Appendix: All Solutions to the Problems
865
initial condition and final condition that x ≡ y ≡ 0, they do not conform to the
requirements of the problem. Moreover when λ > 0, the general solutions are x =
c 1 e
−
√
λt
+ c 2 e
√ λt , y = c 3 e
−
√ λt
+ c 4 e
√ λt , z = c 5 e
−
√ λt
+ c 6 e
√
λt , According to the
initial condition and final condition, there are still x ≡ y ≡ 0, they do not conform
to the requirements of the problem either. Then there must be λ < 0, therefore
⎧
⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎩
x = c 1 sin
−
λT 1
m
t + c 2 cos
−
λT 1
m
t
y = c 3 sin
−
λT 1
m
t + c 4 cos
−
λT 1
m
t
z = c 5 sin
−
λT 1
m
t + c 6 cos
−
λT 1
m
t
By the initial condition, when t 0 = 0, x = y = 0, z = R, we get c 2 = c 4 = 0,
c 6 = R. When t 1 = T 1 , x = y = 0, z = −R, we get
−
λT 1
m
T 1 = π , namely
−
λ T 1
m
=
π
T 1
. Moreover From the constraint condition x
2
+ y
2
+ z
2
= R
2 , we get
c 5 = 0, c
2
1 + c
2
3 = R
2 . Thus, the results are
⎧
⎪ ⎨
⎪ ⎩
x = c 1 sin
π
T 1
t
y = c 3 sin
π
T 1
t
z = R cos
π
T 1
t
(c
2
1 + c
2
3 = R
2
)
The motion path is a meridian.
6.9 Find the extremal curve of the functional J [x, y] =
t 1
t 0
˙
x 2 + ˙
y 2
x−k
dt, where, k is a
constant.
Solution The Euler equation of the functional is
F y −
d
dt
F ˙
y = 0 −
d
dt
˙
y
√
x − k
˙
x 2 + ˙
y 2
= 0
or
˙
y
√
x − k
˙
x 2 + ˙
y 2
= c 1
Squaring the both sides of the above equation and transposing the term, we get
˙
y
2
= c
2
1 (x − k)( ˙
x
2
+ ˙
y
2
)
Let ˙
x = 0, the both sides are divided by ˙
x and arranged, we get
dy
dx
=
˙
y
˙
x
=
c
2
1 (x − k)
1 − c
2
1 (x − k)