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In Hamiltons paper "Ricci flow on surfaces" there is an estimate on $|\nabla R|^2$ which shows that $|\nabla R|^2 \leq C_1 \exp{\frac{rt}{2}}$ for some constant $C_1$. Actually for any solution of the Ricci flow on a surface $|\nabla R|^2$ evolves by $$\frac{\partial}{\partial t}|\nabla R|^2 = \Delta|\nabla R|^2 - 2|\nabla\nabla R|^2 + (4R-3r)|\nabla R|^2.$$ Using the estimate $|R-r|\leq C \exp{rt}$, the above equation implies $$\frac{\partial}{\partial t}|\nabla R|^2 \leq \Delta|\nabla R|^2 - 2|\nabla\nabla R|^2 + (r+4C\exp{rt})|\nabla R|^2.$$ My problem is in the next step where it is stated that taking $t>0$ large enough one can show that $$\frac{\partial}{\partial t}|\nabla R|^2 \leq \Delta|\nabla R|^2 + \frac{r}{2}|\nabla R|^2.$$ After that it is well understood that we can apply maximum principle to get the above estimate. Please anyone help me to understand the last arguement. This case is for $r<0$.

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  • $\begingroup$ It is the case for $r < 0$ indeed. $\endgroup$ Commented Oct 5, 2015 at 18:36
  • $\begingroup$ Is $r$ a constant? $\endgroup$
    – YangMills
    Commented Oct 11, 2015 at 2:04
  • $\begingroup$ yes it is the average scalar curvature and is constant in space-time $\endgroup$ Commented Oct 11, 2015 at 3:39
  • $\begingroup$ then the answer to your question is obvious. For all $t$ large, $4C\exp(rt)$ is smaller than $-r/2$. $\endgroup$
    – YangMills
    Commented Oct 11, 2015 at 3:58

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