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Thread: Stability of a solution

  1. #1
    Apr 2007

    Stability of a solution

    Dear all,
    I have a question on the stability of a solution.

    Let a(t), b(t) and c(t) be continuous functions of t over the interval $\displaystyle [0,\infty)$. Assume (x,y) = $\displaystyle (\phi(t), \psi(t))$ is a solution of the system
    $\displaystyle \dot{x} = -a^2(t)y + b(t), \dot{y} = a^2(t)x + c(t)$

    Show that this solution is stable.

    I rearranged the system to get

    $\displaystyle \frac{d}{dt}\binom{x}{y} = \binom{-a^2(t)y + b(t)}{a^2(t)x + c(t)}
    = \left( \begin{array}{cc} 0 & -a^2(t) \\ a^2(t) & 0 \end{array} \right)\binom{x}{y} + \binom{b(t)}{c(t)}$

    I've only dealt with constant coefficient linear systems before, so I'm having trouble with this question.

    Let $\displaystyle A = \left( \begin{array}{cc} 0 & -a^2(t) \\ a^2(t) & 0 \end{array} \right) $.

    I'm not sure if I can treat A like a constant matrix, i.e. take the determinant of A to be $\displaystyle a^4(t)$ and trace(A) = 0 by treating t as constant. Also, what role does the $\displaystyle \binom{b(t)}{c(t)}$ play here?

    Can someone please help me?

    Thank you.

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  2. #2
    Super Member Rebesques's Avatar
    Jul 2005
    My house.
    You are attempting to use Lyapunov's theorem. Let's find the stationary solution. Solve $\displaystyle 0= -a^2(t)y + b(t), 0= a^2(t)x + c(t)$ to get $\displaystyle y_0(t)= b(t)/a^2(t), \ x_0(t)= -c(t)/a^2(t)$. Goes without saying that $\displaystyle a(t)\neq 0$, or we get a trivial system.

    Now let's rename: $\displaystyle
    \frac{d}{dt}\binom{x}{y} = \binom{f_1(x,y)}{f_2(x,y)}$. By Lyapunov's theorem, if all eigenvalues of the Jacobian matrix $\displaystyle \binom{\frac{\partial f_1}{\partial x} \ \ \frac{\partial f_1}{\partial y}}{\frac{\partial f_2}{\partial x} \ \ \frac{\partial f_2}{\partial y}}$ evaluated at $\displaystyle (x_0(t),y_0(t))$ have non-positive real parts, the solution is stable. But this is just as you say $\displaystyle \left( \begin{array}{cc} 0 & -a^2(t) \\ a^2(t) & 0 \end{array} \right)$, so the characteristic polynomial is $\displaystyle \lambda^2+a^4(t)\Rightarrow Re(\lambda)=0$.
    Last edited by Rebesques; Sep 14th 2007 at 11:16 PM. Reason: old age
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