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 $$\frac{\partial}{\partial t}\nabla R^2 = \Delta\nabla R^2  2\nabla\nabla R^2 + (4R3r)\nabla R^2.$$ Using the estimate $Rr\leq C \exp{rt}$, 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. $\endgroup$– debabrata chakrabortyCommented Oct 5, 2015 at 18:36

$\begingroup$ Is $r$ a constant? $\endgroup$– YangMillsCommented Oct 11, 2015 at 2:04

$\begingroup$ yes it is the average scalar curvature and is constant in spacetime $\endgroup$– debabrata chakrabortyCommented 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 $r/2$. $\endgroup$– YangMillsCommented Oct 11, 2015 at 3:58
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