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Am I right?: solving u" = Au

  1. Nov 12, 2015 #1
    I know how to solve [tex]\frac{d\vec{u}}{dt} = A\vec{u}[/tex], I was just watching a lecture, and the lecturer related that solving that equation is pretty much a direct analogy to [tex]\vec{u} = e^{At}\vec{u}(0)[/tex], in so far as all we need to do after that is understand exactly what it means to take the exponential of a matrix, which is fine...

    My question is this, do I solve [tex]\frac{d^2\vec{u}}{dt^2} = A\vec{u}[/tex] with [tex]\vec{u}=A^{-1}e^{At}\frac{d\vec{u}}{dt}(0)[/tex]?


    OR do I need to go along the lines

    [tex]\frac{d^2\vec{u}}{dt^2} = \lambda^2\vec{u} = A\vec{u}[/tex]

    and solve

    [tex](A - \lambda^2 I)\vec{u}=\vec{0}[/tex]

    Last edited: Nov 12, 2015
  2. jcsd
  3. Nov 13, 2015 #2


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    Science Advisor

    One way to look at it: Introduce a new function v: [itex]\frac{d\vec{u}}{dt}= \vec{v}, \frac{d\vec{v}}{dt}=A\vec{u} [/itex].
  4. Nov 13, 2015 #3


    Staff: Mentor

    I haven't thought about this in quite a while.
    If you had a regular ODE of the form u' = au, or u' - au = 0, the characteristic equation would be r - a = 0, so the solution would be u = Ceat, similar to your first order example above.

    If you had u'' = λ2u, or u'' - λ2u = 0, the characteristic equation would be r2 - λ2 = 0, or r = ±λ. The solution would be u = C1eλt + C2e-λt.

    Your example, with a system of ODEs, will have a solution that is analogous to the above, with exponentials involving the matrix.
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