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2D-picture: focal point or center: direction of rotation


Victor Ivrii:
When considering system
\mathbf{x}'=A\mathbf{x} \qquad \text{with } \
A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}
and discovering that it has two adjoint complex (not real) eigenvalues $\lambda_\pm =\mu \pm i\nu$, one should not only tell, if it is an unstable ($\mu>0$) or stable ($\mu <0$) focal point, or a center ($\mu =0$), but also the direction of rotation.

Observe that, $\lambda_\pm$ are roots of the equation $\lambda ^2- (a+d)\lambda + ad -bc=0$ with the discriminant $D:=(a+d)^2-4(ad -bc))=(a-d)^2+4bc$, and we consider the case $D<0\implies bc <0$. Thus $b$ and $c$ are not $0$ and have opposite signs.

Then, if $b<0$ (and $c>0$) rotation is counter-clockwise, and if $b>0$ (and $c<0$) rotation is clockwise.

Victor Ivrii:
To justify: let us change continuously matrix $A$, so that eigenvalues remain complex conjugate, then it may change from stable to unstable but still remain focal point/center with the same direction of rotation. Then we can make $a=d=0$ and $b=-c=\pm 1$ resulting in the system
x'=b y,\\
$$which solves to $$x=R\cos (bt+ \phi_0),\\
y=R\sin (bt+\phi_0)
or in the polar coordinates
$$r =\mathrm{const},\ \theta = bt +\phi_0$$
which describes a circle with the counter-clockwise (as $b<0$) or clockwise  (as $b>0$) rotation.

It is easy to remember: $x=\cos(t),\ y=\sin(t)$ is counter-clockwise  and satisfies $x'=-y,\ y'=x$.


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