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 ｹ 0 in N and L(0) = 0.
Theorem: Let x*(t) = 0, t ｳ t0 be the zero solution of the homogeneous system x｢ = Ax where x(0) = x0 = 0. Then x*(t) is globally stable for t ｳ 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 ﾌ 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.ｧ
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
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.ｧ
Thus, we have proven that a real n ｴ 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.