Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-79/3/a/solution

In the outer solution, derivative terms are small and , so . To resolve its mismatch with the zero left boundary, set , . The original equation becomes exactly . The left inner variable is therefore , with and as .
Multiplying by gives the first integral of a square-root reaction layer. Matching , fixes its constant:
The positive derivative branch below begins with and approaches the plateau exponentially. Thus the left layer has thickness , not despite the coefficient on the original second derivative. Its small amplitude is essential in the dominant balance.

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