 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-4/13c/b/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let the displayed integrand be

F

. Since it has no explicit dependence on

u

v

\partial_{x} F_{u_{x}} + \partial_{y} F_{u_{y}} = 0, \partial_{x} F_{v_{x}} + \partial_{y} F_{v_{y}} = 0.

F_{u_{x}} = (λ + 2 μ) u_{x} + (λ + μ) v_{y}, F_{u_{y}} = μ u_{y},

F_{v_{x}} = μ v_{x}, F_{v_{y}} = (λ + 2 μ) v_{y} + (λ + μ) u_{x} .

Thus

(λ + 2 μ) u_{xx} + μ u_{yy} + (λ + μ) v_{x y} = 0,

μ v_{xx} + (λ + 2 μ) v_{yy} + (λ + μ) u_{x y} = 0.

Using the Laplacian, gradient, and divergence, these combine into the static Navier-Cauchy equation

μ \nabla^{2} u + (λ + μ) \nabla (\nabla \cdot u) = 0 .

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