act, the first equation in is equivalent to $
\frac{2(\gamma-1)}{a}\eta(t)=\frac{\sigma'}{\sigma}+2\gamma{\kappa}.
$ Put this into the last two equations in , we have $$\left\{
  \begin{array}{rl}
(\sigma\alpha)'=&\sigma'+2\gamma{\kappa}\sigma\\
(\sigma\varphi)'=&\frac{a\sigma}{4}\Big(\frac{\sigma'}{\sigma}+2\gamma{\kappa}\Big)^2,
  \end{array}
\right.$$ Integral above identities on $[0,t]$, we can obtain the explicit expressions of $\alpha(t)$ and $\varphi(t)$ in .

Since $\gamma>1$, $a>0$ and $M$ is closed, the standard parabolic maximum principle in implies $F_{\alpha}\ge0$, that is in Theorem \[pmeGK\].

Let $\varsigma(t)$ be a constant speed geodesic with $\varsigma(t_1)=x_1$ and $\varsigma(t_2)=x_2$ such that $|\dot{\varsigma}(t)|=\frac{d(x_2,x_1)}{t_2-t_1}$. Using differential Harnack estimate and Young inequality, we have

$$\begin{aligned}
v(x_2,t_2)-v(x_1,t_1)=&\int^{t_2}_{t_1}v_t+\langle\nabla v,\dot{\varsigma}(t)\rangle dt\\
\ge&\int^{t_2}_{t_1}\left(\frac1{\alpha(t)}|\nabla v|^2-\frac{\varphi(t)}{\alpha(t)}v-\frac1{\alpha(t)}|\nabla v|^2-\frac{1}{4}\alpha(t)|\dot{\varsigma}(t)|^{2}\right)dt\\
\ge&-v_{max}\int^{t_2}_{t_1}\frac{\varphi(t)}{\alpha(t)}dt
-\frac{1}{4}\frac{d(x_2,x_1)^{2}}{(t_2-t_1)^{2}}\int^{t_2}_{t_1}\alpha(t)dt\end{aligned}$$

and

$$\begin{aligned}
\log\frac{v(x_2,t_2)}{v(x_1,t_1)}
=&\int^{t_2}_{t_1}\left(\frac{d}{dt}\log v(x,t)+\nabla\log v\cdot\dot{\varsigma}(t)\right)dt\\
\ge&\int^{t_2}_{t_1}\left(\frac{1}{\alpha(t)}\Big(|\nabla v|^2-\varphi(t)\Big)
-\frac1{\alpha(t)}|\nabla v|^2-\frac{1}{4}\frac{|\dot{\varsigma}(t)|^{2}}{v_{max}}\alpha(t)\right)dt\\
\ge&-\int^{t_2}_{t_1}\frac{\varphi(t)}{\alpha(t)}dt
-\frac{1}{4}\frac1{v_{max}}\frac{d(x_2,x_1)^{2}}{(t_2-t_1)^{2}}\int^{t_2}_{t_1}\alpha(t)dt.\end{aligned}$$

This finishes the proof of Corollary \[Harnack\].

Since $\alpha(t)>1$, direct calculation implies that

$$\begin{aligned}
\Delta(v^{\beta})=&\beta v^{\beta-1}\left(\Delta v+(\beta-1)\frac{|\nabla v|^2}{v}\right)\\
=&\frac{1}{\alpha (\gamma-1) }\beta v^{\beta-1}\left(\alpha (\gam