Smoothness of solutions of the curve shortening flow given bounded curvature I've been looking at the Lemma 1.5 of The Heat Equation Shrinks Embedded Plane Curves to Round Points (here), where Matthew Grayson proved that
If $\kappa(s,t)$ is bounded for $t\in[0,t_0)$. Then for some $\varepsilon>0$, $C(t)$ exists and is smooth for $t\in [0,t_0+\varepsilon)$.
In the proof he uses the formula 
$${\partial\over\partial t}{\partial\over\partial s}={\partial\over\partial s}{\partial\over\partial t}+\kappa^2{\partial\over\partial s}$$
and also
$${\partial\kappa\over\partial t}={\partial^2\kappa\over\partial s^2}+\kappa^3$$
to obtain that
$${\partial\over\partial t}\left({\partial\kappa\over\partial s}\right)={\partial\over\partial s^2}\left({\partial\kappa\over\partial s}\right)+4\kappa^2\left({\partial\kappa\over\partial s}\right).$$
He claims that this equation bounds the rate of growth of ${\partial\kappa\over\partial s}$ to exponential.
Repeated applications of the first formula yield
$${\partial\over\partial t}\left({\partial^n\kappa\over\partial s^n}\right)={\partial\over\partial s^2}\left({\partial^n\kappa\over\partial s^n}\right)+(n+3)\kappa^2\left({\partial^n\kappa\over\partial s^n}\right)\;+\;\text{previously bounded terms},$$
so he gets the same as before for the $n$-th derivative of $\kappa$.
With this, using again the first formula, it is proved that the curve converges as $t\to t_0$. And similarly, $C(t_0)$ is smooth.
He finally applies the Theorem 1.1 to obtain that $C(t)$ exists and is smooth for some further short time.
I'm struggling to understand the key parts of this proof:


*

*The boundedness of the derivatives is not that obvious to me from the above formulas. I guess he gets an PDE which solution can be bounded by an exponential, but if so I don't really see how.

*From the boundedness of the derivatives of $\kappa$ with respect to $s$ for a fixed $t$, $C(t)$ converges as $t\to t_0$. This is because with $\kappa$ and its derivatives bounded, the curves under the flow must exist for $t\in[0,t_0)$. And the same argument for the smoothness... Am I right?


As you can see I'm kind of lost here, it would be great if anyone could explain to me a little bit these two ideas. Thanks in advance.
 A: $\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$
$\newcommand{\pdk}[3]{\frac{\partial^#3#1}{\partial#2^#3}}$
$\newcommand{\pdd}[3]{\frac{\partial^2#1}{\partial#2\ \partial #3}}$
I worked out this proof and I can answer my own question(s) now.


*

*For my first question on the exponential boundedness of $\pd\kappa s$, let consider
$$F(s,t)=e^{\alpha t}\pd\kappa s$$
and compute its time derivative:
$$\pd F t=\alpha e^{\alpha t}\pd\kappa s+e^{\alpha t}\pdd\kappa s t.$$
Substituting the value of $\pdd\kappa s t$ from the equation obtained previously in the proof,
$$\pd F t= e^{\alpha t}\left(\alpha\pd\kappa s+\pdk\kappa s 3+4\kappa^2\pd\kappa s\right).$$
We want to put this as much in terms of $F$ (rather than $\kappa$) as is possible. We already know $\pd\kappa s= e^{-\alpha t}F$, and differentiating this gives us expressions for the higher derivatives of $\kappa$ with respect to $s$. Substituting these all into $\pd F t$ we get a parabolic equation for $F$,
$$\pd F t =\pdk F s 2+(\alpha+4\kappa^2)F.$$
Since $\kappa$ is bounded we can choose $\alpha$ so that $\alpha+4\kappa^2\leq 0$, so the maximum principle tells us that $F$ achieves its maximum on the parabolic boundary. Then we get


$$\pd\kappa s\leq e^{-\alpha t}\sup_{t=0}F=e^{-\alpha t}\sup_{t=0}\pd\kappa s.$$


*Once we get bounded all derivatives of $\kappa$ with respect to $s$ (plus the hypothesis of $\kappa$ bounded), we obtain that the curvature and its derivatives don't blow up for times $[0,t_0)$, and therefore the curves under the flow are $\mathcal{C}^\infty$. Hence, $C(t_0)$ exists and is smooth.


Hope this can help others to understand what it is done here, and similar computations that show up a lot in geometric analysis.
