Linear Systems of ODE: Nullclines, Eigenvector lines and trajectories

Transcription

1 Linear Systems of ODE: Nullclines, Eigenvector lines and trajectories James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University October 6, 203 Outline NullClines Eigenvector Lines Trajectories

2 Abstract This lecture is about the qualitative analysis of linear systems of ODE. we begin with nullclines and the eigenvector lines. We then discuss trajectories. We can also analyze the solutions to our linear system models for arbitrary ICs using graphical means. We have a lot to cover so let s get started. Let s look at a specific model: x (t) 3 x(t) + 4 x(t) + 2 x(0) x 0 y 0 The set of (x, y) pairs where x 0 is called the nullcline for x; similarly, the points where y 0 is the nullcline for y. The x equation can be set equal to zero to get 3x + 4y 0. This is the same as the straight line y 3/4 x. This straight line divides the x y plane into three pieces: the part where x > 0; the part where x 0; and, the part where x < 0.

3 In this figure, we show the part of the x y plane where x > 0 with one shading and the part where it is negative with another. x > 0 Evaluation Point x 0 x < 0 The x equation for our system is x 3x + 4y. Setting this to 0, we get y 3/4 x whose graph is shown. At the point (0, ), x Hence, every point in the x y plane on this side of the x 0 line will give x > 0. Hence, we have shaded the part of the plane where x > 0 as shown. Similarly, the y equation can be set to 0 to give the equation of the line x + 2y 0. This gives the straight line y /2 x. In the next figure, we show how this line also divided the x y plane into three pieces.

4 I y > 0 Evaluation Point y 0 y < 0 The y equation for our system is y x + 2y. Setting this to 0, we get y /2 x whose graph is shown. At the point (0, ), y Hence, every point in the x y plane on this side of the y 0 line will give y > 0. Hence, we have shaded the part of the plane where y > 0 as shown. The shaded areas can be combined into one figure. In this figure, we divide the x y plane into four regions. In each region, x and y are either positive or negative. Hence, each region can be marked with an ordered pair, (x ±, y ±). (+,+) (+, ) (,+) (, ) The x 3x + 4y and the y x + 2y equations determine four regions in the plane. In each region, the algebraic sign of x and y are shown as an ordered pair. For example, in the region labeled with (+, +), x is positive and y is positive.

5 Homework 52 For these problems, Find the x nullcline and determine the plus and minus regions in the plane. Find the y nullcline and determine the plus and minus regions in the plane. Assemble the two nullclines into one picture showing the four regions that result. 52. x ] (t) ] ] 3 x(t) 3 ] 3 Homework 52 Continued 52.2 ] x (t) ] ] x (t) 2 4 ] ] x (t) ] 2 4

6 Now we add the eigenvector lines. You can verify this system has eigenvalues r 2 and r 2 with associated eigenvectors E ] ], E /4 2. Recall a vector V with components a and b, V ] a b determines a straight line with slope b/a. Hence, these eigenvectors each determine a straight line. The E line has slope /4 and the E 2 line has slope. We can graph these two lines overlaid on the graph shown in the next figure. E y 0 (+, ) x 0 E 2 (+,+) (, ) E 2 x 0 (,+) y 0 E In this figure, we show the sign pairs determined by the x 3x +4y and the y x + 2y equations for all four regions. In addition, the lines corresponding to the eigenvectors E and the dominant E 2 are drawn.

7 Homework 53 For these problems, Find the x nullcline and determine the plus and minus regions in the plane. Find the y nullcline and determine the plus and minus regions in the plane. Assemble the two nullclines into one picture showing the four regions that result. Draw the Eigenvector lines on the same picture. 53. x ] (t) ] ] 3 x(t) 3 ] 3 Homework 53 Continued 53.2 ] x (t) ] ] x (t) 2 4 ] ] x (t) ] 2 4

8 In each of the four regions, we know the algebraic signs x and y. Given an initial condition (x 0, y 0), we can use this information to draw the set of points (x(t), ) corresponding to the solution to our system x (t) 3 x(t) + 4 x(t) + 2 x(0) x 0 y 0. This set of points is called the trajectory corresponding to this solution. The first point on the trajectory is the initial point (x 0, y 0) and the rest of the points follow from the solution. which has the form ] ] ] x(t) a e 2t + b e t. /4 where a and b satisfy the system of equations This can be rewritten as Hence x 0 a + b y 0 (/4)a + b x(t) a e 2t + b e t (/4)a e 2t + b e t. dy dx x (t) ( 2/4)a e 2t + b e t 2a e 2t + b e t

9 When t is large, as long as b is not zero, the terms involving e 2t are negligible and so we have dy dx x (t) b et the slope of E2. b et Hence, when t is large, the slopes of the trajectory approach, the slope of E 2. So, we conclude for large t, for b not zero, the trajectory either parallels the E 2 line or approaches it asymptotically. For an initial condition on the E line, b is zero and dy dx In this case, ( 2/4)a e 2t x (t) 2a e 2t the slope of E. 4 x(t) a e 2t, (/4)a e 2t (x(t), ) (0, 0). along the E line. Conclusions: Given an IC which does not lie on E or E 2, (x(t), ) approaches E 2 as t gets large. Since the second eigenvalue is +, (x(t), ) moves away from the origin (0, 0) as t increases. Note + is the dominant eigenvalue and these trajectories move toward the E 2 line. Given an IC which starts on E 2, (x(t), ) are always on the E 2 line and since the second eigenvalue is, (x(t), ) moves outward from (0, 0) along the E 2 line as t gets large. We need to study the trajectories in the various derivative sign regions next.