RENDICONTI DEL SEMINARIO MATEMATICO

Zero shear viscosity limit in compressible isentropic fluids 45solution of (1), i.e., ρ ǫ satisfies(41) ∂ t b(ρ ǫ ) + [b(ρ ǫ )u ǫ ] x + [b ′ (ρ ǫ )ρ ǫ − b(ρ ǫ )]∂ x u ǫ + b ′ (ρ ǫ ) ρ ǫu ǫx= 0for any b ∈ C 1 (R), b ′ (z) = 0 for z large enough. It is not difficult to verify that onecan take b(z) = z log z in (41) by an approximate argument and the uniform a prioriestimates established for ρ ǫ and u ǫ . Thus, we have(ρ ǫ log ρ ǫ ) t + (u ǫ ρ ǫ log ρ ǫ ) x + ρ ǫ ∂ x u ǫ + ρ ǫu ǫx+ u ǫx ρ ǫ log ρ ǫ = 0.Letting ǫ → 0 in the above equation and making use of (22), (28) and (31), we obtain(39) immediately.Similarly, the limit functions ρ, u are still a renormalized solution to (34).That is, the equation (41) with (ρ ǫ , u ǫ ) replaced by (ρ, u) is still valid, and by anapproximation one can take b(z) = z log z, and hence, the equation (40) holds.Subtraction of (40) from (39) leads to(42) [ρ log ρ−ρ log ρ] t +[u(ρ log ρ−ρ log ρ)] x +ρ log ρ − ρ log ρu = ρu x −ρu x .xOn the other hand, from (38) and the weak lower semicontinuity of convexfunctions, we find that(43) ρu x − ρu x = a ) (ρλ1+γ − ρρ γ ≥ 0, a.e.and(44) ρ log ρ ≥ ρ log ρ, a.e.As in [14, 16], consider a sequence of functions φ m ∈ C ∞ 0(), such that0 ≤ φ m ≤ 1, φ m (x) = 1 for all x such that dist(x,∂) ≥ m −1 ,|∂ x φ m (x)| ≤ 2m and dist(x,∂)|∂ x φ m (x)| ≤ 2 for all x ∈ ,φ m (x) → 1 as m → ∞ for all x ∈ .Notice that by virtue of u ∈ L 2 (0, T; H0 1), |u|[dist(x,∂)]−1 ∈ L 2 (0, T; L 2 ). Therefore,multiplying (42) by xφ m (x) and integrating over (0, t)×, then taking m → ∞,and using (43) and (23), we infer∫∫x(ρ log ρ −ρ log ρ)(x, t)dx ≤ x(ρ log ρ −ρ log ρ)(x, 0) = 0, a.e. t ∈ [0, T],which combined with (44) gives ρ log ρ = ρ log ρ a.e. on Q T . This identity, togetherwith, for example, Theorem 2.11 in [15], yields ρ ǫ → ρ a.e. on Q T , which combinedwith the Egorov theorem proves the lemma.