6.3 Extrema of Functionals with Variable Boundaries and Parametric Forms
381
xϕ x + yϕ y = 0
( 4 )
˙
yϕ x − ˙
xϕ y = 0
( 5 )
Solving for ϕ x from Eq. (5) and substituting it into Eq. (4), we give
y + x
˙
x
˙
y
ϕ y = 0
( 6 )
But ϕ y is arbitrary, thus the transversality condition is
x ˙
x + y ˙
y = 0
( 7 )
Because ϕ(x, y) = 0 is arbitrary, the above equation holds for any variable
boundary.
Problems 6
6.1 Verify whether the values of the following functionals have nothing to do with
the parameter form of the admissible curve.
(1) J [x, y, z] =
t 1
t 0
˙
x 2 + ˙
y 2 + ˙
z 2 dt;
(2) J [x, y] =
t 1
t 0
x ˙
x ˙
y 2 dt;
(3) J [x, y] =
t 1
t 0
(x
2
˙
x + 3x y ˙
y
2
)dt;
(4) J [x, y] =
t 1
t 0
(x ˙
y + y ˙
x
2
)dt;
(5) J [x, y] =
t 1
t 0
(x ˙
y + y ˙
x)
2 dt;
(6) J [x, y] =
t 1
t 0
x ˙
y 2 + y ˙
x 2 dt.
6.2 Prove: The area of the plane figure surrounded by the closed curve Γ
A =
1
2
Γ
(xdy − ydx)
has nothing to do with the parameter form of the curve Γ .
6.3 Find the extremal curve of the functional J [x, y] =
(x 1 ,y 1 )
(0,0)
˙
y
2 −y
2 ˙
x
2
˙
x
dt.
6.4 Find the extremal curve of the functional J [x, y] =
(1,2)
(0,0)
˙
y
2 −3e
˙
y
˙
x ˙
x
2
˙
x
dt.
6.5 Find the extremal curve of the functional
J [x, y]
=
(1,0)
(0,1) (x ˙
y − y ˙
x − 2
˙
x 2 + ˙
y 2 )dt.
6.6 Find the extremal curve of the functional J [x, y] =
π
4
0 ( ˙
x ˙
y + 2x
2
+ 2y
2
)dt,
the boundary conditions are x(0) = y(0) = 0, x(1) = y(1) = 1.