Helfrich Hamiltionian derivation

The Helfrich Hamiltonian describes the bending energy of a membrane (we are leaving out the Gaussian curvature term here).

$$$
F = \int |H|^2
$$$

## Mean curvature

$$H$$ is the mean curvature, which for a height field function $$S(x,y)$$ [is defined](https://en.wikipedia.org/wiki/Mean_curvature) as

$$$
H = \nabla\cdot\left( \frac{(\nabla S,-1)}{\sqrt{1+|\nabla S|^2}} \right)
$$$

Here $$(\nabla S,-1)$$ is the composite vector $$(\frac{\partial S}{\partial x},\frac{\partial S}{\partial y},-1)$$, which we simply write as $$\nabla S - \hat i_z$$, where $$\hat i_z = (0,0,1)$$, thus

$$$
H = \nabla\cdot(\nabla S - \hat i_z)\left(1+|\nabla S|^2\right)^{-\frac12}
$$$

Applying the product rule

$$$
H = \left[\nabla\cdot(\nabla S - \hat i_z)\right]\left(1+|\nabla S|^2\right)^{-\frac12} + 
(\nabla S - \hat i_z)\cdot\nabla\cdot\left(1+|\nabla S|^2\right)^{-\frac12}
$$$

Applying the chain rule

$$$
H = \nabla^2 S \left(1+|\nabla S|^2\right)^{-\frac12} + 
(\nabla S - \hat i_z)\cdot
\left(-\frac12\right)\left(1+|\nabla S|^2\right)^{-\frac32}2(\nabla S)\nabla^2S
$$$

Combining terms

$$$
H = \nabla^2 S \left[\left(1+|\nabla S|^2\right)^{-\frac12} -
(\nabla S - \hat i_z)
\left(1+|\nabla S|^2\right)^{-\frac32}\nabla S\right]
$$$

In a 2D system $$\nabla S \cdot \hat i_z = 0$$, thus

$$$
H = \nabla^2 S \left[\left(1+|\nabla S|^2\right)^{-\frac12} -
|\nabla S|^2
\left(1+|\nabla S|^2\right)^{-\frac32}\right]
$$$

## First variation

For the phase-field equations we need to calculate the functional derivative (or first variation) with respect to the height variable $$S$$ 

$$$
\frac{\delta F}{\delta S} = \underbrace{\frac{\partial F}{\partial S}}_{=0} - \nabla\cdot \frac{\partial F}{\partial\nabla S} - \color{gray}{\nabla^2\cdot \frac{\partial F}{\partial^2\nabla S} + \dots}
$$$

I did not add the higher order derivative terms...

As $$S$$ does not appear underived in $$F$$ the first term of the functional derivative drops out

$$$
\frac{\delta F}{\delta S} = - \nabla\cdot \frac{\partial}{\partial\nabla S} \left(
(\nabla^2 S)^2 \left[\left(1+|\nabla S|^2\right)^{-\frac12} -
|\nabla S|^2
\left(1+|\nabla S|^2\right)^{-\frac32}\right]^2
\right)
$$$ 

which yields 

$$$
\frac{\delta F}{\delta S} = - \nabla\cdot \left(

(\nabla^2 S)^2 
\left(
-\frac{|\nabla S|^2}{\left(1+|\nabla S|^2\right)^\frac32} + \left(1+|\nabla S|^2\right)^{-\frac12}
\right)

\left(
\frac{6|\nabla S|^3}{\left(1+|\nabla S|^2\right)^\frac52} 
- \frac{6\nabla S}{\left(1+|\nabla S|^2\right)^\frac32}
\right)

\right)
$$$ 

Which simplifies to

$$$
\frac{\delta F}{\delta S} = \nabla\cdot \left(
\frac{6(\nabla^2 S)^2\nabla S}{\left(1+|\nabla S|^2\right)^4}
\right)
$$$

Applying the product rule once

$$$
\frac16\frac{\delta F}{\delta S} = 
\left[\nabla\cdot (\nabla^2 S)^2\nabla S\right] \left(1+|\nabla S|^2\right)^{-4}

+(\nabla^2 S)^2\nabla S \left[\nabla\cdot\left(1+|\nabla S|^2\right)^{-4}\right]
$$$

and again, with the chain rule

$$$
\frac16\frac{\delta F}{\delta S} = 
\left[2\nabla^2S\cdot\nabla^3S\cdot\nabla S + (\nabla^2 S)^3\right] \left(1+|\nabla S|^2\right)^{-4}

-(\nabla^2 S)^2\nabla S \left[
8\left(1+|\nabla S|^2\right)^{-5}\nabla S\cdot\nabla^2S
\right]
$$$

simplify

$$$
\frac16\frac{\delta F}{\delta S} = 
\left[2\nabla^2S\cdot\nabla^3S\cdot\nabla S + (\nabla^2 S)^3\right] \left(1+|\nabla S|^2\right)^{-4}

-8(\nabla^2 S)^3|\nabla S|^2 
\left(1+|\nabla S|^2\right)^{-5}
$$$

## Allen-Cahn bending energy equation

We can solve for $$S$$ as a non-conserved order parameter by plugging the result above into an Allen-Cahn equation

$$$
\frac{\partial S}{\partial t} = -6L \left( 
\left[2\nabla^2S\cdot\nabla^3S\cdot\nabla S + (\nabla^2 S)^3\right] \left(1+|\nabla S|^2\right)^{-4}

-8(\nabla^2 S)^3|\nabla S|^2 
\left(1+|\nabla S|^2\right)^{-5}
\right)
$$$

Introducing an additional variable $$\mu = \nabla^2 S$$ we get

$$$
\frac{\partial S}{\partial t} = -6L \left( 
\left[2\mu\cdot\nabla\mu\cdot\nabla S + \mu^3\right] \left(1+|\nabla S|^2\right)^{-4}

-8\mu^3|\nabla S|^2 
\left(1+|\nabla S|^2\right)^{-5}
\right)
$$$

## First variation of the linearized form

Taking 

$$$
F = \int |\nabla^2S|^2
$$$

with the higher order functional derivative terms you get

$$$
\frac{\delta F}{\delta S} = \nabla^2\cdot2\nabla^2S
$$$