822
Appendix: All Solutions to the Problems
δy 1 = δy| x=x 1 + y
(x 1 )δx 1 = δy| x=x 1 +
y
(x 1 )
2e 2x 1
δy 1
(4)
Solving the above equation, we get
δy 1 = δy| x=x 1
2e
x 1
2e x 1 − y (x 1 )
=
2e
x 1
2e x 1 − y δy
x=x 1
(5)
Substituting Eqs. (5) into (3), we obtain
δ J = 2
x 1
0
(y − y
)δydx +
2y
+
y
2
+ y
2
2e 2x 1 − y
δy
x=x 1
(6)
(2) Because x 1 is a variable boundary, δy| x=x 1 = 0, the transversality condition is
2y
+
y
2
+ y
2
2e 2x 1 − y
x=x 1
=
y
2
+ (4e
2x 1 − y
)y
2e 2x 1 − y
x=x 1
= 0
(7)
or
[y
2
+ (4e
2x 1 − y
)y
2
]
x=x 1
= 0
( 8 )
Directly applying Eq. (4.1.25), the transversality condition can also be obtained
[F + (ψ
− y
)F y ]
x=x 1
= [y
2
+ y
2
+ (2e
2x 1 − y
)2y
]
x=x 1
= [y
2
+ (4e
2x 1 − y
)y
]
x=x 1
= 0
( 9 )
Equations (8) and (9) both are the same.
4.8 Let the functional J 1 [y]
=
x 1
x 0
F(x, y, y
)dx and J 2 [y]
=
x 1
x 0
[F(x, y, y
) + P(x, y) + Q(x, y)y
]dx, where, P y = Q x , Prove:
(1) J 1 and J 2 have the same Euler equation; (2) The natural boundary condition
of J 2 is F y + Q = 0.
Proof Obviously, the Euler equation of J 1 is
F y −
d
dx
F y = 0
Taking the first variation to J 2 , and integrating one time to the terms with δy
,
meanwhile taking note that the condition P y = Q x , there is
δ J 2 =
x 1
x 0
(F y δy + F y δy
+ P y δy + Q y y
δy + Qδy
)dx