AEM_3e_Chapter_11

AEM_3e_Chapter_11 - 11 11 Systems of Nonlinear Differential...

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11 11 Systems of Nonlinear Differential Equations EXERCISES 11.1 Autonomous Systems 3. The corresponding plane autonomous system is x 0 = y, y 0 = x 2 y (1 x 3 ) . If ( x, y ) is a critical point, y = 0 and so x 2 y (1 x 3 )= x 2 = 0. Therefore (0 , 0) is the sole critical point. 6. The corresponding plane autonomous system is x 0 = y 0 = x + ²x | x | . If ( x, y ) is a critical point, y = 0 and x + ²x | x | = x ( 1+ ² | x | ) = 0. Hence x =0,1 , 1 . The critical points are (0 , 0), (1 /², 0) and ( 1 /², 0). 9. From x y = 0 we have y = x . Substituting into 3 x 2 4 y = 0 we obtain 3 x 2 4 x = x (3 x 4) = 0. It follows that (0 , 0) and (4 / 3 , 4 / 3) are the critical points of the system. 12. Adding the two equations we obtain 10 15 y/ ( y + 5) = 0. It follows that y = 10, and from 2 x + y +10=0 we can conclude that x = 10. Therefore (10 , 10) is the sole critical point of the system. 15. From x (1 x 2 3 y 2 )=0wehave x =0or x 2 +3 y 2 =1.If x = 0, then substituting into y (3 x 2 3 y 2 ) gives y (3 3 y 2 ) = 0. Therefore y =0,1, 1. Likewise x 2 =1 3 y 2 yields 2 y = 0 so that y = 0 and x 2 3(0) 2 =1. The critical points of the system are therefore (0 , 0), (0 , 1), (0 , 1), (1 , 0), and ( 1 , 0). 18. (a) From Exercises 10 . 2, Problem 6, x = c 1 +2 c 2 e 5 t and y =3 c 1 + c 2 e 5 t , which is not periodic. (b) From X (0) = (3 , 4) it follows that c 1 = c 2 = 1. Therefore x =1+2 e 5 t and y =3+ e 5 t gives y = 1 2 ( x 1)+3. (c) 21. (a) From Exercises 10 . 2, Problem 35, x = c 1 (sin t cos t ) e 4 t + c 2 ( sin t cos t ) e 4 t and y =2 c 1 (cos t ) e 4 t + 2 c 2 (sin t ) e 4 t . Because of the presence of e 4 t , there are no periodic solutions. (b) From X (0) = ( 1 , 2) it follows that c 1 = 1 and c 2 = 0. Therefore x = (sin t cos t ) e 4 t and y = 2(cos t ) e 4 t . 194
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11.2 Stability of Linear Systems (c) 24. Switching to polar coordinates, dr dt = 1 r µ x dx dt + y dy dt = 1 r ( xy x 2 r 2 xy + y 2 r 2 )= r 3 dt = 1 r 2 µ y dx dt + x dy dt = 1 r 2 ( y 2 xyr 2 x 2 + xyr 2 1 . If we use separation of variables, it follows that r = 1 2 t + c 1 and θ = t + c 2 . Since X (0) = (4 , 0), r = 4 and θ = 0 when t = 0. It follows that c 2 = 0 and c 1 = 1 16 . The Fnal solution can be written as r = 4 1 32 t = t. Note that r →∞ as t ( 1 32 ) . Because 0 t 1 32 , the curve is not a spiral. 27. The system has no critical points, so there are no periodic solutions. 30. The system has no critical points, so there are no periodic solutions. EXERCISES 11.2 Stability of Linear Systems 3. (a) All solutions are unstable spirals which become unbounded as t increases. (b) 195
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11.2 Stability of Linear Systems 6. (a) All solutions become unbounded and y = x/ 2 serves as the asymptote. (b) 9. Since ∆ = 41 < 0, we can conclude from Figure 11 . 18 in the text that (0 , 0) is a saddle point. 12. Since ∆ = 1 and τ = 1, τ 2 4∆ = 3 and so from Figure 11 . 18, (0 , 0) is a stable spiral point. 15. Since ∆ = 0 . 01 and τ = 0 . 03, τ 2 4∆ < 0 and so from Figure 11 . 18, (0 , 0) is a stable spiral point.
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AEM_3e_Chapter_11 - 11 11 Systems of Nonlinear Differential...

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