Consider the linear system $$\begin{bmatrix} \dot{x} \\ \dot{y} \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} x \\y \end{bmatrix}$$
Show the critical point is asymptotically stable if $q > 0$ and $p < 0,$ stable if $q > 0$ and $p = 0$ and unstable if $q < 0$ or $p > 0.$ $p$ and $q$ denote the trace and the determinant
of the matrix respectively.
Attempt:
This system is already linear so there is no need to Taylor expand about the critical points to determine the local behaviour. Let $A$ denote the matrix in question, then $$|A- \lambda I| = 0 \Rightarrow \lambda = \frac{p \pm \sqrt{p^2 – 4q}}{2}$$
The nature of each critical point which are given by $x$ and $y$ that satisfy $\dot{x}=\dot{y} = 0$ are $\left(-\frac{a_{12}a_{21}}{a_{11}a_{21}}y_o, y_0\right)$ and depend on the signage and form of the eigenvalues of $A$.
I then took each case separately: Asymptotically stable points occur when $\lambda_i$ are real and negative or of the form of complex conjugates with negative real part $-r \pm i\mu$. My first question is that to determine the conditions on $p$ and $q$ for each case (asymptotically stable, stable or unstable) do I need to just analyse one form of eigenvalues that correspond to that particular nature or analyse them all. For example, $-r \pm i\mu \Rightarrow$ $p < 0$ and $q > 0$, exactly as the question states. Would I then need to show that the same condition holds for the case of the $\lambda_i$ being real and negative? Or is it the case, that since they are already part of the asymptotically stable points, that I do not need to?
There is only one form of eigenvalues corresponding to stable points and I am fine with this.
For unstable, I am a little unsure. The cases are $\lambda_1, \lambda_2 > 0, \lambda_1 = \lambda_2 >0,$
$\lambda_2 < 0 < \lambda_1,$ and $ \lambda_1, \lambda_2 = r \pm i\mu, r > 0$
The last set has to have $p>0$ and $q > 0$, but the question only states $p>0$ or $q<0$. Surely in this case, $q > 0$ otherwise we may not have complex eigenvalues?
Many thanks.
Best Answer
I will do a couple of examples and see if you can do the rest.
Show the critical point is asymptotically stable if $q > 0$ and $p < 0,$ stable if $q > 0$ and $p = 0$ and unstable if $q < 0$ or $p > 0.$ $p$ and $q$ denote the trace and the determinant of the matrix respectively.
Case 1: If $D = (tr A)^2 - 4 \det A > 0$, and $tr(A) > 0$, it means that we have real eigenvalues and an unstable node.
Case 2: If $D = (tr A)^2 - 4 \det A > 0$, $\det A < 0 \rightarrow$ saddle.
Case 3: If $D = (tr A)^2 - 4 \det A < 0$, $\det A >0$, we have complex conjugate eigenvalues. If $tr(A) > 0$, we have an unstable spiral.
Now, can you relate these back to the critical point.