3804 The Borderline Geometric Types

All the other possibilities for the linear system we call borderline types. We will show now that none of them is structurally stable; we begin with the center.
Eigenvalues pure imaginary. Once again we use the eigenvalue with the positive imaginary part: $\lambda = 0 + si, s > 0$. It corresponds to a center: the trajectories are a family of concentric ellipses, centered at the origin. If the coefficients $a, b, c, d$ are changed a little, the eigenvalue $0 + si$ changes a little to $r' + s'i$, where $r' \approx 0, s' \approx s$, and there are three possibilities for the new eigenvalue: $$\begin{aligned} &0+si\rarr r'+s'i:&r'>0&&r'<0&&r'=0\\ &s>0&s'>0&&s'>0&&s'>0\\ &\text{center}&\text{source spiral }&&\text{sink spiral}&&\text{center} \end{aligned}$$ Correspondingly, there are three possibilities for how the geometric picture of the trajectories can change:

Eigenvalues real; one eigenvalue zero. Here $\lambda_1 = 0$, and $\lambda_2 > 0$ or $\lambda_2 < 0$. The general solution to the system has the form ($\alpha_1, \alpha_2$ are the eigenvectors) $$\boldsymbol{x}=c_1\alpha_1+c_2\alpha_2e^{\lambda_2t}$$ If $\lambda_2 < 0$, the geometric picture of its trajectories shows a line of critical points (constant solutions, corresponding to $c_2 = 0$), with all other trajectories being parallel lines ending up (for $t = \infty$) at one of the critical points, as shown below

We continue to assume $\lambda_2 < 0$. As the coefficients of the system change a little, the two eigenvalues change a little also; there are three possibilities, since the eigenvalue $\lambda = 0$ can become positive, negative, or stay zero: $$\begin{aligned} &\lambda_1=0\rarr\lambda_1':&\lambda_1'>0&&\lambda_1'=0&&\lambda_1'<0\\ &\lambda_2<0\rarr\lambda_2':&\lambda_2'<0&&\lambda_2'<0&&\lambda_2'<0\\ &\text{critical line}&\text{unstable saddle}&&\text{critical line}&&\text{sink node} \end{aligned}$$ Here are the corresponding pictures. (The pictures would look the same if we assumed $\lambda_2 > 0$, but the arrows on the trajectories would be reversed.)

One repeated real eigenvalue. Finally, we consider the case where $\lambda_1 = \lambda_2$. Here there are a number of possibilities, depending on whether $\lambda_1$ is positive or negative, and whether the repeated eigenvalue is complete (i.e., has two independent eigenvectors), or defective (i.e., incomplete: only one eigenvector). Let us assume that $\lambda_1 < 0$. We vary the coefficients of the system a little. By the same reasoning as before, the eigenvalues change a little, and by the same reasoning as before, we get as the main possibilities (omitting this time the one where the changed eigenvalue is still repeated): $$\begin{aligned} &\lambda_1<0&\rarr&&\lambda_1'<0&&r+si\\ &\lambda_2<0&\rarr&&\lambda_2'<0&&r-si\\ &\lambda_1=\lambda_2&&&\lambda_1'\neq\lambda_2'&&r\approx\lambda_1,s\approx 0\\ &\text{sink node}&&&\text{sink node}&&\text{sink spiral} \end{aligned}$$ Typical corresponding pictures for the complete case and the defective (incomplete) case are (the last one is left for you to experiment with on the computer screen)

Remarks. Each of these three cases-one eigenvalue zero, pure imaginary eigenvalues, repeated real eigenvalue-has to be looked on as a borderline linear system: altering the coefficients slightly can give it an entirely different geometric type, and in the first two cases, possibly alter its stability as well.