Rewrite equation (1.1) as the following equivalent form:
$$ x'=y,\quad \quad y'=g(x)+p(t). $$
(2.1)
We will apply the PoinacréBirkhoff twist theorem to obtain the existence of the \(2k\pi\)periodic solution for equation (2.1) under the assumption of a weak subquadratic potential. The existence of a 2πperiodic solution then follows from the Massera theorem.
The key point of applying the PoincaréBirkhoff twist theorem is constructing an annulus \(\mathcal {A}\) in the \((x,y)\) phase plane bounded by two starshaped curves \(\gamma^{+}\) and \(\gamma^{}\), such that the solution starting from \(\gamma^{+}\) and \(\gamma^{}\) will move more than and less than j clockwise turns in the time interval \([t_{0}, t_{0}+2k\pi]\), respectively.
When the solution passes through the origin at some time, we cannot compute how many clockwise turns it moves. To avoid this problem, we introduce a modified equation such that if \(z(t)=(x(t),y(t))\) is a solution of the modified equation, then \(z(t_{0})\neq(0,0)\) for some \(t_{0}\) implies that \(z(t)\neq(0,0)\) for all t. Moreover, the twist property of the fixed point obtained by using the PoinacréBirkhoff twist theorem will help us to guarantee that the \(2k\pi\)periodic solution we showed for the modified equation is exactly the \(2k\pi\)periodic solution of the original equation (2.1) for sufficiently large k.
Denote by \(P=\max\{ \vert p(t)\vert \mid t\in[0,2\pi] \}\) and \({\mathcal {B}}_{\mu_{0}}(g)\) the set of all the continuously differentiable functions \(f(x)\) satisfying
$$\bigl\vert f(x)g(x)\bigr\vert \leq\mu_{0},\quad\hbox{for all }x \in{\Bbb {R}}, $$
where \(\mu_{0}\) is sufficient small such that \(\liminf_{\vert x\vert \to\infty} \operatorname {sgn}(x) g(x)>\mu_{0}+P\).
We consider the following modified equation:
$$ x'=\frac{\partial H(t,x,y)}{\partial y},\qquad y'= \frac {\partial H(t,x,y)}{\partial y}, $$
(2.2)
where \(H(t,x,y)=\frac{y^{2}}{2}+K(x^{2}+y^{2})[F(x)xp(t)]+(1K(x^{2}+y^{2}))x^{2}\), \(F(x)=\int_{0}^{x} f(\xi)\,d\xi\) with \(f\in{\mathcal {B}}_{\mu_{0}}(g)\), and \(K\in C^{\infty}(\Bbb {R})\) satisfies that
$$ K\bigl(x^{2}+y^{2}\bigr)=\left \{ \textstyle\begin{array}{l@{\quad}l} 0,& 0 \leq x^{2}+y^{2} \leq R_{1}^{2},\\ 1,& x^{2}+y^{2} \ge R_{2}^{2} , \end{array}\displaystyle \right . $$
with some given constants \(R_{2}>R_{1}>0\).
Let \(z(t;t_{0},z_{0})=(x(t;t_{0},z_{0}),y(t;t_{0},z_{0}))\) be the solution of (2.2) satisfying the initial condition \(z(t_{0};t_{0},z_{0})=z_{0}=(x_{0},y_{0})\). Then we have the following fundamental lemma.
Lemma 2.1
Assume that
g
satisfies
\((g_{1})\)
and
\(f\in{\mathcal {B}}_{\mu_{0}}(g)\)
for sufficiently small
\(\mu_{0}\). Then each initial value problem associated with (2.2) has a unique solution
\(z(t;t_{0},z_{0})=(x(t;t_{0},z_{0}),y(t;t_{0},z_{0}))\)
which is well defined for
\(t\in\Bbb {R}\)
and
\(z_{0}\neq(0,0)\)
implying that
\(z(t;t_{0}, z_{0})\neq(0,0)\)
for all
t. Moreover, for any
\(L>0\), there is
\(E=E(L)>0\)
such that
$$\bigl\vert l(t)l(t_{0})\bigr\vert \leq E \quad\textit{for } \vert tt_{0}\vert \leq L \textit{ and } \vert z_{0} \vert \gg1, $$
where
$$l(t)=\sqrt{\frac {y^{2}(t;t_{0},z_{0})}{2}+G\bigl(x(t;t_{0},z_{0}) \bigr)}\quad\textit{for } t>t_{0}. $$
Proof
f is a continuously differentiable function. The existence and uniqueness of the solution associated to the initial condition is ensured by the existenceuniqueness theorem. Moreover, the solution has continuity with respect to initial conditions. It is easy to see that \(z(t)\equiv(0,0)\) is the solution of (2.2) satisfying the initial condition \(z(t_{0};t_{0},z_{0})=(0,0)\), from which it follows that \(z_{0}\neq(0,0) \Rightarrow z(t;t_{0},z_{0})\neq(0,0)\), ∀t.
Note that equation (2.2) is the same as the equation
$$x'=y,\quad\quad y'=f(x)+p(t) $$
for \(x^{2}+y^{2} \ge R_{2}^{2}\). Then we have, for \(\vert z(t;t_{0},z_{0})\vert \gg1\),
$$\begin{aligned} \bigl\vert l'(t)\bigr\vert =&{\frac{1}{ 2l(t)}\bigl\vert y(t)y'(t)+g\bigl(x(t)\bigr)x'(t)\bigr\vert = \frac{1}{ 2l(t)}\bigl\vert y(t) \bigl(g\bigl(x(t)\bigr)f\bigl(x(t)\bigr)+p(t) \bigr)\bigr\vert } \\ \leq&{ \frac{1}{ 2l(t)}\vert P+\mu_{0}\vert \bigl\vert y(t) \bigr\vert \leq \vert P+\mu_{0}\vert }. \end{aligned}$$
An obvious induction shows that for any \(L>0\), there is \(E=E(L)>0\) such that
$$\bigl\vert l(t)l(t_{0})\bigr\vert \leq E \quad\hbox{for } \vert tt_{0}\vert \leq L \hbox{ and } \vert z_{0} \vert \gg1. $$
Moreover, \(l(t)\rightarrow+\infty\Leftrightarrow \vert z(t)\vert \rightarrow +\infty\). Then the above estimation implies that the global existence of the solution of (2.2) is a consequence of the continuation theorem. The lemma is thus proved. □
If \(z_{0}\neq(0,0)\) we can represent the solution \(z(t;t_{0},z_{0})\) by means of polar coordinates as
$$r(t;t_{0},r_{0},\theta_{0})=\bigl\vert z(t;t_{0},z_{0})\bigr\vert ,\qquad\theta (t;t_{0},r_{0},\theta_{0})=\operatorname {Arg}\bigl(z(t;t_{0},z_{0})\bigr), $$
where \(\vert \cdot \vert \) is Euclidean norm of \({\Bbb {R}}^{2}\). Sometimes we write \(r(t;t_{0},z_{0})=\vert z(t;t_{0},z_{0})\vert \) and \(\theta(t;t_{0},z_{0})=\operatorname {Arg}(z(t;t_{0},z_{0}))\) for ease of notation.
Moreover, we have the following.
Lemma 2.2
Assume that the conditions of Lemma
2.1
hold. Then there is
\(R_{0}>R_{2}>R_{1}>0\), such that
\(\theta' (t;t_{0},r_{0},\theta_{0})<0\)
whenever
\(r(t;t_{0},r_{0},\theta_{0})\ge R_{0}\).
Proof
From \((g_{1})\) and \(\liminf_{\vert x\vert \to\infty} \operatorname {sgn}(x) g(x)>\mu_{0}+P\) we have \(d_{1}>0\) such that \(x(f(x)p(t))>0\) for \(\vert x\vert \ge d_{1}\) and \(t\in\Bbb {R}\). For \(\vert z(t)\vert \ge R_{2}\), we have
$$\theta'(t)=\frac{1}{r^{2}(t)}\bigl(x(t)y'(t)y(t)x'(t) \bigr)=\frac{1}{r^{2}(t)}\bigl(y^{2}(t)+\bigl(f\bigl(x(t)\bigr)p(t) \bigr)x(t)\bigr). $$
Let \(R_{0}\) be a constant such that \(R_{0}\ge2\max\{d_{1},R_{2},\max\{ \vert g(x)\vert : \vert x\vert \leq d_{1}\}+\mu_{0}\}\). Then for \(r(t)\ge R_{0}\), either \(\vert x(t)\vert \ge d_{1}\), and we have \(y^{2}(t)+(f(x(t))p(t))x(t)>0\), or \(\vert x(t)\vert < d_{1}\), which implies \(\vert y(t)\vert >\vert x(t)\vert \) and \(\vert y(t)\vert >1/2r(t)\ge1/2R_{0}\), then \(y^{2}(t)>(f(x(t))p(t))x(t)\). Hence \(\theta'(t)<0\) for \(r(t)\ge R_{0}\).
Lemma 2.2 concludes that \(z(t;t_{0},z_{0})\) moves clockwise around the origin O if \(\vert z(t;t_{0}, z_{0})\vert \ge R_{0}\). □
Lemma 2.3
Assume that the conditions of Lemma
2.1
hold. Then for any solution
\(z(t;t_{0},z_{0})\)
of (2.2) with
\(z_{0}\neq(0,0)\), we have
$$\theta(t_{2};t_{0},r_{0},\theta_{0})  \theta(t_{1};t_{0},r_{0},\theta_{0})< \pi, \quad \forall t_{2}>t_{1}. $$
Proof
Note that in equation (2.2) \(yx'=y^{2}>0\) whenever \(x=0\) and \(y\neq0\). Then for any solution \(z(t;t_{0},z_{0})\) of (2.2) with \(z_{0}\neq(0,0)\), we have \(\theta' (t;t_{0},r_{0},\theta _{0})<0\) whenever \(\theta(t;t_{0},r_{0},\theta_{0})=k\pi+1/2\pi\), \(k\in \Bbb {Z}\). Thus Lemma 2.3 in [7] shows that we have \(\theta (t_{2};t_{0},r_{0},\theta_{0})\theta(t_{1};t_{0},r_{0},\theta_{0})<\pi\) for all \(t_{2}>t_{1}\). This proves the lemma. □
The following lemma is similar to that in [7] with the modified proof as in [7].
Lemma 2.4
Assume that the conditions of Lemma
2.1
hold. Then for any constant
\(j\in{\Bbb {N}}\)
there is
\(K_{j}>R_{0}\)
and there is a continuous increasing function
\(\zeta=\zeta _{j}:[K_{j}, +\infty)\rightarrow{\Bbb {R}}^{+}\), with
\(\zeta_{j} (s)>s\), \(\forall s\geq K_{j}\)
such that the following inference holds for each
\(z(t)\)
as the solution of (2.2):
$$\begin{aligned}& \bigl\vert z(t_{0})\bigr\vert \leq r, \bigl\vert z(t_{1})\bigr\vert \geq\zeta_{j} (r) \textit{ or } \bigl\vert z(t_{0})\bigr\vert \geq\zeta_{j} (r), \bigl\vert z(t_{1})\bigr\vert \leq r \textit{ whenever } r\geq K_{j} \\& \textit{and } \bigl\vert z(t)\bigr\vert \geq R_{0}, \forall t\in [t_{0},t_{1}] \quad\Longrightarrow\quad \\& z(t)\textit{ moves at least }j\textit{ clockwise turns on }[t_{0}, t_{1}]\textit{ around the origin }O. \end{aligned}$$
Proof
Let \(z(t)\) be a solution of (2.2) and suppose that \(z(t)\ge R_{0}\) for each \(t\in I\), with I an interval. Denote
$$\begin{aligned}& H_{\pm}(x,y)=\frac{y^{2}}{2}+G(x)\pm(\mu_{0}+P)x,\quad\quad E_{1}=\bigl\{ (x,y): y\ge0\bigr\} , \\& H_{\pm}(t)=\frac{y^{2}(t)}{2}+G\bigl(x(t)\bigr)\pm(\mu_{0}+P)x(t), \quad\hbox{with } z(t)=\bigl(x(t),y(t)\bigr). \end{aligned}$$
Then
$$\begin{aligned}& \frac{d}{dt}H_{+}(t)=y(t) \bigl(g\bigl(x(t)\bigr)f\bigl(x(t)\bigr)+p(t)+ \mu_{0}+P\bigr)\ge0; \\& \frac{d}{dt}H_{}(t)=y(t) \bigl(g\bigl(x(t)\bigr)f\bigl(x(t)\bigr)+p(t) \mu_{0}P\bigr)\le0, \end{aligned}$$
for \(z(t)\in E_{1}\). Moreover, let \(I_{1}\subset I\) be a nondegenerate interval such that \(z(t)\in E_{1}\) for all \(t\in I_{1}\). Then
$$ H_{+}(t)\ge H_{+}(s),\quad \quad H_{}(t)\le H_{}(s),\quad\hbox{for } \forall t,s\in I_{1}, t>s. $$
(2.3)
From the assumption \((g_{1})\), the definition of f and \(\mu_{0}\) sufficiently small, we have for \(z=(x,y)\),
$$H_{\pm}(x,y)\to+\infty\quad \Leftrightarrow\quad r=\vert z\vert \to+\infty. $$
Then for r sufficiently large, using similar argument as in [7], Lemma 2.7, we have two continuous increasing functions \(L_{\pm }: [r_{1},+\infty)\) with \(r_{1}>R_{0}\) sufficiently large, such that
$$L_{}\bigl(\vert z\vert \bigr)\le H_{\pm}(x,y)\le L_{+}\bigl(\vert z \vert \bigr),\quad\hbox{for }z=(x,y)\in E_{1}, $$
and
$$L_{\pm}(r)\to+\infty,\quad\hbox{for } r\to+\infty. $$
Without loss of generality, we suppose that \(L_{}(r)< r <L_{+}(r)\). Using (2.3) we have
$$L_{+}\bigl(r(t)\bigr)\ge L_{}\bigl(r(s)\bigr),\qquad L_{}\bigl(r(t)\bigr) \le L_{+} \bigl(r(s)\bigr),\quad\hbox{for }\forall t,s\in I_{1}, t>s, $$
from which it follows that
$$ (\xi_{1})^{1}\bigl(r(s)\bigr)\le r(t) \le \xi_{1}\bigl(r(s)\bigr),\quad\hbox{for }\forall t,s\in I_{1}, t>s, $$
(2.4)
where \(\xi_{1}(r)=(L_{})^{1}(L_{+}(r))\) and \((\xi _{1})^{1}(r)=(L_{+})^{1}(L_{}(r))\) with \(r<\xi_{1}(r)\) and \((\xi _{1})^{1}(r)\to+\infty\) for \(r\to+\infty\).
Denote by \(E_{2}=\{ (x,y): y\le0\}\) and \(I_{2}\subset I\) be a nondegenerate interval such that \(z(t)\in E_{2}\) for all \(t\in I_{2}\). We can use a similar argument to above to obtain a continuous increasing function \(\xi_{2}: [r_{2},+\infty)\) with \(r_{2}>R_{0}\) sufficiently large, such that
$$ (\xi_{2})^{1}\bigl(r(s)\bigr)\le r(t) \le \xi_{2}\bigl(r(s)\bigr),\quad\hbox{for }\forall t,s\in I_{2}, t>s. $$
(2.5)
Moreover, \(r<\xi_{2}(r)\) and \((\xi_{2})^{1}(r)\to+\infty\) for \(r\to +\infty\).
Choose a continuous increasing function \(\eta(r)\ge\max\{\xi _{i}(r),i=1,2\}\) with \((\eta)^{1}(r)\to+\infty\) for \(r\to+\infty\). Denote \(\zeta_{1}(r)=\eta\circ\eta\circ\eta(r)\). Let \(K_{1}\) be large enough such that \(K_{1}\ge\zeta_{1}(R_{0})\). Thus (2.4) and (2.5) show that if \(\vert z(t_{0})\vert \ge K_{1}\), then \(\zeta_{1}^{1}(\vert z(t_{0})\vert )\le \vert z(t)\vert \le\zeta_{1}(\vert z(t_{0})\vert )\) before \(z(t)\) completes 1 clockwise turn around the origin O. That is, if \(\vert z(t_{0})\vert \ge K_{1}\), \(\theta (t)\ge\theta(t_{0})2\pi\) and \(\vert z(t)\vert \ge R_{0}\), \(\forall t\in[t_{0},t_{1}]\) then \(\zeta_{1}^{1}(\vert z(t_{0})\vert )\le \vert z(t_{1})\vert \le\zeta_{1}(\vert z(t_{0})\vert )\). In other words, for \(r\ge K_{1}\),
$$\begin{aligned}& \vert z(t_{0})\vert \le r, \vert z(t_{1})\vert \ge\zeta_{1} (r) \hbox{ and } \vert z(t)\vert \ge R_{0}, \forall t\in [t_{0},t_{1}] \\& \hbox{or } \vert z(t_{0})\vert \ge\zeta_{1} (r), \vert z(t_{1})\vert \le r \hbox{ and } \vert z(t)\vert \ge R_{0}, \forall t\in [t_{0},t_{1}] \quad \Longrightarrow\\& z(t)\hbox{ moves at least one clockwise turn on }[t_{0}, t_{1}]\hbox{ around the origin }O. \end{aligned}$$
The lemma is thus proved by induction. □
Let \(T_{j}(t_{0},z_{0})\) denote the minimum time in which the solution \(z(t;t_{0}, z_{0})\) of (2.2) completes j clockwise turns, \(j\in \Bbb {N}\). In the proof of the next lemma we will use Lemma 2.4 to show that \(T_{j}(t_{0},z_{0})\) is well defined for sufficiently large \(z_{0}\). Let \(T^{+}_{j}(t_{0},z_{0})\) denote the minimum time such that if \(T\ge T^{+}_{j}(t_{0},z_{0})\) then the solution \(z(t;t_{0}, z_{0})\) of (2.2) completes at least j clockwise turns in \([t_{0},t_{0}+T]\). Moreover, denote
$$\begin{aligned}& T_{j}^{}(h)=\inf\bigl\{ T_{j}(t_{0}, z_{0}) \mid \forall z_{0}\in \gamma _{h}, \forall f\in{\mathcal {B}}_{\mu_{0}}, \forall t_{0} \bigr\} ; \\& T_{j}^{+}(h)=\sup\bigl\{ T^{+}_{j}(t_{0},z_{0}) \mid \forall z_{0}\in\gamma _{h}, \forall f\in{\mathcal {B}}_{\mu_{0}}, \forall t_{0} \bigr\} . \end{aligned}$$
Then we have the following estimations of the twist properties for the solutions of (2.2).
Lemma 2.5
Assume that the conditions of Lemma
2.1
hold. Then there exists a sequence
\(\{h_{m}^{}\}\)
with
\(\lim_{m\to\infty} h_{m}^{}=+\infty\)
such that
\(T_{j}^{+}(h_{m}^{})<+\infty\). Moreover, if
\((G_{1})\)
holds, then there exists another sequence
\(\{h_{m}^{+}\}\)
with
\(\lim_{m\to\infty} h_{m}^{+}=+\infty\)
such that
\(\lim_{m\to\infty} T_{j}^{}(h_{m}^{+})=+\infty\).
Proof
For sufficiently large h, let
$$\begin{aligned}& r^{+}(h)=\max\bigl\{ \sqrt{x^{2}+y^{2}} : 1/2y^{2}+G(x)=h \bigr\} , \\& r^{}(h)=\min\bigl\{ \sqrt{x^{2}+y^{2}} : 1/2y^{2}+G(x)=h \bigr\} . \end{aligned}$$
Then for for any \(j\in\Bbb {N}\) and \(m\in\Bbb {N}\), let \(r^{(m)}\ge\max\{ m,K_{j+1} \}\) and choose \(h^{}_{m}\) sufficiently large such that \(r^{}(h^{}_{m})\ge\zeta_{j+1}(r^{(m)})\). Let
$$a=\inf\bigl\{ \theta'(t;t_{0},z_{0}) : z_{0}\in\gamma_{h^{}_{m}}, r^{(m)}\leq\bigl\vert z(t;t_{0},z_{0})\bigr\vert \leq\zeta_{j+1} \bigl(r^{+}\bigl(h^{}_{m}\bigr)\bigr) \bigr\} . $$
It is easy to see that
$$a\ge\frac{\inf\{ xg(x)(\mu_{0}+P)\vert x\vert +y^{2} : x^{2}+y^{2}\ge(r^{(m)})^{2} \}}{(\zeta_{j+1}(r^{+}(h^{}_{m})))^{2}}>0. $$
Denote by \(L_{m}= \frac{2(j+1)\pi}{a}\). For any solution \(z(t;t_{0},z_{0})\), \(z_{0}\in\gamma_{h^{}_{m}} \), there are two cases as follows.

(1)
If \(r^{(m)}\leq \vert z(t;t_{0},z_{0})\vert \leq \zeta_{j+1}(r^{+}(h^{}_{m}))\) for \(t\in[t_{0},t_{0}+L_{m}]\), then
$$\theta(t_{0}+L_{m};t_{0},z_{0}) \theta_{0}= \int_{t_{0}}^{t_{0}+L_{m}}\theta '(t)\,dt \leqaL_{m}\leq2(j+1)\pi. $$

(2)
If there exists a time \(t_{1}'\in [t_{0},t_{0}+L_{m}) \) such that \(\vert z(t_{1}';t_{0},z_{0})\vert < r^{(m)}\) or \(\vert z(t_{1}';t_{0},z_{0})\vert >\zeta _{j+1}(r^{+}(h^{}_{m}))\), then we have \(t_{1}\in [t_{0},t_{1}') \) such that \(\vert z(t;t_{0},z_{0})\vert \ge R_{0}\) for \(t\in[t_{0},t_{1}]\) and
$$\bigl\vert z(t_{1};t_{0},z_{0})\bigr\vert < r^{(m)},\qquad \vert z_{0}\vert \ge r^{} \bigl(h^{}_{m}\bigr)\ge\zeta_{j+1}\bigl(r^{(m)}\bigr), $$
or
$$\bigl\vert z(t_{1};t_{0},z_{0})\bigr\vert > \zeta_{j+1}\bigl(r^{+}\bigl(h^{}_{m}\bigr)\bigr),\qquad \vert z_{0}\vert \le r^{+}\bigl(h^{}_{m}\bigr). $$
According to Lemma 2.4, it follows that \(z(t;t_{0},z_{0})\) completes at least \(j+1\) clockwise turns in \([t_{0},t_{1}]\). That is,
$$\theta(t_{1};t_{0},z_{0})\theta_{0} \leq2(j+1)\pi, $$
which implies that
$$\begin{aligned} \theta(t_{0}+L_{m};t_{0},z_{0}) \theta_{0} =& \theta(t_{0}+L_{m};t_{0},z_{0}) \theta(t_{1};t_{0},z_{0})+\theta (t_{1};t_{0},z_{0})\theta_{0} \\ < & \pi2(j+1)\pi=(2j+1)\pi \end{aligned}$$
by using Lemma 2.3.
Hence for both cases, we show that the solution \(z(t;t_{0},z_{0})\), \(z_{0}\in\gamma_{h^{}_{m}} \) completes at least j clockwise turns in \([t_{0},t_{0}+L_{m}]\). Moreover, the above argument shows that for \(T\ge L_{m}\), the solution \(z(t;t_{0},z_{0})\), \(z_{0}\in\gamma_{h^{}_{m}} \) completes at least j clockwise turns in \([t_{0},t_{0}+T]\), from which it follows that \(T_{j}^{+}(h_{m}^{})\leq L_{m}<+\infty\).
By the way, using the same method as employed above we can prove that if \(\vert z_{0}\vert \ge\zeta_{j+1}(K_{j+1})\) then \(T_{j}(t_{0},z_{0})\) is well defined.
On the other hand, from the assumption \((G_{1})\), we have a sequence \(\{x_{m}\}\) with \(x_{m}\to+\infty\) as \(m\to\infty\), such that \(\lim_{m\to+\infty}\frac{G(x_{m})}{x_{m}^{2}}=0\). For any given \(L>0\), choose \(m_{0}\), such that
$$\frac{G(x_{m})}{x_{m}^{2}}\le\frac{1}{9L^{2}},\quad\hbox{and}\quad \sqrt{G(x_{m})}\geq2E, \quad\hbox{for } m\ge m_{0}, $$
where \(E=E(L)\) is defined as in Lemma 2.1.
Consider the solution \(z(t;t_{0},z_{0})\) starting from \(z_{0}\in \gamma_{h^{+}_{m}}\), where
$$h^{+}_{m}=\bigl(\sqrt{G(x_{m})}+E\bigr)^{2}. $$
If \(T_{j}(t_{0},z_{0})\le L\), then Lemma 2.1 implies that
$$G(x_{m})\le\frac{1}{2}y^{2}(t;t_{0},z_{0})+G \bigl(x(t;t_{0},z_{0})\bigr)\le\Bigl(\sqrt {h_{m}^{+}}+E\Bigr)^{2},\quad \hbox{for } t_{0}\le t\le t_{0}+T_{j}(t_{0},z_{0}), $$
which implies that
$$\frac{\vert x'(t;t_{0},z_{0})\vert }{\sqrt{2((\sqrt {h_{m}^{+}}+E)^{2}G(x(t;t_{0},z_{0})))}}\le1, \quad\hbox{for } t_{0}\le t\le t_{0}+T_{j}(t_{0},z_{0}). $$
Note that the solution \(z(t;t_{0},z_{0})\), \(z_{0}\in\gamma_{h^{+}_{m}} \), completes at least j clockwise turns around the origin O in \([t_{0},t_{0}+T_{j}(t_{0},z_{0})]\), then it intersects with the xaxis and yaxis which implies that there exist \(t_{1},t_{2}\in [t_{0},t_{0}+T_{j}(t_{0},z_{0})]\) such that
$$x(t_{1})=0,\qquad y(t_{2})=0\quad\hbox{and}\quad x'(t)=y(t)>0\quad\hbox{for } t\in(t_{1},t_{2}) $$
or
$$y(t_{1})=0,\qquad x(t_{2})=0\quad\hbox{and}\quad x'(t)=y(t)< 0\quad\hbox{for } t\in(t_{1},t_{2}). $$
If \(y(t_{2})=0\) then \(G(x(t_{2}))\ge G(x_{m})\), which implies \(x(t_{2})\ge x_{m}\). If \(y(t_{1})=0\) then \(G(x(t_{1}))\ge G(x_{m})\), which implies \(x(t_{1})\ge x_{m}\). In both cases we have
$$\bigl\vert x(t_{2})x(t_{1})\bigr\vert \ge x_{m}0. $$
Therefore
$$\begin{aligned} T_{j}(t_{0},z_{0}) =& \int_{t_{0}}^{t_{0}+T_{j}(t_{0},z_{0})}\,dt\ge \int _{t_{1}}^{t_{2}}\,dt\ge \biggl\vert \int_{x(t_{1})}^{x(t_{2})}\frac{d s}{\sqrt{2((\sqrt {h_{m}^{+}}+E)^{2}G(s))}}\biggr\vert \\ \ge& \int_{0}^{x_{m}}\frac{d s}{\sqrt {2((\sqrt {h_{m}^{+}}+E)^{2}G(s))}}\ge\frac{1}{\sqrt{2}} \frac{x_{m}}{\sqrt {G(x_{m})}}\cdot\frac{\sqrt{G(x_{m})}}{\sqrt {h_{m}^{+}}+E} \\ =&\frac{1}{\sqrt{2}}\cdot\frac{x_{m}}{\sqrt{G(x_{m})}}\cdot \biggl(1+\frac{2E}{\sqrt{G(x_{m})}} \biggr)^{1}\ge\frac{3L}{2\sqrt {2}}>L. \end{aligned}$$
This is a contradiction. Hence we have proved that \(T_{j}(t_{0},z_{0})>L\) for \(z_{0}\in\gamma_{h^{+}_{m}}\) and \(f\in{\mathcal {B}}_{\mu_{0}}\), which implies that \(T_{j}^{}(h_{m}^{+})\ge L\). Since L is arbitrary we have
$$T_{j}^{}\bigl(h_{m}^{+}\bigr)\to+\infty\quad\hbox{as } m\to \infty. $$
The lemma is thus proved. □
Now we are in the position to prove Theorem 1.2.
Proof of Theorem 1.2
It follows from Lemma 2.5 that there exist sequences \(\{ h^{}_{l}\}\) and \(\{h^{+}_{l}\}\) such that \(\lim_{l\to+\infty}h^{}_{l}=+\infty\), \(\lim_{l\to+\infty }h^{+}_{l}=+\infty\), \(T_{j}^{+}(h^{}_{l})<+\infty\), and \(\lim_{l\to+\infty }T_{j}^{}(h^{+}_{l})=+\infty\). Let \(h_{m}^{}\) be large enough such that \(r^{}(h_{m}^{})\ge K_{j+1}\), \(h^{}_{m}\ge\zeta_{j+1}(R_{0})\) and \(T_{j}(0,z_{0})\) is well defined for every solution \(z(t;z_{0})=(x(t;x_{0},y_{0}),y(t;x_{0},y_{0}))\) starting from \(z_{0}\in\gamma _{h_{m}^{}}\) at \(t=0\). Then there is \(m_{j}^{*}\in{\mathbb {N}}\), such that, for every \(k\ge m_{j}^{*}\), where k primes with j, we can choose \(l=l(m,k)\) sufficiently large, such that \(h_{m}^{}< h_{l}^{+}\) and
$$ T_{j}^{+}\bigl(h_{m}^{}\bigr)< 2k \pi< T_{j}^{}\bigl(h_{l}^{+}\bigr). $$
(2.6)
Denote by \({\mathcal {A}}_{m,l}\) the annulus bounded by \(\gamma_{h_{m}^{}}\) and \(\gamma_{h_{l}^{+}}\). Consider the Poincaré map
$${\mathcal {P}}_{k}:\quad(x_{0},y_{0})\mapsto \bigl(x(2k\pi;x_{0},y_{0}), y(2k\pi;x_{0},y_{0}) \bigr). $$
The uniqueness of the solution to initial value problems for equation (2.2) guarantees that \({\mathcal {P}}_{k}\) is an areapreserving homeomorphism such that \({\mathcal {P}}_{k}(O)=O\). Moreover, \({\mathcal {A}}_{m,l}\) are annuli bounded by strictly starshaped Jordan curves around the origin and the inequality (2.6) implies that \({\mathcal {P}}_{k}\) is boundary twisting on the annulus \({\mathcal {A}}_{m,l}\), that is,
$$\begin{aligned}& \theta(2k\pi;t_{0},z_{0})\theta_{0}< 2j\pi, \quad\hbox{for } z_{0}\in \gamma_{h_{m}^{}}, \\& \theta(2k\pi;t_{0},z_{0})\theta_{0}>2j\pi, \quad\hbox{for } z_{0}\in \gamma_{h_{l}^{+}}. \end{aligned}$$
Hence we can use a recent version of the PoincaréBirkhoff fixed point theorem^{Footnote 1} (Rebelo, [8], Corollary 2) to obtain at least two fixed points of \({\mathcal {P}}_{k}\), \(z_{1},z_{2}\in\mathcal{A}_{m,l}\) with \(z_{1}\neq z_{2}\). These fixed points are initial points of two \(2k\pi\)periodic solutions
$$z_{1}(t)=z(t;z_{1}),\qquad z_{2}(t)=z(t;z_{2}) $$
of (2.2), such that \(z_{1}(t)\) and \(z_{2}(t)\) satisfy
$$\theta(2k\pi;z_{i})\theta(0;z_{i})=2j\pi, \quad i=1,2. $$
We assert that
$$\min\bigl\{ \bigl\vert z_{i}(t)\bigr\vert \mid t\in[0,2k\pi) \bigr\} >R_{2},\quad i=1,2. $$
If the assertion would not hold, then there exists \(t_{1}'\in(0,2k\pi]\) such that \(\vert z_{1}(t_{1}')\vert \leq R_{2}\) (or \(t_{2}'\in(0,2k\pi]\) such that \(\vert z_{2}(t_{2}')\vert \leq R_{2}\)). It follows that there exists \(t_{1}\in(0,t_{1}')\) such that \(\vert z_{1}(0)\vert =\vert z_{1}\vert \ge r^{}(h^{}_{m}) \ge\zeta_{j+1}(K_{j+1})\), \(\vert z_{1}(t_{1})\vert \le K_{j+1}\), and \(\vert z_{1}(t)\vert \ge R_{0}>R_{2}\) for \(t\in[0,t_{1}]\). According to Lemma 2.4, we have
$$\theta(t_{1};z_{1})\theta(0;z_{1})\le2(j+1) \pi. $$
Then Lemma 2.3 implies that
$$\begin{aligned} \theta(2k\pi;z_{1})\theta(0;z_{1}) =&\theta(2k\pi ;z_{1})\theta(t_{1};z_{1})+ \theta(t_{1};z_{1})\theta(0;z_{1}) \\ \leq& \pi2(j+1)\pi=(2j+1)\pi. \end{aligned}$$
This is a contradiction. Hence \(z_{1}(t)>R_{2}\), \(\forall t \in\Bbb {R}\) (or \(z_{2}(t)>R_{2}\), \(\forall t\in \Bbb {R}\)).
Therefore we have proved that \(z_{i}(t)\) is exactly the \(2k\pi\)periodic solution of the equation
$$x'=y,\quad\quad y'=f(x)+p(t) $$
and \(z_{i}(t)\) completes exactly j clockwise turns around the origin O in \([0,2k\pi)\) which implies \(x_{i}(t)\) has exactly 2j zeros in \([0,2k\pi)\), where \(x_{i}(t)=\Pi_{1}(z_{i}(t))\), \(\Pi_{1}(\cdot)\) is the projection for the first component, \(i=1,2\).
Moreover, in the above argument we can choose the same \({\mathcal {A}}_{m,l}\) for any \(f\in{\mathcal {B}}_{\mu_{0}}\), \({\mu_{0}}\) sufficiently small. Further, denote the two \(2k\pi\)periodic solutions obtained above by \(z_{f,1}(t)= z_{1}(t)\) and \(z_{f,2}(t)=z_{2}(t)\). It is easy to find a compact annulus \({\mathcal {C}}_{m,l}\) such that \(z_{f,i}(t)\), \(i=1,2\), are in \({\mathcal {C}}_{m,l}\).
Now consider equation (2.1) with \(g, p: {\mathbb {R}}\to {\mathbb {R}}\) continuous and p 2πleast periodic. We construct the equations
$$x'=y,\quad\quad y'=f_{q}(x)+p(t) $$
with \(f_{q}\in{\mathcal {B}}_{\mu_{0}}\), \(f_{q}\to g\) in \({\mathcal {C}}_{m,l}\) as \(q\to\infty\). As showed previously, we have two sequences of \(2k\pi\)periodic solutions \(z_{1}^{(q)}(t)=z_{f_{q},1}(t)\) and \(z_{2}^{(q)}(t)=z_{f_{q},2}(t)\) for the equation
$$x'=y,\quad\quad y'=f_{q}(x)+p(t) $$
in \({\mathcal {C}}_{m,l}\). Moreover, \(x_{i}^{(q)}(t)=\Pi_{1}(z_{i}^{(q)}(t))\) has exactly 2j zeros in \([0,2k\pi)\), \(i=1,2\). If \(t_{*}\) is the zero time of \(x_{i}^{(q)}(t)\), i.e.
\(x_{i}^{(q)}(t_{*})=0\), then \(\vert x_{i}^{(q)'}(t_{*})\vert \ge c\), where c is a positive constant which is independent of q and i. By a standard compactness argument (see, for example, in [9]), we obtain a \(2k\pi\)periodic solution \(z_{j,k}(t)\) (as the limitation for some subsequence of \(\{z_{1}^{(q)}(t)\}\) and \(\{z_{2}^{(q)}(t)\}\)) for equation (2.1) in \({\mathcal {C}}_{m,l}\). Then we obtain a \(2k\pi\)periodic solution \(x_{j,k}(t)=\Pi_{1}(z_{j,k}(t))\) for equation (1.1). Since \(p(t)\) is 2πleast periodic, \(x_{j,k}(t)\) has exactly 2j zeros in \([0,2k\pi)\) and k is prime with j, and \(x_{j,k}(t)\) has a minimal period \(2k\pi\). Actually, the solution \(z_{j,k}(t)=(r_{k}(t)\cos\theta _{k}(t),r_{k}(t)\sin\theta_{k}(t))\) satisfies
$$ r_{k}(t)>0,\quad \quad\theta_{k}'(t)< 0, \quad\forall t\in\Bbb {R}. $$
(2.7)
Moreover, \(z_{j,k}(t)\) moves exactly j clockwise turns around the origin O in \([0,2k\pi)\). Suppose, by contradiction, that \(z_{j,k}(t)\) has a minimal period \(2l\pi\) with \(l\in\Bbb {N}\) such that \(1\leq l< k\). Then (2.7) implies that \(z_{j,k}(t)\) moves exactly q clockwise turns around the origin O in \([0,2l\pi)\) with \(q\in\Bbb {N}\) such that \(1\leq q< j\). On the other hand, from the additivity property of the rotation we easily see that \(z_{j,k}(t)\) moves exactly kq and lj clockwise turns around the origin O in \([0,2lk\pi)\). It follows that \(kq=lj\). Then
$$k/j=l/q\quad\hbox{with } 1\leq l< k \hbox{ and } 1\leq q< j. $$
This contradicts that k is prime with j. The same argument works for the case when \(j=1\). Hence we have proved that \(z_{j,k}(t)\) (\(x_{j,k}(t)\)) has a minimal period \(2k\pi\).
Theorem 1.2 is thus proved. □