I'm trying understand the article "[Curvature bound for curve shortening flow via distance comparison and a direct proof of Grayson's theorem](https://arxiv.org/pdf/0908.2682.pdf)" by Ben Andrews and Paul Bryan and they stated on the final of the article (precisely, on the proof of corollary 4) that



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> 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 $\tau = \int\_0^t \lambda(t')^2 dt'$.
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I'm trying understand how $\lambda(t)$ was obtained.



A change of parameters used on Curve Shortening Flow

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> Just fixing notation,
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> $\gamma\_1$ is the normalized curve;
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> $\gamma$ is the curve which is a solution for the Curve Shortening Flow (CSF);
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> $t\_1$ is the time parameter of $\gamma\_1$;
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> $t$ is the time parameter of $\gamma$;
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> $k\_1(p,t\_1)$ is the curvature of $\gamma\_1 ( \cdot, t\_1)$ at $p$;
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> $k(p,t)$ is the curvature of $\gamma ( \cdot, t)$ at $p$;
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> $n(p,t)$ is the normal unit vector of $\gamma( \cdot, t)$ at $p$;
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> $n(p,t\_1)$ is the normal unit vector of $\gamma\_1( \cdot, t\_1)$ at $p$
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> According was obtained in the article and writing according the notation that I fixed, we know that
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> $\gamma(p,t) = \lambda(t\_1) \gamma\_1(p,t\_1)$;
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> $\gamma\_1(p,t\_1) = \frac{2 \pi}{L[\gamma ( \cdot , t)]} \gamma(p,t)$ (and we note from these two equations that $\lambda(t\_1) = \frac{L[\gamma ( \cdot , t)]}{2 \pi}$);
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> $k\_1(p, t\_1) = \frac{L[\gamma ( \cdot , t)]}{2 \pi} k(p, t) = \lambda(t\_1) k(p,t)$
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> $\frac{\partial \gamma}{\partial t} = k(p,t) n(p,t)$;
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> $\overline{k^2} = \frac{1}{2 \pi} \int k(p,t)^2 ds(p)$;
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> $\frac{\partial \gamma\_1}{\partial t\_1} = \overline{k^2} \gamma\_1 (p,t\_1) + k\_1(p,t\_1) n(p,t\_1)$.
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> And the next relations are obtained by the Chain Rule and Product's rule for derivatives:
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> $\frac{\partial \gamma}{\partial t\_1} (p,t) = \frac{\partial \gamma}{\partial t} (p,t) \frac{dt}{dt\_1}$;
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> $\frac{\gamma}{t\_1} (p,t) = \lambda'(t\_1) \gamma\_1(p,t\_1) + \lambda(t\_1) \frac{\partial \gamma\_1}{\partial t\_1} (p,t\_1)$.
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> Using all theses relations, we have
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> $\lambda'(t\_1) \gamma\_1(p,t\_1) + \lambda(t\_1) \frac{\partial \gamma\_1}{\partial t\_1} (p,t\_1) = \frac{\partial \gamma}{\partial t} (p,t) \frac{dt}{dt\_1}$
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> $\Longrightarrow \lambda'(t\_1) \gamma\_1(p,t\_1) + \lambda(t\_1) \frac{\partial \gamma\_1}{\partial t\_1} (p,t\_1) = k(p,t) n(p,t) \frac{dt}{dt\_1}$
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> $\Longrightarrow \lambda'(t\_1) \gamma\_1(p,t\_1) + \lambda(t\_1) \left( \overline{k^2} \gamma\_1 (p,t\_1) + k\_1(p,t\_1) n(p,t\_1) \right) = k(p,t) n(p,t) \frac{dt}{dt\_1}$
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> $\Longrightarrow \lambda'(t\_1) \gamma\_1(p,t\_1) + \overline{k^2} \lambda(t\_1) \gamma\_1 (p,t\_1) + \lambda(t\_1) k\_1(p,t\_1) n(p,t\_1) = k(p,t) n(p,t) \frac{dt}{dt\_1}$
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> $\Longrightarrow \lambda'(t\_1) \gamma\_1(p,t\_1) + \overline{k^2} \gamma (p,t) + (\lambda(t\_1))^2 k(p,t) n(p,t) = k(p,t) n(p,t) \frac{dt}{dt\_1}$
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> $\Longrightarrow \lambda'(t\_1) \left( \frac{\gamma(p,t)}{\lambda(t\_1)} \right) + \overline{k^2} \gamma (p,t) + (\lambda(t\_1))^2 k(p,t) n(p,t) = k(p,t) n(p,t) \frac{dt}{dt\_1}$
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> $\Longrightarrow \left( \frac{\lambda'(t\_1)}{\lambda(t\_1)} \right) \gamma(p,t) + \overline{k^2} \gamma (p,t) = k(p,t) n(p,t) \frac{dt}{dt\_1} - (\lambda(t\_1))^2 k(p,t) n(p,t)$
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> $\Longrightarrow \left( \frac{\lambda'(t\_1)}{\lambda(t\_1)} + \overline{k^2} \right) \gamma (p,t) = \left( \frac{dt}{dt\_1} - (\lambda(t\_1))^2 \right) k(p,t) n(p,t)$
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> 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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