I am trying to solve an axisymmetric special plate problem, but stuck at the expansion of the Dirac Delta function $\delta(r)$ in the interval $[0,c]$, consisting of $1$ and orthonormal Bessel functions $J_0(\alpha_kr)$ with weight $r$. $\alpha_k$ is the $k_{th}$ positive root of $J_1(\alpha c)=0$ .
It has been proved in some books that $\{1,J_0(\alpha_kr)\}$ is the eigen-functions corresponding to the eigen-values $\{0,\alpha_k^2\}$ of a singular Sturm-Liouville problem, namely $$ x^2 \frac{d^2X}{dx^2}+x\frac{dX}{dx}+\alpha^2xX=0 $$ and $$ X'(c)=0 $$
(It should be noted that $\alpha_k$ is the $k_{th}$ positive root of $J_1(\alpha c)=0$, instead of $J_0(\alpha c)=0$, since the b.c. forces the derivative to vanish at boundaries. Besides, in my case, because of the other coupled equation's variable-separation solution, the basis funcion set is limited to $\{1,J_0(\alpha_kr)\}$ described above.)
So, I think the funciton set should be complete in $[0,c]$ for any function meeting the above b.c..
But, Mathematica told me that $$ \int_0^c\delta(r)rdr=0 $$ and $$ \int_0^c\delta(r)J_0(\alpha_kr)rdr=0 $$
It seems that the Dirac Delta funciton cannot be expaned on this function set.
Have I done something wrong?
Initial Approach
The series representation for $\delta(x-y)$ defined at the end of the Calculating the coefficients section of the Wikipedia article on Fourier-Bessel series seems to suggest the representation
$$\delta(r)=\frac{2 (c+r)}{(2 c)^2}\ \underset{K\to\infty}{\text{lim}}\ \sum\limits_{k=1}^K \frac{J_{\alpha}\left(\frac{j_{\alpha,k}}{2 c} c\right) J_{\alpha}\left(\frac{j_{\alpha,k}}{2 c} (c+r)\right)}{J_{\alpha+1}\left(j_{\alpha,k}\right){}^2}\tag{1}$$
converges (in a distributional sense) on the interval $(-c,c)$ where $j_{\alpha,k}$ is the $k^{th}$ positive root of $J_{\alpha}(r)=0$.
Figure (1) below illustrates formula (1) for $\delta(r)$ above evaluated at $\alpha=1$, $c=2$, and $K=10$ seems to be converging (in a distributional sense) on the interval $(-c,c)=(-2,2)$.
Figure (1): Illustration of formula (1) for $\delta(r)$ evaluated at $\alpha=1$, $c=2$, and $K=10$
Alternate Approach
The alternate approach is an attempt to derive a formula more consistent with the goals of the question.
Consider the formula
$$\delta'(r-c)=\underset{K\to\infty}{\text{lim}}\ \sum\limits_{n=1}^K a(c,n)\, J_1\left(\frac{j_{1,n}}{2 c} r\right)\tag{2}$$
which I believe converges (in a distributional sense) on the interval $(0, 2 c)$ where $j_{1,k}$ is the $k^{th}$ positive root of $J_1(r)=0$ and
$$a(c,n)=\frac{\int\limits_0^{2 c} r\, \delta'(r-c)\, J_1\left(\frac{j_{1,n}}{2 c} r\right)\, dr}{\frac{1}{2} \left(2 c\, J_{1+1}\left(j_{1,n}\right)\right){}^2}=-\frac{j_{1,n}\, J_0\left(\frac{j_{1,n}}{2}\right)}{4 c^2\, J_2\left(j_{1,n}\right){}^2}\tag{3}.$$
Term-wise integration of formula (2) above using Mathematica results in the following formula
$$\delta(r-c)=\underset{K\to\infty}{\text{lim}}\ \sum\limits_{n=1}^K b(c,n) \left(J_0\left(\frac{j_{1,n}}{2 c} r\right)-1\right)\tag{4}$$
where
$$b(c,n)=\frac{J_0\left(\frac{j_{1,n}}{2}\right)}{2 c\, J_2\left(j_{1,n}\right){}^2}\tag{5}$$
but formula (4) above seems to evaluate with an offset whereas the conjectured formula
$$\delta(r-c)=\frac{1}{2 c}+\underset{K\to\infty}{\text{lim}}\ \sum\limits_{n=1}^K b(c,n)\, J_0\left(\frac{j_{1,n}}{2 c} r\right)\tag{6}$$
seems to converge (in a distributional sense) on the interval $(0,2 c)$ while eliminating this offset.
The conjectured formula (6) above leads to the following conjectured formula
$$\delta(r)=\frac{1}{2 c}+\underset{K\to\infty}{\text{lim}}\ \sum\limits_{n=1}^K b(c,n)\, J_0\left(\frac{j_{1,n}}{2 c} (r+c)\right)\tag{7}$$
which seems to converge (in a distributional sense) on the interval $(-c, c)$ as illustrated in Figure (2) below where $c=\frac{3}{2}$.
Figure (2): Illustration of conjectured formula (7) for $\delta(r)$ evaluated at $K=10$ (blue) and $K=20$ (orange) where $c=\frac{3}{2}$
I'm not sure if the scaling property $\delta(r)=2 c\, \delta(2 c r)$ can be applied to the conjectured formula (7) above for $\delta(r)$ since this has the undesirable effect of decreasing the range of convergence.
Another Conjectured Formula
Consider the following conjectured formula which is based on a conjectured formula user @Charles6 posted in a comment below
$$\delta(x-y)=\frac{2y}{c^2}+\underset{K\to\infty}{\text{lim}}\ \sum\limits_{k=1}^K d(y,c,n)\, J_0\left(\alpha(c,k)\, x\right)\tag{8}$$
where
$$d(y,c,n)=\frac{2 y\, J_0(\alpha(c,k)\, y)}{c^2\, J_0(\alpha(c,k)\, c){}^2}\tag{9}$$
and $\alpha(c,k)$ is the $k^{th}$ positive root of $J_1(c x)$. In other words $\alpha(c,k)=\frac{j_{1,k}}{c}$ where $j_{1,k}$ is the $k^{th}$ positive root of $J_1(x)$.
I believe formula (8) above evaluated at $c=2 y$ is exactly equivalent to formula (6) above evaluated at $c=y$ which is related to the equivalence of $J_0(j_{1,k})^2=J_2(j_{1,k})^2$ where $j_{1,k}$ is the $k^{th}$ positive root of $J_1(x)$.
Formula (8) above (assuming its correct) seems to suggest the following limit representation
$$\delta(x)=\underset{y\to 0}{\text{lim}}\ \left(\frac{2y}{c^2}+\underset{K\to\infty}{\text{lim}}\ \sum\limits_{k=1}^K d(y,c,n)\, J_0\left(\alpha(c,k)\, x\right)\right)\tag{10}$$
which perhaps converges (in a distributional sense) on the interval $(-c, c)$ as $y\to0$ and $K\to\infty$.
Figure (3) below illustrates formula (10) above evaluated at $c=1$, $y=10^{-4}$, and $K=20$.
Figure (3): Illustration of conjectured formula (10) for $\delta(x)$
But something seems wrong with formula (10) since $\int\limits_{-1}^1 \delta(x)\,dx=1$ and there doesn't seem to be enough area under the curve in Figure (3) for it to integrate to $1$ (I believe the area is closer to $\frac{1}{10}$).
Note formula (10) above can be evaluated as
$$\delta(x)=\underset{y\to 0}{\text{lim}}\ \frac{2y}{c^2} \left(1+\underset{K\to\infty}{\text{lim}}\ \sum\limits_{k=1}^K \hat{d}(y,c,n)\, J_0\left(\alpha(c,k)\, x\right)\right)\tag{11}$$
where
$$\hat{d}(y,c,n)=\frac{J_0(\alpha(c,k)\, y)}{J_0(\alpha(c,k)\, c){}^2}\tag{12}$$
so if formula (10) is on the right track, then perhaps the scaling factor $\frac{2y}{c^2}$ in formula (11) above is incorrect and needs to be adjusted.