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I was recently Studying Griffiths Electrodynamics after a long time and there I saw the Laplace equation. Because it was my second time going over Griffiths so I thought maybe I should try to derive the solutions myself just for fun purposes and because I really wanted to apply the trick of Power series to solve differential equations. I started off by writing Laplace equation in Spherical Polar Coordinates (Assuming no dependence on $\phi$) $$\frac{1}{r}\frac{\partial^2}{\partial r^2}\left(rV\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial{\theta}}\left(\sin\theta\frac{\partial V}{\partial\theta}\right)=0$$ This Laplacian was not taken from Griffiths but from Jackson, they Both just differ in the radial part. Now using Method of variable separation, $$V(r,\theta)=\frac{R(r)}{r}\Theta(\theta)$$ $$\frac{r^2}{R(r)}\frac{d^2R}{dr^2}+\frac{1}{\Theta(\theta)\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)=0$$ Now the Angular part is very Obviously just the Legendre Polynomials. Then we are left with the Radial equation which is my main Concern. The book just describes a lazy method and just tells the solution. I wanted to solve the radial part equation using power series method. So my Focus is on the Cauchy's Equation $$\frac{r^2}{R(r)}\frac{d^2R}{dr^2}=l(l+1)$$ So, I assumed a power series $$R(r)=\sum_{k=0}^{\infty}a_kx^k$$ and By substitution we get, $$r^2\frac{d^2}{dr^2}\left(\sum_{k=0}^{\infty}a_kx^k\right)=l(l+1)\sum_{k=0}^{\infty}a_kx^k$$ Which gives $$\sum_{k=0}^{\infty}k(k-1)a_kr^k=\sum_{k=0}^{\infty}l(l+1)a_kx^k$$ Which means $$k(k-1)=l(l+1)=\lambda$$ $$k^2-k-\lambda=0$$ Using Quadratic Formula $$k=\frac{1\pm\sqrt{1+4\lambda}}{2}$$ and by Substituting $\lambda=l(l+1)$ gives $$k=-l,l+1$$ which do indeed Form the radial part solution. But there is a problem with this approach. We know that $l$ is a positive integer and this means that $k=-l$ is not possible since $k>0$. So, we are getting the correct solution but by the wrong method? I don't know what to comment on it. I wondered about it for a while and came to no conclusion. So, can anyone please Identify where I am wrong? It will be really helpful, since the same problem is coming in Cylindrical Coordinates as well.

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You need both solutions in general. Your potential is not uniquely defined by being harmonic, you need to append boundary conditions. If your domain is a ball (includes the origin), this excludes the $V = r^{-l-1}Y_l^m$ solution. Conversely, if your domain is the complement of the ball (includes infinity), then this excludes the $V = r^lY_l^m$ solution. In general if your domain is between two concentric spherical shells you'll need both solutions in order to match the inside and outside shell boundary condition.

For your method, you technically need to exclude the negative solution since you are looking for positive power series. There is no contradiction, since it just shows that your ansatz is not general enough. In your case, it turns out that simple power laws suffices, no need for power series, only you need to let the power to be arbitrary (in some cases, you'll even need complex powers btw). This is how you recover both power laws. In general, a Taylor series won't suffice for singular ODE's, for regular singular points, you'll rather need the Frobenius method and irregular singular points, you'll need to improvise.

The same applies in 2D (and in higher dimensions). The analogue of $l$ is $m\in\mathbb Z$, so that $V \propto e^{im\phi}$. You now have two solutions for the radial dependence: $r^{\pm m}$, and similarly you need to use both depending on your domain. In 2D, it is even more transparent as you can relate it to Laurent series using $z=r^{i\phi}$. If your domain is a disk, then the inner radius of convergence is zero, so your Laurent series is a Taylor series, so you use only positive powers $r^{|m|}$, and the converse applies for the complement of a disk.

The discussion actually generalises in arbitrary dimensions.

Hope this helps.

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  • $\begingroup$ Yeah, I understand your point but my main concern is with the fact that k represent set of positive integers. the solution $k=-l$ should not be possible because k is the index of power series. I am asking, even though it's wrong, why are getting the correct solution? $\endgroup$ Commented May 18 at 11:41
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    $\begingroup$ Why do the powers in the series have to be positive? In fact they don't. You can, and in general do, have a Laurent series that has all positive and negative powers. $\endgroup$
    – mike stone
    Commented May 18 at 12:23

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