346
5 Variational Problems of Conditional Extrema
the moment δy 0 and δy 1 both are arbitrary, if the functional obtains extremum, then
the coefficients of them must be zero, namely
(F y + Φ y 1 )
x=x 1
= 0, (F y − Φ y 0 )
x=x 0
= 0
(5.4.4)
When boundary points (x 0 , y 0 ) and (x 1 , y 1 ) variate respectively on the given
curves y 0 = ϕ(x 0 ) and y 1 = ψ(x 1 ), there are
δy 0 = ϕ
(x 0 )δx 0 , δy 1 = ψ
(x 1 )δx 1
At the moment the expression (5.4.3) can be written as
[F + (ψ
− y
)F y + Φ x1 + Φ y1 ψ
]
x=x1
δx 1 − [F + (ϕ
− y
)F y − Φ x0 − Φ y0 ϕ
]
x=x0
δx 0 = 0 (5.4.5)
Because δx 0 and δx 1 are arbitrary, the coefficients of them should be zero, namely
[F + (ϕ
− y
)F y − Φ x 0 − Φ y 0 ϕ
]
x=x 0
= 0
[F + (ψ
− y
)F y + Φ x 1 + Φ y 1 ψ
]
x=x 1
= 0
(5.4.6)
These are the transversality conditions which variable boundary points should
satisfy. It is thus clear that the natural boundary conditions are the particular cases
of the transversality conditions.
For the functional like
J =
x 1
x 0
F(x, y, y
, y
)dx + Φ(x 0 , y 0 , x 1 , y 1 )
(5.4.7)
using a similar method above, the Euler-Poisson equation can be obtained
F y −
d
dx
F y +
d
2
dx 2 F y = 0
(5.4.8)
The endpoint condition is
F − y
F y −
d
dx
F y
− y
F y + Φ x 1
x=x 1
δx 1 +
F y −
d
dx
F y + Φ y 1
x=x 1
δy 1 + F y
x=x 1
δy
1 = 0
(5.4.9)
and
F − y
F y −
d
dx
F y
− y
F y − Φ x 0
x=x 0
δx 0 +
F y −
d
dx
F y − Φ y 0
x=x 0
δy 0 + F y
x=x 0
δy
0 = 0
(5.4.10)