Re: Finding branches where general solution is possible

*To*: mathgroup at smc.vnet.net*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)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net

> 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