Fundamental solution for a parabolic PDE with costant coefficents

Question

as it is well known, the fundamental solution of the heat equation is the function

$G(t,x)=\frac{1}{(4\pi t)^{n/2}}e^{\frac{|x|^2}{4t}}$,

for all $t>0,x\in\mathbb{R}^n$.

I wonder if exists (and if you have same references) a similar explicit formula for the fundamental solution for a parabolic PDE with constant coefficents.

Answer 1

A parabolic operator with constant coefficients is a linear transformation away from the heat operator, so it is a natural guess that the fundamental solutions should be similar. I will use this idea to find the fundamental solution. (If you just want to see the solution, see the last line.)

Take two positive definite symmetric $n\times n$ matrices $A$ and $Q$. Consider the function $\phi_a(x)=\exp(-\frac1ax^TQx)$ with a parameter $a>0$. A simple calculation gives $$ \partial_i\partial_j\phi_a(x) %= %\partial_i[-\frac2a\phi_a(x)(Qx)_j] %= %\frac4{a^2}\phi_a(x)(Qx)_i(Qx)_j-\frac2a\phi_a(x)Q_{ij} = \frac2a\phi_a(x)(\frac2a(Qx)_i(Qx)_j-Q_{ij}) $$ and $$ \partial_a\phi_a(x)=\frac1{a^2}\phi_a(x)x^TQx. $$ Therefore the function $F(t,x)=t^{-n/2}\phi_{4t}(x)$ satisfies $$ \partial_tF(t,x) = t^{-1}(-\frac n2+\frac1{4t}x^TQx)t^{-n/2}\phi_{4t}(x). $$ Now consider the elliptic second order operator $L=\sum_{ij}A_{ij}\partial_i\partial_j$ — every elliptic second order operator with constant coefficients is of this form. Now $$ L\phi_a(x) %= %A_{ij}\frac2a\phi_a(x)(\frac2a(Qx)_i(Qx)_j-Q_{ij}) %= %\frac2a\phi_a(x)(\frac2a A_{ij}(Qx)_i(Qx)_j-Q_{ij}A_{ij}) = \frac2a\phi_a(x)(\frac2a x^TQAQx-\operatorname{tr}(QA)), $$ so $$ LF(t,x) = t^{-n/2}\frac1{2t}\phi_{4t}(x)(\frac1{2t} x^TQAQx-\operatorname{tr}(QA)). $$ If we assume $Q=A^{-1}$, we get $$ LF(t,x) = t^{-n/2}\frac1{t}\phi_{4t}(x)(\frac1{4t} x^TQx-n/2) = \partial_tF(t,x). $$ We have thus found that $F$ with $Q=A^{-1}$ satisfies the heat equation for $t>0$, so it must be the fundamental solution — up to normalization.

To fix the normalization, we only need to evaluate a Gaussian integral at any fixed time. Suppose the fundamental solution is $cF$ for a constant $c>0$. We should have $$ 1 = \int_{\mathbb R^n}cF(1,x)dx %= %c\int_{\mathbb R^n}\exp(-\frac14x^TA^{-1}x)dx = c\int_{\mathbb R^n}\exp(-\frac14|\sqrt Ax|^2)dx %= %c\int_{\mathbb R^n}\exp(-\frac14|y|^2)d(A^{-1/2}y) = c\det(A)^{-1/2}\int_{\mathbb R^n}\exp(-\frac14|y|^2)dy = c\det(A)^{-1/2}(4\pi)^{n/2}. $$ Therefore the fundamental solution to the operator $L$ given by the matrix $A$ is $$ F(t,x) = \det(A)^{1/2}(4\pi t)^{-n/2}\exp(-\frac1{4t}x^TA^{-1}x). $$ Notice that when $A$ is the identity matrix, this is the usual formula as it should.