Solving Laplace's equation in spherical coordinates

[Happiness]

The angular equation:

##\frac{d}{d\theta}(\sin\theta\,\frac{d\Theta}{d\theta})=-l(l+1)\sin\theta\,\Theta##

Right now, ##l## can be any number.

The solutions are Legendre polynomials in the variable ##\cos\theta##:

##\Theta(\theta)=P_l(\cos\theta)##, where ##l## is a non-negative integer.

Why must ##l## be a non-negative integer?

In the case of the solution to Laplace's equation in cartesian coordinates, ##k=\frac{n\pi}{a}## because we need to satisfy the boundary conditions. How about in the case of spherical coordinates?

Does it mean that if ##l## is not a non-negative integer, but is, say, ##\sqrt3##, then the differential equation can't be solved?

That is, the equation ##\frac{d}{d\theta}(\sin\theta\,\frac{d\Theta}{d\theta})=(-3-\sqrt3)\sin\theta\,\Theta## has no solution?

 

[pasmith]

The constraint that [itex]l[/itex] be a non-negative integer comes from satisfying a boundary condition: we don't want [itex]\Theta[/itex] to blow up when [itex]\theta = 0[/itex] or [itex]\theta = \pi[/itex]. The solutions obtained where [itex]l[/itex] is not a non-negative integer do have that unfortunate property.

 

[Orodruin]

To add to this, you should also be wondering where the second solution went. The differential equation is of order two and should therefore have two independent solutions. The second solution also exists, but blows up at the poles.

 

[Happiness]

This website http://mathworld.wolfram.com/LegendreDifferentialEquation.html explains very well how to solve the angular equation.

To test my understanding, I let ##l=0.4## in http://mathworld.wolfram.com/LegendreDifferentialEquation.html and substitute it into equations (17) and (4), I get

##\Theta=y=a_0(1-\frac{7}{25}x^2-\frac{238}{1875}x^4-...)+a_1(x-\frac{6}{25}x^3-\frac{429}{3125}x^5-...)##, where ##x=\cos\theta##

The series in the bracket following ##a_0## is called the even solution ##y_1##, equation (23). And the series in the bracket following ##a_1## is called the odd solution ##y_2##, equation (24).

The claim is that ##\Theta=a_0\times y_1(x)+a_1\times y_2(x)##, where ##x=\cos\theta##, is always divergent unless ##l## is a non-negative integer. (Seems like a very strong statement.) Is this very difficult to prove mathematically?

When I chose ##l=0.4##, it looks possible that ##\Theta## could be convergent as the coefficients in the series ##y_1## and ##y_2## get smaller and ##-1\leq x\leq1##.

I did the ratio test (on ##y_1## and ##y_2##) using equation (17) but the test is inconclusive since ##\displaystyle\lim_{n\rightarrow +\infty}{|\frac{a_{n+2}}{a_n}|}=1##. (Although I wrote ##a_{n+2}##, it is actually the next term [separately] in the series ##y_1## and ##y_2##, as opposed to being the term after next.)

Edit: My approach in the last paragraph is wrong. Actually it's not even correct to do the ratio test (and any other convergence tests) separately on ##y_1## and ##y_2## because by Riemann series theorem, it's possible for ##y_1## and ##y_2## to be themselves divergent but yet for ##\Theta## to be convergent. Just take ##a_0## and ##a_1## to have opposite signs. (Then ##\Theta## could converge conditionally.)

Then now I'm stuck because it seems that I can only do any convergence test on ##y_1## and ##y_2## separately since I only have the recurrence relation relating ##a_{n+2}## and ##a_n##, equation (17).

 

[Happiness]

The constraint that [itex]l[/itex] be a non-negative integer comes from satisfying a boundary condition: we don't want [itex]\Theta[/itex] to blow up when [itex]\theta = 0[/itex] or [itex]\theta = \pi[/itex]. The solutions obtained where [itex]l[/itex] is not a non-negative integer do have that unfortunate property.

I'm starting to doubt this explanation. I've only read that if the ##l## in the angular equation is a non-negative integer, then the solutions are the Legendre polynomials ##P_l##, which always converge. But I've not seen the claim that the inverse is true, that is, if ##l## is not a non-negative integer, then the solutions diverge, or "blow up".

We have forgotten that the set of Legendre polynomials is complete (for ##-1\leq x \leq 1##). That means that for cases where ##l## is not a non-negative integer, the solutions can always be written as a series involving Legendre polynomials. Then, setting ##l## as a non-negative integer actually does not result in any loss of generality. In other words, it is not that ##l## must be a non-negative integer, but that setting it to be non-negative is just a tool or a technique to get the general solution, which can satisfy all cases.

Am I right?

 

[Orodruin]

Am I right?

No. ##\ell## must be a non-negative integer. Try this for example: http://www.wolframalpha.com/input/?i=LegendreP[0.4,-1]

In the end, it is all about solving a Sturm-Liouville problem and finding the appropriate orthogonal basis to do so. The basis is unique up to linear combinations of functions with the same eigenvalues (in this case ##\ell(\ell+1)##).