Appendix: All Solutions to the Problems
863
Solution 2 The integrand is F = (x ˙
y − y ˙
x) − 2
˙
x 2 + ˙
y 2 , it is easy to see that F
is a positively homogeneous function of order one about ˙
x and ˙
y. taking the partial
derivative of F with respect to ˙
x, we obtain
F ˙
x = −2
˙
x
˙
x 2 + ˙
y 2
Taking the partial to ˙
x again, we get
F ˙
x ˙
x = −2
˙
y
2
( ˙
x 2 + ˙
y 2 )
3
2
If using the Weierstrass form of Euler equation, then
F x ˙
y = 1, F ˙
x y = −1, F 1 =
F ˙
x ˙
x
˙
y 2 = −
2
( ˙
x 2 + ˙
y 2 )
3
2
Substituting the above found result into the curvature formula, we get
1
R
=
F ˙
x y − F x ˙
y
( ˙
x 2 + ˙
y 2 )
3
2 F 1
=
−1 − 1
−2
= 1
According to the boundary conditions, it is a part of the arc length of the unit
circle in the first quadrant.
6.6 Find the extremal curve of the functional J [x, y] =
π
4
0 ( ˙
x ˙
y + 2x
2
+ 2y
2
)dt, the
boundary conditions are x(0) = y(0) = 0, x(1) = y(1) = 1.
Solution The Euler equations of the functional are
4x − ¨
y = 0, 4y − ¨
x = 0
or
y
(4)
− 16y = 0, x
(4)
− 16x = 0
The characteristic equation of the Euler equations is
(r − 2)(r + 2)(r
2
+ 4) = 0
The general solutions of the Euler equations are
x = c 1 sinh 2t + c 2 cosh 2t − c 3 cos 2t − c 4 sin 2t
y = c 1 sinh 2t + c 2 cosh 2t + c 3 cos 2t + c 4 sin 2t