In the construction field, ultra-high performance concrete (UHPC) has drawn remarkable attention owing to its outstanding mechanical properties and durability, with increasingly extensive applications in flat components. Moreover, due to its dense microstructure, UHPC is highly susceptible to explosive spalling at elevated temperatures. In this paper, we comprehensively review the application status of UHPC in this domain, encompassing aspects such as bridge deck overlays, composite slabs, steel-concrete composite systems, joint connections, rehabilitation and strengthening, and thin-walled members. We deeply analyzed the spalling mechanisms of UHPC at high temperatures, mostly including thermal stress and vapor pressure mechanisms, and thoroughly investigated influencing factors such as permeability, heating rate, fiber and aggregate types, specimen size, cooling method, external load, and restraint. Additionally, we summarize effective methods to mitigate fire-induced spalling, such as the application of fire insulation, optimization of curing processes, incorporation of fibers or aggregates, and the utilization of thermal spalling-resistant admixtures. Despite the significant potential of UHPC in flat component applications, numerous challenges persist, including further validation of application feasibility, optimization and improvement of interface performance, in-depth elucidation of spalling mechanisms, research and exploration of new fiber materials, full consideration of the scale effect, and exploration and exploitation of innovative improvement solutions for fire resistance. Researchers should concentrate on addressing these issues to promote the broader and more efficient application of UHPC in the construction field.
Citation: Xiaodong Cheng, Jun Xia, Theofanis Krevaikas, Luigi Di Sarno. A review of the applications of ultra-high performance concrete in flat components and the associated fire-induced spalling risk[J]. AIMS Materials Science, 2025, 12(1): 165-202. doi: 10.3934/matersci.2025010
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In the construction field, ultra-high performance concrete (UHPC) has drawn remarkable attention owing to its outstanding mechanical properties and durability, with increasingly extensive applications in flat components. Moreover, due to its dense microstructure, UHPC is highly susceptible to explosive spalling at elevated temperatures. In this paper, we comprehensively review the application status of UHPC in this domain, encompassing aspects such as bridge deck overlays, composite slabs, steel-concrete composite systems, joint connections, rehabilitation and strengthening, and thin-walled members. We deeply analyzed the spalling mechanisms of UHPC at high temperatures, mostly including thermal stress and vapor pressure mechanisms, and thoroughly investigated influencing factors such as permeability, heating rate, fiber and aggregate types, specimen size, cooling method, external load, and restraint. Additionally, we summarize effective methods to mitigate fire-induced spalling, such as the application of fire insulation, optimization of curing processes, incorporation of fibers or aggregates, and the utilization of thermal spalling-resistant admixtures. Despite the significant potential of UHPC in flat component applications, numerous challenges persist, including further validation of application feasibility, optimization and improvement of interface performance, in-depth elucidation of spalling mechanisms, research and exploration of new fiber materials, full consideration of the scale effect, and exploration and exploitation of innovative improvement solutions for fire resistance. Researchers should concentrate on addressing these issues to promote the broader and more efficient application of UHPC in the construction field.
As it is well-known that the cylindrically symmetric Navier-Stokes equations take the form
ρt+(rρu)rr=0, | (1.1) |
ρ(ut+uur)−ρv2r+Pr=(λ(ru)rr)r−2uμrr, | (1.2) |
ρ(vt+uvr)+ρuvr=(μvr)r+2μvrr−(μv)rr−uvr2, | (1.3) |
ρ(wt+uwr)=(μwr)r+μwrr, | (1.4) |
ρ(e+uer)+P(ru)rr=(κrθr)rr+Q, | (1.5) |
where ρ(r,t) is the density, u(r,t), v(r,t), and w(r,t) are velocities in different directions, θ(r,t) is the temperature, the pressure P and the internal energy e are related with the density and temperature
P=P(ρ,θ)=Rρθande=e(ρ,θ)=cvθ, | (1.6) |
the specific gas constant R and the specific heat at constant volume cv are positive constants, respectively; the symbol Q denotes
Q=λ(ru)2rr2−4μuurr+μw2r+μ(vr−vr)2, | (1.7) |
μ and λ are viscosity coefficients, and κ is the heat conductivity coefficient.
Without loss of generality, we shall consider the system (1.1) with the following initial-boundary data:
{(ρ,u,v,w,θ)|t=0=(ρ0,u0,v0,w0,θ0)(r),0<a≤r≤b<∞,(u,v,w,∂rθ)|r=a=(u,v,w,∂rθ)|r=b=0,t≥0. | (1.8) |
Our main goal is to show the large-time behavior of global solutions to the initial-boundary value problem (1.1)–(1.8) with large initial data. For this purpose, it is convenient to transform the initial-boundary value problem (1.1)–(1.8) into Lagrangian coordinates. We introduce the Lagrangian coordinates (t,x) and denote (˜ρ,˜u,˜v,˜w,˜θ)(t,x)=(ρ,u,v,w,θ)(t,r), where
r=r(t,x)=r0(x)+∫t0u(s,r(s,x))ds, | (1.9) |
and
r0(x):=f−1(x),f(r):=∫rayρ0(y)dz. |
Note that the function f is invertible on [a,b] provided that ρ0(y)>0 for each y∈[a,b] (which will be assumed in Theorem 1.1). Due to (1.1)1 and (1.8), we see
∂∂t∫r(t,x)ayρ(t,y)dy=0. |
Then it is easy to check
∫r(t,x)ayρ(t,y)dy=f(r0(x))=xand∫r(t,1)byρ(t,y)dy=0, | (1.10) |
which translates the domain [0,T]×[a,b] into [0,T]×[0,1]. Hereafter, we denote (˜ρ,˜u,˜v,˜w,˜θ) by (ρ,u,v,w,θ) for simplicity. The identities (1.9) and (1.10) imply
rt(t,x)=u(t,x),rx(t,x)=r−1τ(t,x), | (1.11) |
where τ:=ρ−1 is the specific volume. By means of identities (1.11), system (1.1)–(1.8) is changed to
τt=(ru)x, | (1.12) |
ut−v2r+rPx=r(λ(ru)xτ)x−2uμx, | (1.13) |
vt−uvr=r(μrvxτ)x+2μvx−(μv)x−μτvr2, | (1.14) |
wt=r(μrwxτ)x+μwx, | (1.15) |
et+P(ru)x=(κr2θxτ)x+Q, | (1.16) |
where t>0, x∈Ω=(0,1), P=Rθτ, e=Rθγ−1, and Q=λ(ru)2xτ−4μuux+μr2w2xτ+μτ(rvxτ−vr)2.
Throughout this paper, we assume that μ,λ, and κ are power functions of absolute temperature as follows:
μ=˜μθα,λ=˜λθα,κ=˜κθβ, | (1.17) |
where constants ˜μ,˜λ,˜κ,α, and β are positive constants.
The objective of this paper is to study the global existence and stability of the solutions to an initial-boundary value problem of (1.12)–(1.16) with the initial data:
(τ,u,v,w,θ)(x,0)=(τ0,u0,v0,w0,θ0),x∈(0,1), | (1.18) |
and the boundary conditions:
(u,v,w,θx)(0,t)=(u,v,w,θx)(1,t)=0,t≥0. | (1.19) |
Using Navier-Stokes equations as a model for describing fluid motion has been widely accepted by the physics community. In recent years, some significant progress has been made in the study of Navier-Stokes equations with constant viscosity coefficients. When the initial value has a certain small property and vacuum state does not exist, the global existence, uniqueness, and large-time behavior of the solutions can be easily calculated [1,2,3,4,5,6,7,8]. However, solving the problem of large initial values is very challenging, and the first significant breakthrough was achieved by Lions [9]. Besides, by assuming that the initial value is only sufficiently small in the energy space, Hoff [10,11] confirmed the existence of global weak solutions. In the process of studying fluid motion, a vacuum state is often involved, which makes calculations far more complex. The results in [12,13] indicate the Cauchy problem of Navier-Stokes equations with constant coefficients containing vacuum state is not appropriate. This uncertainty is reflected by the fact that the solutions of the system have no continuous dependence on the initial values. Based on physical considerations, Liu-Xin-Yang [12] studied the Cauchy problem of the Navier-Stokes equations with density dependent viscosity, and proved its local suitability. However, only when the temperature and density change within a suitable range, real fluids can be considered as ideal fluids (viscosity coefficients are constants). In the case of large changes in temperature or density, the viscosity of the real fluid will vary greatly [14].
On the other hand, Navier-Stokes equations can be developed using the Chapman-Enskog expansion of the microscopic particle collision model Boltzman equation. Consequently, it can be determined that the viscosity depends on the temperature. However, compared to the abundant research using classical models, the studies on the physical case using the temperature-dependent viscosity model are lacking. Because the viscosity and heat conductivity are both temperature-dependent, degeneracy and strong non-linearity may appear. Pan-Zhang [15] and Huang-Shi [16] obtained global strong solutions and large-time behavior in bounded domains for one-dimensional Navier-Stokes equations, when α=0 and 0<β<1. The studies of Liu-Yang-Zhao-Zhou [17] and Wan-Wang [18] also acquired global solutions of Navier-Stokes equations in one dimensional and cylindrically symmetrical cases, respectively, with the requirement that |γ−1| was small enough. Wang-Zhao [19] removed the smallness condition of |γ−1|, and established global classical solutions to Navier-Stokes equations in the one-dimensional whole space when μ and κ satisfy:
μ=˜μh(τ)θα,κ=˜κh(τ)θα, |
where α is small enough. In their calculations, the viscosity and heat-conductivity were dependent on temperature and density, and to overcome the difficulties caused by density, the following conditions could not be removed:
‖h(τ)−1τ−1‖L∞(Ω)+‖h(τ)−1τ‖L∞(Ω)≤C. |
This means that estimate of ‖τx‖L2(Ω) can be directly obtained without the upper and lower bounds of density, as long as the coefficient μ−1 or κ−1 appears. However, if h(τ) is constant, then the constants l1=l2=0 and the result of this case cannot be established using the model in [19]. Recently, Sun-Zhang-Zhao [20] considered an initial-boundary value problem of the compressible Navier-Stokes equations for one-dimensional viscous and heat-conducting ideal polytropic fluids with temperature-dependent transport coefficients, and discovered the global-in-time existence of strong solutions. In that paper, the initial data could be large if α≥0 is small and the growth exponent β≥0 is arbitrarily large. It is worth mentioning that the smallness of α>0 depends on the size of the initial data. However, unfortunately the study did not provide a specific relationship between α and the initial data in [20]. Our main results are concluded as follows.
Theorem 1.1. For given positive constants M0,V0>0, assume that
‖(τ0,u0,v0,w0,θ0)‖H2(Ω)≤M0,infx∈(0,1){τ0,θ0}≥V0. | (1.20) |
Then there exist ϵ0>0 and C0 which depend only on β, M0, and V0, such that the initial-boundary value problem (1.12)–(1.19) with 0≤α≤ϵ0:=min{|α1|,|α2|} and β>0 admit a unique global-in-time strong solution (τ,u,v,w,θ) on [0,1]×[0,+∞) satisfying
C−10≤τ(x,t)≤C0,C−11≤θ(x,t)≤C1, |
and
(τ−ˉτ,u,v,w,θ−E0)∈C([0,+∞);H2(Ω)), |
where α1,α2 defined in what follows are dependent only on β, M0, and V0 (see details in (3.2), (3.5), and (3.6)). Moreover, for any t>0, the exponential decay rate is
‖(τ−ˉτ,u,v,w,θ−E0)‖2H1+‖r−ˉr‖2H2≤Ce−γ0t, | (1.21) |
where
ˉτ=∫10τdx,E0=∫10(θ0+u20+v20+w202cv)dx,ˉr=[a2+2ˉτx]12. |
A few remarks are in order.
Remark 1. For k=1,2 and 1≤p≤∞, we adopt the simplified nations for the standard Sobolev space as follows:
‖⋅‖:=‖⋅‖L2(Ω),‖⋅‖k:=‖⋅‖Hk(Ω),‖f‖∞:=maxx∈Ω|f(x)|,‖⋅‖Lp:=‖⋅‖Lp(Ω). |
Remark 2. We remark here that the growth exponent β∈(0,+∞) can be arbitrarily large, and the choice of ϵ0>0 depends only on β, V0, and the H2−norm of the initial data. An outline of this paper is as follows. We devote Section 2 to a discussion of a number of a priori estimates independent of time, which are needed to extend the local solution to all time. Based on the previous estimates, the main results, Theorem 1.1 are proved in Section 3.
Remark 3. In this paper, the positive c, C, and Ci(i=0,1,⋯,16) are some positive constants which depend only on β, M0, and V0, but not on the time t. Furthermore, c and C are different from line to line.
First of all, define
X(t1,t2;m1,m2;N):={(τ,u,v,w,θ)∈C([t1,t2];H2(Ω)),τx∈L2(t1,t2;H1(Ω))(ux,vx,wx,θx)∈L2(t1,t2;H2(Ω)),τt∈C([t1,t2];H1(Ω))∩L2(t1,t2;H1(Ω)),(ut,vt,wt,θt)∈C([t1,t2];L2(Ω))∩L2(t1,t2;H1(Ω)),τ≥m1,θ≥m2,E(t1,t2)≤N2,∀(x,t)∈[0,1]×[t1,t2]}, |
where N, m1, m2, and t1,t2(t2>t1) are constants and
E(t1,t2):=supt1≤t≤t2‖(τx,ux,θx)‖21+‖θt‖2+∫t2t1‖θt‖2dt |
with
θt|t=t1:=1cv[−P(ru)x+(κr2θxτ)x+Q]|t=t1, |
θxt|t=t1:=1cv[−P(ru)x+(κr2θxτ)x+Q]x|t=t1. |
The main purpose of this section is to derive the global t-independent estimates of the solutions (τ,u,v,w,θ)∈X(0,T;m1,m2,N).
We start with the following basic energy estimate.
Lemma 2.1. Assume that the conditions listed in Theorem 1.1 hold. Then there exists a constant 0<ϵ1≤1 depending only on M0 and V0, such that if
m−α2≤2,Nα≤2,αH(m1,m2,N)≤ϵ1, | (2.1) |
where
H(m1,m2,N):=(m1+m2+N+1)5, |
then for T≥0,
∫10ηˆθ(τ,u,v,w,θ)(x,t)dx+∫T0∫10[τu2θ+u2x+w2xτθ+θβθ2xτθ2+τθ(rvxτ−vr)2]dxds≤C, | (2.2) |
where
ηˆθ(τ,u,v,w,θ):=ˆθϕ(τˉτ)+u2+v2+w22+cvˆθϕ(θˆθ),ϕ(z):=z−logz−1. |
Proof. Multiplying (1.12)–(1.16) by Rˆθ(ˉτ−1−τ−1), u, v, w, and (1−ˆθθ−1), respectively, integrating over [0,1], and adding them together, one obtains
ddt∫10ηˆθ(τ,u,v,w,θ)dx+∫10[˜κr2θβθ2xτθ2+Qθ]dx=0, | (2.3) |
where Q=λ(ru)2xτ−4μuux+μr2w2xτ+μτ(rvxτ−vr)2.
Apparently, by means of λ=2μ+λ′, one has
λ(ru)2x−4τμuux=(2μ+λ′)r2u2x+(2μ+λ′)τ2u2r2+2λ′τuux=2μr2u2x+2μτ2u2r2+2μ+3λ′3[rux+τur]2−2μ3[rux+τur]2=23μ(r2u2x+τ2u2r2)+2μ+3λ′3[rux+τur]2+2μ3[rux−τur]2 ≥23μ(r2u2x+τ2u2r2). |
Thus, one has
Q≥Cu2xτ+Cτu2+Cw2xτ+Cτ(rvxτ−vr)2, |
which combined with (2.1) and (2.3) yields
ddt∫10ηˆθ(τ,u,v,w,θ)(t,x)dx+c∫10(τu2θ+u2x+w2xτθ+θβθ2xτθ2+τθ(rvxτ−vr)2)dx≤0. | (2.4) |
Integrating (2.4) over (0,T), we can obtain (2.2) by the initial conditions (τ0,u0,v0,θ0).
Next, by means of Lemma 2.1, we derive the upper and lower bounds of τ.
Lemma 2.2. Assume that the conditions of Lemma 2.1 hold. Then for (x,t)∈Ω×[0,∞),
C−10≤τ(x,t)≤C0. |
Proof. The proof is divided into three steps.
Step 1 (Representation of the formula for τ).
It follows from (1.13) that
(ur)t+u2−v2r2+2uμxr+Px=(λ(lnτ)t)x=λ(lnτ)xt+λx(ru)xτ. |
that is
(uλr)t+g+(λ−1P)x=(lnτ)xt, | (2.5) |
where
g:=u2−v2λr2+2uμxλr−(λ−1)xP−λx(ru)xλτ−(λ−1)tur. |
Integrating (2.5) over [0,t]×[x1(t),x], we have
∫xx1(t)(uλr−u0λ0r0)dξ+∫t0∫xx1(t)gdξds+∫t0λ−1P(x)−λ−1P(x1)ds=lnτ(x,t)−lnτ(x1(t),t)−[lnτ0(x)−lnτ(x1(t),0)], | (2.6) |
where x1(t)∈[0,1] is determined by the following progresses. Next, for convenience, we define
F:=(ru)xτ−λ−1P−∫x0g(ξ)dξ,φ:=∫t0F(x,s)ds+∫x0u0λ0r0dξ. |
It follows from the definitions above that
φx=uλr,φt=F. | (2.7) |
By the definition of F and (1.12), one has
∫t0[λ−1P(x1(t),s)+∫x1(t)0g(ξ,s)dξ]ds=∫t0((ru)xτ−F)(x1(t),s)ds=lnτ(x1(t),t)−lnτ(x1(t),0)−∫t0F(x1(t),s)ds. | (2.8) |
Due to (1.12) and (2.7), we have
(τφ)t−(ruφ)x=τφt−ruφx=τF−u2λ=(ru)x−τPλ−τ∫x0g(ξ)dξ−u2λ. | (2.9) |
Integrating (2.9) over [0,t]×Ω, one has
∫10φτdx=∫10τ0∫x0(u0λ0r0)(ξ)dξdx−∫t0∫10[τλP+τ∫x0gdξ+u2λ]dxds. | (2.10) |
Hence, by virtue of the mean value theorem, there exits x1(t)∈[0,1] such that φ(x1(t),t)=∫10φτdx. By the definition of φ, (2.8), and (2.10), one obtains
∫t0F(x1(t),s)ds=φ(x1(t),t)−∫x1(t)0u0λ0r0(ξ)dξ=∫10τ0∫x0u0λ0r0(ξ)dξdx−∫t0∫10(τλP+τ∫x0gdξ+u2λ)dxds−∫x1(t)0u0λ0r0(ξ)dξ. | (2.11) |
Putting (2.11) into (2.8), it follows that
∫t0(Pλ(x1(t),s)+∫x1(t)0g(ξ,s)dξ)ds=lnτ(x1(t),t)−lnτ(x1(t),0)−∫10τ0∫x0u0λ0r0(ξ)dξdx+∫x1(t)0u0λ0r0(ξ)dξ+∫t0∫10(τλP+τ∫x0gdξ+u2λ)dxds. | (2.12) |
Inserting (2.12) into (2.6), we derive
∫t0Pλds+∫t0∫x0gdξds−∫t0∫10(τλP+τ∫x0gdξ+u2λ)dxds+∫xx1(t)(uλr−u0λ0r0)dξ+∫10τ0∫x0u0λ0r0dξdx−∫x1(t)0u0λ0r0dξ=lnτ−lnτ0. | (2.13) |
Let
g=u2−v2λr2+g1, |
where
g1:=2uμxλr−(λ−1)xP−λx(ru)xλτ−(λ−1)tur. |
It follows from (2.13) that
τ=B−1AD, | (2.14) |
where
A:=exp{∫t0[Pλ(x,s)+∫x0(g1(ξ,s)+u2λr2)dξ+∫10τ∫x0(v2λr2−g1)dξdx]ds},B:=exp{∫t0[∫10(τλP+τ∫x0u2λr2(ξ)dξ+u2λ)dx+∫x0v2λr2dξ]ds},D:=τ0exp{∫10τ0∫x0u0λ0r0dξdx−∫x1(t)0u0λ0r0dξ+∫xx1(t)(uλr−u0λ0r0)(ξ)dξ}. |
By (2.14), one has
τD−1B=A. | (2.15) |
Define that
J:=Pλ(x,s)+∫x0(g1(ξ,s)+u2λr2)dξ+∫10τ∫x0(v2λr2−g1)dξdx. |
Then, multiplying (2.15) by J gives
τD−1BJ=ddtA. |
Since A(0)=1, integrating the above equality over (0,t) about time, one has
τ=DB−1+1λ∫t0B(s)B(t)D(t)D(s)τ[Pλ(x,s)+∫x0(g1(ξ,s)+u2λr2)dξ+∫10τ∫x0(v2λr2−g1)dξdx]ds. | (2.16) |
Step 2 (Lower bound for τ). First of all, by means of (2.1) and (2.2), one has
C−1≤D≤C. | (2.17) |
Next, we estimate B. Employing Jensen's inequality to the convex function ϕ, we have
∫10zdx−log∫10zdx−1≤∫10ϕ(z)dx. | (2.18) |
By (2.18) and Lemma 2.1, one obtains
C−1≤∫10τdx,ˉθ:=∫10θdx≤C, | (2.19) |
which means that
C−1≤∫10τλPdx≤C. | (2.20) |
Hence, by means of the definition of B and (2.20), choosing ε1 suitably small, there exist two constants C1 and C2, such that
ec1t≤B(t)≤ec2t. | (2.21) |
That is,
e−c1(t−s)≤B(s)B(t)≤e−c2(t−s). | (2.22) |
Apparently, by means of (2.1) and (2.19), we deduce
|τ∫10τ∫x0g1dξdx|≤C|α|‖τ‖2∞(‖θ−1‖∞‖θx‖‖u‖+‖θ−ατ−1‖∞‖θx‖+‖θ−1τ−1‖∞‖θx‖‖u‖1+‖θ−1τ−1‖∞‖θt‖‖u‖)≤Cε1. | (2.23) |
Similarly, one also has
‖∫x0g1dξ‖∞≤Cε1. | (2.24) |
Thus, for t≤t0<∞,
τ≥DB−1−Cε1∫t0e−c2(t−s)ds=DB−1−Cε1c2(1−e−c2t)≥Ce−ct0−ε2(1−e−c2t0). |
For a enough large t, we have
infx∈Ωτ(x,t)≥C∫t0B(s)B(t)θds−ε2(1−e−c2t). | (2.25) |
So, we need the estimates of θ and B(s)B(t). By the mean value theorem and (2.19), there exits x2(t)∈[0,1], such that
C−1≤θ(x2(t),t)≤C. | (2.26) |
By Cauchy-Schwarz's inequality and (2.19), one has
|[ln(θ+1)]β2+1−[ln(θ(x2(t),t)+1)]β2+1|=|∫xx2(ln(θ+1))β2θx√τ(θ+1)√τ(ξ)dξ|≤(∫10(ln(θ+1))βθ2xτ(θ+1)2dx)12(∫10τdx)12≤C(∫10θβθ2xτθ2dx)1/2, |
which means that
θ≥C−C∫10θβθ2xτθ2dx. | (2.27) |
By (2.16)–(2.17), (2.23)–(2.24), (2.21), Lemma 2.1, and (2.19), one has
∫10τdx≤Ce−ct+C∫t0B(s)B(t)ds, |
that is
∫t0B(s)B(t)ds≥C−Ce−ct. | (2.28) |
Putting (2.27) into (2.25), by (2.22), (2.28), and Lemma 2.1, for a enough large t, one has
∫t0B(s)B(t)θds≥C∫t0B(s)B(t)(1−∫10θβθ2xτθ2dx)ds≥C−Ce−ct−C(∫t/20+∫tt/2)B(s)B(t)∫10θβθ2xτθ2dxds≥C−Ce−ct−C∫t/20e−c(t−s)∫10θβθ2xτθ2dxds−C∫tt/2∫10θβθ2xτθ2dxds≥C−Ce−ct−Ce−ct/2−C∫tt/2∫10θβθ2xτθ2dxds≥C. | (2.29) |
Inserting (2.29) into (2.25), for a large enough time T0, when t>T0, it follows that
infx∈Ωτ(x,t)≥C. |
Step 3 (Upper bound for τ). By (2.17), (2.22)–(2.24), and Lemma 2.1, one obtains
‖τ‖∞≤C+C∫t0e−c2(t−s)‖τ‖∞(∫10θβθ2xτθ2dx+1)ds, | (2.30) |
where we have used the results
{‖θ‖∞≤C+C‖τ‖∞∫10θβθ2xτθ2dxwhen0<β≤1,‖θ‖∞≤C+C∫10θβθ2xτθ2dxwhen1<β<∞. | (2.31) |
In fact, by Hölder's inequality, for 0<β≤1,
|θ1/2(x,t)−θ1/2(x2(t),t)|≤∫10θ−12θxdx≤‖τ‖1/2∞(∫10θβθ2xτθ2dx)1/2(∫10θ1−βdx)1/2≤‖τ‖1/2∞(∫10θβθ2xτθ2dx)1/2. | (2.32) |
For 1<β<∞,
|θβ/2(x,t)−θβ/2(x2(t),t)|≤∫10θβ/2θxθdx≤(∫10θβθ2xτθ2dx)1/2(∫10τdx)1/2. | (2.33) |
By means of (2.26) and (2.32)–(2.33), we can obtain (2.31).
Thus, the inequality (2.30) combined with Gronwall's inequality and Lemma 2.1 yields that for any t≥0,
supt≥0‖τ(x,t)‖∞≤C. |
However, we cannot get the time-space estimate of vx in Lemma 2.1. To obtain this estimate, we need the following result.
Lemma 2.3. Assume that the conditions listed in Lemma 2.1 hold. Then for any p>0 and T≥0,
∫10θ1−pdx+∫T0∫10(θβθ2xθp+1+θα(u2+u2x+w2x)θp+θατθp(rvxτ−vr)2)dxds≤C. | (2.34) |
Proof. By Lemma 2.1, the result of (2.34) has been established for p=1. In the following steps, we do the estimate for p>0 and p≠1. Multiplying (1.16) by θ−p, integrating over [0,1], and using integration by parts gives
cvp−1ddt∫10θ1−pdx+p∫10˜κr2θβθ2xτθp+1dx+∫10Qθpdx=R∫10θ1−pτ(ru)xdx=R∫10θ1−p−E0τ(ru)xdx+RE0∫10(ru)xτdx. | (2.35) |
Apparently, there exists constant C(p) depending on p such that
|θ1−p−E0|≤C(p)|θ1/2−E1/20|(E1/20+θ12−p). | (2.36) |
By means of (2.35), (2.36), Lemma 2.2, (1.13), and (1.12), we deduce
cvp−1ddt∫10θ1−pdx+p∫10˜κr2θβθ2xτθp+1dx+∫10Qθpdx≤C(p)‖θ1/2−E1/20‖∞∫10(E1/20+θ12−p)(|u|+|ux|)dx+RE0ddt∫10lnτdx≤C(p)‖θ1/2−E1/20‖∞[(∫10u2+u2xθdx)12(∫10θdx)12+(∫10θ1−pdx)12(∫10u2+u2xτθpdx)12]+RE0ddt∫10lnτdx≤C(p)‖θ1/2−E1/20‖2∞+C(p)∫10u2+u2xθdx+δ∫10u2+u2xτθpdx+C(δ,p)‖θ1/2−E1/20‖2∞∫10θ1−pdx+RE0ddt∫10lnτdx. | (2.37) |
Thus, employing the truth of
∫t0‖θ1/2−E1/20‖2∞ds≤C, | (2.38) |
we can conclude from (2.37), Grönwall's inequality, and Lemma 2.2 that (2.34) is correct. In fact,
‖θ1/2−E1/20‖∞≤‖θ1/2−ˉθ1/2‖∞+‖ˉθ1/2−E1/20‖∞. | (2.39) |
By virtue of Lemmas 2.1–2.2 and (2.19), one has
|ˉθζ−Eζ0|=|∫10ddη{[∫10(θ+ηu2+v2+w22cv)dx]ζ}dη|=|ζ∫10[∫10(θ+ηu2+v2+w22cv)dx]ζ−1dη∫10(u,v,w)22dx|≤C‖(u,v,w)‖‖(u,v,w)‖∞≤C∫10|(ux,(vr)x,wx)|dx≤C(∫10[u2xθ+1θ(rvxτ−vr)2+w2xθdx])1/2(∫10θdx)1/2≤C(∫10[u2xθ+1θ(rvxτ−vr)2+w2xθ]dx)1/2, | (2.40) |
where we have used the fact that
\begin{equation*} \left(\frac{v}{r}\right)_x = \frac{\tau}{r^2}\left(\frac{rv_x}{\tau}-\frac{v}{r }\right). \end{equation*} |
For \beta < 1 , it follows from Lemma 2.1 and (2.19) that
\begin{equation} \begin{split} &\|\theta^{1/2}-\bar\theta^{1/2}\|_\infty \\&\quad \leq C\int_0^1\theta^{-\frac{1}{2}}|\theta_x| \mathrm{d} x \\&\quad \leq C\left(\int_0^1\frac{\theta^\beta\theta_x^2}{\theta^2} \mathrm{d} x\right)^{\frac{1}{2}}\left(\int_0^1\theta^{1-\beta} \mathrm{d} x\right)^{\frac{1}{2}} \\&\quad \leq C\left(\int_0^1\frac{\theta^{\beta}\theta_x^2}{\theta^2} \mathrm{d} x\right)^{\frac{1}{2}}. \end{split} \end{equation} | (2.41) |
For 1\leq\beta < \infty ,
\begin{equation} \|\theta^{\frac{1}{2}}-\bar{\theta}^{\frac{1}{2}}\|_\infty \leq C\|\theta^{\frac{\beta}{2}}-\bar{\theta}^{\frac{\beta}{2}}\|_\infty \leq C\int_0^1\theta^{\frac{\beta}{2}-1}|\theta_x| \mathrm{d} x \leq C\left(\int_0^1\frac{\theta^\beta\theta_x^2}{\theta^2} \mathrm{d} x\right)^{\frac{1}{2}}. \end{equation} | (2.42) |
Hence, by (2.39)–(2.42) and Lemmas 2.1–2.2, we can derive (2.38). The proof of Lemma 2.3 is thus complete.
According to Lemmas 2.1–2.3, we can conclude that the following results have been established.
Corollary 2.1. Assume that the conditions listed in Lemma 2.1 hold. Then for -\infty < q < 1 , 0 < p < \infty , and T\geq0 ,
\begin{array}{c} C_1\leq\tau\leq C, \quad C^{-1}\leq\int_0^1\tau \mathrm{d} x\leq C, \quad C^{-1}\leq\int_0^1\theta \mathrm{d} x\leq C, \\ \int_0^1(|\ln\tau|+|\ln\theta|+\theta^q+u^2+v^2+w^2) \mathrm{d} x\leq C_3, \\ \int_0^T\int_0^1\left[(u^2+u_x^2+v^2+v_x^2+w_x^2+\tau_t^2)(1+\theta^{-p})+\frac{\theta^\beta\theta_x^2}{\theta^{1+p}}\right] \mathrm{d} x \mathrm{d} s\leq C. \end{array} | (2.43) |
Here, we have taken p = \alpha in (2.34) to obtain the time-space estimates of v and v_x .
Using the result above, we establish the following estimate about \tau_x .
Lemma 2.4. Assume that the conditions listed in Lemma 2.1 hold. Then for T\geq0 ,
\begin{equation*} \int_0^1\tau_x^2 \mathrm{d} x+\int_0^T\int_0^1\tau_x^2(1+\theta) \mathrm{d} x \mathrm{d} s\leq C_2. \end{equation*} |
Proof. According to the chain rule, one has
\begin{equation} \left(\frac{\lambda\tau_x}{\tau}\right)_t = \left(\frac{\lambda\tau_t}{\tau}\right)_x+\frac{\lambda_\theta}{\tau}(\tau_x\theta_t-\tau_t\theta_x). \end{equation} | (2.44) |
By means of (1.12), (1.13), and (2.44), we have
\begin{equation} \left(\frac{\lambda\tau_x}{\tau}\right)_t = \frac{u_t}{r }+P_x-\frac{v^2}{r^{2}}+\frac{2 u\mu_x}{r} +\frac{\lambda_\theta}{\tau}(\tau_x\theta_t-\tau_t\theta_x). \end{equation} | (2.45) |
Multiplying (2.45) by \frac{\lambda\tau_x}{\tau} , integrating over [0, 1] about x , and using (1.12) and (2.44), we obtain
\begin{equation} \begin{split} &\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\left[\frac{1}{2}\left(\frac{\lambda\tau_x}{\tau}\right)^2-\frac{\lambda u\tau_x}{r \tau}\right] \mathrm{d} x +\int_0^1\frac{R\lambda\theta\tau_x^2}{\tau^3} \mathrm{d} x \\&\quad = \int_0^1\left(\frac{u}{r }\right)_x\frac{\lambda\tau_t}{\tau} \mathrm{d} x +\int_0^1\frac{R\lambda\tau_x\theta_x}{\tau^2} \mathrm{d} x +\int_0^1\frac{\lambda\tau_x( u^2-v^2)}{\tau r^{2}} \mathrm{d} x \\&\qquad +\int_0^1\frac{2 u\mu_x\lambda\tau_x}{r\tau} \mathrm{d} x +\int_0^1\frac{\lambda_\theta}{\tau^2}(\lambda\tau_x-r^{-1}u\tau)(\tau_x\theta_t-\tau_t\theta_x) \mathrm{d} x \\&\quad : = \sum\limits_{i = 1}^{5}I_i. \end{split} \end{equation} | (2.46) |
By Hölder's inequality, (2.1), (1.12), and Corollary 2.1, one has
\begin{equation} I_1 = \int_0^1\left(\frac{u_x}{r }-\frac{\tau u}{r^{3}}\right)\frac{\lambda\tau_t}{\tau} \mathrm{d} x \leq C\|(u, u_x, \tau_t)\|^2\leq C\|(u, u_x)\|^2. \end{equation} | (2.47) |
Using Corollary 2.1 and taking p = \beta , one has
\begin{equation} \int_0^T\int_0^1\frac{\theta_x^2}{\theta} \mathrm{d} x \mathrm{d} s\leq C. \end{equation} | (2.48) |
Hence, we argue the term I_2 as the following
\begin{equation} I_2\leq\delta\int_0^1\frac{\tau_x^2\theta}{\tau^3} \mathrm{d} x +C(\delta)\int_0^1\frac{\theta_x^2}{\theta} \mathrm{d} x. \end{equation} | (2.49) |
By means of integration by parts, Corollary 2.1, and (2.1), one can derive
\begin{equation} \begin{split} I_3& = -\int_0^1\log\tau\left(\frac{\lambda}{r^{2}}(u^2-v^2)\right)_x \mathrm{d} x \\& \leq C\|\ln\tau\|_\infty\int_0^1\Big|\Big(|\alpha|\theta_xu^2, |\alpha|\theta_xv^2, \theta^\alpha u^2, \theta^\alpha v^2, \theta^\alpha u_x^2, \theta^\alpha v_x^2\Big)\Big| \mathrm{d} x \\& \leq C\int_0^1(u^2, v^2, u_x^2, v_x^2) \mathrm{d} x. \end{split} \end{equation} | (2.50) |
By virtue of (2.1), we derive
\begin{equation} I_4\leq C|\alpha|m_2^{-\frac{3}{2}}N\left(\int_0^1u^2 \mathrm{d} x\right)^{1/2}\left(\int_0^1\frac{\tau_x^2\theta}{\tau^3} \mathrm{d} x\right)^{1/2} \leq\delta\int_0^1\frac{\tau_x^2\theta}{\tau^3} \mathrm{d} x+C(\delta)\int_0^1u^2 \mathrm{d} x. \end{equation} | (2.51) |
By means of (2.1), Corollary 2.1, and (1.16), one can deduce
\begin{equation} \begin{split} I_5&\leq C\int_0^1|\alpha||\theta^{-1}|\Big|\Big(\tau_x^2\theta_t, u\tau_x\theta_t, \tau_xu\theta_x, u^2\theta_x, \tau_xu_x\theta_x, uu_x\theta_x\Big)\Big| \mathrm{d} x \\& \leq C|\alpha|m_2^{-2}N\int_0^1\frac{\tau_x^2\theta}{\tau^3} \mathrm{d} x +C|\alpha|m_2^{-\frac{3}{2}}N\left(\int_0^1u^2+u_x^2 \mathrm{d} x\right)^{1/2}\left(\int_0^1\frac{\tau_x^2\theta}{\tau^2} \mathrm{d} x\right)^{1/2} \\&\quad +C|\alpha|m_2^{-1}N\int_0^1u^2+u_x^2 \mathrm{d} x \\& \leq \varepsilon\int_0^1\frac{\tau_x^2\theta}{\tau^3} \mathrm{d} x +C( \varepsilon)\int_0^1u^2+u_x^2 \mathrm{d} x. \end{split} \end{equation} | (2.52) |
Inserting (2.47) and (2.49)–(2.52) into (2.46), and choosing \varepsilon suitable small, we obtain
\begin{equation} \frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\left[\frac{1}{2}\left(\frac{\lambda\tau_x}{\tau}\right)^2-\frac{\lambda u\tau_x}{r \tau}\right] \mathrm{d} x +c\int_0^1\theta\tau_x^2 \mathrm{d} x \leq C\|(u, u_x, \theta_x/\sqrt{\theta}, v, v_x)\|^2. \end{equation} | (2.53) |
Integrating (2.53) over [0, t] , using Cauchy-Schwarz's inequality, (2.48), and Corollary 2.1, for any t\geq0 , one has
\begin{equation} \int_0^1\tau_x^2 \mathrm{d} x+\int_0^t\int_0^1\tau_x^2\theta \mathrm{d} x \mathrm{d} s\leq C. \end{equation} | (2.54) |
By virtue of (2.54), we have
\begin{equation} \begin{split} &\bar\theta\int_0^1\tau_x^2 \mathrm{d} x = \int_0^1\tau_x^2(\bar{\theta}-\theta) \mathrm{d} x+\int_0^1\tau_x^2\theta \mathrm{d} x \\&\quad \leq \frac{\bar{\theta}}{2}\int_0^1\tau_x^2 \mathrm{d} x+\frac{1}{2\bar{\theta}}\|\theta-\bar{\theta}\|_\infty^2\int_0^1\tau_x^2 \mathrm{d} x+\int_0^1\tau_x^2\theta \mathrm{d} x \\&\quad \leq \frac{\bar{\theta}}{2}\int_0^1\tau_x^2 \mathrm{d} x+C\|\theta-\bar{\theta}\|_\infty^2+\int_0^1\tau_x^2\theta \mathrm{d} x. \end{split} \end{equation} | (2.55) |
It follows from (2.19) and (2.48) that
\begin{equation} \int_0^T\|\theta-\bar{\theta}\|_\infty^2 \mathrm{d} s \leq C\int_0^T\int_0^1\frac{\theta_x^2}{\theta} \mathrm{d} x\int_0^1\theta \mathrm{d} x \mathrm{d} t\leq C. \end{equation} | (2.56) |
Thus, it follows from (2.55)–(2.56) that
\begin{equation} \int_0^T\int_0^1\tau_x^2 \mathrm{d} x \mathrm{d} t\leq C. \end{equation} | (2.57) |
The proof of Lemma 2.4 has been completed by (2.54) and (2.57).
Next, based on the estimate of \tau_x , we are devoted to derive the estimates on the first-order derivatives of w_x .
Lemma 2.5. Assume that the conditions listed in Lemma 2.1 hold. Then for T\geq0 ,
\begin{equation*} \int_0^1w_x^2 \mathrm{d} x+\int_0^T\int_0^1w_{xx}^2 \mathrm{d} x \mathrm{d} t\leq C_3. \end{equation*} |
Proof. Multiplying (1.15) by w_{xx} and integrating over [0, 1] about x , we find from (2.1) and Lemma 2.4 that
\begin{equation} \begin{split} &\frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\|w_x\|^2+\int_0^1\frac{\mu r^2w_{xx}^2}{\tau} \mathrm{d} x \\&\quad = -\int_0^1r w_{xx}w_x\left(\frac{\mu r }{\tau}\right)_x \mathrm{d} x-\int_0^1\mu w_{xx}w_x \mathrm{d} x \\&\quad \leq C\int_0^1|w_{xx}w_x|(|\alpha|m_2^{-1}|\theta_x|+1+|\tau_x|) \mathrm{d} x \\&\quad \leq \varepsilon\|w_{xx}\|^2+C( \varepsilon)\|w_x\|^2+C( \varepsilon)\|\tau_x\|^2\|w_x\|_\infty^2 \\&\quad \leq \varepsilon\|w_{xx}\|^2+C( \varepsilon)\|w_x\|^2. \end{split} \end{equation} | (2.58) |
Taking \varepsilon suitably small in (2.58) finds
\begin{equation} \frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\|w_x\|^2 +c\int_0^1w_{xx}^2 \mathrm{d} x \leq C\|w_x\|^2. \end{equation} | (2.59) |
The proof of Lemma 2.5 is complete by integrating (2.59) over (0, t) about time and choosing \varepsilon suitably small.
Based on the above result, we have the following uniform first-order derivatives estimates on the velocity (u, v) .
Lemma 2.6. Assume that the conditions listed in Lemma 2.1 hold. Then for T\geq0 ,
\begin{equation*} \int_0^1(u_x^2+v_x^2+\tau_t^2) \mathrm{d} x+\int_0^T\int_0^1\left(u_{xx}^2+v_{xx}^2+\theta_x^2+u_t^2+v_t^2+w_t^2+\tau_{tx}^2\right) \mathrm{d} x \mathrm{d} t\leq C_4. \end{equation*} |
Proof. Multiplying (1.13) and (1.14) by u_{xx} and v_{xx} , respectively, and integrating over \Omega about x , by integration by parts, one has
\begin{equation} \begin{split} &\frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1(u_x^2+v_x^2) \mathrm{d} x +\int_0^1\frac{r^{2}}{\tau}(\lambda u_{xx}^2+\mu v_{xx}^2) \mathrm{d} x \\&\quad = \int_0^1u_{xx}r P_x \mathrm{d} x +\int_0^1\left(v_{xx}\frac{uv}{r}-u_{xx}\frac{v^2}{r}\right) \mathrm{d} x \\&\qquad -\int_0^1u_{xx}r \left[\left(\frac{\lambda(r u)_x}{\tau}\right)_x-\frac{\lambda r u_{xx}}{\tau}\right] \mathrm{d} x +2 \int_0^1 u\mu_xu_{xx} \mathrm{d} x \\&\qquad -\int_0^1v_{xx}\left[r v_x\left(\frac{\mu r }{\tau}\right)_x+2\mu v_x- (\mu v)_x-\frac{\mu\tau v}{r^{2}}\right] \mathrm{d} x \\&\quad : = \sum\limits_{i = 1}^5II_i. \end{split} \end{equation} | (2.60) |
By Cauchy-Schwarz's inequality, one has
\begin{equation} II_1\leq \varepsilon\|u_{xx}\|^2+C( \varepsilon)\|(\theta_x, \tau_x)\|^2. \end{equation} | (2.61) |
It follows from Sobolev's inequality, the boundary condition of v , and Corollary 2.1, that we have
\begin{equation} II_2\leq \varepsilon\|(u_{xx}, v_{xx})\|^2+C( \varepsilon)\|v\|_\infty^2\|(u, v)\|^2 \leq \varepsilon\|(u_{xx}, v_{xx})\|^2+C( \varepsilon)\|v_x\|^2. \end{equation} | (2.62) |
Direct computation from (2.1) yields
\begin{eqnarray} II_3&\leq& \varepsilon\|u_{xx}\|^2+C( \varepsilon)\int_0^1\Big[\tau_x^2u_x^2+(1+|\alpha|m_2^{-2}\theta_x^2)|(u_x, u\tau_x, u)|^2\Big] \mathrm{d} x\\ &\leq& \varepsilon\|u_{xx}\|^2+C( \varepsilon)\|(u_x, u)\|^2+C( \varepsilon)\|\tau_x\|^2\|(u_x, u)\|_\infty^2\\ &\leq& 2 \varepsilon\|u_{xx}\|^2+C( \varepsilon)\|(u_x, u)\|^2, \end{eqnarray} | (2.63) |
\begin{eqnarray} II_4&\leq& \varepsilon\|u_{xx}\|^2+C( \varepsilon)|\alpha|N^2m_2^{-2}\|u\|^2\leq \varepsilon\|u_{xx}\|^2+C( \varepsilon)\|u\|^2, \end{eqnarray} | (2.64) |
and
\begin{equation} \begin{split} II_5&\leq \varepsilon\|v_{xx}\|^2+C( \varepsilon)\int_0^1\Big[v_x^2(1+|\alpha|m_2^{-2}\theta_x^2+\tau_x^2)+v^2\Big] \mathrm{d} x \\&\leq2 \varepsilon\|v_{xx}\|^2+C( \varepsilon)\|(v_x, v)\|^2. \end{split} \end{equation} | (2.65) |
Putting (2.61)–(2.65) into (2.60) and taking \varepsilon suitably small gives
\begin{equation} \frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1(u_x^2+v_x^2) \mathrm{d} x +c\int_0^1(u_{xx}^2+v_{xx}^2) \mathrm{d} x \leq C\|(\theta_x, \tau_x, v_x, u_x, u, v)\|^2. \end{equation} | (2.66) |
Integrating (2.66) over (0, T) about time, and using Lemma 2.4 and Corollary 2.1, we find
\begin{equation} \int_0^1(u_x^2+v_x^2) \mathrm{d} x+\int_0^T\int_0^1(u_{xx}^2+v_{xx}^2) \mathrm{d} x \mathrm{d} t\leq C+C\int_0^T\int_0^1\theta_x^2 \mathrm{d} x \mathrm{d} t. \end{equation} | (2.67) |
For \beta > 1 , we take p = \beta-1 in (2.43), and then
\begin{equation} \int_0^T\int_0^1\theta_x^2 \mathrm{d} x \mathrm{d} t\leq C. \end{equation} | (2.68) |
Substituting (2.68) into (2.67), it follows for \beta > 1 that
\begin{equation} \int_0^1(u_x^2+v_x^2) \mathrm{d} x+\int_0^T\int_0^1(u_{xx}^2+v_{xx}^2+\theta_x^2) \mathrm{d} x \mathrm{d} t\leq C. \end{equation} | (2.69) |
Next, we need to estimate the L^2(\Omega\times(0, t)) -norm of \theta_x for 0 < \beta\leq1 . We deduce from multiplying (1.16) by \theta^{1-\frac{\beta}{2}} and integration by parts that
\begin{equation} \begin{split} &\frac{2c_v}{4-\beta}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\theta^{2-\frac{\beta}{2}} \mathrm{d} x+\frac{2-\beta}{2}\int_0^1\frac{\tilde{\kappa}r^{2}\theta^{\frac{\beta}{2}}\theta_x^2}{\tau} \mathrm{d} x \\&\quad = -R\int_0^1\frac{\theta^{2-\frac{\beta}{2}}}{\tau}(r u)_x \mathrm{d} x+\int_0^1\theta^{1-\frac{\beta}{2}}Q \mathrm{d} x \\&\quad = R\int_0^1\frac{\bar{\theta}^{2-\frac{\beta}{2}}-\theta^{2-\frac{\beta}{2}}}{\tau}(r u)_x \mathrm{d} x-R\bar{\theta}^{2-\frac{\beta}{2}}\int_0^1\frac{(r u)_x}{\tau} \mathrm{d} x+\int_0^1\theta^{1-\frac{\beta}{2}}Q \mathrm{d} x \\&\quad \leq C\int_0^1\left|\bar{\theta}^{2-\frac{\beta}{2}}-\theta^{2-\frac{\beta}{2}}\right||(u, u_x)| \mathrm{d} x-R\bar{\theta}^{2-\frac{\beta}{2}}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\ln\tau \mathrm{d} x+\int_0^1\theta^{1-\frac{\beta}{2}}Q \mathrm{d} x. \end{split} \end{equation} | (2.70) |
Notice that
\begin{equation} \begin{split} &\int_0^1\left|\bar{\theta}^{2-\frac{\beta}{2}}-\theta^{2-\frac{\beta}{2}}\right||(u, u_x)| \mathrm{d} x \\&\quad \leq C\|\bar{\theta}^{1-\frac{\beta}{4}}-\theta^{1-\frac{\beta}{4}}\|_\infty \left(\int_0^1(1+\theta^{2-\frac{\beta}{2}}) \mathrm{d} x\right)^{1/2}\left(\int_0^1(u^2+u_x^2) \mathrm{d} x\right)^{1/2} \\&\quad \leq C\left(\int_0^1\theta^{-\frac{\beta}{4}}|\theta_x| \mathrm{d} x\right)^2 +C\int_0^1(1+\theta^{2-\frac{\beta}{2}}) \mathrm{d} x\int_0^1(u^2+u_x^2) \mathrm{d} x \\&\quad \leq C\int_0^1\theta^{1-\frac{\beta}{2}} \mathrm{d} x\int_0^1\frac{\theta_x^2}{\theta} \mathrm{d} x +C\int_0^1(1+\theta^{2-\frac{\beta}{2}}) \mathrm{d} x\int_0^1(u^2+u_x^2) \mathrm{d} x \\&\quad \leq C\int_0^1\frac{\theta_x^2}{\theta} \mathrm{d} x +C\int_0^1(1+\theta^{2-\frac{\beta}{2}}) \mathrm{d} x\int_0^1(u^2+u_x^2) \mathrm{d} x, \end{split} \end{equation} | (2.71) |
and
\begin{equation} \begin{split} &\int_0^1\theta^{1-\frac{\beta}{2}}Q \mathrm{d} x \\&\quad \leq C\left( \|\bar{\theta}^{1-\frac{\beta}{2}}-\theta^{1-\frac{\beta}{2}}\|_\infty +1\right)\int_0^1(u^2+u_x^2+v^2+v_x^2+w_x^2) \mathrm{d} x \\&\quad \leq C\int_0^1\theta^{-\frac{\beta}{2}}|\theta_x| \mathrm{d} x\int_0^1(u_x^2+v_x^2+w_x^2) \mathrm{d} x+C\int_0^1(u_x^2+v_x^2+w_x^2) \mathrm{d} x \\&\quad \leq C\int_0^1(\theta^{-\frac{1}{2}}+\theta^{\frac{\beta}{4}})|\theta_x| \mathrm{d} x\int_0^1(u_x^2+v_x^2+w_x^2) \mathrm{d} x+C\int_0^1(u_x^2+v_x^2+w_x^2) \mathrm{d} x \\&\quad \leq \varepsilon\int_0^1\theta^{\frac{\beta}{2}}\theta_x^2 \mathrm{d} x+C( \varepsilon)\left(\int_0^1(u_x^2+v_x^2+w_x^2) \mathrm{d} x\right)^2 \\&\qquad +C\int_0^1\left(\frac{\theta_x^2}{\theta}+u_x^2+v_x^2+w_x^2\right) \mathrm{d} x. \end{split} \end{equation} | (2.72) |
We can conclude from (2.70)–(2.72) that
\begin{equation*} \int_0^1\theta^{2-\frac{\beta}{2}} \mathrm{d} x +\int_0^T\int_0^1\theta^{\beta/2}\theta_x^2 \mathrm{d} x \mathrm{d} t \leq C+C\int_0^T \left(\int_0^1 (u_x^2+v_x^2+w_x^2) \mathrm{d} x\right)^2 \mathrm{d} s, \end{equation*} |
which combined with Young's inequality and Corollary 2.1 yields
\begin{equation} \begin{split} &\int_0^T\int_0^1\theta_x^2 \mathrm{d} x \mathrm{d} t \\&\quad\leq C\int_0^T\int_0^1\frac{\theta^\beta\theta_x^2}{\theta^2} \mathrm{d} x \mathrm{d} s+C\int_0^T\int_0^1\theta^{\beta/2}\theta_x^2 \mathrm{d} x \mathrm{d} s \\&\quad \leq C+C\int_0^T\left(\int_0^1(u_x^2+v_x^2+w_x^2) \mathrm{d} x\right)^2 \mathrm{d} t. \end{split} \end{equation} | (2.73) |
By means of Lemma 2.5, (2.67), and (2.73), we find for 0 < \beta\leq 1 ,
\begin{equation} \int_0^1(u_x^2+v_x^2) \mathrm{d} x+\int_0^T\int_0^1(u_{xx}^2+v_{xx}^2+\theta_x^2) \mathrm{d} x \mathrm{d} t\leq C. \end{equation} | (2.74) |
By virtue of (1.12)–(1.16), (2.1), Corollary 2.1, Lemma 2.4, (2.69), and (2.74), it follows that
\begin{equation} \int_0^1\tau_t^2 \mathrm{d} x+\int_0^T\int_0^1(u_t^2+v_t^2+w_t^2+\tau_{tx}^2) \mathrm{d} x \mathrm{d} s\leq C. \end{equation} | (2.75) |
To obtain the first-order derivative estimate of the temperature, we need to first establish the uniform upper and lower bounds of \theta .
Lemma 2.7. Assume that the conditions listed in Lemma 2.1 hold. Then for T\geq0 ,
\begin{equation*} C_1^{-1}\leq\theta\leq C_1. \end{equation*} |
Proof. First of all, multiplying (1.16) by \theta , and integrating over [0, 1] about x , yields
\begin{equation} \begin{split} &\frac{c_v}{2}\frac{ \mathrm{d} }{ \mathrm{d} t}\int_{0}^{1}\theta^2 \mathrm{d} x + \int_{0}^{1}\frac{\tilde{\kappa}r^{2}\theta^\beta\theta_x^2}{\tau} \mathrm{d} x \\&\quad = \int_0^1\theta Q \mathrm{d} x-R\int_0^1\frac{\theta^2(r u)_x}{\tau} \mathrm{d} x \\&\quad \leq C\|{(u, u_x, v, v_x, w_x)}\|_\infty^2\int_0^1\theta \mathrm{d} x+\|u_x\|_\infty^2\int_0^1\theta^2 \mathrm{d} x. \end{split} \end{equation} | (2.76) |
Applying Gronwall's inequality to (2.76), we can obtain
\begin{equation} \int_{0}^{1}\theta^2 \mathrm{d} x + \int_{0}^{T}\int_{0}^{1}\theta^\beta\theta_x^2 \mathrm{d} x \mathrm{d} t\leq C. \end{equation} | (2.77) |
Based on the estimate above, we can get the bound of \int_{0}^{1}\theta^\beta\theta_x^2 \mathrm{d} x which will be used to obtain the upper bound of \theta . Multiplying (1.16) by \theta^\beta \theta_t and integrating over (0, 1) about x , it follows that
\begin{equation} c_v\int_{0}^{1}\theta^\beta\theta_t^2 \mathrm{d} x +R \int_{0}^{1}\frac{\theta^{\beta+1}\theta_t(ru)_x}{\tau} \mathrm{d} x -\int_0^1\theta^\beta\theta_t Q \mathrm{d} x = \int_0^1\left(\frac{\tilde{\kappa}r^{2}\theta^{\beta}\theta_x}{\tau}\right)_x\theta^\beta\theta_t \mathrm{d} x. \end{equation} | (2.78) |
By integration by parts, one has
\begin{equation} \begin{split} &\int_0^1\left(\frac{\tilde{\kappa}r^{2}\theta^{\beta}\theta_x}{\tau}\right)_x\theta^\beta\theta_t \mathrm{d} x \\&\quad = -\int_0^1\frac{\tilde{\kappa}r^{2}\theta^{\beta}\theta_x}{\tau}\left(\theta^\beta\theta_x\right)_t \mathrm{d} x \\&\quad = -\frac{\tilde{\kappa}}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\frac{r^{2}}{\tau}(\theta^\beta\theta_x)^2 \mathrm{d} x +\frac{\tilde{\kappa}}{2}\int_0^1\left(\frac{2ru}{\tau}-\frac{ru}{\tau}-\frac{r^3u_x}{\tau^2}\right)(\theta^\beta\theta_x)^2 \mathrm{d} x. \end{split} \end{equation} | (2.79) |
Inserting (2.79) into (2.78), we can deduce that
\begin{equation} \begin{split} &\frac{\tilde{\kappa}}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\frac{r^{2}}{\tau}(\theta^\beta\theta_x)^2 \mathrm{d} x +c_v\int_0^1\theta^\beta\theta_t^2 \mathrm{d} x \\&\quad = -R\int_0^1\frac{\theta^{\beta+1}\theta_t(r u)_x}{\tau} \mathrm{d} x +\int_0^1\theta^\beta\theta_tQ \mathrm{d} x +\frac{\tilde{\kappa}}{2}\int_0^1\left(\frac{ru}{\tau}-\frac{r^3u_x}{\tau^2}\right)(\theta^\beta\theta_x)^2 \mathrm{d} x \\&\quad \leq \frac{c_v}{2}\int_0^1\theta^\beta\theta_t^2 \mathrm{d} x +C\int_0^1\theta^{\beta+2}(u^2+u_x^2) \mathrm{d} x \\&\qquad +C\int_0^1\theta^\beta\Big(u^4+u_x^4+v^4+v_x^4+w_x^4\Big) \mathrm{d} x +C\|(u, u_x)\|_\infty\int_0^1(\theta^\beta\theta_x)^2 \mathrm{d} x \\&\quad \leq \frac{c_v}{2}\int_0^1\theta^\beta\theta_t^2 \mathrm{d} x +C\|(u^2, u_x^2, u^4, u_x^4, v^4, v_x^4, w_x^4)\|_\infty +C\left(\int_0^1\theta^\beta\theta_t^2 \mathrm{d} x\right)^2. \end{split} \end{equation} | (2.80) |
By Sobolev's inequality, Corollary 2.1, and Lemmas 2.5–2.6, one can find that
\begin{equation} \int_0^T\|(u^2, u_x^2, u^4, u_x^4, v^4, v_x^4, w_x^4)\|_\infty \mathrm{d} s\leq C. \end{equation} | (2.81) |
By virtue of (2.80), Grönwall's inequality, and (2.81), we can obtain
\begin{equation} \int_0^1(\theta^\beta\theta_x)^2 \mathrm{d} x+\int_0^T\int_0^1\theta^\beta\theta_t^2 \mathrm{d} x \mathrm{d} s\leq C. \end{equation} | (2.82) |
Thanks to (2.82), it follows that
\begin{equation} \|\theta^{\beta+1}-\bar{\theta}^{\beta+1}\|_\infty\leq C\left(\int_0^1(\theta^\beta\theta_x)^2 \mathrm{d} x\right)^{\frac{1}{2}}\leq C. \end{equation} | (2.83) |
That is, for t\geq0 ,
\begin{equation} \|\theta\|_\infty\leq C. \end{equation} | (2.84) |
Thanks to (2.77) and (2.84), one has
\begin{equation} \begin{split} &\int_0^T\int_0^1(\theta^{\beta+1}-\bar{\theta}^{\beta+1})^2 \mathrm{d} x \mathrm{d} t \\&\quad \leq \int_0^T\int_0^1\theta^{2\beta}\theta_x^2 \mathrm{d} x \mathrm{d} t \\&\quad \leq C\sup\limits_{0\leq t\leq T}\|\theta\|_\infty^\beta\int_0^T\int_0^1\theta^{\beta}\theta_x^2 \mathrm{d} x \mathrm{d} t \\&\quad \leq C. \end{split} \end{equation} | (2.85) |
Combining (2.83) and (2.84), one has
\begin{equation} \begin{split} &\int_0^T\left|\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1(\theta^{\beta+1}-\bar{\theta}^{\beta+1})^2 \mathrm{d} x\right| \mathrm{d} t \\&\quad \leq C\int_0^T\int_0^1(\theta^{\beta+1}-\bar{\theta}^{\beta+1})^2 \mathrm{d} x \mathrm{d} t + C\int_0^T\|\theta^\beta\theta_t\|^2 \mathrm{d} t \\&\quad \leq C\sup\limits_{0\leq t\leq T}\|\theta\|_\infty^\beta\int_0^T\int_0^1\theta^{\beta}\theta_x^2 \mathrm{d} x \mathrm{d} t \\&\quad \leq C. \end{split} \end{equation} | (2.86) |
So, from (2.83), (2.85), and (2.86), one has
\lim\limits_{t\rightarrow +\infty}\int_0^1(\theta^{\beta+1}-\bar{\theta}^{\beta+1})^2 \mathrm{d} x = 0. |
From (2.83), when t\rightarrow +\infty ,
\|(\theta^{\beta+1}-\bar{\theta}^{\beta+1})\|_\infty^2\leq C\|(\theta^{\beta+1}-\bar{\theta}^{\beta+1})\| \|\theta^\beta\theta_x\|\rightarrow 0, |
and we can obtain that there exists some time T_0\gg1 such that when t > T_0 ,
\begin{equation} \theta(x, t)\geq \frac{\gamma_1}{2}. \end{equation} | (2.87) |
Fixing T_0 in (2.87), multiplying (1.16) by \theta^{-p} , p > 2 , and integrating over [0, 1] about x yield
\begin{equation*} \begin{split} &\frac{c_v}{p-1}\frac{ \mathrm{d}}{ \mathrm{d} t}\|\theta^{-1}\|_{p-1}^{p-1} + p\int_0^1\frac{\tilde{\kappa}r^2\theta^\beta\theta_x^2}{\tau\theta^{p+1}} \mathrm{d} x +\int_0^1\frac{Q}{\theta^p} \mathrm{d} x \\&\quad = R\int_0^1\frac{\theta}{\tau \theta^p}(ru)_x \mathrm{d} x \\&\quad \leq \frac{1}{2}\int_0^1\frac{u^2+u_x^2}{\tau \theta^p} \mathrm{d} x +C\|\theta^{-1}\|_{L^{p-1}}^{p-2}. \end{split} \end{equation*} |
Hence,
\frac{ \mathrm{d}}{ \mathrm{d} t}\|\theta^{-1}\|_{L^{p-1}}\leq C, |
where C is a generic positive constant independent of p . Thus, integrating the above inequality over (0, t) and letting p\rightarrow \infty , we arrive that
\theta^{-1}(x, t)\leq C(T_0+1)\leftrightarrow \theta (x, t) \geq [C(T_0+1)]^{-1}, \forall (x, t)\in [0, 1]\times [T_0, +\infty). |
The proof of Lemma 2.7 is complete.
Lemma 2.8. Assume that the conditions listed in Lemma 2.1 hold. Then for T\geq0 ,
\begin{equation*} \int_0^1\theta_x^2 \mathrm{d} x+\int_0^T\int_0^1(\theta_{xx}^2+\theta_t^2) \mathrm{d} x \mathrm{d} s\leq C_5. \end{equation*} |
Proof. Multiplying (1.16) by \theta_{xx} , integrating over [0, 1] on x , and by Hölder's, Poincaré's, and Cauchy-Schwarz's inequalities, Corollary 2.1, Lemma 2.4, and Lemma 2.7, we have
\begin{equation} \begin{split} &\frac{c_v}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1 \theta_x^2 \mathrm{d} x +\int_0^1\frac{\kappa r^{2} \theta_{xx}^2}{\tau} \mathrm{d} x \\&\quad = \int_0^1 \theta_{xx}\left[\frac{R \theta}{\tau}(r u)_x- \theta_x\left(\frac{\kappa r^{2}}{\tau}\right)_x-Q\right] \mathrm{d} x \\&\quad \leq \varepsilon\int_0^1 \theta_{xx}^2 \mathrm{d} x+C( \varepsilon)\int_0^1\left[ \theta^2(r u)_x^2- \theta_x^2\left(\frac{\kappa r^{2}}{\tau}\right)_x^2-Q^2\right] \mathrm{d} x \\&\quad \leq \varepsilon\int_0^1 \theta_{xx}^2 \mathrm{d} x+C( \varepsilon)\|u_x\|^2\|\theta\|_\infty^2 +C( \varepsilon)\| \theta_x\|^2+C( \varepsilon)\|\theta_x\|_\infty^2\|\tau_x\|^2 \\&\qquad +C( \varepsilon)\int_0^1(u^4+v^4+u_x^4+v_x^4+w_x^4) \mathrm{d} x \\&\quad \leq \varepsilon\| \theta_{xx}\|^2 +C( \varepsilon)\Big(\|u_x\|^2 +\| \theta_x\|^2+\| \theta_x\|\| \theta_{xx}\| +\|u\|^2\|u\|^2+\|v\|_\infty^2\|v\|^2\Big) \\&\qquad +C( \varepsilon)\Big(\|u_x\|_\infty^2\|u_x\|^2 +\|v_x\|_\infty^2\|v_x\|^2 +\|w_x\|_\infty^2\|w_x\|^2\Big) \\&\quad \leq \varepsilon\| \theta_{xx}\|^2+ C( \varepsilon)\|(u_x, v_x, w_x, u_{xx}, v_{xx}, w_{xx})\|^2+C( \varepsilon)\| \theta_x\|^2. \end{split} \end{equation} | (2.88) |
Choosing \varepsilon suitably small in (2.88) gives
\begin{equation} \frac{c_v}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1 \theta_x^2 \mathrm{d} x +c\int_0^1 \theta_{xx}^2 \mathrm{d} x \leq C\|(u_x, v_x, w_x)\|_1^2+C\| \theta_x\|^2. \end{equation} | (2.89) |
Integrating (2.89) and using Lemmas 2.5–2.6, one has
\begin{equation} \| \theta_x(t)\|^2+\int_0^T\| \theta_{xx}\|^2 \mathrm{d} s\leq C. \end{equation} | (2.90) |
Hence, similar to (2.75), by means of (1.16), Corollary 2.1, Lemmas 2.4–2.7, and (2.90), one can deduce that
\begin{equation*} \int_0^T\int_0^1\theta_t^2 \mathrm{d} x \mathrm{d} t\leq C. \end{equation*} |
Next, we derive the second-order derivatives estimates of (\tau, u, v, w, \theta) .
Lemma 2.9. Assume that the conditions listed in Lemma 2.1 hold. Then for T\geq0 ,
\begin{equation*} \begin{split} &\int_0^1(u_t^2+v_t^2+w_t^2+\theta_t^2+u_{xx}^2+v_{xx}^2+w_{xx}^2+\theta_{xx}^2+\tau_{xt}^2) \mathrm{d} x\\ &\quad+\int_0^T\int_0^1(u_{xt}^2+\tau_{tt}^2+v_{xt}^2+w_{xt}^2+\theta_{xt}^2) \mathrm{d} x \mathrm{d} s\leq C_6. \end{split} \end{equation*} |
Proof. Applying \partial_t to (1.13) and multiplying by u_t in L^2 , one has
\begin{equation} \begin{split} &\frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1u_t^2 \mathrm{d} x+\int_0^1\frac{\tilde\lambda r^2\theta^\alpha u_{xt}^2}{\tau} \mathrm{d} x \\&\quad = \int_0^1ru_{xt}\left[P_t-\left(\frac{\lambda}{\tau}\right)_t(ru)_x-\frac{\lambda}{\tau}\Big((ru)_{xt}-ru_{xt}\Big)\right] \mathrm{d} x-\int_0^1r_xu_t\left[\frac{\lambda(ru)_x}{\tau}\right]_t \mathrm{d} x \\&\qquad +\int_0^1u_t\left[\left(\frac{v^2}{r}\right)_t-r_tP_x+r_xP_t+r_t\left(\frac{\lambda(ru)_x}{\tau}\right)_x-2(u\mu_x)_t\right] \mathrm{d} x \\&\quad : = \sum\limits_{i = 1}^3III_i. \end{split} \end{equation} | (2.91) |
Applying \partial_t to (1.14) and multiplying by v_t in L^2 , one has
\begin{equation} \begin{split} &\frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1v_t^2 \mathrm{d} x +\int_0^1\frac{\tilde\mu r^2\theta^\alpha v_{xt}^2}{\tau} \mathrm{d} x \\&\quad = \int_0^1\bigg\{v_t\left[2\Big(\mu v_x\Big)_t-r_x\left(\frac{\mu rv_x}{\tau}\right)_t-\Big(\mu v\Big)_{xt}\right] -rv_{xt}v_x\left[\frac{\mu r}{\tau}\right]_t\bigg\} \mathrm{d} x \\&\qquad +\int_0^1v_t\left[r_t\left(\frac{\mu rv_x}{\tau}\right)_x-\left(\frac{uv}{r}\right)_t -\left(\frac{\mu\tau v}{r^2}\right)_t\right] \mathrm{d} x \\&\quad : = \sum\limits_{i = 4}^5III_i. \end{split} \end{equation} | (2.92) |
Applying \partial_t to (1.15) and multiplying by w_t in L^2 , one has
\begin{equation} \begin{split} &\frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1w_t^2 \mathrm{d} x +\int_0^1\frac{\tilde\mu r^2\theta^\alpha w_{xt}^2}{\tau} \mathrm{d} x \\&\quad = \int_0^1w_tr_t\left(\frac{\mu rw_x}{\tau}\right)_x \mathrm{d} x -\int_0^1\bigg\{rw_xw_{xt}\left(\frac{\mu r}{\tau}\right)_t+w_t\left[r_x\left(\frac{\mu rw_x}{\tau}\right)_t-\Big(\mu w_x\Big)_t\right]\bigg\} \mathrm{d} x \\&\quad : = \sum\limits_{i = 6}^7III_i. \end{split} \end{equation} | (2.93) |
Adding (2.91)–(2.93) together, we get
\begin{equation} \frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1(u_t^2+v_t^2+w_t^2) \mathrm{d} x +\int_0^1\frac{\tilde\lambda r^2\theta^\alpha u_{xt}^2+\tilde\mu r^2\theta^\alpha v_{xt}^2+\tilde\mu r^2\theta^\alpha w_{xt}^2}{\tau} \mathrm{d} x = \sum\limits_{i = 1}^7III_i. \end{equation} | (2.94) |
Before the computations of III_1 to III_7 , we need to keep in mind the following facts:
\begin{eqnarray*} &&\|(u, v, w)\|_\infty \leq C, \\ &&a\leq r\leq b, \quad r_x = r^{-1}\tau, \quad r_t = u, \quad r_{tx} = u_x, \\ &&C^{-1}\leq\tau\leq C, \quad C^{-1}\leq\theta\leq C, \\ &&|(ru)_x|\leq C|(u, u_x)|, \quad |(ru)_{xt}-ru_{xt}|\leq C|(u^2, u_t, uu_x)|, \\ &&|(ru)_{xt}|\leq C|(u^2, u_t, uu_x, u_{xt})|. \end{eqnarray*} |
Then, by Hölder's, Sobolev's, and Cauchy-Schwarz's inequalities, one has
\begin{equation} \begin{split} III_1&\leq C\int_0^1|u_{xt}||(\theta_t, \tau_t, \theta_tu_x, \tau_tu_x, u, u_t, u_x)| \mathrm{d} x \\& \leq \varepsilon\|u_{xt}\|^2+C( \varepsilon)\|(\theta_t, \tau_t, u, u_t)\|^2+C( \varepsilon)\|(\theta_t, \tau_t)\|_\infty^2\|u_x\|^2 \\& \leq \varepsilon\|u_{xt}\|^2+\delta\|\theta_{xt}\|^2 +C( \varepsilon, \delta)\|(\theta_t, \tau_t, u, u_t, \tau_{xt})\|^2, \end{split} \end{equation} | (2.95) |
and
\begin{equation} \begin{split} III_2&\leq C\int_0^1|u_t||(\theta_t, \theta_tu_x, u^2, u_t, u_x, u_{xt}, \tau_t, \tau_t^2)| \mathrm{d} x \\&\leq \varepsilon\|u_{xt}\|^2 +C( \varepsilon)\|(u_t, \theta_t, u, u_x, \tau_t)\|^2 + \varepsilon\|\theta_t\|_\infty^2\|u_x\|^2+C( \varepsilon)\|\tau_t\|_\infty^2\|\tau_t\|^2 \\& \leq \varepsilon\|u_{xt}\|^2+\delta\|\theta_{xt}\|^2 +C( \varepsilon, \delta)\|(u_t, \theta_t, u, u_x, \tau_t, \tau_{xt})\|^2. \end{split} \end{equation} | (2.96) |
By virtue of (1.13), one has
\begin{equation*} \left|\left(\frac{\lambda(ru)_x}{\tau}\right)_x\right| \leq C \Big|\Big(u_t, v^2, \theta_x, \tau_x\Big)\Big|. \end{equation*} |
Thus, it follows from Hölder's, Sobolev's, and Cauchy-Schwarz's inequalities that
\begin{equation} \begin{split} III_3&\leq C\int_0^1|u_t||(v_t, v^2, \theta_x, \tau_x, \tau_t, \theta_t, u_t, u_t\theta_x, \theta_x\theta_t, \theta_{xt})| \mathrm{d} x\\&\leq \varepsilon\|\theta_{xt}\|^2+C( \varepsilon)\|(v_t, u_t, v, \theta_x, \theta_t, \theta_{t}, \tau_x, \tau_t)\|^2 + \varepsilon\|(u_t, \theta_t)\|_\infty^2\|\theta_x\|^2\\ &\leq \varepsilon\|(u_{xt}, \theta_{xt})\|^2 +C( \varepsilon)\|(v_t, u_t, \theta_t, \tau_t, v, \theta_x, \tau_x)\|^2, \end{split} \end{equation} | (2.97) |
and
\begin{equation} \begin{split} III_4&\leq C\int_0^1|v_t||(\theta_tv_x, v_x, v_{xt}, v_x\tau_t, \theta_x\theta_t, \theta_{xt}, \theta_xv_t)| \mathrm{d} x +C\int_0^1|v_{xt}v_x||(\theta_t, v, \tau_t)| \mathrm{d} x \\&\leq \frac{ \varepsilon}{2}\|(v_{xt}, \theta_{xt})\|^2 +C( \varepsilon)\|(v_t, v_x)\|^2+C( \varepsilon)\|(\theta_t, \tau_t, v_t)\|_\infty^2\|(v_x, \theta_x)\|^2 \\&\leq \varepsilon\|(v_{xt}, \theta_{xt})\|^2 +C( \varepsilon)\|(v_t, v_x, \theta_t, \tau_t, \tau_{tx})\|^2. \end{split} \end{equation} | (2.98) |
It follows from (1.14) that
\begin{equation*} \left|\left(\frac{\mu rv_x}{\tau}\right)_x\right| \leq C|(v_t, v, v_x, \theta_x)|. \end{equation*} |
Then
\begin{equation} III_5\leq C\int_0^1|v_t||(v_t, v, \theta_x, v_x, u_t, \theta_t, \tau_t)| \mathrm{d} x \leq C\|(v_t, v, \theta_x, v_x, u_t, \theta_t, \tau_t)\|^2. \end{equation} | (2.99) |
By virtue of (1.15), we can obtain
\begin{equation} III_6\leq C\int_0^1|w_t||u|\left|\left(\frac{\mu rw_x}{\tau}\right)_x\right| \mathrm{d} x \leq C\int_0^1|w_t|\left|(w_t, w_x)\right| \mathrm{d} x \leq C\|(w_t, w_x)\|^2, \end{equation} | (2.100) |
and
\begin{equation} \begin{split} III_7&\leq C\int_0^1|w_{xt}|\Big|(w_x\theta_t, w_x\tau_t, w_x)\Big| +|w_t|\Big|(\theta_tw_x, \tau_tw_x, w_x, w_{xt})\Big| \mathrm{d} x \\&\leq \varepsilon\|w_{xt}\|^2+C( \varepsilon)\|(w_x, w_t)\|^2+C( \varepsilon)\|(\theta_t, \tau_t)\|_\infty^2\|w_x\|^2 \\&\leq \varepsilon\|(w_{xt}, \theta_{xt})\|^2 +C( \varepsilon)\|(w_x, w_t, \theta_t, \tau_t, \tau_{xt})\|^2. \end{split} \end{equation} | (2.101) |
Putting (2.95)–(2.101) into (2.94) gives
\begin{equation} \begin{split} &\frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\|(u_t, v_t, w_t)\|^2+c\|(u_{xt}, v_{xt}, w_{xt})\|^2 \\&\quad\leq \varepsilon\|(u_{xt}, v_{xt}, w_{xt}, \theta_{xt})\|^2+C( \varepsilon)\|(u_t, v_t, w_t, \theta_t, w_x, \tau_x, \theta_x)\|^2 +C( \varepsilon)\|(\tau_t, u, v)\|_1^2. \end{split} \end{equation} | (2.102) |
Applying \partial_t to (1.16) and multiplying by \theta_t in L^2 , it follows that
\begin{equation} \begin{split} &\frac{c_v}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\theta_t^2 \mathrm{d} x +\int_0^1\frac{\tilde{\kappa}\theta^\beta r^2\theta_{xt}^2}{\tau} \mathrm{d} x \\&\quad = \int_0^1\theta_t\left[Q_t-\Big(P(ru)_x\Big)_t\right]-\theta_x\theta_{xt}\left(\frac{\kappa r^2}{\tau}\right)_t \mathrm{d} x. \end{split} \end{equation} | (2.103) |
First of all, by means of the definition of Q , one has
\begin{equation} \begin{split} |\theta_tQ_t| \leq& C|\theta_t||(u, \tau_t, u_t, u_x, u_{xt}, u_x\tau_t, u_xu_t, u_x^2, u_xu_{xt})| \\&+C|\theta_t||(\theta_t, \theta_tu_x^2, \tau_tu_x^2, \theta_tw_x^2, w_x^2, \tau_tw_x^2, w_xw_{xt}, \theta_tv_x^2, \tau_tv_x^2)| \\&+C|\theta_t||(v_x^2, v_{xt}, \tau_tv_x, v_t, v, v_x, v_xv_{xt}, v_xv_t)| \\\leq& C( \varepsilon)|(\theta_t, u, \tau_t, u_t, u_x, w_x, v_x, v_t, v, v_x)|^2+ \varepsilon|(u_{xt}, w_{xt}, v_{xt})|^2 \\&+C( \varepsilon)|(\tau_t, u_t, u_x)|^2|(u_x, w_x, v_x)|^2 +C( \varepsilon)|\theta_t|^2|(u_x, v_x, w_x)|^2. \end{split} \end{equation} | (2.104) |
Using (2.104) and Sobolev's inequality, we can derive from (2.103) that
\begin{equation} \begin{split} &\frac{c_v}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\theta_t^2 \mathrm{d} x +\int_0^1\frac{\tilde{\kappa}\theta^\beta r^2\theta_{xt}^2}{\tau} \mathrm{d} x \\&\quad\leq C\int_0^1\left(|\theta_t||(Q_t, \theta_t, \tau_t, \theta_tu_x, \tau_tu_x, u, u_t, u_x, u_{xt})| +|\theta_x||(\theta_{xt}\theta_t, \theta_{xt}, \theta_{xt}\tau_t)|\right) \mathrm{d} x \\&\quad \leq C( \varepsilon)\|(\theta_t, u, \tau_t, u_t, u_x, \theta_x, w_x, v_x, v_t, v, v_x)\|^2 + \varepsilon\|(u_{xt}, w_{xt}, v_{xt}, \theta_{xt})\|^2 \\&\qquad +C( \varepsilon)\|(\tau_t, \theta_t, u_t, u_x)\|_\infty^2\|(u_x, w_x, v_x, \theta_x)\|^2 +C( \varepsilon)\|\tau_t\|_\infty^2\|\theta_t\|^2 \\&\quad \leq C( \varepsilon)\|(\theta_t, u, \tau_t, u_t, u_x, w_x, v_x, v_t, v, \tau_{tx}, u_{xx})\|^2 + \varepsilon\|(u_{xt}, w_{xt}, v_{xt}, \theta_{xt})\|^2 \\&\qquad +C( \varepsilon)\|(u_x, v_x, w_x)\|_{1}^2\|\theta_t\|^2 +C( \varepsilon)\|\tau_t\|_{1}^2\|\theta_t\|^2. \end{split} \end{equation} | (2.105) |
Adding (2.102) to (2.105) and choosing \varepsilon > 0 suitably small, it follows that
\begin{equation} \begin{split} &\frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\|(\sqrt{c_v}\theta_t, u_t, v_t, w_t)\|^2 +c\|(u_{xt}, v_{xt}, w_{xt}, \theta_{xt})\|^2 \\&\quad\leq C\|(u_t, v_t, w_t, \theta_t, w_x, \tau_x, \theta_x)\|^2 +C\|(u_x, \tau_t, v)\|_{1}^2 \\&\qquad +C\|(u_x, v_x, w_x)\|_{1}^2\|\theta_t\|^2 +C\|\tau_t\|_{1}^2\|\theta_t\|^2. \end{split} \end{equation} | (2.106) |
By means of (2.106) and Grönwall's inequality, we deduce
\begin{equation} \|(u_t, v_t, w_t, \theta_t)\|^2+\int_0^T\|(u_{xt}, v_{xt}, w_{xt}, \theta_{xt})\|^2 \mathrm{d} s\leq C. \end{equation} | (2.107) |
According to (1.13), one has
\begin{equation*} \frac{\lambda r^2u_{xx}}{\tau} = u_t-\frac{v^2}{r}+rP_x+2u\mu_x-r\left[\left(\frac{\lambda(ru)_x}{\tau}\right)_x-\frac{\lambda ru_{xx}}{\tau}\right], \end{equation*} |
which means that
\begin{equation*} |u_{xx}|\leq C|(u_t, v, \theta_x, \tau_x, \theta_xu_x, \tau_xu_x, u, u_x)|. \end{equation*} |
Hence, by means of (2.107), we obtain
\begin{equation*} \|u_{xx}\|^2\leq C. \end{equation*} |
Similarly, use the equations (1.12)–(1.16), we also can derive
\begin{equation} \|(v_{xx}, w_{xx}, \theta_{xx}, \tau_{tx})\|^2+\int_0^T\|\tau_{tt}\|^2 \mathrm{d} s\leq C_7. \end{equation} | (2.108) |
Here, we omit the details of (2.108). The proof of Lemma 2.9 is complete.
Lemma 2.10. Assume that the conditions listed in Lemma 2.1 hold. Then for T\geq0 ,
\begin{equation*} \int_0^1\tau_{xx}^2 \mathrm{d} x+\int_0^T\int_0^1(\tau_{xx}^2+\tau_{xxt}^2+u_{xxx}^2+v_{xxx}^2+w_{xxx}^2+\theta_{xxx}^2) \mathrm{d} x \mathrm{d} s\leq C_7. \end{equation*} |
Proof. Apply \partial_x to (2.45) and multiply by (\lambda\tau_x/\tau)_x in L^2 to get
\begin{equation} \begin{split} &\frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\left(\frac{\lambda\tau_x}{\tau}\right)_x^2 \mathrm{d} x +\int_0^1\frac{R\theta}{\lambda\tau}\left(\frac{\lambda\tau_x}{\tau}\right)_x^2 \mathrm{d} x \\&\quad = \int_0^1\left(\frac{\lambda\tau_x}{\tau}\right)_x\left[\frac{\lambda_\theta}{\tau}(\tau_x\theta_t-\tau_t\theta_x)\right]_x \mathrm{d} x +\int_0^1\left(\frac{\lambda\tau_x}{\tau}\right)_x\left(\frac{u_t}{r}-\frac{v^2}{r^2}+\frac{2u\mu_x}{r}\right)_x \mathrm{d} x \\&\qquad -R\int_0^1\left(\frac{\lambda\tau_x}{\tau}\right)_x\left[\frac{2\theta_x\tau_x}{\tau^2}-\frac{\theta_{xx}}{\tau} -\frac{2\theta\tau_x^2}{\tau^3}-\frac{\theta\tau_x}{\lambda\tau}\left(\frac{\lambda}{\tau}\right)_x\right] \mathrm{d} x \\&\quad \leq C( \varepsilon)\int_0^1\Big[ |(\tau_x, \theta_x)|^2|(\tau_x\theta_t, \tau_t\theta_x)|^2 +|(\tau_{xx}\theta_t, \tau_x\theta_{xt}, \tau_{tx}\theta_x, \tau_t\theta_{xx})|^2 \Big] \mathrm{d} x \\&\qquad +C( \varepsilon)\int_0^1\Big[ |(u_{xt}, u_t, v_x, v, u_x\theta_x, \theta_{xx}, \theta_x^2, \theta_x)|^2 +|(\theta_{xx}, \tau_x^2, \theta_x\tau_x)|^2\Big] \mathrm{d} x \\&\qquad + \varepsilon\int_0^1\left(\frac{\lambda\tau_x}{\tau}\right)_x^2 \mathrm{d} x \\&\quad : = \sum\limits_{i = 8}^9III_i+ \varepsilon\int_0^1\left(\frac{\lambda\tau_x}{\tau}\right)_x^2 \mathrm{d} x, \end{split} \end{equation} | (2.109) |
where the following fact has been used:
\begin{equation*} \begin{split} \left(\frac{\theta}{\tau}\right)_{xx} & = \frac{\theta_{xx}}{\tau}-2\frac{\theta_x\tau_x}{\tau^2}+2\frac{\theta\tau_x^2}{\tau^3}-\frac{\theta\tau_{xx}}{\tau^2} \\& = \frac{\theta_{xx}}{\tau}-2\frac{\theta_x\tau_x}{\tau^2}+2\frac{\theta\tau_x^2}{\tau^3} -\frac{\theta}{\lambda\tau}\left[\left(\frac{\lambda\tau_x}{\tau}\right)_x-\frac{\lambda_x\tau_x}{\tau} +\frac{\lambda\tau_x^2}{\tau^2}\right]. \end{split} \end{equation*} |
By Sobolev's inequality and Lemmas 2.6–2.9, we have
\begin{equation} \begin{split} III_8 \leq& C( \varepsilon)\Big( \|(\tau_x, \theta_x)\|_\infty^4\|(\tau_t, \theta_t)\|^2 +\|\theta_t\|_\infty^2\|\tau_{xx}\|^2 \\& +\|\tau_x\|_\infty^2\|\theta_{xt}\|^2 +\|\tau_t\|_\infty^2\|\theta_{xx}\|^2 +\|\theta_x\|_\infty^2\|\tau_{xt}\|^2 \Big) \\\leq& C( \varepsilon) \Big( \|(\tau_x, \theta_x)\|^4 +\|(\tau_x, \theta_x)\|^2\|(\tau_{xx}, \theta_{xx})\|^2 \\&+\|\theta_t\|_{1}^2\|\tau_{xx}\|^2 +\|\theta_{xt}\|^2\|\tau_x\|^2 +\|\tau_t\|_{1}^2+\|\theta_x\|_{1}^2 \Big) \\\leq& C( \varepsilon)\Big(\|(\tau_x, \theta_x, \tau_t)\|^2+\|\theta_t\|^2\|\tau_{xx}\|^2\Big), \end{split} \end{equation} | (2.110) |
and
\begin{equation} \begin{split} III_9 &\leq C( \varepsilon)\|(u_{xt}, u_t, v_x, v, \theta_{xx}, \theta_x)\|^2 +C( \varepsilon)\|(u_x, \theta_x, \tau_x)\|_\infty^2\|(\theta_x, \tau_x)\|^2 \\&\leq \varepsilon \|\tau_{xx}\|^2+C( \varepsilon)\|(u_{xt}, u_t, v_x, v, \theta_{xx}, \theta_x, \tau_x)\|^2. \end{split} \end{equation} | (2.111) |
Noting that
\begin{equation*} |\tau_{xx}|\leq C\left|\left(\frac{\lambda\tau_x}{\tau}\right)_x\right| +C\Big|\Big(\theta_x\tau_x, \tau_x^2\Big)\Big|, \end{equation*} |
we can derive from Sobolev's inequality and Lemma 2.4 that
\begin{equation*} \begin{split} \|\tau_{xx}\|^2&\leq C\left\|\left(\frac{\lambda\tau_x}{\tau}\right)_x\right\|^2 +C\|\theta_x\|_\infty^2\|\tau_x\|^2+C\|\tau_x\|_4^4 \\&\leq C\left\|\left(\frac{\lambda\tau_x}{\tau}\right)_x\right\|^2 +C\|\theta_x\|_{1}^2+C\|\tau_x\|^4+C\|\tau_x\|^3\|\tau_{xx}\|. \end{split} \end{equation*} |
So, it follows from Cauchy-Schwarz's inequality and Lemma 2.4 that
\begin{equation} \|\tau_{xx}\|^2\leq C\|\theta_{x}\|_{1}^2+C\|\tau_x\|^2+C\left\|\left(\frac{\lambda\tau_x}{\tau}\right)_x\right\|^2. \end{equation} | (2.112) |
Taking \varepsilon suitably small, putting (2.110)–(2.111) into (2.109), and using Lemmas 2.4, 2.8, and 2.9, we find
\begin{equation} \begin{split} &\frac{1}{2}\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\left(\frac{\lambda\tau_x}{\tau}\right)_x^2 \mathrm{d} x +c\int_0^1\left(\frac{\lambda\tau_x}{\tau}\right)_x^2 \mathrm{d} x \\&\quad\leq C\|(\tau_t, \theta_t, \theta_{x}, u_t, v)\|_{1}^2+C\|\tau_x\|^2+C\|\theta_{t}\|_{1}^2\left\|\left(\frac{\lambda\tau_x}{\tau}\right)_x\right\|^2. \end{split} \end{equation} | (2.113) |
By (2.113), Grönwall's inequality, Corollary 2.1, and Lemmas 2.6 and 2.8–2.9, one obtains
\begin{equation} \int_0^1\left(\frac{\lambda\tau_x}{\tau}\right)_x^2 \mathrm{d} x +\int_0^T\int_0^1\left(\frac{\lambda\tau_x}{\tau}\right)_x^2 \mathrm{d} x \mathrm{d} s\leq C. \end{equation} | (2.114) |
It follows from (2.112) and (2.114) that
\begin{equation} \|\tau_{xx}\|^2+\int_0^T\|\tau_{xx}\|^2 \mathrm{d} s\leq C. \end{equation} | (2.115) |
Letting \partial_x act on (1.13) gives
\begin{equation} \begin{split} &\frac{\tilde\lambda\theta^2 r^2u_{xxx}}{\tau}+r_x\left(\frac{\lambda\tau_t}{\tau}\right)_x +r\left[\left(\frac{\lambda}{\tau}\right)_x\tau_t\right]_x+r\left(\frac{\lambda}{\tau}\right)_x(ru)_{xx} \\&\quad = u_{xt}-\left(\frac{v^2}{r}\right)_x+(rP_x)_x+2\Big(u\mu_x\Big)_x-\frac{r\lambda}{\tau}[(ru)_{xxx}-ru_{xxx}]. \end{split} \end{equation} | (2.116) |
It follows from (2.115) and (2.116) that
\begin{equation*} \begin{split} &\int_0^T\|u_{xxx}\|^2 \mathrm{d} s \\&\quad\leq C\int_0^T\|(\theta_x\tau_t, \tau_{xt}, \tau_t\tau_x)\|^2 \mathrm{d} s +C\int_0^T\|(\theta_x^2\tau_t, \theta_{xx}\tau_t, \theta_x\tau_{tx}, \theta_x\tau_t\tau_x)\|^2 \mathrm{d} s \\&\qquad+C\int_0^T\|(\tau_{xx}\tau_t, \tau_x\tau_{tx}, \tau_x^2\tau_t)\|^2 \mathrm{d} s +C\int_0^T\|(\theta_x, \theta_x\tau_x, \theta_xu_x, \theta_xu_{xx})\|^2 \mathrm{d} s \\&\qquad+C\int_0^T\|(u_{xt}, v_x, v, \theta_x, \tau_x, \theta_{xx}, \theta_x\tau_x, \tau_{xx})\|^2 \mathrm{d} s +C\int_0^T\|(u_x\theta_x, \theta_x^2, \theta_{xx})\|^2 \mathrm{d} s \\&\qquad+C\int_0^T\|(u, \tau_x, \tau_{xx}, u_x, \tau_xu_x, u_{xx})\|^2 \mathrm{d} s \\&\quad\leq C\int_0^T\|(\tau_t, \tau_{xx}, \tau_{tx}, \theta_x, u_x, u_{xt}, v_x, v, \theta_{xx}, u, \tau_x, u_{xx})\|^2 \mathrm{d} s \\&\quad \leq C, \end{split} \end{equation*} |
where the following fact has been used:
\begin{equation*} \|(\theta_x, \tau_x, \theta_x^2, \tau_t, \theta_x\tau_t, \tau_x^2)\|_\infty \leq C+C\|(\theta_x, \tau_t, \tau_x)\|_{1}^2 \leq C. \end{equation*} |
Similarly, using (1.14)–(1.15), we also have
\begin{equation*} \int_0^T\|(v_{xxx}, w_{xxx})\|^2 \mathrm{d} s\leq C. \end{equation*} |
Letting \partial_x act on (1.16) gives
\begin{equation} \frac{\tilde{\kappa}\theta^\beta r^2\theta_{xxx}}{\tau} = c_v\theta_{xt}+(P\tau_t)_x-\left(\frac{\kappa r^2}{\tau}\right)_{xx}\theta_x-2\left(\frac{\kappa r^2}{ \tau}\right)_x\theta_{xx}-Q_x. \end{equation} | (2.117) |
It follows from (2.114) and (2.117) that
\begin{equation} \begin{split} &\int_0^T\|\theta_{xxx}\|^2 \mathrm{d} s \\&\quad\leq C\int_0^T\|(\theta_{xt}, \theta_x\tau_t, \tau_x\tau_t, \tau_{tx})\|^2 \mathrm{d} s \\&\qquad +C\int_0^T\|(\theta_x^3, \theta_{xx}\theta_x, \theta_x^2, \theta_x^2\tau_x, \theta_x\tau_x, \theta_x\tau_x^2, \theta_x\tau_{xx})\|^2 \mathrm{d} s \\&\qquad +C\int_0^T\|(\theta_{xx}, \tau_x\theta_{xx})\|^2+\|Q_x\|^2 \mathrm{d} s. \end{split} \end{equation} | (2.118) |
By the definition of Q , one has
\begin{equation} \begin{split} &\int_0^T\|Q_x\|^2 \mathrm{d} s \\&\quad \leq C\int_0^T\|(\theta_x, \theta_xu_x, u, \tau_x, u_x, u_{xx}, u_x\tau_x, u_x^2, u_xu_{xx})\|^2 \mathrm{d} s \\&\qquad +C\int_0^T\|(\theta_xw_x^2, w_x^2, w_xw_{xx}, w_x^2\tau_x)\|^2 \mathrm{d} s \\&\qquad +C\int_0^T\|(\theta_xv_x^2, \tau_xv_x^2, v_x^2, v_xv_{xx}, v_x^2\tau_x, v_x, v_{xx}, v_x\tau_x, v)\|^2 \mathrm{d} s. \end{split} \end{equation} | (2.119) |
Since the following estimates have been obtained:
\begin{equation*} \|(\theta_x, \tau_x, u_x, w_x, v_x)\|_\infty\leq C\|(\theta_x, \tau_x, u_x, w_x, v_x)\|_{1}\leq C, \end{equation*} |
putting (2.119) into (2.118) yields
\begin{equation*} \begin{split} &\int_0^T\|\theta_{xxx}\|^2 \mathrm{d} s \\&\quad \leq C\int_0^T\|(\theta_{xt}, \tau_t, \tau_{xt}, \theta_x, \theta_{xx}, \tau_x, \tau_{xx}, u_x, u, u_{xx}, w_x, w_{xx}, v_x, v_{xx}, v)\|^2 \mathrm{d} s \\&\quad\leq C. \end{split} \end{equation*} |
The proof of Lemma 2.10 is complete.
With all a priori estimates from Section 2 at hand, we can complete the proof of Theorem 1.1 . For this purpose, it will be shown that the existence and uniqueness of local solutions to the initial-boundary value problem (1.12)–(1.19) can be obtained by using the Banach theorem and the contractivity of the operator defined by the linearization of the problem on a small time interval.
Lemma 3.1. Letting (1.20) hold, then there exists T_0 = T_0(V_0, V_0, M_0) > 0 , depending only on \beta , V_0 , and M_0 , such that the initial boundary value problem (1.12)–(1.19) has a unique solution (\tau, u, v, w, \theta)\in X(0, T_0;\frac{1}{2}V_0, \frac{1}{2}V_0, 2M_0) .
Proof of Theorem 1.1: First, to prove Theorem 1.1, according to (1.20), one has
\begin{eqnarray*} &&\tau_0\geq V_0, \theta_0\geq V_0, \qquad \forall x\in \Omega, \\&&\|(\tau_0, u_0, v_0, w_0, \theta_0)\|_{H^2}\leq M_0. \end{eqnarray*} |
Combined with Lemma 3.1, there exists t_1 = T_0(V_0, V_0, M_0) such that (\tau, u, v, w, \theta)\in X(0, t_1;\frac{1}{2}V_0, \frac{1}{2}V_0, 2M_0).
We find the positive constant |\alpha| \leq \alpha_1 , where \alpha_1 satisfies
\begin{equation} \left(\frac{1}{2}V_0\right)^{-|\alpha_1|}\leq 2, \qquad (2M_0)^{|\alpha_1|}\leq 2, \qquad |\alpha_1|H(\frac{1}{2}V_0, \frac{1}{2}V_0, 2M_0)\leq \epsilon_1, \end{equation} | (3.1) |
where \epsilon_1 is chosen in Lemma 2.1. That means that one can choose
\begin{equation} |\alpha_1|: = \min\left\{\frac{\ln 2}{|\ln 2-\ln V_0|}, \frac{\ln 2}{|\ln 2+\ln M_0|}, \epsilon_1 H^{-1}\left(\frac{1}{2}V_0, \frac{1}{2}V_0, 2M_0\right)\right\}. \end{equation} | (3.2) |
One deduces from Lemmas 2.1–2.10 with T = t_1 that for each t\in [0, t_1] , the local solution (\tau, u, v, w, \theta) satisfies
\begin{equation} C_0^{-1}\leq v(x, t)\leq C_0, \qquad C_1^{-1}\leq \theta(x, t)\leq C_1, \qquad x\in (0, 1), \end{equation} | (3.3) |
and
\begin{equation} \sup\limits_{0\leq t\leq t_1}\|(\tau, u, v, w, \theta)\|_{2}^2+\int_{0}^{t_1}\|\theta_{t}\|^2 \mathrm{d} t\leq C_8^2, \end{equation} | (3.4) |
where C_i(i = 2, \cdots, 7) is chosen in Section 2 and C_8^2: = \sum_{i = 2}^{7}C_i . It follows from Lemma 2.9 and Lemma 2.10 that (\tau, u, v, w, \theta)\in C([0, T); H^2). If one takes (\tau, u, v, w, \theta)(\cdot, t_1) as the initial data and applies Lemma 3.1 again, the local solution (\tau, u, v, w, \theta) can be extended to the time interval [t_1, t_1+t_2] with t_2(C_0, C_1, C_8) such that (\tau, u, v, w, \theta)\in X(t_1, t_1+t_2; \frac{1}{2}C_0, \frac{1}{2}C_1, \frac{1}{2}C_8). Moreover, for all (x, t)\in [0, 1]\times [0, t_1+t_2] , one gets
\begin{equation*} \frac{1}{2}C_0\leq v(x, t), \qquad \frac{1}{2}C_1\leq \theta(x, t), \end{equation*} |
and
\begin{equation*} \sup\limits_{t_1\leq t\leq t_1+t_2}\|(\tau, u, v, w, \theta)\|_{2}^2+\int_{t_1}^{t_1+t_2}\|\theta_{t}\|^2 \mathrm{d} t\leq 4C_8^2, \end{equation*} |
which combined with (3.3) and (3.4) implies that for all t\in [0, t_1+t_2],
\begin{equation*} \frac{1}{2}C_0\leq v(x, t), \qquad \frac{1}{2}C_1\leq \theta(x, t), \end{equation*} |
\begin{equation*} \sup\limits_{0\leq t\leq t_1+t_2}\|(\tau, u, v, w, \theta)\|_{2}^2+\int_{0}^{t_1+t_2}\|\theta_{t}\|^2 \mathrm{d} t\leq 5C_8^2. \end{equation*} |
Take \alpha\leq \min \{\alpha_1, \alpha_2\}, where \alpha_i(i = 1, 2) are positive constants satisfying (3.1) and
\begin{equation*} \left(\frac{1}{2}C_0\right)^{-\alpha_2}\leq 2, \qquad (\sqrt{5}C_8)^{\alpha_2}\leq 2, \qquad \alpha_2H(\frac{1}{2}C_0, \frac{1}{2}C_1, \sqrt{5}C_8)\leq \epsilon_1, \end{equation*} |
where the value of \epsilon_1 is chosen in Lemma 2.1. That means that we can choose
\begin{equation} |\alpha_2|: = \min\left\{\frac{\ln 2}{|\ln 2-\ln C_0|}, \frac{\ln 2}{|\ln \sqrt{5}+\ln C_8|}, \epsilon_1 H^{-1}\left(\frac{1}{2}C_0, \frac{1}{2}C_1, \sqrt{5}C_8\right)\right\}. \end{equation} | (3.5) |
Then one can employ Lemmas 2.1–2.10 with T = t_1+t_2 to infer the local solution (\tau, u, v, w, \theta) satisfying (3.3) and (3.4).
Choosing
\begin{equation} \epsilon_0 = \min \{\alpha_1, \alpha_2\}, \end{equation} | (3.6) |
and repeating the above procedure, one can extend the solution (\tau, u, v, w, \theta) step-by-step to a global one provided that |\alpha|\leq \epsilon_0 . Furthermore,
\begin{equation*} \|(\tau, u, v, w, \theta)\|_{H^2}^2+\int_{0}^{+\infty}\left[\|(u_x, v_x, w_x, \theta_{x})\|^2+\|\tau\|^2\right] \mathrm{d} t\leq C_9^2, \end{equation*} |
from which we derive that the solution (\tau, u, v, w, \theta)\in X(0, +\infty; C_0, C_1, C_9).
The large-time behavior (1.21) follows from Lemmas 2.4–2.10 by using a standard argument [21].
First, thanks to (1.15), (2.1), (2.43), (2.55), (2.62), (2.73), Corollary 2.1, and Lemmas 2.4–2.10, taking \hat{\theta} = E_0 , one has
\begin{eqnarray} &&\frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\eta_{E_0}(\tau, u, v, w, \theta) \mathrm{d} x +c_1\|(u, v)\|_{1}^2+c_1\|(w_x, \theta_x)\|^2\leq0, \end{eqnarray} | (3.7) |
\begin{eqnarray} && \frac{ \mathrm{d}}{ \mathrm{d} t}\int_0^1\left[\frac{1}{2}\left(\frac{\lambda\tau_x}{\tau}\right)^2-\frac{\lambda u\tau_x}{r \tau}\right] \mathrm{d} x +c_2\|\tau_x\|^2 \leq C_{10}\|(u, u_x, \theta_x, v, v_x)\|^2, \end{eqnarray} | (3.8) |
\begin{eqnarray} && \frac{ \mathrm{d}}{ \mathrm{d} t}\|(u_x, v_x, w_x)\|^2 +c_3\|(u_{xx}, v_{xx}, w_{xx})\|^2 \leq C_{11}\|(\theta_x, \tau_x, v_x, u_x, u, v, w_x)\|^2, \end{eqnarray} | (3.9) |
\begin{eqnarray} && \frac{ \mathrm{d}}{ \mathrm{d} t}\| \theta_x\|^2 +c_4\| \theta_{xx}\|^2 \leq C_{12}\|(u_x, v_x, w_x)\|_{1}^2+C_{12}\| \theta_x\|^2. \end{eqnarray} | (3.10) |
By Cauchy-Schwarz's inequality, one has
\begin{equation} \left|\frac{\lambda u\tau_x}{r \tau}\right|\leq \frac{1}{4}\left(\frac{\lambda\tau_x}{\tau}\right)^2+C\|u\|^2. \end{equation} | (3.11) |
Hence, by means of (3.11), Poincaré's inequalities, Corollary 2.1, and Lemma 2.7, one can deduce
\begin{equation*} c\|\tau_x\|^2-C_{13}\|u\|^2\leq\int_0^1\left[\frac{1}{2}\left(\frac{\lambda\tau_x}{\tau}\right)^2-\frac{\lambda u\tau_x}{r \tau}\right] \mathrm{d} x\leq C\|(\tau_x, u_x)\|^2. \end{equation*} |
Multiplying (3.7)–(3.10) by C_{14} , C_{15} , and C_{16} , respectively, and adding them together with (3.10), one has
\begin{equation} \frac{ \mathrm{d}}{ \mathrm{d} t}\mathcal{A} +c\|(u_x, v_x, w_x, \theta_x)\|_{H^1}^2+c\|\tau_x\|^2\leq 0, \end{equation} | (3.12) |
where we have defined
\begin{equation*} \mathcal{A}: = \int_0^1C_{14}\eta_{E_0}(\tau, u, v, w, \theta)+C_{15}\left[\frac{1}{2}\left(\frac{\lambda\tau_x}{\tau}\right)^2-\frac{\lambda u\tau_x}{r \tau}\right] \mathrm{d} x+C_{16}\|(u_x, v_x, w_x)\|^2+\|\theta_x\|^2, \end{equation*} |
and chosen constants C_{14} > C_{15} > C_{16} > 0 suitably large such that
c_1C_{14}-C_{10}C_{15}-C_{11}C_{16}-C_{12} > 0, |
c_2C_{15}-C_{11}C_{16}-C_{12} > 0, |
c_3C_{16}-C_{12} > 0. |
Taking \frac{C_{14}}{2} > C_{13} and using Poincaré's inequality gives
\begin{equation} c\|(\tau-\bar\tau, u, v, w, \theta-E_0)\|^2\leq\mathcal{A}\leq C\|(u_x, v_x, w_x, \theta_x)\|_1^2+C\|\tau_x\|^2, \end{equation} | (3.13) |
where we have used the facts
\begin{equation*} \|\theta-E_0\|^2\leq C\int_0^1|\theta-\bar\theta|^2 \mathrm{d} x+C\|(u, v, w)\|^2\leq C\|(\theta_x, u_x, v_x, w_x)\|^2. \end{equation*} |
By means of (3.12) and (3.13), we can derive that
\begin{equation} \|(\tau-\bar\tau, u, v, w, \theta-E_0)(t)\|_{H^1( \Omega)}^2\leq C\text{e}^{-ct}. \end{equation} | (3.14) |
By means of \bar r , one has
\begin{equation} r^2-\bar r^2 = 2\int_0^x\tau-\bar\tau \mathrm{d}\xi. \end{equation} | (3.15) |
By means of (3.14) and (3.15), we have
\begin{equation*} \|r-\bar r\|_2^2\leq C\text{e}^{-ct}. \end{equation*} |
The proof is thus complete.
Dandan Song: Writing-original draft, Writing-review & editing, Supervision, Formal Analysis; Xiaokui Zhao: Writing-review & editing, Methodology, Supervision.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors are grateful to the referees for their helpful suggestions and comments on the manuscript. This work was supported by the NNSFC (Grant No. 12101200), the Doctoral Scientific Research Foundation of Henan Polytechnic University (No. B2021-53) and the China Postdoctoral Science Foundation (Grant No. 2022M721035).
The authors declare there is no conflict of interest.
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