In this paper, we study the large time behavior of the solution for one-dimensional compressible micropolar fluid model with large initial data. This model describes micro-rotational motions and spin inertia which is commonly used in the suspensions, animal blood, and liquid crystal. We get the uniform positive lower and upper bounds of the density and temperature independent of both space and time. In particular, we also obtain the asymptotic behavior of the micro-rotation velocity.
Citation: Haibo Cui, Junpei Gao, Lei Yao. Asymptotic behavior of the one-dimensional compressible micropolar fluid model[J]. Electronic Research Archive, 2021, 29(2): 2063-2075. doi: 10.3934/era.2020105
[1] | Haibo Cui, Junpei Gao, Lei Yao . Asymptotic behavior of the one-dimensional compressible micropolar fluid model. Electronic Research Archive, 2021, 29(2): 2063-2075. doi: 10.3934/era.2020105 |
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In this paper, we study the large time behavior of the solution for one-dimensional compressible micropolar fluid model with large initial data. This model describes micro-rotational motions and spin inertia which is commonly used in the suspensions, animal blood, and liquid crystal. We get the uniform positive lower and upper bounds of the density and temperature independent of both space and time. In particular, we also obtain the asymptotic behavior of the micro-rotation velocity.
In this paper, we consider the one-dimensional compressible micropolar fluid model in Lagrange coordinates:
{vt=ux,ut+Px=μ(uxv)x,wt+νvw=λ(wxv)x,(e+u22)t+(Pu)x=(μuuxv+kθxv)x+νvw2+λw2xv,e=cvθ,P=Rθv, | (1.1) |
where
When
When
In this paper, motivated by the result of [13], we get asymptotic behavior of the one-dimensional compressible micropolar fluid model with large initial data. We supplement the system (1.1) with the initial condition
(v,u,w,θ)|t=0=(v0,u0,w0,θ0), | (1.2) |
and three types of far-field and boundary conditions:
(1) Cauchy problem
Ω=R, lim|x|→∞(v(x,t),u(x,t),w(x,t),θ(x,t))=(1,0,0,1),t>0; | (1.3) |
(2) Boundary and far-field conditions for
u(0,t)=0, θx(0,t)=0, limx→∞(v(x,t),u(x,t),w(x,t),θ(x,t))=(1,0,0,1),t>0; | (1.4) |
(3) Boundary and far-field conditions for
u(0,t)=0, θ(0,t)=1, limx→∞(v(x,t),u(x,t),w(x,t),θ(x,t))=(1,0,0,1),t>0. | (1.5) |
We now state the main result of this paper as follows:
Theorem 1.1. Consider the initial value problem given by (1.1)-(1.3), and initial-boundary value problem given by (1.1), (1.2), (1.4) and (1.1), (1.2), (1.5). Assume that
1C1≤v(x,t)≤C1, | (1.6) |
1C≤θ(x,t)≤C, | (1.7) |
and for any
limt→∞(∥(v−1,u,w,θ−1)(t)∥Lp(Ω)+∥(vx,ux,wx,θx)(t)∥L2(Ω))=0. |
Remark 1. Compared with the work [13], we have to deal with the strong couple of the micro-rotation velocity
Lemma 2.1. (Energy inequality) Under the conditions of Theorem 1.1, the following inequality holds
sup0≤t<∞∫Ω(12u2+12w2+R(v−lnv−1)+cv(θ−lnθ−1))dx+∫∞0∫Ω(μu2xvθ+κθ2xvθ2+λw2xvθ+νvw2θ)dxds≤C0, | (2.1) |
0<α1≤∫i+1ivdx≤α2,0<α1≤∫i+1iθdx≤α2, | (2.2) |
and there are
α1≤v(ai(t),t)≤α2,α1≤θ(bi(t),t)≤α2, | (2.3) |
where
Proof. The above estimates can been obtained by the similar arguments as that in [10].
Lemma 2.2. Under the conditions of Theorem 1.1, there holds
sup0≤t<∞∫Ω12w2dx+∫∞0∫Ω(λw2xv+νvw2)dxds≤∫Ω12w20dx, | (2.4) |
and
sup0≤t<∞∫Ω14w4dx+∫∞0∫Ω(λw2w2xv+νvw4)dxds≤∫Ω14w40dx. | (2.5) |
Proof. Multiplying
Multiplying
In this subsection, the specific volume
Lemma 2.3. Under the hypotheses of Theorem 1.1, it holds that
0<C1−1≤v(x,t)≤C1. | (2.6) |
For more detailed proof of Lemma 2.3, refer to [5,10,11] and references therein.
Lemma 2.4. Under the conditions of Theorem 1.1, let
∫Ω(|θ−1|2+u2θ+w2θ+u4)dx+∫T0∫Ω(θ2x+u2xθ+w2xθ+w2θ+u2u2x)dxds≤C. | (2.7) |
Proof. Motivated by [13], the proof will be divided into three steps.
Step 1. At first, for any
Ωa(t)≜{x∈Ω|θ(x,t)>a}, |
and we derive from (2.1) that
sup0≤t<∞∫Ωa(t)θ≤C(a)sup0≤t<∞∫Ω(θ−lnθ−1)≤C(a). |
Using
cvθt+Rθuxv=(κθxv)x+μu2xv+νvw2+λw2xv. | (2.8) |
Multiplying (2.8) by
cv2∫Ω(θ−2)2+dx+κ∫T0∫Ω2(s)θ2xvdxds=cv2∫Ω(θ0−2)2+dx−R∫T0∫Ωθuxv(θ−2)+dxds+μ∫T0∫Ωu2xv(θ−2)+dxds+λ∫T0∫Ωw2xv(θ−2)+dxds+ν∫T0∫Ωvw2(θ−2)+dxds. | (2.9) |
Next, multiplying
∫Ωu2(θ−2)+dx+2μ∫T0∫Ωu2xv(θ−2)+dxds=∫Ωu20(θ0−2)+dx+2R∫T0∫Ωθuxv(θ−2)+dxds+2R∫T0∫Ω2(s)θuθxvdxds−2μ∫T0∫Ω2(s)uuxθxvdxds+∫T0∫Ω2(s)u2θsdxds. | (2.10) |
Then multiplying
∫Ωw2(θ−2)+dx+2λ∫T0∫Ωw2xv(θ−2)+dxds+2ν∫T0∫Ωvw2(θ−2)+dxds=∫Ωw20(θ0−2)+dx−2λ∫T0∫Ω2(s)wwxθxvdxds+∫T0∫Ω2(s)w2θsdxds. |
Adding the above three integral expressions together and using (2.8) to get
∫Ω(cv2(θ−2)2++u2(θ−2)++w2(θ−2)+)dx+κ∫T0∫Ω2(s)θ2xvdxds+μ∫T0∫Ωu2xv(θ−2)+dxds+λ∫T0∫Ωw2xv(θ−2)+dxds+ν∫T0∫Ωvw2(θ−2)+dxds=∫Ω(cv2(θ0−2)2++u20(θ0−2)++w20(θ0−2)+)dx+R∫T0∫Ωθuxv(θ−2)+dxds+2R∫T0∫Ω2(s)θuθxvdxds−2μ∫T0∫Ω2(s)uuxθxvdxds+1cv∫T0∫Ω2(s)u2(μu2xv−Rθuxv)dxds+κcv∫T0∫Ω2(s)u2(θxv)xdxdsλcv∫T0∫Ω2(s)u2w2xvdxds+νcv∫T0∫Ω2(s)vu2w2dxds+λcv∫T0∫Ω2(s)w2w2xvdxds+1cv∫T0∫Ω2(s)w2(μu2xv−Rθuxv)dxds+κcv∫T0∫Ω2(s)w2(θxv)xdxds+νcv∫T0∫Ω2(s)vw4dxds−2λ∫T0∫Ω2(s)wwxθxvdxds=:∫Ω(cv2(θ0−2)2++u20(θ0−2)++w20(θ0−2)+)dx+12∑j=1Ij. | (2.11) |
Now the main task is to estimate
For
|I1|=R|∫T0∫Ωθuxv(θ−2)+dxds|≤ϵ∫T0∫Ω2(s)u2xv(θ−2)+dxds+C(ϵ)∫T0∫Ωθ2(θ−2)+dxds |
≤ϵ∫T0∫Ω2(s)u2xv(θ−2)+dxds+C(ϵ)∫T0supΩ(θ−32)2+ds. |
It follows from Cauchy inequality and (2.6) that for any
|I2|=R|∫T0∫Ω2(s)θuθxvdxds|≤ϵ∫T0∫Ω2(s)θ2xdxds+C(ϵ)∫T0∫Ω2(s)u2θ2dxds≤ϵ∫T0∫Ωθ2xdxds+C(ϵ)∫T0supΩ(θ−32)2+ds, |
where we have used the fact that
∫T0∫Ω2(s)u2θ2dxds≤16∫T0∫Ωu2(θ−32)2+dxds≤C∫T0supΩ(θ−32)2+ds. | (2.12) |
Then for
|I3|=2μ|∫T0∫Ω2(s)uuxθxvdxds|≤ϵ∫T0∫Ωθ2xdxds+C(ϵ)∫T0∫Ωu2u2xdxds. |
For
|I4|≤C∫T0∫Ωu2u2xdxds+C∫T0supΩ(θ−32)2+ds. |
For any
χη(θ)={0, θ≤2,θ−2η,2≤θ≤η+2,2,θ≥η+2. |
For
I5=κcv∫T0∫Ωlimη→0χη(θ)u2(θxv)xdxds=limη→0κcv∫T0∫Ω(−2χη(θ)uuxθxv−χ′η(θ)u2θ2xv)dxds≤ϵ∫T0∫Ωθ2xdxds+C(ϵ)∫T0∫Ωu2u2xdx. |
It follows from Cauchy inequality that
|I6|+|I7|≤C∫T0∫Ωu2w2xdxds+C∫T0∫Ωu2w2dxds. |
For
|I8|≤C∫T0∫Ωu2xw2dxds+C∫T0∫Ω2(s)w2θ2dxds≤C∫T0∫Ωu2xw2dxds+C∫T0supΩ(θ−32)2+ds. |
For
I9=κcv∫T0∫Ωlimη→0χη(θ)w2(θxv)xdxds=limη→0κcv∫T0∫Ω(−2χη(θ)wwxθxv−χ′η(θ)w2θ2xv)dxds≤ϵ∫T0∫Ω2(s)θ2xdxds+C∫T0∫Ωw2w2xdxds≤ϵ∫T0∫Ω2(s)θ2xdxds+C. |
For the terms
|I10|+|I11|≤C, |
and
|I12|≤ϵ∫T0∫Ωθ2xdxds+C. |
Noticing that
∫T0∫Ω(θ2x+u2xθ+w2xθ+w2θ)dxds=∫T0∫Ω3(s)(θ2x+u2xθ+w2xθ+w2θ)dxds+∫T0∫Ω−Ω3(s)(θ2x+u2xθ+w2xθ+w2θ)dxds≤3∫T0∫Ω3(s)(θ2x+u2x(θ−2)++w2x(θ−2)++w2(θ−2)+)dxds+C∫T0∫Ω(μu2xvθ+κθ2xvθ2+λw2xvθ+νvw2θ)dxds≤C∫T0∫Ω2(s)(κθ2xv+μu2xv(θ−2)++λw2xv(θ−2)++νvw2(θ−2)+)dxds+C. |
Then
∫Ω(cv2(θ−2)2++u2(θ−2)++w2(θ−2)+)dx+∫T0∫Ω(θ2x+u2xθ+w2xθ+w2θ)dxds≤C+C∫T0∫Ωu2u2xdx+C∫T0supΩ(θ−32)2+ds+C∫T0∫Ωu2w2xdxds+C∫T0∫Ωu2w2dxds+C∫T0∫Ωu2xw2dx. | (2.13) |
Step 2. Multiplying
∫Ωu2w2dx+2λ∫T0∫Ωu2w2xvdxds+2ν∫T0∫Ωvu2w2dxds=∫Ωu20w20dx−4λ∫T0∫Ωuuxwwxvdxds+2∫T0∫Ωuusw2dxds. |
Using
∫Ωu2w2dx+2λ∫T0∫Ωu2w2xvdxds+2ν∫T0∫Ωvu2w2dxds+2μ∫T0∫Ωu2xw2vdxds=∫Ωu20w20dx−4λ∫T0∫Ωuuxwwxvdxds−4μ∫T0∫Ωuuxwwxvdxds |
+2R∫T0∫Ωθuxw2vdxds+4R∫T0∫Ωuθwwxvdxds=:∫Ωu20w20dx+4∑j=1Jj. | (2.14) |
What's more, it follows from (2.1) and (2.6) that for any
sup0≤t≤T∫Ω(v−1)2dx+sup0≤t≤T∫Ω∖Ωα(θ−1)2dx≤Csup0≤t≤T∫Ω(v−lnv−1)dx+Csup0≤t≤T∫Ω(θ−lnθ−1)dx≤C. | (2.15) |
Then we can estimate
|J1|+|J2|≤C∫T0∫Ωu2u2xdxds+C∫T0∫Ωw2w2xdxds≤C∫T0∫Ωu2u2xdxds+C, |
|J3|≤μ∫T0∫Ωu2xw2vdx+C∫T0∫Ω2(s)w2θ2dxds≤μ∫T0∫Ωu2xw2vdx+C∫T0∫Ω2(s)w2θ2dxds+C∫T0∫Ω−Ω2(s)w2θ2dxds≤μ∫T0∫Ωu2xw2vdx+C∫T0supΩ(θ−32)2+ds+C, |
and
|J4|≤λ∫T0∫Ωu2w2xvdx+C∫T0∫Ω2(s)w2θ2dxds+C∫T0∫Ω−Ω2(s)w2θ2dxds≤λ∫T0∫Ωu2w2xvdx+C∫T0supΩ(θ−32)2+ds+C. |
Therefore, we have
∫Ωu2w2dx+∫T0∫Ω(u2w2x+u2w2+u2xw2)dxds≤C+C∫T0∫Ωu2u2xdxds+C∫T0supΩ(θ−32)2+ds. | (2.16) |
And similarly,
sup0≤t<∞∫Ωu4dx+∫∞0∫Ωu2u2xdxds≤C+Cδ∫T0∫Ωθu2xdxds+C∫T0supΩ(θ−32)2+ds. |
Adding all the above inequalities, one obtains
∫Ω(cv2(θ−2)2++u2(θ−2)++w2(θ−2)++u4)dx+∫T0∫Ω(θ2x+u2xθ+w2xθ+w2θ+u2u2x)dxds≤C+C∫T0supΩ(θ−32)2+ds. | (2.17) |
Step 3. We estimate the last term on the right hand side of (2.17).
∫T0supΩ(θ−32)2+ds≤C(ϵ)+ϵ∫T0∫Ωθ2xdxds. | (2.18) |
Then
∫Ω(cv2(θ−2)2++u2(θ−2)++w2(θ−2)++u4)dx+∫T0∫Ω(θ2x+u2xθ+w2xθ+w2θ+u2u2x)dxds≤C. | (2.19) |
The proof of Lemma 3.4 is complete.
Lemma 2.5. There exists a positive constant
sup0≤t≤T∫Ω(v2x+u2x+θ2x+w2x)dx+∫T0∫Ω(θv2x+u2xx+θ2xx+w2xx)dxds≤C. | (2.20) |
Moreover,
supΩ×[0,T]θ(x,t)≤C. | (2.21) |
Proof. The process will be divided into five steps:
Step 1. Integrating
μ2ddt∫Ω(vxv)2dx+R∫Ωθv2xv3dx=ddt∫Ωuvxvdx+R∫Ωvxθxv2dx+∫Ωu2xvdx. |
Integrating the above equality on
∫Ω(vxv)2dx+R∫T0∫Ωθv2xv3dxds≤C+∫Ωuvxvdx+C∫T0∫Ω(vxθxv2+u2xv)dxds≤C+δ∫Ω(vxv)2dx+C(δ)∫Ωu2dx+δ∫T0∫Ωθv2xdxds+C(δ)∫T0∫Ωθ2xθdxds. |
It follows from Cauchy inequality, (1.6), (2.1), (2.6), (2.7) and (2.18) that
sup0≤t≤T∫Ωv2xdx+∫T0∫Ωθv2xdxds≤C. | (2.22) |
Step 2. Integrating
12ddt∫Ωu2xdx+μ∫Ωu2xxvdx=μ∫Ωuxvxuxxv2dx−R∫Ωuxxθxvdx+R∫Ωθvxuxxv2dx. | (2.23) |
Using (2.7), (2.22) and Sobolev inequality, we get
∫T0∣μ∫Ωuxvxuxxv2dx−R∫Ωuxxθxvdx+R∫Ωθvxuxxv2dx∣ds≤μ4∫T0∫Ωu2xxvdxds+C∫T0∫Ωu2xv2xdxds+C∫T0∫Ωθ2xdxds+C∫T0∫Ωθ2v2xdxds≤C+μ4∫T0∫Ωu2xxvdxds+CsupΩ×[0,T]θ∫T0∫Ωθv2xdxds+C∫T0∥ux∥2L∞(Ω)ds≤C+μ2∫T0∫Ωu2xxvdxds+CsupΩ×[0,T]θ, | (2.24) |
then by using (2.23), one has
sup0≤t≤T∫Ωu2xdx+∫T0∫Ωu2xxdxds≤C+CsupΩ×[0,T]θ. | (2.25) |
Step 3. Integrating
12ddt∫Ωθ2xdx+κ∫Ωθ2xxvdx=κ∫Ωθxvxθxxv2dx−μ∫Ωu2xθxxvdx+R∫Ωθuxθxxvdx−∫Ωνvw2θxxdx−λ∫Ωw2xθxxvdx, | (2.26) |
then using Cauchy inequality, (2.7), (2.22), (2.25), Sobolev inequality, we get
∫T0∣κ∫Ωθxvxθxxv2dx−μ∫Ωu2xθxxvdx+R∫Ωθuxθxxv−∫Ωνvw2θxxdx−λ∫Ωw2xθxxvdx∣ds≤C∫T0∥θxx∥L2(Ω)∥θx∥L∞(Ω)∥vx∥L2(Ω)ds+C∫T0∥θxx∥L2(Ω)∥ux∥L∞(Ω)∥ux∥L2(Ω)ds+C∫T0∥θxx∥L2(Ω)∥θ∥L∞(Ω)∥ux∥L2(Ω)ds+C∫T0∥θxx∥L2(Ω)ds≤C∫T0∥θxx∥L2(Ω)∥θx∥12L2(Ω)∥θxx∥12L2(Ω)∥vx∥L2(Ω)ds+C∫T0∥θxx∥L2(Ω)∥ux∥H1(Ω)∥ux∥L2(Ω)ds+C∫T0∥θxx∥L2(Ω)∥ux∥H1(Ω)∥θ∥L∞(Ω)ds+C∫T0∥θxx∥L2(Ω)ds≤κ4∫T0∫Ωθ2xxvdxds+C+CsupΩ×[0,T]θ3. | (2.27) |
Hence we obtain by using (2.26) that
sup0≤t≤T∫Ωθ2xdx+∫T0∫Ωθ2xxdxds≤C+CsupΩ×[0,T]θ3. | (2.28) |
Step 4. Integrating
12ddt∫Ωw2xdx+λ∫Ωw2xxvdx=λ∫Ωwxvxwxxv2dx−ν∫Ωvwwxxdx, | (2.29) |
using (2.1), (2.22) and Lemma 2.2, one has
∫T0∣λ∫Ωwxvxwxxv2dx−ν∫Ωvwwxxdx∣ds≤λ4∫T0∫Ωw2xxvdxds+C. | (2.30) |
Then, by using (2.29), we get
sup0≤t≤T∫Ωw2xdx+∫T0∫Ωw2xxdxds≤C. | (2.31) |
Step 5. Using Sobolev inequality and (2.7), for any
∥θ−1∥2C(Ω)≤C∥θ−1∥L2(Ω)∥θx∥L2(Ω)≤C∥θx∥L2(Ω). | (2.32) |
Combining (2.28), yields
supΩ×[0,T](θ−1)2≤C+CmaxˉΩ×[0,T]θ32. |
It implies that there exists a positive constant
θ(x,t)≤C. | (2.33) |
Hence, putting together (2.22), (2.25), (2.28), (2.31) and (2.33), we finish the proof of Lemma 2.5.
Lemma 2.6. Under the conditions of Theorem 1.1, it holds that
limt→∞(∥(v−1,u,w,θ−1)(t)∥Lp(Ω)+∥(vx,ux,wx,θx)(t)∥L2(Ω))=0, | (2.34) |
for any
C−12≤θ(x,t)≤C2,for(x,t)∈ˉΩ×[0,∞). | (2.35) |
Proof. First, it follows from (2.7), (2.20), (2.23), (2.24), (2.26), (2.27), (2.29) and (2.30) that
∫∞0(∥ux∥2L2(Ω)+|ddt∥ux∥2L2(Ω)|)dt+∫∞0(∥θx∥2L2(Ω)+|ddt∥θx∥2L2(Ω)|)dt+∫∞0(∥wx∥2L2(Ω)+|ddt∥wx∥2L2(Ω)|)dt≤C, |
which directly gives
limt→∞(∥ux∥L2(Ω)+∥θx∥L2(Ω)+∥wx∥L2(Ω))=0. | (2.36) |
By applying (2.36) to (2.32), we get
limt→∞∥θ−1∥C(ˉΩ)=0. |
Hence, there exists some
12≤θ≤32,for any(x,t)∈ˉΩ×[T∗,∞). | (2.37) |
Combining (2.20), leads to
∫∞T∗∥vx∥2L2(Ω)ds≤C. | (2.38) |
Then combining
∫∞T∗∣ddt∥vx∥2L2(Ω)∣ds=2∫∞T∗∣∫Ωuxxvxdx∣ds≤∫∞T∗∫Ωu2xxdxds+∫∞T∗∫Ωv2xdxds≤C, |
which together with (2.38) implies
limt→∞∥vx∥L2(Ω)=0. | (2.39) |
Therefore, combining (2.1), (2.7), (2.36), (2.39) and (2.15), we can get (2.34).
Finally, we will establish the lower bound of
C−13e−C3t≤θ,,for any(x,t)∈ˉΩ×[0,∞), |
which together with (2.37), yield
C−13e−C3T∗≤θ. |
Combining (2.33) and (2.35), we choose
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