Fundamental solution for a parabolic PDE with costant coefficents 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.
 A: 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.
