# [SOLVED]Integrating delta/Bessel function

#### dwsmith

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

#### dwsmith

##### Well-known member
\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: