E209A: Analysis and Control of Nonlinear Systems Problem Set 2 Solutions

Transcription

1 E9A: Analysis and Control of Nonlinear Systems Problem Set Solutions Gabe Hoffmann Stanford University Winter 5 Problem : The method of isoclines. We consider the system The equilibria are found by setting the state derivatives to zero. Combining these results, we find that So, we have equilibria at (,) and (,). ẋ = x x x () ẋ = x + x () ẋ = x x x = x (x x ) = (3) ẋ = x + x = x = x (4) x ( x ) = x = or x = (5) Now, on the x axis, x =. So, on the x axis, ẋ = x x x =. Therefore, the x axis is invariant. The slope dx dx is found as follows. dx dx = ( ) dx dt ( dx dt ) = x + x x x x (6) So, dx dx when the denominator goes to zero while the numerator remains finite. Factoring the denominator, we see that it is (x )(x x ). So, when x = or x = x, dx dx. Now, to find the isoclines on which dx dx = c for finite c. For c =, from (6), we find that the isocline is x + x =. So, c = on x = x (7)

2 For c =.5, from (6), we find that the isocline is x +x x x x =.5. So, c =.5 on x = x x (8) For c =, from (6), we find that the isocline is x + x = x x x. So, c = on x = and x = (9) For c =, from (6), we find that the isocline is x + x = (x x x ). So, c = on x = x x () The next part of the problem requires the curves to be sketched with the associated slopes on top of the isoclines. These drawings should be done by hand. The exact curves are plotted in Figure..5.5 x x Figure : Isoclines for Problem Finally, we need to conjecture the phase portrait from this information. The trajectories are found by tracing curves through the slopes that we plotted on top of the isoclines. In order to establish the direction of these trajectories, we need to look at the differential equations at some of the isoclines to determine the direction of the derivative field. This plot should be made by hand in your solutions. It is plotted by computer in Figure

3 x = x x x x = x + x.5.5 x x Figure : Isoclines and phase portrait for Problem 3

4 Problem : The pumping heart. The Van der Pol oscillator equations are ẋ = v µ(x 3 /3 x) () v = x () where x is the muscle fiber length, v is the stimulous, and µ > is a parameter. The equilibria are found by setting the state derivatives to zero. ẋ = v µ(x 3 /3 x) = (3) v = x = (4) Combining these two equations, we see that the only equilibrium is at (,). Next, we need to find the stability of this equilibrium. To do so, we linearize about it. µ(x ) µ = (5) (,) Next, we determine the eigenvalues of the Jacobian. µ λ det (Df λi) = det λ = λ µλ + = (6) Thus, the eigenvalues are at λ = µ ± µ 4 So, when µ >, there are two positive real eigenvalues, and the equilibrium is an unstable node. When µ =, there are two identical eigenvalues greater than zero, so the equilibrium is an improper unstable node. When < µ <, there are two complex eigenvalues with positive real parts, so the equilibrium is an unstable focus. See the figures at the end of this problem for plots at small and large µ. Finally, we need to show that in the transition from long fibers to short fibers, the contraction happens slowly at first, but that at a high enough stimulus the fibers contract suddenly to push the blood all throughout the body. We cannot directly read from the phase portrait how sudden the contraction is, because it s the plot of v versus x, not x versus time. Although you can see some time information from the vector field plot of f(x,x ) which we typically plot along with the phase portrait in pplane, we are supposd to show that the contraction happens slowly at first, and then suddenly at a high stimulus. From the phase plot, we can find out how v and x change - the system exhibits a limit cycle. Let s look at ẋ = v µ(x 3 /3 x) (8) (7) 4

5 In the upper right corner of the limit cycle, at the point where ẋ =, v = µ(x 3 /3 x), so the muscle fiber length is constant. The extension is positive, so the equation v = x says that v will continue decreasing. As v decreases, ẋ will transition to a slightly negative value. Thus, with x decreasing slightly, the equation µ(x 3 /3 x) decreases as well. This decrease balances out the decrease in v, moderating the change in ẋ, until the cubic equation reaches a local minimum, found by d ( ) x 3 /3 x = dx (9) x = x = ± () The positive solution corresponds to the minimum. So when x = is reached, the cubic equation turns around, and x begins to decrease very increasingly rapidly. It is no longer dependent on the magnitude of v increasing. Thus, the fiber contracts increasingly rapidly until the maximum of the cubic equation is reached, and the rate decreases again, and eventually goes to zero. We begin the same process all over again, symmetric about the origin. x = v mu (x 3 /3 x) v = x mu = 4 3 v x Figure 3: Phase portrait with µ =. This is an unstable focus. 5

6 x = v mu (x 3 /3 x) v = x mu = 4 3 v x Figure 4: Phase portrait with µ =. This is an improper unstable node. x = v mu (x 3 /3 x) v = x mu = v x Figure 5: Phase portrait with µ = 4. This is an unstable node. 6

7 Problem 3: Modification of Duffing s equation. The modified Duffing equation is First, we find its equilibria. ẋ = x () ẋ = x x 3 δx + x x () ẋ = = x (3) ẋ = = x ( x ) (4) So, we have equilibria at (,), (,), and (,). Next, we linearize about the equilibria. Around (,), So the characteristic equation is det 3x + x x x δ λ δ λ δ xe (5) (6) = λ + δλ = (7) Therefore, the eigenvalues at (,) are λ = δ± δ +4, a saddle for any value of δ. Around (,), δ So the characteristic equation is λ det = λ + (δ )λ + = (9) δ λ Therefore, the eigenvalues at (,) are λ = (δ )± (δ ) 8, which will have stability varying with δ. The result for (-,) is the same. The stability regions are (8) δ < Unstable Node (3) δ = Unstable Improper Node (3) < δ < Unstable Focus (3) δ = Center in linearized system (33) < δ < + Stable Focus (34) δ = + Stable Improper Node (35) δ > + Stable Node (36) 7

8 Now, to apply Bendixson s theorem, we evaluate the divergence of our vector field. div(f) = f x + f x = + ( δ) + x = x δ (37) So, δ is never identically zero on any region in R except on the lines where δ = x. The line cannot contain an orbit, the first condition to say that orbits are possible is not met. What about the second condition? From (37), we know that div(f) changes sign on the lines where δ = x, except when δ =. Thus, by Bendixson s theorem, orbits could exist that cross these lines where a sign change occurs. They could not occur in the region between the lines alone, our outside the lines alone. To conjecture plausible phase portraits, we return to the eigenvalue expressions. In the regions defined by (3) and (3), the line x = ± δ doesn t exist, so no closed orbits can exist. The phase portrait can then be conjectured and sketched by hand in the neighborhood of the equilibrium points. It shouldn t have any closed orbits. The actual phase portrait is shown in Figure 6. x = x x = x x 3 delta x + x x delta = sqrt().5.5 x x Figure 6: Phase portrait for δ =. By Bendixson s theorem, there can t be any closed orbits. The region defined by (3) will be similar to the region defined by (3) and (3 when δ, except the equilibria at (±,) change from nodes to foci. When δ >, a the lines x = ± δ exist between each equilibrium in region (3). Thus, orbits could exist crossing any combination of these lines, as long as they enclose an equilibrium point as well, and satisfy index theory. Again, sketches just need to show guesses of where closed orbits could occur, and the behavior around the equilibrium. Figures 7, 8 and 9 show interesting actual phase portraits in this domain. In the region defined by (33), the lines x = ± δ lie on the equilibria at (±,). So, the points could be nonlinear centers. Again, a closed orbit could exist across the lines. Sketches need to 8

9 x = x x = x x 3 delta x + x x delta = x x Figure 7: Phase portrait for δ =.5, with red lines indicating where the divergence changes sign. By Bendixson s theorem, there can be closed orbits crossing the lines x = ±.5. It turns out that there aren t in this case. show guesses of what the phase portrait could be. It turns out that the equilibria at (±,) are very strange, and not exactly nonlinear centers, as shown in Figure. In the regions defined by (34), (35), and (36), the lines x = ± δ no longer separate the equilibria. Again, a closed orbit could exist across the lines. Once again, sketches just need to show guesses of what the phase portrait could be. One example of a real phase portrait is given in Figure. 9

10 x = x x = x x 3 delta x + x x delta = x x Figure 8: Phase portrait for δ =.8. By Bendixson s theorem, there can be closed orbits crossing the lines x = ±.8. It turns out that there is one closed orbit encompassing all three equilibria. x = x x = x x 3 delta x + x x delta = x x Figure 9: Phase portrait for δ =.9. By Bendixson s theorem, there can be closed orbits crossing the lines x = ±.9. Increasing δ by just., the system goes from having orbit to having 3 orbits. Note that all orbits cross at least one of the sign change lines.

11 x = x x = x x 3 delta x + x x delta =.5.5 x x Figure : Phase portrait for δ =. By Bendixson s theorem, there can be closed orbits crossing the lines x = ±. It turns out that there is one closed orbit again, encompassing all 3 equilibria. x = x x = x x 3 delta x + x x delta =..5.5 x x Figure : Phase portrait for δ =.. By Bendixson s theorem, there can be closed orbits crossing the lines x = ±.. It turns out that there is a closed orbit, surrounding all 3 equilibria.

12 Problem 4: First integrals. We consider the differential equation θ = sin θ (38) First, we put it in state model form with x = θ and x = θ. Then, ẋ = x (39) ẋ = sin x (4) The equilibrium points are then found by setting the derivatives to zero. ẋ = = x x = (4) ẋ = = sin x x = sinx = x = π 6 + πn, 5π 6 + πn (4) Thus, the equilibria are at ( π 6 + πn,) and (5π 6 + πn,). Next, we linearize the system. cos x (43) At ( 5π 6 + πn,), the Jacobian is 3 (44) So the eigenvalues are λ = ±3 /4, thus this point is a saddle. At ( π 6 + πn,), the Jacobian is 3 (45) So the eigenvalues are λ = ±3 /4 j, thus this point is a center in the linearized system. Now we ask if this is a nonlinear center. We use the first integral. If we can find a scalar function V (x,x ) has a time derivative of zero along the system trajectories, then that function is constant along the trajectories. Hence, its level curves would define the trajectories. Let s find such a scalar function for this system. So, one solution is finding V such that d dt V (x,x ) = V x x t + V x x t = V x ẋ + V x ẋ = { V x = ẋ V x = ẋ (46)

13 Then ẋ ẋ ẋ ẋ =, and the function V would satisfy the differntial equation. As many in the class correctly observed, this function V is called a Hamiltonian function. So, plugging in our system dynamics, we see that { V x = sin x V (47) x = x Integrating the first equation, V (x,x ) = sin x dx (48) = cos x x + g (x ) (49) Integrating the second equation, V (x,x ) = x dx (5) = x + g (x ) (5) Combining these results, we see that V = x cos x x + C (5) where C is a constant of integration. It can be anything without effecting our use of V, so let s set it to zero. Now, looking at the region around ( π 6 + πn,), define an arbitrary point very close to (π 6,), ( π 6 + ǫ,ǫ ), where ǫ,ǫ. Set V (ǫ,ǫ ) = c, where c is some arbitrary constant that the function V evaluates to at that point. Then, V (ǫ,ǫ ) = c (53) ǫ cos ( π 6 + ǫ ) ( π 6 + ǫ ) = c (54) ǫ (cos π 6 cos ǫ sin π ) 6 sin ǫ ǫ = c (55) ǫ 3cos ǫ + sin ǫ ǫ = c (56) ǫ ( 3 sin ǫ ) + sin ǫ ǫ = c (57) Now, using the fact that ǫ, we know that sin ǫ ǫ and sin ǫ ǫ. So, ǫ 3 + ( ) ǫ 3 + ǫ ǫ = c (58) 3 ǫ + ǫ = c (59) Thus, as we approach the equilibria, the level curves of V are a family of ellipses, which form closed orbits around the equilibria. Thus, the equillibria do correspond to nonlinear centers. At this point, the solution requires a hand drawn sketch of the phase portrait. The exact plot of the phase portrait is shown in Figure. (6) 3

14 x = x x = sin(x ) 6 4 x x Figure : Phase portrait for Problem 4. 4

15 Problem 5: Reaction-diffusion equations. The governing equations are First, we find the equilibria. Subtracting (64) from (63), we obtain Substituting this back into (63), we find that So, the equilibria are at (, ), (, ) and (, ). ẋ = (x x ) + x ( x ) (6) ẋ = (x x ) + x ( x ) (6) ẋ = = (x x ) + x ( x ) (63) ẋ = = (x x ) + x ( x ) (64) 4(x x ) + (x x ) + x 3 x3 ( ) = (65) (x x ) 3 + x + x x + x = }{{} (66) always positive x = x (67) x ( x ) = (68) Now, to find the stability, we linearize the system. First, we find the Jacobian. 3x 3x (69) So, at (,), The characteristic equation is given by λ det(df λi) = det λ So the eigenvalues are λ = 3,. Thus, this equilibrium is a saddle. At (,) and (-,-), 4 4 The characteristic equation is given by 4 λ det(df λi) = det 4 λ (7) = (λ + ) 4 = (7) (7) = (λ + 4) 4 = (73) 5

16 So the eigenvalues are λ =, 6. Thus, these equilibria are stable nodes. To find out about limit cycles, we use Bendixson s theorem to see if any closed orbits exist. div(f) = f x + f x = + 3x + 3x = 3(x + x ) < (74) So, because the divergence is always negative, it can never have a sign change, or be equivalently zero. Therefore, no closed orbits can exist in this system, so neither can a limit cycle. 6

17 Problem 6: Hamiltonian systems. The equation of motion is ẍ x + x 3 = (75) First, we put it in state model form with x = x and x = ẋ. Then, ẋ = x (76) ẋ = x x 3 (77) We linearize the system about (±,). 3x = (78) So the eigenvalues are λ = ± j, thus this point is a center in the linearized system, but by Hartman-Grobman, we don t know about the stability of the nonlinear system. Next, we look at (8) in the problem statement to determine the Hamiltonian function. Combining these two equations, we find that ẋ = H x = x H = x + g (x ) (79) ẋ = H x = x x 3 H = 4 x4 x + g (x ) (8) H = 4 x4 x + x + C (8) As in Problem 4, we can set C to zero, because we don t care about the constant offset of H here. Now, we need to determine what Ḣ is along the trajectories of the system. Ḣ = H x x t + H x x t = H H + H ( H ) = (8) x x x x Therefore, H is constant along a trajectory of the system. In other words, a contour of constant H corresponds to a trajectory on the phase portrait. Now, looking at (±,), let s focus on (,). Because the system is symmetric about the origin, what we derive for (,) applies to (-,) as well. Take a point ( + ǫ,ǫ ) which is very close to (,). That is, ǫ and ǫ are very small. Evaluating H at that point, we find the function for a constant contour of H. H = 4 ( + ǫ ) 4 + ǫ ( + ǫ ) (83) = 4 ( + 4ǫ + 6ǫ + 4ǫ3 + ǫ4 ) ( + ǫ + ǫ ) + ǫ (84) = ǫ + ǫ ǫ4 + ǫ 4 (85) 7

18 Getting rid of higher order terms, and setting H to a constant c, we obtain H = ǫ + ǫ = c (86) Which is the equation of an ellipse. Thus, as we approach the equilibrium point, the level curves of H approach an ellipse centered on the equlibrium point. Therefore, there are closed orbits around the equilibrium points (±,). 8