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Differential equation system, got stuck in a physics problem

  1. Sep 11, 2007 #1
    hi guys,
    I'm solving a pretty complex problem: calculating a trajectory of a charged particle in a custom magnetic field. I arrive to the point where this very nice equation system blocks my way :P


    hope you can help me somehow
  2. jcsd
  3. Sep 11, 2007 #2


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    It's not a valid link, there's no picture in it. I'm using Firefox as a webbrowser.
  4. Sep 12, 2007 #3
    Strange... But if you copy and paste it, it works 100%
  5. Sep 12, 2007 #4


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    As I understand it, the equation is
    [tex]m x'''(t) = \frac{q z''(t) y(t)}{r(t)}, m y'''(t) = \frac{q z''(t) x(t)}{r(t)}, m z'''(t) = \frac{ q\left( x'(t) x(t) + y'(t) y(t) \right) }{ r(t) }[/tex]
    (the last [itex]y'''(t)[/itex] in the original should be a [itex]z'''(t)[/itex] I presume), where
    [tex]r(t) = \sqrt{ x(t)^2 + y(t)^2 [/tex].

    This looks very complicated. Are you sure it is correct? I would either try polar or spherical coordinates (especially if there is some kind of symmetry in the system), or otherwise to first solve [itex]x(r(t))[/itex] or at least the functions in a more convenient variable than [itex]t[/itex]. But actually, I have no idea :confused:
  6. Sep 12, 2007 #5
    Yeah it's complicated indeed... and you got that right with [tex]z'''(t)[/tex]

    I actually know what it should look like when it's solved (graphically i mean) and there is some pretty serious symmetry in the system... I'll give a shot with the cilindrical coordinates, but there will be much work at it especially because I have to reconstruct the field function with polar coordinates (mabye esier... who knows...)

    anyway if someone has any ideas how to solve this (also with maple or similar), I'd relly apreciate it

    as soon as i get the cilindrical set of equations, I'll post them.

    thanks, ramses

    Edit: ah yeah forgot to say it's an equation system, so three equations together
  7. Sep 12, 2007 #6


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    How did you get third derivatives? Kinematic equations always involve acceleration, the second derivative.
  8. Sep 12, 2007 #7
    :surprised ... you're right... i made some confusion added a derivation grade for acceleration and velocity XD ... so, here are the correct ones:

    m x(t)'' = q z(t)' \frac{x(t)}{r(t)}\\
    m y(t)'' = q z(t)' \frac{y(t)}{r(t)}\\
    m z(t)'' = -\frac{q}{r(t)}(x(t)' x(t) + y(t)' y(t))\\


    [tex]r(t)=\sqrt{x(t)^2 + y(t)^2}[/tex]

    still working on those polar ones

    thanks for your interest

    (P.S. I finally realized that fourmulaes can be written with latex!! XD)
  9. Sep 13, 2007 #8
    So, as

    [tex]r^{\prime}(t) = \frac{1}{2} \frac{(2 x(t) x^{\prime}(t) + 2 y(t) y^{\prime}(t))}{\sqrt{x^2(t) + y^2(t)}} = \frac{x(t)x^{\prime}(t) + y(t) y^{\prime}(t)}{\sqrt{x^2(t) + y^2(t)}}[/tex]

    You can write

    [tex]m z^{\prime \prime}(t) = -q r^{\prime}(t) [/tex]

    Hence, for z you get

    [tex]z^{\prime}(t) = \kappa - \frac{q}{m} r(t) [/tex]

    where [tex]\kappa[/tex] is a constant.

    Now, for [tex]\kappa = 0[/tex], the solution is straightforward. You will then get

    [tex]x^{\prime \prime} = -(\frac{q}{m})^2 x(t) [/tex]


    [tex]y^{\prime \prime} = -(\frac{q}{m})^2 y(t) [/tex]

    But in the more general case, [tex]\kappa[/tex] may not be taken as 0, so you'll get

    [tex]x^{\prime \prime}(t) = \frac{q}{m} x(t) (\frac{\kappa}{r(t)} - \frac{q}{m})[/tex]


    [tex]y^{\prime \prime}(t) = \frac{q}{m} y(t) (\frac{\kappa}{r(t)} - \frac{q}{m})[/tex]
    Last edited: Sep 13, 2007
  10. Sep 13, 2007 #9
    :bugeye: you got it! Well thanks very much!

    just for curiosity are you a math student?
  11. Sep 13, 2007 #10
    I've only solved it for [tex]\kappa =0[/tex]. For [tex]\kappa \ne 0[/tex], it is still a nasty mess.

    And, no, I'm not a student.
  12. Sep 13, 2007 #11
    It's still a great advance, because I can set the initial conditions for z velocity (so [tex]z(t)'[/tex]) to be 0 forcing the constant to be zero. So I have the problem solved for this specifical case. But as you say it still needs work for the more general case...

    Hope you didn't take me thinking you were a student as offensive, I just thought that you were closely involved on math so supposed that. If any offense was taken I apologise. Thanks again for this partial solution
  13. Sep 13, 2007 #12
    No worries, mate. :cool:
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