I have a linear set of equation,
$\frac{dx(t)}{dt}= 4 \frac{1-a^{-1}}{a} y(t) - 8 (1-a^{-1}) x(t)$
and
$\frac{dy(t)}{dt} = b x(t) - \frac{b}{a} y(t)$
with initial conditions $x(0)=x_0$ and $y(0)=y_0$
After solving these i get following set of solutions
$x(\tau) = A_1 e^{-\lambda_1 \tau} + B_1 e^{-\lambda_2 \tau}$ and $y (\tau) = A_2 e^{-\lambda_1 \tau} + B_2 e^{-\lambda_2 \tau}$
Where, $\lambda_1 = -\frac{8-8a-b - \kappa}{2 a}, \lambda_2 = -\frac{8-8a-b + \kappa}{2 a}$
and
$A_1 = \frac{x(0) a \left(\kappa+8a-b-8\right)-8 (a-1) y(0)}{2 a \kappa} \\ B_1 = \frac{ x(0) a \left(\kappa-8a+b+8\right) + 8 (a-1) y(0)}{2 a \kappa} \\ A_2 = \frac{\left(\kappa-8a+b+8\right) y(0)-2 b x(0) a}{2 \kappa} \\ B_2 = \frac{\left(\kappa+8a-b-8\right) y(0)+2 b x(0) a}{2 \kappa} \\ \kappa \equiv \sqrt{b^2+64 (a-1)^2}$
Now in the limit $a \rightarrow 1^{+}$ I get solutions that gives correct equilibrium (i.e. $x(\infty)=y(\infty)=0$) as well as consistent initial conditions.
But in the limit $a \rightarrow \infty$ I get weird solution, where my solution for $y(t)$ does not go to right equilibrium (i.e. $y(\infty) \neq0$).
I dont understand why i get different types of behavior in two different limiting cases.
Thanks.
With $a→∞$ the system reduces to \begin{align} \dot x &= -8 x\\ \dot y &= bx \end{align} and has the solution \begin{align} x(t) &= x_0e^{-8t}\\ y(t) &= \frac{b}{8}x_0(1-e^{-8t})+y_0 \end{align} and indeed $y(\infty)=\frac{b}{8}x_0+y_0$ is not necessarily zero.
With the new insight into the negative eigenvalues one can see that $$ λ_1=\frac{8(a-1)+b+\sqrt{64(a-1)^2+b^2}}{2a} $$ is always positive going from $b$ to $8$ for a from $1$ to $∞$. $$ λ_2=\frac{8(a-1)+b-\sqrt{64(a-1)^2+b^2}}{2a} =\frac{8(1-a^{-1})b}{8(a-1)+b+\sqrt{64(a-1)^2+b^2}} $$ is always positive, with limits $0$ at $a=1$ and $a\to∞$.
Which means that the stationary point $(x,y)=(0,0)$ is always stable except in the limit cases which are limit cases. Stability there depends on higher order terms, or here their absence.