Stability of Linear Systems

For linear systems with constant coefficients, stability is completely determined by the eigenvalues of the coefficient matrix. This provides a complete and computable stability theory.

Eigenvalue Criterion

Theorem7.3Stability criterion for linear systems

For the system $\mathbf{x}' = A\mathbf{x}$ with constant matrix $A$ :

The origin is asymptotically stable if and only if all eigenvalues have $\text{Re}(\lambda_j) < 0$ .
The origin is stable (not asymptotically) if and only if all eigenvalues satisfy $\text{Re}(\lambda_j) \leq 0$ and those with $\text{Re}(\lambda_j) = 0$ are semisimple (geometric multiplicity equals algebraic multiplicity).
The origin is unstable if any eigenvalue has $\text{Re}(\lambda_j) > 0$ , or if an eigenvalue with $\text{Re}(\lambda_j) = 0$ is not semisimple.

ExampleStability analysis

$A = \begin{pmatrix}-1 & 1\\0 & -2\end{pmatrix}$ : eigenvalues $-1, -2$ , both negative. Asymptotically stable.

$A = \begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$ : eigenvalues $\pm i$ , pure imaginary, semisimple. Stable (center), not asymptotically stable.

$A = \begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}$ : eigenvalue $0$ with algebraic multiplicity $2$ , geometric multiplicity $1$ . Unstable ( $x(t) = x_0 + y_0 t$ grows).

The Routh-Hurwitz Criterion

Definition7.3Hurwitz matrix

For a polynomial $p(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_0$ , the Hurwitz matrix is constructed from the coefficients. All roots have negative real parts if and only if all leading principal minors of the Hurwitz matrix are positive.

Theorem7.4Routh-Hurwitz criterion

For a polynomial $p(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_0$ , all roots satisfy $\text{Re}(\lambda) < 0$ if and only if:

$n = 1$ : $a_0 > 0$ .
$n = 2$ : $a_1 > 0$ and $a_0 > 0$ .
$n = 3$ : $a_2 > 0$ , $a_0 > 0$ , and $a_2 a_1 > a_0$ .

ExampleApplying Routh-Hurwitz

Is $\lambda^3 + 3\lambda^2 + 4\lambda + 2 = 0$ stable? Check: $a_2 = 3 > 0$ , $a_0 = 2 > 0$ , and $a_2 a_1 = 12 > 2 = a_0$ . Yes, all roots have $\text{Re}(\lambda) < 0$ .

Is $\lambda^3 + \lambda^2 + 2\lambda + 8 = 0$ stable? $a_2 a_1 = 2 < 8 = a_0$ . No, the system is unstable.

Exponential Stability

Definition7.4Exponential stability

The equilibrium $\mathbf{x}_0 = \mathbf{0}$ is exponentially stable if there exist $C, \alpha > 0$ such that $\|\mathbf{x}(t)\| \leq Ce^{-\alpha t}\|\mathbf{x}(0)\|$ for all $t \geq 0$ and all initial conditions.

RemarkLinear systems and exponential stability

For constant-coefficient linear systems, asymptotic stability and exponential stability are equivalent. The decay rate $\alpha$ can be taken as any number less than $\min_j(-\text{Re}(\lambda_j))$ . For nonautonomous or nonlinear systems, the distinction matters.