Laplace's Equation in Spherical Coordinates

Question

The general solution of the Laplace equation in spherical coordinates is (independant of $\phi$):

$$V(r,\theta ) = \sum ^{\infty} _{l=0} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l (\cos \theta ))$$

An example (Griffiths EM, 3rd edition, example 3.6):

The potential $V_0 (\theta )$ is specified on the surface of a hollow sphere of radius $R$. Find the potential inside the sphere.

Here $B_l$ must be zero because it would blow up at the origin, this is clear, but the next step loses me.

The Legendre polynomials constitute a complete set of functions on the interval $-1 \le x \le 1$ $(0\le \theta \le \pi)$

$$\int ^1 _{-1} P_l (x) P_{l'} (x) dx = \int ^\pi _0 P_l (\cos \theta ) P_{l'} (\cos \theta ) \sin \theta d\theta.$$

Because they are orthogonal functions if $l\ne l'$ the integral is zero, how ever if they are equal, we get an answer of $\frac{2}{2L+1}$, where did this answer come from? Griffiths skips this step.

Why was $\sin$ introduced into the second integral?

I think the Legendre polynomial is tripping me up. How are these dealt with under integrals?

Answer 1

To prove that Legendre polynomials are orthogonal, i.e. $$ \int_{-1}^1 P_n(x)P_m(x)\,\text{d}x = \frac{2\,\delta_{m,n}}{2n+1} $$ you can use Rodrigues' formula (eq 3.62 in Griffiths): $$ P_n(x) = \frac{1}{2^n n!}\left(\frac{\text{d}}{\text{d}x}\right)^n (x^2-1)^n. $$ We get $$ I_{mn} = \int_{-1}^1 P_n(x)P_m(x)\,\text{d}x = \int_{-1}^1 \frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^m \!X^m\,\left(\frac{\text{d}}{\text{d}x}\right)^n \!X^n\,\text{d}x, $$ where I used the short-hand notation $X = x^2-1$. Without loss of generality, we can assume that $n\geqslant m$. Let us integrate this by parts: $$ \begin{multline} I_{mn} = \left[\frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^m \!X^m \,\left(\frac{\text{d}}{\text{d}x}\right)^{n-1} \!X^n\right]_{-1}^1 \\- \int_{-1}^1 \frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^{m+1} \!X^m\,\left(\frac{\text{d}}{\text{d}x}\right)^{n-1} \!X^n\,\text{d}x. \end{multline} $$ However, since the derivative $$ \left(\frac{\text{d}}{\text{d}x}\right)^{n-1} (x^2-1)^n $$ will contain a factor $(x^2-1)$, the integrated part will vanish, so that only the integral remains: $$ I_{mn} = -\int_{-1}^1 \frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^{m+1} \!X^m\,\left(\frac{\text{d}}{\text{d}x}\right)^{n-1} \!X^n\,\text{d}x. $$ In this manner, we can integrate by parts $n$ times, until we obtain $$ I_{mn} = (-1)^n\int_{-1}^1 \frac{(x^2-1)^n}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^{m+n} \!(x^2-1)^m\,\text{d}x. $$ Now, $$ (x^2-1)^m = x^{2m} - mx^{2m-2} + \ldots $$ So that, if $n>m$, $$ \left(\frac{\text{d}}{\text{d}x}\right)^{m+n} \!(x^2-1)^m = 0, $$ and $I_{mn}=0$.

On the other hand, if $n=m$, we get $$ \left(\frac{\text{d}}{\text{d}x}\right)^{n+n} \!(x^2-1)^n = (2n)! $$ and $$ I_{nn} = (-1)^n\int_{-1}^1 \frac{(2n)!}{2^{2n} n!\,n!}(x^2-1)^n\,\text{d}x. $$ Substituting $x$ with $y = (x+1)/2$, we can rewrite this as $$ I_{nn} = 2\int_0^1 \frac{(2n)!}{n!\,n!}y^n(1-y)^n\,\text{d}y, $$ which is a beta function: $$ I_{nn} = 2\frac{(2n)!}{n!\,n!}B(n+1,n+1) = 2\frac{(2n)!}{n!\,n!}\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)} = 2\frac{(2n)!}{(2n+1)!} = \frac{2}{2n+1}, $$ and this concludes the proof.

With $x=\cos\theta$, you can write it as $$ \begin{multline} \int_{-1}^1 P_n(x)P_m(x)\,\text{d}x = \int_{-1}^1 P_n(\cos\theta)P_m(\cos\theta)\,\text{d}\cos\theta \\= \int_{0}^\pi P_n(\cos\theta)P_m(\cos\theta)\sin\theta\,\text{d}\theta. \end{multline} $$ Because of the orthogonality property, you can find the coefficients by integrating $V(r,\theta)$ with each Legendre polynomial: $$ \int_{0}^\pi V(r,\theta)P_l(\cos\theta)\sin\theta\,\text{d}\theta = \frac{2}{2l+1}\left(A_lr^l + \frac{B_l}{r^{l+1}}\right). $$

Answer 2

Why was $\sin$ introduced into the second integral?

Looks like $x = \cos\theta$. Note, then, that $dx= -\sin\theta d\theta$. This also seems to match the new limits of integrations.

Answer 3

For actually working out problems in Griffiths, it is sufficient to take the orthogonality relation as given, as suresh's comment suggests.

If you want to go a tiny bit deeper, you can take them Legendre polynomials as given in either Griffiths or Wikipedia, and plug them into the orthogonality integral, and actually work out the integral yourself to confirm the orthogonality for a few members of the sequence. Obviously, you can't do that for all the (infinite) members of the sequence, so that isn't really a formal proof of the relation. But it might help you see how the polynomials work.

The real reason the Legendre polynomials follow that normalization relation is because the Legendre equation is a form of the [Sturm-Lioville] equation. In Sturm-Liouville theory, you can write the orthogonality relation in terms of the $p(x), q(x)$, and $w(x)$ functions. You can then identify what those functions actually are for the Legendre equation, and directly plug them in to get the orthogonality relation that Griffiths cites. The actual Sturm-Liouville theory is typically left for math methods books like Arfken to explain.

The $\sin\theta$ factor comes from the substitution $x = \cos\theta, dx = \sin\theta\ d\theta$. This is just a matter of switching between the $x$ variable (easier to work with in the mathematical theory) and the $\theta$ variable that corresponds to a physical variable of interest.