Regularity to the spherically symmetric compressible Navier–Stokes equations with density-dependent viscosity

Regularity to the spherically symmetric compressible Navier–Stokes equations with... College of Mathematics and This paper is concerned with the dynamics for the compressible Navier–Stokes Statistics, North China University of Water Resources and Electric Power, equations with density-dependent viscosity in bounded annular domains in R .Inthe Zhengzhou, P.R. China paper, we shall analyze the spherical symmetric model and establish the regularity in 2 4 H and H under certain assumptions imposed on the initial data. Keywords: Navier–Stokes equations; Density-dependent viscosity; Regularity; Spherical symmetry 1 Introduction It is well known that the compressible isentropic Navier–Stokes equations which describe the motion of compressible fluids can be written in Eulerian coordinates as ρ + div(ρU)=0, (1.1) (ρU) + div(ρU ⊗ U) – div μ(ρ)D(U) – ∇ λ(ρ)divU + ∇P(ρ)=0, (1.2) where ρ(x, t), U(x, t)and P(ρ)= ρ (γ > 1) stand for the fluid density, velocity and pressure, respectively, and the strain tensor is given by ∇U +(∇U) D(U)= . (1.3) The Lamé viscosity coefficients μ(ρ), λ(ρ) satisfy the natural restrictions μ(ρ)>0, μ(ρ)+ N λ(ρ) ≥ 0. (1.4) For simplicity of the presentation, we consider only the viscosity terms μ(ρ)= ρ, λ(ρ)=0 and D(U)= ∇U.Then(1.1)–(1.2)become ρ + div(ρU)=0, (1.5) (ρU) + div(ρU ⊗ U)+ ∇P(ρ) – div(ρ∇U)=0. (1.6) © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Huang and Lian Boundary Value Problems (2018) 2018:85 Page 2 of 13 We are concerned with the spherically symmetric solutions of system (1.5)–(1.6)in bounded annular domains G = {x ∈ R ,0 < a < |x| < b <+∞}.Tothisend,wedenote |x| = r, ρ(x, t)= ρ(r, t), U(x, t)= u(r, t) , (1.7) which leads to the following system of equations for r >0: 2ρu ρ +(ρu) + = 0, (1.8) t r 2ρu 2u 2 γ (ρu) + ρu + ρ + –(ρu) – ρ = 0. (1.9) t r r r We shall consider problem (1.8)–(1.9)inthe region G subject to the initial data (ρ, u)(r,0) = (ρ , u )(r), r ∈ [a, b], (1.10) 0 0 and the boundary condition u(a, t)= u(b, t)=0, t ∈ [0, T]. (1.11) First we find it convenient to transfer problems (1.8)–(1.11) into that in Lagrangian co- ordinates and draw the desired results. We introduce the following coordinate transfor- mation: x(r, t)= ρ(s, τ)s ds, t = τ, (1.12) then the boundaries r = a and r = b become b b 2 2 x =0, x = ρ(s, τ)s ds = ρ (s)s ds, (1.13) a a where ρ (s)s ds is the total initial mass and, without loss of generality, we can normalize it to 1. So in terms of Lagrangian coordinates, the domain G becomes  = (0, 1). The relations between Lagrangian and Eulerian coordinates are satisfied by ∂x ∂x 2 2 = ρr , = ρur . (1.14) ∂r ∂t The initial boundary value problem (1.8)–(1.11)are changedto 2 2 ρ + ρ r u = 0, (1.15) u 2ρ u t x γ 2 2 + ρ = ρ r u – , (1.16) 2 x x x r r (ρ, u)(x,0) = (ρ , u )(x), x ∈ [0, 1], (1.17) 0 0 u(0, t)= u(1, t)=0, t > 0. (1.18) Huang and Lian Boundary Value Problems (2018) 2018:85 Page 3 of 13 Much progress was achieved recently on the compressible Navier–Stokes equations with density-dependent viscosity coefficient. Firstly let us recall some well-known results as re- gards the one-dimensional compressible isentropic Navier–Stokes equations with the flow density being connected with the infinite vacuum, [19, 24, 25] for the local well-posedness and the global existence of weak solutions to an initial boundary value problem with the viscous gas being connected vacuum states with jump discontinuities, [9, 10]for theglobal behavior with the initial density being piecewise smooth; [28–31, 34] for the local exis- tence, the global existence, the asymptotic behavior and the uniqueness of weak solutions with a viscous gas being connected vacuum states with continuous density. In spatial multi-dimension, there is a huge literature as regards the global existence, the regularity and the asymptotic behavior of a solution to system (1.1)–(1.2) with constant viscosity, we refer the reader to [2, 3, 11–14, 16, 20–23, 33] and the references therein. For the 3-D flow of a compressible fluid with cylindrical symmetry, the global existence and the large-time behavior of generalized solutions have been proved in [1, 4, 5, 7, 15, 23, 26, 27, 32] for the isentropic and the nonisentropic cases. The corresponding study of the regularity of a solution for any given initial datum has been carried out in [17]. For the 3- D flow of compressible fluid with spherical symmetry, there are some interesting results, [18] for the global well-posedness of classical solutions with large oscillations and vacuum; [33] for the global existence and uniqueness of the weak solution without a solid core; [14] for the structure of the solution; [21] for the global existence of the exterior problem and the initial boundary value problem. Besides, we would like to refer to [6, 8]asregards the existence and regularity of solutions for micropolar fluid with spherical symmetry in the three-dimensional case. In the paper, we shall analyze the spherical symmetric model and focus on the initial boundary problem of an isentropic compressible fluid. We show the regularity in H and H under certain assumptions imposed on the initial data. The notation in this paper will be as follows: 1,2 p m,p 1 1,2 1 L , 1 ≤ p ≤ +∞, W , m ∈ N, H = W , H = W denote the usual (Sobolev) 0 0 spaces on [0, 1]. To denote various constants, we use C (i = 1,2,4) to denote the generic positive constant depending only on the H norm of initial datum (ρ , u ), 0 0 min ρ (x) and variable t, respectively. In addition, · denotes the norm in the x∈[0,1] 0 space L . The basic assumption of this paper is the following: inf ρ > ρ, (1.19) [0,1] for some constant ρ >0. 2 2 Theorem 1.1 Let γ >1. Assume that the initial data (ρ , u ) ∈ H () × H () and (1.19) 0 0 2 2 hold, then there exists a unique generalized global solution (ρ(t), u(t)) ∈ (H ()) to the problem (1.15)–(1.18) verifying that, for any T >0, ∞ 2 2 2 ρ ∈ L [0, T], H () ∩ L [0, T], H () , (1.20) ∞ 2 2 3 u ∈ L [0, T], H () ∩ L [0, T], H () , (1.21) ∞ 2 2 1 u ∈ L [0, T], L () ∩ L [0, T], H () . (1.22) t Huang and Lian Boundary Value Problems (2018) 2018:85 Page 4 of 13 Theorem 1.2 Let γ >1. Assume that the initial data satisfies (1.19) and (ρ , u ) ∈ H () × 0 0 4 4 2 H (), then there exists a unique generalized global solution (ρ(t), u(t)) ∈ (H ()) to the problem (1.15)–(1.18) verifying that, for any T >0, ∞ 4 2 4 ρ ∈ L [0, T], H () ∩ L [0, T], H () , (1.23) ∞ 4 2 4 u ∈ L [0, T], H () ∩ L [0, T], H () , (1.24) ∞ 2 2 3 u ∈ L [0, T], H () ∩ L [0, T], H () , (1.25) ∞ 2 2 1 u ∈ L [0, T], L () ∩ L [0, T], H () . (1.26) tt Corollary 1.1 Under assumptions of Theorem 1.2,(1.23)–(1.24) imply (ρ(t), u(t)) is the classical solution verifying that, for any t >0, ρ(t) + u(t) ≤ C . (1.27) 3+1/2 3+1/2 C C 2 Proof of Theorem 1.1 This section is devoted to deriving the estimates of the solutions to prove Theorem 1.1 which will be presented in a sequence of lemmas. We begin with the following lemma. Lemma 2.1 (Theorem 2.2 in [21]) Under the assumptions in Theorem 1.1, then there exist positive constants ρ >0 and ρ >0 so that the unique global solution (ρ(t), u(t)) to problem (1.15)–(1.18) exists and satisfies, for any T >0, 0< ρ ≤ ρ(x, t) ≤ ρ , (2.1) 1 t 1 2 2 2 2 2 2 2 u +(ρ – ρ¯) + u + u + ρ (x, t) dx + ρ + u x t x x x 0 0 0 2 2 2 + u + u + u (x, s) dx ds ≤ C , t ∈ [0, T], (2.2) t xx xt where ρ¯ = ρ(s, t)s ds. b–a a Lemma 2.2 Under the conditions in Theorem 1.1, for any T >0, 2 2 ρ (t) + ρ (s) ds ≤ C , t ∈ [0, T], (2.3) xx xx 2 t 1 u (t) + u (x, s) dx ds ≤ C , t ∈ [0, T]. (2.4) xx 2 xxx 0 0 Proof We infer from (1.16)that u 2ρ u t x γ –1 2 2 2 2 =–γρ ρ +2ρρ r u + ρ r u +4rr u +2r u +2rr u – . (2.5) x x xx x x xx 2 x r r Multiplying (2.5)by u in L (), we deduce xx 1 1 2 2 2 γ –1 2 ρ r u dx = u + γρ ρ –2ρρ r u xx x x xx 0 0 2ρ u 2 2 – ρ 4rr u +2r u +2rr u + dx. (2.6) x x xx r Huang and Lian Boundary Value Problems (2018) 2018:85 Page 5 of 13 Using Young’s inequality, Sobolev’s embedding theorem, and Lemma 2.1,wededucefrom (2.6)that 1 1 1 2 2 2 2 2 2 u dx ≤ C u + ρ + ρ u dx + u dx xx t x x x xx 0 0 0 1 1 2 2 2 ≤ C + C u  ρ + u dx 2 1 x L x xx 0 0 ≤ C + u dx, xx whence u dx ≤ C . (2.7) xx Differentiating (1.16)withrespect to x, and exploiting (1.15), we have u 2ρ u t x γ 2 2 = –ρ + ρ r u – , (2.8) x xx r r x x which gives γ –1 ρ + γρ ρ = E (x, t) (2.9) txx xx 0 with 2ρ ur 2ρ u +2ρ u 2u r u x x xx x x t x tx γ –2 2 E (x, t)= – – γ (γ –1)ρ ρ + – . 2 3 2 r r r r Multiplying (2.9)by ρ , integrating the resultant over [0, 1], we deduce xx γ –1 2 ρ (t) + γρ ρ dx xx xx dt 2ρ ur 2ρ u +2ρ u 2u r u x x xx x x t x tx γ –2 2 = – – γ (γ –1)ρ ρ + – ρ dx xx 2 3 2 r r r r 2 2 ≤ C ρ  u  + u r  + ρ + ρ u  + ρ ur  –2 ρ dx. 1 xx tx t x x x x x x xx Using the Young inequality and the interpolation inequality to the above inequality, then we get 2 2 2 2 2 ρ (t) ≤ C u  + u  + ρ  + u  + u xx 1 tx t x x dt 2 2 + C ρ  1+ u . (2.10) 1 xx Integrating (2.10)withrespect to t, using initial condition ρ ∈ H and Lemma 2.1,we have 2 2 ρ (t) ≤ C + C ρ (s) ds, ∀t ∈ [0, T]. xx 2 1 xx Then using the Gronwall inequality to the above inequality, we can get (2.3). Huang and Lian Boundary Value Problems (2018) 2018:85 Page 6 of 13 Differentiating (1.16)withrespect to x, we can obtain 2 2 2 2 2 u =– (4ruρ r +2r ρ u +2ρρ 2rr u + r u +4ρρ 2r u xxx x x xx x x x x x x 2 2 ρ r 2 2 2 2 +4rr u +2rr u + r u + ρ 6r r u +6r u +6rr u + r u x x xx xx x xx x xx xx 2ρ u +2ρ u xx x x 2 2 + ρ 6r r u +6r u +6rr u +6rr u +2rr u + x xx x xx x x xx xxx 2ρ ur u u r x x tx t x γ –1 γ –2 2 – + – + γρ ρ + γ (γ –1)ρ ρ . (2.11) xx 2 2 3 r r r Integrating (2.11)over x and t, applying the embedding theorem, Lemma 2.1,(2.7)and (2.3), we conclude, for any t ∈ [0, T], t 1 t 1 2 2 4 2 4 2 2 2 2 2 2 u dx ds ≤ C u + ρ + ρ + ρ u + ρ u + ρ u + u (x, s) dx ds xxx tx x xx x x x x xx x t 0 0 0 0 t 1 t 2 2 4 2 2 ≤ C + C u  ∞ ρ + ρ dx ds + C ρ  ∞ u  ds 1 1 x 1 x xx L xx x L 0 0 0 ≤ C , which, along with (2.7), gives (2.4). The proof is complete. Proof of Theorem 1.1 Clearly, (1.20)–(1.22) follow from Lemmas 2.1–2.2.Thiscompletes the proof of Theorem 1.1. 3 Proof of Theorem 1.2 In this section, we shall complete the proof of Theorem 1.2.Tothisend,weassumethatin this sectionthatall assumptionsinTheorem 1.2 hold. We begin with the following lemma. Lemma 3.1 The following estimate holds for any T >0: t t 2 2 2 u (t) + u (s) ds ≤ C + C u (s) ds, t ∈ [0, T]. (3.1) tt ttx 4 2 txx 0 0 Proof We easily infer from (1.16)and (2.1)–(2.4)that u (t) ≤ C ρ (t) + u (t) + u (t) . (3.2) t 1 x x xx Differentiating (1.16)withrespect to x and using Lemmas 2.1–2.2,wehave u (t) ≤ C ρ (t) + ρ (t) + u (t) + u (t) + u (t) , (3.3) tx 2 x xx x xx xxx or u (t) ≤ C ρ (t) + ρ (t) + u (t) + u (t) + u (t) . (3.4) xxx 2 x xx x xx tx Huang and Lian Boundary Value Problems (2018) 2018:85 Page 7 of 13 Differentiating (1.16)withrespect to x twice and using the Cauchy–Schwarz inequality, we have u (t) ≤ C ρ (t) + u (t) , (3.5) txx 2 x 2 x 3 H H u (t) ≤ C ρ (t) + u (t) + u (t) . (3.6) xxxx 2 x 2 x 2 txx H H After differentiating (1.16)withrespect to t,using (3.2)–(3.3)and (3.5), we can get u (t) ≤ C u (t) + u (t) + u (t) + ρ (t) + u (t) + u (t) (3.7) tt 2 tx txx t x x xx ≤ C ρ (t) + u (t) . (3.8) 2 x 2 x 3 H H Now differentiating (1.16)withrespect to t twice, multiplying the resulting equation by ( ) in L (), and using integration by parts and (1.18), we conclude 2 tt 1 1 1 d u u γ 2 2 dx =– –ρ + ρ r u dx x tt 2 2 2 dt r r 0 0 tt ttx 1 2 2ρ u u u – – dx =: A + A . (3.9) 1 2 3 2 r r r tt tt We use (2.1)–(2.4), (3.2)–(3.8), (1.16) and the interpolation inequality to deduce that, for any small ε ∈ (0, 1), γ 2 2 A =– –ρ + ρ r u dx x tt 2 ttx 2 2 2 ≤ – ρ u dx + C ρ u  + ρ u + ρ u  + ρ u  + ρ u  + u  + u 2 tt x x t t t x t xt x t ttx t + u u  + u  + u  u  + u  + u u  + u  + u  + u  + u x t tx tt ttx tt x t tx t x 2 2 2 2 2 ≤ – ρ u dx + ε u (t) + C ρ (t) + ρ (t) ttx 2 tt 1 t ttx 2 2 2 2 + u (t) + u (t) + u (t) + ρ (t) t 1 tt x 1 x 1 H H H 2 2 2 2 –1 ≤ –C u (t) + C u (t) + ρ (t) + u (t) , (3.10) ttx 2 x x t 2 2 1 2 H H H 2ρ u u u A =– – dx 3 2 r r r tt tt 2 2 ≤ C ρ  + ρ u  + ρ u  + ρ u + u + u  + u  + ρ u 2 ttx tx t x tt tx tt t xt 3 3 2 + ρ u  + ρ u + u  + u u  + u  + u r  + ur  + ur x t x t tt t t t tt 2 2 2 2 2 2 ≤ C u (t) + u (t) + u (t) + u (t) + ρ (t) + u (t) . (3.11) 2 t tt txx tx x x Integrating (3.9)withrespect to t,and using(2.1)–(2.4), initial condition (1.16)and (3.10)– (3.11), then we obtain (3.1). Lemma 3.2 For any T >0 and ε ∈ (0, 1), the following estimate holds: t t 2 2 2 –1 2 u (t) + u (s) ≤ C + C ε u (s) ds, t ∈ [0, T]. (3.12) tx txx 4 ttx 0 0 Huang and Lian Boundary Value Problems (2018) 2018:85 Page 8 of 13 Proof Differentiating (1.16)withrespect to t and x, then multiplying the resultant by ( ) tx in L [0, 1], and integrating by parts, we know that 1 2 1 d u dx 2 dt r tx u u γ 2 2 γ 2 2 = –ρ + ρ r u – –ρ + ρ r u dx x tx x tx 2 2 r r tx 0 txx 1 2 u 2ρ u u + – dx 3 2 r r r tx tx = B (x, t)+ B (t)+ B (t), (3.13) 0 1 2 where γ 2 2 B (x, t)= –ρ + ρ r u , x tx tx 0 γ 2 2 B (t)=– –ρ + ρ r u dx, x tx txx 1 2 u 2ρ u u B (t)= – dx. 3 2 r r r tx tx Now using Young’s inequality several times, and employing (2.1)–(2.4)and theinterpo- lation inequality, after some calculation, we have, for any ε ∈ (0, 1), B (x, t) ≤ C ρ  ∞ + ρ ρ  ∞ + ρ ρ u  ∞ + ρ u  ∞ + ρ u  ∞ 0 2 tx L t x L x t x L tx x L t xx L + u  ∞ + u  ∞ u  ∞ txx L tx L tx L 1 1 1 1 1 1 2 2 2 2 2 2 ≤ C u  2 + ρ  1 + u  u  + u  u  u  u 2 x H x H tx txx txx txxx tx txx –1 2 2 2 –6 2 ≤ C ε u  + u  + C ε u  + u  2 + ρ  1 , (3.14) txx txxx 2 tx x H x H 1 1 2 2 2 2 2 2 B (t) ≤ – ρ u dx + ε ρ u dx + C ρ (t) + ρ (t) 1 2 tx 1 x 1 txx txx H H 0 0 2 2 2 + u (t) + u (t) + u (t) , (3.15) x 1 t tx B (t) ≤ C u u  + u  + u  + u  + u + ρ  + ρ u  + ρ u 2 2 x t tx x t txx tx x xx t + ρ u  + ρ  + ρ u  + ρ  + ρ u  + ρ  u  + u x tx xx x x tx x t x tx x 2 2 2 2 2 ≤ C u  + u  + ρ  + ρ  + ρ 2 x 1 t 1 x 1 tx txx H H H 2 2 2 ≤ C u  + u  + ρ  . (3.16) 2 x 2 t 1 x 1 H H H Differentiating (1.16)withrespect to x and t, and using Lemmas 2.1–2.2 and (3.7)–(3.8), we conclude u (t) ≤ C u (t) + ρ (t) + u (t) + u (t) . (3.17) txxx 2 x 2 x 1 tx 1 ttx H H H We integrate (3.13)withrespect to t,use (3.3), (3.14)–(3.16) and Lemmas 2.1–2.2 to obtain (3.12). The proof is complete.  Huang and Lian Boundary Value Problems (2018) 2018:85 Page 9 of 13 Lemma 3.3 The following estimates hold for any T >0: 2 2 2 2 u (t) + u (t) + u  + u  (s) ≤ C , t ∈ [0, T], (3.18) tt tx txx ttx 4 2 2 2 2 ρ (t) + u (t) + ρ  + u  (s) ds ≤ C , t ∈ [0, T]. (3.19) xxx xxx xxx xxxx 4 Proof We insert (3.1)into(3.12)and pick ε small enough to get (3.18). Differentiating (2.9)withrespect to x,wehave γ –1 ρ + γρ ρ = E (x, t), (3.20) txxx xxx 1 where γ –2 E (x, t)= E (x, t)– γ (γ –1)ρ ρ ρ . 1 0x x xx Taking into account estimate (3.18), from (2.1)–(2.4)wecan get E (t) ≤ C E  + ρ ρ 1 2 0x x xx ≤ C ρ u  + ρ  + ρ u  + ρ  + ρ u  + ρ ρ 2 x x xx xx x xxx x xx xx x + u  + u ρ  + u tx t x txx ≤ C ρ (t) + u (t) + u (t) , 2 x x t 2 2 2 H H H which, along with Lemmas 2.1–2.2 and (3.18), implies t t 2 2 E (s) ds ≤ C + C ρ (s) ds. (3.21) 1 4 2 xxx 0 0 After multiplying (3.20)by ρ in L [0, 1], we deduce xxx 1 d 2 γ –1 2 ρ  + γ ρ ρ dx ≤ C E (t) , (3.22) xxx 1 1 xxx 2 dt which implies ρ  ≤ C E (t) . (3.23) xxx 1 1 dt Integrating (3.23)withrespect to t,using (3.21), we conclude ρ  ≤ C + C ρ (s) ds, xxx 4 2 xxx which, after applying the Gronwall inequality and (3.22), yields 2 2 ρ (t) + ρ (s) ds ≤ C , ∀t ∈ [0, T]. (3.24) xxx xxx 4 0 Huang and Lian Boundary Value Problems (2018) 2018:85 Page 10 of 13 By (3.4), (3.6), (3.18) and Lemmas 2.1–2.2,weconclude 2 2 u (t) + u (s) ds ≤ C , ∀t ∈ [0, T], xxx xxxx 4 which, along with (3.24), gives (3.19). The proof is complete. Lemma 3.4 The following estimates hold for any T >0: 2 2 ρ (t) + ρ (s) ds ≤ C , t ∈ [0, T], (3.25) xxxx xxxx 4 2 2 2 u (t) + u (t) + u (s) ds ≤ C , t ∈ [0, T]. (3.26) txx xxxx xxxxx 4 Proof Differentiating (1.16)withrespect to t, and using Lemmas 2.1–2.2 and Lemma 3.3, we can get u (t) ≤ C u (t) + ρ (t) + u (t) + u (t) txx 1 t 1 x x 1 tt H H ≤ C , ∀t ∈ [0, T]. (3.27) Differentiating (3.20)withrespect to x,wehave γ –1 ρ + γρ ρ = E (x, t), (3.28) txxxx xxxx 2 where E (x, t)= E (x, t)– γ (γ –1)ρ ρ (3.29) 2 1x x xxx and γ –2 E (x, t)= E (x, t)– γ (γ –1) ρ ρ ρ . 1x 0xx x xx An easy calculation with the interpolation inequality, (3.2)–(3.8) and Lemmas 2.1–2.2 and Lemma 3.3 gives E (t) ≤ C ρ u  + ρ u  + ρ u  + ρ + u r  + u r  ++u 0x 2 x x xxx x xx xx tx x t xx txx ≤ C ρ  3 + u  2 + u  + u  , (3.30) 2 x H x H tx txx E (t) ≤ C ρ  + ρ u  + ρ u  + ρ u  + ρ u 0xx 2 xxxx xx x x xx xxx x xx xx + u  + u u  + u txx tx x txxx ≤ C ρ  3 + u  2 + u  2 , (3.31) 2 x x tx H H H 2 2 E (t) ≤ C E (t) + ρ ρ + ρ + ρ ρ 1x 2 0xx xx x xxx x xx ≤ C ρ  3 + u  2 + u  2  . (3.32) 2 x x tx H H H Taking into account estimate (3.30)–(3.32), from (3.29), we obtain E (t) ≤ C u (t) + ρ (t) + u (t) . (3.33) 2 4 x 3 x 3 tx 2 H H H Huang and Lian Boundary Value Problems (2018) 2018:85 Page 11 of 13 Inserting (3.17)into(3.33), and integrating (3.33)withrespect to t, using Lemmas 2.1– 2.2 and Lemma 3.3,wehave t t 2 2 E (s) ds ≤ C + C ρ (s) ds, ∀t ∈ [0, T]. (3.34) 2 4 2 xxxx 0 0 Multiplying (3.28)by ρ in L [0, 1], we can get xxxx 1 d 2 γ –1 2 ρ  + γ ρ ρ dx ≤ C E (t) , (3.35) xxxx 2 2 xxxx 2 dt which implies ρ  ≤ C E (t) . (3.36) xxxx 2 2 dt Integrating (3.36)withrespect to t,using (3.34), we conclude ρ  ≤ C + C ρ (s) ds, t ∈ [0, T] (3.37) xxxx 4 2 xxxx which, after using Gronwall’s inequality, yields ρ (t) ≤ C , t ∈ [0, T]. (3.38) xxxx 4 Thus, we can obtain (3.25)byvirtueof(3.37)–(3.38). By (3.6), (3.17), (3.25), Lemmas 2.1–2.2 and Lemma 3.3,wehave 2 2 u (t) + u (s) ds ≤ C , t ∈ [0, T]. (3.39) xxxx txxx 4 On the other hand, we differentiate (1.16)withrespect to x three times, use Lemmas 2.1– 2.2 and (3.25), (3.27) to conclude, for any t ∈ [0, T], u (t) ≤ C u (t) + u (t) + ρ (t) . (3.40) xxxxx 4 txxx x 3 x 3 H H Thus we deduce from (3.38)–(3.40)that u (s) ds ≤ C , ∀t ∈ [0, T] xxxxx 4 which, along with (3.39), gives (3.26). This completes the proof of the lemma. Proof of Theorem 1.2 Applying Lemmas 2.1–2.2 and Lemmas 3.1–3.4, we readily get es- timates (1.23)–(1.26). This completes the proof of Theorem 1.2. 4Conclusions In this paper, we have established the regularity of global solutions for the spherically sym- metric compressible fluid with density-dependent viscosity in H and H . The biggest dif- 2 4 ference from other papers is that our domain has spherical symmetry and the viscosity coefficients are density dependent. Huang and Lian Boundary Value Problems (2018) 2018:85 Page 12 of 13 Acknowledgements The authors would like to thank the referees for their useful suggestions, which have significantly improved the paper. Funding This research was supported by the NSFC (No. 11501199) and the Natural Science Foundation of Henan Province (No. 18B110010). Availability of data and materials Data sharing not applicable to this article as no datasets were generated or analyzed during the current study. Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors carried out the proofs and conceived the study. All authors read and approved the final manuscript. Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 27 February 2018 Accepted: 24 May 2018 References 1. Aoyama, R., Kagei, Y.: Large time behavior of solutions to the compressible Navier–Stokes equations around a parallel flow in a cylindrical domain. Nonlinear Anal. 127, 362–396 (2015) 2. Bresch, D., Desjardins, B.: Existence of global weak solutions for a 2D viscous shallow water equations and convergence to the quasi-geostrophic model. Commun. Math. Phys. 238, 211–223 (2003) 3. 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Yang, T., Zhao, H.: A vacuum problem for the one-dimensional compressible Navier–Stokes equations with density-dependent viscosity. J. Differ. Equ. 184, 163–184 (2002) 31. Yang, T., Zhu, C.: Compressible Navier–Stokes equations with degenerate viscosity coefficient and vacuum. Commun. Math. Phys. 230, 329–363 (2002) 32. Yao, L., Zhang, T., Zhu, C.: Boundary layers for compressible Navier–Stokes equations with density-dependent cylindrical symmetry. Ann. Inst. Henri Poincaré, Anal. Non Linéaire 28, 677–709 (2011) 33. Zhang, T., Fang, D.: Global behavior of spherically symmetric Navier–Stokes equations with density-dependent viscosity. J. Differ. Equ. 236, 293–341 (2007) 34. Zhu, C.: Asymptotic behavior of compressible Navier–Stokes equations with density-dependent viscosity and vacuum. Commun. Math. Phys. 293, 279–299 (2010) http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Boundary Value Problems Springer Journals

Regularity to the spherically symmetric compressible Navier–Stokes equations with density-dependent viscosity

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College of Mathematics and This paper is concerned with the dynamics for the compressible Navier–Stokes Statistics, North China University of Water Resources and Electric Power, equations with density-dependent viscosity in bounded annular domains in R .Inthe Zhengzhou, P.R. China paper, we shall analyze the spherical symmetric model and establish the regularity in 2 4 H and H under certain assumptions imposed on the initial data. Keywords: Navier–Stokes equations; Density-dependent viscosity; Regularity; Spherical symmetry 1 Introduction It is well known that the compressible isentropic Navier–Stokes equations which describe the motion of compressible fluids can be written in Eulerian coordinates as ρ + div(ρU)=0, (1.1) (ρU) + div(ρU ⊗ U) – div μ(ρ)D(U) – ∇ λ(ρ)divU + ∇P(ρ)=0, (1.2) where ρ(x, t), U(x, t)and P(ρ)= ρ (γ > 1) stand for the fluid density, velocity and pressure, respectively, and the strain tensor is given by ∇U +(∇U) D(U)= . (1.3) The Lamé viscosity coefficients μ(ρ), λ(ρ) satisfy the natural restrictions μ(ρ)>0, μ(ρ)+ N λ(ρ) ≥ 0. (1.4) For simplicity of the presentation, we consider only the viscosity terms μ(ρ)= ρ, λ(ρ)=0 and D(U)= ∇U.Then(1.1)–(1.2)become ρ + div(ρU)=0, (1.5) (ρU) + div(ρU ⊗ U)+ ∇P(ρ) – div(ρ∇U)=0. (1.6) © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Huang and Lian Boundary Value Problems (2018) 2018:85 Page 2 of 13 We are concerned with the spherically symmetric solutions of system (1.5)–(1.6)in bounded annular domains G = {x ∈ R ,0 < a < |x| < b <+∞}.Tothisend,wedenote |x| = r, ρ(x, t)= ρ(r, t), U(x, t)= u(r, t) , (1.7) which leads to the following system of equations for r >0: 2ρu ρ +(ρu) + = 0, (1.8) t r 2ρu 2u 2 γ (ρu) + ρu + ρ + –(ρu) – ρ = 0. (1.9) t r r r We shall consider problem (1.8)–(1.9)inthe region G subject to the initial data (ρ, u)(r,0) = (ρ , u )(r), r ∈ [a, b], (1.10) 0 0 and the boundary condition u(a, t)= u(b, t)=0, t ∈ [0, T]. (1.11) First we find it convenient to transfer problems (1.8)–(1.11) into that in Lagrangian co- ordinates and draw the desired results. We introduce the following coordinate transfor- mation: x(r, t)= ρ(s, τ)s ds, t = τ, (1.12) then the boundaries r = a and r = b become b b 2 2 x =0, x = ρ(s, τ)s ds = ρ (s)s ds, (1.13) a a where ρ (s)s ds is the total initial mass and, without loss of generality, we can normalize it to 1. So in terms of Lagrangian coordinates, the domain G becomes  = (0, 1). The relations between Lagrangian and Eulerian coordinates are satisfied by ∂x ∂x 2 2 = ρr , = ρur . (1.14) ∂r ∂t The initial boundary value problem (1.8)–(1.11)are changedto 2 2 ρ + ρ r u = 0, (1.15) u 2ρ u t x γ 2 2 + ρ = ρ r u – , (1.16) 2 x x x r r (ρ, u)(x,0) = (ρ , u )(x), x ∈ [0, 1], (1.17) 0 0 u(0, t)= u(1, t)=0, t > 0. (1.18) Huang and Lian Boundary Value Problems (2018) 2018:85 Page 3 of 13 Much progress was achieved recently on the compressible Navier–Stokes equations with density-dependent viscosity coefficient. Firstly let us recall some well-known results as re- gards the one-dimensional compressible isentropic Navier–Stokes equations with the flow density being connected with the infinite vacuum, [19, 24, 25] for the local well-posedness and the global existence of weak solutions to an initial boundary value problem with the viscous gas being connected vacuum states with jump discontinuities, [9, 10]for theglobal behavior with the initial density being piecewise smooth; [28–31, 34] for the local exis- tence, the global existence, the asymptotic behavior and the uniqueness of weak solutions with a viscous gas being connected vacuum states with continuous density. In spatial multi-dimension, there is a huge literature as regards the global existence, the regularity and the asymptotic behavior of a solution to system (1.1)–(1.2) with constant viscosity, we refer the reader to [2, 3, 11–14, 16, 20–23, 33] and the references therein. For the 3-D flow of a compressible fluid with cylindrical symmetry, the global existence and the large-time behavior of generalized solutions have been proved in [1, 4, 5, 7, 15, 23, 26, 27, 32] for the isentropic and the nonisentropic cases. The corresponding study of the regularity of a solution for any given initial datum has been carried out in [17]. For the 3- D flow of compressible fluid with spherical symmetry, there are some interesting results, [18] for the global well-posedness of classical solutions with large oscillations and vacuum; [33] for the global existence and uniqueness of the weak solution without a solid core; [14] for the structure of the solution; [21] for the global existence of the exterior problem and the initial boundary value problem. Besides, we would like to refer to [6, 8]asregards the existence and regularity of solutions for micropolar fluid with spherical symmetry in the three-dimensional case. In the paper, we shall analyze the spherical symmetric model and focus on the initial boundary problem of an isentropic compressible fluid. We show the regularity in H and H under certain assumptions imposed on the initial data. The notation in this paper will be as follows: 1,2 p m,p 1 1,2 1 L , 1 ≤ p ≤ +∞, W , m ∈ N, H = W , H = W denote the usual (Sobolev) 0 0 spaces on [0, 1]. To denote various constants, we use C (i = 1,2,4) to denote the generic positive constant depending only on the H norm of initial datum (ρ , u ), 0 0 min ρ (x) and variable t, respectively. In addition, · denotes the norm in the x∈[0,1] 0 space L . The basic assumption of this paper is the following: inf ρ > ρ, (1.19) [0,1] for some constant ρ >0. 2 2 Theorem 1.1 Let γ >1. Assume that the initial data (ρ , u ) ∈ H () × H () and (1.19) 0 0 2 2 hold, then there exists a unique generalized global solution (ρ(t), u(t)) ∈ (H ()) to the problem (1.15)–(1.18) verifying that, for any T >0, ∞ 2 2 2 ρ ∈ L [0, T], H () ∩ L [0, T], H () , (1.20) ∞ 2 2 3 u ∈ L [0, T], H () ∩ L [0, T], H () , (1.21) ∞ 2 2 1 u ∈ L [0, T], L () ∩ L [0, T], H () . (1.22) t Huang and Lian Boundary Value Problems (2018) 2018:85 Page 4 of 13 Theorem 1.2 Let γ >1. Assume that the initial data satisfies (1.19) and (ρ , u ) ∈ H () × 0 0 4 4 2 H (), then there exists a unique generalized global solution (ρ(t), u(t)) ∈ (H ()) to the problem (1.15)–(1.18) verifying that, for any T >0, ∞ 4 2 4 ρ ∈ L [0, T], H () ∩ L [0, T], H () , (1.23) ∞ 4 2 4 u ∈ L [0, T], H () ∩ L [0, T], H () , (1.24) ∞ 2 2 3 u ∈ L [0, T], H () ∩ L [0, T], H () , (1.25) ∞ 2 2 1 u ∈ L [0, T], L () ∩ L [0, T], H () . (1.26) tt Corollary 1.1 Under assumptions of Theorem 1.2,(1.23)–(1.24) imply (ρ(t), u(t)) is the classical solution verifying that, for any t >0, ρ(t) + u(t) ≤ C . (1.27) 3+1/2 3+1/2 C C 2 Proof of Theorem 1.1 This section is devoted to deriving the estimates of the solutions to prove Theorem 1.1 which will be presented in a sequence of lemmas. We begin with the following lemma. Lemma 2.1 (Theorem 2.2 in [21]) Under the assumptions in Theorem 1.1, then there exist positive constants ρ >0 and ρ >0 so that the unique global solution (ρ(t), u(t)) to problem (1.15)–(1.18) exists and satisfies, for any T >0, 0< ρ ≤ ρ(x, t) ≤ ρ , (2.1) 1 t 1 2 2 2 2 2 2 2 u +(ρ – ρ¯) + u + u + ρ (x, t) dx + ρ + u x t x x x 0 0 0 2 2 2 + u + u + u (x, s) dx ds ≤ C , t ∈ [0, T], (2.2) t xx xt where ρ¯ = ρ(s, t)s ds. b–a a Lemma 2.2 Under the conditions in Theorem 1.1, for any T >0, 2 2 ρ (t) + ρ (s) ds ≤ C , t ∈ [0, T], (2.3) xx xx 2 t 1 u (t) + u (x, s) dx ds ≤ C , t ∈ [0, T]. (2.4) xx 2 xxx 0 0 Proof We infer from (1.16)that u 2ρ u t x γ –1 2 2 2 2 =–γρ ρ +2ρρ r u + ρ r u +4rr u +2r u +2rr u – . (2.5) x x xx x x xx 2 x r r Multiplying (2.5)by u in L (), we deduce xx 1 1 2 2 2 γ –1 2 ρ r u dx = u + γρ ρ –2ρρ r u xx x x xx 0 0 2ρ u 2 2 – ρ 4rr u +2r u +2rr u + dx. (2.6) x x xx r Huang and Lian Boundary Value Problems (2018) 2018:85 Page 5 of 13 Using Young’s inequality, Sobolev’s embedding theorem, and Lemma 2.1,wededucefrom (2.6)that 1 1 1 2 2 2 2 2 2 u dx ≤ C u + ρ + ρ u dx + u dx xx t x x x xx 0 0 0 1 1 2 2 2 ≤ C + C u  ρ + u dx 2 1 x L x xx 0 0 ≤ C + u dx, xx whence u dx ≤ C . (2.7) xx Differentiating (1.16)withrespect to x, and exploiting (1.15), we have u 2ρ u t x γ 2 2 = –ρ + ρ r u – , (2.8) x xx r r x x which gives γ –1 ρ + γρ ρ = E (x, t) (2.9) txx xx 0 with 2ρ ur 2ρ u +2ρ u 2u r u x x xx x x t x tx γ –2 2 E (x, t)= – – γ (γ –1)ρ ρ + – . 2 3 2 r r r r Multiplying (2.9)by ρ , integrating the resultant over [0, 1], we deduce xx γ –1 2 ρ (t) + γρ ρ dx xx xx dt 2ρ ur 2ρ u +2ρ u 2u r u x x xx x x t x tx γ –2 2 = – – γ (γ –1)ρ ρ + – ρ dx xx 2 3 2 r r r r 2 2 ≤ C ρ  u  + u r  + ρ + ρ u  + ρ ur  –2 ρ dx. 1 xx tx t x x x x x x xx Using the Young inequality and the interpolation inequality to the above inequality, then we get 2 2 2 2 2 ρ (t) ≤ C u  + u  + ρ  + u  + u xx 1 tx t x x dt 2 2 + C ρ  1+ u . (2.10) 1 xx Integrating (2.10)withrespect to t, using initial condition ρ ∈ H and Lemma 2.1,we have 2 2 ρ (t) ≤ C + C ρ (s) ds, ∀t ∈ [0, T]. xx 2 1 xx Then using the Gronwall inequality to the above inequality, we can get (2.3). Huang and Lian Boundary Value Problems (2018) 2018:85 Page 6 of 13 Differentiating (1.16)withrespect to x, we can obtain 2 2 2 2 2 u =– (4ruρ r +2r ρ u +2ρρ 2rr u + r u +4ρρ 2r u xxx x x xx x x x x x x 2 2 ρ r 2 2 2 2 +4rr u +2rr u + r u + ρ 6r r u +6r u +6rr u + r u x x xx xx x xx x xx xx 2ρ u +2ρ u xx x x 2 2 + ρ 6r r u +6r u +6rr u +6rr u +2rr u + x xx x xx x x xx xxx 2ρ ur u u r x x tx t x γ –1 γ –2 2 – + – + γρ ρ + γ (γ –1)ρ ρ . (2.11) xx 2 2 3 r r r Integrating (2.11)over x and t, applying the embedding theorem, Lemma 2.1,(2.7)and (2.3), we conclude, for any t ∈ [0, T], t 1 t 1 2 2 4 2 4 2 2 2 2 2 2 u dx ds ≤ C u + ρ + ρ + ρ u + ρ u + ρ u + u (x, s) dx ds xxx tx x xx x x x x xx x t 0 0 0 0 t 1 t 2 2 4 2 2 ≤ C + C u  ∞ ρ + ρ dx ds + C ρ  ∞ u  ds 1 1 x 1 x xx L xx x L 0 0 0 ≤ C , which, along with (2.7), gives (2.4). The proof is complete. Proof of Theorem 1.1 Clearly, (1.20)–(1.22) follow from Lemmas 2.1–2.2.Thiscompletes the proof of Theorem 1.1. 3 Proof of Theorem 1.2 In this section, we shall complete the proof of Theorem 1.2.Tothisend,weassumethatin this sectionthatall assumptionsinTheorem 1.2 hold. We begin with the following lemma. Lemma 3.1 The following estimate holds for any T >0: t t 2 2 2 u (t) + u (s) ds ≤ C + C u (s) ds, t ∈ [0, T]. (3.1) tt ttx 4 2 txx 0 0 Proof We easily infer from (1.16)and (2.1)–(2.4)that u (t) ≤ C ρ (t) + u (t) + u (t) . (3.2) t 1 x x xx Differentiating (1.16)withrespect to x and using Lemmas 2.1–2.2,wehave u (t) ≤ C ρ (t) + ρ (t) + u (t) + u (t) + u (t) , (3.3) tx 2 x xx x xx xxx or u (t) ≤ C ρ (t) + ρ (t) + u (t) + u (t) + u (t) . (3.4) xxx 2 x xx x xx tx Huang and Lian Boundary Value Problems (2018) 2018:85 Page 7 of 13 Differentiating (1.16)withrespect to x twice and using the Cauchy–Schwarz inequality, we have u (t) ≤ C ρ (t) + u (t) , (3.5) txx 2 x 2 x 3 H H u (t) ≤ C ρ (t) + u (t) + u (t) . (3.6) xxxx 2 x 2 x 2 txx H H After differentiating (1.16)withrespect to t,using (3.2)–(3.3)and (3.5), we can get u (t) ≤ C u (t) + u (t) + u (t) + ρ (t) + u (t) + u (t) (3.7) tt 2 tx txx t x x xx ≤ C ρ (t) + u (t) . (3.8) 2 x 2 x 3 H H Now differentiating (1.16)withrespect to t twice, multiplying the resulting equation by ( ) in L (), and using integration by parts and (1.18), we conclude 2 tt 1 1 1 d u u γ 2 2 dx =– –ρ + ρ r u dx x tt 2 2 2 dt r r 0 0 tt ttx 1 2 2ρ u u u – – dx =: A + A . (3.9) 1 2 3 2 r r r tt tt We use (2.1)–(2.4), (3.2)–(3.8), (1.16) and the interpolation inequality to deduce that, for any small ε ∈ (0, 1), γ 2 2 A =– –ρ + ρ r u dx x tt 2 ttx 2 2 2 ≤ – ρ u dx + C ρ u  + ρ u + ρ u  + ρ u  + ρ u  + u  + u 2 tt x x t t t x t xt x t ttx t + u u  + u  + u  u  + u  + u u  + u  + u  + u  + u x t tx tt ttx tt x t tx t x 2 2 2 2 2 ≤ – ρ u dx + ε u (t) + C ρ (t) + ρ (t) ttx 2 tt 1 t ttx 2 2 2 2 + u (t) + u (t) + u (t) + ρ (t) t 1 tt x 1 x 1 H H H 2 2 2 2 –1 ≤ –C u (t) + C u (t) + ρ (t) + u (t) , (3.10) ttx 2 x x t 2 2 1 2 H H H 2ρ u u u A =– – dx 3 2 r r r tt tt 2 2 ≤ C ρ  + ρ u  + ρ u  + ρ u + u + u  + u  + ρ u 2 ttx tx t x tt tx tt t xt 3 3 2 + ρ u  + ρ u + u  + u u  + u  + u r  + ur  + ur x t x t tt t t t tt 2 2 2 2 2 2 ≤ C u (t) + u (t) + u (t) + u (t) + ρ (t) + u (t) . (3.11) 2 t tt txx tx x x Integrating (3.9)withrespect to t,and using(2.1)–(2.4), initial condition (1.16)and (3.10)– (3.11), then we obtain (3.1). Lemma 3.2 For any T >0 and ε ∈ (0, 1), the following estimate holds: t t 2 2 2 –1 2 u (t) + u (s) ≤ C + C ε u (s) ds, t ∈ [0, T]. (3.12) tx txx 4 ttx 0 0 Huang and Lian Boundary Value Problems (2018) 2018:85 Page 8 of 13 Proof Differentiating (1.16)withrespect to t and x, then multiplying the resultant by ( ) tx in L [0, 1], and integrating by parts, we know that 1 2 1 d u dx 2 dt r tx u u γ 2 2 γ 2 2 = –ρ + ρ r u – –ρ + ρ r u dx x tx x tx 2 2 r r tx 0 txx 1 2 u 2ρ u u + – dx 3 2 r r r tx tx = B (x, t)+ B (t)+ B (t), (3.13) 0 1 2 where γ 2 2 B (x, t)= –ρ + ρ r u , x tx tx 0 γ 2 2 B (t)=– –ρ + ρ r u dx, x tx txx 1 2 u 2ρ u u B (t)= – dx. 3 2 r r r tx tx Now using Young’s inequality several times, and employing (2.1)–(2.4)and theinterpo- lation inequality, after some calculation, we have, for any ε ∈ (0, 1), B (x, t) ≤ C ρ  ∞ + ρ ρ  ∞ + ρ ρ u  ∞ + ρ u  ∞ + ρ u  ∞ 0 2 tx L t x L x t x L tx x L t xx L + u  ∞ + u  ∞ u  ∞ txx L tx L tx L 1 1 1 1 1 1 2 2 2 2 2 2 ≤ C u  2 + ρ  1 + u  u  + u  u  u  u 2 x H x H tx txx txx txxx tx txx –1 2 2 2 –6 2 ≤ C ε u  + u  + C ε u  + u  2 + ρ  1 , (3.14) txx txxx 2 tx x H x H 1 1 2 2 2 2 2 2 B (t) ≤ – ρ u dx + ε ρ u dx + C ρ (t) + ρ (t) 1 2 tx 1 x 1 txx txx H H 0 0 2 2 2 + u (t) + u (t) + u (t) , (3.15) x 1 t tx B (t) ≤ C u u  + u  + u  + u  + u + ρ  + ρ u  + ρ u 2 2 x t tx x t txx tx x xx t + ρ u  + ρ  + ρ u  + ρ  + ρ u  + ρ  u  + u x tx xx x x tx x t x tx x 2 2 2 2 2 ≤ C u  + u  + ρ  + ρ  + ρ 2 x 1 t 1 x 1 tx txx H H H 2 2 2 ≤ C u  + u  + ρ  . (3.16) 2 x 2 t 1 x 1 H H H Differentiating (1.16)withrespect to x and t, and using Lemmas 2.1–2.2 and (3.7)–(3.8), we conclude u (t) ≤ C u (t) + ρ (t) + u (t) + u (t) . (3.17) txxx 2 x 2 x 1 tx 1 ttx H H H We integrate (3.13)withrespect to t,use (3.3), (3.14)–(3.16) and Lemmas 2.1–2.2 to obtain (3.12). The proof is complete.  Huang and Lian Boundary Value Problems (2018) 2018:85 Page 9 of 13 Lemma 3.3 The following estimates hold for any T >0: 2 2 2 2 u (t) + u (t) + u  + u  (s) ≤ C , t ∈ [0, T], (3.18) tt tx txx ttx 4 2 2 2 2 ρ (t) + u (t) + ρ  + u  (s) ds ≤ C , t ∈ [0, T]. (3.19) xxx xxx xxx xxxx 4 Proof We insert (3.1)into(3.12)and pick ε small enough to get (3.18). Differentiating (2.9)withrespect to x,wehave γ –1 ρ + γρ ρ = E (x, t), (3.20) txxx xxx 1 where γ –2 E (x, t)= E (x, t)– γ (γ –1)ρ ρ ρ . 1 0x x xx Taking into account estimate (3.18), from (2.1)–(2.4)wecan get E (t) ≤ C E  + ρ ρ 1 2 0x x xx ≤ C ρ u  + ρ  + ρ u  + ρ  + ρ u  + ρ ρ 2 x x xx xx x xxx x xx xx x + u  + u ρ  + u tx t x txx ≤ C ρ (t) + u (t) + u (t) , 2 x x t 2 2 2 H H H which, along with Lemmas 2.1–2.2 and (3.18), implies t t 2 2 E (s) ds ≤ C + C ρ (s) ds. (3.21) 1 4 2 xxx 0 0 After multiplying (3.20)by ρ in L [0, 1], we deduce xxx 1 d 2 γ –1 2 ρ  + γ ρ ρ dx ≤ C E (t) , (3.22) xxx 1 1 xxx 2 dt which implies ρ  ≤ C E (t) . (3.23) xxx 1 1 dt Integrating (3.23)withrespect to t,using (3.21), we conclude ρ  ≤ C + C ρ (s) ds, xxx 4 2 xxx which, after applying the Gronwall inequality and (3.22), yields 2 2 ρ (t) + ρ (s) ds ≤ C , ∀t ∈ [0, T]. (3.24) xxx xxx 4 0 Huang and Lian Boundary Value Problems (2018) 2018:85 Page 10 of 13 By (3.4), (3.6), (3.18) and Lemmas 2.1–2.2,weconclude 2 2 u (t) + u (s) ds ≤ C , ∀t ∈ [0, T], xxx xxxx 4 which, along with (3.24), gives (3.19). The proof is complete. Lemma 3.4 The following estimates hold for any T >0: 2 2 ρ (t) + ρ (s) ds ≤ C , t ∈ [0, T], (3.25) xxxx xxxx 4 2 2 2 u (t) + u (t) + u (s) ds ≤ C , t ∈ [0, T]. (3.26) txx xxxx xxxxx 4 Proof Differentiating (1.16)withrespect to t, and using Lemmas 2.1–2.2 and Lemma 3.3, we can get u (t) ≤ C u (t) + ρ (t) + u (t) + u (t) txx 1 t 1 x x 1 tt H H ≤ C , ∀t ∈ [0, T]. (3.27) Differentiating (3.20)withrespect to x,wehave γ –1 ρ + γρ ρ = E (x, t), (3.28) txxxx xxxx 2 where E (x, t)= E (x, t)– γ (γ –1)ρ ρ (3.29) 2 1x x xxx and γ –2 E (x, t)= E (x, t)– γ (γ –1) ρ ρ ρ . 1x 0xx x xx An easy calculation with the interpolation inequality, (3.2)–(3.8) and Lemmas 2.1–2.2 and Lemma 3.3 gives E (t) ≤ C ρ u  + ρ u  + ρ u  + ρ + u r  + u r  ++u 0x 2 x x xxx x xx xx tx x t xx txx ≤ C ρ  3 + u  2 + u  + u  , (3.30) 2 x H x H tx txx E (t) ≤ C ρ  + ρ u  + ρ u  + ρ u  + ρ u 0xx 2 xxxx xx x x xx xxx x xx xx + u  + u u  + u txx tx x txxx ≤ C ρ  3 + u  2 + u  2 , (3.31) 2 x x tx H H H 2 2 E (t) ≤ C E (t) + ρ ρ + ρ + ρ ρ 1x 2 0xx xx x xxx x xx ≤ C ρ  3 + u  2 + u  2  . (3.32) 2 x x tx H H H Taking into account estimate (3.30)–(3.32), from (3.29), we obtain E (t) ≤ C u (t) + ρ (t) + u (t) . (3.33) 2 4 x 3 x 3 tx 2 H H H Huang and Lian Boundary Value Problems (2018) 2018:85 Page 11 of 13 Inserting (3.17)into(3.33), and integrating (3.33)withrespect to t, using Lemmas 2.1– 2.2 and Lemma 3.3,wehave t t 2 2 E (s) ds ≤ C + C ρ (s) ds, ∀t ∈ [0, T]. (3.34) 2 4 2 xxxx 0 0 Multiplying (3.28)by ρ in L [0, 1], we can get xxxx 1 d 2 γ –1 2 ρ  + γ ρ ρ dx ≤ C E (t) , (3.35) xxxx 2 2 xxxx 2 dt which implies ρ  ≤ C E (t) . (3.36) xxxx 2 2 dt Integrating (3.36)withrespect to t,using (3.34), we conclude ρ  ≤ C + C ρ (s) ds, t ∈ [0, T] (3.37) xxxx 4 2 xxxx which, after using Gronwall’s inequality, yields ρ (t) ≤ C , t ∈ [0, T]. (3.38) xxxx 4 Thus, we can obtain (3.25)byvirtueof(3.37)–(3.38). By (3.6), (3.17), (3.25), Lemmas 2.1–2.2 and Lemma 3.3,wehave 2 2 u (t) + u (s) ds ≤ C , t ∈ [0, T]. (3.39) xxxx txxx 4 On the other hand, we differentiate (1.16)withrespect to x three times, use Lemmas 2.1– 2.2 and (3.25), (3.27) to conclude, for any t ∈ [0, T], u (t) ≤ C u (t) + u (t) + ρ (t) . (3.40) xxxxx 4 txxx x 3 x 3 H H Thus we deduce from (3.38)–(3.40)that u (s) ds ≤ C , ∀t ∈ [0, T] xxxxx 4 which, along with (3.39), gives (3.26). This completes the proof of the lemma. Proof of Theorem 1.2 Applying Lemmas 2.1–2.2 and Lemmas 3.1–3.4, we readily get es- timates (1.23)–(1.26). This completes the proof of Theorem 1.2. 4Conclusions In this paper, we have established the regularity of global solutions for the spherically sym- metric compressible fluid with density-dependent viscosity in H and H . The biggest dif- 2 4 ference from other papers is that our domain has spherical symmetry and the viscosity coefficients are density dependent. Huang and Lian Boundary Value Problems (2018) 2018:85 Page 12 of 13 Acknowledgements The authors would like to thank the referees for their useful suggestions, which have significantly improved the paper. Funding This research was supported by the NSFC (No. 11501199) and the Natural Science Foundation of Henan Province (No. 18B110010). 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Boundary Value ProblemsSpringer Journals

Published: May 30, 2018

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