University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2021
Views:

385

Pages:

1
Author:

Elaina Collier

Review of Linearization in Calculus In Calculus I, the linearization of y = f (x) at a point (a, f (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. It is very difficult to analyze a general non-linear system of differential equations, so we will “linearize” the system at the equilibria, and we know how to analyze linear systems. In Calculus III, we learn how to linearize z = f (x, y) at a point (a, b, f (a, b))- we get the tangent plane: L(x, y) = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b) From this, you can guess what we do for w = f (x, y, z) at (a, b, c, f (a, b, c)): L(x, y, z) = f (a, b, c) + fx (a, b, c)(x − a) + fy (a, b, c)(y − b) + fz (a, b, c)(z − c) and so on. Vector Fields and Linearization Given a vector field F(x, y), we can also linearize the whole by linearizing each function. For example, if f (x, y) F(x, y) = g(x, y) Then the linearization at x = a, y = b is given by: f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b) L(x, y) = g(a, b) + gx (a, b)(x − a) + gy (a, b)(y − b) This is typically written in matrix-vector form: fx (a, b) fy (a, b) (x − a) f (a, b) L(x, y) = + gx (a, b) gy (a, b) (y − b) g(a, b) The matrix of first partial derivatives has a special name- It is the Jacobian matrix for our vector fieldYou might have heard of this matrix in Calc III since it comes up in optimization problems. Homework: Linearization and some review 1. Find all solutions to each system of equations. (a) (2 + x)(y − x) = 0 (4 − x)(y + x) = 0 (b) x + x2 + y 2 y − xy =0 =0 2. Linearize each function about the given point. (a) f (x) = x2 + 2x + 1 + cos(x) at x = 0 (b) f (x, y) = x − 2y + x2 + exy at x = 1, y = 0. −x + y + 2xy (c) F(x, y) = at (x = 0, y = 0). −4x − y + x2 − y 2 (1 + x) sin(y) at (x = 0, y = 0). (d) F(x, y) = 1 − x − cos(y) 1 (c) x − x2 − xy 3y − xy − 2y 2 =0 =0

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