# Stability of a solution

• September 6th 2007, 03:24 AM
lancer6238
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 $[0,\infty)$. Assume (x,y) = $(\phi(t), \psi(t))$ is a solution of the system
$\dot{x} = -a^2(t)y + b(t), \dot{y} = a^2(t)x + c(t)$

Show that this solution is stable.

I rearranged the system to get

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

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

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

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

You are attempting to use Lyapunov's theorem. Let's find the stationary solution. Solve $0= -a^2(t)y + b(t), 0= a^2(t)x + c(t)$ to get $y_0(t)= b(t)/a^2(t), \ x_0(t)= -c(t)/a^2(t)$. Goes without saying that $a(t)\neq 0$, or we get a trivial system.
Now let's rename: $
. By Lyapunov's theorem, if all eigenvalues of the Jacobian matrix $\binom{\frac{\partial f_1}{\partial x} \ \ \frac{\partial f_1}{\partial y}}{\frac{\partial f_2}{\partial x} \ \ \frac{\partial f_2}{\partial y}}$ evaluated at $(x_0(t),y_0(t))$ have non-positive real parts, the solution is stable. But this is just as you say $\left( \begin{array}{cc} 0 & -a^2(t) \\ a^2(t) & 0 \end{array} \right)$, so the characteristic polynomial is $\lambda^2+a^4(t)\Rightarrow Re(\lambda)=0$.