# Time derivative of area under curve shortening flow

I asked the same question but get no answer in other place. Here is the following.

For a compact Riemannian surface $$\Sigma$$. For an initial embedded closed curve $$\gamma_0$$ in $$\Sigma$$, a family $$\gamma_t$$ $$(0\leq t is parametrized by $$\begin{equation} F : S^{1} \times[0, T) \rightarrow \Sigma, \end{equation}$$ it is called a curve shortening flow, if $$\begin{equation} \frac{\partial}{\partial t} F(\theta, t)=-\kappa_{t}(F(\theta, t)) v_{t}(F(\theta, t)) \end{equation}$$ P. Topping states on page 51 in  that $$\begin{equation} \frac{d A_{t}}{d t}=-\int_{\gamma_{t}} \kappa_{t}. ~~~(1) \end{equation}$$ where $$A_t$$ is the area of the set bounded by the curve $$\gamma_t$$.

I know how to derive this for $$\Sigma=\mathbb{R}^2$$. How to prove (1) for $$\Sigma$$ being a surface? Thank you very much.

References

 P. Topping, Mean curvature flow and geometric inequalities, J. Reine Angew. Math. 503, 47-61, 1998

• It suffices to do this in a coordinate chart containing the curve. You could for example choose coordinates so that the curve is the graph of a function. Then grind away at the calculation. Aug 5, 2019 at 13:00

A late answer if the OP is still interested in a solution. In $$\mathbb{R}^2$$, the area $$A$$ enclosed by the closed embedded curve $$\gamma \colon I \subset \mathbb{R} \to \mathbb{R}^2$$ is given by (using the Green's identity) $$A = \frac 12\oint x\,dy - y\,dx = \frac 12\int_I \left(x\frac{dy}{dz} - y\frac{dy}{dz}\right)dz = \frac 12\int_I RX\cdot \frac{dX}{dz},$$ where $$R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},$$ $$X = (x,y)$$. Thus, we deduce that \begin{align*} \frac{d}{dt}A &= \frac 12\int_I RX\cdot \frac{dX}{dz} \\ &= \frac 12\int_I \left(R\frac{dX}{dt}\cdot\frac{dX}{dz} + RX\frac{d^2 X}{dzdt}\right) dz \\ &= \frac 12\int_I \left(R\frac{dX}{dt}\cdot\frac{dX}{dz} - R\frac{dX}{dz}\frac{dX}{dt}\right) dz, \end{align*} where the last identity follows from integration by parts. Now recall that under the MCF, we have $$\frac{dX}{dt} = \kappa N$$ for $$N$$ being a unit normal vector, and we also have $$\frac{dX}{dz} = T\,\frac{ds}{dz}$$ for $$T$$ being the unit tangent vector (here $$\frac{ds}{dz} = \|\frac{dX}{dz}\|$$ measures the speed of the curve). Note that $$RN = -T$$ and $$RT =N$$. Finally, putting all these together, you will arrive at $$\frac{d}{dt}A = -\int_{\gamma} \kappa \, ds,$$ which is the advertised conclusion .
Remark: The last display, i.e., $$\int_{\gamma} \kappa$$, will equal to $$2\pi$$ thanks to Gauss-Bonnet formula (if the enclosed region is compact and convex).