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Nonlinear Control Sy.. - Free

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A. 7. CHAPTER 10 331<br />

Equations (A.62) and (A.63) can be rewritten as follows:<br />

By the Jabobi identity we have that<br />

or<br />

Similarly<br />

LfTi=LfTi+1, (A.64)<br />

L9T1 = L9T2 = = L9Tn_1 = 0 L9Tn # 0 (A.65)<br />

VT1[f,g] _ V(L9T1)f -V(LfT1)g<br />

0-L9T2=0<br />

VTladfg = 0.<br />

OT1adfg = 0 (A.66)<br />

VT1ad f-1g # 0. (A.67)<br />

We now claim that (A.66)-(A.67) imply that the vector fields g, adf g, , adf-1g are linearly<br />

independent. To see this, we use a contradiction argument. Assume that (A.66)<br />

(A.67) are satisfied but the g, adfg, , adf-1g are not all linearly independent. Then, for<br />

some i < n - 1, there exist scalar functions Al (X), A2(x), , .i_1(x) such that<br />

and then<br />

i-1<br />

adfg = Ak adfg<br />

k=0<br />

n-2<br />

adf-19= adfg<br />

k=n-i-1<br />

and taking account of (A.66). we conclude that<br />

n-2<br />

OT1adf 1g = > Ak VT1 adf g = 0<br />

k=n-i-l<br />

which contradicts (A.67). This proves that (i) is satisfied. To prove that the second property<br />

is satisfied notice that (A.66) can be written as follows<br />

VT1 [g(x), adfg(x), ... , adf 2g(x)] = 0 (A.68)<br />

that is, there exist T1 whose partial derivatives satisfy (A.68). Hence A is completely<br />

integrable and must be involutive by the Frobenius theorem.

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