Local Linear Analysis of Nonlinear Autonomous DEs 2019

Local Linear Analysis of Nonlinear Autonomous DEs In Calculus I, the linearization of y = f (x) at a point x = a was defined to be the tangent line to f at a: L(x) = f (a) + f 0 (a)(x − a) We are going to do something similar here- The point at which we analyze the ODE will be at the equilibria. We will then “linearize” about the equilibria and classify them using the Poincare diagram. Local Linearization Local linear analysis is the process by which we analyze a nonlinear system of differential equations about its equilibrium solutions (also known as critical points or fixed points). Given a system of nonlinear autonomous DEs: x0 = f (x, y) y 0 = g(x, y) or x0 (t) = F (x) we first find the equilibrium solutions by setting the derivatives to zero, then solve simultaneously, the system: f (x, y) = 0 g(x, y) = 0 Given an equilibrium, say x = a, y = b, the linearization of the system at the point (a, b) is the following matrix, also called the Jacobian matrix of F . " x y #0 " = fx (a, b) fy (a, b) gx (a, b) gy (a, b) #" x−a y−b # We can then use the Poincaré Diagram to determine the local behavior. We must use some caution in the case of centers and degenerate nodes, however. Because the linearization is an approximation of the true solution, the actual solutions are of a slightly perturbed system. This means that while the linearization gives a center, the true solution may be a center or a spiral (we would use a computer simulation to see what we actually get). Example 1: Competing Species Suppose we have two populations that are competing for similar resources, like rabbits (x(t)) and hamsters (y(t)). It seems reasonable to suppose that: In the absence of the other, each population is modeled by a population model with an environmental threshold (what we called the logistic model. Back in Chapter 2, that was (book’s notation on p 81): y 0 = r(1 − y/k)y = ay − by 2 so that y = 0 is unstable and y = k is stable. 1 To simplify matters, we assume some constants: x0 = x − x2 y 0 = 34 y − y 2 Now, our second assumption will be: The rate of change of populations (down) will be proportional to the number of interactions between the populations. For example, if there are 3 rabbits and 2 hamsters, there are a total of 6 possible rabbit-hamster interactions possible. Our equations become: x0 = x − x2 − xy y 0 = 34 y − y 2 − 12 xy The analysis proceeds by getting the equilibria (a.k.a. critical points): From the first equation, either x = 0 or x = −y + 1: • x = 0 in the second equation: y  3 4  − y = 0 so that y = 0 or y = 3/4. • If x = −y + 1 in the first, then the second equation becomes: 3 1 1 1 y − y − (−y + 1) = y − y =0 4 2 4 2     Therefore, y = 0 (and x = 0, but we’ve counted that one), or y = 1/2 (then x = 1/2, too). We have 4 equilibrium solutions: (0, 0), (1, 0), (0, 3/4), (1/2, 1/2) Now we linearize the system about each equilibrium solution to determine its stability. First, the matrix of partial derivatives is: " 1 − 2x − y −x −0.5y 0.75 − 2y − 0.5x # Evaluating this at each of the critical points (in order) gives us: " 1 0 0 0.75 # " −1 −1 0 0.25 # " 0.25 0 −.375 −.75 # " −.5 −.5 −.25 −.5 # Using the Poincaré Diagram, we see that the origin is indeed a SOURCE, the equilibria on the x− and y− axes are SADDLES, and the point of intersection of the two lines is a SINK. Putting these together, we can look at the direction field to examine the global behavior. From this we see that if both of the initial populations are not zero, the model predicts that all solutions will tend to the sink at (1/2, 1/2)- We might call this peaceful coexistence. 2 Figure 1: The Competing Species model, with peaceful coexistence. Almost all solutions tend to the equilibrium (1/2, 1/2). In the second case, War! Example 2: We now repeat the analysis for a slightly different system: x0 = x(1 − x − y) = x − x2 − xy 0 y = y(0.5 − 0.25y − 0.75x) = 0.5y − 0.25y 2 − 0.75xy The critical points have changed slightly: (0, 0), (1, 0), (0, 2), (1/2, 1/2) The matrix of partial derivatives is: " 1 − 2x − y −x −.75y 0.5 − 0.5y − 0.75x # Evaluating this at each of the critical points (in order) gives us: " 1 0 0 0.5 # " −1 −1 0 −0.25 # " −1 0 −1.5 −.5 # " −.5 −.5 −.375 −.125 # Using the Poincaré Diagram, we see that the origin is still a SOURCE, the equilibria on the x− and y− axes are still SADDLES. The big difference is that the point of intersection of the two lines now produces a SADDLE. We will recall that a saddle point in UNSTABLE- this means that for just about any initial condition, one of the two species will die off. Notice how critical these parameters are to the general outcome- This has important public policy implications! Example 3: Consider the following system: x0 = x − 0.5xy = x(1 − 0.5y) y 0 = −0.75y + 0.25xy = y(−0.75 + 0.25x) Analyze the behavior of the solutions of the system by using local linearization. SOLUTION: We always get equilibria first. In this case, from the first equation we get 3 • x = 0: From the second, we must have y = 0. • y = 2: From the second, x = 3. We have only two equilibria for this system, (0, 0) and (3, 2). Now compute the “Jacobian matrix” of partial derivatives: " fx (a, b) fy (a, b) gx (a, b) gy (a, b) # " ⇒ 1 − 0.5y −0.5x 0.25y −0.75 + 0.25x # Linearizing about the two equilibrium gives (in order): " 1 0 0 −0.75 # " 0 −1.5 0.5 0 # In the first case, the trace is 1/4 and the determinant is −3/4. By the Poincare diagram, the origin is a SADDLE. At the point (3, 2), the trace is 0 and the determinant is positive: We have a CENTER. We should check the direction field to verify our analysis. In the figure below, we first show the (x, y) plane, then in the next figure, we plot x(t) and y(t) versus t. Figure 2: The direction field (top) and a plot of x(t) and y(t) as functions of time (bottom). They both reveal the existence of periodic solutions. Side remark: This set of equations is called “Predator-Prey”- Can you tell which variable is the predator, and which is the prey? 4