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I'm trying understand the article "Curvature bound for curve shortening flow via distance comparison and a direct proof of Grayson's theorem" by Ben Andrews and Paul Bryan and they stated on the final of the article (precisely, on the proof of corollary 4) that

The un-normalized equation can be recovered from the normalized one by setting $\lambda(t) = \frac{L[\widetilde{F}_0]}{2 \pi} \exp \left( - \int_0^t \overline{k^2}(t') dt' \right)$, and $\widetilde{F}(p, \tau) = \lambda(t) F(p,t)$, where $\tau = \int_0^t \lambda(t')^2 dt'$.

I'm trying understand how $\lambda(t)$ was obtained.

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Just fixing notation,

$\gamma_1$ is the normalized curve;

$\gamma$ is the curve which is a solution for the Curve Shortening Flow (CSF);

$t_1$ is the time parameter of $\gamma_1$;

$t$ is the time parameter of $\gamma$;

$k_1(p,t_1)$ is the curvature of $\gamma_1 ( \cdot, t_1)$ at $p$;

$k(p,t)$ is the curvature of $\gamma ( \cdot, t)$ at $p$;

$n(p,t)$ is the normal unit vector of $\gamma( \cdot, t)$ at $p$;

$n(p,t_1)$ is the normal unit vector of $\gamma_1( \cdot, t_1)$ at $p$

According was obtained in the article and writing according the notation that I fixed, we know that

$\gamma(p,t) = \lambda(t_1) \gamma_1(p,t_1)$;

$\gamma_1(p,t_1) = \frac{2 \pi}{L[\gamma ( \cdot , t)]} \gamma(p,t)$ (and we note from these two equations that $\lambda(t_1) = \frac{L[\gamma ( \cdot , t)]}{2 \pi}$);

$k_1(p, t_1) = \frac{L[\gamma ( \cdot , t)]}{2 \pi} k(p, t) = \lambda(t_1) k(p,t)$

$\frac{\partial \gamma}{\partial t} = k(p,t) n(p,t)$;

$\overline{k^2} = \frac{1}{2 \pi} \int k(p,t)^2 ds(p)$;

$\frac{\partial \gamma_1}{\partial t_1} = \overline{k^2} \gamma_1 (p,t_1) + k_1(p,t_1) n(p,t_1)$.

And the next relations are obtained by the Chain Rule and Product's rule for derivatives:

$\frac{\partial \gamma}{\partial t_1} (p,t) = \frac{\partial \gamma}{\partial t} (p,t) \frac{dt}{dt_1}$;

$\frac{\gamma}{t_1} (p,t) = \lambda'(t_1) \gamma_1(p,t_1) + \lambda(t_1) \frac{\partial \gamma_1}{\partial t_1} (p,t_1)$.

Using all theses relations, we have

$\lambda'(t_1) \gamma_1(p,t_1) + \lambda(t_1) \frac{\partial \gamma_1}{\partial t_1} (p,t_1) = \frac{\partial \gamma}{\partial t} (p,t) \frac{dt}{dt_1}$

$\Longrightarrow \lambda'(t_1) \gamma_1(p,t_1) + \lambda(t_1) \frac{\partial \gamma_1}{\partial t_1} (p,t_1) = k(p,t) n(p,t) \frac{dt}{dt_1}$

$\Longrightarrow \lambda'(t_1) \gamma_1(p,t_1) + \lambda(t_1) \left( \overline{k^2} \gamma_1 (p,t_1) + k_1(p,t_1) n(p,t_1) \right) = k(p,t) n(p,t) \frac{dt}{dt_1}$

$\Longrightarrow \lambda'(t_1) \gamma_1(p,t_1) + \overline{k^2} \lambda(t_1) \gamma_1 (p,t_1) + \lambda(t_1) k_1(p,t_1) n(p,t_1) = k(p,t) n(p,t) \frac{dt}{dt_1}$

$\Longrightarrow \lambda'(t_1) \gamma_1(p,t_1) + \overline{k^2} \gamma (p,t) + (\lambda(t_1))^2 k(p,t) n(p,t) = k(p,t) n(p,t) \frac{dt}{dt_1}$

$\Longrightarrow \lambda'(t_1) \left( \frac{\gamma(p,t)}{\lambda(t_1)} \right) + \overline{k^2} \gamma (p,t) + (\lambda(t_1))^2 k(p,t) n(p,t) = k(p,t) n(p,t) \frac{dt}{dt_1}$

$\Longrightarrow \left( \frac{\lambda'(t_1)}{\lambda(t_1)} \right) \gamma(p,t) + \overline{k^2} \gamma (p,t) = k(p,t) n(p,t) \frac{dt}{dt_1} - (\lambda(t_1))^2 k(p,t) n(p,t)$

$\Longrightarrow \left( \frac{\lambda'(t_1)}{\lambda(t_1)} + \overline{k^2} \right) \gamma (p,t) = \left( \frac{dt}{dt_1} - (\lambda(t_1))^2 \right) k(p,t) n(p,t)$

In order to obtain $\lambda(t_1) = \frac{L[\gamma( \cdot, 0)]}{2 \pi} \exp \left( - \int_0^{t_1} \overline{k^2}(t') dt' \right)$ (using my notation) as in the article, we need $\frac{dt}{dt_1} = (\lambda(t_1))^2$, therefore, $t = \int_0^{t_1} \lambda(t')^2 dt'$ (using my notation).

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