Appendix: All Solutions to the Problems
853
du =
− sin ϕ sin ϕ−cos ϕ cos ϕ
sin
2 ϕ
dϕ = −
dϕ
sin
2 ϕ
u
=
du
dθ
=
− sin ϕ sin ϕ−cos ϕ cos ϕ
sin
2 ϕ
dϕ = −
ϕ
sin
2 ϕ
du
= −
dϕ
sin
2 ϕ−2ϕ
sin ϕ cos ϕdϕ
sin
4 ϕ
= −
dϕ
sin ϕ−2ϕ
cos ϕdϕ
sin
3 ϕ
u
=
du
dθ
=
d
2 u
dθ 2 = −
ϕ
sin ϕ−2ϕ
2 cos ϕ
sin
3 ϕ
Substituting the above equation into the Euler equation, we get
u
+ u = 0
Integrating, we get
u = c 1 sin θ + c 2 cos θ
or
cot ϕ = c 1 sin θ + c 2 cos θ
Thus the general solution is ϕ = arccot(c 1 sin θ + c 2 cos θ), r = R, where the
integral constants are determined by the boundary conditions A and B.
It is followed from the boundary condition A(R, 0, 0) that the endpoint is on the
x axis, ϕ =
π
2
, θ = 0, we get c 2 = 0. It is followed from the boundary condition
B
0,
√
2
2
R,
√
2
2
R
that the endpoint is on the Oyz plane, and is on the angular bisector
of the Oyz plane, ϕ =
π
4
, θ =
π
2
, we get c 1 = 1. Thus the extremal curve is
ϕ = arccot(sin θ), r = R.
Squaring the both sides of cot ϕ = sin θ , we get
cot
2
ϕ =
cos
2
ϕ
sin
2
ϕ
=
1 − sin
2
ϕ
sin
2
ϕ
= sin
2
θ
Solve for sin
2
ϕ from the above expression
sin
2
ϕ =
1
1 + sin
2
θ
Differentiating the both ends of cot ϕ = sin θ , we get
− sin
2 ϕ−cos
2 ϕ
sin
2 ϕ
dϕ = −
dϕ
sin
2 ϕ
= cos θ dθ
ϕ
= − sin
2
ϕ cos θ =
cos θ
1+sin
2 θ
ϕ
2
+ sin
2
ϕ =
cos
2 θ+1+sin
2 θ
(1+sin
2 θ) 2
=
2
(1+sin
2 θ) 2
Substituting the above equation into the expression of the functional, the geodesic
line from point A to point B is