464
8 Direct Methods of Variational Problems
y n (x) =
n
k=1
a k (1 − x)x
k
Taking n = 1, 2 to calculate the approximate value. First of all consider the case
of n = 1, at this point we have
y 1 (x) = a 1 (1 − x)x
Substituting the above expression into the functional, we give
J [a 1 ] =
1
0
[a
2
1 (1 − 2x)
2
− a
2
1 (1 − x)
2
− 2a 1 (1 − x)x
2
]dx =
3
10
a
2
1 −
1
6
a 1
Putting
∂ J
∂a 1
= 0, namely
3
5
a 1 −
1
6
= 0
we get
a 1 =
5
18
Therefore there is
y 1 (x) =
5
18
(1 − x)x
Then consider the case of n = 2, at the point let
y 2 (x) = a 1 (1 − x)x + a 2 (1 − x)x
2
= x(1 − x)(a 1 + a 2 x)
Substituting the above expression into the functional, and putting
∂ J
∂a 1
= 0,
∂ J
∂a 2
=
0, we get the simultaneous equations
⎧
⎪ ⎨
⎪ ⎩
3
10
a 1 +
3
20
a 2 =
1
12
3
20
a 1 +
13
105
a 2 =
1
20
Solving the equations, we obtain
a 1 =
71
369
, a 2 =
7
41