By Theorem
2.1, we need only to prove this theorem for
\(s\geq 3\). According to Lemma
3.1, there exists
\(\xi (t)\in \mathbb{S}\) such that
$$\begin{aligned} I_{1}(t)=u_{x}\bigl(t,\xi (t)\bigr)=\inf _{x\in \mathbb{S}}u_{x}(t,x), \quad t \in [0,T). \end{aligned}$$
(40)
Since
\(q(t,\cdot )\) defined by (
14) is a diffeomorphism of the circle for any
\(t\in [0,T)\), we know that there exists
\(x_{1}(t) \in \mathbb{S}\) such that
$$\begin{aligned} q\bigl(t,x_{1}(t)\bigr)=\xi (t), \quad t\in [0,T). \end{aligned}$$
(41)
Then (
38) and (
40) imply that
$$\begin{aligned} I_{1}(0)=u_{x}\bigl(0,\xi (0)\bigr)=\inf _{x\in \mathbb{S}}u_{0,x}(x)=u_{0,x}(x _{1}), \quad t\in [0,T). \end{aligned}$$
(42)
Therefore we can choose
\(\xi (0)=x_{1}\) and
$$\begin{aligned} \rho_{0}\bigl(\xi (0)\bigr)=\rho_{0}(x_{1})=0, \quad t\in [0,T). \end{aligned}$$
(43)
Using Lemma
3.3, we have
$$\begin{aligned} \rho \bigl(t,-q\bigl(t,x_{1}(t)\bigr)\bigr)=\rho \bigl(t,-\xi (t) \bigr)=0, \quad \forall t\in [0,T). \end{aligned}$$
(44)
On the other hand, due to
\(\sup_{x\in \mathbb{S}}(v_{x}(t,x))=- \inf_{x\in \mathbb{S}}(-v_{x}(t,x))\), we similarly define
$$\begin{aligned} I_{2}(t)=u_{x}\bigl(t,\eta (t)\bigr)=\sup _{x\in \mathbb{S}}u_{x}(t,x), \quad t \in [0,T). \end{aligned}$$
(45)
There exists
\(x_{2}(t)\in \mathbb{S}\) such that
$$\begin{aligned} q\bigl(t,x_{2}(t)\bigr)=\eta (t), \quad t\in [0,T). \end{aligned}$$
(46)
Moreover, we have
$$\begin{aligned} \rho \bigl(t,-q\bigl(t,x_{2}(t)\bigr)\bigr)=\rho \bigl(t,-\eta (t) \bigr)=0, \quad \forall t\in [0,T). \end{aligned}$$
(47)
Differentiating the first equation of system (
7) with respect to
x yields
$$\begin{aligned} u_{tx}-\sigma u_{x}^{2}-(\sigma u+ \gamma_{1})u_{xx}=A^{-1}\partial ^{2}_{x}\biggl(2\mu_{0}u+\frac{\sigma }{2}u^{2}_{x}+ \frac{1}{2}\rho^{2}\biggr) . \end{aligned}$$
(48)
Using (
10), we get
$$\begin{aligned} u_{tx}-(\sigma u+\gamma_{1})u_{xx}= \frac{\sigma }{2}u_{x}^{2}+2\mu _{0}A^{-1} \partial^{2}_{x}u-\frac{1}{2}\rho^{2} + \int^{1}_{0}\frac{ \sigma }{2}u_{x}^{2}+ \frac{1}{2}\rho^{2}\,dx . \end{aligned}$$
(49)
Recalling the definitions of
\(I_{i}\) (
\(i=1,2\)), we get
$$\begin{aligned} \frac{dI_{1}}{dt}&=\frac{\sigma }{2}I^{2}_{1}+2 \mu_{0}\partial^{2} _{x}A^{-1}u+ \int^{1}_{0}\biggl(\frac{\sigma }{2}u^{2}_{x}+ \frac{1}{2}\rho ^{2}\biggr)\,dx \\ & =\frac{\sigma }{2}I^{2}_{1}+2\mu_{0} \int^{1}_{0}g(y)u_{xx}\bigl(t,\xi (t)-y \bigr)\,dy+ \int^{1}_{0}\biggl(\frac{\sigma }{2}u^{2}_{x}+ \frac{1}{2}\rho^{2}\biggr)\,dx \end{aligned}$$
(50)
and
$$\begin{aligned} \frac{dI_{2}}{dt}&=\frac{\sigma }{2}I^{2}_{2}+2 \mu_{0}\partial^{2} _{x}A^{-1}u+ \int^{1}_{0}\biggl(\frac{\sigma }{2}u^{2}_{x}+ \frac{1}{2}\rho ^{2}\biggr)\,dx \\ &=\frac{\sigma }{2}I^{2}_{2}+2\mu_{0} \int^{1}_{0}g(y)u_{xx}\bigl(t,\eta (t)-y \bigr)\,dy+ \int^{1}_{0}\biggl(\frac{\sigma }{2}u^{2}_{x}+ \frac{1}{2}\rho^{2}\biggr)\,dx . \end{aligned}$$
(51)
We notice that
\(g(x)=\frac{1}{2}(x-\frac{1}{2})^{2}+\frac{23}{24}\) is continuous on
\(\mathbb{S}\), decreasing on
\([0,\frac{1}{2}]\), and increasing on
\([\frac{1}{2},1]\). Therefore, we have
$$\begin{aligned} & \biggl\vert \int^{1}_{0}g(y)u_{xx}\bigl(t,\xi (t)-y \bigr)\,dy \biggr\vert \\ &\quad = \biggl\vert \int^{\frac{1}{2}}_{0}g(y)u_{xx}\bigl(t,\xi (t)-y \bigr)\,dy \biggr\vert + \biggl\vert \int ^{1}_{\frac{1}{2}}g(y)u_{xx}\bigl(t,\xi (t)-y \bigr)\,dy \biggr\vert . \end{aligned}$$
(52)
From equality (
11), we deduce
$$\begin{aligned} & \biggl\vert \int^{\frac{1}{2}}_{0}g(y)u_{xx}\bigl(t,\xi (t)-y \bigr)\,dy \biggr\vert \\ &\quad \leq \biggl\vert g(0) \int^{\zeta }_{0}u_{xx}\bigl(t,\xi (t)-y\bigr) \,dy \biggr\vert + \biggl\vert g\biggl( \frac{1}{2}\biggr) \int^{\frac{1}{2}}_{\zeta }u_{xx}\bigl(t,\xi (t)-y\bigr) \,dy \biggr\vert \\ &\quad =\frac{13}{12} \bigl\vert u_{x}\bigl(t,\xi (t) \bigr)-u_{x}\bigl(t,\xi (t)-\zeta \bigr) \bigr\vert + \frac{23}{24} \biggl\vert u_{x}\biggl(t,\xi (t)- \frac{1}{2}\biggr)-u_{x}\bigl(t,\xi (t)-\zeta \bigr) \biggr\vert \\ &\quad \leq \frac{49}{24}\bigl(I_{2}(t)-I_{1}(t)\bigr) . \end{aligned}$$
(53)
In an analogous way, we get
$$\begin{aligned} \biggl\vert \int^{1}_{\frac{1}{2}}g(y)u_{xx}\bigl(t,\xi (t)-y \bigr)\,dy \biggr\vert \leq \frac{49}{24}\bigl(I_{2}(t)-I_{1}(t) \bigr) . \end{aligned}$$
(54)
Thus, by (
53) and (
54), it implies
$$\begin{aligned} \biggl\vert \int^{1}_{0}g(y)u_{xx}\bigl(t,\eta (t)-y \bigr)\,dy \biggr\vert \leq \frac{49}{12}\bigl(I _{2}(t)-I_{1}(t) \bigr) . \end{aligned}$$
(55)
If
\(\mu_{0}\leq 0\) and
\(\sigma \geq 1\), from (
55), (
50), and (
51) we deduce that, for
a.e.
\(t\in (0,T)\),
$$\begin{aligned} \frac{dI_{1}}{dt} \geq \frac{\sigma }{2}I^{2}_{1}+ \frac{49}{6}\mu_{0}(I _{2}-I_{1})+ \frac{1}{2}\mu^{2}_{1} \end{aligned}$$
(56)
and
$$\begin{aligned} \frac{dI_{2}}{dt} &\geq \frac{\sigma }{2}I^{2}_{2}+ \frac{49}{6}\mu _{0}(I_{2}-I_{1})+ \frac{1}{2}\mu^{2}_{1} \\ & =\frac{\sigma }{2}I^{2}_{2}-\frac{49}{6} \mu_{0}(I_{2}+I_{1})+ \frac{49}{3} \mu_{0}I_{2}+\frac{1}{2}\mu^{2}_{1} . \end{aligned}$$
(57)
Summing up (
56) and (
57) results in
$$\begin{aligned} \frac{d(I_{1}+I_{2})}{dt} &\geq \frac{\sigma }{2}\bigl(I^{2}_{1}+I^{2} _{2}\bigr)+\frac{49}{3}\mu_{0}(I_{2}-I_{1})+ \mu^{2}_{1} \\ & =\frac{\sigma }{2}\bigl(I^{2}_{1}+I^{2}_{2} \bigr)+\frac{49}{3}\mu_{0}(I_{2}+I _{1})- \frac{98}{3}\mu_{0}I_{1}+\mu^{2}_{1} . \end{aligned}$$
(58)
From the assumption of Theorem
4.1
\(I_{0}(0)+I_{2}(0)\geq - \frac{98}{3}\mu_{0}+2\mu_{1}\), we now claim that, for all
\(t\in T\),
$$\begin{aligned} (I_{1}+I_{2}) (t)\geq -\frac{98}{3} \mu_{0}+2\mu_{1} . \end{aligned}$$
(59)
Let
\(I(t)=(I_{1}+I_{2})(t)+\frac{98}{3}\mu_{0}-2\mu_{1}\). Then we claim that
\(I(t)\geq 0\). It is observed that
\(I(t)\) is continuous on
\([0,T)\). Assume that
\(I(t)\geq 0\) is not valid, then there is
\(t_{0}\in (0,T)\) such that
\(I(t_{0})<0\). Let
\(t_{1}=\max \{t< t_{0}: I(t)=0 \}\). Then
\(I(t_{1})=0\) and
\(I'(t_{1})<0\), namely
$$\begin{aligned} (I_{1}+I_{2}) (t_{1})=-\frac{98}{3} \mu_{0}+2\mu_{1} \end{aligned}$$
(60)
and
$$\begin{aligned} I'(t)=\bigl(I'_{1}+I'_{2} \bigr) (t_{1})< 0 . \end{aligned}$$
(61)
Due to
$$\begin{aligned} I_{2}(t_{1})\geq \frac{1}{2}(I_{1}+I_{2}) (t_{1})=-\frac{98}{6}\mu_{0}+ \mu_{1} \end{aligned}$$
(62)
and
$$\begin{aligned} I_{1}(t_{1})=-\frac{98}{3}\mu_{0}+2 \mu_{1}-I_{2}(t_{1}) . \end{aligned}$$
(63)
Thus, we get
$$\begin{aligned} I'(t_{1})&=I'_{1}(t_{1})+I'_{2}(t_{1}) \\ & \geq \frac{\sigma }{2}\bigl(I^{2}_{1}+I^{2}_{2} \bigr)-\frac{49}{3}\mu_{0}(I_{2}+I _{1})+ \frac{98}{3}\mu_{0}I_{2}+\mu^{2}_{1} \\ & =\frac{\sigma }{2}I^{2}_{2}+\frac{\sigma }{2}\biggl(- \frac{98}{3}\mu_{0}+2 \mu_{1}-I_{2}(t_{1}) \biggr)^{2}-\frac{49}{3}\mu_{0}\biggl(- \frac{98}{3}\mu_{0}+2 \mu_{1}\biggr) \\ & \quad {}+\frac{98}{3}\mu_{0}I_{2}(t_{1})+ \mu^{2}_{1} \\ & =\sigma I^{2}_{2}-\sigma \biggl(-\frac{98}{3} \mu_{0}+2\mu_{1}\biggr)I_{2}+ \frac{98}{3} \mu_{0}I_{2}+\frac{\sigma }{2}\biggl(-\frac{98}{3} \mu_{0}+2\mu _{1}\biggr)^{2} \\ & \quad {}-\frac{49}{3}\mu_{0}\biggl(-\frac{98}{3} \mu_{0}+2\mu_{1}\biggr)+\mu^{2}_{1} \\ & =\sigma \biggl[I_{2}-\frac{\sigma (-\frac{98}{3}\mu_{0}+2\mu_{1})- \frac{98}{3}\mu_{0}}{2\sigma }\biggr]^{2} - \frac{[\sigma (-\frac{98}{3}\mu _{0}+2\mu_{1})-\frac{98}{3}\mu_{0}]^{2}}{4\sigma } \\ & \quad {} +\frac{\sigma }{2}\biggl(-\frac{98}{3}\mu_{0}+2 \mu_{1}\biggr)^{2} - \frac{49}{3}\mu_{0} \biggl(-\frac{98}{3}\mu_{0}+2\mu_{1}\biggr)+ \mu^{2}_{1} \\ & =\sigma \biggl[I_{2}-\frac{\sigma (-\frac{98}{3}\mu_{0}+2\mu_{1})- \frac{98}{3}\mu_{0}}{2\sigma }\biggr]^{2}+ \frac{\sigma }{4}\biggl(-\frac{98}{3}\mu _{0}+2 \mu_{1}\biggr)^{2} \\ & \quad {} -\frac{1}{4\sigma }\biggl(\frac{98}{3}\mu_{0} \biggr)^{2}+\mu^{2}_{1} \\ & >0, \end{aligned}$$
(64)
which gives rise to a contradiction with (
61). Therefore, (
59) is true.
$$\begin{aligned} &\frac{d(I_{2}(t)+\frac{49}{3\sigma }\mu_{0})}{dt} \\ &\quad =\frac{dI_{2}}{dt} \\ &\quad \geq \frac{\sigma }{2}I^{2}_{2}-\frac{49}{6} \mu_{0}(I_{2}+I_{1})+ \frac{49}{3} \mu_{0}I_{2}+\frac{1}{2}\mu^{2}_{1} \\ &\quad \geq \frac{\sigma }{2}I^{2}_{2}-\frac{49}{6} \mu_{0}\biggl(-\frac{98}{3}\mu _{0}+2 \mu_{1}\biggr)+\frac{49}{3}\mu_{0}I_{2}+ \frac{1}{2}\mu^{2}_{1} \\ &\quad =\frac{\sigma }{2}\biggl(I_{2}(t)+\frac{49}{3\sigma } \mu_{0}\biggr)^{2}-\frac{1}{2 \sigma }\biggl( \frac{49}{3}\mu_{0}\biggr)^{2}+\biggl( \frac{49}{3}\mu_{0}\biggr)^{2} - \frac{49}{3} \mu_{0}\mu_{1}+\frac{1}{2}\mu^{2}_{1} \\ &\quad \geq \frac{\sigma }{2}\biggl(I_{2}(t)+\frac{49}{3\sigma } \mu_{0}\biggr)^{2}. \end{aligned}$$
(65)
Since
\(I_{2}(t)\) is locally Lipschitz on
\((0,T)\), we have that
\(\frac{1}{(I_{2}(t)+\frac{49}{3\sigma }\mu_{0})}\) is also locally Lipschitz on
\((0,T)\), then
\(\frac{1}{(I_{2}(t)+\frac{49}{3\sigma }\mu _{0})}\) is absolutely continuous on
\((0,T)\).