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Appendix: All Solutions to the Problems
Because δy and δy
are arbitrary, the Euler equation is
( py
)
− (qy
)
+ r y − s = 0
The natural boundary conditions are
py
x 1
x 0
= 0, (qy
− py
)
x 1
x 0
= 0
4.6 Let the functional J [y] =
1
0 F(x, y, y
)dx, determine the Euler equation and
natural boundary conditions of the following cases.
(1) F = y
2
+ yy
+ y
2 ; (2) F = x y
2
− yy
+ y;
(3) F = y
2
+ k
2 cos y; (4) F = a(x)y
2
+ b(x)y
2 .
Solution
(1) The Euler equation of the functional is
y
+ 2y − 2y
− y
= 0
or
y
− y = 0
The natural boundary condition is
F y
1
0
= (2y
+ y)
1
0
= 0
(2) The Euler equation of the functional is
−y
+ 1 − 2y
− 2x y
+ y
= 0
or
2x y
+ 2y
− 1 = 0
The natural boundary condition is
F y
1
0
= (2x y
− y)
1
0
= 0
(3) The Euler equation of the functional is
−k
2 sin y − 2y
= 0
or