428
7 Variational Principles
= · · · =
n
k=0
(−1)
2k
x 1
x 0
p k (x)[y
(k)
]
2 dx =
n
k=0
x 1
x 0
p k (x)[y
(k)
]
2 dx (4)
Since p k (x) ∈ C
k
[x 0 , x 1 ] and p k (x) 0, (T y, y) 0, it can be known that T is
a positive operator, the found functional is
J 2n [y] = (T y, y) − 2(y, f ) =
x 1
x 0
n
k=0
p k (x)[y
(k)
]
2
− 2 f (x)y
dx
(5)
Example 7.7.3 Let the second differential equation be
−
d
dx
p(x)
du
dx
+ r (x)u − λu = 0
where, λ is a parameter, p(x) 0 and r (x) − λ 0 are both continuous functions.
The boundary conditions are
u
(x 0 ) − au(x 0 ) = 0, u
(x 1 ) + bu(x 1 ) = 0
where, a, b are constants and not zero. Find the variational expression of the
differential equation.
Solution The problem is the case equivalent to y = u, q = r − λ, α = γ = 1, x,
σ = b, A = B = 0 in Eq. (7.4.4) and the boundary conditions (7.7.3), from the
functional (7.7.20) the corresponding variational expression of the above problem
can immediately be written as
J [y] =
x 1
x 0
( pu
2
+ ru
2
− λu
2
)dx + ap(x 0 )u
2
(x 0 ) + bp(x 1 )u
2
(x 1 )
7.8 Variational Problems of Equivalence with Boundary
Value Problem of Self Conjugate Partial Differential
Equation
The variational problems of elliptic partial differential equations are discussed in this
section. The equation like
−∇ · ( p∇u) + qu = f (x, y, z ∈ V )
(7.8.1)
is called the general elliptic (differential) equation. It has the following three kinds
of boundary conditions