Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 359 / 1 / b / ii / Solution

Integrating the exact Galerkin energy identity gives

\frac{1}{2} ∣ u_{m} (t) ∣^{2} + ν \int_{0}^{t} ∥ u_{m} (τ) ∥^{2} d τ = \frac{1}{2} ∣ P_{m} u_{0} ∣^{2} \leq \frac{1}{2} ∣ u_{0} ∣^{2} .

Thus one may take

K_{0} (T) = ∣ u_{0} ∣, K_{1} (T) = \frac{∣ u _{0} ∣}{2 ν}

for the first two requested bounds; these constants happen not to grow with

T

.

The Galerkin equation and contractivity of

P_{m}

on

V^{'}

give

\frac{d u _{m}}{d t}_{V^{'}} \leq ν ∥ A u_{m} ∥_{V^{'}} + ∥ B (P_{N} u_{m}, u_{m}) ∥_{V^{'}} .

Since

∥ A u_{m} ∥_{V^{'}} = ∥ u_{m} ∥

, part a gives

\frac{d u _{m}}{d t}_{V^{'}} \leq (ν + c λ_{N}^{1/4} ∣ u_{m} ∣) ∥ u_{m} ∥ \leq (ν + c λ_{N}^{1/4} ∣ u_{0} ∣) ∥ u_{m} ∥.

Consequently

\frac{d u _{m}}{d t}_{L^{2} (0, T; V^{'})} \leq K_{0}^{'} (T) := (ν + c λ_{N}^{1/4} ∣ u_{0} ∣) \frac{∣ u _{0} ∣}{2 ν} .

All three constants are independent of

m

.

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