Autonomous Systems of ODEs#

Definition: Autonomous System of ODEs

A system of ODEs \(\boldsymbol{F}\) is autonomous if \(\boldsymbol{F}\) does not depend explicitly on its first argument (the independent variable \(x\)).

First-Order#

Definition: Stationary Point

We say that \(\boldsymbol{v} \in \mathbb{R}^n\) is a stationary point of the autonomous system

\[\boldsymbol{y}' = f(\boldsymbol{y})\]

if \(f(\boldsymbol{v}) = \boldsymbol{0}\).

Definition: Attractivity

We say that \(\boldsymbol{v}\) is attractive if there exists some \(\delta \gt 0\) such that each solution \(\boldsymbol{y}\) with \(||\boldsymbol{y}(0) - \boldsymbol{v}|| \lt \delta\) approaches \(\boldsymbol{v}\) at \(+\infty\):

\[\lim_{t \to \infty} \boldsymbol{y}(t) = \boldsymbol{v}\]

Definition: Stability

We say that \(\boldsymbol{v}\) is stable if, for each \(\varepsilon \gt 0\), there exists some \(\delta \gt 0\) such that for each solution \(\boldsymbol{y}\) with \(||\boldsymbol{y}(0) - \boldsymbol{v}|| \lt \delta\), we have \(||\boldsymbol{y}(t) - \boldsymbol{v}|| \lt \varepsilon\) for all \(t \ge 0\).

If \(\boldsymbol{v}\) is not stable, then we call it unstable.

Definition: Asymptotic Stability

We say that \(\boldsymbol{v}\) is asymptotically stable if it is both attractive and stable

Theorem: Attractivity \(\implies\) Stability with One Equation

Consider the autonomous system with just a single ordinary differential equation

\[y' = f(y)\]

and let \(v \in \mathbb{R}\) be a stationary point.

If \(v\) is attractive, then it is also stable.

Proof

TODO

Theorem: Attractivity \(\implies\) Stability with One Equation

Consider the autonomous system with just a single ordinary differential equation

\[y' = f(y)\]

and let \(v \in \mathbb{R}\) be a stationary point.

If \(v\) is attractive, then it is also stable.

Proof

TODO

Theorem

Consider the autonomous system

\[\boldsymbol{y}' = f(\boldsymbol{y})\]

with a continuously differentiable real vector field \(f: \mathbb{R}^n \to \mathbb{R}^n\) and a stationary point \(\boldsymbol{v} \in \mathbb{R}^n\). Let \(\lambda_1, \dotsc, \lambda_n \in \mathbb{C}\) be the eigenvalues of the Jacobian matrix \(\boldsymbol{J}_f(\boldsymbol{v})\) as a complex matrix.

If \(\operatorname{Re}(\lambda_i) \lt 0\) for all \(i \in \{1, \dotsc, n\}\), then \(\boldsymbol{v}\) is asymptotically stable.

If there exists some \(j \in \{1, \dotsc, n\}\) with \(\operatorname{Re}(\lambda_j) \gt 0\), then \(\boldsymbol{v}\) is unstable.

Example

Consider the following autonomous system:

\[\boldsymbol{y}' = \begin{bmatrix} y_1 + y_2 + 2 \\ y_2 - y_1^2 + 4 \end{bmatrix}\]

To find the stationary points, we solve

\[\begin{bmatrix} y_1 + y_2 + 2 \\ y_2 - y_1^2 + 4 \end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}\]

and obtain:

\[\boldsymbol{v} = \begin{bmatrix} -2 \\ 0 \end{bmatrix} \qquad \boldsymbol{w} = \begin{bmatrix} 1 \\ -3 \end{bmatrix}\]

We get the following Jacobian matrices:

\[\boldsymbol{J}_f (\boldsymbol{v}) = \begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix} \qquad \boldsymbol{J}_f (\boldsymbol{w}) = \begin{bmatrix} 1 & 1 \\ -2 & 1 \end{bmatrix}\]

We see that \(\boldsymbol{J}_f (\boldsymbol{v})\) has the eigenvalues \(-1\) and \(3\) and since \(\operatorname{Re}(3) = 3 \gt 0\), we know that \(\boldsymbol{v}\) must be unstable.

We see that \(\boldsymbol{J}_f (\boldsymbol{w})\) has the eigenvalues \(1 \pm \mathrm{i}\sqrt{2}\), and since \(\operatorname{Re}(1 + \mathrm{i}\sqrt{2}) = 1 \gt 0\), we know that \(\boldsymbol{w}\) must also be unstable.

Proof

TODO

Theorem

Consider the autonomous system

\[\boldsymbol{y}' = \boldsymbol{A}\boldsymbol{y} + \boldsymbol{b}\]

with a real matrix \(\boldsymbol{A} \in \mathbb{R}^{n \times n}\) and \(\boldsymbol{b} \in \mathbb{R}^n\) and let \(\boldsymbol{v} \in \mathbb{R}^n\) be a stationary point. Let \(\lambda_1, \dotsc, \lambda_n \in \mathbb{C}\) be the eigenvalues of \(\boldsymbol{A}\) as a complex matrix.

Then \(\boldsymbol{v}\) is asymptotically stable if and only if \(\operatorname{Re}(\lambda_i) \lt 0\) for all \(i \in \{1, \dotsc, n\}\).

If \(\operatorname{Re}(\lambda_j) \gt 0\) for some \(j \in \{1, \dotsc, n\}\), then \(\boldsymbol{v}\) is unstable.

Example

Consider the following autonomous system:

\[\boldsymbol{y}' = \underset{\boldsymbol{A}}{\underbrace{\begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix}}}\boldsymbol{y}\]

We see that \(\boldsymbol{v} = \boldsymbol{0}\) is a stationary point.

The eigenvalues of \(\boldsymbol{A}\) are \(-1\) and \(-2\). Therefore, \(\boldsymbol{v}\) is asymptotically stable.

Example

Consider the following autonomous system:

\[\boldsymbol{y}' = \underset{\boldsymbol{A}}{\underbrace{\begin{bmatrix} 1 & 0 \\ 0 & -2 \end{bmatrix}}}\boldsymbol{y}\]

We see that \(\boldsymbol{v} = \boldsymbol{0}\) is a stationary point.

The eigenvalues of \(\boldsymbol{A}\) are \(1\) and \(-2\). Therefore, \(\boldsymbol{v}\) is unstable, since \(\operatorname{Re}(1) = 1 \gt 0\).

Example

Consider the following autonomous system:

\[\boldsymbol{y}' = \underset{\boldsymbol{A}}{\underbrace{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}}}\boldsymbol{y}\]

We see that \(\boldsymbol{v} = \boldsymbol{0}\) is a stationary point.

The eigenvalues of \(\boldsymbol{A}\) are \(1\) and \(0\). Therefore, \(\boldsymbol{v}\) is unstable, since \(\operatorname{Re}(1) = 1 \gt 0\).

Proof

TODO