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##### MAT244--Lectures & Home Assignments / Re: 2D-picture: focal point or center: direction of rotation

« Last post by**Victor Ivrii**on

*»*

**Today**at 03:44:37 AMTo 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,\\

y'=-bx

$$which solves to $$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 $x'=-y,\ y'=x$.

$$

x'=b y,\\

y'=-bx

$$which solves to $$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 $x'=-y,\ y'=x$.