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 $\gamma_t$ $(0\leq t<T)$ 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 [1] 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.


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

  • $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Aug 5, 2019 at 13:00

1 Answer 1


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).

  • $\begingroup$ (and even if the region is not convex) $\endgroup$
    – alesia
    Jun 20, 2021 at 17:49
  • $\begingroup$ @alesia thanks for the comment! $\endgroup$
    – Fei Cao
    Jun 20, 2021 at 19:25

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