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Re: Finding branches where general solution is possible

  • To: mathgroup at smc.vnet.net
  • Subject: [mg131847] Re: Finding branches where general solution is possible
  • From: "Dr. Wolfgang Hintze" <weh at snafu.de>
  • Date: Thu, 17 Oct 2013 00:16:14 -0400 (EDT)
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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?
> 
> 
> 
> 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.>>

Version 8 of Mathematica has no difficulty in DSolve-ing the equation generally.
The solution is given in terms of InverseFunction as follows:

In[1]:= eq = 
 Derivative[2][s][t] == 
  Cot[s[t]]*(1 + Derivative[1][s][t])*(1 + 2*Derivative[1][s][t])

Out[1]= Derivative[2][s][t] == 
 Cot[s[t]]*(1 + Derivative[1][s][t])*(1 + 2*Derivative[1][s][t])

In[2]:= DSolve[eq, s[t], t]

Out[2]= {
   {s[t] -> 
   InverseFunction[
     2*(-((I*E^C[1]*Cos[#1/2]^3*
              Log[2*I*E^C[1]*Cos[#1] - 
                Sqrt[2]*Cos[#1/2]^2*
                 Sqrt[(-2 + E^(2*C[1]) - E^(2*C[1])*Cos[2*#1])*
                   Sec[#1/2]^4]]*
                          
              Sqrt[(-(2 - E^(2*C[1]) + E^(2*C[1])*Cos[2*#1]))*
                Sec[#1/2]^4]*Sin[#1/2])/
            Sqrt[(-E^(2*C[1]))*(2 - E^(2*C[1]) + 
                E^(2*C[1])*Cos[2*#1])*Sin[#1]^2]) - #1/2) & ][
    t + C[2]]}, 
   {s[t] -> 
   InverseFunction[-2*(-((I*E^C[1]*Cos[#1/2]^3*
              Log[2*I*E^C[1]*Cos[#1] - 
                Sqrt[2]*Cos[#1/2]^2*
                 Sqrt[(-2 + E^(2*C[1]) - E^(2*C[1])*Cos[2*#1])*
                   Sec[#1/2]^4]]*
                          
              Sqrt[(-(2 - E^(2*C[1]) + E^(2*C[1])*Cos[2*#1]))*
                Sec[#1/2]^4]*Sin[#1/2])/
            Sqrt[(-E^(2*C[1]))*(2 - E^(2*C[1]) + 
                E^(2*C[1])*Cos[2*#1])*Sin[#1]^2]) + #1/2) & ][
    t + C[2]]}}

Best regards,
Wolfgang



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