RENDICONTI DEL SEMINARIO MATEMATICO

Zero shear viscosity limit in compressible isentropic fluids 47First, we show the strong convergence of w ǫ . Multiplying (54) by 2w ǫ φ withφ ∈ C ∞ (Q T ), φ ≥ 0 and φ(·, T) = 0, then integrating over (0, T)×, integrating byparts and using the boundary conditions (56), we obtain∫ T0∫∫wǫ 2 φ tdydt +∫ T ∫= 2ǫ0 w0 2 φ(y, 0)dy[x 2 ρ ǫ (∂ y w ǫ ) 2 φ + x 2 ρ ǫ ∂ y w ǫ φ y w ǫ]dydt.Transforming this identity into the Eulerian coordinates, we see that∫ T0∫= 2ǫ∫xρ ǫ wǫ 2 (φ t + u ǫ φ x )dxdt +∫ T0∫xρ 0 w0 2 φ(x, 0)dx[x(∂ x w ǫ ) 2 φ + x∂ x w ǫ φ x w ǫ]dxdt.(57)Letting ǫ → 0 in the above equation, and using (26), (31) and (45), we find that∫ T0∫∫xρw 2 (φ t + uφ x )dxdt + xρ 0 w0 2 φ(x, 0)dx = 2〈ǫxw2 x ,φ〉 ≥ 0,where ǫxw 2 x , the weak limit of ǫx(∂ xw ǫ ) 2 in the space of signed Radon measures onQ T , is a nonnegative Radon measure on Q T . By transforming (57) into the Lagrangiancoordinates (z, t), the inequality (57) in the variables (z, t) reads(58)∫ T0∫∫w 2 φ t dzdt + w0 2 φ(z, 0)dz ≥ 0.On the other hand, transforming (49) into the Lagrangian coordinates (calculatedin the weak form), testing then the resulting equation with 2wφ, we get(59)∫ T0∫∫w 2 φ t dzdt + w0 2 φ(z, 0)dz = 0.Subtracting (59) from (58) and noticing that φ t can be nonpositive and arbitrary,we find that w 2 (z, t) ≤ w 2 (z, t), a.e. in Q T . Hence, w 2 (z, t) = w 2 (z, t) a.e. in Q T ,which implies(60) w ǫ → w strongly in L 2 (Q T ).As a consequence of (60) and (59), we see that the left hand side of (57) in theLagrangian coordinates (z, t) is equal to zero, therefore,〈ǫxw 2 x ,φ〉 = 0, ∀ φ ∈ C∞ (Q T ), φ ≥ 0 in Q T , φ(·, T) = 0.