LECTURE 31 Apr 15 Linearization of autonomous nonlinear systems Sec 10 1 in Polking Boggess and Arnold Goal analyze a nonlinear autonomous system near its equilibrium points by linearizing the system 1 dimensional case x0 f x equilibrium at x0 0 so f 0 0 nonlinear DE 0 Examples x x0 x0 x0 linearized DE 2 x x 2x x3 4x5 4 sin x e7x 1 0 x x0 x0 x0 x 2x 4x 7x method remove higher degree terms using sin x x using e7x 1 7x Note each of the nonlinear DEs has an equilibrium at x 0 since right side equals 0 at x 0 Method we linearize the DE by removing all terms on the right side that are quadratic or cubic etc leaving only the linear term on the right In other words replace the right side with its tangent line approximation around the equilibrium point x0 0 Why linearize The phase line is the same for the nonlinear DE and its linearization for x near 0 and so their solutions should behave similarly close to the equilibrium And we know how to solve the linear DE it has exponential growth decay 0 1 1 0 0 2 e g nonlinear DE x x x has equilibrium points at x 0 1 The linearization x0 x has equilibrium point at x 0 a 0 Phase lines show behavior of solutions will be similar when x is near 0 x0 x x2 x0 x 1 dimensional case x0 f x equilibrium at x0 R so f x0 0 Method replace f x by tangent line approximation f x0 f 0 x0 x x0 Equilibrium has f x0 0 so linearization reduces to x0 f 0 x0 x x0 Remember f 0 x0 is some actual number while x t is a function nonlinear DE Examples a a 2 0 x0 2 x x 4 x0 sin x 1 2 x0 e2x e2 a 2 6 1 2 linearized DE 0 x 4 x 2 x0 cos 6 x 6 x0 2e2 x 1 a 0 u0 4u u0 cos 6 u u0 2e2 u The shifted variable u x x0 has equilibrium point at u0 0 Phase lines show behavior of solutions for x near 2 will be similar to behavior of linearized system when u is near 0 2 x0 x2 4 shifted DE for u x x0 x0 4 x 2 u0 4u 2 dimensional case Examples x0 f x y y 0 g x y equilibrium at x0 y0 0 0 so f 0 0 0 nonlinear system x0 2x x3 3 sin y y 0 x ey 1 x0 2xey x3 5y y 0 x 7y cos x xy x0 f x y y 0 g x y linearized system x0 2x 3y y 0 x y x0 2x 5y y 0 x 7y f x0 f x 0 0 x y 0 0 y g g y 0 x 0 0 x y 0 0 y method remove higher degree terms using sin y y and ey 1 y using ey 1 y and cos x 1 Method we linearize the system by removing all terms on the right side that are quadratic or cubic etc including all products of variables leaving only the linear terms on the right That is replace the right side with its tangent plane approximation around the equilibrium point x0 y0 0 0 Why linearize The phase portraits are similar for the nonlinear DE and its linearization for x y near the origin and so the solutions of the two systems should behave similarly close to the equilibrium And we know how to solve the homogeneous linear system 2 dimensional case x0 f x y y 0 g x y equilibrium at x0 y0 R2 so f x0 y0 0 Method replace f x y and g x y by their tangent plane approximations and then shift the variables with u x x0 and v y y0 to move the equilibrium to the origin The linearized and shifted system is u0 v0 where J f x x0 y0 g x x0 y0 f x x0 y0 u g x x0 y0 u f y x0 y0 v g y x0 y0 v or u 0 J u f y x0 y0 g y x0 y0 is the Jacobian matrix at x0 y0 and u u v Remember the partial derivatives are evaluated at the equilibrium point and so they are actual numbers while u t and v t are the functions to be solved for Exercise Competing species model with carrying capacities x0 4 1 x y x y 0 4 4 7x 3y y Find an equilibrium point x0 y0 in the first quadrant x0 0 y0 0 Compute the partial derivatives for the Jacobian matrix evaluate x0 y0 and write down the linearized system Classify the phase portrait of the linearized system For an equilibrium with x0 0 and y0 0 we need 1 x0 y0 0 and 4 7x0 3y0 0 Solving gives 4 1 2x y 4x x0 1 4 y0 3 4 Compute J and evaluate at x0 1 4 y0 3 4 to 28y 4 4 7x 6y 1 1 1 1 get J So the linearized system is u 0 u This matrix has trace T 10 and 21 9 21 9 determinant D 12 The determinant is negative so the phase portrait for u is a saddle around the origin Page 2