In physics it is often of interest to solve Laplace equation \begin{equation} \nabla^2 V = f , \end{equation} where $f$ is a distribution, e.g. the Dirac delta function for a point charge, the Dirac surface distribution for a surface charge, or a rapidly decaying function for a volume distribution of charge.

If we consider the limited case where $f$ is a Schwartz function we may solve the equation above by first finding the Green's function which is \begin{equation} G(\mathbf r) = \frac{1}{4 \pi | \mathbf r |} \end{equation} and it is then possible to find the solution as \begin{equation} V(\mathbf r) = \int G(\mathbf r - \mathbf r') f(\mathbf r') \mathrm{d}^3 r' = \int G(\mathbf r') f(\mathbf r - \mathbf r') \mathrm{d}^3 y . \end{equation} However, it is my understanding that the Green's function should be considered as a (tempered) distribution and that this equation can then be viewed as the distribution $G$ acting on the function $f_{\mathbf r}$ (which is translated by $\mathbf r$).

However, this does not work if $f$ is instead a distribution since we then would have a multiplication of distributions. Can someone please guide me to how to properly interpret the above in terms of distributions? The closest I have come to an answer is to perhaps consider the solution $V(\mathbf r)$ as a convolution between two distributions, but I cannot find any good sources for this even though I have searched quite extensively.


1 Answer 1


While multiplication of distributions is not well-defined, convolution of distributions is well-defined provided that at least one of the two is compactly supported. The Green's function on the full space is not compactly supported, so you can only make sense of $\Delta u = f$ this way for compactly supported distributions $f$. In this case the solution is the distributional convolution of $G$ with $f$. In the setting of "regular" distributions (i.e. continuous linear functionals on $C^\infty_c$, as opposed to the Schwarz class $\mathcal{S}$), this is defined purely through the equality:


for all $C^\infty_c$ functions $h$, which is analogous to associativity of multiplication. This only means anything once we've defined convolution of a distribution with a test function, but that's easy enough: $(f*h)(x)=\langle f_y,h(x-y) \rangle$. From here it follows that $\langle f,h \rangle = (f*(h \circ m))(0)$, where $m(x)=-x$. From here we get an equation defining $G*f$:

$$\langle G*f,h \rangle = ((G*f)*(h \circ m))(0)=(G*(f*(h \circ m)))(0).$$

Now we rewrite the outer convolution using the definition:

$$\langle G*f,h \rangle = \langle G_y,(f*(h \circ m))(0-y) \rangle = \langle G,(f*(h \circ m)) \circ m \rangle.$$

Now we have written things in terms of just applying $G$ to a function, so it is reasonably clearly defined. You can find more details along these lines in distribution theory notes like http://www.math.chalmers.se/~hasse/distributioner_eng.pdf

  • $\begingroup$ In physics it is (almost) always the case that the source term has bounded support so this should solve most problems I have! Thanks. $\endgroup$ Sep 20, 2016 at 19:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .