Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Systems of Linear Homogenous Differential equations with Constant Coefficients

  1. Mar 6, 2010 #1
    Hello,
    I am looking at different ways to solve Systems of Linear Homogenous Differential equations with Constant Coefficients that is [tex]\acute{x}=Ax[/tex] (x and x' are vectors A is a matix) then the solutions are [tex]x= \xi e^{\lambda t}[/tex] where [tex]\xi[/tex] are the eigenvectors and [tex]\lambda[/tex] the eigenvalues of A and the general solution is the sum of all the eigenvectors with constants inserted.
    i.e [tex] x= c_{1}\xi^{(1)} e^{\lambda t}+c_{2}\xi^{(2)} e^{\lambda t}[/tex]

    The problem I have with this is that I can't figure out how to get mixing since surely if n=2, say, then x=(x1,x2) but to get x1 you are just adding weighted amounts of x1 doesn't ever couple to x2 i.e [tex]x_{1} = c_{1}\xi^{(1)}_1 e^{\lambda t}+c_{2}\xi^{(2)}_{1} e^{\lambda t}[/tex]I assume I am being idiot but if someone could point out where I am going wrong that would be brilliant.

    Thanks very much,

    P.S I normally solve it like this http://physics.ucsc.edu/~peter/114A/coupled_fol.pdf [Broken] if you know any links to other ways to solve them I would be grateful.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 7, 2010 #2
    If you have :
    d X/dt = AX where A is a diagonalisable matrix and X a column vector.
    U^-1AU=D where D is the diagonal matrix : diag(lambda_1...lambda_n)

    d X/dt = AX <=>
    d X/dt = UDU^-1X <=>
    U^-1dX/dt = DU^-1X <=>
    dY/dt = D Y with Y =U^-1X provided that A is time independant. The mixing comes from the fact the eigendirections are Y and not X : you need the transfer matrix.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook