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A set of two 2nd order linear ode

  1. Jan 20, 2010 #1
    I'm doing an undergrad research job..
    I have encountered the following coupled 2nd order linear ODE with constant coefficients a, b, c, d...
    [tex] \begin{align} \frac{d^2 y_1}{d x^2} + a^2y_1 & = -c y_2 \\
    \frac{d^2 y_2}{d x^2} - b^2y_2 & = -d y_1 \end{align} [/tex]

    In addition, I would like to impose constraints
    [tex] \frac{d y_1}{d x}|_{x=0} = \frac{d y_2}{d x}_{x=0} = 0 [/tex]
    (y_1, y_2 are even in x)

    if c were zero..the solution would be simple..
    [tex] \begin{align} y_1 & = A \cos(ax) \\
    y_2 & = B \cosh(bx) + C \cos(ax) \end{align} [/tex]
    but i cannot proceed further when c is not zero..
    did I fall into the silly trap??
    plz help me find the solutions..
     
  2. jcsd
  3. Jan 20, 2010 #2
    Re-write this vector form:

    [tex]
    \vec{y}''=A\vec{y}
    [/tex]

    with y=[y1 y2]
    and A=[-a^2 -c; -d b^2 ]

    Like in an ordinary constant coefficient ODE, we suggest solutions of the form
    [tex]
    \vec{y}=\vec{v}e^{sx}
    [/tex]

    Substituting into the equation gives:

    [tex]
    s^{2} \vec{v}e^{sx}=A\vec{v}e^{sx}
    [/tex]

    or

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

    For this algebric equation to have a non-trivial solution, the determinant of the matrix must equal to zero. You might recognize this as searching for eigenvalues, only here the eigenvalues are [tex]\lambda=s^{2}[/tex].

    In the most simple case, you'll get two real positive eigenvalues with one eigenvector which,
    for which you will have four basic solutions:

    [tex]\vec{y}=K_{11}\vec{v_{1}}e^{\sqrt{\lambda_{1}}x}+K_{12}\vec{v_{1}}e^{-\sqrt{\lambda_{1}}x}+K_{21}\vec{v_{2}}e^{\sqrt{\lambda_{2}}x}+K_{22}\vec{v_{2}}e^{-\sqrt{\lambda_{2}}x}[/tex]

    Off course other cases, will need more algebric manipulation, to pull out a real solution (f.e. if you get complex eigenvalues, your solutions will be periodic)
     
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