5.3 Isoperimetric Problems
339
Making the auxiliary functional
J
∗
[y] =
x 1
x 0
(y
1 + y 2 + λ
1 + y 2 )dx
(5.5)
Since H = y
1 + y 2 + λ
1 + y 2 does not contain x, there exists the first
integral
(y + λ)
1 + y 2 − y
(y + λ)y
1 + y 2
= c 1
(5.6)
Simplifying Eq. (6), we get
y + λ = c 1
1 + y 2
(5.7)
Let y
= sinh t, substituting it into the Eq. (7), we obtain
y = c 1
1 + sinh
2 t − λ = c 1 cosh t − λ
(5.8)
Differentiating the expression (8), there is
dx =
dy
y =
c 1 sinh tdt
sinh t
= c 1 dt
(5.9)
Integrating Eq. (9), we obtain
x = c 1 t + c 2
(5.10)
Making use of the expression (10), t in the expression (8) of y is eliminated, we
get
y = c 1 cosh
x − c 2
c 1
− λ
(5.11)
This is the general equation of a catenary, as shown in Fig. 5.2. Where, The
arbitrary constants c 1 , c 2 and λ can be determined by the boundary conditions
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
y 0 = c 1 cosh
x 0 − c 2
c 1
− λ
y 1 = c 1 cosh
x 1 − c 2
c 1
− λ
(5.12)
and the constraint condition
339
Making the auxiliary functional
J
∗
[y] =
x 1
x 0
(y
1 + y 2 + λ
1 + y 2 )dx
(5.5)
Since H = y
1 + y 2 + λ
1 + y 2 does not contain x, there exists the first
integral
(y + λ)
1 + y 2 − y
(y + λ)y
1 + y 2
= c 1
(5.6)
Simplifying Eq. (6), we get
y + λ = c 1
1 + y 2
(5.7)
Let y
= sinh t, substituting it into the Eq. (7), we obtain
y = c 1
1 + sinh
2 t − λ = c 1 cosh t − λ
(5.8)
Differentiating the expression (8), there is
dx =
dy
y =
c 1 sinh tdt
sinh t
= c 1 dt
(5.9)
Integrating Eq. (9), we obtain
x = c 1 t + c 2
(5.10)
Making use of the expression (10), t in the expression (8) of y is eliminated, we
get
y = c 1 cosh
x − c 2
c 1
− λ
(5.11)
This is the general equation of a catenary, as shown in Fig. 5.2. Where, The
arbitrary constants c 1 , c 2 and λ can be determined by the boundary conditions
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
y 0 = c 1 cosh
x 0 − c 2
c 1
− λ
y 1 = c 1 cosh
x 1 − c 2
c 1
− λ
(5.12)
and the constraint condition