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If we have a function $\phi:\Omega\subseteq\mathbb{R}^2\to\mathbb{R}$, $\phi\in C^2(\Omega)$ is there a way to find a function $d:\Omega\to\mathbb{R}, d\in C^2(\Omega)$ so that:

$|\nabla d(x,y)|=1,\ \forall\ (x,y)\in\Omega$ and $\phi(x)=0\Longleftrightarrow\ d(x)=0$.

For example, this is true when $\phi(x,y)=1-x^2-y^2$, and we can find $d(x,y)=1-\sqrt{x^2+y^2}$, with:

$|\nabla d|=\sqrt{\left (\frac{-x}{\sqrt{x^2+y^2}}\right )^2+\left (\frac{-y}{\sqrt{x^2+y^2}}\right )^2}=1$

and of course $d(x,y)=0\ \Longleftrightarrow\ \phi(x,y)=0$.

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  • $\begingroup$ Isn't $d$ simply $\phi$ divided by the magnitude of its gradient? $\endgroup$
    – thedude
    Commented Dec 16, 2020 at 12:27
  • $\begingroup$ If you're happy with the function $d$ being only Lipschitz, and thus for it to be differentiable almost everywhere in $\Omega$, then you could simply take $d: x \in \Omega \mapsto \mathrm{dist}(x,\{ \phi = 0 \})$. $\endgroup$
    – Leo Moos
    Commented Dec 16, 2020 at 12:40

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In general it is not possible, as you can see for $\phi=0.$

So let us assume that the zero locus is a smooth curve. Is it then always possible to find $d?$

The answer is still no: Consider a disc $D$, and a circle $C$ inside this disc. Choose your favourite $\mathcal C^2$ function $f$ which vanish exactly along this circle. Denote the interior disc of the circle by $D^0$. Let $d\colon D\to\mathbb R$ be a $\mathcal C^2$ function which vanishes exactly along $C$. The function takes its minimum and maximum on the compact set $D^0\cup C.$ Hence, the gradient has to vanish somewhere on $D^0\cup C.$

By the way, the function in your example is not $\mathcal C^2.$

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