Solution
ID: past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2010/a/1/iii/solution
Past exam of the perturbation methods and applied complex variables course of the University of Oxford 2010 A 1 iii Solution by
Codex 0 2026-10-31
For the left inner coordinate , the equation becomes . At leading order the solution that matches a bounded outer expansion is ; the growing exponential is excluded. The boundary value gives . Since is finite, the possible derivative must vanish, so and .
At the next order, . Its boundary value is zero, giving , while its physical derivative at zero is . Hence , and the left inner expansion is .
At the right endpoint put . The equation is . Matching the outer expansion at one and imposing giveAt the next order, . The solution with no growing exponential, the correct matching polynomial, and isIndeed the operator sends to and sends to .
Add the two decaying inner corrections to the outer solution, subtracting their common matching parts. Since , the composite asymptotic expansion isIt reproduces all three boundary conditions to the retained algebraic orders, with exponentially small interactions between endpoints. The right endpoint has an boundary layer in . The left correction has only amplitude in , but its derivative is and is necessary for the derivative boundary condition. Thus the left layer is a weak derivative boundary layer; if “genuine” refers specifically to an order-one change in the function itself, only the right layer is genuine in that sense. Both nonzero corrections belong in the expansion through .
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