solprogram3

solprogram3 - Solutions to Program 3 Foundations of...

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Solutions to Program 3 Foundations of Computational Math 1 Fall 2010 1 General Task In assignment you will implement solvers for nonlinear systems and apply them to a specific system of 2 nonlinear equations. You will also analyze some of your iterations to show that the must converge. 2 Submission of Results Expected results comprise: 1. A document giving the analyses requested below and the describing the results of the experiments required below. 2. The source code, makefiles, and instructions on how to compile and execute your code including the math department’s machine used if applicable. 3. Code documentation should be included in each routine. These results should be emailed to [email protected] by 11:59PM on the due date above. You may be asked to demonstrate your code if your document does not completely convince me that you tested your code sufficiently. 3 Specific Tasks 3.1 The Equations Consider the system of equations in R 2 ξ 2 + η 2 = 4 e ξ + η = 1 Show that the system has two solutions in R 2 , one with ξ > 0 and η < 0 and one with ξ < 0 and η > 0. Solution: It is easy to see that the two equations are a circle of radius 2 center at the origin and exponential opening down and to the left. The exponential, passes through the origin, has an asymptote at η = 1 as ξ → -∞ , and is such that η → -∞ as ξ → ∞ . Therefore, there must be two intersection points: one with ξ > 0 and η < 0 and one with ξ < 0 and η > 0. 1

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The leftmost one with ξ < 0 and η > 0 can be isolated further using the exponential’s horizontal asymptote and the finite extent of the circle and the intersection of the asymptote with the circle. It must satisfy - 2 < ξ < - 3 and 0 < η < 1 The rightmost one with ξ r > 0 and η r < 0 can be isolated further using the finite extent of the circle and the intersection of the exponential with the line η = - 2, i.e., ξ = ln 3. It must satisfy 0 < ξ r < ln 3 1 . 09 and - 2 < η r < 0 Also note that if we set η = - 2 we have ln(1 + 2) < 2 so the exponential is still inside the circle and we have 0 < ξ r < ln 3 1 . 09 and - 2 < η r < - 2 These bounds could be refined but that amounts to using a bisection-like method to solve the problem analytically. The regions could also be used to choose potential initial points for the iterations below but we will consider those choices more widely over R 2 . 3.2 Iteration 1 Consider the following function and iteration: G 1 ( x ) = γ 1 ( ξ, η ) γ 2 ( ξ, η ) = ln(1 - η ) - 4 - ξ 2 ξ k +1 η k +1 = γ 1 ( ξ k , η k ) γ 2 ( ξ k , η k ) = ln(1 - η k ) - 4 - ξ 2 k 1. Note that all iterates must remain real. Consider where in the R 2 a value of ξ k or η k would cause complex values to appear on the next iteration.
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