Lyapunov's Method

Blake's Newton


Let L(x1, x2, .., xn) be a scalar function of n components of x. L(x) is positive definite in a neighborhood N of the origin if L(x) > 0 for all x 0 in N and L(0) = 0.

Theorem: Let x*(t) = 0, t t0 be the zero solution of the homogeneous system x = Ax where x(0) = x0 = 0. Then x*(t) is globally stable for t t0 if there exists L(x) with the following properties in some neighborhood N of 0:

(i) L(x) and its partial derivatives are continuous.

(ii) L(x) is positive definite, or L(x) > 0.

(iii) dL(x)/dt is negative definite, or dL(x)/dt < 0.

Proof: By (ii) the quadratic form L(x) exhibits an ellipsoid curve. By (iii), the ellipsoid curve shrinks to zero. Choose e > 0 such that Ne N above. Any half-path starting in Ne remains in it because L(x) is a quadrative form (by (ii)) which exhibits an ellipsoid curve that is continuous as well as its partial derivatives (by (i)). The same holds for every sufficiently small e > 0 and hence for every sufficiently small neighborhood of the origin. The zero solution is therefore globally stable.

Theorem: (Lyapunov's Second Method) The system (dx/dt) = Ax is globally stable if and only if for some positive definite matrix W, the equation:

A H + HA = -W

has a positive definite matrix H.

Proof: If for some positive definite matrix W, the equation A H + HA = -W has a positive definite matrix H, let us show that (dx/dt) = Ax is globally stable. Since H is positive definite, then L(x) = x Hx is positive definite (where x is now the transpose of x and not the time derivative), i.e. L(x) > 0. Also, L(x) positive definite implies that V(x) and its partial derivatives are continuous. Differentiating L(x), then:

dL(x)/dt = (dx /dt)Hx + x H(dx/dt)

or, as dx/dt = Ax:

dL(x)/dt = (Ax) Hx + x HAx

= x A Hx + x HAx

= x (A H + HA)x

thus, as A H + HA = -W:

dL(x)/dt = x (-W)x

W positive definite implies -W is negative definite, thus:

dL(x)/dt = x (-W)x < 0

Finally, we can note that (i) L(x) and its partial derivatives are continuous; (ii) V(x) is positive definite; (iii) dL(x)/dt is negative definite. As a result, dx/dt is globally stable according to our previous theorem.

Conversely, if dx/dt = Ax is stable, let us show that for some positive definite matrix W, the equation A H + HA = -W has a positive definite matrix H. dx/dt = Ax stable implies all the eigenvalues of A are negative, i.e. l < 0 for any eigenvalue l of A. Now, as l x = Ax, then (Ax) = (l x) , which implies x A = l x . Thus, premultiplying A H + AH by x and postmultiplying it by x, we obtain:

x (A H + HA)x = x (-W)x


x A Hx + x HAx = x (-W)x

or substiting in l x and l x:

l x Hx + x Hl x = x (-W)x

or simply:

2l x Hx = x (-W)x

As -W is negative definite, then x (-W)x < 0, thus 2l x Hx < 0. As l < 0 by the assumption of stability, then it must be that x Hx > 0, or H is a positive definite matrix.

Thus, we have proven that a real n n matrix A is a stable matrix if and only if there exists a symmetric positive definite matrix H such that A H + HA is negative definite. In practice, we can choose W = I and solve for H in the equation A H + HA = -I. The solution has the form H = a (A )-1A-1 + b I where a and b are constants. Thus, choosing a Lyapunov function, L(x) = x Hx, we can use this solution to determine H.

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