133 the pendulum equation x sin x 0 can be integrated

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1.33. The pendulum equation ¨ x + sin x = 0 can be integrated once to give the equation of the phase paths in the form 1 2 ˙ x 2 cos x = C = − cos a , ( i ) using the condition that x = a when ˙ x = 0. The origin is a centre about which the paths are symmetric in both the x and y = ˙ x axes. Without loss of generality assume that t = 0 initially. The pendulum completes the first cycle when x = 2 π . From (i) the quarter period is K = K 0 d t = 1 2 a 0 d x ( cos x cos a) = 1 2 a 0 d x ( sin 2 ( 1 / 2 )a sin 2 ( 1 / 2 )x) . Now apply the substitution sin 1 2 x = sin 1 2 a sin φ so that the limits are replaced by φ = 0 and φ = 1 2 π . Since 1 2 cos 1 2 x d x d φ = sin 1 2 a cos φ , then K(β) = 1 2 π 0 d φ ( 1 β sin 2 φ) . For small β , expand the integrand in powers of β using the binomial expansion so that K(β) = ( 1 2 0 1 + 1 2 β sin 2 φ + 3 8 β 2 sin 4 φ + · · · d φ = 1 2 π + 1 8 πβ + 9 128 β 2 π + · · ·
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48 Nonlinear ordinary differential equations: problems and solutions Now expand β in powers of a : β = sin 2 1 2 a = 1 2 a 1 48 a 3 + O(a 5 ) . Finally T = 4 K(β) = 2 π 1 + 1 4 1 4 a 2 1 48 a 4 + 9 1024 a 4 + O(a 6 ) = 2 π 1 + 1 16 a 2 + 11 3072 a 4 + O(a 6 ) . as a 0. 1.34 Repeat Problem 1.33 with the equation ¨ x + x εx 3 = 0 (ε > 0 ) , and show that T = 4 2 ( 2 εa 2 ) K(β) , β = εa 2 2 εa 2 , and that T = 2 π 1 + 3 8 εa 2 + 57 256 ε 2 a 4 + O(ε 3 a 6 ) as εa 2 0. 1.34. The damped equation ¨ x + x x 3 = 0 has phase paths given by 1 2 ˙ x 2 = 1 4 x 4 1 2 x 2 + C = 1 4 x 4 1 2 x 2 1 4 a 4 + 1 2 a 2 , = 1 2 (x 2 a 2 )( x 2 + a 2 2 ) assuming that x = a when ˙ x = 0. The equation has equilibrium points at x = 0 and x = ± 1 / . Oscillations about the origin (which is a centre) occur if a < 1. In this case the period T is given by T = 4 2 a 0 d x (a 2 x 2 ) ( 2 a 2 x 2 ) = 4 2 ( 1 / 2 0 d φ ( 2 a 2 a 2 sin 2 φ) (substituting x = sin φ ) = 4 2 ( 2 a 2 ) K(β)
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1 : Second-order differential equations in the phase plane 49 where β = a 2 /( 2 a 2 ) and K(β) = ( 1 / 2 0 d φ ( 1 β sin 2 φ) . From the previous problem, with µ = a 2 , T = 2 π 2 ( 2 µ) 1 + µ 4 ( 2 µ) + 9 µ 2 64 ( 2 µ) 2 + O(µ 3 ) = 2 π 2 ( 2 µ) 1 + 1 8 µ 1 + 1 2 µ + 9 µ 2 256 + O(µ 3 ) = 2 π 1 + µ 4 + 3 µ 2 32 1 + µ 8 + 25 µ 2 256 + O(µ 3 ) = 2 π 1 + 3 µ 8 + 57 µ 2 256 + O(µ 3 ) as µ 0. 1.35 Show that the equations of the form ¨ x + g(x) ˙ x 2 + h(x) = 0 are effectively conserva- tive. (Find a transformation of x which puts the equations into the usual conservative form. Compare with NODE, eqn (1.59).) 1.35. The significant feature of the equation ¨ x + g(x) ˙ x 2 + h(x) = 0 ( i ) is the ˙ x 2 term. Let z = f (x) , where f (x) is twice differentiable and it is assumed that z = f (x) can be uniquely inverted into x = f 1 (z) . Differentiating ˙ z = f (x) ˙ x , ¨ z = f (x) ¨ x + f (x) ˙ x 2 . Therefore ˙ x = ˙ z f (x) , ¨ x = ¨ z f (x) f (x) ˙ x 2 f (x) = ¨ z f (x) f (x) ˙ z 2 f (x) 3 .
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