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Appendix: All Solutions to the Problems
The Euler equation of the functional is
d
dt
F ˙
y =
(2 − 3e
˙
y
˙
x )( ˙
x ¨
y − ¨
x ˙
y)
˙
x 2
= 0
Because the denominator can not be zero, there is 2−3e
˙
y
˙
x = 0 or ˙
x ¨
y− ¨
x ˙
y = 0. The
solution in the former round brackets does not conform to the boundary conditions,
casting out. Finding the solution in the latter round brackets, let ˙
y = p, ˙
x = q, there
are ˙
p = ¨
y, ˙
q = ¨
x, separating the variables, we get
d p
p
=
dq
q
Integrating, we get
ln
p
q
= ln
˙
y
˙
x
= ln y
= ln c 1
or
y
= c 1
Integrating again, we get y = c 1 x + c 2 . From the boundary conditions y(0) = 0,
y(1) = 2, we get c 2 = 0, c 1 = 2, so the extremal curve is y = 2x.
6.5 Find the extremal curve of the functional J [x, y]
=
(1,0)
(0,1) (x ˙
y − y ˙
x − 2
˙
x 2 + ˙
y 2 )dt.
Solution 1 Let F = (x ˙
y − y ˙
x) − 2
˙
x 2 + ˙
y 2 , the Euler equation of the functional
is
˙
y + ˙
y + 2
d
dt
˙
x
˙
x 2 + ˙
y 2
= 0
or
˙
y
1 +
¨
x ˙
y − ˙
x ¨
y
( ˙
x 2 + ˙
y 2 )
3
2
= 0
In order to make the above equation identically hold, or ˙
y = 0, or ˙
x
2
+ ˙
y
2
=
˙
x ¨
y − ¨
x ˙
y = 1. If ˙
y = 0, then y =constant is not the desired solution. Therefore it
should choose ˙
x
2
+ ˙
y
2
= 1, obviously its solution is a unit circle of x
2
+ y
2
= 1,
taking note that the boundary conditions, it should be a part of the arc length of the
unit circle in the first quadrant.