Timeline for How to pick out harmonics based on boundary conditions?
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| Oct 9, 2013 at 8:40 | comment | added | user6818 | @IgorKhavkine How do you get that "p" right? Thats the question. Picking $\lambda$ to that specific value that you stated doesn't immediately get the asymptotics right. | |
| Oct 9, 2013 at 7:26 | comment | added | Igor Khavkine | That's the thing. As far as I can tell, the 'new' differential equation is the same as the old one (up to identifying $a$ with $\lambda$ using the formula I gave). So I just don't see where the puzzle is. | |
| Oct 8, 2013 at 19:38 | comment | added | user6818 | @IgorKhavkine The asymptotics I wrote down are for $z \rightarrow 0$. Is there anyway that can be used to pick out which of these satisfy the new differential equation and also go as $z^p$ near $z=0$? | |
| Oct 8, 2013 at 13:42 | comment | added | Igor Khavkine | I'm sorry, I'm not sure I can confirm your counting, as it would take me quite some time to go through the details. Also, I may have misunderstood your question. The asymptotic formulas you gave are for $z\to \infty$, not $z\to 0$. Right? Then you can ignore my comment. Still, if you can follow the original argument that leads to the $z\to \infty$ asymptotics, the same method should work for $z\to 0$. I'm afraid I might not be more helpful beyond this suggestion. | |
| Oct 8, 2013 at 0:05 | comment | added | user6818 | @IgorKhavkine I am not getting you :) This value of $\lambda$ that you state only ensures that the function, $h^{\lambda l m \sigma}_{i_1 i_2 \dots i_s}$ is a solution of $(-\nabla ^2 +a)$, but how do I ensure the $z^p$ behaviour near $z=0$? (...and can you also kindly confirm or correct the counting of the solutions as stated in my first point?..) | |
| Oct 7, 2013 at 9:58 | comment | added | Igor Khavkine | If you set $a = -(\lambda^2+\rho^2+s)$, you get $\lambda = \sqrt{-(a+\rho^2+s)}$ (recall that $\rho$ is fixed by $n$). It seems that you already have the formula for the $z$-asymptotic behavior of solutions in terms of $\lambda$ and hence in terms of $a$ as well. Is that not sufficient? | |
| Oct 7, 2013 at 8:28 | history | edited | Willie Wong |
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| Oct 7, 2013 at 1:20 | review | Close votes | |||
| Oct 8, 2013 at 13:56 | |||||
| Oct 7, 2013 at 0:47 | history | edited | user6818 | CC BY-SA 3.0 |
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| Oct 7, 2013 at 0:35 | history | asked | user6818 | CC BY-SA 3.0 |