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5 Variational Problems of Conditional Extrema
Fig. 5.2 Catenary
y
x
A (x 0 , y 0 )
0
B (x 1 , y 1 )
x 1
x 0
1 + y 2 dx =
x 1
x 0
cosh
x − c 2
c 1
dx = c 1
sinh
x 1 − c 2
c 1
− sinh
x 0 − c 2
c 1
= L
(5.13)
When L = 2a sinh 1 and y 0 (−a) = y 1 (a) = a cosh 1, it can solve for c 1 = a and
c 2 = λ = 0. Thus the catenary equation is
y = a cosh
x
a
(5.14)
Example 5.3.4 In a plane, in all smooth closed curves of length l, find the curve that
can round into the largest area.
Solution Because l is the closed curve, adopting the parameter form, let x = x(s),
y = y(s), there s is the arc length. The question comes down to under the condition
of the circumference
L =
s
0
x + y ds
(5.1)
find the extremum of the functional
J =
1
2
s
0
(x y
− x
y)ds
(5.2)
Making the auxiliary functional
J
∗
=
s
0
1
2
(x y
− x
y) + λ
x + y
ds =
s
0
H ds
(5.3)
where
H =
1
2
(x y
− x
y) + λ
x + y
(5.4)
5 Variational Problems of Conditional Extrema
Fig. 5.2 Catenary
y
x
A (x 0 , y 0 )
0
B (x 1 , y 1 )
x 1
x 0
1 + y 2 dx =
x 1
x 0
cosh
x − c 2
c 1
dx = c 1
sinh
x 1 − c 2
c 1
− sinh
x 0 − c 2
c 1
= L
(5.13)
When L = 2a sinh 1 and y 0 (−a) = y 1 (a) = a cosh 1, it can solve for c 1 = a and
c 2 = λ = 0. Thus the catenary equation is
y = a cosh
x
a
(5.14)
Example 5.3.4 In a plane, in all smooth closed curves of length l, find the curve that
can round into the largest area.
Solution Because l is the closed curve, adopting the parameter form, let x = x(s),
y = y(s), there s is the arc length. The question comes down to under the condition
of the circumference
L =
s
0
x + y ds
(5.1)
find the extremum of the functional
J =
1
2
s
0
(x y
− x
y)ds
(5.2)
Making the auxiliary functional
J
∗
=
s
0
1
2
(x y
− x
y) + λ
x + y
ds =
s
0
H ds
(5.3)
where
H =
1
2
(x y
− x
y) + λ
x + y
(5.4)