Timeline for Second order ODE
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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May 6, 2014 at 15:09 | comment | added | user37929 | @MichaelRenardy thank you for this reference. | |
May 6, 2014 at 15:09 | vote | accept | CommunityBot | ||
May 6, 2014 at 9:50 | comment | added | Michael Renardy | You can get rid of this term by an elementary transformation like $v(s)=\exp(\alpha s)w(s)$ for an appropriate $\alpha$. There is a book by Ronveaux on the Heun equation, it discusses reduction to standard form, among other things. | |
May 6, 2014 at 7:35 | comment | added | user37929 | @MichaelRenardy of course there is a missing parenthesis in the middle term of the last term, sorry for that. | |
May 5, 2014 at 21:05 | comment | added | user37929 | @MichaelRenardy so I get: $u(t)=v(s) \Rightarrow u′(t)=v′(s)2t \Rightarrow u′′(t)=v′′(s)4s+2v′(s) $This should give me the ODE: $v''(s) + \frac{1}{2} \left(\frac{1}{s} + \frac{1}{s-1} \right)v'(s) + \left( n \beta \frac{2s-1}{s(1-s)} + \beta^2 \frac{2s-1)^2}{s(1-s)} +C \right)v(s)=0$. Now in the equation you mention at dlmf.nist.gov/31.12 there is no term containing $s^2$ in the nominator in the last term. So does this mean, that I have to take $C$ such that the $s^2$ does not appear? How do I see that by doing so, I don't throw away solutions? | |
May 5, 2014 at 13:42 | comment | added | Michael Renardy | Substitute $s=t^2$ and $u(t)=v(s)$ or, respectively $u(t)=tv(s)$. | |
May 5, 2014 at 13:32 | comment | added | user37929 | @MichaelRenardy could you give me a hint/reference how such a separation is done? Cause I have difficulties to see the relationship betwen the Heun's function and my equation. How do I use the symmetry? | |
May 5, 2014 at 7:18 | comment | added | Robert Israel | Specifically, the solution Maple 18 finds is $$ u \left( t \right) ={ a_1}\,{{\rm e}^{\beta\,\sqrt {n}{t}^{2}}}{ \it HeunC} \left( 2\,\beta\,\sqrt {n},-1/2,-1/2,-\beta/2,3/8-1/4\,n{ \beta}^{2}-C/4+\beta/4,{t}^{2} \right) +{a_2}\,{{\rm e}^{\beta\, \sqrt {n}{t}^{2}}}{\it HeunC} \left( 2\,\beta\,\sqrt {n},1/2,-1/2,- \beta/2,3/8-1/4\,n{\beta}^{2}-C/4+\beta/4,{t}^{2} \right) t $$ | |
May 5, 2014 at 6:56 | vote | accept | CommunityBot | ||
May 6, 2014 at 7:33 | |||||
May 5, 2014 at 3:12 | history | answered | Michael Renardy | CC BY-SA 3.0 |