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I tried to derive Euler-Lagrange from a functional:

$E(\phi) = \int_{\Omega} |\nabla \phi | \delta(\phi) dx$

where $\phi$ is real-valued function depending on $x$. I denote $F(\phi, \nabla\phi) = |\nabla \phi | \delta(\phi)$ and I tried to follow the procedure of deriving Euler-Lagrange as to compute:

$\frac{\partial F}{\partial \phi}-\sum_{i=1}^{m+1} \frac{\partial}{\partial x_i}\left(\frac{\partial F}{\partial \phi_{x_i}}\right)=0$

Some academic papers just directly give the result Euler-Lagrange: $\delta(\phi)\cdot \text{div}\left( \frac{\nabla \phi}{|\nabla \phi|}\right)$

Does anyone know how this result is derived? Thank you.

Teddy
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1 Answers1

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Given $F(\phi, \nabla \phi) = |\nabla \phi| \delta(\phi)$, the Euler-Lagrange can be written as:

(Eq. 1): $\frac{\partial F}{\partial \phi} - \nabla\cdot\left(\frac{\partial F}{\partial \nabla \phi}\right) = |\nabla \phi| \delta'(\phi) - \nabla\cdot\left(\delta(\phi)\cdot \frac{\nabla \phi}{|\nabla \phi|} \right)$

From the product rule of divergence, the second term is

$\nabla\cdot\left(\delta(\phi)\cdot \frac{\nabla \phi}{|\nabla \phi|} \right) = \nabla \delta(\phi) \cdot \frac{\nabla \phi}{|\nabla \phi|} + \delta(\phi)\cdot \nabla\cdot\left(\frac{\nabla \phi}{|\nabla \phi|} \right)$

Since $\nabla\delta(\phi) = \delta'(\phi)\cdot \nabla\phi$, then this term becomes

$\nabla \delta(\phi) \cdot \frac{\nabla \phi}{|\nabla \phi|} + \delta(\phi)\cdot \nabla\cdot\left(\frac{\nabla \phi}{|\nabla \phi|} \right) = \delta'(\phi) \cdot \frac{\nabla \phi \cdot \nabla \phi}{|\nabla \phi|} + \delta(\phi)\cdot \nabla\cdot\left(\frac{\nabla \phi}{|\nabla \phi|} \right) = \delta'(\phi)\cdot|\nabla\phi| + \delta(\phi)\cdot \nabla\cdot\left(\frac{\nabla \phi}{|\nabla \phi|} \right)$

Plugging this result into above equation 1 yields:

$\frac{\partial F}{\partial \phi} - \nabla\cdot\left(\frac{\partial F}{\partial \nabla \phi}\right) = |\nabla \phi| \delta'(\phi) - \delta'(\phi)\cdot|\nabla\phi| - \delta(\phi)\cdot \nabla\cdot\left(\frac{\nabla \phi}{|\nabla \phi|} \right) = - \delta(\phi)\cdot \nabla\cdot\left(\frac{\nabla \phi}{|\nabla \phi|} \right)$

Hence, the Euler-Lagrange can be derived as:

$\delta(\phi)\cdot \nabla\cdot\left(\frac{\nabla \phi}{|\nabla \phi|} \right) = 0$

Teddy
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