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E209A: Analysis and Control of Nonlinear Systems Problem Set 3 Solutions Michael Vitus Stanford University Winter 2007 Problem 1: Planar phase portraits. Part a Figure 1: Problem 1a This phase portrait is correct. Consider the closed contour shown in Figure 2. For the contour, which is positively oriented in the counter-clockwise direction, we ﬁnd that 1 If (D) = dθf (x) = 1 (1) 2π J The only equilibrium point in D is an unstable node. Therefore, If (xi ) = 1 (2) which is consistent with the equation (1). Figure 2: Problem 1a - contour 1 Figure 3: Problem 1b Figure 4: Problem 1b - contour Part b This phase portrait is impossible. Consider the closed contour shown in Figure 4. For the contour, which is positively oriented in the counter-clockwise direction, we ﬁnd that 1 If (D) = dθf (x) = 1 (3) 2π J Inside the contour, there are two equilibria, which are both foci. Therefore, If (xi ) = 2 (4) which is inconsistent with equation (3). Figure 5 shows the correct phase portrait in which the right middle arrow has been reversed. Figure 5: Problem 1b - ﬁxed Part c This phase portrait is impossible. Consider the closed contour shown in Figure 7. For the contour, which is positively oriented in the counter-clockwise direction, we ﬁnd that 1 If (D) = dθf (x) = 1 (5) 2π J Inside the contour, there are 8 equilibria of which 4 are saddles. Therefore, If (xi ) = 0 (6) 2 Figure 6: Problem 1c Figure 7: Problem 1c - contour which is inconsistent with Eqn. (5). To correct the phase portrait, we need to add an equilibrium point inside the middle circle. This equilibrium can be either an unstable/stable node, unstable/stable focus or a center. Problem 2 The governing equations are ˙ x1 = x2 (7) x2 = −x1 + x2 1 − x2 − 2x2 ˙ 1 2 The equilibrium of the system is (0, 0). To analyze the system, lets deﬁne the function 1 2 V (x) = x + x2 (8) 2 1 2 which is proportional to the squared distance from a point (x1 , x2 ) to the origin. Now we take the derivative of V (x) w.r.t. time along the system’s trajectory ˙ V (x) = x1 x1 + x2 x2 = x1 x2 + x1 −x1 + x2 1 − x2 − 2x2 ˙ ˙ 1 2 ˙ (9) V (x) = x2 1 − x2 − 2x2 2 1 2 ˙ The purpose of V (x) is to show how the system’s trajectories are evolving w.r.t. the contours of V (x). If V ˙ ˙ (x) < 0 then the trajectories of the system are moving closer to (0, 0). If V (x) > 0 then the trajectories of the system are moving away from (0, 0). Let’s deﬁne a region such that 1 ≤ x2 + x2 ≤ 1 1 2 (10) 2 3 Let’s ﬁrst analyze the outer boundary in which x2 + x2 = 1, plugging this into Eqn. (9) and simplifying we 1 2 obtain V (x) = −x4 ≤ 0 2 (11) which shows that trajectories either enter or orbit the region. 1 Now let’s analyze the inner boundary in which x2 + x2 = 2 , plugging this into Eqn. (9) and simplifying we 1 2 obtain V (x) = x2 x2 ≥ 0 1 2 (12) which shows that trajectories either enter or orbit the region. Therefore we have shown that the annulus is invariant. By Poincare’-Bendixson’s theorem, there must be either a closed orbit or an equilibrium point inside, but the only equilibrium point is at (0, 0) which is outside the annulus. Therefore, the annulus must contain a closed orbit. Problem 3 Part (i) The governing equations are x1 = −x1 + x2 ˙ (13) ˙ x2 = g (x1 ) + ax2 , where a is a constant, a = 1 Applying Bendixson’s theorem, ∂f1 ∂f2 div(f ) = + = −1 + a (14) ∂x1 ∂x2 2 which is neither zero nor changes sign in all of so there are no closed orbits or limit cycles. Part (ii) The governing equations are ˙ x1 = x1 x2 (15) ˙ x2 = x2 Solving for the equiibria results in the x1 -axis being an equilibrium set. We know that a closed orbit cannot cross the x1 -axis, therefore it must be either be in the upper or lower halves of the plane. Since there are no equilibrium points other than the x1 -axis, we can conclude that there are no closed orbits as a result from index theory that a closed orbit must enclose an equilibrium point. Problem 4 The governing equation is x + x (1 − x) + 2x = 0 ¨ ˙ (16) ˙ Letting x1 = x and x2 = x results in ˙ x1 = x2 (17) x2 = −x1 (1 − x1 ) − 2x2 ˙ 4 Part (a) Solving for the equilibria yields (0, 0) and (1, 0). To determine the stability of the equilibrium points, we will need to linearize the state model about them. The Jacobian is: 0 1 Df = (18) 2x1 − 1 −2 Substituting in for the equilibrium point at (0, 0): 0 1 Df |(0,0) = (19) −1 −2 The eigenvalues of Eqn. 19 are λ = −1, −1 which is a stable node. Substituting in for the equilibrium point at (1, 0) yields 0 1 Df |(1,0) = (20) 1 −2 √ The eigenvalues of Eqn. 20 are λ = −1 ± 2 which is a saddle. Part (b) For this part of the problem we need to show that the region in Figure 8 is positively invariant. Figure 8: Problem 4 - Invariant Region • On x2 = 0, the governing equations are ˙ x1 = 0 (21) x2 = −x1 (1 − x1 ) ˙ for x1 ∈ [0, 1], x2 ≤ 0. ˙ • On x1 = 1, the governing equations are ˙ x1 = x2 (22) x2 = −2x2 ˙ for x2 ∈ [0, −1], x1 ≤ 0 and x2 ≥ 0. ˙ ˙ 5 T • On x2 = −x1 , the inward pointing normal on this line with respect to K is [1 1] . Therefore, the T magnitude of f along the vector [1 1] is x2 [1 1] = x2 − x1 (1 − x1 ) − 2x2 = x2 1 (23) −x1 (1 − x1 ) − 2x2 for x2 = −x1 , x2 ≥ 0. Since the magnitude is greater than zero, we know that the trajectories are 1 pointing in toward K along the line x2 = −x1 . Therefore, the region K is positively invariant. Part (c) By Poincare’-Bendixson, all trajectories inside K converge to either a closed orbit or an equilibrium point. ˙ There are no closed orbits in K since x1 < 0 in K, or by index theory - if there were a closed orbit it would surround an equilibrium point and there are no equilibrium points in the interior of K. Therefore, all trajectories not at (1, 0) converge to (0, 0). Problem 5 Model 1 The governing equations of the system are x = ax − bxy ˙ (24) y = −dy + cxy ˙ The equilibria of the system are: (0, 0) and d , a . To determine the stability of the equilibrium points, we c b will need to linearize the state model about them. The Jacobian is: a − by −bx Df = (25) cy −d + cx For (0, 0): a 0 Df |(0,0) = (26) 0 −d which has eigenvalues of λ = a, −d and therefore is a saddle point. d a For c, b : −db 0 Df |( d , a ) = ac c (27) c b b 0 √ which has eigenvalues of λ = ± adj. By the Hartman-Grobman theorem, we can’t trust the linearization since the eigenvalues are on the jω-axis. As you will see from the simulation, Figure 9, the equilibrium point at d , a is indeed a center. The intuition c b behind this is that any perturbations from steady state will cause cyclic variations. Consider when sardines are low and sharks are high, the sharks will start to die oﬀ because their food source is low. Since the sharks are dying oﬀ, the sardines will start to increase in population because there are less predators around. With a large sardine population and low shark population, the sharks will ﬂourish and increase in number which in turn decreases the sardine population. Now this cyclic trend will repeat. 6 Figure 9: Problem 5 - Model 1 7 Model 2 The governing equations of the system are x = (a − by − λx)x ˙ (28) y = (−d + cx − µy)y ˙ The equilibrium points of the system are: a bd+aµ ac−dλ (0, 0) λ, 0 bc+λµ , bc+λµ (29) To determine the stability of the equilibrium points, we will need to linearize the state model about them. The Jacobian is: a − by − 2λx −bx Df = (30) cy −d + cx − 2µy For (0, 0): a 0 Df = (31) 0 −d which has eigenvalues of λ = a, −d and therefore is a saddle point. a For λ, 0 : −a − ba λ Df = (32) 0 −d + ca λ which has eigenvalues of λ = −a, ac − d. If ac > λd then it is a saddle. If ac ≤ λd then it is a stable node. λ For bd+aµ , ac−dλ : bc+λµ bc+λµ If ac < λd, then the equilibrium point will be negative which can’t happen. If ac ≥ λd, then the linearization is: a − bye − 2λxe −bxe Df(xe ,ye ) = (33) cye −d + cxe − 2µye bd+aµ ac−dλ Plugging in (xe , ye ) = bc+λµ , bc+λµ and simplifying the expression we obtain: −λxe −bxe Df(xe ,ye ) = (34) cye −µye The characteristic equation is: (s + λxe ) (s + µye ) + bcxe ye = 0 (35) which simpliﬁes to s2 + (λxe + µye ) s + (λµ + bc) xe ye = 0 (36) Deﬁning β = λxe + µye (37) α = (λµ + bc) xe ye Since xe > 0, ye > 0, and λ, µ, b > 0, β>0 α>0 (38) Therefore, the equilibrium will always be stable, and if α2 − 4β > 0 stable node α2 − 4β < 0 stable focus (39) α2 − 4β = 0 improper stable node 8 Figure 10 depicts an example solution. The plot depicts that there always exists a stable population of sharks and sardines. No matter what the initial populations are, they always reach the stable equilibrium. Figure 10: Problem 5 - Model 2 Problem 6 The governing equations are 4x1 x2 x1 = a − x1 − ˙ 1+x2 1 x2 (40) x2 = bx1 1 − ˙ 1+x12 Part (a) Solve for the equilibrium points. 4x1 x2 a − x1 − =0 (41) 1 + x2 1 x2 bx1 1 − =0 (42) 1 + x2 1 x2 from Eqn. (42), x1 = 0 or 1+x2 = 1. 1 If x1 = 0, then from Eqn. (41) a = 0 which is a contradiction. x2 If 1+x2 = 1, then from Eqn. (41) a − x1 − 4x1 = 0 which yields x1 = a . 5 1 Therefore there is a unique equilibrium at a a 2 ,1 + (43) 5 5 To determine the stability of the equilibrium points, we will need to linearize the state model about them. The Jacobian is: 4x2 8x2 x2 −4x1 −1 − 1+x2 + 1 2 2 1+x1 1 (1+x2 ) 1 Df = (44) x2 2bx2 x2 −bx1 b 1 − 1+x2 + 1 2 1+x2 1 1+x2 ) ( 1 1 9 simplifying yields, 1 −5 + 3x2 1 −4x1 Df = (45) 1 + x2 1 2bx2 1 −bx1 a a 2 Evaluating the jacobian at 5,1 + 5 is a 2 1 −5 + 3 5 −4 a 5 Df = 2 (46) a 2 2b a −b a 1+ 5 5 5 The characteristic equation of Eqn. 46 is a 2 a a a 2 s2 + 5 − 3 +b s + 5b 1+ =0 (47) 5 5 5 5 a 2 This if we can choose a value for b such that 5 − 3 5 + ba 5 < 0 then the equilibrium point is unstable. Solving for this value of b yields, a 25 b<3 − (48) 5 a Part (b) Now we need to show the invariance of the region M = (x1 , x2 ) |x1 ≥ 0, x2 ≥ 0, x1 ≤ a, x2 ≤ 1 + a2 . • For x1 = 0, x1 = a > 0 ˙ • For x2 = 0, x2 = bx1 > 0 if x1 > 0 and at (0, 0), x1 > 0 ˙ ˙ • For x1 = a, x1 < 0 for x2 > 0 and at (a, 0), x2 > 0 ˙ ˙ • For x2 = 1 + a2 , x2 < 0 if 0 < x1 < a ˙ Therefore the region M is invariant. Since M is a closed and bounded region, and for b < 3 a − 25 contains only one unstable equilibrium point 5 a there must be a closed orbit inside M by the Poincare’-Bendixson theorem. Problem 7 Part (a) The governing equations are x1 = σ(x2 − x1 ) ˙ x2 = (1 + λ − x3 )x1 − x2 ˙ (49) x3 = x1 x2 − bx3 ˙ where σ, λ, and b are positive constants. ˙ ˙ To solve for the equilibrium points, ﬁrst set x1 = 0 which yields x1 = x2 . Plugging that into x2 = 0, √ yields x3 = λ. Finally, plugging that into x3 = 0 gives x1 = x2 = ± bλ. Also, by inspection (0, 0, 0) is an ˙ equilibrium point. Therefore the three equilibrium points are: √ √ √ √ (0, 0, 0) ( bλ, bλ, λ) (− bλ, − bλ, λ) (50) 10 To determine the stability of the equilibrium points, we will need to linearize the state model about them. The Jacobian is −σ σ 0 Df = 1 + λ − x3 −1 −x1 (51) x2 x1 −b For (0, 0, 0): −σ σ 0 Df |(0,0,0) = 1+λ −1 0 (52) 0 0 −b The characteristic equation of Eqn. 52 is: s3 + (b + σ + 1) s2 + (σ b − λ σ + b) s − bλ σ = 0 (53) Since the constant coeﬃcient is negative we can conclude that the equilibrium point at (0, 0, 0) is always unstable, which comes from the necessary condition for Routh stability. √ √ For ( bλ, bλ, λ): −σ s+σ 0 √ Df |(√bλ,√bλ,λ) = −1 s + 1 − bλ (54) √ √ bλ bλ s + b The characteristic equation of Eqn. 54 is: s3 + (b + σ + 1) s2 + (bλ + σ b + b) s + 2 bλ σ = 0 (55) To determine the stability we can construct the Routh array. 1 b(λ + σ + 1) b+σ+1 2bλσ (b + σ + 1)(bλ + bσ + b) − 2bλσ (56) b+σ+1 2bλσ In order for the system to be stable, all of the coeﬃcients in the ﬁrst column have to be positive. Considering the third coeﬃcient from the top: (b + σ + 1)(bλ + bσ + b) − 2bλσ >0 (57) b+σ+1 After Simplifying: (b + 1 − σ)λ + σ 2 + (2 + b)σ + (b + 1) > 0 (58) Now factor the σ terms: (b + 1 − σ)λ + (σ + 1)(σ + b + 1) > 0 (59) Finally, solving for λ: (σ + 1)(σ + b + 1) λ> (60) (σ − b − 1) Also, since we know that λ is positive, (σ − b − 1) > 0 which simpliﬁes to σ >b+1 (61) 11 Figure 11: Problem 7 Part (b) Figure 11 shows the projection on the (x1 , x2 ) plane of a trajectory of the Lorentz equations with σ = 10, λ = 24, and b = 2 with initial condition of (5, 5, 5). Part (c) 1 Diﬀerentiating the function V (x1 , x2 , x3 ) = 2 x2 + x2 + (x3 − a)2 : 1 2 ˙ V = x1 x1 + x2 x2 + (x3 − a)x3 ˙ ˙ ˙ (62) Now plugging in the state equations and simplifying gives us: ˙ V = −σx2 − x2 − bx3 (x3 − a) (63) 1 2 From this point there are two ways to solve it. Method 1 In this method, we will try to show that ˙ V ≤ −αV + β (64) where α and β are positive constants. Completing the square for x3 in Eqn. (63) results in: ˙ V = − σx2 + x2 + 2 x2 − 2ax3 + a2 + 2 bx2 − 1 ba2 b 1 1 2 3 3 2 ˙ 2 (65) V = − σx + x + (x3 − a) + bx + 1 ba2 2 1 2 b 2 2 1 2 2 3 2 ˙ ˙ Since we are trying to bound V we can drop the 1 bx2 since it will always be positive and will only make V 2 3 smaller, therefore Eqn. 65 becomes: ˙ b 2 1 V ≤ − σx2 + x2 + (x3 − a) 1 2 + ba2 (66) 2 2 12 From this we can easily verify that 1 2 β= ba (67) 2 In order to determine α, we need to compare the coeﬃcients in Eqn. 66 and V (x). Through this comparison we ﬁnd that α = min(2σ, 2, b) (68) ˙ Since Eqn. 64 holds, we can then ﬁnd the region in which V is always less than zero. −αV + β < 0 β (69) V >α ˙ Therefore, V < 0 when V > β α. β No trajectory can leave the region given by V ≤ α and any trajectory starting outside of this region must either enter it or asymptotically approach its boundary. Method 2 This method is more of a geometric argument. By completing the square for x3 in Eqn. 63 we obtain, a 2 1 ˙ V = − σx2 + x2 + b x3 − 1 2 + ba2 (70) 2 4 Realizing that the right hand side of Eqn. 70 represents an ellipse in which we will now consider the region bounded by: a 2 1 2 σx2 + x2 + b x3 − 1 2 = ba (71) 2 4 ˙ which is an ellipse center at 0, 0, a . On this surface, V = 0. 2 a 2 1 ˙ Outside the region, σx2 + x2 + b x3 − 1 2 2 > 4 ba2 , we can show that V < 0. Therefore, we can conclude that all trajectories are bounded within the region: a 2 1 2 σx2 + x2 + b x3 − 1 2 ≤ ba (72) 2 4 13

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