Engineering Optimization: Theory and Practice, Fourth Edition

680 Optimal Control and Optimality Criteria Methods

is a function of the two variablesxandu, we can write the Euler–Lagrange equations

[withu 1 =x,u′ 1 = ∂x/∂t= ̇x,u 2 = uandu′ 2 = ∂u/∂t= ̇uin Eq. (12.10)] as

∂F

∂x

−

d

dt

(

∂F

∂x ̇

)

= 0 (12.28)

∂F

∂u

−

d

dt

(

∂F

∂u ̇

)

= 0 (12.29)

In view of relation (12.27), Eqs. (12.28) and (12.29) can be expressed as

∂f 0

∂x

+λ

∂f

∂x

+ ̇λ= 0 (12.30)

∂f

∂u

+λ

∂f

∂u

= 0 (12.31)

A new functionalH, called theHamiltonian, is defined as

H=f 0 + λf (12.32)

and Eqs. (12.30) and (12.31) can be rewritten as

−

∂H

∂x

=λ ̇ (12.33)

∂H

∂u

= 0 (12.34)

Equations (12.33) and (12.34) represent two first-order differential equations. The inte-

gration of these equations leads to two constants whose values can be found from the

known boundary conditions of the problem. If two boundary conditions are specified

asx(0)= k 1 and x(T )=k 2 , the two integration constants can be evaluated without

any difficulty. On the other hand, if only one boundary condition is specified as, say,

x(0)= k 1 , the free-end condition is used as∂F /∂x ̇=0 orλ=0 att=T.

Example 12.4Find the optimal controluthat makes the functional

J=

∫ 1

0

(x^2 +u^2 ) dt (E 1 )

stationary with

x ̇=u (E 2 )

andx(0)=1. Note that the value ofxis not specified att=1.

SOLUTION The Hamiltonian can be expressed as

H=f 0 + λu=x^2 +u^2 + λu (E 3 )

and Eqs. (12.33) and (12.34) give

− 2 x=λ ̇ (E 4 )

2 u+λ= 0 (E 5 )