Appendix: All Solutions to the Problems
847
y = 3
(y 1 + y 0 )t 1 − 2(x 1 − x 0 )
t
3
1
t
2
+ 2
3(x 1 − x 0 ) − t 1 (y 1 + 2y 0 )
t
2
1
t + y 0
The extremal function is
u = 6
(y 1 + y 0 )t 1 − 2(x 1 − x 0 )
kt
3
1
t + 2
3(x 1 − x 0 ) − t 1 (y 1 + 2y 0 )
kt
2
1
In the first group of boundary conditions, there are
x =
2x 0
t
3
1
t
3
−
3
t
2
1
t
2
+ x 0 , y =
6x 0
t
3
1
t
2
−
6x 0
t
2
1
t, u = x 1
=
6x 0 (2t − t 1 )
kt
3
1
In the second group of boundary conditions, there are
x = −t
3
+ 2t
2
, y = −3t
2
+ 4t, u =
−6t + 4
k
5.9 Find the geodesic line on the circular cylindrical surface z =
√
1 − x 2 .
Solution Let the curve joining two points A(x 0 , y 0 , z 0 ) and B(x 1 , y 1 , z 1 ) on the
circular cylindrical surface be
y = y(x)
z = z(x)
(x 0 ≤ x ≤ x 1 )
The length of the curve
J [y, z] =
x 1
x 0
1 + y 2 + z 2 dx
Making the auxiliary functional
J
∗
[y, z] =
x 1
x 0
[
1 + y 2 + z 2 + λ(x)(z −
1 − x 2 )]dx
The Euler equations of the functional are
⎧
⎨
⎩
d
dx
y
√
1+y 2 +z 2
= 0
d
dx
z
√
1+y 2 +z 2
= λ(x)
Integrating we get