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Show that the area of the inside region of a simple closed curve in the plane along the curve shortening flow is given by \begin{equation*} A(\tau) = -2\pi\tau + A(\tau = 0) \end{equation*}

As a hint, I'm supposed to first show the (more general) formula for the area; \begin{equation*} A = -\frac{1}{2} \int_\gamma \left( \gamma_\tau \cdot N_{\gamma_\tau} \right) \|\dot{\gamma}_\tau \| \text{d}t \end{equation*}

I know it has to do with using Green's theorem (Green's identity?); \begin{equation} \int_\gamma \left( \frac{\partial g}{\partial x}-\frac{\partial f}{\partial y} \right) dxdy = \int_\gamma f(x,y)dx + g(x,y)dy \end{equation} (As it appears in the book I'm following). I have a pretty clear outline of where to go after the above result (the area by integration), but I'm unsure how to get started using Green's Theorem and the Curve-Shortening Flow identities.

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    $\begingroup$ Your area formula by integration can be derived directly from the Green's formula, setting f(x,y)=y, g(x,y)=-x. (So the formula is nothing to do with the curve shorting flow) Then after differentiating the formula by the time parameter $\tau$ and applyng the curve shorteninng flow to the terms in the integrant, you might need the fact the total curvature of the simple closed curve is $2\pi$. $\endgroup$ Commented Mar 1, 2025 at 19:06
  • $\begingroup$ Do you mean the area as; $$\begin{equation} A(\gamma) = \frac{1}{2} \int_0^T \left(x\dot{y}-y\dot{x}\right) dt \end{equation}$$? I don't see how $||\dot{\gamma}_\tau||$ gets fitted into this identity. $\endgroup$ Commented Mar 2, 2025 at 0:12
  • $\begingroup$ P.s. I thought $t$, and not $\tau$ was the time parameter? $\endgroup$ Commented Mar 2, 2025 at 1:15
  • $\begingroup$ $v=(\dot y,-\dot x)$ is a normal vector with length $\vert \dot\gamma\vert$. So unit normal vector $N_\gamma$ is given by $N_\gamma=v/\vert\dot\gamma\vert$. $\tau$ is the parameter for evolution by curve shortening flow and $t$ is the parameter of the evolving curve. Anyhow it is not the question of research interest. $\endgroup$ Commented Mar 2, 2025 at 8:06

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