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Solve A-B*x^3-C/r^2 = 0

  1. Mar 12, 2008 #1
    Hi all,

    I'm dealing with a central force problem (V(x) = k*x^3 potential) and I am stuck solving for x in the equation:
    [tex]
    A-Bx^3-\frac{C}{x^2}=0
    [/tex]

    Tried to do it using the symbolic toolbox in matlab but I get a "can't find closed form solution" error message. Anyone know the solution to the problem or how to obtain one?
     
  2. jcsd
  3. Mar 12, 2008 #2

    HallsofIvy

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    Multiply through by x2 and you have Ax2- Bx5- C= 0 or Bx5- Ax2+ C= 0. For some A, B, C, you might be able to find a solution but there is no general formula for fifth degree equations- some have solutions that cannot be written as radicals. How did you get that equation?
     
    Last edited: Mar 12, 2008
  4. Mar 12, 2008 #3
    The equation arises when I try to find r_min & r_max of a particle's trajectory in a V(r) = k*r^3 central force potential. The full equation of motion is described in this post: https://www.physicsforums.com/showthread.php?t=221342 :)

    You bring up a very good point that solutions exists only for certain values of A,B, & C because these parameters relate to the energy, strength of the potential, angular momentum and mass of the particle. It is easy to see that if the particle has too large an angular momentum or the energy is too high then it is not bound by this potential. Therefore, physically r_min & r_max do not exist.

    Is it possible to solve the equation with a given constraint? My math skills are pathetic at best so is it possible to get mathematica or matlab to do it? :)

    Btw, what do you mean by "some have solutions that cannot be written as radicals"?
     
  5. Mar 12, 2008 #4

    Mute

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    A radical is the root of a number (square root, cube root, etc.). In general, polynomials of degree five or higher do not have solutions that can be written down in terms of roots, so except in special cases you have very little hope of writing down the solutions symbolically (at least in terms of familiar numbers or radicals - maybe in terms of some special functions, but that's not really going to be much better).
     
  6. Mar 13, 2008 #5
    I think his teaching must have just done a bong and gave them all the question for a laugh which they could not answer.....
     
  7. Mar 13, 2008 #6
    Being handed unsolvable problems by your supervisor is all part of the fun of research. :) But anyway, everything can be solved numerically ;)
     
  8. Mar 13, 2008 #7
    There are 5 solutions. 4 of which are imaginary. The real one is:
    x -> -(B^(-1/5)) * (-A^2+C)^(1/5)

    The rest involve factors of imaginary "i" to the n/5th.
     
  9. Mar 13, 2008 #8

    HallsofIvy

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    If A= 2, B= 1, C= 5, that gives x= (1-1/5(-4+5))1/4= 1
    which does NOT satisfy 2- x3- 5/x2= 0.
     
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