In this paper, we investigate an initial boundary value problem of a planar magnetohydrodynamics system with temperature-dependent viscosity, heat conductivity, and resistivity. When all of the relative coefficients mentioned above are power functions of temperature, the existence and uniqueness of a global-in-time non-vacuum strong solutions are proved under some special assumptions. At the same time, we obtain the nonlinear exponential stability of the solution. In fact, the initial data could be large if the power of viscosity is small enough.
Citation: Dandan Song, Xiaokui Zhao. Large time behavior of strong solution to the magnetohydrodynamics system with temperature-dependent viscosity, heat-conductivity, and resistivity[J]. Electronic Research Archive, 2025, 33(2): 938-972. doi: 10.3934/era.2025043
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In this paper, we investigate an initial boundary value problem of a planar magnetohydrodynamics system with temperature-dependent viscosity, heat conductivity, and resistivity. When all of the relative coefficients mentioned above are power functions of temperature, the existence and uniqueness of a global-in-time non-vacuum strong solutions are proved under some special assumptions. At the same time, we obtain the nonlinear exponential stability of the solution. In fact, the initial data could be large if the power of viscosity is small enough.
The governing equations of a planar magnetohydrodynamics (MHD) compressible flow can be written in Lagrange variable form as:
vt=ux, | (1.1) |
ut+(P+12|b|2)x=(μuxv)x, | (1.2) |
wt−bx=(λwxv)x, | (1.3) |
(vb)t−wx=(νbxv)x, | (1.4) |
(e+u2+|w|2+v|b|22)t+(u(P+12|b|2)−w⋅b)x=(κθxv+μuuxv+λw⋅wxv+νb⋅bxv)x. | (1.5) |
Here t>0 represents the time, and x∈Ω=(0,1) denotes the Lagrange mass coordinate. The unknown functions v(x,t)>0, u(x,t), w=(w1(x,t),w2(x,t)), b=(b1(x,t),b2(x,t)), θ(x,t)>0, e, and P are the specific volume of the gas, longitudinal velocity, transverse velocity, transverse magnetic field, absolute temperature, internal energy, and pressure, respectively. μ and λ are the viscosity of the flow, ν is the resistivity, and κ is the heat conductivity.
In this paper, we consider the MHD flow of a perfect gas. Thus, P and e satisfy:
P=Rθvande=Cvθ+Const. | (1.6) |
Here R>0 denotes the specific gas constant, and Cv>0 stands for the heat capacity at constant volume. It is assumed that μ, λ, ν, and κ satisfy
μ=˜μθα,λ=˜λθα,ν=˜νθα,andκ=˜κθβ, | (1.7) |
which contain the positive constants ˜μ>0, ˜λ>0, ˜ν>0, ˜κ>0, α>0, and β>0.
The systems (1.1)–(1.7) are supplemented with initial conditions
(v,u,w,b,θ)(x,0)=(v0,u0,w0,b0,θ0), | (1.8) |
and boundary ones
(u,w,b,θx)|∂Ω=0. | (1.9) |
Obviously, the initial data (1.8) should be compatible with the boundary conditions (1.9).
When w=b=0, the Eqs (1.1)–(1.5) are converted into the compressible Navier–Stokes equation, which can be derived from Boltzmann's equation, assuming that the space and time scales are larger than all inherent scale–lengths, such as the Debye length or the gyro-radii of the charged particles [1,2,3,4,5]. Also, one can deduce from the Chapman–Enskog expansion for the first level of approximation in kinetic theory that the viscosity μ and λ may depend on the temperature or the density (see Chapman and Cowling [6]). Experimental results [7] show that the transport coefficients μ and κ vary according to gas temperature and density at very high temperatures and densities.
The central point of magnetohydrodynamics theory is that conductive fluids can support magnetic fields. Li and Shang [8] proved the existence and uniqueness of the global-in-time classical solution to the initial-boundary value problem when the viscosity, resistivity, and heat conductivity depend on the specific volume v and the temperature θ. In that paper, the coefficients are assumed to be proportional to h(v)θα, where h(v) is a non-degenerate and smooth function satisfying some natural conditions, and the absolute value of the exponent α is sufficiently small. It's worth noting that Li and Shang considered the planar compressible magnetohydrodynamic system for the domain [0,1]×R2. Besides, Huang et al. [9] proved the large-time behavior of strong solutions to equations of compressible planar magnetohydrodynamic flow with the heat conductivity is the power function of temperature. Similar results can be observed in various other reports [10,11,12,13,14,15,16].
Recently, Sun et al. [17] verified the existence and uniqueness of a global-in-time non-vacuum strong solution to a one-dimensional compressible Navier–Stokes system for a viscous and heat-conducting ideal polytropic gas. It was assumed that the viscosity μ and heat conductivity κ depend on temperature θ with μ(θ)=θα and κ(θ)=θβ for sufficiently small α>0 and arbitrary β≥0.
Before presenting our main results, we need to provide some explanations of the symbols first. Throughout this paper, the positive general constant C will be different in different lines. For 1≤p≤∞, and integer k≥0, we adopt the simplified notations for the standard Sobolev space as follows:
Lp=Lp(Ω),Wk,p=Wk,p(Ω),Hk=Wk,2(Ω). |
Without loss of generality, we assume that ˜λ=˜μ=˜ν=˜κ=R=cv=1, and
∫10v0dx=1,∫10(θ0+u20+|w0|2+v0|b0|22)dx=1. | (1.10) |
Inspired above, we have the following main results.
Theorem 1.1. For given positive constants M0>0 and V0>0. Assume that
‖(v0,u0,w0,b0,θ0)‖H2≤M0,infx∈[0,1]{v0,θ0}≥V0>0. |
Then there exist ϵ0>0, C0>0 and C1>0 which depend only on β, M0, V0, such that the initial boundary value problem (1.1)–(1.9) with 0≤α≤ε0 (see (3.5)) admits a unique global-in-time strong solution (v,u,w,b,θ)(x,t) on [0,1]×[0,+∞) satisfying
C−10≤v(x,t)≤C0,C1≤θ(x,t)≤C−11, | (1.11) |
and
{(v,u,w,b,θ)∈C([0,+∞);H2),vx∈L2(0,+∞;H1),(ux,wx,bx,θx)∈L2(0,+∞;H2). |
Furthermore, for any t≥0, it holds that
‖(v−1,u,w,b,θ−1)‖H1≤Ce−η0t, |
where C,η0>0 are some positive constants.
Remark 1.1. From the view of physics, the resistance is a function of temperature (e.g., [18]). This implies that our results is physical. Dou et al.'s 2021 study[19], published in Scientific Reports, delves into a variety of issues, including Enhanced Oil Recovery, where the technology can potentially improve the extraction of residual oil from oil fields. The equations presented in our study, especially those related to magnetic force distribution and the relationship between magnetic force and displacement, are pivotal for understanding and optimizing these applications. They assist in predicting the behavior of magnetic foams under various conditions, which is crucial for designing effective systems in the aforementioned fields.
In this paper, we try to use the framework of Li and Liang [20] to prove the global wellposedness of the solution. It should be emphasized that the key step is to derive the time–independent positive lower and upper bounds of specific volume and temperature. The foremost obstacles lie in the strong non-linearities caused by the temperature-dependent viscosity, resistivity, and heat-conductivity from Eq (1.7). Fortunately, these nonlinear terms are involved with μx, λx, νx, μt, λt, or νt which can be controlled by the smallness of α. With the help of upper and lower bounds of the specific volume, we can then estimate the higher-order derivatives of the solutions, and the upper and lower bounds of the temperature.
The rest of this paper is organized as follows. Section 2 is devoted to a discussion of a number of a priori estimates independent of time, which are required to extend the local solution to the time global. Based on the estimates given in Section 2, the main results of Theorem 1.1 are established in Section 3.
For constants N,m1,m2, and T, define
X(0,T;m1,m2,N):={(v,u,θ,w,b):(v−1,u,w,b,θ−1)∈C([0,T];H2),vx∈L2(0,T;H1),(ux,θx,wx,bx)∈L2(0,T;H2),vt∈C([0,T];H1),(ut,θt,wt,bt)∈L2(0,T;H1),v(x,t)≥m1,θ(x,t)≥m2,E(0,T)≤N2,∀(x,t)∈[0,1]×[0,T]}, |
where
E(0,T):=sup0≤t≤T‖(u,w,b,vx,θx)‖2H1+∫T0‖θt‖2L2dt. |
The main purpose of this section is to derive certain t-dependent a priori estimates for the variables (v,u,θ,w,b) in the function space X(0,T;m1,m2,N), relevant to the initial boundary value problem (1.1)–(1.9) for T>0 and 0<mi≤1(i=1,2), 2≤N<+∞. It follows from Sobolev's inequality that
m1≤v(x,t)≤2N,m2≤θ(x,t)≤2N,for∀(x,t)∈[0,1]×[0,T]. |
Firstly, let us derive the time-independent lower and upper bounds of v.
Lemma 2.1. Assume that the conditions listed in Theorem 1.1 hold; then there exists a constant 0<ϵ1≤1 depending only on β, M0, and V0, such that if
m−α2≤2,Nα≤2,αH(m1,m2,N)≤ϵ1, | (2.1) |
where
H(m1,m2,N):=(m−11+m−12+N+1)8. |
Then for (x,t)∈[0,1]×[0,T],
C−10≤v(x,t)≤C0. | (2.2) |
Here (and in what follow), C0,Ci(i=1,2,⋯,10), and C denote some generic positive constants depending only on β, ‖(v0,u0,w0,b0,θ0)‖H2, infx∈(0,1)v0(x), and infx∈(0,1)θ0(x).
Proof. The proof is divided into four steps.
Step 1. (Basic energy estimate)
According to (1.1), (1.5), (1.9), and (1.10), for t>0, one has
∫10v(x,t)dx=1,∫10(θ+u2+|w|2+v|b|22)(x,t)dx=1. | (2.3) |
In light of (2.1), it is deduced that
‖θα+θ−α‖L∞([0,1]×[0,T])≤m−α2+(2N)α≤2+4=6. | (2.4) |
Simplifying (1.5) gives
θt+θvux=(θβθxv)x+θα(u2x+|wx|2+|bx|2)v. | (2.5) |
Then, multiplying (1.1)–(1.4), and (2.5) by (1−v−1), u, w, b, and (1−θ−1), respectively, and integrating them over [0,1]×[0,T], together with (2.4) gives
sup0≤t≤T∫10(u2+|w|2+v|b|22+(v−lnv)+(θ−lnθ))dx+∫T0W(s)ds≤E0, | (2.6) |
where
W(t)=∫10(θαθ2xvθ2+θα(u2x+|wx|2+|bx|2)vθ)(x,t)dx, | (2.7) |
and E0 is the initial total entropy defined by
E0=∫10(u20+|w0|2+v|b0|22+(v0−lnv0)+(θ0−lnθ0))dx. |
Step 2. (Representation formula for v)
First, (1.2) can be written as
ut+(P+12|b|2)x=μ(lnv)xt+μxuxv, |
that is,
(uμ)t+g+(μ−1(P+12|b|2))x=(lnv)xt, | (2.8) |
where
g=−(μ−1)tu−(μ−1)x(P+12|b|2)−μxuxμv. |
Integrating (2.8) over [0,t]×[x1(t),x], it follows
∫xx1(t)(uμ−u0μ0)dξ+∫10∫xx1(t)gdξds+∫t0(P+12|b|2μ−P+12|b|2μ(x1))ds=lnv(x,t)−lnv(x1(t),t)−[lnv0(x)−lnv(x1(t),0)], |
where x1(t)∈[0,1] is determined by the following steps and μ0=μ(θ0). Moreover, for ease of notation, define
F=uxv−μ−1(P+12|b|2)−∫x0g(ζ)dζ,φ=∫t0F(x,s)ds+∫x0u0μ0dζ. |
Based on the above definitions that
φx=uμ,φt=F. | (2.9) |
It is easy to show that
∫t0[μ−1(P+12|b|2)(x1(t),s)+∫x1(t)0g(ξ,s)dξ]ds=∫t0(uxμ−F)(x1(t),s)ds=∫t0[(lnv)t−F](x1(t),s)ds=lnv(x1(t),t)−lnv(x1(t),0)−∫t0F(x1(t),s)ds. | (2.10) |
With the help of (1.1) and (2.9) that
(vφ)t−(uφ)x=vφt−uφx=vF−u2μ=ux−vμ(P+12|b|2)−v∫x0g(ξ)dξ−u2μ. | (2.11) |
Integrating (2.11) over [0,t]×[0,1], it follows
∫10vφdx=∫10v0∫x0u0μ0dξdx−∫t0∫10[vμ(P+12|b|2)+v∫x0g(ξ)dξ+u2μ]dxds. | (2.12) |
Hence, according to the mean value theorem, there exists x1(t)∈[0,1] such that φ(x1(t),t)=∫10vφdx. By the definition of φ, (2.9) and (2.12), one obtains
∫t0F(x1(t),s)ds=φ(x1(t),t)−∫x1(t)0u0μ0dξ=∫10v0∫x0u0μ0dξdx−∫t0∫10[vμ(P+12|b|2)+v∫x0g(ξ)dξ+u2μ]dxds−∫x1(t)0u0μ0dξ. | (2.13) |
Substituting (2.13) into (2.10) gives
∫t0[μ−1(P+12|b|2)(x1(t),s)+∫x1(t)0g(ξ,s)dξ]ds=lnv(x1(t),t)−lnv(x1(t),0)−∫10v0∫x0u0μ0dξdx+∫x1(t)0u0μ0dξ+∫t0∫10[vμ(P+12|b|2)+v∫x0g(ξ)dξ+u2μ]dxds. | (2.14) |
Moreover, substituting (2.14) into (2.9),
∫t0μ−1(P+12|b|2)ds+∫t0∫x0gdξds−∫t0∫10[vμ(P+12|b|2)+v∫x0g(ξ)dξ+u2μ]dxds+∫xx1(t)(uμ−u0μ0)dξ+∫10v0∫x0u0μ0dξdx−∫x1(t)0u0μ0dξ=lnv(x,t)−lnv(x,0). | (2.15) |
It follows from (2.15) that
v=B−1AD, | (2.16) |
where
A:=exp{∫t0[μ−1(P+12|b|2)+∫x0gdζ]ds};B:=exp{∫t0∫10[vμ(P+12|b|2)+v∫x0g(ζ)dζ+u2μ]dxds};D:=v0exp{∫xx1(t)(uμ−u0μ0)dζ+∫10v0∫x0u0μ0dζdx−∫x1(t)0u0μ0dζ}. |
From (2.16), one has
vD−1B=A. | (2.17) |
Furthermore, we define
J:=1μ(P+12|b|2)+∫x0dζ. |
Then, multiplying (2.17) by J, we have
vD−1BJ=∂∂tA. |
Since A(0)=1, integrating the above equality over (0,t) with respect to time, one obtains
v=DB−1+∫t0B(s)B(t)⋅D(t)D(s)⋅v[1μ(P+12|b|2)+∫x0gdξ]ds. | (2.18) |
Step 3. (Lower bound for v)
Applying Jensen's inequality to the convex function θ−lnθ leads to
∫10θdx−ln∫10θdx≤∫10(θ−lnθ)dx, |
which, together with (2.3) and (2.6), leads to
ˉθ(t)=∫10θ(x,t)dx∈[α1,1], | (2.19) |
where 0<α1<α2 are two roots of
x−lnx=e0. |
The expression of D and
∫xx1(t)uμdξ≤∫10θ−αudx≤m−2α2+∫10u2dx≤C, |
imply that
C−1≤D≤C. | (2.20) |
Next, we will estimate B. It follows from (2.3) and (2.4) that
∫10(vμ(P+12|b|2)+u2μ)dx=∫10θ−α(θ+12v|b|2+u2)dx≤m−α2∫10(θ+12v|b|2+u2)dx≤4, | (2.21) |
and
∫10(vμ(P+12|b|2)+u2μ)dx≥∫10θ−α(θ+12v|b|2+u2)dx≥(2N)−α∫10θdx≥α14. | (2.22) |
On the other hand, by the expression of g, there exists a sufficiently small ϵ>0 such that
∫10∫x0g(ξ)dξ=∫10v∫x0αθ−α−1(θtu+θx(P+12|b|2)−θαθxuxv)dξdx≤αm−α−12∫10v∫10(θtu+m−11θθx+12θx|b|2−m−α2m−11θxux)dξdx≤αm−α−12(‖θt‖2L2+‖u‖2L2+m−11‖θx‖L2‖θ‖L2+‖θx‖L2‖ux‖L2)≤ϵt. | (2.23) |
Putting (2.21)–(2.23) into the expression of B, we will find that there exist C2,C3, such that
eC2t≤B(t)≤eC3t. |
That means
e−C2(t−s)≤B(s)B(t)≤e−C3(t−s). |
Thus, for 0≤t<t0, one has
v≥DB−1−Cε∫t0e−C2(t−s)ds≥Ce−Ct0−Cε(1−e−C2t0). |
For large enough t>t0, it follows
infx∈Ωv(x,t)≥C∫t0B(s)B(t)θ1−αds−Cε(1−e−C3t). | (2.24) |
Therefore, one needs the estimates of θ and B(s)B(t). By the mean value theorem and (2.3), there exists x2(t)∈[0,1], such that C−1≤θ(x2(t),t)≤C. Based on Cauchy-Schwarz's inequality, one has
|[ln(θ+1)]β2+1−[ln(θ(x2(t),t)+1)]β2+1|=|∫xx2(ln(θ+1))β⋅θx√v(θ+1)⋅√v(ζ)dζ|≤C(∫10(ln(θ+1))βθ2xvθ2dx)12⋅(∫10vdx)12≤C(∫10(ln(θ+1))βθ2xvθ2dx)12, |
which implies
θ≥C−CW(t). |
From (2.18), (2.20), (2.21), and (2.23), one has
∫t0B(s)B(t)θ1−αds≥∫t0B(s)B(t)(1−∫10θβθ2xvθ2dx)ds≥C−e−Ct−C{∫t20B(s)B(t)∫10θβθ2xvθ2dxds+∫tt2B(s)B(t)∫10θβθ2xvθ2dxds}≥C−e−Ct−Ce−C2t−C∫tt2∫10θβθ2xvθ2dxds≥C. | (2.25) |
For the large enough time T0, when t>T0, plugging (2.25) into (2.24) gives
infx∈Ωv(x,t)≥C. |
Step 4. (Upper bound for v)
According to Holder's inequality, for 0<β≤1, one has
|θ12(x,t)−θ12(x2(t),t)|≤∫10θ−12⋅θxdx≤‖v‖12∞⋅(∫10θβθ2xvθ2dx)12⋅(∫10θ1−βdx)12≤‖v‖12∞⋅(∫10θβθ2xvθ2dx)12. | (2.26) |
That means
θ(x,t)≤C+‖v‖∞∫10θβθ2xvθ2dx. | (2.27) |
For 1<β<∞, one has
|θβ2(x,t)−θβ2(x2(t),t)|≤∫10θβ2−1⋅θxθdx≤(∫10θβθ2xvθ2dx)12⋅(∫10θ1−βdx)12≤(∫10θβθ2xvθ2dx)12, |
which means
θ(x,t)≤C+∫10θβθ2xvθ2dx. | (2.28) |
Then the standard calculations give
maxx∈[0,1]|b|2(x,t)≤C∫10|b⋅bx|dx≤C∫10θα|bx|2vθdx+C∫10vθ1−α|b|2dx≤C∫10θα|bx|2vθdx+C. | (2.29) |
It follows from the expression of v and (2.26)–(2.29) that
‖v‖∞≤Ce−Ct+C∫t0e−C(t−s)((1+‖v‖∞)∫10θβθ2xvθ2dx+∫10θα|bx|2vθdx)ds. |
By using Grönwall's inequality, one has
‖v‖∞≤C. |
Up to now, the proof of Lemma 2.1 has been finished.
Lemma 2.2. Assume that the conditions listed in Lemma 2.1 hold; then for any p>0, there exists some positive constant C(p) such that
sup0≤t≤T∫10θ1−pdx+∫T0∫10(θβθ2xθp+1+u2x+|wx|2+|bx|2θp)dxdt≤C(p). | (2.30) |
Proof. From (2.6), we see that (2.30) holds for p=1. Then we assume p>0 and p≠1. Multiplying (2.5) by θ−p and integrating by parts, one can arrive at
1p−1(∫10θ1−pdx)t+p∫10θβθ2xvθp+1dx+∫10θα(u2x+|wx|2+|bx|2)vθpdx=∫10(θ1−p−1)uxvdx+∫10uxvdx≤C(p)∫10|θ12−1|(θ12−p+1)|ux|dx+∫10vtvdx≤C(p)maxx∈[0,1]|θ12−1|∫10(θ12−p+1)|ux|dx+∫10(lnv)tdx≤C(p)maxx∈[0,1]|θ12−1|(∫10(θ12−p)|ux|dx+∫10|ux|dx)+(∫10lnvdx)t≤C(p)maxx∈[0,1]|θ12−1|(∫10vθ1−pθαdx)12(∫10θαu2xvθpdx)12+C(p)maxx∈[0,1]|θ12−1|(∫10u2xvθdx)12(∫10vθdx)12+(∫10lnvdx)t≤12∫10θαu2xvθpdx+C(p)maxx∈[0,1]|θ12−1|(∫10vθ1−pθαdx+1)+(∫10lnvdx)t. | (2.31) |
Moreover, it follows from (2.3) and (2.19) that
α1≤∫10θdx≤∫10(θ+ηθα(u2+|w|2+v|b|2)2)dx≤1. |
For any real number q, it follows from (2.2) and (2.4) that
|1−ˉθq|=|∫10ddη(∫10(θ+ηθα(u2+|w|2+v|b|2)2)dx)qdη|=|∫10q(∫10θ+ηθα(u2+|w|2+v|b|2)2dx)q−1dη⋅∫10θα(u2+|w|2+v|b|2)2dx|≤C(q)maxx∈[0,1](|u|+|w|+|b|)(∫10(u2+|w|2+v|b|2)dx)12≤C(q)∫10(|ux|+|wx|+|bx|)dx≤C(q)(∫10θα(u2x+|wx|2+|bx|2)vθdx)12(∫10vθ1−αdx)≤C(q)W12(t). | (2.32) |
After that, for β∈(0,1), it follows from (2.27), (2.28), and (2.19) that
maxx∈[0,1]|θ12−1|≤maxx∈[0,1]|θ12−ˉθ12|+maxx∈[0,1]|ˉθ12−1|≤C∫10θ−12|θx|dx+CW12(t)≤C(∫10θβθ2xvθ2dx)12⋅(∫10vθ1−βdx)12+CW12(t)≤CW12(t), | (2.33) |
when β≥1, one has
maxx∈[0,1]|θ12−1|≤maxx∈[0,1]|θ12−ˉθ12|+maxx∈[0,1]|ˉθ12−1|≤maxx∈[0,1]|θβ2−ˉθβ2|+CW12(t)≤C∫10θβ2−1|θx|dx+CW12(t)≤C(∫10θβθ2xvθ2dx)12(∫10vθ1−βdx)12+CW12(t)≤CW12(t). | (2.34) |
Therefore, for β>0, it follows from (2.33), (2.34), and (2.6) that
∫T0maxx∈[0,1](θ12−1)2dt≤C. | (2.35) |
Finally, we see that for p∈[0,1], one has
∫10θ1−pdx≤∫10θdx+1≤C. |
And for β≥0, it follows from (2.33), (2.34), and (2.6) that
sup0≤t<∞∫10|lnv|dx≤C. | (2.36) |
As a result, according to (2.33), (2.34), (2.7), (2.35), and Grönwall's inequality, we derive (2.30) from (2.31), which finishes the proof of Lemma 2.2.
Lemma 2.3. Assume that the conditions listed in Lemma 2.1 hold; then for all T>0,
sup0≤t≤T∫10v2xdx+∫T0∫10(v2x(1+θ)+u2x+|wx|2+|bx|2)dxdt≤C4. | (2.37) |
Proof. First, integrating (2.5) over [0,1]×[0,t], by (2.36), one has
∫T0∫10θα(u2x+|wx|2+|bx|2)vdxdt=∫10θdx−∫10θ0dx+∫T0∫10θ−1vuxdxdt+∫10lnvdx−∫10lnv0dx≤12∫T0∫10θαu2xvdxdt+C∫T0∫10(θ−1)2vθαdxdt+C≤12∫T0∫10θαu2xvdxdt+C∫T0maxx∈[0,1]|θ12−1|2∫10(θ12+1)2dxdt+C≤12∫T0∫10θαu2xvdxdt+C∫T0maxx∈[0,1]|θ12−1|2dt+C≤12∫T0∫10θαu2xvdxdt+C∫T0W(t)dt+C≤C, |
thus it follows from (2.2) and (2.4) that
∫T0∫10(u2x+|wx|2+|bx|2)dxdt≤C. | (2.38) |
Next, since
(θαvxv)t=θα(vxv)t+αθα−1θtvxv=θα(vtv)x+αθα−1θtvxv=(θαvtv)x−αθα−1θxvtv+αθα−1θtvxv=(θαvtv)x+αθα−1v(vxθt−θxvt), |
the momentum equation (1.2) can be rewritten as
(u−θαvxv)t=−(θv+12|b|2)x−αθα−1v(vxθt−θxvt). | (2.39) |
Multiplying (2.39) by (u−θαvxv) and integrating it over [0,1]×[0,t] yields that for any t∈[0,T],
12∫10(u−θαvxv)2dx−12∫10(u−θαvxv)2(x,0)dx=∫T0∫10(θvxv2−θxv−b⋅bx)⋅(u−θαvxv)dxdt−∫T0∫10αθα−1v(vxθt−θxvt)(u−θαvxv)dxdt=−∫T0∫10(θα+1v2xv3)dxdt+∫T0∫10θuvxv2dxdt−∫T0∫10θxv(u−θαvxv)dxdt−∫T0∫10b⋅bx(u−θαvxv)dxdt−∫T0∫10αθα−1v(vxθt−θxvt)(u−θαvxv)dxdt=−∫T0∫10(θα+1v2xv3)dxdt+4∑i=1Ii. | (2.40) |
Each Ii(i=1,2,3,4) can be estimated as follows. First, based on Cauchy's inequality, we have
|I1|=|∫T0∫10θuvxv2dxdt|≤18∫T0∫10θα+1v2xv3dxdt+C∫T0∫10θ1−αu2vdxdt≤18∫T0∫10θα+1v2xv3dxdt+C, | (2.41) |
where it has been used
∫T0∫10θ1−αu2vdxdt≤C∫T0∫10θu2dxdt≤C∫T0maxx∈[0,1]|u|2∫10θdxdt≤C∫T0maxx∈[0,1]|u|2dt≤C∫T0∫10u2xdxdt≤C. |
Next, by using (2.4), (2.6), and (2.30) with p=β, it follows
|I2|=|∫T0∫10θxv(u−θαvxv)dxdt|≤18∫T0∫10θα+1v2xv3dxdt+C∫T0∫10θ1−αu2vdxdt+C∫T0∫10θαθ2xvθdxdt≤18∫T0∫10θα+1v2xv3dxdt+C. | (2.42) |
Combining (2.42) with Cauchy's inequality leads to
|I3|=|∫T0∫10b⋅bx(u−θαvxv)dxdt|=∫T0∫10(|bx|2+|b|2(u−θαvxv)2)dxdt≤C∫T0W(t)∫10(u−θαvxv)2dxdt+C. | (2.43) |
Rewriting (2.5) as
θt=−θβθxvxv2+βθβ−1θ2xv+θβθxxv+θα(u2x+|wx|2+|bx|2)v−θvux. |
We set
Y=−(uv−θαvx)(vxθt−θxvt)=θαv2xθt−uvvxθt−θαθxuxvx+uvθxux=uvθxux+(θαθt+uθβθxv)v2x−(βθβ−1uθ2x+θβuθxx+θαu(u2x+|wx|+|bx|2)−θuux+θαθxux)vx=uvθxux+R1v2x+R2vx, | (2.44) |
where
R1:=θαθt+uθβθxv, |
and
R2:=−(βθβ−1uθ2x+θβuθxx+θαu(u2x+|wx|+|bx|2)−θuux+θαθxux). |
Then from (2.44), one has
|I4|=|∫T0∫10αθα−1v2(uvθxux+R1v2x+R2vx)dxdt|≤|∫T0∫10αθα−1uθxuxvdxdt|+|∫T0∫10αθα−1v2R1v2xdxdt|+|∫T0∫10αθα−1v2R2vxdxdt|:=3∑i=1Ji. | (2.45) |
Each Ji(i=1,2,3) can be estimated as follows. First, by means of Cauchy's inequality
J1≤∫T0‖αθ1+α−β2‖∞(∫10θβθ2xvθ2dx+∫10θαu2xvθdx)dt≤C. | (2.46) |
According to the definition of R1, one has
J2=|∫T0∫10(αθα−1v2xv2⋅uθβθxv+αθα−1v2xv2⋅θαθt)dxdt|=|∫T0∫10αθα+β−1uv2xθxv3dxdt|+|∫T0∫10αθ2α−1v2xθtv2dxdt|≤∫T0‖αθα+β−12vxuv‖∞(∫10θα+1v2xv3dx+∫10θβθ2xvθ2dx)dt+∫T0‖αθ32(α−1)vx√v‖∞(∫10θα+1v2xv3dx+∫10θ2tdx)dt≤18∫T0∫10θα+1v2xv3dxdt+C∫T0∫10θβθ2xvθ2dxdt+C‖α‖∞∫T0∫10θ2tdxdt≤18∫T0∫10θα+1v2xv3dxdt+C. | (2.47) |
It follows from the definition of R2 that
R22≤C(θ2αθ2xu2x+u2β2θ2β−2θ4x+θ2βu2θ2xx+u2θ2αu4x+u2θ2α|wx|4+u2θ2α|bx|4+u2xθ2α+2), |
then, one has
J3=|∫T0∫10αθα−1v2R2vxdxdt|≤18∫T0∫10θα+1v2xv3dxdt+C∫T0∫10α2θα−3R22vdxdt≤18∫T0∫10θα+1v2xv3dxdt+C. | (2.48) |
Inserting (2.46)–(2.48) into (2.45) gives
|I4|≤14∫T0∫10θα+1v2xv3dxdt+C. | (2.49) |
Putting (2.41)–(2.43), and (2.49) into (2.40), combining Grönwall's inequality gives
∫10(u−vxv)2dx+∫T0∫10θα+1v2xv3dxdt≤C. |
Note that
∫10(u−vxv)2dx=∫10(u2−2uvxv+v2xv2)2dx=∫10u2dx+∫10v2xv2dx−2∫10uvxvdx, |
that means
∫10u2dx+∫10v2xv2dx+∫T0∫10θα+1v2xv3dxdt≤C+2∫10uvxvdx≤12∫10v2xv2dx+C∫10u2dx+C≤C. |
On the other hand, it follows from (2.33) and (2.34) that
∫10v2xdx=∫10v2x(1−θ)dx+∫10θv2xdx≤Cmaxx∈[0,1]|θ12−1|2∫10v2xdx+∫10θv2xdx≤CW(t)∫10v2xdx+∫10θv2xdx. |
Together with (2.38), the proof of Lemma 2.3 has been completed.
Lemma 2.4. Assume that the conditions listed in Lemma 2.1 hold; then for all T>0, one has
sup0≤t≤T∫10(|bx|2+|wx|2)dx+∫T0∫10(|bt|2+|wt|2+|bxx|2+|wxx|2)dxdt≤C5. | (2.50) |
Proof. First, rewrite (1.3) as
wt=θαwxxv−θαwxvxv2+αθα−1θxwxv+bx. | (2.51) |
Multiplying (2.51) by wxx and integrating over [0,1]×[0,T], one obtains:
12∫10|wx|2dx+∫T0∫10θα|wxx|2vdxdt=∫T0∫10θαvxwx⋅wxxv2dxdt+∫T0∫10αθα−1wx⋅wxxvdxdt+∫T0∫10bx⋅wxxdxdt=3∑i=1Ii. | (2.52) |
Each Ii(i=1,2,3) is estimated as follows. From Cauchy's inequality and (2.4), it shows
I1=∫T0∫10θαvxwx⋅wxxv2dxdt≤18∫T0∫10θα|wxx|2vdxdt+C∫T0∫10θα|wx|2v2xv3dxdt≤18∫T0∫10θα|wxx|2vdxdt+C∫T0maxx∈[0,1]|wx|2∫10θαv2xv3dxdt≤18∫T0∫10θα|wxx|2vdxdt+C∫T0maxx∈[0,1]|wx|2dt≤14∫T0∫10θα|wxx|2vdxdt+C, | (2.53) |
where it has been used
∫T0maxx∈[0,1]|wx|2dt≤∫T0∫10|wx|⋅|wxx|dxdt+C∫T0∫10|wx|2dxdt≤18∫T0∫10θα|wxx|2vdxdt+C∫T0∫10|wx|2dxdt≤18∫T0∫10θα|wxx|2vdxdt+C. | (2.54) |
Next, from the a priori assumption, one has
I2=−∫T0∫10αθα−1θxwx⋅wxxvdxdt≤18∫T0∫10θα|wxx|2vdxdt+C∫T0∫10α2θα−2θ2x|wx|2vdxdt≤18∫T0∫10θα|wxx|2vdxdt+C∫T0maxx∈[0,1]|wx|2⋅|α|2∫10θα−2θ2xvdxdt≤18∫T0∫10θα|wxx|2vdxdt+C∫T0maxx∈[0,1]|wx|2⋅|α|2⋅mα−22⋅‖θ2x‖2L2dt≤18∫T0∫10θα|wxx|2vdxdt+C∫T0maxx∈[0,1]|wx|2⋅|α|2⋅H(m1,m2,N)dt≤18∫T0∫10θα|wxx|2vdxdt+C∫T0maxx∈[0,1]|wx|2dt≤14∫T0∫10θα|wxx|2vdxdt+C. | (2.55) |
Furthermore, according to (2.37), one has
I3=∫T0∫10bx⋅wxxdxdt≤18∫T0∫10θα|wxx|2vdxdt+C∫T0∫10|bx|2θ−αvdxdt≤18∫T0∫10θα|wxx|2vdxdt+C. | (2.56) |
Substituting (2.53), (2.55), and (2.56) into (2.52), one has
∫10|wx|2dx+∫T0∫10|wxx|2dxdt≤C. | (2.57) |
Combining (2.51) with (2.54) gives
∫T0∫10|wt|2dxdt≤∫T0∫10(θ2αw2xxv2+θ2α|wx|2v2xv4+α2θ2α−2θ2x|wx|2v2+|bx|2)dxdt≤C∫T0maxx∈[0,1]|wx|2dt+C≤C. | (2.58) |
Next, rewrite (1.4) as
bt=θαbxxv2−θαbxvxv3+αθα−1θxbxv2−uxbv+wxv. | (2.59) |
Multiplying (2.59) by bxx and integrating the result over [0,1]×[0,T] yields
12∫10|bx|2dx+∫T0∫10θα|bxx|2v2dxdt=−∫T0∫10αθα−1θxbx⋅bxxv2dxdt+∫T0∫10θαvxbx⋅bxxv3dxdt+∫T0∫10uxb⋅bxxvdxdt−∫T0∫10wx⋅bxxvdxdt:=4∑i=1Ji. | (2.60) |
Each Ji(i=1,2,3) is estimated as follows. From Cauchy's inequality and (2.4), one has
J1=−∫T0∫10αθα−1θxbx⋅bxxv2dxdt≤18∫T0∫10θα|bxx|2v2dxdt+C∫T0∫10α2θα−2θ2x|bx|2v2dxdt≤18∫T0∫10θα|bxx|2v2dxdt+C∫T0maxx∈[0,1]|bx|2⋅|α|2∫10θα−2θ2xv2dxdt≤18∫T0∫10θα|bxx|2v2dxdt+C∫T0maxx∈[0,1]|bx|2⋅|α|2⋅mα−22⋅‖θ2x‖2L2dt≤18∫T0∫10θα|wxx|2v2dxdt+C∫T0maxx∈[0,1]|bx|2⋅|α|2⋅H(m1,m2,N)dt≤18∫T0∫10θα|bxx|2v2dxdt+C∫T0maxx∈[0,1]|bx|2dt≤18∫T0∫10θα|bxx|2v2dxdt+C, | (2.61) |
where it has been used
∫T0maxx∈[0,1]|bx|2dt≤∫T0∫10|bx|⋅|bxx|dxdt+C∫T0∫10|bx|2dxdt≤18∫T0∫10θα|bxx|2v2dxdt+C∫T0∫10|bx|2dxdt≤18∫T0∫10θα|bxx|2v2dxdt+C. | (2.62) |
It follows from (2.2), (2.4), (2.37), and (2.62) that
J2=∫T0∫10θαvxbx⋅bxxv3dxdt≤18∫T0∫10θα|bxx|2v2dxdt+C∫T0maxx∈[0,1]|bx|2∫10θαv2xv4dxdt≤18∫T0∫10θα|bxx|2v2dxdt+C∫T0maxx∈[0,1]|bx|2dt≤14∫T0∫10θα|bxx|2v2dxdt+C. | (2.63) |
From (2.2)–(2.4), one has
J3=∫T0∫10uxb⋅bxxvdxdt≤18∫T0∫10θα|bxx|2v2dxdt+max(x,t)∈[0,1]×[0,T]|b|2⋅C∫T0∫10u2xdxdt≤18∫T0∫10θα|bxx|2v2dxdt+max(x,t)∈[0,1]×[0,T]|b|2≤18∫T0∫10θα|bxx|2v2dxdt+Csup0<t<T∫10|b|⋅|bx|dx≤18∫T0∫10θα|bxx|2v2dxdt+18sup0<t<T|bx|2+C. | (2.64) |
According to (2.37), it follows
J4=∫T0∫10wx⋅bxxvdxdt≤18∫T0∫10θα|bxx|2v2dxdt+C∫T0∫10|wx|2dxdt≤18∫T0∫10θα|bxx|2v2dxdt+C. | (2.65) |
Inserting (2.61), (2.63)–(2.65) into (2.60), which implies
sup0<t<T∫10|bx|2dx+∫T0∫10|bxx|2dxdt≤C. | (2.66) |
From (2.59), one obtains
∫T0∫10|bt|2dxdt≤C∫T0∫10(θ2α|bxx|2v4+α2θ2α−2θ2x|bx|2v4+θ2α|bx|2v2xv6+u2x|b|2v2+|wx|2v2)dxdt≤C. | (2.67) |
Therefore, it follows from (2.57), (2.58), (2.66), and (2.67) that (2.50) is correct. Then the Lemma 2.4 has been proved.
Lemma 2.5. Assume that the conditions listed in Lemma 2.1 hold; then for all T>0,
∫T0∫10θ2xdxdt≤C6. |
Proof. For the case of β>1, setting p=β−1 in (2.30) will give
∫T0∫10θ2xdxdt≤C. | (2.68) |
For 0<β≤1, multiplying (2.5) by θ1−β2 and integrating by parts, it gives
24−β(∫10θ2−β2dx)t+2−β2∫10θβ2θ2xvdx=−∫10θ2−β2uxvdx+∫10θα+1−β2(u2x+|wx|2+|bx|2)vdx=∫10(ˉθ2−β2−θ2−β2)uxvdx+∫10(1−ˉθ2−β2)uxvdx−∫10uxvdx+∫10θα+1−β2(u2x+|wx|2+|bx|2)vdx:=4∑i=1Ii. | (2.69) |
Each Ii(i=1,2,3,4) can be estimated as follows. First, by (2.7), one has
I1=∫10(ˉθ2−β2−θ2−β2)uxvdx=∫10|ˉθ1−β4−θ1−β4|(ˉθ1−β4+θ1−β4)|ux|dx≤Cmaxx∈[0,1]|ˉθ1−β4−θ1−β4|(∫10(θ2−β2+1)dx)12(∫10u2xdx)12≤18maxx∈[0,1]|ˉθ1−β4−θ1−β4|+C∫10(θ2−β2+1)dx∫10u2xdx≤18(∫10θ−β4|θx|dx)2+C∫10(θ2−β2+1)dx∫10u2xdx≤18∫10θβ2θ2xvdx+C∫10θβθ2xvθ2dx+C∫10(θ2−β2+1)dx∫10u2xdx≤18∫10θβ2θ2xvdx+CW(t)+C∫10θ2−β2dx∫10u2xdx+C∫10u2xdx. | (2.70) |
According to (2.32), one obtains
I2=∫10(1−ˉθ2−β2)uxvdx≤maxx∈[0,1]|1−ˉθ2−β2|∫10uxdx≤maxx∈[0,1]|1−ˉθ2−β2|2+C∫10u2xdx≤CW(t)+C∫10u2xdx. | (2.71) |
It follows
I3=−∫10uxvdx≤C∫10u2xdx≤C. | (2.72) |
Next, it follows from (2.4) that
I4=∫10θα+1−β2(u2x+|wx|2+|bx|2)vdx≤Cmaxx∈[0,1](|θ1−β2−ˉθ1−β2|+1)∫10(u2x+|wx|2+|bx|2)dx≤18∫10θ−β2|θx|dx∫10(u2x+|wx|2+|bx|2)dx+C∫10(u2x+|wx|2+|bx|2)dx≤18∫10θβ2θ2xvdx+C∫10θβθ2xvθ2dx+C∫10(u2x+|wx|2+|bx|2)dx+C(∫10u2xdx)2+C(∫10|wx|2dx)2+C(∫10|bx|2dx)2≤18∫10θβ2θ2xvdx+CW(t)+C∫10(u2x+|wx|2+|bx|2)dx+C(∫10u2xdx)2+C(∫10|wx|2dx)2+C(∫10|bx|2dx)2. | (2.73) |
Substituting (2.70)–(2.73) into (2.69), integrating on [0,T], and combining (2.37) and Grönwall's inequality, one has when 0<β≤1,
∫T0∫10θ2xdxdt≤C. | (2.74) |
Then from (2.68) and (2.74) the proof of Lemma 2.5 has ended.
Lemma 2.6. Assume that the conditions listed in Lemma 2.1 hold; then for all T>0, one has
sup0≤t≤T∫10u2xdx+∫T0∫10u2xxdxdt≤C7. | (2.75) |
Proof. Rewrite (1.2) as
ut=θαuxxv−θαuxvxv2+αθα−1θxuxv−θxv+θvxv2−b⋅bx. | (2.76) |
Multiplying (2.76) by uxx and integrating the result over [0,1]×[0,T], it shows
12∫10u2xdx+∫T0∫10θαu2xxvdxdt≤∫T0∫10θαvxuxuxxv2dxdt−∫T0∫10αθα−1θxuxuxxvdxdt+∫T0∫10θxuxxvdxdt−∫T0∫10θvxuxxv2dxdt+∫T0∫10b⋅bxuxxdxdt=5∑i=1Ji. | (2.77) |
Each Ji can be estimated as follows. First, according to (2.37), one has
J1=∫T0∫10θαvxuxuxxv2dxdt≤18∫T0∫10θαu2xxvdxdt+C∫T0maxx∈[0,1]|ux|2∫10v2xdxdt≤18∫T0∫10θαu2xxvdxdt+C∫T0maxx∈[0,1]|ux|2dt≤14∫T0∫10θαu2xxvdxdt+C, | (2.78) |
where it has been used
∫T0maxx∈[0,1]|ux|2dt≤∫T0∫10|ux|⋅|uxx|dxdt+C∫T0∫10u2xdxdt≤18∫T0∫10θαu2xxvdxdt+C∫T0∫10u2xdxdt≤18∫T0∫10θαu2xxvdxdt+C. | (2.79) |
Secondly, combining (2.30) with p=β+1−α, one obtains
J2=−∫T0∫10αθα−1θxuxuxxvdxdt≤18∫T0∫10θαu2xxvdxdt+C∫T0maxx∈[0,1]|ux|2⋅|α|2mα−22∫10θ2xdxdt≤18∫T0∫10θαu2xxvdxdt+C∫T0maxx∈[0,1]|ux|2dt≤14∫10θαu2xxvdx+C. | (2.80) |
Next, from (2.2) and (2.4), one has
J3=∫T0∫10θxuxxvdxdt≤18∫T0∫10θαu2xxvdxdt+C∫T0∫10θ2xdxdt≤18∫T0∫10θαu2xxvdxdt+C. | (2.81) |
Furthermore, from (2.37), one has
J4=−∫T0∫10θvxuxxv2dxdt≤18∫T0∫10θαu2xxvdxdt+C∫T0∫10θ2v2xdx≤18∫T0∫10θαu2xxvdxdt+C∫T0(maxx∈[0,1](θ−ˉθ)2+1)∫10v2xdxdt≤18∫T0∫10θαu2xxvdxdt+C∫T0maxx∈[0,1](θ−ˉθ)2dt+C∫T0∫10v2xdxdt≤18∫T0∫10θαu2xxvdx+C∫T0∫10θ2xdxdt≤18∫T0∫10θαu2xxvdx+C, | (2.82) |
where we have used
∫T0maxx∈[0,1]|θ−ˉθ|2dt≤C. |
In fact, for 0<β<2,
∫T0maxx∈[0,1]|θ−ˉθ|2dt≤C∫T0∫10|θx|2dxdt≤C, |
and for β≥2,
∫T0maxx∈[0,1]|θ−ˉθ|2dt≤C∫T0maxx∈[0,1]|θβ2−ˉθβ2|2dt≤C∫T0∫10θβ−2|θx|2dxdt≤C(∫T0∫10θβθ2xvθ2dxdt)12(∫T0∫10vdxdt)12≤C. |
Finally, it follows from (2.62) that
J5=∫T0∫10b⋅bxuxxdxdt≤18∫T0∫10θαu2xxvdxdt+C∫T0maxx∈[0,1]|bx|2∫10v|b|2dxdt≤18∫T0∫10θαu2xxvdxdt+C∫T0maxx∈[0,1]|bx|2dt≤18∫10θαu2xxvdx+C. | (2.83) |
Substituting (2.78) and (2.80)–(2.83) into (2.77) gives
∫10u2xdx+∫T0∫10u2xxdxdt≤C. | (2.84) |
On the other hand, from (1.2), one has
|ut|2≤C(u2xx+v2xu2x+α2θ2α−2θ2xu2x+θ2x+θ2v2x+|b|2|bx|2), |
and
∫T0∫10u2tdxdt≤C. |
Combining this with (2.84), the proof of Lemma 2.6 has been finished.
Lemma 2.7. Assume that the conditions listed in Lemma 2.1 hold; then for all T>0,
C1≤θ(x,t)≤C−11, | (2.85) |
sup0≤t≤T∫10θ2xdx+∫T0∫10(θ2t+θ2xx)dxdt≤C8. | (2.86) |
Proof. Multiplying (2.5) by θ gives
12ddt∫10θ2dx+∫10θβθ2xvdx=∫10(1−θ2)uxvdx−(∫10lnvdx)t+∫10θα+1(u2x+|wx|2+|bx|2)vdx≤Cmaxx∈[0,1](|1−θ|2+u2x+|wx|2+|bx|2)−(∫10lnvdx)t. |
It follows from (2.32), (2.74), and (2.6) that
∫T0maxx∈[0,1](θ−1)2dt≤C∫T0maxx∈[0,1](θ−ˉθ)2dt+C∫T0maxx∈[0,1](ˉθ−1)2dt≤C∫T0∫10θ|θx|dxdt+C∫T0V(t)dt≤∫T0θ2xdxdt+C∫T0∫10θ2dxdt+C∫T0V(t)dt≤C∫T0∫10θ2dxdt+C. |
By combining this with (2.3), (2.37), and Grönwall's inequality, one has
∫T0∫10θβθ2xdxdt≤C. | (2.87) |
Next, multiplying (2.5) by θβθt and integrating it over (0,1), by (2.4), one has
12ddt∫10(θβθx)2vdx+∫10θβθ2tdx=−12∫10(θβθx)2v2uxdx−∫10θβ+1θtuxvdx+∫10θα+βθt(u2x+|wx|2+|bx|2)vdx≤12∫10θβθ2tdx+Cmaxx∈[0,1]|ux|θβ2∫10θ3β2θ2xv2dx+C∫10θβ+2u2xdx+∫10θβ(u4x+|wx|4+|bx|4)dx=12∫10θβθ2tdx+3∑i=1Ii. | (2.88) |
Moreover, each Ii(i=1,2,3) can be estimated as follows. First,
I1=Cmaxx∈[0,1]|ux|θβ2∫10θ3β2θ2xv2dx≤Cmaxx∈[0,1]u2xθβ+C(∫10θ32βθ2xdx)2≤Cmaxx∈[0,1]u2xmaxx∈[0,1](1+θ2β+2)+C∫10θβθ2xdx⋅∫10θ2βθ2xdx. | (2.89) |
Second,
I2≤C∫10(1+θ2β+2)u2xdx≤Cmaxx∈[0,1]u2xmaxx∈[0,1](1+θ2β+2). | (2.90) |
Finally,
I3≤Cmaxx∈[0,1](u4x+|wx|4+|bx|4)maxx∈[0,1](1+θ2β+2). |
Substituting (2.88)–(2.90) into (2.87) gives
12ddt∫10(θβθx)2vdx+∫10θβθ2tdx≤Cmaxx∈[0,1](u2x+u4x+|wx|4+|bx|4)maxx∈[0,1](1+θ2β+2)+C∫10θβθ2xdx∫10θ2βθ2xdx. |
Direct calculations yield
maxx∈[0,1]|θβ+1−ˉθβ+1|≤C+C∫10θ2βθ2xdx, | (2.91) |
and
maxx∈[0,1](1+θ2β+2)≤maxx∈[0,1](1+θβ+1)2≤C(∫10θβ|θx|dx)2≤C∫10θ2βθ2xdx. |
From this and (2.91), and integrating over [0,T], together with Grönwall's inequality, one has
sup0≤t≤T∫10θ2βθ2xdx+∫T0∫10θβθ2tdxdt≤C. | (2.92) |
Combining with (2.91), one has
max(x,t)∈[0,1]×[0,T]θ(x,t)≤C. | (2.93) |
On the one hand, (2.93) gives
∫T0∫10(θβ+1−ˉθβ+1)2dxdt≤C∫T0∫10θ2βθ2xdxdt≤C. | (2.94) |
Together with (2.4), (2.37), (2.91), and (2.92), one has
∫T0|ddt∫10(θβ+1−ˉθβ+1)2dx|dt≤C∫T0∫10(θβ+1−ˉθβ+1)2dxdt+C∫T0∫10(θ2βθ2t+ˉθ2t)dxdt≤C∫T0∫10u2xdxdt+C≤C. | (2.95) |
Combining (2.37), (2.93), and (2.94) leads to
limt→+∞∫10(θβ+1−ˉθβ+1)2dx=0. |
Then combining (2.91) gives
maxx∈[0,1](θβ+1−ˉθβ+1)4≤∫10(θβ+1−ˉθβ+1)2dx∫10θ2βθ2xdx→0,ast→+∞. | (2.96) |
Therefore, it follows from (2.19) and (2.95) that there exists some T0, such that
θ(x,t)≥γ12, |
for all (x,t)∈[0,1]×[T0,+∞).
On the other hand, for p>2, multiplying (2.5) by 1θp, one has
1p−1(∫10(1θ)p−1dx)t+∫10μu2xvθpdx≤∫10uxvθp−1dx≤12∫10μu2xvθpdx+12∫101μvθp−2dx. |
That means
‖1θ‖p−2Lp−1ddt‖1θ‖Lp−1≤C‖1θ‖p−2Lp−2≤C‖1θ‖p−2Lp−1, | (2.97) |
where the positive constant C independent of p and T. (2.97) gives
sup0<t<T‖θ−1‖Lp−1≤C(T+1). |
Letting p→+∞, there exists a positive constant C1≤γ12 such that
θ(x,t)≥C1, |
for all (x,t)∈[0,1]×[0,T0]. Combining this, (2.96), and (2.92) yields that for all (x,t)∈[0,1]×[0,+∞),
C1≤θ≤C−11. | (2.98) |
Together with (2.91), one has
sup0≤t≤T∫10θ2xdx+∫T0∫10θ2tdxdt≤C. | (2.99) |
Finally, it follows from (2.5) that
θβθxxv=θt+θvux−βθβ−1θ2xv+θβvxθxv2−θα(u2x+|wx|2+|bx|2)v, |
from this and (2.37), (2.97), (2.98) yields
(2.100) |
Combining with (2.98)–(2.100), the proof of Lemma 2.7 has been finished.
Lemma 2.8. Assume that the conditions listed in Lemma 2.1 hold; then for all , one has
(2.101) |
Proof. First, differentiating (1.2) with respect to shows
Multiplying the above equation by , one obtains after integration by parts,
(2.102) |
where in the last inequality it has been used
Next, differentiating (1.3) with respect to shows
Multiplying the above equation by , one also gets after integration by parts,
(2.103) |
Similarly, differentiating (1.4) with respect to shows
Multiplying the above by , one gets after integration by parts that
(2.104) |
At the end, differentiating (2.5) with respect to shows
Multiplying the above by and integrating by parts, one obtains
(2.105) |
According to (2.86), one has
Combining (2.102)–(2.105) and Grönwall's inequality, we deduce
(2.106) |
Finally, we rewrite (1.2) as
It follows from (2.106), (2.85), (2.86) and (2.37) that
(2.107) |
Similarly, rewriting (2.5) as
Using (2.5), (2.106), (2.85), (2.86), (2.37), and (2.107), one has
(2.108) |
Next, rewriting (1.3) as
Then it follows that
(2.109) |
At the end, rewriting (1.4) as
thus, one has
(2.110) |
Combining (2.106)–(2.110), the proof of Lemma 2.8 has been proved.
Lemma 2.9. Assume that the conditions listed in Lemma (2.1) hold; then for all , one has
(2.111) |
Proof. First, differentiating (1.2) with respect to gives
(2.112) |
Multiplying (2.112) by , and integrating it over , we arrive at
Together with (2.37), (2.75), (2.85), (2.86), (2.101), and Grönwall's inequality, one has
That means
(2.113) |
Furthermore, (2.112) can be written as
Together with (2.37), (2.75), (2.85), (2.86), (2.101), and (2.113), one has
(2.114) |
Next, differentiating (2.5) with respect to , it shows
Thus, one has
(2.115) |
Similarly, differentiating (1.4) with respect to , one has
It implies that
(2.116) |
Finally, differentiating (1.3) with respect to gives
Then, one has
(2.117) |
Combining with (2.113)–(2.117), we obtain (2.111). The Lemma 2.9 has been proved.
With all the a priori estimates in Section 2 at hand, we will 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 (1.1)–(1.9), which 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 the (1.10) holds, then there exists , depending only on , and , such that the initial boundary value problem (1.1)–(1.9) has a unique solution .
Proof of Theorem 1.1 First, using Lemma 3.1, the problem (1.1)–(1.9) has a unique solution , where .
For the positive constants with being small enough such that
(3.1) |
where is chosen in Lemma 2.1, one deduces from Lemmas 2.1–2.9 with that the solution satisfies
(3.2) |
and
(3.3) |
where is chosen in Section 2, and . It follows from Lemmas 2.8 and 2.9 that . If one takes as the initial data and applies Lemma 3.1 again, the local solution can be extended to the time interval with . Moreover, one obtains
and
Combining with (3.2), (3.3) implies
and
(3.4) |
Taking , where is chosen in (3.1) and is chosen to be such that
where is chosen in Lemma 2.1. Then one can employ Lemmas 2.1–2.9 with to infer the local solution satisfies (3.2) and (3.3).
Thus, choosing
(3.5) |
and repeating the above procedure, one can then extend the solution step by step to a global one provided that . Furthermore, one derives the initial boundary value problem (1.1)–(1.9) has a unique global solution satisfying (3.2) and (3.3). Moreover, .
The large-time behavior (1.11) follows from Lemmas 2.3–2.9 by using a standard argument (see Reference [21]).
First, similar to (2.6), multiplying (1.1) by , (1.2) by , (1.3) by , (1.4) by , (2.5) by and adding them altogether, integrating the resultant equality over , one has after using (2.2) and (2.85) that
where (and in what follows) and denote some generic positive constants depending only on and .
By means of (2.87), (2.2), (2.85), (2.86), (2.101) and Sobolev's inequality, one obtains
(3.6) |
Next, multiplying (2.39) by and integrating the resultant equality over , using (2.2), (2.85), (2.50), (2.75), and Poincare's inequality yields that
(3.7) |
By the virtue of (2.50), (2.75), and (2.54), one obtains that
(3.8) |
Furthermore, adding (3.5) multiplied by , (2.7) multiplied by , and (3.4) multiplied by , which satisfy
From (3.6) and choosing suitably small, it follows
where
By using Cauchy–Schwarz's inequality, one obtains
which along with Poincare's inequality, yields that
Finally, using Poincare's inequality and (3.7) implies that
where it has been used the conservation of energy implied by (1.5), (2.30), and the following fact:
By means of (3.6) and (3.8), one obtains that
Thus, the proof of Theorem 1.1 has been completed.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors are grateful to the anonymous referees for their helpful comments and valuable suggestions, which have improved the presentation of the manuscript. This work was supported by the Jiangxi Provincial Natural Science Foundation (Grant Nos. 20232BAB211001, 20242BAB25006).
The authors declare there are no conflicts of interest.
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