In order to study the existence of \(2kT\)-periodic solutions to system (2.1), we firstly study some properties of all possible \(2kT\)-periodic solutions to the following system:
$$ \left \{ \textstyle\begin{array}{l} u'(t)=\lambda\phi(v(t))=\lambda\frac{\varphi _{q}(v(t))}{\sqrt{1-|\varphi_{q}(v(t))|^{2}}}, \\ v'(t)=-\lambda F(t,\varphi(v(t)))-\lambda G(t,u(t-\tau(t)))+\lambda e_{k}(t),\quad \lambda\in(0,1], \end{array}\displaystyle \right . $$
(3.1)
where \((u_{k}, v_{k})^{\top}\in Z_{k}\subset X_{k}\). For each \(k\in\mathbb{N}\) and all \(\lambda\in(0,1]\). Let
$$\Delta=\bigl\{ \omega=(u,v)^{\top}\in X_{k}: L\omega=\lambda N \omega,\lambda \in(0,1]\bigr\} . $$
This means that Δ represents the set of all the possible \(2kT\)-periodic solutions to (3.1).
Theorem 3.1
Assume that assumptions (H1)-(H3) hold, \(\frac{\alpha}{1+\sigma_{0}}> \frac{m_{1}\beta\sqrt{1-\sigma_{1}}+\sqrt {2}\beta^{2}\|\tau\|_{0}}{m_{0}(1-\sigma_{1})}\), and
$$ \biggl[ \frac{ \beta\|\tau\|_{0}d_{0}d_{1}}{ T\sqrt {2(1-\sigma_{1})}}+ \frac{m_{1}d_{0}d_{1}+Ad_{0}}{2T} \biggr]^{\frac{1}{q}} + \frac{\sqrt{T}d_{0}\beta+\sqrt{T(1-\sigma_{1})}(m_{1} d_{1}+A) }{\sqrt {2(1-\sigma_{1})}}< 1, $$
where
$$d_{0}:=\frac{A (1-\sigma_{1}) (1+\sigma_{0})(m_{0}+m_{1})+\sqrt{2}A\beta \|\tau\|_{0}(1+\sigma_{0})\sqrt{1-\sigma_{1}}}{ \alpha m_{0} (1-\sigma_{1}) - m_{1}\beta(1+\sigma_{0})\sqrt{1-\sigma _{1}}-\sqrt{2}\beta^{2} \|\tau\|_{0}(1+\sigma_{0})} $$
and
$$d_{1}:=\frac{\beta d_{0}}{m_{0}\sqrt{1-\sigma_{1}}}+\frac{A}{m_{0}}. $$
Then, for each
\(k\in\mathbb{N}\), if
\((u, v)^{\top}\in\Delta\), there are positive constants
\(\rho_{1}\), \(\rho_{2}\), \(\rho_{3}\), \(\rho_{4}\), \(A_{1}\), \(A_{2}\), \(A_{3}\), and
\(A_{4}\), which are independent of
k
and
λ, such that
$$\begin{aligned}& \|u\|_{0}\leq\rho_{1},\qquad \|v\|_{0}\leq\rho _{2}< 1,\qquad \bigl\| u'\bigr\| _{0}\leq \rho_{3},\qquad \bigl\| v'\bigr\| _{0}\leq \rho_{4}, \\& \|u\|_{2}\le A_{1},\qquad \bigl\| u'\bigr\| _{2} \le A_{2},\qquad \|v\|_{p}\le A_{3}, \qquad \bigl\| v'\bigr\| _{2}\le A_{4}. \end{aligned}$$
Proof
For each \(k \in\mathbb{N}\), if \((u, v)^{\top}\in\Delta\), then \((u(t),v(t))^{\top}\) satisfies (3.1). Multiplying the second equation of (3.1) by \(u(t)\) and integrating from \(-kT\) to kT, we have
$$\begin{aligned}& \int_{-kT}^{kT} \bigl\langle u'(t),v(t) \bigr\rangle \,dt \\& \quad = -\int_{-kT}^{kT}\bigl\langle u(t),v'(t) \bigr\rangle \,dt \\& \quad = \lambda \int_{-kT}^{kT} \biggl\langle u(t),F \biggl(t,\frac {u'(t)}{\lambda}\biggr)\biggr\rangle \,dt \\& \qquad {}+\lambda \int _{-kT}^{kT} \bigl\langle u(t),G\bigl(t,u\bigl(t- \tau(t)\bigr)\bigr) \bigr\rangle \,dt -\lambda \int_{-kT}^{kT}\bigl\langle u(t),e_{k}(t)\bigr\rangle \,dt \\& \quad = \lambda \int_{-kT}^{kT} \biggl\langle u(t),F \biggl(t,\frac {u'(t)}{\lambda}\biggr) \biggr\rangle \,dt \\& \qquad {}+\lambda \int _{-kT}^{kT} \bigl\langle u(t)-u\bigl(t-\tau(t) \bigr),G\bigl(t,u\bigl(t-\tau(t)\bigr)\bigr) \bigr\rangle \,dt \\& \qquad {}+\lambda \int_{-kT}^{kT} \bigl\langle u\bigl(t- \tau (t)\bigr),G\bigl(t,u\bigl(t-\tau(t)\bigr)\bigr)\bigr\rangle \,dt-\lambda \int _{-kT}^{kT} \bigl\langle u(t),e_{k}(t) \bigr\rangle \,dt, \end{aligned}$$
which combining with (H1) and (H2) gives
$$\begin{aligned}& \int_{-kT}^{kT} \frac{\vert v(t)\vert ^{q}}{\sqrt{1-\vert \varphi _{q}(v(t))\vert ^{2}}}\,dt+ \alpha \int _{-kT}^{kT}\bigl\vert u\bigl(t-\tau (t)\bigr)\bigr\vert ^{2}\,dt \\& \quad \leq \frac{m_{1}}{\lambda} \int_{-kT}^{kT}\bigl\vert u(t)\bigr\vert \bigl\vert u'(t)\bigr\vert \,dt+\beta \int_{-kT}^{kT}\bigl\vert u(t)-u\bigl(t-\tau(t) \bigr)\bigr\vert \bigl\vert u\bigl(t-\tau(t)\bigr)\bigr\vert \,dt \\& \qquad {}+ \int_{-kT}^{kT}\bigl\vert u(t)\bigr\vert \bigl\vert e_{k}(t)\bigr\vert \,dt. \end{aligned}$$
(3.2)
Furthermore,
$$ \int_{-kT}^{kT}\bigl\vert u\bigl(t-\tau(t)\bigr) \bigr\vert ^{2}\,dt= \int_{-kT-\tau(-kT)}^{kT-\tau(kT)} \frac{1}{1-\tau'(\mu(s))}\bigl\vert u(s)\bigr\vert ^{2}\, ds. $$
It follows from Lemma 2.4 that
$$ \int_{-kT-\tau(-kT)}^{kT-\tau(kT)}\frac{1}{1-\tau '(\mu(s))}\bigl\vert u(s) \bigr\vert ^{2}\, ds= \int_{-kT }^{kT } \frac{1}{1-\tau'(\mu(s))}\bigl\vert u(s)\bigr\vert ^{2}\, ds. $$
By Remark 2.1, we have
$$ \frac{1}{1+\sigma_{0}}\| u\|^{2}_{2}\leq \int _{-kT }^{kT }\frac{1}{1-\tau'(\mu(s))}\bigl\vert u(s)\bigr\vert ^{2}\, ds \leq\frac{1}{1-\sigma_{1}}\| u\|^{2}_{2}. $$
(3.3)
Substituting (3.3) into (3.2) and combining with \(\frac {|v(t)|^{q}}{\sqrt{1-|\varphi_{q}(v(t))|^{2}}}>|v(t)|^{q}\), we get
$$\begin{aligned}& \int_{-kT}^{kT} \bigl\vert v(t)\bigr\vert ^{q}\, dt+\frac{\alpha}{1+\sigma_{0}}\int_{-kT}^{kT} \bigl\vert u(t)\bigr\vert ^{2}\,dt \\& \quad \leq \frac{m_{1}}{\lambda} \biggl(\int_{-kT}^{kT} \bigl\vert u(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \biggl( \int_{-kT}^{kT} \bigl\vert u'(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} +\beta \biggl(\int _{-kT}^{kT} \bigl\vert u\bigl(t-\tau(t)\bigr)\bigr\vert ^{2}\,dt \biggr)^{\frac {1}{2}} \\& \qquad {}\times \biggl(\int_{-kT}^{kT} \bigl\vert u(t)-u\bigl(t-\tau(t)\bigr)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} + \biggl(\int_{-kT}^{kT} \bigl\vert e_{k}(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \biggl( \int_{-kT}^{kT} \bigl\vert u(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}. \end{aligned}$$
By applying Lemma 2.2 and (3.3), we see that
$$\begin{aligned}& \int_{-kT}^{kT} \bigl\vert v(t)\bigr\vert ^{q}\, dt+\frac{\alpha}{1+\sigma _{0}} \int_{-kT}^{kT} \bigl\vert u(t)\bigr\vert ^{2}\,dt \\& \quad \leq \frac{m_{1}}{\lambda} \biggl( \int_{-kT}^{kT} \bigl\vert u(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \biggl( \int_{-kT}^{kT} \bigl\vert u'(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} +\frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt{1-\sigma _{1}}} \biggl( \int _{-kT}^{kT} \bigl\vert u(t)\bigr\vert ^{2}\,dt \biggr)^{\frac {1}{2}} \\& \qquad {}\times \biggl( \int_{-kT}^{kT} \bigl\vert u'(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} + \biggl( \int_{-kT}^{kT} \bigl\vert e_{k}(t)\bigr\vert ^{2}\,dt \biggr)^{\frac {1}{2}} \biggl( \int_{-kT}^{kT} \bigl\vert u(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}, \end{aligned}$$
i.e.,
$$ \| v\|^{q}_{q}+\frac{\alpha}{1+\sigma_{0}}\| u\|^{2}_{2} \leq\frac{m_{1}}{\lambda}\| u\|_{2}\bigl\Vert u'\bigr\Vert _{2} +\frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt{1-\sigma _{1}}}\| u\|_{2}\bigl\Vert u'\bigr\Vert _{2} +A\| u\|_{2}. $$
This implies that
$$ \| v\|^{q}_{q} \leq\frac{m_{1}}{\lambda}\| u\|_{2} \bigl\Vert u'\bigr\Vert _{2} +\frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt{1-\sigma _{1}}}\| u \|_{2}\bigl\Vert u'\bigr\Vert _{2} +A\| u \|_{2} $$
(3.4)
and
$$ \frac{\alpha}{1+\sigma_{0}}\| u\|^{2}_{2} \leq\frac{m_{1}}{\lambda}\| u \|_{2}\bigl\Vert u'\bigr\Vert _{2} + \frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt{1-\sigma _{1}}}\| u\|_{2}\bigl\Vert u'\bigr\Vert _{2} +A\| u\|_{2}. $$
(3.5)
Multiplying the second equation of (3.1) by \(u'(t)\) and integrating from \(-kT\) to kT, we have
$$\begin{aligned} 0 =& \lambda\int_{-kT}^{kT}\biggl\langle \frac{\varphi _{q}(v(t))}{\sqrt{1-|\varphi_{q}(v(t))|^{2}}},v'(t) \biggr\rangle \,dt= \int _{-kT}^{kT}\bigl\langle u'(t),v'(t) \bigr\rangle \,dt \\ =&-\lambda \int_{-kT}^{kT} \biggl\langle u'(t),F\biggl(t,\frac {u'(t)}{\lambda}\biggr) \biggr\rangle \,dt-\lambda \int_{-kT}^{kT} \bigl\langle u'(t),G \bigl(t,u\bigl(t-\tau(t)\bigr)\bigr) \bigr\rangle \,dt \\ &{}+\lambda \int_{-kT}^{kT} \bigl\langle u'(t),e_{k}(t)\bigr\rangle \,dt. \end{aligned}$$
Combining with (H1), (H2), and (3.3), we get
$$\begin{aligned}& m_{0} \int_{-kT}^{kT} \bigl\vert u'(t)\bigr\vert ^{2}\,dt \\& \quad \leq\biggl\vert \lambda^{2} \int_{-kT}^{kT} \biggl\langle \frac {u'(t)}{\lambda},F\biggl(t,\frac{u'(t)}{\lambda}\biggr) \biggr\rangle \,dt\biggr\vert \\& \quad \leq\lambda\beta \int_{-kT}^{kT} \bigl\vert u'(t)\bigr\vert \bigl\vert u\bigl(t-\tau (t)\bigr)\bigr\vert \,dt+\lambda \int_{-kT}^{kT}\bigl\vert u'(t)\bigr\vert \bigl\vert e_{k}(t)\bigr\vert \,dt \\& \quad \leq\frac{\lambda\beta}{\sqrt{1-\sigma_{1}}} \bigl\Vert u'\bigr\Vert _{2}\| u\|_{2}+\lambda A\bigl\Vert u'\bigr\Vert _{2}, \end{aligned}$$
which results in
$$ \bigl\Vert u'\bigr\Vert _{2}\leq\frac{\lambda\beta}{m_{0}\sqrt{1-\sigma _{1}}}\| u \|_{2}+\frac{ \lambda A}{m_{0}}\leq\frac{ \beta }{m_{0}\sqrt{1-\sigma_{1}}}\| u \|_{2}+ \frac{ A}{m_{0}}. $$
(3.6)
Substituting (3.6) into (3.5), we obtain
$$\begin{aligned} \frac{\alpha}{1+\sigma_{0}}\| u\|^{2}_{2} \leq&\frac{m_{1}}{\lambda}\| u \|_{2} \biggl(\frac{\lambda \beta}{m_{0}\sqrt{1-\sigma_{1}}}\| u \|_{2}+\frac{\lambda A}{m_{0}} \biggr) \\ &{} +\frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt{1-\sigma _{1}}}\| u\|_{2} \biggl(\frac{\beta}{m_{0}\sqrt{1-\sigma_{1}}}\| u \| _{2}+\frac{A}{m_{0}} \biggr)+A\| u\|_{2}. \end{aligned}$$
It follows from \(\frac{\alpha}{1+\sigma_{0}}> \frac{m_{1}\beta\sqrt {1-\sigma_{1}}+\sqrt{2}\beta^{2}\|\tau\| _{0}}{m_{0}(1-\sigma_{1})}\) that
$$\begin{aligned} \| u\|_{2}&\leq\frac{A (1-\sigma_{1}) (1+\sigma _{0})(m_{0}+m_{1})+\sqrt{2}A\beta \|\tau\|_{0}(1+\sigma _{0})\sqrt{1-\sigma_{1}}}{ \alpha m_{0} (1-\sigma_{1}) - m_{1}\beta(1+\sigma_{0})\sqrt{1-\sigma _{1}}-\sqrt{2}\beta^{2} \|\tau\|_{0}(1+\sigma_{0})} \\ &:=d_{0}. \end{aligned}$$
(3.7)
Substituting (3.7) into (3.6), we get
$$ \bigl\Vert u'\bigr\Vert _{2}\leq\frac{ \lambda d_{0}\beta}{m_{0}\sqrt {1-\sigma_{1}}}+ \frac{ \lambda A}{m_{0}}, $$
(3.8)
i.e.,
$$ \bigl\Vert u'\bigr\Vert _{2}\leq\frac{ d_{0}\beta}{m_{0}\sqrt{1-\sigma _{1}}}+ \frac{ A}{m_{0}}:=d_{1}. $$
(3.9)
Substituting (3.7), (3.8), and (3.9) into (3.4), we have
$$\begin{aligned} \| v\|^{q}_{q} &\leq\frac{m_{1}}{\lambda}\| u\|_{2} \bigl\Vert u'\bigr\Vert _{2} +\frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt{1-\sigma _{1}}}\| u \|_{2}\bigl\Vert u'\bigr\Vert _{2} +A\| u \|_{2} \\ &\leq\frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt {1-\sigma_{1}}}d_{0}d_{1}+ m_{1} d_{0}d_{1}+Ad_{0}. \end{aligned}$$
(3.10)
Moreover, it follows from Lemma 2.3 that
$$\begin{aligned} \bigl\vert u(t)\bigr\vert &\leq(2T)^{-\frac{1}{2}}\biggl( \int _{t-kT}^{t+kT}\bigl\vert u(s)\bigr\vert ^{2}\, ds\biggr)^{\frac{1}{2}}+T(2T)^{-\frac {1}{2}}\biggl( \int _{t-kT}^{t+kT}\bigl\vert u'(s)\bigr\vert ^{2}\, ds\biggr)^{\frac{1}{2}} \\ &=(2T)^{-\frac{1}{2}}\biggl( \int_{-kT}^{kT}\bigl\vert u(s)\bigr\vert ^{2} \, ds\biggr)^{\frac {1}{2}}+T(2T)^{-\frac{1}{2}}\biggl( \int_{-kT}^{kT}\bigl\vert u'(s)\bigr\vert ^{2}\, ds\biggr)^{\frac{1}{2}}, \end{aligned}$$
which combining with (3.7) and (3.9) yields
$$\bigl\vert u(t)\bigr\vert \leq(2T)^{-\frac{1}{2}}d_{0} +T(2T)^{-\frac{1}{2}}d_{1}:=\rho_{1}, \quad \mbox{for all } t \in R, $$
and then
$$ \|u\|_{0}=\max_{t\in[-kT,kT]}\bigl\vert u(t)\bigr\vert \leq\rho_{1}. $$
(3.11)
Clearly, \(\rho_{1}\) is independent of k and λ.
Multiplying the second equation of (3.1) by \(v'(t)\) and integrating from \(-kT\) to kT, in view of (H1) and (H2), we have
$$\begin{aligned} \int_{-kT}^{kT} \bigl\vert v'(t) \bigr\vert ^{2} \,dt =&-\lambda \int_{-kT}^{kT} \biggl\langle v'(t),F\biggl(t,\frac{u'(t)}{\lambda}\biggr)\biggr\rangle \,dt-\lambda \int_{-kT}^{kT} \bigl\langle v'(t),G\bigl(t,u\bigl(t-\tau (t)\bigr)\bigr)\bigr\rangle \,dt \\ &{}+\lambda \int_{-kT}^{kT} \bigl\langle v'(t),e_{k}(t)\bigr\rangle \,dt \\ \leq&\frac{m_{1}}{\lambda} \int_{-kT}^{kT}\bigl\vert v'(t)\bigr\vert \bigl\vert u'(t)\bigr\vert \,dt+\beta \int_{-kT}^{kT}\bigl\vert v'(t)\bigr\vert \bigl\vert u\bigl(t-\tau (t)\bigr)\bigr\vert \,dt \\ &{}+ \int_{-kT}^{kT}\bigl\vert v'(t)\bigr\vert \bigl\vert e_{k}(t)\bigr\vert \,dt. \end{aligned}$$
By applying the Hölder inequality and (3.3), we have
$$\bigl\Vert v'\bigr\Vert _{2}\leq \frac{m_{1}}{\lambda} \bigl\Vert u'\bigr\Vert _{2}+\frac{\beta}{\sqrt{1-\sigma_{1}}} \| u\| _{2}+A . $$
By (3.7), (3.8), and (3.9), we have
$$\begin{aligned} \bigl\Vert v'\bigr\Vert _{2} &\leq\frac{m_{1}}{\lambda} \bigl\Vert u'\bigr\Vert _{2}+\frac{\beta}{\sqrt{1-\sigma_{1}}} \| u\| _{2}+A \\ & \leq\frac{m_{1}}{\lambda} \biggl(\frac{ \lambda d_{0}\beta}{m_{0}\sqrt {1-\sigma_{1}}}+\frac{ \lambda A}{m_{0}}\biggr)+ \frac{\beta}{\sqrt{1-\sigma _{1}}} \| u\|_{2}+A \\ & = m_{1} \biggl(\frac{ d_{0}\beta}{m_{0}\sqrt{1-\sigma_{1}}}+\frac{ A}{m_{0}}\biggr)+ \frac{\beta}{\sqrt{1-\sigma_{1}}} \| u\|_{2}+A \\ & \leq\frac{d_{0}\beta}{\sqrt{1-\sigma_{1}}}+ m_{1} d_{1}+A. \end{aligned}$$
(3.12)
By applying Lemma 2.3 again and combining with (3.10) and (3.12), we get
$$\begin{aligned} \bigl\vert v(t)\bigr\vert &\leq(2T)^{-\frac{1}{q}} \biggl( \int _{t-kT}^{t+kT}\bigl\vert v(s)\bigr\vert ^{q}\, ds \biggr)^{\frac{1}{q}}+T(2T)^{-\frac {1}{2}} \biggl( \int_{t-kT}^{t+kT}\bigl\vert v'(s)\bigr\vert ^{2}\, ds \biggr)^{\frac{1}{2}} \\ &=(2T)^{-\frac{1}{q}} \biggl( \int_{-kT}^{kT}\bigl\vert v(s)\bigr\vert ^{q} \, ds \biggr)^{\frac{1}{q}}+T(2T)^{-\frac{1}{2}} \biggl( \int_{-kT}^{kT} \bigl\vert v'(s)\bigr\vert ^{2}\, ds \biggr)^{\frac{1}{2}} \\ &\leq(2T)^{-\frac{1}{q}} \biggl( \frac{\sqrt{2}\beta\|\tau \|_{0}}{\sqrt{1-\sigma_{1}}}d_{0}d_{1}+ m_{1} d_{0}d_{1}+Ad_{0} \biggr)^{\frac {1}{q}}+T(2T)^{-\frac{1}{2}} \biggl(\frac{d_{0}\beta}{\sqrt{1-\sigma _{1}}}+ m_{1} d_{1}+A \biggr) \\ &= \biggl[ \frac{ \beta\|\tau\|_{0}d_{0}d_{1}}{ T\sqrt {2(1-\sigma_{1})}}+ \frac{m_{1}d_{0}d_{1}+Ad_{0}}{2T} \biggr]^{\frac{1}{q}} + \frac{\sqrt{T}d_{0}\beta+\sqrt{T}(m_{1} d_{1}+A)\sqrt{1-\sigma _{1}}}{\sqrt{2(1-\sigma_{1})}} \\ &:=\rho_{2}. \end{aligned}$$
Since
$$ \biggl[ \frac{ \beta\|\tau\|_{0}d_{0}d_{1}}{ T\sqrt {2(1-\sigma_{1})}}+ \frac{m_{1}d_{0}d_{1}+Ad_{0}}{2T} \biggr]^{\frac{1}{q}} + \frac{\sqrt{T}d_{0}\beta+\sqrt{T}(m_{1} d_{1}+A)\sqrt{1-\sigma _{1}}}{\sqrt{2(1-\sigma_{1})}}< 1, $$
we have
$$ \|v\|_{0}=\max_{t\in[-kT,kT]}\bigl\vert v(t)\bigr\vert \leq\rho_{2}< 1. $$
(3.13)
Clearly, \(\rho_{2}\) is independent of k and λ.
Furthermore, it follows from (3.1) that
$$\begin{aligned} \bigl\Vert u'\bigr\Vert _{0}&=\max _{t\in[-kT,kT]}\bigl\vert u'(t)\bigr\vert =\max _{t\in [-kT,kT]}\lambda\frac{\varphi_{q}(v(t))}{\sqrt{1-(\varphi _{q}(v(t)))^{2}}} \\ &\leq\frac{\rho^{q-1}_{2}}{\sqrt{1-\rho^{2q-2}_{2}}}: =\rho_{3}. \end{aligned}$$
(3.14)
Clearly, \(\rho_{3}\) is independent of k and λ.
Define \(F_{\rho_{3}}=\max_{|x|\leq\rho_{3},t\in[0,T]}|F(t,x)|\) and \(G_{\rho_{1}}=\max_{|y|\leq\rho_{1},t\in[0,T]}|G(t,y)|\), then from the second equation of (3.1), we get
$$ \bigl\Vert v'\bigr\Vert _{0}=\max _{t\in[-kT,kT]}\bigl\vert v'(t)\bigr\vert \leq F_{\rho_{3}}+G_{\rho _{1}}+A:=\rho_{4}. $$
(3.15)
\(\rho_{4}\) is also independent of k and λ. Therefore, from (3.7), (3.9), (3.10), (3.11), (3.12), (3.13), (3.14), and (3.15), we know that all the conclusions of Theorem 3.1 hold. □
Theorem 3.2
Assume that the conditions of Theorem
3.1
are satisfied. Then, for each
\(k\in\mathbb{N}\), system (3.1) has at least one
\(2kT\)-periodic solution
\((u_{k}(t), v_{k}(t))^{\top}\)
in
\(\Delta\subset X_{k}\)
such that
$$\begin{aligned}& \|u_{k}\|_{0}\leq\rho_{1},\qquad \|v_{k}\|_{0}\leq\rho _{2}< 1,\qquad \bigl\Vert u_{k}'\bigr\Vert _{0}\leq \rho_{3},\qquad \bigl\Vert v_{k}'\bigr\Vert _{0}\leq\rho_{4}, \\& \|u_{k}\|_{2}\le A_{1},\qquad \bigl\Vert u_{k}'\bigr\Vert _{2}\le A_{2}, \qquad \|v_{k}\|_{p}\le A_{3},\qquad \bigl\Vert v_{k}'\bigr\Vert _{2}\le A_{4}, \end{aligned}$$
where
\(\rho_{1}\), \(\rho_{2}\), \(\rho_{3}\), \(\rho_{4}\), \(A_{1}\), \(A_{2}\), \(A_{3}\), and
\(A_{4}\)
are constants defined by Theorem
3.1.
Proof
In order to use Lemma 2.1, for each \(k \in\mathbb{N}\), we consider the following system:
$$ \left \{ \textstyle\begin{array}{l} u'(t)=\lambda\varphi(v(t))=\lambda\frac{\varphi _{q}(v(t))}{\sqrt{1-|\varphi_{q}(v(t))|^{2}}}, \\ v'(t)=-\lambda F(t, \varphi(v(t)))-\lambda G(t,u(t-\tau(t)))+\lambda e_{k}(t),\quad \lambda\in(0,1), \end{array}\displaystyle \right . $$
(3.16)
where \(v(t)=\varphi_{p}(\frac{\frac{u'(t)}{\lambda}}{\sqrt {1+|\frac{u'(t)}{\lambda}|^{2}}})\). Let \(\Omega_{1}\subset X_{k}\) represents the set of all possible \(2kT\)-periodic solutions of (3.16). Since \((0,1)\subset(0, 1]\), then \(\Omega_{1}\subset\Delta\), where Δ is defined by Theorem 3.1. If \((u,v)^{\top}\in\Omega_{1}\), by using Theorem 3.1, we have
$$\|u\|_{0}\leq\rho_{1},\qquad \bigl\Vert u' \bigr\Vert _{0}\leq\rho_{3},\qquad \|v\|_{0} \leq\rho _{2}< 1,\qquad \bigl\Vert v'\bigr\Vert _{0}\leq\rho_{4}. $$
Define \(\Omega_{2}=\{\omega=(u,v)^{\top}\in \ker L,QN \omega=0\}\). If \((u,v)^{\top}\in\Omega_{2}\), then \((u,v)^{\top}=(a_{1},a_{2})^{\top}\in \mathbb{R}^{2}\) (constant vector) such that
$$\left \{ \textstyle\begin{array}{l} \int_{-kT}^{kT}\frac{\varphi _{q}(a_{2})}{\sqrt{1-|\varphi_{q}(a_{2})|^{2}}}\,dt=0, \\ \int_{-kT}^{kT}[-F(t,\frac{\varphi_{q}(a_{2})}{\sqrt {1-|\varphi_{q}(a_{2})|^{2}}})-G(t,a_{1})+e_{k}(t)]\,dt=0, \end{array}\displaystyle \right . $$
i.e.,
$$ \left \{ \textstyle\begin{array}{l} a_{2}=0, \\ \int_{-kT}^{kT}[-F(t,0)-G(t,a_{1})+e_{k}(t)]\,dt=0. \end{array}\displaystyle \right . $$
(3.17)
Multiplying the second equation of (3.17) by \(a_{1}\) and combining with (H1) and (H2), we have
$$\begin{aligned} 2kT\alpha|a_{1}|^{2}&\leq \int_{-kT}^{kT} \bigl\vert F(t,0)\bigr\vert |a_{1}|\,dt+ \int_{-kT}^{kT}|a_{1}| \bigl\vert e_{k}(t)\bigr\vert \,dt \\ &\leq2kT |a_{1}|A. \end{aligned}$$
Thus,
$$|a_{1}|\leq\frac{A}{\alpha}:=\varpi. $$
Now, if we define \(\Omega= \{\omega=(u, v)^{\top}\in X_{k},\|u\|_{0}< \rho _{1}+ \varpi, \|v\|_{0} < \frac{1+\rho_{2}}{2}<1\}\), it is easy to see that \(\Omega_{1}\cup\Omega_{2}\subset\Omega\). So, condition (h1) and condition (h2) of Lemma 2.1 are satisfied. In order to verify the condition (h3) of Lemma 2.1, define
$$H(\omega,\mu):\bigl(\Omega\cap\mathbb{R}^{2}\bigr)\times[0,1] \rightarrow \mathbb{R}\mbox{:} \quad H(\omega,\mu)=\mu\omega+(1-\mu) JQN (\omega), $$
where \(J : \operatorname{Im}Q\rightarrow\ker L \) is a linear isomorphism, \(J (u, v)=(v, u)^{\top}\). From assumption (H1), we have
$$\omega^{\top}H(\omega,\mu)\neq0,\quad \forall(\omega,\mu)\in\partial \Omega\cap\mathbb{R}^{2}\times[0,1]. $$
Hence,
$$\begin{aligned} \deg\bigl\{ JQN ,\Omega\cap\mathbb{R}^{2},0\bigr\} &=\deg\bigl\{ H( \omega ,0),\Omega\cap \mathbb{R}^{2} ,{ 0}\bigr\} \\ &=\deg\bigl\{ H(\omega,1),\Omega\cap\mathbb{R}^{2},0\bigr\} \\ &\neq0. \end{aligned}$$
Thus, the condition (h3) of Lemma 2.1 is also satisfied. Therefore, by using Lemma 2.1, we can see that (2.1) has a \(2kT\)-periodic solution \((u_{k}, v_{k})^{\top}\in\overline{\Omega}\). Clearly, \(u_{k}\) is a \(2kT\)-periodic solution to (1.5), and \((u_{k}, v_{k})^{\top}\) must be in Δ for the case of \(\lambda= 1\). Thus, by using Theorem 3.1, we have
$$\begin{aligned}& \|u_{k}\|_{0}\leq\rho_{1},\qquad \|v_{k}\|_{0}\leq\rho _{2}< 1,\qquad \bigl\Vert u_{k}'\bigr\Vert _{0}\leq \rho_{3},\qquad \bigl\Vert v_{k}'\bigr\Vert _{0}\leq\rho_{4}, \\& \|u_{k}\|_{2}\le A_{1},\qquad \bigl\Vert u_{k}'\bigr\Vert _{2}\le A_{2}, \qquad \|v_{k}\|_{p}\le A_{3},\qquad \bigl\Vert v_{k}'\bigr\Vert _{2}\le A_{4}. \end{aligned}$$
Hence, all the conclusions of Theorem 3.2 hold. □
Theorem 3.3
Suppose that the conditions in Theorem
3.1
hold, then (1.4) has a nontrivial homoclinic solution.
Proof
From Theorem 3.2, we see that for each \(k\in\mathbb{N}\), there exists a \(2kT\)-periodic solution \((u_{k},v_{k})^{\top}\) to (2.1) with \((u_{k},v_{k})^{\top}\in X_{k}\) and
$$ \|u_{k}\|_{0}\leq\rho_{1},\qquad \|v_{k}\|_{0}\leq\rho_{2}< 1,\qquad \bigl\Vert u'_{k}\bigr\Vert _{0}\leq\rho _{3},\qquad \bigl\Vert v'_{k}\bigr\Vert _{0}\leq\rho_{4}, $$
(3.18)
where \(\rho_{1}\), \(\rho_{2}\), \(\rho_{3}\), \(\rho_{4}\) are constants independent of \(k \in\mathbb{N}\). Equation (3.18) together with Lemma 2.5 shows that there are a function \(w_{0}:=(u_{0},u_{0})^{\top}\in C(\mathbb {R},\mathbb{R}^{2n})\) and a subsequence \(\{(u_{k_{j}},v_{k_{j}})^{\top}\}\) of \(\{(u_{k},v_{k})^{\top}\}_{k\in\mathbb{N}}\) such that for each interval \([a,b]\subset\mathbb{R}\), \(u_{k_{j}}(t)\rightarrow u_{0}(t)\), and \(v_{k_{j}}(t)\rightarrow v_{0}(t)\) uniformly on \([a,b]\). Below, we will show that \((u_{0}(t),v_{0}(t))^{\top}\) is just a homoclinic solution to (1.4).
Since \((u_{k}(t),v_{k}(t))^{\top}\) is a \(2kT\)-periodic solution of (2.1), it follows that
$$ \left \{ \textstyle\begin{array}{l} u_{k}'(t)= \phi(v_{k}(t))= \frac{\varphi _{q}(v_{k}(t))}{\sqrt{1-|\varphi_{q}(v_{k}(t))|^{2}}}, \\ v_{k}'(t)=-F(t,\varphi(v_{k}(t)))-G(t,u_{k}(t-\tau(t)))+e_{k}(t). \end{array}\displaystyle \right . $$
(3.19)
For all \(a,b\in R\) with \(a < b \), there must be a positive integer \(j_{0}\) such that for \(j>j_{0}\), \([-k_{j}T,k_{j}T-\varepsilon_{0}] \supset[a-\|\tau\|_{0}, b+\|\tau\|_{0}]\). So for \(j>j_{0}\), from (1.6) and (3.19) we see that
$$\left \{ \textstyle\begin{array}{l} u_{k_{j}}'(t)=\varphi(y_{k_{j}}(t))=\frac{\varphi_{q}(v_{k_{j}}(t))}{\sqrt {1-|\varphi_{q}(v_{k_{j}}(t))|^{2}}}, \\ v_{k_{j}}'(t)=-F(t,\varphi(y_{k_{j}}(t)))-G(t,u_{k_{j}}(t-\tau(t)))+e(t),\quad t\in(a,b), \end{array}\displaystyle \right . $$
which results in
$$ u_{k_{j}}'(t)=\frac{\varphi_{q}(v_{k_{j}}(t))}{\sqrt{1-|\varphi _{q}(v_{k_{j}}(t))|^{2}}}\rightarrow \frac{\varphi_{q}(v_{0}(t))}{\sqrt {1-|\varphi_{q}(v_{0}(t))|^{2}}} $$
(3.20)
and
$$\begin{aligned} v_{k_{j}}'(t)&=-F\bigl(t,\varphi \bigl(v_{k_{j}}(t) \bigr)\bigr)-G\bigl(t,u_{k_{j}}\bigl(t-\tau(t)\bigr)\bigr)+e(t) \\ &\rightarrow -F\bigl(t,\varphi\bigl(v_{0}(t)\bigr)\bigr)-G \bigl(t,u_{0}\bigl(t-\tau(t)\bigr)\bigr)+e(t) \end{aligned}$$
(3.21)
uniformly for \(t\in[a,b]\) as \(j\rightarrow+\infty\). Since \(u_{k_{j}}(t)\rightarrow u_{0}(t)\) and \(u_{k_{j}}(t)\) is continuously differentiable for \(t\in(a,b)\), it follows that
$$u_{k_{j}}'(t)\rightarrow u_{0}'(t) \quad \mbox{uniformly for } t\in[a,b] \mbox{ as } j\rightarrow+\infty, $$
which together with (3.20) yields
$$u_{0}'(t)=\frac{\varphi_{q}(v_{0}(t))}{\sqrt{1-|\varphi _{q}(v_{0}(t))|^{2}}},\quad t\in(a,b). $$
Similarly, by (3.21) we have
$$v_{0}'(t)=-F\bigl(t,\varphi\bigl(v_{0}(t) \bigr)\bigr)-G\bigl(t,u_{0}\bigl(t-\tau(t)\bigr)\bigr)+e(t),\quad t \in(a,b). $$
Considering a, b to be two arbitrary constants with \(a< b\), it is easy to see that \((u_{0}(t),v_{0}(t))^{\top}\), \(t\in R\), is a solution to the following equation:
$$\left \{ \textstyle\begin{array}{l} u'(t)= \phi(v(t))= \frac{\varphi_{q}(v(t))}{\sqrt {1-|\varphi_{q}(v(t))|^{2}}}, \\ v'(t)=-F(t,\varphi(v(t)))-G(t,u(t-\tau(t)))+e(t), \end{array}\displaystyle \right . $$
i.e.,
$$ \left \{ \textstyle\begin{array}{l} u_{0}'(t)= \phi(v_{0}(t))= \frac{\varphi _{q}(v_{0}(t))}{\sqrt{1-|\varphi_{q}(v_{0}(t))|^{2}}}, \\ v_{0}'(t)=-F(t,\varphi(v_{0}(t)))-G(t,u_{0}(t-\tau(t)))+e(t). \end{array}\displaystyle \right . $$
(3.22)
Now, we will prove \(u_{0}(t)\rightarrow0\) and \(u_{0}'(t)\rightarrow0\) as \(|t|\rightarrow+\infty\).
Since
$$\begin{aligned} \begin{aligned} \int^{+\infty}_{-\infty}\bigl(\bigl\vert u_{0}(t)\bigr\vert ^{2}+\bigl\vert u'_{0}(t) \bigr\vert ^{2}\bigr)\,dt&=\lim_{i\rightarrow+\infty}\int ^{iT}_{-iT}\bigl(\bigl\vert u_{0}(t) \bigr\vert ^{2}+\bigl\vert u'_{0}(t)\bigr\vert ^{2}\bigr)\,dt \\ &=\lim_{i\rightarrow+\infty}\lim_{j\rightarrow+\infty}\int ^{iT}_{-iT}\bigl(\bigl\vert u_{0}(t) \bigr\vert ^{2}+\bigl\vert u'_{0}(t)\bigr\vert ^{2}\bigr)\,dt. \end{aligned} \end{aligned}$$
By using the conclusion of Theorem 3.2, we have
$$\int^{iT}_{-iT}\bigl(\bigl\vert u_{k_{j}}(t)\bigr\vert ^{2}+\bigl\vert u'_{k_{j}}(t) \bigr\vert ^{2}\bigr)\,dt\leq\int^{k_{j}T}_{-k_{j}T} \bigl(\bigl\vert u_{k_{j}}(t)\bigr\vert ^{2}+\bigl\vert u'_{k_{j}}(t)\bigr\vert ^{2}\bigr)\,dt \leq A_{0}^{2} + A_{1}^{2}. $$
Let \(i\rightarrow+\infty\) and \(j \rightarrow+\infty\), we have
$$\int^{+\infty}_{-\infty}\bigl(\bigl\vert u_{0}(t)\bigr\vert ^{2}+\bigl\vert u'_{0}(t) \bigr\vert ^{2}\bigr)\,dt\leq A_{0}^{2}+A_{1}^{2}, $$
and then
$$\int_{\vert t\vert \geq r}\bigl(\bigl\vert u_{0}(t)\bigr\vert ^{2}+\bigl\vert u'_{0}(t)\bigr\vert ^{2}\bigr)\,dt \rightarrow 0 $$
as \(r\rightarrow+\infty\). So by using Lemma 2.3, we obtain
$$\begin{aligned} \bigl\vert u_{0}(t)\bigr\vert &\leq(2T)^{-\frac{1}{2}} \biggl( \int^{t+T}_{t-T}\bigl\vert u_{0}(s)\bigr\vert ^{l_{1}+1}\,ds \biggr)^{\frac{1}{2}}+T(2T)^{-\frac{1}{2}} \biggl( \int^{t+T}_{t-T}\bigl\vert u_{0}'(s) \bigr\vert ^{2}\,ds \biggr)^{\frac{1}{2}} \\ &\leq \bigl[(2T)^{-\frac{1}{2}}+T(2T)^{-\frac{1}{2}} \bigr] \biggl[ \biggl(\int ^{t+T}_{t-T}\bigl\vert x(s)\bigr\vert ^{2}\,ds \biggr)^{1/(2)}+ \biggl(\int^{t+T}_{t-T} \bigl\vert u_{0}'(s)\bigr\vert ^{2}\,ds \biggr)^{\frac{1}{2}} \biggr] \\ &\rightarrow0 \quad \mbox{as } \vert t\vert \rightarrow+\infty, \end{aligned}$$
which implies that
$$ u_{0}(t)\rightarrow0 \quad \mbox{as } |t|\rightarrow+ \infty. $$
(3.23)
Similarly, we can prove that
$$v_{0}(t)\rightarrow0 \quad \mbox{as } |t|\rightarrow+\infty, $$
which together with the first equation of (3.22) gives
$$ u_{0}'(t)\rightarrow0 \quad \mbox{as } |t|\rightarrow+ \infty. $$
(3.24)
It is easy to see from (3.22) that \(u_{0}(t)\) is a solution for (1.4). Thus, by (3.23) and (3.24), \(u_{0}(t)\) is just a homoclinic solution to (1.4). Clearly, \(u_{0}(t)\not\equiv0\), otherwise, \(e(t)\equiv0\), which contradicts assumption (H3). Hence, the conclusion of Theorem 3.3 holds. □
Remark 3.1
Obviously, the prescribed mean curvature equations studied in [11, 12, 15, 16] are special cases of (1.4). This implies that the main result in this paper is essentially new.