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Hindawi Publishing Corporation Boundary Value Problems Volume 2010, Article ID 796065, 21 pages doi:10.1155/2010/796065 Research Article One-Dimensional Compressible Viscous Micropolar Fluid Model: Stabilization of the Solution for the Cauchy Problem ´ Nermina Mujakovic Department of Mathematics, University of Rijeka, Omladinska 14, 51000 Rijeka, Croatia Correspondence should be addressed to Nermina Mujakovi´ , mujakovic@inet.hr c Received November 2009; Revised 24 May 2010; Accepted June 2010 Academic Editor: Salim Messaoudi Copyright q 2010 Nermina Mujakovi´ This is an open access article distributed under the c Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited We consider the Cauchy problem for nonstationary 1D flow of a compressible viscous and heat-conducting micropolar fluid, assuming that it is in the thermodynamical sense perfect and polytropic This problem has a unique generalized solution on R× 0, T for each T > Supposing that the initial functions are small perturbations of the constants we derive a priori estimates for the solution independent of T , which we use in proving of the stabilization of the solution Introduction In this paper we consider the Cauchy problem for nonstationary 1D flow of a compressible viscous and heat-conducting micropolar fluid It is assumed that the fluid is thermodynamically perfect and polytropic The same model has been considered in 1, , where the global-in-time existence and uniqueness for the generalized solution of the problem on R× 0, T , T > 0, are proved Using the results from 1, we can also easily conclude that the mass density and temperature are strictly positive Stabilization of the solution of the Cauchy problem for the classical fluid where microrotation is equal to zero has been considered in 4, In was analyzed the Holder ă continuous solution In is considered the special case of our problem We use here some ideas of Kanel’ and the results from 1, as well Assuming that the initial functions are small perturbations of the constants, we first derive a priori estimates for the solution independent of T In the second part of the work we analyze the behavior of the solution as T → ∞ In the last part we prove that the solution of our problem converges uniformly on R to a stationary one 2 Boundary Value Problems The case of nonhomogeneous boundary conditions for velocity and microrotation which is called in gas dynamics “problem on piston” is considered in Statement of the Problem and the Main Result Let ρ, v, ω, and θ denote, respectively, the mass density, velocity, microrotation velocity, and temperature of the fluid in the Lagrangean description The problem which we consider has the formulation as follows : ∂ρ ∂t ∂v ∂t ρ ρ ∂θ ∂t −Kρ2 θ ∂v ∂x ∂ω ∂t ρ2 ρ2 ∂v ∂x 0, 2.1 −K ∂ ρθ , ∂x 2.2 ∂ ∂ω ρ ∂x ∂x −ω , 2.3 ∂ ∂v ρ ∂x ∂x A ρ ∂v ∂x ρ2 ∂ω ∂x ω2 Dρ ∂θ ∂ ρ ∂x ∂x 2.4 in R × R , where K, A, and D are positive constants Equations 2.1 – 2.4 are, respectively, local forms of the conservation laws for the mass, momentum, momentum moment, and energy We take the following nonhomogeneous initial conditions: ρ x, ρ0 x , v x, v0 x , ω x, ω0 x , θ x, 2.5 θ0 x for x ∈ R, where ρ0 , v0 , ω0 , and θ0 are given functions We assume that there exist m, M ∈ R , such that m ≤ ρ0 x ≤ M, m ≤ θ0 x ≤ M, x ∈ R 2.6 In the previous papers 1, we proved that for ρ0 − 1, v0 , ω0 , θ0 − ∈ H R 2.7 the problem 2.1 – 2.5 has, for each T ∈ R , a unique generalized solution: x, t −→ ρ, v, ω, θ x, t x, t ∈ Π R × 0, T , 2.8 Boundary Value Problems with the following properties: ρ − ∈ L∞ 0, T ; H R ∩ H1 Π , 2.9 v, ω, θ − ∈ L∞ 0, T ; H R ∩ H Π ∩ L2 0, T ; H R Using the results from 1, we can easily conclude that θ, ρ > in Π 2.10 We denote by Bk R , k ∈ N0 , the Banach space Bk R 0, ≤ n ≤ k , u ∈ Ck R : lim |Dn u x | |x| → ∞ 2.11 where Dn is nth derivative; the norm is defined by u sup sup|Dn u x | Bk R 2.12 x∈R n≤k From Sobolev’s embedding theorem 7, Chapter IV and the theory of vector-valued distributions 8, pages 467–480 one can conclude that from 2.9 one has ρ − ∈ L∞ 0, T ; B0 R v, ω, θ − ∈ L2 0, T ; B1 R ∩C ∩C 0, T ; L2 R , 2.13 ∩ L∞ 0, T ; B0 R , 0, T ; H R 2.14 and hence v, ω, θ − ∈ C 0, T ; B0 R , ρ ∈ L∞ Π 2.15 From 2.7 and 2.6 it is easy to see that there exist the constants E1 , E2 , E3 , M1 ∈ R , M1 > 1, such that R v0 dx 2A R ω0 dx K R 2 R R 1 − ln − dx ρ0 ρ0 v02 ω A ρ dx ρ0 R θ02 dx v0 ln E2 , dx ≤ E3 , ρ0 sup θ0 x < M1 |x| it holds that 2A v2 dx R R t θ − ln θ − dx ω2 dx ρ θ R R ∂v ∂x 1 − ln − dx ρ ρ K R ρ θ ∂ω ∂x ω2 ρθ ρ D θ ∂θ ∂x 3.1 dx dτ E1 , where E1 is defined by 2.16 Proof Multiplying 2.1 , 2.2 , 2.3 , and 2.4 , respectively, by Kρ−1 − ρ−1 , v, A−1 ρ−1 ω, and ρ−1 − θ−1 , integrating by parts over R and over 0, t , and taking into account 2.13 and 2.14 , after addition of the obtained equations we easily get equality 3.1 independently of t If we multiply 2.2 by ∂/∂x ln 1/ρ , integrate it over R and 0, t , and use some equalities and inequalities which hold by 2.1 and 2.13 – 2.15 together with Young’s inequality we get, as in , the following formula: ρ2 R ∂ ∂x ≤ ρ dx K v2 dx R R K t R ρ θ t θρ3 R ∂θ ∂x ∂ ∂x ρ dx dτ t dx dτ ρ R ∂v ∂x dx dτ R v0 ln dx ρ0 3.2 ρ dx ρ0 for each t > Using 3.1 , 2.16 , 2.18 , and 2.21 we get easily ρ2 R ∂ ∂x ρ dx K t θρ3 R ∂ ∂x ρ dx dτ ≤ K1 θ t , 3.3 Boundary Value Problems where θ t sup x,τ ∈R× 0,t θ x, τ , 2μE1 K1 θ t E3 E1 θ t 3.4 As in , we introduce the increasing function η ψ η eξ − − ξ dξ 3.5 that satisfies the following inequality: ψ ln ρ ≤ R 1 − − ln dx ρ ρ 1/2 ρ R ∂ ∂x ρ 1/2 dx ≤ K2 θ t , 3.6 where K2 θ t 2E1 2μ K We can easily conclude that there exist the quantities η θ t η θ t eη − − η dη η1 θ t 1/2 E3 E1 θ t 3.7 < and η1 θ t eη − − η dη > 0, such that K2 θ t 3.8 Comparing 3.8 and 3.6 we obtain, as in 5, Lemma 3.2 , the following result Lemma 3.2 For each t > there exist the strictly positive quantities u1 exp η1 θ t exp η θ t and u1 such that u1 ≤ ρ−1 x, τ ≤ u1 , x, τ ∈ R × 0, t Now we find out some estimates for the derivatives of the functions v, ω, and θ 3.9 Boundary Value Problems Lemma 3.3 For each t > it holds that 2 ∂v ∂x R 6u1/2 Du1 D 12 R ∂ω ∂x ρ − K3 θ τ dx R ∂v ∂x dx dτ 3.10 2 − K4 θ τ dx R ∂ω ∂x dx dτ ∂2 θ ∂x2 R dx ∂v ∂x t ∂θ ∂x R t t 3u1 Du1 2 ∂ω A ∂x dx dτ ≤ K5 θ t , where 3.11 Du1 , 1/2 12u1 A1/2 E1 48K1 θ t θ t E1 E1 K θ t D , 1/2 24u1 E1 K4 θ t K5 θ t Du1 K3 θ t u1 u1 3.12 u1 u1 2Kθ t K1 θ t 9θ t 3.13 3θ t 3AE1 D E1 θ t u2 1 u1 u1 u2 E2 Proof Multiplying 2.2 , 2.3 , and 2.4 , respectively, by −∂2 v/∂x2 , −A−1 ρ−1 ∂2 ω/∂x2 , and −ρ−1 ∂2 θ/∂x2 , integrating over R× 0, t , and using the following equality: − t R ∂v ∂2 v dx dτ ∂t ∂x2 ∂v ∂x R dx|t 3.14 that is satisfied for the functions θ and ω as well, after addition of the obtained equalities we find that R ∂v ∂x t ρ R ∂ω A ∂x ∂2 ω ∂x2 ∂θ ∂x 2 dx|t t dx dτ D ρ R t ρ ∂2 θ ∂x2 R ∂2 v ∂x2 dx dτ dx dτ Boundary Value Problems t − R t − K ρ R ∂ρ ∂ω ∂ ω dx dτ ∂x ∂x ∂x2 t ∂v ∂x ρ R t −D R R t − t ∂ρ ∂v ∂2 v dx dτ ∂x ∂x ∂x2 K R ω∂ ω dx dτ ρ ∂x2 t ∂ω ∂x ρ R θ R t ∂2 θ dx dτ − ∂x2 t ∂θ ∂2 v dx dτ ∂x ∂x2 K ρθ R ∂2 θ dx dτ − ∂x2 ∂ρ ∂2 v dx dτ ∂x ∂x2 ∂v ∂2 θ dx dτ ∂x ∂x2 t R ω2 ∂2 θ dx dτ ρ ∂x2 ∂ρ ∂θ ∂2 θ dx dτ ∂x ∂x ∂x2 3.15 Using 3.3 , 3.9 , the inequality ∂v ∂x ≤2 R 1/2 ∂v ∂x ⎛ ∂2 v ∂x2 ⎝ R ⎞1/2 ⎠ 3.16 , that holds for the functions ∂θ/∂x, ∂ω/∂x, v, and ω as well, and applying Young’s inequality with a sufficiently small parameter on the right-hand side of 3.15 , we find, similarly as in , the following estimates: t R ≤ ∂ρ ∂v ∂2 v dx dτ ∂x ∂x ∂x2 t R 2u1/2 ≤ u1 ∂ρ ∂x ρ t R ρ2 ∂v ∂x dx dτ ∂ρ ∂x dx R t R ∂v ∂x ∂2 v ∂x2 ρ dx dτ 1/2 ⎛ ⎝ dx ∂2 v ∂x2 ρ R ⎞1/2 dx⎠ dτ 3.17 t ≤ 16 ≤ 16 ρ R t R R ∂2 v ∂x2 t ρ ∂v ∂x2 ∂2 v ∂x2 ρ 2 dx dτ dx dτ 16K1 θ t dx dτ 16K1 θ t t u1 u2 u1 u1 ∂v ∂x R t θ t R dx dτ ρ θ ∂v ∂x dx dτ, Boundary Value Problems t K ρ R ∂θ ∂2 v dx dτ ∂x ∂x2 ∂θ ∂x ρ R t t ≤ K2 ρ θ2 ≤ K u1 θ t u1 t K θ R ≤K t θ2 ρ R ≤K θ t θρ R t R ≤ R dx dτ, R dx dτ t dx dτ ∂2 v ∂x2 ρ R ∂ω ∂x2 R ∂2 ω ∂x2 t ρ dx dτ ∂ρ ∂x ∂2 v ∂x2 ρ 3.19 R dx dτ, t R ∂ω ∂x dx R ∂2 ω ∂x2 ρ dx dτ 1/2 ⎛ ⎝ dx ρ R ⎞1/2 ∂2 ω ∂x2 dx⎠ dτ 3.20 ∂2 ω ∂x2 R ∂ω ∂x ρ2 R ρ ρ dx dτ ⎛ ⎜ 2⎝ dx dτ 8K1 θ t u1/2 u1 dx dτ 8K1 θ t ⎞2 ⎟ ⎠ t u1 u1 ∂ω ∂x R t θ t R ρ θ dx dτ ∂ω ∂x dx dτ, ω ∂2 ω dx dτ ρ ∂x2 t ≤ t ≤ R 3.18 ∂2 v ∂x2 ρ t ρ ∂ ∂x t t dx dτ t dx dτ dx dτ R ∂ρ ∂x ρ 2u1/2 ≤ u1 t R ∂ρ ∂ω ∂2 ω dx dτ ∂x ∂x ∂x2 ≤ ≤ ∂2 v ∂x2 ρ ∂θ ∂x ∂ρ ∂x t R t dx dτ ∂ρ ∂2 v dx dτ ∂x ∂x2 t u2 θ R ω2 dx dτ ρ3 t t R t ∂2 ω ∂x2 ρ ω2 dx dτ ρθ R t ρ R dx dτ ∂2 ω ∂x2 3.21 dx dτ, 10 Boundary Value Problems t K ρθ R ∂v ∂2 θ dx dτ ∂x ∂x2 t 3K ≤ 2D t 3K θ t ≤ 2D t ρ R ∂v ∂x t ≤ 2D 3u1/2 ≤ Du1 t dx dτ R D t dx dτ R 3.22 ∂2 θ ∂x2 ρ ∂2 θ ∂x2 ρ ρ R dx dτ, R dx dτ ρ dx dτ ⎞1/2 ∂2 v ∂x2 R ∂v ∂x ∂2 θ ∂x2 ⎛ ⎝ dx dx⎠ dτ t 16 ∂2 v ∂x2 ρ D R t ρ R ∂2 θ ∂x2 dx dτ dx dτ ∂2 θ ∂x2 R t D 3/2 t ρ dx dτ ∂v ∂x R t D dx dτ ∂v ∂x t D ∂2 θ dx dτ ∂x2 3u1/2 Du1 ≤4 R ρ θ ∂v ρ ∂x R ∂v θ ρ ∂x R dx dτ, 3.23 t ρ R ≤ ∂ω ∂x t 2D ∂ω ∂x ρ R 3u1/2 ≤ Du1 ≤2 ∂2 θ dx dτ ∂x2 t D t ρ R ∂2 θ ∂x2 ρ R ρ R ∂ω ∂x R ∂2 θ ∂x2 ⎛ ⎝ dx t 3/2 t D dx dτ ∂ω ∂x R 3u1/2 Du1 dx dτ dx dτ ⎞1/2 ∂2 ω ∂x2 dx⎠ t ρ R D dτ ∂2 ω ∂x2 t ρ R ∂2 θ ∂x2 dx dτ dx dτ dx dτ, 3.24 Boundary Value Problems t R ω2 ∂2 θ dx dτ ρ ∂x2 t ≤ 2D ≤ ≤ 11 R 6AE1 u3 D ω4 dx dτ ρ3 t ω2 dx 6AE1 u3 D t R 1/2 R 1/2 ∂ω ∂x ρ ω2 x dτ ρθ dx dτ dx t u1 θ t D dτ ρ θ R ∂ω ∂x t ∂2 θ ∂x2 ρ R dx dτ dx dτ ∂2 θ ∂x2 ρ R R θ t 2u1 ∂2 θ ∂x2 ρ R t D t D dx dτ, 3.25 t D R ∂ρ ∂θ ∂2 θ dx dτ ∂x ∂x ∂x2 t 3D ≤ R 3Du1/2 ≤ u1 t ⎝ ρ dx dτ ρ R ⎞1/2 ∂θ ∂x dx⎠ R ∂2 θ ∂x2 dx dτ 1/2 ρ2 dx R ∂ρ ∂x dx dτ dx dτ ⎛ ⎜ ≤⎝ t D ∂2 θ ∂x2 R 2 ∂2 θ ∂x2 R ρ ∂θ ∂x ⎛ t D ∂ρ ∂x ρ 12DK1 θ t θ t u1 u1 ⎞2 ⎟ ⎠ D t R ρ θ2 ∂θ ∂x dx dτ D t ρ R ∂2 θ ∂x2 dx dτ 3.26 Inserting 3.17 – 3.26 into 3.15 and using estimates 3.1 , 3.3 , and 2.17 we obtain R ∂v ∂x ≤ K5 θ t ∂ω A ∂x t ρ R ∂2 ω ∂x2 6u1/2 Du1 ∂θ ∂x t R dx D 12 dx dτ 2 ∂v ∂x t ρ R t ∂2 v ∂x2 ρ R ∂2 θ ∂x2 dx dτ dx dτ dx dτ 2 3u1/2 Du1 t R ∂ω ∂x dx dτ, 3.27 12 Boundary Value Problems where K5 θ t is defined by 3.10 We also use the following inequality: ∂v ∂x R ∂2 v − v dx ≤ u1/2 R ∂x dx v2 dx R ρ ⎞1/2 dx⎠ 3.28 ⎞1/2 ∂v ∂x2 ρ ⎝ ∂2 v ∂x2 R 1/2 ⎝ ⎛ R ⎛ ≤ 2E1 u1 1/2 dx⎠ that is satisfied for the function ∂ω/∂x as well Therefore we have ρ R ∂ω ∂x2 ρ R ∂2 v ∂x2 ∂v ∂x −1 dx ≥ 2E1 u1 R dx R , ∂ω ∂x −1 dx ≥ 2AE1 u1 2 dx 3.29 Inserting 3.29 into 3.27 we obtain R ∂v ∂x 2 ∂ω A ∂x 3u1/2 Du1 6u1/2 Du1 t ∂θ ∂x R t R ∂ω ∂x ∂v ∂x D dx 12 ⎛ t ρ ⎝ dx 2 dx R dxdτ Du1 Du1 1/2 24u1 E1 ∂ω ∂x − 1/2 12u1 A1/2 E1 ⎛ ⎝ ∂2 θ ∂x2 R − R ∂v ∂x 2 ⎞ dx⎠dτ ⎞ dx⎠dτ ≤ K5 θ t , 3.30 and 3.10 is satisfied In the continuation we use the above results and the conditions of Theorem 2.1 Similarly as in 4, , we derive the estimates for the solution ρ, v, ω, θ of problem 2.1 – 2.7 , defined by 2.8 – 2.10 in the domain Π R× 0, T , for arbitrary T > Taking into account assumption 2.19 and the fact that θ ∈ C Π see 2.15 we have the following alternatives: either sup θ x, t x,t ∈Π θ T ≤ M1 , 3.31 or there exists t1 , < t1 < T , such that θ t < M1 for ≤ t < t1 , θ t1 M1 3.32 Boundary Value Problems 13 Now we assume that 3.32 is satisfied and we will show later that because of the choice E1 , E2 , E3 , and M1 the conditions of Theorem 2.1 , the property 3.32 is impossible Because K2 θ t , defined by 3.7 , increases with increasing θ t , we can easily conclude that < K2 M1 K2 θ t and K2 M1 for ≤ t < t1 , 3.33 E5 Therefore, comparing 2.20 and 3.8 , we obtain u < u1 θ t , u > u1 θ t , 3.34 where u, u, and u1 θ t , u1 θ t are defined by 2.22 and Lemma 3.2 It is important to point out that the quantities K3 θ t and K4 θ t , defined by 3.11 - 3.12 , decrease with M1 they become increasing θ t and for θ t1 Du K3 M1 , 1/2 24uE1 K4 M1 Du 1/2 12uA1/2 E1 3.35 Now, using these facts we will obtain the estimates for R ∂ω/∂x dx and ∂v/∂x dx on 0, t1 Taking into account the assumptions 2.23 and 2.24 of Theorem 2.1 R and the following inclusion see 2.14 : ∂v ∂ω , ∈C ∂x ∂x 0, T ; L2 R , 3.36 we have again the following two alternatives: either R ∂v ∂x x, t dx ≤ K3 M1 , R ∂ω ∂x x, t dx ≤ K4 M1 for t ∈ 0, t1 , 3.37 or there exists t2 , < t2 < t1 , such that ∂ω/∂x and ∂v/∂x or conversely have the following properties: R R R ∂ω ∂x x, t dx < K4 M1 3.38 for t2 < t1 , 3.39 for ≤ t ≤ t2 3.40 ∂ω ∂x ∂v ∂x for ≤ t < t2 , x, t2 dx K4 M1 x, t dx ≤ K3 M1 14 Boundary Value Problems We assume that 3.38 – 3.40 are satisfied Then we have K4 M1 < K4 θ t , θ t < M1 , for t ∈ 0, t2 Using 3.41 from 3.10 , for t R ∂ω ∂x K3 M1 < K3 θ t 3.41 t2 , we obtain x, t dx ≤ 2AK5 θ t2 , ≤ t ≤ t2 3.42 Since K5 θ t , defined by 3.13 , increases with the increase of θ t , it holds that K5 θ t ≤ K5 M1 , t ∈ 0, t2 3.43 t ∈ 0, t2 3.44 Using condition 2.25 we get 2AK5 θ t < K4 M1 , and conclude that ∂ω ∂x R x, t2 dx < K4 M1 3.45 This inequality contradicts 3.39 Consequently, the only case possible is when t2 t1 , 3.46 and then for ∂ω/∂x 3.37 is satisfied If in 3.38 – 3.40 the functions ∂ω/∂x and ∂v/∂x exchange positions, using assumption 2.25 , in the same way as above we obtain that the function ∂v/∂x satisfies the inequality R ∂v ∂x x, t dx ≤ K3 M1 , t ∈ 0, t1 3.47 With the help of 3.37 from 3.10 we can easily conclude that R ∂θ ∂x x, t dx < 2K5 M1 for < t ≤ t1 3.48 Now, as in , we introduce the function Ψ by Ψ θ x, t θ x,t s − − ln s ds 3.49 Boundary Value Problems 15 Taking into account that from 2.14 follows that θ x, t → as |x| → ∞, we have Ψ θ x, t −→ as |x| −→ ∞ 3.50 Consequently, ψ θ x, t ≤ ψ θ x, t θ x,t d ψ s ds ds x θ x, t − − ln θ x, t −∞ ≤ 3.51 ∂θ x, t dx ∂x θ x, t − − ln θ x, t dx 1/2 R ∂θ ∂x R 1/2 x, t dx Using 3.32 , 3.48 , and 3.1 from 3.51 we get max 0≤θ x,t ≤M1 ψ θ t1 ψ θ x, t ψ M1 ≤ 2K5 M1 E1 1/2 , 3.52 or M1 s − − ln s ds − 2K5 M1 E1 1/2 ≤ 3.53 Since this inequality contradicts 2.25 , it remains to assume that t1 following lemma T Hence we have the Lemma 3.4 For each T > it holds that θ x, t ≤ M1 , R ∂ω ∂x R ∂v ∂x R ∂θ ∂x x, t ∈ Π, 3.54 x, t dx ≤ K4 M1 , ≤ t ≤ T, 3.55 x, t dx ≤ K3 M1 , ≤ t ≤ T, 3.56 x, t dx ≤ 2K5 M1 , ≤ t ≤ T Proof These conclusions follow from 3.32 , 3.37 , and 3.48 directly 3.57 16 Boundary Value Problems Lemma 3.5 It holds that 0 0, 3.58 , 3.59 , 3.60 x, t ∈ Π, 3.61 where u and u are defined by 2.20 – 2.22 and a constant h depends only on the data of problem 2.1 – 2.5 Proof Because the quantity u1 θ t in Lemma 3.2 decreases with increasing θ t while u1 θ t increases, it follows, in the same way as in , from 3.9 and 3.54 that 3.58 is satisfied Using the inequalities v2 x −∞ v ∂v dx ≤ ∂x v2 dx 1/2 R ∂v ∂x R 1/2 dx , 3.62 ω2 x −∞ ω ∂ω dx ≤ ∂x ω2 dx 1/2 R ∂ω ∂x R 1/2 dx and estimations 3.1 , 3.55 , and 3.56 we get immediately 3.59 and 3.60 From 3.50 , 3.53 , 3.56 , and 3.1 we have, as in , for θ x, t ≤ that θ x,t s − − ln s ds ≤ 2K5 M1 E1 1/2 < s − − ln s ds 3.63 holds because of 2.25 Hence we conclude that there exists the constant h > such that θ x, t ≥ h Boundary Value Problems 17 Remark 3.6 Using the properties of the functions u1 exp η θ t and u1 exp η1 θ t defined in Lemma 3.2, from 3.55 - 3.56 and 3.59 - 3.60 we get the following estimates: ∂ω ∂x R R ∂v ∂x dx ≤ 1/2 24E1 sup |v x, t | ≤ 3.64 exp − 2λ t , exp − D2 72 x,t ∈Π exp{−2λ t }, 1/4 D 18 −η θ t D x,t ∈Π η1 θ t exp{−2λ t }, 1/2 12A1/2 E1 sup |ω x, t | ≤ where λ t D dx ≤ 2λ t , 1/4 > Lemma 3.7 For each T > it holds that R ∂v ∂x R ∂θ ∂x R ∂ω ∂x T T T T R T ∂ρ ∂x R T 0 3.66 dx dτ ≤ K8 , 3.67 3.68 dx dτ ≤ K10 , dx ≤ K11 , R R ∂2 v ∂x2 R ∂2 ω ∂x2 T dx dτ ≤ K7 , 3.69 ∂2 θ ∂x2 T 3.65 ω2 dx dτ ≤ K9 , ρθ ∂ρ ∂x R dx dτ ≤ K6 , t ∈ 0, T , 3.70 dx dτ ≤ K12 , 3.71 dx dτ ≤ K13 , 3.72 2 dx dτ ≤ K14 , where the constants K6 , K7 , , K14 ∈ R are independent of T 3.73 18 Boundary Value Problems Proof Taking into account 3.58 , 3.61 , and 3.54 from 3.1 , 3.3 , 3.10 , and 3.27 we get all above estimates Proof of Theorem 2.1 In the following we use the results of Section The conclusions of Theorem 2.1 are immediate consequences of the following lemmas Lemma 4.1 It holds that ∂v ∂x R ∂ω ∂x x, t dx −→ 0, R ∂θ ∂x x, t dx −→ 0, R x, t dx −→ 0, 4.1 when t → ∞ Proof Let ε > be arbitrary With the help of 3.65 – 3.69 we conclude that there exists t0 > such that t t0 ∂v ∂x R t t0 ∂θ ∂x dx dτ < ε, R t t0 ∂ω ∂x dx dτ < ε, R dx dτ < ε, 4.2 t t0 t R ω dx dτ < ε, ρθ t0 R ∂ρ ∂x dx dτ < ε, for t > t0 , and ∂v ∂x R x, t0 dx < ε, R ∂θ ∂x ∂ω ∂x x, t0 dx < ε, R x, t0 dx < ε Similarly to 3.10 , we have R ∂v ∂x ∂ω A ∂x 6u1/2 Du 3u Du t t0 R t t0 R ∂θ ∂x dx ∂ω ∂x t ρ t0 R ∂2 θ ∂x2 2 ∂v ∂x D 12 K3 θ τ dx − R 2 dx K4 θ τ − R dx dτ ∂v ∂x ∂ω ∂x dx dτ dx dτ 4.3 Boundary Value Problems ≤ 2 ∂v ∂x R 19 16K1 θ t ⎛ ⎜ 2⎝ t u u2 t0 8K1 θ t u1/2 u t 3K 2D t0 ∂ω A ∂x t ⎟ ⎠ R ∂v ∂x R t0 t0 dx dτ ∂ω ∂x 3AE1 u D t t0 ∂ω ∂x R dx dτ R R θ2 ρ dx dτ ∂ρ ∂x ⎜ ⎝ t u2 θ t dx dτ t0 3AE1 θ t u3 Du ⎛ t K2 dx dτ ∂θ ∂x ρ dx dτ ⎞2 t0 θ2 ρ t K2 x, t0 dx ∂v ∂x R ∂θ ∂x t t0 R R ω2 dx dτ ρθ ω2 dx dτ ρθ 12DK1 θ t u1/2 u ⎞2 ⎟ ⎠ t t0 ∂θ ∂x R dx dτ 4.4 Taking into account 3.54 , 3.58 , 3.61 , and 4.2 – 4.3 from 4.4 we obtain ∂v ∂x R ∂v A ∂x ∂θ ∂x dx ≤ K15 ε for t > t0 , 4.5 where K15 depends only on the data of our problem and does not depend on t0 Hence relations 4.1 hold Lemma 4.2 It holds that v x, t −→ 0, ω x, t −→ 0, θ x, t −→ 4.6 when t → ∞, uniformly with respect to all x, x ∈ R Proof We have see 3.51 and 3.62 v2 x, t ≤ 1/2 R v2 x, t dx R ω2 x, t ≤ ω2 x, t dx R ψ θ x, t ≤ ∂v ∂x R ∂ω ∂x 1/2 θ x, t − − ln θ x, t dx R 1/2 , x, t dx 1/2 1/2 R ∂θ ∂x 4.7 , x, t dx 1/2 dx 20 Boundary Value Problems Taking into account 3.1 from 4.7 we get ∂v ∂x 1/2 v x, t ≤ 2E1 R ω2 x, t ≤ 2AE1 1/2 R 1/2 ≤ E1 ψ θ x, t R 1/2 ∂ω ∂x , x, t dx 1/2 1/2 ∂θ ∂x 4.8 , x, t dx dx Using 4.1 and property 3.50 of the function ψ we can easily obtain that 4.6 holds Now, as in , we analyze the behavior of the function ρ as t → ∞ From 3.69 and 3.72 we conclude that for ε > there exists t0 > such that t t0 R ∂ρ ∂x t dx dτ < ε, t0 ∂2 v ∂x2 R dx dτ < ε, R ∂ρ ∂x x, t0 dx < ε 4.9 for t > t0 Deriving 2.1 with respect to x, multiplying by ∂/∂x 1/ρ , and integrating over R and t0 , t , after using 3.58 and Young’s inequality, we obtain u4 R ∂ρ ∂x dx ≤ u2 t t0 u4 R R ∂ρ ∂x ∂ρ ∂x 2 dx dτ u2 t t0 R ∂2 v ∂x2 dx dτ 4.10 x, t0 dx With the help of 4.9 we get easily the following result Lemma 4.3 It holds that R ∂ρ ∂x x, t dx −→ 4.11 when t → ∞ Similarly as for the function ψ θ , we have ψ 1/ρ ρ s − − ln s ds ≤ R 1 − − ln dx ρ ρ 1/2 R ρ2 ∂ρ ∂x 4.12 1/2 dx Boundary Value Problems 21 Taking into account 3.1 from 4.12 one has ψ ρ ≤ K −1 E1 1/2 u R ∂ρ ∂x 1/2 dx 4.13 Using 4.11 we obtain the following conclusion Lemma 4.4 It holds that ρ x, t −→ 4.14 when t → ∞, uniformly with respect to x ∈ R References N Mujakovi´ , “One-dimensional flow of a compressible viscous micropolar fluid: the Cauchy c problem,” Mathematical Communications, vol 10, no 1, pp 1–14, 2005 N Mujakovi´ , “Uniqueness of a solution of the Cauchy problem for one-dimensional compressible c viscous micropolar fluid model,” Applied 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Numerical Methods for Science and Technology Vol 5, Springer, Berlin, Germany, 1992  ... fluid: the Cauchy c problem,” Mathematical Communications, vol 10, no 1, pp 1–14, 2005 N Mujakovi´ , “Uniqueness of a solution of the Cauchy problem for one-dimensional compressible c viscous micropolar. .. 3.27 we get all above estimates Proof of Theorem 2.1 In the following we use the results of Section The conclusions of Theorem 2.1 are immediate consequences of the following lemmas Lemma 4.1 It... 1979 Russian ı N Mujakovi´ and I Draˇ i´ , ? ?The Cauchy problem for one-dimensional flow of a compressible viscous c zc fluid: stabilization of the solution, ” Glasnik Matematiˇ ki In press c N