# 2 to nonlinear problems y f x y 0 glya 0 axb gr

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Unformatted text preview: ting these results into (7.3.1) and neglecting the O( y2) terms y + y = f (x y) y + f (x y) 0 or 0 y y = f (x y) y + f (x y) ; y : 0 0 y This has the form of a linear system y = A(x) y + b(x) (7.3.3a) 0 with A(x) = f (x y) y b(x) = f (x y) ; y : 0 Linearizing the left boundary condition yields gL(y(a) + y(a)) = gL(y(a)) + gLy (y(a)) y(a) = 0: 18 (7.3.3b) This has the form of the linear boundary condition L y(a) = l (7.3.4a) with L = gLy (y(a)) l = ;gL (y(a)): (7.3.4b) At the right end, we have R y(b) = r (7.3.5a) with R = gRy (y(b)) r = ;gR (y(b)): (7.3.5b) The linearized system (7.3.3 - 7.3.5) may be solved by iteration using the procedures of Section 7.2 thus, 1. Beginning with an initial guess y(0) (x), 0 x 1, 2. Solve the linear system y ( ) = A( ) (x) y( ) + b( ) (x) 0 L( ) y( ) (a) = l( ) R( ) y( )(b) = r( ) where A( )(x) = f (x Y( ) ), etc. y 3. After each iteration, set y( +1) = y( ) + y( ) and repeat the procedure until convergence. The procedure is awkward since interpolation...
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## This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.

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