Fundamental solution of elliptic pde with constant coefficient It is known that the fundamental solution of Laplace's equation on $\mathbb{R^n}$ is 
\begin{align}
\Gamma(x)=\begin{cases}
\frac{\log |x|}{2\pi}, n=2\\
\frac{1}{n(2-n)\nu_n}|x|^{2-n}, n\ge 3.
\end{cases}
\end{align}
where $\nu_n$ is the volume of the unit ball in $\mathbb{R}^n, n\ge 3$. Can we find  a fundamental solution of the following equation
\begin{align}
\sum_{i=1}^n \sum_{j=1}^n c_{ij} u_{x_ix_j}(x)=0
\end{align}
where $c_{ij}'s$ are constants? When $c_{ij}=\begin{cases} 1, i=j\\
0, i\ne j
\end{cases}$, we recover Laplace equation. Is there any book discussing this type of question?
 A: Sure, you can find it with some calculus tricks.  You didn't explicitly state it in your post, but in the title you say the PDE is elliptic, so I am going to assume ellipticity for the coefficient matrix $C$.  This gives us two things: that $C = C^T$ and that $C \ge \lambda I$ for some $\lambda >0$.
Since $C$ is symmetric and positive definite we can write $C = A^{-1} (A^{-1})^T$ for another positive definite symmetric matrix $A$.  This decomposition can be achieved using the fact that $C$ is diagonalizable over an orthonormal basis and the eigenvalues are strictly positive (no smaller than $\lambda$, actually).
Suppose now that functions $u$ and $f$ are given.  Define $v(x) = u(Ax)$.  A direct calculation then shows that 
$$
\Delta u = f \Leftrightarrow \sum_{ij} C_{ij} \partial_i \partial_j v(x) = f(Ax).
$$
Now, given a function $g$ we define $f(x) = g(A^{-1}x)$ to see that $f(Ax) = g(x)$.  Then
$$
u(x) = \int_{\mathbb{R}^n} \Gamma(y) f(x-y) dy = \int_{\mathbb{R}^n} \Gamma(y) g(A^{-1}(x-y)) dy
$$
solves $\Delta u = f$.  Using the above, we set 
$$
v(x) = u(Ax) = \int_{\mathbb{R}^n} \Gamma(y) g(x-A^{-1}y) dy = \int_{\mathbb{R}^n} \Gamma(A(x-z)) (\det A ) g(z) dz
$$
and note that 
$$
\sum_{ij} C_{ij} \partial_i \partial_j v(x) = f(Ax) = g(x).
$$
From this we see that $v$ solves the desired PDE and so the fundamental solution is 
$$
\Psi(z) = \Gamma(Az) \det{A}
$$
