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IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 43, NO. 2, FEBRUARY 1998 261<br />

By using (21) of , (18) on M, and by imposing<br />

1<br />

sup x<br />

fr(x)g = 1 R (36)<br />

we get the following inequalities:<br />

jy 0 (x)j 1)<br />

jy0(x)j 1 aj([y 0 (x)])j (37)<br />

a<br />

1 jy 0 (x)j )<br />

jy0(x)j(jyj+j 6 (x)j+r(x))<br />

1 1 j([y 0 (x)])j (38)<br />

<br />

max<br />

<br />

R a ; 2 b<br />

<br />

a + b(jyj + j6 (x)j)<br />

2<br />

1j([y 0 (x)])j: (39)<br />

This yields, for all (x; y)<br />

jy 0 (x)j 1 max R<br />

a ; 2 M (y; 6 (x))<br />

b<br />

1j([y 0 (x)])j: (40)<br />

Also, we have<br />

minfR; jy 0 (x)jg R minf1; jy0(x)jg<br />

R j([y 0 (x)])j: (41)<br />

<br />

B. Proof of Theorem 1<br />

1) Proof of Global Stability: We propose a Lyapunov function<br />

W (x; y) which belongs to the family of Lyapunov functions described<br />

in [3] and is “flattened” inside the set A 0 , as proposed in<br />

[1]<br />

W (x; y) :=V(x)+ K(y; 6 (x)); when (x; y) 2 A 6<br />

0; when (x; y) 2 A 0 (42)<br />

where K is given by (14). One can verify that W is C 1 , positive<br />

definite, and proper. We next compute W _ in each of the two sets<br />

A 6 and A 0 .<br />

In the set A 0 we obtain, using A2)<br />

_W (x; y) =L f V(x)+L g V(x)y0(x): (43)<br />

Therefore, _W (x; y) is negative definite on A 0 , regardless of the<br />

value of the control variable u.<br />

In the set A 6 we obtain, with u = (x; y) given by (6)<br />

_W (x; y) 0(x)+T(x; y) 0 [y 0 6 (x)]<br />

1 M (y; 6 (x))([y 0 (x)]) (44)<br />

where M is from (17) and<br />

T (x; y) =[y0 6 (x)][L g V (x) +M(y; 6 (x))h(x; y)<br />

0 (a +2bj 6 (x)j)(L f 6 (x) +L g 6 (x)y)]:<br />

It remains to determine the negativeness of _W (x; y) on the set A 6 .<br />

<strong>For</strong> this we observe that by completing the squares and using (30),<br />

(34), and (35), we get, for all (x; y) 2 A 6<br />

T (x; y) 1 3 (x) +jy06 (x)j<br />

1[(c 2 1(c 0 + a + b) 2 + c 0 c 1 + c 2 2(a + b) 2<br />

+(c 1 +c 2 +c 3)(a + b))jy 0 (x)j<br />

+ c 2 M (y; 6 (x)) minfR; jy 0 (x)jg] (45)<br />

where c 3 is given by the boundedness of L g 6 and 6 . Then, with<br />

(40) and (41), we get more simply<br />

T (x; y) 1 3 (x) +jy06 (x)jM(y; 6 (x)) 1j([y 0 (x)])j<br />

1 1 <br />

[c 2 1(c 0 + a + b) 2 + c 0 c 1<br />

+ c 2 2(a + b) 2 +(c 1 +c 2 +c 3 )(a + b)]<br />

1 max R a ; 2 + c 2R : (46)<br />

b<br />

So by imposing that be large enough, we finally arrive at<br />

_W (x; y) 0 2 (x)0 1jy06 (x)j<br />

3 2<br />

1M(y; 6 (x))j([y 0 (x)])j (47)<br />

for all (x; y) 2 A 6 . This proves the negative definiteness of W _ (x; y)<br />

on A 6 .<br />

2) Construction of the Function : With the properties of the<br />

function defined in (32), we can construct a C 1 positive definite<br />

function (x; y) such that<br />

(x; y) 1 j([y 0 (x)])j; when (x; y) 2 A6<br />

4<br />

1<br />

4 (r(x)); when (x; y) 2 (48)<br />

A0<br />

and furthermore is constant outside some compact set. With this<br />

choice for we obtain<br />

where<br />

S(x; y) =<br />

ju 0 (x; y)j (x; y) =) _ W 0S(x; y) (49)<br />

2<br />

3 (x)+ a jy06 (x)j<br />

4<br />

1j([y 0 (x)])j; when (x; y) 2 A 6 (50)<br />

(x); when (x; y) 2 A 0 .<br />

3) Construction of the Functions U and : We define U as<br />

U (x; y) := ln[1 + W (x; y)] (51)<br />

where W is given by (42). We now show that there exists a function<br />

(x; y) such that<br />

• is C 1 and positive definite;<br />

• for all (x; y) 2 IR n 2 IR<br />

S(x; y)<br />

(x; y) <br />

1+W(x; y) ; (52)<br />

• there exist CIR n 2 IR, a compact neighborhood of the origin,<br />

and a constant >0such that for all (x; y) 2C<br />

(x; y) [(x) +(y0(x)) 2 ]: (53)<br />

First, because s 2 s(s) when jsj 1, and (33) holds, there is some<br />

compact neighborhood C of the origin such that for all (x; y) 2C<br />

j([y 0 (x)])j jy 0 (x)j;<br />

1<br />

8 ar(x)2 1 3 (x):<br />

(54)<br />

Let us now bound S(x; y) from below. We begin by observing from<br />

(54) that on A 6 \C<br />

a<br />

4 jy06 (x)jj([y 0 (x)])j<br />

1 4 a[(y 0 (x))2 0 r(x)jy 0 (x)j] (55)<br />

1 4 a[ 1 2 (y 0 (x))2 0 1 2 r(x)2 ] (56)<br />

1 8 a(y 0 (x))2 0 1 3 (x): (57)<br />

Furthermore, because jy 0 (x)j r(x)on A 0 , we have from (54)<br />

that<br />

1<br />

3 (x) 1 8 a(y 0 (x))2 (58)

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