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I'm looking to find the derivative of a convolution integral of the following form:

\begin{equation} \frac{d}{dr}((G(r,t)*f(t)) = \frac{d}{dr} (\int_{-\infty}^{\infty} G(r,t-\tau)f(\tau) d\tau) \end{equation}.

Here, $G(r,t)$ is the 2D free space Greens function for the wave equation defined as: $ G(r,t) =\frac{1}{2\pi} \frac{H(t-r/c)}{\sqrt{t^2 - r^2/c^2}}$, where $H$ is a Heaviside and c is the wave speed constant. And $f(t)$ is a smooth function, I currently have let $f(t) = e^{-\alpha(t-t0)^2}$, ($\alpha$ is constant) and have computed the convolution integral numerically,

\begin{equation} G(r,t)*f(t) = \int_{0}^{max(0,tn-r/c)} \frac{e^{-\alpha(\tau-t0)^2}}{\sqrt{(t-\tau)^2 - r^2/c^2}}d\tau. \end{equation}

I was wondering if i could compute the derivative of the above convolution integral as $\frac{dG(r,t)}{dr}*f(t)$, then complete the convolution integral numerically as before but with $\frac{dG}{dr}$ or if there are some complications im not taking into account.

Thanks in advance.

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  • $\begingroup$ the derivative of $\partial G/\partial r$ has a delta function contribution $\delta(t-r/c)(t^2-r^2/c^2)^{-1/2}$, which diverges. $\endgroup$ Nov 26, 2019 at 18:41
  • $\begingroup$ So perhaps, using a finite difference method to solve this problem would be better. $\endgroup$ Nov 28, 2019 at 16:15
  • $\begingroup$ if it diverges, then the finite difference method will not converge; there is no meaningful answer, I'm afraid. $\endgroup$ Nov 28, 2019 at 18:00

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