PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 63 (77), 1998, 102–114

A NEW PROOF OF A THEOREM OF BELOV

Ljubica S. Velimirović

Communicated by Mileva Prvanović

Abstract. Belov in [2] gave necessary and sufficient condition for rotational surface generated by a special quadrangular meridian, to be rigid. Belov’s theorem disproved the hypothesis of Boyarski that each toroid rotational surface with convex meridian is rigid. We give another proof of Belov’s theorem. The field of infinitesimal bendings is determined, the rotational field is obtained too. The method, used here, can be applied in a case of every rotational surface generated by a simple polygon

[6] .

0. Introduction

One of the basic problems of infinitesimal bendings theory in E_{3} is to point out nonrigid surfaces, and, if possible, to determine the infinitesimal bending field and the field of rotations.

It is known [3] that a circular torus is rigid. Among surfaces topologically equivalent to the torus Belov [2] pointed out a class of nonrigid toroids with quadrangular meridian of a special form, which can be convex or nonconvex.

We give a new proof of Belov’s theorem by using matrices. Such a proof of Belov’s theorem makes possible to determine the infinitesimal bending field and the rotational field when the surface is nonrigid. The Belov’s proof is only a proof of the existence of the bending field. Using this procedure we have proved the rigidity of toroid rotational surface with triangular meridian, not containing a side orthogonal to axis of rotation (see [5]). This procedure is also applied in the case of a toroid with a meridian in the form of a simple polygon [6].

We quote some well-known definitions and properties that we shall use afterwards. At first, we define infinitesimal deformation of a surface, and then infinitesimal bending, as a particular case ([1], [3], [4]).

AMS Subject Classification (1991): Primary 53A05 Kew Words: Infinitesimal bending, rigidity, fields of bendings and rotations.

Supported by Grant 04M03D of RFNS through Math. Inst. SANU.

Definition 1. Let S be a regular surface of the class C^{m} (m ≥ 3), defined in a vector form

S : r̄ = r̄(u, v),

included in a family of surfaces

(0.1)  S_{ε} : r̄_{ε}(u, v) = r̄(u, v) + εz̄(u, v),

where ε(ε → 0), u, v ∈ R^{1} and r̄_{0}(u, v) = r̄(u, v). It is said that the surfaces S_{ε} are infinitesimal deformations of the surface S.

Definition 2. The surfaces S_{ε} are infinitesimal bendings of a surface S if for line elements of these surfaces we have

(0.2)  ds_{ε}^{2} − ds^{2} = O(ε).

Infinitesimal bending can be defined in another manner, too, which follows from the following theorem:

Theorem 1. [1,3,4] The condition (0.2) is equivalent to each of the conditions

ds_{ε} − ds = O(ε), (0.3) dr̄dz̄ = 0, (0.4)

\frac{∂s_{ε}}{∂ε} |_{ε=0} = 0, (0.5)

r̄_{u}z̄_{u} = r̄_{v}z̄_{v} = r̄_{u}z̄_{v} + r̄_{v}z̄_{u} = 0. (0.6)

Definition 3. A vector field z̄(u, v), defined at points of the surface S satisfying the conditions of Definition 2, i.e., any of the conditions (0.2)–(0.6), is an infinitesimal bending field of the surface S.

The field

(0.7)  z̄ = ā × r̄ + b̄,

where ā, b̄ are constant vectors and × denotes a vector product, satisfies identically (0.4). It is known [3] that the vector z̄, in this case, induces only a motion of the surface as a solid body, without intrinsic deformations.

Definition 4. The field z̄ of the form (0.7) is a trivial infinitesimal bending field or infinitesimal motion field.

If z̄ = \overline{const}, then we have a particular case of (0.7).

Definition 5. If a surface possesses only a trivial infinitesimal bending field, then it is rigid, otherwise it is nonrigid.

1. A new proof of Belov’s theorem

Theorem 2. (Belov [2]) The quadrangle B, with apexes A(−1, b), B(0, b + c_{1}), C(1, b), D(0, b − c_{2}), rotates around u-axis of the coordinate system u0ρ. A necessary and sufficient condition for nonrigidity of the toroid rotational surface generated by the meridian B is

(1.1)  1/c_{2} − 1/c_{1} = k^{2}/b,

where k ≥ 2 is an integer.

Proof. Designating ρ on sides AB,…, DA with ρ_{(1)}, ρ_{(2)}, ρ_{(3)}, ρ_{(4)}, we get

(1.2)  \begin{matrix}ρ_{(1)} = b + c_{1} (u + 1) , & ρ_{(2)} = b + c_{1} (1 − u) , \\ ρ_{(3)} = b + c_{2} (u − 1) , & ρ_{(4)} = b − c_{2} (u + 1) \end{matrix}

If ē is the ort of the u-axis, ā(v) the ort of ρ-axis, v the angle between ā and the plane of initial position of meridian ρ = ρ (u), then [1, p. 90] the rotational surface generated by the meridian ρ = ρ (u) at the coordinate system with orthonormal base ē, ā (v) , ā′ (v) is:

(1.3)  r̄(u, v) = uē + ρ(u)ā(v).

It is known, [1, p. 91], that for every k ∈ {2, 3,… } there is a field of infinitesimal bendings

(1.4) z̄(u, v) = z̄_{k}(u, v) = [φ_{k}(u)e^{ikv} + φ̃_{k}(u)e^{−ikv}] ē

+ [ψ_{k}(u)e^{ikv} + ψ̃_{k}(u)e^{−ikv}]ā(v) + [χ_{k}(u)e^{ikv} + χ̃_{k}(u)e^{−ikv}] ā′(v)

for the surface (1.3), where e.g., φ̃_{k} (u) is the conjugate value for φ_{k}(u). The functions ψ_{k} (u) and χ_{k} (u) satisfy the same equation

(1.5)  ρ(u)λ′′ (u) + (k^{2} − 1)ρ′′ (u) λ (u) = 0,

and also the equations

(1.6)  \begin{matrix}φ′_{k}(u) + ρ′(u)ψ′_{k}(u) = 0, ψ_{k} (u) + ikχ_{k} (u) = 0, \\ ikφ_{k} (u) + ρ′ (u) [ikψ_{k} (u) − χ_{k} (u)] + ρ (u) χ′_{k} (u) = 0. \end{matrix}

are valid.

At turning points u = σ of the meridian ρ = ρ(u) we have [1, p. 112, eq. (15)]:

(1.7) ρ(σ)[ψ′_{k}(σ + 0) − ψ′_{k}(σ − 0)] + (k^{2} − 1)ψ_{k}(σ)[ρ′(σ + 0) − ρ′(σ − 0)] = 0,

supposing that the field z̄_{k}(u) is continuous at these points. We have analogous equation for χ_{k}(u).

Omitting the index k, we designate ψ_{k}(u) by ψ(u) and by ψ_{(1)}(u),…, ψ_{(4)}(u) the corresponding values on the sides AB,…, DA respectively. According to (1.5) and (1.2) we have the linearity of the functions ψ_{(i)}(u), i.e.,

(1.8)  ψ_{(i)}(u) = M_{i}u + N_{i}, (i = 1,…, 4).

From the continuity of the functions ψ_{(i)}(u) at the apexes we have

ψ_{(1)}(−1) = ψ_{(4)}(−1), ψ_{(1)}(0) = ψ_{(2)}(0), ψ_{(2)}(1) = ψ_{(3)}(1), ψ_{(3)}(0) = ψ_{(4)}(0),

from which one gets

(1.9) −M_{1} + N_{1} = −M_{4} + N_{4}, N_{1} = N_{2}, M_{2} + N_{2} = M_{3} + N_{3}, N_{3} = N_{4}.

At the apex A, according to (1.7), we will replace ψ′_{k}(σ − 0) with ψ′_{(4)}(−1) and ψ′_{k}(σ + 0) with ψ′_{(1)}(−1). Analogously at the apex C we will replace ψ′_{k}(σ − 0) with ψ′_{(2)}(1) and ψ′_{k}(σ + 0) with ψ′_{(3)}(1). In such a way we have at A:

ρ(−1)[ψ′_{(1)}(−1) − ψ′_{(4)}(−1)] + (k^{2} − 1)ψ_{(1)}(−1)[ρ′_{1}(−1) − ρ′_{4}(−1)] = 0,

i.e.,

(1.10 a)  b(M_{1} − M_{4}) + (k^{2} − 1)(N_{1} − M_{1})(c_{1} + c_{2}) = 0,

and for the apexes B, C, D:

(b + c_{1})(M_{2} − M_{1}) − 2(k^{2} − 1)c_{1}N_{1} = 0, (1.10 b)

b(M_{3} − M_{2}) + (k^{2} − 1)(c_{1} + c_{2})(M_{2} + N_{2}) = 0, (1.10 c)

(b − c_{2})(M_{4} − M_{3}) − 2(k^{2} − 1)c_{2}N_{3} = 0. (1.10 d)

We consider equations (1.9) and (1.10) as a system with respect to M_{i}, N_{i} (i = 1,…, 4). At first, let us consider (1.9) as a system with respect to N_{i}. The extended matrix of this system is

N_{1} N_{2} N_{3} N_{4}

(1.11) P = [ \begin{matrix}1 & 0 & 0 & −1 & ⋮ & M_{1} − M_{4} \\ 1 & −1 & 0 & 0 & ⋮ & 0 \\ 0 & 1 & −1 & 0 & ⋮ & −M_{2} + M_{3} \\ 0 & 0 & 1 & −1 & ⋮ & 0 \end{matrix} ] ∼

N_{1} N_{2} N_{3} N_{4}

∼ [ \begin{matrix}1 & 0 & 0 & −1 & ⋮ & M_{1} − M_{4} \\ 0 & −1 & 0 & 1 & ⋮ & −M_{1} + M_{4} \\ 0 & 0 & −1 & 1 & ⋮ & −M_{1} − M_{2} + M_{3} + M_{4} \\ 0 & 0 & 0 & 0 & ⋮ & −M_{1} − M_{2} + M_{3} + M_{4} \end{matrix} ],

where M is the matrix of the system (1.9) (with respect to N_{i} formed by first four columns of P). The system (1.9) is compatible if and only if rank P = rank M, i.e.

(1.12)  −M_{1} − M_{2} + M_{3} + M_{4} = 0 ⇔ M_{4} = M_{1} + M_{2} − M_{3}.

From (1.11,12) we get a reduced system,

N_{1} − N_{4} = M_{1} − M_{4}, −N_{2} + N_{4} = −M_{1} + M_{4}, −N_{3} + N_{4} = 0

and substituting M_{4} from (1.12), we get

(1.13)  N_{1} = N_{2} = −M_{2} + M_{3} + N_{4}, N_{3} = N_{4}.

Substituting (1.12,13) at (1.10) we get a homogeneous system with unknowns M_{1}, M_{2}, M_{3}, N_{4}:

−(c_{1} + c_{2})(k^{2} − 1)M_{1} − [(c_{1} + c_{2})(k^{2} − 1) + b]M_{2}
+ [(c_{1} + c_{2})(k^{2} − 1) + b]M_{3} + (c_{1} + c_{2})(k^{2} − 1)N_{4} = 0,

−(b + c_{1})M_{1} + [2c_{1}(k^{2} − 1) + b + c_{1}]M_{2} − 2c_{1}(k^{2} − 1)M_{3} − 2c_{1}(k^{2} − 1)N_{4} = 0,

−bM_{2} + [(c_{1} + c_{2})(k^{2} − 1) + b]M_{3} + (c_{1} + c_{2})(k^{2} − 1)N_{4} = 0,

(b − c_{2})M_{1} + (b − c_{2})M_{2} + 2(c_{2} − b)M_{3} − 2c_{2}(k^{2} − 1)N_{4} = 0. (1.14)

The system (1.14) has nontrivial solutions if and only if the rank of the matrix N of this system is less then 4. We have to investigate the conditions under which that is valid. We have

N_{4} M_{3} M_{2} M_{1}

N ∼ [ \begin{matrix}a_{11} & a_{12} & a_{13} & a_{14} \\ 0 & a_{22} & a_{23} & a_{24} \\ 0 & 0 & 1 & 1 \\ 0 & 0 & X & Y \end{matrix} ]

where

\begin{matrix}a & _{11} = (c_{1} + c_{2})(k^{2} − 1), a_{12} = (c_{1} + c_{2})(k^{2} − 1) + b, \\ a & _{13} = (c_{1} + c_{2})(1 − k^{2}) − b, a_{14} = (c_{1} + c_{2})(1 − k^{2}), \\ a & _{22} = \frac{2bc_{1}}{c_{1} + c_{2}} , a_{23} = b + c_{1} − \frac{2bc_{1}}{c_{1} + c_{2}} , \\ a & _{24} = 2c_{1}(1 − k^{2}) − b − c_{1}, \\ X & = −b + \frac{(c_{1} + c_{2})(b + c_{1}) − bc_{2}}{c_{1}} − \frac{(c_{1} + c_{2})(b + c_{1})c_{2}k^{2}}{bc_{1}} , \\ Y & = \frac{(c_{1} + c_{2})b}{c_{1}} + \frac{(2c_{1}k^{2} − c_{1} + b)(c_{1} + c_{2})(c_{2}k^{2} − b)}{bc_{1}}. \end{matrix}

Let us consider three possible cases:

1. For X = 0 and Y ≠ 0 or X ≠ 0 and Y = 0 the system has only trivial solution and the surface is rigid.

2. X = Y = 0. From X = 0 we have

(1.15)  k^{2}c_{2}(b + c_{1}) = bc_{1}.

Substituting k^{2} from this equation into the equation Y = 0, we have

(1.16)  b(c_{2} − c_{1}) + c_{1}c_{2} = 0.

For c_{1} = c_{2} we get c_{1}c_{2} = 0, but then we have c_{1} = c_{2} = 0, which is impossible. If we find b from (1.16) and substitute in (1.15), we get k = 1, which is impossible too, because k ∈ {2, 3,… }. Therefore, for X = Y = 0 the surface is rigid.

3. If X ≠ 0 and Y ≠ 0, we get

N_{4} M_{3} M_{2} M_{1}

(1.17) N ∼ [ \begin{matrix}a_{11} & a_{12} & a_{13} & a_{14} \\ 0 & a_{22} & a_{23} & a_{24} \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & Y − X \end{matrix} ]

where

Y − X = \frac{2k^{2}}{bc_{1}} (c_{1} + c_{2})(c_{1}c_{2}k^{2} + bc_{2} − bc_{1}).

The system (1.14) has nontrivial solutions, i.e., the surface is nonrigid if and only if

Y − X = 0 ⇔ c_{1}c_{2}k^{2} + bc_{2} − bc_{1} = 0 ⇔ 1/c_{2} − 1/c_{1} = k^{2}/b,

i.e., when (1.1) is valid. The theorem is proved.

In what follows the surface satisfying the conditions of the Theorem 2 will be named Belov’s surface.

2. Determination of the infinitesimal bending field

In order to determine the bending field z̄(u, v) by virtue of (1.4), we have to find the functions φ_{(i)}(u), ψ_{(i)}(u), χ_{(i)}(u), i = 1,…, 4. The following lemma is valid:

Lemma 1. The functions φ_{(i)}(u), ψ_{(i)}(u), χ_{(i)}(u), i = 1,…, 4, where the index in the brackets is related respectively to the sides AB, BC, CD, DA of the meridian, have the values

(2.1.1)  \begin{aligned}φ_{(1)}(u) & = −c_{1}M_{1}u, ψ_{(1)}(u) = (u − P)M_{1}, \\ χ_{(1)}(u) & = \frac{i}{k} (u − P)M_{1}, \end{aligned}

(2.1.2)  \begin{aligned}φ_{(2)}(u) & = −c_{1}M_{1}u, ψ_{(2)}(u) = −(u + P)M_{1}, \\ χ_{(2)}(u) & = −\frac{i}{k} (u + P)M_{1}, \end{aligned}

(2.1.3)  \begin{aligned}φ_{(3)}(u) & = −c_{1}M_{1}u, ψ_{(3)}(u) = [\frac{c_{1}}{c_{2}} (u − 1) − (P + 1)] M_{1}, \\ χ_{(3)}(u) & = \frac{i}{k} [\frac{c_{1}}{c_{2}} (u − 1) − (P + 1)] M_{1}, \end{aligned}

(2.1.4)  \begin{aligned}φ_{(4)}(u) & = −c_{1}M_{1}u, ψ_{(4)}(u) = − [\frac{c_{1}}{c_{2}} (u + 1) + P + 1] M_{1}, \\ χ_{(4)}(u) & = −\frac{i}{k} [\frac{c_{1}}{c_{2}} (u + 1) + P + 1] M_{1}, \end{aligned}

where is M_{1} ≠ 0 an arbitrary constant, and

(2.2)  P = \frac{b + c_{1}}{c_{1}(k^{2} − 1)} = \frac{bc_{2} + c_{1}c_{2}}{bc_{1} − bc_{2} − c_{1}c_{2}} ,

Proof. According to (1.17) for Y-X=0 we can get reduced system

(2.3)  \begin{matrix}(c_{1} + c_{2})(k^{2} − 1)N_{4} + [(c_{1} + c_{2})(k^{2} − 1) + b]M_{3} \\ + [(c_{1} + c_{2})(1 − k^{2}) − b]M_{2} + (c_{1} + c_{2})(1 − k^{2})M_{1} = 0, \\ \frac{2bc_{1}}{c_{1} + c_{2}} M_{3} + (b + c_{1} − \frac{2bc_{1}}{c_{1} + c_{2}} )M_{2} + [2c_{1}(1 − k^{2}) − b − c_{1}]M_{1} = 0, \\ M_{2} + M_{1} = 0. \end{matrix}

From (2.3) and (1.13) we have

(2.4)  \begin{matrix}M_{2} = −M_{1}, M_{3} = (\frac{c_{2}}{c_{1}} + \frac{c_{1} + c_{2}}{b} k^{2})M_{1}, \\ N_{4} = − [\frac{c_{2}}{c_{1}} + \frac{c_{1} + c_{2}}{b} k^{2} + \frac{k^{2}}{k^{2} − 1} + \frac{b}{c_{1}(k^{2} − 1)}] M_{1} = N_{3}. \end{matrix}

Further, from (1.12,13), we get

(2.5)  M_{4} = −(\frac{c_{2}}{c_{1}} + \frac{c_{1} + c_{2}}{b} k^{2})M_{1}, N_{1} = N_{2} = −\frac{b + c_{1}}{c_{1}(k^{2} − 1)}M_{1}.

From (1.1) we obtain

k^{2} = \frac{(c_{1} − c_{2})b}{c_{1}c_{2}}, \frac{c_{2}}{c_{1}} + \frac{c_{1} + c_{2}}{b} k^{2} = \frac{c_{1}}{c_{2}} , (2.6)

\frac{c_{2}^{2}}{c_{1}} + \frac{c_{1}c_{2} + c_{2}^{2}}{b} k^{2} = c_{1}, \frac{k^{2}}{k^{2} − 1} = \frac{b(c_{1} − c_{2})}{bc_{1} − bc_{2} − c_{1}c_{2}} , (2.7)

\frac{b}{c_{1}(k^{2} − 1)} = \frac{bc_{2}}{bc_{1} − bc_{2} − c_{1}c_{2}} . (2.8)

Using (2.6), we obtain

\frac{b + c_{1}}{c_{1}(k^{2} − 1)} = \frac{bc_{2} + c_{1}c_{2}}{bc_{1} − bc_{2} − c_{1}c_{2}} ⇒ (2.2),

and from (2.4)–(2.8), we get

(2.9)  \begin{matrix}M_{2} = −M_{1}, M_{3} = \frac{c_{1}}{c_{2}} M_{1}, \\ N_{4} = N_{3} = − (\frac{c_{1}}{c_{2}} + \frac{bc_{1}}{bc_{1} − bc_{2} − c_{1}c_{2}} ) M_{1} = − (\frac{c_{1}}{c_{2}} + P + 1) M_{1}. \end{matrix}

Further, by virtue of (1.12,13), (2.2) we have

(2.10)  M_{4} = −\frac{c_{1}}{c_{2}} M_{1}, N_{1} = N_{2} = −P M_{1}.

From (2.9,10) the magnitudes M_{i} N_{i} (i = 1,…, 4) are expressed by M_{1} (undetermined constant). Consequently, from (1.8) one gets the values of ψ_{(i)}(u), given in (2.1).

The functions ψ_{(i)}(u) = ψ_{(i)k}(u) are defined by (2.1, 2) for k ∈ {2, 3,… }. According to (1.6) we get the functions φ_{(i)k}(u) = φ_{(i)}(u), χ_{(i)k}(u) = χ_{(i)}(u), given at (2.1). The Lemma is proved.

As we mentioned, the problem of infinitesimal bendings can be considered solved only if the infinitesimal bendings field z̄(u, v) is determined, as in this case by virtue of (0.1) the deformed surface S_{ϵ} can be presented. In this case we have:

Theorem 3. The infinitesimal bendings field z̄(u, v) of Belov’s toroid is defined by the equations

(2.11.1)

z̄_{(1)k}(u, v) = − 2c_{1}M_{1}u cos kv ē + 2 (u − P) M_{1} cos kvā(v)

+ 2 (P − u) \frac{M_{1}}{k} sin kvā′(v), u ∈ [−1, 0], v ∈ [0, 2π],

(2.11.2)

z̄_{(2)k}(u, v) = − 2c_{1}M_{1}u cos kv ē − 2 (u + P) M_{1} cos kv ā(v)

+ 2 (P + u) \frac{M_{1}}{k} sin kv ā′(v), u ∈ [0, 1], v ∈ [0, 2π],

(2.11.3)

z̄_{(3)k}(u, v) = − 2c_{1}M_{1}u cos kv ē + 2\frac{c_{1}}{c_{2}} (u − Q) M_{1} cos kv ā(v)

+ 2\frac{c_{1}}{kc_{2}} (Q − u) M_{1} sin kvā′(v), u ∈ [0, 1], v ∈ [0, 2π],

(2.11.4)

z̄_{(4)k}(u, v) = − 2c_{1}M_{1}u cos kv ē − 2\frac{c_{1}}{c_{2}} (u + Q) M_{1} cos kv ā(v)

+ 2\frac{c_{1}}{kc_{2}} (Q + u) M_{1} sin kv ā′(v), u ∈ [−1, 0], v ∈ [0, 2π],

where M_{1} is an undetermined constant, P is given by (2.2), the parameters b, c_{1}, c_{2}, k ∈ {2, 3,… } satisfy (1.1) and

Q = \frac{bc_{1} − c_{1}c_{2}}{bc_{1} − bc_{2} − c_{1}c_{2}} .

The index in the brackets (i = 1,…, 4) is related to the parts of the surface obtained by the rotation of the sides AB,…, DA of the Belov’s quadrangle respectively.

Proof. According to (1.4) with respect to (2.1.1–4), where, for example,

φ_{k}e^{ikv} + φ̃_{k}e^{−ikv} = 2 Re(φ_{k}) cos kv − 2 Im(φ_{k}) sin kv,
φ_{k} = Re(φ_{k}) + i Im(φ_{k}),

we obtain (2.11.1–4).

3. A field of rotations

In the study of infinitesimal bendings of surfaces an important role plays the field of rotations ȳ(u, v) (see [1], [4]), related to z̄(u, v), r̄(u, v) by the equation

(3.1)  dz̄ = ȳ × dr̄.

As given in [3, p. 331], the rotational field ȳ, in the coordinate system with base the r̄_{1}, r̄_{2}, ν̄, has coordinates

y^{1} = \frac{1}{\sqrt{a}}(b_{12}z^{1}+b_{22}z^{2}+ \frac{∂z_{0}}{∂u^{2}} ), y^{2} = −\frac{1}{\sqrt{a}}(b_{11}z^{1}+b_{12}z^{2}+ \frac{∂z_{0}}{∂u^{1}} ), (3.1′a, b)

y_{0} = \frac{1}{2\sqrt{a}}(\frac{∂z_{2}}{∂u^{1}} − \frac{∂z_{1}}{∂u^{2}} ), (3.1′ c)

where u^{1} = u, u^{2} = v, z̄ = z^{i}r̄_{i} + z_{0}ν̄, ȳ = y^{i}r̄_{i} + y_{0}ν̄, a = det(a_{ij}), a_{ij}, b_{ij} are the first and the second fundamental tensor of the surface, r̄_{i} = \frac{∂r̄}{∂u^{i}}, ν̄ is the unit normal vector of the surface.

By virtue of (2.11) we see that the field z̄ is determined in the frame ē, ā, ā′, and from (3.1′) it follows that we need the coordinates of this field in the frame r̄_{1}, r̄_{2}, ν̄. We shall prove the needy lemma for the part S_{(1)} of the surface. One can prove it analogously for the other parts, too.

Lemma 2. The infinitesimal bendings field z̄ for the part S_{(1)} of the Belov’s surface in the frame r̄_{1}, r̄_{2}, ν̄ has the contravariant coordinates

z_{(1)}^{1} = −\frac{2c_{1}P M_{1}}{p} cos kv, z_{(1)}^{2} = (P − u) \frac{2M_{1} sin kv}{k(c_{1}u + b + c_{1})}, (3.2 a,b)

z_{0(1)} = 2M_{1} [\frac{P}{\sqrt{p}} − \sqrt{p}u] cos kv. (3.2 c)

Proof. From (1.2,3) we get for S_{(1)}

(3.3.1)  \begin{matrix}r̄_{(1)}(u, v) = uē + ρ_{(1)}(u)ā(v) = uē + (c_{1}u + b + c_{1})ā(v), \\ r̄_{(1)u} = ē + c_{1}ā(v), r̄_{(1)v} = (c_{1}u + b + c_{1})ā′(v), \end{matrix}

(3.4.1)  ν̄_{(1)} = \frac{(c_{1}u + b + c_{1})(c_{1}ē − ā)}{|c_{1}u + b + c_{1}| |c_{1}ē − ā|} = \frac{c_{1}ē − ā(v)}{\sqrt{p}}, p = (c_{1})^{2} + 1,

as c_{1}u + b + c_{1} > 0 (c_{1}u + b + c_{1} ≤ 0 ⇔ u ≤ −\frac{b}{c_{1}} − 1 < −1, which is impossible, because in the considered case |u| ≤ 1, b, c_{1} > 0).

According to (3.3,4) we have

\begin{aligned}z̄_{(1)k} & = z_{(1)}^{1}r̄_{(1)u} + z_{(1)}^{2}r̄_{(1)v} + z_{0(1)}ν̄_{(1)} \\ & = [z_{(1)}^{1} + \frac{c_{1}z_{0(1)}}{\sqrt{p}} ]ē + [c_{1}z_{(1)}^{1} − \frac{z_{0(1)}}{\sqrt{p}} ]ā(v) + z_{(1)}^{2}(c_{1}u + b + c_{1})ā′(v). \end{aligned}

112 Velimirović

Comparing this equation with (2.11.1), we get equations from which we get the values (3.2).

We can prove now the theorem related to the determination of the field of rotations ȳ.

Theorem 4. The field ȳ(u, v) of infinitesimal rotations of Belov’s toroid is defined by equations

ȳ_{(i)k} = y_{(i)}^{1}r̄_{(i)u} + y_{(i)}^{2}r̄_{(i)v} + y_{0(i)}ν̄_{(i)}, i = 1,…, 4

where

y_{(1)}^{1} = \frac{2M_{1} sin kv}{k(c_{1}u + b + c_{1})[(c_{1})^{2} + 1]}\{uk^{2}[(c_{1})^{2} + 1] − u − \frac{b + c_{1}}{c_{1}} \}, (3.5.1a)

y_{(1)}^{2} = \frac{2M_{1} cos kv}{c_{1}u + b + c_{1}} , (3.5.1b)

y_{0(1)} = \frac{−2M_{1} sin kv}{k\sqrt{(c_{1})^{2} + 1}}, u ∈ [−1, 0], v ∈ [0, 2π], (3.5.1c)

y_{(2)}^{1} = \frac{2M_{1} sin kv}{k(−c_{1}u + b + c_{1})[(c_{1})^{2} + 1]}\{u − uk^{2}[(c_{1})^{2} + 1] − \frac{b + c_{1}}{c_{1}} \}, (3.5.2a)

y_{(2)}^{2} = \frac{2M_{1} cos kv}{c_{1}u − b − c_{1}} , (3.5.2b)

y_{0(2)} = \frac{2M_{1} sin kv}{k\sqrt{(c_{1})^{2} + 1}}, u ∈ [0, 1], v ∈ [0, 2π], (3.5.2c)

y_{(3)}^{1} = \frac{2M_{1}c_{1} sin kv}{kc_{2}(c_{2}u + b − c_{2})[(c_{2})^{2} + 1]}\{uk^{2}[(c_{2})^{2} + 1] − u − \frac{b − c_{2}}{c_{2}} \}, (3.5.3a)

y_{(3)}^{2} = \frac{2M_{1}c_{1} cos kv}{c_{2}(c_{2}u + b − c_{2})}, (3.5.3b)

y_{0(3)} = \frac{−2M_{1}c_{1} sin kv}{kc_{2}\sqrt{(c_{1})^{2} + 1}}, u ∈ [0, 1], v ∈ [0, 2π], (3.5.3c)

y_{(4)}^{1} = \frac{2M_{1}c_{1} sin kv}{kc_{2}(−c_{2}u + b − c_{2})[(c_{2})^{2} + 1]}\{u − uk^{2}[(c_{1})^{2} + 1] − \frac{b − c_{2}}{c_{2}} \}, (3.5.4a)

y_{(4)}^{2} = \frac{2M_{1}c_{1} cos kv}{c_{2}(c_{2}u − b + c_{2})}, (3.5.4b)

y_{0(4)} = \frac{2M_{1}c_{1} sin kv}{kc_{2}\sqrt{(c_{2})^{2} + 1}}, u ∈ [−1, 0], v ∈ [0, 2π] (3.5.4c)

and the indices in the brackets are related to the parts of the surface obtained by rotation of the sides AB, BC, CD, DA of Belov’s quadrangle, M_{1} is an arbitrary constant.

Proof. From (3.3.1) we have r̄_{(1)uu} = 0, r̄_{(1)uv} = c_{1}ā′(v), r̄_{(1)vv} = (c_{1}u + b + c_{1})ā′′(v) = −(c_{1}u + b + c_{1})ā(v), and for the coefficients of the second fundamental form of the surface b_{ij} = r̄_{ij}ν̄ we have

b_{(1)11} = b_{(1)12} = 0, b_{(1)22} = \frac{1}{\sqrt{p}}(c_{1}u + b + c_{1}).

According to (3.1′a, b) and (3.2) one obtains

y_{(1)}^{1} = \frac{2M_{1} sin kv}{k\sqrt{a_{(1)}p}} [u (k^{2}p − 1) − \frac{b + c_{1}}{c_{1}} ] , y_{(1)}^{2} = \frac{2M_{1} cos kv}{c_{1}u + b + c_{1}}

and because of

a_{(1)} = (c_{1}u + b + c_{1})^{2} p, p = (c_{1})^{2} + 1,

we obtain the first two equations in (3.5.1).

In order to find y_{0(1)} according to (3.1′c), first of all we have to find z_{1(1)}, z_{2(1)}, where z_{i} = a_{ip}z^{p}. Based on (3.3.1), we have for S_{(1)}

a_{11} = (c_{1})^{2} + 1, a_{12} = 0, a_{22} = (c_{1}u + b + c_{1})^{2},

and from (3.2) we have

z_{1(1)} = −2M_{1}P c_{1} cos kv, z_{2(1)} = 2M_{1}(c_{1}u + b + c_{1}) (P − u) \frac{sin kv}{k} ,

\frac{∂z_{1(1)}}{∂u^{2}} = \frac{∂z_{1(1)}}{∂v} = 2M_{1}kP c_{1} sin kv, \frac{∂z_{2}}{∂u^{1}} = (P c_{1} − 2c_{1}u − b − c_{1})\frac{sin kv}{k}

and substituting into (3.1′c) we get the third equation at (3.5.1).

The field of rotation for S_{(1)} is

ȳ_{(1)k} = y_{(1)}^{1}r̄_{(1)u} + y_{(1)}^{2}r̄_{(1)v} + y_{0(1)}ν̄_{(1)}, u ∈ [−1, 0], v ∈ [0, 2π],

where y_{(1)}^{1}, y_{(1)}^{2}, y_{0(1)} are given by (3.5.1a−c). The same field ȳ_{(1)k} in the coordinate system with the basis ē, ā(v), ā′(v) one gets according to (3.3.1), (3.4.1). In the same way we get components of the field of rotations for S_{(2)} S_{(3)} and S_{(4)}. For S_{(2)} we have:

(3.3.2)  \begin{matrix}r̄_{(2)}(u, v) = uē + ρ_{(2)}(u)ā(v) = uē + (−c_{1}u + b + c_{1})ā(v), \\ r̄_{(2)u} = ē − c_{1}ā(v), r̄_{(2)v} = (−c_{1}u + b + c_{1})ā′(v), \end{matrix}

(3.4.2)  ν̄_{(2)} = −\frac{c_{1}ē + ā}{\sqrt{(c_{1})^{2} + 1}},

y_{(2)}^{1} = \frac{2M_{1} sin kv}{k(−c_{1}u + b + c_{1})[(c_{1})^{2} + 1]}\{u − uk^{2}[(c_{1})^{2} + 1] − \frac{b + c_{1}}{c_{1}} \}, (3.5.2a)

y_{(2)}^{2} = \frac{2M_{1} cos kv}{c_{1}u − b − c_{1}} , (3.5.2b)

(3.5.2c)  y_{0(2)} = \frac{2M_{1} sin kv}{k\sqrt{(c_{1})^{2} + 1}},

ȳ_{(2)k} = y_{(2)}^{1}r̄_{(2)u} + y_{(2)}^{2}r̄_{(2)v} + y_{0(2)}ν̄_{(2)}, u ∈ [−1, 0], v ∈ [0, 2π],

where y_{(2)}^{1}, y_{(2)}^{2}, y_{0(2)} are given by (3.5.2). For S_{(3)}:

\begin{matrix}r̄_{(3)}(u, v) = uē + ρ_{(3)}(u)ā(v) = uē + (c_{2}u + b − c_{2})ā(v), \\ r̄_{(3)u} = ē + c_{2}ā(v), r̄_{(3)v} = (c_{2}u + b − c_{2})ā′(v), \end{matrix}

ν̄_{(3)} = \frac{c_{2}ē − ā(v)}{\sqrt{p}}, q = (c_{2})^{2} + 1,

ȳ_{(3)k} = y_{(3)}^{1}r̄_{(3)u} + y_{(3)}^{2}r̄_{(3)v} + y_{0(3)}ν̄_{(3)}, u ∈ [0, 1], v ∈ [0, 2π],

where y_{(3)}^{1}, y_{(3)}^{2}, y_{0(3)} are given by (3.5.3).

For S_{(4)}:

\begin{matrix}r̄_{(4)}(u, v) = uē + ρ_{(4)}(u)ā(v) = uē + (−c_{2}u + b − c_{2})ā(v), \\ r̄_{(4)u} = ē − c_{2}ā(v), r̄_{(4)v} = (−c_{2}u + b − c_{2})ā′(v), \end{matrix}

ν̄_{(4)} = −\frac{c_{2}ē − ā(v)}{\sqrt{q}}, q = (c_{2})^{2} + 1,

ȳ_{(4)k} = y_{(4)}^{1}r̄_{(4)u} + y_{(4)}^{2}r̄_{(4)v} + y_{0(4)}ν̄_{(4)}, u ∈ [−1, 0], v ∈ [0, 2π],

where y_{(4)}^{1}, y_{(4)}^{2}, y_{0(4)} are given by (3.5.4).

References

[1] S. E. Kon-Fossen, Nekotorye voprosy differencial’noi geometrii v celom, Fizmatgiz, Moskva, 1959.

[2] K. M. Belov, O beskonechno malyh izgibaniyah toroobraznoi poverhnosti vrashceniya, Sib. Mat. Ž. 9 (1968), 490–494.

[3] I. N. Vekua, Obobshcennye analiticheskie funkcii, Nauka, Moskva, 1988.

[4] N. V. Efimov, Kachestvennye voprosy teorii deformacii poverhnostei, Uspehi Mat. Nauk. 3 (1948), 47–158.

[5] Lj. S. Velimirović, On the infinitesimal rigidity of a class of toroid surfaces of rotation, Zbornik radova Prirodno-matematičkog fakulteta u Kragujevcu 16 (1994), 123–130.

[6] Lj. S. Velimirović, Beskonechno maly izgibaniya toroobraznoi poverhnostei vrascheniya s mnogougolnym meridianom, to appear in Izvestiya VUZ-Matematika.

Gra - devinski fakultet (Received 01 09 1997) 18000 Niš (Revised 23 03 1998) Yugoslavia