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