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Solve a system of nonhomogeneous DEs

  1. Mar 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Solve
    $$\ddot {\vec a}=A\vec a+B\dot{ \vec a}+\vec F$$

    if A and B are known matrices and F is a constant vector.

    2. Relevant equations


    3. The attempt at a solution

    My plan was to define ##\vec b=\dot{\vec a}## to move from second order to first order system ##\dot{\vec b}=A\vec a+B\vec b+\vec F##

    But now... well, I have got absolutely no idea how to continue! I wanted to use theorem 23 on page 580 in chapter 11.7 in this work: http://www.math.utah.edu/~gustafso/2250systems-de.pdf BUT I have got a ##A\vec a ## term... which... wel... I don't know what to do with it!
     
    Last edited by a moderator: Mar 25, 2016
  2. jcsd
  3. Mar 25, 2016 #2

    Ray Vickson

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    If ##A## is invertible there is no reason to appeal to Theorem 23; that is just "overkill". Think first how you would deal with the scalar DE ##x'' = A x + B x' + c## when ##A, B, c## are constants and ##A \neq 0##.
     
  4. Mar 25, 2016 #3
    In case of a scalar DE i would rewrite your equation to $${x}''-B{x}'-Ax=c$$
    Than the solution of a complementary part are roots of this characteristic equation $$D^2-BD-A=0$$ meaning $$D_{1,2}=\frac B 2 \pm \frac 1 2 \sqrt{B^2+4A}$$ therefore $$x_{hom}(t)=C_1e^{(\frac B 2 + \frac 1 2 \sqrt{B^2+4A})t}+C_2e^{(\frac B 2 - \frac 1 2 \sqrt{B^2+4A})t}.$$
    To get the particular solution we could guess it. Or not? Lets say $$x_{part}=C_0$$
    Right?
     
  5. Mar 25, 2016 #4

    Ray Vickson

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    If ##A,B,C## are constants and ##A \neq 0## the DE ##x'' = Ax + Bx' + C## is the same as the homogeneous DE
    [tex] (x+C/A)'' = A(x+C/A) + B(x+C/A)' . [/tex]

    In your case it is not completely clear what to do in the case that ##A## is not invertible and the linear system ##Ar = F## has no solution.

    BTW: my answer assumes that you know how to deal with the homogeneous system, and are just confused what to do about the non-homogeneous system (but with a constant "forcing" term).
     
  6. Mar 25, 2016 #5
    If I understand you correctly, than in my case of system of equations and assuming A is invertible $${(x-A^{-1}C)}''=A(x-A^{-1}C)+B{(x-A^{-1}C)}'$$ with ##x## being a vector ##\vec x=(x(t),y(t))## or did I not understand you correctly?

    Your assumption is wrong, to be honest. I am sure I used to know how to do this with matrices, yet I can't seem to remember, nor can I find any good paper with a nice explanation and an example.

    However, if I define ##\vec y=(\vec x-A^{-1}C)## than the equation above can be written $${\vec y}''-B{\vec y}'-A\vec y=0.$$

    Now all I was able to figure out from several papers is that it makes sense to define ##\vec u ={\vec y}'## to reduce the order of the differential equation. Doing so brings me to $$ { \vec u}'-B\vec u-A\vec y=0.$$

    Now you can safely assume that the ##-A\vec y## term is confusing me. Without it I would find the eigenvalues and eigenvectors of a matrix B. With that term I have no idea!
     
  7. Mar 25, 2016 #6

    Ray Vickson

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    The standard method is to re-write your system as
    [tex] \vec{y}' = \vec{u}, \; \; \vec{u}' = A \vec{y} + B \vec{u}[/tex]
    or
    [tex] \frac{d}{dt}\pmatrix{y \\ u} = \pmatrix{0 & I \\ A & B} \pmatrix{y\\u} [/tex]
    This is of the form ##\vec{z}' = {\mathbf M} \vec{z} ##.
     
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