Re: Analytical solution possible?

*To*: mathgroup at smc.vnet.net*Subject*: [mg131988] Re: Analytical solution possible?*From*: Roland Franzius <roland.franzius at uos.de>*Date*: Tue, 12 Nov 2013 00:14:30 -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*: <l5po5s$7md$1@smc.vnet.net>

Am 11.11.2013 05:58, schrieb Narasimham: > A numerical solution is possible for the following ODE, but its analytical solution is sought. It goes on indefinitely without indication about inverse functions or messages. > > Thanks in advance for any suggested workaround. > > NoISQL = {SI'[t] == (R[t] - a)/(2 a - R[t])/R[t]^2,R'[t] == Cot[SI[t]]/R[t]}; DSolve[NoISQL, {SI, R}, t]; Standard eleimination of dt between the two equations yields an equation for the orbit with variables R,S separated Eliminate[ {dR/dt == Cot[S]/R, dS/dt == (R - a)/(R^2 (2 a - R))} /. {R -> r + a}, dt] 2 dS Cot[S] == dr (1/(a - r) - 1/(a + r)) && a - r != 0 && a + r!= 0 Integrate[(1/(a - r) - 1/(a + r)), r] == Integrate[2 Cot[S], S] + C[1] -Log[a^2 - r^2] == 2 Log[Sin[S]] + C[1] where in the real domain the arguments of both Log's have to be taken as absolute values. -Log[Abs[a^2 - r^2]] == 2 Log[Abs[ C[1] Sin[S]] The orbits are then given by r -> Sqrt[a^2 - C[1] Csc[S]^2] Finally, replacement of S by R in R' yields an integrable differential equation for t Solve[ R * dR/dt == Sqrt[ -((-2 a R + R^2 + C[1])/C[1])], dt] -- Roland Franzius