336
5 Variational Problems of Conditional Extrema
Making the auxiliary functional
J
∗
=
+∞
−∞
H dx =
+∞
−∞
{− p(x) ln[kp(x)] + λ 1 p(x) + λ 1 x
2 p(x)}dx
(5.4)
The Euler equation is
− ln[kp(x)] − 1 + λ 1 + λ 2 x
2
= 0
(5.5)
Solving for p(x) from Eq. (5), we give
p(x) =
1
k
e
λ 1 −1+λ 2 x
2
(5.6)
Substituting the expression (6) into both integrals of the expression (2) respectively
we give
+∞
−∞
p(x)dx =
1
k
e
λ 1 −1
+∞
−∞
e
λ 2 x
2 dx = 1
(5.7)
+∞
−∞
x
2 p(x)dx =
1
k
e
λ 1 −1
+∞
−∞
x
2 e
λ 2 x
2 dx =
−1
2kλ 2
e
λ 1 −1
+∞
−∞
e
λ 2 x
2 dx = σ
2
(5.8)
Solving for λ 2 from the expression (7) and the expression (8), we get
λ 2 = −
1
2σ 2
(5.9)
Substituting the expression (9) into the last integral of the expression (8), we
obtain
+∞
−∞
e
λ 2 x
2 dx =
+∞
−∞
e
−
1
2σ 2 x
2
dx
(5.10)
Let ξ =
x
√
2σ
, there is dx =
√
2σ dξ , and when x → −∞, ξ → −∞, when
x → +∞, ξ → +∞, thus the expression (10) may be written as
+∞
−∞
e
λ 2 x
2 dx =
+∞
−∞
e
−
1
2σ 2 x
2
dx =
√
2σ
+∞
−∞
e
−ξ
2 dξ
(5.11)
Let I (a) =
a
0 e
−ξ
2 dξ , where a is any finite positive number, thus
[I (a)]
2
=
a
0
e
−ξ
2 dξ
a
0
e
−η
2 dη =
¨
D
e
−(ξ
2 +η
2 ) dξ dη
(5.12)