- #1
sir_manning
- 66
- 0
Good afternoon
I have a question regarding the limits on the orthogonality integral of Legendre Polynomicals:
[tex]\int_{-1}^1 P_l(u)P_{l'}du = 2/(2l+1)[/tex]
I am in the middle of a question involving the solution of Laplace's equation inside a hemisphere, which means that for the usual [tex]u=cos\theta[/tex], the limits will be from 0 to 1 instead. So after solving for my boundary conditions and multiplying by [tex]P_{l'}[/tex], I have:
[tex]P_{l'}(u)C = A_{l}\sum{P_{l}(u)P_{l'}(u)}[/tex]
The next step is to integrate each side. I want to use the orthogonality integral above, but since I have to integrate the LHS from 0 to 1, won't I have to do the same for the RHS? Or can I integrate from -1 to 1 on each side, and then change the limit on the LHS because it isn't defined for 0 to 1. I've already done the latter and the result seems to work out, but I'm simply wondering if it is justified or not.
I have a question regarding the limits on the orthogonality integral of Legendre Polynomicals:
[tex]\int_{-1}^1 P_l(u)P_{l'}du = 2/(2l+1)[/tex]
I am in the middle of a question involving the solution of Laplace's equation inside a hemisphere, which means that for the usual [tex]u=cos\theta[/tex], the limits will be from 0 to 1 instead. So after solving for my boundary conditions and multiplying by [tex]P_{l'}[/tex], I have:
[tex]P_{l'}(u)C = A_{l}\sum{P_{l}(u)P_{l'}(u)}[/tex]
The next step is to integrate each side. I want to use the orthogonality integral above, but since I have to integrate the LHS from 0 to 1, won't I have to do the same for the RHS? Or can I integrate from -1 to 1 on each side, and then change the limit on the LHS because it isn't defined for 0 to 1. I've already done the latter and the result seems to work out, but I'm simply wondering if it is justified or not.