The aim of this section is to prove that the Poincaré map associated to Eq. (1.2) fits the hypotheses of Theorem 2.2 exposed in Sect. 2.1.
Multiplying both sides of Eq. (1.2) by \(\sigma(h)(t )\), we have
$$ \bigl(\sigma(h) (t )u'\bigr)'+\omega \sigma(h) (t )\sin u=\sigma(h) (t )p(t). $$
(4.1)
Note that Eq. (4.1) is equivalent to the planar system
$$ \textstyle\begin{cases} u'=f_{1}(t,v)=\sigma(h)(t )v, \\ v'=f_{2}(t,u)=\omega\sigma(h)(t )\sin u+\sigma(h)(t )p(t). \end{cases} $$
(4.2)
Let \((u,v)^{\top}=(u(t,\theta,r),v(t,\theta,r))^{\top}\) be the solution of the system (4.2) satisfying the initial condition
$$ \textstyle\begin{cases} u(0)=\theta,\\ v(0)=r. \end{cases} $$
(4.3)
It is easy to verify that there exist two positive constants \(q_{1}\) and \(q_{2}\) such that
$$ \sqrt{f^{2}_{1}(t,v)+f^{2}_{2}(t,u)} \leq q_{1}\sqrt{u^{2}+v^{2}}+q_{2}, $$
which guarantees that the solution \((u,v)^{\top}\) of the initial value problem (4.2)–(4.3) is unique and globally defined. Then we can define the Poincaré map associated to the system (4.2) as
$$\begin{aligned} S(\theta,r) =\bigl(Q(\theta,r),P(\theta,r)\bigr) =\bigl(u(T, \theta,r),v(T,\theta,r)\bigr). \end{aligned}$$
Obviously, the fixed points of the Poincaré map S correspond to the Tperiodic solutions of the system (4.2). It follows from 2πperiodicity of the function sinu and the uniqueness of \((u(t,\theta,r),v(t,\theta,r))^{\top}\) that
$$ u(t,\theta+2\pi,r)=u( t,\theta,r)+2\pi, \qquad v(t,\theta+2\pi,r)=v(t, \theta,r). $$
(4.4)
Then we have
$$ Q(\theta+2\pi,r)=Q(\theta,r)+2\pi,\qquad P(\theta+2\pi,r)=P(\theta,r), $$
which implies that the Poincaré map S is defined on the cylinder. Based on the theorem of differentiability with respect to the initial conditions, it is easy to see that the Poincaré map \(S\in C^{2}(A)\). Since \((u(t,\theta,r),v(t,\theta,r))^{\top}\) is unique and globally defined, the Poincaré map S is a diffeomorphism of A. The isotopy to the identity is given by the flow
$$\begin{aligned} \Psi_{\lambda}(\theta,r)&=\Psi\bigl((1\lambda)T,\theta,r\bigr) \\ &=\bigl(x\bigl((1\lambda)T, \theta,r\bigr),y\bigl((1\lambda)T,\theta,r\bigr) \bigr),\quad \lambda\in[0,1]. \end{aligned}$$
Note that \(\Psi_{0}(\theta,r)=S(\theta,r)\), \(\Psi_{1}(\theta ,r)=(\theta,r)\) and this isotopy is valid on the cylinder. Therefore, the Poincaré map \(S\in\mathbb{\varepsilon}^{2}(A)\).
Theorem 4.1
Assume that
$$ \int_{0}^{T}\sigma(h) (s)p(s)\,\mathrm{ d}s=0. $$
(4.5)
Then Eq. (1.2) has at least two geometrically distinct
Tperiodic solutions, and at least one of them is unstable.
Proof
In order to apply Theorem 2.2, we need to prove that the Poincaré map S is exact symplectic and satisfies the boundary twist condition (2.7).
Let us first prove that the Poincaré map S is exact symplectic. Consider the \({\mathbb {C}}^{1}\) function
$$\begin{aligned} V(\theta,r)={}& \int_{0}^{T} \biggl[\frac{ (\sigma(h)(t)u'(t,\theta,r) )^{2}}{2} +\omega \sigma(h) (t)\cos u(t,\theta,r) \biggr]\,\mathrm{ d} t \\ &{}+ \int_{0}^{T}\sigma(h) (t)p(t)u(t,\theta,r)\, \mathrm{ d} t. \end{aligned}$$
By (4.4) and (4.5), we have
$$\begin{aligned} V(\theta+2\pi,r) ={}& \int_{0}^{T} \biggl[\frac{ (\sigma(h)(t)u'(t,\theta+2\pi,r) )^{2}}{2} +\omega \sigma(h) (t)\cos u(t,\theta+2\pi,r) \biggr]\,\mathrm{ d} t \\ &{}+ \int_{0}^{T}\sigma(h) (t)p(t)u(t,\theta+2\pi,r)\, \mathrm{ d} t \\ ={}& \int_{0}^{T} \biggl[\frac{(\sigma(h)(t)u'(t,\theta,r))^{2}}{2} +\omega \sigma(h) (t)\cos u(t,\theta,r) \biggr]\,\mathrm{ d} t \\ &{}+ \int_{0}^{T}\sigma(h) (t)p(t)u(t,\theta,r)\, \mathrm{ d}t+2\pi \int _{0}^{T}\sigma(h) (t)p(t)\,\mathrm{ d}t \\ ={}&V(\theta,r). \end{aligned}$$
Let us compute the partial derivatives of \(V(\theta,r)\)
$$\begin{aligned} V_{\theta}(\theta,r) ={}& \int_{0}^{T}\sigma(h) (t) \biggl[u'(t, \theta,r)\frac{\partial u'(t,\theta,r)}{\partial\theta}\omega\sin u(t,\theta,r)\frac {\partial u(t,\theta,r)}{\partial\theta} \biggr]\, \mathrm{ d} t \\ &{}+ \int_{0}^{T}\sigma(h) (t)p(t)\frac{\partial u(t,\theta,r)}{\partial \theta}\, \mathrm{ d} t. \end{aligned}$$
Then, by the second equation of the system (4.2), we have
$$ V_{\theta}(\theta,r) = \int_{0}^{T}\sigma(h) (t) \biggl[u'(t, \theta,r)\frac{\partial u'(t,\theta,r)}{\partial\theta}+v'(t,\theta,r)\frac{\partial u(t,\theta,r)}{\partial\theta} \biggr] \,\mathrm{ d} t. $$
(4.6)
Integrating by parts and using the first equation of the system (4.2), we have
$$\begin{aligned} &\int_{0}^{T}v'(t,\theta,r) \frac{\partial u(t,\theta,r)}{\partial \theta}\,\mathrm{ d}t \\ &\quad = \biggl(\frac{\partial u(t,\theta,r)}{\partial\theta}v(t,\theta ,r) \biggr)\Big^{T}_{0}  \int_{0}^{T}\frac{\partial u'(t,\theta,r)}{\partial\theta }v(t,\theta,r) \, \mathrm{ d} t \\ &\quad =v(T,\theta,r)\frac{\partial u(T,\theta,r)}{\partial\theta }r \int_{0}^{T}\sigma(h) (t)u'(t, \theta,r)\frac{\partial u'(t,\theta ,r)}{\partial\theta} \,\mathrm{ d} t. \end{aligned}$$
Substituting the above equality into (4.6) gives
$$ V_{\theta}(\theta,r) =v(T,\theta,r)\frac{\partial u(T,\theta,r)}{\partial\theta}r. $$
(4.7)
Analogously, we have
$$ V_{r}(\theta,r) = v(T,\theta,r)\frac{\partial u(T,\theta,r)}{\partial r}. $$
(4.8)
By (4.7) and (4.8), we have
$$\begin{aligned} \mathrm{ d}V={}&V_{\theta}(\theta,r)\,\mathrm{ d}\theta+ V_{r}(\theta,r) \,\mathrm{ d}r \\ ={}& \biggl[v(T,\theta,r)\frac{\partial u(T,\theta,r)}{\partial\theta }r \biggr]\,\mathrm{ d}\theta+v(T, \theta,r)\frac{\partial u(T,\theta ,r)}{\partial r}\,\mathrm{ d}r \\ ={}&v(T,\theta,r) \biggl[\frac{\partial u(T,\theta,r)}{\partial\theta }\,\mathrm{ d}\theta+\frac{\partial u(T,\theta,r)}{\partial r} \,\mathrm{ d}r \biggr]r\,\mathrm{ d}\theta \\ ={}&v(T,\theta,r)\,\mathrm{ d}u(T,\theta,r)r\,\mathrm{ d}\theta \\ ={}&P\,\mathrm{ d}Qr\,\mathrm{ d}\theta, \end{aligned}$$
which means that the Poincaré map \(S(\theta,r)\) is exact symplectic.
We next prove that the Poincaré map S satisfies the boundary twist condition (2.7). Integrating the second equation of the system (4.2) from 0 to t with \(t\in[0,t]\), we obtain
$$\begin{aligned} v(t) &=r+ \int_{0}^{t} \bigl[\omega\sigma(h) (s)\sin u+ \sigma (h) (s)p(s) \bigr]\,\mathrm{ d}s \\ &\geq r \int_{0}^{T}\bigl(\omega\sigma(h) (s)+ \bigl\vert \sigma(h) (s)p(s) \bigr\vert \bigr)\,\mathrm{ d}s \\ &= rT\bigl(\omega\overline{\sigma(h)}+\overline{ \bigl\vert \sigma(h)p \bigr\vert }\bigr) , \end{aligned}$$
where \(\overline{\xi}=\frac{1}{T}\int_{0}^{T}\xi(s)\,\mathrm{ d}s\). Thus we can find a positive constant \(\rho_{1}\geq T(\omega\overline {\sigma(h)}+\overline{\sigma(h)p})>0\) such that \(v(t)>0\) if \(r>\rho_{1}\), \(\forall t\in[0,T]\). By the first equation of (4.2), we know that
$$ u'(t)=\sigma(h) (t)v(t)>0, $$
which means that u is increasing for \(t\in[0,T]\). So we can choose a positive constant ρ with \(\rho>\rho_{1}\), then we have
$$ Q(\theta,\rho)\theta=u(T,\theta,\rho)u(0,\theta,\rho)>0. $$
By a standard compactness argument, we can conclude that there exists \(\epsilon>0\) such that
$$ Q(\theta,\rho)\theta>\epsilon, \quad \theta\in[0,2\pi). $$
Analogously, we have
$$ Q(\theta,\rho)\theta< \epsilon, \quad \theta\in[0,2\pi). $$
In order to apply Theorem 2.2, we take \(A={\mathbb {R}}\times[\rho,\rho]\). By the solutions of the system (4.2) are globally defined, one can find a larger B such that \(S(A)\subset \operatorname{int}B\). Since the righthand side of the system (4.2) is analytic with respect to the variables \((u,v)\), the Poincaré map S is also analytic, as follows from the analytic dependence on the initial conditions.
Up to now, all the conditions of Theorem 2.2 are satisfied, thus we see that the Poincaré map
$$S(\theta,r)=\bigl(Q(\theta,r),P(\theta,r)\bigr)=\bigl(u(T,\theta,r),v(T, \theta,r)\bigr) $$
has at least two fixed points, and at least one of them is unstable. That is, Eq. (1.2) has at least two geometrically distinct Tperiodic solutions and at least one of them is unstable. □
Now we consider the existence of the socalled Tperiodic solutions with winding number of Eq. (1.2), i.e., solutions u such that
$$ u(t+T)=u(t)+2N\pi,\quad N\in {\mathbb {Z}}, \forall t\in {\mathbb {R}}. $$
Such solutions are also called running solutions. Obviously, we get the usual Tperiodic solutions when \(N=0\).
Let u be a Tperiodic solution of Eq. (1.2) with winding number N. Taking the change of variables
$$ y(t)=u(t)\frac{2N\pi}{T} t. $$
Obviously, we have
$$\begin{aligned} y(t+T) &=u(t+T)\frac{2N\pi}{T}(t+T) \\ &=u(t)\frac{2N\pi}{T} t \\ &=y(t), \end{aligned}$$
which implies that Tperiodic solutions with winding number N of Eq. (1.2) correspond to Tperiodic solutions of the equation
$$ y''+h(t)y'+\omega\sin \biggl(y+\frac{2N\pi}{T}t \biggr)=p(t)+\frac {2N\pi}{T}h(t). $$
(4.9)
Proceeding as in the proof of Theorem 4.1, we can prove the following result.
Theorem 4.2
Assume that (4.5) holds and
$$ \int_{0}^{T}h(s)\sigma(h) (s)\,\mathrm{ d}s=0 . $$
(4.10)
Then for every integer
N, Eq. (1.2) has at least two geometrically distinct
Tperiodic solutions with winding number
N, and at least one of them is unstable.
Let u is a periodic solution of (1.2) and take its minimal period \(\tau>0\). Since T is the minimal period of p, we have \(\tau =kT, k\in {\mathbb {N}}\mbox{ with } k\geq1\). Therefore, if \(k=1\), x is a harmonic (periodic) solution of Eq. (1.2); if \(k\in {\mathbb {N}}\mbox{ with } k>1\), u is a subharmonic solution of Eq. (1.2).
Finally, we study the existence of korder subharmonic solutions with winding number N of Eq. (1.2), that is,
$$ u(t+kT)=u(t)+2N\pi,\quad k\in {\mathbb {N}}\mbox{ with } k>1, \forall t\in {\mathbb {R}}. $$
(4.11)
Theorem 4.3
Assume that (4.5) and (4.10) hold. Then, for each couple of relatively prime natural numbers
N
and
k, Eq. (1.2) has at least two geometrically distinct
korder subharmonic solutions with winding number
N
and
kT
is the minimal period. Moreover, at least one of them is unstable.
Proof
By Theorem 4.2, with T replaced by kT, we see that Eq. (1.2) has at least two geometrically distinct korder subharmonic solutions with winding number N. Moreover, at least one of them is unstable. We only need to verify that kT is the minimal period of periodic solutions with winding number N for Eq. (1.2).
Conversely, suppose that mT is the minimal period, where \(m\in\{ 1,2,\ldots,k1\}\). So there exists an integer i with \(i\neq0\), such that
$$ u(t+mT)=u(t)+2i\pi,\quad \forall t\in {\mathbb {R}}. $$
(4.12)
Notice that there exist positive integers \(l_{1}\) and \(l_{2}\) such that
$$ l_{1}m=l_{2}k. $$
(4.13)
By (4.12), we have
$$ u(t+l_{1}mT)=u(t)+2l_{1}i\pi,\quad \forall t\in {\mathbb {R}}. $$
By (4.11), we have
$$ u(t+l_{2}kT)=u(t)+2l_{2}N\pi,\quad \forall t\in {\mathbb {R}}. $$
According to the above two equalities and the uniqueness of the solution u, we have
$$ l_{1}i=l_{2}N. $$
(4.14)
By (4.13) and (4.14), we have
$$ \frac{N}{k}=\frac{i}{m}, $$
which is impossible because N and k are relatively prime and i is a nonzero integer and \(m\in\{1,2,\ldots,k1\}\). □
Example 4.4
Consider the following damped pendulum equation:
$$ u''\frac{2\alpha\sin t}{1+\alpha\cos t}u'+ \omega\sin u=\frac {2\alpha\sin t}{1+\alpha\cos t}, $$
(4.15)
where α and ω are positive constants with \(\alpha<1\). Then the following conclusions hold:

(I)
Equation (4.15) has at least two geometrically distinct 2πperiodic solutions, and at least one of them is unstable.

(II)
For every integer N, Eq. (4.15) has at least two geometrically distinct 2πperiodic solutions with winding number N, and at least one of them is unstable.

(III)
For each couple of relatively prime natural numbers N and k, Eq. (4.15) has at least two geometrically distinct korder subharmonic solutions with winding number N and kT is the minimal period. Moreover, at least one of them is unstable.
Proof
Equation (4.15) can be regarded as a problem of the form Eq. (1.2), where
$$ h(t)=\frac{2\alpha\sin t}{1+\alpha\cos t}, \qquad p(t)=\frac{2\alpha \sin t}{1+\alpha\cos t}. $$
By calculating, we have
$$\begin{aligned} &\int_{0}^{2\pi}p(s)\sigma(h) (s)\,\mathrm{ d}s= \int_{0}^{T}\frac{2\alpha \sin t}{1+\alpha\cos t}\cdot \biggl( \frac{1+\alpha\cos t}{1+\alpha } \biggr)^{2}\,\mathrm{ d}t=0, \\ &\int_{0}^{2\pi}h(s)\sigma(h) (s)\,\mathrm{ d}s= \int_{0}^{T}\frac{2\alpha \sin t}{1+\alpha\cos t}\cdot \biggl( \frac{1+\alpha\cos t}{1+\alpha } \biggr)^{2}\,\mathrm{ d}t=0. \end{aligned}$$
Now the results (I)–(III) follow directly from Theorems 4.1–4.3. □