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# 1 x0 x x1 u1 u1 u2 u1r 1 1 the integration

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Unformatted text preview: pendent basis for the solution. The point x1 is determined as part of the solution process but, for the present, let's assume that it is known. New initial conditions U2(x1 ) are determined by orthogonalizing the columns of U1 (x1). We'll subsequently show that this is equivalent to nding an upper triangular matrix P1 such that U2(x1 )P1 = U1 (x1): We also select new initial conditions v2 (x1) for the particular solution that are in the orthogonal complement of U2 (x1 ). Again, we'll postpone the details and note that this is equivalent to determining a vector w1 such that v2 (x1 ) = v1(x1 ) ; U2(x1 )w1 : The time integration can now continue by solving U2 = AU2 x1 < x < x2 0 11 v2 = Av2 + b x1 < x < x2 : The quantities U2 (x1 ) and v2(x1 ) are used as initial conditions. The process continues to x2 where the columns of U2 (x2 ) fail to satisfy a linear independence test. 0 The general procedure is summarized as follows. 1. Determine U1 (a) and v1 (a) as solutions of LU1(a) = 0 (7.2.5a) Lv1(a) = l: (7.2.5b) 2. Beginning with j = 1, solve the IVPs Uj = AUj xj...
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