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 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

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