Re: Finding branches where general solution is possible

*To*: mathgroup at smc.vnet.net*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)*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*References*: <l525d7$9f9$1@smc.vnet.net>

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