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From what I understand the Dirac's Delta derivatives have the meaning $$\int_{-\infty}^{\infty}\delta^{(k)}(x)\phi(x)dx=(-1)^k\int_{-\infty}^{\infty}\delta(x)\phi^{(k)}(x)dx$$ Assuming, of course that the function $\phi$ is differentiable up to order $k$. If that's true, you can then say $$\phi^{(k)}(x_0)=(-1)^k\int_{-\infty}^{\infty}\delta^{(k)}(x-x_0)\phi(x)dx$$ Is this correct? And also, could it be useful in any circumstance?

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    $\begingroup$ Sure. I don't know what you mean when you say "useful", though. $\endgroup$ – Ian May 31 '16 at 17:13
  • $\begingroup$ It follows from the definition. $\endgroup$ – copper.hat May 31 '16 at 17:25
  • $\begingroup$ @Ian, useful means that if you knew this you could do something better or more easily than if you didn´t $\endgroup$ – moonshine May 31 '16 at 18:10
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It is correct provided that one understand the notation $\int_{-\infty}^\infty\delta(x)\phi(x)\ dx$ correctly. This is by no means an integral and $\delta$ is not a function with real variables.

Also, one should specify in what function space is the function $\phi$ so that your identities would make sense.


Any distribution $T$ is infinitely differentiable in the sense of distribution by the definition $$ \langle \partial^\alpha T,\varphi\rangle:=(-1)^{|\alpha|}\langle \partial^\alpha\varphi,T\rangle. $$

However, it is not true that $T$ has arbitrary order weak derivative. For instance, the Heaviside step function is not weakly differentiable but its distributional derivative is the delta function.


I have never seen the word "derivable" used in that way before. (Due to change of OP)


In the theory of electromagnetism, the first derivative of the delta function represents a point magnetic dipole situated at the origin. As for the higher order distributional derivatives, can I say that it is useful for demonstrating an example of distributions of which the $k$-th order derivative can be explicitly calculated?

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  • $\begingroup$ I'm sorry, English is not my first language, I've learnt maths mostly in Spanish :(. What would be the word for "a function that can be derived"? $\endgroup$ – moonshine May 31 '16 at 18:08
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    $\begingroup$ en.wikipedia.org/wiki/Differentiable_function $\endgroup$ – Jack May 31 '16 at 18:12
  • $\begingroup$ I've never heard "weak derivative" as distinct from "distributional derivative" before. $\endgroup$ – Ian May 31 '16 at 18:24
  • $\begingroup$ @Ian terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions $\endgroup$ – Jack May 31 '16 at 18:26
  • $\begingroup$ Fair enough. I still have usually seen the opposite convention, where "weak derivative" and "distributional derivative" are synonymous. We can still talk about the weak derivative "being" a locally integrable function where (from this perspective) we really mean that it, as a distribution, can be represented as integration against a locally integrable function. $\endgroup$ – Ian May 31 '16 at 18:29

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