“…The next task is then to return to Eulerian coordinates, where the solution (u(t), µ(t)) for each positive time t satisfies u(t) ∈ H 1 (R), as well as being a weak global solution of (4), and µ(t) is a nonnegative Radon measure such that µ ac (t) = u 2 x (t, ·) d x. When u is a smooth solution, µ = µ ac , but, at a blow-up time t c , the singular part of µ, which we denote µ s , accounts for the singular part of the energy, as we have lim t↑t c ((u 2 (t, x) + u 2 x (t, x)) d x) = µ s (t c ) + (u 2 (t c , x) + u 2 x (t c , x)) d x.…”