7.7 Variational Problems of Equivalence with Boundary Value Problem …
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(5) The Hermite equation
d
dx
(e
−x
2 y
) + 2ne
−x
2 y = 0
(7.7.10)
(6) The associated Laguerre equation
d
dx
(e
−x x
n+1 y
) +
a −
n + 1
2
e
−x x
n y = 0
(7.7.11)
If introducing the operator T , namely
T = −
d
dx
p(x)
d
dx
+ q(x)
(7.7.12)
where, the domain of definition of T is D, then Eq. (7.4.4) is the operator equation,
and it can be written as
T y = −
d
dx
[ p(x)y
] + q(x)y = f (x)
(7.7.13)
For Eq. (7.4.4), constructing the quadratic functional
J [y] = (T y, y) − 2(y, f ) =
x 1
x 0
[−( py
)
+ qy − 2 f ]ydx
(7.7.14)
The property of the operator T is discussed below. According to the boundary
conditions (7.7.3), let α = 0, γ = 0, using integration by pats, there is
(T y, z) =
x 1
x 0
[−( py
)
+ qy)]zdx =
x 1
x 0
( py
z
+ qyz)dx − py
z
x 1
x 0
=
x 1
x 0
( py
z
+ qyz)dx +
A + βy
α
pz
x=x 0
+
σ y − B
γ
pz
x=x 1
(7.7.15)
If A = B = 0, transposing y, z in the expression (7.7.15), the form on the right end
does not change, namely which is symmetric for y, z ∈ D. Hence (T y, z) = (y, T z),
namely T is a symmetric operator. And in the expression (7.7.15), put z = y and get
(T y, y) =
x 1
x 0
( py
2
+ qy
2
)dx +
β
α
p(x 0 )y
2
(x 0 ) +
σ
γ
p(x 1 )y
2
(x 1 )
(7.7.16)
According to the assumption p 0, q 0, α, β, γ , σ are nonnegative, therefore
(T y, y) 0, namely T is a positive definite operator.