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Linear 1st order coupled DEs

  1. Apr 11, 2010 #1
    This has got me really stumped. I've started out with an equation for the differential of a 2x2 matrix:

    [tex]
    \frac{d[N]}{dz} = f(z)[E][D] - \{[W],[N]\} - c\sigma_1[N] - d\sigma_2[N]
    [/tex]

    where all terms in [] are matrices, {} denotes anti-commutator, and the [tex]\sigma[/tex]'s are the Pauli matrices. Everything but c and d are functions of z, but they're all known functions - only N is unknown.
    I know physically that this should describe equations of motion for the matrix N in which the off-diagonal terms are exponentially damped. I think it's sensible to split the equation into equations for the diagonal and off-diagonal elements. If I'm right this results in equations of the form:

    [tex]
    \frac{dN_{\alpha\alpha}}{dz} = f(z)(DE)_{\alpha\alpha} - W_{\alpha\beta}N_{\beta\alpha} - W_{\beta\alpha}N_{\alpha\beta} - 2W_{\alpha\alpha}N_{\alpha\alpha}
    [/tex]

    and

    [tex]
    \frac{dN_{\alpha\beta}}{dz} = f(z)(DE)_{\alpha\beta} - Tr[N]W_{\alpha\beta} - Tr[W]N_{\alpha\beta} - cN_{\alpha\beta} - d(\sigma_2)_{\alpha\beta}N_{\alpha\beta} [/tex]

    where [tex]\alpha \neq \beta[/tex]. These are what I have to try to solve. And I'm stumped. I'm hoping someone can give me tips on solving coupled differential equations, stuff of the form:
    [tex]
    \frac{dy}{dz} = f(z) + g(z)y + h(z)x [/tex]

    [tex]
    \frac{dx}{dz} = u(z) + v(z)x + w(z)y
    [/tex]

    Anyone got any ideas on this kind of thing? Any kind of help would be really appreciated.

    (of course i suppose i'm really trying to solve:
    [tex]
    \frac{dy_1}{dz} + a(z)b(y) = c(z) + d(z)u(x)[/tex]
    [tex]
    \frac{dy_2}{dz} + e(z)f(y) = g(z) + h(z)v(x)[/tex]
    [tex]
    \frac{dx}{dz} + p(z)q(x) = r(z) + s(z)t(y)
    [/tex]
    :uhh:
     
    Last edited: Apr 11, 2010
  2. jcsd
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