- Research
- Open Access
New positive periodic solutions to singular Rayleigh prescribed mean curvature equations
- Zhiyan Li1Email author and
- Weigao Ge2
- Received: 31 December 2016
- Accepted: 12 April 2017
- Published: 24 April 2017
Abstract
This paper is concerned with the existence of positive periodic solutions for the prescribed mean curvature Rayleigh equations with a singularity. Our results are based on the Manásevich-Mawhin continuation theorem. The results to be obtained here extend the existing ones in the literature. Moreover, an example is given to illustrate the applicability of our results.
Keywords
- positive periodic solution
- Manásevich-Mawhin continuation theorem
- prescribed mean curvature Rayleigh equation
- singularity
MSC
- 34K13
- 34C25
1 Introduction
The Rayleigh equation arises from many applied fields, such as the physics, mechanics, and engineering technique fields. So, it is meaningful and necessary to study the periodic solutions for the Rayleigh equation. In 1977, Gaines and Mawhin [1] discussed the existence of solutions for the following Rayleigh equation: By applying the continuation theorems, Gaines and Mawhin proved that the Rayleigh equation can support periodic solutions.
$$u''(t) + f\bigl(u'(t)\bigr) + g\bigl(t, u(t)\bigr)=0. $$
In recent years, the prescribed mean curvature equation and its modified forms have been studied widely since they arise from some certain problems associated with differential geometry and physics such as combustible gas dynamics; see [2–5] and the references therein. Due to the wide range of application background of the prescribed mean curvature equations, many researchers have worked on the existence of periodic solutions for the prescribed mean curvature equations. For the related papers, we refer the reader to [6–12].
On this basis of work of Gaines and Mawhin [1], some researchers discussed the existence of periodic solutions to some types of prescribed mean curvature Rayleigh equations; see [13, 14] and the references therein. For example, by using Mawhin’s continuation theorem in the coincidence degree theory, Li et al. [14] considered the periodic solutions for the following prescribed mean curvature Rayleigh equation: where τ, \(e\in C(\mathbb {R},\mathbb {R})\) are T-periodic, and f, \(g\in C(\mathbb {R}\times \mathbb {R},\mathbb {R})\) are T-periodic in the first argument, \(T>0\) is a constant.
$$ \biggl(\frac{x'(t)}{\sqrt{1+(x'(t))^{2}}} \biggr)'+f\bigl(t, x'(t) \bigr)+g\bigl(t,x\bigl(t-\tau(t)\bigr)\bigr)=e(t), $$
(1.1)
Singular equations appear in a great deal of physical models and play an important role in the differential equations. Recently, Lu and Kong in [15] extended the prescribed mean curvature Liénard equations to the singular case and studied the positive periodic solutions for the following prescribed mean curvature Liénard equation with a singularity: where \(\sigma=kT\), \(k=1,2,\ldots, n\), f and \(g: (0,+\infty )\rightarrow {\mathbb {R}}\) are continuous functions, g can be singular at \(u=0\), i.e., \(g(u)\) can be unbounded as \(u\rightarrow0^{+}\). \(e(t)\) is T-periodic with \(\int_{0}^{T} e(t)\,dt=0\). In order to establish the existence result of positive periodic solutions, the authors gave the following conditions:
By applying the Manásevich-Mawhin continuation theorem, the authors proved that equation (1.2) has at least one positive T-periodic solution.
$$ \biggl(\frac{u'(t)}{\sqrt{1+(u'(t))^{2}}} \biggr)'+f\bigl(u(t)\bigr)u'(t)+g \bigl(u(t-\sigma)\bigr)=e(t), $$
(1.2)
- [\(A_{1}\)]:
-
There exist positive constants \(D_{1}\) and \(D_{2}\) with \(D_{1}< D_{2}\) such that
(1) For each positive continuous T-periodic function \(x(t)\) satisfying \(\int_{0}^{T} g( x(t))\,dt=0\), there exists a positive point \(\tau\in [0,T]\) such that$$D_{1}\leq x(\tau)\leq D_{2}; $$(2) \(g(x)<0\) for all \(x\in(0,D_{1})\) and \(g(x)>0\) for all \(x>D_{2}\).
- [\(A_{2}\)]:
-
\(g(x(t))=g_{1}(x(t))+g_{0}(x(t))\), where \(g_{1}: (0,+\infty )\rightarrow {\mathbb {R}}\) is a continuous function and
(1) There exist positive constants \(m_{0}\) and \(m_{1}\) such that \(g(x)\leq m_{0}x+m_{1}\);
(2) \(\displaystyle\int_{0}^{1} g_{0}(x)\,dx=-\infty\).
- [\(A_{3}\)]:
-
There exist positive constants γ, \(c_{0} \), \(c_{1} \) such that \(\gamma< f(x)\leq c_{0} \vert x \vert +c_{1}\).
Based on Lu and Kong in [15], Chen and Kong [16] further study the existence of positive periodic solutions for a prescribed mean curvature p-Laplacian equation with a singularity of repulsive type and a time-varying delay where \(g: [0,T]\times(0,+\infty)\times(0,+\infty)\rightarrow {\mathbb {R}}\) is a continuous function. g can be singular at \(u=0\), i.e., g can be unbounded as \(u\rightarrow0^{+}\). \(\tau, p\in(\mathbb {R}, \mathbb {R})\) are T-periodic with \(\int_{0}^{T} p(t)\,dt=0\), β is a constant.
$$\biggl(\varphi_{p} \biggl(\frac{x'(t)}{\sqrt{1+(x'(t))^{2}}} \biggr) \biggr)'+\beta x'(t)+g\bigl(t, x(t),x\bigl(t-\tau(t) \bigr)\bigr)=p(t), $$
Compared with the results in the literature, the prescribed mean curvature Rayleigh equations with singular effects have been scarcely studied.
Inspired by the above facts, in this paper, we further consider the following prescribed mean curvature Rayleigh equations with a singularity: where \(f\in C([0,T]\times \mathbb {R}, \mathbb {R})\) is a T-periodic function about t and \(f (t,0)=0\), \(g: (0, +\infty) \rightarrow \mathbb {R}\) is a continuous function and has a strong singularity at the origin:
\(e\in C([0,T], \mathbb {R})\) is a T-periodic function, \(0\leq\tau< T\) and τ is a constant. By means of the Manásevich-Mawhin continuation theorem, we prove that (1.3) has at least one positive T-periodic solution.
$$ \biggl(\frac{u'(t)}{\sqrt{1+(u'(t))^{2}}} \biggr)'+f\bigl(t,u'(t) \bigr)+g\bigl(u(t-\tau)\bigr)=e(t), $$
(1.3)
$$ \lim_{x\rightarrow0^{+}} \int_{x}^{1}g(s)\,ds=+\infty. $$
(1.4)
Remark 1.1
The theorem and methods used to obtain the periodic solutions to (1.1) in [14] and [13] can be applied to the (1.3) if there is no singularity in (1.3). So, we extend the prescribed mean curvature Rayleigh equations to the singular case.
Remark 1.2
If \(x\in C^{1}(\mathbb {R}, \mathbb {R})\) with T-periodic, then \(f(x)x'\) in equation (1.2) satisfies \(\int_{0}^{T}f(x(t))x'(t)\,dt = 0\), which is crucial to obtain the priori bounds of T-periodic solutions for equation (1.2). However, the first order derivative term in equation (1.3) is \(f(t,x')\). Generally, \(\int_{0}^{T}f(t,x'(t))\,dt = 0\) does not hold. For example, let us define then it is easy to see that \(\int_{0}^{T} (3- \sin^{2} 8t )x'(t)\,dt\neq0\) for some \(x\in C^{1}(\mathbb {R},\mathbb {R})\). This implies that our method to complete estimate the priori bounds for all T-periodic solutions to equation (1.3) is different from the corresponding ones.
$$f\bigl(t,x'(t)\bigr)= \bigl(3- \sin^{2} 8t \bigr)x'(t), $$
Remark 1.3
From [15] and [16], the conditions composed on \(e(t)\) and \(p(t)\) are \(\int_{0}^{T} e(t)\,dt=0\) and \(\int _{0}^{T} p(t)\,dt=0\). But, in this paper, it is unnecessary. For example, let us define then it is easy to see that \(\int_{0}^{T}\frac{e^{\cos^{2} 8t}}{12} \,dt\neq0\). So, our results can be more general.
$$e(t)=\frac{e^{\cos^{2} 8t}}{12}, $$
2 Preliminary
Throughout this paper, for any T-periodic continuous function \(u(t)\), we always use the notations as follows:
$$\Vert u \Vert _{2} = \biggl( \int_{0}^{T} \bigl\vert u(t) \bigr\vert ^{2} \,dt \biggr)^{1/2}\quad\mbox{and}\quad \Vert u \Vert _{0} = \max_{t\in[0,T ]} \bigl\vert u(t) \bigr\vert . $$
In order to use Lemma 2.1, let us consider the problem Obviously, if \((u(t),v(t))^{\top}\) is a solution of (2.1), then \(u(t)\) is a solution of (1.3).
$$ \textstyle\begin{cases} u'(t)=\phi(v(t))=\frac{v(t)}{\sqrt{1-v^{2}(t)}},\\ v'(t)=-f(t,\phi(v(t))) -g(u(t-\tau))+e(t). \end{cases} $$
(2.1)
Lemma 2.1
[17]
Assume that there exist positive constants \(E_{1}\), \(E_{2}\), \(E_{3}\) with \(E_{1}< E_{2}\) such that the following conditions hold:
(1) for each
\(\lambda\in(0, 1]\), each possible positive
T-periodic solution
\(x=(u,v)^{\top}\)
to the system
satisfies the inequalities
\(E_{1}< u(t)<E_{2}\)
and
\(\Vert u' \Vert _{0} < E_{3}\)
for all
\(t\in[0, T]\).
$$\textstyle\begin{cases} u'(t)=\lambda\phi(v(t))=\lambda\frac{v(t)}{\sqrt{1-v^{2}(t)}},\\ v'(t)=-\lambda f(t, \phi(v(t))) -\lambda g(u(t-\tau))+\lambda e(t), \end{cases} $$
(2) Each possible solution
C
to the equation
\(g(C)-\frac{1}{T}\int_{0}^{T}e(t)\,dt=0\)
satisfies
$$E_{1}< C < E_{2}. $$
(3) We have \(( g(E_{1})- \frac{1}{T}\int_{0}^{T}e(t)\,dt ) ( g(E_{2})- \frac{1}{T}\int_{0}^{T}e(t)\,dt )<0\).
Then equation (1.3) has at least one positive T-periodic solution.
Lemma 2.2
[18]
Let
\(u(t)\)
be a continuously differentiable
T-periodic function. Then, for any
\(t_{0} \in[0, T]\),
$$\biggl( \int_{0}^{T} \bigl\vert u(t) \bigr\vert ^{2}\,dt \biggr)^{1/2}\leq\frac{T}{\pi} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{1/2}+\sqrt{T} \bigl\vert u(t_{0}) \bigr\vert . $$
For the sake of convenience, we list the following assumptions:
- (H1)There exist constants \(0< d_{1}< d_{2}\) such that$$g(x)-e(t)>0,\quad \forall x \in(0, d_{1}),\quad\mbox{and}\quad g(x)-e(t)< 0,\quad \forall x \in (d_{2}, +\infty), t\in[0,T]. $$
- (H2)There exist positive constants \(m_{0}\) and \(m_{1}\) such that$$\bigl\vert g(x) \bigr\vert \leq m_{0} x+m_{1},\quad \forall x \in(0, +\infty). $$
- (H3)There exists a positive constant a such that$$f(t,x)x\geq a \vert x \vert ^{2},\quad \forall(t,x)\in[0,T]\times \mathbb {R}. $$
- (H4)There exist positive constants β and γ such that$$f(t,x) \leq \beta \vert x \vert +\gamma,\quad\forall(t,x)\in[0,T]\times \mathbb {R}. $$
3 Main results
In this section, we will consider the existence of positive periodic solution for (1.3) with a singularity.
First of all, we embed equation (1.3) into the following equation family with a parameter \(\lambda\in(0, 1]\):
$$ \textstyle\begin{cases} u'(t)=\lambda\phi(v(t))=\lambda\frac{v(t)}{\sqrt{1-v^{2}(t)}},\\ v'(t)=-\lambda f(t, \phi(v(t))) -\lambda g(u(t-\tau))+\lambda e(t). \end{cases} $$
(3.1)
Theorem 3.1
Suppose the conditions (H1)-(H4) hold, \(a\pi> m_{0}T \)
and
where
\(A_{0}= \frac{\pi\sqrt{T}(m_{0}d_{2}+m_{1}+ \Vert e \Vert _{0})}{a\pi-m_{0}T}\), \(A_{1}=d_{2}+A_{0}\), then there exist positive constants
\(A_{1}\), \(A_{2}\), \(A_{3}\)
and
\(A_{4}\), which are independent of
λ
such that
where
\(x=(u,v)^{\top}\)
is any solution to equation (3.1), \(\lambda\in(0,1]\).
$$\beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T< 1, $$
$$A_{2}\leq u(t)\leq A_{1},\qquad \bigl\Vert u' \bigr\Vert _{0} \leq A_{3},\qquad \Vert v \Vert _{0} \leq A_{4}, $$
Proof
Let t̅, \(\underline{t}\), respectively, be the global maximum point and global minimum point \(u(t)\) on \([0,T]\); then \(u'(\overline{t})=0\) and \(u'(\underline {t})=0\). We claim that In fact, if (3.2) does not hold, then there exists \(\varepsilon >0\) such that \(v'(t)< 0\) for \(t \in(\underline{t}-\varepsilon, \underline {t}+\varepsilon)\). Therefore, \(v(t)\) is strictly decreasing for \(t \in(\underline {t}-\varepsilon, \underline{t }+\varepsilon)\). Thus, from the first equation of (3.1), we can see that \(u'(t)\) is strictly decreasing for \(t \in(\underline{t}-\varepsilon, \underline{t}+\varepsilon)\). This contradicts the definition of \(\underline{t}\). Therefore, (3.2) is true. From the second equation of (3.1), (3.2) and \(f(t,0)=0\), we have In a similar way, we get It follows from (H1), (3.3) and (3.4) that Thus, we can see that there exists a point \(t_{0}\in[0,T]\) such that
$$ v'(\underline{t})\geq0. $$
(3.2)
$$ g\bigl(u(\underline{t}-\tau)\bigr)-e(\underline{t})\leq0. $$
(3.3)
$$ g\bigl(u(\overline{t}-\tau)\bigr)-e(\overline{t})\geq0. $$
(3.4)
$$\begin{aligned} u(\underline{t}-\tau)\geq d_{1} \quad\mbox{and}\quad u(\overline{t}-\tau)\leq d_{2}. \end{aligned}$$
$$ d_{1}\leq u(t_{0}) \leq d_{2}. $$
(3.5)
Multiplying the second equation of (3.1) by \(u'(t)\) and integrating over the interval \([0, T]\), we have which together with (H2) and (H3) gives By using Lemma 2.2 and (3.5), we have i.e., Since \(a\pi> m_{0}T \), we have By means of the Hölder inequality, (3.5) and (3.6), we have Clearly, \(A_{1}\) is independent of λ.
$$\begin{aligned} 0={}& \int_{0}^{T}v'(t)u'(t)\,dt= \lambda \int_{0}^{T}\frac{v(t)}{\sqrt {1-v^{2}(t)}}\cdot v'(t)\,dt \\ ={}&{-}\lambda \int_{0}^{T}f \biggl(t,\frac{u'(t)}{\lambda} \biggr)u'(t)\,dt-\lambda \int_{0}^{T}g\bigl(u(t-\tau)\bigr)u'(t)\,dt \\ &{}+\lambda \int_{0}^{T} e(t)u'(t)\,dt, \end{aligned}$$
$$\begin{aligned} &a \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \\ &\quad \leq \lambda \int_{0}^{T} \bigl\vert g\bigl(u(t-\tau )\bigr) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt +\lambda \int_{0}^{T} \bigl\vert e(t) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad \leq\lambda\cdot \int_{0}^{T} \bigl(m_{0} \bigl\vert u(t-\tau) \bigr\vert +m_{1} \bigr) \bigl\vert u'(t) \bigr\vert \,dt +\lambda\cdot \Vert e \Vert _{0} \sqrt{T}\cdot \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \\ &\quad\leq\lambda\cdot m_{0} \biggl( \int_{0}^{T} \bigl\vert u(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} +\lambda\cdot m_{1} \sqrt{T} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \\ &\qquad{}+\lambda\cdot \Vert e \Vert _{0} \sqrt{T}\cdot \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2}. \end{aligned}$$
$$\begin{aligned} &a \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \\ &\quad\leq\lambda\cdot m_{0} \biggl[ \frac{T}{\pi} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{1/2}+\sqrt{T}d_{2} \biggr] \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \\ &\qquad{}+\lambda\cdot m_{1} \sqrt{T} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} +\lambda\cdot \Vert e \Vert _{0} \sqrt{T}\cdot \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \\ &\quad= \lambda\cdot\frac{ m_{0}T}{\pi} \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2}\,dt + \lambda\cdot m_{0} d_{2} \sqrt{T} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \\ &\qquad{}+\lambda\cdot m_{1} \sqrt{T} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} +\lambda\cdot \Vert e \Vert _{0} \sqrt{T}\cdot \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2}, \end{aligned}$$
$$\begin{aligned} a \bigl\Vert u' \bigr\Vert ^{2}_{2} ={}& a \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \\ \leq{}& \frac{ m_{0}T}{\pi} \bigl\Vert u' \bigr\Vert ^{2}_{2} + m_{0}d_{2} \sqrt{T} \bigl\Vert u' \bigr\Vert _{2} + m_{1} \sqrt{T} \bigl\Vert u' \bigr\Vert _{2} + \Vert e \Vert _{0} \sqrt{T} \bigl\Vert u' \bigr\Vert _{2}. \end{aligned}$$
$$ \bigl\Vert u' \bigr\Vert _{2} \leq \frac{\pi\sqrt{T}(m_{0}d_{2}+m_{1}+ \Vert e \Vert _{0})}{a\pi -m_{0}T}:=A_{0}. $$
(3.6)
$$\begin{aligned} \Vert u \Vert _{0} =&\max _{t\in[0,T]} \bigl\vert u(t) \bigr\vert \leq\max _{t\in [0,T]} \biggl\vert u(t_{0})+ \int_{t_{0}}^{t}u'(s)\,ds \biggr\vert \\ \leq& d_{2}+ \int_{0}^{T} \bigl\vert u'(s) \bigr\vert \,ds\leq d_{2}+ \sqrt{T} \bigl\Vert u' \bigr\Vert _{2} \\ \leq& d_{2}+\frac{\pi T(m_{0}d_{2}+m_{1}+ \Vert e \Vert _{0})}{a\pi-m_{0}T}:= A_{1}. \end{aligned}$$
(3.7)
From the second equation of (3.1), we have Furthermore, from (3.7) and (H2), we get Substituting (3.9) into (3.8) and by using (H4), (3.6) and (3.7), we can obtain Integrating the first equation of (3.1) on the interval \([0,T]\), we have Then we can see that there exists \(\eta\in[0,T]\) such that \(v(\eta )=0\). It implies that which together with (3.10) yields Since \(\beta\sqrt{T} A_{0}+ ( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma)T<1\), we have Clearly, \(A_{4}\) is independent of λ.
$$\begin{aligned} \int_{0}^{T} \bigl\vert v'(t) \bigr\vert \,dt \leq& \lambda \int_{0}^{T} \biggl\vert f \biggl(t, \frac{u'(t)}{\lambda} \biggr) \biggr\vert \,dt+\lambda \int_{0}^{T} \bigl\vert g\bigl(u(t-\tau)\bigr) \bigr\vert \,dt \\ &{}+\lambda \int_{0}^{T} \bigl\vert e(t) \bigr\vert \,dt. \end{aligned}$$
(3.8)
$$\begin{aligned} \int_{0}^{T} \bigl\vert g\bigl(u(t-\tau)\bigr) \bigr\vert \,dt \leq& \int_{0}^{T} \bigl[ m_{0}u(t- \tau)+m_{1} \bigr] \,dt \\ \leq& m_{0}T \cdot \Vert u \Vert _{0}+ m_{1}T \\ \leq& m_{0} A_{1}T+ m_{1}T. \end{aligned}$$
(3.9)
$$\begin{aligned} \int_{0}^{T} \bigl\vert v'(t) \bigr\vert \,dt \leq&\lambda \int_{0}^{T} \biggl\vert f \biggl(t, \frac{u'(t)}{\lambda} \biggr) \biggr\vert \,dt+\lambda \int_{0}^{T} \bigl\vert g\bigl(u(t-\tau)\bigr) \bigr\vert \,dt +\lambda \int_{0}^{T} \bigl\vert e(t) \bigr\vert \,dt \\ \leq&\lambda \int_{0}^{T} \biggl( \beta \biggl\vert \frac{u'(t)}{\lambda } \biggr\vert +\gamma \biggr) \,dt + m_{0} A_{1}T+ m_{1}T + \Vert e \Vert _{0}T \\ \leq&\beta \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt+\gamma T + m_{0} A_{1}T+ m_{1}T + \Vert e \Vert _{0}T \\ \leq&\beta\sqrt{T} \bigl\Vert u' \bigr\Vert _{2}+ \gamma T + m_{0} A_{1}T+ m_{1}T + \Vert e \Vert _{0}T \\ \leq& \beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T. \end{aligned}$$
(3.10)
$$\int_{0}^{T}u'(t)\,dt= \int_{0}^{T} \frac{v(t)}{\sqrt{1-v^{2}(t)}}\,dt=0. $$
$$\bigl\vert v(t) \bigr\vert = \biggl\vert \int_{\eta}^{t} v'(s)\,ds+v(\eta) \biggr\vert \leq \int _{0}^{T} \bigl\vert v'(s) \bigr\vert \,ds, $$
$$\begin{aligned} \bigl\vert v(t) \bigr\vert \leq& \int_{0}^{T} \bigl\vert v'(s) \bigr\vert \,ds \\ \leq& \beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T :=A_{4}. \end{aligned}$$
(3.11)
$$ \begin{aligned} \Vert v \Vert _{0}=\max _{t\in[0,T]} \bigl\vert v(t) \bigr\vert \leq A_{4}< 1. \end{aligned} $$
(3.12)
From the first equation of (3.1), we can see that Clearly, \(A_{3}\) is independent of λ.
$$ \bigl\Vert u' \bigr\Vert _{0} \leq \lambda\max_{t\in[0,T]}\frac{ \vert v(t) \vert }{\sqrt {1-v^{2}(t)}}\leq \lambda \cdot\frac{A_{4}}{1-A_{4}^{2}} \leq \frac{A_{4}}{1-A_{4}^{2}}:=A_{3}. $$
(3.13)
In the following, we will prove that there exists a positive constant \(A_{2}\) which is dependent of λ such that Indeed, it follows from the second equation of (3.1) that Multiplying both sides of (3.15) by \(u'(t)\) and integrating on \([\xi, t]\), here \(\xi\in[0, T]\), we get then Furthermore, by (3.10) and (3.13) we obtain By using (H4) and (3.13), we have Substituting (3.17) and (3.18) into (3.16), we obtain i.e., From the strong force condition (1.4), we know that (3.14) holds. Therefore, from (3.7), (3.12), (3.13) and (3.14), we can see that the proof of Theorem 3.1 is now completed. □
$$ u(t)\geq A_{2}. $$
(3.14)
$$ v'(t+\tau)=-\lambda f \biggl(t+\tau,\frac{u'(t+\tau)}{\lambda} \biggr)- \lambda g\bigl(u(t)\bigr)+\lambda e(t+\tau). $$
(3.15)
$$\begin{aligned} \lambda \int_{u(\xi)}^{u(t)}g_{0}(u)\,du={}& \lambda \int_{\xi }^{t}g_{0}\bigl(u(s) \bigr)u'(s)\,ds \\ ={}& - \int_{\xi}^{t}v'(t+\tau)u'(t)\,dt -\lambda \int_{\xi}^{t} f \biggl(t+\tau,\frac{u'(t+\tau)}{\lambda} \biggr) u'(t)\,dt \\ &{} +\lambda \int_{\xi}^{t} e(t+\tau)u'(t)\,dt; \end{aligned}$$
$$\begin{aligned} \lambda \biggl\vert \int_{u(\xi)}^{u(t)}g_{0}(u)\,du \biggr\vert =& \int_{0}^{T} \bigl\vert v'(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &{}+\lambda \int_{0}^{T} \biggl\vert f \biggl(t+\tau, \frac{u'(t+\tau )}{\lambda} \biggr) \biggr\vert \cdot \bigl\vert u'(t) \bigr\vert \,dt \\ &{}+\lambda \int_{0}^{T} \bigl\vert e(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt. \end{aligned}$$
(3.16)
$$\begin{aligned} \int_{0}^{T} \bigl\vert v'(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \leq& \bigl\Vert u' \bigr\Vert _{0}\cdot \int _{0}^{T} \bigl\vert v'(t+\tau) \bigr\vert \,dt \\ \leq&\lambda \cdot\frac{A_{4}}{1-A_{4}^{2}} \cdot \bigl[\beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T \bigr]. \end{aligned}$$
(3.17)
$$\begin{aligned} & \int_{0}^{T} \biggl\vert f \biggl(t+\tau, \frac{u'(t+\tau)}{\lambda} \biggr) \biggr\vert \cdot \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad\leq \int_{0}^{T} \biggl( \beta \biggl\vert \frac{u'(t+\tau)}{\lambda} \biggr\vert +\gamma \biggr) \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad\leq\beta \int_{0}^{T} \biggl\vert \frac{u'(t+\tau)}{\lambda} \biggr\vert \cdot \bigl\vert u'(t) \bigr\vert \,dt+\gamma \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad\leq\frac{\beta T}{\lambda} \cdot \bigl\Vert u' \bigr\Vert ^{2}_{0}+\gamma T\cdot \bigl\Vert u' \bigr\Vert _{0} \\ &\quad\leq\frac{\beta T}{\lambda} \cdot \biggl(\lambda \cdot\frac {A_{4}}{1-A_{4}^{2}} \biggr)^{2}+\gamma T\cdot\lambda \cdot\frac {A_{4}}{1-A_{4}^{2}} \\ &\quad\leq \beta T \cdot\lambda \cdot \biggl(\frac{A_{4}}{1-A_{4}^{2}} \biggr)^{2}+ \gamma T\cdot\lambda \cdot\frac{A_{4}}{1-A_{4}^{2}}. \end{aligned}$$
(3.18)
$$\begin{aligned} \lambda \biggl\vert \int_{u(\xi)}^{u(t)}g_{0}(u)\,du \biggr\vert \leq& \int_{0}^{T} \bigl\vert v'(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt +\lambda \int_{0}^{T} \biggl\vert f \biggl(t+\tau, \frac{u'(t+\tau )}{\lambda} \biggr) \biggr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &{}+\lambda \int_{0}^{T} \bigl\vert e(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ \leq&\lambda \cdot\frac{A_{4}}{1-A_{4}^{2}} \cdot \bigl[ \beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T \bigr] \\ &{}+\beta T \cdot\lambda^{2} \cdot \biggl(\frac{A_{4}}{1-A_{4}^{2}} \biggr)^{2}+\gamma T\cdot\lambda^{2} \cdot\frac{A_{4}}{1-A_{4}^{2}} + \lambda^{2} \cdot \Vert e \Vert _{0} \cdot \biggl( \frac{A_{4}}{1-A_{4}^{2}} \biggr) \end{aligned}$$
$$\begin{aligned} \biggl\vert \int_{u(\xi)}^{u(t)}g_{0}(u)\,du \biggr\vert \leq{}& \frac {A_{4}}{1-A_{4}^{2}} \cdot \bigl[\beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T \bigr] \\ &{}+\beta T \cdot \biggl(\frac{A_{4}}{1-A_{4}^{2}} \biggr)^{2} +\gamma T \cdot \frac{A_{4}}{1-A_{4}^{2}} + \Vert e \Vert _{0} \cdot \frac{A_{4}}{1-A_{4}^{2}}. \end{aligned}$$
Theorem 3.2
Assume that the conditions in Theorem 3.1 hold, then equation (1.3) has at least one positive T-periodic solution.
Proof
Define It follows from (3.5), (3.7), (3.13) and (3.14) that Then we can see that the condition (1) of Lemma 2.1 is satisfied.
$$0< E_{1}< \min\{d_{1}, A_{2}\},\qquad E_{2}> \max\{d_{2}, A_{1}\},\qquad E_{3}>A_{3}. $$
$$ E_{1}< u(t)< E_{2},\qquad \bigl\Vert u' \bigr\Vert _{0}< E_{3}. $$
(3.19)
For a possible solution C to the equation it is easy to see that \(E_{1}< C < E_{2}\) is satisfied. Thus, the condition (2) of Lemma 2.1 is satisfied.
$$g(C)-\frac{1}{T} \int_{0}^{T}e(t)\,dt=0, $$
Finally, we prove that the condition (3) of Lemma 2.1 is also satisfied. In fact, from (H1), we have and which implies that the condition (3) of Lemma 2.1 is also satisfied. Therefore, by application of Lemma 2.1, we conclude that (1.3) has at least one positive T-periodic solution. □
$$g(E_{1})-\frac{1}{T} \int_{0}^{T}e(t)\,dt>0, $$
$$g(E_{2})-\frac{1}{T} \int_{0}^{T}e(t)\,dt< 0, $$
4 Example
Consider the following prescribed mean curvature Rayleigh equations with a singularity:
$$ \begin{aligned} & \biggl(\frac{u'(t)}{\sqrt{1+(u'(t))^{2}}} \biggr)'+ \bigl(3-\sin^{2} 8t \bigr)u'(t)-\frac{1}{6u(t-1)}+ \frac{u(t-1)}{4}= \frac{e^{\cos^{2} 8t}}{12}. \end{aligned} $$
(4.1)
Conclusion
Problem (4.1) has at least one positive \(\pi /8\)-periodic solution.
Proof
Corresponding to equation (1.3), we have It is easy to see that (H1)-(H4) hold if we choose Moreover, \(a\pi> m_{0}T \) and then we have Hence, the conditions of Theorem 3.1 are satisfied. Therefore, by Theorem 3.2, we can see that equation (4.1) has at least one positive \(\pi/8\)-periodic solution. □
$$\begin{aligned} &{f\bigl(t,u'(t)\bigr)= \bigl(3- \sin^{2} 8t \bigr)u'(t),\qquad e(t)=\frac{e^{\cos^{2} 8t}}{12},} \\ &{g\bigl(u(t-1)\bigr)=-\frac{1}{6u(t-1)}+\frac{u(t-1)}{4},\qquad T= \frac{\pi}{8}.} \end{aligned}$$
$$d_{2}=1,\qquad m_{0}=\frac{1}{4},\qquad m_{1}= \frac{1}{6},\qquad a=2,\qquad \beta=3,\qquad \gamma=\frac{1}{8}. $$
$$A_{0}= \frac{\pi\sqrt{T}(m_{0}d_{2}+m_{1}+ \Vert e \Vert _{0})}{a\pi-m_{0}T}\approx 0.15915,\qquad A_{1}=d_{2}+A_{0} \approx1.15915, $$
$$\begin{aligned} \beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T\approx0.56026< 1. \end{aligned}$$
Declarations
Acknowledgements
The authors are grateful for the referee’s helpful suggestions and comments. This work is supported by the Fundamental Research Funds for the Central Universities (Grant No. 2016B07514, No. 2015B27914).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), 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
(1)
Department of Mathematics and Physics, Hohai University, Campus Changzhou, 213022, China
(2)
Department of Mathematics, Beijing Institute of Technology, Beijing, 100081, China
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