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Amin
Posts: 1967
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Re: Analytic ODE solution possible for this one?

January 7th, 2018, 5:19 pm

Yes, using my method you would have to take a time step that would correspond to number of terms of the series expansion included in the solution. You could, however, use a far larger step than most of the numerical methods. If you could recognize the series expansion of the solution, you can find the corresponding closed form solution.
What makes you think that the new method is not general?
 
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tw
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Re: Analytic ODE solution possible for this one?

January 7th, 2018, 6:34 pm

Amin wrote:
Yes, using my method you would have to take a time step that would correspond to number of terms of the series expansion included in the solution. You could, however, use a far larger step than most of the numerical methods. If you could recognize the series expansion of the solution, you can find the corresponding closed form solution.
What makes you think that the new method is not general?

Well I quickly read through it and their proposed ansatz solution seemed to work just fine. However the algebra look fairly hardcore, I was summoning up the energy to verify it, when I scanned to the bottom of the paper and they plotted the analytic solution versus a numerical integration and another
perturbation approach. 
Just on one of the final oscillations before it decays away, the analytic solution deviates away and has a nonzero asymptotic solution.
I will read it a bit more carefully.I like the way if gives an explicit  form for the decay of the frequency of the oscillations through the "omega" functions which go as the argument of the Jacobi elliptic functions (which is very much of interest to me). Need to rework the whole thing for the care of a constant force acting on it. 
 
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Cuchulainn
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Re: Analytic ODE solution possible for this one?

January 8th, 2018, 2:11 pm

tw, 
You obviously have your reasons for an explicit solution but you also mention asymptotic solutions. The Lyapunov method might be useful?

https://en.wikipedia.org/wiki/Lyapunov_stability

and it has an example for Dr. Pol's equation.

This link compares the approaches
https://books.google.nl/books?id=XejcAw ... ns&f=false



// BTW I see Jacobi is in Boost

http://www.boost.org/doc/libs/1_65_1/li ... bi_cd.html
 
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ppauper
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Re: Analytic ODE solution possible for this one?

June 13th, 2018, 5:42 pm

I've got a pair of coupled ODEs. I can write down a general form for one in terms of a parameter [$]n=0,1,2,\cdots[$]
Can anyone square the circle for me and write a general form for the other?

[$]f_{n}''+2\zeta f_{n}'-2nf_{n}=0[$]
and
[$]g_{n}''-2\zeta g_{n}'-2(n+1)g_{n}=-4f_{n}'[$]
with [$]f_{n}(\zeta)=\frac{1}{n!(2i)^{n}}H_{n}(i\zeta)[$]
(Hermite polynomial) so [$]f_{n+1}'(\zeta)=f_{n}(\zeta)[$].
what's [$]g_{n}(\zeta)[$] ?

edit: latex fixed
I'm looking for a general polynomial solution
Last edited by ppauper on June 13th, 2018, 7:41 pm
 
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Alan
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Re: Analytic ODE solution possible for this one?

June 13th, 2018, 6:45 pm

ppauper wrote:
I've got a pair of coupled ODEs. I can write down a general form for one in terms of a parameter $n=0,1,2,\cdots$
Can anyone square the circle for me and write a general form for the other?

\[$]f_{n}''+2\zeta f_{n}'-2nf_{n}=0[$]
and
\[$]g_{n}''-2\zeta g_{n}'-2(n+1)g_{n}=-4f_{n}'[$]
\with $f_{n}(\zeta)=\frac{1}{n!(2i)^{n}}H_{n}(i\zeta)$
(Hermite polynomial) so $f_{n+1}'(\zeta)=f_{n}(\zeta)$.
what's $g_{n}(\zeta)[$] ?

A Green function solution might be what you are looking for:
[$] g_{n}(\zeta) = 4 \int G_{n}(\zeta,\zeta') f_{n}'(\zeta') \, d \zeta'[$].
The Green function is constructed in a standard way from homogeneous solutions to your second ODE.
 
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ppauper
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Re: Analytic ODE solution possible for this one?

June 13th, 2018, 9:42 pm

variation of parameters
I tried it but it's not that helpful
The two homogeneous solutions are [$]g_{a}=e^{\zeta^{2}}f[$] and [$]g_{b}=e^{\zeta^{2}}f\int\frac{e^{-\zeta^{2}}d\zeta}{f(\zeta)^{2}}[$] with Wronskian [$]W=g_{a}g_{b}'-g_{a}'g_{b}=e^{\zeta^{2}}[$]

variation of parameters gives something  like [$]g=\pi^{1/2}e^{\zeta^{2}}{\rm erf}(\zeta)f(\zeta) [$] and for the application I need polynomial solutions (which is what you get if you do it case by case as it were)
Obviously the difference is in the second homogeneous solution and I need to evaluate
[$]\int\frac{e^{-\zeta^{2}}d\zeta}{H(n,i\zeta)^{2}}[$] where [$]H(n,\zeta)[$] is a Hermite polynomial

I have polynomial solutions for specific values of [$]n[$] but I'd like a master solution that covers all values of [$]n[$]
 
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Alan
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Re: Analytic ODE solution possible for this one?

June 14th, 2018, 2:47 am

Another suggestion is to post the problem that led to these ODE's. Maybe someone will see a more direct approach to that original problem.
 
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ppauper
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Re: Analytic ODE solution possible for this one?

June 18th, 2018, 11:51 pm

it's for a series solution of the Black-Scholes PDE.
I've got a maple code for spitting out as many of the [$]g_{n}[$] as I care to, but it would be nice to have a general expression


I've got the solution down to a double sum
[$]g_{n}=
-\sum_{m=0}^{n}\frac{1}{m!(n-m)!}x^{n-m}[$]
[$]\times\left[x^{m-1}
+\frac{(m-1)x^{m-3}}{2}
+\frac{(m-1)(m-3)x^{m-5}}{2^{2}}
+\frac{(m-1)(m-3)(m-5)x^{m-7}}{2^{3}}
+\cdots\right][$]
the polynomial inside the square brackets terminates when the exponent becomes negative
That polynomial
[$]x^{m-1}
+\frac{(m-1)x^{m-3}}{2}
+\frac{(m-1)(m-3)x^{m-5}}{2^{2}}
+\frac{(m-1)(m-3)(m-5)x^{m-7}}{2^{3}}
+\cdots[$]
looks so simple there's got to be a good chance it's a known function
It can be written very compactly using th double factorial
In terms of actual factorials (versus double factorials) it gets messier with separate expressions needed for even and odd powers

It took a lot of work, Cauchy formula for repeated integration along with the integral form of the Hermite polynomials
 
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Alan
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Re: Analytic ODE solution possible for this one?

June 19th, 2018, 10:43 pm

ppauper wrote:
it's for a series solution of the Black-Scholes PDE.
I've got a maple code for spitting out as many of the [$]g_{n}[$] as I care to, but it would be nice to have a general expression


I've got the solution down to a double sum
[$]g_{n}=
-\sum_{m=0}^{n}\frac{1}{m!(n-m)!}x^{n-m}[$]
[$]\times\left[x^{m-1}
+\frac{(m-1)x^{m-3}}{2}
+\frac{(m-1)(m-3)x^{m-5}}{2^{2}}
+\frac{(m-1)(m-3)(m-5)x^{m-7}}{2^{3}}
+\cdots\right][$]
the polynomial inside the square brackets terminates when the exponent becomes negative
That polynomial
[$]x^{m-1}
+\frac{(m-1)x^{m-3}}{2}
+\frac{(m-1)(m-3)x^{m-5}}{2^{2}}
+\frac{(m-1)(m-3)(m-5)x^{m-7}}{2^{3}}
+\cdots[$]
looks so simple there's got to be a good chance it's a known function
It can be written very compactly using th double factorial
In terms of actual factorials (versus double factorials) it gets messier with separate expressions needed for even and odd powers

It took a lot of work, Cauchy formula for repeated integration along with the integral form of the Hermite polynomials

 I don't understand when that polynomial inside the large brackets terminates. 
"when the exponent becomes negative" is confusing as there are various exponents.

Please write out *all* the terms for [$]n = 0,1,2,3,4[$]. (Don't use any [$]\cdots[$]). That should clear it up. Once I understand it, I'll see if Mathematica can turn it into a known function.
 
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ppauper
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Re: Analytic ODE solution possible for this one?

June 20th, 2018, 5:20 am

it's [$]h_{m}=\sum_{0\le k\le (m-1)/2}\frac{(m-1)!!x^{m-1-2k}}{(m-1-2k)!!}[$] with [$]k[$] an integer
[$]0[$] for [$]m=0[$]
[$]1[$] for [$]m=1[$] 
[$]x[$] for [$]m=2[$]
[$]x^{2}+1[$] for [$]m=3[$]
[$]x^{3}+\frac{3x}{2}[$] for [$]m=4[$]
[$]x^{4}+2x^{2}+2[$] for [$]m=5[$]
[$]x^{5}+\frac{5x^{3}}{2}+\frac{15x}{4}[$]
 
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Alan
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Re: Analytic ODE solution possible for this one?

June 20th, 2018, 3:12 pm

ppauper wrote:
it's [$]h_{m}=\sum_{0\le k\le (m-1)/2}\frac{(m-1)!!x^{m-1-2k}}{(m-1-2k)!!}[$] with [$]k[$] an integer
[$]0[$] for [$]m=0[$]
[$]1[$] for [$]m=1[$] 
[$]x[$] for [$]m=2[$]
[$]x^{2}+1[$] for [$]m=3[$]
[$]x^{3}+\frac{3x}{2}[$] for [$]m=4[$]
[$]x^{4}+2x^{2}+2[$] for [$]m=5[$]
[$]x^{5}+\frac{5x^{3}}{2}+\frac{15x}{4}[$]

Success. So inserting the missing 2^k, and letting m = 2 J + 1, I defined

H[J_] := Sum[(2 J)!! x^(2 J - 2 k)/(2^k (2 J - 2 k)!!), {k, 0, J}]

Then,  with  FullSimplify[H[J]], Mathematica found

[$] H(J) = e^{x^2} \Gamma(1+J,x^2)[$],
using the incomplete Gamma function. 

I'll leave the case of even m to you.
 
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ppauper
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Re: Analytic ODE solution possible for this one?

June 23rd, 2018, 6:48 am

thanks
it looks like the best I'm going to get is Kummer hypergeometric functions
If I go back to the original equation and turn it into a 4th order equation, I find a general solution
[$]\left[c_{1}+c_{2}{\rm erf}(x)\right]M\left(\frac{n}{2}+\frac{1}{2},\frac{1}{2}.x^{2}\right)
+\left[c_{3}+c_{4}{\rm erf}(x)\right]M\left(\frac{n}{2}+1,\frac{3}{2}.x^{2}\right)[$]
some of the coefficients vanish:
2 vanish when [$]n[$] is odd and (a different) 2 vanish when [$]n[$] is even
The bad news is that while there are nice expressions for [$]M\left(m+\frac{1}{2},\frac{1}{2}.x^{2}\right)[$] and [$]M\left(m+\frac{1}{2},\frac{3}{2}.x^{2}\right)[$],
the expressions for [$]M\left(m,\frac{1}{2}.x^{2}\right)[$] and [$]M\left(m,\frac{3}{2}.x^{2}\right)[$] all seem to be lengthy and unpleasant and worse than what I had already

On a side note, a rant at maple.
I grew up using Kummer functions as my degenerate hypergeometric functions, and early versions of maple used them.
Then for no apparent reason, maple switched to Whittaker functions, including the (old old) version of maple on my laptop.
It's annoying as whenever maple spits them out as a solution to something, I have to grab Abramowitz & Stegun and convert them back to the Kummer functions that we all know and love. Maple should switch back to Kummer functions if they haven't already








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