1 The relation between a second order linear ode and a system of two rst order linear odes
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1 Math 1280 Spring, The relation between a second order linear ode and a system of two rst order linear odes In Chapter 3 of the text you learn to solve some second order linear ode's, such as x 00 + x = 0 (1) Then, in chapter 7, you studied systems of two rst order linear ode's, such as y 0 = x (2) In fact, you saw that (1) and (2) are completely equivalent to each other. Suppose, for example, that (x (t) ; y (t)) solves (2). Then by dierentiating the rst equation of (2), we get x 00 = y 0 Then use the second equation of (2) to get x 00 = x; or x 00 + x = 0 Similarly, if x (t) solves (1) ; and we set y (t) = x 0 ; then the pair of functions (x (t) ; y (t)) solves the system (2) This can be used to help draw phase planes. Consider the system y 0 = x 2y (3) Dierentiating the rst equation and using the second, we get or You learned to solve this previously. x 00 = y 0 = x 2y = x 2x 0 ; x x + x = 0 Substituting x = e rt ; we get r 2 + 2r + 1 = 0 (r + 1) 2 = 0; so r 1 = r 2 = 1 Therefore one solution is x 1 (t) = e t. You learned in chapter 3 that a second solution is given by x 2 (t) = te t 1
2 Converting to a system, we have x 0 1 (t) = e t = y 1 (t) Hence one solution to (3) is (x 1 (t) ; y 1 (t)) = e t ; e t For a second solution, we have x 2 (t) = te t ; x 0 2 (t) = e t te t = y 2 (t). Thus, a second solution of (3) is (x 2 (t) ; y 2 (t)) = te t ; e t te t We now wish to graph each of these in the phase plane. The rst is easy, because we easily see that y = x The second is not so easy to see, but the computer can do it easily. The following graph contains both trajectories. The following graph lls in some more of the trajectories. These will be further explained in class. See also case (3b) in section 9.1 of the text. Note that the matrix here is It turns out that the eigenvalues are r 1 = r 2 = 1; and there is only one linearly 1 independent eigenvector, say Observe that the two trajectories which are 1 straight lines above are on the line pointing in the direction of this vector. (One in x < 0; the other in x > 0) 2
3 2 Nonlinear phase planes Usually, a system x 0 = f (x; y) y 0 = g (x; y) is nonlinear, and can't be solved. We will learn various techniques for guring out the phase plane. The rst step is almost always to look for the equilibrium points. (The text calls these \critical points".) This is a matter of solving some equations, and can be tricky. Go over example 1 on page 502. Here is another example y 0 = y + x x 3 From the rst equation we see that y = 0 at any equilibrium point, and from the second, we see that there are three equilibrium points (0; 0), (1; 0) ; and ( 1; 0). Now we use PPlane to try to understand the phase plane. In doing so, we must be sure that the \window" includes the equilibrium points. We can use tools from the menu at the top of the phase plane window to show the equilibrium points clearly, and to choose a useful window. You may nd other useful tools by browsing along the top set of buttons. We will discuss the results in class. In the next section we discuss a very important analytical method for analyzing the phase plane for nonlinear systems. 3 The linearization of a nonlinear system around an equilibrium point In this section I will use the alternative notation for partial derivatives f x (x; y) (x; Consider a general system of two autonomous equations in two unknowns x 0 = f (x; y) y 0 = g (x; y) (4) 3
4 Suppose that (x 0 ; y 0 ) is an equilibrium point for (4). Then the \linearization of (4) around (x 0 ; y 0 ) is the linear system where u 0 = Au fx (x 0 ; y 0 ) f y (x 0 ; y 0 ) g x (x 0 ; y 0 ) g y (x 0 ; y 0 ) Notice that I use a dierent letter for the unknown function for the linearized system. The text does this also { see Example 3 in section 9.3. Here is a dierent example Clearly, in this example, y 0 = x + x 3 (5) f (x; y) = y g (x; y) = x + x 3 First we look for equilibrium points. It is easy to see that the only equilibrium point is (0; 0) We then nd the needed partial derivatives f x (x; y) = 0; f y (x; y) = 1 g x (x; y) = 1 + 3x 3 ; g y (x; y) = 0 The matrix A is therefore and the linearized system is ; u 0 = v v 0 = u Note that the solutions of the linear system do not involve sine and cosine. Instead, the general solution is e t e t u (t) = c 1 e t + c 2 e t (6) From this we can draw the phase plane. 4
5 We now wish to see what the relation between this picture is and the phase plane for the original system, (5). The picture below can easily be seen using PPlane. Can we see any similarity to the phase plane above? Notice that the \window" here is 10 x 10; 10 y 10 Let's zoom in. Now we can see a greater similarity. The window is 1 x 1; 1 y 1 We say that the linearized system gives \local" information about the phase plane, in some small region around the equilibrium point. The concept of linearization is one of the most important in applied mathematics. We will look at quite a few examples. Here is another one. y 0 = x y 3 (7) The only equilibrium point is (x 0 ; y 0 ) = (0; 0). We linearize around this point f x (x y) = 0; f y (x; y) = 1 g x (x; y) = 1; g y (x; y) = 3y 2 5
6 The linearized system is familiar We did the phase plane in class u 0 = v v 0 = u This is a \center". Now let's do the original system. (Will be passed out in class.) We nd spirals, not circles. In this case, it turns out, the linearized system is not a good predictor of the actual behavior of the system. We have to decide when it is useful, and when it is not. There is a very useful way of showing that the system (7) is a spiral and not a center. Recall that for the linear system, we know that u 2 + v 2 = c We can try to determine if, for (7), x 2 + y 2 is also constant. We do this by dierentiating and using the dierential equations. We get d dt x (t) 2 + y (t) 2 = 2xx 0 + 2yy 0 = 2xy + 2y x y 3 = 2y 4 This is negative, except when y = 0 So the quantity x (t) 2 + y (t) 2 is decreasing. Since x 2 +y 2 is the square of the distance from (x; y) to (0; 0), this shows that we get a stable spiral, and not a center for this system. (Well, some more has to be said, but I'll say that in class.) 6
7 4 Homework Due at the beginning of class on Wednesday, January 20. section 9.1, 9 section 9.2, 5, 7 section 9.3, 5, 10 7
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