# Analytical testcase for 2D/3D anisotropic Diffusion (Heat Kernel)

I want to verify and compare different Discretizations of the anisotropic diffusion equation in 2D / 3D.

In order to both test the timestepping and the spatial discretisations I had a look at using the heat kernels as an analytical solution to diffusion equations. The one dimensional heat kernel (wiki-link) looks like this: $$u(x,t) = \frac{1}{(4\pi t)^{d/2}} e^{-x^2/4t}\$$ and solves following diffusion equation of dimension $d$ analytically: $$\frac{\partial u}{\partial t} = -\Delta u$$

The idea is that for an analytical solution I can verify my convergence both spatially and temporal. It is important in my case, that the diffusion is anisotropic, and might not be aligned with the underlying grids.

$$\frac{\partial u}{\partial t} = \nabla (D \nabla u)$$

In order to expand the analytical solution to my case, where the diffusion is not scalar and unity, but a full (symmetric positive definite) tensor D, I had a look at multivariate gaußian distributions (wiki-link).

My question is the following:

1. Is there somewhere in the literature a heat kernel for anisotropic (but homogeneous) diffusion which I can use to verify my numerics?

By analogy I would come up with something like:

$$u(\vec{x},\vec{\mu},t) = \frac{1}{(4\pi~ |\mathbf{D}| t)^{d/2}} \exp \large(\frac{-(\vec{x}-\vec{\mu})^{T} \mathbf{D}^{-1}(\vec{x}-\vec{\mu})}{4~|\mathbf{D}|~t}\large).$$

1. Is the above equation a proper analytical solution of the anisotropic diffusion equation?

I am not sure if the exponent $d/2$ is correct for the $|D|$. I am aware that the question might touch both physics, and numerics, but I thought this might be useful to other people trying to verify their codes.

Any hints or help is appreciated!

• I'd start by looking for a coordinate transformation (using the eigenvalues and eigenvectors of $D$) which transforms your anisotropic diffusion into isotropic diffusion. Then you can transform the classical heat kernel solution back into the original anisotropic space. But I haven't worked out the details. – user3883 Jul 20 '18 at 10:46
• This may be a stupid comment, but have you thought about just manufacturing an analytical solution using the method of manufactured solutions? This is generally what I do for anisotropic equations... – BlaB Aug 22 '18 at 14:07

Since Laplacian is an elliptic operator you are looking for the Cholesky decomposition of the assumed constant diffusion matrix $D$: $$D = LL^T$$
Therefore the parabolic equation may be written as: $$\partial_t u=-\nabla^T(D\nabla u)=-(L^T\nabla)^T(L^T\nabla u)=-\tilde{\nabla}^T(\tilde{\nabla} u)=-\tilde{\nabla}^2u$$
Naming $\tilde{\nabla}=L^T\nabla$ and therefore $\tilde{x}=L^{-1}x$. This last PDE is totally decoupled (using Fourier transform and separation of variables).
The solution for the unnormalised heat kernel will be (using $\tilde{x}=L^{-1}x$): $$\Phi=\prod_{n=1}^d\frac{\exp{(-\tilde{x}_n^2/(4t))}}{\sqrt{4\pi t}}=\frac{\exp{(-\tilde{x}^T\tilde{x}/(4t))}}{(4\pi t)^{d/2}}=\frac{\exp{(-x^TD^{-1}x/(4t))}}{(4\pi t)^{d/2}}$$
• There is one step missing here in getting kernel in $x$ space. The solution involves a convolution so you have to transform the $d$-dimensional measure $d \tilde{x}$ to $dx$. In $x$-space the kernel becomes $\exp(-x^\top D^{-1}x/(4t))/(4\pi t |D|)^{d/2}$. – cfdlab Aug 22 '18 at 11:00