424
7 Variational Principles
The expression (7.7.15) and expression (7.7.16) are obtained in the cases of α = 0,
γ = 0. If α = γ = 0, the terms with α and γ do not occur in the expressions, at this
time whether or not A and B are zero, T is a symmetric positive definite operator.
To sum up, when A = B = 0 or α = γ = 0, the self-conjugate differential
operator T given by the problem is a symmetric positive definite operator. Thus
Eq. (7.4.4) is the Euler equation of the quadratic functional
J [y] =
x 1
x 0
( py
2
+ qy
2
− 2y f )dx
(7.7.17)
When α = 0, γ = 0, taking the first and second variations to the functional
(7.7.17), there are
δ J [y] = 2
x 1
x 0
( py
δy
+ qyδy − f δy)dx = 2
x 1
x 0
( py
dδy + qyδydx − f δydx)
= 2 py
δy
x 1
x 0
+ 2
x 1
x 0
−
d
dx
( py
) + qy − f
δydx
(7.7.18)
δ
2 J [y] = 2
x 1
x 0
[ p(δy
)
2
+ q(δy)
2
]dx 0
(7.7.19)
If y = y(x) is the solution of the differential (7.7.2), then at y = y(x 0 ) and y =
y(x 1 ) there are δ J = 2 p(x 0 )y
(x 0 )δy(x 0 ), δ J = 2 p(x 1 )y
(x 1 )δy(x 1 ) respectively,
using the boundary conditions (7.7.3), we get
δ J = 2 p(x 1 )δy(x 1 )
B − σ y(x 1 )
γ
− 2 p(x 0 )δy(x 0 )
A + βy(x 0 )
α
At this point, δ J usually is not equal to zero, the functional (7.7.17) is not the
required functional, it would need to be modified. Considering the functional
J B = −p(x 1 )
2By(x 1 ) − σ y
2
(x 1 )
γ
+ p(x 0 )
2 Ay(x 0 ) + βy
2
(x 0 )
α
The first variation is
δ J B = −p(x 1 )
2Bδy(x 1 ) − 2σ y(x 1 )δy(x 1 )
γ
+ p(x 0 )
2 Aδy(x 0 ) + 2βy(x 0 )δy(x 0 )
α
= −2 p(x 1 )δy(x 1 )
B − σ y(x 1 )
γ
+ 2 p(x 0 )δy(x 0 )
A + βy(x 0 )
α
= −δ J
The second variation is
δ
2 J B = −p(x 1 )δ
2 2By(x 1 ) − σ y
2
(x 1 )
γ
+ p(x 0 )δ
2 2 Ay(x 0 ) + βy
2
(x 0 )
α