Solution

ID: past-exam-of-the-viscous-flow-course-of-the-university-of-oxford/2023/1/b/iii/solution

Subtract the particular plane Poiseuille flow profile from the full rectangular duct flow. The remainder is a harmonic function:
The data are even in and . The boundary value problem is unique: if two solutions differed by , Green's first identity and the zero Dirichlet boundary condition would give
Thus is constant, and its zero boundary values force . Reflection of a solution in either coordinate axis therefore gives the same solution. We may seek even separated eigenfunctions.
Use separation of variables with
Each summand solves the Laplace equation, vanishes at , and has the required evenness. The hyperbolic cosine ratio equals one at either side wall. The functions have orthogonality relation
They are the even Dirichlet eigenfunctions on , so their Fourier series represents the even boundary data. Projecting that data onto the th eigenfunction yields the Fourier coefficient
For completeness, two integrations by parts give the quadratic arch expansion in half-integer cosine modes: the displayed integral is , agreeing with the identity supplied in the paper.
Adding the particular solution gives
The coefficients are absolutely summable and the hyperbolic cosine ratio is at most one throughout the closed rectangle, giving uniform convergence and the prescribed boundary values. On compact subsets of the interior, the hyperbolic cosine ratio supplies exponential decay in , justifying differentiation of the series and the Laplace equation. The side-wall correction slows the plane Poiseuille flow enough to satisfy all four no-slip boundary conditions.

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