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ALGORITHMS OF ISOMETRIC SURFACES CONSTRUCTIONS
G. Tsitsiashvili, M. Osipova
Russia, Vladivostok, Radio str. 7, IAM FEB RAS e-mail: guram@iam.dvo.ru , mao1975@ list.ru
In this paper a problem of a construction of isometric surfaces is considered. It is important in different technical applications. This problem usually is solved for surfaces with constant Gauss curvature using so called "gold" Gauss theorem. But in practical applications it is interesting to consider more wide class of surfaces. Here this problem is solved for surfaces of cylinder, conical and spherical types and their generalizations using a criterion of isometric based on equalities of coefficients of first quadratic forms.
1. INTRODUCTION
In the differential geometry a concept of the isometric is introduced as follows. Define a surface in three dimensional space E3 as a set described by equalities
X = x (Mj, u2), Y = y (ul3 u2), Z = z (ul3 u2), (ul3 u2 )e G ^ E2, where the functions x, y, z on the domain G are smooth. Say that the surfaces
51 =|X = x (m1, u2 ), Y = y1 (m1, u2), Z = z1 (m1, u2), (u1, u2 )e G},
52 ={x = X2 (^ U2 ) , Y = y2 (U1, U2 ) , z = Z2 (U1, U2 ) , (U1, U2 )g G}
are isometric if for any pair of points on the surface S1 (a length of a geodesic curve connected these points) coincides with a distance between appropriate points on the surface S2. So the surface S1 may be transformed into the surface S2 without a dilation and a compression. Last condition has important significance in a solution of modern technical problems including reliability theory also [2].
From the Gauss theorem [1] the surfaces S1, S2 are isometric if and only if coefficients of their first quadratic forms coincide:
4, = A,, i,J = 1,2,
(1)
where
(
Ak -
Ai,j -
dxk
du,
du.
dzt du
(
' /
dxk dyk dz>
dU: V j
i, j, k -1,2.
du . du .
j j j
Here " • " is the scalar product operation. But in a literature a problem of the isometric is solved mainly for surfaces with constant Gauss curvature [3, p.126] using so called the "gold" Gauss theorem. This restriction is sufficiently inconvenient in an applied sense and it is worthy to use a direct application of the condition (1) and a search of more general relations for isometric surfaces. In this paper these relations are constructed for surfaces of cylinder, conical and spherical types and for some their generalizations.
2. SURFACE OF CYLINDER TYPE
Define the first quadrant S1 in the space E2 by the equalities x = u1, yi = 0, z1 = u2 and a surface of a cylinder type S2 by equalities
x2 = a (u1), y2 = b (u1), z2 = u2. Coefficients of the first quadratic form for the surface S1 satisfy the relations A1j = Si j where Si j, i, j = 1,2, is the Kronecker symbol. For the surface S2 it is not difficult to obtain the equalities
Ai22 = A21 = 0, A22 = 1, A21 =(a (U1 ))2 + (b (u ))2 = 1. So for the isometric of the surfaces S1, S2 it is sufficient to fulfill relations
(a'(u1 ))2 < 1, (b'(u1 ))2 = 1 -(a'(u1 ))2. (2)
Consequently to solve considered problem for the cylinder surface S2 it is sufficient using the function a (t) to find the function b (t) from the equality (2).
Example 1. A partial case of this problem solution is the pair of functions
a (u1 ) = cos u1, b (u1 ) = sin u1. (3)
If in the first quadrant of E2 we separate a stripe with sides not parallel to coordinate axes u1, u2, then from the equalities (3) it is possible to transform this stripe into spiral surface enveloped around the cylinder S2 .
3. SURFACE OF CONICAL TYPE
Assume that the angle S2 in the space E2 is described by the equalities x = u2 c°s u\, yj = u2 sin uj. Define the surface of a conical type S2 by the formulas
x2 = U2a (Uj ) , y2 = u2b (U ) , Z2 = U2c (Uj ).
Coefficients of the first quadratic form for the surface S2 satisfy the formulas
Aj,j = U2 , Aj,2 = ^2,1 = 0 , ^2,2 = 1 . And coefficients of the first quadratic form for the surface S2 satisfy the relations
r
Aj2j = u2 ((a ' (uj ))2 + (V (uj ))2 + (c' (uj ))2 ) , Aj22 = A^ = U2(a2 (Uj ) + b2 (Uj ) + c2 (ux )) ,
A22 = a2 (u2) + V (U2) + c2 (U2). So to fulfill the equalities in (j) it is sufficient to satisfy the formulas
a2 ( U2 ) + V ( U2 ) + c2 ( U2 ) = j, (4)
(a' (Uj ))2 +(V (uj ))2 +(c' (uj ))2 = j, (5)
Seek a solution of this problem as follows
a (u2 ) = cos F cos G, b (u2 ) = sin F sin G, c (uj ) = sin G, F = F (5), G = G (5), 5 = 5 (u2 )
where F (5), G(5) are arbitrary smooth functions of an argument 5 and 5 (u1 ) is a smooth function of the argument u1 . Such a choice of the functions a, b, c leads to the equality (4)
independently on a choice of the function 5 . Simple calculations show that to fulfill the formulas (5) it is sufficient to satisfy the condition
1 = ( 5' (U ))2 [(F'(5))2 +(G' (5))
that leads to the formula
u1 = ±j^(F' (5))2 cos2 G (5) + (G' (5))2 d5. (6)
Example 2. In a partial case it is possible to take real a and put
G(5)= a, F(5) = 5, 5(u1 )= u1
and so
x2 = u2 cos u1 cos a , y2 = u2 cos u1 cos a , z2 = sin a .
(7)
The relations (7) define usual conical surface and the formulas (6) specify a wide class of surfaces isometric to an angle in the space E2.
4. SURFACE OG SPHERICAL TYPE
Suppose that S1 is defined as follows
x1 = a (u2) cos u1, y1 = a (u2) sin u1, z1 = b (u2), (a' (u2 ))2 +(b' (u2 ))2 = 1. A1,1 = a (U2 ) , A1,2 = ^2,1 = 0 , ^2,2 = 1 . Assume that for some real y the condition (a' (u2 ))2 < 1 is true. Define the surface S2 by the equalities
x2 = ya (u2) cos — , y2 = ya (u2) sin —, z2 = B (u2).
Then the functions
A21 = a2 (U2), a22 = 4,1 = 0, 4,2 = (B' (U2 ))2 + y2 (a (U2 ))2.
So the surfaces S1, S2 are isometric if
( b' (U2 ))2+y2 (a (U2 ))2 = 1,
consequently
B(u2) = jV1"y2 (a'(u2))2du2.
Example 3. Assume that the equality a (u2) = cos u2 is true and so the surface S1 is a sphere with unity radius and with a cut along a meridian. Then the surface S2 may coincide with a bending of a sphere by a type [3,Figure 3]: spindled surface of a rotating with a self overlap and etc Consider possible generalizations of isometric surface with a spherical type. Suppose that smooth functions F ( 5 ), G (5) satisfy the condition
(F (5 ))2 +((G (5 ))2 = 1 (8)
and f (u1, u2) is smooth function. Define the surface S1 as follows
1 = F (f («1, u2)), y1 = G (f (u1. u2)), Z1 = R (u2),
X =
then the functions
(9)
A =
df ( u2)
v du1 j
, A,2 = A2,1 =
df (^ u2 ) df (u^ u2 )
du1 du2
Take a real number y and define the surface S2 by the equalities
f f (u,, u2 C
x2 = yF 2) , y2 = yG
A =
A2,2 "
df (U1, u2 )
V du2 j
+
(R ' (u2 ))2.
y j
f (U1, u2 )
A
^ 2 y j
, z2 = R (u2 ) .
(10)
Then the functions
(
A2 = A1,1 =
df (U1, u2 )
2
V du1 j
, A 2 = A 2 = , A1,2 A2,1
df (U1, u2 ) df (^ u2 )
22
A2 = A2,2 "
T +(R (u2))
V du2 j
2
du1 du2
and so the relation (1) is true. Consequently the surfaces S1, S2 defined by the formulas (8) - (10) are isometric.
2
REFERENCES
1. Andreev G.N. 2007. Additional chapters of geometry. Differential geometry of curved surfaces. Moscow.: MGIU. (In Russian).
2. Ushakov I.A. 2009. History of science through prism of inspiration. Glamour of geometry. Book 3. Moscow: URSS. (In Russian).
3. Fomenko V.T. 1998. Bending of surfaces// Soros educational journal. Vol. 5. P. 122-127. (In Russian).
