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Existence of solutions for a shear thickening fluid-particle system with non-Newtonian potential

Yukun Song
/ Fengming Liu
Published Online: 2018-06-27 | DOI: https://doi.org/10.1515/math-2018-0064

Abstract

This paper is concerned with a compressible shear thickening fluid-particle interaction model for the evolution of particles dispersed in a viscous non-Newtonian fluid. Taking the influence of non-Newtonian gravitational potential into consideration, the existence and uniqueness of strong solutions are established.

MSC 2010: 76A05; 76A10

1 Introduction

We consider a compressible non-Newtonian fluid-particle interaction model which reads as follows $ρt+(ρu)x=0,(ρu)t+(ρu2)x+ρΨx−λ[(ux2+μ1)p−22ux]x+(P+η)x=−ηΦx,(x,t)∈ΩT(|Ψx|q−2Ψx)x=4πg(ρ−1|Ω|∫Ωρdx),ηt+[η(u−Φx)]x=ηxx$(1)

with the initial and boundary conditions $(ρ,u,η)|t=0=(ρ0,u0,η0),x∈Ω,u|∂Ω=Ψ|∂Ω=0,t∈[0,T],$(2)

and the no-flux condition for the density of particles $(ηx+ηΦx)|∂Ω=0,t∈[0,T].$(3)

where ρ, u, η, P(ρ) = γ denote the fluid density, velocity, the density of particle in the mixture and pressure respectively, Ψ denotes the non-Newtonian gravitational potential and the given function Φ(x) denotes the external potential. a > 0, γ > 1, μ1 > 0, p > 2, 1 < q < 2, λ > 0 is the viscosity coefficient and β ≠ 0 is a constant. Ω is a one-dimensional bounded interval, for simplicity we only consider Ω = (0, 1), ΩT = Ω × [0, T].

In fact, there are extensive studies concerning the theory of strong and weak solutions for the multi-dimensional fluid-particle interaction models for the newtonian case. In [1], Carrillo et al. discussed the global existence and asymptotic behavior of the weak solutions for a fluid-particle interaction model. Subsequently, Fang et al. [2] obtained the global classical solution in dimension one. In dimension three, Ballew and Trivisa [3, 4] established the global existence of weak solutions and the existence of weakly dissipative solutions under reasonable physical assumptions on the initial data. In addition, Constantin and Masmoudi [5] obtained the global existence of weak solutions for a coupled incompressible fluid-particle interaction model in 2D case followed the spirit of reference [6].

The non-Newtonian fluid is an important type of fluid because of its immense applications in many fields of engineering fluid mechanics such as inks, paints, jet fuels etc., and biological fluids such as blood (see [7]). Many researchers turned to the study of this type of fluid under different conditions both theoretically and experimentally. For details, we refer the readers to [8, 9, 10, 11, 12] and the references therein. To our knowledge, there seems to be a very few mathematical results for the case of the fluid-interaction model systems with non-Newtonian gravitational potential. There are still no existence results to problem (1)-(3) when p > 2, 1 < q < 2 which describes that the motion of the compressible viscous isentropic gas flow is driven by a non-Newtonian gravitational force.

We are interested in the existence and uniqueness of strong solutions on a one dimensional bounded domain. The strong nonlinearity of (1) bring us new difficulties in getting the upper bound of ρ and the method used in [2] is not suitable for us. Motivated by the work of Cho et al. [13, 14] on Navier-Stokes equations, we establish local existence and uniqueness of strong solutions by the iteration techniques.

Throughout the paper we assume that a = λ = 1. In the following sections, we will use simplified notations for standard Sobolev spaces and Bochner spaces, such as $\begin{array}{}{L}^{p}={L}^{p}\left(\mathit{\Omega }\right),{H}_{0}^{1}={H}_{0}^{1}\left(\mathit{\Omega }\right),C\left(0,T;{H}^{1}\right)\end{array}$ = C(0, T;H1(Ω)).

We state the definition of strong solution as follows:

Definition 1.1

The (ρ, u, Φ, η) is called a strong solution to the initial boundary value problem (1)-(3), if the following conditions are satisfied:

1. $ρ∈L∞(0,T∗;H1(Ω)),u∈L∞(0,T∗;W01,p(Ω)∩H2(Ω)),η∈L∞(0,T∗;H2(Ω)),Ψ∈L∞(0,T∗;H2(Ω)),ρt∈L∞(0,T∗;L2(Ω)),ut∈L2(0,T∗;H01(Ω)),Ψt∈L∞(0,T∗H1(Ω)),ρut∈L∞(0,T∗;L2(Ω)),ηt∈L∞(0,T∗;L2(Ω)),((ux2+μ1)p−22ux)x∈L2(0,T∗;L2(Ω))$

2. For all φL(0, T*;H1(Ω)), φtL(0, T*;L2(Ω)), for a.e. t ∈ (0, T), we have $∫Ωρφ(x,t)dx−∫0t∫Ω(ρφt+ρuφx)(x,s)dxds=∫Ωρ0φ(x,0)dx$(4)

3. For all $\begin{array}{}\varphi \in {L}^{\mathrm{\infty }}\left(0,{T}_{\ast };{W}_{0}^{1,p}\left(\mathit{\Omega }\right)\cap {H}^{2}\left(\mathit{\Omega }\right)\right),{\varphi }_{t}\in {L}^{2}\left(0,{T}_{\ast };{H}_{0}^{1}\left(\mathit{\Omega }\right)\right),\end{array}$ for a.e. t ∈ (0, T), we have $∫Ωρuϕ(x,t)dx−∫0t∫Ω{ρu2ϕx−ρΨxϕ−λ(ux2+μ1)p−22uxϕx+(P+η)ϕx+ηΦxϕ}(x,s)dxds=∫Ωρ0u0ϕ(x,0)dx$(5)

4. For all ϑL(0, T*;H2(Ω)), ϑtL(0, T*H1(Ω)), for a.e. t∈ (0, T), we have $−∫0t∫Ω|Ψx|q−2Ψxϑx(x,s)dxds=∫0t∫Ω4πg(ρ−1|Ω|∫Ωρdx)ϑ(x,0)dxds$(6)

5. For all ψL(0, T*;H2(Ω)), ψtL(0, T*;L2(Ω)), for a.e. t ∈ (0, T), we have $∫Ωηψ(x,t)dx−∫0t∫Ω[η(u−Φx)−ηx]ψx(x,s)dxds=∫Ωη0ψ(x,0)dx$(7)

1.1 Main results

Theorem 1.2

Let μ1 > 0 be a positive constant and ΦC2(Ω), and assume that the initial data (ρ0, u0, η0) satisfy the following conditions $0≤ρ0∈H1(Ω),u0∈H01(Ω)∩H2(Ω),η0∈H2(Ω)$

and the compatibility condition $−[(u0x2+μ1)p−22u0x]x+(P(ρ0)+η0)x+η0Φx=ρ012(g+βΦx),$(8)

for some gL2(Ω). Then there exist a T*∈(0, +∞) and a unique strong solution (ρ, u, η) to (1)-(3) such that $ρ∈L∞(0,T∗;H1(Ω)),ρt∈L∞(0,T∗;L2(Ω)),u∈L∞(0,T∗;W01,p(Ω)∩H2(Ω)),ut∈L2(0,T∗;H01(Ω)),η∈L∞(0,T∗;H2(Ω)),ηt∈L∞(0,T∗;L2(Ω)),Ψ∈L∞(0,T∗;H2(Ω)),Ψt∈L∞(0,T∗H1(Ω)),ρut∈L∞(0,T∗;L2(Ω)),((ux2+μ1)p−22ux)x∈L2(0,T∗;L2(Ω)).$(9)

2 A priori Estimates for Smooth Solutions

In this section, we will prove the local existence of strong solutions. By virtue of the continuity equation (1)1, we deduce the conservation of mass $∫Ωρ(t)dx=∫Ωρ0dx:=m0,(t>0,m0>0).$

Provided that (ρ, u, η) is a smooth solution of (1)-(3) and ρ0δ, where 0 < δ ≪ 1 is a positive number. We denote by $\begin{array}{}{M}_{0}=1+{\mu }_{1}+{\mu }_{1}^{-1}+|{\rho }_{0}{|}_{{H}^{1}}+|g{|}_{{L}^{2}},\end{array}$ and introduce an auxiliary function $Z(t)=sup0≤s≤t(1+|u(s)|W01,p+|ρ(s)|H1+|ηt(s)|L2+|η(s)|H1+|ρut(s)|L2).$

Then we estimate each term of Z(t) in terms of some integrals of Z(t), apply arguments of Gronwall-type and thus prove that Z(t) is locally bounded.

2.1 Estimate for $\begin{array}{}|u{|}_{{W}_{0}^{1,p}}\end{array}$

By using (1)1, we rewrite the (1)2 as $ρut+ρuux+ρΨx−[(ux2+μ1)p−22ux]x+(P+η)x=−ηΦx.$(10)

Multiplying (10) by ut, integrating (by parts) over ΩT, we have $∬ΩTρ|ut|2dxds+∬ΩT(ux2+μ1)p−22uxuxtdxds=−∬ΩT(ρuux+ρΨx+Px+ηx+ηΦx)utdxds.$(11)

We deal with each term as follows: $∫Ω(ux2+μ1)p−22uxuxtdx=12∫Ω(ux2+μ1)p−22(ux2)tdx=12ddt∫Ω(∫0ux2(s+μ1)p−22ds)dx∫0ux2(s+μ1)p−22ds=∫μ1ux2+μ1tp−22dt=2p[(ux2+μ1)p2−μ1p2]≥2p|ux|p−2pμ1p2$ $−∬ΩTPxutdxds=∬ΩTPuxtdxds=ddt∬ΩTPuxdxds−∬ΩTPtuxdxds.$

Since from (1)1 we get $Pt=−γPux−Pxu$(12) $−∬ΩT(ηx+ηΦx)utdxds=ddt∬ΩT(ηx+ηΦx)udxds−∬ΩT(ηx+ηΦx)tudxds$ $−∬ΩT(ηx+ηΦx)tudxds=∬ΩTηt(ux−Φxu)dxds=−∬ΩT[ηx−η(u−Φx)](ux−Φxu)xdxds.$

Substituting the above into (11), we obtain $∫0t|ρut(s)|L22ds+1p∫Ω|ux(t)|pdx≤C+∫Ω|Pux|dx+∬ΩT(|ρuuxut|+|ρΨxut|+|γPux2|+|Pxuux|)dxds+∬ΩT(|ηxuxx|+|ηxΦxux|+|ηxΦxxu|+|ηuuxx|+|ηu2Φxx|+|ηuΦxux|+|ηΦxuxx|+|ηΦxΦxxu|+|ηΦx2ux|)dxds.$

Using Young’s inequality, we obtain $∫0t|ρut(s)|L22ds+|ux(t)|Lpp≤C+C∫0t(|ρ|L∞|u|L∞2|ux|Lp2+|ρ|L∞|Ψxx|L22+|P|L∞|ux|Lp2+|Px|L2|u|L∞|ux|Lp+|ηx|L2|uxx|L2+|ηx|L2|ux|Lp+|ηx|L2|u|L∞+|η|L∞|u|L∞|uxx|L2+|η|L∞|u|L∞2+|η|L∞|u|L∞|ux|Lp+|η|L∞|uxx|L2+|η|L∞|u|L∞+|η|L∞|ux|Lp)ds+C|P(t)|L2pp−1.$(13)

On the other hand, multiplying (1)3 by Ψ and integrating over Ω, we get $∫Ω|Ψx|qdx=−∫Ω(|Ψx|q−2Ψx)xΨdx=−4πg(∫ΩρΨdx−m0∫ΩΨdx)≤8πgm0|Ψ|L∞≤8πgm0|Ψx|Lq≤1q|Φx|Lqq+1p(8π.gm0)p$

Then we have $∫Ω|Ψx|qdx≤C(m0),1

Differentiating (1)3 with respect to x, multiplying it by Ψx and integrating over Ω, we have $∫Ω(|Ψx|q−2Ψx)xΨxxdx=−4πg∫ΩρxΨxdx.$

By virtue of $∫Ω(|Ψx|q−2Ψx)xΨxxdx=∫Ω(|Ψx|q−4[(q−2)Ψx2+Ψx2]Ψxx2dx≥C∫Ω|Ψx|q−2Ψxx2dx≥C|Ψx|L∞q−2|Ψxx|L22≥C|Ψxx|L2q−2|Ψxx|L22≥C|Ψxx|L2q$

and $−4πg∫ΩρxΨxdx≤C∫Ω|ρx||Ψx|dx≤C|ρx|L2p+C|Ψx|L2q≤C|ρx|L2p+C(ε)|Ψxx|L2q.$

Therefore, $|Ψxx|L2≤CZpq(t).$(14)

We deal with the term of |uxx|L2. Notice that $|[(ux2+μ1)p−22ux]x|≥μ1p−22|uxx|.$

Then $|uxx|≤C|ρut+ρuux+ρΨx+(P+η)x+ηΦx|.$

Taking the above inequality by L2 norm, we get $|uxx|L2≤C|ρut+ρuux+ρΨx+(P+η)x+ηΦx|L2≤C(|ρ|L∞12|ρut|L2+|ρ|L∞|u|L∞|ux|L2+|ρ|L2|Ψxx|L2+|Px|L2+|ηx|L2+|η|L2|Φx|L2).$

Hence, we deduce that $|uxx|L2≤CZmax(pq+1,3)(t).$(15)

Moreover, using (1)1, we have $|P(t)|L2pp−1≤∫Ω|P(t)|2dx=∫Ω|P(0)|2dx+∫0t∂∂s(∫ΩP(s)2dx)ds≤∫Ω|P(0)|2dx+2∫0t∫ΩP(s)γργ−1(−ρxu−ρux)dxds≤C+C∫0t|P|L∞|ρ|L∞γ−1|ρ|H1|ux|Lpds≤C(1+∫0tZ2γ+1(s)ds).$(16)

Combining (13)-(16), yields $∫0t|ρut(s)|L22(s)ds+|ux(t)|Lpp≤C(1+∫0tZmax(2pq,2γ+3)(s)ds),$(17)

where C is a positive constant, depending only on M0.

2.2 Estimate for |ρ|H1

From (1)3, taking it by L2 norm, we get $|ηxx|L2≤|ηt+(η(u−Φx))x|L2≤|ηt|L2+|ηx|L2|u|L∞+C|ηx|L2+|η|H1|ux|Lp+C|η|H1≤CZ2(t).$(18)

Multiplying (1)1 by ρ, integrating over Ω, we have $12ddt∫Ω|ρ|2ds+∫Ω(ρu)xρdx=0.$

Integrating by parts, using Sobolev inequality, we deduce that $ddt|ρ(t)|L22≤∫Ω|ux||ρ|2dx≤|uxx|L2|ρ|L22.$(19)

Differentiating (1)1 with respect to x, and multiplying it by ρx, integrating over Ω, using Sobolev inequality, we have $ddt∫Ω|ρx|2dx=−∫Ω[32ux(ρx)2+ρρxuxx](t)dx≤C[|ux|L∞|ρx|L22+|ρ|L∞|ρx|L2|uxx|L2]≤C|ρ|H12|uxx|L2.$(20)

From (19) and (20), by Gronwall’s inequality, it follows that $sup0≤t≤T|ρ(t)|H12≤|ρ0|H12exp⁡{C∫0t|uxx|L2ds}≤Cexp⁡(C∫0tZmax(pq+1,3)(s)ds).$(21)

Besides, by using (1)1, we can also get the following estimates: $|ρt(t)|L2≤|ρx(t)|L2|u(t)|L∞+|ρ(t)|L∞|ux(t)|L2≤CZ2(t).$(22)

2.3 Estimate for |ηt|L2 and |η|H1

Multiplying (1)4 by η, integrating the resulting equation over ΩT, using the boundary conditions (3), Young’s inequality, we have $∫0t|ηx(s)|L22ds+12|η(t)|L22≤∬ΩT(|ηuηx|+|ηΦxηx|)dxds≤14∫0t|ηx(s)|L22ds+C∫0t|ux|Lp2|η|H12ds+C∫0t|η|H12+C≤14∫0t|ηx(s)|L22ds+C(1+∫0tZ4(t)ds).$(23)

Multiplying (1)4 by ηt, integrating (by parts) over ΩT, using the boundary conditions (3), Young’s inequality, we have $∫0t|ηt(s)|L22ds+12|ηx(t)|L22≤∬ΩT|η(u−Φx)ηxt|dxds≤14∫0t|ηxt(s)|L22ds+C∫0t|η|H12|ux|Lp2ds+C∫0t|η|H12ds+C≤14∫0t|ηxt(s)|L22ds+C(1+∫0tZ4(t)ds).$(24)

Differentiating (1)4 with respect to t, multiplying the resulting equation by ηt, integrating (by parts) over ΩT, we get $∫0t|ηxt(s)|L22ds+12|ηt(t)|L22=∬ΩT(η(u−Φx))tηxtdxds≤C+∬ΩT(|ηtuηxt|+|ηtΦxηxt|+|ηxutηt|+|ηuxtηt|)dxds≤C(1+∫0t(|ηt|L22||ux|Lp2+|ηt|L22+|ηx|L22|ηt|L22+|η|H12|ηt|L22)dx)+12∫0t|ηxt|L22+12∫0t|uxt|L22≤C(1+∫0tZ2γ+6(s)ds).$(25)

Combining (23)-(25), we get $|η|H12+|ηt|L22+∫0t(|ηx|L22+|ηt|L22+|ηxt|L22)(s)ds≤C(1+∫0tZ2γ+6(s)ds).$(26)

2.4 Estimate for $\begin{array}{}|\sqrt{\rho }{u}_{t}{|}_{{L}^{2}}\end{array}$

Differentiating equation (10) with respect to t, multiplying the result equation by ut, and integrating it over Ω with respect to x, we have $12ddt∫Ωρ|ut|2dx+∫Ω[(ux2+μ1)p−22ux]tuxtdx=∫Ω[(ρu)x(ut2+uuxut+Ψxut)−ρuxut2−ρΨxtut−(P+η)tuxt−ηtΦxut]dx.$(27)

Note that $[(ux2+μ1)p−22ux]tuxt=(ux2+μ1)p−42(p−1)(ux2+μ1)uxt2≥μ1p−22uxt2.$

Combining (12), (27) can be rewritten into $12ddt∫Ωρ|ut|2dx+∫Ω|uxt|2dx≤2∫Ωρ|u||ut||uxt|dx+∫Ωρ|u||ux|2|ut|dx+∫Ωρ|u|2|uxx||ut|dx+∫Ωρ|u|2|ux||uxt|dx+∫Ωρ|u||Ψxx||ut|dx+∫Ωρ|u||Ψx||uxt|dx+∫Ωρ|ux||ut|2dx+∫Ωγ|P||ux||uxt|dx+∫Ω|Px||u||uxt|dx+∫Ω|ηt||uxt|dx+∫Ω|ηt||Φx||ut|dx+∫Ωρ|Ψxt||ut|dx=∑j=112Ij.$(28)

By using Sobolev inequality, Hölder inequality and Young’s inequality, (14), (15), we estimate each term of Ij as follows $I1=2∫Ωρ|u||ut||uxt|dx≤2|ρ|L∞12|u|L∞|ρut|L2|uxt|L2≤CZ5(t)+17|uxt|L22I2=∫Ωρ|u||ux|2|ut|dx≤|ρ|L∞12|u|L2|ux|Lp2|ρut|L2≤CZ5(t)I3=∫Ωρ|u|2|uxx||ut|dx≤|ρ|L∞12|u|L∞2|uxx|L2|ρut|L2≤CZmax(pq+5,7)(t)I4=∫Ωρ|u|2|ux||uxt|dx≤|ρ|L∞|u|L∞2|ux|L2|uxt|L2≤CZ8(t)+17|uxt|L22I5=∫Ωρ|u||Ψxx||ut|dx≤|ρ|L∞12|u|L∞|Ψxx|L2|ρut|L2≤CZqp+3(t)I6=∫Ωρ|u||Ψx||uxt|dx≤|ρ|L∞|u|L∞|Ψxx|L2|uxt|L2≤CZ2pq+4(t)+17|uxt|L22I7=∫Ωρ|ux||ut|2dx≤|ux|L∞|ρut|L22≤CZmax(pq+3,5)(t)I8=∫Ωγ|P||ux||uxt|dx≤C|P|L∞|ux|Lp|uxt|L2≤CZ2γ+2(t)+17|uxt|L22I9=∫Ω|Px||u||uxt|dx≤|Px|L2|u|L∞|uxt|L2≤CZ2γ+2(t)+17|uxt|L22I10=∫Ω|ηt||uxt|dx≤|ηt|L2|uxt|L2≤CZ2(t)+17|uxt|L22I11=∫Ω|ηt||Φx||ut|dx≤|ηt|L2|Φx|L2|ut|L∞≤CZ2(t)+17|uxt|L22I12=∫Ωρ|Ψxt||ut|dx≤C|ρ|L∞12|Ψxt|L2|ρut|L2,$

where C is a positive constant, depending only on M0.

Next, we deal with the term |Ψxt|L2 of I12. Differentiating (1)3 with respect to t, multiplying it by Ψt, integrating over Ω and using Young’s inequality, we obtain $∫Ω(|Ψx|q−2Ψx)tΨxtdx=−4πg∫ΩρtΨtdx.$

By virtue of $∫Ω(|Ψx|q−2Ψx)tΨxtdx=∫Ω(|Ψx|q−4[(q−2)Ψx2+Ψx2]Ψxt2dx≥C∫Ω|Ψx|q−2Ψxt2dx≥C|Ψx|L∞q−2|Ψxt|L22≥C|Ψxx|L2q−2|Ψxt|L22$

and $−4πg∫ΩρtΨtdx≤C∫Ω|ρt||Ψt|dx≤C|ρt|L22+C|Ψt|L22≤C|ρt|L22+C(ε)|Ψxt|L22,$

then $\begin{array}{}|{\mathit{\Psi }}_{xt}{|}_{{L}^{2}}\le C{Z}^{2-\frac{p\left(2-q\right)}{q}}\left(t\right).\end{array}$ Therefore, $I12=∫Ωρ|Ψxt||ut|dx≤C|ρ|L∞12|Ψxt|L2|ρut|L2≤Z4+p(2−q)q(t).$

Substituting Ij(j = 1, 2, …, 12) into (28), and integrating over (τ, t) ⊂ (0, T) on the time variable, we have $|ρut(t)|L22+∫τt|uxt|L22(s)ds≤|ρut(τ)|L22+C∫τtZmax(2pq+4,8)(s)ds.$(29)

To obtain the estimate of $\begin{array}{}|\sqrt{\rho }{u}_{t}\left(t\right){|}_{{L}^{2}}^{2},\end{array}$ we need to estimate $\begin{array}{}\underset{\tau \to 0}{lim}|\sqrt{\rho }{u}_{t}\left(\tau \right){|}_{{L}^{2}}^{2}.\end{array}$ Multiplying (10) by ut and integrating over Ω, we have $∫Ωρ|ut|2dx≤2∫Ω(ρ|u|2|ux|2+ρ|Ψx|2+ρ−1|−[(ux2+μ1)p−22ux]x+(P+η)x+ηΦx|2)dx.$

According to the smoothness of (ρ, u, η), we obtain $limτ→0∫Ω(ρ|u|2|ux|2+ρ|Ψx|2+ρ−1|−[(ux2+μ1)p−22ux]x+(P+η)x+ηΦx|2)dx=∫Ω(ρ|u0|2|u0x|2+ρ0|Ψx|2+ρ0−1|−[(u0x2+μ1)p−22u0x]x+(P0+η0)x+η0Φx|2)dx≤|ρ0|L∞|u0|L∞2|u0x|L22+|ρ0|L∞|Ψx|L22+|g|L22+β|Φx|L22≤C.$

Therefore, taking a limit on τ in (29), as τ → 0, we conclude that $|ρut(t)|L22+∫0t|uxt|L22(s)ds≤C(1+∫0tZmax(2pq+4,8)(s)ds),$(30)

where C is a positive constant, depending only on M0.

Combining the estimates of (15), (18), (21), (22), (17), (26), (30) and the definition of Z(t), we conclude that $Z(t)≤C~exp⁡(C~~∫0tZmax(2pq+4,8)(s)ds),$(31)

where , c͂͂ are positive constants, depending only on M0. This means that there exist a time T1 > 0 and a constant C > 0, such that $esssup0≤t≤T1(|ρ|H1+|u|W01,p∩H2+|η|H2+|ηt|L2+|ρut|L2+|ρt|L2)+∫0T1(|ρut|L22+|uxt|L22+|ηx|L22+|ηt|L22+|ηxt|L22)ds≤C.$(32)

3 Proof of the Main Theorem

In this section, our proof will be based on the usual iteration argument and some ideas developed in [13, 14]. Precisely, we construct the approximate solutions, by using the iterative scheme, inductively, as follows: first define u0 = 0 and assuming that uk−1 was defined for k ≥ 1, let ρk, uk, ηk be the unique smooth solution to the following problems: $ρtk+ρxkuk−1+ρkuxk−1=0ρkutk+ρkuk−1uxk+ρkΨxk−[((uxk)2+μ1)p−22uxk]x+Pxk+ηxk=−ηkΦx(|Ψxk|q−2Ψxk)x=4πg(ρk−m0)ηtk+(ηk(uk−1−Φx))x=ηxxk$

with the initial and boundary conditions $(ρk,uk,ηk)|t=0=(ρ0,u0,η0)uk|∂Ω=(ηxk+ηkΦx)|∂Ω=0$

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 $esssup0≤t≤T1(|ρk|H1+|uk|W01,p∩H2+|ηk|H2+|ηtk|L2+|ρkutk|L2+|ρtk|L2)+∫0T1(|ρkutk|L22+|uxtk|L22+|ηxk|L22+|ηtk|L22+|ηxtk|L22)ds≤C,$(33)

where C is a generic constant depending only on M0, but independent of k.

In addition, we first find ρk from the initial problem $ρtk+uk−1ρxk+uxk−1ρk=0,andρk|t=0=ρ0$

with smooth function uk−1, obviously, there is a unique solution ρk to the above problem and also by a standard argument, we could obtain that $ρk(x,t)≥δexp⁡[−∫0T1|uxk−1(.,s)|L∞ds]>0,for all t∈(0,T1).$

Next, we have to prove that the approximate solution (ρk, uk, ηk) converges to a solution to the original problem (1) in a strong sense. To this end, let us define $ρ¯k+1=ρk+1−ρk,u¯k+1=uk+1−uk,Ψ¯k+1=Ψk+1−Ψk,η¯k+1=ηk+1−ηk,$

then we can verify that the functions ρk+1, uk+1, ηk+1 satisfy the system of equations $ρ¯tk+1+(ρ¯k+1uk)x+(ρku¯k)x=0$(34) $ρk+1u¯tk+1+ρk+1uku¯xk+1−([((uxk+1)2+μ1)p−22uxk+1]x−[((uxk)2+μ1)p−22uxk]x)=−ρ¯k+1(utk+ukuxk+Ψxk+1)−(Pk+1−Pk)x+ρk(u¯kuxk−Ψ¯xk+1)−η¯xk+1−η¯k+1Φx$(35) $(|Ψxk+1|q−2Ψxk+1)x−(|Ψxk|q−2Ψxk)x=4πgρ¯k+1$(36) $η¯tk+1+(ηku¯k)x+(η¯k+1(uk−Φx))x=η¯xxk+1.$(37)

Multiplying (34) by ρk+1, integrating over Ω and using Young’s inequality, we obtain $ddt|ρ¯k+1|L22≤C|ρ¯k+1|L22|uxk|L∞+|ρk|H1|u¯xk|L2|ρ¯k+1|L2≤C|uxxk|L2|ρ¯k+1|L22+Cζ|ρk|H12|ρ¯k+1|L22+ζ|u¯xk|L22≤Cζ|ρ¯k+1|L22+ζ|u¯xk|L22,$(38)

where Cζ is a positive constant, depending on M0 and ζ for all t < T1 and k ≥ 1.

Multiplying (35) by uk+1, integrating over Ω and using Young’s inequality, we obtain $12ddt∫Ωρk+1|u¯k+1|2dx+∫Ω([((uxk+1)2+μ1)p−22uxk+1]x−[((uxk)2+μ1)p−22uxk]x)u¯k+1dx≤C∫Ω(|ρ¯k+1|(|utk|+|ukuxk|+|Ψxk+1|)|u¯k+1|+|Pxk+1−Pxk||u¯k+1|+|ρk|u¯k||uxk||u¯k+1|+|ρk||Ψ¯xk+1||u¯k+1|+|η¯xk+1||u¯k+1|+|η¯k+1Φx||u¯k+1|)dx.$(39)

Let $σ(s)=(s2+μ1)p−22s,$

then $σ′(s)=((s2+μ1)p−22s)′=(s2+μ1)p−42((p−1)s2+μ1)≥μ1p−22.$

We estimate the second term of (39) as follows $∫Ω([((uxk+1)2+μ1)p−22uxk+1]x−[((uxk)2+μ1)p−22uxk]x)u¯k+1dx=∫Ω∫01σ′(θuxk+1+(1−θ)uxk)dθ|u¯xk+1|2dx≥μ1p−22∫Ω|u¯xk+1|2dx.$(40)

Similarly, multiplying (36) by Ψk+1, integrating over Ω, we get $∫Ω[(|Ψxk+1|q−2Ψxk+1)x−(|Ψxk|q−2Ψxk)x]Ψ¯k+1dx=4πg∫Ωρ¯k+1Ψ¯k+1dx,$(41)

since $∫Ω[|Ψxk+1|q−2Ψxk+1−|Ψxk|q−2Ψxk]Ψ¯xk+1dx=(q−1)∫Ω(∫01|θΨxk+1+(1−θ)Ψxk|q−2dθ)(Ψ¯xk+1)2dx$

and $∫01|θΨxk+1+(1−θ)Ψxk|q−2dθ=∫011|θΨxk+1+(1−θ)Ψxk|2−qdθ≥∫011(|Ψxk+1|+|Ψxk|2−q)dθ=1(|Ψxk+1|+|Ψxk|)2−q.$

Then $∫Ω[|Ψxk+1|q−2Ψxk+1−|Ψxk|q−2Ψxk]Ψ¯xk+1dx≥1(|Ψxk+1(t)|L∞+|Ψxk(t)|L∞)2−q∫Ω(Ψ¯xk+1)2dx.$

That means (41) turns into $∫Ω(Ψ¯xk+1)2dx≤C|ρ¯k+1|L22.$(42)

Substituting (40) and (42) into (39), using Young’s inequality, yields $ddt∫Ωρk+1|u¯k+1|2dx+∫Ω|u¯xk+1|2dx≤C(|ρ¯k+1|L2|uxtk|L2|u¯xk+1|L2+|ρ¯k+1|L2|uxk|Lp|uxxk|L2|u¯xk+1|L2+|ρ¯k+1|L2|Ψxk+1|L2|u¯xk+1|L2+|Pk+1−Pk|L2|u¯xk+1|L2+|ρk|L212|ρku¯k|L2|uxxk|L2|u¯xk+1|L2+|ρk|L∞|Ψ¯xk+1|L2|u¯xk+1|L2+|η¯k+1|L2|u¯xk+1|L2+|η¯k+1|L2|u¯xk+1|L2)≤Bζ(t)|ρ¯k+1|L22+C(|ρku¯k|L22+|η¯k+1|L22)+ζ|u¯xk+1|L22,$(43)

where $\begin{array}{}{B}_{\zeta }\left(t\right)=C\left(1+|{u}_{xt}^{k}\left(t\right){|}_{{L}^{2}}^{2}\right),\end{array}$ for all tT1 and k ≥ 1. Using (33) we derive $∫0tBζ(s)ds≤C+Ct.$

Multiplying (37) by ηk+1, integrating over Ω, using (33) and Young’s inequality, we have $12ddt∫Ω|η¯k+1|2dx+∫Ω|η¯xk+1|2dx≤∫Ω|η¯k+1||uk−Φx||η¯xk+1|dx+∫Ω(|ηk||u¯k|)x|η¯k+1|dx≤|η¯k+1|L2|uk−Φx|L∞|η¯xk+1|L2+|ηxk|L2|u¯k|L∞|η¯k+1|L2+|ηk|L∞|u¯xk|L2|η¯k+1|L2≤Cζ|η¯k+1|L22+ζ|η¯xk+1|L22+ζ|u¯xk|L22.$(44)

Collecting (38), (43) and (44), we obtain $ddt(|ρ¯k+1(t)|L22+|ρk+1u¯k+1(t)|L22+|η¯k+1(t)|L22)+|u¯xk+1(t)|L22+|η¯xk+1|L22≤Eζ(t)|ρ¯k+1(t)|L22+C|ρku¯k|L22+Cζ|η¯k+1|L22+ζ|u¯xk|L22,$(45)

where Eζ(t) depends only on Bζ(t) and Cζ, for all tT1 and k ≥ 1. Using (33), we have $∫0tEζ(s)ds≤C+Cζt.$

Integrating (45) over (0, t) ⊂ (0, T1) with respect to t, using Gronwall’s inequality, we have $|ρ¯k+1(t)|L22+|ρk+1u¯k+1(t)|L22+|η¯k+1(t)|L22+∫0t|u¯xk+1(t)|L22ds+∫0t|η¯xk+1|L22ds≤Cexp⁡(Cζt)∫0t(|ρku¯k(s)|L22+|u¯xk(s)|L22)ds.$(46)

From the above recursive relation, choose ζ > 0 and 0 < T* < T1 such that C exp $\begin{array}{}\left({C}_{\zeta }{T}_{\ast }\right)<\frac{1}{2},\end{array}$ using Gronwall’s inequality, we deduce that $∑k=1K[sup0≤t≤T∗(|ρ¯k+1(t)|L22+|ρk+1u¯k+1(t)|L22+|η¯k+1(t)|L22dt+∫0T∗|u¯xk+1(t)|L22+∫0T∗|η¯xk+1(t)|L22dt](47)

Since all of the constants do not depend on δ, as k → ∞, we conclude that sequence (ρk, uk, ηk) converges to a limit (ρδ, uδ, ηδ) in the following convergence $ρ→ρδin L∞(0,T∗;L2(Ω)),$(48) $u→uδinL∞(0,T∗;L2(Ω))∩L2(0,T∗;H01(Ω)),$(49) $η→ηδinL∞(0,T∗;L2(Ω))∩L2(0,T∗;H1(Ω)),$(50)

and there also holds $esssup0≤t≤T1(|ρδ|H1+|uδ|W01,p∩H2+|ηδ|H2+|ηtδ|L2+|ρδutδ|L2+|ρtδ|L2)+∫0T∗(|ρδutδ|L22+|uxtδ|L22+|ηxδ|L22+|ηtδ|L22+|ηxtδ|L22)ds≤C.$(51)

For each small δ > 0, let $\begin{array}{}{\rho }_{0}^{\delta }={J}_{\delta }\ast {\rho }_{0}+\delta ,{J}_{\delta }\end{array}$ is a mollifier on Ω, and $\begin{array}{}{u}_{0}^{\delta }\in {H}_{0}^{1}\left(\mathit{\Omega }\right)\cap {H}^{2}\left(\mathit{\Omega }\right)\end{array}$ is a smooth solution of the boundary value problem $−[((u0xδ)2+μ1)p−22u0xδ]x+(P(ρ0δ)+η0δ)x+η0δΦx=(ρ0δ)12(gδ+βΦx),u0δ(0)=u0δ(1)=0,$(52)

where $\begin{array}{}{g}^{\delta }\in {C}_{0}^{\mathrm{\infty }}\end{array}$ and satisfies $\begin{array}{}|{g}^{\delta }{|}_{{L}^{2}}\le |g{|}_{{L}^{2}},\underset{\delta \to {0}^{+}}{lim}|{g}^{\delta }-g{|}_{{L}^{2}}=0.\end{array}$

We deduce that (ρδ, uδ, ηδ) is a solution of the following initial boundary value problem $ρt+(ρu)x=0,(ρu)t+(ρu2)x+ρΨx−λ[(ux2+μ1)p−22ux]x+(P+η)x=−ηΦx,(|Ψx|q−2Ψx)x=4πg(ρ−1|Ω|∫Ωρdx),ηt+(η(u−Φx))x=ηxx,(ρ,u,η)|t=0=(ρ0δ,u0δ,η0δ),u|∂Ω=Ψ|∂Ω=(ηx+ηΦx)|∂Ω=0,$

where $\begin{array}{}{\rho }_{0}^{\delta }\ge \delta ,p>2,1

By the proof of Lemma 2.3 in [11], there exists a subsequence $\begin{array}{}\left\{{u}_{0}^{{\delta }_{j}}\right\}\text{\hspace{0.17em}of\hspace{0.17em}}\left\{{u}_{0}^{\delta }\right\},\text{\hspace{0.17em}as\hspace{0.17em}}{\delta }_{j}\to {0}^{+},{u}_{0}^{\delta }\to {u}_{0}\text{\hspace{0.17em}in\hspace{0.17em}}{H}_{0}^{1}\left(\mathit{\Omega }\right)\end{array}$$\begin{array}{}{H}^{2}\left(\mathit{\Omega }\right),-\left(|{u}_{0x}^{{\delta }_{j}}{|}^{p-2}{u}_{0x}^{{\delta }_{j}}{\right)}_{x}\to -\left(|{u}_{0x}{|}^{p-2}{u}_{0x}{\right)}_{x}\end{array}$ in L2(Ω), Hence, u0 satisfies the compatibility condition (8) of Theorem 1.2. By virtue of the lower semi-continuity of various norms, we deduce that (ρ, u, η) satisfies the following uniform estimate $esssup0≤t≤T1(|ρ|H1+|u|W01,p∩H2+|η|H2+|ηt|L2+|ρut|L2+|ρt|L2)+∫0T∗(|ρut|L22+|uxt|L22+|ηx|L22+|ηt|L22+|ηxt|L22)ds≤C,$(53)

where C is a positive constant, depending only on M0.

The uniqueness of solution can be obtained by the same method as the above proof of convergence, we omit the details here. This completes the proof.

Acknowledgement

The authors would like to thank the anonymous referees for their valuable suggestions and comments which improved the presentation of the paper. This work was supported by the National Natural Science Foundation of China (Nos.11526105; 11572146), the founds of education department of Liaoning Province (Nos.JQL201715411; JQL201715409).

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Accepted: 2018-05-06

Published Online: 2018-06-27

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 704–717, ISSN (Online) 2391-5455,

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