370
6 Variational Problems in Parametric Forms
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F x −
d
dt
F ˙
x = ( ˙
x F x ˙
x + ˙
y F x ˙
y ) − ( ˙
x F x ˙
x + ˙
y F y ˙
x + ¨
x F ˙
x ˙
x + ¨
y F ˙
x ˙
y ) =
˙
y[F x ˙
y − F ˙
x y + ( ˙
x ¨
y − ¨
x ˙
y)F 1 ] = 0
F y −
d
dt
F ˙
y = ( ˙
x F y ˙
x + ˙
y F y ˙
y ) − ( ˙
x F x ˙
y + ˙
y F y ˙
y + ¨
x F ˙
x ˙
y + ¨
y F ˙
y ˙
y ) =
− ˙
x[F x ˙
y − F ˙
x y + ( ˙
x ¨
y − ¨
x ˙
y)F 1 ] = 0
(6.2.7)
Because ˙
x and ˙
y can not be zero simultaneously, Eqs. (6.2.1) are equivalent to the
following equation
F x ˙
y − F ˙
x y + ( ˙
x ¨
y − ¨
x ˙
y)F 1 = 0
(6.2.8)
Equation (6.2.8) is called the Weierstrass form of Euler equation.
The derivative of y with respect to x is written in the form of parameters, there are
y
=
dy
dx
=
dy
dt
dx
dt
=
˙
y
˙
x
(6.2.9)
y
=
d
2 y
dx 2 =
˙
x ¨
y
dt
dx
− ˙
y ¨
x
dt
dx
˙
x 2
=
˙
x ¨
y − ˙
y ¨
x
˙
x 3
(6.2.10)
In addition, the curvature radius R of the extremal curve can be represented as
−
1
R
=
y
(1 + y 2 )
3
2
=
˙
x ¨
y − ˙
y ¨
x
( ˙
x 2 + ˙
y 2 )
3
2
(6.2.11)
Substituting Eq. (6.2.11) into Eq. (6.2.8), then R can be written as
−
1
R
=
F ˙
x y − F x ˙
y
( ˙
x 2 + ˙
y 2 )
3
2 F 1
(6.2.12)
Equation (6.2.12) is also called the Weierstrass form of Euler equation.
Example 6.2.1 In all closed curves surrounding a certain area, find the curve that
the length is minimum.
Solution Let x = x(t), y = y(t), t 0 t t 1 be the equation of any closed curve.
In the Green formula, taking P = −y, Q = x, the area of the domain D can be given
A =
¨
D
dxdy =
1
2
Γ
(xdy − ydx) =
1
2
t 1
t 0
(x ˙
y − ˙
x y)dt
(1)
After removing the constant
1
2
in the functional (1), which does not affect the
property of the problem discussed.
The length of the curve is