Linear dynamical system

Solution of linear dynamical systems

If the initial vector ${\displaystyle {\mathbf{x}}_0\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\mathbf{x}(t=0)}$ is aligned with a right eigenvector ${\displaystyle {\mathbf{r}}_k}$ of the matrix ${\displaystyle \mathbf{A}}$, the dynamics are simple

${\displaystyle \frac{d}{dt}\mathbf{x}(t)=\mathbf{A}{\mathbf{r}}_k={\lambda }_k{\mathbf{r}}_k}$

where ${\displaystyle {\lambda }_k}$ is the corresponding eigenvalue; the solution of this equation is

${\displaystyle \mathbf{x}(t)={\mathbf{r}}_k{e}^{{\lambda }_{k}t}}$

as may be confirmed by substitution.

If ${\displaystyle \mathbf{A}}$ is diagonalizable, then any vector in an ${\displaystyle N}$-dimensional space can be represented by a linear combination of the right and left eigenvectors (denoted ${\displaystyle {\mathbf{l}}_k}$) of the matrix ${\displaystyle \mathbf{A}}$.

${\displaystyle {\mathbf{x}}_0=\sum _{k=1}^N\left({\mathbf{l}}_k\cdot {\mathbf{x}}_0\right){\mathbf{r}}_k}$

Therefore, the general solution for ${\displaystyle \mathbf{x}(t)}$ is a linear combination of the individual solutions for the right eigenvectors

${\displaystyle \mathbf{x}(t)=\sum _{k=1}^n\left({\mathbf{l}}_k\cdot {\mathbf{x}}_0\right){\mathbf{r}}_k{e}^{{\lambda }_{k}t}}$

Similar considerations apply to the discrete mappings.

Classification in two dimensions

The roots of the characteristic polynomial det(A - λI) are the eigenvalues of A. The sign and relation of these roots, ${\displaystyle {\lambda }_n}$, to each other may be used to determine the stability of the dynamical system

${\displaystyle \frac{d}{dt}\mathbf{x}(t)=\mathbf{A}\mathbf{x}(t).}$

For a 2-dimensional system, the characteristic polynomial is of the form ${\displaystyle {\lambda }^2-\tau \lambda +\mathrm{\Delta }=0}$ where ${\displaystyle \tau }$ is the trace and ${\displaystyle \mathrm{\Delta }}$ is the determinant of A. Thus the two roots are in the form:

${\displaystyle {\lambda }_1=\frac{\tau +\sqrt{{\tau }^2-4\mathrm{\Delta }}}{2}}$

${\displaystyle {\lambda }_2=\frac{\tau -\sqrt{{\tau }^2-4\mathrm{\Delta }}}{2}}$,

and ${\displaystyle \mathrm{\Delta }={\lambda }_1{\lambda }_2}$ and ${\displaystyle \tau ={\lambda }_1+{\lambda }_2}$. Thus if ${\displaystyle \mathrm{\Delta }<0}$ then the eigenvalues are of opposite sign, and the fixed point is a saddle. If ${\displaystyle \mathrm{\Delta }>0}$ then the eigenvalues are of the same sign. Therefore, if ${\displaystyle \tau >0}$ both are positive and the point is unstable, and if ${\displaystyle \tau <0}$ then both are negative and the point is stable. The discriminant will tell you if the point is nodal or spiral (i.e. if the eigenvalues are real or complex).