672 Optimal Control and Optimality Criteria Methods
Figure 12.2 Curve of minimum time of
descent.
Since potential energy is converted to kinetic energy as the particle moves down the
path, we can write
1
2 mν
(^2) = mgx
Hence
dt=
[
1 +(y′)^2
2 gx
] 1 / 2
dx (E 1 )
and the integral to be stationary is
t=
∫xB
0
[
1 +(y′)^2
2 gx
] 1 / 2
dx (E 2 )
The integrand is a function ofxandy′and so is a special case of Eq. (12.1). Using
the Euler–Lagrange equation,
d
dx
(
∂F
∂y′
)
−
∂F
∂y
=0 withF=
[
1 +(y′)^2
2 gx
] 1 / 2
we obtain
d
dx
(
y′
{x[1+(y′)^2 ]}^1 /^2
)
= 0
Integratingyields
y′=
dy
dx
=
(
C 1 x
1 −C 1 x
) 1 / 2
(E 3 )
whereC 1 is a constant of integration. The ordinary differential equation (E 3 ) yields on
integration the solution to the problem as
y(x)=C 1 sin−^1 (x/C 1 )−( 2 C 1 x−x^2 )^1 /^2 +C 2 (E 4 )
Example 12.2 Design of a Solid Body of Revolution for Minimum Drag Next we
consider the problem of determining the shape of a solid body of revolution for mini-
mum drag. In the general case, the forces exerted on a solid body translating in a fluid