Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 64 / 4 / b / Solution

Self-adjointness gives $⟨\mathbf{\xi },F\mathbf{\xi }⟩=⟨F\mathbf{\xi },\mathbf{\xi }⟩=⟨\mathbf{\xi },F\mathbf{\xi }{⟩}^{\ast }$, soThe integrated expression from part (a) also writes it explicitly asThis is the quadratic change in potential energy associated with the perturbation, including pressure, magnetic and imposed-gravity restoring effects. Its sign need not be positive term by term.

The perturbation kinetic energy is $T=\frac{1}{2}⟨\dot{\mathbf{\xi }},\dot{\mathbf{\xi }}⟩$, and the total quadratic perturbation energy isSince the equilibrium and $F$ are time independent, differentiating and using the self-adjoint operator identity givesThus $\boxed{dE/dt=0}$ along the linearized motion. This conserved kinetic-plus-restoring energy is the basis of the magnetohydrodynamic energy principle.