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# EDO 9th Boyce Cap9 Resolução

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0 x g(s)ds > 0 for x < 0 . Therefore V (x, y) is positive definite. The rate of change of V along any trajectory is V\u2d9 = Vx dx dt + Vy dy dt = g(x) · (y) + y [\u2212g(x)\u2212 y] = \u2212y2 . It follows that V\u2d9 (x, y) is only negative semidefinite. Hence the origin is a stable critical point. (b) Given V (x, y) = 1 2 y2 + 1 2 y sin(x) + \u222b x 0 sin(s)ds , It is easy to see that V (0 , 0) = 0 . The rate of change of V along any trajectory is V\u2d9 = Vx dx dt + Vy dy dt = [ sin x+ y 2 cos x ] (y) + [ y + 1 2 sin x ] [\u2212 sin x\u2212 y] = = 1 2 y2 cos x\u2212 1 2 sin2 x\u2212 y 2 sin x\u2212 y2 . For \u2212pi/2 < x < pi/2 , we can write sin x = x\u2212 \u3b1x3/6 and cos x = 1\u2212 \u3b2 x2/2 , in which \u3b1 = \u3b1(x) , \u3b2 = \u3b2(x) . Note that 0 < \u3b1 , \u3b2 < 1 . Then V\u2d9 (x , y) = y2 2 (1\u2212 \u3b2 x 2 2 )\u2212 1 2 (x\u2212 \u3b1x 3 6 )2 \u2212 y 2 (x\u2212 \u3b1x 3 6 )\u2212 y2 . 9.6 507 Using polar coordinates, V\u2d9 (r , \u3b8) = \u2212r 2 2 [1 + sin \u3b8 cos \u3b8 + h(r , \u3b8)] = \u2212r 2 2 [ 1 + 1 2 sin 2\u3b8 + h(r , \u3b8) ] . It is easy to show that |h(r , \u3b8)| \u2264 1 2 r2 + 1 72 r4. So if r is sufficiently small, then |h(r , \u3b8)| < 1/2 and \u2223\u2223 12 sin 2\u3b8 + h(r , \u3b8)\u2223\u2223 < 1 . Hence V\u2d9 (x , y) is negative definite. Now we show that V (x, y) is positive definite. Since g(u) = sin u , V (x, y) = 1 2 y2 + 1 2 y sin(x) + 1\u2212 cos x . This time we set cos x = 1\u2212 x 2 2 + \u3b3 x4 24 . Note that 0 < \u3b3 < 1 for \u2212pi/2 < x < pi/2 . Converting to polar coordinates, V (r, \u3b8) = r2 2 [ 1 + sin \u3b8 cos \u3b8 \u2212 r 2 12 sin \u3b8 cos3 \u3b8 \u2212 \u3b3 r 2 24 cos4 \u3b8 ] = r2 2 [ 1 + 1 2 sin 2\u3b8 \u2212 r 2 12 sin \u3b8 cos3 \u3b8 \u2212 \u3b3 r 2 24 cos4 \u3b8 ] . Now \u2212 r 2 12 sin \u3b8 cos3 \u3b8 \u2212 \u3b3 r 2 24 cos4 \u3b8 > \u22121 8 for r < 1 . It follows that when r > 0 , V (r, \u3b8) > r2 2 [ 7 8 + 1 2 sin 2\u3b8 ] \u2265 3 r 2 16 > 0 . Therefore V (x, y) is indeed positive definite, and by Theorem 9.6.1 , the origin is an asymptotically stable critical point. 12.(a) We consider the linear system( x y )\u2032 = ( a11 a12 a21 a22 )( x y ) . Let V (x, y) = Ax2 +Bxy + Cy2, in which A = \u2212a 2 21 + a 2 22 + (a11a22 \u2212 a12a21) 2\u2206 B = a12a22 + a11a21 \u2206 C = \u2212a 2 11 + a 2 12 + (a11a22 \u2212 a12a21) 2\u2206 , 508 Chapter 9. Nonlinear Differential Equations and Stability and \u2206 = (a11 + a22)(a11a22 \u2212 a12a21). Based on the hypothesis, the coefficients A and B are negative. Therefore, except for the origin, V (x, y) is negative on each of the coordinate axes. Along each trajectory, V\u2d9 = (2Ax+By)(a11 x+ a12 y) + (2Cy +Bx)(a21 x+ a22 y) = \u2212x2 \u2212 y2. Hence V\u2d9 (x, y) is negative definite. Theorem 9.6.2 asserts that the origin is an unstable critical point. (b) We now consider the system( x y )\u2032 = ( a11 a12 a21 a22 )( x y ) + ( F1(x , y) G1(x , y) ) , in which F1(x , y)/r \u2192 0 and G1(x , y)/r \u2192 0 as r \u2192 0 . Let V (x, y) = Ax2 +Bxy + Cy2, in which A = a221 + a 2 22 + (a11a22 \u2212 a12a21) 2\u2206 B = \u2212a12a22 + a11a21 \u2206 C = a211 + a 2 12 + (a11a22 \u2212 a12a21) 2\u2206 , and \u2206 = (a11 + a22)(a11a22 \u2212 a12a21). Based on the hypothesis, A ,B > 0 . Ex- cept for the origin, V (x, y) is positive on each of the coordinate axes. Along each trajectory, V\u2d9 = x2 + y2 + (2Ax+By)F1(x , y) + (2Cy +Bx)G1(x , y) . Converting to polar coordinates, for r 6= 0 , V\u2d9 = = r2 + r(2A cos \u3b8 +B sin \u3b8)F1 + r(2C sin \u3b8 +B cos \u3b8)G1 = r2 + r2 [ (2A cos \u3b8 +B sin \u3b8) F1 r + (2C sin \u3b8 +B cos \u3b8) G1 r ] . Since the system is almost linear, there is an R such that\u2223\u2223\u2223\u2223(2A cos \u3b8 +B sin \u3b8) F1r + (2C sin \u3b8 +B cos \u3b8) G1r \u2223\u2223\u2223\u2223 < 12 , and hence (2A cos \u3b8 +B sin \u3b8) F1 r + (2C sin \u3b8 +B cos \u3b8) G1 r > \u22121 2 for r < R . It follows that V\u2d9 > 1 2 r2 as long as 0 < r < R . Hence V\u2d9 is positive definite on the domain D = { (x , y) |x2 + y2 < R2} . By Theorem 9.6.2, the origin is an unstable critical point. 9.7 509 9.7 3. The equilibrium solutions of the ODE dr dt = r(r \u2212 1)(r \u2212 3) are given by r1 = 0 , r2 = 1 and r3 = 3 . Note that dr dt > 0 for 0 < r < 1 and r > 3 ; dr dt < 0 for 1 < r < 3 . r = 0 corresponds to an unstable critical point. The equilibrium solution r2 = 1 is asymptotically stable, whereas the equilibrium solution r3 = 3 is unstable. Since the critical values are isolated, a limit cycle is given by r = 1 , \u3b8 = t+ t0 which is asymptotically stable. Another periodic solution is found to be r = 3 , \u3b8 = t+ t0 which is unstable. 5. The equilibrium solutions of the ODE dr dt = sin pir are given by r = n , n = 0 , 1 , 2 , . . . . Based on the sign of r \u2032 in the neighborhood of each critical value, the equilibrium solutions r = 2k , k = 1 , 2 , . . . correspond to unstable periodic solutions, with \u3b8 = t+ t0 . The equilibrium solutions r = 2k + 1 , k = 0 , 1 , 2 , . . . correspond to stable limit cycles, with \u3b8 = t+ t0 . The solution r = 0 represents an unstable critical point. 6. The equilibrium solutions of the ODE dr dt = r|r \u2212 2|(r \u2212 3) are given by r1 = 0 , r2 = 2 and r3 = 3 . Note that dr dt < 0 for 0 < r < 3 ; dr dt > 0 for r > 3 . r = 0 corresponds to an asymptotically stable critical point. The equilibrium so- lution r2 = 2 is semistable, whereas the equilibrium solution r3 = 3 is unstable. Since the critical values are isolated, a semistable limit cycle is given by r = 2 , \u3b8 = \u2212t+ t0. Another periodic solution is found to be r = 3 , \u3b8 = \u2212t+ t0 which is unstable. 510 Chapter 9. Nonlinear Differential Equations and Stability 10. Given F (x , y) = a11 x+ a12 y and G(x , y) = a21 x+ a22 y , it follows that Fx +Gy = a11 + a22 . Based on the hypothesis, Fx +Gy is either positive or negative on the entire plane. By Theorem 9.7.2, the system cannot have a nontrivial periodic solution. 12. Given that F (x , y) = \u22122x\u2212 3y \u2212 xy2 and G(x , y) = y + x3 \u2212 x2y , Fx +Gy = \u22121\u2212 x2 \u2212 y2. Since Fx +Gy < 0 on the entire plane, Theorem 9.7.2 asserts that the system cannot have a nontrivial periodic solution. 14.(a) Based on the given graphs, the following table shows the estimated values: µ = 0.2 T \u2248 6.29 µ = 1.0 T \u2248 6.66 µ = 5.0 T \u2248 11.60 (b) The initial conditions were chosen as x(0) = 2 , y(0) = 0 . T \u2248 6.38 . T \u2248 7.65 . 9.7 511 T \u2248 8.86 . T \u2248 10.25 . (c) The period, T , appears to be a quadratic function of µ . 15.(a) Setting x = u and y = u \u2032, we obtain the system of equations dx dt = y dy dt = \u2212x+ µ(1\u2212 1 3 y2)y . 512 Chapter 9. Nonlinear Differential Equations and Stability (b) Evidently, y = 0 . It follows that x = 0 . Hence the only critical point of the system is at (0 , 0). The components of the vector field are infinitely differentiable everywhere. Therefore the system is almost linear. The Jacobian matrix of the vector field is J = ( 0 1 \u22121 µ\u2212 µy2 ) . At the critical point (0 , 0), the coefficient matrix of the linearized system is J(0 , 0) = ( 0 1 \u22121 µ ) , with eigenvalues r1,2 = µ 2 ± 1 2 \u221a µ2 \u2212 4 . If µ = 0 , the equation reduces to the ODE for a simple harmonic oscillator. For the case 0 < µ < 2 , the eigenvalues are complex, and the critical point is an unstable spiral. For µ \u2265 2 , the eigenvalues are real, and the origin is an unstable node. (c) The initial conditions were chosen as x(0) = 2 , y(0) = 0 . A \u2248 2.16 and T \u2248 6.65 . (d) A \u2248 2.00 and T \u2248 6.30 . 9.7 513 A \u2248 2.04 and T \u2248 6.38 . A \u2248 2.6 and T \u2248 7.62 . A \u2248 4.37 and T \u2248 11.61 . (e) A T µ = 0.2 2.00 6.30 µ = 0.5 2.04 6.38 µ = 1.0 2.16 6.65 µ = 2.0 2.6 7.62 µ = 5.0 4.37 11.61 514 Chapter 9. Nonlinear Differential Equations and Stability 16.(a) The critical points are solutions of the algebraic system µx+ y \u2212 x(x2 + y2) = 0 \u2212x+ µ y \u2212 y(x2 + y2)