On Free Boundary Problem for the Non-Newtonian Shear Thickening Fluids

WANG Shu and YUAN Fang,2,?

1 College of Applied Sciences,Beijing University of Technology,Beijing 100124,China.

2 China Everbright Bank,Beijing 100054,China.

Abstract. The aim of this paper is to explore the free boundary problem for the Non-Newtonian shear thickening fluids. These fluids not only have vacuum,but also have strong nonlinear properties. In this paper, a class of approximate solutions is first constructed,and some uniform estimates are obtained for these approximate solutions.Finally, the existence of free boundary problem solutions is proved by these uniform estimates.

Key Words: On free boundary problem;the non-Newtonian shear thickening fluids;vacuum.

1 Introduction

It is well known that the non-newtonian shear thickening flows can be described by the following equations(for example,see[1-6])

wherep>2,A>0,μ>0 andγ>1 are some given positive constants,andρ,u,ργrepresent the density,velocity and pressure for the non-Newtonian fluids,respectively.

We assume that the initial densityρ0is some given nonnegative function satisfying suppρ0=[a0,b0]for some constantsa0andb0,and||ρ0||L1(a0,b0)=1.Letx=a(t)andx=b(t)represent the free boundary which is the interface between fluid and vacuum, and then haveρ(a(t),t)=ρ(b(t),t)=0,anda′(t)=u(a(t),t)witha(0)=a0,andb′(t)=u(b(t),t)withb(0)=b0.

We introduce the Lagrange coordinate transformation

Clearly, the left dividing line Γ0:x=a(t) for the interface is a straight line Γ0:y=0 in Lagrange coordinates. In addition,in the right dividing line Γ1:x=b(t)for the interface,we have

Therefore, the right dividing line Γ1:x=b(t) for the interface is a straight line Γ1:y=1 in Lagrange coordinates. In particular, in Lagrange coordinates, the original equations(1.1)-(1.2)are transformed into the following equations

This paper is to solve the above equations(1.5)-(1.6)inQS≡(0,1)×(0,S)(S>0)with the following initial condition

and the following boundary condition

where the initial densityρ0=ρ0(y)and the initial velocityu0=u0(y)have the following properties[A1]-[A3]:

[A1]The initial densityρ0∈C(?∞,+∞)∩C1(0,1)satisfies

[A2]The initial velocityu0∈C3(?∞,+∞)satisfiesu0y(0)=u0y(1)=0.

[A3]The initial value(ρ0,u0)also has the following property:

Our main results are the following theorems.

Theorem 1.1.Let p>2and γ>1, and assume that[A1]-[A3]hold. Then there is a positive number S0∈(0,1)such that, the initial-boundary problem(1.5)-(1.8)has at least one solution(ρ,u)=(ρ(y,s),u(y,s))for(y,s)∈QS0. In particular, the solution(ρ,u)also has the following properties:

(i)There exist two positive constantsμ1andμ2depending only on A,p,γ and M0such that

for almost all(y,s)∈QS0.

(ii)The solution(ρ,u)has the following regularity:

(iii)For almost all(y,s)∈QS0,the solution(ρ,u)=(ρ(y,s),u(y,s))satisfies Eqs.(1.5)-(1.6).

(iv)For almost all s∈(0,S0),the solution(ρ,u)satisfies initial conditions(1.7)in the following sense:

whereμ3is a positive constant depending only on A,p,μ,γ and M0.

(v)For almost all(y,s)∈QS0, the solution(ρ,u)satisfies boundary condition(1.8)in the following sense:

whereμ4is a positive constant depending only on A,p,μ,γ and M0.

We shall prove Theorem 1.1 in Section 4. In order to prove Theorem 1.1, we need some Lemmas in Sections 2-3.

2 Fundamental lemmas

In order to prove our results,we need following Lemmas.

Lemma 2.1.Let0

for all r∈(?∞,+∞). Then we have

In addition,for any q>0,we also have

where C1is a positive constant depending only on q.

Proof.The conclusions(2.2)-(2.3)of Lemma 2.1 can be obtained by direct calculation. For(2.4),we have two cases:(i)0

To prove(2.4)in the case: (i)0

By(2.2),we have

for|r|≤?. For|r|≥?,by(2.5),we compute

Combining the above inequality with(2.6) we have(2.4) in the case 0

Lemma 2.2.We define

for all r∈(?∞,+∞)and all η∈(0,1),where j∈C∞0(?∞,+∞)is a nonnegative function satisfying

In addition,we also define

where m=1/4,and

Then we have

In addition,for any ?∈(0,?0),we have

where ?0∈(0,1/2)and C2∈(1,+∞)are some positive constants depending only on p and M0.

Proof.From(2.7)-(2.10),the conclusion(2.11)-(2.13)can be obtained by direct calculation.Therefore,the details of the proofs for(2.11)-(2.13)are omitted.

To prove(2.14). In fact,by(2.7)-(2.10),fory∈[0,1],we compute

By the above inequality,we can obtain(2.14)and(2.15). Thus the proof of Lemma 2.2 is completed.

Lemma 2.3.Let ?∈(0,1). We denote

for all y∈(?∞,+∞),and then have

In addition,we also have

where C3is a positive constant depending only on M0.

Proof.From (2.16), the conclusions (2.17)-(2.18) can be obtained by direct calculation.Therefore, the details of the proof for Lemma 2.3 are omitted. Thus the proof of Lemma 2.3 is completed.

Lemma 2.4.Assume that{hn(s):n=1,2,···}is a sequence of nonnegative continuous functions satisfying the following inequalities

for all s∈(0,1)and all n=1,2,···, where C4and C5are some given nonnegative real numbers.Then we have

for all s∈(0,1)and all n=1,2,···,where C6=sups∈(0,1)h1(s).

Proof.Applying the mathematical induction method,we immediately get

for alls∈(0,1) and alln=1,2,···. This implies (2.19). Thus the proof of Lemma 2.4 is completed.

Lemma 2.5.We define a function

for all r∈(?∞,+∞). Then we have

where ν1and ν2are some positive constants depending only onμand p.

Proof.From (2.20), the conclusions (2.21)-(2.22) can be obtained by direct calculation.Therefore, the details of the proof of Lemma 2.5 are omitted. Thus the proof of Lemma 2.5 is completed.

3 The constructions and uniform estimates of the approximate solutions

Byρ?0defined by Lemma 2.2 andu?0defined by Lemma 2.3, we construct a sequence of the approximate solutions as follows.

Step 1. We defineρ0=ρ?0and then consider the following initial-boundary problem

By[7],the initial-boundary problem(3.1)has a unique smooth solutionu1=u1(y,s).

Step 2. We consider the following initial value problem

Clearly,the initial value problem(3.2)has a smooth solutionρ1=ρ1(y,s).

Step 3. We consider the following initial-boundary problem

By[7], the initial boundary problem (3.3) has a unique smooth solutionu2=u2(y,s). In addition,we also consider the following initial value problem

Clearly,the initial value problem(3.4)also has a smooth solutionρ2=ρ2(y,s).

Repeating the above process we can find a sequence{(ρn,un)}∞n=1of the approximate solutions,which are smooth and satisfy the following equations

with initial conditions

and boundary conditions

where

Using[A1]and(2.14),by(3.5),we have

Next, we shall find some uniform estimates of approximate solutionsWe have the following lemmas.

Lemma 3.1.Let p>2. For any positive integer k,we define

where

Then,for all n=1,···,k,and all(y,s)∈[0,1]×[0,Sk],and ?∈(0,Sk],we have

whereμ1,μ2and C7are some positive constants depending only on A,p,γ and M0.

Proof.By(3.13)-(3.14),using Lemma 2.2,we compute

Similarly to the above inequality,we also haveρn≥C?1(ρ0(y)+?m).Therefore,we have(3.16). From(3.16),by[A3],we get(3.17). Thus the proof of Lemma 3.1 is completed.

Lemma 3.2.Let p>2and denote

whereμ1and C1are defined by Lemma3.1and Lemma2.1,respectively. Then,for all n=1,···,k,and all(y,s)∈[0,1]×[0,S1k],and ?∈(0,S1k),we have

where C8is a positive constant depending only on A,p,μ,γ and M0.

Proof.By Lemmas 2.1 and 3.1,we compute

which implies that

Then,forn=1,···,kand?∈(0,S1k),by(3.21),we have

By(3.21)and(3.22),using Lemma 3.1 we get

In addition,by Lemma 3.1,we compute

Combining the above inequality with(3.22)-(3.23) we have (3.19)-(3.20). Thus the proof of Lemma 3.2 is completed.

Lemma 3.3.Let p>2and γ>1. For all n=1,···,k,and all(y,s)∈[0,1]×[0,Sk],and ?∈(0,S1k),we have

where C9is a positive constant depending only on A,p,μand γ.

Proof.By Lemmas 2.1 and 3.1,we compute

which implies(3.25). By Lemma 2.5 and Lemma 3.1,we compute

which implies(3.25). Thus the proof of Lemma 3.3 is completed.

Lemma 3.4.Let11. For all n=1,2,···,k, and all(y,s)∈[0,1]×[0,S1k], and ?∈(0,S1k],we have

where C10is a positive constant depending only on A,p,μ,γ and M0.

Proof.By(3.5)-(3.6)and(3.9)-(3.11),we compute

which implies

where

By(3.25),for alls∈(0,1),we have

We now calculate the items on both sides of(3.31). By(3.8)and(3.11),we have

By(3.12)and(3.32),using Lemma 2.2 we get

By(3.9)-(3.11)and(3.32)-(3.33),we obtain

This implies

Similarly to(3.35),we also have

Applying(3.35)-(3.36),we compute

Using Young’s inequality we compute

Combining(3.37)-(3.38)with(3.31)we conclude that

We now calculate the two items on the right side of(3.39). First,applying Lemma 2.1,Lemma 2.5 and Lemma 3.1,by(3.28)-(3.30),we compute

which implies

By(3.40)and[A3],using Lemma 3.1 we get

fors∈[0,S1k],whereCis a positive constant depending only onA,p,μ,γandM0.

Finally, let us calculate the second item on the right side of(3.39). Using(3.9)-(3.11),we compute we compute

By the above inequality and(3.7),using Lemma 2.2-2.3,we compute

By the above inequality,using Lemma 2.2 we have

for alls∈(0,S1k), whereCis a positive constant depending only onA,p,μ,γandM0.Combining (3.41)-(3.42) with (3.39) we get (3.26). Thus the proof of Lemma 3.4 is completed.

Lemma 3.5.Let p>2and γ>1. For all n=1,···,k,and all(y,s)∈[0,1]×[0,S1k],and ?∈(0,S1k],we have

where C11is a positive constant depending only on A,p,γ,μand M0.

Proof.Applying Lemma 3.4,by(3.34),we compute

which implies

On the other hand,using Lemmas 3.1 and 3.3,we have

In addition,by(3.9)and(3.44)-(3.45),we get

By(3.46),using Lemma 2.5 we get

By(3.47),using Lemma 3.2 we get

By(3.48),using Lemma 3.1 we have

Using(3.44)-(3.49)we have(3.43). Thus the proof of Lemma 3.5 is completed.

Lemma 3.6.Let p>2and γ>1. For all n=1,···,k,and all(y,s)∈[0,1]×[0,S1k],and ?∈(0,S1k],we have

where C12is a positive constant depending only on A,p,μ,γ and M0.

Proof.By(3.5),using Lemmas 3.1 and 3.5,for alln=1,···,k,and all(y,s)∈(0,1)×(0,S1k),and?∈(0,S1k),we have

In addition,by(3.5),for alln=1,···,k,and alls∈(0,S1k),and?∈(0,S1k),we also have

By[A3]and Lemma 2.2,we have

Using Schwarz’s inequality and applying Lemmas 2.5,3.1 and 3.5,by(3.9),we compute

Using the above inequality,by(3.52)-(3.53),we conclude that

for alls∈(0,S1k), whereC13is a positive constant depending only onA,p,μ,γandM0.Applying Gronwall’s inequality,by(3.54),we have

By the above inequality and(3.54),we conclude that

for alln=1,2,···, and alls∈(0,S1k), whereCis a positive constant depending only onA,p,μ,γandM0. Applying Lemma 2.4,by(3.51)and(3.55),we get(3.50). Thus the proof of Lemma 3.6 is completed.

Lemma 3.7.Let p>2and γ>1. For all n=1,···,k,and all(y,s)∈[0,1]×(0,S1k],and ?∈(0,S1k],we have

where C14is a positive constant depending only on A,p,γ and M0.

Proof.By(3.6)and[A3],using Lemmas 2.3 and 3.5,we compute

which implies

By Lemmas 3.1 and 3.5,we get

By(3.6)and Lemma 3.4,we have

By(3.58)and[A3],we have

Combining(3.57)-(3.60)we get(3.56). Thus the proof of Lemma 3.7 is completed.

Lemma 3.8.Let p>2and γ>1. For all n=1,···,k,and all(y,s)∈[0,1]×[0,S1k],and ?∈(0,S1k],we have

where C15is a positive constant depending only on A,p,μ,γ and M0.

Proof.From Lemmas 2.2-2.3 and Lemmas 3.6-3.7,we compute

which implies(3.61). Thus the proof of Lemma 3.8 is completed.

Lemma 3.9.Let p>2and γ>1. For all n=1,···,k,and all(y,s)∈[0,1]×[0,S1k],and ?∈(0,S1k],we have

where C16is a positive constant depending only on A,p,μ,γ and M0.

Proof.Applying Lemmas 2.5 and 3.5,we compute

By the above inequality and Lemma 3.3,we get

On the other hand,by(3.9)-(3.10),using Lemma 3.4 and Lemma 3.6 we have

which implies

By(3.63)-(3.64),we have

Similar to(3.65),we also have

Combining(3.65)-(3.66)and applying Lemma 3.1,we get(3.62).Thus the proof of Lemma 3.9 is completed.

4 The proof of Theorem 1.1

In order to prove Theorem 1.1 we need the following lemmas.

Lemma 4.1.Let p>2and γ>1,and denote

where ?1is defined by Lemma3.2, α is defined by(3.15), C7is defined by Lemma3.1, C11is defined by Lemma3.5. Then,for any positive integer k,we have

Proof.For any given positive integerk, by(3.13)-(3.15), we only have two cases: Case I:Sk≥?1,Case II:Sk∈(0,?1).

We now prove(4.2)in the Cases I and Case II,respectively.

Case I:Sk≥?1.

In this case,we have(4.2),and then Lemma 4.1 in the Case I is proved.

Case II:Sk∈(0,?1).

In this case,by(3.14)and(3.18),we have

Applying Lemmas 3.1 and 3.5,by the above equation,we get

which implies(4.2). Therefore,Lemma 4.1 in the Case II is also proved. Combining Case I with Case II,we have(4.2)and then the proof of Lemma 4.1 is completed.

Lemma 4.2.Let p>2and γ>1. Then,for S0defined by Lemma4.1,there exist

and ρ?∈L∞(0,S0;L2(0,1))such that

strongly in L∞(0,S0;L2(0,1))∩L2(0,S0;H1(0,1))as n→∞,and

strongly in L∞(0,S0;L2(0,1))as n→∞. In addition,we have

weakly in L2(QS0)as n→∞. In particular,for almost all(y,s)∈QS0,we also have

whereμ1,μ2,C7are defined by Lemma3.1,C12and C14are defined by Lemma3.6and Lemma3.7,respectively.

Proof.Denote

By(3.5)-(3.6),we have

which implies

By the above equation,we have

Applying Young’s inequality and using Lemmas 2.1, 2.5 and 3.5, by (3.9)-(3.11), we compute

which implies

Combining(4.10)-(4.11)with(3.7)we get

This implies that

for alls∈[0,S0],whereC17is a positive constant independent ofn. Applying Gronwall’s inequality,by(4.12),we obtain

By(4.12)-(4.13)we have

for alls∈(0,S0),whereC18is a positive constant independent ofn. Applying Lemma 2.5,by(4.14),we get

where

Applying Lemma 3.1,by(4.9)and(4.15)-(4.16),we conclude that

for all positive integern, whereCis a positive constant independent ofn. Combining(4.14)with(4.17)we get

for alls∈[0,S0], whereCis a positive constant independent ofn. This implies that, the sequence{ρn}∞n=1is a Cauchy’s sequence inL∞(0,S0;L2(0,1)). Therefore,we have (4.4).Similarly,we also have(4.3). In addition,applying Lemma 3.4,Lemmas 3.7-3.8,by(4.3)-(4.4),we have(4.5)-(4.8). Thus the proof of Lemma 4.2 is completed.

Lemma 4.3.Let p>2and γ>1. Then,for S0defined by Lemma4.1,we have

strongly in L2(QS0)as n→∞, where R?=AGγ?(ρ?), F?=Ψ??R? andΓ?=G?(ρ?)u?y. In addition,we also have

weakly in L2(QS0)as n→∞. In particular,we also have

for almost all(y,s)∈QS0,where C20is a positive constant depending only on A,p,μ,γ and M0.

Proof.By Lemma 2.4,we compute

By the above inequality,applying Lemma 4.2, we get.Similarly,we also have

Therefore we have(4.18). In addition,from the lower half continuity of the norm,using Lemma 3.5, by (4.18), we get (4.20). Using Lemma 3.4 and Lemma 3.6-3.7, by (3.9), we have

whereCis a positive constant depending only onA,p,μ,γandM0. In addition, by the above inequality and (4.19), from the weak lower half continuity of the norm, we get(4.21). Thus the proof of Lemma 4.3 is completed.

Lemma 4.4.Let p>2and γ>1. Then,for S0defined by Lemma4.1,there exist(ρ,u)∈L∞(QS0)and a subsequencesuch that

strongly in L2(QS0)as ?=?j→0+. In addition,we also have

weakly in L2(QS0)as ?=?j→0+,and

weakly inas ?=?j→0+. In particular, for almost all(y,s)∈QS0, we alsohave

where μ1,μ2and C7are defined by Lemma3.1; C12and C14are defined Lemma3.6and Lemma3.7,respectively.

Proof.From the lower half continuity of the norm, applying Sobolev’s imbedding theorem(see,e.g.,[8]), by Lemma 4.2, we have(4.22)-(4.27). Thus the proof of Lemma 4.4 is completed.

Lemma 4.5.Let p>2and γ>1. Then,for S0defined by Lemma4.1,we have

weakly in L2(QS0)as ?=?j→0+. In particular,we also have

for almost all(y,s)∈QS0,where C11is defined by Lemma3.5.

Proof.We define

for allr∈(0,1). By[A1]and(4.28),for anyr∈(0,1),we have

For any givenand for allν∈(0,1),by(4.30)-(4.31),applying Lemmas 4.3-4.4,we compute

which implies that

for allν∈(0,1), whereCis a positive constant independent of?andν. In the above inequality,letting?=?j→0+andν→0+in turn,using Lemma 4.4,we get

for allφ∈L2(QS0). This implies(4.28). From the weak lower half continuity of the norm,by(4.20)and(4.28),we get(4.29). Thus the proof of Lemma 4.5 is completed.

Lemma 4.6.Let p>2and γ>1. Then,for S0defined by Lemma4.1,we have

strongly in L2(QS0)as ?=?j→0+.

Proof.For any givenδ∈(0,1/4),we define a cutoff functionξδ∈C∞0(?∞,+∞)such that

whereC21is an absolute constant independent ofδ. Applying Lemma 2.5, Lemmas 4.2-4.5,by(4.33),for any givenδ∈(0,1/4),we compute

which implies that

By(4.34),using Lemma 2.5,we compute

Combining the above inequality with(4.34)we get

whereCis a positive constant independent of?andδ. In the above inequality, letting?=?j→0+andδ→0+in turn,using Lemmas 4.4-4.5,we get

which implies(4.32). Thus the proof of Lemma 4.6 is completed.

Lemma 4.7.Let p>2and γ>1. Then,for S0defined by Lemma4.1,we have

strongly in L2(QS0)as ?=?j→0+. In addition,we also have

weakly in L2(QS0)as ?=?j→0+. In particular,we also have

for almost all(y,s)∈QS0,where C20is defined by Lemma4.3.

Proof.Applying Lemma 2.1,Lemmas 4.2-4.3,forξδdefined by(4.33),we compute

which implies

whereCis a positive constant independent of?andδ. In(4.38), letting?=?j→0+andδ→0+in turn,by Lemma 4.6,we get

Using Lemmas 4.3-4.4,we compute

By the above inequality and(4.39),we have

Similar to(4.40),we also have

Combining(4.41)with(4.39)-(4.40) we have(4.35). From the weak lower half continuity of the norm, by Lemma 4.3 and (4.35), we have (4.36)-(4.37). Thus the proof of Lemma 4.7 is completed.

Now,let us prove the Theorem 1.1. In fact,the conclusions(i)-(ii)of Theorem 1.1 can be obtained by using Lemma 4.4. To prove the conclusion(iii)of Theorem 1.1,we choose anyφ∈L2(QS0),by(3.6),we have

Lettingn→∞in the above equation,by Lemma 4.2-4.3,we get

In addition,letting?=?j→0+in the above equation,by Lemma 4.4 and Lemma 4.7,we get

for allφ∈L2(QS0). This implies

for almost all(y,s)∈QS0. Similarly,we also have

for almost all(y,s)∈QS0. Combining(4.42)-(4.43) we get the conclusion(iii)of Theorem 1.1. The conclusions(iv)-(v)can be obtained by Lemmas 3.8-3.9 and Lemmas 4.2 and 4.4,and the details are omitted here. Thus the proof of Theorem 1.1 is completed.

Acknowledgement

This work is supposed by NSFC(no.11771031 and no.11531010)China.