# (p.281) Appendix A Stability

# (p.281) Appendix A Stability

The study of system stability may be found in textbooks on differential equations (Gourmelin and Wadi, 2009) or dynamical systems (Strogatz 1994, Jean, 2011). For general aspects on differential calculus see (Cartan 1985).

# Differential equation

Let us consider a differential equation,

where $f:U\subset {\mathbb{R}}^{N}\to {\mathbb{R}}^{N}$ is a continuously differentiable function defined on an open set *U*. A solution $\mathbf{\text{Y}}:I\to U$ is a function differentiable on an interval $I\subset \mathbb{R}$ which verifies eqn (A.1) on *I*. A maximal solution is a solution defined on the largest possible interval. The Cauchy–Lipschitz theorem states the existence and uniqueness of a maximal solution for any initial condition $\mathbf{\text{Y}}(0)={\mathbf{\text{Y}}}_{0}\in U$. A point ${\mathbf{\text{Y}}}^{\ast}\in U$ is an equilibrium if $f({\mathbf{\text{Y}}}^{\ast})=0$.

The differential equation is said to be linear if *f* is a matrix **A**,

For all linear systems, the origin is an equilibrium (but not necessarily the only one). For any ${\mathbf{\text{Y}}}_{0}\in {\mathbb{R}}^{N}$, the unique solution of eqn (A.2) verifying $\mathbf{\text{Y}}(0)={\mathbf{\text{Y}}}_{0}$ is

and is defined on $I=\mathbb{R}$. Let us recall that the exponential of matrix **A** is defined by the series $\sum _{n=0}^{\mathrm{\infty}}{\mathbf{\text{A}}}^{n}/n!$.

Example A.1

In Chapter 2, we introduced the following

N-degrees-of-freedom system,(A.4)$$\mathbf{\text{M}}\ddot{\mathbf{\text{X}}}+(\mathbf{\text{C}}+\mathbf{\text{G}})\dot{\mathbf{\text{X}}}+\mathbf{\text{K}}\mathbf{\text{X}}=0$$where

M,K,C, andGare respectively the mass, stiffness, damping, and gyroscopic matrices. They are all real-valued andM,Kare symmetric and positive-definite, i.e. ${\mathbf{\text{M}}}^{T}=\mathbf{\text{M}},\text{}{\mathbf{\text{K}}}^{T}=\mathbf{\text{K}},\text{}\mathbf{\text{M}}0$, and $\mathbf{\text{K}}>0$. The gyroscopic matrix is antisymmetric ${\mathbf{\text{G}}}^{T}=-\mathbf{\text{G}}$ and $\mathbf{\text{C}}=\mathbf{\text{diag}}({c}_{1},\dots ,{c}_{N})$ is diagonal with non-negative entries ${c}_{i}\ge 0$ hence ${\mathbf{\text{C}}}^{T}=\mathbf{\text{C}}$. If all ${c}_{i}>0$ thenCis positive-definite but only semipositive when at least one ${c}_{i}=0$.The matrices

MandKare invertible since they are diagonalizable (real-valued and symmetric) and all their eigenvalues are non-zero.By introducing the state vector,

(A.5)$$\mathbf{\text{Y}}=\left(\begin{array}{l}\mathbf{\text{X}}\\ \dot{\mathbf{\text{X}}}\end{array}\right)$$which has values in ${\mathbb{R}}^{2N}$, the governing equation (A.4) takes the form (A.2) with

(A.6)$$A=\left(\begin{array}{cc}0& I\\ -{M}^{-1}K& -{M}^{-1}(C+G)\end{array}\right)$$All properties of system (A.4) are embodied in the matrix

A.

# (p.282) Stability

The stability of an equilibrium is defined in the sense of Lyapunov by the following.

Definition A.1

The equilibrium ${\mathbf{\text{Y}}}^{\ast}$ is stable if for any $\u03f5>0$ there exists $\mathrm{\delta}>0$

such that(A.7)$$\mathrm{i}\mathrm{f}\phantom{\rule{1em}{0ex}}\parallel \mathbf{\text{Y}}(0)-{\mathbf{\text{Y}}}^{\ast}\parallel <\mathrm{\delta}\phantom{\rule{1em}{0ex}}\text{then}\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\ge 0\phantom{\rule{1em}{0ex}}\parallel \mathbf{\text{Y}}(t)-{\mathbf{\text{Y}}}^{\ast}\parallel <\u03f5$$

Stability means that any solution starting in a neighbourhood of equilibrium remains indefinitely in a neighbourhood of equilibrium (Fig. A.1). A stronger definition is the asymptotic stability.

Definition A.2

The equilibrium ${\mathbf{\text{Y}}}^{\ast}$ is said to be attracting if there exists $\mathrm{\delta}>0$

such that(A.8)$$\text{if}\phantom{\rule{1em}{0ex}}\parallel \mathbf{\text{Y}}(0)-{\mathbf{\text{Y}}}^{\ast}\parallel <\mathrm{\delta}\phantom{\rule{1em}{0ex}}\text{then}\phantom{\rule{1em}{0ex}}\underset{t\to \mathrm{\infty}}{lim}\mathbf{\text{Y}}(t)={\mathbf{\text{Y}}}^{\ast}$$An equilibrium ${\mathbf{\text{Y}}}^{\ast}$

is asymptotically stable if it is both stable and attracting.

Asymptotic stability means that any solution starting in a neighbourhood of equilibrium tends to equilibrium by remaining arbitrarily close to it for all time (Fig. A.1). Asymptotic stability trivially implies stability but the converse is not true. An equilibrium may be attracting but not stable. An example of such a pathological case is given by the differential equation $\dot{\mathrm{\theta}}=1-cos\mathrm{\theta}$ and the equilibrium ${\mathrm{\theta}}^{\ast}=0$. For all initial conditions ${\mathrm{\theta}}_{0}\ne 0$, the solution is $\mathrm{\theta}(t)=2arctan\left[1/\left(1/tan({\mathrm{\theta}}_{0}/2)-t\right)\right]$. The point 0 is attracting since $lim\mathrm{\theta}(t)=0$ when $t\to \mathrm{\infty}$ for any ${\mathrm{\theta}}_{0}$. But 0 is not stable since $lim\mathrm{\theta}(t)=\mathrm{\pi}$ when $t\to 1/arctan({\mathrm{\theta}}_{0}/2)$. The solution cannot be confined to the neighbourhood of 0.

(p.283)
In the linear case, stability depends on the position of eigenvalues of **A** in the complex plane. The following theorem is established by discussing the solution (A.3).

Theorem A.1

Let us consider a matrix

Aand the linear differential equation (A.2). The origin is an equilibrium. We note ${\mathrm{\lambda}}_{i},\text{}i=1,\dots ,N$ the complex eigenvalues ofAandE_{i}the related eigenspaces. Then,

• the origin is stable if and only if $\text{Re}({\mathrm{\lambda}}_{i})\le 0$ for all

iand when $\text{Re}({\mathrm{\lambda}}_{i})=0$ the dimension ofE_{i}is equal to the multiplicity of ${\mathrm{\lambda}}_{i}$ in the characteristic polynomial; and• the origin is asymptotically stable if and only if $\text{Re}({\mathrm{\lambda}}_{i})<0$ for all

i.

Although this criterion only applies to linear systems, it may be also useful for non-linear systems by linearizing the differential equation about equilibrium. But the result is weakened. The following theorem is known as the first Lyapunov method.

Theorem A.2

Let $\mathbf{\text{D}}f$ be the Jacobian matrix of a continuously differentiable function $f:U\to {\mathbb{R}}^{N}$ at equilibrium ${\mathbf{\text{Y}}}^{\ast}$ and let ${\mathrm{\lambda}}_{i},\text{}i=1,\dots ,N$ be its complex eigenvalues. Then

• if $\text{Re}({\mathrm{\lambda}}_{i})<0$ for all

ithen ${\mathbf{\text{Y}}}^{\ast}$ is asymptotically stable; and• if $\text{Re}({\mathrm{\lambda}}_{i})>0$ for at least one

ithen ${\mathbf{\text{Y}}}^{\ast}$ is not stable.

When $\text{Re}({\mathrm{\lambda}}_{i})=0$ for one or more eigenvalues, one cannot conclude. For instance, the differential equation $\dot{y}=-{y}^{3}$ has $Df=0$ at ${y}^{\ast}=0$. Theorem 2 does not apply. However, the origin is asymptotically stable since the solution is $y(t)=\text{sgn}(\phantom{\rule{thinmathspace}{0ex}}{y}_{0})/\sqrt{2t+1/{y}_{0}^{2}}$ for $t\ge 0$ and goes to zero when $t\to \mathrm{\infty}$. Conversely, the differential equation $\dot{\mathrm{\theta}}=1-cos\mathrm{\theta}$ has also $Df=0$ at ${\mathrm{\theta}}^{\ast}=0$ but we have previously seen that 0 is not stable.

Example A.2

Let us discuss the stability of eqn (A.4) with respect to ${Y}^{\ast}=0$. We have seen that

Ais given by eqn (A.6). Stability is therefore related to the eigenvalues ofA, that is the roots of the characteristic polynomial,$$\begin{array}{rcl}det\left(\mathbf{\text{A}}-\mathrm{\lambda}\mathbf{\text{I}}\right)& =& det\left(-\mathrm{\lambda}\mathbf{\text{I}}\right)det\left(-{\mathbf{\text{M}}}^{-1}(\mathbf{\text{C}}+\mathbf{\text{G}})-\mathrm{\lambda}\mathbf{\text{I}}\right)-det\left(-{\mathbf{\text{M}}}^{-1}\mathbf{\text{K}}\right)det\left(\mathbf{\text{I}}\right)\\ & =& (det\mathbf{\text{M}}{)}^{-1}det\left({\mathrm{\lambda}}^{2}\mathbf{\text{M}}+\mathrm{\lambda}(\mathbf{\text{C}}+\mathbf{\text{G}})+\mathbf{\text{K}}\right)\end{array}$$Thus, the system stability depends on the position of zeros of the polynomial,

(A.9)$$\mathrm{\Delta}(\mathrm{\lambda})=det\left({\mathrm{\lambda}}^{2}\mathbf{\text{M}}+\mathrm{\lambda}(\mathbf{\text{C}}+\mathbf{\text{G}})+\mathbf{\text{K}}\right)$$with respect to the axis $\text{Re}(z)=0$ in the complex plane.

# (p.284) Lyapunov functions

To study stability in the general case, we introduce the concept of Lyapunov functions.

Definition A.3

A continuously differentiable function

Vdefined in an open setUsuch that ${Y}^{\ast}\in U\subset {\mathbb{R}}^{n}$ is said to be a Lyapunov function if

1. $V({\mathbf{\text{Y}}}^{\ast})=0$

2. $V(\mathbf{\text{Y}})>0$ for all $\mathbf{\text{Y}}\ne {\mathbf{\text{Y}}}^{\ast}$

3. $\dot{\mathbf{\text{V}}}=f.\mathrm{\nabla}V(\mathbf{\text{Y}})\le 0$

If, furthermore, $\dot{V}(\mathbf{\text{Y}})<0$ for all $\mathbf{\text{Y}}\ne {\mathbf{\text{Y}}}^{\ast}$ then

Vis a strict Lyapunov function.

In the above definition, $\mathrm{\nabla}V$ is the gradient of *V* defined in *U*. This is a vector oriented in the direction of increasing values of *V*. The velocity $\dot{\mathbf{\text{Y}}}=f(\mathbf{\text{Y}})$ is a vector tangential to the trajectory at point **Y**. The scalar product $f.\mathrm{\nabla}V$ is therefore the time derivative of *V* along trajectories. The third condition imposes that trajectories are oriented toward lower levels of *V* (Fig. A.2).

We may now enunciate Lyapunov’s second theorem on stability.

Theorem A.3

Let us consider the differential equation (A.1) and an equilibrium point ${\mathbf{\text{Y}}}^{\ast}$. If there exists a Lyapunov function in a neighbourhood of ${\mathbf{\text{Y}}}^{\ast}$ then ${\mathbf{\text{Y}}}^{\ast}$ is stable. If there exists a strict Lyapunov function in a neighbourhood of ${\mathbf{\text{Y}}}^{\ast}$ then ${\mathbf{\text{Y}}}^{\ast}$ is asymptotically stable.

A common application of Lyapunov functions is concerning the energy. The theorem then states that if the energy is non-increasing (no energy is injected), the equilibrium is stable. Furthermore, if the system dissipates *at all times*, the energy is a strict Lyapunov function and the equilibrium is asymptotically stable.

Example A.3

To establish the stability of system (A.4), we consider the Lyapunov function $V(Y)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times {Y}^{T}PY$ where

(A.10)$$\mathbf{\text{P}}=\left(\begin{array}{cc}\mathbf{\text{K}}& \mathbf{\text{0}}\\ \mathbf{\text{0}}& \mathbf{\text{M}}\end{array}\right)$$Since

MandKare symmetric and positive-definite,Pis also symmetric and positive-definite. The first two conditions of a Lyapunov function are therefore fulfilled. By developing the product ${\mathbf{\text{Y}}}^{T}\mathbf{\text{P}}\mathbf{\text{Y}}$, we obtain(A.11)$$V(\mathbf{\text{Y}})=\frac{1}{2}{\mathbf{\text{X}}}^{T}\mathbf{\text{K}}\mathbf{\text{X}}+\frac{1}{2}{\dot{\mathbf{\text{X}}}}^{T}\mathbf{\text{M}}\dot{\mathbf{\text{X}}}$$so that $V(\mathbf{\text{Y}})$ is simply the energy of system.

To check the third condition, we calculate $\dot{V}(\mathbf{\text{Y}})$:

(A.12)$$\dot{V}(\mathbf{\text{Y}})=\frac{1}{2}{\dot{\mathbf{\text{Y}}}}^{T}\mathbf{\text{P}}\mathbf{\text{Y}}+\frac{1}{2}{\mathbf{\text{Y}}}^{T}\mathbf{\text{P}}\dot{\mathbf{\text{Y}}}=\frac{1}{2}{\mathbf{\text{Y}}}^{T}\left({\mathbf{\text{A}}}^{T}\mathbf{\text{P}}+\mathbf{\text{P}}\mathbf{\text{A}}\right)\mathbf{\text{Y}}$$By denoting $Q=-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \left({A}^{T}P+PA\right)$, a mere calculation gives

(A.13)$$\mathbf{\text{Q}}=\left(\begin{array}{cc}\mathbf{\text{0}}& \mathbf{\text{0}}\\ \mathbf{\text{0}}& \mathbf{\text{C}}\end{array}\right)$$and $\dot{V}(\mathbf{\text{Y}})=-{\dot{\mathbf{\text{X}}}}^{T}\mathbf{\text{C}}\dot{\mathbf{\text{X}}}$. This equality is nothing other than the energy balance. Since

Cis semipositive, we obtain $\dot{V}(\mathbf{\text{Y}})\le 0$ and the third condition is also fulfilled. By applying the Lyapunov theorem, we conclude that system (A.4) is stable about the origin, that is all zeros of the polynomial $\mathrm{\Delta}(\mathrm{\lambda})$ are in the half-plane $\text{Re}(z)\le 0$. But, unfortunately, $\mathbf{\text{Q}}$ is not positive definite andVis not a strict Lyapunov function. Therefore, we cannot conclude asymptotic stability. To do that, we need a stronger result.

# (p.285) Invariance principle

We now introduce an important extension of Lyapunov’s second method. Standard references are (Krasovskii 1963), and (LaSalle 1960).

Definition A.4

Let us consider a differential equation $\dot{\mathbf{\text{Y}}}=f(\mathbf{\text{Y}})$ with an equilibrium ${\mathbf{\text{Y}}}^{\ast}\in U$. A set $\mathcal{M}$ is termed

if any solution starting from $\mathcal{M}$ is entirely in $\mathcal{M}$, i.e. $\mathbf{\text{Y}}(0)\in \mathcal{M}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathbf{\text{Y}}(t)\in \mathcal{M}$ for allinvariantt, and termedif $\mathbf{\text{Y}}(0)\in \mathcal{M}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathbf{\text{Y}}(t)\in \mathcal{M}$ for all $t\ge 0$.positively invariant

Theorem A.4

Let

Vbe a continuously differentiable function onUsuch that $\dot{V}(\mathbf{\text{Y}})\le 0$. LetSbe the set of all points inUsuch that $\dot{V}(\mathbf{\text{Y}})=0$. If $\mathcal{M}$ is the largest invariant set inSthen every solution $\mathbf{\text{Y}}(t)$ bounded for $t>0$ approaches $\mathcal{M}$ as $t\to \mathrm{\infty}$, i.e.\ $\underset{t\to \mathrm{\infty}}{lim}d(\mathbf{\text{Y}}(t),\mathcal{M})=0$.

(p.286)
Let us choose *V* as the energy of a dissipating system. If no energy is provided to the system, the function *V* is non-increasing and any trajectory tends to a state which does not dissipate.

Theorem A.5

Let

Vbe a Lyapunov function onUandΩa compact neighbourhood of ${\mathbf{\text{Y}}}^{\ast}$. IfΩis positively invariant and if the only solution $\mathbf{\text{Y}}(t)$ contained in the set $S=\left\{\mathbf{\text{Y}}\in U,\dot{V}(\mathbf{\text{Y}})=0\right\}$ is the trivial solution $\mathbf{\text{Y}}(t)={\mathbf{\text{Y}}}^{\ast}$, then ${\mathbf{\text{Y}}}^{\ast}$ is asymptotically stable. Furthermore all points inΩtend to ${\mathbf{\text{Y}}}^{\ast}$.

What is interesting about this theorem is that asymptotic stability is obtained with a non-strict Lyapunov function.

Finally, let us mention this last result.

Theorem A.6

Let

Vbe a Lyapunov function on ${\mathbb{R}}^{N}$ such that $\underset{\parallel \mathbf{\text{Y}}\parallel \to \mathrm{\infty}}{lim}V(\mathbf{\text{Y}})=\mathrm{\infty}$. If the only solution $\mathbf{\text{Y}}(t)$ contained in the set $S=\left\{\mathbf{\text{Y}}\in {\mathbb{R}}^{N},\dot{V}(\mathbf{\text{Y}})=0\right\}$ is the trivial solution $\mathbf{\text{Y}}(t)={\mathbf{\text{Y}}}^{\ast}$, then ${\mathbf{\text{Y}}}^{\ast}$ is globally asymptotically stable.

Globally asymptotic stability means that all solutions tend to ${\mathbf{\text{Y}}}^{\ast}$, not only those starting in a neighbourhood of ${\mathbf{\text{Y}}}^{\ast}$.

Example A.4

Let us consider the special case of system (A.4) when

Cis positive-definite (all resonators are damped). The condition $\dot{V}(\mathbf{\text{Y}})=-{\dot{\mathbf{\text{X}}}}^{T}\mathbf{\text{C}}\dot{\mathbf{\text{X}}}=0$ implies that $\dot{\mathbf{\text{X}}}=0$. Then, the setSreduces to $S=\left\{\mathbf{\text{Y}}=(\mathbf{\text{X}},0)\right\}$. A trajectory contained inSmust verify $\dot{\mathbf{\text{X}}}\equiv 0$ at any time. Therefore $\ddot{\mathbf{\text{X}}}\equiv 0$ and by substitution into eqn (A.4), we get $\mathbf{\text{X}}\equiv 0$ and finally $\mathbf{\text{Y}}\equiv 0$. Applying the invariance principle, we conclude that system (A.4) is asymptotically stable. In particular, the zeros ofΔare all in the half-plane $\text{Re}(z)<0$.

Example A.5

Let us examine an example with a semipositive damping matrix

C. The system is shown in Fig. A.3. The set of governing equations is(A.14)$$\{\begin{array}{l}{m}_{1}{\ddot{X}}_{1}+{c}_{1}{\dot{X}}_{1}+({k}_{1}+K){X}_{1}-K{X}_{2}=0\hfill \\ {m}_{2}{\ddot{X}}_{2}+({k}_{2}+K){X}_{2}-K{X}_{1}=0\hfill \end{array}$$The mass, stiffness, and damping matrices are

(A.15)$$M=\left(\begin{array}{cc}{m}_{1}& 0\\ 0& {m}_{2}\end{array}\right)\text{}K=\left(\begin{array}{cc}{k}_{1}+K& -K\\ -K& {k}_{2}+K\end{array}\right)\text{}C=\left(\begin{array}{cc}{c}_{1}& 0\\ 0& 0\end{array}\right)$$Matrix

Cis only semipositive because resonator 2 is undamped. The system energy is(A.16)$$V=\frac{1}{2}{m}_{1}{\dot{X}}_{1}^{2}+\frac{1}{2}{m}_{2}{\dot{X}}_{2}^{2}+\frac{1}{2}({k}_{1}+K){X}_{1}^{2}-K{X}_{1}{X}_{2}+\frac{1}{2}({k}_{2}+K){X}_{2}^{2}$$is positive-definite. The time derivative of the energy is

(A.17)$$\dot{V}=-{c}_{1}{\dot{X}}_{1}^{2}$$and is seminegative. Hence $S=\left\{Y=({X}_{1},{X}_{2},{\dot{X}}_{1},{\dot{X}}_{2})\in {\mathbb{R}}^{4}/{\dot{X}}_{1}=0\right\}$. If a solution ${X}_{1}(t),{X}_{2}(t)$ stays in

S, then ${\dot{X}}_{1}\equiv 0$ and therefore ${\ddot{X}}_{1}\equiv 0$. The first governing equation gives $({k}_{1}+K){X}_{1}-K{X}_{2}=0$. Furthermore,X_{1}being constant in time,X_{2}is also constant and therefore ${\ddot{X}}_{2}=0$. The second governing equation gives $-K{X}_{1}+({k}_{2}+K){X}_{2}=0$. In a matrix form this reads $\mathbf{\text{K}}X=0$ and sinceKis invertible, $X=0$. Applying the invariance principle, we conclude that the system is asymptotically stable and therefore all zeros of $\mathrm{\Delta}(\mathrm{\lambda})$ are in the half-plane $\text{Re}(z)<0$.A direct method to check this last result would be to expand

Δ,(A.18)$$\mathrm{\Delta}(\mathrm{\lambda})=\left[{m}_{1}{\mathrm{\lambda}}^{2}+{c}_{1}\mathrm{\lambda}+({k}_{1}+K)\right]\left[{m}_{2}{\mathrm{\lambda}}^{2}+({k}_{2}+K)\right]-{K}^{2}$$and to calculate the four complex roots. But this way is evidently more tedious.

(p.287) In most cases, the stability of a system reduces to localizing the roots of the characteristic polynomial $\mathrm{\Delta}(\mathrm{\lambda})$ of $\mathbf{\text{A}}$. A real polynomial $\mathrm{\Delta}(\mathrm{\lambda})\in \mathbb{R}[X]$ is said to be Hurwitz stable if all its roots are located in the left half-plane $\text{Re}(z)<0$. A first simple property of Hurwitz stable polynomials is the following.

Theorem A.7

Let $\mathrm{\Delta}(\mathrm{\lambda})\in \mathbb{R}[X]$. If $\mathrm{\Delta}(\mathrm{\lambda})$ is Hurwitz stable then all its coefficients have the same sign.

This property is easy to prove by remarking that since $\mathrm{\Delta}(\mathrm{\lambda})\in \mathbb{R}[X]$ its complex roots appear in pairs ${\text{\lambda}}_{j},\text{}{\overline{\lambda}}_{j}$. The factorized form of $\mathrm{\Delta}(\mathrm{\lambda})$,

(p.288)
where *α* is the leading coefficient of $\mathrm{\Delta}(\mathrm{\lambda})$, shows a product of polynomials whose coefficients are all positive. Developing can only give non-negative coefficients.

The converse is generally false. For instance, the polynomial ${\mathrm{\lambda}}^{3}+{\mathrm{\lambda}}^{2}+\mathrm{\lambda}+6=(\mathrm{\lambda}+2)({\mathrm{\lambda}}^{2}-\mathrm{\lambda}+3)$ has positive coefficients but two complex roots in the half-plane $\text{Re}(z)>0$.

The Routh–Hurwitz algorithm provides a criterion to recognize a Hurwitz stable polynomial.

Theorem A.8

Given ${a}_{n}{\mathrm{\lambda}}^{n}+{a}_{n\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}1}{\mathrm{\lambda}}^{n\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}1}+\dots +{a}_{0}$ where

a_{i}are real coefficients, we construct the Routh array,$$\begin{array}{llll}{a}_{n}\phantom{\rule{thinmathspace}{0ex}}& {a}_{n-2}\phantom{\rule{thinmathspace}{0ex}}& {a}_{n-4}\phantom{\rule{thinmathspace}{0ex}}& \dots \\ {a}_{n-1}\phantom{\rule{thinmathspace}{0ex}}& {a}_{n-3}\phantom{\rule{thinmathspace}{0ex}}& {a}_{n-5}\phantom{\rule{thinmathspace}{0ex}}& \dots \\ {l}_{n-2,1}\phantom{\rule{thinmathspace}{0ex}}& {l}_{n-2,2}\phantom{\rule{thinmathspace}{0ex}}& {l}_{n-2,3}\phantom{\rule{thinmathspace}{0ex}}& \dots \\ \vdots \\ {l}_{0,1}\end{array}$$completed by zeros on the right. The third and subsequent rows are calculated by the recursive rule,

$${l}_{k,i}=\left[{l}_{k+1,i}{l}_{k+2,i+1}-{l}_{k+2,1}{l}_{k+1,i+1}\right]/{l}_{k+1,1}$$If all coefficients in the left column are positive (respectively negative), then $\mathrm{\Delta}(\mathrm{\lambda})$ is Hurwitz stable.

We can now revisit the last example.

Example A.6

The characteristic polynomial (A.18), $\mathrm{\Delta}(\mathrm{\lambda})\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{m}_{1}{m}_{2}{\mathrm{\lambda}}^{4}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{m}_{2}{c}_{1}{\mathrm{\lambda}}^{3}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\left[{m}_{1}({k}_{2}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}K)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{m}_{2}({k}_{1}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}K)\right]{\mathrm{\lambda}}^{2}+{c}_{1}({k}_{2}+K)\mathrm{\lambda}+{k}_{1}{k}_{2}+K({k}_{1}+{k}_{2})$ leads to the Routh array,

$$\begin{array}{llllll}{m}_{1}{m}_{2}\hfill & {m}_{1}({k}_{2}+K)+{m}_{2}({k}_{1}+K)\hfill & \hfill & {k}_{1}{k}_{2}+K({k}_{1}+{k}_{2})\hfill & \hfill & 0\hfill \\ {m}_{2}{c}_{1}\hfill & {c}_{1}({k}_{2}+K)\hfill & \hfill & 0\hfill & \hfill & 0\hfill \\ {m}_{2}({k}_{1}+K)\hfill & {k}_{1}{k}_{2}+K({k}_{1}+{k}_{2})\hfill & \hfill & 0\hfill & \hfill & 0\hfill \\ {c}_{1}{K}^{2}/({k}_{1}+K)\hfill & 0\hfill & \hfill & 0\hfill & \hfill & 0\hfill \\ {k}_{1}{k}_{2}+K({k}_{1}+{k}_{2})\hfill & 0\hfill & \hfill & 0\hfill & \hfill & 0\hfill \end{array}\text{}$$All coefficients of the left column are positive. The polynomial is Routh stable and the system is asymptotically stable.