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Re: Finding branches where general solution is possible

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  • Subject: [mg131940] Re: Finding branches where general solution is possible
  • From: Roland Franzius <roland.franzius at uos.de>
  • Date: Mon, 4 Nov 2013 23:16:39 -0500 (EST)
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Am 02.11.2013 07:17, schrieb Narasimham:
>> 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]

ops, yes.

So we take

  dv/dt dt  /((1 + v) (1 + 2 v))  = Cot[x] (dt/dx) dx

or

  dv  v /((1 + v) (1 + 2 v)) =  Cot[x] dx

which yields once integrated with a first constant of integration k

solution[k_] := Solve[ Log[(1 + (dx/dt))^2] - Log[1 + 2 (dx/dt)] == 
Log[k Sin[x]^2] dt]

Solutions  readily may be given for special values of k, eg  k=1 which 
is complex or k=-2.

By time translation invariance, the second constant of integration is of 
course t+C[2].

Since the two log functions with factors 1 and 1/2, winding around two 
different infinite branch points it is of course left to the experienced 
user to identify his special branch for a solution to a given problem.

-- 

Roland Franzius



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