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Homoclinic solutions for n-dimensional prescribed mean curvature p-Laplacian equations

Boundary Value Problems20152015:105

  • Received: 25 February 2015
  • Accepted: 28 May 2015
  • Published:


In this paper, a n-dimensional prescribed mean curvature Rayleigh p-Laplacian equation with a deviating argument, \((\varphi_{p}(\frac{u'(t)}{\sqrt{1+|u'(t)|^{2}}}))'+F(t,u'(t))+G(t,u(t-\tau(t)))=e(t)\), is studied. By means of Mawhin’s continuation theorem and some analysis methods, a new result on the existence of homoclinic solutions for the equation is obtained. Our research enriches the contents of prescribed mean curvature equations.


  • homoclinic solution
  • Mawhin’s continuation theorem
  • n-dimensional
  • prescribed mean curvature
  • p-Laplacian

1 Introduction

In recent years, the existence of homoclinic solutions has been studied widely for the Hamiltonian systems and the p-Laplacian systems (see [14] and the references therein). For example, in [1], Lzydorek and Janczewska studied the existence of homoclinic solutions for second-order Hamiltonian system in the following form:
$$\ddot{q}+V_{q}(t,q)=f(t), $$
where \(q\in\mathbb{R}^{n} \) and \(V\in C^{1}(\mathbb{R}\times\mathbb {R}^{n},\mathbb{R})\), \(V(t,q)=-K(t,q)+W(t,q) \) is T-periodic with respect to t. Lu in [4] studied the existence of homoclinic solutions for a second-order p-Laplacian differential system with delay
$$ \frac{d}{dt}\bigl[\varphi_{p}\bigl(u'(t)\bigr) \bigr]+\frac{d}{dt} \nabla F\bigl(u(t)\bigr)+\nabla G\bigl(u(t)\bigr)+\nabla H\bigl(u\bigl(t-\gamma(t)\bigr)\bigr)=e(t), $$
where \(p\in(1,+\infty)\), \(\varphi_{p}:\mathbb{R}^{n}\rightarrow\mathbb {R}^{n}\), \(\varphi_{p}(u)=(|u_{1}|^{p-2}u_{1}, |u_{2}|^{p-2}u_{2},\ldots,|u_{n}|^{p-2}u_{n})\) for \(u\neq0=(0,0,\ldots,0)\) and \(\varphi_{p}(0)=(0,0,\ldots,0)\), \(F\in C^{2}(\mathbb{R}^{n},\mathbb{R})\), \(G, H\in C^{1}(\mathbb{R}^{n},\mathbb{R})\), \(e\in C(\mathbb{R}, \mathbb{R}^{n})\), and \(\gamma(t)\) is a continuous T-periodic function with \(\gamma(t)\geq0\); T is a given constant.
In the recent past, the prescribed mean curvature equation
$$\biggl(\frac{u'(t)}{\sqrt{1+(u'(t))^{2}}}\biggr)'=f\bigl(u(t)\bigr), $$
and its modified forms, which arises from some problems associated to differential geometry and combustible gas dynamics, were studied extensively [510]. Also, we note that the existence of periodic solutions for the prescribed curvature mean equations has attracted much attention from researchers. For example, Feng in [11] studied the problem of the existence of periodic solution for a prescribed mean curvature Liénard equation
$$ \biggl(\frac{u'(t)}{\sqrt{1+(u'(t))^{2}}}\biggr)'+f\bigl(u(t)\bigr)u'(t)+g \bigl(t,u\bigl(t-\tau (t)\bigr)\bigr)=e(t), $$
where \(\tau,e \in C(\mathbb{R},\mathbb{R}) \) are T-periodic, and \(g\in C(\mathbb{R}\times\mathbb{R},\mathbb{R})\) is T-periodic in the first argument, \(T>0\) is a constant. Aiming to apply Mawhin’s continuation theorem, Feng made (1.1) equivalent to the following system through the transformation \(v(t)=\frac{u'(t)}{\sqrt{1+(u'(t))^{2}}}\):
$$ \left \{ \textstyle\begin{array}{l} u'(t)=\varphi(v(t))=\frac{v(t)}{\sqrt {1-v^{2}(t)}}, \\ v'(t)=-f(t,\varphi(v(t)))-g(t, u(t-\tau(t)))+e(t). \end{array}\displaystyle \right . $$
Li in [12] further studied the existence of periodic solutions for a prescribed mean curvature Rayleigh equation of the form
$$ \biggl(\frac{u'(t)}{\sqrt{1+(u'(t))^{2}}}\biggr)'+f\bigl(t,u'(t) \bigr)+g\bigl(t,u\bigl(t-\tau(t)\bigr)\bigr)=e(t), $$
and Wang in [13] discussed the following boundary valued problem:
$$ \left \{ \textstyle\begin{array}{l} (\varphi_{p}(\frac{x'(t)}{\sqrt {1+(x'(t))^{2}}}))'+f(t,x'(t))+g(t,x(t-\tau(t)))=e(t), \\ x_{1}(0)=x_{1}(\omega),\qquad x_{2}(0)=x_{2}(\omega), \end{array}\displaystyle \right . $$
where \(p>1\) and \(\varphi_{p}: \mathbb{R}\rightarrow\mathbb{R}\) is given by \(\varphi_{p}(s)=|s|^{p-2}s\) for \(s\neq0\) and \(\varphi _{p}(0)=0\), \(g\in C(\mathbb{R}^{2},\mathbb{R})\), \(e, \tau\in C(\mathbb {R},\mathbb{R})\), \(g(t+\omega,x)=g(t,x)\), \(f(t+\omega,x)=f(t,x)\), \(f(t,0)=0\), \(e(t+\omega)=e(t)\) and \(\tau(t+\omega)=\tau(t)\). By using a similar transformation in [11], (1.2) is converted to the following system:
$$ \left \{ \textstyle\begin{array}{l} x_{1}'(t)=\phi(x_{2}(t))=\frac{\varphi_{q}(x_{2}(t))}{\sqrt {1-\varphi^{2}_{q}(x_{2}(t))}}, \\ x_{2}'(t)=-f(t,\frac{\varphi_{q}(x_{2}(t))}{\sqrt{1-\varphi ^{2}_{q}(x_{2}(t))}})-g(t,x_{1}(t-\tau(t)))+e(t), \\ x_{1}(0)=x_{1}(\omega), \qquad x_{2}(0)=x_{2}(\omega). \end{array}\displaystyle \right . $$
Under the conditions imposed on f and g such as
$$ f(t,x)\geq a|x|^{r},\quad \forall(t,x)\in\mathbb{R}^{2} $$
$$ g(t,x)-e(t)\geq-m_{1}|x|-m_{2},\quad \forall t\in\mathbb{R}, x>d, $$
where \(a, r\geq1\); \(m_{1}\) and \(m_{2}\) are positive constants, the author found that (1.2) has at least one periodic solution. It is easy to see from the first equation of (1.3) that the function \(\varphi _{q}(x_{2}(t))\) must satisfy \(\max_{t\in[0,T]}|\varphi _{q}(x_{2}(t))|<1\). This implies that the open and bounded set Ω associated to Mawhin’s continuation theorem must satisfy \(\overline {\Omega} \in\{(x_{1}; x_{2})^{\top}\in X : \| x_{1}\| _{\infty} < M;\| x_{2}\|_{\infty} < 1\}\). Thus, there must be a constant \(\rho\in(0,1)\) such that \(\Omega\in\{(x_{1}; x_{2})^{\top}\in X : \| x_{1}\|_{\infty} < M;\| x_{2}\|_{\infty} < \rho\}\). But in [13], the author obtained \(\Omega= \{(x_{1}; x_{2})^{\top}\in X : \| x_{1}\|_{\infty} < M_{1};\| x_{2}\|_{\infty } <M_{2}\} \) and there was no proof as regards \(M_{2}<1\). A similar problem also occurred in [14].
Inspired by the above fact, the aim of this paper is to investigate the existence of homoclinic solution to the following n-dimensional prescribed mean curvature equation with a deviating argument:
$$ \biggl(\varphi_{p}\biggl(\frac{u'(t)}{\sqrt{1+|u'(t)|^{2}}}\biggr) \biggr)'+F\bigl(t,u'(t)\bigr)+G\bigl(t,u\bigl(t-\tau(t) \bigr)\bigr)=e(t), $$
where \(p\in(1,+\infty)\), \(\varphi_{p}:\mathbb{R}^{n}\rightarrow\mathbb {R}^{n}\), \(\varphi_{p}(u)=(|u_{1}|^{p-2}u_{1}, |u_{2}|^{p-2}u_{2},\ldots,|u_{n}|^{p-2}u_{n})\) for \(u\neq0=(0,0,\ldots,0)\) and \(\varphi_{p}(0)=(0,0,\ldots,0)\), \(F\in C(\mathbb{R}\times\mathbb {R}^{n};\mathbb{R}^{n})\), \(G\in C (\mathbb{R}\times\mathbb {R}^{n};\mathbb{R}^{n} )\), \(e\in C(\mathbb{R};\mathbb{R}^{n})\), \(\tau(t)\) is a continuous T-periodic function and \(T > 0\) is given constant.
In order to study the homoclinic solution for (1.4), firstly, like in [14, 15] and [16], the existence of a homoclinic solution for (1.4) is obtained as a limit of a certain sequence of \(2kT\)-periodic solutions for the following equation:
$$ \biggl(\varphi_{p}\biggl(\frac{u'(t)}{\sqrt{1+|u'(t)|^{2}}}\biggr) \biggr)'+F\bigl(t,u'(t)\bigr)+G\bigl(t,u\bigl(t-\tau(t) \bigr)\bigr)=e_{k}(t), $$
where \(k\in\mathbb{N}\). \(e_{k} : \mathbb{R}\rightarrow\mathbb{R}\) is a \(2kT\)-periodic function such that
$$ e_{k}(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} e(t),&t\in[-kT,kT-\varepsilon_{0}), \\ e(kT-\varepsilon_{0})+\frac{e(-kT)-e(kT-\varepsilon_{0})}{\varepsilon _{0}}(t-kT+\varepsilon_{0}), &t\in[kT-\varepsilon_{0},kT], \end{array}\displaystyle \right . $$
where \(\varepsilon_{0}\in(0,T)\) is a constant independent of k. Obviously, for each \(k\in\mathbb{N}\), from (1.6) we observe that \(e_{k}\in C(\mathbb{R},\mathbb{R}^{n})\) with \(e_{k}(t+2kT)\equiv e_{k}(t)\). In this paper, the approach for solving the \(2kT\)-periodic solutions to (1.5) is based on Mawhin’s continuation theorem [17], which is different from the corresponding ones in [14] associated to critical point theory.

The rest of this paper organized as follows. In Section 2, we state some necessary definitions and lemmas. In Section 3, we prove the main result.

2 Preliminaries

First of all, we give the definition of the homoclinic solution. A solution \(u(t)\) is named homoclinic (to 0) if \(u(t)\rightarrow0\) and \(u'(t)\rightarrow0\) as \(|t|\rightarrow+\infty\). In addition, if \(u\neq0\), then u is called a nontrivial homoclinic solution.

In the following, we recall some notations and lemmas, which are important for proving our main result.

Throughout this paper, \(\|\cdot\|\) will denote the Euclidean norm on \(\mathbb{R}^{n}\) and \(\langle\cdot,\cdot\rangle: \mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}\) denote the standard inner product.

For each \(k\in\mathbb{N}\), define
$$\begin{aligned}& C_{2kT}=\bigl\{ u|u\in C\bigl(\mathbb{R},\mathbb{R}^{n} \bigr),u(t+2kT)\equiv u(t)\bigr\} , \\& C^{1}_{2kT}=\bigl\{ u|u\in C^{1}\bigl(\mathbb{R}, \mathbb{R}^{n}\bigr),u(t+2kT)\equiv u(t)\bigr\} \end{aligned}$$
$$ C^{2}_{2kT}=\bigl\{ u|u\in C^{2}\bigl(\mathbb{R}, \mathbb{R}^{n}\bigr),u(t+2kT)\equiv u(t)\bigr\} . $$
If the norm of \(C_{2kT}\), \(C^{1}_{2kT}\), and \(C^{2}_{2kT}\) is defined by \(\|\cdot\|_{C_{2kT}}=\|\cdot\|_{0}\), \(\|x\|_{C^{1}_{2kT}}=\max\{\| x\|_{0}, \| x'\|_{0}\}\), and \(\|x\|_{C^{2}_{2kT}}=\max\{\| x\| _{0}, \| x'\|_{0}, \| x''\|_{0}\}\), respectively, then \(C_{2kT}\), \(C^{1}_{2kT}\), and \(C^{2}_{2kT}\) are all Banach spaces.

Moreover, for any \(\psi\in C_{2kT}\), define \(\|\psi \|_{r}=( \int_{-kT}^{kT}|\psi(t)|^{r}\, dt)^{\frac{1}{r}}\), where \(r\in(1,+\infty)\).

In order to use Mawhin’s continuation theorem, we first recall it.

Let X and Y be two Banach spaces, a linear operator \(L: D(L)\subset X \rightarrow Y\) is said to be a Fredholm operator of index zero provided that
  1. (a)

    ImL is a closed subset of Y,

  2. (b)

    \(\dim\ker L=\operatorname{codim} \operatorname{Im}L<\infty\).

Let \(\Omega\subset X\) be an open and bounded set, and let \(L: D(L)\subset X \rightarrow Y\) be a Fredholm operator of index zero. This means that there are continuous linear projectors \(P: X\rightarrow X\) and \(Q: Y\rightarrow Y\) such that \(\operatorname{Im}P =\ker L\), \(\ker Q=\operatorname{Im}L\), \(X=\ker L\oplus\ker P\) and \(Y=\operatorname{Im} L\oplus \operatorname{Im} Q\). Obviously, \(L: {D}(L)\cap\ker P\rightarrow\operatorname{Im} L\) has its right inverse. Let \(K_{P}: \operatorname{Im} L\rightarrow D(L)\cap\ker P\) be the right inverse of \(L: D(L)\cap\ker P\rightarrow\operatorname{Im} L\). A continuous operator \(N:\Omega\subset X\rightarrow Y\) is said to be L-compact in \(\overline{\Omega}\) provided that
  1. (c)

    \(K_{p}(I-Q)N(\overline{\Omega})\) is a relative compact set of X,

  2. (d)

    \(QN(\overline{\Omega})\) is a bounded set of Y.


Lemma 2.1


Let X and Y be two real Banach spaces, Ω be an open and bounded subset of X, \(L: D(L)\subset X \rightarrow Y\) be a Fredholm operator of index zero and the operator \(N:\overline{\Omega}\subset X\rightarrow Y\) be L-compact in \(\overline{\Omega}\). In addition, if the following conditions hold:

\(Lx\neq\lambda Nx\), \(\forall(x,\lambda)\in\partial\Omega \times(0,1)\);


\(QNx\neq0\), \(\forall x\in\ker L\cap\partial\Omega\);


\(\deg\{ JQN,\Omega\cap\ker L,0\}\neq0\), where \(J: \operatorname{ Im}Q\rightarrow\ker L\) is a homeomorphism.

Then \(Lx=Nx\) has at least one solution in \(D(L)\cap\overline{\Omega}\).

Lemma 2.2


Let \(0<\alpha<T\) be a constant, \(\tau\in C(\mathbb{R},\mathbb{R})\) be a T-periodic function and \(\max_{t\in[0,T]}|\tau(t)|=\alpha\), then for all \(u\in C^{1}(\mathbb{R},\mathbb{R})\) with \(u(t+T)\equiv u(t)\), we have
$$\int_{0}^{T}\bigl\vert u(t)-u\bigl(t-\tau(t) \bigr)\bigr\vert ^{2}\, dt\leq2\alpha ^{2} \int _{0}^{T}\bigl\vert u'(t)\bigr\vert ^{2}\, dt. $$

Lemma 2.3


If \(u : \mathbb{R}\rightarrow\mathbb{R}\) is continuously differentiable on \(\mathbb{R}\), \(a>0\), \(\mu>1\), and \(p>1\) are constants, then for every \(t\in\mathbb{R}\), the following inequality holds:
$$ \bigl\vert u(t)\bigr\vert \leq(2a)^{-\frac{1}{\mu}}\biggl( \int _{t-a}^{t+a}\bigl\vert u(s)\bigr\vert ^{\mu}\, ds\biggr)^{\frac{1}{\mu}}+a(2a)^{-\frac {1}{p}}\biggl( \int_{t-a}^{t+a}\bigl\vert u'(s)\bigr\vert ^{p}\, ds\biggr)^{\frac{1}{p}}. $$

Lemma 2.4


Suppose \(\tau\in C^{1}( {\mathbb{R},\mathbb {R}})\) with \(\tau(t+\omega)\equiv\tau(t)\) and \(\tau'(t)<1\), \(\forall t\in[0,\omega]\). Then the function \(t-\tau(t)\) has an inverse \(\mu(t)\) satisfying \(\mu\in C( {\mathbb{R},\mathbb{R}})\) with \(\mu(t+\omega)\equiv\mu(t)+\omega\), \(\forall t\in[0,\omega ]\).

Throughout this paper, besides τ being a periodic function with period T, we suppose in addition that \(\tau\in C^{1}( {\mathbb {R},\mathbb{R}})\) with \(\tau'(t)<1\), \(\forall t\in[0,T]\).

Remark 2.1

From the above assumption, one can find from Lemma 2.4 that the function \((t-\tau(t))\) has an inverse denoted by \(\mu(t)\). Define \(\sigma_{0}=-\min_{t\in[0,T]}\tau'(t)\), \(\sigma _{1}=\max_{t\in[0,T]}\tau'(t)\) and \(\|\tau\| _{0}=\max_{t\in[0,T]}|\tau(t)|\). Clearly, \(\sigma_{0}\geq0\) and \(0\leq\sigma_{1}<1\).

Lemma 2.5


Let \(u_{k}\in C^{2}_{2kT}\) be a \(2kT\)-periodic function for each \(k\in\mathbb{N}\) with
$$|u_{k}|_{0}\leq A_{0}, \qquad \bigl\vert u'_{k}\bigr\vert _{0}\leq A_{1}, \qquad \bigl\vert u''_{k}\bigr\vert _{0}\leq A_{2}, $$
where \(A_{0}\), \(A_{1}\), and \(A_{2}\) are constants independent of \(k\in \mathbb{N}\). Then there exists a function \(u\in C^{1}(\mathbb {R},\mathbb{R}^{n})\) such that for each interval \([c,d]\subset\mathbb {R}\), there is a subsequence \(\{u_{k_{j}}\}\) of \(\{u_{k}\}_{k\in\mathbb {N}}\) with \(u'_{k_{j}}(t)\rightarrow u'_{0}(t)\) uniformly on \([c,d]\).
Equation (1.5) is equivalent to the following system:
$$ \left \{ \textstyle\begin{array}{l} u'(t)= \phi(v(t))= \frac{\varphi_{q}(v(t))}{\sqrt {1-|\varphi_{q}(v(t))|^{2}}}, \\ v'(t)=-F(t,\varphi(v(t)))-G(t,u(t-\tau(t)))+e_{k}(t), \end{array}\displaystyle \right . $$
where \(\varphi_{q}(s)=|s|^{q-2}s\), \(\frac{1}{p}+\frac{1}{q}=1\), \(v(t)= \varphi_{p}(\frac{u'(t)}{\sqrt{1+|u'(t)|^{2}}})=\phi^{-1}(u'(t))\).
$$ X_{k}=Y_{k}=\bigl\{ \omega=\bigl(u(t),v(t) \bigr)^{\top}: u\in C_{2kT}, v\in C_{2kT}\bigr\} , $$
and the norm \(\|\omega\|_{X_{k}}=\|\omega\|_{Y_{k}}=\max\{\| u\|_{2kT},\| v\|_{2kT}\}\). Obviously, \(X_{k}\) and \(Y_{k}\) are Banach spaces.
Now we define the operator
$$ L:D(L)\subset X_{k}\rightarrow Y_{k},\qquad L\omega= \omega'=\bigl(u'(t),v'(t) \bigr)^{\top}, $$
where \(D(L)=\{\omega|\omega=(u(t),v(t))^{\top}: u\in C_{2kT}^{1}, u\in C_{2kT}^{1}\}\). Let
$$ Z_{k}=\bigl\{ \omega=\bigl(u(t),v(t)\bigr)^{\top}\in X_{k}: v\in C(R,B_{k})\bigr\} , $$
where \(B_{k}=\{x\in\mathbb{R}^{n}:|x|<1\}\). The nonlinear operator
$$N:\overline{\Omega}\subset Z_{k}\rightarrow Y_{k} $$
is defined as
$$N\omega= \biggl(\frac{\varphi_{q}(v(t))}{\sqrt{1-|\varphi _{q}(v(t))|^{2}}},-F\biggl(t,\frac{\varphi_{q}(v(t))}{\sqrt{1-|\varphi _{q}(v(t))|^{2}}}\biggr)-G\bigl(t,u \bigl(t-\tau(t)\bigr)\bigr) +e_{k}(t) \biggr)^{\top}, $$
where Ω is an open bounded subset of \(Z_{k}\). Clearly, the problem of the existence of a \(2kT\)-periodic solution to (2.1) is equivalent to the problem of the existence of a solution in \(\overline {\Omega}\) for the equation \(L\omega=N\omega\).

By simple calculating, we have \(\ker L =\mathbb{R}^{2n}\) and \(\operatorname{Im}L =\{z\in Y_{k}, \int_{0}^{2kT}z(s)\, ds=0\}\). Therefore, L is a Fredholm operator of index zero.

$$ P:X_{k}\rightarrow\ker L,\qquad P\omega=\frac{1}{2kT} \int _{0}^{2kT}\omega(s)\, ds $$
$$ Q:Y_{k}\rightarrow\operatorname{Im}Q,\qquad Qz=\frac{1}{2kT} \int _{0}^{2kT}z(s)\, ds. $$
If we define \({K}_{p}=L|^{-1}_{\operatorname{Ker}L\cap D(L)}\), then it is easy to see that
$$ (K_{p}z) (t) = \int_{0}^{2kT}G_{k}(t,s)z(s) \, ds, $$
$$ G_{k}(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{s-2kT}{2kT},&0\leq t\leq s ; \\ \frac{s}{2kT}, &s\leq t\leq2kT. \end{array}\displaystyle \right . $$
For all \(\overline{\Omega}\) such that \(\overline{\Omega}\subset Z_{k}\subset X_{k}\), we can see that \(K_{p}(I-Q)N (\overline{\Omega})\) is a relative compact set of \(X_{k}\) and \({QN}(\overline{\Omega})\) is a bounded set of \(Y_{k}\), so the operator N is L-compact in \(\overline{\Omega}\).
For the sake of convenience, we list the following assumptions:
There are two constants \(m_{0}>0\) and \(m_{1}>0\) such that
$$\bigl\langle x,F(t,x)\bigr\rangle \leq-m_{0}|x|^{2}\quad \mbox{and}\quad \bigl\vert F(t,x)\bigr\vert \leq m_{1}|x|,\quad \mbox{for all } (t,x)\in\mathbb{R}\times\mathbb{R}^{n}. $$
There are two constants \(\alpha>0\) and \(\beta>0\) such that
$$\bigl\langle x,G(t,x)\bigr\rangle \leq-\alpha|x|^{2} \quad \mbox{and} \quad \bigl\vert G(t,x) \bigr\vert \leq \beta|x|,\quad \mbox{for all } (t,x) \in\mathbb{R}\times\mathbb{R}^{n}. $$
\(e\in C(\mathbb{R},\mathbb{R}^{n})\) is a bounded function with \(e(t)\neq0=(0,0,\ldots,0)^{\top}\) and
$$A:= \biggl(\int_{R}\bigl\vert e(t)\bigr\vert ^{2}\, dt \biggr)^{\frac{1}{2}}+\sup_{t\in R}\bigl\vert e(t)\bigr\vert < +\infty. $$

Remark 2.2

From (1.6), we can see that \(|e_{k}(t)|\leq\sup_{t\in R}|e(t)|\). So if (H3) holds, for each \(k\in\mathbb{N}\), \(( \int_{ -kT}^{kT} |e(t)|^{2}\, dt)^{\frac{1}{2}}< A\).

3 Main results

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:
$$ \left \{ \textstyle\begin{array}{l} u'(t)=\lambda\phi(v(t))=\lambda\frac{\varphi _{q}(v(t))}{\sqrt{1-|\varphi_{q}(v(t))|^{2}}}, \\ v'(t)=-\lambda F(t,\varphi(v(t)))-\lambda G(t,u(t-\tau(t)))+\lambda e_{k}(t),\quad \lambda\in(0,1], \end{array}\displaystyle \right . $$
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
$$ \biggl[ \frac{ \beta\|\tau\|_{0}d_{0}d_{1}}{ T\sqrt {2(1-\sigma_{1})}}+ \frac{m_{1}d_{0}d_{1}+Ad_{0}}{2T} \biggr]^{\frac{1}{q}} + \frac{\sqrt{T}d_{0}\beta+\sqrt{T(1-\sigma_{1})}(m_{1} d_{1}+A) }{\sqrt {2(1-\sigma_{1})}}< 1, $$
$$d_{0}:=\frac{A (1-\sigma_{1}) (1+\sigma_{0})(m_{0}+m_{1})+\sqrt{2}A\beta \|\tau\|_{0}(1+\sigma_{0})\sqrt{1-\sigma_{1}}}{ \alpha m_{0} (1-\sigma_{1}) - m_{1}\beta(1+\sigma_{0})\sqrt{1-\sigma _{1}}-\sqrt{2}\beta^{2} \|\tau\|_{0}(1+\sigma_{0})} $$
$$d_{1}:=\frac{\beta d_{0}}{m_{0}\sqrt{1-\sigma_{1}}}+\frac{A}{m_{0}}. $$
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
$$\begin{aligned}& \|u\|_{0}\leq\rho_{1},\qquad \|v\|_{0}\leq\rho _{2}< 1,\qquad \bigl\| u'\bigr\| _{0}\leq \rho_{3},\qquad \bigl\| v'\bigr\| _{0}\leq \rho_{4}, \\& \|u\|_{2}\le A_{1},\qquad \bigl\| u'\bigr\| _{2} \le A_{2},\qquad \|v\|_{p}\le A_{3}, \qquad \bigl\| v'\bigr\| _{2}\le A_{4}. \end{aligned}$$


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
$$\begin{aligned}& \int_{-kT}^{kT} \frac{\vert v(t)\vert ^{q}}{\sqrt{1-\vert \varphi _{q}(v(t))\vert ^{2}}}\,dt+ \alpha \int _{-kT}^{kT}\bigl\vert u\bigl(t-\tau (t)\bigr)\bigr\vert ^{2}\,dt \\& \quad \leq \frac{m_{1}}{\lambda} \int_{-kT}^{kT}\bigl\vert u(t)\bigr\vert \bigl\vert u'(t)\bigr\vert \,dt+\beta \int_{-kT}^{kT}\bigl\vert u(t)-u\bigl(t-\tau(t) \bigr)\bigr\vert \bigl\vert u\bigl(t-\tau(t)\bigr)\bigr\vert \,dt \\& \qquad {}+ \int_{-kT}^{kT}\bigl\vert u(t)\bigr\vert \bigl\vert e_{k}(t)\bigr\vert \,dt. \end{aligned}$$
$$ \int_{-kT}^{kT}\bigl\vert u\bigl(t-\tau(t)\bigr) \bigr\vert ^{2}\,dt= \int_{-kT-\tau(-kT)}^{kT-\tau(kT)} \frac{1}{1-\tau'(\mu(s))}\bigl\vert u(s)\bigr\vert ^{2}\, ds. $$
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
$$ \frac{1}{1+\sigma_{0}}\| u\|^{2}_{2}\leq \int _{-kT }^{kT }\frac{1}{1-\tau'(\mu(s))}\bigl\vert u(s)\bigr\vert ^{2}\, ds \leq\frac{1}{1-\sigma_{1}}\| u\|^{2}_{2}. $$
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
$$\begin{aligned}& \int_{-kT}^{kT} \bigl\vert v(t)\bigr\vert ^{q}\, dt+\frac{\alpha}{1+\sigma _{0}} \int_{-kT}^{kT} \bigl\vert u(t)\bigr\vert ^{2}\,dt \\& \quad \leq \frac{m_{1}}{\lambda} \biggl( \int_{-kT}^{kT} \bigl\vert u(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \biggl( \int_{-kT}^{kT} \bigl\vert u'(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} +\frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt{1-\sigma _{1}}} \biggl( \int _{-kT}^{kT} \bigl\vert u(t)\bigr\vert ^{2}\,dt \biggr)^{\frac {1}{2}} \\& \qquad {}\times \biggl( \int_{-kT}^{kT} \bigl\vert u'(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} + \biggl( \int_{-kT}^{kT} \bigl\vert e_{k}(t)\bigr\vert ^{2}\,dt \biggr)^{\frac {1}{2}} \biggl( \int_{-kT}^{kT} \bigl\vert u(t)\bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$ \| v\|^{q}_{q}+\frac{\alpha}{1+\sigma_{0}}\| u\|^{2}_{2} \leq\frac{m_{1}}{\lambda}\| u\|_{2}\bigl\Vert u'\bigr\Vert _{2} +\frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt{1-\sigma _{1}}}\| u\|_{2}\bigl\Vert u'\bigr\Vert _{2} +A\| u\|_{2}. $$
This implies that
$$ \| v\|^{q}_{q} \leq\frac{m_{1}}{\lambda}\| u\|_{2} \bigl\Vert u'\bigr\Vert _{2} +\frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt{1-\sigma _{1}}}\| u \|_{2}\bigl\Vert u'\bigr\Vert _{2} +A\| u \|_{2} $$
$$ \frac{\alpha}{1+\sigma_{0}}\| u\|^{2}_{2} \leq\frac{m_{1}}{\lambda}\| u \|_{2}\bigl\Vert u'\bigr\Vert _{2} + \frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt{1-\sigma _{1}}}\| u\|_{2}\bigl\Vert u'\bigr\Vert _{2} +A\| u\|_{2}. $$
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
$$ \bigl\Vert u'\bigr\Vert _{2}\leq\frac{\lambda\beta}{m_{0}\sqrt{1-\sigma _{1}}}\| u \|_{2}+\frac{ \lambda A}{m_{0}}\leq\frac{ \beta }{m_{0}\sqrt{1-\sigma_{1}}}\| u \|_{2}+ \frac{ A}{m_{0}}. $$
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
$$\begin{aligned} \| u\|_{2}&\leq\frac{A (1-\sigma_{1}) (1+\sigma _{0})(m_{0}+m_{1})+\sqrt{2}A\beta \|\tau\|_{0}(1+\sigma _{0})\sqrt{1-\sigma_{1}}}{ \alpha m_{0} (1-\sigma_{1}) - m_{1}\beta(1+\sigma_{0})\sqrt{1-\sigma _{1}}-\sqrt{2}\beta^{2} \|\tau\|_{0}(1+\sigma_{0})} \\ &:=d_{0}. \end{aligned}$$
Substituting (3.7) into (3.6), we get
$$ \bigl\Vert u'\bigr\Vert _{2}\leq\frac{ \lambda d_{0}\beta}{m_{0}\sqrt {1-\sigma_{1}}}+ \frac{ \lambda A}{m_{0}}, $$
$$ \bigl\Vert u'\bigr\Vert _{2}\leq\frac{ d_{0}\beta}{m_{0}\sqrt{1-\sigma _{1}}}+ \frac{ A}{m_{0}}:=d_{1}. $$
Substituting (3.7), (3.8), and (3.9) into (3.4), we have
$$\begin{aligned} \| v\|^{q}_{q} &\leq\frac{m_{1}}{\lambda}\| u\|_{2} \bigl\Vert u'\bigr\Vert _{2} +\frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt{1-\sigma _{1}}}\| u \|_{2}\bigl\Vert u'\bigr\Vert _{2} +A\| u \|_{2} \\ &\leq\frac{\sqrt{2}\beta\|\tau\|_{0}}{\sqrt {1-\sigma_{1}}}d_{0}d_{1}+ m_{1} d_{0}d_{1}+Ad_{0}. \end{aligned}$$
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
$$ \|u\|_{0}=\max_{t\in[-kT,kT]}\bigl\vert u(t)\bigr\vert \leq\rho_{1}. $$
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
$$\begin{aligned} \bigl\Vert v'\bigr\Vert _{2} &\leq\frac{m_{1}}{\lambda} \bigl\Vert u'\bigr\Vert _{2}+\frac{\beta}{\sqrt{1-\sigma_{1}}} \| u\| _{2}+A \\ & \leq\frac{m_{1}}{\lambda} \biggl(\frac{ \lambda d_{0}\beta}{m_{0}\sqrt {1-\sigma_{1}}}+\frac{ \lambda A}{m_{0}}\biggr)+ \frac{\beta}{\sqrt{1-\sigma _{1}}} \| u\|_{2}+A \\ & = m_{1} \biggl(\frac{ d_{0}\beta}{m_{0}\sqrt{1-\sigma_{1}}}+\frac{ A}{m_{0}}\biggr)+ \frac{\beta}{\sqrt{1-\sigma_{1}}} \| u\|_{2}+A \\ & \leq\frac{d_{0}\beta}{\sqrt{1-\sigma_{1}}}+ m_{1} d_{1}+A. \end{aligned}$$
By applying Lemma 2.3 again and combining with (3.10) and (3.12), we get
$$\begin{aligned} \bigl\vert v(t)\bigr\vert &\leq(2T)^{-\frac{1}{q}} \biggl( \int _{t-kT}^{t+kT}\bigl\vert v(s)\bigr\vert ^{q}\, ds \biggr)^{\frac{1}{q}}+T(2T)^{-\frac {1}{2}} \biggl( \int_{t-kT}^{t+kT}\bigl\vert v'(s)\bigr\vert ^{2}\, ds \biggr)^{\frac{1}{2}} \\ &=(2T)^{-\frac{1}{q}} \biggl( \int_{-kT}^{kT}\bigl\vert v(s)\bigr\vert ^{q} \, ds \biggr)^{\frac{1}{q}}+T(2T)^{-\frac{1}{2}} \biggl( \int_{-kT}^{kT} \bigl\vert v'(s)\bigr\vert ^{2}\, ds \biggr)^{\frac{1}{2}} \\ &\leq(2T)^{-\frac{1}{q}} \biggl( \frac{\sqrt{2}\beta\|\tau \|_{0}}{\sqrt{1-\sigma_{1}}}d_{0}d_{1}+ m_{1} d_{0}d_{1}+Ad_{0} \biggr)^{\frac {1}{q}}+T(2T)^{-\frac{1}{2}} \biggl(\frac{d_{0}\beta}{\sqrt{1-\sigma _{1}}}+ m_{1} d_{1}+A \biggr) \\ &= \biggl[ \frac{ \beta\|\tau\|_{0}d_{0}d_{1}}{ T\sqrt {2(1-\sigma_{1})}}+ \frac{m_{1}d_{0}d_{1}+Ad_{0}}{2T} \biggr]^{\frac{1}{q}} + \frac{\sqrt{T}d_{0}\beta+\sqrt{T}(m_{1} d_{1}+A)\sqrt{1-\sigma _{1}}}{\sqrt{2(1-\sigma_{1})}} \\ &:=\rho_{2}. \end{aligned}$$
$$ \biggl[ \frac{ \beta\|\tau\|_{0}d_{0}d_{1}}{ T\sqrt {2(1-\sigma_{1})}}+ \frac{m_{1}d_{0}d_{1}+Ad_{0}}{2T} \biggr]^{\frac{1}{q}} + \frac{\sqrt{T}d_{0}\beta+\sqrt{T}(m_{1} d_{1}+A)\sqrt{1-\sigma _{1}}}{\sqrt{2(1-\sigma_{1})}}< 1, $$
we have
$$ \|v\|_{0}=\max_{t\in[-kT,kT]}\bigl\vert v(t)\bigr\vert \leq\rho_{2}< 1. $$
Clearly, \(\rho_{2}\) is independent of k and λ.
Furthermore, it follows from (3.1) that
$$\begin{aligned} \bigl\Vert u'\bigr\Vert _{0}&=\max _{t\in[-kT,kT]}\bigl\vert u'(t)\bigr\vert =\max _{t\in [-kT,kT]}\lambda\frac{\varphi_{q}(v(t))}{\sqrt{1-(\varphi _{q}(v(t)))^{2}}} \\ &\leq\frac{\rho^{q-1}_{2}}{\sqrt{1-\rho^{2q-2}_{2}}}: =\rho_{3}. \end{aligned}$$
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
$$ \bigl\Vert v'\bigr\Vert _{0}=\max _{t\in[-kT,kT]}\bigl\vert v'(t)\bigr\vert \leq F_{\rho_{3}}+G_{\rho _{1}}+A:=\rho_{4}. $$
\(\rho_{4}\) 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.


In order to use Lemma 2.1, for each \(k \in\mathbb{N}\), we consider the following system:
$$ \left \{ \textstyle\begin{array}{l} u'(t)=\lambda\varphi(v(t))=\lambda\frac{\varphi _{q}(v(t))}{\sqrt{1-|\varphi_{q}(v(t))|^{2}}}, \\ v'(t)=-\lambda F(t, \varphi(v(t)))-\lambda G(t,u(t-\tau(t)))+\lambda e_{k}(t),\quad \lambda\in(0,1), \end{array}\displaystyle \right . $$
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
$$\left \{ \textstyle\begin{array}{l} \int_{-kT}^{kT}\frac{\varphi _{q}(a_{2})}{\sqrt{1-|\varphi_{q}(a_{2})|^{2}}}\,dt=0, \\ \int_{-kT}^{kT}[-F(t,\frac{\varphi_{q}(a_{2})}{\sqrt {1-|\varphi_{q}(a_{2})|^{2}}})-G(t,a_{1})+e_{k}(t)]\,dt=0, \end{array}\displaystyle \right . $$
$$ \left \{ \textstyle\begin{array}{l} a_{2}=0, \\ \int_{-kT}^{kT}[-F(t,0)-G(t,a_{1})+e_{k}(t)]\,dt=0. \end{array}\displaystyle \right . $$
Multiplying the second equation of (3.17) by \(a_{1}\) and combining with (H1) and (H2), we have
$$\begin{aligned} 2kT\alpha|a_{1}|^{2}&\leq \int_{-kT}^{kT} \bigl\vert F(t,0)\bigr\vert |a_{1}|\,dt+ \int_{-kT}^{kT}|a_{1}| \bigl\vert e_{k}(t)\bigr\vert \,dt \\ &\leq2kT |a_{1}|A. \end{aligned}$$
$$|a_{1}|\leq\frac{A}{\alpha}:=\varpi. $$
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
$$\omega^{\top}H(\omega,\mu)\neq0,\quad \forall(\omega,\mu)\in\partial \Omega\cap\mathbb{R}^{2}\times[0,1]. $$
$$\begin{aligned} \deg\bigl\{ JQN ,\Omega\cap\mathbb{R}^{2},0\bigr\} &=\deg\bigl\{ H( \omega ,0),\Omega\cap \mathbb{R}^{2} ,{ 0}\bigr\} \\ &=\deg\bigl\{ H(\omega,1),\Omega\cap\mathbb{R}^{2},0\bigr\} \\ &\neq0. \end{aligned}$$
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.


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
$$ \|u_{k}\|_{0}\leq\rho_{1},\qquad \|v_{k}\|_{0}\leq\rho_{2}< 1,\qquad \bigl\Vert u'_{k}\bigr\Vert _{0}\leq\rho _{3},\qquad \bigl\Vert v'_{k}\bigr\Vert _{0}\leq\rho_{4}, $$
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
$$ \left \{ \textstyle\begin{array}{l} u_{k}'(t)= \phi(v_{k}(t))= \frac{\varphi _{q}(v_{k}(t))}{\sqrt{1-|\varphi_{q}(v_{k}(t))|^{2}}}, \\ v_{k}'(t)=-F(t,\varphi(v_{k}(t)))-G(t,u_{k}(t-\tau(t)))+e_{k}(t). \end{array}\displaystyle \right . $$
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
$$ u_{k_{j}}'(t)=\frac{\varphi_{q}(v_{k_{j}}(t))}{\sqrt{1-|\varphi _{q}(v_{k_{j}}(t))|^{2}}}\rightarrow \frac{\varphi_{q}(v_{0}(t))}{\sqrt {1-|\varphi_{q}(v_{0}(t))|^{2}}} $$
$$\begin{aligned} v_{k_{j}}'(t)&=-F\bigl(t,\varphi \bigl(v_{k_{j}}(t) \bigr)\bigr)-G\bigl(t,u_{k_{j}}\bigl(t-\tau(t)\bigr)\bigr)+e(t) \\ &\rightarrow -F\bigl(t,\varphi\bigl(v_{0}(t)\bigr)\bigr)-G \bigl(t,u_{0}\bigl(t-\tau(t)\bigr)\bigr)+e(t) \end{aligned}$$
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:
$$\left \{ \textstyle\begin{array}{l} u'(t)= \phi(v(t))= \frac{\varphi_{q}(v(t))}{\sqrt {1-|\varphi_{q}(v(t))|^{2}}}, \\ v'(t)=-F(t,\varphi(v(t)))-G(t,u(t-\tau(t)))+e(t), \end{array}\displaystyle \right . $$
$$ \left \{ \textstyle\begin{array}{l} u_{0}'(t)= \phi(v_{0}(t))= \frac{\varphi _{q}(v_{0}(t))}{\sqrt{1-|\varphi_{q}(v_{0}(t))|^{2}}}, \\ v_{0}'(t)=-F(t,\varphi(v_{0}(t)))-G(t,u_{0}(t-\tau(t)))+e(t). \end{array}\displaystyle \right . $$
Now, we will prove \(u_{0}(t)\rightarrow0\) and \(u_{0}'(t)\rightarrow0\) as \(|t|\rightarrow+\infty\).
$$\begin{aligned} \begin{aligned} \int^{+\infty}_{-\infty}\bigl(\bigl\vert u_{0}(t)\bigr\vert ^{2}+\bigl\vert u'_{0}(t) \bigr\vert ^{2}\bigr)\,dt&=\lim_{i\rightarrow+\infty}\int ^{iT}_{-iT}\bigl(\bigl\vert u_{0}(t) \bigr\vert ^{2}+\bigl\vert u'_{0}(t)\bigr\vert ^{2}\bigr)\,dt \\ &=\lim_{i\rightarrow+\infty}\lim_{j\rightarrow+\infty}\int ^{iT}_{-iT}\bigl(\bigl\vert u_{0}(t) \bigr\vert ^{2}+\bigl\vert u'_{0}(t)\bigr\vert ^{2}\bigr)\,dt. \end{aligned} \end{aligned}$$
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
$$ u_{0}(t)\rightarrow0 \quad \mbox{as } |t|\rightarrow+ \infty. $$
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
$$ u_{0}'(t)\rightarrow0 \quad \mbox{as } |t|\rightarrow+ \infty. $$
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.



The authors express their thanks to the referee for his (or her) valuable suggestions. The research was supported by the National Natural Science Foundation of China (Grant No. 11271197).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (, which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

College of Mathematics and Statistics, Nanjing University of Information Science and Technology, Nanjing, China
Department of Mathematics, Anhui Normal University, Wuhu, 241000, China


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