Extinction time of a 2-sphere under Ricci Flow So I was trying to pick up some differential geometry on my own and decided to try and solve Ricci Flow for a 2-sphere. Unless restrictions are imposed on the system,  the surface will collapse and vanish. My knowledge on DG isn't yet very robust yet so instead I decided to discard the volume restrictions and calculate the extinction time $t_\text{x}$ instead. Here is my approach to the final result:
Starting from the Ricci Flow PDE $\partial_tg_{ij}=-2R_{ij}$ I simplified it to the 2D case where $R_{ij}=Kg_{ij}$ for Gaussian curvature $K$ according to what I have covered so far. Thus the PDE becomes $\partial_tg_{ij}=-Kg_{ij}$.
Now for a sphere, I found the first fundamental form to be given by$$E=a^2sin^2(v)$$$$G=a^2$$$$F=0$$and the Gaussian curvature to be $K=1/a^2$. Another reason why I chose to start with this example was to keep it as simple as possible with constant curvatures and sparse a $g_{ij}$.
Now the next step is the part that I think may be kind of sketchy. Given that $K$ is uniform throughout during each time-step and due to the normals being radial, then I took a guess that a sphere at $t=0$ will produce another sphere at time $t=0+dt$ therefore the only thing that has any time dependence within $g_{ij}$ must be $a$ so I just set it arbitrarily to be $a(t)$.
Now from the RF equation and the first fundamental form, after using the fact that only $a$ depends on $t$ and simplifying I got only one equation$$a\dot{a}=-1 \implies a(t)=\sqrt{a_0^2-2t} \implies t_\text{x}=\frac{a_0^2}{2}\text{ at }a=0$$
Does this make any sense? If not (which I think is the most likely possibility) what am I missing that I should look up?
It's a shame that there aren't many online examples regarding RF on surfaces. I tried to look for a solution but I couldn't find any, and when I worked it out the answer just seemed too "simple" so to speak
 A: Your reasoning works! Another way to achieve that is to guess that since the initial curvature $K_0$ is constant, the flow will only deform the metric by rescaling it, so that $g(t) = r(t)g_0$ for some function $r(t)$ with $r(0) = 1$. The curvature of this metric is $K(t) = r^{-1}(t)K_0$. The (simplified) equation then reads 
$$-K_0g_0 = -K(t)g(t) = \partial_t g(t) = r'(t)g_0,$$
so that $r(t) = 1-tK_0$. This shows that for any surface with constant curvature $K_0$, we get a solution $g(t) = (1-tK_0)g_0$. We see that indeed, for a sphere where $K_0 \equiv 1$, the time of extinction is $1$ while for a flat torus, the flow exists for all time (it doesn't change the metric at all). We also see that if $K_0 \equiv -1$ on a higher genus surface, the flow exists for all time and just expands the metric indefinitely. In fact, one could extend the definition of the flow in the above examples to $(-\infty,1)$ for the sphere, to $(-\infty,+\infty)$ for the flat torus, and to $(-1,+\infty)$ for the higher genus surfaces!
By the way, this same reasoning works for an initial metric that is Einstein ($\text{Ric}(g) = \lambda g$ for some $\lambda \in \mathbb{R}$) in any dimension and yields the solution $g(t) = (1-2\lambda t)g_0$.
