Proof. Consider the following homotopy equation with periodic boundary condition

(2.6)\begin{equation} ( \phi_{p}(u'))'+\lambda \hbar\varepsilon\phi_{p}(u')+\lambda a_{+}(t)\phi_{p}(u^{+})-\lambda a_{-}(t)\phi_{p}(u^{-})+\lambda W_{0}(t,u)=\lambda e(t),\end{equation}

where $\lambda \in (0,1)$.

First, if $u\in C_{T}^{1}$ is a solution of (2.6), we claim that there exist positive constants $b_{i},i=1,2$ such that $\|u\|_{\infty }\leq b_{1}+b_{2}\|u'\|_{{1}}$. Let $t_{1}$ and $t_{2}$ represent minimum and maximum points respectively. Thus, we know that

\[ \phi_{p}(u'(t_{1}))=0,(\phi_{p}(u'(t_{1})))'\geq 0\ \mbox{and}\ \phi_{p}(u'(t_{2}))=0,(\phi_{p}(u'(t_{2})))'\leq 0. \]

Thus, we need to consider the following three cases.

\begin{align*} & (A1):u(t_1 ) \le 0,u(t_2 ) \ge 0, \\ & (A2):u(t_1 ) \ge 0,u(t_2 ) \ge 0, \\ & (A3):u(t_1 ) \le 0,u(t_2 ) \le 0. \end{align*}

For the case (A1), it is clear that $\|u\|_{\infty }\leq \|u'\|_{{1}}$. For the case (A2), from (2.6), we have

\[ a_{+}(t_{1})\phi_{p}(u^{+}(t_{1}))+W_{0}(t_{1},u(t_{1})))\leq e(t_{1})\leq \|e\|_{\infty} \]

and

\[ a_{+}(t_{2})\phi_{p}(u^{+}(t_{2}))+W_{0}(t_{2},u(t_{2})))\geq e(t_{2})\geq{-}\|e\|_{\infty}. \]

Thus, we know that

\[ \{ a_{+}(t)\phi_{p}(u^{+}(t))+W_{0}(t,u(t)):t\in [0,T]\}\bigcap [-\|e\|_{\infty},\|e\|_{\infty}]\neq \emptyset, \]

i.e., there exists $t_{3}\in [0,T]$ such that

\[ |a_{+}(t_{3})\phi_{p}(u^{+}(t_{3}))+W_{0}(t_{3},u(t_{3}))|\leq \|e\|_{\infty}. \]

From $(H2)$, we can obtain that $|u(t_{3})|\leq a_{3}.$ Thus, the claim is proved. By the similar way to the case $(A2)$, the case $(A3)$ is also true. Here, we omit the detail.

Multiplying the both sides of (2.6) by $u'$ and integrating from $0$ to $T$, we have

(2.7)\begin{align} & \hbar\varepsilon\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s+\int_{0}^{T} a(s)\phi_{p}(u^{+}(s))u'(s)\,{\rm d}s-\int_{0}^{T} b(s)\phi_{p}(u^{-}(s))u'(s)\,{\rm d}s\nonumber\\ & \quad+\int_{0}^{T}g_{1}(s,u(s))u'(s)\,{\rm d}s=\int_{0}^{T}e(s)u'(s)\,{\rm d}s. \end{align}

By the definition of $\hbar$, we get two cases as follows.

\begin{align*} & (i):\ \varepsilon\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\leq{-}\int_{0}^{T}g_{1}(s,u(s))u'(s)\,{\rm d}s+\int_{0}^{T}e(s)u'(s)\,{\rm d}s.\\ & (ii):\ \varepsilon\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\leq\int_{0}^{T}g_{1}(s,u(s))u'(s)\,{\rm d}s-\int_{0}^{T}e(s)u'(s)\,{\rm d}s. \end{align*}

Moreover, by $(H1)$ and the basic inequality $(x+y)^{r}\leq 2^{r}(x^{r}+y^{r}), x,y,r>0$, we have

\begin{align*} & \int_{0}^{T}|g_{1}(s,u(s))u'(s)|\,{\rm d}s\\ & \quad \leq \left(\int_{0}^{T}|g_{1}(s,u(s))|^{q}\,{\rm d}s\right)^{\frac{1}{q}} \left(\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \quad \leq \left(\int_{0}^{T}(a_{1}(t)+a_{2}(t)|u(s)|^{\theta})^{q}\,{\rm d}s\right)^{\frac{1}{q}} \left(\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \quad \leq \left[2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}+2^{\frac{1}{q}+1}\|a_{2}\|_{\infty}\left(\int_{0}^{T}|u(s)|^{\theta q}\,{\rm d}s\right)^{\frac{1}{q}}\right]\left(\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \quad \leq \left(2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}+2^{\frac{2}{q}+1+\theta}\|a_{2}\|_{\infty}b_{1}^{\theta}T^{\frac{1}{q}}\right)\left(\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \qquad+2^{\frac{2}{q}+1+\theta}\|a_{2}\|_{\infty}b_{2}^{\theta}T^{\frac{1}{q}}\left(\int_{0}^{T}|u'(s)|\,{\rm d}s\right)^{\theta}\left(\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \quad \leq \left(2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}+2^{\frac{2}{q}+1+\theta}\|a_{2}\|_{\infty}b_{1}^{\theta}T^{\frac{1}{q}}\right)\left(\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \qquad+2^{\frac{2}{q}+1+\theta}\|a_{2}\|_{\infty}b_{2}^{\theta}T^{\frac{1+\theta}{q}}\left(\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\right)^{\frac{1+\theta}{p}}. \end{align*}

Thus, from (i) or (ii), we have

\begin{align*} & \varepsilon\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\nonumber\\ & \leq \left.\left(2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}+2^{\frac{2}{q}+1+\theta}\|a_{2}\|_{\infty}b_{1}^{\theta}T^{\frac{1}{q}}+\int_{0}^{T}|e(s)|^{q}\right)^{\frac{1}{q}}\right)\left(\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \quad+ 2^{\frac{2}{q}+1+\theta}\|a_{2}\|_{\infty}b_{2}^{\theta}T^{\frac{1+\theta}{q}}\left(\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\right)^{\frac{1+\theta}{p}}. \end{align*}

So, we can get

(2.8)\begin{align} \varepsilon \left(\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{q}}& \leq 2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}+2^{\frac{2}{q}+1+\theta}\|a_{2}\|_{\infty}b_{1}^{\theta}T^{\frac{1}{q}}+\left(\int_{0}^{T}|e(s)|^{q}\right)^{\frac{1}{q}} \nonumber\\ & \quad+2^{\frac{2}{q}+1+\theta}\|a_{2}\|_{\infty}b_{2}^{\theta}T^{\frac{1+\theta}{q}}\left(\int_{0}^{T}|u'(s)|^{p}\,{\rm d}s\right)^{\frac{\theta}{p}}. \end{align}

Noting that $\theta \in [0,p-1)$, we claim that $\|u'\|_{{p}}$ is bounded, i.e, there exists a positive constant $D$ such that $\|u'\|_{{p}}\leq D$. Otherwise, we can find the sequence $\{u'_{n}(t)\}_{n\in \mathbb {Z^{+}}}$ which satisfies (2.8) such that

\[ \|u'_{n}\|_{{p}}\to \infty,\text{as } n\to \infty. \]

Moreover, multiplying (2.8) by $\frac {1}{\|u'_{n}\|_{{p}}^{\theta }}$, since $\theta \in [0,p-1)$, we have

\begin{align*} & \varepsilon \|u'_{n}\|_{{p}}^{p-1-\theta}\nonumber\\ & \quad\leq \frac{2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}+2^{\frac{2}{q}+1+\theta}\|a_{2}\|_{\infty}b_{1}^{\theta}T^{\frac{1}{q}}+\left(\int_{0}^{T}|e(s)|^{q}\right)^{\frac{1}{q}}}{\|u'_{n}\|_{{p}}^{\theta}}\\ & \qquad +2^{\frac{2}{q}+1+\theta}\|a_{2}\|_{\infty}b_{2}^{\theta}T^{\frac{1+\theta}{q}}, \end{align*}

which implies a contradiction as $n\to \infty$. Thus, the claim is true. Therefore, we have

\[ \|u\|_{\infty}\leq b_{1}+b_{2}\|u'\|_{{1}}\leq b_{1}+b_{2}T^{\frac{1}{q}}\|u'\|_{{p}}\leq b_{1}+b_{2}T^{\frac{1}{q}}D:=\overline{D}. \]

On the other hand, (2.6) is equivalent to

(2.9)\begin{equation} ( \phi_{p}(u'(t))e^{\lambda \hbar\varepsilon t})'=\lambda \left[{-}a_{+}(t)\phi_{p}(u^{+})+ a_{-}(t)\phi_{p}(u^{-})-W_{0}(t,u)+ e(t)\right]e^{\lambda \hbar\varepsilon t}.\end{equation}

Since $u(0)=u(T)$, there exists $t^{*}\in (0,T)$ such that $u'(t^{*})=0$. Integrating from $t^{*}$ to $t$ on (2.9), we have

\begin{align*} phi_{p}(u')& =\lambda e^{-\lambda \hbar\varepsilon t}\int_{t^{*}}^{t}({-}a_{+}(s)\phi_{p}(u^{+}(s))+ a_{-}(s)\phi_{p}(u^{-}(s))\\ & \quad -W_{0}(s,u(s))+ e(s))e^{\lambda \hbar\varepsilon s}\,{\rm d}s. \end{align*}

Thus, we can obtain

(2.10)\begin{equation} |u'(t)|^{p-1}\leq e^{2\varepsilon T}(Ta_{+,M}\overline{D}^{p-1}+Ta_{-,M}\overline{D}^{p-1}+lT+T\|e\|_{\infty}):=K^{p-1},\end{equation}

where $l={\max }_{|u|\leq \overline {D},0\leq t\leq T}|W_{0}(t,u(t))|$. It implies that $\|u'\|_{\infty }\leq K.$

Let

\[ \Omega=\{u(t)\in C^{1}_{T}:\|u\|\leq M+1\}, \]

where $M>K+\overline {D}$ such that $a_{+}(t)\phi _{p}(M+1)+W_{0}(t,M+1)-e(t)>0$ and $a_{-}(t)\phi _{p}(-M-1)+W_{0}(t,-M-1)-e(t)<0$. Clearly, the problem (2.4) has no solution on $\mathbb {R}\cap \partial \Omega$. Moreover, we can get

\begin{align*} & F(M+1)=\frac{1}{T}\int_{0}^{T}-(a_{+}(s)\phi_{p}(M+1)+W_{0}(s,M+1)-e(s))\,{\rm d}s<0,\\ & F({-}M-1)=\frac{1}{T}\int_{0}^{T}-(a_{-}(s)\phi_{p}({-}M-1)+W_{0}(s,-M-1)-e(s))\,{\rm d}s>0. \end{align*}

Thus, $F(a)$ has no solution on $\mathbb {R}\cap \partial \Omega$. Making a homotopy

\begin{align*} H(u,\lambda)& =\lambda u+(1-\lambda)\frac{1}{T}\int_{0}^{T}({-}a_{+}(s)\phi_{p}(u^{+}(s))+ a_{-}(s)\phi_{p}(u^{-}(s))\\ & \quad -W_{0}(s,u(s))+ e(s))\,{\rm d}s, \end{align*}

where $u\in \mathbb {R}\cap \partial \Omega$, we have

\begin{align*} uH(u,\lambda)& =\lambda u^{2}+(1-\lambda)\frac{1}{T}u\int_{0}^{T}({-}a_{+}(s)\phi_{p}(u^{+}(s))+ a_{-}(s)\phi_{p}(u^{-}(s))\\ & \quad -W_{0}(s,u(s))+ e(s))\,{\rm d}s<0. \end{align*}

Based on the homotopic transformation, we have

\[ \mbox{deg}_{B}(F,\mathbb{R}\cap\partial\Omega,0)=\mbox{deg}_{B}(u,\mathbb{R}\cap\partial\Omega,0)\neq 0. \]

Thus, the problem (2.4) has at least one solution.

Proof. First, we can obtain the priori estimate that if $u\in C_{T}^{1}$ is a solution of the problem (2.5), then we have $|u(t)|\leq L,$ where $L$ is a positive constant and independent of $\lambda$. Otherwise, we can find solutions $\{u_{n}(t)\}$ of the problem (2.5) corresponding with $\lambda =\lambda _{n}$, namely, $u_{n}(t)$ satisfies the following equation

(2.11)\begin{align} & ( \phi_{p}(u_{n}'))'+\lambda_{n}\hbar \varepsilon\phi_{p}(u_{n}')+\lambda_{n} a_{+}(t)\phi_{p}(u_{n}^{+})\notag\\ & \quad -\lambda_{n} a_{-}(t)\phi_{p}(u_{n}^{-})+\lambda_{n} W_{1}(t,u_{n},u'_{n})=\lambda_{n} e(t)\end{align}

and periodic boundary condition such that

\[ \|u_{n}\|_{\infty}\to \infty,\ \mbox{as}\ n\to \infty. \]

For any $u \in C_{T}^{1}\setminus \{0\}$, let $z_{n}(t)=\frac {u_{n}(t)}{\left \| u_{n} \right \|_{\infty }}$, so $\|z_{n}\|_{\infty }=1$. Moreover, multiplying (2.11) by $\frac {u_{n}'}{\left \| u_{n} \right \|^{p}_{\infty }}$ and integrating from 0 to $T$, we have

(2.12)\begin{align} & \hbar\varepsilon\int_{0}^{T}|z_{n}'(s)|^{p}\,{\rm d}s+\int_{0}^{T} a(s)\phi_{p}(z_{n}^{+}(s))z_{n}'(s)\,{\rm d}s-\int_{0}^{T} b(s)\phi_{p}(z_{n}^{-}(s))z_{n}'(s)\,{\rm d}s\nonumber\\ & \quad+\frac{1}{{\|u_{n}\|}_{\infty}^{p-1}}\int_{0}^{T}\overline{g}_{1}(s,u'_{n}(s))z_{n}'(s)\,{\rm d}s=\frac{1}{{\|u_{n}\|}_{\infty}^{p-1}}\int_{0}^{T}e(s)z_{n}'(s)\,{\rm d}s. \end{align}

By definition of $\hbar$, we get two cases as follows.

\begin{align*} (i):\ \varepsilon\int_{0}^{T}|z_{n}'(s)|^{p}\,{\rm d}s& \leq{-}\frac{1}{{\|u_{n}\|}_{\infty}^{p-1}}\int_{0}^{T}\overline{g}_{1}(s,u'_{n}(s))z_{n}'(s)\,{\rm d}s\notag\\ & \quad +\frac{1}{{\|u_{n}\|}_{\infty}^{p-1}}\int_{0}^{T}e(s)z_{n}'(s)\,{\rm d}s.\\ (ii):\ \varepsilon\int_{0}^{T}|z_{n}'(s)|^{p}\,{\rm d}s& \leq \frac{1}{{\|u_{n}\|}_{\infty}^{p-1}}\int_{0}^{T}\overline{g}_{1}(s,u'_{n}(s))z_{n}'(s)\,{\rm d}s\notag\\ & \quad -\frac{1}{{\|u_{n}\|}_{\infty}^{p-1}}\int_{0}^{T}e(s)z_{n}'(s)\,{\rm d}s. \end{align*}

By $(H3)$, we can get

\begin{align*} & \frac{1}{{\|u_{n}\|}_{\infty}^{p-1}}\int_{0}^{T}\overline{g}_{1}(s,u'_{n}(s))z_{n}'(s)\,{\rm d}s\\ & \quad\leq\frac{1}{{\|u_{n}\|}_{\infty}^{p-1}}\left(\int_{0}^{T}|\overline{g}_{1}(s,u'_{n}(s))|^{q}\,{\rm d}s\right)^{\frac{1}{q}} \left(\int_{0}^{T}|z'_{n}(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \quad\leq\frac{1}{{\|u_{n}\|}_{\infty}^{p-1}}\left(\int_{0}^{T}(a_{1}(t)+a_{2}(t)|u'_{n}(s)|^{\theta})^{q}\,{\rm d}s\right)^{\frac{1}{q}} \left(\int_{0}^{T}|z'_{n}(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}} \\ & \quad\leq\frac{2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}}{{\|u_{n}\|}_{\infty}^{p-1}}\left(\int_{0}^{T}|z'_{n}(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \qquad +\frac{2^{\frac{1}{q}+1}\|a_{2}\|_{\infty}}{{\|u_{n}\|}_{\infty}^{p-\theta-1}}\left(\int_{0}^{T}|z'_{n}(s)|^{\theta q}\,{\rm d}s\right)^{\frac{1}{q}}\left(\int_{0}^{T}|z'_{n}(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \quad\leq\frac{2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}}{{\|u_{n}\|}_{\infty}^{p-1}}\left(\int_{0}^{T}|z'_{n}(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \qquad +\frac{2^{\frac{1}{q}+1}\|a_{2}\|_{\infty}T^{\frac{p-\theta-1}{p}}}{{\|u_{n}\|}_{\infty}^{p-\theta-1}}\left(\int_{0}^{T}|z'_{n}(s)|^{p}\,{\rm d}s\right)^{\frac{1+\theta}{p}} . \end{align*}

Thus, from (i) or (ii), one has

\begin{align*} \varepsilon\int_{0}^{T}|z_{n}'(s)|^{p}\,{\rm d}s& \leq \frac{2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}}{{\|u_{n}\|}_{\infty}^{p-1}}\left(\int_{0}^{T}|z'_{n}(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\\ & \quad +\frac{2^{\frac{1}{q}+1}\|a_{2}\|_{\infty}T^{\frac{p-\theta-1}{p}}}{{\|u_{n}\|}_{\infty}^{p-\theta-1}}\left(\int_{0}^{T}|z'_{n}(s)|^{p}\,{\rm d}s\right)^{\frac{1+\theta}{p}}\\ & \quad+ \frac{1}{{\|u_{n}\|}_{\infty}^{p-1}} \left(\int_{0}^{T}|e(s)|^{q}\right)^{\frac{1}{q}}\left(\int_{0}^{T}|z_{n}'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}, \end{align*}

which yields that

(2.13)\begin{align} \varepsilon\left(\int_{0}^{T}|z_{n}'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{q}}& \leq \frac{2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}}{{\|u_{n}\|}_{\infty}^{p-1}}+\frac{2^{\frac{1}{q}+1}\|a_{2}\|_{\infty}T^{\frac{p-\theta-1}{p}}}{{\|u_{n}\|}_{\infty}^{p-\theta-1}}\left(\int_{0}^{T}|z'_{n}(s)|^{p}\,{\rm d}s\right)^{\frac{\theta}{p}}\nonumber\\ & \quad+ \frac{1}{{\|u_{n}\|}_{\infty}^{p-1}} \left(\int_{0}^{T}|e(s)|^{q}\right)^{\frac{1}{q}}. \end{align}

We claim that $\|z_{n}'\|_{{p}}\to 0$ as $n\to \infty$. If not, we have the following two cases.

\begin{align*} & (B1)\ \|z_{n}'\|_{{p}}\to c\ \mbox{as}\ n\to \infty,\ \mbox{where}\ c \ \mbox{is a positive constant};\\ & (B2)\ \|z_{n}'\|_{{p}}\to \infty\ \mbox{as}\ n\to \infty. \end{align*}

For the case $(B1)$, from (2.13), we have

\begin{align*} & \left(\int_{0}^{T}|z_{n}'(s)|^{p}\,{\rm d}s\right)^{\frac{\theta}{p}}\left[\varepsilon\left(\int_{0}^{T}|z_{n}'(s)|^{p}\,{\rm d}s\right)^{\frac{p-\theta-1}{p}}-\frac{2^{\frac{1}{q}+1}\|a_{2}\|_{\infty}T^{\frac{p-\theta-1}{p}}}{{\|u_{n}\|}_{\infty}^{p-\theta-1}}\right]\\ & \quad\leq \frac{2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}}{{\|u_{n}\|}_{\infty}^{p-1}}+ \frac{1}{{\|u_{n}\|}_{\infty}^{p-1}} \left(\int_{0}^{T}|e(s)|^{q}\right)^{\frac{1}{q}}, \end{align*}

which implies that $\varepsilon c^{p-1}\leq 0$ as $n\to \infty$. This is impossible. Clearly, for the case $(B2)$, we can also get a contradiction. Thus, $\|z_{n}'\|_{{p}}\to 0$ as $n\to \infty$. Moreover, for any $t\in [0,T]$, by Hölder inequality, we have

\[ |z_{n}(t)-z_{n}(0)|\leq T^{\frac{1}{q}}\left(\int_{0}^{T}|z_{n}'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\to 0,\ \mbox{as}\ n \to \infty. \]

Since $\|z_{n}\|=1$, we can obtain

\[ \lim_{n\to \infty} z_{n}(t)={\pm} 1,\quad \forall t\in [0,T]\Rightarrow \lim_{n\to \infty} u_{n}(t)=\lim_{n\to \infty} \|u_{n}\|_{\infty}z_{n}(t)={\pm} \infty. \]

Thus, we have

(2.14)\begin{equation} \lim_{n\to \infty} a_{{\pm}}(t)\phi_{p}(u_{n}^{{\pm}}(t))={\pm} \infty,\ \forall t\in [0,T].\end{equation}

Similarly, we have

(2.15)\begin{align} \varepsilon\left(\int_{0}^{T}|u_{n}'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{q}}& \leq 2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}+2^{\frac{1}{q}+1}\|a_{2}\|_{\infty}T^{\frac{p-\theta-1}{p}}\left(\int_{0}^{T}|u'_{n}(s)|^{p}\,{\rm d}s\right)^{\frac{\theta}{p}}\nonumber\\ & \quad+ \left(\int_{0}^{T}|e(s)|^{q}\right)^{\frac{1}{q}}. \end{align}

Clearly, $\|u'_{n}\|_{{p}}$ is bounded, i.e, there exists a positive constant $\overline {L}$ such that ${\|u'_{n}\|_{{p}}\leq \overline {L}}$. Otherwise,

\[ \|u'_{n}\|_{{p}}\to \infty,\ \mbox{as}\ n\to \infty. \]

From (2.15), we have

(2.16)\begin{equation} \varepsilon\|u_{n}\|_{{p}}^{\frac{p-\theta-1}{p}}\leq \frac{2^{\frac{1}{q}+1}\|a_{1}\|_{\infty}T^{\frac{1}{q}}+ \left(\int_{0}^{T}|e(s)|^{q}\right)^{\frac{1}{q}}}{\|u_{n}\|_{{p}}^{\frac{\theta}{p}}}+2^{\frac{1}{q}+1}\|a_{2}\|_{\infty}T^{\frac{p-\theta-1}{p}},\end{equation}

which implies a contradiction. Thus, $\|u'_{n}\|_{{p}}$ is bounded. Moreover, we can also get

(2.17)\begin{align} \left|\int_{0}^{T}\phi_{p}(u_{n}'(s))\,{\rm d}s\right|& \leq \int_{0}^{T}|u_{n}'(s)|^{p-1}\,{\rm d}s\leq T^{\frac{1}{p}}\left(\int_{0}^{T}|u_{n}'(s)|^{p}\,{\rm d}s\right)^{\frac{1}{q}}\nonumber\\ & \leq T^{\frac{1}{p}} \overline{L}^{p-1}, \end{align}

and

(2.18)\begin{align} \left|\int_{0}^{T}\overline{g}_{1}(s,u_{n}'(s))\,{\rm d}s\right|& \leq \|a_{1}\|_{\infty}T+\|a_{2}\|_{\infty}\int_{0}^{T}|u'_{n}|\,{\rm d}s\nonumber\\ & \leq\|a_{1}\|_{\infty}T+\|a_{2}\|_{\infty}T^{\frac{1}{q}}\left(\int_{0}^{T}|u'_{n}|^{p}\,{\rm d}s\right)^{\frac{1}{p}}\nonumber\\ & \leq\|a_{1}\|_{\infty}T+\|a_{2}\|_{\infty}T^{\frac{1}{q}}\overline{L}. \end{align}

By $(H4)$, we have

(2.19)\begin{align} & \text{if }\lim_{n\to \infty} u_{n}(t)=\infty,\ \overline{g}_{0}(u_{n})\geq \varrho{|u_{n}|}^{\gamma}\Rightarrow \overline{g}_{0}(u_{n})\to \infty \ \mbox{as}\ n \to \infty \ \mbox{or}\nonumber\\ & \text{if }\lim_{n\to \infty} u_{n}(t)={-}\infty,\ \overline{g}_{0}(u_{n})\leq{-} \varrho{|u_{n}|}^{\gamma}\Rightarrow \overline{g}_{0}(u_{n})\to -\infty \ \mbox{as}\ n \to \infty, \end{align}

which implies that

(2.20)\begin{equation} \lim_{n\to\infty}\left|\int_{0}^{T}(e(s)- a_{+}(s)\phi_{p}(u_{n}^{+}(s))+a_{-}(s)\phi_{p}(u_{n}^{-}(s))\,{\rm d}s\right|=\infty.\end{equation}

However, integrating from 0 to $T$ on (2.11), we have

\begin{align*} & \hbar\varepsilon\int_{0}^{T}\phi_{p}(u_{n}'(s))\,{\rm d}s+\int_{0}^{T} a_{+}(s)\phi_{p}(u_{n}^{+}(s))\,{\rm d}s-\int_{0}^{T} a_{-}(s)\phi_{p}(u_{n}^{-}(s))\,{\rm d}s\nonumber\\ & \quad+ \int_{0}^{T}W_{1}(s,u_{n}(s),u'_{n}(s))\,{\rm d}s =\int_{0}^{T}e(s)\,{\rm d}s. \end{align*}

This implies a contradiction for $n$ large enough. Therefore, $|u(t)|\leq L$. By the similar way to lemma 2.2, we can obtain that there exists a positive constant $\Theta$ such that $|u'(t)|\leq \Theta$. Moreover, $\mbox {deg}_{B}(F,\mathbb {R}\cap \partial \Omega,0)\neq 0$. Thus, the problem (2.5) has at least one solution.

Proof. In view of lemma 2.3, we claim that if $\{u_n(t)\}$ are solutions of the problem (2.4) corresponding to $\varepsilon = \{\varepsilon _n\}$, where $\varepsilon _n\to 0$ as $n\to \infty$, then there exist constants $\widetilde {D},\ \widetilde {K}>0$, independent of $\varepsilon _n$, such that for $t\in [0,T]$,

\[ |u_{n}(t)|\leq \widetilde{D},\ |u_{n}'(t)|\leq \widetilde{K}. \]

Otherwise,

\[ \|u_{n}\|_{\infty}\to \infty\ \mbox{as}\ n\to \infty. \]

For any $u \in C_{T}^{1}\setminus \{0\}$, let $z_{n}(t)=\frac {u_{n}(t)}{\left \| u_{n} \right \|_{\infty }}$, so $\|z_{n}\|_{\infty }=1$. Moreover, dividing the problem (2.4) by $\left \| u_{n} \right \|_{\infty }^{p-1}$, we have

(2.21)\begin{align} & ( \phi_{p}(z_{n}'))'+\hbar \varepsilon_{n}\phi_{p}(z_{n}')+ a_{+}(t)\phi_{p}(z_{n}^{+})- a_{-}(t)\phi_{p}(z_{n}^{-})+ \frac{g_{0}(u_{n})}{\left\| u_{n} \right\|_{\infty}^{p-1}}\nonumber\\ & \quad- \frac{g_{1}(t,u_{n}(t))}{\left\| u_{n} \right\|_{\infty}^{p-1}}= \frac{e(t)}{\left\| u_{n} \right\|_{\infty}^{p-1}}, \end{align}

which is equivalent to

(2.22)\begin{align} & ( \phi_{p}(z_{n}')e^{\varepsilon_{n}\hbar t})'+ e^{\varepsilon_{n} \hbar t}a_{+}(t)\phi_{p}(z_{n}^{+})- e^{\varepsilon_{n}\hbar t}a_{-}(t)\phi_{p}(z_{n}^{-})+ e^{\varepsilon_{n}\hbar t} \frac{g_{0}(u_{n})}{\left\| u_{n} \right\|_{\infty}^{p-1}}\nonumber\\ & \quad- e^{\varepsilon_{n}\hbar t}\frac{g_{1}(t,u_{n}(t))}{\left\| u_{n} \right\|_{\infty}^{p-1}}= e^{\varepsilon_{n}\hbar t} \frac{e(t)}{\left\| u_{n} \right\|_{\infty}^{p-1}}. \end{align}

Since $z_{n}(0)=z_{n}(T)$, there exists $t_{*}\in (0,T)$ such that $z_{n}'(t_{*})=0$. Thus, we have

\begin{align*} | \phi_{p}(z_{n}')|e^{\varepsilon_{n}\hbar t}& \leq\int_{t_{*}}^{t}\left(\left|a_{+}(s)\phi_{p}(z_{n}^{+})\right|+ \left|a_{-}(t)\phi_{p}(z_{n}^{-})\right|+\left|\frac{g_{0}(u_{n})}{\left\| u_{n} \right\|_{\infty}^{p-1}}\right|\right.\\ & \left. \quad +\left| \frac{g_{1}(t,u_{n}(t))}{\left\| u_{n} \right\|_{\infty}^{p-1}}\right|+\left|\frac{e(s)}{\left\| u_{n} \right\|_{\infty}^{p-1}}\right|\right)e^{\varepsilon_{n}\hbar s}\,{\rm d}s. \end{align*}

From $\varepsilon _{n} \to 0$ as $n\to \infty$, there exists a positive constant $d_{1}$ such that $e^{\varepsilon _{n}\hbar t}\leq d_{1}$. Moreover, we have

(2.23)\begin{align} \int_{t_{*}}^{t} \left| \frac{g_{1}(t,u_{n}(t))}{\left\| u_{n} \right\|_{\infty}^{p-1}}e^{\varepsilon_{n}\hbar s}\,{\rm d}s\right|& \leq \frac{d_{1}}{\left\| u_{n} \right\|_{\infty}^{p-1}}\int_{0}^{T} | g_{1}(s,u_{n}(s))|\,{\rm d}s\nonumber\\ & \leq \frac{d_{1}}{\left\| u_{n} \right\|_{\infty}^{p-1}}\left(\|a_{1}\|_{\infty}T+\|a_{2}\|_{\infty}\int_{0}^{T} |u_{n}(s)|^{\theta}\,{\rm d}s\right)\nonumber\\ & \leq \frac{d_{1}T\|a_{1}\|_{\infty}}{\left\| u_{n} \right\|_{\infty}^{p-1}}+ \frac{d_{1} T \|a_{2}\|_{\infty}}{\left\| u_{n} \right\|_{\infty}^{p-\theta-1}}, \end{align}

which implies that there exists a constant $M_{1}>0$ which is independent of $n$ such that

(2.24)\begin{equation} \int_{t_{*}}^{t} \left| \frac{g_{1}(t,u_{n}(t))}{\left\| u_{n} \right\|_{\infty}^{p-1}}e^{\varepsilon_{n}\hbar s}\,{\rm d}s\right|\leq M_{1}.\end{equation}

Based on $\|z_{n}\|=1$ and $(H5)$, for any $t \in [0,T]$, there exists a constant $M_{2}>0$ which is independent of $n$ such that

\[ \int_{t_{*}}^{t}(\left|a_{+}(s)\phi_{p}(z_{n}^{+})\right|+ \left|a_{-}(s)\phi_{p}(z_{n}^{-})\right|+\left|\frac{g_{0}(u_{n})}{\left\| u_{n} \right\|_{\infty}^{p-1}}\right|)e^{\varepsilon_{n}\hbar s}\,{\rm d}s\leq M_{2}, \]

which together with $e\in C([0,T])$ and the monotonicity of $\phi _{p}(\cdot )$ yield that

(2.25)\begin{equation} |\phi_{p}(z_{n}'(t))|\leq M_{3},\ \mbox{i.e.},\ |z_{n}'(t)|\leq M_{3}^{q-1},\end{equation}

where $M_{3}>0$ is a constant and independent of $n$. Thus, from the definition of $z_{n}(t)$, it follows that $z_{n}(t)$ is uniformly bounded. By the standard argument, we can get that $z_{n}(t)$ is also equicontinuous. Moreover, from (2.22), we can also find a constant $M_{4}>0$ which is independent of $n$ such that

\[ |( \phi_{p}(z_{n}'(t)))'|\leq M_{4}. \]

Furthermore, $\forall t_{1},\ t_{2}\in [0,1]$, assuming that $t_{1}\leq t_{2},$ for any $n\in \mathbb {N}$, we have

\begin{align*} | \phi_{p}(z_{n}'(t_{2}))-\phi_{p}(z_{n}'(t_{1}))|& \leq \int_{t_{1}}^{t_{2}}|( \phi_{p}(z_{n}'(t)))'\,{\rm d}s|\\ & \leq M_{4}|t_{2}-t_{1}| \to 0\ \mbox{uniformly as}\ t_{1}\to t_{2}, \end{align*}

which together with the monotonicity and uniformly continuity of $\phi _{p}(\cdot )$ on $[-M_{3}^{q-1},M_{3}^{q-1}]$ yield that $z_{n}'(t)$ is also uniformly bounded and equicontinuous. By Arzelá–Ascoli theorem, there exists a subsequence of $\{z_{n}\}$ (without loss of generality, take its subsequence) and $z(t)\in C_{T}^{1}$ such that for any $t\in [0,T]$,

(2.26)\begin{equation} \mathop {\lim }\limits_{n \to \infty } z_n (t) = z(t),\mathop {\lim }\limits_{n \to \infty } z_n' (t) = z'(t).\end{equation}

Thus, there exists a $\widetilde {t}\in [0,T]$ such that $z(\widetilde {t})\neq 0.$ Multiplying both sides of (2.4) by $\frac {u_{n}'(t)}{\|u_{n}\|^{p}}$ and integrating from $\widetilde {t}$ to $t$, one has

(2.27)\begin{align} & \int_{\widetilde{t}}^{t}( \phi_{p}(z_{n}'))'z_{n}'\,{\rm d}s\,{+}\,\varepsilon_{n}\hbar\int_{\widetilde{t}}^{t}|z_{n}|^{p}\,{\rm d}s\,{+}\int_{\widetilde{t}}^{t} a_{+}(s)\phi_{p}(z_{n}^{+})z_{n}'\,{\rm d}s\,{-}\int_{\widetilde{t}}^{t} a_{-}(s)\phi_{p}(z_{n}^{-})z_{n}'\,{\rm d}s\nonumber\\ & \quad+ \frac{G_{0}(u_{n}(t))-G_{0}(u_{n}(\widetilde{t}))}{\left\| u_{n} \right\|_{\infty}^{p}}- \int_{\widetilde{t}}^{t}\frac{ g_{1}(s,u_{n}(s))z_{n}'}{\left\| u_{n} \right\|_{\infty}^{p-1}}\,{\rm d}s=\int_{\widetilde{t}}^{t}\frac{e(s)z_{n}'}{\left\| u_{n} \right\|_{\infty}^{p-1}}\,{\rm d}s, \end{align}

Following $(H4)$, we know that

\[ \mathop {\lim }\limits_{|u| \to \infty } \frac{{pG_{0}(u)}}{{|u|^{p} }} = \eta. \]

For any $\epsilon >0$, there exists a constant $l_{1}>0$ such that

\[ \left|\frac{{pG_{0}(u)}}{{|u|^{p} }}-\eta\right|\leq \epsilon,\ |u|\geq l_{1},\ \mbox{i.e.}, \left|G_{0}(u)-\frac{\eta|u|^{p}}{p}\right|\leq \epsilon\frac{|u|^{p}}{p},\ |u|\geq l_{1}. \]

By the continuity of $G_{0}$, define

\[ l_{2}:=\max_{|u|\leq l_{l}}\left|G_{0}(u)-\frac{\eta|u|^{p}}{p}\right|. \]

Thus, we have

\[ \left|G_{0}(u_{n})-\frac{\eta|u_{n}|^{p}}{p}\right|\leq \epsilon\frac{|u_{n}|^{p}}{p}+l_{2}\Rightarrow \left|\frac{G_{0}(u_{n})}{\left\| u_{n} \right\|_{\infty}^{p}}-\frac{\eta|z_{n}|^{p}}{p}\right|\leq \epsilon+\frac{l_{2}}{\|u_{n}\|^{p}}. \]

Taking limit in the above equality, by arbitrariness of $\epsilon$, we can obtain

(2.28)\begin{equation} \lim_{n\to\infty}\frac{G_{0}(u_{n})}{\left\| u_{n} \right\|_{\infty}^{p}}=\frac{\eta|z|^{p}}{p}.\end{equation}

Moreover, by (2.24) and (2.25), we have

(2.29)\begin{equation} \int_{\widetilde{t}}^{t}\frac{ g_{1}(s,u_{n}(s))z_{n}'(s)}{\left\| u_{n} \right\|_{\infty}^{p-1}}\,{\rm d}s, \int_{\widetilde{t}}^{t}\frac{e(s)z_{n}'}{\left\| u_{n} \right\|_{\infty}^{p-1}}\,{\rm d}s\to 0,\ \mbox{as}\ n\to \infty.\end{equation}

Furthermore, we know that

(2.30)\begin{equation} \int_{\widetilde{t}}^{t}( \phi_{p}(z_{n}'(s)))'z_{n}'(s)\,{\rm d}s=\frac{1}{q}|\phi_{p}(z_{n}'(t))|^{q}-\frac{1}{q}|\phi_{p}(z_{n}'(\widetilde{t}))|^{q}.\end{equation}

By (2.28)–(2.30), taking limit $n\to \infty$ in (2.27), we have

(2.31)\begin{align} & \frac{1}{q}|\phi_{p}(z'(t))|^{q}+\int_{\widetilde{t}}^{t} a_{+}(s)\phi_{p}(z^{+}(s))z'(s)\,{\rm d}s\notag\\ & \quad -\int_{\widetilde{t}}^{t} a_{-}(s)\phi_{p}(z^{-}(s))z'(s)\,{\rm d}s+\frac{\eta|z(t)|^{p}}{p}=\sigma,\end{align}

where

\[ \sigma=\frac{1}{q}|\phi_{p}(z'(\widetilde{t}))|^{q}+\frac{\eta|z(\widetilde{t})|^{p}}{p}>0. \]

Now, we claim that $z'(t)$ has only limited zero points on $[0, T].$ If not, assume that $z'(t)$ has unlimited zero points $\{\mu _{i}\}_{i\in \mathbb {N}}$ on $[0, T].$ Without loss of generality, there exists a $\mu _{*}$ such that

\[ \lim_{i\to \infty}\mu_{i}=\mu_{*}. \]

Following the continuity of $z'(t)$, we have $z'(\mu _{*})=0$ and

(2.32)\begin{equation} \int_{\widetilde{t}}^{\mu_{*}} a_{+}(s)\phi_{p}(z^{+}(s))z'(s)\,{\rm d}s-\int_{\widetilde{t}}^{\mu_{*}} a_{-}(s)\phi_{p}(z^{-}(s))z'(s)\,{\rm d}s+\frac{\eta|z(\mu_{*})|^{p}}{p}=\sigma.\end{equation}

By the above equality, we can get $z(\mu _{*})\neq 0$. In fact, if $z(\mu _{*})=0$, we have

(2.33)\begin{equation} \int_{\widetilde{t}}^{\mu_{*}} a_{+}(s)\phi_{p}(z^{+}(s))z'(s)\,{\rm d}s-\int_{\widetilde{t}}^{\mu_{*}} a_{-}(s)\phi_{p}(z^{-}(s))z'(s)\,{\rm d}s=\sigma.\end{equation}

By choosing a suitable radius $r_{1}$ and $r_{2}$ such that $\widetilde {t}$ belonging to the left neighbourhood of $(\mu _{*}-r_{1},\mu _{*})$ or the right neighbourhood of $(\mu _{*},\mu _{*}+r_{2})$, we just need to consider the following four cases.

\begin{align*} & (C1)\ z(t)\leq 0\ z'(t)\geq 0,\ t\in [\widetilde{t}, \mu_{*}],\\ & (C2)\ z(t)\geq 0\ z'(t)\leq 0,\ t\in [\widetilde{t}, \mu_{*}],\\ & (C3)\ z(t)\geq 0\ z'(t)\geq 0,\ t\in [\mu_{*},\widetilde{t}],\\ & (C4)\ z(t)\leq 0\ z'(t)\leq 0,\ t\in [\mu_{*},\widetilde{t}]. \end{align*}

For the case $(C1)$, we can obtain

\begin{align*} 0<\frac{1}{q}|\phi_{p}(z'(\widetilde{t}))|^{q}+\frac{\eta|z(\widetilde{t})|^{p}}{p}=\sigma& \leq a_{-,L}\int_{\widetilde{t}}^{\mu_{*}} \phi_{p}(z(s))z'(s)\,{\rm d}s\\ & ={-}\frac{a_{-,L}}{p}|z(\widetilde{t})|^{p}<0, \end{align*}

which is impossible. For the case $(C2)$, it follows

\begin{align*} 0<\frac{1}{q}|\phi_{p}(z'(\widetilde{t}))|^{q}+\frac{\eta|z(\widetilde{t})|^{p}}{p}=\sigma& \leq a_{+,L}\int_{\widetilde{t}}^{\mu_{*}} \phi_{p}(z(s))z'(s)\,{\rm d}s\\ & ={-}\frac{a_{+,L}}{p}|z(\widetilde{t})|^{p}<0, \end{align*}

which implies a contradiction. Moreover, by the same way, we can also obtain a contradiction under the case of $(C3)$ or $(C4)$. Thus, $z(\mu _{*})\neq 0$. Without loss of generality, assume that $z(\mu _{*})>0$. According to its continuity, there exist constants $\xi,\nu >0$ such that

\[ z(t)>\xi,\quad t\in [\mu_{*}-\nu,\mu_{*}+\nu]. \]

Choosing $n$ large enough, we can also get

\[ z_{n}(t)>\xi,\quad t\in [\mu_{*}-\nu,\mu_{*}+\nu]. \]

Since $\mu _{*}$ is a limiting point of $\{\mu _{i}\}$, so we can find two zero points called $\mu ^{*},\ \mu ^{**}$ of $z'(t)$ in $[\mu _{*}-\nu,\mu _{*}+\nu ]$. Integrating from $\mu ^{*}$ to $\mu ^{**}$ on (2.27), taking limit $n\to \infty$, we can obtain

(2.34)\begin{equation} \lim_{n\to \infty} \left(\int_{\mu^{*}}^{\mu^{**}} \frac{g_{0}(u_{n}(s))}{\left\| u_{n} \right\|_{\infty}^{p-1}}e^{\varepsilon_{n}\hbar s}\,{\rm d}s+\int_{\mu^{*}}^{\mu^{**}} e^{\varepsilon_{n}\hbar s} a_{+}(s)\phi_{p}(z_{n}(s))\,{\rm d}s\right)=0.\end{equation}

On the other hand, for $t\in [\mu ^{*}, \mu ^{**}]$, choose $n$ large enough such that

\[ u_{n}(t)\geq z_{n}(t)\|u\|_{\infty}\geq \xi\|u\|_{\infty}\geq d. \]

Then, we have

\[ \frac{g_{0}(u_{n}(t))}{\left\| u_{n} \right\|^{p-1}}=\frac{g_{0}(u_{n}(t))}{\phi_{p}(u_{n}(t))}\phi_{p}(z_{n}(t))\geq\delta \phi_{p}(\xi). \]

Thus, we can obtain

(2.35)\begin{align} & \lim_{n\to \infty}\left(\int_{\mu^{*}}^{\mu^{**}} \frac{g_{0}(u_{n}(s))}{\left\| u_{n} \right\|^{p-1}}e^{\varepsilon_{n} \hbar s}\,{\rm d}s+\int_{\mu^{*}}^{\mu^{**}} e^{\varepsilon_{n}\hbar s}a_{+}(s)\phi_{p}(z(s))\,{\rm d}s\right)\nonumber\\ & \quad\geq (a_{{+},L}+\delta)\phi_{p}(\xi)(\mu^{**}-\mu^{*})>0, \end{align}

which contradicts to (2.34). Thus, $z'(t)$ has only limited zero points on $[0, T].$ Differentiating (2.31), one has

\[ z'(t)\left(( \phi_{p}(z'(t)))'+a(t)\phi_{p}(z^{+}(t))-b(t)\phi_{p}(z^{-}(t))+\eta\phi_{p}(z(t))\right)=0. \]

Since $z'(t)$ has only limited zero points on $[0, T]$, we have

(2.36)\begin{equation} (\phi_{p}(z'(t)))'+a(t)\phi_{p}(z^{+}(t))-b(t)\phi_{p}(z^{-}(t))+\eta\phi_{p}(z(t))=0, \end{equation}

which together with $\eta \in (0,\infty )\setminus \bigcup _{k=0}^{\infty }[\lambda _{2k}^{min},\lambda _{2k}^{max}]$ yield $z(t)\equiv 0$ for $t\in [0,T]$. In fact, if there exists a $z\in C_{T}^{1}$ such that (2.36) is satisfied on a subset $\Delta \subset [0,T]$ and $z\equiv 0$ on $[0,T]\setminus \Delta,$ it must be an eigenvector corresponding to a half-eigenvalue, which implies a contradiction. Therefore, $z(t)\equiv 0$. However, it contradicts to $\|z\|_{\infty }=1$. Thus, there exists a constant $\widetilde {D} >0$ which is independent of $\varepsilon _n$ such that $|u_{n}(t)|\leq \widetilde {D}$.

Since $u_{n}(0)=u_{n}(T)$, there exists $t_{n}\in (0,T)$ such that $u_{n}'(t_{n})=0$. By the same way to lemma 2.2, we can also get that there exists a constant $\widetilde {K} >0$ that is independent of $\varepsilon _n$ such that $|u'_{n}(t)|\leq \widetilde {K}$. By Arzelá–Ascoli theorem (without loss of generality, take its subsequence), there exist $u\in C_{T}^{1}$ and $t_{0}\in [0,T]$ such that $\mathop {\lim }\limits _{n \to \infty } t_n = t_{0}$,

(2.37)\begin{equation} \mathop {\lim }\limits_{n \to \infty } u_n (t) = u(t),\quad \mathop {\lim }\limits_{n \to \infty } u_n' (t) = u'(t),\ \forall t\in [0,T].\end{equation}

Taking limit $n\to \infty$ in the following equation

\[ \phi_{p}(u_{n}')e^{\varepsilon_{n} \hbar t}=\int_{t_{n}}^{t}\left({-}a_{+}(s)\phi_{p}(u_{n}^{+})+ a_{-}\phi_{p}(u_{n}^{-}) -W_{0}(s,u(s))+e(s)\right)e^{\varepsilon_{n} \hbar s}\,{\rm d}s, \]

we have

\[ \phi_{p}(u')=\int_{t_{0}}^{t}\left({-}a_{+}(s)\phi_{p}(u^{+})+ a_{-}\phi_{p}(u^{-}) -W_{0}(s,u(s))+e(s)\right)\,{\rm d}s, \]

which implies that for all $t\in [0,T]$

\[ (\phi_{p}(u'(t)))'+a_{+}(t)\phi_{p}(u^{+}(t))-a_{-}(t)\phi_{p}(u^{-}(t))+W_{0}(t,u(t))=e(t) \]

and $u(0) = u(T),u'(0) = u'(T).$ Thus, the problem (1.1) has at least one periodic solution.

Proof. Based on lemma 2.4, by similar way to theorem 2.6, the conclusion of theorem 2.8 is also true. Here, we omit the detail.