Rewrite equation (1.1) as the following equivalent form:
$$ x'=y,\quad \quad y'=-g(x)+p(t). $$
(2.1)
We will apply the Poinacré-Birkhoff twist theorem to obtain the existence of the \(2k\pi\)-periodic solution for equation (2.1) under the assumption of a weak sub-quadratic potential. The existence of a 2π-periodic solution then follows from the Massera theorem.
The key point of applying the Poincaré-Birkhoff twist theorem is constructing an annulus \(\mathcal {A}\) in the \((x,y)\) phase plane bounded by two star-shaped curves \(\gamma^{+}\) and \(\gamma^{-}\), such that the solution starting from \(\gamma^{+}\) and \(\gamma^{-}\) will move more than and less than j clock-wise 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 clock-wise 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 Poinacré-Birkhoff 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 \(H(t,x,y)=\frac{y^{2}}{2}+K(x^{2}+y^{2})[F(x)-xp(t)]+(1-K(x^{2}+y^{2}))x^{2}\), \(F(x)=\int_{0}^{x} f(\xi)\,d\xi\) 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
$$\bigl\vert l(t)-l(t_{0})\bigr\vert \leq E \quad\textit{for } \vert t-t_{0}\vert \leq L \textit{ and } \vert z_{0} \vert \gg1, $$
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 existence-uniqueness 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
$$\bigl\vert l(t)-l(t_{0})\bigr\vert \leq E \quad\hbox{for } \vert t-t_{0}\vert \leq L \hbox{ and } \vert z_{0} \vert \gg1. $$
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 clock-wise 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):
$$\begin{aligned}& \bigl\vert z(t_{0})\bigr\vert \leq r, \bigl\vert z(t_{1})\bigr\vert \geq\zeta_{j} (r) \textit{ or } \bigl\vert z(t_{0})\bigr\vert \geq\zeta_{j} (r), \bigl\vert z(t_{1})\bigr\vert \leq r \textit{ whenever } r\geq K_{j} \\& \textit{and } \bigl\vert z(t)\bigr\vert \geq R_{0}, \forall t\in [t_{0},t_{1}] \quad\Longrightarrow\quad \\& z(t)\textit{ moves at least }j\textit{ clock-wise turns on }[t_{0}, t_{1}]\textit{ around the origin }O. \end{aligned}$$
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 clock-wise 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 \(r\ge K_{1}\),
$$\begin{aligned}& \vert z(t_{0})\vert \le r, \vert z(t_{1})\vert \ge\zeta_{1} (r) \hbox{ and } \vert z(t)\vert \ge R_{0}, \forall t\in [t_{0},t_{1}] \\& \hbox{or } \vert z(t_{0})\vert \ge\zeta_{1} (r), \vert z(t_{1})\vert \le r \hbox{ and } \vert z(t)\vert \ge R_{0}, \forall t\in [t_{0},t_{1}] \quad \Longrightarrow\\& z(t)\hbox{ moves at least one clock-wise turn on }[t_{0}, t_{1}]\hbox{ around the origin }O. \end{aligned}$$
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 clock-wise 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 clock-wise 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
$$\theta(t_{0}+L_{m};t_{0},z_{0})- \theta_{0}= \int_{t_{0}}^{t_{0}+L_{m}}\theta '(t)\,dt \leq-aL_{m}\leq-2(j+1)\pi. $$
-
(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 \(j+1\) clock-wise turns in \([t_{0},t_{1}]\). That is,
$$\theta(t_{1};t_{0},z_{0})-\theta_{0} \leq-2(j+1)\pi, $$
which implies that
$$\begin{aligned} \theta(t_{0}+L_{m};t_{0},z_{0})- \theta_{0} =& \theta(t_{0}+L_{m};t_{0},z_{0})- \theta(t_{1};t_{0},z_{0})+\theta (t_{1};t_{0},z_{0})-\theta_{0} \\ < & \pi-2(j+1)\pi=-(2j+1)\pi \end{aligned}$$
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 clock-wise 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 clock-wise 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 clock-wise turns around the origin O in \([t_{0},t_{0}+T_{j}(t_{0},z_{0})]\), then it intersects with the x-axis and y-axis 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 area-preserving homeomorphism such that \({\mathcal {P}}_{k}(O)=O\). Moreover, \({\mathcal {A}}_{m,l}\) are annuli bounded by strictly star-shaped 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 Poincaré-Birkhoff fixed point theoremFootnote 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
$$\theta(t_{1};z_{1})-\theta(0;z_{1})\le-2(j+1) \pi. $$
Then Lemma 2.3 implies that
$$\begin{aligned} \theta(2k\pi;z_{1})-\theta(0;z_{1}) =&\theta(2k\pi ;z_{1})-\theta(t_{1};z_{1})+ \theta(t_{1};z_{1})-\theta(0;z_{1}) \\ \leq& \pi-2(j+1)\pi=-(2j+1)\pi. \end{aligned}$$
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 clock-wise 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 2π-least 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 2π-least 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 clock-wise 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 clock-wise 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 clock-wise 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. □