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Re: Finding branches where general solution is possible

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  • Subject: [mg131936] Re: Finding branches where general solution is possible
  • From: Narasimham <mathma18 at gmail.com>
  • Date: Sat, 2 Nov 2013 02:27:03 -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]
> 
Integrate[1/((1 + v) (2 + v)), v] == Integrate[Cot[x], x]

Integrate[1/((1 + v) (2 + v)), v] == Integrate[Cot[x]/v, x]

Regards
Narasimham



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