Citation: Paola Goatin, Sheila Scialanga. Well-posedness and finite volume approximations of the LWR traffic flow model with non-local velocity[J]. Networks and Heterogeneous Media, 2016, 11(1): 107-121. doi: 10.3934/nhm.2016.11.107
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Fluid-particle interaction model arises in many practical applications, and is of primary importance in the sedimentation analysis of disperse suspensions of particles in fluids. This model is one of the commonly used models nowadays in biotechnology, medicine, mineral processing and chemical engineering [27]-[25]. Usually, the fluid flow is governed by the Navier-Stokes equations for a compressible fluid while the evolution of the particle densities is given by the Smoluchowski equation [4], the system has the form:
{ρt+div(ρu)=0,(ρu)t+div(ρu⊗u)+∇(P(ρ)+η)−μΔu−λ∇divu=−(η+βρ)∇Φ,ηt+div(η(u−∇Φ))−Δη=0, | (1) |
where
There are many kinds of literatures on the study of the existence and behavior of solutions to Navier-Stokes equations (See [1]-[17]). Taking system (1) as an example, Carrillo
Despite the important progress, there are few results of non-Newtonian fluid-particle interaction model. As we know, the Navier Stokes equations are generally accepted as a right governing equations for the compressible or incompressible motion of viscous fluids, which is usually described as
{ρt+div(ρu)=0,(ρu)t+div(ρu⊗u)−div(Γ)+∇P=ρf, |
where
Eij(∇u)=∂ui∂xj+∂uj∂xi, |
is the rate of strain. If the relation between the stress and rate of strain is linear, namely,
Γ=μ(∂ui∂xj+∂uj∂xi)q, |
for
Γij=(μ0+μ1|E(∇xu)|p−2)Eij(∇xu). |
For
Non-Newtonian fluid flows are frequently encountered in many physical and industrial processes [8,9], such as porous flows of oils and gases [7], biological fluid flows of blood [30], saliva and mucus, penetration grouting of cement mortar and mixing of massive particles and fluids in drug production [13]. The possible appearance of the vacuum is one of the major difficulties when trying to prove the existence and strong regularity results. On the other hand, the constitutive behavior of non-Newtonian fluid flow is usually more complex and highly non-linear, which may bring more difficulties to study such flows.
In recent years, there has been many research in the field of non-Newtonian flows, both theoretically and experimentally (see [14]-[26]). For example, in [14], Guo and Zhu studied the partial regularity of the generalized solutions to an incompressible monopolar non-Newtonian fluids. In [32], the trajectory attractor and global attractor for an autonomous non-Newtonian fluid in dimension two was studied. The existence and uniqueness of solutions for non-Newtonian fluids were established in [29] by applying Ladyzhenskaya's viscous stress tensor model.
In this paper, followed by Ladyzhenskaya's model of non-Newtonian fluid, we consider the following system
{ρt+(ρu)x=0,(ρu)t+(ρu2)x+ρΨx−λ(|ux|p−2ux)x+(P+η)x=−ηΦx,(x,t)∈ΩT(|Ψx|q−2Ψx)x=4πg(ρ−1|Ω|∫Ωρdx),ηt+(η(u−Φx))x=ηxx, | (2) |
with the initial and boundary conditions
{(ρ,u,η)|t=0=(ρ0,u0,η0),x∈Ω,u|∂Ω=Ψ|∂Ω=0,t∈[0,T], | (3) |
and the no-flux condition for the density of particles
(ηx+ηΦx)|∂Ω=0,t∈[0,T], | (4) |
where
The system describes a compressible shear thinning fluid-particle interaction system for the evolution of particles dispersed in a viscous non-Newtonian fluid and the particle is driven by non-Newtonian gravitational potential. To our knowledge, there still no existence results for (2)-(4) when
We state the definition of strong solution as follows:
Definition 1.1. The
(ⅰ)
ρ∈L∞(0,T∗;H1(Ω)),u∈L∞(0,T∗;W1,p0(Ω)∩H2(Ω)),Ψ∈L∞(0,T∗;H2(Ω)),η∈L∞(0,T∗;H2(Ω)),ρt∈L∞(0,T∗;L2(Ω)),ut∈L2(0,T∗;H10(Ω)),Ψt∈L∞(0,T∗;H1(Ω)),ηt∈L∞(0,T∗;L2(Ω)),√ρut∈L∞(0,T∗;L2(Ω)),(|ux|p−2ux)x∈C(0,T∗;L2(Ω)). |
(ⅱ) For all
∫Ωρϕ(x,t)dx−∫t0∫Ω(ρϕt+ρuϕx)(x,s)dxds=∫Ωρ0ϕ(x,0)dx, | (5) |
(ⅲ) For all
∫Ωρuφ(x,t)dx−∫t0∫Ω{ρuφt+ρu2φx−ρΨxφ−λ|ux|p−2uxφx+(P+η)φx−ηΦxφ}(x,s)dxds=∫Ωρ0u0φ(x,0)dx, | (6) |
(ⅳ) For all
−∫t0∫Ω|Ψx|q−2Ψxψx(x,s)dxds=∫t0∫Ω4πg(ρ−1|Ω|∫Ωρdx)ψ(x,0)dxds, | (7) |
(ⅴ) For all
∫Ωηϑ(x,t)dx−∫t0∫Ω[η(u−Φx)−ηx]ϑx(x,s)dxds=∫Ωη0ϑ(x,0)dx. | (8) |
The main result of this paper is stated in the following theorem.
Theorem 1.2. Let
0≤ρ0∈H1(Ω),u0∈H10(Ω)∩H2(Ω),η0∈H2(Ω), |
and the compatibility condition
−(|u0x|p−2u0x)x+(P(ρ0)+η0)x+η0Φx=ρ120(g+Φx), | (9) |
for some
ρ∈L∞(0,T∗;H1(Ω)),u∈L∞(0,T∗;W1,p0(Ω)∩H2(Ω)),Ψ∈L∞(0,T∗;H2(Ω)),η∈L∞(0,T∗;H2(Ω)),ρt∈L∞(0,T∗;L2(Ω)),ut∈L2(0,T∗;H10(Ω)),Ψt∈L∞(0,T∗;H1(Ω)),ηt∈L∞(0,T∗;L2(Ω)),√ρut∈L∞(0,T∗;L2(Ω)),(|ux|p−2ux)x∈C(0,T∗;L2(Ω)). |
Remark 1. By using exactly the similar argument, we can prove the result also hold for the case
In this section, we will prove the local existence of strong solutions. From the continuity equation
∫Ωρ(t)dx=∫Ωρ0dx:=m0,(t>0,m0>0) |
Because equation
ρt+(ρu)x=0, | (10) |
(ρu)t+(ρu2)x+ρΨx−[(εu2x+1u2x+ε)2−p2ux]x+(P+η)x=−ηΦx, | (11) |
[(ϵΨ2x+1Ψ2x+ϵ)2−q2Ψx]x=4πg(ρ−m0), | (12) |
ηt+(η(u−Φx))x=ηxx, | (13) |
with the initial and boundary conditions.
(ρ,u,η)|t=0=(ρ0,u0,η0),x∈Ω, | (14) |
u|∂Ω=Ψ|∂Ω=(ηx+ηΦx)|∂Ω=0,t∈[0,T], | (15) |
and
{−[(εu20x+1u20x+ε)2−p2u0x]x+(P(ρ0)+η0)x+η0Φx=ρ120(g+Φx),u0|∂Ω=0. | (16) |
Provided that
We first get the estimate of
{−[(εu20x+1u20x+ε)2−p2u0x]x+(P(ρ0)+η0)x+η0Φx=ρ120(g+Φx),u0|∂Ω=0. | (16) |
Then
|u0xx|L2≤1p−1|(u20x+εεu20x+1)1−p2|L∞|(P(ρ0)+η0)x+η0Φx−ρ120(g+Φx)|L2≤1p−1(|u0x|2L∞+1)1−p2(|(P(ρ0)+η0)x+η0Φx−ρ120(g+Φx)|L2)≤1p−1(|u0xx|2L2+1)1−p2(|Px(ρ0)|L2+|η0x|L2+|η0|L∞|Φx|L2+|ρ0|12L∞|g|L2+|ρ0|12L∞|Φx|L2). |
Applying Young's inequality, we have
|u0xx|L2≤C(|Px(ρ0)|L2+|η0x|L2+|η0|L∞|Φx|L2+|ρ0|12L∞|g|L2+|ρ0|12L∞|Φx|L2)1p−1≤C, |
thus
|u0|L∞+|u0x|L∞+|u0xx|L2≤C, | (17) |
where
Next, we introduce an auxiliary function
Z(t)=sup0≤s≤t(1+|ρ(s)|H1+|u(s)|W1,p0+|√ρut(s)|L2+|ηt(s)|L2+|η(s)|H1). |
We will derive some useful estimate to each term of
In order to prove the main Theorem, we first give some useful lemmas for later use.
Lemma 2.1. Let
{−[(ε(uε0x)2+1(uε0x)2+ε)2−p2uε0x]x+(P(ρ0)+η0)x+η0Φx=ρ120(g+Φx),uε0(0)=uε0(1)=0. | (18) |
Then there are a subsequence
uεj0→u0inH10(Ω)∩H2(Ω),[(εj(uεj0x)2+1(uεj0x)2+εj)2−p2uεj0x]x→(|u0x|p−2u0x)xinL2(Ω). |
Proof. According to (18), we have
uεj0→u0inH10(Ω)∩H2(Ω),[(εj(uεj0x)2+1(uεj0x)2+εj)2−p2uεj0x]x→(|u0x|p−2u0x)xinL2(Ω). |
Taking it by the
|uε0xx|L2≤|(ε(uε0x)2+1(uε0x)2+ε)1−p2|L∞|(P(ρ0)+η0)x+η0Φx+ρ120(g+Φx)|L2≤(|uε0x|2L∞+1)1−p2|(P(ρ0)+η0)x+η0Φx+ρ120(g+Φx)|L2, |
then
|uε0xx|L2≤C(1+|(P(ρ0)+η0)x+η0Φx+ρ120(g+Φx)|L2)1p−1≤C. | (19) |
Therefore, by the above inequality, as
uεj0→u0inC32(Ω),uεj0xx→u0xxinL2(Ω)weakly. |
Thus, we can obtain
|uεi0x−uεj0x|L∞(Ω)<α1. |
Now, we prove that
Let
|uεi0x−uεj0x|L∞(Ω)<α1. |
For all
|uεi0xx−uεj0xx|L2(Ω)≤|ϕi−ϕj|L∞(Ω)|(P(ρ0)+η0)x+η0Φx−ρ120(g+Φx)|L2(Ω). |
With the assumption, we can obtain
|(P(ρ0)+η0)x+η0Φx−ρ120(g+Φx)|L2(Ω)≤C, |
where
|ϕi−ϕj|L∞(Ω)≤|∫10ϕ′(θ(uεi0x)2+(1−θ)(uεj0x)2)dθ((uεi0x)2−(uεj0x)2)|L∞(Ω), | (20) |
where
By the simple calculation, we can get
ϕ′(s)≤2p−1(1+s−p2), |
where
|ϕi−ϕj|L∞(Ω)≤2p−1|(1+∫10(θ(uεi0x)2+(1−θ)(uεj0x)2)dθ)((uεi0x)2−(uεj0x)2)|L∞(Ω)≤2p−1|uεi0x−uεj0x|L∞(Ω)|uεi0x+uεj0x|L∞(Ω)+4(2−p)(p−1)|uεi0x−uεj0x|2−p2L∞(Ω)|uεi0x+uεj0x|2−p2L∞(Ω)≤α. |
Substituting this into (18), we have
|uεi0xx−uεj0xx|L∞(Ω)<α, |
then there is a subsequence
{uεj0xx}→χinL2(Ω). |
By the uniqueness of the weak convergence, we have
χ={uε0xx}. |
Since
[(εj(uεj0x)2+1(uεj0x)2+εj)2−p2uεj0x]x→(|u0x|p−2u0x)xinL2(Ω). |
This completes the proof of Lemma 2.1.
Lemma 2.2.
sup0≤t≤T|ρ(t)|2H1≤Cexp(C∫t0Z6γ(3p−4)(q−1)(s)ds), | (21) |
where
Proof. We estimates for
[(εu2x+1u2x+ε)2−p2ux]x=ρut+ρuux+ρΨx+(P+η)x+ηΦx. |
We note that
|uxx|≤1p−1(u2x+ε)1−p2|ρut+ρuux+ρΨx+(P+η)x+ηΦx|≤1p−1(|ux|2−p+1)|ρut+ρuux+ρΨx+(P+η)x+ηΦx|. |
Taking it by the
|uxx|p−1L2≤C(1+|ρut|L2+|ρuux|L2+|ρΨx|L2+|(P+η)x|L2+|ηΦx|L2)≤C(1+|ρ|12L∞|√ρut|L2+|ρ|L∞|u|L∞|ux|p2Lp|ux|1−p2L∞+|ρ|γ−1L∞|ρx|L2+|ηx|L2+|η|L∞|Φx|L2+|ρ|L2|Ψxx|L2)≤C[1+|ρ|12L∞|√ρut|L2+(|ρ|L∞|u|L∞|ux|p2Lp)2(p−1)3p−4+|ρ|γ−1L∞|ρx|L2+|ηx|L2+|η|L∞|Φx|L2+|ρ|L2|Ψxx|L2]+12|uxx|p−1L2. | (22) |
On the other hand, by
|Ψxx|≤1q−1(|Ψx|2−q+1)|4πg(ρ−m0)|. |
Taking it by
|Ψxx|L2≤CZ1q−1(t). | (23) |
This implies that
|uxx|L2≤CZmax{qq−1,(p−1)(4+p)3p−4γ}(t)≤CZ6γ(3p−4)(q−1)(t). | (24) |
By (13), taking it by the
|ηxx|L2≤|ηt+(η(u−Φx))x|L2≤|ηt|L2+|ηx|L2|u|L∞+|ηx|L2|Φx|L∞+|η|L2|uxx|L2+|η|L∞|Φxx|L2≤CZ6γ+2(3p−4)(q−1)(t). | (25) |
Multiplying (10) by
12ddt∫Ω|ρ|2ds+∫Ω(ρu)xρdx=0. |
Integrating it by parts, using Sobolev inequality, we obtain
ddt|ρ(t)|2L2≤∫Ω|ux||ρ|2dx≤|uxx|L2|ρ|2L2. | (26) |
Differentiating
ddt∫Ω|ρx|2dx=−∫Ω[32ux(ρx)2+ρρxuxx](t)dx≤C[|ux|L∞|ρx|2L2+|ρ|L∞|ρx|L2|uxx|L2]≤C|ρ|2H1|uxx|L2. | (27) |
From (26) and (27) and the Gronwall's inequality, then lemma 2.2 holds.
Lemma 2.3.
|η|2H1+|ηt|2L2+∫t0(|ηx|2L2+|ηt|2L2+|ηxt|2L2)(s)ds≤C(1+∫t0Z4(s)ds), | (28) |
where
Proof. Multiplying
∫t0|ηx(s)|2L2ds+12|η(t)|2L2≤∬ΩT(|ηuηx|+|ηΦxηx|)dxds≤14∫t0|ηx(s)|2L2ds+C∫t0|ux|2Lp|η|2H1ds+C∫t0|η|2H1ds+C≤14∫t0|ηx(s)|2L2ds+C(1+∫t0Z4(t)ds). | (29) |
Multiplying
∫t0|ηt(s)|2L2ds+12|ηx(t)|2L2≤∬ΩT|η(u−Φx)ηxt|dxds≤14∫t0|ηxt(s)|2L2ds+C∫t0|η|2H1|ux|2Lpds+C∫t0|η|2H1ds+C≤14∫t0|ηxt(s)|2L2ds+C(1+∫t0Z4(t)ds). | (30) |
Differentiating
∫t0|ηxt(s)|2L2ds+12|ηt(t)|2L2=∬ΩT(η(u−Φx))tηxtdxds≤C+∬ΩT(|ηtuηxt|+|ηtΦxηxt|+|ηxutηt|+|ηuxtηt|)dxds≤C(1+∫t0(|ηt|2L2||ux|2Lp+|ηt|2L2+|ηx|2L2|ηt|2L2+|η|2H1|ηt|2L2)dx)+12∫t0|ηxt|2L2+12∫t0|uxt|2L2≤C(1+∫t0Z4(s)ds). | (31) |
Combining (29)-(31), we obtain the desired estimate of Lemma 2.3.
Lemma 2.4.
∫t0|√ρut(s)|2L2(s)ds+|ux(t)|pLp≤C(1+∫t0Z10+4γ(3p−4)(q−1)(s)ds), | (32) |
where
Proof. Using (10), we rewritten the (11) as
ρut+(ρu)ux+ρΨx−[(εu2x+1u2x+ε)2−p2ux]x+(P+η)x=−ηΦx. | (33) |
Multiplying (33) by
∬ΩTρ|ut|2dxds+∬ΩT(εu2x+1u2x+ε)2−p2uxuxtdxds=−∬ΩT(ρuux+ρΨx+Px+ηx+ηΦx)utdxds. | (34) |
We deal with each term as follows:
∫Ω(εu2x+1u2x+ε)2−p2uxuxtdx=12∫Ω(εu2x+1u2x+ε)2−p2(u2x)tdx=12ddt∫Ω(∫u2x0(εs+1s+ε)2−p2ds)dx, |
∫u2x0(εs+1s+ε)2−p2ds≥∫u2x0(s+1)2−p2ds=2p[(u2x+1)p2−1], |
−∬ΩTPxutdxds=∬ΩTPuxtdxds=ddt∬ΩTPuxdxds−∬ΩTPtuxdxds. |
By virtue of
Pt=−γPux−Pxu,−∬ΩTηxutdxds=∬ΩTηuxtdxds=ddt∬ΩTηuxdxds−∬ΩTηtuxdxds.−∬ΩTηΦxutdxds=−ddt∬ΩTηΦxudxds+∬ΩTηtΦxudxds. | (35) |
Substituting the above into (34), using Sobolev inequality and Young's inequality, we have
∫t0|√ρut(s)|2L2ds+|ux(t)|pLp≤∬ΩT(|ρuuxut|+|ρΨxut|+|γPu2x|+|Pxuux|+|ηtux|+|ηtΦxu|)dxds+∫Ω(|Pux|+|ηux|+|ηΦxu|)dx+C≤C+∫t0(|ρ|12L∞|u|L∞|ux|p2Lp|ux|1−p2L∞|√ρut|L2+|ρ|12L∞|Ψx|L∞|√ρut|L2)ds+∫t0(γ|P|L2|ux|p2Lp|ux|1−p2L∞|uxx|L2+aγ|ρ|γ−1L∞|ρx|L2|u|L∞|ux|L∞+|ηt|L2|ux|p2Lp|ux|1−p2L∞+|ηt|L2|Φx|L2|u|L∞)ds+|P|Lpp−1|ux|Lp+|η|Lpp−1|ux|Lp+|η|Lpp−1|Φx|Lp|u|L∞≤C(1+∫t0(|ρ|L∞|ux|2+pLp|uxx|2−pL2+|ρ|H1|Ψxx|2L2+|P|L∞|ux|p2Lp|uxx|2−p2L2+|ρ|γ−1L∞|ρx|L2|ux|Lp|uxx|L2+|ηt|L2|ux|p2Lp|uxx|1−p2L2+|ηt|L2|ux|Lp)ds)+|P|pp−1Lpp−1+|η|pp−1Lpp−1+12∫t0|√ρut(s)|2L2ds+12|ux(t)|pLp. | (36) |
To estimate (36), combining (35) we have the following estimates
∫Ω|P(t)|pp−1dx=∫Ω|P(0)|pp−1dx+∫t0∂∂s(∫ΩP(s)pp−1dx)ds≤∫Ω|P(0)|pp−1dx+pp−1∫t0∫Ωaγργ−1P(s)1p−1(−ρxu−ρux)dxds≤C+C∫t0|ρ|γ−1L∞|P|1p−1L∞|ρ|H1|ux|Lpds≤C(1+∫t0Zγp−1+γ+1(s)ds), | (37) |
In exactly the same way, we also have
∫Ω|η(t)|pp−1dx≤C(1+∫t0Z1p−1+1(s)ds), | (38) |
which, together with (36) and (37), implies (32) holds.
Lemma 2.5.
|√ρut(t)|2L2+∫t0|uxt|2L2(s)ds≤C(1+∫t0Z26γ(3p−4)(q−1)(s)ds), | (39) |
where
Proof. Differentiating equation
12ddt∫Ωρ|ut|2dx+∫Ω[(εu2x+1u2x+ε)2−p2ux]tuxtdx=∫Ω[(ρu)x(u2t+uuxut+Ψxut)−ρuxu2t+(P+η)tuxt−ηtΦxut−ρΨxtut]dx. | (40) |
Note that
∫Ω[(εu2x+1u2x+ε)2−p2ux]tuxtdx=∫Ω[(εu2x+1u2x+ε)−p2ux](εu2x+1)(u2x+ε)−(2−p)(1−ε2)u2x(u2x+ε)2u2xtdx≥(p−1)∫Ω(u2x+1)p−22|uxt|2dx, | (41) |
Let
ω=(u2x+1)p−24, |
from (24), it follows that
|ω−1|L∞=|(u2x+1)2−p4|L∞≤C(|uxx|2−p2L2+1)≤CZ2γ(3p−4)(q−1)(t). |
Combining (35), (40) can be rewritten into
ddt∫Ω|ρ|ut|2dx+∫Ω|ωuxt|2dx≤2∫Ωρ|u||ut||uxt|dx+∫Ωρ|u||ux|2|ut|dx+∫Ω|ρx||u|2|ux||ut|dx+∫Ω|ρx||u||Ψx||ut|dx+∫Ωρ|ux||Ψx||ut|dx+∫Ωρ|ux||ut|2dx+∫ΩγP|ux||uxt|dx+∫Ω|Px||u||uxt|dx+∫Ω|ηt||uxt|dx+∫Ω|ηt||Φx||ut|dx+∫Ωρ|Ψxt||ut|dx=11∑j=1Ij. | (42) |
Using Sobolev inequality, Young's inequality,
ddt∫Ω|ρ|ut|2dx+∫Ω|ωuxt|2dx≤2∫Ωρ|u||ut||uxt|dx+∫Ωρ|u||ux|2|ut|dx+∫Ω|ρx||u|2|ux||ut|dx+∫Ω|ρx||u||Ψx||ut|dx+∫Ωρ|ux||Ψx||ut|dx+∫Ωρ|ux||ut|2dx+∫ΩγP|ux||uxt|dx+∫Ω|Px||u||uxt|dx+∫Ω|ηt||uxt|dx+∫Ω|ηt||Φx||ut|dx+∫Ωρ|Ψxt||ut|dx=11∑j=1Ij. | (42) |
ddt∫Ω|ρ|ut|2dx+∫Ω|ωuxt|2dx≤2∫Ωρ|u||ut||uxt|dx+∫Ωρ|u||ux|2|ut|dx+∫Ω|ρx||u|2|ux||ut|dx+∫Ω|ρx||u||Ψx||ut|dx+∫Ωρ|ux||Ψx||ut|dx+∫Ωρ|ux||ut|2dx+∫ΩγP|ux||uxt|dx+∫Ω|Px||u||uxt|dx+∫Ω|ηt||uxt|dx+∫Ω|ηt||Φx||ut|dx+∫Ωρ|Ψxt||ut|dx=11∑j=1Ij. | (42) |
In order to estimate
∫Ω[(ϵΨ2x+1Ψ2x+ϵ)2−q2Ψx]tΨxtdx=−4πg∫Ω(ρu)xΨtdx, | (43) |
and
∫Ω[(ϵΨ2x+1Ψ2x+ϵ)2−q2Ψx]tΨxtdx≥(q−1)∫Ω(Ψ2x+1)q−22|Ψxt|2dx. | (44) |
Let
βq=(Ψ2x+1)q−24 |
then
|(βq)−1|L∞=|(Ψ2x+1)2−q4|L∞≤C(|Ψxx|2−q2L2+1)≤CZ2−q2(q−1)(t). |
Then (43) can be rewritten into
∫Ω|βqΨxt|2dx≤C∫Ω(ρu)Ψxtdx≤C|ρ|L2|u|L∞|βqΨxt|L2|(βq)−1|L∞. |
Using Young's inequality, combining the above estimates we deduce that
I11≤|ρ|12L∞|√ρut|L2|βqΨxt|L2|(βq)−1|L∞≤CZ5q−32(q−1)(t). |
Substituting
|√ρut(t)|2L2+∫t0|ωuxt|2L2(s)ds≤|√ρut(τ)|2L2+∫t0Z26γ(3p−4)(q−1)(s)ds. | (45) |
To obtain the estimate of
∫Ωρ|ut|2dx≤2∫Ω(ρ|u|2|ux|2+ρ|Ψx|2+ρ−1|−[(εu2x+1u2x+ε)2−p2ux]x+(P+η)x+ηΦx|2)dx. |
According to the smoothness of
limτ→0∫Ω(ρ|u|2|ux|2+ρ|Ψx|2+ρ−1|−[(εu2x+1u2x+ε)2−p2ux]x+(P+η)x+ηΦx|2)dx=∫Ω(ρ0|u0|2|u0x|2+ρ0|Ψx|2+ρ−10|−[(εu20x+1u20x+ε)2−p2u0x]x+(P0+η0)x+η0Φx|2)dx≤|ρ0|L∞|u0|2L∞|u0x|2L2+|ρ0|L∞|Ψx|2+|g|2L2+|Φx|2L2≤C. |
Then, taking a limit on
|√ρut(t)|2L2+∫t0|uxt|2L2(s)ds≤C(1+∫t0Z26γ(3p−4)(q−1)(s)ds), | (46) |
This complete the proof of Lemma 2.5.
With the help of Lemma 2.2 to Lemma 2.5, and the definition of
Z(t)≤Cexp(˜C∫t0Z26γ(3p−4)(q−1)(s)ds), | (47) |
where
esssup0≤t≤T1(|ρ|H1+|u|W1,p0∩H2+|η|H2+|ηt|L2+|√ρut|L2+|ρt|L2)+∫T10(|√ρut|2L2+|uxt|2L2+|ηx|2L2+|ηt|2L2+|ηxt|2L2)ds≤C, | (48) |
where
In this section, the existence of strong solutions can be established by a standard argument. We construct the approximate solutions by using the iterative scheme, derive uniform bounds and thus obtain solutions of the original problem by passing to the limit. Our proof will be based on the usual iteration argument and some ideas developed in [10]. Precisely, we first define
ρkt+ρkxuk−1+ρkuk−1x=0, | (49) |
ρkukt+ρkuk−1ukx+ρkΨkx+Lpuk+Pkx+ηkx=−ηkΦx, | (50) |
LqΨk=4πg(ρk−m0), | (51) |
ηkt+(ηk(uk−1−Φx))x=ηkxx, | (52) |
with the initial and boundary conditions
(ρk,uk,ηk)|t=0=(ρ0,u0,η0), | (53) |
uk|∂Ω=(ηkx+ηkΦx)|∂Ω=0, | (54) |
where
Lpθk=−[(ε(θkx)2+1(θkx)2+ε)2−p2θkx]x. |
With the process, the nonlinear coupled system has been deduced into a sequence of decoupled problems and each problem admits a smooth solution. And the following estimates hold
esssup0≤t≤T1(|ρk|H1+|uk|W1,p0∩H2+|ηk|H2+|ηkt|L2+|√ρkukt|L2+|ρkt|L2)+∫T10(|√ρkukt|2L2+|ukxt|2L2+|ηkx|2L2+|ηkt|2L2+|ηkxt|2L2)ds≤C, | (55) |
where
In addition, we first find
ρkt+uk−1ρkx+uk−1xρk=0, |
ρk|t=0=ρ0, |
with smooth function
ρk(x,t)≥δexp[−∫T10|uk−1x(.,s)|L∞ds]>0,for all t∈(0,T1). |
Next, we will prove the approximate solution
ˉρk+1=ρk+1−ρk,ˉuk+1=uk+1−uk,ˉηk+1=ηk+1−ηk,ˉΨk+1=Ψk+1−Ψk. |
By a direct calculation, we can verify that the functions
ˉρk+1t+(ˉρk+1uk)x+(ρkˉuk)x=0, | (56) |
ρk+1ˉuk+1t+ρk+1ukˉuk+1x+(Lpuk+1−Lpuk)=−ˉρk+1(ukt+ukukx+Ψk+1x)−(Pk+1−Pk)x−ˉηk+1x+ρk(ˉukukx−ˉΨk+1x)−ˉηk+1Φx, | (57) |
LqΨk+1−LqΨk=4πgˉρk+1, | (58) |
ˉηk+1t+(ηkˉuk)x+(ˉηk+1(uk−Φx))x=ˉηk+1xx. | (59) |
Multiplying (56) by
ddt|ˉρk+1|2L2≤C|ˉρk+1|2L2|ukx|L∞+|ρk|H1|ˉukx|L2|ˉρk+1|L2≤C|ukxx|L2|ˉρk+1|2L2+Cξ|ρk|2H1|ˉρk+1|2L2+ξ|ˉukx|2L2≤Cξ|ˉρk+1|2L2+ξ|ˉukx|2L2, | (60) |
where
Multiplying (57) by
12ddt∫Ωρk+1|ˉuk+1|2dx+∫Ω(Lpuk+1−Lpuk)ˉuk+1dx≤C∫Ω[|ˉρk+1|(|ukt|+|ukukx|+|Ψk+1x|)+|Pk+1x−Pkx|+|ˉηk+1x|+|ρk|ˉuk||ukx|+|ρk||ˉΨk+1x|+|ˉηk+1Φx|]|ˉuk+1|dx≤C(|ˉρk+1|L2|ukxt|L2|ˉuk+1x|L2+|ˉρk+1|L2|ukx|Lp|ukxx|L2|ˉuk+1x|L2+|ˉρk+1|L2|Ψk+1x|L2|ˉuk+1x|L2+|Pk+1−Pk|L2|ˉuk+1x|L2+|ˉηk+1|L2|ˉuk+1x|L2+|ρk|12L2|√ρkˉuk|L2|ukxx|L2|ˉuk+1x|L2+|ρk|H1|ˉΨk+1x|L2|ˉuk+1x|L2+|ˉηk+1|L2|ˉuk+1x|L2). | (61) |
Let
σ(s)=(εs2+1s2+ε)2−p2s, |
then
σ′(s)=(εs2+1s2+ε)−p2(εs2+1)(s2+ε)−(2−p)(1−ε2)s2(s2+ε)2≥p−1(s2+ε)2−p2. |
To estimate the second term of (61), we have
∫Ω(Lpuk+1−Lpuk)ˉuk+1dx=∫Ω∫10σ′(θuk+1x+(1−θ)ukx)dθ|ˉuk+1x|2dx≥∫Ω[∫10dθ|θuk+1x+(1−θ)ukx|2−pL∞+1](ˉuk+1x)2≥C−1∫Ω|ˉuk+1x|2dx. | (62) |
On the other hand, multiplying (58) by
∫Ω(LqΨk+1−LqΨk)ˉΨk+1dx=4πg∫Ωˉρk+1ˉΨk+1dx. | (63) |
Since
∫Ω(LqΨk+1−LqΨk)ˉΨk+1xdx=(q−1)∫Ω(∫10|θΨk+1x+(1−θ)Ψkx|q−2dθ)(ˉΨk+1x)2dx, |
and
∫10|θΨk+1x+(1−θ)Ψkx|q−2dθ=∫101|θΨk+1x+(1−θ)Ψkx|2−qdθ≥∫101(|Ψk+1x|+|Ψkx|2−q)dθ=1(|Ψk+1x|+|Ψkx|)2−q, |
then
∫Ω[|Ψk+1x|q−2Ψk+1x−|Ψkx|q−2Ψkx]ˉΨk+1xdx≥1(|Ψk+1x(t)|L∞+|Ψkx(t)|L∞)2−q∫Ω(ˉΨk+1x)2dx, |
which implies
∫Ω(ˉΨk+1x)2dx≤C|ˉρk+1|2L2. | (64) |
From (55), (62) and (64), (61) can be re-written as
ddt∫Ωρk+1|ˉuk+1|2dx+C−1∫Ω|ˉuk+1x|2dx≤Bξ(t)|ˉρk+1|2L2+C(|√ρkˉuk|2L2+|ˉηk+1|2L2)+ξ|ˉuk+1x|2L2, | (65) |
where
∫t0Bξ(s)ds≤C+Ct. |
Multiplying (59) by
12ddt∫Ω|ˉηk+1|2dx+∫Ω|ˉηk+1x|2dx≤∫Ω|ˉηk+1||uk−Φx||ˉηk+1x|dx+∫Ω(|ηk||ˉuk|)x|ˉηk+1|dx≤|ˉηk+1|L2|uk−Φx|L∞|ˉηk+1x|L2+|ηkx|L2|ˉuk|L∞|ˉηk+1|L2+|ηk|L∞|ˉukx|L2|ˉηk+1|L2≤Cξ|ˉηk+1|2L2+ξ|ˉηk+1x|2L2+ξ|ˉukx|2L2. | (66) |
Combining (60), (65) and (66), we have
ddt(|ˉρk+1(t)|2L2+|√ρk+1ˉuk+1(t)|2L2+|ˉηk+1(t)|2L2)+|ˉuk+1x(t)|2L2+|ˉηk+1x|2L2≤Eξ(t)|ˉρk+1(t)|2L2+C|√ρkˉuk|2L2+Cξ|ˉηk+1|2L2+ξ|ˉukx|2L2, | (67) |
where
∫t0Eξ(s)ds≤C+Cξt. |
Integrating (67) over
|ˉρk+1(t)|2L2+|√ρk+1ˉuk+1(t)|2L2+|ˉηk+1(t)|2L2+∫t0|ˉuk+1x(t)|2L2ds+∫t0|ˉηk+1x|2L2ds≤Cexp(Cξt)∫t0(|√ρkˉuk(s)|2L2+|ˉukx(s)|2L2)ds. | (68) |
From the above recursive relation, choose
K∑k=1[sup0≤t≤T∗(|ˉρk+1(t)|2L2+|√ρk+1ˉuk+1(t)|2L2+|ˉηk+1(t)|2L2dt+∫T∗0|ˉuk+1x(t)|2L2+∫T∗0|ˉηk+1x(t)|2L2dt]<C, | (69) |
where
Therefore, as
ρk→ρεin L∞(0,T∗;L2(Ω)), | (70) |
uk→uεin L∞(0,T∗;L2(Ω))∩L2(0,T∗;H10(Ω)), | (71) |
ηk→ηεin L∞(0,T∗;L2(Ω))∩L2(0,T∗;H1(Ω)). | (72) |
By virtue of the lower semi-continuity of various norms, we deduce from the uniform estimate (55) that
esssup0≤t≤T1(|ρε|H1+|uε|W1,p0∩H2+|ηε|H2+|ηεt|L2+|√ρεuεt|L2+|ρεt|L2)+∫T∗0(|√ρεuεt|2L2+|uεxt|2L2+|ηεx|2L2+|ηεt|2L2+|ηεxt|2L2)ds≤C. | (73) |
Since all of the constants are independent of
ρε→ρδin L∞(0,T∗;L2(Ω)), | (74) |
uε→uδin L∞(0,T∗;L2(Ω))∩L2(0,T∗;H10(Ω)), | (75) |
ηε→ηδin L∞(0,T∗;L2(Ω))∩L2(0,T∗;H1(Ω)), | (76) |
and there also holds
esssup0≤t≤T1(|ρδ|H1+|uδ|W1,p0∩H2+|ηδ|H2+|ηδt|L2+|√ρδuδt|L2+|ρδt|L2)+∫T∗0(|√ρδuδt|2L2+|uδxt|2L2+|ηδx|2L2+|ηδt|2L2+|ηδxt|2L2)ds≤C. | (77) |
For each small
{Lpuδ0+(P(ρδ0)+ηδ0)x+ηδ0Φx=(ρδ0)12(gδ+Φx),uδ0|∂Ω=0, | (78) |
where
We deduce that
{ρt+(ρu)x=0,(ρu)t+(ρu2)x+ρΨx−λ(|ux|p−2ux)x+(P+η)x=−ηΦx,(|Ψx|q−2Ψx)x=4πg(ρ−1|Ω|∫Ωρdx),ηt+(η(u−Φx))x=ηxx,(ρ,u,η)|t=0=(ρδ0,uδ0,ηδ0),u|∂Ω=(ηx+ηΦx)|∂Ω=0, |
where
By the proof of Lemma 2.1, there exists a subsequence
esssup0≤t≤T1(|ρ|H1+|u|W1,p0∩H2+|η|H2+|ηt|L2+|√ρut|L2+|ρt|L2)+∫T∗0(|√ρut|2L2+|uxt|2L2+|ηx|2L2+|ηt|2L2+|ηxt|2L2)ds≤C, | (79) |
where
The authors would like to thank the anonymous referees for their valuable suggestions.
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