Fundamental solution of elliptic pde with constant coefficient

Question

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?

Answer 1

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