Theorem 3.1
If (H1) holds, then equation (1.6) has a positive
T-periodic solution if and only if
\(\overline{h}<0\).
Proof
Suppose that equation (1.6) has a positive T-periodic solution \(y(t)\), then
$$ y''+f(y)y'- \frac{\alpha(t)}{y^{\mu}}=h(t). $$
(3.1)
Integrating (3.1) on the interval \([0,T]\), and by using
$$\int_{0}^{T}y''(s)\,ds= \int_{0}^{T}f\bigl(y(s)\bigr)y'(s) \,ds=0, $$
we have
$$\int_{0}^{T} \frac{\alpha(s)}{y^{\mu}(s)}\,ds=-T\overline {h}, $$
which together with the assumption of \(\alpha(t)\ge0\) and \(y(t)>0\) for all \(t\in[0,T]\) gives a necessary condition for the existence of a positive T-periodic solution of equation (1.6): \(\overline {h}<0\).
Below, we will show the proof of sufficiency. In order to do it, suppose that \(\overline {h}<0\), and let u be an arbitrary positive T-periodic solution of (2.1). Then
$$ u''+\lambda f(u)u'-\lambda \frac{\alpha(t)}{u^{\mu}}=\lambda h(t),\quad \lambda\in(0,1]. $$
(3.2)
Integrating (3.2) over the interval \([0,T]\), we have
$$ \int^{T}_{0}\frac{\alpha(t)}{u^{\mu}}\,dt=- \int^{T}_{0} h(t)\,dt=-T\overline{h}. $$
(3.3)
Due to the fact that \(\alpha(t)\) is non-negative, \(\frac{1}{u_{M}^{\mu }}\int_{0}^{T}\alpha(t)\,dt\leq\int^{T}_{0}\frac{\alpha(t)}{u^{\mu}(t)}\,dt\leq\frac{1}{u_{m}^{\mu}}\int_{0}^{T}\alpha(t)\,dt\), where \(u_{m}\), \(u_{M}\) are the global minimum and maximum, respectively, of u. Then there is a point \(\eta\in[0,T]\) such that
$$ \frac{1}{u^{\mu}(\eta)} \int^{T}_{0}\alpha(t)\,dt=-T\overline{h}, $$
which results in
$$ T\overline{\alpha}=-u^{\mu}(\eta)T\overline{h}, $$
and then
$$ u(\eta)=\biggl(-\frac{\overline{\alpha}}{\overline{h}}\biggr)^{\frac{1}{\mu}}. $$
(3.4)
Multiplying (3.2) with \(u(t)\), and integrating it over the interval \([0,T]\), we obtain
$$ \int^{T}_{0}u''(t)u(t) \,dt-\lambda \int^{T}_{0}\frac{\alpha(t)u(t)}{u^{\mu }(t)}\,dt=\lambda \int^{T}_{0}h(t)u(t)\,dt. $$
By using \(\int^{T}_{0}u''(t)u(t)\,dt=-\int_{0}^{T}|u'(t)|^{2}\,dt\), we have
$$ \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt+\lambda \int^{T}_{0} \alpha(t)u^{1-\mu }(t)\,dt=- \lambda \int^{T}_{0}h(t)u(t)\,dt, $$
which together with the fact of \(\alpha(t)\ge0\) for all \(t\in[0,T]\) and \(\alpha(t)\not\equiv0\) gives
$$ \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt< \int^{T}_{0} \bigl\vert h(t) \bigr\vert u(t) \,dt\le \biggl( \int ^{T}_{0}h^{2}(t)d \biggr)^{\frac{1}{2}} \biggl( \int^{T}_{0}u^{2}(t)\,dt \biggr)^{\frac{1}{2}}. $$
(3.5)
By Lemma 2.2, we have
$$ \biggl( \int^{T}_{0}u^{2}(t)\,dt \biggr)^{\frac{1}{2}}< \frac{T}{\pi} \biggl( \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}+\sqrt{T}u(\eta). $$
It follows from (3.4) that
$$ \biggl( \int^{T}_{0}u^{2}(t)\,dt \biggr)^{\frac{1}{2}}< \frac{T}{\pi} \biggl( \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}+\sqrt{T} \biggl(- \frac{\overline {\alpha}}{\overline{h}} \biggr)^{\frac{1}{\mu}}. $$
Substituting it into (3.5), we have
$$\begin{aligned} \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt&< \biggl( \int^{T}_{0}h^{2}(t)\,dt \biggr)^{\frac {1}{2}} \biggl[\frac{T}{\pi} \biggl( \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} +\sqrt{T} \biggl(- \frac{\overline{\alpha}}{\overline{h}} \biggr)^{\frac{1}{\mu }} \biggr] \\ &=\frac{T}{\pi} \biggl( \int^{T}_{0}h^{2}(t)\,dt \biggr)^{\frac{1}{2}} \biggl( \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} +T^{\frac{1}{2}} \biggl(- \frac{\overline{\alpha}}{\overline{h}} \biggr)^{\frac {1}{\mu}} \biggl( \int^{T}_{0}h^{2}(t)\,dt \biggr)^{\frac{1}{2}}, \end{aligned}$$
by the inequality \(X^{2}-AX-B<0\), we can obtain \(X< A+B^{\frac{1}{2}}\). Let \(X= (\int^{T}_{0}|u'(t)|^{2}\,dt )^{\frac{1}{2}}\), \(A=\frac{T}{\pi } (\int^{T}_{0}h^{2}(t)\,dt )^{\frac{1}{2}}\), \(B=T^{\frac{1}{2}} (-\frac{\overline{\alpha}}{\overline{h}} )^{\frac{1}{\mu}} (\int ^{T}_{0}h^{2}(t)\,dt )^{\frac{1}{2}}\), we have
$$ \biggl( \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}< \frac{T}{\pi}\|h\| _{2}+T^{\frac{1}{4}} \biggl(-\frac{\overline{\alpha}}{\overline{h}} \biggr)^{\frac{1}{2\mu}} \|h\|_{2}^{\frac{1}{2}}:=\rho_{1}. $$
(3.6)
Combined (3.4) with (3.6), we have
$$ u(t)\le u(\eta)+\sqrt{T} \bigl\Vert u' \bigr\Vert _{2}< \biggl(-\frac{\overline{\alpha }}{\overline{h}} \biggr)^{\frac{1}{\mu}}+T^{\frac{1}{2}} \rho_{1}:=M. $$
(3.7)
Let \(t_{1}\), \(t_{2}\) be the maximum point and the minimum point of \(u(t)\) on \([0,T]\), respectively, then
$$ \int_{t_{1}}^{t_{2}}u''(t)\,dt+ \lambda \int _{t_{1}}^{t_{2}}f\bigl(u(t)\bigr)u'(t) \,dt-\lambda \int_{t_{1}}^{t_{2}}\frac{\alpha (t)}{u^{\mu}(t)}\,dt=\lambda \int_{t_{1}}^{t_{2}}h(t)\,dt, $$
which together with \(u'(t_{1})=u'(t_{2})=0\) yields
$$ F\bigl(u(t_{2})\bigr)-F\bigl(u(t_{1})\bigr)= \int_{t_{1}}^{t_{2}}\frac{\alpha(t)}{u^{\mu }(t)}\,dt+ \int_{t_{1}}^{t_{2}}h(t)\,dt, $$
where \(F(x)=\int_{1}^{x}f(s)\,ds\), and then
$$ \bigl\vert F\bigl(u(t_{2})\bigr) \bigr\vert \leq \bigl\vert F \bigl(u(t_{1})\bigr) \bigr\vert + \int_{0}^{T}\frac{\alpha(t)}{u^{\mu }(t)}\,dt+ \int_{0}^{T} \bigl\vert h(t) \bigr\vert \,dt. $$
It follows from (3.4) and (3.7) that
$$\begin{aligned} \bigl\vert F\bigl(u(t_{2})\bigr) \bigr\vert &\leq\max _{(-\frac{\overline{\alpha}}{\overline {h}})^{\frac{1}{\mu}}\leq z\leq M} \bigl\vert F(z) \bigr\vert -T\overline{h}+T \overline{|h|} \\ &\leq\max_{(-\frac{\overline{\alpha}}{\overline{h}})^{\frac{1}{\mu }}\leq z\leq M} \bigl\vert F(z) \bigr\vert +2T \overline{h}_{-}. \end{aligned}$$
(3.8)
It is easy to see from (H1) that there is a constant \(\gamma_{0}>0\), such that
$$\bigl\vert F(z) \bigr\vert = \biggl\vert \int_{1}^{z}f(s)\,ds \biggr\vert >\max _{(-\frac{\overline{\alpha}}{\overline {h}})^{\frac{1}{\mu}}\leq z\leq M} \bigl\vert F(z) \bigr\vert +2T\overline{h}_{-} \quad \mbox{for } z\in(0,\gamma_{0}]. $$
By (3.8), we have
$$ u(t_{2})>\gamma_{0}, $$
and then
$$ u(t)>\gamma_{0}\quad \mbox{for all } t\in[0,T]. $$
(3.9)
Next, if u attains its maximum over \([0,T]\) at \(t_{1}\in[0,T]\), then \(u'(t_{1})=0\) and we see from (3.2) that
$$ u'(t)=\lambda \int_{t_{1}}^{t}\biggl[-f(u)u'+ \frac{\alpha(t)}{u^{\mu}}+h(t)\biggr]\,dt\quad \mbox{for all }t\in[t_{1},t_{1}+T]. $$
Therefore,
$$\begin{aligned} \bigl\vert u'(t) \bigr\vert & \leq\lambda \bigl\vert F \bigl(u(t)\bigr)-F\bigl(u(t_{1})\bigr) \bigr\vert + \lambda \int _{t_{1}}^{t_{1}+T} \frac{\alpha(t)}{u^{\mu}(t)}\,dt+ \lambda \int_{t_{1}}^{t_{1}+T} h_{+}(s)\,ds \\ &\leq2 \lambda\max_{\gamma_{0} \leq u\leq M} \bigl\vert F(u) \bigr\vert + \lambda \int _{0}^{T}\frac{\alpha(t)}{u^{\mu}(t)}\,dt+T \overline{h}_{+}. \end{aligned}$$
(3.10)
Substituting (3.3) into (3.10), we have
$$ \bigl\vert u'(t) \bigr\vert \leq2 \max_{\gamma_{0} \leq u\leq M} \bigl\vert F(u) \bigr\vert - T\overline {h}+T\overline{h}_{+}:=M^{*} \quad \mbox{for all } t\in[0,T] $$
and then
$$ \bigl\vert u'(t) \bigr\vert \leq M^{*} \quad \mbox{for all } t\in[0,T]. $$
(3.11)
From (3.7), (3.9), (3.11) and Remark 2.1, we can choose \(M_{0}:=\min\{\gamma_{0}, D_{1}\}\) where \(D_{1}\) is determined in Remark 2.1, \(M_{1}=M\) and \(M_{2}=M^{*}\) such that all the conditions of Lemma 2.1 are satisfied. Thus, by using Lemma 2.1, we see that equation (1.6) has at least one positive T-periodic solution. The proof is complete. □
Theorem 3.2
Suppose that
\(\overline {h}<0\)
and (H2) holds. Then equation (1.6) has at least one positive
T-periodic solution.
Proof
Suppose that u is an arbitrary positive T-periodic solution of equation (2.1), then
$$ u''+\lambda f(u)u'-\lambda \frac{\alpha(t)}{u^{\mu}}=\lambda h(t),\quad \lambda\in(0,1]. $$
(3.12)
Similar to the proof of (3.4) and (3.6), we find that there is a point \(\xi\in[0,T]\) such that
$$ u(\xi)= \biggl(-\frac{\overline{\alpha}}{\overline{h}} \biggr)^{\frac{1}{\mu}} $$
(3.13)
and
$$ \biggl( \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}< \frac{T}{\pi}\|h\| _{2}+T^{\frac{1}{4}} \biggl(-\frac{\overline{\alpha}}{\overline{h}} \biggr)^{\frac{1}{2\mu}} \|h\|_{2}^{\frac{1}{2}}:=\rho_{1}. $$
(3.14)
In view of the inequality
$$\begin{aligned} u(t)&=u(\xi)+ \int^{t}_{\xi}u'(s)\,ds\leq u(\xi)+ \int^{\xi+T}_{\xi } \bigl\vert u'(s) \bigr\vert \,ds \\ &=u(\xi)+ \int^{T}_{0} \bigl\vert u'(s) \bigr\vert \,ds,\quad t\in[\xi,\xi+T], \end{aligned}$$
and by using (3.13), together with Schwarz inequality, we have
$$ \max_{t\in[0,T]}u(t)\leq \biggl(-\frac{\overline{\alpha}}{\overline{h}} \biggr)^{\frac{1}{\mu}}+T^{\frac{1}{2}} \biggl( \int^{T}_{0} \bigl\vert u'(s) \bigr\vert ^{2}\,ds \biggr)^{\frac{1}{2}} $$
(3.15)
and
$$ \min_{t\in[0,T]}u(t)\geq u(\xi)- \biggl\vert \int^{t}_{\xi}u'(s)\,ds \biggr\vert \geq \biggl(-\frac {\overline{\alpha}}{\overline{h}} \biggr)^{\frac{1}{\mu}}-T^{\frac {1}{2}} \biggl( \int^{T}_{0} \bigl\vert u'(s) \bigr\vert ^{2}\,ds \biggr)^{\frac{1}{2}}. $$
(3.16)
Substituting (3.14) into (3.15) and (3.16), respectively, we have
$$ \max_{t\in[0,T]}u(t)\leq \biggl(-\frac{\overline{\alpha}}{\overline{h}} \biggr)^{\frac{1}{\mu}}+T^{\frac{1}{2}} \biggl[\frac{T}{\pi}\|h \|_{2} +T^{\frac{1}{4}} \biggl(-\frac{\overline{\alpha}}{\overline{h}} \biggr)^{\frac {1}{2\mu}}\|h\|_{2}^{\frac{1}{2}} \biggr] $$
(3.17)
and
$$ \min_{t\in[0,T]}u(t)\geq \biggl(-\frac{\overline{\alpha}}{\overline {h}} \biggr)^{\frac{1}{\mu}}-T^{\frac{1}{2}} \biggl[\frac{T}{\pi}\|h \|_{2} +T^{\frac{1}{4}} \biggl(-\frac{\overline{\alpha}}{\overline{h}} \biggr)^{\frac {1}{2\mu}}\|h\|_{2}^{\frac{1}{2}} \biggr]. $$
(3.18)
The rest of the proof works almost analogously to the corresponding ones of Theorem 3.1. □
Example 3.1
Considering the following equation:
$$ x''(t)+\frac{x'(t)}{x^{2}(t)}- \frac{\sin^{2}t}{x^{\frac{2}{3}}}=-1+\cos t. $$
(3.19)
Corresponding to equation (1.6), we have \(f(x)=\frac{1}{x^{2}}\), \(\mu=\frac{2}{3}\), \(\alpha(t)=\sin^{2}t\), \(h(t)=-1+\cos t\), and then \(T=2\pi\), \(\int_{0}^{1}f(s)\,ds=+\infty\). This implies that assumption (H1) holds. Since \(\overline{h}=-1<0\), by using Theorem 3.1, we find that equation (3.19) has at least one positive 2π-periodic solution.
Example 3.2
Considering the following equation:
$$ x''(t)+\frac{x'(t)}{x^{\frac{1}{2}}(t)}- \frac{\sin^{2}8t}{x^{\frac {3}{4}}}=-\cos^{2}8t. $$
(3.20)
Corresponding to equation (1.6), f can be regarded as \(f(x)=\frac{1}{x^{\frac{1}{2}}}\), \(\mu=\frac{3}{4}\), \(\alpha(t)=\sin ^{2}8t\) and \(h(t)=-\cos^{2}8t\). Since \(\int_{0}^{1}f(s)\,ds=2\), it follows that assumption (H1) does not hold. This implies that Theorem 3.1 cannot be used to study the existence of periodic solutions to (3.20). But, by simple calculating, we can verify that
$$ -\frac{\overline{h}}{\overline{\alpha}}=1,\qquad \|h\|^{2}_{2}= \frac{3T}{8}, $$
where \(T=\frac{\pi}{8}\), and then
$$ \biggl(-\frac{\overline{\alpha}}{\overline{h}}\biggr)^{\frac{1}{\mu}}-T^{\frac {1}{2}}\biggl[ \frac{T}{\pi}\|h\|_{2} +\biggl(T^{\frac{1}{2}}\biggl(- \frac{\overline{\alpha}}{\overline{h}}\biggr)^{\frac{1}{\mu }}\|h\|_{2}\biggr)^{\frac{1}{2}} \biggr]= 1-\biggl(\frac{\sqrt{6}\pi}{256}+\frac{\sqrt[4]{6}\pi}{16}\biggr)>0, $$
which implies that assumption (H2) holds. Thus, by using Theorem 3.2, we find that (3.20) has at least one positive \(\frac{\pi}{8}\)-periodic solution.
Remark
The above two examples can neither be studied by using the results in [31, 32, 34] and [35], since \(f(x)\) in (3.19) and in (3.20) are all singular at \(x=0\), nor be studied by using the results in [33], since the restoring force terms of \(\frac{\sin^{2}t}{x^{\frac{2}{3}}}\) in (3.19) and \(\frac{\sin^{2}8t}{x^{\frac{3}{4}}}\) in (3.20) have weak singularities at \(x=0\).