844
Appendix: All Solutions to the Problems
θ
2
1 + R 2 θ 2 = c
Thus solving for θ
2
=
c
1−cR 2 = c
2
1 , namely θ
= c 1 is a constant, integrating we
get θ = c 1 z + c 2 . The general solution is θ = c 1 z + c 2 , r = R, where, the constants
c 1 , c 2 are determined by the boundary conditions.
Solution 2 In the cylindrical coordinates r, θ , z, (ds)
2
= (Rdθ)
2
+ (dz)
2 , the found
geodesic line is the following functional
J =
θ 1
θ 0
R 2 + z 2 dθ
Because the circular cylindrical surface directly chooses the cylindrical coordinates, it is necessary to make the auxiliary functional, and because the integrand of
the functional is only the function of z
, the general solution of the Euler equation is
a straight line z = c 1 θ + c 2 , r = R. It follows that the latter method of solution is
relatively simple.
From the boundary conditions y = 0, z = 0, y = R, z = R, we get c 2 = 0,
c 1 =
2
π
, so the extremal curve is z =
2
π
θ. The length of geodesic line is
J =
π
2
0
R 2 +
2
π
2
dθ =
π
2
R 2 +
2
π
2
=
1
2
(π R) 2 + 4
5.6 Under the isoperimetric condition
x 1
x 0
ydx = a, find the extremal curve of the
functional J [y] =
x 1
x 0
y
2 dx, there a is a constant.
Solution Making the auxiliary functional J
∗
[y] =
x 1
x 0
(y
2
+ λy)dx, The Euler
equation of the functional is
λ − 2y
= 0
Integrating we get y =
λ
4
x
2
+ c 1 x + c 2 , where c 1 , c 2 and λ are determined by the
boundary conditions and the isoperimetric condition.
5.7 Under the isoperimetric condition
x 1
x 0
r (x)y
2 dx = 1 and the boundary conditions
y(x 0 ) = y(x 1 ) = 0, write the differential equation of the extremal curve for the
functional J [y] =
x 1
x 0
[ p(x)y
2
+ q(x)y
2
]dx, where, p(x), q(x), r (x) ∈ C
1
[x 0 , x 1 ]
are all the known functions.
Solution Making the auxiliary functional
J
∗
[y] =
x 1
x 0
[ p(x)y
2
+ q(x)y
2
+ λr (x)y
2
]dx