Using the substitution θ z z β in 220223 leads to 1

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Using the substitution θ = ¨ Z ˙ Z β in (2.20)–(2.23) leads to, 1 2 ... Z ˙ Z 1 4 ¨ Z 2 + ˙ Z 2 ( σ 2 + + σ 2 × ω 2 ) + κ + 2 k β 2 2 ˙ Z 2 2 = 0 , (2.24) ˙ Z ˙ σ + + ¨ + p ˙ Z = 0 , (2.25) ˙ Z ˙ σ × + ¨ × q ˙ Z = 0 , (2.26) ˙ Z ˙ ω + ¨ = 0 . (2.27) Solving for σ + , σ × and ω , respectively, from (2.25)–(2.27), and substituting in (2.24) followed by one further time differentiation yields, .... Z + 2 α 2 ¨ Z + α 1 Z = D, (2.28) where α 1 = 4( p 2 + q 2 ), α 2 = κ + 2 k β 2 2 , and D = 4( pE + qF ) with E and F as constants of integration. Now, one can write the general solution of the ESR variables as, θ = ¨ Z ˙ Z β, σ + = pZ + E ˙ Z , σ × = qZ + F ˙ Z , ω = G ˙ Z , (2.29) where G is also a constant of integration, and Z ( t ) is the solution of the differential equation (2.28). This scheme of solution of course leads to more number of constants of integration than conditions available from the initial values of the ESR variables. Therefore, one has to also ensure the satisfaction of the individual equations in (2.20)–(2.23), which yields appropriate number of constraint equations required for the solution of all the constants of integration involved. It is evident from (2.28) that the solution of Z ( t ) is dependent on the values of α 1 and α 2 , which are in turn dependent on the curvature, stiffness and damping Int. J. Geom. Methods Mod. Phys. 2009.06:645-666. Downloaded from by UNIVERSITY OF CINCINNATI on 04/01/15. For personal use only.
652 A. Dasgupta, H. Nandan & S. Kar parameter values. The solutions for all possible cases are straightforward, though may be tedious. Therefore, for the purpose of illustration, we consider only some special cases in detail in the rest of this section. The nature of solutions for all possible cases are discussed and summarized separately later. Case I. k = k + = k × = 0 and β = 0 We start with the case without stiffness and damping. Here, we have some simplifications in (2.28) with α 1 = 1 and α 2 = κ . The solutions for the spherical and hyperbolic geometries are presented below. Spherical geometry ( κ = 1) The exact solutions for the ESR variables in this case are obtained as follows. The solution for the equation for Z ( t ) is obtained as: Z ( t ) = ( A 1 + B 2 + B 1 t ) sin t + ( B 1 A 2 B 2 t ) cos t 2 E. (2.30) From this expression, using the above-mentioned definitions for the ESR variables, we obtain: θ = ( A 2 + B 1 + B 2 t ) cos t + ( B 2 A 1 B 1 t ) sin t ( A 1 + B 1 t ) cos t + ( A 2 + B 2 t ) sin t , (2.31) σ + = 2 E − { ( A 1 + B 2 + B 1 t ) sin t + ( B 1 A 2 B 2 t ) cos t } 2[( A 1 + B 1 t ) cos t + ( A 2 + B 2 t ) sin t ] , (2.32) σ × = F [( A 1 + B 1 t ) cos t + ( A 2 + B 2 t ) sin t ] , (2.33) ω = G [( A 1 + B 1 t ) cos t + ( A 2 + B 2 t ) sin t ] (2.34) where A 1 , B 1 , A 2 , B 2 , E , F and G are the integration constants and will also be used throughout in the text for the analytical solutions of other cases as well. These can be defined in terms of the initial conditions on the ESR variables at t = 0 which are represented as θ 0 , σ +0 , σ × 0 and ω 0 . The integration constants are calculated by putting back the solutions (2.31)–(2.34) in Eq. (2.9) and using the initial conditions on the ESR variables as, A 1 = 4 ( θ 2 0 + 4 σ 2 +0 ) 4 I 0 , A 2 = A 1 ( θ 0 + 2 σ +0 ) 2 , B 1 = A 1 ( θ 0
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