Linear Systems of ODEs#
Theorem: Basis for Solutions with
Consider the following linear system of ODEs
\[\boldsymbol{y}'(t) = \boldsymbol{A} \boldsymbol{y}(t)\]
with the real matrix \(\boldsymbol{A} \in \mathbb{R}^{n \times n}\).
If \(\boldsymbol{A}\) is diagonalizable as a complex matrix, then a basis for its solution space can be generated as follows:
- Each real eigenvalue \(\lambda\) with a corresponding eigenvector \(\boldsymbol{v}\) yields the basis element \(\mathrm{e}^{\lambda t}\boldsymbol{v}\).
- Each pair of complex conjugate eigenvalues \(\lambda = \alpha \pm \beta \mathrm{i}\) with corresponding eigenvectors \(\boldsymbol{v} = \boldsymbol{a} \pm \boldsymbol{b}\mathrm{i}\) yields the basis elements \(\mathrm{e}^{\alpha t} (\cos(\beta t) \boldsymbol{a} - \sin (\beta t) \boldsymbol{b})\) and \(\mathrm{e}^{\alpha t} (\sin(\beta t) \boldsymbol{a} + \cos (\beta t) \boldsymbol{b})\).
Example
Consider the following linear system of ODEs:
\[\boldsymbol{y}'(t) = \begin{bmatrix}1 & -2 \\ 2 & 1\end{bmatrix} \boldsymbol{y}\]
We have:
\[\boldsymbol{A} = \begin{bmatrix}1 & -2 \\ 2 & 1\end{bmatrix}\]
This is diagonalizable as a complex matrix with the following eigenvalues:
\[\lambda = 1 + 2\mathrm{i} \qquad \overline{\lambda} = 1 - 2\mathrm{i}\]
We thus get \(\alpha = 1\) and \(\beta = 2\). One eigenvector of \(\lambda\) is the following:
\[\boldsymbol{v} = \begin{bmatrix} \mathrm{i} \\ 1 \end{bmatrix} = \begin{bmatrix}0 \\ 1\end{bmatrix} + \mathrm{i} \cdot \begin{bmatrix}1 \\ 0\end{bmatrix}\]
We thus get:
\[\boldsymbol{a} = \begin{bmatrix}0 \\ 1\end{bmatrix} \qquad \boldsymbol{b} = \begin{bmatrix}1 \\ 0\end{bmatrix}\]
Therefore, we obtain the following basis:
\[\left\{ \mathrm{e}^{\alpha t} (\cos(\beta t) \boldsymbol{a} - \sin (\beta t) \boldsymbol{b}), \mathrm{e}^{\alpha t} (\sin(\beta t) \boldsymbol{a} + \cos (\beta t) \boldsymbol{b}) \right\}\]
\[\left\{ \mathrm{e}^{t} \left(\cos (2t) \begin{bmatrix}0 \\ 1\end{bmatrix} - \sin(2t) \begin{bmatrix}1 \\ 0\end{bmatrix}\right), \mathrm{e}^{t} \left(\sin (2t) \begin{bmatrix}0 \\ 1\end{bmatrix} + \cos(2t) \begin{bmatrix}1 \\ 0\end{bmatrix}\right) \right\}\]
\[\left\{ \mathrm{e}^t \begin{bmatrix}-\sin(2t) \\ \cos (2t)\end{bmatrix}, \mathrm{e}^t \begin{bmatrix}\cos (2t) \\ \sin(2t)\end{bmatrix}\right\}\]
The general solution is thus the following:
\[\boldsymbol{y}(t) = c_1 \mathrm{e}^t \begin{bmatrix}-\sin(2t) \\ \cos (2t)\end{bmatrix} + c_2 \mathrm{e}^t \begin{bmatrix}\cos (2t) \\ \sin(2t)\end{bmatrix} \qquad c_1, c_2 \in \mathbb{R}\]
Proof
TODO