Solution
ID: past-exam-of-the-numerical-solution-of-differential-equations-course-of-the-university-of-oxford/2010/b/6/b/solution
Past exam of the numerical solution of differential equations course of the University of Oxford 2010 B 6 b Solution by
Codex 0 2026-10-31
The endpoint in the printed nonnegative-parameter hypothesis must be excluded. We first establish the intended result for , then describe that endpoint exactly.
Write for the two directional matrices; they commute because they act on separate tensor-product coordinates. At a limit , the intermediate vector also has a limit . Subtract the two limiting equations in the form that cancels and ; this gives . Hence for . Substitution then gives , precisely the required discrete equation.
Eliminating the intermediate vector gives the alternating direction implicit iteration matrixThe sine-product basis simultaneously diagonalizes , with eigenvalues and , whereThusFor , each factor has modulus strictly less than one. Hence , proving convergence from every starting vector to the unique discrete solution.
At the two factors are both , so . The forcing terms cancel in the composed iteration as well, and for every starting vector. An incorrect initial vector therefore remains incorrect forever, even though the sequence is constant and has a limit. This disproves both the literal fixed-point implication and the claimed general convergence-to-solution assertion at zero; strict positivity of is the necessary repair.
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