This section is devoted to the proof of the three theorems enunciated in the introduction of this work.

First consider the case $p>2$ for which we prove that the functional *ϕ* satisfies the Palais-Smale condition.

**Lemma 11** *Suppose that* $p>2$ *and conditions* (H1) *and* (H2) *hold*. *Then the functional* $\varphi :X\to \mathbb{R}$ *satisfies condition* (PS).

*Proof* Let

$({u}_{k})\subset X$ and

$C>0$ be such that

$|\varphi ({u}_{k})|\le C,\phantom{\rule{2em}{0ex}}{\varphi}^{\prime}({u}_{k})\to 0,\phantom{\rule{1em}{0ex}}k\to \mathrm{\infty}.$

Then we have

$C\ge \frac{1}{2}{\parallel {u}_{k}\parallel}_{X}^{2}+\sum _{j=1}^{n}({G}_{j}({u}_{k}^{\prime}({t}_{j}))+{H}_{j}({u}_{k}({t}_{j})))-\frac{1}{p}{\int}_{0}^{T}c(t){|{u}_{k}(t)|}^{p}\phantom{\rule{0.2em}{0ex}}dt$

(14)

and

$\left|\u3008{\varphi}^{\prime}({u}_{k}),v\u3009\right|\le {\parallel v\parallel}_{X},\phantom{\rule{1em}{0ex}}\mathrm{\forall}v\in X,$

(15)

for all sufficiently large

*k*,

$k>N$. Taking

$v={u}_{k}$ in (15), we have for

$k>N$$|{\parallel {u}_{k}\parallel}_{X}^{2}+\sum _{j=1}^{n}({g}_{j}({u}_{k}^{\prime}({t}_{j})){u}_{k}^{\prime}({t}_{j})+{h}_{j}({u}_{k}({t}_{j})){u}_{k}({t}_{j}))-{\int}_{0}^{T}c(t){|{u}_{k}(t)|}^{p}\phantom{\rule{0.2em}{0ex}}dt|\le {\parallel {u}_{k}\parallel}_{X}.$

In particular,

$\frac{{\parallel {u}_{k}\parallel}_{X}}{p}\ge -\frac{{\parallel {u}_{k}\parallel}_{X}^{2}}{p}-\frac{1}{p}\sum _{j=1}^{n}({g}_{j}({u}_{k}^{\prime}({t}_{j})){u}_{k}^{\prime}({t}_{j})+{h}_{j}({u}_{k}({t}_{j})){u}_{k}({t}_{j}))+\frac{1}{p}{\int}_{0}^{T}c(t){|{u}_{k}(t)|}^{p}\phantom{\rule{0.2em}{0ex}}dt.$

Adding the last inequality with (14), by assumption (H2), we obtain

$C+\frac{{\parallel {u}_{k}\parallel}_{X}}{p}\ge \frac{p-2}{2p}{\parallel {u}_{k}\parallel}_{X}^{2},$

which implies that $({u}_{k})$ is a bounded sequence in *X*.

Then, by the compact inclusion

$X\subset {C}^{1}([0,T])$, it follows that, up to a subsequence,

${u}_{k}\rightharpoonup u$ weakly in

*X* and

${u}_{k}\to u$ strongly in

${C}^{1}([0,T])$. As a consequence, from the inequality

$\left|\u3008{\varphi}^{\prime}({u}_{k})-{\varphi}^{\prime}(u),{u}_{k}-u\u3009\right|\le \parallel {\varphi}^{\prime}({u}_{k})\parallel \parallel {u}_{k}-u\parallel +\left|\u3008{\varphi}^{\prime}(u),{u}_{k}-u\u3009\right|$

it follows that

$\u3008{\varphi}^{\prime}({u}_{k})-{\varphi}^{\prime}(u),{u}_{k}-u\u3009\to 0$

(16)

and

$\begin{array}{c}\sum _{j=1}^{n}({h}_{j}({u}_{k}({t}_{j}))-{h}_{j}(u({t}_{j})))({u}_{k}({t}_{j})-u({t}_{j}))\to 0,\hfill \\ \sum _{j=1}^{n}({g}_{j}({u}_{k}^{\prime}({t}_{j}))-{g}_{j}({u}^{\prime}({t}_{j})))({u}_{k}^{\prime}({t}_{j})-{u}^{\prime}({t}_{j}))\to 0,\hfill \\ {\int}_{0}^{T}c(t)({|{u}_{k}|}^{p-2}{u}_{k}-{|u|}^{p-2}u)({u}_{k}-u)\phantom{\rule{0.2em}{0ex}}dt\to 0.\hfill \end{array}$

Then by (16) it follows that

${\parallel {u}_{k}-u\parallel}_{X}^{2}\to 0,$

*i.e.*, ${u}_{k}\to u$ strongly in *X*, which completes the proof. □

Now, we are in a position to prove the main results of this paper.

*Proof of Theorem 2* We find by (H1) and (8) that the following inequalities are valid for every

$u\in X$:

$\begin{array}{rcl}\varphi (u)& \ge & \frac{1}{2}{\parallel u\parallel}_{X}^{2}-\frac{{c}_{2}T}{p}{\parallel u\parallel}_{\mathrm{\infty}}^{p}\\ \ge & \frac{1}{2}{\parallel u\parallel}_{X}^{2}-\frac{{c}_{2}T{M}^{p}}{p}{\parallel u\parallel}_{X}^{p}.\end{array}$

It is evident that this last expression is strictly positive when

${\parallel u\parallel}_{X}=\rho $, with

*ρ* small enough. Next, let

${u}_{0}(t)=sin(\frac{\pi t}{T})\in X$ and

${u}_{\lambda}(t)=\lambda {u}_{0}(t)$, with

$\lambda >0$. Then, by (H2) and (3), we have

$\varphi ({u}_{\lambda})\le \frac{{\lambda}^{2}}{2}{\parallel {u}_{0}\parallel}_{X}^{2}+C(2n+\sum _{j=1}^{n}({\lambda}^{{\gamma}_{j}}+{\lambda}^{{\sigma}_{j}}))-\frac{{\lambda}^{p}}{p}{\int}_{0}^{T}c(t){|{u}_{0}|}^{p}\phantom{\rule{0.2em}{0ex}}dt,$

where $C=max\{{C}_{j},{K}_{j}:1\le j\le n\}$.

Since $p>2$, we conclude that $\varphi ({u}_{\lambda})<0$ for sufficiently large *λ*. According to the mountain-pass Theorem 9, together with Lemmas 11 and 7, we deduce that there exists a nonzero classical solution of the problem (P). □

Now consider the case $1<p<2$. In the next result we prove that the Palais-Smale condition is also valid.

**Lemma 12** *Suppose that* $1<p<2$ *and conditions* (H1) *and* (H3) *hold*. *Then the functional* $\varphi :X\to \mathbb{R}$ *is bounded from below and satisfies condition* (PS).

*Proof* By

$1<p<2$, conditions (H1), (H3), and inequality (8), it follows that the functional

*ϕ* is bounded from below:

$\begin{array}{rcl}\varphi (u)& \ge & \frac{1}{2}{\parallel u\parallel}_{X}^{2}-\frac{{c}_{2}T}{p}{\parallel u\parallel}_{\mathrm{\infty}}^{p}\ge \frac{1}{2}{\parallel u\parallel}_{X}^{2}-\frac{{c}_{2}T{M}^{p}}{p}{\parallel u\parallel}_{X}^{p}\\ \ge & \frac{p-2}{2p}{\left({c}_{2}T{M}^{p}\right)}^{2/(2-p)}.\end{array}$

(17)

Further, if $({u}_{k})$ is a (PS) sequence, by (17) it follows that $({u}_{k})$ is a bounded sequence in *X*. Then, as in Lemma 11, we conclude that $({u}_{k})$ has a convergent subsequence. □

Now we are in a position to prove the next existence result for the problem (P).

*Proof of Theorem 3* By assumption, we know that

${g}_{j}$ and

${h}_{j}$ are odd functions. So

${G}_{j}$ are

${H}_{j}$ are even functions and the functional

*ϕ* is even. By Lemma 12 we know that

*ϕ* is bounded from below and satisfies condition (PS). Let

$m\in \mathbb{N}$,

$m\ge 3$ be a natural number and define, for any

$\rho >0$ fixed, the set

${K}_{\rho}^{m}=\{\sum _{j=1}^{m}{\lambda}_{j}sin\left(\frac{j\pi t}{T}\right):\sum _{j=1}^{m}{\lambda}_{j}^{2}={\rho}^{2}\}\subset X.$

${K}_{\rho}^{m}$ is homeomorphic to

${\mathbb{S}}^{m-1}$ by the odd mapping defined as

$H:{K}_{\rho}^{m}\to {\mathbb{S}}^{m-1}$$H(\sum _{j=1}^{m}{\lambda}_{j}sin\left(\frac{j\pi t}{T}\right))=(-\frac{{\lambda}_{1}}{\rho},-\frac{{\lambda}_{2}}{\rho},\dots ,-\frac{{\lambda}_{m}}{\rho}).$

Moreover, for

$w={\sum}_{j=1}^{m}{\lambda}_{j}sin(\frac{j\pi t}{T})\in {K}_{\rho}^{m}$, the following inequalities hold:

$\rho \sqrt{\frac{T{b}_{1}}{2}}\le {\parallel w\parallel}_{X}\le \rho m\sqrt{\frac{T}{2}({\left(\frac{m\pi}{T}\right)}^{4}+a{\left(\frac{m\pi}{T}\right)}^{2}+{b}_{2})}.$

(18)

Clearly

${K}_{\rho}^{m}$ is a subset of the

*m*-dimensional subspace

${X}_{m}=sp\{sin\left(\frac{\pi t}{T}\right),\dots ,sin\left(\frac{m\pi t}{T}\right)\}\subset X,$

and there exist positive constants

${C}_{1}(m)$ and

${C}_{2}(m)$, such that

${C}_{1}(m){\parallel w\parallel}_{{X}_{m}}\le {\parallel w\parallel}_{{L}^{p}}\le {C}_{2}(m){\parallel w\parallel}_{{X}_{m}},$

(19)

where ${\parallel \cdot \parallel}_{{X}_{m}}$ is the induced norm of ${\parallel \cdot \parallel}_{X}$ on ${X}_{m}$.

Arguing as in [[

15], pp.16-18], one can prove that there exists

$\epsilon =\epsilon (m)>0$, such that

$meas\{t\in [0,T]:c(t){|u(t)|}^{p}\ge \epsilon {\parallel u\parallel}_{X}^{p},u\in {X}_{m}\mathrm{\setminus}\{0\}\}\ge \epsilon .$

(20)

Denote

${\mathrm{\Omega}}_{u}=\{t\in [0,T]:c(t){|u(t)|}^{p}\ge \epsilon {\parallel u\parallel}_{X}^{p}\}.$

By (H3) we see that for every

$w\in {K}_{\rho}^{m}$,

$w={\sum}_{k=1}^{m}{\lambda}_{k}sin(\frac{k\pi t}{T})$, the following inequalities are fulfilled:

$\begin{array}{rcl}\left|{G}_{j}({w}^{\prime}({t}_{j}))\right|& \le & {A}_{j}{|{w}^{\prime}({t}_{j})|}^{2}\le {A}_{j}{\left(\sum _{k=1}^{m}\frac{k\pi |{\lambda}_{k}|}{T}|cos\left(\frac{k\pi {t}_{j}}{T}\right)|\right)}^{2}\\ \le & {A}_{j}{\left(\frac{m\pi}{T}\right)}^{2}{(\sum _{k=1}^{m}|{\lambda}_{k}|)}^{2}\le {A}_{j}{m}^{3}{\left(\frac{\pi}{T}\right)}^{2}\left(\sum _{k=1}^{m}{\lambda}_{k}^{2}\right)\end{array}$

(21)

and

$\begin{array}{rcl}{H}_{j}(w({t}_{j}))& \le & {B}_{j}{|w({t}_{j})|}^{2}\le {B}_{j}{(\sum _{k=1}^{m}|{\lambda}_{k}||sin\left(\frac{k\pi {t}_{j}}{T}\right)|)}^{2}\\ \le & {B}_{j}{(\sum _{k=1}^{m}|{\lambda}_{k}|)}^{2}\le {B}_{j}m\left(\sum _{k=1}^{m}{\lambda}_{k}^{2}\right).\end{array}$

Denote

$C=max\{{A}_{j}{m}^{3}{(\frac{\pi}{T})}^{2},{B}_{j}m:1\le j\le n\}$. Then by (18)-(21) we have

$\begin{array}{rcl}\varphi (w)& =& \frac{1}{2}{\parallel w\parallel}_{X}^{2}+\sum _{j=1}^{n}({G}_{j}({w}^{\prime}({t}_{j}))+{H}_{j}(w({t}_{j})))-\frac{1}{p}{\int}_{0}^{T}c(t){|w|}^{p}\phantom{\rule{0.2em}{0ex}}dt\\ \le & \frac{1}{2}{\parallel w\parallel}_{{X}_{m}}^{2}+2nC\sum _{j=1}^{m}{\lambda}_{j}^{2}-\frac{\epsilon}{p}{\parallel w\parallel}_{{L}^{p}}^{p}meas{\mathrm{\Omega}}_{w}\\ =& \frac{1}{2}{\parallel w\parallel}_{{X}_{m}}^{2}+2nC{\rho}^{2}-\frac{\epsilon}{p}{\parallel w\parallel}_{{L}^{p}}^{p}meas{\mathrm{\Omega}}_{w}\\ \le & (\frac{1}{2}+\frac{4nC}{T{b}_{1}}){\parallel w\parallel}_{{X}_{m}}^{2}-\frac{{\epsilon}^{2}}{p}{\parallel w\parallel}_{{L}^{p}}^{p}\\ \le & K{\parallel w\parallel}_{{X}_{m}}^{2}-\frac{{\epsilon}^{2}{C}_{1}^{p}(m)}{p}{\parallel w\parallel}_{{X}_{m}}^{p},\end{array}$

where $K=\frac{1}{2}+\frac{4nC}{T{b}_{1}}$.

By the last inequality, it follows that

$\varphi (w)<0$ if

${\parallel w\parallel}_{{X}_{m}}<{(\frac{{\epsilon}^{2}{C}_{1}^{p}(m)}{pK})}^{1/(2-p)}$. Then, by (18), choosing

$\rho <{\left(\frac{{\epsilon}^{2}{C}_{1}^{p}(m)}{pK}\right)}^{1/(2-p)}{\left(m\sqrt{\frac{T}{2}({\left(\frac{m\pi}{T}\right)}^{4}+a{\left(\frac{m\pi}{T}\right)}^{2}+{b}_{2})}\right)}^{-1}$

we obtain $\varphi (w)<0$ for any $w\in {K}_{\rho}^{m}$.

By Clarke’s Theorem 10, there exist at least *m* pairs of different critical points of the functional *ϕ*. Since *m* is arbitrary, there exist infinitely many solutions of the problem (P), which concludes the proof. □

Concerning the problem (P

_{1}), one can introduce similarly the notions of classical and weak solutions. In this case it is not difficult to verify that the weak solutions are critical points of the functional

${\varphi}_{1}:X\to \mathbb{R}$ defined as

$\begin{array}{rcl}{\varphi}_{1}(u)& =& \frac{1}{2}{\int}_{0}^{T}({u}^{\prime \prime 2}+a{u}^{\prime 2})\phantom{\rule{0.2em}{0ex}}dt-\frac{1}{2}{\int}_{0}^{T}b(t){u}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ +\sum _{j=1}^{n}({G}_{j}({u}^{\prime}({t}_{j}))+{H}_{j}(u({t}_{j})))+\frac{1}{p}{\int}_{0}^{T}c(t){|u|}^{p}\phantom{\rule{0.2em}{0ex}}dt.\end{array}$

(22)

*Proof of Theorem 4* By the Poincaré inequalities (7) we find that $\u2980u{\u2980}^{2}={\int}_{0}^{T}({u}^{\prime \prime 2}+a{u}^{\prime 2})\phantom{\rule{0.2em}{0ex}}dt$ is an equivalent norm to ${\parallel \cdot \parallel}_{X}$ in *X* and the functional ${I}_{1}(u)=\frac{1}{2}\u2980u{\u2980}^{2}$ is convex.

Since the functional

${I}_{2}(u)=-\frac{1}{2}{\int}_{0}^{T}b(t){u}^{2}\phantom{\rule{0.2em}{0ex}}dt+\sum _{j=1}^{n}({G}_{j}({u}^{\prime}({t}_{j}))+{H}_{j}(u({t}_{j})))+\frac{1}{p}{\int}_{0}^{T}c(t){|u|}^{p}\phantom{\rule{0.2em}{0ex}}dt$

is sequentially weakly continuous, from the fact that the inclusion $X\subset {C}^{1}([0,T])$ is compact, we deduce that the functional ${\varphi}_{1}:X\to \mathbb{R}$ is weakly lower semi-continuous.

Next, let us see that

${\varphi}_{1}:X\to \mathbb{R}$ is bounded from below:

$\begin{array}{rcl}{\varphi}_{1}(u)& \ge & \frac{1}{2}\u2980u{\u2980}^{2}+{\int}_{0}^{T}\frac{1}{p}c(t){|u|}^{p}-\frac{1}{2}b(t){u}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ \ge & \frac{1}{2}\u2980u{\u2980}^{2}+{\int}_{0}^{T}(\frac{1}{p}{c}_{1}{|u|}^{p}-\frac{1}{2}{b}_{2}{u}^{2})\phantom{\rule{0.2em}{0ex}}dt\ge CT,\end{array}$

(23)

where

$C=min\{\frac{1}{p}{c}_{1}{t}^{p}-\frac{1}{2}{b}_{2}{t}^{2}:t\ge 0\}=\frac{2-p}{p}{b}_{2}{\left(\frac{{b}_{2}}{{c}_{1}}\right)}^{2/(p-2)}.$

Then, by Theorem 8, there exists a minimizer of ${\varphi}_{1}$, which is a critical point of ${\varphi}_{1}$.

Let

*u* be a weak solution of (P

_{1}),

*i.e.*, a critical point of

${\varphi}_{1}$. Then

$\begin{array}{r}{\int}_{0}^{T}({u}^{\prime \prime 2}+a{u}^{\prime 2}-b(t){u}^{2})\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}+\sum _{j=1}^{n}({g}_{j}({u}^{\prime}({t}_{j})){u}^{\prime}({t}_{j})+{h}_{j}(u({t}_{j}))u({t}_{j}))+{\int}_{0}^{T}c(t){|u|}^{p}\phantom{\rule{0.2em}{0ex}}dt=0.\end{array}$

(24)

If

$0<T\le {T}_{2}=\pi \sqrt{\frac{a+\sqrt{{a}^{2}+4{b}_{2}}}{2{b}_{2}}}$ then

${(\frac{\pi}{T})}^{4}+a{(\frac{\pi}{T})}^{2}-{b}_{2}\ge 0$. Suppose that

*u* is a nonzero solution and

$0<T\le {T}_{2}$. By (H2′), (7), and (24) it follows that

$\begin{array}{rcl}0& >& -{\int}_{0}^{T}c(t){|u|}^{p}\phantom{\rule{0.2em}{0ex}}dt={\int}_{0}^{T}({u}^{\prime \prime 2}+a{u}^{\prime 2}-b(t){u}^{2})\phantom{\rule{0.2em}{0ex}}dt\\ +\sum _{j=1}^{n}({g}_{j}({u}^{\prime}({t}_{j})){u}^{\prime}({t}_{j})+{h}_{j}(u({t}_{j}))u({t}_{j}))\\ \ge & {\int}_{0}^{T}({\left(\frac{\pi}{T}\right)}^{4}+a{\left(\frac{\pi}{T}\right)}^{2}-{b}_{2}){u}^{2}\phantom{\rule{0.2em}{0ex}}dt\ge 0,\end{array}$

which is a contradiction. Then, for $0<T\le {T}_{2}$, the problem (P_{1}) has only the zero solution.

Suppose now that $T\ge {T}_{1}=\pi \sqrt{\frac{a+\sqrt{{a}^{2}+4{b}_{1}}}{2{b}_{1}}}$.

Take

${u}_{\epsilon}(t)=\epsilon sin(\frac{\pi t}{T})\in X$,

$\epsilon >0$. Then

${\varphi}_{1}({u}_{\epsilon})\le \frac{{\epsilon}^{2}}{2}({\left(\frac{\pi}{T}\right)}^{4}+a{\left(\frac{\pi}{T}\right)}^{2}-{b}_{1})+D\left(\sum _{j=1}^{n}({\epsilon}^{{\gamma}_{j}}+{\epsilon}^{{\sigma}_{j}})\right)+\frac{{c}_{2}T}{p}{\epsilon}^{p},$

(25)

where $D=max\{{d}_{j},{e}_{j}:1\le j\le n\}$. For $T>{T}_{1}=\pi \sqrt{\frac{a+\sqrt{{a}^{2}+4{b}_{1}}}{2{b}_{1}}}$ it follows that ${(\frac{\pi}{T})}^{4}+a{(\frac{\pi}{T})}^{2}-{b}_{1}<0$. Then, since ${\gamma}_{j},{\sigma}_{j}\in (2,p)$, by (25) it follows that ${\varphi}_{1}({u}_{\epsilon})<0$ for sufficiently small $\epsilon >0$. In consequence we show that $min\{{\varphi}_{1}(u):u\in X\}<0$. So we ensure the existence of a nonzero minimizer of ${\varphi}_{1}$, which completes the proof of Theorem 4. □