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ELEC4631 Chap.5 Lyapunov Stability II: The Lyapunov Equation and Linear Stability Tests

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Chapter 5 of 13 · ELEC4631

Lyapunov Stability II: The Lyapunov Equation and Linear Stability Tests

For x-dot=Ax, asymptotic stability has exact spectral and quadratic-certificate characterisations. A is Hurwitz when every eigenvalue lies in the open left half-plane; then the origin is globally asymptotically stable. Axis eigenvalues create the subtle case: Lyapunov stability requires no right-half-plane eigenvalues and semisimple Jordan structure for every imaginary-axis eigenvalue. Internal stability must be distinguished from BIBO stability because a cancelled uncontrollable or unobservable mode may disappear from the transfer function. A quadratic V=x^TPx has derivative x^T(A^TP+PA)x. For Hurwitz A and any symmetric Q>0, the Lyapunov equation A^TP+PA=-Q has a unique symmetric P>0. Linearisation carries the same spectral language to nonlinear equilibria, but only locally and with an inconclusive boundary when the Jacobian has critical axis eigenvalues.

In this chapter

What this chapter covers

  • 01Hurwitz and LTI asymptotic stability
  • 02Axis eigenvalues and semisimple multiplicity
  • 03Internal versus BIBO stability
  • 04Quadratic Lyapunov functions
  • 05Solving A^TP+PA=-Q
  • 06First indirect method for nonlinear systems
  • 07Decay-rate Lyapunov inequalities
Worked example · free

Solve and verify a two-state Lyapunov equation

Q [6 marks]. AskSia-authored practice weighting: 6 marks. Let A=diag(-1,-2) and Q=I. Solve A^TP+PA=-Q for symmetric P=[[p11,p12],[p12,p22]], verify P>0 and conclude stability.
  • expandSubstitution gives [[-2p11,-3p12],[-3p12,-4p22]]=-I.
  • entryMatch the (1,1) entry: p11=1/2.
  • entryMatch off-diagonal entries: p12=0.
  • entryMatch the (2,2) entry: p22=1/4.
  • certificateP=diag(1/2,1/4) is symmetric positive definite because both diagonal eigenvalues are positive.
  • conclusionV-dot=-x^Tx<0 for x!=0; hence the LTI origin is globally asymptotically stable.
P=diag(1/2,1/4). It is positive definite and satisfies the full residual A^TP+PA+I=0. Therefore V=x^TPx is a global quadratic Lyapunov function and the origin is globally asymptotically stable.
Sia tip — Solve only independent symmetric entries, but verify the complete matrix equation afterwards.
Glossary

Key terms

Hurwitz
Every eigenvalue of A has strictly negative real part; equivalent to asymptotic stability of continuous-time x-dot=Ax.
Semisimple eigenvalue
An eigenvalue whose geometric multiplicity equals algebraic multiplicity, so its Jordan blocks have size one.
Internal stability
Stability of every state trajectory under the internal realisation, including modes that may be hidden from the transfer function.
BIBO stability
Every bounded input produces bounded output under the zero-state input-output map; it can coexist with a hidden unstable internal mode.
Lyapunov equation
A^TP+PA=-Q. For Hurwitz A and Q>0 it has a unique P>0.
First indirect method
Use the Jacobian spectrum at a nonlinear equilibrium for a local conclusion; critical axis eigenvalues make the test inconclusive.
FAQ

Lyapunov Stability II: The Lyapunov Equation and Linear Stability Tests FAQ

Are imaginary-axis eigenvalues always unstable?

No. With no RHP eigenvalues, semisimple imaginary-axis modes can yield Lyapunov stability but not asymptotic decay. A nontrivial Jordan block on the axis creates polynomial growth and instability. State multiplicity conditions rather than using the location alone.

Why can a stable transfer function hide an unstable state?

The transfer function contains only modes reachable from the input and visible at the output. An uncontrollable or unobservable unstable eigenvalue can cancel algebraically. BIBO stability of the reduced input-output map therefore does not certify internal stability of a nonminimal realisation.

May I choose any Q in the Lyapunov equation?

For the standard Hurwitz theorem, choose any symmetric Q>0; the resulting unique solution P is positive definite. A convenient Q such as I simplifies arithmetic. If A is not Hurwitz, a positive Q will not yield the required positive-definite P certificate.

What does linearisation prove?

A Hurwitz Jacobian proves local asymptotic stability of the nonlinear equilibrium; an RHP eigenvalue proves instability. Axis eigenvalues are inconclusive because higher-order nonlinear terms may dominate. The result is local, unlike the global LTI statement for a fixed Hurwitz A.

How do spectral and Lyapunov-equation answers differ?

A spectrum question can be answered quickly once every eigenvalue and any axis multiplicity are known. It classifies the fixed LTI dynamics but does not produce an explicit energy geometry. A Lyapunov-equation question asks for a symmetric P, its positive-definiteness evidence, the matrix residual and the resulting V-dot. It produces a reusable certificate that later supports decay rates and convex synthesis. Do not mix their steps: solving eigenvalues does not substitute for the requested P, and a numerical P without verifying A^TP+PA=-Q does not complete the certificate. When both are available, use the spectrum to predict whether positive P should exist and use the residual to catch arithmetic. Explain why an LTI Hurwitz conclusion is global, whereas a Jacobian result transplanted to a nonlinear model is local.

What is the safest imaginary-axis classification routine?

List every eigenvalue with algebraic and geometric multiplicity. If any real part is positive, the equilibrium is unstable. If all real parts are negative, it is asymptotically stable. If none is positive and at least one lies on the axis, check that every axis eigenvalue is semisimple. Semisimple axis modes can remain bounded and give Lyapunov stability, but not asymptotic decay; a defective axis mode creates polynomial growth and instability. Zero is an imaginary-axis eigenvalue and follows the same rule. Do not call the system marginally stable from a root plot alone. Connect the multiplicity test to the Jordan form or to the appearance of t e^{j omega t}, because that is the actual unbounded mechanism.

How can I create an independent check without software dependence?

For a two-state A, compare eigenvalue sum and product with trace and determinant. After solving a Lyapunov equation, use principal minors to establish P>0 and multiply A^TP+PA by hand to recover -Q. For a linearisation, substitute the equilibrium before taking the Jacobian and state which higher-order terms were discarded. Software can then test eigenvalues or residuals on a fresh case, but the written proof remains complete without it. Asking an AI tutor to perturb A while preserving Hurwitz stability is useful: predict how P changes qualitatively, solve the new symmetric equations and request a residual check only after your result is fixed.

Study strategy

Exam move

Drill the continuous-time spectrum classification as a complete sentence, including the imaginary-axis multiplicity clause. Then solve Lyapunov equations with a symmetric unknown and a simple Q, checking P through principal minors or eigenvalues and checking the full residual. Pair each transfer-function problem with a question about hidden modes and minimality so internal and BIBO language never collapse. For nonlinear systems, compute the equilibrium and Jacobian first, then label the indirect-method outcome as local or inconclusive. Use a rate inequality only after the base stability certificate is clear; the additional term strengthens the derivative bound rather than replacing positive definiteness of P.

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