7.7 Variational Problems of Equivalence with Boundary Value Problem …
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or
(e
p 1
p 0
dx y
)
+
p 2
p 0
e
p 1
p 0
dx y = 0,
Putting
p = e
p 1
p 0
dx , q =
p 2
p 0
e
p 1
p 0
dx
then we yield
( py
)
+ qy = 0
Example 7.7.2 Let self-conjugate ordinary differential equation be
n
k=0
(−1)
k d
k
dx k [ p k (x)y
(k)
] = f (x)
(1)
The boundary conditions are
y
(k)
(x 0 ) = y
(k)
(x 1 ) = 0 (k = 0, 1, · · · , n − 1)
(2)
Try to establish the equivalent functional of the boundary value problem.
Solution Putting T y =
n
k=0 (−1)
k d
k
dx k [ p k (x)y
(k)
] = f (x), then there is
(T y, y) =
n
k=0
(−1)
k
x 1
x 0
y
d
k
dx k [ p k (x)y
(k)
]dx
(3)
Applying integration by parts, and taking note that the boundary condition given
in the problem, we obtained
(T y, y) =
n
k=0
(−1)
k
x 1
x 0
y
d
k
dx k [ p k (x)y
(k)
]dx
=
n
k=0
(−1)
k
d
k−1
dx k−1 [ p k (x)y
(k)
]y
x 1
x 0
−
x 1
x 0
y
d
k−1
dx k−1 [ p k (x)y
(k)
]dx
=
n
k=0
(−1)
k
−
d
k−2
dx k−2 [ p k (x)y
(k)
]y
x 1
x 0
+
x 1
x 0
y
d
k−2
dx k−2 [ p k (x)y
(k)
]dx
= · · · =
n
k=0
(−1)
k
(−1)
i
x 1
x 0
y
(i) d
k−i
dx k−i [ p k (x)y
(k)
]dx