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Re: Finding branches where general solution is possible

  • To: mathgroup at smc.vnet.net
  • Subject: [mg131891] Re: Finding branches where general solution is possible
  • From: "Dr. Wolfgang Hintze" <weh at snafu.de>
  • Date: Mon, 28 Oct 2013 00:39:28 -0400 (EDT)
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  • References: <l4a4cr$af9$1@smc.vnet.net>

Am Donnerstag, 24. Oktober 2013 05:32:43 UTC+2 schrieb Narasimham:
> > 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?
>
> > > 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]) -
>
> > -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]) + 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]) -
>
> > ])*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]) -
>
> > -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]) + 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]) +
>
> > ])*Sin[#1]^2]) + #1/2) & ][
>
> >     t + C[2]]}}
>
> >
>
> > Best regards,
>
> > Wolfgang
>
>
>
>
>
> Thanks Wolfgang, I am having difficulty in simplifying this Inverse perio=
dic function for a given boundary condition at t = 0 for #1. E.g., can I =
get a closed form/general expression for s as a function of t, at least for=
 boundary condition s[0] == Pi/2 ? ... so as to be able to plot the num=
erical case given from that expression ?
>
>
>
> Regards
>
> Narasimham

I'm always having some Problems with InverseFunction.

In this case I proceeded as in the following related example

(* here is an expression including InverseFunction *)

In[4]:= ss =
  InverseFunction[(1/
        2)*((a*Log[
            Sqrt[2]*Sqrt[-a^2]*Cos[#1] +
             Sqrt[-2 + a^2 - a^2*Cos[2*#1]]])/
         Sqrt[-a^2] + #1) & ][-(t/2) + C[1]];

(* the next line inverts the function *)

In[2]:= t[s_] := -2*First[List @@ ss[[0]]][s]

(* and here comes the inverted function for further use *)

In[3]:= t[s]

Out[3]= -s - (a*
    Log[Sqrt[2]*Sqrt[-a^2]*Cos[s] + Sqrt[-2 + a^2 - a^2*Cos[2*s]]])/
  Sqrt[-a^2]

Hope this helps.

Best regards,
Wolfgang



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