The Computable Differential Equation Lecture ... - Bruce E. Shapiro

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154 CHAPTER 8. BOUNDARY VALUE PROBLEMS We will modify the Euler program slightly to take the number of mesh points and the two boundary points – rather than the step size – as parameter. This is because we want to force the problem to stop at the second boundary point. The following code will return a single Euler solution for the variable u (we don’t really care about v since it was not in the original problem). EulerIVP[f , ya , a , b , n ] := Module[{ t, y, h, solution}, h = (b - a)/n; solution = {{a, ya[[1]]}}; y = ya; For[t = a, t < b, t += h, y = y + h*f[t, y]; AppendTo[solution, {t + h, y[[1]]}]; ]; Return[solution]; ] Suppose we want to estimate the solution using a mesh with 100 intervals. Then our first guess would be obtained with f[t , u , v ] := {v, u/(t*t) - v/t} EulerIVP[f, 1, 0, 1, 2., 100] which will return {{1, 1}, {1.01,1}, {1.02, 1.0001},...,{2.,1.25041}} where the intervening values have been omitted. The very last value is what we are interested in: our first guess gives u(2) (0) = 1.25041. Our goal is to iterate on this value: if we have two guesses at c and the corresponding two guess at u(2), we can use linear interpolation to make a third guess: m (i) = u(b)(i) − u(b) (i−1) c (i) − c (i−1) (8.9) c (i+1) = c (i) + u(b) − u(b)(i) m (8.10) The problem is that we don’t have a good second estimate for c, so we pick one arbitrarily, namely c (1) = c (0) + 1 = 1 (8.11) Then returns EulerIVP[f, 1, 1, 1, 2., 100] {{1, 1}, {1.01,1.01}, {1.02, 1.02},...,{2.,2}} Math 582B, Spring 2007 California State University Northridge c○2007, B.E.Shapiro Last revised: May 23, 2007
CHAPTER 8. BOUNDARY VALUE PROBLEMS 155 Thus using u(b) (0) = 1.25041, u(b) (1) = 2, with our values for c (0) and c (1) we calculate Next gives m (1) = 2 − 1.25041 1 − 0 EulerIVP[f, 1, -0.33406, 1, 2., 100] = 0.74959 (8.12) c (2) = 1 + 1 − 2 = −0.33406 (8.13) 0.74959 {{1, 1}, {1.01,0.9966594}, {1.02, 0.993452206},..., {2.,1.0000059174753426}} which shows convergence at the boundary value to better than one part in 10 5 . The three estimates are illustrated in figure 8.1. Now we put it all together and automate the shooting process. Shoot[f , ua , ub , a , b , n , MaxIterations :5, tolerance :10 −6 ] := Module[{c, ya, ubObserved, ubObservedOld, cold, m}, c = (ub - ua)/(b - a); (* first guess *) ya = {ua, c}; s = EulerIVP[f, ya, a, b, n]; (* first shot *) ubObserved = Last[Last[s]]; cold = c; c = c + 1; (* second guess *) Do[ ubObservedOld = ubObserved; ya = ua, c; s = EulerIVP[f, ya, a, b, n]; (* next shot *) ubObserved = Last[Last[s]]; If[Abs[ubObserved - ubObservedOld] < tolerance, Break[]]; m = (ubObserved - ubObservedOld)/(c - cold) (* slope *) cold = c; c = cold + (ub - ubObserved)/m; (* next guess at c *); , {MaxIterations} ]; Return[s]; ]; The problem with this implementation is that it can miss solutions, if the BVP has multiple solutions. To fix this we will have to replace the linear slope correction with a full Newton root finder. At this point it is not yet clear what we are finding the root of, so we will need to develop the theory somewhat first before we can return to the shooting algorithm. c○2007, B.E.Shapiro Last revised: May 23, 2007 Math 582B, Spring 2007 California State University Northridge
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The Computable Differential Equatio
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Contents 1 Classifying The Problem
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CONTENTS v Timeline on Computable D
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Chapter 1 Classifying The Problem 1
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CHAPTER 1. CLASSIFYING THE PROBLEM
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Chapter 2 Successive Approximations
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CHAPTER 2. SUCCESSIVE APPROXIMATION
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Chapter 3 Approximate Solutions 3.1
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