Appendix: All Solutions to the Problems
821
2y
+ k
2 sin y = 0
The natural boundary condition is
F y
1
0
= 2y
1
0
= y
1
0
= 0
(4) The Euler equation of the functional is
2b(x)y − 2a
(x)y
− 2a(x)y
= 0
or
a(x)y
+ a
(x)y
− b(x)y = 0
The natural boundary condition is
F y
1
0
= 2a(x)y
1
0
= a(x)y
1
0
= 0
4.7 Let the functional J [y] =
x 1
0 (y
2
+ y
2
)dx, the boundary conditions are y(0) =
0, y(x 1 ) = e
2x 1 . Find: (1) The variation δ J ; (2) The transversality condition.
Solution
(1) F = y
2
+ y
2 , F y = 2y
. according to the endpoint condition y(x 1 ) = ψ(x 1 ) =
e
2x 1 , taking the variation to it, there is
δy 1 = 2e
2x 1 δx 1 = 2y(x 1 )δx 1 = 2ψ(x 1 )δx 1
(1)
or
δx 1 =
δy 1
2e 2x 1
(2)
The first variation of the functional is
δ J =
x 1
0
F y −
d
dx
F y
δydx + (F − y
F y )
x=x 1
δx 1 + F y
x=x 1
δy 1
= 2
x 1
0
(y − y
)δydx +
y
2
− y
2
2e x 1
x=x 1
δy 1 + 2y
x=x 1
δy 1
= 2
x 1
0
(y − y
)δydx +
2y
+
y
2
− y
2
2e 2x 1
x=x 1
δy 1
(3)
Furthermore according to Eq. (4.1.16), there is