In this section we present our main results.
Given two nonnegative constants
c,
d such that
$\overline{q}{c}^{p}\ne \overline{r}{d}^{p}$, where
$\overline{q}=\frac{1}{p}{\left(\frac{{q}_{0}}{ba}\right)}^{p1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\overline{r}=\frac{1}{p}{\left(\frac{2}{ba}\right)}^{p1}[{\parallel q\parallel}_{\mathrm{\infty}}+\frac{p+2}{p+1}{\left(\frac{ba}{2}\right)}^{p}{\parallel r\parallel}_{\mathrm{\infty}}],$
put
${a}_{d}(c):=\frac{{\int}_{a}^{b}{max}_{\xi \le c}F(x,\xi )\phantom{\rule{0.2em}{0ex}}dx{\int}_{\frac{a+b}{2}}^{b}F(x,d)\phantom{\rule{0.2em}{0ex}}dx}{\overline{q}{c}^{p}\overline{r}{d}^{p}}.$
Theorem 3.1 Under the following conditions:
 (i)
there exist three constants ${c}_{1}$,
${c}_{2}$,
d,
with ${\left(\frac{{q}_{0}^{p2}}{{2}^{p1}}\right)}^{\frac{1}{p}}{c}_{1}<d<{\left(\frac{{q}_{0}^{p1}}{{2}^{p1}}\right)}^{\frac{1}{p}}\frac{1}{{[{\parallel q\parallel}_{\mathrm{\infty}}+\frac{p+2}{p+1}{(\frac{ba}{2})}^{p}{\parallel r\parallel}_{\mathrm{\infty}}]}^{\frac{1}{p}}}{c}_{2},$
(3.1)
such that
${a}_{d}({c}_{2})<{a}_{d}({c}_{1}),$
and ${\int}_{a}^{\frac{a+b}{2}}F(x,t)\phantom{\rule{0.2em}{0ex}}dx\ge 0$ $\mathrm{\forall}t\in [0,d]$;
 (ii)
there exist
$\nu >p$
and
$R>0$
such that
For each $\lambda \in \phantom{\rule{0.2em}{0ex}}]\frac{1}{{a}_{d}({c}_{1})},\frac{1}{{a}_{d}({c}_{2})}[$, problem (P) admits at least two nontrivial weak solutions ${\overline{u}}_{1}$, ${\overline{u}}_{2}$, with ${\overline{u}}_{1}$ such that ${(\frac{{q}_{0}}{ba})}^{\frac{p1}{p}}{c}_{1}<\parallel {\overline{u}}_{1}\parallel <{(\frac{{q}_{0}}{ba})}^{\frac{p1}{p}}{c}_{2}$.
Proof The proof of this theorem is divided into two steps. In the first part, by applying Theorem 2.1, we prove the existence of a local minimum for the functional $\mathrm{\Phi}\lambda \mathrm{\Psi}$, where $\mathrm{\Phi}(u)$ and $\mathrm{\Psi}(u)$ are functionals given in (2.7) for all $u\in X$. Obviously, Φ and Ψ satisfy all regularity assumptions requested in Theorem 2.1, and the critical points in X of the functional $\mathrm{\Phi}\lambda \mathrm{\Psi}$ are exactly the weak solutions of problem (P). To this end, we verify condition (2.4) of Theorem 2.1.
Define the following function
${u}_{0}\in X$, by setting
${u}_{0}(x)=\{\begin{array}{cc}\frac{2d}{ba}(xa)\hfill & \text{if}x\in [a,\frac{a+b}{2}[,\hfill \\ d\hfill & \text{if}x\in [\frac{a+b}{2},b],\hfill \end{array}$
and estimate
$\mathrm{\Psi}({u}_{0})$ and
$\mathrm{\Phi}({u}_{0})$ as follows:
$\mathrm{\Psi}({u}_{0})\ge {\int}_{\frac{a+b}{2}}^{b}F(x,d)\phantom{\rule{0.2em}{0ex}}dx,$
(3.2)
and
$\mathrm{\Phi}({u}_{0})\le \frac{1}{p}{\left(\frac{2}{ba}\right)}^{p1}[{\parallel q\parallel}_{\mathrm{\infty}}+\frac{p+2}{p+1}{\left(\frac{ba}{2}\right)}^{p}{\parallel r\parallel}_{\mathrm{\infty}}]{d}^{p}.$
(3.3)
Fix ${c}_{1}$, d, ${c}_{2}$ satisfying (3.1) and put ${r}_{1}=\frac{1}{p}{(\frac{{q}_{0}}{ba})}^{p1}{c}_{1}^{p}$ and ${r}_{2}=\frac{1}{p}{(\frac{{q}_{0}}{ba})}^{p1}{c}_{2}^{p}$.
From (3.1), one has ${r}_{1}<\mathrm{\Phi}({u}_{0})<{r}_{2}$.
Moreover, for all
$u\in X$ such that
$u\in {\mathrm{\Phi}}^{1}(]\mathrm{\infty},{r}_{2}[)$, one has
$\mathrm{\Psi}(u)={\int}_{a}^{b}F(x,u(x))\phantom{\rule{0.2em}{0ex}}dx\le {\int}_{a}^{b}\underset{\xi \le {c}_{2}}{max}F(x,\xi )\phantom{\rule{0.2em}{0ex}}dx.$
(3.4)
Hence,
$\underset{u\in {\mathrm{\Phi}}^{1}(]\mathrm{\infty},{r}_{2}[)}{sup}\mathrm{\Psi}(u)\le {\int}_{a}^{b}\underset{\xi \le {c}_{2}}{max}F(x,\xi )\phantom{\rule{0.2em}{0ex}}dx.$
(3.5)
Now, arguing as before, we obtain
$\underset{u\in {\mathrm{\Phi}}^{1}(]\mathrm{\infty},{r}_{1}])}{sup}\mathrm{\Psi}(u)\le {\int}_{a}^{b}\underset{\xi \le {c}_{1}}{max}F(x,\xi )\phantom{\rule{0.2em}{0ex}}dx.$
(3.6)
From hypothesis (i) and bearing in mind (3.3), (3.2), (3.5), and (3.6), we obtain
$\beta ({r}_{1},{r}_{2})<{\rho}_{2}({r}_{1},{r}_{2}).$
From Theorem 2.1, for each
$\lambda \in \phantom{\rule{0.2em}{0ex}}]\frac{1}{{a}_{d}({c}_{1})},\frac{1}{{a}_{d}({c}_{2})}[$,
$\mathrm{\Phi}\lambda \mathrm{\Psi}$ admits at least one critical point
${\overline{u}}_{1}$ which is a local minimum such that
${\left(\frac{{q}_{0}}{ba}\right)}^{\frac{p1}{p}}{c}_{1}<\parallel {\overline{u}}_{1}\parallel <{\left(\frac{{q}_{0}}{ba}\right)}^{\frac{p1}{p}}{c}_{2}.$
Now, we prove the existence of the second local minimum distinct from the first one. To this end, we must show that the functional $\mathrm{\Phi}\lambda \mathrm{\Psi}$ satisfies the hypotheses of the mountain pass theorem.
Clearly, the functional $\mathrm{\Phi}\lambda \mathrm{\Psi}$ is of class ${C}^{1}$ and $(\mathrm{\Phi}\lambda \mathrm{\Psi})(0)=0$.
From the first part of the proof, we can assume that ${\overline{u}}_{1}$ is a strict local minimum for $\mathrm{\Phi}\lambda \mathrm{\Psi}$ in X. Therefore, there is $\rho >0$ such that ${inf}_{\parallel u{u}_{1}\parallel =\rho}(\mathrm{\Phi}\lambda \mathrm{\Psi})(u)>(\mathrm{\Phi}\lambda \mathrm{\Psi})({\overline{u}}_{1})$, so condition [[13], (${I}_{1}$), Theorem 2.2] is verified.
Now, choosing any
$u\in X\setminus \{0\}$, from (ii) one has
$\begin{array}{rcl}(\mathrm{\Phi}\lambda \mathrm{\Psi})(tu)& =& \frac{1}{p}{\parallel tu\parallel}^{p}\lambda {\int}_{a}^{b}F(x,tu(x))\phantom{\rule{0.2em}{0ex}}dx\\ \le & \frac{{t}^{p}}{p}{\parallel u\parallel}^{p}\lambda {t}^{\mu}{a}_{3}{\int}_{a}^{b}{u}^{\mu}+{a}_{4}(ba)\to \mathrm{\infty}\end{array}$
as $t\to +\mathrm{\infty}$, so condition [[13], (${I}_{2}$), Theorem 2.2] is verified. Moreover, by standard computations, $\mathrm{\Phi}\lambda \mathrm{\Psi}$ satisfies the PalaisSmale condition. Hence, the classical theorem of Ambrosetti and Rabinowitz ensures a critical point ${\overline{u}}_{2}$ of $\mathrm{\Phi}\lambda \mathrm{\Psi}$ such that $(\mathrm{\Phi}\lambda \mathrm{\Psi})({\overline{u}}_{2})>(\mathrm{\Phi}\lambda \mathrm{\Psi})({\overline{u}}_{1})$. So, ${\overline{u}}_{1}$ and ${\overline{u}}_{2}$ are two distinct weak solutions of (P) and the proof is complete. □
Remark 3.1 We observe that in literature the existence of at least one nontrivial solution for differential problems is obtained associating to the classical AmbrosettiRabinowitz condition a hypothesis on the nonlinear term of type $f(x,t)=o(t)$ as $t\to 0$. This implies that the problem possesses also the trivial solution $u\equiv 0$. In Theorem 3.1, we find a nontrivial solution of the problem that actually is a proper local minimum of the EulerLagrange functional associated to the problem different from zero.
Now, we present an application of Theorem 2.2 which we will use to obtain multiple solutions.
Theorem 3.2 Assume that there exist two constants $\overline{c}$,
$\overline{d}$,
with ${(\frac{{q}_{0}^{p2}}{{2}^{p1}})}^{\frac{1}{p}}\overline{c}<\overline{d}$,
such that ${\int}_{a}^{b}\underset{\xi \le \overline{c}}{max}F(x,\xi )\phantom{\rule{0.2em}{0ex}}dx<{\int}_{\frac{a+b}{2}}^{b}F(x,\overline{d})\phantom{\rule{0.2em}{0ex}}dx,$
(3.7)
and
$\underset{\xi \to +\mathrm{\infty}}{lim\hspace{0.17em}sup}\frac{F(x,\xi )}{{\xi }^{p}}\le 0\phantom{\rule{1em}{0ex}}\mathit{\text{uniformly in}}X.$
(3.8)
Then,
for each $\lambda >\tilde{\lambda}$,
where $\tilde{\lambda}=\frac{\overline{r}{\overline{d}}^{p}\overline{q}{\overline{c}}^{p}}{{\int}_{\frac{a+b}{2}}^{b}F(x,\overline{d})\phantom{\rule{0.2em}{0ex}}dx{\int}_{a}^{b}{max}_{\xi \le \overline{c}}F(x,\xi )\phantom{\rule{0.2em}{0ex}}dx},$
problem (P) admits at least one nontrivial weak solution $\tilde{u}$ such that $\parallel \tilde{u}\parallel >\frac{\overline{c}}{{(\frac{{q}_{0}}{ba})}^{\frac{p1}{p}}}$.
Proof The functionals Φ and Ψ satisfy all regularity assumptions requested in Theorem 2.2. Moreover, by standard computations, condition (3.8) implies that
$\mathrm{\Phi}\lambda \mathrm{\Psi}$,
$\lambda >0$, is coercive. So, our aim is to verify condition (2.5) of Theorem 2.2. To this end, put
$\begin{array}{c}r=\frac{{\overline{c}}^{p}}{p}{\left(\frac{{q}_{0}}{ba}\right)}^{p1},\phantom{\rule{1em}{0ex}}\text{and}\hfill \\ {u}_{0}(x)=\{\begin{array}{cc}\frac{2\overline{d}}{ba}(xa)\hfill & \text{if}x\in [a,\frac{a+b}{2}[,\hfill \\ \overline{d}\hfill & \text{if}x\in [\frac{a+b}{2},b].\hfill \end{array}\hfill \end{array}$
Arguing as in the proof of Theorem 3.1, we obtain that
$\rho (r)\ge \frac{{\int}_{\frac{a+b}{2}}^{b}F(x,\overline{d})\phantom{\rule{0.2em}{0ex}}dx{\int}_{a}^{b}{max}_{\xi \le \overline{c}}F(x,\xi )\phantom{\rule{0.2em}{0ex}}dx}{\frac{1}{p}{(\frac{2}{ba})}^{p1}{\overline{d}}^{p}[{\parallel q\parallel}_{\mathrm{\infty}}+\frac{p+2}{p+1}{(\frac{ba}{2})}^{p}{\parallel r\parallel}_{\mathrm{\infty}}]\frac{1}{p}{(\frac{{q}_{0}}{ba})}^{p1}{\overline{c}}^{p}}.$
So, from our assumption, it follows that $\rho (r)>0$.
Hence, from Theorem 2.2 for each $\lambda >\tilde{\lambda}$, the functional $\mathrm{\Phi}\lambda \mathrm{\Psi}$ admits at least one local minimum $\tilde{u}$ such that $\parallel \tilde{u}\parallel >\frac{\overline{c}}{{(\frac{{q}_{0}}{ba})}^{\frac{p1}{p}}}$ and our conclusion is achieved. □
Remark 3.2 We point out that the same statement of above given result can be obtained by using a classical direct methods theorem (see [14]), but in addition we get the location of the solution, hence in particular the solution is nontrivial.
Now, we point out some results when the nonlinear term is with separable variables. To be precise, let

$\alpha \in {L}^{1}([a,b])$ such that $\alpha (x)\ge 0$ a.e. $x\in [a,b]$, $\alpha \not\equiv 0$, and

$g:\mathbb{R}\to \mathbb{R}$ be a nonnegative continuous function,
consider the following boundary value problem:
We observe that the following results give the existence of multiple nonnegative solutions since the nonlinear term is supposed to be nonnegative. In order to justify what has been said above, we point out the following weak maximum principle.
Lemma 3.1 Suppose that $\overline{u}\in X$ is a weak solution of problem (P1), then $\overline{u}$ is nonnegative.
Proof We claim that a weak solution
$\overline{u}$ is nonnegative. In fact, arguing by a contradiction and setting
$A=\{x\in [a,b]:\overline{u}(x)<0\}$, one has
$A\ne \mathrm{\varnothing}$. Put
${\overline{u}}^{}=min\{\overline{u},0\}$, one has
${\overline{u}}^{}\in X$ (see, for instance, [[
15], Lemma 7.6]). So, taking into account that
$\overline{u}$ is a weak solution and by choosing
$v={\overline{u}}^{}$, one has
${\int}_{A}{\overline{u}}^{\prime}(x){\overline{u}}^{\prime}(x)\phantom{\rule{0.2em}{0ex}}dx=\lambda {\int}_{A}\alpha (x)g(\overline{u}(x))\overline{u}(x)\phantom{\rule{0.2em}{0ex}}dx\le 0,$
that is, ${\parallel \overline{u}\parallel}_{{W}^{1,2}(A)}=0$ which is absurd. Hence, our claim is proved. □
Corollary 3.1
Assume that
(i′)
there exist three nonnegative constants ${c}_{1}$,
${c}_{2}$,
d,
with ${c}_{1}<\sqrt{2}d<{c}_{2}$,
such that $\frac{{\parallel \alpha \parallel}_{{L}^{1}([a,b])}G({c}_{2}){\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(d)}{{c}_{2}^{2}2{d}^{2}}<\frac{{\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(d){\parallel \alpha \parallel}_{{L}^{1}([a,b])}G({c}_{1})}{2{d}^{2}{c}_{1}^{2}};$
(3.9)
(ii′)
there exist $\nu >2$ and $R>0$ such that $0<\nu G(t)\le tg(t),\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}t\ge R.$
Then,
for each $\lambda \in \mathrm{\Lambda}$,
where $\begin{array}{rcl}\mathrm{\Lambda}& =& ]\frac{2{d}^{2}{c}_{1}^{2}}{2(ba)[{\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(d){\parallel \alpha \parallel}_{{L}^{1}([a,b])}G({c}_{1})]},\\ \frac{{c}_{2}^{2}2{d}^{2}}{2(ba)[{\parallel \alpha \parallel}_{{L}^{1}([a,b])}G({c}_{2}){\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(d)]}[,\end{array}$
problem (P1) admits at least two nonnegative weak solutions ${\overline{u}}_{1}$ and ${\overline{u}}_{2}$ such that $\frac{1}{\sqrt{ba}}{c}_{1}<\parallel {\overline{u}}_{1}\parallel <\frac{1}{\sqrt{ba}}{c}_{2}$.
Theorem 3.3 Assume that there exist two positive constants c,
d,
with $\sqrt{2}d<c$,
such that $\frac{G(c)}{{c}^{2}}<\left(\frac{{\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}}{2{\parallel \alpha \parallel}_{{L}^{1}([a,b])}}\right)\frac{G(d)}{{d}^{2}}.$
(3.10)
Further,
suppose that there exist $\nu >2$ and $R>0$ such that $0<\nu G(t)\le tg(t),\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}t\ge R.$
Then, for each $\lambda \in \phantom{\rule{0.2em}{0ex}}]\frac{2{d}^{2}}{2(ba){\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(d)},\frac{{c}^{2}}{2(ba){\parallel \alpha \parallel}_{{L}^{1}([a,b])}G(c)}[$, problem (P1) admits at least two nonnegative weak solutions.
Proof Our aim is to apply Corollary 3.1. To this end, we pick
${c}_{1}=0$ and
${c}_{2}=c$. From (3.10), one has
$\begin{array}{c}\frac{{\parallel \alpha \parallel}_{{L}^{1}([a,b])}G({c}_{2}){\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(d)}{{c}_{2}^{2}2{d}^{2}}\hfill \\ \phantom{\rule{1em}{0ex}}<\frac{\frac{1}{{c}^{2}}[{c}^{2}{\parallel \alpha \parallel}_{{L}^{1}([a,b])}G(c)2{d}^{2}{\parallel \alpha \parallel}_{{L}^{1}([a,b])}G(c)]}{{c}^{2}2{d}^{2}}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{{\parallel \alpha \parallel}_{{L}^{1}([a,b])}G(c)}{{c}^{2}}.\hfill \end{array}$
On the other hand, one has
$\frac{{\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(d){\parallel \alpha \parallel}_{{L}^{1}([a,b])}G({c}_{1})}{2{d}^{2}{c}_{1}^{2}}=\frac{{\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(d)}{2{d}^{2}}.$
Hence, from Corollary 3.1 and taking (2.6) into account, the conclusion follows. □
A further consequence of Theorem 3.1 is the following result.
Theorem 3.4
Assume that
$\underset{t\to {0}^{+}}{lim}\frac{g(t)}{t}=+\mathrm{\infty},$
(3.11)
and there are constants $\mu >2$ and $R>0$ such that,
for all $\xi \ge R$,
one has Then, for each $\lambda \in \phantom{\rule{0.2em}{0ex}}]0,{\lambda}^{\ast}[$, where ${\lambda}^{\ast}=\frac{1}{2(ba){\parallel \alpha \parallel}_{{L}^{1}([a,b])}}{sup}_{c>0}\frac{{c}^{2}}{G(c)}$, problem (P1) admits at least two nonnegative weak solutions.
Proof Fix $\lambda \in \phantom{\rule{0.2em}{0ex}}]0,{\lambda}^{\ast}[$. Then there is $c>0$ such that $\lambda <\frac{1}{2(ba){\parallel \alpha \parallel}_{{L}^{1}([a,b])}}\frac{{c}^{2}}{G(c)}$. From (3.11), there is $d<\frac{c}{\sqrt{2}}$ such that $\frac{2(ba){\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(d)}{2{d}^{2}}>\frac{1}{\lambda}$. Hence, Theorem 3.3 ensures the conclusion. □
Next, as a consequence of Theorems 3.3 and 3.2, the following theorem of the existence of three solutions is obtained.
Theorem 3.5
Assume that
$\underset{\xi \to +\mathrm{\infty}}{lim\hspace{0.17em}sup}\frac{G(\xi )}{{\xi}^{2}}\le 0.$
(3.12)
Moreover,
assume that there exist four positive constants c,
d,
$\overline{c}$,
$\overline{d}$,
with $\sqrt{2}d<c\le \overline{c}<\sqrt{2}\overline{d}$,
such that (3.10),
$G(\overline{c})<\left(\frac{{\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}}{{\parallel \alpha \parallel}_{{L}^{1}([a,b])}}\right)\frac{1}{2}G(\overline{d}),$
and
$\frac{G(c)}{{c}^{2}}<\frac{{\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(\overline{d}){\parallel \alpha \parallel}_{{L}^{1}([a,b])}G(\overline{c})}{\frac{1}{ba}{\overline{d}}^{2}\frac{1}{2(ba)}{\overline{c}}^{2}}$
(3.13)
are satisfied.
Then, for each $\lambda \in \mathrm{\Lambda}=\phantom{\rule{0.2em}{0ex}}]max\{\tilde{\lambda},\frac{2{d}^{2}}{2(ba){\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(d)}\},\frac{{c}^{2}}{2(ba){\parallel \alpha \parallel}_{{L}^{1}([a,b])}G(c)}[$, problem (P1) admits at least three weak nonnegative solutions.
Proof First, we observe that $\mathrm{\Lambda}\ne \mathrm{\varnothing}$ owing to (3.13). Next, fix $\lambda \in \mathrm{\Lambda}$. Theorem 3.3 ensures a nontrivial weak solution $\overline{u}$ such that $\parallel \overline{u}\parallel <c$ which is a local minimum for the associated functional $\mathrm{\Phi}\lambda \mathrm{\Psi}$, as well as Theorem 3.2 guarantees a nontrivial weak solution $\tilde{u}$ such that $\parallel \tilde{u}\parallel >c$ which is a local minimum for $\mathrm{\Phi}\lambda \mathrm{\Psi}$. Hence, the mountain pass theorem as given by Pucci and Serrin (see [2]) ensures the conclusion. □
Further,
assume that there exist two positive constants $\overline{c}$,
$\overline{d}$,
with $\overline{c}<\sqrt{2}\overline{d}$,
such that $\frac{G(\overline{c})}{{\overline{c}}^{2}}<\left(\frac{{\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}}{2{\parallel \alpha \parallel}_{{L}^{1}([a,b])}}\right)\frac{G(\overline{d})}{{\overline{d}}^{2}}.$
(3.16)
Then, for each $\lambda \in \phantom{\rule{0.2em}{0ex}}]\frac{2{\overline{d}}^{2}}{2(ba){\parallel \alpha \parallel}_{{L}^{1}([\frac{a+b}{2},b])}G(\overline{d})},\frac{{\overline{c}}^{2}}{2(ba){\parallel \alpha \parallel}_{{L}^{1}([a,b])}G(\overline{c})}[$, problem (P1) admits at least three weak nonnegative solutions.
Proof Clearly, (3.15) implies (3.8). Moreover, by choosing d small enough and $c=\overline{c}$, simple computations show that (3.14) implies (3.10). Finally, from (3.16) we get (3.7) and also (3.13). Hence, Theorem 3.5 ensures the conclusion. □
Finally, we present two examples of problems that admit multiple solutions owing to Theorems 3.4 and 3.6.
Example 3.1 Owing to Theorem 3.4, for each
$\lambda \in \phantom{\rule{0.2em}{0ex}}]0,\frac{1}{2}[$, the problem
$\{\begin{array}{c}{u}^{\u2033}=\lambda ({u}^{4}+1)\phantom{\rule{1em}{0ex}}\text{in}]0,1[,\hfill \\ u(0)={u}^{\prime}(1)=0\hfill \end{array}$
admits at least two nonnegative solutions. In fact, one has ${lim}_{u\to {0}^{+}}\frac{g(u)}{u}={lim}_{u\to {0}^{+}}\frac{{u}^{4}+1}{u}=+\mathrm{\infty}$ and (AR) is satisfied as a simple computation shows. Moreover, one has ${\lambda}^{\ast}=\frac{1}{2(ba){\parallel \alpha \parallel}_{1}}{sup}_{c>0}\frac{{c}^{2}}{G(c)}=\frac{1}{2}$.
Example 3.2 Consider the following problem:
It has three nonnegative solutions. In fact, let
$g:\mathbb{R}\to \mathbb{R}$ be a function defined as
$g(u)=\{\begin{array}{cc}(\frac{{u}^{7}}{{e}^{u}}+1)\hfill & \text{if}u\ge 0,\hfill \\ 1\hfill & \text{if}u0.\hfill \end{array}$
Owing to Theorem 3.6, the following problem
admits three nonnegative classical solutions. In fact, one has
$\underset{u\to {0}^{+}}{lim}\frac{g(u)}{u}=\underset{u\to {0}^{+}}{lim}\frac{(\frac{{u}^{7}}{{e}^{u}}+1)}{u}=+\mathrm{\infty}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{u\to +\mathrm{\infty}}{lim}\frac{g(u)}{u}=\underset{u\to +\mathrm{\infty}}{lim}\frac{(\frac{{u}^{7}}{{e}^{u}}+1)}{u}=0.$
Moreover, taking into account that $G(u)=u\frac{{\sum}_{i=0}^{7}\frac{7!}{i!}{u}^{i}}{{e}^{u}}+7!$, by choosing $\overline{c}=2$ and $\overline{d}=9$, one has $\frac{G(2)}{{2}^{2}}<\frac{1}{4}\frac{G(9)}{{9}^{2}}$ and $\frac{2\phantom{\rule{0.2em}{0ex}}{9}^{2}}{G(9)}<\frac{1}{10}<\frac{{2}^{2}}{2G(2)}$.
So, it is clear that any nonnegative solution u of problem (${P}_{g}$) is also a solution of problem (${P}_{E}$).