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
$$\beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T< 1, $$
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
$$A_{2}\leq u(t)\leq A_{1},\qquad \bigl\Vert u' \bigr\Vert _{0} \leq A_{3},\qquad \Vert v \Vert _{0} \leq A_{4}, $$
where
\(x=(u,v)^{\top}\)
is any solution to equation (3.1), \(\lambda\in(0,1]\).
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
$$ v'(\underline{t})\geq0. $$
(3.2)
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
$$ g\bigl(u(\underline{t}-\tau)\bigr)-e(\underline{t})\leq0. $$
(3.3)
In a similar way, we get
$$ g\bigl(u(\overline{t}-\tau)\bigr)-e(\overline{t})\geq0. $$
(3.4)
It follows from (H1), (3.3) and (3.4) that
$$\begin{aligned} u(\underline{t}-\tau)\geq d_{1} \quad\mbox{and}\quad u(\overline{t}-\tau)\leq d_{2}. \end{aligned}$$
Thus, we can see that there exists a point \(t_{0}\in[0,T]\) such that
$$ d_{1}\leq u(t_{0}) \leq d_{2}. $$
(3.5)
Multiplying the second equation of (3.1) by \(u'(t)\) and integrating over the interval \([0, T]\), we have
$$\begin{aligned} 0={}& \int_{0}^{T}v'(t)u'(t)\,dt= \lambda \int_{0}^{T}\frac{v(t)}{\sqrt {1-v^{2}(t)}}\cdot v'(t)\,dt \\ ={}&{-}\lambda \int_{0}^{T}f \biggl(t,\frac{u'(t)}{\lambda} \biggr)u'(t)\,dt-\lambda \int_{0}^{T}g\bigl(u(t-\tau)\bigr)u'(t)\,dt \\ &{}+\lambda \int_{0}^{T} e(t)u'(t)\,dt, \end{aligned}$$
which together with (H2) and (H3) gives
$$\begin{aligned} &a \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \\ &\quad \leq \lambda \int_{0}^{T} \bigl\vert g\bigl(u(t-\tau )\bigr) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt +\lambda \int_{0}^{T} \bigl\vert e(t) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad \leq\lambda\cdot \int_{0}^{T} \bigl(m_{0} \bigl\vert u(t-\tau) \bigr\vert +m_{1} \bigr) \bigl\vert u'(t) \bigr\vert \,dt +\lambda\cdot \Vert e \Vert _{0} \sqrt{T}\cdot \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \\ &\quad\leq\lambda\cdot m_{0} \biggl( \int_{0}^{T} \bigl\vert u(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} +\lambda\cdot m_{1} \sqrt{T} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \\ &\qquad{}+\lambda\cdot \Vert e \Vert _{0} \sqrt{T}\cdot \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2}. \end{aligned}$$
By using Lemma 2.2 and (3.5), we have
$$\begin{aligned} &a \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \\ &\quad\leq\lambda\cdot m_{0} \biggl[ \frac{T}{\pi} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{1/2}+\sqrt{T}d_{2} \biggr] \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \\ &\qquad{}+\lambda\cdot m_{1} \sqrt{T} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} +\lambda\cdot \Vert e \Vert _{0} \sqrt{T}\cdot \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \\ &\quad= \lambda\cdot\frac{ m_{0}T}{\pi} \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2}\,dt + \lambda\cdot m_{0} d_{2} \sqrt{T} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} \\ &\qquad{}+\lambda\cdot m_{1} \sqrt{T} \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2} +\lambda\cdot \Vert e \Vert _{0} \sqrt{T}\cdot \biggl( \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2} \,dt \biggr)^{1/2}, \end{aligned}$$
i.e.,
$$\begin{aligned} a \bigl\Vert u' \bigr\Vert ^{2}_{2} ={}& a \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \\ \leq{}& \frac{ m_{0}T}{\pi} \bigl\Vert u' \bigr\Vert ^{2}_{2} + m_{0}d_{2} \sqrt{T} \bigl\Vert u' \bigr\Vert _{2} + m_{1} \sqrt{T} \bigl\Vert u' \bigr\Vert _{2} + \Vert e \Vert _{0} \sqrt{T} \bigl\Vert u' \bigr\Vert _{2}. \end{aligned}$$
Since \(a\pi> m_{0}T \), we have
$$ \bigl\Vert u' \bigr\Vert _{2} \leq \frac{\pi\sqrt{T}(m_{0}d_{2}+m_{1}+ \Vert e \Vert _{0})}{a\pi -m_{0}T}:=A_{0}. $$
(3.6)
By means of the Hölder inequality, (3.5) and (3.6), we have
$$\begin{aligned} \Vert u \Vert _{0} =&\max _{t\in[0,T]} \bigl\vert u(t) \bigr\vert \leq\max _{t\in [0,T]} \biggl\vert u(t_{0})+ \int_{t_{0}}^{t}u'(s)\,ds \biggr\vert \\ \leq& d_{2}+ \int_{0}^{T} \bigl\vert u'(s) \bigr\vert \,ds\leq d_{2}+ \sqrt{T} \bigl\Vert u' \bigr\Vert _{2} \\ \leq& d_{2}+\frac{\pi T(m_{0}d_{2}+m_{1}+ \Vert e \Vert _{0})}{a\pi-m_{0}T}:= A_{1}. \end{aligned}$$
(3.7)
Clearly, \(A_{1}\) is independent of λ.
From the second equation of (3.1), we have
$$\begin{aligned} \int_{0}^{T} \bigl\vert v'(t) \bigr\vert \,dt \leq& \lambda \int_{0}^{T} \biggl\vert f \biggl(t, \frac{u'(t)}{\lambda} \biggr) \biggr\vert \,dt+\lambda \int_{0}^{T} \bigl\vert g\bigl(u(t-\tau)\bigr) \bigr\vert \,dt \\ &{}+\lambda \int_{0}^{T} \bigl\vert e(t) \bigr\vert \,dt. \end{aligned}$$
(3.8)
Furthermore, from (3.7) and (H2), we get
$$\begin{aligned} \int_{0}^{T} \bigl\vert g\bigl(u(t-\tau)\bigr) \bigr\vert \,dt \leq& \int_{0}^{T} \bigl[ m_{0}u(t- \tau)+m_{1} \bigr] \,dt \\ \leq& m_{0}T \cdot \Vert u \Vert _{0}+ m_{1}T \\ \leq& m_{0} A_{1}T+ m_{1}T. \end{aligned}$$
(3.9)
Substituting (3.9) into (3.8) and by using (H4), (3.6) and (3.7), we can obtain
$$\begin{aligned} \int_{0}^{T} \bigl\vert v'(t) \bigr\vert \,dt \leq&\lambda \int_{0}^{T} \biggl\vert f \biggl(t, \frac{u'(t)}{\lambda} \biggr) \biggr\vert \,dt+\lambda \int_{0}^{T} \bigl\vert g\bigl(u(t-\tau)\bigr) \bigr\vert \,dt +\lambda \int_{0}^{T} \bigl\vert e(t) \bigr\vert \,dt \\ \leq&\lambda \int_{0}^{T} \biggl( \beta \biggl\vert \frac{u'(t)}{\lambda } \biggr\vert +\gamma \biggr) \,dt + m_{0} A_{1}T+ m_{1}T + \Vert e \Vert _{0}T \\ \leq&\beta \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt+\gamma T + m_{0} A_{1}T+ m_{1}T + \Vert e \Vert _{0}T \\ \leq&\beta\sqrt{T} \bigl\Vert u' \bigr\Vert _{2}+ \gamma T + m_{0} A_{1}T+ m_{1}T + \Vert e \Vert _{0}T \\ \leq& \beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T. \end{aligned}$$
(3.10)
Integrating the first equation of (3.1) on the interval \([0,T]\), we have
$$\int_{0}^{T}u'(t)\,dt= \int_{0}^{T} \frac{v(t)}{\sqrt{1-v^{2}(t)}}\,dt=0. $$
Then we can see that there exists \(\eta\in[0,T]\) such that \(v(\eta )=0\). It implies that
$$\bigl\vert v(t) \bigr\vert = \biggl\vert \int_{\eta}^{t} v'(s)\,ds+v(\eta) \biggr\vert \leq \int _{0}^{T} \bigl\vert v'(s) \bigr\vert \,ds, $$
which together with (3.10) yields
$$\begin{aligned} \bigl\vert v(t) \bigr\vert \leq& \int_{0}^{T} \bigl\vert v'(s) \bigr\vert \,ds \\ \leq& \beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T :=A_{4}. \end{aligned}$$
(3.11)
Since \(\beta\sqrt{T} A_{0}+ ( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma)T<1\), we have
$$ \begin{aligned} \Vert v \Vert _{0}=\max _{t\in[0,T]} \bigl\vert v(t) \bigr\vert \leq A_{4}< 1. \end{aligned} $$
(3.12)
Clearly, \(A_{4}\) is independent of λ.
From the first equation of (3.1), we can see that
$$ \bigl\Vert u' \bigr\Vert _{0} \leq \lambda\max_{t\in[0,T]}\frac{ \vert v(t) \vert }{\sqrt {1-v^{2}(t)}}\leq \lambda \cdot\frac{A_{4}}{1-A_{4}^{2}} \leq \frac{A_{4}}{1-A_{4}^{2}}:=A_{3}. $$
(3.13)
Clearly, \(A_{3}\) is independent of λ.
In the following, we will prove that there exists a positive constant \(A_{2}\) which is dependent of λ such that
$$ u(t)\geq A_{2}. $$
(3.14)
Indeed, it follows from the second equation of (3.1) that
$$ v'(t+\tau)=-\lambda f \biggl(t+\tau,\frac{u'(t+\tau)}{\lambda} \biggr)- \lambda g\bigl(u(t)\bigr)+\lambda e(t+\tau). $$
(3.15)
Multiplying both sides of (3.15) by \(u'(t)\) and integrating on \([\xi, t]\), here \(\xi\in[0, T]\), we get
$$\begin{aligned} \lambda \int_{u(\xi)}^{u(t)}g_{0}(u)\,du={}& \lambda \int_{\xi }^{t}g_{0}\bigl(u(s) \bigr)u'(s)\,ds \\ ={}& - \int_{\xi}^{t}v'(t+\tau)u'(t)\,dt -\lambda \int_{\xi}^{t} f \biggl(t+\tau,\frac{u'(t+\tau)}{\lambda} \biggr) u'(t)\,dt \\ &{} +\lambda \int_{\xi}^{t} e(t+\tau)u'(t)\,dt; \end{aligned}$$
then
$$\begin{aligned} \lambda \biggl\vert \int_{u(\xi)}^{u(t)}g_{0}(u)\,du \biggr\vert =& \int_{0}^{T} \bigl\vert v'(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &{}+\lambda \int_{0}^{T} \biggl\vert f \biggl(t+\tau, \frac{u'(t+\tau )}{\lambda} \biggr) \biggr\vert \cdot \bigl\vert u'(t) \bigr\vert \,dt \\ &{}+\lambda \int_{0}^{T} \bigl\vert e(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt. \end{aligned}$$
(3.16)
Furthermore, by (3.10) and (3.13) we obtain
$$\begin{aligned} \int_{0}^{T} \bigl\vert v'(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \leq& \bigl\Vert u' \bigr\Vert _{0}\cdot \int _{0}^{T} \bigl\vert v'(t+\tau) \bigr\vert \,dt \\ \leq&\lambda \cdot\frac{A_{4}}{1-A_{4}^{2}} \cdot \bigl[\beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T \bigr]. \end{aligned}$$
(3.17)
By using (H4) and (3.13), we have
$$\begin{aligned} & \int_{0}^{T} \biggl\vert f \biggl(t+\tau, \frac{u'(t+\tau)}{\lambda} \biggr) \biggr\vert \cdot \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad\leq \int_{0}^{T} \biggl( \beta \biggl\vert \frac{u'(t+\tau)}{\lambda} \biggr\vert +\gamma \biggr) \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad\leq\beta \int_{0}^{T} \biggl\vert \frac{u'(t+\tau)}{\lambda} \biggr\vert \cdot \bigl\vert u'(t) \bigr\vert \,dt+\gamma \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad\leq\frac{\beta T}{\lambda} \cdot \bigl\Vert u' \bigr\Vert ^{2}_{0}+\gamma T\cdot \bigl\Vert u' \bigr\Vert _{0} \\ &\quad\leq\frac{\beta T}{\lambda} \cdot \biggl(\lambda \cdot\frac {A_{4}}{1-A_{4}^{2}} \biggr)^{2}+\gamma T\cdot\lambda \cdot\frac {A_{4}}{1-A_{4}^{2}} \\ &\quad\leq \beta T \cdot\lambda \cdot \biggl(\frac{A_{4}}{1-A_{4}^{2}} \biggr)^{2}+ \gamma T\cdot\lambda \cdot\frac{A_{4}}{1-A_{4}^{2}}. \end{aligned}$$
(3.18)
Substituting (3.17) and (3.18) into (3.16), we obtain
$$\begin{aligned} \lambda \biggl\vert \int_{u(\xi)}^{u(t)}g_{0}(u)\,du \biggr\vert \leq& \int_{0}^{T} \bigl\vert v'(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt +\lambda \int_{0}^{T} \biggl\vert f \biggl(t+\tau, \frac{u'(t+\tau )}{\lambda} \biggr) \biggr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &{}+\lambda \int_{0}^{T} \bigl\vert e(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ \leq&\lambda \cdot\frac{A_{4}}{1-A_{4}^{2}} \cdot \bigl[ \beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T \bigr] \\ &{}+\beta T \cdot\lambda^{2} \cdot \biggl(\frac{A_{4}}{1-A_{4}^{2}} \biggr)^{2}+\gamma T\cdot\lambda^{2} \cdot\frac{A_{4}}{1-A_{4}^{2}} + \lambda^{2} \cdot \Vert e \Vert _{0} \cdot \biggl( \frac{A_{4}}{1-A_{4}^{2}} \biggr) \end{aligned}$$
i.e.,
$$\begin{aligned} \biggl\vert \int_{u(\xi)}^{u(t)}g_{0}(u)\,du \biggr\vert \leq{}& \frac {A_{4}}{1-A_{4}^{2}} \cdot \bigl[\beta\sqrt{T} A_{0}+ \bigl( m_{0} A_{1}+ m_{1}+ \Vert e \Vert _{0}+\gamma\bigr)T \bigr] \\ &{}+\beta T \cdot \biggl(\frac{A_{4}}{1-A_{4}^{2}} \biggr)^{2} +\gamma T \cdot \frac{A_{4}}{1-A_{4}^{2}} + \Vert e \Vert _{0} \cdot \frac{A_{4}}{1-A_{4}^{2}}. \end{aligned}$$
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. □
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
$$0< E_{1}< \min\{d_{1}, A_{2}\},\qquad E_{2}> \max\{d_{2}, A_{1}\},\qquad E_{3}>A_{3}. $$
It follows from (3.5), (3.7), (3.13) and (3.14) that
$$ E_{1}< u(t)< E_{2},\qquad \bigl\Vert u' \bigr\Vert _{0}< E_{3}. $$
(3.19)
Then we can see that the condition (1) of Lemma 2.1 is satisfied.
For a possible solution C to the equation
$$g(C)-\frac{1}{T} \int_{0}^{T}e(t)\,dt=0, $$
it is easy to see that \(E_{1}< C < E_{2}\) is satisfied. Thus, the condition (2) of Lemma 2.1 is satisfied.
Finally, we prove that the condition (3) of Lemma 2.1 is also satisfied. In fact, from (H1), we have
$$g(E_{1})-\frac{1}{T} \int_{0}^{T}e(t)\,dt>0, $$
and
$$g(E_{2})-\frac{1}{T} \int_{0}^{T}e(t)\,dt< 0, $$
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. □