Laplacian of the squared distance function I'm having trouble to understand some properties of the squared distance function assigned to a curve which moves by mean curvature.
We assume that a smooth mean curvature flow of an embedded curve $\Gamma_t$ starting from $\Gamma_0$ exists up to some time $T>0$. We can express $\Gamma_t$ in terms of a parametrization $\gamma(.,t)$ such that $\Gamma_t = \gamma([0,1],t)$, where $\gamma : [0,1] \times [0,T) \rightarrow \mathbb{R}^3$, and it satisfies $\frac{\partial}{\partial t}\gamma(\theta,t)=\kappa_{\gamma(\theta,t)}N_{\gamma(\theta,t)}$ and $\gamma([0,1],0)=\Gamma_0$. Now for $\sigma >0$ we define the (truncated) squared distance function $\phi_\sigma(x,t):=\frac{1}{2} \left\{
\begin{array}{ll}
d(x,\Gamma_t)^2 & \textrm{if} \; d(x,\Gamma_t)<\sigma \\
4\sigma^2 & \, \textrm{if} \; d(x,\Gamma_t)>2\sigma \\
\end{array}
\right.$
It is stated that $\partial_t \phi_\sigma = 0$ and $\Delta\phi_\sigma = 2$ on $\Gamma_t$ without further explanation. I never took a differential geometry class thus I'm no expert on this topic and this isn't obvious to me. I assume the first assumption is valid because if we calculate the time derivative of the squared distance function, we are still left with a term $d(x,\Gamma_t)$ because of the chain rule, and this term is $0$. But the equality with the Laplacian I don't understand. Is it simple calculation or is it another argument based on the mean curvature flow? Any help is appreciated.
 A: First of all, since you consider only points on $\Gamma_t$, we may assume that $\phi_\sigma = \frac{1}{2} d(\cdot, \Gamma_t)^2$. 
Let $x\in \Gamma_t$ and let $x(s)$ be a curve so that $x(s) \in \Gamma_s$ and $x(t) =x$. Then $\phi_\sigma (x(s), s) = 0$ by definition. By Chain rule, at $s=t$
$$\nabla \phi_\sigma \cdot \dot x(t) + \partial _t \phi_\sigma (x, t) = 0.$$
Thus it suffices to show that $\nabla \phi_\sigma (x, t) = 0$. This is easy: first, since $\phi_\sigma =0$ on $\Gamma_t$, 
$$\nabla \phi_\sigma (x,t) \cdot \frac{\partial \gamma_t}{\partial \theta} = 0$$
since $\frac{\partial \gamma_t}{\partial \theta}$ is tangential to $\Gamma_t$. Next, let $\nu$ be any normal vector to $\Gamma_t$ at $x$. Then 
$$ \phi_\sigma ( x+ s\nu , t) = \frac{1}{2} s^2$$
when $|s|$ is small enough. Thus by Chain rule again, 
$$\nabla \phi_\sigma (x, t) \cdot \nu =\frac{d}{ds}\bigg|_{s=0} \phi_\sigma (x+s\nu, t)= 0$$
and so $\nabla \phi_\sigma (x, t) = 0$. Note that we did not use the mean curvature flow equation. 
The equality $\Delta \phi_\sigma = 2$ is in some sense easier, since it involves only $\Gamma_t$ (and so again is independent of the mean curvature flow equation). Let $e_1(\theta), e_2(\theta) $ be two normal vector fields to $\Gamma_t$. Then for $s_1, s_2$ small, 
$$\tag{1} \phi_\sigma (\gamma_t(\theta) + s_1 e_1(\theta) + s_2 e_2(\theta)) = \frac{1}{2} (s_1^2 + s_2^2).$$
Then 
\begin{align*}
\Delta \phi_\sigma (x, t) &= \frac{\partial^2 \phi_\sigma}{\partial \theta^2} +\frac{\partial^2 \phi_\sigma}{\partial s_1^2} +\frac{\partial^2 \phi_\sigma}{\partial s_2^2} + X \cdot \nabla \phi_\sigma \\
&= 0 + 1+1+0 =2
\end{align*}
The term $X$ is not important since we know $\nabla \phi_\sigma (x, t) = 0$. 
Remark: To see that (1) holds in a neighborhood $U$ of $\Gamma_t$, first one can find a neighborhood $U$ of $\Gamma_t$ so that for all $y\in U$, there is an unique $\theta$ so that 
$$ |y-\gamma(\theta) | = d(y, \Gamma_t).$$
Since the ball in $\mathbb R^3$ with center $y$ and radius $|y-\gamma(\theta) | $ touches $\Gamma_t$ at one point $\gamma (\theta)$, $y-\theta$ is normal to $\Gamma_t$. Thus there are $s_1, s_2$ so that 
$$ y-\gamma(\theta) = s_1 e_1(\theta) + s_2e_2(\theta).$$
Thus $y = \gamma(\theta) + s_1 e_1(\theta) + s_2e_2(\theta)$ and 
\begin{align*}
\phi_\sigma (y, t) &= \frac{1}{2} d(y, \Gamma_t)^2 \\
&= \frac{1}{2} |y-\gamma(\theta)|^2 \\
&=\frac{1}{2} |s_1 e_1(\theta) + s_2e_2(\theta)|^2\\
&= \frac{1}{2} (s_1^2 + s_2^2)^2. 
\end{align*}
