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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)
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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



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