# 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?

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 $$\Psi(z) = \Gamma(Az) \det{A}$$