 Solution

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

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

For the displacement gradient tensor specified in the question,

\nabla u : \nabla u^{T} = tr (\nabla u \nabla u^{T}) = u_{x}^{2} + u_{y}^{2} + v_{x}^{2} + v_{y}^{2},

while the divergence is

\nabla \cdot u = u_{x} + v_{y}

\frac{μ}{2} (u_{x}^{2} + u_{y}^{2} + v_{x}^{2} + v_{y}^{2}) + \frac{λ + μ}{2} (u_{x} + v_{y})^{2} .

Expanding the square and collecting terms yields

J = \iint_{Ω} [(\frac{λ}{2} + μ) (u_{x}^{2} + v_{y}^{2}) + \frac{μ}{2} (u_{y}^{2} + v_{x}^{2}) + (λ + μ) u_{x} v_{y}] d x d y .

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