g(x(tx
n+1(t) = const and xn+1(t1) =
1)): (4.15)
t1
Using the conditions above we get the following:
t1Z
g(x(t1)) = [xn+1(t)]dt: (4.16)
t0
Hence we could reformulate the cost function by merging both integrals into one
t1Z
C(u) = [xn+1(t) + f0(x(t);u(t))]dt: (4.17)
t0
We have a new state which is n + 1 dimensional and the terminal set X1will be given as
 g(x(t
X 1))1 = (x1;x2;:::;xn;xn+1) : xn+1(t) = : (4.18)t
1
We will write it in the form
X1 = f(x1;x2;:::;xn;xn+1) : (x1;x2;:::;xn;xn+1) = 0g; (4.19)
where
(x1;x2;:::;xn;xn+1) = g(x(t1))txn+1: (4.20)
From equation (4.15), we have
(t1) = (1(t1);:::;n(t1);n+1(t1))
= r (x1(t1);:::;xn(t1);xn+1(t1)
 @g @g
=  ;:::; ;t@x
1 @x
1 = 0:
n
20
Page 1  |  Page 2  |  Page 3  |  Page 4  |  Page 5  |  Page 6  |  Page 7  |  Page 8  |  Page 9  |  Page 10  |  Page 11  |  Page 12  |  Page 13  |  Page 14  |  Page 15  |  Page 16  |  Page 17  |  Page 18  |  Page 19  |  Page 20  |  Page 21  |  Page 22  |  Page 23  |  Page 24  |  Page 25  |  Page 26  |  Page 27  |  Page 28  |  Page 29  |  Page 30  |  Page 31  |  Page 32  |  Page 33  |  Page 34  |  Page 35  |  Page 36  |  Page 37  |  Page 38  |  Page 39  |  Page 40  |  Page 41  |  Page 42  |  Page 43  |  Page 44  |  Page 45  |  Page 46  |  Page 47  |  Page 48  |  Page 49  |  Page 50  |  Page 51  |  Page 52  |  Page 53  |  Page 54  |  Page 55  |  Page 56