Theorem 3.1
Assume that
\([H_{1}]\)-\([H_{3}]\)
hold. Then equation (1.10) has at least one positive
T-periodic solution.
Proof
Firstly, we will show that there exist \(N_{1}>0\) and \(N_{2}>0\) such that each positive T-periodic solution \(x(t)\) of equation (2.3) satisfying
$$ x(t)< N_{1} \quad \mbox{and}\quad \bigl|x'(t)\bigr| < N_{2}\quad \mbox{for all } t\in [0,T]. $$
(3.1)
Suppose that x is an arbitrary positive T-periodic solution of equation (2.3). Then
$$ x''+\lambda f\bigl(x' \bigr)+\lambda \varphi (t)x-\frac{\lambda }{x^{\alpha }}= \lambda p(t),\quad \lambda \in (0,1]. $$
(3.2)
This implies that \(x\in \Omega \). So by Lemma 2.2 there exists a point \(t_{0}\in [0,T]\) such that
and then
$$ \vert x\vert _{\infty }\leq M_{0}+T^{\frac{n}{n+1}} \biggl( \int^{T}_{0}\bigl\vert x'(s)\bigr\vert ^{n+1}\,ds \biggr)^{\frac{1}{n+1}}. $$
(3.3)
Integrating (3.2) over the interval \([0,T]\), we get
$$ \int^{T}_{0}f\bigl(x'(t)\bigr)\,dt+ \int^{T}_{0}\varphi (t)x(t)\,dt- \int^{T}_{0}\frac{1}{x ^{\alpha }(t)}\,dt= \int^{T}_{0}p(t)\,dt. $$
(3.4)
On the other hand, similarly to the proof of (2.11), we have
$$\begin{aligned} \int^{T}_{0}\bigl\vert x'(t)\bigr\vert ^{n+1}\,dt \leq& \sigma^{-1}\vert x\vert _{\infty }\Vert \varphi \Vert _{ \frac{n+1}{n}} \biggl( \int^{T}_{0}\bigl\vert x'(t)\bigr\vert ^{n+1}\,dt \biggr)^{\frac{1}{n+1}} \\ &{}+\sigma^{-1}\Vert p\Vert _{\frac{n+1}{n}} \biggl( \int^{T}_{0}\bigl\vert x'(t)\bigr\vert ^{n+1}\,dt \biggr)^{\frac{1}{n+1}}. \end{aligned}$$
(3.5)
Substituting (3.3) into (3.5), we have
$$\begin{aligned}& \int^{T}_{0}\bigl\vert x'(t)\bigr\vert ^{n+1}\,dt \\& \quad \leq \sigma^{-1}\Vert \varphi \Vert _{\frac{n+1}{n}}T^{\frac{n}{n+1}} \biggl( \int^{T}_{0}\bigl\vert x'(t)\bigr\vert ^{n+1}\,dt \biggr)^{\frac{2}{n+1}} \\& \qquad {} +\bigl(\sigma^{-1}\Vert \varphi \Vert _{\frac{n+1}{n}}M_{0}+ \sigma^{-1}\Vert p\Vert _{ \frac{n+1}{n}} \bigr) \biggl( \int^{T}_{0}\bigl\vert x'(t)\bigr\vert ^{n+1}\,dt \biggr)^{ \frac{1}{n+1}}. \end{aligned}$$
(3.6)
According to (3.6), we list two cases.
- Case 1::
-
If \(n>1\), then there exists \(\rho_{0}>0\) such that \((\int ^{T}_{0}\vert x'(t)\vert ^{n+1}\,dt)^{\frac{1}{n+1}}\leq \rho_{0}\);
- Case 2::
-
If \(n=1\), then by assumption \([H_{3}]\) there exists \(\rho_{1}>0\) such that \((\int^{T}_{0}\vert x'(t)\vert ^{2}\,dt)^{ \frac{1}{2}}\leq \rho_{1}\).
Letting \(\rho =\max \{\rho_{0},\rho_{1}\}\), it follows from Case 1 or Case 2 that
$$ \biggl( \int^{T}_{0}\bigl\vert x'(t)\bigr\vert ^{n+1}\,dt \biggr)^{\frac{1}{n+1}}\leq \rho, $$
(3.7)
and according to (3.3), we have
$$ x(t)\leq M_{0}+T^{\frac{n}{n+1}}\rho:=N_{1} \quad \mbox{for all } t\in [0,T]. $$
(3.8)
Clearly, there is a point \(t_{2}\in [0,T]\) such that \(x'(t_{2})=0\). Multiplying both sides of (3.2) by \(x'(t)\) and integrating it over the interval \([t_{2},t]\), we get
$$\begin{aligned}& \int^{t}_{t_{2}}x''(t)x'(t)\,dt \\& \quad =\lambda \int^{t}_{t_{2}}\biggl[-f\bigl(x'(t) \bigr)x'(t)-\varphi (t)x(t)x'(t)+\frac{x'(t)}{x ^{\alpha }(t)}+p(t)x'(t) \biggr]\,dt \\& \qquad \mbox{for all } t\in [t_{2},t_{2}+T], \end{aligned}$$
and then
$$\begin{aligned} \frac{\vert x'(t)\vert ^{2}}{2} \leq& \lambda \bigl\vert x'\bigr\vert _{\infty } \biggl[\vert x\vert _{\infty } \int^{t_{2}+T}_{t_{2}}\bigl\vert \varphi (t)\bigr\vert \,dt+ \int^{t_{2}+T}_{t_{2}}\frac{1}{x ^{\alpha }(t)}\,dt+ \int^{t_{2}+T}_{t_{2}}\bigl\vert p(t)\bigr\vert \,dt \biggr] \\ =&\lambda \bigl\vert x'\bigr\vert _{\infty } \biggl[ \vert x\vert _{\infty } \int^{T}_{0}\bigl\vert \varphi (t)\bigr\vert \,dt+ \int^{T}_{0}\frac{1}{x^{\alpha }(t)}\,dt+ \int^{T}_{0}\bigl\vert p(t)\bigr\vert \,dt \biggr] \\ =&\lambda \bigl\vert x'\bigr\vert _{\infty } \biggl[N_{1}T\overline{\vert \varphi \vert }+ \int^{T} _{0}\frac{1}{x^{\alpha }(t)}\,dt+T\overline{\vert p\vert } \biggr]\quad \mbox{for all } t \in [t_{2},t_{2}+T]. \end{aligned}$$
(3.9)
Since
$$\bigl\vert x'\bigr\vert _{\infty }=\max _{t\in [0,T]}\bigl\vert x'(t)\bigr\vert =\max _{t\in [t_{2},t_{2}+T]}\bigl\vert x'(t)\bigr\vert , $$
it follows from (3.9) that
$$\frac{\vert x'\vert ^{2}_{\infty }}{2}\leq \lambda \bigl\vert x'\bigr\vert _{\infty } \biggl[N_{1}T\overline{\vert \varphi \vert }+ \int^{T}_{0}\frac{1}{x^{\alpha }(t)}\,dt+T\overline{\vert p \vert } \biggr], $$
that is,
$$\frac{\vert x'\vert _{\infty }}{2}\leq \lambda \biggl[N_{1}T\overline{\vert \varphi \vert }+ \int^{T}_{0}\frac{1}{x^{\alpha }(t)}\,dt+T\overline{\vert p \vert } \biggr], $$
which implies that
$$ \frac{\vert x'(t)\vert }{2}\le \frac{\vert x'\vert _{\infty }}{2}\le \lambda \biggl[N_{1}T\overline{\vert \varphi \vert }+ \int^{T}_{0}\frac{1}{x^{\alpha }(t)}\,dt+T\overline{\vert p \vert } \biggr]\quad \mbox{for all } t\in [0,T]. $$
(3.10)
On the other hand, from (3.4) and condition (2.1) in \([H_{1}]\) we have
$$\begin{aligned} \int^{T}_{0}\frac{1}{x^{\alpha }(t)}\,dt =& \int^{T}_{0}f\bigl(x'(t)\bigr)\,dt+ \int ^{T}_{0}\varphi (t)x(t)\,dt- \int^{T}_{0}p(t)\,dt \\ \leq& L \int^{T}_{0}\bigl\vert x'(t)\bigr\vert \,dt+N_{1}T\overline{\vert \varphi \vert }+T \overline{\vert p\vert } \\ \leq& L\rho T^{\frac{n}{n+1}}+N_{1}T\overline{\vert \varphi \vert }+T \overline{\vert p\vert }, \end{aligned}$$
where ρ is determined in (3.7). Substituting this formula into (3.10), we obtain
$$ \bigl\vert x'(t)\bigr\vert \leq \lambda \bigl[2L \rho T^{\frac{n}{n+1}}+4N_{1}T\overline{\vert \varphi \vert }+4T \overline{\vert p\vert } \bigr]:=\lambda N_{2}\quad \mbox{for all } t \in [0,T]. $$
(3.11)
So we have
$$ \bigl\vert x'(t)\bigr\vert \leq N_{2} \quad \mbox{for all } t\in [0,T]. $$
(3.12)
We further show that there exists a constant \(\gamma_{0}\in (0,\gamma)\) such that each positive T= periodic solution of (2.3) satisfies
$$ x(t)>\gamma_{0}\quad \mbox{for all } t\in [0,T]. $$
(3.13)
In fact, suppose that \(x(t)\) is an arbitrary positive T-periodic solution of (2.3). Then
$$ x''+\lambda f\bigl(x' \bigr)+\lambda \varphi (t)x-\frac{\lambda }{x^{\alpha }}= \lambda p(t), \quad \lambda \in (0,1]. $$
(3.14)
By Lemma 2.3 we see that there is a point \(t_{1}\in [0,T]\) such that
$$x(t_{1})\geq \gamma. $$
For \(t\in [t_{1},t_{1}+T]\), multiplying both sides of (3.14) with \(x'(t)\) and integrating it over the interval \([t_{1},t]\) (or \([t,t_{1}]\)), we get
$$\frac{\vert x'(t)\vert ^{2}}{2}-\frac{\vert x'(t_{1})\vert ^{2}}{2}+\lambda \int^{t}_{t _{1}}f\bigl(x' \bigr)x'\,dt=\lambda \int^{t}_{t_{1}}\frac{1}{x^{\alpha }}x'\,dt - \lambda \int^{t}_{t_{1}}\varphi (t)xx'\,dt+\lambda \int^{t}_{t_{1}}p(t)x'\,dt, $$
which results in
$$\begin{aligned}& \lambda \int^{x(t)}_{x(t_{1})}\frac{1}{s^{\alpha }}\,ds \\& \quad = \frac{\vert x'(t)\vert ^{2}}{2}-\frac{\vert x'(t_{1})\vert ^{2}}{2}+\lambda \int^{t} _{t_{1}}f\bigl(x'(s) \bigr)x'(s)\,ds +\lambda \int^{t}_{t_{1}}\varphi (s)x(s)x'(s)\,ds- \lambda \int^{t}_{t_{1}}p(s)x'(s)\,ds, \end{aligned}$$
that is,
$$\begin{aligned} \lambda \int^{x(t_{1})}_{x(t)}\frac{1}{s^{\alpha }}\,ds = &- \frac{\vert x'(t)\vert ^{2}}{2}+\frac{\vert x'(t_{1})\vert ^{2}}{2}-\lambda \int^{t}_{t _{1}}f\bigl(x'(s) \bigr)x'(s)\,ds \\ &{} -\lambda \int^{t}_{t_{1}}\varphi (s)x(s)x'(s)\,ds+ \lambda \int^{t} _{t_{1}}p(s)x'(s)\,ds. \end{aligned}$$
According to (2.2) in \([H_{1}]\), we get \(\int^{t}_{t_{1}}f(x'(s))x'(s)\,ds \ge 0\). Thus, it follows from the last formula that
$$\begin{aligned} \lambda \int^{x(t_{1})}_{x(t)}\frac{1}{s^{\alpha }}\,ds \le& - \frac{\vert x'(t)\vert ^{2}}{2}+\frac{\vert x'(t_{1})\vert ^{2}}{2}-\lambda \int^{t}_{t _{1}}\varphi (s)x(s)x'(s)\,ds+ \lambda \int^{t}_{t_{1}}p(s)x'(s)\,ds \\ \le& \bigl\vert x'\bigr\vert _{\infty }^{2}+ \lambda \int_{0}^{T}\bigl\vert \varphi (s)x(s)x'(s)\bigr\vert \,ds+ \lambda \int^{T}_{0}\bigl\vert p(s)x'(s)\bigr\vert \,ds, \end{aligned}$$
which, together with (3.8) and (3.11), yields
$$\lambda \int^{x(t_{1})}_{x(t)}\frac{1}{s^{\alpha }}\,ds\le \lambda^{2} N ^{2}_{2}+\lambda^{2} N_{1}N_{2}T\overline{\vert \varphi \vert }+ \lambda^{2} N _{2}T\overline{\vert p\vert }, $$
that is,
$$ \int^{x(t_{1})}_{x(t)}\frac{1}{s^{\alpha }}\,ds\leq N^{2}_{2}+ N_{1}N _{2}T\overline{ \vert \varphi \vert }+ N_{2}T\overline{\vert p\vert }:=N_{3}. $$
(3.15)
Since \(\alpha \ge 1\), it follows that there exists \(\gamma_{0}\in (0, \gamma)\) such that
$$\int^{\gamma }_{\eta }\frac{1}{x^{\alpha }(t)}\,dt>N_{3} \quad \mbox{for all } \eta \in (0,\gamma_{0}), $$
which, together with (3.15), implies that
$$x(t)>\gamma_{0}\quad \mbox{for all } t\in [0,T]. $$
So (3.13) holds.
Let \(n_{0}=\min \{D_{1},\gamma_{0}\}\) and \(n_{1}\in (N_{1}+D_{2},+ \infty)\) be two constants. Then from (3.8), (3.12), and (3.13) we see that each possible positive T-periodic solution x to (2.3) satisfies
$$n_{0}< x(t)< n_{1},\qquad \bigl\vert x'(t)\bigr\vert < N_{2}. $$
This implies that condition 1 and condition 2 of Lemma 2.1 hold. In addition, from Remark 2.1 we can infer that
$$\frac{1}{c^{\alpha }}-c\overline{\varphi }+\overline{p}>0 \quad \mbox{for } c \in (0,n_{0}] $$
and
$$\frac{1}{c^{\alpha }}-c\overline{\varphi }+\overline{p}< 0 \quad \mbox{for } c \in [n_{1},+\infty), $$
which results in
$$\biggl(\frac{1}{n^{\alpha }_{0}}-n_{0}\overline{\varphi }+\overline{p} \biggr) \biggl(\frac{1}{n^{\alpha }_{1}}-n_{1}\overline{\varphi }+ \overline{p} \biggr)< 0. $$
Therefore, condition 3 of Lemma 2.1 holds. Thus, by Lemma 2.1 we see that equation (1.10) has at least one positive T-periodic solution. The proof is complete. □
Example 3.1
Consider the equation
$$ x''(t)+10x'(t)- \frac{(x'(t))^{3}}{1+(x'(t))^{2}}+a(1+2\sin t)x(t)-\frac{1}{x ^{2}(t)}=\cos t, $$
(3.16)
where \(a\in (0,\infty)\). Corresponding to (1.10), we see that \(f(x)=10x-\frac{x^{3}}{1+x^{2}}\), \(\varphi (t)=a(1+2\sin t)\), \(p(t)=\cos t\), and \(T=2\pi \).
Firstly, from (3.16) we see that \(f(0)=0\) and
$$\overline{\varphi_{+}}=\frac{1}{T} \int_{0}^{T}\varphi_{+}(t)\,dt= \frac{\frac{2 \pi }{3}+\sqrt{3}}{\pi }a,\qquad \overline{\varphi_{-}}=\frac{1}{T}\varphi _{-}(t)\,dt=\frac{-\frac{\pi }{3}+\sqrt{3}}{\pi }a. $$
Obviously, \([H_{2}]\) is satisfied. Secondly, integrating \(f(x')\) over the internal \([0,T]\), we get
$$\begin{aligned} \biggl\vert \int^{T}_{0}f\bigl(x'\bigr)\,dt \biggr\vert =& \biggl\vert \int^{T}_{0}\biggl[10x'(t)- \frac{(x'(t))^{3}}{1+(x'(t))^{2}}\biggr]\,dt \biggr\vert \\ =& \biggl\vert - \int^{T}_{0}\frac{(x'(t))^{3}}{1+(x'(t))^{2}}\,dt \biggr\vert \\ =& \biggl\vert \int^{T}_{0}\frac{\vert x'(t)\vert ^{3}}{1+(x'(t))^{2}}\,dt \biggr\vert \\ \leq& \int^{T}_{0}\bigl\vert x'(t)\bigr\vert \,dt, \end{aligned}$$
which implies that we can chose \(L=1\) such that assumption \([H_{1}]\) holds. Besides, from
$$yf(y)=10y^{2}-\frac{y^{4}}{1+y^{2}} \geq 9y^{2} $$
we see that the constant σ can be chosen as \(\sigma =9\) such that assumption \([H_{1}]\) is satisfied. Last, let \(L=1\), \(\sigma =9\), \(n=1\). Then we get
$$\begin{aligned}& 1-\frac{LT^{\frac{-1}{2}}+T^{\frac{1}{2}}\overline{\varphi_{+}}}{ \sigma (\overline{\varphi_{+}}-\overline{\varphi_{-}})} \Vert \varphi \Vert _{2}=1-\frac{ \sqrt{3}}{9}-\frac{18+4\sqrt{3}\pi }{27}a>0, \\& 1-\sigma^{-1}\Vert \varphi \Vert _{2}T^{\frac{1}{2}}=1- \frac{2\pi }{3 \sqrt{3}}a>0. \end{aligned}$$
If
$$a< \frac{27-3\sqrt{3}}{18+4\sqrt{3}\pi }, $$
then \([H_{3}]\) holds. Thus, by Theorem 3.1 we have that equation (3.16) has at least one positive 2π-periodic solution.