Let \(v=\frac{p}{p-1}u^{p-1}\). From (1.1), by simple calculations, it is easy to see that
$$ v_{t}=(p-1)v\Delta v+|\nabla v|^{2}. $$
(2.1)
We define
$$w=\frac{|\nabla v|^{2}}{v^{\beta}}, $$
where β is a constant to be determined. Then we have
$$\begin{aligned}& w_{t}= \frac{2v_{i}v_{it}}{v^{\beta}}-\beta\frac{v_{i}^{2}v_{t}}{v^{\beta+1}} \\& \hphantom{w_{t}}= 2\frac{v_{i}[(p-1)v\Delta v+|\nabla v|^{2}]_{i}}{v^{\beta}}-\beta\frac{v_{i}^{2}[(p-1)v\Delta v+|\nabla v|^{2}]}{v^{\beta+1}} \\& \hphantom{w_{t}}= 2(p-1)\frac{v_{i}^{2}v_{jj}}{v^{\beta}}+2(p-1)\frac{v_{i}v_{jji}}{v^{\beta-1}} +4\frac{v_{ij}v_{i}v_{j}}{v^{\beta}} -\beta(p-1)\frac{v_{i}^{2}v_{jj}}{v^{\beta}}-\beta\frac{v_{i}^{2}v_{j}^{2} }{v^{\beta+1}}, \end{aligned}$$
(2.2)
$$\begin{aligned}& w_{j}=\frac{2v_{i}v_{ij}}{v^{\beta}}-\beta\frac{v_{i}^{2}v_{j}}{v^{\beta+1}}, \end{aligned}$$
(2.3)
and hence
$$ w_{jj}=\frac{2v_{ij}^{2}}{v^{\beta}}+\frac{2v_{i}v_{ijj}}{v^{\beta}} -4\beta \frac{v_{ij}v_{i}v_{j}}{v^{\beta+1}}-\beta\frac{v_{i}^{2}v_{jj}}{v^{\beta+1}} +\beta(\beta+1)\frac{v_{i}^{2}v_{j}^{2}}{v^{\beta+2}}. $$
(2.4)
By virtue of (2.2) and (2.4),
$$\begin{aligned} (p-1)v\Delta w-w_{t} =&2(p-1)\frac{v_{ij}^{2}}{v^{\beta-1}}+2(p-1) \frac {v_{i}v_{ijj}}{v^{\beta-1}} -4\beta(p-1)\frac{v_{i}v_{j}v_{ij}}{v^{\beta}} \\ &{}-\beta(p-1)\frac{v_{i}^{2}v_{jj}}{v^{\beta}}+\beta(\beta+1) (p-1)\frac {v_{i}^{2}v_{j}^{2}}{v^{\beta+1}} \\ &{}-2(p-1)\frac{v_{i}^{2}v_{jj}}{v^{\beta}}-2(p-1)\frac{v_{i}v_{jji}}{v^{\beta -1}}-4\frac{v_{ij}v_{i}v_{j}}{ v^{\beta}} \\ &{}+\beta(p-1) \frac{v_{i}^{2}v_{jj}}{v^{\beta}}+\beta\frac{v_{i}^{2}v_{j}^{2}}{v^{\beta+1}} \\ =&2(p-1)\frac{v_{ij}^{2}}{v^{\beta-1}}-2(p-1)\frac{v_{i}^{2}v_{jj}}{v^{\beta}} +2(p-1)\frac{R_{ij}v_{i}v_{j}}{v^{\beta-1}} \\ &{}-4\bigl[1+\beta(p-1)\bigr]\frac{v_{ij}v_{i}v_{j}}{v^{\beta}} +\beta\bigl[(\beta+1) (p-1)+1 \bigr]\frac{v_{i}^{2}v_{j}^{2}}{v^{\beta+1}}, \end{aligned}$$
(2.5)
where, in the second equality, we use the Ricci formula: \(v_{ijj}-v_{jji}=R_{ij}v_{j}\).
In order to prove Theorem 1.1, we need the following lemma (cf. Lemma A.1 in [11]).
Lemma 2.1
Let
\(A=(a_{ij})\)
be a nonzero
\(n\times n\)
symmetric matrix. Then for
\(a, b\in\mathbb{R}\),
$$ \max_{A\in\mathcal{S}(n);|e|=1} \biggl(\frac{aA+b (\operatorname{ tr}A) I_{n}}{|A|}(e,e) \biggr)^{2}=(a+b)^{2}+(n-1)b^{2}, $$
(2.6)
where
\(I_{n}\)
is an identity matrix.
Notice that
$$ \nabla v\nabla w =2\frac{v_{ij}v_{i}v_{j}}{v^{\beta}}-\beta\frac{v_{i}^{2}v_{j}^{2}}{v^{\beta+1}}. $$
(2.7)
It follows from (2.5) that, for any constant ε,
$$\begin{aligned}& (p-1) v\Delta w-w_{t}+\varepsilon\nabla v\nabla w \\& \quad= 2(p-1)\frac{v_{ij}^{2}}{v^{\beta-1}}-2(p-1)\frac{v_{i}^{2}v_{jj}}{v^{\beta}} +2(p-1)\frac{R_{ij}v_{i}v_{j}}{v^{\beta-1}} \\& \quad\quad{}+\bigl[2\varepsilon-4\bigl(1+\beta(p-1)\bigr)\bigr]\frac{v_{ij}v_{i}v_{j}}{v^{\beta}} +\beta \bigl[(\beta+1) (p-1)+1-\varepsilon\bigr]\frac{v_{i}^{2}v_{j}^{2}}{v^{\beta+1}} \\& \quad= 2(p-1)\frac{|A|^{2}}{v^{\beta-1}}-2(p-1)\frac{\operatorname{tr}A}{|A|}w|A| +2(p-1)v\operatorname{Ric}(e,e)w \\& \quad\quad{}+\bigl[2\varepsilon-4\bigl(1+\beta(p-1)\bigr)\bigr]\frac{A(e,e)}{|A|}w|A| + \beta\bigl[(\beta+1) (p-1)+1-\varepsilon\bigr]v^{\beta-1}w^{2} \\& \quad= 2(p-1)\frac{|A|^{2}}{v^{\beta-1}}+ \biggl(\bigl[ 2\varepsilon-4\bigl(1+\beta(p-1)\bigr) \bigr]\frac{A(e,e)}{|A|}-2(p-1)\frac{\operatorname{tr}A}{|A|} \biggr)w|A| \\& \quad\quad{}+2(p-1)v\operatorname{Ric}(e,e)w+\beta\bigl[(\beta+1) (p-1)+1-\varepsilon \bigr]v^{\beta-1}w^{2} \\& \quad= 2(p-1) \biggl[\frac{|A|}{v^{\frac{\beta-1}{2}}}+\frac{1}{4(p-1)} \biggl(\bigl[ 2 \varepsilon-4\bigl(1+\beta(p-1)\bigr)\bigr]\frac{A(e,e)}{|A|}-2(p-1) \frac{\operatorname{tr}A}{|A|} \biggr)wv^{\frac{\beta-1}{2}} \biggr]^{2} \\& \quad\quad{}-\frac{1}{8(p-1)} \biggl(\bigl[ 2\varepsilon-4\bigl(1+\beta(p-1)\bigr)\bigr] \frac{A(e,e)}{|A|}-2(p-1)\frac{\operatorname{tr}A}{|A|} \biggr)^{2}v^{\beta-1}w^{2} \\& \quad\quad{}+2(p-1)v\operatorname{Ric}(e,e)w+\beta\bigl[(\beta+1) (p-1)+1-\varepsilon \bigr]v^{\beta-1}w^{2} \\& \quad\geq -\frac{1}{8(p-1)} \biggl|\bigl[ 2\varepsilon-4\bigl(1+\beta(p-1)\bigr)\bigr] \frac{A(e,e)}{|A|}-2(p-1)\frac{(\operatorname{tr}A)I_{n}(e,e)}{|A|} \biggr|^{2}v^{\beta-1}w^{2} \\& \quad\quad{}+2(p-1)v\operatorname{Ric}(e,e)w+\beta\bigl[(\beta+1) (p-1)+1-\varepsilon \bigr]v^{\beta-1}w^{2}, \end{aligned}$$
(2.8)
where \(A_{ij}=(v_{ij})\) and \(e=\nabla v/|\nabla v|\). By virtue of Lemma 2.1, we have
$$\begin{aligned}& (p-1) v\Delta w-w_{t}+\varepsilon\nabla v\nabla w \\& \quad\geq -\frac{1}{8(p-1)} \biggl|\bigl[ 2\varepsilon-4\bigl(1+\beta(p-1)\bigr)\bigr] \frac{A(e,e)}{|A|}-2(p-1)\frac{(\operatorname{tr}A)I_{n}(e,e)}{|A|} \biggr|^{2}v^{\beta-1}w^{2} \\& \quad\quad{}+2(p-1)v\operatorname{Ric}(e,e)w+\beta\bigl[(\beta+1) (p-1)+1-\varepsilon \bigr]v^{\beta-1}w^{2} \\& \quad\geq -\frac{1}{8(p-1)} \bigl(\bigl[2\varepsilon-4\bigl(1+\beta (p-1) \bigr)-2(p-1)\bigr]^{2}+4(n-1) (p-1)^{2} \\& \quad\quad{}-8(p-1)\beta\bigl[(\beta+1) (p-1)+1-\varepsilon\bigr] \bigr)v^{\beta -1}w^{2}+2(p-1)v\operatorname{Ric}(e,e)w \\& \quad= -\frac{1}{8(p-1)}f(\beta,\varepsilon)v^{\beta-1}w^{2}+2(p-1)v\operatorname{Ric}(e,e)w, \end{aligned}$$
(2.9)
where
$$f(\beta,\varepsilon)=\bigl[2\varepsilon-4\bigl(1+\beta(p-1)\bigr)-2(p-1) \bigr]^{2}+4(n-1) (p-1)^{2} -8(p-1)\beta\bigl[(\beta+1) (p-1)+1-\varepsilon\bigr]. $$
For the purpose of showing that the coefficient of \(v^{\beta-1}w^{2}\) is positive, we minimize the function \(f(\beta,\varepsilon)\) by letting
and
$$\beta=-\frac{p-\varepsilon+2}{2(p-1)}=-\frac{1}{p-1}, $$
such that
$$f=-4+4(n-1) (p-1)^{2}. $$
Then (2.9) becomes
$$\begin{aligned} (p-1)v\Delta w-w_{t}+p\nabla v\nabla w \geq& \frac{1-(n-1)(p-1)^{2}}{2(p-1)}v^{\beta-1}w^{2}-2(p-1)Kvw \\ =&\alpha v^{\beta-1}w^{2}-2(p-1)Kvw, \end{aligned}$$
(2.10)
where \(\alpha=\frac{1-(n-1)(p-1)^{2}}{2(p-1)}\).
We first recall the well-known smooth cutoff function ψ which originated with Li and Yau [12], satisfying the following:
-
(1)
The cutoff function ψ satisfies \(\psi=\psi(d(x,x_{0}),t)\equiv\psi(r,t)\) and \(\psi(r,t)=1\) in \(Q_{R/2,T/2}\) with \(0\leq\psi\leq1\).
-
(2)
The function ψ is decreasing as a radial function in the spatial variables.
-
(3)
\(|\partial_{r}\psi|/\psi^{a}\leq C_{a}/R\) and \(|\partial_{r}^{2}\psi|/\psi^{a}\leq C_{a}/R^{2}\) for \(a\in(0,1)\).
-
(4)
\(|\partial_{t}\psi|/\psi^{\frac{1}{2}}\leq C/T\).
By virtue of (2.10), we have
$$\begin{aligned}& \bigl[(p-1)v\Delta-\partial_{t}\bigr](\psi w) \\& \quad= \psi\bigl[(p-1)v\Delta-\partial_{t}\bigr]w+(p-1)v w\Delta\psi-w \psi_{t} \\& \quad\quad{}+2(p-1)v\frac{1}{\psi}\nabla\psi\nabla(\psi w)-2(p-1)vw\frac{|\nabla \psi|^{2}}{\psi} \\& \quad\geq \alpha v^{\beta-1}\psi w^{2}-2(p-1)Kv\psi w-p\nabla v\nabla( \psi w) \\& \quad\quad{}+pw\nabla v\nabla\psi+(p-1)v w\Delta\psi-w\psi_{t} \\& \quad\quad{}+2(p-1)v\frac{1}{\psi}\nabla\psi\nabla(\psi w)-2(p-1)vw\frac{|\nabla\psi|^{2}}{\psi}. \end{aligned}$$
(2.11)
Next we will apply the maximum principle to ψw in a closed set. Assume ψw achieves its maximum at the point \((x_{1},t_{1})\) and assume \((\psi w)(x_{1},t_{1})>0\) (otherwise the proof is trivial), which implies \(t_{1}>0\). Then at the point \((x_{1},t_{1})\)
$$(\Delta-\partial_{t}) (\psi w)\leq0, \quad\nabla(\psi w)=0, $$
and (2.11) becomes
$$\begin{aligned} \alpha v^{\beta-1}\psi w^{2} \leq&-pw\nabla v\nabla \psi+2(p-1)vw\frac{|\nabla\psi|^{2}}{\psi} \\ &{}-(p-1)v w\Delta\psi+w\psi_{t}+2(p-1)Kv\psi w. \end{aligned}$$
(2.12)
That is,
$$\begin{aligned} 2\psi w^{2} \leq&-p\gamma v^{1-\beta} w\nabla v \nabla\psi+2(p-1)\gamma v^{2-\beta}w\frac{|\nabla\psi|^{2}}{\psi} \\ &{}-(p-1)\gamma v^{2-\beta} w\Delta\psi+\gamma v^{1-\beta} w \psi_{t}+2(p-1)\gamma Kv^{2-\beta}\psi w, \end{aligned}$$
(2.13)
where \(\alpha=\frac{2}{\gamma}\) and \(\gamma=\frac{1-(n-1)(p-1)^{2}}{4(p-1)}\).
It have been shown in [1] (see equations (2.6)-(2.10) in [1]) that
$$\begin{aligned} -p\gamma v^{1-\beta} w\nabla v\nabla\psi\leq \frac{1}{4}\psi w^{2}+CM^{4-2\beta}\frac{1}{R^{4}}, \end{aligned}$$
(2.14)
where we used the fact that \(0< v\leq M\) and \(\beta\leq2\),
$$\begin{aligned}& 2(p-1)\gamma v^{2-\beta}w\frac{|\nabla\psi|^{2}}{\psi}\leq \frac{1}{4}\psi w^{2}+CM^{4-2\beta}\frac{1}{R^{4}}, \end{aligned}$$
(2.15)
$$\begin{aligned}& -(p-1)\gamma v^{2-\beta} w\Delta\psi\leq\frac{1}{4} \psi w^{2}+CM^{4-2\beta} \biggl(\frac{1}{R^{4}}+K \frac{1}{R^{2}} \biggr), \end{aligned}$$
(2.16)
$$\begin{aligned}& \gamma v^{1-\beta} w\psi_{t}\leq\frac{1}{4} \psi w^{2}+CM^{4-2\beta}\frac{1}{T^{2}}, \end{aligned}$$
(2.17)
$$\begin{aligned}& 2(p-1)\gamma Kv^{2-\beta}\psi w\leq\frac{1}{4}\psi w^{2}+CM^{4-2\beta}K^{2}. \end{aligned}$$
(2.18)
Substituting (2.14)-(2.18) into (2.13), we obtain
$$ 2\psi w^{2}\leq \frac{5}{4}\psi w^{2}+CM^{4-2\beta} \biggl(\frac{1}{R^{4}}+\frac{1}{T^{2}}+K^{2} \biggr), $$
(2.19)
which gives at the point \((x_{1},t_{1})\)
$$ \psi w^{2}\leq CM^{4-2\beta} \biggl( \frac{1}{R^{4}}+\frac{1}{T^{2}}+K^{2} \biggr). $$
(2.20)
Therefore, for all \((x,t)\in Q_{R,T}\),
$$\begin{aligned} \bigl(\psi^{2} w^{2}\bigr) (x,t) \leq&\bigl( \psi^{2} w^{2}\bigr) (x_{1},t_{1}) \leq \bigl(\psi w^{2}\bigr) (x_{1},t_{1}) \\ \leq&CM^{4-2\beta} \biggl(\frac{1}{R^{4}}+\frac{1}{T^{2}}+K^{2} \biggr). \end{aligned}$$
Notice that \(\psi=1\) in \(Q_{R/2,T/2}\) and \(w=\frac{|\nabla v|^{2}}{v^{\beta}}\). Hence, we have
$$ \frac{|\nabla v|}{v^{\frac{\beta}{2}}}(x,t)\leq CM^{1-\frac{\beta}{2}}\biggl(\sqrt{K} + \frac{1}{R}+\frac{1}{\sqrt{T}}\biggr). $$
(2.21)
One concludes the proof of Theorem 1.1 by letting \(\beta=-\frac{1}{p-1}\).