Mathematica 9 is now available
Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2013

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Finding branches where general solution is possible

  • To: mathgroup at smc.vnet.net
  • 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)
  • 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: <l3ljoe$jt5$1@smc.vnet.net> <l3no2j$ov6$1@smc.vnet.net>

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?
> 
> > 
> 
> > 
> 
> > 
> 
> > 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

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




  • Prev by Date: Re: Help with Manipulate
  • Next by Date: why the mantissas used below are all roots of powers of 10
  • Previous by thread: Re: Finding branches where general solution is possible
  • Next by thread: Re: Finding branches where general solution is possible