4
$\begingroup$

I'm trying to solve the 6.2 problem of Jackson's Classical Electrodynamics textbook. At some point, to get the desired solution, I have to exchange a derivative applied to a Dirac delta-function with the integral operator:

$$\int_{\mathbb{R^3}} \frac{\partial \delta(\mathbf{x}-\mathbf{x_0}(t))}{\partial t}f(\mathbf{x})\,d^3x=\frac{\partial}{\partial t}\int_{\mathbb{R^3}} \delta(\mathbf{x}-\mathbf{x_0}(t))f(\mathbf{x})\,d^3x=\frac{\partial}{\partial t}f(\mathbf{x_0}(t))$$

Under which hypothesis can I do something like that (i.e. exchange the order of differentiation and integration)? I expect that known theorems of real analysis do not apply in this case, since the $\delta$ is not even a proper function.

$\endgroup$
  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Feb 19 '17 at 11:23
  • $\begingroup$ @Qmechanic, I think it's a transversal question, since this result is needed to obtain the Feynman-Heaviside electromagnetical fields. $\endgroup$ – Alessandro Zunino Feb 19 '17 at 11:27
4
$\begingroup$

By definition, if $T$ is a distribution, then $$\langle \partial_x T, f \rangle := - \langle T, \partial_x f \rangle\tag{1}$$ for every test function $f=f(x)$ in $C_0^\infty(\mathbb R^n)$ (or also $C^\infty(\mathbb R^n)$ if $T$ has compact support as the delta function). Here the derivative is just a bit more complicated. However, since the chain rule for taking the derivative of distributions composed with smooth functions is valid also for distributions (it is a general theorem) we have $$\partial_t \delta(x-x_0(t)) = \frac{dx_0}{dt}|_{x_0}\cdot \nabla_{x_0} \delta(x-x_0(t)) = -\frac{dx_0}{dt}|_{x}\cdot \nabla_x \delta(x-x_0(t))\:. \tag{2}$$ As a consequence, for every function $f \in C^\infty(\mathbb R^n)$, applying (2), $$\int \partial_t \delta(x-x_0(t)) f(x) d^nx = -\int \frac{dx_0}{dt}|_{x_0}\cdot \nabla_x \delta(x-x_0(t)) f(x) d^nx$$ $$= -\frac{dx_0}{dt}|_{x_0}\cdot\int \nabla_x\delta(x-x_0(t)) f(x) d^nx \:.$$ Applying (1) $$\int \partial_t \delta(x-x_0(t)) f(x) d^nx = + \frac{dx_0}{dt}|_{x_0}\cdot \int \delta(x-x_0(t)) \nabla_x f(x) d^nx$$ $$= \frac{dx_0}{dt}|_{x_0}\cdot \nabla_x f(x)|_{x_0(t)} = \frac{d}{dt}f(x_0(t))\:.$$

$\endgroup$
  • $\begingroup$ Thank you very much, this looks like a great answer, but it's not completely clear to me. Can you please explain better (2)? It looks like you used a sort of chain rule for distributions, but how does it works exactly? And why you could exchange x_0 with x in the gradient operator? $\endgroup$ – Alessandro Zunino Feb 19 '17 at 11:03
  • $\begingroup$ Yes, that is nothing but the standard chain rule which is assumed to be valid also for distributions if the internal function ($x_0$) is smooth. Regardind the second issue, I used the obvious fact that taking the derivative with respect to $x$ or to $-x_0$ is the same as they appear summed in the argument. This is again an application of chain rule. $\endgroup$ – Valter Moretti Feb 19 '17 at 11:51
  • $\begingroup$ Thank you again. Just a last question: can the hypothesis on the smoothness of f(x) be relaxed? The particular function I'm working with has simple poles, so it's not a member of C^\infty(R^n). $\endgroup$ – Alessandro Zunino Feb 19 '17 at 15:41
  • $\begingroup$ If you want to stick to distribution theory you cannot relax that hypothesis (also using Hoermander's machinery to define multiplication of distribution, since the so called wavefront set of the delta is too large at the singular point). However the delta is also a measure, so in principle you may relax the smoothness requirement and construct an ad-hoc theory... $\endgroup$ – Valter Moretti Feb 19 '17 at 16:23
  • $\begingroup$ If the poles of your function are far from the singularity of $\delta$ you can simply ignore them... $\endgroup$ – Valter Moretti Feb 19 '17 at 16:24

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.