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A set of nonlinear coupled ODE's

  1. Nov 2, 2006 #1
    Hi,

    I'm looking for an analytical solution to this set of coupled equations:

    [tex]U^{'}_{i}(x)=c (f_i(x)-(\vec{U}\cdot\vec{f})U_i(x))[/tex]

    Where the vectors are 3 dimentional, c is constant and f is a vector of given functions:

    [tex]f_i(x) , i\in 1..3[/tex]

    There probably isn't one but I thought I'd try anyway.

    thanks
     
    Last edited: Nov 2, 2006
  2. jcsd
  3. Nov 3, 2006 #2
    right now it looks like you have 1 equation in two or more unknowns, which will be impossible to solve.
     
  4. Nov 3, 2006 #3
    Not really

    There are 3 equations for each component of [tex]\vec{U}(x)[/tex] since the index i goes between 1..3. For example the first equation can be explicitly written as:
    [tex]U_1^{'}(x)=c(f_1(x)-U_1(f_1 U_1+f_2 U_2+f_3 U_3))[/tex]
    As for the unknows, that doesn't mean you can't formulate a general solution. consider the equation:
    [tex]U_i^{'}(x)=f_iU_i(x)[/tex] (no summation)
    whose solution is:
    [tex]U_i(x)=\exp{\left(\int_0^x f_i(t)dt\right)}[/tex]
     
    Last edited: Nov 3, 2006
  5. Nov 3, 2006 #4
    I think if we knew something more about the U's or the f's it may be possible to get somewhere, otherwise I wouldn't know where to begin.

    Is there anything more to know?
     
  6. Nov 3, 2006 #5
    I wasn't sure what kind of summation notation you were using, but after playing with it for a few minutes I was able to move the u1's and f1's to one side, so if you really need that analytical solution I would guess that its there, as you could wirte

    G(f1,u1,u1')=-F2u2-F3u3
    G(f2,u2,u2')=-F1u1-F3u3
    G(f3,u3,u3')=-F1u1-F2u2
    (note I used the same G as it would be the same seperation procedure)
    which can then be written as the matrix equation

    G=FU
    then you might be able to play with it some more to get the G's and u's together in order to make 3 first orderdifferential equations, but I don't know, but you would have to assume that det(F) does not equal 0.

    hmm one thing I do notice there though is that if G=(FU)^t than it would become three ordinary differential equations in 1 variable. does anybody know of an identity that will let you do that?

    if you do come up with a general solution post it, i'd love to see it. also out of curiosity whats it for?

    EDIT:edited for reasons of gross inaccuracy
     
    Last edited: Nov 3, 2006
  7. Nov 4, 2006 #6
    You are right Matthew the f's and u's have some very neat properties which I left out because a lot of mathematicians are not familiar with the Minkovsky product which is the appropriate product for unified space time. Let's denote it by star, such that:
    [tex]\vec{f}(x)\star\vec{g}(x)=f_1g_1-f_2g_2-f_3g_3[/tex]
    Note the diffrence in sign because this is Minkovsky space not Euclidian space. So in fact in my original equation you should replace Euclidiean product [tex]\vec{U}(x)\cdot\vec{f}(x)[/tex] with [tex]\vec{U}(x)\star\vec{f}(x)[/tex]. This of course doesn't really shed more light on the problem, however The U's are unit vectors under the Minkovsky product and the f's are null vectors. Namely:
    [tex]\vec{U}(x)\star\vec{U}(x)=1[/tex]
    [tex]\vec{f}(x)\star\vec{f}(x)=0[/tex]
    Note that for Euclidian vectors null vectors are always trivial, because a zero norm implies that the vector is the zero vector. This is not true for Minkovsky space vectors, that can be non trivial if they have a zero norm under the Minkovsky product.

    Luke: These equations are related to the dynamics of confined quarks within mesons or Baryons.
     
    Last edited: Nov 4, 2006
  8. Nov 5, 2006 #7
    hmm well i you ha the other function, and the initial conditions you could use a lplace or fourier transform to get the answer, but /i do't think you'll be able to form a general solution
     
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