Let L(x_{1}, x_{2}, .., x_{n}) 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: Letx*(t) = 0, t ｳ t_{0}be the zero solution of the homogeneous systemx｢=Axwherex(0) =x_{0}=0. Thenx*(t) is globally stable for t ｳ t_{0}if there exists L(x) with the following properties in some neighborhood N of0:(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) =Axis globally stable if and only if for some positive definite matrixW, the equation:

A｢ H+HA= -Whas 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 (d**x/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 d**x**/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, d**x**/dt is globally stable according to our previous theorem.

Conversely, if d**x**/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**. d**x**/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

or:

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.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｢ **)^{-1}**A**^{-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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