 Solution

ID: past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2011/2/c/solution

μ / ρ = c_{T}^{2}

and

(λ + μ) / ρ = c_{L}^{2} - c_{T}^{2}

, it becomes

u_{tt} = c_{T}^{2} Δ u + (c_{L}^{2} - c_{T}^{2}) \nabla div u .

Equivalently, using the vector calculus identity

Δ u = \nabla div u - \nabla \times (\nabla \times u)

u_{tt} = c_{L}^{2} \nabla div u - c_{T}^{2} \nabla \times (\nabla \times u) .

The divergence and curl consequently propagate at

c_{L}

and

c_{T}

, respectively.

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