Legendre's ODE: Fundamental Solution for L = 2, 3, 4...

[member 428835]

Hi PF!

I'm wondering what the fundamental solution is for this ODE
$$
f''(x)+\cot (x) f'(x) + \left( 2-\frac{L^2}{\sin(x)} \right) f(x) = 0 : L = 2,3,4...
$$

I know one solution is $$
(\cos(x)+L)\left(\frac{1-\cos(s)}{1+\cos(s)}\right)^{L/2}
$$
but I don't know the other. Mathematica isn't much help here, as it only gives me a solution but not for ##L=1## (Legendre Polynomials) but not a general solution for ##L\geq 2##. Any help is very appreciated!

 

[eys_physics]

Are you sure that your equation is correct?
If it instead would be
##f''(x)+\cot(x)f'(x)+(2-\frac{L^2}{\sin(x)^2})f(x)=0##,
it is the associated Legendre equation.

 

[member 428835]

Are you sure that your equation is correct?
If it instead would be
##f''(x)+\cot(x)f'(x)+(2-\frac{L^2}{\sin(x)^2})f(x)=0##,
it is the associated Legendre equation.

Thanks, you're correct, I made a typo. Ughhh such a dumb mistake!

 

[member 428835]

Are you sure that your equation is correct?
If it instead would be
##f''(x)+\cot(x)f'(x)+(2-\frac{L^2}{\sin(x)^2})f(x)=0##,
it is the associated Legendre equation.

So I'm thinking about this, and if ##L \geq 2## then ##P_1^L = 0## identically. Thus, we only have one solution for the ##L \geq 2## case. But the ODE is second order, so we must be missing something. Clearly ##Q_1^L(\cos x)## is one solution, but what would the second be?

 

[eys_physics]

Well, there are non-trivial solutions. But, for ##L\geq 2## they are the so-called Legendre functions, see https://en.wikipedia.org/wiki/Legendre_function .

 

[member 428835]

Well, there are non-trivial solutions. But, for ##L\geq 2## they are the so-called Legendre functions, see https://en.wikipedia.org/wiki/Legendre_function .

I saw this just before you posted it., but thanks a bunch!