5.3 Isoperimetric Problems
341
The various partial derivatives of H are
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ H
∂ x
=
1
2
y
,
∂ H
∂ y
= −
1
2
x
∂ H
∂ x = −
1
2
y +
λx
x 2 + y 2
∂ H
∂ y =
1
2
x +
λy
x 2 + y 2
(5.5)
Since the parameter s is the arc length, there is x
2
+ y
2
= 1, Eq. (5) can be
written as
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ H
∂ x
=
1
2
y
,
∂ H
∂ y
= −
1
2
x
∂ H
∂ x = −
1
2
y + λx
∂ H
∂ y =
1
2
x + λy
(5.6)
The Euler equation is
y
− λx
= 0
x
+ λy
= 0
(5.7)
Integrating once, we get
y − λx
= c 1
x + λy
= c 2
(5.8)
Eliminating λ in Eq. (8), we get
(x − c 2 )dx + (y − c 1 )dy = 0
(5.9)
Integrating Eq. (9), to yield
(x − c 2 )
2
+ (y − c 1 )
2
= c
2
3
(5.10)
This is a class of circles, the radii are c 3 , the centers of the circle are (c 2 ,c 1 ). It
can be verified that the Lagrange multiplier λ is the radii c 3 . In fact, eliminating c 1 ,
c 2 from the Eqs. (8) and (10), to yield
(x − c 2 )
2
+ (y − c 1 )
2
= λ
2
(x
2
+ y
2
) = λ
2
= c
2
3
(5.11)
therefore λ = c 3 .