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Engineering Mathematics Class #10 Chapter 4 ( Part2)

Engineering Mathematics Class #10 Chapter 4 ( Part2). Sheng-Fang Huang. Five Types of Critical Points. Improper Node. The system The two eigenvectors are x (1) = [1 1] T and – x (2) = [1 -1] T. Proper Node. The system

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Engineering Mathematics Class #10 Chapter 4 ( Part2)

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  1. Engineering Mathematics Class #10 Chapter 4 (Part2) Sheng-Fang Huang

  2. Five Types of Critical Points

  3. Improper Node • The system • The two eigenvectors are x(1) = [1 1]T and –x(2) = [1 -1]T.

  4. Proper Node • The system has a proper node at the origin. Its characteristic equation (1 –λ)2 = 0 has the root λ = 1. Any x ≠0 is an eigenvector, and we can take [1 0]T and [0 1]T. Hence a general solution is

  5. Fig. 82. Trajectories of the system (10) (Proper node)

  6. Saddle Point • The system has a saddle point at the origin. Its characteristic equation (1 –λ)(–1 –λ) = 0 has the roots λ1 = 1 and λ2 = –1. For λ1= 1 an eigenvector [1 0]T is obtained from the second row of (A –λI)x = 0, that is, 0x1 + (–1 – 1)x2 = 0. For λ2 = –1 the first row gives [0 1]T. Hence a general

  7. Fig.83. Trajectories of the system (11) (Saddle point)

  8. Canter • The system (12) has a center at the origin. The characteristic equation λ2 + 4 = 0 gives the eigenvalues 2i and –2i. For 2i an eigenvector follows from the first equation –2ix1 + x2 = 0 of (A –λI)x = 0, say, [1 2i]T. For λ = –2i that equation is –(–2i)x1 + x2 = 0 and gives, say, [1 –2i]T. Hence a complex general solution is (12*)

  9. Rewrite the given equations in the form y'1 = y2, 4y1 = –y'2; then the product of the left sides must equal the product of the right sides,

  10. Fig.84.Trajectories of the system (12) (Center)

  11. Spiral Point • The system has a spiral point at the origin, as we shall see. The characteristic equation is λ2 + 2λ+ 2 = 0. It gives the eigenvalues –1 + i and –1 – i. Corresponding eigenvectors are obtained from (–1 –λ)x1 + x2 = 0. For λ = –1 + i this becomes –ix1 + x2 = 0 and we can take [1 i]T as an eigenvector. This gives the complex general solution

  12. Accordingly, we start again from the beginning and instead of that rather lengthy systematic calculation we use a shortcut. We multiply the first equation in (13) by y1, the second by y2, and add, obtaining y1y'1 + y2y'2 = –(y12 + y22). We now introduce polar coordinates r, t, where r2 = y12 + y22. Differentiating this with respect to t gives 2rr' = 2y1y'1 + 2y2y'2. Hence the previous equation can be written rr' = –r2, Thus, r' = –r, dr/r = –dt, ln│r│ = –t +c*, r =ce-t.

  13. Fig. 85. Trajectories of the system (13) (Spiral point)

  14. 4.4 Criteria for Critical Points. Stability • We continue our discussion of homogeneous linear systems with constant coefficients (1)

  15. (3) We also recall from Sec. 4.3 that there are various types of critical points, and we shall now see how these types are related to the eigenvalues. The latter are solutions λ=λ1 and λ2 of the characteristic equation (4)

  16. This is a quadratic equation λ2 – pλ +q = 0 with coefficients p, q and discriminant Δ given by p =a11 + a22, q = det A = a11a22 – a12a21, Δ = p2 – 4q. From calculus we know that the solutions of this equation are (6)

  17. Table 4.1 Eigenvalue Criteria for Critical Points

  18. DEFINITION Stable A critical point P0 of (1) is called stable if all trajectories of (1) that at some instant are close to P0 remain close to P0 at all future times Stability

  19. Fig.89.Stable critical point P0 of (1) (The trajectory initiating at P1 stays in the disk of radius ε.)

  20. DEFINITION Unstable, Stable and Attractive P0 is called unstable if P0 is not stable. P0 is called stable and attractive if P0 is stable and every trajectory that has a point in Dδapproaches P0 as t →∞. Stability

  21. Fig.90.Stable and attractive critical point P0 of (1)

  22. Table 4.2 Stability Criteria for Critical Points

  23. Stability Chart

  24. Example 1: • In Example 1, Sec. 4.3, we have y' = y, p = –6, q = 8, Δ = 4, a node by Table 4.1(a), which is stable and attractive by Table 4.2(a).

  25. Example 2:Free Motions of a Mass on a Spring • What kind of critical point does my" + cy' + ky = 0 in Sec. 2.4 have? • Solution.Division by m gives y'' = –(k/m)y – (c/m)y'. To get a system, set y1 = y, y2 = y'. Then y'2 = y'' = –(k/m)y1 – (c/m)y2. Hence

  26. We see that p = –c/m, q =k/m, Δ = (c/m)2 – 4k/m. From Tables 4.1 and 4.2, we obtain the following results. Note that in the last three cases the discriminant Δ plays an essential role. • No damping. c = 0, p = 0, q > 0, a center. • Underdamping. c2 < 4mk, p < 0, q > 0, Δ < 0, a stable and attractive spiral point. • Critical damping. c2 = 4mk, p < 0, q > 0, Δ = 0, a stable and attractive node. • Overdamping. c2 > 4mk, p < 0, q > 0, Δ > 0, a stable and attractive node.

  27. 4.5 Qualitative Methods for nonlinear Systems • Qualitative methods are methods of obtaining qualitative information on solutions without actually solving a system. • These methods are particularly valuable for systems whose solution by analytic methods is difficult or impossible.

  28. Phase plane method is a kind of qualitative method. • In this section, we will extend phase plane method from linear system to nonlinear systems y'1 = f1(y1, y2) (1) y' = f(y), thus y'2 = f2(y1, y2) • We assume that (1) is autonomous. • Constant coefficients; the independent variable t does not occur explicitly.

  29. As a convenience, each time we first move the critical point P0: (a, b) to be considered to the origin (0, 0). This can be done by a translation which moves P0 to (0, 0). Thus we can assume P0 to be the origin (0, 0), and for simplicity we continue to write y1, y2 (instead of ˜y1, ˜y2).

  30. Linearization of Nonlinear Systems • How to determine the kind and stability of a critical point P0: (0, 0) of nonlinear system? • Linearization • Rewrite (1) as y'1 = a11y1 + a12y2 + h1(y1, y2) (2) y' = Ay + h(y), thus y'2 = a21y1 + a22y2 + h2(y1, y2) A is constant (independent of t). One can prove that:

  31. THEOREM 1 Linearization If ƒ1 and ƒ2 in (1) are continuous and have continuous partial derivatives in a neighborhood of the critical point P0: (0, 0), and if detA≠ 0 in (2), then the kind and stability of the critical point of (1) are the same as those of the linearized system y'1 = a11y1 + a12y2 (3)y' = Ay, thus y'2 = a21y1 + a22y2 Exceptions occur if A has equal or pure imaginary eigenvalues; then (1) may have the same kind of critical point as (3) or a spiral point.

  32. Example 1: Free Undamped Pendulum. • A pendulum consisting of a body of mass m and a rod of length L. Determine the locations and types of the critical points. Assume that the mass of the rod and air resistance are negligible.

  33. Example 1: Free Undamped Pendulum. • Solution. Step 1. Setting up the mathematical model. • Let θ denote the angular displacement, measured counterclockwise from the equilibrium position. The weight of the bob is mg (g the acceleration of gravity). • The mathematical model is: mLθ'' +mg sinθ = 0. Dividing this by mL, we have (4) θ'' + k sinθ = 0

  34. Step 2. Critical points (0, 0), ±(2π, 0), ±(4π, 0), ‥‥ , Linearization. To obtain a system of ODEs, we set θ =y1, θ' = y2. Then from (4) we obtain a nonlinear system (1) of the form y'1 = ƒ1(y1, y2) = y2 (4*) y'2 = ƒ2(y1, y2) = –k sin y1.

  35. The right sides are both zero when y2 = 0 and sin y1 = 0. This gives infinitely many critical points (nπ, 0), where n = 0, ±1, ±2, ‥‥. We consider (0, 0). Since the Maclaurin series is the linearized system at (0, 0) is

  36. To apply our criteria in Sec. 4.4: p =a11 + a22 = 0 q = det A = k = g/L (> 0) Δ = p2 – 4q = –4k. Thus, we conclude that (0, 0) is a center, which is always stable. Since sinθ = sin y1 is periodic with period 2π, the critical points (nπ, 0), n = ±2, ±4, ‥‥ , are all centers.

  37. Step 3. Critical points ± (π, 0), ±(3π, 0), ±(5π, 0), ‥‥, Linearization. • We now consider the critical point (π, 0), setting θ–π = y1 and (θ–π)' = θ' =y2. Then in (4), the linearized system at (π, 0) is now

  38. We see that p = 0, q = –k (< 0) Δ = –4q = 4k. • Hence, this gives a saddle point, which is always unstable. • Because of periodicity, the critical points (nπ, 0), n = ±1, ±3, ‥‥, are all saddle points.

  39. Fig.92.Example 1 (C will be explained in Example 4.)

  40. Example 2: Linearization of the Damped Pendulum Equation • Now we add a damping term cθ' (damping proportional to the angular velocity) to equation (4), so that it becomes (5) θ'' + cθ' +k sinθ = 0 where k > 0 and c ≥ 0. Setting θ =y1, θ' =y2, as before, we obtain the nonlinear system (use θ'' =y'2 ) y'1 = y2 y'2 = –k sin y1 – cy2.

  41. Critical points as before, namely, (0, 0), (±π, 0), (±2π, 0), ‥‥. We consider (0, 0). Linearizing sin y1≈y1 as in Example 1, we get the linearized system at (0, 0) (6) p = q = Δ =

  42. We now consider the critical point (π, 0). We set θ–π= y1, (θ–π)' = θ' =y2 and linearize sinθ = sin (y1 + π) = –sin y1≈ –y1. This gives the new linearized system at (π, 0) (6*)

  43. p =a11 + a22 = –c, q = det A = –k Δ = p2 – 4q =c2 + 4k. This gives the following results for the critical point at (π, 0). • No damping. • c = 0, p = 0, q < 0, Δ > 0, a saddle point. • Damping. • c > 0, p < 0, q < 0, Δ > 0, a saddle point. Since sin y1 is periodic with period 2π, the critical points (±2π, 0), (±4π, 0), ‥‥are of the same type as (0, 0), and the critical points (–π, 0), (±3π, 0), ‥‥are of the same type as (π, 0), so that our task is finished.

  44. Fig.93.Trajectories in the phase plane for the damped pendulum in Example 2

  45. Lotka–Volterra Population ModelExample 3: Predator–Prey Population Model • This model concerns two species, say, rabbits and foxes, and the foxes prey on the rabbits. • Step 1. Setting up the model. We assume the following. 1.Rabbits have unlimited food supply. Hence if there were no foxes, their number y1(t) would grow exponentially, y1'= ay1 2.Actually, y1 is decreased because of the kill by foxes, say, at a rate proportional to y1y2, where y2(t) is the number of foxes. Hence y1'= ay1 – by1y2, where a > 0 and b > 0.

  46. 3.If there were no rabbits, then y2(t) would exponentially decrease to zero, y2'= –ly2. However, y2 is increased by a rate proportional to the number of encounters between predator and prey; together we have y2'= –ly2 + ky1y2, where k > 0 and l > 0. This gives the Lotka–Volterra system y1'= ƒ1(y1, y2) = ay1 – by1y2 (7) y2' = ƒ2(y1, y2) = ky1y2 – ly2 .

  47. Step 2. Critical point (0, 0), Linearization. We see from (7) that the critical points are the solutions of (7*) ƒ1(y1, y2) = y1(a –by2) = 0, ƒ2(y1, y2) = y2(ky1 – l) = 0. The solutions are (y1, y2) = (0, 0) and (l/k, a/b). We consider (0, 0). Dropping –by1y2 and ky1y2 from (7) gives the linearized system Its eigenvalues are λ1 = a > 0 and λ2 = –l < 0. They have opposite signs, so that we get a saddle point.

  48. Step 3. Critical point (l /k, a/b), Linearization. We set Then the critical point (l/k, a/b) corresponds to = (0, 0). Since we obtain from (7)

  49. Dropping the two nonlinear terms and , we have the linearized system (7**)

  50. Fig.94.Ecological equilibrium point and trajectory of the linearized Lotka–Volterra system (7**)

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