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Differential equation with cross product

  1. Feb 21, 2007 #1
    ok, i don't know what to do with something like this:

    (d^2R/dt^2 ) + (dR/dt) x B = 0

    where the capitals are vectors (sorry i suck at latex). R is a position vector in x-y plane and B is in the z-direction.

    do i split this into equations for x and y directions separately and solve them? for x-direction i would get

    d^2x/dt^2 + (dy/dt)|B| = 0

    but how do i solve this? i'm pretty sure all the DEs i've ever solved had, for example, d^2x/dt^2 and dx/dt in them, but not d^2x/dt^2 and dy/dt.

    or is there a quick way of doing it without having to expand the cross product?

    any help would be well appreciated!!
     
  2. jcsd
  3. Feb 21, 2007 #2

    arildno

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    This is a SYSTEM of differential equations that you can rewrite as:
    [tex]\frac{d\vec{U}}{dt}=A\vec{U}, \vec{U}=\frac{d\vec{R}}{dt}[/tex]
    and A is a matrix.
    If A has constant coefficients, then the system is readily solvable with eigen-vector decomposition.

    Question:
    Did you get this equation from a physical problem with a Coriolis term?
    Just curious..

    EDIT:
    Insofar as B is constant, you may gain two decoupled 3.order diff.eq's in R (i.e 2.order in U).
     
    Last edited: Feb 21, 2007
  4. Feb 21, 2007 #3

    dextercioby

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    Maybe [itex] \vec{B} [/itex] is the magnetic field and he chose units such as q=1...
     
  5. Feb 21, 2007 #4
    damn, i wish i was more proficient with matrices and stuff. i might have to do a bit of quick revision.

    the equation is just electron motion in a magnetic field without the constants.

    my friend says that from the equation

    d^2x/dt^2 + (dy/dt)|B| = 0

    you just integrate to get

    dx/dt + y|B| = constant

    that doesn't seem quite right to me...is it? if it is that's all i need for now.
     
  6. Feb 21, 2007 #5

    J77

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    Is B a constant, or a function of x and y?
     
  7. Feb 21, 2007 #6

    HallsofIvy

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    Yes, writing R as <x, y, 0> you get
    [tex]\frac{d^2x}{dt^2}+ B\frac{dy}{dt}= 0[/tex]
    and
    [tex]\frac{d^2y}{dt^2}- B\frac{dx}{dt}= 0[/itex]

    You can, as Arildno suggested, introduce [itex]u= \frac{dx}{dt}[/itex] and [itex]v= \frac{dy}{dt}[/itex] and write this as a system of 4 first order differential equations.

    Another way to handle it is this: differentiate the first equation again to get
    [tex]\frac{d^3x}{dt^3}+ B\frac{d^2y}{dt^2}= 0[/itex]
    and use the second equation to substitute for the second derivative of y
    [tex]\frac{d^3x}{dt^3}+ B^2\frac{dx}{dt}= 0[/tex]
    That's easy to solve.

    Once you know x(t), you can use the first equation to solve for [itex]\frac{dy}{dt}[/itex] and integrate once more to find y(t).

    (I just noticed J77's comment. I am assuming here that B is a constant.)
     
    Last edited: Feb 21, 2007
  8. Feb 21, 2007 #7
    awesome, thanks. and yes B is constant.
     
  9. Feb 21, 2007 #8

    arildno

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    You can certainly use your friend's trick in the case of constant B. It doesn't help you an overly lot, though, compared to other techniques mentioned.
     
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