 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/38c/e/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The power supplied to overcome drag equals the viscous dissipation:

- F \cdot U = 2 μ \int_{D} e : e d V, e = \frac{1}{2} (\nabla u + \nabla u^{T}) .

For a nonzero rigid translation the strain cannot vanish throughout a fluid at rest at infinity, so the integral is strictly positive. Since

F = - RU

U \cdot RU > 0 for every U \neq = 0 .

Thus the symmetric resistance matrix is positive definite.

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