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The integral equation for (causal) convolution is given by

$$y(t) = \int_{-\infty}^{t} K(t - \tau) x(\tau) d\tau$$

Can one write an equivalent differential equation for general well-behaved kernel $K$?

Since convolution with an exponential kernel is a solution to a linear ODE with with time-dependent input (e.g. see 2nd example here), the transformation is possible for some kernels. If it turns out that there are well-behaved kernels for which the transformation is not possible, it would be great to address the question on determining the set of kernels for which the transformation is possible.

EDIT: I have found a related post, where respondents argue that this operation is indeed impossible for general kernels. I will comment further when I have fully understood if the outlined arguments apply to my case, where causality is enforced by the limits of integration.

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    $\begingroup$ Presumably you have tried differentiating both sides with respect to $t$? $\endgroup$ May 12, 2020 at 0:15
  • $\begingroup$ @paulgarrett yes, the problem is the derivative of the kernel itself. I am 100% convinced that this is a very well known problem and that it has been addressed before, just that naive googling was unsuccessful $\endgroup$ May 12, 2020 at 0:17
  • $\begingroup$ I don't have much useful to say... except maybe that viewing this as "convolution" is a bit misleading, since the upper bound on the limit is $t$, not $+\infty$. So the actual kernel for the operator is not symmetric, etc. But I do not have useful keywords to give... $\endgroup$ May 12, 2020 at 0:22
  • $\begingroup$ @paulgarrett That is why I have called it causal convolution. Present of variable $y$ may not depend on future of variable $x$ $\endgroup$ May 12, 2020 at 0:28
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    $\begingroup$ Typically, if you go to the Fourier or Laplace transformed equation, you can treat the transformed parameter (say the frequency for Fourier) as the differential operator. If the transformed kernel is a rational function of the frequency, you can get a differential equation. $\endgroup$
    – user619894
    Jun 3, 2020 at 20:41

1 Answer 1

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Let me elaborate on my comment:

rewrite $y(t) = \int_{-\infty}^{t} K(t - \tau) x(\tau) d\tau$ as

$y(t) = \int_{-\infty}^{\infty} \Theta(t-\tau)K(t - \tau) x(\tau) d\tau$ for $\Theta(x)$ the Heavyside function.

Call $\Theta(x) K(x) = R(x)$.

Then a Fourier transform yields:

$\hat y(\omega) = \hat R(\omega) \hat x(\omega) $ . If $\hat R(\omega) = {P(\omega)\over Q(\omega)} $ then

$Q(\omega) \hat y(\omega) = R(\omega) \hat x(\omega)$

In the inverse Fourier we may interpret $i\omega\rightarrow {d\over dt}$ to get $Q( {d\over dt}) y(t) = R( {d\over dt})x(t) $ , a differential equation.

For example :

$y(t) = \int_{-\infty}^{t} e^{-k(t - \tau)} x(\tau) d\tau$ yields

$\hat y(\omega) = {1\over i\omega +k} \hat x(\omega)$

$\rightarrow$

$({d \over dt} +k)y(t)=x(t)$

I am not claiming that only kernels whose transforms are polynominals of $\omega$ can be transformed into differential equations, but it is quite a broad family.

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  • $\begingroup$ Ok, this answer really contributes. I'll wait a few days just in case somebody offers the general solution, then award you the bounty. $\endgroup$ Jun 4, 2020 at 13:04
  • $\begingroup$ Maybe ask yourself the reverse question: Given a differential equation, when is the solution a convolution of the inhomogeneous term? $\endgroup$
    – user619894
    Jun 4, 2020 at 13:06
  • $\begingroup$ One more question: If I take a rational function and perform partial fractions, I will get a polynomial + terms like a/(A - w) and a/(A - iw). The inverse-transform of a polytnomial is a sum of derivatives of dirac deltas, and the transforms of the other terms are exponentials of complex functions. Thus, in time domain, transformable kernels include linear combinations of dirac delta functions and derivatives, exponentials, sines and cosines. Is this correct? $\endgroup$ Jun 4, 2020 at 13:12
  • $\begingroup$ I will ponder a bit over your last question... $\endgroup$ Jun 4, 2020 at 13:12
  • $\begingroup$ I didn't follow the delta function part. Are you worried about a polynomial of $\omega$ multiplying the $x(t)$ ? $\endgroup$
    – user619894
    Jun 4, 2020 at 13:19

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