# Nature of stability of critical points

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.

• Observing that for $2 \times 2$ matrix $A$ characteristic polynomial is $\lambda^2 - {\rm tr} A \cdot \lambda + {\rm det} A$, ${\rm tr} A = \lambda_1 + \lambda_2$, ${\rm det} A = \lambda_1 \cdot \lambda_2$ tells you how to easily project all you know about eigenvalues to coefficients of characteristic poly and matrix trace/determinant. Oct 26 '13 at 12:46

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.

• Stable if $q > 0$ and $p = 0$, we have: $\lambda_{1,2} = \pm i \sqrt{q} \rightarrow$ stable center.
• Unstable if $q < 0$ or $p > 0$, we have three cases here:

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.

• Can you do the case $q > 0$ and $p < 0$? (You have a stable node or stable spiral.)

Now, can you relate these back to the critical point.

• In case 2 of unstable, is $q < 0$ just to ensure that $\sqrt{p^2-4q} \in \mathbb{R}$? For a stable node, $\lambda_i < 0$ I see that this implies $p < 0$ , however I don't see why $q>0$. The conditions on a stable spiral are clear to me. Thanks.
– CAF
Oct 26 '13 at 20:33
• When $q < 0$, this guarantees that the sqrt term is positive. What you get from that is two real eigenvalues. with one negative and one positive. The reason for that is that we have $p \pm sqrt{p^2 + 4q}$ and the latter term is greater than $p$ (see why?), hence a pos and neg eigenvalue, which is a saddle. Oct 26 '13 at 21:35
• @CAF: When $q>0$, we have that $p^2-4q$ can be pos or neg. When negative, we have imaginary eigenvalues with $p<0$ which is a stable spiral. When positive, we have the stable node case. Also recall that we can have the degenerate cases, but those are not asked. Regards Oct 26 '13 at 21:39
• @CAF: Another thing yo can try is to choose numbers for the matrix that satisfy all of these, so you can see how they actually materialize. Regards Oct 26 '13 at 21:57
• Many thanks, it is clear.
– CAF
Oct 26 '13 at 22:10