Appendix: All Solutions to the Problems
883
Substituting them into the above functional, we get
J =
2
1
xa
2
(4x
2
− 12x + 9) −
x
2
− 1
x
a
2
(x
4
− 6x
3
+ 13x
2
− 12x + 4)
−2x
2 a(x
2
− 3x + 2)
dx = a
2
(2.3 − 4 ln 2) +
23
30
a
Deriving a and putting it be zero, we get
∂ J
∂a
= −2a(2.3 − 4 ln 2) +
23
30
= 0
Solve for a
a =
23
60(2.3 − 4 ln 2)
= −0.811135169
Therefore, the approximate solution can be obtained
y =
23(x − 1)(x − 2)
60(2.3 − 4 ln 2)
= −0.811135169(x − 1)(x − 2)
The integrand is F = x y
2
−
x
2 −1
x
y
2
− 2x
2 y, the Euler equation of the functional
is
−2x y
− 2y
− 2
x
2
− 1
x
y − 2x
2
= 0
or
x
2 y
+ x y
+ (x
2
− 1)y + x
3
= 0
or
y
+
y
x
+
1 −
1
x 2
y + x = 0
The solution of the Euler equation is y = 3.6072J 1 (x)+0.75195N 1 (x)−x, where
J 1 and N 1 are the first-order Bessel function and the first-order Neumann function.
8.6 Using the Ritz method to find the approximate solution of the minimum problem
of the functional J [y] =
2
0 (y
2
+ y
2
+ 2x y)dx, and do a comparison with the
accurate solution, the boundary conditions are y(0) = 0, y(2) = 0.
Hint: The approximate solution can be searched in the form of y n (x) = x(2 −
x)(a 0 + a 1 x + · · · + a n x
n
), and calculating the values of n = 0 and n = 1.
Solution Let F = y
2
+ y
2
+ 2x y, the Euler equation of the functional is