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The Dirac-delta function has the property, $$ x \delta'(x) = -\delta(x) $$

If I consider an ODE of the following form, $$ \frac{df}{dx} = \frac{-1}{x} f(x) $$ here we can show that $f(x)$ satisfies the above property and therefore $f(x) = \delta(x)$. However, if I solve the above ODE using separability, $$ \int \frac{df}{f} = -\int \frac{dx}{x} $$ I get a result of the kind $f(x) \sim 1/x$. How do I reconcile between these two seemingly contradictory solutions.

user35952
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1 Answers1

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Some obstacles to your wish:

  1. The solution of an ODE is always a continuously differentiable function.

  2. $ xf'(x)+f(x)=0 $ is not defined as ODE at $x=0$ (but to both sides of it). So how any generalized solution behaves at $x=0$ is not covered by ODE theory.

  3. There are different ways how one can try to extend a solution to $x=0$, but most will involve that the extension be at least continuous.

One can try to modify the equation slightly so that the new equation is regular, such as $$ (x^2+\varepsilon^2)f'(x)+xf(x)=0\implies f(x)=\frac{c}{\sqrt{x^2+ε^2}}. $$ But this solution does not have a finite area under the curve, so it can not be used to construct an approximation of the Dirac delta.

Lutz Lehmann
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  • So, in what sense is $x\delta'(x) = -\delta(x)$ true? – user35952 Dec 12 '22 at 11:50
  • As a distribution. Take any compactly supported test function $\phi$, then $$(xδ0')[\phi]=δ_0'[x\phi]=-δ_0[(x\phi)']=-[x\phi'+\phi]{x=0}=-\phi(0)=-\delta_0[\phi].$$ As for the ODE solution, $f(x)=0$ is a valid solution for $x>0$ and $x<0$, so you can still get $\delta$ as a generalized solution in the weak sense. But it is not "organic" as ODE solution. – Lutz Lehmann Dec 12 '22 at 12:06
  • Thank you, as a physicist it was not clear to me that this rigorously doesn't converge to such a solution until you pointed it out. Further, can we make complete rigorous theory in which I can consider generalized functions and distributions as well in ODE? A side note, I think if you add the above the arguments in the answer, it might also be useful to someone who comes across the answer at a later time. – user35952 Dec 12 '22 at 12:10
  • There is the theory by Fillippov, replace everything non-smooth by smooth approximations in the distributional sense and filter out those cases where the result, in some limit, is the same independent of the approximation. /// One can probably use other perturbations to get approximations of delta, I would think that $2ε^2\cosh(x/ε)f'(x)+ε\sinh(x/ε)f(x)$ extends the previous attempt leading to $f(x)=\frac{c}{\sqrt{\cosh(x/ε)}}$ which should have a finite area due to the exponential decay towards infinity. – Lutz Lehmann Dec 12 '22 at 12:23
  • A question from an amateur : Is this because $\delta(x)$ isn't mathematically a function technically? – Hersh Dec 12 '22 at 12:41
  • Yes, technically $\delta(x)$ is a distribution not a function. – user35952 Dec 12 '22 at 12:48
  • @LutzLehmann Thank you, that seems interesting. Can you share some resources about the theory by Fillippov? – user35952 Dec 12 '22 at 12:49
  • Not more than you would find with an internet search. I got to this name in the context of multi-phase systems, sliding mode etc. mostly as a general theory to support the more special solutions there. – Lutz Lehmann Dec 12 '22 at 13:00
  • The above idea does not work, $2ε^2\cosh(x/ε)$ does not converge to $x^2$, in the worst possible way. One could try again with $x^2\cosh(ε/x)$, etc., but that does really not look nice. – Lutz Lehmann Dec 12 '22 at 13:07
  • THIS ANSWER might be of interest. – Mark Viola Dec 12 '22 at 16:00