Re: Finding branches where general solution is possible
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- Subject: [mg131908] Re: Finding branches where general solution is possible
- From: Roland Franzius <roland.franzius at uos.de>
- Date: Mon, 28 Oct 2013 23:24:56 -0400 (EDT)
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Am 16.10.2013 10:46, schrieb Narasimham:
> For following function with period 2 Pi in which branches is it possible to get a general solution?
>
> Regards
> Narasimham
>
> DSolve[{si''[th] Tan[si[th]]==(1+si'[th]) (1+2si'[th]),si[0]==Pi/4 },si,th]
> NDSolve[{si''[th] Tan[si[th]]==(1+si'[th]) (1+2si'[th]),si'[0]==0,si[0]==Pi/4},si,{th,0,6Pi}];
> SI[u_]=si[u]/.First[%];Plot[SI[th],{th,0,6Pi}]
>
> DSolve::bvnul: For some branches of the general solution, the given boundary conditions lead to an empty solution.>>
>
The standard method of separation of variables leads to elliptic integrals.
You may fiddle around with the following result on the Riemann surface
of the arctan function family living on a doubly infinity branched
surface around the poles of the integrand 1/(1+z^2 ) at z=+-I
as a sum of two logarithms.
Setting si=x and si' = v
v' /((1 + v) (1 + 2 v)) = Cot[x]
In: Integrate[1/((1 + v) (2 + v)), v] == Integrate[Cot[x], x]
Out: Log[1 + v] - Log[2 + v] == Log[Sin[x]]
In: D[ Log[1 + v] - Log[2 + v], v] // Together
Out: 1/((1 + v) (2 + v))
The true genral indefinite integrals are of course
Log[Abs[(1 + v)/(1 + 2 v)]] == Log[Abs[Sin[x]]] + C[1]
or exponentiated on both sides
(1 + v)/(1 + 2 v) = C[1] Sin[x] v != {-1/2,-1}
In: Solve[(1 + (dx/dt))/(1 + 2 (dx/dt)) == C[1] Sin[x], dt]
Out: {{dt -> (dx - 2 dx C[1] Sin[x])/(-1 + C[1] Sin[x])}}
In: t -> C[2] + Integrate[(1 - 2 C[1] Sin[x])/(-1 + C[1] Sin[x]), x]
Out: t -> C[2] -2 x -
(2 ArcTan[ (C[1] - Tan[x/2])/Sqrt[1 - C[1]^2] ]) / Sqrt[1 - C[1]^2]
--
Roland Franzius