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For the level set method, $\phi(\vec{x},t)$ is the level set function in 3D and the level set $\phi(\vec{x},t) = 0$ forms the interface. For evolving $\phi$ the derivation says to imagine a particle $\textbf{x}(t)$ on the surface, then we differentiate with respect to t

$\frac{d}{dt}(\phi(\vec{x}, t)=0)$

Using chain rule we get

$\frac{\partial \phi}{\partial t} + \triangledown \phi \cdot \frac{d\vec{x}}{dt} = 0$

Then this equation is turned into

$\frac{\partial \phi}{\partial t} + \triangledown \phi \cdot \vec{V} = 0$

I guess $\vec{V}$ is the speed, and its elements are to replace $dx$, $dy$, .. that are elements of $d{\vec{x}}$. Hence we can substitute our own displacement and receive a result.

What I do not understand is they continue like this:

Seperate $\vec{V}$ into normal and tangential components

$\frac{\partial \phi}{\partial t} + \triangledown \phi \cdot (\vec{V_N}\vec{N} + \vec{V_T}\vec{T}) = 0$

Then since $\vec{N} = \frac{\triangledown \phi}{|\triangledown \phi |}$

We get

$\frac{\partial \phi}{\partial t} + V_N \cdot |\triangledown \phi| = 0$

What happened to $V_T$? Also how can the dot product of $\triangledown \phi$ with $\frac{\triangledown \phi}{|\triangledown \phi |}$ result in $|\triangledown \phi|$ ?


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$\frac{\partial \phi}{\partial t} + \triangledown \phi \cdot (\vec{V_N}\vec{N} + \vec{V_T}\vec{T}) = 0$ <== I think that the $\vec{N}$ and $\vec{T}$ should be removed in this equation. – Thomas Rot Feb 17 '11 at 13:10
up vote 4 down vote accepted

Since $\nabla \phi$ is normal to the surface, its inner product with a vector tangential to the surface is zero. Furthermore $\nabla \phi \cdot \nabla \phi=|\nabla\phi|^2$, hence the second identity: $$\nabla\phi\cdot\frac{\nabla\phi}{|\nabla\phi|}=\frac{|\nabla\phi|^2}{| \nabla\phi|}=|\nabla\phi|$$.

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