E209A: Analysis and Control of Nonlinear Systems Problem Set 3 Solutions

Transcription

1 E09A: Analysis and Control of Nonlinear Systems Problem Set 3 Solutions Michael Vitus Stanford University Winter 007 Problem : Planar phase portraits. Part a Figure : Problem a This phase portrait is correct. Consider the closed contour shown in Figure. For the contour, which is positively oriented in the counter-clockwise direction, we find that I f D) = dθ f x) = ) π The only equilibrium point in D is an unstable node. Therefore, If x i ) = ) J which is consistent with the equation ). Figure : Problem a - contour

2 Figure 3: Problem b Part b Figure 4: Problem b - contour 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 I f D) = dθ f x) = 3) π Inside the contour, there are two equilibria, which are both foci. Therefore, If x i ) = 4) J which is inconsistent with equation 3). Figure shows the correct phase portrait in which the right middle arrow has been reversed. Figure : Problem b - 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 I f D) = dθ f x) = ) π Inside the contour, there are 8 equilibria of which 4 are saddles. Therefore, If x i ) = 0 6) J

3 Figure 6: Problem c Figure 7: Problem c - contour which is inconsistent with Eqn. ). 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 The governing equations are x = x ) x = x + x x x 7) The equilibrium of the system is 0, 0). To analyze the system, lets define the function V x) = x + x ) which is proportional to the squared distance from a point x, x ) to the origin. Now we take the derivative of V x) w.r.t. time along the system s trajectory [ )] V x) = x ẋ + x ẋ = x x + x x + x x x V x) = x x x) 9) 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). 8) Let s define a region such that x + x 0) 3

4 Let s first analyze the outer boundary in which x + x =, plugging this into Eqn. 9) and simplifying we obtain V x) = x 4 0 ) which shows that trajectories either enter or orbit the region. Now let s analyze the inner boundary in which x + x =, plugging this into Eqn. 9) and simplifying we obtain V x) = x x 0 ) 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 ẋ = x + x ẋ = g x ) + ax, where a is a constant, a 3) Applying Bendixson s theorem, divf) = f x + f x = + a 4) which is neither zero nor changes sign in all of R so there are no closed orbits or limit cycles. Part ii) The governing equations are ẋ = x x ẋ = x ) Solving for the equiibria results in the x -axis being an equilibrium set. We know that a closed orbit cannot cross the x -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 x -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 Letting x = x and x = ẋ results in ẍ + x x) + ẋ = 0 6) ẋ = x ẋ = x x ) x 7) 4

5 Part a) Solving for the equilibria yields 0, 0) and, 0). To determine the stability of the equilibrium points, we will need to linearize the state model about them. The Jacobian is: [ ] 0 Df = 8) x Substituting in for the equilibrium point at 0, 0): 0 Df 0,0) = The eigenvalues of Eqn. 9 are λ =, which is a stable node. 9) Substituting in for the equilibrium point at, 0) yields 0 Df,0) = 0) The eigenvalues of Eqn. 0 are λ = ± 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 x = 0, the governing equations are for x [0, ], ẋ 0. ẋ = 0 ẋ = x x ) ) On x =, the governing equations are for x [0, ], ẋ 0 and ẋ 0. ẋ = x ẋ = x )

6 On x = x, the inward pointing normal on this line with respect to K is [ ] T. Therefore, the magnitude of f along the vector [ ] T is [ ] x [ ] = x x x ) x x x ) x = x 3) for x = x, x 0. Since the magnitude is greater than zero, we know that the trajectories are pointing in toward K along the line x = x. 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 ẋ < 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, 0) converge to 0, 0). Problem Model The governing equations of the system are ẋ = ax bxy ẏ = dy + cxy 4) The equilibria of the system are: 0, 0) and d c, ) a b. To determine the stability of the equilibrium points, we will need to linearize the state model about them. The Jacobian is: a by bx Df = ) cy d + cx For 0, 0): Df 0,0) = [ a 0 0 d which has eigenvalues of λ = a, d and therefore is a saddle point. ] 6) For d c, a b ) : 0 Df d c, a b ) = db c ac 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 c, a b ) is indeed a center. The intuition 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. 7) 6

7 Figure 9: Problem - Model 7

8 Model The governing equations of the system are The equilibrium points of the system are: 0, 0) ẋ = a by λx)x ẏ = d + cx µy)y ) a λ, 0) bd+aµ bc+λµ, ac dλ bc+λµ To determine the stability of the equilibrium points, we will need to linearize the state model about them. The Jacobian is: a by λx bx Df = 30) cy d + cx µy For 0, 0): Df = [ a 0 0 d which has eigenvalues of λ = a, d and therefore is a saddle point. For a λ, 0) : which has eigenvalues of λ = a, ac λ ] 8) 9) 3) a ba Df = λ 0 d + ca 3) λ d. If ac > λd then it is a saddle. If ac λd then it is a stable node. ) For bd+aµ bc+λµ, ac dλ bc+λµ : If ac < λd, then the equilibrium point will be negative which can t happen. If ac λd, then the linearization is: Plugging in x e, y e ) = The characteristic equation is: which simplifies to Defining Since x e > 0, y e > 0, and λ, µ, b > 0, a bye λx Df xe,y e) = e bx e cy e d + cx e µy e ) bd+aµ bc+λµ, ac dλ bc+λµ and simplifying the expression we obtain: λxe bx Df xe,y e) = e cy e µy e Therefore, the equilibrium will always be stable, and if 33) 34) s + λx e ) s + µy e ) + bcx e y e = 0 3) s + λx e + µy e ) s + λµ + bc) x e y e = 0 36) α 4β > 0 α 4β < 0 α 4β = 0 β = λx e + µy e α = λµ + bc) x e y e 37) β > 0 α > 0 38) stable node stable focus improper stable node 39) 8

9 Figure 0 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 0: Problem - Model Problem 6 The governing equations are ẋ = a x 4xx +x ) 40) ẋ = bx x +x Part a) Solve for the equilibrium points. x from Eqn. 4), x = 0 or =. +x If x = 0, then from Eqn. 4) a = 0 which is a contradiction. x If =, then from Eqn. 4) a x +x 4x = 0 which yields x = a. Therefore there is a unique equilibrium at a a ) ), + a x 4x x + x = 0 4) bx x ) + x = 0 4) To determine the stability of the equilibrium points, we will need to linearize the state model about them. The Jacobian is: 4x + 8x +x x 4x +x Df = ) +x ) b x + bx +x x bx 44) +x ) +x 43) 9

10 simplifying yields, Df = + x a Evaluating the jacobian at, + ) ) a is Df = [ + ) a The characteristic equation of Eqn. 46 is a ) s a b This if we can choose a value for b such that Solving for this value of b yields, + 3x 4x bx bx 3 a + 3 a ) 4 a b a ) b a ] 4) 46) ) s + b a a ) ) + = 0 47) ) ) + b a < 0 then the equilibrium point is unstable. b < 3 a a 48) Part b) Now we need to show the invariance of the region M = { x, x ) x 0, x 0, x a, x + a }. For x = 0, ẋ = a > 0 For x = 0, ẋ = bx > 0 if x > 0 and at 0, 0), ẋ > 0 For x = a, ẋ < 0 for x > 0 and at a, 0), ẋ > 0 For x = + a, ẋ < 0 if 0 < x < a Therefore the region M is invariant. Since M is a closed and bounded region, and for b < 3 a a contains only one unstable equilibrium point there must be a closed orbit inside M by the Poincare -Bendixson theorem. Problem 7 Part a) The governing equations are where σ, λ, and b are positive constants. ẋ = σx x ) ẋ = + λ x 3 )x x ẋ 3 = x x bx 3 49) To solve for the equilibrium points, first set ẋ = 0 which yields x = x. Plugging that into ẋ = 0, yields x 3 = λ. Finally, plugging that into ẋ 3 = 0 gives x = x = ± 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) 0) 0

11 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 = + λ x 3 x ) x x b For 0, 0, 0): The characteristic equation of Eqn. is: Df 0,0,0) = + λ 0 σ σ b ) s 3 + b + σ + ) s + σ b λ σ + b) s bλ σ = 0 3) 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): Df bλ, bλ,λ) = s + σ σ 0 s + b λ b λ b λ s + b 4) The characteristic equation of Eqn. 4 is: s 3 + b + σ + ) s + bλ + σ b + b) s + bλ σ = 0 ) To determine the stability we can construct the Routh array. bλ + σ + ) b + σ + bλσ b + σ + )bλ + bσ + b) bλσ b + σ + 6) bλσ 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: After Simplifying: Now factor the σ terms: Finally, solving for λ: b + σ + )bλ + bσ + b) bλσ b + σ + > 0 7) b + σ)λ + σ + + b)σ + b + ) > 0 8) b + σ)λ + σ + )σ + b + ) > 0 9) λ > σ + )σ + b + ) σ b ) 60) Also, since we know that λ is positive, σ b ) > 0 which simplifies to σ > b + 6)

12 Figure : Problem 7 Part b) Figure shows the projection on the x, x ) plane of a trajectory of the Lorentz equations with σ = 0, λ = 4, and b = with initial condition of,, ). Part c) Differentiating the function V x, x, x 3 ) = x + x + x 3 a) ) : Now plugging in the state equations and simplifying gives us: From this point there are two ways to solve it. Method V = x ẋ + x ẋ + x 3 a)ẋ 3 6) V = σx x bx 3 x 3 a) 63) In this method, we will try to show that V αv + β 64) where α and β are positive constants. Completing the square for x 3 in Eqn. 63) results in: V = σx + x + b x 3 ax 3 + a ) + ) bx 3 ba) V = σx + x + b x 3 a) + bx 3 + 6) ba Since we are trying to bound V we can drop the bx 3 since it will always be positive and will only make V smaller, therefore Eqn. 6 becomes: V σx + x + b ) x 3 a) + ba 66)

13 From this we can easily verify that β = ba 67) In order to determine α, we need to compare the coefficients in Eqn. 66 and V x). Through this comparison we find that α = minσ,, b) 68) Since Eqn. 64 holds, we can then find the region in which V is always less than zero. αv + β < 0 V > β α 69) Therefore, V < 0 when V > β α. No trajectory can leave the region given by V β α and any trajectory starting outside of this region must either enter it or asymptotically approach its boundary. Method This method is more of a geometric argument. By completing the square for x 3 in Eqn. 63 we obtain, V = σx + x + b x 3 a ) ) + 4 ba 70) Realizing that the right hand side of Eqn. 70 represents an ellipse in which we will now consider the region bounded by: which is an ellipse center at 0, 0, a ). On this surface, V = 0. σx + x + b x 3 a ) = 4 ba 7) Outside the region, σx + x + b x 3 a ) > 4 ba, we can show that V < 0. Therefore, we can conclude that all trajectories are bounded within the region: σx + x + b x 3 a ) 4 ba 7) 3