application of Theorem 2.2. Compared to the conditions in Theorem 3.2, here the numbers ${w}_{1}$, ${\tau}_{1}$ and ${\tau}_{4}$ have different ranges and condition (P) is also different. Note that in the proof of Theorem 3.3 the functionals *ψ* and Θ are chosen differently from those in Theorem 3.2.

**Theorem 3.3** *Let* $\delta \in (0,\frac{1}{2})$ *be fixed*.

*Let* (C1)-(C4)

*hold*.

*Assume that there exist numbers* ${\tau}_{j}$,

$1\le j\le 4$,

*with* $\delta \le {\tau}_{1}\le {\tau}_{2}<\frac{1}{2}<{\tau}_{3}\le {\tau}_{4}\le 1-\delta $

*such that* (C7)

*and* (C8)

*hold*.

*Suppose that there exist numbers* ${w}_{i}$,

$1\le i\le 5$,

*with* $\{\begin{array}{l}0<{w}_{1}\le \frac{2\delta \theta {w}_{2}}{\pi}<{w}_{2}<{w}_{3}<\frac{\pi {w}_{3}}{2\delta \theta}\le {w}_{4}\le {w}_{5}<\frac{q{w}_{2}}{{p}_{3}},\\ {\tau}_{1}{w}_{5}+({\tau}_{4}-{\tau}_{1}){w}_{2}<{w}_{3}(\frac{1}{2}-{\tau}_{2})\end{array}$

*such that the following hold*:

- (P)
$f(u,v)<\frac{1}{{p}_{2}}({w}_{2}-\frac{{w}_{5}{p}_{3}}{q})$ *for* $u\in [0,{\tau}_{1}{w}_{5}+({\tau}_{4}-{\tau}_{1}){w}_{2}]$ *and* $v\in [{w}_{1},{w}_{2}]$;

- (Q)
$f(u,v)\le \frac{{w}_{5}}{q}$ *for* $u,v\in [0,{w}_{5}]$;

- (R)
$f(u,v)>\frac{{w}_{3}}{{p}_{1}}$ *for* $u\in [{w}_{3}(\frac{1}{2}-{\tau}_{2}),{w}_{4}(\frac{1}{2}-{\tau}_{2})]$ *and* $v\in [{w}_{3},{w}_{4}]$.

*Then we have conclusions* (a) *and* (b) *of Theorem * 3.2.

*Proof* To apply Theorem 2.2, we shall define the following functionals on the cone

*C* (see (3.8)):

$\{\begin{array}{l}\gamma (x)=\parallel x\parallel ,\\ \psi (x)={min}_{t\in [{\tau}_{1},{\tau}_{4}]}x(t),\\ \beta (x)={max}_{t\in [{\tau}_{1},{\tau}_{4}]}x(t),\\ \alpha (u)={min}_{t\in [{\tau}_{2},{\tau}_{3}]}x(t),\\ \mathrm{\Theta}(u)={max}_{t\in [{\tau}_{2},{\tau}_{3}]}x(t).\end{array}$

(3.37)

As in the proof of Theorem 3.2, using (Q) and Lemma 3.3 we can show that $S:\overline{P}(\gamma ,{w}_{5})\to \overline{P}(\gamma ,{w}_{5})$.

Next, to see that condition (a) of Theorem 2.2 is fulfilled, we use (R) and a similar argument as in the proof of Theorem 3.2.

We shall now prove that condition (b) of Theorem 2.2 is satisfied. Note that

$x(t)=\frac{1}{2}({w}_{1}+{w}_{2})\in \{x\in Q(\gamma ,\beta ,\psi ,{w}_{1},{w}_{2},{w}_{5})\mid \beta (u)<{w}_{2}\}\ne \mathrm{\varnothing}.$

Let

$x\in Q(\gamma ,\beta ,\psi ,{w}_{1},{w}_{2},{w}_{5})$. Then we have

$\psi (x)\ge {w}_{1}$,

$\beta (x)\le {w}_{2}$ and

$\gamma (x)\le {w}_{5}$ which imply

$x(s)\in [{w}_{1},{w}_{2}],\phantom{\rule{1em}{0ex}}s\in [{\tau}_{1},{\tau}_{4}]\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x(s)\in [0,{w}_{5}],\phantom{\rule{1em}{0ex}}s\in [0,1]$

(3.38)

and also (3.30). In view of (3.9), (3.38), (3.30), (C8), (P) and (Q), we obtain, as in the proof of Theorem 3.2, that $\beta (Sx)<{w}_{2}$. Therefore, condition (b) of Theorem 2.2 is fulfilled.

Next, using a similar argument as in the proof of Theorem 3.2, we see that condition (c) of Theorem 2.2 is met.

Finally, we shall verify that condition (d) of Theorem 2.2 is fulfilled. Let

$x\in C$. It is clear that

$\beta (Sx)=\underset{t\in [{\tau}_{1},{\tau}_{4}]}{max}Sx(t)\le \parallel Sx\parallel .$

(3.39)

Noting that

*S* maps

*C* into

*C*, we find

$\psi (Sx)=\underset{t\in [{\tau}_{1},{\tau}_{4}]}{min}Sx(t)\ge \underset{t\in [\delta ,1-\delta ]}{min}Sx(t)\ge \frac{2\delta \theta}{\pi}\parallel Sx\parallel .$

(3.40)

A combination of (3.39) and (3.40) gives

$\psi (Sx)\ge \frac{2\delta \theta}{\pi}\beta (Sx),\phantom{\rule{1em}{0ex}}x\in C.$

(3.41)

Let

$x\in Q(\gamma ,\beta ,{w}_{2},{w}_{5})$ with

$\psi (Sx)<{w}_{1}$. Then (3.41) and the inequality

${w}_{1}\le \frac{2\delta \theta {w}_{2}}{\pi}$ lead to

$\beta (Sx)\le \frac{\pi}{2\delta \theta}\psi (Sx)<\frac{\pi}{2\delta \theta}{w}_{1}\le \frac{\pi}{2\delta \theta}\frac{2\delta \theta}{\pi}{w}_{2}={w}_{2}.$

Thus, $\beta (Sx)<{w}_{2}$ for all $x\in Q(\gamma ,\beta ,{w}_{2},{w}_{5})$ with $\psi (Sx)<{w}_{1}$.

Conclusion (a) now follows from Theorem 2.2 immediately, while conclusion (b) is similarly obtained as in Theorem 3.2. □