By Xi-Ping Zhu

ISBN-10: 0821833111

ISBN-13: 9780821833117

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Y' I i f SLY_ 1( and the fact that the third component of P is its projection onto n, we find that K = AIA2 for any vector field n. 2 with (8) and (9), we see that the mean curvatures of the first and second kinds coincide, while the total curvatures of the first and second 5)2 kinds are different in the non-holonomic case: K - KI = 4 In the theory of surfaces it is well known that H2 > K. The generalization of this inequality to the case of an arbitrary vector field is the following: (10) We may state (10) in the special system of coordinates constructed above with respect to which {I = C2 = fax, = 0.

Introduce the following notations for them: du dv = rl. The area element of F is dS = R(1 - kR cos v) dudv. The following relation holds: 11 dS = [kR cos v1- R2(ng + s))'`] du dv. Hence, the integral of streamline normal curvature over F has the form r J F 2i 21T ! 1 rJR2(`et jdS=JJ Rkcosvdudv- F +'1)2dudv. (5) 0 0 0 0 The first integral on the right is zero because R and k are independent of v. Using (4) and (5) we can write l J KdV = G 2x I ff R2( 0 0 21 , )2 dude. VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE 37 FIGURE 10 As the first fundamental form may be transformed to I= (1 -kRcosv)2 2+R2(s +rt)' the integrand on the right is less then 1/2.

VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE 37 FIGURE 10 As the first fundamental form may be transformed to I= (1 -kRcosv)2 2+R2(s +rt)' the integrand on the right is less then 1/2. So, we have already proved the following statement. Theorem If the tubular surface F with the axial line of length I is an invariant surface of the vector field a then the integral of curvature K of the field n over a domain bounded by F is non-negative and does not exceed irl. 0< fKdV

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