Re: Finding branches where general solution is possible
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- Subject: [mg131864] Re: Finding branches where general solution is possible
- From: "Dr. Wolfgang Hintze" <weh at snafu.de>
- Date: Sat, 19 Oct 2013 04:12:16 -0400 (EDT)
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Am Donnerstag, 17. Oktober 2013 06:12:03 UTC+2 schrieb Dr. Wolfgang Hintze:
> Am Mittwoch, 16. Oktober 2013 10:46:06 UTC+2 schrieb Narasimham:
>
> > For following function with period 2 Pi in which branches is it possible to get a general solution?
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> > Regards
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> >
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> > Narasimham
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> > DSolve[{si''[th] Tan[si[th]]==(1+si'[th]) (1+2si'[th]),si[0]==Pi/4 },si,th]
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> >
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> > NDSolve[{si''[th] Tan[si[th]]==(1+si'[th]) (1+2si'[th]),si'[0]==0,si[0]==Pi/4},si,{th,0,6Pi}];
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> >
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> > SI[u_]=si[u]/.First[%];Plot[SI[th],{th,0,6Pi}]
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> > DSolve::bvnul: For some branches of the general solution, the given boundary conditions lead to an empty solution.>>
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> Version 8 of Mathematica has no difficulty in DSolve-ing the equation generally.
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> The solution is given in terms of InverseFunction as follows:
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>
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> In[1]:= eq =
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> Derivative[2][s][t] ==
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> Cot[s[t]]*(1 + Derivative[1][s][t])*(1 + 2*Derivative[1][s][t])
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>
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> Out[1]= Derivative[2][s][t] ==
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> Cot[s[t]]*(1 + Derivative[1][s][t])*(1 + 2*Derivative[1][s][t])
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>
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> In[2]:= DSolve[eq, s[t], t]
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>
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> Out[2]= {
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> {s[t] ->
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> InverseFunction[
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> 2*(-((I*E^C[1]*Cos[#1/2]^3*
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> Log[2*I*E^C[1]*Cos[#1] -
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> Sqrt[2]*Cos[#1/2]^2*
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> Sqrt[(-2 + E^(2*C[1]) - E^(2*C[1])*Cos[2*#1])*
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> Sec[#1/2]^4]]*
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>
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> Sqrt[(-(2 - E^(2*C[1]) + E^(2*C[1])*Cos[2*#1]))*
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> Sec[#1/2]^4]*Sin[#1/2])/
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> Sqrt[(-E^(2*C[1]))*(2 - E^(2*C[1]) +
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> E^(2*C[1])*Cos[2*#1])*Sin[#1]^2]) - #1/2) & ][
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> t + C[2]]},
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> {s[t] ->
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> InverseFunction[-2*(-((I*E^C[1]*Cos[#1/2]^3*
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> Log[2*I*E^C[1]*Cos[#1] -
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> Sqrt[2]*Cos[#1/2]^2*
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> Sqrt[(-2 + E^(2*C[1]) - E^(2*C[1])*Cos[2*#1])*
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> Sec[#1/2]^4]]*
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>
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> Sqrt[(-(2 - E^(2*C[1]) + E^(2*C[1])*Cos[2*#1]))*
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> Sec[#1/2]^4]*Sin[#1/2])/
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> Sqrt[(-E^(2*C[1]))*(2 - E^(2*C[1]) +
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> E^(2*C[1])*Cos[2*#1])*Sin[#1]^2]) + #1/2) & ][
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> t + C[2]]}}
>
>
>
> Best regards,
>
> Wolfgang
In order to make things (which Mathematica "knows") more tranparent consider that the differential equation is of the form
s''(t) = f(s(t)) g(s'(t))
Here variables can be separated as follows.
Letting s' = u and using s as the indepent variable instead of t we have
du/dt = du/ds ds/dt = du/ds u
and the differential equation becomes
du/ds u = f(s) g(u)
or
du u/g(u) = ds f(s)
This can be integrated once and gives (possibly implicitly) u as a function h of s and the constant C[1].
But this is another differential equation
ds/dt = h(s,C[1])
which can be integrated to give
Int(1/h(s,C[1])) = t+C[2]
This is then the general solution (two constants of integration) in the inverted form t = t(s).
Best regards,
Wolfgang