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Dy/dt = iy

  1. Aug 4, 2015 #1
    Does a first order system with a solution that contains the complex number i have any meaning?

    e.g. dy/dt = iy

    therefore y = e^(it) = isin(t) + cos(t)

    Follow-up question. Can I breakdown a second order system into two cascaded first order systems that have complex roots?
  2. jcsd
  3. Aug 5, 2015 #2
    Yes, I think it could be, just for the differentiation in complex numbers.
  4. Aug 5, 2015 #3


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    Exactly what do you mean by "meaning"? A differential equation involving complex numbers has every bit as much meaning as an algebraic equation involving complex numbers or, indeed, as complex numbers themselves.

    If dy/dt= iy then [itex]y= Ce^{it}[/itex] where C can be any complex number.

    The second order equation, [itex]d^2y/dt^2+ y= 0[/itex] can be written as a pair of first order differential equations by defining [itex]x= dy/dt[/itex] so that [itex]d^2y/dt^2+ y= dx/dt+ y= 0[/itex] and [itex]dy/dt= x[/itex].

    You ask about "two cascaded first order systems that have complex roots". First the "system" is the pair of first order equations. Second, the system does not have "complex roots", its characteristic equation has.

    We can write the example above, [itex]dx/dt= -y[/itex] and [itex]dy/dt= x[/itex] as the matrix equation
    [tex]\frac{d}{dt}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}[/tex]

    That matrix has "eigenvalue equation" (the characteristic equation of the system of differential equations)
    [tex]\left|\begin{array}{cc}-\lambda & -1 \\ 1 & -\lambda \end{array}\right|= \lambda^2+ 1= 0[/tex]
    which has roots [itex]\lambda= \pm i[/itex].

    So the matrix has eigenvalues i and -i and corresponding eigenvectors (1, -i) and (1, i). In particular, if we construct the matrix [itex]B= \begin{bmatrix}1 & 1 \\ -i & i\end{bmatrix}[/itex], having those eigenvectors as columns, which has inverse matrix [itex]B^{-1}= \begin{bmatrix}1/2 & i/2 \\ 1/2 & -i/2\end{bmatrix}[/itex]
    Then the equation dX/dt= Ax becomes [itex]d(BX)/dt= (BAB^{-1})(BX)[/itex] which is
    [tex]\frac{d}{dt}\begin{bmatrix}u \\ v\end{bmatrix}= \begin{bmatrix}i & 0 \\ 0 & -i\end{bmatrix}\begin{bmatrix}u \\ v \end{bmatrix}[/tex]
    where u and v are defined by [itex]\begin{bmatrix}u \\ v \end{bmatrix}= \begin{bmatrix}1 & 1 \\ -i & i \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}[/itex] so that u= x+ y and v= -ix+ iy.

    That last system reduces to the two uncoupled equations du/dt= iu and dv/dt= -iv so that we have [itex]u(t)= Ce^{it}[/itex] and [itex]v(t)= De^{-it}[/itex]. Since u= x+ y and v= -ix+ iy, [itex]x= (1/2)u+ (1/2)i v= C' e^{it}+ D' e^{-it}[/itex] and [itex]y= (1/2)u- (1/2)iv= C'e^{it}- D'e^{-it}[/itex] where C'= (1/2)C and D'= (1/2)Di.
    Last edited by a moderator: Aug 5, 2015
  5. Aug 9, 2015 #4
    Thank you. Makes sense.
  6. Aug 10, 2015 #5
    I should also add that when I said "meaning" I was thinking of a physical interpretation. For example a first order system with a real answer corresponds to exponential decay or growth. A second order second system with a real answer is usually some sort of harmonic oscillator. As far as I can tell, an answer to ODE that contain complex numbers don't appear to have a simple physical meaning.

    Also, when I said cascaded I think convolved would have been a better word to use. As far as I understand, which was shown with the system representation, any second order system can be decomposed as two convolved first order ODE. I'm not sure of uses of this expect to solve ODE's and to look at stability. Can it be used in control systems or DSP?
  7. Aug 10, 2015 #6

    jim mcnamara

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    Staff: Mentor

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