[ALGEBRA] Stability of systems

This page uses the roots of complex polynomials to describe how linear systems evolve. A discrete system is built from modes $r^n$ and a continuous system from modes $e^{rt}$ , where $r$ is a characteristic root; a zero root needs separate treatment. The modulus $|r|$ decides whether a discrete mode decays and the real part $\operatorname{Re}(r)$ decides the same for a continuous mode.

In the mode calculations, $r$ denotes a characteristic root, not the polar radius used in polar form.

General discrete systems

[DEFINITION]

Complex polynomial

A complex polynomial is a function $p:\mathbb C\to\mathbb C$ of the form

p(z)=a_0+a_1z+a_2z^2+\cdots+a_nz^n, \qquad a_0,\ldots,a_n\in\mathbb C,

for a non-negative integer $n$ .

[DEFINITION]

Polar form

For $z\ne0$ , the polar form of $z$ is

z=r(\cos\theta+i\sin\theta)=re^{i\theta}, \qquad r=|z|>0,

where $\theta$ is an argument of $z$ . The expressions $a+bi$ and $re^{i\theta}$ are two writings of the same complex number.

[DEFINITION]

Discrete-time system

A discrete-time system describes values $x_n$ at times $n=0,1,2,\ldots$ through a recurrence

x_{n+1}=f(n,x_n,x_{n-1},\ldots).

A first-order system uses $x_{n+1}=f(n,x_n)$ and needs one initial value; a second-order system uses $x_{n+1}=f(n,x_n,x_{n-1})$ and needs two.

The simplest linear case is the first-order recurrence

x_{n+1}=a x_n.

Repeated substitution gives $x_n=a^n x_0$ for $n\ge1$ , where $x_0$ is the initial value. So $|a|<1$ gives decay, $|a|>1$ gives growth and $|a|=1$ gives boundary behaviour, such as a constant or oscillating mode that does not decay. Compound interest is the case $a=1+r$ , so its long-term behaviour depends only on the powers of one number.

Nonlinear recurrences fall outside the characteristic-root method.

[EXAMPLE]

A nonlinear non-example: logistic iteration

For the parameterised logistic recurrence

x_{n+1}=r x_n(1-x_n),\qquad x_0=\frac13,

the first terms for $r=2$ are

x_1=\frac49,\qquad x_2=2\left(\frac49\right)\left(1-\frac49\right) =\frac{40}{81}.

The factor $1-x_n$ depends on the current value, so this recurrence is not linear and has no characteristic polynomial of the kind used below. The delayed logistic recurrence

x_{n+1}=rx_n(1-x_{n-1}),\qquad x_0=\frac15,\quad x_1=\frac14

requires both $x_0$ and $x_1$ before it can be iterated.

[routine]

[1]:Discrete-time system

Linear recurrences and characteristic modes

[DEFINITION]

Discrete linear system

A homogeneous second-order discrete linear system has the form

a x_{n+1}+b x_n+c x_{n-1}=0, \qquad n\ge1.

Here $a,b,c\in\mathbb C$ , $a\ne0$ and the coefficients are constant, so a characteristic-root calculation can replace the recurrence by an algebraic equation.

[METHOD]

Find discrete modes from a characteristic equation

For $a\ne0$ and

a x_{n+1}+b x_n+c x_{n-1}=0,

try a non-trivial solution $x_n=K r^n$ with $K\ne0$ and $r\ne0$ . Substitution gives

K r^{n-1}(ar^2+br+c)=0.

For this nonzero mode, the prefactor is nonzero, so its characteristic equation is

ar^2+br+c=0.

The zero-root case, if it occurs, is not covered by this $r^{n-1}$ factorisation and must be checked directly in the recurrence. When the non-zero roots $r_1,r_2$ are distinct, the general solution is

x_n=K_1r_1^n+K_2r_2^n.

If there is a repeated non-zero root $r$ , the general solution is instead

x_n=(K_1+K_2n)r^n.

The expression $ar^2+br+c$ is a complex polynomial. Its non-zero roots are exactly the numbers for which the proposed exponential mode satisfies the substitution. The factor $n$ in the repeated-root solution matters for stability: even when $|r|=1$ , the term $nr^n$ is unbounded.

If $c=0$ , then zero is a characteristic root and the recurrence reduces to $a x_{n+1}+b x_n=0$ for $n\ge1$ . Here $x_0$ and $x_1$ are independent initial values. For $b\ne0$ , the later terms are $x_n=x_1(-b/a)^{n-1}$ for $n\ge1$ ; for $b=0$ , all terms from $x_2$ onward are zero. This handles the zero-root case without using $0^0$ .

Discrete stability

[DEFINITION]

Discrete stability

For the linear systems considered here, a system is stable when every solution satisfies $|x_n|\to0$ as $n\to\infty$ and unstable when some solution is unbounded.

Here stable means asymptotically stable. A system whose solutions stay bounded but do not all decay is a boundary case, neither stable nor unstable under this convention.

Informally, “stable” sometimes means only “bounded”. The stricter convention stops a unit-modulus mode from being mistaken for a decaying one.

[THEOREM]

Discrete stability from root moduli

Write a characteristic root in polar form as

r_j=\rho_j e^{i\theta_j}.

Then

|r_j^n|=\rho_j^n.

Thus $\rho_j<1$ gives a decaying mode, $\rho_j>1$ gives an unbounded mode and $\rho_j=1$ gives a non-decaying mode.

For the whole system, every solution decays precisely when all characteristic roots satisfy $|r_j|<1$ . A root with $|r_j|>1$ supplies an unbounded solution. If all roots satisfy $|r_j|\le1$ and a unit-modulus root is present, its multiplicity decides the remaining case: when each such root is simple, every solution is bounded but some do not decay; a repeated unit-modulus root instead produces an unbounded term $nr_j^n$ .

[EXAMPLE]

Test a recurrence

Consider

x_{n+1}=x_n+x_{n-1}.

The characteristic equation is

r^2-r-1=0,

whose roots are

r_1=\frac{1+\sqrt5}{2}, \qquad r_2=\frac{1-\sqrt5}{2}.

Since $|r_1|>1$ , the corresponding mode grows. The system is unstable. The second root has modulus less than $1$ , but one growing mode is enough: choosing initial data with a non-zero coefficient of $r_1^n$ produces an unbounded solution.

[routine]

[1]:Discrete stability from root moduli

This page keeps the mode calculation scalar. For vector states, eigenvalues take over the role of characteristic roots once a matrix system is diagonalised: in matrix powers, in Markov systems and, in continuous time, in linear ODE systems. The equations differ: a recurrence gives a polynomial in $r$ directly, while a matrix $A$ uses $p_A(\lambda)=\det(A-\lambda I)$ in the eigenvalue test.

Continuous systems and characteristic modes

This section can be read now or deferred until after second-order constant-coefficient equations in Calculus. The trial-function calculation it needs is stated below.

[DEFINITION]

Continuous-time system

A continuous-time system describes a function $x(t)$ through a differential equation with $t\in\mathbb R$ . For $a,b,c\in\mathbb C$ with $a\ne0$ , consider the second-order homogeneous equation

a x''(t)+b x'(t)+cx(t)=0.

The exponential trial function $x(t)=Ke^{rt}$ leads to the algebraic equation

ar^2+br+c=0.

[1]:Find continuous modes from a characteristic equation

[METHOD]

Find continuous modes from a characteristic equation

For distinct roots $r_1,r_2$ of $ar^2+br+c=0$ , the solution has the form

x(t)=K_1e^{r_1t}+K_2e^{r_2t}.

For a repeated root $r$ , the solution has the form

x(t)=(K_1+K_2t)e^{rt}.

If $r=\alpha+i\beta$ , then

e^{rt}=e^{\alpha t}e^{i\beta t}.

The factor $e^{\alpha t}$ controls growth or decay, while $e^{i\beta t}$ supplies oscillation. When the coefficients and required solution are real, a conjugate pair $\alpha\pm i\beta$ (with $\beta\ne0$ ) combines into

e^{\alpha t}\bigl(A\cos(\beta t)+B\sin(\beta t)\bigr), \qquad A,B\in\mathbb R.

The constant-coefficient ODE method in Calculus is the same calculation, with repeated roots treated in more detail. The damped oscillator below shows each case.

[1]:Continuous-time system
[2]:Complex polynomial
[3]:Continuous stability from real parts

Continuous stability

[DEFINITION]

Continuous stability

For the linear systems considered here, a continuous system is stable when every solution decays to zero as $t\to\infty$ and unstable when some solution is unbounded.

As in the discrete case, this is asymptotic stability; bounded non-decaying solutions form a boundary case.

[1]:Continuous stability from real parts

[THEOREM]

Continuous stability from real parts

For a characteristic root $r=\alpha+i\beta$ ,

|e^{rt}|=e^{\alpha t}.

Therefore $\alpha<0$ gives decay, $\alpha>0$ gives an unbounded mode and, for a simple root, $\alpha=0$ gives a non-decaying oscillatory or constant mode. A repeated root on the imaginary axis introduces a term $t e^{rt}$ , whose magnitude is unbounded.

For the whole system, every solution decays precisely when all characteristic roots have negative real part. A root with positive real part supplies an unbounded solution. If no root has positive real part but a root lies on the imaginary axis and each such root is simple, every solution is bounded but some do not decay. A repeated root there gives an unbounded solution.

[EXAMPLE]

Classify a damped oscillator

For

m x''+c x'+kx=0,

the roots satisfy

mr^2+cr+k=0.

Take $m=k=1$ and vary the damping $c$ .

If $c>2$ , the roots are distinct negative real numbers: the motion is overdamped and decays without oscillating.
If $c=2$ , the repeated root is $r=-1$ , so $(K_1+K_2t)e^{-t}\to0$ : the motion is critically damped and decays.
If $0<c<2$ , the roots are a complex conjugate pair with negative real part: the motion is underdamped and oscillates while its amplitude decays.
If $c=0$ , the roots are $r=\pm i$ , so the motion is periodic rather than decaying.
If $-2<c<0$ , the conjugate roots have positive real part $-c/2$ , so oscillations grow and the system is unstable.

[routine]

[1]:Continuous stability from real parts

Stability of systems

Complex polynomial

Polar form

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