Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 6 / 3 / f / Solution

The orthogonal projection satisfies $\Pi L=\Pi (\Pi -I)=0$, hence $d\Pi f_t/dt=0$ and $\Pi f_t=\Pi f_0=\rho (f_0)M$. Set $h_t=f_t-\Pi f_0$. Then $\Pi h_t=0$ and ${\dot{h}}_t=L{h}_t$. Crucially, applying the energy identity of part (e) to $h_t$ givesSolving this scalar equation and taking square roots provesThis argument uses (e) explicitly, rather than bypassing the requested energy method with an explicit solution formula.