MHB Integrating delta/Bessel function

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The discussion focuses on solving an equation involving delta and Bessel functions, specifically integrating a delta function to derive coefficients for a series solution. The delta function is used to evaluate integrals, leading to expressions for coefficients C_{mn} and D_{mn} that depend on the values of the Bessel functions at specific points. Participants are seeking clarification on the integration process and the correct formulation of the coefficients. There are attempts to compute these coefficients using numerical integration in a programming environment, but errors were encountered. The conversation emphasizes the importance of correctly applying the properties of the delta function in the context of Bessel function integrals.
Dustinsfl
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\begin{alignat*}{3}
u_t(r,\theta,0) & = & \delta(\mathbf{x} - \mathbf{x}_0) & = & \delta(r - r_0, \theta - \theta_0)
\end{alignat*}
$$
\int_A\delta(\mathbf{x} - \mathbf{x}_0)f(r,\theta)dA = \int_0^{2\pi}\int_0^a\delta(r - r_0, \theta - \theta_0)f(r,\theta)rdrd\theta = f(\mathbf{x}_0)
$$

How do I solve this?
$$
u_t(r,\theta,0) = \frac{c}{a}\sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn} \left(z_{mn}\frac{r}{a}\right)z_{mn} \left[C_{mn}\cos m\theta + D_{mn}\sin m\theta\right] = \delta(r - r_0,\theta - \theta_0).
$$
 
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dwsmith said:
\begin{alignat*}{3}
u_t(r,\theta,0) & = & \delta(\mathbf{x} - \mathbf{x}_0) & = & \delta(r - r_0, \theta - \theta_0)
\end{alignat*}
$$
\int_A\delta(\mathbf{x} - \mathbf{x}_0)f(r,\theta)dA = \int_0^{2\pi}\int_0^a\delta(r - r_0, \theta - \theta_0)f(r,\theta)rdrd\theta = f(\mathbf{x}_0)
$$

How do I solve this?
$$
u_t(r,\theta,0) = \frac{c}{a}\sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn} \left(z_{mn}\frac{r}{a}\right)z_{mn} \left[C_{mn}\cos m\theta + D_{mn}\sin m\theta\right] = \delta(r - r_0,\theta - \theta_0).
$$

$$
C_{mn} = \frac{2 \int_0^a \int_0^{2\pi} r \delta(r - r_0,\theta - \theta_0) \mathcal{J}_{mn} \left(z_{mn}\frac{r}{a}\right) \cos m\theta d\theta dr}{z_{mn}ac\pi \mathcal{J}_{m+1}^2(z_{mn})} = \frac{2 \mathcal{J}_{mn} \left(z_{mn}\frac{r_0}{a}\right) \cos m\theta_0}{z_{mn}ac\pi \mathcal{J}_{m+1}^2(z_{mn})}\quad m\neq 0
$$
$$
C_{0n} = \frac{\int_0^a \int_0^{2\pi} r \delta(r - r_0,\theta - \theta_0) \mathcal{J}_{0n} \left(z_{0n}\frac{r}{a}\right)d\theta dr}{z_{0n}ac\pi \mathcal{J}_{1}^2(z_{0n})} = \frac{\mathcal{J}_{0n} \left(z_{0n}\frac{r_0}{a}\right) }{z_{0n}ac\pi \mathcal{J}_{1}^2(z_{0n})}
$$
$$
D_{mn} = \frac{2 \int_0^a \int_0^{2\pi} r \delta(r - r_0,\theta - \theta_0) \mathcal{J}_{mn} \left(z_{mn}\frac{r}{a}\right) \sin m\theta d\theta dr}{z_{mn}ac\pi \mathcal{J}_{m+1}^2(z_{mn})} = \frac{2 \mathcal{J}_{mn} \left(z_{mn}\frac{r_0}{a}\right) \sin m\theta_0}{z_{mn}ac\pi \mathcal{J}_{m+1}^2(z_{mn})}
$$
If this isn't correct, how do I integrate this with the delta?
I tried the below but it errored.
Code:
ClearAll["Global`*"]
z = Table[N[BesselJZero[m, n]], {n, 1, 20}, {m, 0, 5}];
a=c=1;
z // TableForm

A = Table[
   2*NIntegrate[
      DiracDelta[
        r - Subscript[r, 0], \[Theta] - Subscript[\[Theta], 0]]*r*
       Cos[m*\[Theta]]*BesselJ[m, z[[m, n]]*r/a], {\[Theta], 0, 
       2*Pi}, {r, 0, a}]/(Pi*c*a*BesselJ[m + 1, z[[m, n]]*r/a]^2), {m,
     1, 5}, {n, 1, 20}];
 
Last edited:
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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