2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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:

$$(G*f)*h=G*(f*h)$$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ In physics it is (almost) always the case that the source term has bounded support so this should solve most problems I have! Thanks. $\endgroup$ – JezuzStardust Sep 20 '16 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.