Now, we present the necessary definitions from the theory of cones in Banach spaces.
Definition 1 Let E be a real Banach space. A nonempty convex closed set P\subset E is said to be a cone provided that

(i)
ku\in P for all u\in P and all k\ge 0, and

(ii)
u,u\in P implies u=0.
Note that every cone P\subset E induces an ordering in E given by x\le y if yx\in P.
Definition 2 A map Φ is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E if \mathrm{\Phi}:P\to {\mathbb{R}}_{+} is continuous and
\mathrm{\Phi}(tx+(1t)y)\ge t\mathrm{\Phi}(x)+(1t)\mathrm{\Phi}(y)
for all x,y\in P and t\in [0,1].
Similarly, we say the map φ is a nonnegative continuous convex functional on a cone P of a real Banach space E if \phi :P\to {\mathbb{R}}_{+} is continuous and
\phi (tx+(1t)y)\le t\phi (x)+(1t)\phi (y)
for all x,y\in P and t\in [0,1].
Definition 3 An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.
Let φ and Θ be nonnegative continuous convex functionals on P, let Φ be a nonnegative continuous concave functional on P, and let Ψ be a nonnegative continuous functional on P. Then, for positive numbers a, b, c, d, we define the following sets:
We will use the following fixed point theorem of Avery and Peterson to establish multiple positive solutions to problem (1).
Theorem 1 (see [1])
Let P be a cone in a real Banach space E. Let φ and Θ be nonnegative continuous convex functionals on P, let Φ be a nonnegative continuous concave functional on P, and let Ψ be a nonnegative continuous functional on P satisfying \mathrm{\Psi}(kx)\le k\mathrm{\Psi}(x) for 0\le k\le 1 such that for some positive numbers \overline{M} and d,
\mathrm{\Phi}(x)\le \mathrm{\Psi}(x)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\parallel x\parallel \le \overline{M}\phi (x)
for all x\in \overline{P(\phi ,d)}. Suppose
T:\overline{P(\phi ,d)}\to \overline{P(\phi ,d)}
is completely continuous and there exist positive numbers a, b, c with a<b such that
(S_{1}): \{x\in P(\phi ,\mathrm{\Theta},\mathrm{\Phi},b,c,d):\mathrm{\Phi}(x)>b\}\ne 0 and \mathrm{\Phi}(Tx)>b for x\in P(\phi ,\mathrm{\Theta},\mathrm{\Phi},b,c,d),
(S_{2}): \mathrm{\Phi}(Tx)>b for x\in P(\phi ,\mathrm{\Phi},b,d) with \mathrm{\Theta}(Tx)>c,
(S_{3}): 0\notin R(\phi ,\mathrm{\Psi},a,d) and \mathrm{\Psi}(Tx)<a for x\in R(\phi ,\mathrm{\Psi},a,d) with \mathrm{\Psi}(x)=a.
Then
T
has at least three fixed points
{x}_{1},{x}_{2},{x}_{3}\in \overline{P(\phi ,d)}
such that
and
\mathrm{\Psi}({x}_{3})<a.
We apply Theorem 1 with the cone K instead of P and let {\overline{P}}_{r}=\{x\in K:\parallel x\parallel \le r\}. Now, we define the nonnegative continuous concave functional Φ on K by
\mathrm{\Phi}(x)=\underset{[0,\eta ]}{min}x(t).
Note that \mathrm{\Phi}(x)\le {\parallel x\parallel}_{1}. Put \mathrm{\Psi}(x)=\mathrm{\Theta}(x)={\parallel x\parallel}_{1},\phi (x)={\parallel {x}^{\prime}\parallel}_{1}.
Now, we can formulate the main result of this section.
Theorem 2 Let the assumptions H_{1}H_{4} hold with \xi =0, \gamma >0. Let \alpha (t)\le t, t\in J. In addition, we assume that there exist positive constants a, b, c, d, M, a<b and such that
with
and
(A_{1}): f(t,u,v)\le \frac{d}{\mu} for (t,u,v)\in J\times [0,Md]\times [d,d],
(A_{2}): f(t,u,v)\ge \frac{b}{L} for (t,u,v)\in [0,\eta ]\times [b,\frac{b}{\rho}]\times [d,d],
(A_{3}): f(t,u,v)\le \frac{a}{\mu} for (t,u,v)\in J\times [0,a]\times [d,d].
Then problem (1) has at least three nonnegative solutions {x}_{1}, {x}_{2}, {x}_{3} satisfying {\parallel {x}_{i}^{\prime}\parallel}_{1}\le d, i=1,2,3,
b\le \mathrm{\Phi}({x}_{1}),\phantom{\rule{2em}{0ex}}a<{\parallel {x}_{2}\parallel}_{1}\phantom{\rule{1em}{0ex}}\mathit{\text{with}}\phantom{\rule{0.1em}{0ex}}\mathrm{\Phi}({x}_{2})<b
and {\parallel {x}_{3}\parallel}_{1}<a.
Proof Basing on the definitions of T, we see that T\overline{P} is equicontinuous on J, so T is completely continuous.
Let x\in \overline{P(\phi ,d)}, so \phi (x)={\parallel {x}^{\prime}\parallel}_{1}\le d. By Lemma 1, {\parallel x\parallel}_{1}\le Md, so 0\le x(t)\le Md, t\in J. Assumption (A_{1}) implies f(t,x(\alpha (t)),{x}^{\prime}(\beta (t)))\le \frac{d}{\mu}.
Moreover, in view of (7),
Combining it, we have
\begin{array}{rcl}\phi (Tx)& =& \underset{[0,1]}{max}{(Tx)}^{\prime}(t)\\ \le & \frac{1}{\mathrm{\Delta}}(1{B}_{1}){\lambda}_{1}[Fx]+\frac{1}{\mathrm{\Delta}}(1\gamma {A}_{1}){\lambda}_{2}[Fx]+\underset{[0,1]}{max}{(Fx)}^{\prime}(t)\\ \le & \frac{d}{\mu}(\frac{1}{\mathrm{\Delta}}(1{B}_{1}){D}_{1}+\frac{1}{\mathrm{\Delta}}(1\gamma {A}_{1}){D}_{2}+{D}_{3})<d.\end{array}
This proves that T:\overline{P(\phi ,d)}\to \overline{P(\phi ,d)}.
Now, we need to show that condition (S_{1}) is satisfied. Take
{x}_{0}(t)=\frac{1}{2}(b+\frac{b}{\rho}),\phantom{\rule{1em}{0ex}}t\in J.
Then {x}_{0}(t)>0, t\in J, and
for p(t)=1, t\in J. Moreover,
This proves that
\{{x}_{0}\in P(\phi ,\mathrm{\Theta},\mathrm{\Phi},b,\frac{b}{\rho},d):b<\mathrm{\Phi}({x}_{0})\}\ne \mathrm{\varnothing}.
Let b\le x(t)\le \frac{b}{\rho} for t\in [0,\eta ]. Then 0\le \alpha (t)\le t\le \eta for t\in [0,\eta ], so b\le x(\alpha (t))\le \frac{b}{\rho}, t\in [0,\eta ]. Assumption (A_{2}) implies f(t,x(\alpha (t)),{x}^{\prime}(\beta (t)))\ge \frac{b}{L}. Hence,
Moreover,
It yields
\begin{array}{rcl}\mathrm{\Phi}(Tx)& =& \underset{[0,\eta ]}{min}(Tx)(t)=min((Tx)(0),(Tx)(\eta ))\ge \gamma (Tx)(\eta )\\ =& \frac{\gamma}{\mathrm{\Delta}}[1{B}_{2}(1{B}_{1})\eta ]{\lambda}_{1}[Fx]+\frac{\gamma}{\mathrm{\Delta}}[\eta \gamma +{A}_{2}+(1\gamma {A}_{1})\eta ]{\lambda}_{2}[Fx]\\ +\frac{\gamma}{\delta}{\int}_{0}^{1}G(\eta ,s)h(s)f(s,x(\alpha (s)),{x}^{\prime}(\beta (s)))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \frac{b\gamma}{L}(\frac{1}{\mathrm{\Delta}}([1{B}_{2}(1{B}_{1})\eta ]{D}_{1}+[\eta \gamma +{A}_{2}+(1\gamma {A}_{1})\eta ]{D}_{2})+\frac{{D}_{4}}{\delta})\\ >& b.\end{array}
This proves that condition (S_{1}) holds.
Now, we need to prove that condition (S_{2}) is satisfied. Take x\in P(\phi ,\mathrm{\Phi},b,d) and {\parallel Tx\parallel}_{1}>\frac{b}{\rho}=c. Then
\mathrm{\Phi}(Tx)=\underset{[0,\eta ]}{min}(Tx)(t)\ge \rho {\parallel Tx\parallel}_{1}>\rho \frac{b}{\rho}=b,
so condition (S_{2}) holds.
Indeed, \phi (0)=0<a, so 0\notin R(\phi ,\mathrm{\Psi},a,d). Suppose that x\in R(\phi ,\mathrm{\Psi},a,d) with \mathrm{\Psi}(x)={\parallel x\parallel}_{1}=a. Note that G(t,s)\le G(s,s), t\in J. Then
\begin{array}{rcl}{\parallel Fx\parallel}_{1}& \le & \frac{\gamma}{\delta}{\int}_{0}^{1}G(\eta ,s)h(s)f(s,x(\alpha (s)),{x}^{\prime}(\beta (s)))\phantom{\rule{0.2em}{0ex}}ds\\ +{\int}_{0}^{1}G(s,s)h(s)f(s,x(\alpha (s)),{x}^{\prime}(\beta (s)))\phantom{\rule{0.2em}{0ex}}ds\\ \le & \frac{a}{\mu}[\frac{\gamma}{\delta}{\int}_{0}^{1}G(\eta ,s)h(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{1}G(s,s)h(s)\phantom{\rule{0.2em}{0ex}}ds]\\ \le & \frac{a}{\mu}{D}_{5}\end{array}
and finally,
\begin{array}{rcl}\mathrm{\Psi}(Tx)& =& \underset{t\in J}{max}(Tx)(t)\\ \le & \frac{1}{\mathrm{\Delta}}[1{B}_{2}]{\lambda}_{1}[Fx]+\frac{1}{\mathrm{\Delta}}[\eta \gamma +{A}_{2}+1\gamma {A}_{1}]{\lambda}_{2}[Fx]+{\parallel Fx\parallel}_{1}\\ \le & \frac{a}{\mu}(\frac{1}{\mathrm{\Delta}}([1{B}_{2}]{D}_{1}+[\eta \gamma +{A}_{2}+1\gamma {A}_{1}]{D}_{2})+{D}_{5})\\ <& a.\end{array}
This shows that condition (S_{3}) is satisfied.
Since all the conditions of Theorem 1 are satisfied, problem (1) has at least three nonnegative solutions {x}_{1}, {x}_{2}, {x}_{3} such that \parallel {x}_{i}^{\prime}\parallel \le d for i=1,2,3, and
b\le \underset{[0,\eta ]}{min}{x}_{1}(t),\phantom{\rule{2em}{0ex}}a<{\parallel {x}_{2}\parallel}_{1}\phantom{\rule{1em}{0ex}}\text{with}\underset{[0,\eta ]}{min}{x}_{2}(t)b,\phantom{\rule{2em}{0ex}}{\parallel {x}_{3}\parallel}_{1}a.
This ends the proof. □
Example Consider the following problem:
\{\begin{array}{c}{x}^{\u2033}(t)+hf(t,x(\alpha (t)),{x}^{\prime}(\beta (t)))=0,\phantom{\rule{1em}{0ex}}t\in (0,1),\hfill \\ x(0)=\frac{1}{4}x(\frac{1}{2}),\phantom{\rule{2em}{0ex}}x(1)=\frac{1}{2}{\int}_{0}^{1}x(t)(7t2)\phantom{\rule{0.2em}{0ex}}dt,\hfill \end{array}
(8)
where
f(t,u,v)=\{\begin{array}{cc}\frac{1}{100}cost+{(\frac{v}{20,000})}^{2},\hfill & (t,u,v)\in [0,1]\times [0,1]\times [d,d],\hfill \\ \frac{1}{100}cost+2(u1)+{(\frac{v}{20,000})}^{2},\hfill & (t,u,v)\in [0,1]\times [1,16]\times [d,d],\hfill \\ \frac{1}{100}cost+30+{(\frac{v}{20,000})}^{2},\hfill & (t,v)\in [0,1]\times [d,d],u\ge 16\hfill \end{array}
with d=2,000. For example, we can take \alpha (t)=\overline{\rho}t, \beta (t)=\sqrt{t} on J with fixed \overline{\rho}\in (0,1). Indeed, f\in C([0,1]\times {\mathbb{R}}_{+}\times [d,d],{\mathbb{R}}_{+}), \gamma =\frac{1}{4}, \eta =\frac{1}{2}, h(t)=h>0, \xi =0 and
{\lambda}_{1}[x]=0,\phantom{\rule{2em}{0ex}}{\lambda}_{2}[x]=\frac{1}{2}{\int}_{0}^{1}x(t)(7t2)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{2em}{0ex}}\rho =\frac{1}{8}.
Note that dB(t)=\frac{1}{2}(7t2)\phantom{\rule{0.2em}{0ex}}dt, so the measure changes the sign on J. Moreover,
so the assumption H_{4} holds; see Remark 4. Next,
Put a=1, b=2, h=30, then c=16, \mu >37.18, L<1.94. Let \mu =40, L=1. Then
and
f(t,u,v)\le \frac{1}{100}+30+{\left(\frac{2,000}{20,000}\right)}^{2}=30.02<50=\frac{d}{\mu}
for (t,u,v)\in [0,1]\times [0,2d]\times [d,d].
All the assumptions of Theorem 2 hold, so problem (8) has at least three positive solutions.
Remark 5 We can also construct an example in which, for example, {\lambda}_{1}[x]={\int}_{0}^{1}x(t)(3t1)\phantom{\rule{0.2em}{0ex}}dt to use the results of Remark 4. Note that also this measure changes the sign.