Expansions of the delta functions, of following type which are called completeness relations is very useful in many problems in physics.
$\delta(x-y) = \sum_{n=0}^\infty P_n(x)P_n(y) \frac{2n+1}{2} $
Can similar expansions be written for derivatives of the delta functions ? (I am interested in 1st and 2nd derivatives). Something like:
$D^n\delta(x-y) = \sum_{n=0}^\infty P_n(x)P_n(y) A_n $
I was trying to construct expansions for the derivatives of the delta function on the surface of a sphere in 3 dimensions.
$\delta(\cos\theta-1)$
with respect to the chord distance $2\sin\frac{\theta}{2}$. Ultimate aim being to write it in terms of Spherical Harmonics.
($\theta$ here refer to the angle subtended by a pair of points at the center)
Two approaches suggesting that this cannot be done, but that there will be "non-diagonal" terms:
Assuming that we can just transfer the formulas from $L^2[-1,1]$ functions to distributions, we would have $$ D^m\delta(x-y) = \sum_{k=0}^{\infty} a_k(y) \, P_k(x), $$ where $$ a_k(y) = \frac{2k+1}{2} \int_{-1}^{1} D^m\delta(x-y)\,P_k(x)\,dx = \frac{2k+1}{2} (-1)^m P_k^{(m)}(y). $$ Thus we get $$ D^m\delta(x-y) = \sum_{k=0}^{\infty} \frac{2k+1}{2} (-1)^m P_k^{(m)}(y) \, P_k(x). $$ But then we should expand $P_k^{(m)}(y)$ as a series in $P_\ell(y)$ which makes our final result not be diagonal.
We would expect that $$ D^m\delta(x-y) = \sum_{n=0}^{\infty} \frac{2n+1}{2} P_n^{(m)}(x) \, P_n(y) $$ and that $P_n^{(m)}$ can be expanded in Legendre polynomials, $$ P_n^{(m)}(x) = \sum_{k=0}^{\infty} c_{m,k} \, P_k(x) $$ and thus $$ D^m\delta(x-y) = \sum_{n=0}^{\infty} \frac{2n+1}{2} \left(\sum_{k=0}^{\infty} c_{m,k} \, P_k(x)\right) \, P_n(y) = \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{2n+1}{2} c_{m,k} \, P_k(x) \, P_n(y) $$