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Bogdan
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Working with the level set method introduced by Osher & Sethian in shape optimization I came across a simple question that I did not succeed to prove. It mainly asserts that the perimeter of the zero level set is continuous with respect to the level set function. The continuity of the area is true (see here Lebesgue measure of a neighbourhood of a curve). It is well known that the perimeter is a lower semicontinuous functional, so one inequality is checked.

Here is my question:

Let $\Omega\subset\mathbb{R}^2$ be a bounded set with smooth boundary, $\phi_n:\overline\Omega\to\mathbb{R},\ n\geq 1$ be a sequence of functions such that:

$\bullet\ \phi_n\to \phi\ \text{in}\ C^1(\overline{\Omega})$ i.e. $\lim\limits_{n\to\infty} \sup_{x\in\overline{\Omega}}|\phi_n(x)-\phi(x)|+\sup_{x\in\overline{\Omega}} |\nabla\phi_n(x)-\nabla\phi(x)|=0$

$\bullet$ $\nabla\phi_n\neq 0$ on $\phi_n=0$ and $\nabla\phi\neq 0$ on $\phi=0$

$\bullet$ $\{\phi=0\}\Subset\Omega$

Then how can we prove that:

$$\lim\limits_{n\to\infty} \int_{\phi_n=0} 1 \ d\sigma=\int_{\phi=0} 1\ d\sigma$$

?

For the continuity of the perimeter on level sets with respect to the level you can see other posts here: Continuity of Hausdorff measure on level sets, Continuity of surface integrals on level sets

Working with the level set method introduced by Osher & Sethian in shape optimization I came across a simple question that I did not succeed to prove. It mainly asserts that the perimeter of the zero level set is continuous with respect to the level set function. The continuity of the area is true (see here Lebesgue measure of a neighbourhood of a curve). It is well known that the perimeter is a lower semicontinuous functional, so one inequality is checked.

Here is my question:

Let $\Omega\subset\mathbb{R}^2$ be a bounded set with smooth boundary, $\phi_n:\overline\Omega\to\mathbb{R},\ n\geq 1$ be a sequence of functions such that:

$\bullet\ \phi_n\to \phi\ \text{in}\ C^1(\overline{\Omega})$ i.e. $\lim\limits_{n\to\infty} \sup_{x\in\overline{\Omega}}|\phi_n(x)-\phi(x)|+\sup_{x\in\overline{\Omega}} |\nabla\phi_n(x)-\nabla\phi(x)|=0$

$\bullet$ $\nabla\phi_n\neq 0$ on $\phi_n=0$ and $\nabla\phi\neq 0$ on $\phi=0$

Then how can we prove that:

$$\lim\limits_{n\to\infty} \int_{\phi_n=0} 1 \ d\sigma=\int_{\phi=0} 1\ d\sigma$$

?

For the continuity of the perimeter on level sets with respect to the level you can see other posts here: Continuity of Hausdorff measure on level sets, Continuity of surface integrals on level sets

Working with the level set method introduced by Osher & Sethian in shape optimization I came across a simple question that I did not succeed to prove. It mainly asserts that the perimeter of the zero level set is continuous with respect to the level set function. The continuity of the area is true (see here Lebesgue measure of a neighbourhood of a curve). It is well known that the perimeter is a lower semicontinuous functional, so one inequality is checked.

Here is my question:

Let $\Omega\subset\mathbb{R}^2$ be a bounded set with smooth boundary, $\phi_n:\overline\Omega\to\mathbb{R},\ n\geq 1$ be a sequence of functions such that:

$\bullet\ \phi_n\to \phi\ \text{in}\ C^1(\overline{\Omega})$ i.e. $\lim\limits_{n\to\infty} \sup_{x\in\overline{\Omega}}|\phi_n(x)-\phi(x)|+\sup_{x\in\overline{\Omega}} |\nabla\phi_n(x)-\nabla\phi(x)|=0$

$\bullet$ $\nabla\phi_n\neq 0$ on $\phi_n=0$ and $\nabla\phi\neq 0$ on $\phi=0$

$\bullet$ $\{\phi=0\}\Subset\Omega$

Then how can we prove that:

$$\lim\limits_{n\to\infty} \int_{\phi_n=0} 1 \ d\sigma=\int_{\phi=0} 1\ d\sigma$$

?

For the continuity of the perimeter on level sets with respect to the level you can see other posts here: Continuity of Hausdorff measure on level sets, Continuity of surface integrals on level sets

Source Link
Bogdan
  • 1.8k
  • 9
  • 16

Continuity of the perimeter of level sets w.r.t. level function

Working with the level set method introduced by Osher & Sethian in shape optimization I came across a simple question that I did not succeed to prove. It mainly asserts that the perimeter of the zero level set is continuous with respect to the level set function. The continuity of the area is true (see here Lebesgue measure of a neighbourhood of a curve). It is well known that the perimeter is a lower semicontinuous functional, so one inequality is checked.

Here is my question:

Let $\Omega\subset\mathbb{R}^2$ be a bounded set with smooth boundary, $\phi_n:\overline\Omega\to\mathbb{R},\ n\geq 1$ be a sequence of functions such that:

$\bullet\ \phi_n\to \phi\ \text{in}\ C^1(\overline{\Omega})$ i.e. $\lim\limits_{n\to\infty} \sup_{x\in\overline{\Omega}}|\phi_n(x)-\phi(x)|+\sup_{x\in\overline{\Omega}} |\nabla\phi_n(x)-\nabla\phi(x)|=0$

$\bullet$ $\nabla\phi_n\neq 0$ on $\phi_n=0$ and $\nabla\phi\neq 0$ on $\phi=0$

Then how can we prove that:

$$\lim\limits_{n\to\infty} \int_{\phi_n=0} 1 \ d\sigma=\int_{\phi=0} 1\ d\sigma$$

?

For the continuity of the perimeter on level sets with respect to the level you can see other posts here: Continuity of Hausdorff measure on level sets, Continuity of surface integrals on level sets