Take $P$ to be a differential operator on vector-valued functions. That is, it takes $v:\mathbb{R}^3\rightarrow\mathbb{R}^3$ and outputs $P[v]:\mathbb{R}^3\rightarrow\mathbb{R}^3$, e.g. the curl operator or the vector Laplacian.

Suppose $\Omega\subseteq\mathbb{R}^3$ is a compact region with a smooth surface as its boundary. I'd like to write the solution of the following Dirichlet problem as a boundary integral: $$ \begin{array}{rl} P[v](x)\equiv 0&\forall x\in\mathrm{int}\,\Omega\\ v\textrm{ prescribed} &\forall x\in\partial\Omega \end{array} $$ That is, I need the Green's function of $P$. I'd like to solve for it in closed-form, so I assume the Green's function should be over $\mathbb{R}^3$ rather than just $\Omega$, similar to $G(x,y)=-\frac{1}{4\pi\|x-y\|_2}$ for the scalar Laplacian.

I have two related questions:

  • For scalar PDE $Q[f]\equiv0$, the Green's function is defined as the solution of $Q[f](x)=\delta(x)$. Is there an analogous definition of a delta function that makes sense in the vector-valued case?
  • If I use the Fourier transform for vector-valued functions e.g. discussed in this document, what would be the Fourier transform of the delta function?

Note: There's some relationship to this question, but I'm hoping for a more general answer.


I'm not sure in what generality and rigor you are expecting an answer; you may want to look at the theory of pseudodifferential operators on vector bundles. But perhaps the following will be more relevant to you:

It helps to think of $ P $ as an operator with matrix coefficients or equally as a matrix of differential operators.

The identity operator $ \mathcal{I} $ defined by $ \mathcal{I}[v] = v $ has the kernel $ I \; \delta(r) $ where $ I: \mathbb{R}^3 \rightarrow \mathbb{R}^3 $ is the identity matrix, $$ \mathcal{I}[v](r) = \int \delta(r-r') \; I \cdot v(r') \; d^3r' = \int \delta(r-r') \; I \cdot v(r') \; d^3r' = \int \delta(r-r') \; v(r') \; d^3r' = v(r). $$

Since $ I $ is a constant, it just factors out of the fourier transform, so that transform of the kernel of $\mathcal{I}$ is just $ I \; e^{i k \cdot r } $.

Let's look at those examples you had. The vector laplacian is $$ \begin{pmatrix} \Delta & 0 & 0 \\ 0 & \Delta & 0 \\ 0 & 0 & \Delta \end{pmatrix}, $$ and in the fourier basis $$ ( k_x^2 + k_y^2 + k_z^2 ) \; I, $$ so it is easy to invert, and after inverse fourier transform will give you the usual Green's function.

Curl is $$ \begin{pmatrix} 0 & -\partial_z & \partial_y \\ \partial_z & 0 & - \partial_x \\ -\partial_y & \partial_x & 0\end{pmatrix}. $$ In fourier basis $$ C = i \begin{pmatrix} 0 & -k_z & k_y \\ k_z & 0 & -k_x \\ - k_y & k_x & 0 \end{pmatrix} $$ This one is tricky to invert because it has a kernel (gradients) and cokernel (div of curl is zero). You can check though that $$ C \cdot \left( \frac{1}{k_x^2 + k_y^2 + k_z^2} C \right) $$ is the identity when restricted to subspace $ \left\{ v \; \big| \; v \cdot k = 0 \right\} $, which is precisely the subspace of divergence free fields. Inverse fourier transform to obtain $$ \frac{1}{(x^2+ y^2 + z^2)^{3/2}} \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix}, $$ which is a weird way of writing the kernel $$ \frac{(r - r')}{|r-r'|^3} \times \cdot $$ that you will recognize from the Biot-Savart law.

(Sorry for sloppy notation, missing $ -1 $s and $ \pi $s, etc.)

  • $\begingroup$ Thanks! This confirms that what we were doing for our PDE is correct ---- turns out I was just computing the inverse Fourier transform incorrectly...oops! Lost a week to my calculus mishap. Your idea of thinking like a matrix inverse is a handy "formal" trick! $\endgroup$ – Justin Solomon Apr 4 '17 at 19:42
  • $\begingroup$ I'm glad I could help! $\endgroup$ – user81327 Apr 4 '17 at 23:46

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