Routh–Hurwitz stability criterion

Remarks

The rows of $H$ are $(a_1, a_3, a_5, \ldots)$, $(a_0, a_2, a_4, \ldots)$, $(0, a_1, a_3, \ldots)$, $(0, a_0, a_2, \ldots)$, and so on. One proof counts the roots of $p$ in the right half-plane by following the argument of $p(i\omega)$ as $\omega$ runs over $\R$. Routh's array computes the same information by a Euclidean algorithm on the even and odd parts of $p$.

For $n = 2$ and $p(s) = s^2 + a_1 s + a_2$, the conditions are $a_1 > 0$ and $a_1 a_2 > 0$, that is, $a_1 > 0$ and $a_2 > 0$. For $n = 3$, they are $a_1 > 0$, $a_1 a_2 - a_0 a_3 > 0$ and $a_3 > 0$, since $\Delta_3 = a_3 \Delta_2$. Positive coefficients are necessary but not sufficient when $n \ge 3$. For $p(s) = s^3 + s^2 + s + 1 = (s + 1)(s^2 + 1)$, all coefficients are positive, but $\Delta_2 = 1 - 1 = 0$, and indeed $\pm i$ are roots.

Going deeper

For a student

Applied to the characteristic polynomial of a real matrix $A$, the criterion tests whether $A$ is Hurwitz-stable. With the linearisation of a nonlinear system, it gives a test for the stability of an equilibrium that uses only arithmetic on the entries of the Jacobian matrix. It is the standard tool in control engineering for stability with respect to parameters, since the conditions are explicit inequalities. The analogous test for roots inside the unit disc, used for discrete-time systems, is the Schur–Cohn or Jury test.