Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How is this a 4th order ODE?

  1. Jan 31, 2010 #1
    I am not interested in the solution, but I am curious what makes: uiv = u, a 4th-order ordinary differential equation. 'i' is the square-root of -1, v is some element of the reals, and the differentiating variable is x.


    cheers
     
  2. jcsd
  3. Feb 1, 2010 #2
    Really interesting! Thanks for bringing this to my attention!

    I'm not familiar with this problem, so I'm not offering an answer, but here's what I did:

    First restate

    u^i*v = e^(ln(u)*i*v)=u

    If you take the x derivative of both sides, you get

    i*e^(ln(u)*iv)*(u'v/u + v'ln(u))=u'

    recognize that the first part is u by the original eqn, then divide both sides by u',

    iv +iv' u/u' = 1

    That guy is separable, so

    v'/(1-iv) = u'/(u*ln(u)) = a set of constants

    Looks like a first order equation to me.

    But I'm wondering if maybe that i allows higher orders? Like the way that i^i = (-e^(i*2*pi*n)*i/2) = -e^(pi*n) for any integer value of n? But that trick only works because i has an absolute value of 1, which isn't necessarily true of u or v. Also, I have no idea why n would stop at 4.

    Where did you hear it was a 4th order ODE?
     
  4. Feb 1, 2010 #3
    My text book. :D
    Birkhoff, Rota, Ordinary Differnential Equations (4th), 1989, pg 73 Example 1
     
  5. Feb 1, 2010 #4
    This is not a differential equation at all. It's simply

    [tex]u^{iv-1} = 1[/tex]

    [tex](iv-1) \ln (re^{i\phi}) = 0 [/tex]

    [tex] (iv-1) (i\phi + \ln{r}) = 0[/itex]

    which has solutions: (r=1, phi=0 => u=1) when v!=0, and any u otherwise.
     
  6. Feb 1, 2010 #5
    Hi Hamster,

    My textbook explicitly states that it is a 4th-order DE and the purpose of the example is to build a basis of solutions to the DE.
     
  7. Feb 1, 2010 #6
    The only way it could be a DE is if 'iv' denotes fourth derivative (Roman "4") rather than i times v:

    [tex]u'''' = u[/tex]
     
  8. Feb 1, 2010 #7
    That is it. The book uses z = μ + iυ (nu) to represent a complex number and upon closer inspection (holding the book up to my face) the exponent is iv (in cursive, go figure.).

    Thanks all!
     
  9. Feb 1, 2010 #8
    Hahaha, oh man that's hilarious.
     
  10. Feb 4, 2010 #9
    :rofl:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How is this a 4th order ODE?
  1. 4th oder ODE (Replies: 5)

Loading...