338
5 Variational Problems of Conditional Extrema
+∞
−∞
e
−ξ
2 dξ =
√ π
(5.18)
Substituting the expression (18) into the expression (11), then substituting the
expression (9) and the expression (11) into the expression (8), to yield
e
λ 1 −1
=
k
√
2πσ
(5.19)
Substituting the expression (9) and the expression (19) into the expression (100),
the density of the probability distribution is
p(x) =
1
√
2πσ
e
−
x 2
2σ 2
(5.20)
It conforms to the normal distribution.
Example 5.3.3 Find a suspension line of the barycenter being the lowest, which
passes through fixed points A, B, the length is L, let the linear density ρ be a constant.
And find the equation of the curve when L = 2a sinh 1 and y 0 (−a) = y 1 (a) =
a cosh 1.
Solution Let the ordinate of the suspension line be y(x), and pass through two points
A(x 0 , y 0 ), B(x 1 , y 1 ), the length of the suspension line is
L =
x 1
x 0
1 + y 2 dx
(5.1)
When the suspension line is in the equilibrium position, its barycenter should be
at the lowest position, it can be represented as
y c =
1
Lρ
s 1
s 0
ρyds =
1
L
s 1
s 0
yds =
1
L
x 1
x 0
y
1 + y 2 dx
(5.2)
Or to consider the problem from the aspects of gravity doing work. The work
done for gravity is
W =
s 1
s 0
ρgyds =
x 1
x 0
ρgy
1 + y 2 dx = ρg
x 1
x 0
y
1 + y 2 dx
(5.3)
Because L, ρ and g are all constants, they have not impact on the variation of the
functional, the work done for the barycenter and gravity can be substituted by the
following functional
J [y] =
x 1
x 0
y
1 + y 2 dx
(5.4)