Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 331 / 2 / b / iv

Table of contents
- Solution iv

Solution

After the pressure enforces incompressibility, let

L

denote the displayed linear operator. Integration by parts gives

⟨ Φ_{i}, L Φ_{j} ⟩ = - σ \int_{V} \nabla u_{i} : \nabla u_{j} d V - σ Ra \int_{V} \nabla θ_{i} \cdot \nabla θ_{j} d V + σ Ra \int_{V} (w_{i} θ_{j} + θ_{i} w_{j}) d V .

The pressure terms vanish by incompressibility and the boundary conditions. The expression is symmetric under

i \leftrightarrow j

, so

⟨ Φ_{i}, L Φ_{j} ⟩ = ⟨ L Φ_{i}, Φ_{j} ⟩ .

Thus

L

is a self-adjoint operator in the energy inner product and hence a normal operator. Its orthogonal eigenmodes cannot generate non-normal transient growth.

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