At first, we consider the auxiliary autonomous system

$$ x'=ay^{+}-by^{-}, \qquad y'=-g(x). $$

(2.1)

The orbits of system (2.1) are curves determined by the equation

$$ \Gamma_{c}{:}\quad \frac{1}{2}a{y^{+}}^{2}+ \frac{1}{2}b{y^{-}}^{2}+G(x)=c,$$

(2.2)

where *c* is an arbitrary constant. We can easily prove the following lemma.

### Lemma 2.1

*Assume that condition*
\((h_{1})\)
*holds*. *Then there exists a constant*
\(c_{0}>0\)
*such that*, *for any*
\(c>c_{0}\), \(\Gamma_{c}\)
*is a closed curve which is star*-*shaped around the origin*
*O*.

From Lemma 2.1 we know that, for \(c\geq c_{0}\), each \(\Gamma_{c}\) intersects with the *x*-axis at two points \((w(c), 0)\) and \((d(c), 0)\), where \(w(c)\) and \(d(c)\) are continuous and satisfy

$$w(c)< 0< d(c),\qquad G\bigl(w(c)\bigr)=G\bigl(d(c)\bigr)=c. $$

Let \((x_{c}(t), y_{c}(t))\) be the solution of system (2.1) lying on the curve \(\Gamma_{c}\) with \(c\geq c_{0}\). Obviously, \((x_{c}(t), y_{c}(t))\) is periodic. Let us denote by \(T(c)\) the least period of \((x_{c}(t), y_{c}(t))\). From the first equation of (2.1) and (2.2) we have that

$$T(c)=\biggl(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\biggr) \int _{w(c)}^{d(c)}\frac{ds}{\sqrt{2(c-G(s))}}. $$

By the definition, \(T(c)\) is continuous for \(c\geq c_{0}\).

Now we perform some phase-plane analysis for system (1.1)′. Let \((x(t), y(t))=(x(t, x_{0}, y_{0}), y(t, x_{0}, y_{0}))\) be the solution of system (1.1)′ satisfying the initial condition

$$x(0)=x_{0}, \qquad y(0)=y_{0}. $$

### Lemma 2.2

*Assume that conditions*
\((h_{i})\) (\(i=1,2,3\)) *hold*. *Then each solution*
\((x(t), y(t))\)
*of system* (1.1)′ *exists uniquely on the whole*
*t*-*axis*.

### Proof

The proof follows directly from the fact that the nonlinearities are locally Lipschitz continuous and all have at most linear growth. □

According to Lemma 2.2, the Poincaré map \(P: \mathbf{R}^{2}\to \mathbf{R}^{2}\) is well defined by

$$P: (x_{0}, y_{0})\to(x_{1}, y_{1})= \bigl(x(2\pi, x_{0}, y_{0}), y(2\pi, x_{0}, y_{0})\bigr). $$

Clearly, the Poincaré map *P* is an area-preserving homeomorphism. The fixed points of *P* correspond to the 2*π* periodic solutions of system (1.1)′.

Now, we take the polar coordinates transformation \(x=r\cos\theta\), \(y=r\sin\theta\) to system (1.1)′. Under this transformation, system (1.1)′ becomes

$$ \left\{\textstyle\begin{array}{ll} \frac{dr}{dt}=&r(a\sin^{+}\theta-b\sin^{-}\theta)\cos \theta-g(r\cos \theta)\sin\theta+p_{1}(t, r\sin\theta)\cos\theta \\ &{}+p_{2}(t, r\cos\theta)\sin\theta, \\ \frac{d\theta}{dt}=&-(a\sin^{+}\theta-b\sin^{-}\theta)\sin\theta - \frac{1}{r}g(r\cos\theta)\cos\theta -\frac{1}{r}p_{1}(t, r \sin\theta)\sin\theta \\ &{}+\frac{1}{r}p_{2}(t, r\cos\theta)\cos\theta. \end{array}\displaystyle \right.$$

(2.3)

Denote by \((r(t), \theta(t))=(r(t, r_{0}, \theta_{0}), \theta(t, r_{0}, \theta_{0}))\) the solution of (2.3) with the initial value

$$r(0)=r_{0}, \qquad\theta(0)=\theta_{0}, $$

with \(x_{0}=r_{0}\cos\theta_{0}\), \(y_{0}=r_{0}\sin\theta_{0}\). Clearly, the Poincaré map *P* can be written in the polar coordinate form \(P: (r_{0}, \theta_{0})\to(r^{*}, \theta^{*})\) with

$$r^{*}=r(2\pi, r_{0}, \theta_{0}), \qquad\theta^{*}=\theta(2\pi, r_{0}, \theta_{0})+2l\pi, $$

where *l* is an arbitrary integer.

Applying the polar coordinate transformation \(x=\rho\cos\varphi\), \(y=\rho\sin\varphi\) to system (2.1), we get

$$ \left\{\textstyle\begin{array}{l} \frac{d\rho}{dt}=\rho(a\sin^{+}\varphi-b\sin^{-}\varphi)\cos \varphi-g(\rho\cos\varphi)\sin\varphi, \\ \frac{d\varphi}{dt}=-(a\sin^{+}\varphi-b\sin^{-}\varphi)\sin\varphi - \frac{1}{\rho}g(\rho\cos\varphi)\cos\varphi. \end{array}\displaystyle \right.$$

(2.4)

Denote by \((\rho(t), \varphi(t))=(\rho(t, \rho_{0}, \varphi_{0}), \varphi(t, \rho_{0}, \varphi_{0}))\) the solution of (2.4) satisfying the initial value

$$\rho(0)=\rho_{0}, \qquad\varphi(0)=\varphi_{0}. $$

Using conditions \((h_{i})\) (\(i=1,2,3\)), it is not hard to prove the following lemma.

### Lemma 2.3

*Assume that conditions*
\((h_{i})\) (\(i=1,2,3\)) *hold*. *Then there exist constants*
\(\gamma>1\)
*and*
\(R_{0}>0\)
*such that*

$$\frac{r_{0}}{\gamma}\leq r(t)\leq\gamma r_{0}, \quad \forall t\in[0, 2\pi], r_{0}\geq R_{0}. $$

*In particular*, *under conditions*
\((h_{i})\) (\(i=1, 2\)), \(\rho(t)\)
*satisfies the inequality*

$$\frac{\rho_{0}}{\gamma}\leq\rho(t)\leq\gamma\rho_{0}, \quad \forall t\in[0, 2\pi], \rho_{0}\geq R_{0}. $$

### Lemma 2.4

*Assume that conditions*
\((h_{i})\) (\(i=1,2,3\)) *hold*, *and let*

$$\Phi(r_{0}, \theta_{0})=\varphi(2\pi, r_{0}, \theta_{0})-\theta_{0}; \qquad \Theta(r_{0}, \theta_{0})=\theta(2\pi, r_{0}, \theta_{0})- \theta_{0}. $$

*Then*, *for any sufficiently small*
*ε*, *there exists a positive constant*
*ζ*
*such that*

$$\big|\Phi(r_{0}, \theta_{0})-\Theta(r_{0}, \theta_{0})\big|\leq\varepsilon\quad\textit{for } r_{0}\geq\zeta. $$

### Proof

Let \((\bar{x}(t), \bar{y}(t))=(\bar{x}(t, x_{0}, y_{0}), \bar{y}(t, x_{0}, y_{0}))\) be the solution of (2.1) with \((\bar{x}(0), \bar{y}(0))=(x_{0}, y_{0})\). It is noted that \((x(t), y(t))=(x(t, x_{0}, y_{0}), y(t, x_{0}, y_{0}))\) is a solution of system (1.1)′ with \((x(0), y(0))=(x_{0}, y_{0})\). Set

$$\begin{aligned}& u(t)=u(t, x_{0}, y_{0})=x(t, x_{0}, y_{0})-\bar{x}(t, x_{0}, y_{0}),\\& v(t)=v(t, x_{0}, y_{0})=y(t, x_{0}, y_{0})-\bar{y}(t, x_{0}, y_{0}). \end{aligned}$$

Then we have

$$\begin{aligned}& \frac{du(t)}{dt}=a\bigl[y^{+}(t)-\bar{y}^{+}(t)\bigr]-b\bigl[y^{-}(t)-\bar{y}^{-}(t)\bigr]+p_{1}\bigl(t, y(t)\bigr), \\& \frac{dv(t)}{dt}=-\bigl[g\bigl(x(t)\bigr)-g\bigl(\bar{x}(t)\bigr)\bigr]+p_{2}\bigl(t, x(t)\bigr). \end{aligned}$$

Let \(d(t)=\sqrt{u^{2}(t)+v^{2}(t)}\). Then we get

$$d'(t)\leq\delta d(t)+\big|p_{1}\bigl(t, y(t) \bigr)\big|+\big|p_{2}\bigl(t, x(t)\bigr)\big| $$

with \(\delta=\frac{1}{2}(\mu+L)\), \(\mu=\max\{a, b\}\). From condition \((h_{3})\) we have that, for any sufficiently small \(\eta>0\), there exists \(c_{\eta}>0\) such that

$$\big|p_{1}(t, y)\big|\leq\eta|y|+c_{\eta},\quad \forall(t,y)\in \mathbf{R}^{2} $$

and

$$\big|p_{2}(t, x)\big|\leq\eta|x|+c_{\eta}, \quad\forall(t,x)\in \mathbf{R}^{2}. $$

Therefore, we obtain

$$d'(t)\leq\delta d(t)+\eta\bigl(\big|x(t)\big|+\big|y(t)\big|\bigr)+2c_{\eta}. $$

Solving this inequality, we get

$$d(t)\leq\eta e^{2\pi\delta} \int_{0}^{2\pi}\bigl(\big|x(t)\big|+\big|y(t)\big|\bigr)\,dt+A_{\eta} \leq \sqrt{2}\eta e^{2\pi\delta} \int_{0}^{2\pi}\sqrt{x^{2}(t)+y^{2}(t)}\,dt+A_{\eta}, $$

where \(A_{\eta}=\frac{2c_{\eta} e^{2\pi\delta}}{\delta}\). It follows from Lemma 2.3 that, for \(t\in[0, 2\pi]\),

$$d(t)\leq\eta\beta r_{0}+A_{\eta}, $$

where \(\beta=2\sqrt{2}\pi\gamma e^{2\pi\delta}\). Write \(\psi(t)=\psi(t, r_{0}, \theta_{0})=\varphi(t, r_{0}, \theta_{0})-\theta(t, r_{0},\theta_{0})\). Clearly, if \(|\psi(t)|<\pi\), then \(\psi(t)\) is just the angle between the vectors \((x(t), y(t))\) and \((\bar{x} (t), \bar{y}(t))\). Hence, we have

$$\cos\psi(t)=\frac{r^{2}(t)+\rho^{2}(t)-d^{2}(t)}{2r(t)\rho(t)}\geq 1-\frac{d^{2}(t)}{2r(t)\rho(t)}. $$

It follows that

$$\sin^{2}\frac{\psi(t)}{2}\leq\frac{d^{2}(t)}{4r(t)\rho(t)}. $$

According to Lemma 2.3, we have that if *η* is sufficiently small and \(r_{0}\) is large enough, then

$$\bigg|\sin\frac{\psi(t)}{2}\bigg|\leq\frac{d(t)}{2\sqrt{r(t)\rho(t)}} \leq\frac{\gamma(\eta\beta r_{0}+A_{\eta})}{2r_{0}}\leq \frac{\varepsilon}{4}. $$

Since \(\psi(0)=0\) and \(\psi(t)\) varies continuously as *t* increases from 0 to 2*π*, we have

$$\big|\psi(t)\big|\leq4\bigg|\sin\frac{\psi(t)}{2}\bigg|\leq\varepsilon. $$

Consequently, we have that there exists \(\zeta>0\) such that, for \(r_{0}\geq\zeta\),

$$\big|\Phi(r_{0}, \theta_{0})-\Theta(r_{0}, \theta_{0})\big|\leq\varepsilon, \quad \forall r_{0}\geq\zeta. $$

□

### Lemma 2.5

*Assume that conditions*
\((h_{i})\) (\(i=1,2\)) *and*
\((\tau)\)
*hold*. *Then there exists a constant*
\(\omega>0\)
*such that*, *for*
\(t\in\mathbf{R}\)
*and*
*k*
*large enough*,

$$\varphi'(t)\leq-\omega, $$

*with*
\((\rho_{0}\cos\varphi_{0}, \rho_{0}\sin\varphi_{0})\in\Gamma_{a_{k}}\)
*or*
\((\rho_{0}\cos\varphi_{0}, \rho_{0}\sin\varphi_{0})\in\Gamma_{b_{k}}\).

### Proof

From the definition of \(T(c)\) and condition \((\tau)\) we know that, for each \(k\in\mathbf{N}\),

$$T(a_{k})\leq\frac{2\pi}{n}-m\sigma; \qquad T(b_{k})\geq \frac{2\pi}{n}+m\sigma. $$

In what follows, without loss of generality, we assume that the sequence \(\{T(b_{k})\}\) is bounded. Otherwise, we can replace the sequence \(\{T(b_{k})\}\) with a bounded one because \(T(c)\) is continuous for *c* large enough. We shall only deal with the first case, and the second one can be proved similarly. Let us set

$$d_{k}=d(a_{k}), \qquad w_{k}=w(a_{k}). $$

Obviously, \(d_{k}\to+\infty\), \(w_{k}\to-\infty\) as \(k\to\infty\). Next, we prove that there exist two positive constants \(\nu_{i}\) (\(i=1, 2\)) such that

$$\liminf_{k\to\infty}\frac{g(d_{k})}{d_{k}}=\nu_{1}; \qquad \liminf_{k\to\infty}\frac{g(w_{k})}{w_{k}}=\nu_{2}. $$

Assume by contradiction that

$$\liminf_{k\to\infty}\frac{g(d_{k})}{d_{k}}=0. $$

Then there exists a subsequence of \(\{d_{k}\}\) (we still denote it by \(\{d_{k}\}\)) such that

$$\lim_{k\to\infty}\frac{g(d_{k})}{d_{k}}=0. $$

Set

$$\frac{g(d_{k})}{d_{k}}=\varepsilon_{k}. $$

We have that \(\varepsilon_{k}\to0\) as \(k\to\infty\). From condition \((h_{2})\) we know that, for \(0< x\leq d_{k}\),

$$\begin{aligned} \biggl\vert \frac{g(x)}{x}-\varepsilon_{k} \biggr\vert &= \biggl\vert \frac {g(x)}{x}-\frac{g(d_{k})}{d_{k}} \biggr\vert \leq \frac{|g(d_{k})-g(x)|}{x}+\frac{|g(d_{k})(d_{k}-x)|}{d_{k} x} \\&\leq \frac{L(d_{k}-x)}{x}+\biggl(L+\frac{|g(0)|}{d_{k}}\biggr)\frac{(d_{k}-x)}{x}. \end{aligned} $$

For simplicity, we assume \(g(0)=0\). Then we get that, for \(0< x\leq d_{k}\),

$$ \biggl\vert \frac{g(x)}{x}-\varepsilon_{k} \biggr\vert \leq \frac{2L(d_{k}-x)}{x}. $$

(2.5)

Consequently, we have that, for \(0\leq x\leq d_{k}\),

$$g(x)\leq\varepsilon_{k} x+2L(d_{k}-x). $$

It follows that, for \(0\leq x\leq d_{k}\),

$$\begin{aligned} G(d_{k})-G(x)&= \int_{x}^{d_{k}}g(s)\,ds\leq \int_{x}^{d_{k}}\bigl[\varepsilon_{k} x+2L(d_{k}-x)\bigr]\,dx \\&=\frac{1}{2}\varepsilon_{k} \bigl(d_{k}^{2}-x^{2} \bigr)+L(d_{k}-x)^{2}. \end{aligned} $$

From the definition of *T* we have that

$$\begin{aligned} T(a_{k})&= m \int_{w_{k}}^{d_{k}}\frac{dx}{\sqrt{2(G(d_{k})-G(x))}}\geq m \int_{0}^{d_{k}}\frac{dx}{\sqrt{2(G(d_{k})-G(x))}} \\&\geq m \int_{0}^{d_{k}}\frac{dx}{\sqrt{\varepsilon_{k}(d_{k}^{2}-x^{2})+2L(d_{k}-x)^{2}}} \\&=m \int_{0}^{1}\frac{dt}{\sqrt{\varepsilon_{k}(1-t^{2})+2L(1-t)^{2}}}. \end{aligned} $$

Since

$$\liminf_{k\to\infty} \int_{0}^{1}\frac{dt}{\sqrt{\varepsilon _{k}(1-t^{2})+2L(1-t)^{2}}}\geq \frac{1}{\sqrt{2L}} \int_{0}^{1}\frac{dt}{1-t}=+\infty, $$

we have that

$$\lim_{k\to\infty}T(a_{k})=+\infty. $$

This is a contradiction because \(T(a_{k})\) is a bounded sequence. Therefore, there exists a constant \(\nu_{1}>0\) such that

$$ \liminf_{k\to\infty}\frac{g(d_{k})}{d_{k}}=\nu_{1}. $$

(2.6)

Similarly, there exists \(\nu_{2}>0\) such that

$$\liminf_{k\to\infty}\frac{g(w_{k})}{w_{k}}=\nu_{2}. $$

From condition \((h_{1})\) and (2.5), (2.6) we know that there exists sufficiently small \(\varepsilon_{0}>0\) such that, for *k* large enough and \(x\in[(1-\varepsilon_{0})d_{k}, d_{k}]\),

$$ \frac{g(x)}{x}\geq\frac{g(d_{k})}{d_{k}}-\frac{2L(d_{k}-x)}{x}\geq \frac{1}{2}\nu_{1}-\frac{2\varepsilon_{0} L}{1-\varepsilon_{0}}\geq\frac{1}{4} \nu_{1}. $$

(2.7)

Therefore, if \(\rho(t)\cos\varphi(t)\in[(1-\varepsilon_{0})d_{k}, d_{k}]\), then we have

$$\varphi'(t)\leq -\bigl(a\sin^{+}\varphi(t)-b\sin^{-}\varphi(t)\bigr) \sin\varphi(t)-\frac {1}{4}\nu_{1} \cos^{2}\varphi(t) \leq-\omega_{1}, $$

where \(\omega_{1}=\min\{a, b, \frac{1}{4}\nu_{1}\}\). Next, we deal with the case \(\rho(t)\cos\varphi(t)\in[0, (1-\varepsilon_{0})d_{k}]\). Set \(x_{k}=(1-\varepsilon_{0})d_{k}\). Assume that the line \(x=x_{k}\) intersects with the curve \(\Gamma_{a_{k}}\) at two points \((x_{k}, y_{k}^{+})\) and \((x_{k}, y_{k}^{-})\) with \(y_{k}^{-}<0<y_{k}^{+}\). Then we have

$$\frac{1}{2}a{y_{k}^{+}}^{2}+G(x_{k})=a_{k}=G(d_{k}), \qquad \frac{1}{2}b{y_{k}^{-}}^{2}+G(x_{k})=a_{k}=G(d_{k}). $$

Therefore, we get

$$y_{k}^{+}=\sqrt{\frac{2}{a}\bigl(G(d_{k})-G\bigl((1- \varepsilon_{0})d_{k}\bigr)\bigr)}, \qquad y_{k}^{-}=- \sqrt{\frac{2}{b}\bigl(G(d_{k})-G\bigl((1-\varepsilon_{0})d_{k} \bigr)\bigr)}. $$

From (2.7) we have

$$\frac{y_{k}^{+}}{x_{k}}=\frac{\sqrt{\frac{2}{a}(G(d_{k})-G((1-\varepsilon _{0})d_{k}))}}{(1-\varepsilon_{0})d_{k}} \geq\sqrt{\frac{\varepsilon_{0}\nu_{1}}{2a(1-\varepsilon_{0})}} $$

and

$$\frac{y_{k}^{-}}{x_{k}}=-\frac{\sqrt{\frac{2}{b}(G(d_{k})-G((1-\varepsilon _{0})d_{k}))}}{(1-\varepsilon_{0})d_{k}} \leq-\sqrt{\frac{\varepsilon_{0}\nu_{1}}{2b(1-\varepsilon_{0})}}. $$

Set

$$\beta^{+}=\arctan \sqrt{\frac{\varepsilon_{0}\nu_{1}}{2a(1-\varepsilon_{0})}}, \qquad \beta^{-}=\arctan \sqrt{ \frac{\varepsilon_{0}\nu_{1}}{2b(1-\varepsilon_{0})}}. $$

From condition \((h_{1})\) we know that there exists \(A>0\) such that \(g(x)\geq0\) for \(x\geq A\). If \(\rho(t)\cos\varphi(t)\in[A, (1-\varepsilon_{0})d_{k}]\) and \(\rho(t)\sin\varphi(t)\geq0\), then we have

$$\varphi'(t)\leq-a \sin^{2}\beta^{+}-\frac{1}{\rho}g( \rho\cos\varphi)\cos\varphi\leq-a \sin^{2}\beta^{+}. $$

If \(\rho(t)\cos\varphi(t)\in[0, A]\) and \(\rho(t)\sin\varphi (t)\geq 0\), then we have that, for \(\rho_{0}\) large enough,

$$\varphi'(t)\leq-a \sin^{2}\beta^{+}-\frac{1}{\rho}g( \rho\cos\varphi)\cos\varphi\leq -\frac{1}{2}a \sin^{2}\beta^{+}. $$

Similarly, we have that, if \(\rho(t)\cos\varphi(t)\in[A, (1-\varepsilon_{0})d_{k}]\) and \(\rho(t)\sin\varphi(t)\leq0\), then we have

$$\varphi'(t)\leq-b \sin^{2}\beta^{-}-\frac{1}{\rho}g( \rho\cos\varphi)\cos\varphi\leq-b \sin^{2}\beta^{-}. $$

If \(\rho(t)\cos\varphi(t)\in[0, A]\) and \(\rho(t)\sin\varphi (t)\leq 0\), then we have that, for \(\rho_{0}\) large enough,

$$\varphi'(t)\leq-b \sin^{2}\beta^{-}-\frac{1}{\rho}g( \rho\cos\varphi)\cos\varphi\leq -\frac{1}{2}b \sin^{2}\beta^{-}. $$

In conclusion, we have proved that there exists \(\omega_{2}>0\) such that

$$\varphi'(t)\leq-\omega_{2}, $$

with \((\rho_{0}\cos\varphi_{0}, \rho_{0}\sin\varphi_{0})\in\Gamma_{a_{k}}\), \(\rho(t)\cos\varphi(t)\geq0\) and *k* large enough. Similarly, we can prove that there exists \(\omega_{2}'>0\) such that

$$\varphi'(t)\leq-\omega_{2}', $$

with \((\rho_{0}\cos\varphi_{0}, \rho_{0}\sin\varphi_{0})\in\Gamma_{a_{k}}\), \(\rho(t)\cos\varphi(t)\leq0\) and *k* large enough. Let us set \(\omega=\min\{\omega_{2}, \omega_{2}'\}\). Then we have that

$$\varphi'(t)\leq-\omega, $$

with \((\rho_{0}\cos\varphi_{0}, \rho_{0}\sin\varphi_{0})\in\Gamma_{a_{k}}\) and *k* large enough. □

### Lemma 2.6

*Assume that conditions*
\((h_{i})\) (\(i=1,2\)) *and*
\((\tau)\)
*hold*, *and let*
\(\Phi(\rho_{0}, \varphi_{0})=\varphi(2\pi, \rho_{0}, \varphi_{0})-\varphi_{0}\). *Then there exist two positive constants*
*δ*
*and*
\(\varrho_{0}\)
*such that*

$$\begin{aligned}& \Phi(\rho_{0}, \varphi_{0})< -2n\pi-\delta,\qquad(\rho_{0}\cos\varphi_{0}, \rho_{0}\sin\varphi_{0})\in\Gamma_{a_{k}},\quad a_{k}\geq\varrho_{0};\\& \Phi(\rho_{0}, \varphi_{0})>-2n\pi+\delta,\qquad(\rho_{0}\cos\varphi_{0}, \rho_{0}\sin\varphi_{0})\in\Gamma_{b_{k}},\quad b_{k}\geq\varrho_{0}. \end{aligned}$$

### Proof

From Lemma 2.5 we have that there exists \(\varrho_{0}>0\) such that, for \(a_{k}\geq\varrho_{0}\) or \(b_{k}\geq\varrho_{0}\),

$$\varphi'(t)\leq-\omega, \qquad(\rho_{0}\cos \varphi_{0}, \rho_{0}\sin\varphi_{0})\in \Gamma_{a_{k}}\quad\text{or}\quad (\rho_{0}\cos\varphi_{0}, \rho_{0}\sin\varphi_{0})\in\Gamma_{b_{k}}. $$

Write \(\Phi(\rho_{0}, \varphi_{0})=-2l\pi-\phi\), where \(l\geq0\) is an integer, \(0\leq\phi<2\pi\). Let us denote by \(t_{\phi}\) the time for \(\varphi(t)\) to decrease from \(-2l\pi\) to \(-2l\pi-\phi\). If \((\rho_{0}\cos\varphi_{0}, \rho_{0}\sin\varphi_{0})\in\Gamma_{a_{k}}\), then we have

$$2\pi=l T(a_{k})+t_{\phi}=lm\tau(a_{k})+t_{\phi}. $$

Since \(t_{\phi}\leq T(a_{k})=m\tau(a_{k})\), we have

$$2\pi=lm\tau(a_{k})+t_{\phi}\leq(l+1)m\tau(a_{k})\leq (l+1) \biggl(\frac{2\pi}{n}-m\sigma\biggr). $$

It follows that \(l\geq n\). If \(l\geq n+1\), then we have

$$\Phi(\rho_{0}, \varphi_{0})=-2l\pi-\phi\leq-2(n+1)\pi. $$

If \(l=n\), then we have

$$t_{\phi}=2\pi-nm\tau(a_{k})\geq2\pi-n\biggl( \frac{2\pi}{n}-m\sigma\biggr)\geq nm\sigma. $$

Therefore, we get

$$-\phi= \int_{nT(a_{k})}^{nT(a_{k})+t_{\phi}}\varphi'(t)\,dt\leq -nm \sigma\omega. $$

Furthermore,

$$\Phi(\rho_{0}, \varphi_{0})\leq-2n\pi-nm\sigma\omega. $$

Set \(\delta=\min\{2\pi, nm\sigma\omega\}\). Then we have

$$\Phi(\rho_{0}, \varphi_{0})\leq-2n\pi-\delta, \qquad ( \rho_{0}\cos\varphi_{0}, \rho_{0}\sin \varphi_{0})\in\Gamma_{a_{k}}, \quad a_{k}\geq \varrho_{0}. $$

Similarly, we can prove

$$\Phi(\rho_{0}, \varphi_{0})\geq-2n\pi+\delta, \qquad ( \rho_{0}\cos\varphi_{0}, \rho_{0}\sin \varphi_{0})\in\Gamma_{b_{k}},\quad b_{k}\geq \varrho_{0}. $$

□