RENDICONTI DEL SEMINARIO MATEMATICO

40 J. Fan - S. Jiangwhich are independent of ǫ .We start with the following conservation identity which is obtained by multiplying(1) by x, integrating the resulting equation over (0, t) × and using the boundaryconditions (5):∫∫(14)xρ ǫ (x, t)dx = xρ 0 (x)dx.The following lemma gives an elementary energy estimate which is proved in[17] by multiplying the system (2)–(4) by (u ǫ ,v ǫ ,w ǫ ) in L 2 ((0, t) × ) and using (1).(15)LEMMA 1. The following energy estimate holds.∫ [ xsupt∈[0,T] 2 ρ ǫ(u 2 ǫ + v2 ǫ + w2 ǫ ) + axγ − 1 ργ ǫ∫ T ∫+0](x, t)dx[x (λ + 2ǫ)(∂ x u ǫ ) 2 + ǫ(∂ x v ǫ ) 2 + ǫ(∂ x w ǫ ) 2+(λ + 2ǫ) u2 ǫx 2 + ǫ v2 ǫx 2 ]dxdt ≤ C.whereAs in [17], we rewrite the equation (2) in the form[((ρ ǫ u ǫ ) t + ρ ǫ u 2 ǫ + P ǫ − (λ + 2ǫ) ∂ x u ǫ + u ) ]ǫ+ σ ǫxσ ǫ :=It is easy to see that by (15),∫ xρ ǫ (u 2 ǫ − v2 ǫ )r 1z(16) ‖σ ǫ ‖ L ∞ (Q T ) ≤ C.ϕ ǫ (t, x) :=Introducing the function∫ t0{ ((λ + 2ǫ)∂ x u ǫ + u ǫxone has((17) ∂ x ϕ ǫ = ρ ǫ u ǫ , ∂ t ϕ ǫ = (λ + 2ǫ)dz, P ǫ := aρ γ ǫ .x= 0,) } ∫ x− ρu 2 ǫ − P ǫ − σ ǫ (x,τ)dτ + ρ 0 u 0 dξ,r 1∂ x u ǫ + u ǫx)− ρ ǫ u 2 ǫ − P ǫ − σ ǫ .Observe that by virtue of (15), (17), the Cauchy-Schwarz inequality and (14),‖∂ x ϕ ǫ ‖ L ∞ (0,T;L 1 ) ≤ ‖ρ ǫu ǫ ‖ L ∞ (0,T;L 1 )≤ C‖ √ xρ ǫ u ǫ ‖ L ∞ (0,T;L 2 ) ‖√ xρ ǫ ‖ L ∞ (0,T;L 2 )≤ C