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Let $w=w(x,t)$, $x\in\mathbb{R},\ t\in[0,T)$ be a classical solution of the graph curve shortening flow (call (A))

$$ \begin{cases} w_t=\frac {w_{xx}}{1+w_x^2}\quad &\text{ in } \mathbb{R}\times(0,T),\\\\ w(x,0)=w_0(x) \quad &\text{ on } \mathbb{R}. \end{cases} $$

Here, $w_0(x)$ is a globally Lipschitz function on $\mathbb{R}$. Then, we easily see that $W(x,z,t)=w(x,t)-z$ is a classical solution to the level-set mean curvature flow (call (B))

$$ \begin{cases} W_t= \textrm{tr}\left\{\left(I_2-\widehat{DW}\otimes\widehat{DW}\right)D^2W\right\}\quad &\text{ in } \mathbb{R}\times(0,T),\\\\ W(x,0)=w_0(x)-z \quad &\text{ on } \mathbb{R}, \end{cases} $$

where $\widehat{DW}=\frac{DW}{|DW|}$, the gradient $DW$, the Hessian $D^2W$ are with respect to the variable $(x,z)$.

My question is the following; can we say that $W(x,z,t)=w(x,t)-z$ is a viscosity solution to (B) if $w(x,t)$ is a viscosity solution to (A)?

What we've just checked is the classical solution version, and the question is about viscosity solutions.

Thanks!

JJW
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  • What have you tried? I would start with the test function definition of viscosity solution. – Jeff Dec 19 '22 at 14:04
  • @Jeff Yes, I've tried with the definition of viscosity solutions with test functions. Unless test functions $\phi$ have second-order derivatives $\phi_{xz}=\phi_{zx}=0$ at a "tested point," it was unclear to me to check with the definition. – JJW Dec 20 '22 at 21:20
  • You can always assume test functions are smooth (i.e., infinitely differentiable). There is a standard argument for showing this equivalence (it is likely in the Evans & Spruck paper on mean curvature motion). – Jeff Dec 20 '22 at 21:49

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