Solving the radial part of Poisson's equation via power series method

Question

We're solving Poisson's equation in Electrodynamics, and one ODE we must solve is: $$\dfrac{d^2 R}{dr^2}+\dfrac{2}{r}\dfrac{dR}{dr}-\dfrac{l(l+1)}{r^2}R=0$$ where $l\in\mathbb{R}$. I already know that the solution is in the form $R=Ar^l+\dfrac{B}{r^{l+1}}$ but I want to solve it via power series (instead of the miracle Ansatz). I assumed that the solution takes the form: $$R=\sum_{i=0}^{\infty}a_ir^i$$ such that when I plug this in the ODE I get: $$\sum_{i=0}^{\infty}\left[i(i-1)+a_i+2ia_i-l(l+1) \right]r^{i-2}=0$$ and so $$a_i=\dfrac{l(l+1)}{i(i+1)}$$ My solution is then $$R=l(l+1)\sum_{i=0}^{\infty}\dfrac{r^i}{i(i+1)}=l(l+1)\left[\dfrac{1}{2}r+\dfrac{1}{6}r^2+\dfrac{1}{12}r^3+\dfrac{1}{20}r^4+\dots \right]$$ This solution doesn't seem to match with the actual solution. Where is my mistake?

Answer 1

The problem is your third equation. The result of plugging your guess into the original ode gives you

\begin{eqnarray} \sum_i a_i\left[i(i-1) +2i -l(l+1) \right]r^{i-2} = 0 \end{eqnarray}

Answer 2

Hint:)

For differential equation $$\dfrac{d^2 R}{dr^2}+\dfrac{2}{r}\dfrac{dR}{dr}-\dfrac{\ell(\ell+1)}{r^2}R=0$$ we see $p(x)=\dfrac{2}{r}$ and $q(x)=-\dfrac{\ell(\ell+1)}{r^2}$ (comparison with $y''+p(r)y'+q(r)y=0$) which show $r=0$ is regular point of this equation, then $$p_0=\lim_{r\to0}rp(r)=2~~~,~~~q_0=\lim_{r\to0}r^2q(r)=-\ell(\ell+1)$$ so the Ferobenius characteristic equation $\lambda(\lambda+1)+p_0\lambda+q_0=0$ is $$\lambda(\lambda+1)+2\lambda-\ell(\ell+1)=0$$ hence $\lambda=\ell$ and $\lambda=-\ell-1$, and two answers must be of the form $$R=x^\ell\sum_{r=0}^\infty a_nr^2~~~~~,~~~~~R=x^{-\ell-1}\sum_{r=0}^\infty a_nr^2$$ so you must consider these series as solutions.

Answer 3

Let's put the equation in the form $$ r^2R''(r)+r R'(r)-l(l+1)R(r)=0\tag 1 $$ We look for a series solution of the form (Frobenius Method) $$ R(r)=\sum_{n=0}^{\infty}a_nr^{n+s}\tag 2 $$ Substituting $(2)$ in $(1)$ we have $$ \sum_{n=0}^{\infty}(n+s)(n+s-1)a_nr^{n+s}+2\sum_{n=0}^{\infty}(n+s)a_nr^{n+s}-l(l+1)\sum_{n=0}^{\infty}a_nr^{n+s}=0\tag 3 $$ We then re-order the sums to contain the same power of $r$. We find the indicial equation by setting $n=0$ in all the sums of $(3)$ that bear the lowest exponent of $r$. Since all the sums in $(3)$ are to the same power of $r$ (i.e, $r^{n+s}$), all the sums contribute to the indicial equation and we obtain for $a_0\ne 0$: $$ s(s-1)+2s-l(l+1)=0\tag 4 $$ and solving $(4)$ we find $$ s=l\quad\text{and}\quad s=-(l+1)\tag 5 $$ We next find the recursion relation. Note that since all the sums already involve the same exponent of $r$, there is no need to do any re-indexing. That means that all the coefficients will remain expressed as $a_n$, and so we can write immediately: $$ \Big[(n+s)(n+s-1)+2(n+s)-l(l+1)\Big]a_n=0\tag 6 $$ Equation $(6)$ tells us that either the bracketed expression vanishes or the value of $a_n$ vanishes. We know that the bracketed expression vanishes for $n=0$. However, for any other value of $n$, the bracketed expression cannot vanish, meaning that $a_n = 0$ for all values of $n > 0$. This means the only term in the series expansion is the $a_0$ term, so our solution for equation $(1)$ becomes simply:

$$ R(r)=Ar^l+Br^{-(l+1)} $$