Nonlinearly Exponential Stability of Compressible Navier-Stokes System with Degenerate Heat-Conductivity
aa r X i v : . [ m a t h . A P ] S e p Nonlinearly Exponential Stability of CompressibleNavier-Stokes System with Degenerate Heat-Conductivity ∗ Bin Huang, Xiaoding Shi † Department of Mathematics, Faculty of Science,Beijing University of Chemical Technology,Beijing 100029, P. R. China
Abstract
We study the large-time behavior of strong solutions to the one-dimensional,compressible Navier-Stokes system for a viscous and heat conducting ideal poly-tropic gas, when the viscosity is constant and the heat conductivity is proportionalto a positive power of the temperature. Both the specific volume and the tem-perature are proved to be bounded from below and above independently of time.Moreover, it is shown that the global solution is nonlinearly exponentially stableas time tends to infinity. Note that the conditions imposed on the initial data arethe same as those of the constant heat conductivity case ([Kazhikhov-Shelukhin.J. Appl. Math. Mech. 41 (1977); Kazhikhov. Boundary Value Problems for Hy-drodynamical Equations, 50 (1981)] and can be arbitrarily large. Therefore, ourresult can be regarded as a natural generalization of the Kazhikhov’s ones for theconstant heat conductivity case to the degenerate and nonlinear one.
Keywords: Compressible Navier-Stokes system; Degenerate heat-conductivity; Strongsolutions; Nonlinearly exponential stabilityMath Subject Classification: 35Q35; 76N10.
We consider the compressible Navier-Stokes system, describing the one-dimensionalmotion of a viscous heat-conducting gas, written in the Lagrange variables (see [5, 28]) v t = u x , (1.1) u t + P x = (cid:16) µ u x v (cid:17) x , (1.2) (cid:18) e + u (cid:19) t + ( P u ) x = (cid:18) κ θ x v + µ uu x v (cid:19) x , (1.3)where t > x ∈ (0 ,
1) denotes the Lagrange mass coordinate, and the unknownfunctions v > , u, θ > , e > , and P are, respectively, the specific volume of the gas, ∗ Partially supported by NNSFC 11671027 and 11471321. † Email addresses: [email protected](B. Huang), [email protected] (X. Shi) P and e satisfy P = Rθ/v, e = c v θ + const. , (1.4)where both specific gas constant R and heat capacity at constant volume c v are positiveconstants. For µ and κ, we consider the case where µ and κ are proportional to (possiblydifferent) powers of θ : µ = ˜ µθ α , κ = ˜ κθ β , (1.5)with constants ˜ µ, ˜ κ > α, β ≥ . The system (1.1)-(1.5) is supplemented with the initial conditions:( v, u, θ )( x, t = 0) = ( v , u , θ )( x ) , x ∈ (0 , , (1.6)and boundary ones: u (0 , t ) = u (1 , t ) = 0 , θ x (0 , t ) = θ x (1 , t ) = 0 , t ≥ . (1.7)One can deduce from the Chapman-Enskog expansion for the first level of approx-imation in kinetic theory that the viscosity µ and heat conductivity κ are functionsof temperature alone ( [6, 7]). In particular, if the intermolecular potential varies as r − a , with intermolecular distance r , then µ and κ are both proportional to the power( a + 4) / (2 a ) of the temperature, that is, (1.5) holds with α = β = ( a + 4) / (2 a ) . Indeed,for Maxwellian molecules ( a = 4) , the dependence is linear, while for elastic spheres( a → ∞ ) , the dependence is like θ / . For constant coefficients ( α = β = 0) and large initial data, Kazhikhov and Shelukhin[18] first obtained the global existence of solutions in bounded domains. From then on,significant progress has been made on the mathematical aspect of the initial boundaryvalue problems, see [1–4, 11, 12, 15] and the references therein. Moreover, much efforthas been made to generalize this approach to other cases and in particular to modelssatisfying (1.5), which in fact has proved to be challenging especially for temperaturedependence on µ. Motivated by the fact that in the case of isentropic flow a temperaturedependence in the viscosity translates into a density dependence, there is a body ofliterature (see [4, 8–10, 16, 25] and the references therein) studying the case that µ isindependent of θ , and heat conductivity is allowed to depend on temperature in aspecial way with a positive lower bound and balanced with corresponding constitutionrelations.When it comes to the physical case (1.5) with α = β , there is few results partiallybecause of the possible degeneracy and strong nonlinearity in viscosity and heat diffu-sion introduced in such relations. As a first step in this direction, Jenssen-Karper [13]proved the global existence of a weak solution to (1.1)–(1.7) under the assumption that α = 0 and β ∈ (0 , / . Later, for α = 0 and β ∈ (0 , ∞ ) , Pan-Zhang [26] obtain theglobal strong solution under the condition that( v , u , θ ) ∈ H × H × H . (1.8)Concerning the large-time behavior of the strong solutions to (1.1)-(1.7), Kazhikhov[17] (see also [1–4, 19–21, 23, 27] among others) first obtains that for the case that α = β = 0 , the strong solution is nonlinearly exponentially stable as time tends toinfinity. However, it should be mentioned here that the methods used there relies2eavily on the non-degeneracy of the heat conductivity κ and cannot be applied directlyto the degenerate and nonlinear case ( α = 0 , β > α = 0 and β > , the global strong solutions obtainedby [26] are indeed asymptotically stable as time tends to infinity. Moreover, we willimprove the results of [26] by relaxing their assumptions on the initial data. Then westate our main result as follows. Theorem 1.1
Suppose that α = 0 , β > , (1.9) and that the initial data ( v , u , θ ) satisfies ( v , θ ) ∈ H (0 , , u ∈ H (0 , , (1.10) and inf x ∈ (0 , v ( x ) > , inf x ∈ (0 , θ ( x ) > . (1.11) Then, the initial-boundary-value problem (1.1) - (1.7) has a unique strong solution ( v, u, θ ) satisfying v, θ ∈ L ∞ (0 , ∞ ; H (0 , , u ∈ L ∞ (0 , ∞ ; H (0 , ,v t ∈ L ∞ (0 , ∞ ; L (0 , ∩ L (0 , ∞ ; H (0 , ,u x , θ x , u t , θ t , v xt , u xx , θ xx ∈ L ((0 , × (0 , ∞ )) . (1.12) Moreover, there exists some positive constants C and η > such that for any ( x, t ) ∈ (0 , × (0 , ∞ ) , C − ≤ v ( x, t ) ≤ C, C − ≤ θ ( x, t ) ≤ C, (1.13) and that for any t > , (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) v − Z v dx, u, θ − Z θ dx (cid:19) ( · , t ) (cid:13)(cid:13)(cid:13)(cid:13) H (0 , ≤ Ce − η t . (1.14)A few remarks are in order. Remark 1.1
For α = β = 0 , under the conditions (1.10) and (1.11) , Kazhikhov andShelukhin [18] first obtained existence of global strong solutions to the initial-boundary-value problem (1.1) - (1.7) . Later, Kazhikhov [17] further proves that the strong solutionis nonlinearly exponentially stable as time tends to infinity. Therefore, our Theorem1.1 can be regarded as a natural generalization of the classical results [17, 18] to thedegenerate and nonlinear case that α = 0 , β > . Remark 1.2
As far as the existence of global strong solutions is concerned, our resultalso improves Pan and Zhang’s result [26] where they need the initial data satisfy (1.8) which are stronger than (1.10) . We now make some comments on the analysis of this paper. The key step to study thelarge-time behavior of the global strong solutions is to get the time-independent lowerand upper bounds of both v and θ (see (2.12) and (2.50)). Compared with [17, 23], themain difficulty comes from the degeneracy and nonlinearity of the heat conductivitybecause of β > . Hence, to obtain (2.50), some new ideas are needed. The keyobservations are as follows: First, after using the standard energetic estimate (see (2.2))3nd modifying the idea due to Kazhikhov [17], we obtain that the specific volume v is bounded from above and below time-independently (see (2.12)). Then, although itseems difficult to obtain the uniform lower bound of θ at first, after observing that (see(2.28)) Z T max x ∈ [0 , (cid:12)(cid:12)(cid:12)(cid:12) θ / ( x, t ) − Z θ / ( x, t ) dx (cid:12)(cid:12)(cid:12)(cid:12) dt ≤ C, we prove that the L ∞ (0 , ∞ ; L p )-norm of θ − is bounded (see (2.20)), which in turn notonly implies that (see (2.32) and (2.34)) Z T max x ∈ [0 , (cid:12)(cid:12)(cid:12)(cid:12) θ ( x, t ) − Z θ ( x, t ) dx (cid:12)(cid:12)(cid:12)(cid:12) dt + Z T Z u x dxdt ≤ C, but also yields that the L ((0 , × (0 , T ))-norm of θ x is bounded provided β > β ∈ (0 , , we find that the L ((0 , × (0 , T ))-norm of θ x can bebounded by the L (0 , T ; L (0 , u x which plays an important role in obtainingthe uniform bound on L ((0 , × (0 , T ))-norm of both θ x and u xx (see Lemma 2.8) for β ∈ (0 , We first state the following the local existence result which can be proved by using theprinciple of compressed mappings (c.f. [14, 22, 29]).
Lemma 2.1
Let (1.9) - (1.11) hold. Then there exists some T > such that the initial-boundary-value problem (1.1) - (1.7) has a unique strong solution ( v, u, θ ) satisfying v, θ ∈ L ∞ (0 , T ; H (0 , , u ∈ L ∞ (0 , T ; H (0 , ,v t ∈ L ∞ (0 , T ; L (0 , ∩ L (0 , T ; H (0 , ,u t , θ t , v xt , u xx , θ xx ∈ L ((0 , × (0 , T )) . Then, the a priori estimates (see (2.12), (2.30), (2.42), (2.50), and (2.65) below) wherethe constants depend only on the data of the problem make it possible to continue thelocal solution to the whole interval [0 , ∞ ) and finish the proof of Theorem 1.1.Next, without loss of generality, we assume that ˜ µ = ˜ κ = R = c v = 1 and that Z v dx = 1 , Z (cid:18) u θ (cid:19) dx = 1 . (2.1)Motivated by the second law of thermodynamics, one has the following standard ener-getic estimate embodying the dissipative effects of viscosity and thermal diffusion. Lemma 2.2
It holds that sup ≤ t< ∞ Z (cid:18) u v − ln v ) + ( θ − ln θ ) (cid:19) dx + Z t V ( s ) ds ≤ E , (2.2) where V ( t ) , Z (cid:18) θ β θ x vθ + u x vθ (cid:19) ( x, t ) dx, nd E , Z (cid:18) u v − ln v ) + ( θ − ln θ ) (cid:19) dx. Proof.
It follows from (1.1), (1.3), (1.7), and (2.1) that for t > Z v ( x, t ) dx = 1 , Z (cid:18) u θ (cid:19) ( x, t ) dx = 1 . (2.3)Noticing that the energy equation (1.3) can be written as θ t + θv u x = (cid:18) θ β θ x v (cid:19) x + u x v , (2.4)multiplying (1.1) by 1 − v − , (1.2) by u , (2.4) by 1 − θ − , and adding them altogether,we get (cid:0) u / v − ln v ) + ( θ − ln θ ) (cid:1) t + u x vθ + θ β θ x vθ = (cid:18) uu x v − uθv (cid:19) x + u x + (cid:18)(cid:0) − θ − (cid:1) θ β θ x v (cid:19) x , which together with (1.7) yields (2.2) and finishes the proof of Lemma 2.2. ✷ Next, we derive the following representation of v which is essential in obtaining thetime-independent upper and lower bounds of v . Lemma 2.3
We have the following expression of vv ( x, t ) = D ( x, t ) Y ( t ) + Z t D ( x, t ) Y ( t ) θ ( x, τ ) D ( x, τ ) Y ( τ ) dτ, (2.5) where D ( x, t ) = v ( x ) exp (cid:26)Z x ( u ( y, t ) − u ( y )) dy (cid:27) × exp (cid:26) − Z v Z x udydx + Z v Z x u dydx (cid:27) , (2.6) and Y ( t ) = exp (cid:26) − Z t Z (cid:0) u + θ (cid:1) dxds (cid:27) . (2.7) Proof.
First, denoting by σ , u x v − θv , (2.8)it follows from (1.2) that (cid:18)Z x udy (cid:19) t = σ − σ (0 , t ) , (2.9)which implies vσ (0 , t ) = vσ − v (cid:18)Z x udy (cid:19) t . x over (0 ,
1) together with (2.3) and (2.8) yields σ (0 , t ) = Z vσdx − Z v (cid:18)Z x udy (cid:19) t dx = Z ( u x − θ ) dx − (cid:18)Z v Z x udydx (cid:19) t + Z u x Z x udydx = − (cid:18)Z v Z x udydx (cid:19) t − Z (cid:0) θ + u (cid:1) dx. (2.10)Next, since u x = v t , we have σ = (ln v ) t − θv , which together with (2.9) and (2.10) gives (cid:18)Z x udy (cid:19) t = (ln v ) t − θv + (cid:18)Z v Z x udydx (cid:19) t + Z (cid:0) θ + u (cid:1) dx. Integrating this over (0 , t ) leads to v ( x, t ) = D ( x, t ) Y ( t ) exp (cid:26)Z t θv ds (cid:27) , (2.11)with D ( x, t ) and Y ( t ) as in (2.6) and (2.7) respectively.Finally, denoting by g = Z t θv ds, we have by using (2.11) g t = θ ( x, t ) v ( x, t ) = θ ( x, t ) D ( x, t ) Y ( t ) exp { g } , which gives exp { g } = 1 + Z t θ ( x, τ ) D ( x, τ ) Y ( τ ) dτ. Putting this into (2.11) leads to (2.5). ✷ With Lemmas 2.3 and 2.2 at hand, we are in a position to prove the time-independentupper and lower bounds of v . Lemma 2.4
For any ( x, t ) ∈ [0 , × [0 , + ∞ ) , it holds C − ≤ v ( x, t ) ≤ C, (2.12) where (and in what follows) C denotes some generic positive constant depending onlyon β, k ( v − , u , θ − k H (0 , , inf x ∈ [0 , v ( x ) , and inf x ∈ [0 , θ ( x ) . roof. First, since the function x − ln x is convex, Jensen’s inequality gives Z θdx − ln Z θdx ≤ Z ( θ − ln θ ) dx, which together with (2.2) and (2.3) leads to¯ θ ( t ) , Z θ ( x, t ) dx ∈ [ α , , (2.13)where 0 < α < α are two roots of x − ln x = E . Next, both (2.3) and Cauchy’s inequality imply (cid:12)(cid:12)(cid:12)(cid:12)Z v Z x udydx (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z v (cid:12)(cid:12)(cid:12)(cid:12)Z x udy (cid:12)(cid:12)(cid:12)(cid:12) dx ≤ Z v (cid:18)Z u dy (cid:19) / dx ≤ C, which combined with (2.6) gives C − ≤ D ( x, t ) ≤ C. (2.14)Furthermore, one deduces from (2.3) that1 ≤ Z (cid:0) u + θ (cid:1) dx ≤ , which yields that for any 0 ≤ τ < t < ∞ ,e − t ≤ Y ( t ) ≤ , e − t − τ ) ≤ Y ( t ) Y ( τ ) ≤ e − ( t − τ ) . (2.15)Next, it follows from (2.3) that (cid:12)(cid:12)(cid:12) θ β +12 ( x, t ) − ¯ θ β +12 ( t ) (cid:12)(cid:12)(cid:12) ≤ β + 12 (cid:18)Z θ β θ x θ v dx (cid:19) / (cid:18)Z θvdx (cid:19) / ≤ CV / ( t ) max x ∈ [0 , v / ( x, t ) , which together with (2.13) leads to α − CV ( t ) max x ∈ [0 , v ( x, t ) ≤ θ ( x, t ) ≤ C + CV ( t ) max x ∈ [0 , v ( x, t ) , (2.16)for all ( x, t ) ∈ [0 , × [0 , ∞ ) . Next, it follows from (2.5), (2.14), (2.15), and (2.16) that v ( x, t ) ≤ C + C Z t e − t + τ max x ∈ [0 , θ ( x, τ ) dτ ≤ C + C Z t e − t + τ (cid:18) V ( τ ) max x ∈ [0 , v ( x, τ ) (cid:19) dτ ≤ C + C Z t V ( τ ) max x ∈ [0 , v ( x, τ ) dτ, v ( x, t ) ≤ C (2.17)for all ( x, t ) ∈ [0 , × [0 , + ∞ ) . Combining this with (2.5), (2.14), (2.15), and (2.16)yields that v ( x, t ) ≥ C Z t e − t − τ ) min x ∈ [0 , θ ( x, τ ) dτ ≥ C Z t e − t − τ ) (cid:16) α − CV ( τ ) (cid:17) dτ ≥ Cα − Cα e − t − C Z t e − t − τ ) V ( τ ) dτ. (2.18)Noticing that Z t e − t − τ ) V ( τ ) dτ = Z t/ e − t − τ ) V ( τ ) dτ + Z tt/ e − t − τ ) V ( τ ) dτ ≤ e − t Z ∞ V ( τ ) dτ + Z tt/ V ( τ ) dτ → , as t → ∞ , we deduce from (2.18) that there exists some ˜ T > v ( x, t ) ≥ Cα
16 (2.19)for all ( x, t ) ∈ [0 , × [ ˜ T , + ∞ ) . Finally, using (2.11), (2.14), and (2.15), we obtain that there exists some positiveconstant C such that v ( x, t ) ≥ C − for all ( x, t ) ∈ [0 , × [0 , ˜ T ] . Combining this, (2.19), and (2.17) gives (2.12) and finishesthe proof of Lemma 2.4. ✷ To obtain the uniform (with respect to time) lower bound of the temperature, weneed the following time-independent bound on the L ∞ (0 , T ; L p )-norm of θ − . Lemma 2.5
For any p > , there exists some positive constant C ( p ) such that sup ≤ t ≤ T Z θ − p dx + Z T Z θ β θ x θ p +1 dxdt + Z T Z u x θ p dxdt ≤ C ( p ) . (2.20) Proof.
First, it follows from (2.2) that (2.20) holds for p = 1 . p = 1 , multiplying (2.4) by 1 /θ p and integration by parts gives1 p − (cid:18)Z θ − p dx (cid:19) t + p Z θ β θ x vθ p +1 dx + Z u x vθ p dx = Z (cid:0) θ − p − (cid:1) u x v dx + Z u x v dx ≤ C ( p ) Z (cid:12)(cid:12)(cid:12) θ − (cid:12)(cid:12)(cid:12) (cid:16) θ − p (cid:17) | u x | dx + (cid:18)Z ln vdx (cid:19) t ≤ C ( p ) max x ∈ [0 , (cid:12)(cid:12)(cid:12) θ − (cid:12)(cid:12)(cid:12) Z | u x | dx + (cid:18)Z θ − p dx (cid:19) / (cid:18)Z u x vθ p dx (cid:19) / ! + (cid:18)Z ln vdx (cid:19) t ≤ C ( p ) max x ∈ [0 , (cid:12)(cid:12)(cid:12) θ − (cid:12)(cid:12)(cid:12) + C ( p ) (cid:18)Z | u x | dx (cid:19) + 12 Z u x vθ p dx + C ( p ) max x ∈ [0 , (cid:12)(cid:12)(cid:12) θ − (cid:12)(cid:12)(cid:12) Z θ − p dx + (cid:18)Z ln vdx (cid:19) t . (2.21)Next, we claim that for any real number q, there exists a positive constant C ( q ) suchthat 1 − ¯ θ q ≤ C ( q ) V / ( t ) . (2.22)Indeed, standard calculation gives1 − ¯ θ q = Z ddη (cid:18)(cid:18)Z ( θ + η u dx (cid:19) q (cid:19) dη = q Z (cid:18)Z ( θ + η u dx (cid:19) q − dη · Z u dx ≤ C ( q ) Z u dx, (2.23)where in the last inequality we have used α ≤ Z θdx ≤ Z ( θ + η u dx ≤ , due to (2.3) and (2.13). Furthermore, Z u dx ≤ max x ∈ [0 , | u | (cid:18)Z u dx (cid:19) / ≤ C Z | u x | dx, (2.24)and Z | u x | dx ≤ (cid:18)Z u x vθ dx (cid:19) / (cid:18)Z vθdx (cid:19) / ≤ C (cid:18)Z u x vθ dx (cid:19) / ≤ CV / ( t ) . (2.25)Combining all these estimates (2.23)–(2.25) gives (2.22).9hen, it follows from (2.22) and (2.13) that for β ∈ (0 , , max x ∈ [0 , (cid:12)(cid:12)(cid:12) θ − (cid:12)(cid:12)(cid:12) ≤ max x ∈ [0 , (cid:12)(cid:12)(cid:12) θ − ¯ θ (cid:12)(cid:12)(cid:12) + max x ∈ [0 , (cid:12)(cid:12)(cid:12) ¯ θ − (cid:12)(cid:12)(cid:12) ≤ C Z θ − | θ x | dx + CV / ( t ) ≤ C (cid:18)Z θ β − θ x dx (cid:19) / (cid:18)Z θ − β dx (cid:19) / + CV / ( t ) ≤ CV / ( t ) , (2.26)and that for β ≥ , max x ∈ [0 , (cid:12)(cid:12)(cid:12) θ − (cid:12)(cid:12)(cid:12) ≤ max x ∈ [0 , (cid:12)(cid:12)(cid:12) θ − ¯ θ (cid:12)(cid:12)(cid:12) + max x ∈ [0 , (cid:12)(cid:12)(cid:12) ¯ θ − (cid:12)(cid:12)(cid:12) ≤ C max x ∈ [0 , (cid:12)(cid:12)(cid:12) θ β − ¯ θ β (cid:12)(cid:12)(cid:12) + CV / ( t ) ≤ C Z θ β − | θ x | dx + CV / ( t ) ≤ CV / ( t ) . (2.27)It thus follows from (2.26), (2.27), and (2.2) that Z T max x ∈ [0 , (cid:12)(cid:12)(cid:12) θ − (cid:12)(cid:12)(cid:12) dt ≤ C. (2.28)Finally, noticing that for p ∈ (0 , , Z θ − p dx ≤ Z θdx + 1 ≤ C, and that both (2.2) and (2.3) implysup ≤ t< ∞ Z | ln v | dx ≤ C, (2.29)after using (2.25), (2.2), (2.28), and Gronwall’s inequality, we obtain (2.20) from (2.21)and finish the proof of Lemma 2.5. ✷ Next, using Lemma 2.5, we have the following estimate on the L ∞ (0 , T ; L )-norm of v x . Lemma 2.6
There exists a positive constant C such that sup ≤ t ≤ T Z v x dx + Z T Z v x ( θ + 1) dxdt ≤ C, (2.30) for any T ≥ . Proof.
First, choosing p = β in (2.20) gives Z T Z θ − θ x dxdt ≤ C, (2.31)10hich together with (2.13) implies Z T max x ∈ [0 , (cid:0) θ ( x, t ) − ¯ θ ( t ) (cid:1) dt ≤ C Z T Z θ − θ x dx Z θdxdt ≤ C. (2.32)Next, integrating the momentum equation (1.2) multiplied by u over [0 ,
1] withrespect to x, we obtain after integrating by parts12 (cid:18)Z u dx (cid:19) t + Z u x v dx = Z θv u x dx = Z (cid:0) θ − ¯ θ (cid:1) v u x dx + (cid:0) ¯ θ − (cid:1) Z u x v dx + Z u x v dx ≤ C max x ∈ [0 , (cid:0) θ − ¯ θ (cid:1) + CV ( t ) + (cid:18)Z ln vdx (cid:19) t , (2.33)where in the last inequality we have used (2.22) and (2.25). Combining this with (2.2),(2.29), and (2.32) yields Z T Z u x dxdt ≤ C. (2.34)Next, using (1.1), we rewrite momentum equation (1.2) as (cid:16) v x v (cid:17) t = u t + (cid:18) θv (cid:19) x , (2.35)due to (cid:16) v t v (cid:17) x = (cid:16) v x v (cid:17) t . Multiplying (2.35) by v x v leads to12 (cid:20)(cid:16) v x v (cid:17) (cid:21) t = v x v u t + v x v (cid:18) θv (cid:19) x = (cid:16) v x v u (cid:17) t − u (ln v ) xt + v x θ x v − v x θv = (cid:16) v x v u (cid:17) t − [ u (ln v ) t ] x + u x v + v x θ x v − v x θv . (2.36)Integrating (2.36) over [0 , × [0 , T ] , one hassup ≤ t ≤ T Z (cid:20) (cid:16) v x v (cid:17) − v x v u (cid:21) dx + Z T Z v x θv dxdt ≤ C + Z T Z v x θ x v dxdt ≤ C + 12 Z T Z v x θv dxdt + C Z T Z θ − θ x dxdt ≤ C + 12 Z T Z v x θv dxdt, ≤ t ≤ T Z v x dx + Z T Z v x θdxdt ≤ C, (2.37)due to the following simply fact: Z v x v udx ≤ Z (cid:16) v x v (cid:17) dx + C. Finally, it follows from (2.37) that¯ θ Z v x dx = Z v x (cid:0) ¯ θ − θ (cid:1) dx + Z v x θdx ≤ ¯ θ Z v x dx + 12¯ θ Z v x (cid:0) θ − ¯ θ (cid:1) dx + Z v x θdx ≤ ¯ θ Z v x dx + C max x ∈ [0 , (cid:0) θ − ¯ θ (cid:1) + Z v x θdx, which together with (2.32) and (2.37) leads to Z T Z v x dxdt ≤ C. Combining this with (2.37) gives (2.30) and finishes the proof of Lemma 2.6. ✷ For further uses, we need the following estimate on the L ((0 , × (0 , T ))-norm of θ x for β ∈ (0 , . Lemma 2.7 If < β ≤ , there exists a positive constant C such that Z T Z θ x dxdt ≤ C + C Z T (cid:18)Z u x dx (cid:19) dt, (2.38) for any T > . Proof.
Multiplying (2.4) by θ − β and integration by parts gives24 − β (cid:18)Z θ − β dx (cid:19) t + (2 − β )2 Z θ β θ x v dx = − Z θ − β v u x dx + Z θ − β u x v dx = Z (cid:16) ¯ θ − β − θ − β (cid:17) v u x dx + (cid:16) − ¯ θ − β (cid:17) Z u x v dx − Z u x v dx + Z θ − β u x v dx ≤ C Z (cid:12)(cid:12)(cid:12) θ − β − ¯ θ − β (cid:12)(cid:12)(cid:12) | u x | dx + CV ( t ) − (cid:18)Z ln vdx (cid:19) t + Z θ − β u x v dx, (2.39)12here in the last inequality we have used (2.22). Direct calculation yields that Z (cid:12)(cid:12)(cid:12) θ − β − ¯ θ − β (cid:12)(cid:12)(cid:12) | u x | dx ≤ C max x ∈ [0 , (cid:12)(cid:12)(cid:12) θ − β − ¯ θ − β (cid:12)(cid:12)(cid:12) (cid:18)Z (cid:16) θ − β + 1 (cid:17) dx (cid:19) / (cid:18)Z u x dx (cid:19) / ≤ C (cid:18)Z θ − β | θ x | dx (cid:19) + C Z (cid:16) θ − β + 1 (cid:17) dx Z u x dx ≤ C Z θ − β dx Z θ − θ x dx + C Z (cid:16) θ − β + 1 (cid:17) dx Z u x dx ≤ C Z θ − θ x dx + C Z (cid:16) θ − β + 1 (cid:17) dx Z u x dx, (2.40)and that for any δ > Z θ − β u x v dx ≤ C (cid:18) max x ∈ [0 , (cid:12)(cid:12)(cid:12) θ − β − ¯ θ − β (cid:12)(cid:12)(cid:12) + 1 (cid:19) Z u x dx ≤ C Z θ − β | θ x | dx Z u x dx + C Z u x dx ≤ δ Z (cid:16) θ − + θ β (cid:17) θ x dx + C ( δ ) (cid:18)Z u x dx (cid:19) + C Z u x dx. (2.41)Putting (2.40) and (2.41) into (2.39), choosing δ suitably small, and using (2.34), (2.31),and the Gronwall inequality, one obtains Z θ − β/ dx + Z T Z θ β/ θ x dxdt ≤ C + C Z T (cid:18)Z u x dx (cid:19) dt, which together with (2.2) implies Z T Z θ x dxdt ≤ Z T Z (cid:16) θ β − + θ β/ (cid:17) θ x dxdt ≤ C + Z T Z θ β/ θ x dxdt ≤ C + C Z T (cid:18)Z u x dx (cid:19) dt. This gives (2.38) and finishes the proof of Lemma 2.7. ✷ Then, we have the following uniform estimate on the L ((0 , × (0 , T ))-norm of u t and u xx . Lemma 2.8
There exists a positive constant C such that sup ≤ t ≤ T Z u x dx + Z T Z ( u t + u xx ) dxdt ≤ C. (2.42) for any T ≥ . roof. First, we rewrite the momentum equation (1.2) as u t − u xx v = − u x v x v − θ x v + θv x v . (2.43)Multiplying both sides of (2.43) by u xx and integrating the resultant equality in x over[0 ,
1] lead to 12 ddt Z u x dx + Z u xx v dx ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z u x v x v u xx dx (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z θ x v u xx dx (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z θv x v u xx dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z u xx v dx + C Z (cid:0) u x v x + v x θ + θ x (cid:1) dx. (2.44)Direct computation yields that for any δ > , Z (cid:0) u x v x + v x θ + θ x (cid:1) dx ≤ C (cid:18) max x ∈ [0 , u x + max x ∈ [0 , (cid:0) θ − ¯ θ (cid:1) + 1 (cid:19) Z v x dx + Z θ x dx ≤ C max x ∈ [0 , u x + C max x ∈ [0 , (cid:0) θ − ¯ θ (cid:1) + C Z v x dx + Z θ x dx ≤ δ Z u xx dx + C ( δ ) Z u x dx + C max x ∈ [0 , (cid:0) θ − ¯ θ (cid:1) + C Z v x dx + Z θ x dx, (2.45)where in the last inequality we have usedmax x ∈ [0 , u x ≤ Z (cid:12)(cid:12)(cid:0) u x (cid:1) x (cid:12)(cid:12) dx ≤ (cid:18)Z u xx dx (cid:19) / (cid:18)Z u x dx (cid:19) / ≤ δ Z u xx dx + C ( δ ) Z u x dx, (2.46)due to R u x dx = 0 . Putting (2.45) into (2.44) and choosing δ suitably small yields Z u x dx + Z T Z u xx dxdt ≤ C + C Z T Z θ x dxdt, (2.47)due to (2.32), (2.34), and (2.30).Next, on the one hand, if β > , choosing p = β − Z T Z θ x dxdt ≤ C, (2.48)which along with (2.47) givessup ≤ t ≤ T Z u x dx + Z T Z u xx dxdt + Z T Z θ x dxdt ≤ C. (2.49)14n the other hand, if β ∈ (0 , , it follows from (2.47), (2.38), (2.34), and Gronwall’sinequality that (2.49) still holds.Finally, it follows from (2.43), (2.49), (2.45), (2.32), (2.34), and (2.30) that Z T Z u t dxdt ≤ C, which together with (2.49) gives (2.42) and finishes the proof of Lemma 2.8. ✷ Now, we can prove the uniform lower and upper bounds of the temperature θ. Lemma 2.9
There exists a positive constant C such that for any ( x, t ) ∈ [0 , × [0 , T ] C − ≤ θ ( x, t ) ≤ C. (2.50) Proof.
First, for p > β + 1 , multiplying (2.4) by θ p − and integrating the resultantequality in x over (0 ,
1) leads to1 p (cid:18)Z θ p dx (cid:19) t + ( p − Z θ p + β − θ x v dx = Z θ p − u x v dx − Z θ p u x v dx ≤ C max x ∈ [0 , u x Z θ p − dx + C Z | ( θ p − u x | dx − Z u x v dx ≤ C max x ∈ [0 , u x Z θ p − dx + C Z | θ − | (cid:0) θ p − + 1 (cid:1) | u x | dx − Z u x v dx ≤ C max x ∈ [0 , (cid:0) u x + | θ − | (cid:1) (cid:18) Z θ p dx (cid:19) − (cid:18)Z ln vdx (cid:19) t . (2.51)It follows from (2.49), (2.46), and (2.34) that Z T max x ∈ [0 , u x dt + Z T Z θ x dxdt ≤ C, (2.52)which together with (2.22) shows Z T max x ∈ [0 , | θ − | dt ≤ C Z T (cid:18) max x ∈ [0 , (cid:12)(cid:12) θ − ¯ θ (cid:12)(cid:12) + max x ∈ [0 , (cid:12)(cid:12) ¯ θ − (cid:12)(cid:12) (cid:19) dt ≤ C Z T Z θ x dxdt + C Z T V ( t ) dt ≤ C. (2.53)Combining (2.51)-(2.53) with the Gronwall inequality givessup ≤ t ≤ T Z θ p dx + Z T Z θ p + β − θ x dxdt ≤ C ( p ) . (2.54)Next, multiplying (2.4) by θ β θ t and integrating the resultant equality over (0 , Z θ β θ t dx + Z θ β +1 θ t u x v dx = Z θ β θ t (cid:18) θ β θ x v (cid:19) x dx + Z θ β θ t u x v dx = − Z θ β θ x v (cid:16) θ β θ t (cid:17) x dx + Z θ β θ t u x v dx = − Z θ β θ x v (cid:16) θ β θ x (cid:17) t dx + Z θ β θ t u x v dx = − Z (cid:0) ( θ β θ x ) (cid:1) t v dx + Z θ β θ t u x v dx = − (cid:18)Z ( θ β θ x ) v dx (cid:19) t − Z ( θ β θ x ) u x v dx + Z θ β θ t u x v dx, which gives Z θ β θ t dx + 12 (cid:18)Z ( θ β θ x ) v dx (cid:19) t = − Z ( θ β θ x ) u x v dx − Z θ β +1 θ t u x v dx + Z θ β θ t u x v dx ≤ C max x ∈ [0 , | u x | Z (cid:16) θ β θ x (cid:17) dx + 12 Z θ β θ t dx + C Z θ β +2 u x dx + C Z θ β u x dx ≤ C (cid:18)Z (cid:16) θ β θ x (cid:17) dx (cid:19) + C max x ∈ [0 , u x + C max x ∈ [0 , u x + 12 Z θ β θ t dx (2.55)due to (2.54).Next, it follows from (2.46) and (2.42) that Z T max x ∈ [0 , u x dt ≤ C, (2.56)which together with (2.55), the Gronwall inequality, (2.52), and (2.54) leads tosup ≤ t ≤ T Z (cid:16) θ β θ x (cid:17) dx + Z T Z θ β θ t dxdt ≤ C. (2.57)This in particular givesmax x ∈ [0 , (cid:12)(cid:12)(cid:12) θ β +1 − ¯ θ β +1 (cid:12)(cid:12)(cid:12) ≤ ( β + 1) Z θ β | θ x | dx ≤ C (cid:18)Z (cid:16) θ β θ x (cid:17) dx (cid:19) / ≤ C, which implies that for all ( x, t ) ∈ [0 , × [0 , ∞ ) ,θ ( x, t ) ≤ C. (2.58)16ext, it follows from (2.22) that Z T Z (cid:16) θ β +2 − (cid:17) dxdt ≤ Z T Z (cid:16) θ β +2 − ¯ θ β +2 (cid:17) dxdt + C Z T V ( t ) dt ≤ C Z T (cid:18)Z θ β +1 | θ x | dx (cid:19) dt + C ≤ C Z T Z θ β − θ x dxdt + C ≤ C, (2.59)where in the last inequality we have used (2.58). Combining this, (2.57), and (2.58) inparticular gives Z T (cid:12)(cid:12)(cid:12)(cid:12)(cid:18)Z (cid:16) θ β +2 − (cid:17) dx (cid:19) t (cid:12)(cid:12)(cid:12)(cid:12) dt = 2 Z T (cid:12)(cid:12)(cid:12)(cid:12)Z (cid:16) θ β +2 − (cid:17) (cid:16) θ β +2 (cid:17) t dx (cid:12)(cid:12)(cid:12)(cid:12) dt ≤ C Z T Z (cid:16) θ β +2 − (cid:17) dxdt + C Z T Z θ β +2 θ t dxdt ≤ C + C Z T Z θ β θ t dxdt ≤ C, which together with (2.59) leads tolim t →∞ Z (cid:16) θ β +2 − (cid:17) dx = 0 . (2.60)Then, we claim that lim t →∞ (cid:0) − ¯ θ (cid:1) = 0 , (2.61)which combined with (2.60) giveslim t →∞ Z (cid:16) θ β +2 − ¯ θ β +2 (cid:17) dx = 0 . (2.62)It thus follows from (2.57) and (2.58) thatmax x ∈ [0 , (cid:16) θ β +2 − ¯ θ β +2 (cid:17) ≤ C Z (cid:12)(cid:12)(cid:12) θ β +2 − ¯ θ β +2 (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) ( θ β +2 ) x (cid:12)(cid:12)(cid:12) dx ≤ C (cid:18)Z (cid:16) θ β +2 − ¯ θ β +2 (cid:17) dx (cid:19) / (cid:18)Z (cid:16) θ β θ x (cid:17) dx (cid:19) / ≤ C (cid:18)Z (cid:16) θ β +2 − ¯ θ β +2 (cid:17) dx (cid:19) / , which together with (2.62) and (2.61) implies that there exists some T > θ ( x, t ) ≥ / , (2.63)17or all ( x, t ) ∈ [0 , × [ T , ∞ ) . Moreover, it follows from [26, Lemma 2.2] that thereexists some constant C ≥ θ ( x, t ) ≥ C − , for all ( x, t ) ∈ [0 , × [0 , T ] . Combining this, (2.63), and (2.58) gives (2.50).Finally, it remains to prove (2.61). Indeed, it follows from (2.33) and (2.58) that12 (cid:18)Z u dx (cid:19) t + Z u x v dx = Z θv u x dx ≤ C Z | u x | dx, which yields that there exists some constant C such that for any N > s, t ∈ [ N, N + 1] Z u ( x, t ) dx − Z u ( x, s ) dx ≤ C Z N +1 N Z | u x | dxdτ. Integrating this with respect to s over ( N, N + 1) and using (2.24) yields thatsup N ≤ t ≤ N +1 Z u ( x, t ) dx ≤ C (cid:18)Z N +1 N Z u x dxdτ (cid:19) / . (2.64)Letting N → ∞ in (2.64) giveslim t →∞ Z u ( x, t ) dx = 0 , due to (2.34). Combining this with (2.3) gives (2.61) and finishes the proof of Lemma2.9. ✷ Next, we have the following uniform estimate on the L ((0 , × (0 , T ))-norm of θ t and θ xx . Lemma 2.10
There exists a positive constant C such that sup ≤ t ≤ T Z θ x dx + Z T Z (cid:0) θ t + θ xx (cid:1) dxdt ≤ C. (2.65) Proof.
First, both (2.50) and (2.57) lead tosup ≤ t ≤ T Z θ x dx + Z T Z θ t dxdt ≤ C. (2.66)Next, it follows from (2.4) that (cid:0) θ β θ x (cid:1) x v = θ β θ x v x v − u x v + θu x v + θ t , which together with (2.30), (2.50), (2.34), (2.56), and (2.66) gives Z T Z (cid:12)(cid:12)(cid:12)(cid:16) θ β θ x (cid:17) x (cid:12)(cid:12)(cid:12) dxdt ≤ C Z T max x ∈ [0 , (cid:16) θ β θ x (cid:17) Z v x dxdt + C ≤ C Z T max x ∈ [0 , (cid:16) θ β θ x (cid:17) dt + C. (2.67)18ince θ x (0 , t ) = 0 , we get by (2.52) and (2.50), Z T max x ∈ [0 , (cid:16) θ β θ x (cid:17) dt ≤ Z T Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:18)(cid:16) θ β θ x (cid:17) (cid:19) x (cid:12)(cid:12)(cid:12)(cid:12) dxdt ≤ C ( δ ) + δ Z T Z (cid:12)(cid:12)(cid:12)(cid:16) θ β θ x (cid:17) x (cid:12)(cid:12)(cid:12) dxdt, which together with (2.67) and (2.50) implies Z T max x ∈ [0 , θ x dt + Z T Z (cid:12)(cid:12)(cid:12)(cid:16) θ β θ x (cid:17) x (cid:12)(cid:12)(cid:12) dxdt ≤ C. (2.68)Finally, since θ xx = (cid:0) θ β θ x (cid:1) x θ β − βθ x θ , it follows from (2.68), (2.50), and (2.57) that Z T Z θ xx dxdt ≤ C Z T Z (cid:12)(cid:12)(cid:12)(cid:16) θ β θ x (cid:17) x (cid:12)(cid:12)(cid:12) dxdt + C Z T max x θ x Z θ x dxdt ≤ C + C sup ≤ t ≤ T Z θ x dx Z T max x ∈ [0 , θ x dt ≤ C, which together with (2.66) gives (2.65) and finishes the proof of Lemma 2.10. ✷ Finally, we have the following nonlinearly exponential stability of the strong solu-tions.
Lemma 2.11
There exist some positive constants C and η both depending only on β, k ( v − , u , θ − k H (0 , , inf x ∈ [0 , v ( x ) , and inf x ∈ [0 , θ ( x ) such that k ( v − , u, θ − · , t ) k H (0 , ≤ Ce − η t . (2.69) Proof.
Noticing that all the constants C in Lemmas 2.6, 2.8, and 2.10 are independentof T, we have Z ∞ (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ddt k v x ( · , t ) k L (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) ddt k u x ( · , t ) k L (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) ddt k θ x ( · , t ) k L (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) dt ≤ C, (2.70)where we have used Z u x u xt dx = − Z u t u xx dx. It thus follows from (2.70) thatlim t →∞ k ( v x , u x , θ x )( · , t ) k L (0 , = 0 , which in particular implieslim t →∞ k ( v − , u, θ − · , t ) k H (0 , = 0 . Therefore, since we know that the temperature remains bounded from above and belowindependently of time and the solution becomes small in H -norm for large time t , wecan conclude the solution decays to the constant state exponentially as t → ∞ , that is,(2.69) holds (c.f. [24]). ✷ eferenceseferences