# 352sect17 - Applied Math 352 Autumn Quarter, 2007 R. J....

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Unformatted text preview: Applied Math 352 Autumn Quarter, 2007 R. J. LeVeque 17 Newton’s Method for Nonlinear Systems In Section 12.2 we studied Newton’s method for finding a zero of a function f ( x ) of a single variable x , which is a solution of the equation f ( x ) = 0. This is one equation in one unknown, but generally f ( x ) is a nonlinear function. The idea of Newton’s method is to linearize the equation about the current approximation x ( n ) and solve the resulting linear equation for the next approximation x ( n +1) . Now suppose we have a system of m equations in m unknowns. If the equations are all linear then we can write this in the form Ax- b = ¯ 0 for some matrix A and vector b , and solve the system using the backslash operator. If the equations are nonlinear, then we can use the same idea as for one equation. Now x ( n ) ∈ lR m will be a vector representing an approximation to the solution x * ∈ lR m and we will replace the nonlinear equations by a linearization about x ( n ) that will result in a linear system of m equations. Solving the linear system will give us a new approximation x ( n +1) . A general system of m equations in m unknowns can be written as f ( x ) = ¯ where f is a function mapping lR m to lR m . The i th component of f ( x ) is a function f i ( x ) mapping the vector x to a single real number. Example 17.1. Figure 17.1 shows two circles: one with radius 3 centered at the origin and one of radius 2 centered at ( x 1 , x 2 ) = (1 , 2). Suppose we want to find a point where the circles intersect. Then we need to find a solution ( x * 1 , x * 2 ) to the two equations x 2 1 + x 2 2 = 9 ( x 1- 1) 2 + ( x 2- 2) 2 = 4 . (17.1) This system can be written as f ( x ) = ¯ 0 where f ( x ) = f 1 ( x ) f 2 ( x ) = x 2 1 + x 2 2- 9 ( x 1- 1) 2 + ( x 2- 2) 2- 4 . (17.2) This is a nonlinear function since the components of f cannot be expressed as linear combina- tions of the components of x . For this particular nonlinear system of equations, we can eliminate one of the unknowns fairly easily, reducing the problem to one nonlinear equation in a single unknown that can be solved using Newton’s method. For example, from the first equation of (17.1) we find x 2 = ± q 9- x 2 1 . (17.3) Choosing the + or- sign corresponds to restricting our attention to the upper or lower half...
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352sect17 - Applied Math 352 Autumn Quarter, 2007 R. J....

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