Theorem 1
Assume that [H1]-[H3] hold. Then equation (1.1) has at leat one positive
T-periodic solution.
Proof
Firstly, we will show that there exist \(M_{1},M_{2}\) with \(M_{1}>k^{*}M\) and \(M_{2}>0\) such that each positive T-periodic solution \(u(t)\) of equation (2.1) satisfies the inequalities
$$ u(t)< M_{1},\qquad \bigl|u'(t)\bigr|< M_{2}, \quad\mbox{for all } t\in[0,T]. $$
(3.1)
In fact, if u is an arbitrary positive T-periodic solution of equation (2.1), then
$$ u''+\lambda f(u)u'-\lambda g(u)+\lambda\varphi(t)u=\lambda h(t),\quad \lambda\in(0,1]. $$
(3.2)
This implies \(u\in \varOmega \). So by using Lemma 3 that there is a point \(t_{0}\in[0,T]\) such that
$$ u(t_{0})\le k^{*}M, $$
(3.3)
and then
$$ |u|_{\infty}\le k^{*}M+T^{1/2} \biggl( \int _{0}^{T}\bigl|u'(s)\bigr|^{2}\,ds \biggr)^{1/2}. $$
(3.4)
Integrating (3.2) over the interval \([0,T]\), we have
$$ - \int_{0}^{T} g\bigl(u(t)\bigr)\,dt+ \int_{0}^{T}\varphi (t)u(t)\,dt= \int_{0}^{T}h(t)\,dt. $$
(3.5)
Since \(g(x)\rightarrow+\infty\) as \(x\rightarrow0^{+}\), we see from (3.5) that there is a point \(t_{1}\in[0,T]\) such that
$$ u(t_{1})\ge\gamma, $$
(3.6)
where \(\gamma< k^{*}M\) is a positive constant, which is independent of \(\lambda\in(0,1]\). Similar to the proof of (2.9), we have
$$\begin{aligned} & \biggl( \int_{0}^{T}\bigl|u'(t)\bigr|^{2}\,dt \biggr)^{1/2} \\ &\quad< |\varphi _{+}|_{\infty}^{1/2} \biggl( \int_{0}^{T}\bigl|u(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{2}}+ \biggl( \int_{0}^{T}\bigl|u(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{4}} \biggl( \int_{0}^{T}\bigl|h_{-}(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{4}}. \end{aligned}$$
(3.7)
By using Lemma 2, we have
$$ \biggl( \int_{0}^{T}\bigl|u(s)\bigr|^{2}\,ds \biggr)^{1/2}\le \frac{T}{\pi} \biggl( \int_{0}^{T}\bigl\vert u'(s)\bigr\vert ^{2}\,ds \biggr)^{1/2}+\sqrt{T}\bigl\vert u(t_{0})\bigr\vert , $$
(3.8)
where \(t_{0}\) is determined in (3.3). Substituting (3.7) into (3.8), we have
$$\begin{aligned} & \biggl( \int_{0}^{T}\bigl|u(t)\bigr|^{2}\,dt \biggr)^{1/2} \\ &\quad< \frac{T}{\pi} \biggl[|\varphi_{+}|_{\infty}^{1/2} \biggl( \int _{0}^{T}\bigl|u(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{2}}+ \biggl( \int_{0}^{T}\bigl|u(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{4}} \biggl( \int_{0}^{T}\bigl|h_{-}(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{4}} \biggr] \\ &\qquad{} +T^{\frac{1}{2}}k^{*}M \\ &\quad=\frac{T}{\pi}|\varphi_{+}|_{\infty}^{1/2} \biggl( \int_{0}^{T}\bigl|u(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{2}}+\frac{T}{\pi} \biggl( \int_{0}^{T}\bigl|h_{-}(t)\bigr|^{2}\,dt \biggr)^{\frac {1}{4}} \biggl( \int_{0}^{T}\bigl|u(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{4}}+ T^{\frac{1}{2}}k^{*}M, \end{aligned}$$
which results in
$$\begin{aligned} & \biggl(1- \frac{T}{\pi}|\varphi_{+}|_{\infty }^{1/2} \biggr) \biggl( \int_{0}^{T}\bigl|u(t)\bigr|^{2}\,dt \biggr)^{1/2} \\ &\quad< \frac{T}{\pi} \biggl( \int_{0}^{T}\bigl|h_{-}(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{4}} \biggl( \int_{0}^{T}\bigl|u(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{4}}+ T^{\frac{1}{2}}k^{*}M. \end{aligned}$$
(3.9)
Since \(\frac{T}{\pi}|\varphi_{+}|_{\infty}^{1/2}<\sigma_{1}\in(0,1)\), it follows from (3.9) that there is a constant \(\rho>0\), which is independent of \(\lambda\in(0,1]\), such that
$$\biggl( \int_{0}^{T}\bigl|u(t)\bigr|^{2}\,dt \biggr)^{1/2}< \rho, $$
and then by (3.7), we have
$$\biggl( \int_{0}^{T}\bigl|u'(t)\bigr|^{2}\,dt \biggr)^{1/2}< |\varphi_{+}|_{\infty}^{1/2}\rho + \biggl( \int_{0}^{T}\bigl|h_{-}(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{4}}\rho^{1/2}. $$
It follows from (3.4) that
$$|u|_{\infty}< k^{*}M+T^{1/2}|\varphi_{+}|_{\infty}^{1/2} \rho+(T\rho )^{1/2} \biggl( \int_{0}^{T}\bigl|h_{-}(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{4}}:=M_{1}, $$
i.e.,
$$ u(t)< M_{1}, \quad\mbox{for all } t\in[0,T]. $$
(3.10)
Now, if u attains its maximum over \([0,T]\) at \(t_{2}\in[0,T]\), then \(u'(t_{2})=0\) and we deduce from (3.2) that
$$u'(t)=\lambda \int_{t_{2}}^{t}\bigl[-f(u)u'+g(u)- \varphi(t)u+h(t)\bigr]\,dt $$
for all \(t\in[t_{2},t_{2}+T]\). Thus, if \(F'=f\), then
$$\begin{aligned} \bigl\vert u'(t)\bigr\vert \leq&\lambda\bigl|F \bigl(u(t)\bigr)-F\bigl(u(t_{2})\bigr)\bigr|+\lambda \int _{t_{2}}^{t_{2}+T}g\bigl(u(t)\bigr)\,dt \\ &{}+\lambda \int_{t_{2}}^{t_{2}+T}\bigl|\varphi (s)\bigr|u(s)\,ds+\lambda \int_{t_{2}}^{t_{2}+T}\bigl|h(s)\bigr|\,ds \\ \leq& 2\lambda\max_{0\leq u\leq M_{1}}\bigl|F(u)\bigr|+\lambda \int _{0}^{T}g\bigl(u(s)\bigr)\,ds+\lambda T \overline{|\varphi|}|u|_{\infty}+\lambda T\overline{|h|}. \end{aligned}$$
(3.11)
From (3.2), we see that
$$\begin{aligned} \int_{0}^{T}g\bigl(u(s)\bigr)\,ds =& \int_{0}^{T}\varphi(t)u(t)\,dt-T\bar {h} \\ \le&T\overline{\varphi_{+}}|u|_{\infty}+T\overline{h_{-}}. \end{aligned}$$
It follows from (3.10) and (3.11) that
$$\begin{aligned} &\bigl\vert u'(t)\bigr\vert \leq2\lambda \Bigl( \max_{0\leq u\leq M_{1}}\bigl\vert F(u)\bigr\vert + T\overline{| \varphi|}|u|_{\infty}+T\overline{|h|} \Bigr) \\ &\quad< 2\lambda \Bigl( \max_{0\leq u\leq M_{1}}\bigl|F(u)\bigr|+ M_{1}T \overline{|\varphi |}+T\overline{|h|} \Bigr) \\ &\quad:=\lambda M_{2},\quad t\in[0,T], \end{aligned}$$
(3.12)
and then
$$ \bigl\vert u'(t)\bigr\vert < M_{2}, \quad\mbox{for all } t\in[0,T]. $$
(3.13)
Equations (3.10) and (3.13) imply that (3.1) holds.
Below, we will show that there exists a constant \(\gamma_{0}\in (0,\gamma)\), such that each positive T-periodic solution of equation (2.1) satisfies
$$ u(t)>\gamma_{0} \quad\mbox{for all } t\in[0,T]. $$
(3.14)
Suppose that \(u(t)\) is an arbitrary positive T-periodic solution of equation (2.1), then
$$ u''+\lambda f(u)u'-\lambda g(u)+\lambda\varphi(t)u=\lambda h(t),\quad \lambda\in(0,1]. $$
(3.15)
Let \(t_{1}\) be determined in (3.6). Multiplying (3.15) by \(u'(t)\) and integrating it over the interval \([t_{1},t]\) (or \([t,t_{1}]\)), we get
$$\frac{|u'(t)|^{2}}{2}-\frac{|u'(t_{1})|^{2}}{2}+\lambda \int _{t_{1}}^{t}f(u) \bigl(u' \bigr)^{2}\,dt=\lambda \int_{t_{1}}^{t}g(u)u'\,dt-\lambda \int _{t_{1}}^{t}\varphi(t)uu'\,dt+\lambda \int_{t_{1}}^{t} h(t)u'\,dt, $$
which yields the estimate
$$\begin{aligned} \lambda\biggl\vert \int_{u(t)}^{u(t_{1})}g(s)\,ds\biggr\vert \leq{}& \frac {|u'(t)|^{2}}{2}+\frac{|u'(t_{1})|^{2}}{2}+\lambda \int _{0}^{T}\bigl\vert f(u)\bigr\vert \bigl(u'\bigr)^{2}\,dt \\ &{}+\lambda \int_{0}^{T}\bigl\vert \varphi (t)uu' \bigr\vert \,dt+\lambda \int_{0}^{T} \bigl\vert h(t)u'\bigr\vert \,dt. \end{aligned}$$
From (3.10) and (3.12), we get
$$\lambda\biggl\vert \int_{u(t)}^{u(t_{1})}g(s)\,ds\biggr\vert \leq\lambda M_{2}^{2}+\lambda\max_{0\leq u\leq M_{1}}\bigl\vert f(u)\bigr\vert TM_{2}^{2}+\lambda M_{1}M_{2}T \overline{|\varphi|}+\lambda M_{2}T\overline{|h|}, $$
which gives
$$ \biggl\vert \int_{u(t)}^{u(t_{1})}g(s)\,ds\biggr\vert \leq M_{3}, \quad\mbox{for all } t\in[t_{1},t_{1}+T], $$
(3.16)
with
$$M_{3}= M_{2}^{2}+\max_{0\leq u\leq M_{1}} \bigl\vert f(u)\bigr\vert TM_{2}^{2}+ M_{1}M_{2}T \overline {|\varphi|}+ M_{2}T\overline{|h|}. $$
From [H3] there exists \(\gamma_{0}\in(0,\gamma)\) such that
$$ \int_{\eta}^{\gamma}g(u)\,du>M_{3}, \quad\mbox{for all } \eta\in(0,\gamma_{0}]. $$
(3.17)
Therefore, if there is a \(t^{*}\in[t_{1},t_{1}+T]\) such that \(u(t^{*})\le\gamma _{0}\), then from (3.17) we get
$$\int_{u(t^{*})}^{\gamma}g(s)\,ds> M_{3}, $$
which contradicts (3.16). This contradiction gives that \(u(t)>\gamma_{0}\) for all \(t\in[0,T]\). So (3.14) holds. Let \(m_{0}=\min\{D_{1},\gamma_{0}\}\) and \(m_{1}\in(M_{1}+D_{2}, +\infty)\) be two constants, then from (3.1) and (3.14), we see that each possible positive T-periodic solution u to equation (2.1) satisfies
$$m_{0}< u(t)< m_{1},\quad \bigl\vert u'(t) \bigr\vert < M_{2}. $$
This implies that condition 1 and condition 2 of Lemma 1 are satisfied. Also, we can deduce from Remark 1 that
$$g(c)-\bar{\varphi}c+\bar{h}>0, \quad\mbox{for } c\in(0, m_{0}] $$
and
$$g(c)-\bar{\varphi}c+\bar{h}< 0, \quad\mbox{for } c\in[ m_{1},+\infty), $$
which results in
$$\bigl(g(m_{0})-\bar{\varphi}m_{0}+\bar{h} \bigr) \bigl(g(m_{1})-\bar {\varphi}m_{1}+\bar{h} \bigr)< 0. $$
So condition 3 of Lemma 1 holds. By using Lemma 1, we see that equation (1.1) has at least one positive T-periodic solution. The proof is complete. □
Let us consider the equation
$$ x''+f(x)x'- \frac{1}{x^{\gamma}}+\varphi(t)x=h(t), $$
(3.18)
where \(f : [0,+\infty)\rightarrow R\) is an arbitrary continuous function, \(\varphi,h: R\rightarrow R\) are T-periodic functions with \(h\in L^{1}([0,T], R)\) and \(\varphi\in C([0,T], R)\), and the sign of the function φ is allowed to change for \(t\in[0,T]\), \(\gamma \ge1\) is a constant. Corresponding to equation (1.1), \(g(x)=\frac{1}{x^{\gamma}}\). For this case, \(g(x)\rightarrow+\infty\) as \(x\rightarrow0^{+}\), and assumptions [H2]-[H3] are satisfied. Thus, by using Theorem 1, we have the following results.
Corollary 1
Assume that the function
\(\varphi(t)\)
satisfies the following conditions:
$$\int_{0}^{T}\varphi_{+}(s)\,ds>0, \qquad \sigma:= \frac{\int_{0}^{T}\varphi_{-}(s)\,ds}{\int _{0}^{T}\varphi_{+}(s)\,ds}\in[0,1) $$
and
$$\sigma_{1}:=\frac{T}{\pi}|\varphi_{+}|_{\infty}^{1/2}+ \frac{T^{\frac {1}{2}} (\int_{0}^{T}\varphi_{-}(s)^{2}\,ds )^{\frac{1}{2}}}{\int _{0}^{T}\varphi_{+}(s)\,ds}\in(0,1). $$
Then, equation (3.18) possesses at least one positive
T-periodic solution.
Remark 2
Corresponding to equation (1.4) and equation (1.5), the function \(g(t,x)\) associated to equation (3.18) can be regarded as
$$ g(t,u)=-\frac{1}{u^{\gamma}}+\varphi(t)u-h(t),\quad (t,u)\in[0,T] \times(0,+\infty). $$
(3.19)
For the case of \(\varphi(t)\ge0\) for all \(t\in[0,T]\), we see that if x is a positive T-periodic continuous function satisfying \(\int _{0}^{T} g(t,x(t))\,dt=0\), then
$$ \int_{0}^{T}\frac{1}{x^{\gamma}(t)}\,dt= \int _{0}^{T}\varphi(t)x(t)\,dt- \int_{0}^{T}h(t)\,dt. $$
(3.20)
By applying the integral mean value theorem to the term \(\int _{0}^{T}\varphi(t)x(t)\,dt\) in equation (3.20), one can easily verify that \(g(t,u)\) determined in (3.19) satisfies the balance condition (h1). However, if the sign of the function \(\varphi(t)\) is changeable for \(t\in[0,T]\), then it is unclear from (3.20) whether the balance condition (h1) is satisfied. For this case, the main results of [18, 19] cannot be applied to equation (3.18).
Corollary 2
Assume that the function
\(\varphi(t)\)
satisfies
\(\varphi(t)\ge0\)
for all
\(t\in[0,T]\)
with
\(\int_{0}^{T}\varphi(s)\,ds>0\), and
$$|\varphi|_{\infty}< \biggl(\frac{\pi}{T} \biggr)^{2}. $$
Then, equation (3.18) possesses at least one positive
T-periodic solution.
Example 1
Consider the following equation:
$$ x''(t)+f\bigl(x(t)\bigr)x'(t)- \frac {1}{x^{2}(t)}+a(1+2\sin2t)x(t)=\cos2t, $$
(3.21)
where f is an arbitrary continuous function, \(a\in(0,+\infty)\) is a constant. Corresponding to equation (3.18), we have \(\gamma=2\), \(\varphi(t)=a(1+2\sin2t)\) and \(h(t)=\cos2t\), \(T=\pi\). By simply calculating, we can verify that
$$\begin{aligned}& \int_{0}^{T}\varphi_{+}(t)\,dt=\biggl( \frac{2\pi}{3}+\frac {3}{2}\biggr)a,\qquad \int_{0}^{T}\varphi_{-}(t)\,dt=\biggl( \frac{3}{2}-\frac{\pi}{3}\biggr)a, \\& \int_{0}^{T}\bigl(\varphi_{-}(t)\bigr)^{2}\,dt= \frac{3\pi a}{2}, \end{aligned}$$
and then
$$\sigma:=\frac{\int_{0}^{T}\varphi_{-}(s)\,ds}{\int_{0}^{T}\varphi_{+}(s)\,ds}=\frac {9-2\pi}{4\pi+9}\in(0,1) $$
and
$$\sigma_{1}:=\frac{T}{\pi}|\varphi_{+}|_{\infty}^{1/2}+ \frac{T^{\frac {1}{2}} (\int_{0}^{T}\varphi_{-}(s)^{2}\,ds )^{\frac{1}{2}}}{\int _{0}^{T}\varphi_{+}(s)\,ds}=\sqrt{3a}+\frac{3\pi\sqrt{6}}{4\pi+9}. $$
Thus, if \(0< a<\frac{1}{3} (\frac{4\pi+9-3\pi\sqrt{6}}{4\pi+9} )^{2}\), then \(\sigma_{1}\in(0,1)\). By using Corollary 1, we see that equation (3.21) has at least one positive π-periodic solution.
Remark 3
Since the sign of \(\varphi(t)=1+2\sin t\) is changed for \(t\in[0,T]\), whether the right inequality of (1.6) in the balance condition (h1) is satisfied remains unclear. So the conclusion of the example cannot be obtained by using the main results in [18, 19].