# 0 the general procedure is summarized as follows 1

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Unformatted text preview: 1 < x < xj 0 (7.2.6a) ; vj = Avj + b 0 xj 1 < x < xj : ; (7.2.6b) Let the solution of the BVP have the form y(x) = Uj (x)cj + vj (x) xj 1 < x < xj : ; (7.2.6c) 3. At a failure of a linear independence test, compute new initial conditions Uj+1(xj ) and vj+1(xj ) by solving Uj+1(xj )Pj = Uj (xj ) (7.2.7a) vj+1(xj ) = vj (xj ) + Uj+1(xj )w : j (7.2.7b) 4. Repeat the previous two steps until reaching the terminal point xN = b. The BVP solution at b is y(b) = UN (b)cN + vN (b): 5. Using the terminal condition Ry(b) = RUN (b)cN + RvN (b) = r determine cN as the solution of RUN (b)cN = RvN (b) ; r 12 (7.2.8) 6. Once cN has been determined, the remaining cj , j = 1 2 : : : N ; 1, are determined by enforcing solution continuity at the orthogonalization points, i.e., Uj (xj )cj + vj (xj ) = Uj+1(xj )cj+1 + vj+1(xj ): This relation can be simpli ed by the use of (7.2.7) to Uj+1(xj )Pj cj + vj (xj ) = Uj+1(xj )cj+1 + vj (xj ) + Uj+1(xj )wj or Uj+1(xj ) Pj cj ; cj+1 ; wj ] = 0: Since Pj is non-singular, we may take the solution of this system as Pj cj = cj+1 + wj j...
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