Denote by deg the Leray-Schauder degree. To prove Theorem 1.1, we need the following results.

**Lemma 2.1** [20]

*Let* Ω *be a bounded open region in a real Banach space* *X*. *Assume that* $K:\overline{\mathrm{\Omega}}\to \mathbb{R}$ *is completely continuous and* $p\notin (I-K)(\partial \mathrm{\Omega})$. *Then the equation* $(I-K)(x)=p$ *has a solution in* Ω *if* $deg(I-K,\mathrm{\Omega},p)\ne 0$.

**Lemma 2.2** (Borsuk Theorem [20])

*Assume that* *X* *is a real Banach space*. *Let* Ω *be a symmetric bounded open region with* $\theta \in \mathrm{\Omega}$. *Assume that* $K:\overline{\mathrm{\Omega}}\to \mathbb{R}$ *is completely continuous and odd with* $\theta \notin (I-K)(\partial \mathrm{\Omega})$. *Then* $deg(I-K,\mathrm{\Omega},\theta )$ *is odd*.

*Proof of Theorem 1.1* Take

$({\lambda}_{+},{\lambda}_{-})\in [{p}_{1},{p}_{2}]\times [{q}_{1},{q}_{2}]$. Consider the following homotopy problem:

$\{\begin{array}{c}-{({\varphi}_{p}({u}^{\prime}))}^{\prime}=(1-\mu )({\lambda}_{+}{({u}^{+})}^{p-1}-{\lambda}_{-}{({u}^{-})}^{p-1})+\mu f(t,u)\equiv {f}_{\mu}(t,u),\hfill \\ u(0)=u(2\pi ),\phantom{\rule{2em}{0ex}}{u}^{\prime}(0)={u}^{\prime}(2\pi ),\hfill \end{array}$

(2.1)

where $\mu \in [0,1]$.

By (1.3) and the regularity arguments, it follows that

$u\in {C}^{1}[0,2\pi ]$, and furthermore there exists

$a,b\in {\mathbb{R}}^{+}$ such that, if

*u* is a solution of problem (2.1), then

${\parallel u\parallel}_{{C}^{1}}\le a{\parallel u\parallel}_{\mathrm{\infty}}+b.$

(2.2)

In what follows, we shall prove that there exists

$C>0$ independent of

$\mu \in [0,1]$ such that

${\parallel u\parallel}_{\mathrm{\infty}}\le C$ for all possible solution

$u(t)$ of (2.1). Assume by contradiction that there exist a sequence of number

$\{{\mu}_{n}\}\subset [0,1]$ and corresponding solutions

$\{{u}_{n}\}$ of (2.1) such that

${\parallel {u}_{n}\parallel}_{\mathrm{\infty}}\to +\mathrm{\infty}\phantom{\rule{1em}{0ex}}\text{as}n\to +\mathrm{\infty}.$

(2.3)

Set

${z}_{n}=\frac{{u}_{n}}{{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}}$. Obviously,

${\parallel {z}_{n}\parallel}_{\mathrm{\infty}}=1$. Define

${\alpha}_{n}(t)=\{\begin{array}{cc}\frac{f(t,{u}_{n})}{{|{u}_{n}|}^{p-2}{u}_{n}},\hfill & {u}_{n}(t)>M,\hfill \\ 0,\hfill & {u}_{n}(t)\le M\hfill \end{array}$

and

${\beta}_{n}(t)=\{\begin{array}{cc}\frac{f(t,{u}_{n})}{{|{u}_{n}|}^{p-2}{u}_{n}},\hfill & {u}_{n}(t)<-M,\hfill \\ 0,\hfill & {u}_{n}(t)\ge -M.\hfill \end{array}$

By (1.3), there exists

${M}_{0}>0$ such that

$|{\alpha}_{n}(t)|,|{\beta}_{n}(t)|\le {M}_{0},\phantom{\rule{1em}{0ex}}\text{a.e.}t\in [0,2\pi ].$

Then there exist

${\alpha}_{0},{\beta}_{0}\in {L}^{\mathrm{\infty}}(0,2\pi )$ such that

${\alpha}_{n}(t)\stackrel{\ast}{\rightharpoonup}{\alpha}_{0}(t),\phantom{\rule{2em}{0ex}}{\beta}_{n}(t)\stackrel{\ast}{\rightharpoonup}{\beta}_{0}(t)\phantom{\rule{1em}{0ex}}\text{in}{L}^{\mathrm{\infty}}(0,2\pi ).$

(2.4)

In addition, using (1.3) and the regularity arguments, there exists

${M}_{1}>0$ such that, for each

*n*, we have

${\parallel {z}_{n}\parallel}_{{C}^{1}}\le {M}_{1}$, and thus there exists

${z}_{0}\in {C}^{1}[0,2\pi ]$ such that, passing to a subsequence if possible,

${z}_{n}\to {z}_{0}\phantom{\rule{1em}{0ex}}\text{in}{C}^{1}[0,2\pi ].$

(2.5)

Clearly,

${\parallel {z}_{0}\parallel}_{\mathrm{\infty}}=1$. In view of

$\{{\mu}_{n}\}\subset [0,1]$, there exists

${\mu}_{0}\in [0,1]$ such that, passing to a subsequence if possible,

${\mu}_{n}\to {\mu}_{0}\phantom{\rule{1em}{0ex}}\text{as}n\to +\mathrm{\infty}.$

(2.6)

Note that for

$\mu =0$, problem (2.1) has only the trivial solution, it follows that

${\mu}_{0}\in (0,1]$. Denote

$\overline{\alpha}(t)=(1-{\mu}_{0}){\lambda}_{+}+{\mu}_{0}{\alpha}_{0}(t)$,

$\overline{\beta}(t)=(1-{\mu}_{0}){\lambda}_{-}+{\mu}_{0}{\beta}_{0}(t)$. It is easily seen that

${z}_{0}$ is a nontrivial solution of the following problem:

$\{\begin{array}{c}-{({\varphi}_{p}({z}_{0}^{\prime}))}^{\prime}=\overline{\alpha}(t){({z}_{0}^{+})}^{p-1}-\overline{\beta}(t){({z}_{0}^{-})}^{p-1},\hfill \\ {z}_{0}(0)={z}_{0}(2\pi ),\phantom{\rule{2em}{0ex}}{z}_{0}^{\prime}(0)={z}_{0}^{\prime}(2\pi ).\hfill \end{array}$

(2.7)

We now distinguish three cases:

- (i)
${z}_{0}$ changes sign in $[0,2\pi ]$;

- (ii)
${z}_{0}(t)\ge 0$, $\mathrm{\forall}t\in [0,2\pi ]$;

- (iii)
${z}_{0}(t)\le 0$, $\mathrm{\forall}t\in [0,2\pi ]$.

In the following, it will be shown that each case leads to a contradiction.

Then, as

$n\to +\mathrm{\infty}$, we get

In addition, as shown in [

11], we have

$|{I}^{0}|=0$. Define

${\eta}_{+}(t)=\{\begin{array}{cc}\eta (t),\hfill & t\in {I}^{+},\hfill \\ {\alpha}_{0}(t),\hfill & t\in {I}^{-},\hfill \end{array}$

and

${\eta}_{-}(t)=\{\begin{array}{cc}{\beta}_{0}(t),\hfill & t\in {I}^{+},\hfill \\ \eta (t),\hfill & t\in {I}^{-}.\hfill \end{array}$

By (1.4) and (2.4), it follows that

${\alpha}_{0}(t)\equiv {\eta}_{+}(t),\phantom{\rule{2em}{0ex}}{\beta}_{0}(t)\equiv {\eta}_{-}(t),\phantom{\rule{1em}{0ex}}\text{a.e.}t\in [0,2\pi ].$

Thus,

${z}_{0}$ satisfies

$\{\begin{array}{c}-{({\varphi}_{p}({z}_{0}^{\prime}))}^{\prime}=\tilde{\alpha}(t){({z}_{0}^{+})}^{p-1}-\tilde{\beta}(t){({z}_{0}^{-})}^{p-1},\hfill \\ {z}_{0}(0)={z}_{0}(2\pi ),\phantom{\rule{2em}{0ex}}{z}_{0}^{\prime}(0)={z}_{0}^{\prime}(2\pi ).\hfill \end{array}$

(2.8)

Here, $\tilde{\alpha}(t)=(1-{\mu}_{0}){\lambda}_{+}+{\mu}_{0}{\eta}_{+}(t)$, $\tilde{\beta}(t)=(1-{\mu}_{0}){\lambda}_{-}+{\mu}_{0}{\eta}_{-}(t)$.

Now we prove that there exist

$\overline{n}\in {\mathbb{Z}}^{+}$ and

$0<{\kappa}_{1}<1<{\kappa}_{2}$ such that

${\kappa}_{1}\le \frac{max{u}_{n}}{-min{u}_{n}}\le {\kappa}_{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall}n\ge \overline{n}.$

(2.9)

In fact, if not, we assume, by contradiction, that there exists a subsequence of

$\{{u}_{n}\}$, we still denote it as

$\{{u}_{n}\}$ with

$max{u}_{n}\to \mathrm{\infty}$ and

$min{u}_{n}\to -\mathrm{\infty}$, such that

$\frac{max{u}_{n}}{-min{u}_{n}}\to 0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\frac{max{u}_{n}}{-min{u}_{n}}\to +\mathrm{\infty}.$

Combing with (2.5),

${\parallel {z}_{0}\parallel}_{\mathrm{\infty}}=1$ and the fact that

${z}_{0}$ changes sign, we obtain

$max\frac{{u}_{n}}{{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}}/(-min\frac{{u}_{n}}{{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}})\to \frac{max{z}_{0}}{-min{z}_{0}}>0.$

A contradiction. Hence, (2.9) holds.

For any

$(t,\mu )\in [0,2\pi ]\times [0,1]$, define

and

${\overline{F}}_{1}(t,s,\mu )={\int}_{0}^{s}{\overline{f}}_{1}(t,\tau ,\mu )\phantom{\rule{0.2em}{0ex}}d\tau ,\phantom{\rule{2em}{0ex}}{\overline{F}}_{2}(t,r,\mu )={\int}_{0}^{r}{\overline{f}}_{2}(t,\tau ,\mu )\phantom{\rule{0.2em}{0ex}}d\tau .$

Denote

${s}_{n}=max{u}_{n}$,

${r}_{n}=min{u}_{n}$. Then by (2.9) it follows that

${s}_{n}\to +\mathrm{\infty}$ and

${r}_{n}\to -\mathrm{\infty}$. Taking

${t}_{n}$ such that

${u}_{n}({t}_{n})={s}_{n}$,

${t}_{n}^{0}$ is the nearest point satisfying

${t}_{n}^{0}<{t}_{n}$ and

${u}_{n}({t}_{n}^{0})=0$. Since

${t}_{n}^{0},{t}_{n}\in [0,2\pi ]$, there exist

${\overline{t}}^{0},\overline{t}\in [0,2\pi ]$ such that

${t}_{n}^{0}\to {\overline{t}}^{0},\phantom{\rule{2em}{0ex}}{t}_{n}\to \overline{t}\phantom{\rule{1em}{0ex}}\text{as}n\to +\mathrm{\infty}.$

(2.10)

By (2.5), we obtain

${z}_{0}({\overline{t}}^{0})=0$,

${z}_{0}(\overline{t})={max}_{t\in [0,2\pi ]}{z}_{0}(t)$. Note that

${\parallel {u}_{n}\parallel}_{\mathrm{\infty}}\to +\mathrm{\infty}$, we have

${u}_{n}(t)\to +\mathrm{\infty}$,

$\mathrm{\forall}t\in ({\overline{t}}^{0},\overline{t})$. Hence, together with

${\mu}_{n}\to {\mu}_{0}$ and (1.4), there exist subsequences of

$\{{u}_{n}\}$ and

$\{{\mu}_{n}\}$, we still denote them as

$\{{u}_{n}\}$ and

$\{{\mu}_{n}\}$, such that, for a.e.

$t\in [0,2\pi ]$,

$\frac{{\overline{f}}_{1}(t,{u}_{n}(\tau ),{\mu}_{n})}{{({u}_{n}(\tau ))}^{p-1}}\to 0,\phantom{\rule{1em}{0ex}}\text{a.e.}\tau \in ({\overline{t}}^{0},\overline{t}).$

Using (1.3), for a.e.

$t\in [0,2\pi ]$,

$\{\frac{{\overline{f}}_{1}(t,{u}_{n}(\tau ),{\mu}_{n})}{{({u}_{n}(\tau ))}^{p-1}}\}$ is uniformly bounded with respect to

$\tau \in ({\overline{t}}^{0},\overline{t})$, we obtain by the Lebesgue dominated convergence theorem that

${\int}_{{\overline{t}}^{0}}^{\overline{t}}\left|\frac{{\overline{f}}_{1}(t,{u}_{n}(\tau ),{\mu}_{n})}{{({u}_{n}(\tau ))}^{p-1}}\right|\phantom{\rule{0.2em}{0ex}}d\tau \to 0,\phantom{\rule{1em}{0ex}}\text{uniformly for a.e.}t\in [0,2\pi ].$

By (1.4) and (2.2), we get

In view of (2.11), we obtain that

$\left|\frac{p{\overline{F}}_{1}(t,{s}_{n},{\mu}_{n})}{{s}_{n}^{p}}\right|\to 0$

(2.12)

holds uniformly for a.e.

$t\in [0,2\pi ]$. Similarly,

$\left|\frac{p{\overline{F}}_{2}(t,{r}_{n},{\mu}_{n})}{{|{r}_{n}|}^{p}}\right|\to 0$

(2.13)

holds uniformly for a.e. $t\in [0,2\pi ]$.

On the other hand, for

$\{{s}_{n}\}$,

$\{{r}_{n}\}$ satisfying (2.12)-(2.13), denoting

we obtain by (1.5)-(1.6) that

${p}_{1}\le {\xi}_{1}(t)\le {q}_{1},\phantom{\rule{2em}{0ex}}{p}_{2}\le {\xi}_{2}(t)\le {q}_{2},\phantom{\rule{1em}{0ex}}\text{a.e.}t\in [0,2\pi ].$

(2.14)

Using

${\mu}_{n}\to {\mu}_{0}$, we have

We claim that there exists subinterval

${I}_{1}\subset [0,2\pi ]$ with

$|{I}_{1}|>0$ such that

${\xi}_{1}(t)-{\eta}_{+}(t)\ne 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in {I}_{1},$

(2.17)

or subinterval

${I}_{2}\subset [0,2\pi ]$ with

$|{I}_{2}|>0$ such that

${\xi}_{2}(t)-{\eta}_{-}(t)\ne 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in {I}_{2}.$

(2.18)

Indeed, if not, we assume that

${\eta}_{+}(t)={\xi}_{1}(t)$,

${\eta}_{-}(t)={\xi}_{2}(t)$, a.e.

$t\in [0,2\pi ]$. Together with the choosing of

${\lambda}_{+}$,

${\lambda}_{-}$ and (2.14), we get

${p}_{1}\le \tilde{\alpha}(t)\le {q}_{1},\phantom{\rule{2em}{0ex}}{p}_{2}\le \tilde{\beta}(t)\le {q}_{2},\phantom{\rule{1em}{0ex}}\text{a.e.}t\in [0,2\pi ].$

Then by (1.7), it follows that ${z}_{0}\equiv 0$. A contradiction. Combining (2.12)-(2.13) with (2.15)-(2.18), we obtain a contradiction.

Case (ii). In this case, we have

${s}_{n}\to +\mathrm{\infty}\text{and}\{{r}_{n}\}\text{is uniformly bounded}.$

Using similar arguments as in Case (i), by (1.4) and (2.4) it follows that

${\alpha}_{0}(t)\equiv {\eta}_{+}(t)$,

$\mathrm{\forall}t\in [0,2\pi ]$. Taking

${\overline{f}}^{+}={\overline{f}}_{1}$,

${\overline{F}}^{+}={\overline{F}}_{1}$, a.e.

$t\in [0,2\pi ]$. We can see that there exists subsequence of

$\{{s}_{n}\}$, which is still denoted by

$\{{s}_{n}\}$, such that

$\left|\frac{p{\overline{F}}^{+}(t,{s}_{n},{\mu}_{n})}{{s}_{n}^{p}}\right|\to 0$

(2.19)

holds uniformly for a.e.

$t\in [0,2\pi ]$. On the other hand, for

$\{{s}_{n}\}$ satisfying (2.19), denoting

${\xi}^{+}(t)=\underset{n\to +\mathrm{\infty}}{lim\hspace{0.17em}inf}\frac{pF(t,{s}_{n})}{{|{s}_{n}|}^{p}},$

we obtain by (1.5) that

${p}_{1}\le {\xi}^{+}(t)\le {q}_{1},\phantom{\rule{1em}{0ex}}\text{a.e.}t\in [0,2\pi ].$

(2.20)

Using

${\mu}_{n}\to {\mu}_{0}$, we have

We shall show that there exists subinterval

${I}_{+}\subset [0,2\pi ]$ with

$|{I}_{+}|$ such that

${\xi}^{+}(t)-{\eta}_{+}(t)\ne 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in {I}_{+}.$

(2.22)

In fact, if not, we assume that

${\eta}_{+}(t)={\xi}^{+}(t)$, a.e.

$t\in [0,2\pi ]$. By the choosing of

${\lambda}_{+}$ and (2.20), we get

${p}_{1}\le \tilde{\alpha}(t)\le {q}_{1}$, a.e.

$t\in [0,2\pi ]$. Thus,

${z}_{0}$ is a nontrivial solution of the following problem:

$\{\begin{array}{c}-{({\varphi}_{p}({z}_{0}^{\prime}))}^{\prime}=\tilde{\alpha}(t){z}_{0}^{p-1},\hfill \\ {z}_{0}(0)={z}_{0}(2\pi ),\phantom{\rule{2em}{0ex}}{z}_{0}^{\prime}(0)={z}_{0}^{\prime}(2\pi ).\hfill \end{array}$

(2.23)

Taking 1 as test function in problem (2.23), we get

$0={\int}_{0}^{2\pi}\tilde{\alpha}(t){z}_{0}^{p-1}\phantom{\rule{0.2em}{0ex}}dt.$

(2.24)

By $\tilde{\alpha}(t)\ge {p}_{1}>0$ for a.e. $t\in [0,2\pi ]$, it follows that ${z}_{0}(t)=0$ for a.e. $t\in [0,2\pi ]$, which is contrary to that ${\parallel {z}_{0}\parallel}_{\mathrm{\infty}}=1$. Hence, (2.22) holds. Clearly, (2.21)-(2.22) contradict (2.19).

Case (iii). In this case, ${r}_{n}\to -\mathrm{\infty}$ and $\{{s}_{n}\}$ is uniformly bounded. Similar arguments as in Case (ii) imply a contradiction.

In a word, (2.3) cannot hold, and hence by (2.2) there exists

$C>0$ independent of

$\mu \in [0,1]$ such that, if

*u* is a solution of problem (2.1), then

${\parallel u\parallel}_{{C}^{1}}\le C.$

(2.25)

Note that, for each

$h\in {L}^{\mathrm{\infty}}(0,2\pi )$, the problem

$\{\begin{array}{c}-{({\varphi}_{p}({u}^{\prime}))}^{\prime}+{\varphi}_{p}(u)=h(t),\hfill \\ u(0)=u(2\pi ),\phantom{\rule{2em}{0ex}}{u}^{\prime}(0)={u}^{\prime}(2\pi )\hfill \end{array}$

(2.26)

has a unique solution

${G}_{p}(h)\in {C}^{1}[0,2\pi ]$. Clearly, the operator

${G}_{p}$ seen as an operator from

$C[0,2\pi ]$ into

${C}^{1}[0,2\pi ]$ is completely continuous. Define

$\psi :{C}^{1}[0,2\pi ]\to C[0,2\pi ]$ by

$\psi (u)(t)=f(t,u(t))$. Then solving problem (1.1) is equivalent to finding solutions in

${C}^{1}[0,2\pi ]$ of the equation

Let

$(\alpha ,\beta )\in [{p}_{1},{q}_{1}]\times [{p}_{1},{q}_{2}]$. Define the operator

${T}_{\alpha ,\beta}:{C}^{1}[0,2\pi ]\to {C}^{1}[0,2\pi ]$ by

${T}_{\alpha ,\beta}(u)={G}_{p}({\varphi}_{p}(u)+\alpha {({u}^{+})}^{p-1}-\beta {({u}^{-})}^{p-1})$. Denote

${B}_{R}=\{u\in {C}^{1}[0,2\pi ]:{\parallel u\parallel}_{{C}^{1}}<R,R\in \mathbb{R}\}$. Clearly,

$deg(I-{T}_{\alpha ,\beta},{B}_{R},0)$ is well defined for all

$R>0$. Owing to

${\lambda}_{+}\cdot {\lambda}_{-}>0$, there is a continuous curve

$\alpha (\tau )$,

$\beta (\tau )$,

$\tau \in [0,1]$, whose image is in

${\mathbb{R}}^{2}\setminus {\mathrm{\Sigma}}_{p}$ and

$(\lambda ,\lambda )\in \mathbb{R}\setminus {\mathrm{\Sigma}}_{p}$ such that

$(\alpha (0),\beta (0))=({\lambda}_{+},{\lambda}_{-})$,

$(\alpha (1),\beta (1))=(\lambda ,\lambda )$. From the invariance property of Leray-Schauder degree under compact homotopies, it follows that the degree

$deg(I-{T}_{\alpha (\tau ),\beta (\tau )},{B}_{R},0)$ is constant for

$\tau \in [0,1]$. Obviously, the operator

${T}_{\lambda ,\lambda}$ is odd. By the Borsuk’s theorem, it follows that

$deg(I-{T}_{\lambda ,\lambda},{B}_{R},0)\ne 0$ for all

$R>0$. Thus,

$deg(I-{T}_{{\lambda}_{+},{\lambda}_{-}},{B}_{R},0)\ne 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}R>0.$

Consider the following homotopy:

$H(\mu ,u)={G}_{p}({\varphi}_{p}(u)+(1-\mu )({\lambda}_{+}{\left({u}^{+}\right)}^{p-1}-{\lambda}_{-}{\left({u}^{-}\right)}^{p-1})+\mu \psi (u)),$

for

$(\mu ,u)\in [0,1]\times {C}^{1}[0,2\pi ]$. By (2.25), we can see that there exists

${R}_{0}>0$ such that

$H(\mu ,u)\ne u,\phantom{\rule{1em}{0ex}}\mathrm{\forall}\mu \in [0,1],\mathrm{\forall}u\in \partial {B}_{{R}_{0}}.$

From the invariance property of Leray-Schauder degree, it follows that

$\begin{array}{rcl}deg(I-H(1,\cdot ),{B}_{{R}_{0}},0)& =& deg(I-H(0,\cdot ),{B}_{{R}_{0}},0)\\ =& deg(I-{T}_{{\lambda}_{+},{\lambda}_{-}},{B}_{{R}_{0}},0)\ne 0.\end{array}$

Hence, problem (1.1) has a solution. The proof is complete. □