An ill-posed problem in hydrodynamic stability of multi-layer Hele-Shaw flow
aa r X i v : . [ phy s i c s . f l u - dyn ] A ug An ill-posed problem in hydrodynamic stability ofmulti-layer Hele-Shaw flow
Gelu Pa¸sa
Simion Stoilow Institute of Mathematics of Romanian AcademyCalea Grivit¸ei 21, Bucharest 010702, Romania
Abstract
An useful approximation for the displacement of two immiscible fluids in aporous medium is the Hele-Shaw model. We consider several liquids with dif-ferent constant viscosities, inserted between the displacing fluids. The linearstability analysis of this model leads us to an ill-posed problem. The growthrates (in time) of the perturbations exist iff some compatibility conditions onthe interfaces are verified. We prove that these conditions cannot be fulfilled.
Keywords:
Hele-Shaw cells displacements; Constant viscosity fluids.
1. Introduction
The Stokes flow in a rectangular Hele-Shaw cell is studied in [1], [4],[5]. In [9] has been shown that the sharp interface, which exists betweentwo immiscible fluids in a Hele-Shaw cell (or an equivalent porous medium),becomes unstable if the displacing fluid is less viscous.Some experimental and numerical results (see [6] and references therein)have shown that an intermediate liquid with a continuous variable increasingviscosity in the flow direction, inserted between the displacing fluids, canminimize the Saffman - Taylor instability. A theoretical model for this three-layer flow was first considered in [2], [3]. An optimal intermediate viscosity,which minimizes the instability, was obtained by using a numerical procedure.In a large number of papers was considered the multi-layer Hele-Shawflow: the injection of a sequence of fluids, with different viscosities, in a
Email address: [email protected] (Gelu Pa¸sa)
Preprint submitted to Elsevier August 31, 2020 omogeneous equivalent porous medium. Some instability control seems beobtained by using this flow model - see [7], [8] and references therein.We consider the following problem: can we use a sequence of liquids with constant viscosities, inserted between the initial fluids, in order to minimizethe Saffman-Taylor instability ?In this paper we give a negative answer for the above poblem. The anal-ysis of the linear stability of this model leads us to an ill-posed problem.The growth rates (in time) of the perturbations are obtained by using usualboundary conditions on the interfaces, based on the Laplace-Young law. Weshow that these conditions, related also to the amplitude of the velocity per-turbations in each layer, cannot be fulfilled. Thus the growth rates can notexist.
2. The flow model
In [2], [3] was proposed an “optimal policy” in order to minimize theSaffman-Taylor instability, which appears when a less viscous fluid is dis-placing a more viscous one in a Hele-Shaw cell. The point is to consideran intermediate region, filled by a given amount of polymer-solute, with avariable continuous viscosity. This “three-layer” structure is moving with thevelocity, say, U , of the displacing fluid far upstream. The cell is parallel withthe plane xOy and the flow is in the positive direction Ox . A basic solutionis considered, with two straight interfaces. The length of intermediate region(between interfaces) is constant. The viscosity µ ( x ) of the intermediate liq-uid is invertible with respect to the concentration of the polymer-solute. Thecontinuity equation for the polymer-solute is used to get the equation of the(a priori) unknown viscosity µ ( x ) in the intermediate region: µ t + uµ x + vµ y = 0 , (1)where ( u, v ) are the velocity components and the indices t, x, y denote thepartial derivatives with respect to time and x, y . In [2], [3] were considereda viscosity jump and a surface tension only on the second interface , betweenthe intermediate region and the displaced fluid. On the first interface, theviscosity was continuous and the surface tension was missing. A linear stabil-ity analysis of the basic slolution was performed and an “optimal” viscositywas obtained by using a numerical procedure. This optimal viscosity givesus the minimum values of the growth rates (in time) of perturbations.2f the intermediate region is large enough, the viscosity is continuous andsurface tensions are missing on both interfaces, then there exist a variableviscosity which leads us to an arbitrary small (positive) growth rates of per-turbations - see [6].In this paper we consider a basic solution with several constant intermedi-ate viscosities. The surface tensions exist on all interfaces. As in [2], [3], theLaplace-Young law is used to get the boundary conditions on the interfaces(which depends also on the amplitudes of the perturbations of the horizon-thal velocity). We show that these conditions cannot be fulfilled. Thus thegrowth rates can not exist.We recall below the model given in [2], [3], with variable intermediateviscosity and use it to get the flow model with constant intermediate viscosity.The following three-layer basic flow with the intermediate region x ∈ ( U t + a, U t + b ) is considered u = U, v = 0; x L = U t + a, x R = U t + b ; P x = − µ G U, P y = 0; µ G = µ L , x < x L , µ G = µ R , x > x R ,µ G = µ ( x − U t ) , x ∈ ( x L , x R ) , (2)where P is the basic pressure, µ G is the basic viscosity, µ L , µ R are the constantviscosities of the displacing and displaced fluids. The last relation is obtainedby using the relations (1) with u = U, v = 0. We use the moving referenceframe x = x − U t . Thus the intermediate region becomes the segment ( a, b ),in which µ G = µ ( x ). Without loss of generality we can consider b < . The stability system for small perturbations, obtained in [2], is linear.Thus the following perturbation u ′ of the horizontal velocity is considered u ′ ( x, y, t ) = f ( x )[cos( ky ) + i sin( ky )] e σt , k ≥ , i = √− , (3)where f ( x ) is the amplitude, σ is the growth constant and k are the wavenumbers. The cross derivation of the perturbed pressures leads us to theamplitude equation (see the relation (2.17) of [2], where f is denoted by ψ and µ is denoted by µ ) − ( µf x ) x + k µf = 1 σ U k f µ x , ∀ x / ∈ { a, b } . (4)3he viscosity is constant outside the intermediate region, thus it follows − f xx + k f = 0 , x / ∈ ( a, b ); − ( µf x ) x + k µf = 1 σ U k f µ x , ∀ x ∈ ( a, b ) . (5)The perturbations must decay to zero in the far field and f is continuous,therefore we get f ( x ) = f ( a ) e k ( x − a ) , ∀ x ≤ a ; f ( x ) = f ( b ) e − k ( x − b ) , ∀ x ≥ b. (6)As we mentioned above, he boundary conditions are obtained by usingthe Laplace - Young law. It means the pressure jump (across the interfaces)should equal the surface tension T times the curvature and the horizontalvelocity should be continuous.Let’s consider two surface tensions T ( a ) , T ( b ) and two viscosity jumps at x = a and x = b . Thus f x could be discontinuous in a, b . The right and leftlimit values at the points a, b are denoted by the superscripts + , − . Just likein [2], page 84 formula (2.27), we get: µ − ( a ) f − x ( a ) − µ + ( a ) f + x ( a ) = kE ( a ) σ f ( a ) ,µ − ( b ) f − x ( b ) − µ + ( b ) f + x ( b ) = kE ( b ) σ f ( b ) ,E ( a ) := kU [ µ + ( a ) − µ L ] − k T ( a ) ,E ( b ) := kU [ µ R − µ − ( b )] − k T ( b ) . (7)In both relations (7) we have the same eigenvalue(s), then the unknown amplitude f must verify the compatibility relation kE ( a ) f ( a ) µ − ( a ) f − x ( a ) − µ + ( a ) f + x ( a ) = kE ( b ) f ( b ) µ − ( b ) f − x ( b ) − µ + ( b ) f + x ( b ) . (8)The stability of the basic solution (2) is governed by the system (5) - (8).4 . The non existence of the growth rates We consider a constant intermediate viscosity µ = µ s.t.0 < µ L < µ < µ R ,µ − ( a ) = µ L , µ + ( a ) = µ = µ − ( b ) , µ + ( b ) = µ R . (9)Thus both viscosity jumps are positive in the flow direction. As µ is constantinside the intermediate region, from (5) we obtain − f xx + k f = 0 , ∀ x ∈ ( a, b ) . (10)The stability system for the flow with constant viscosity (9) is given by theequations (6)-(10).A crucial difference from the Gorell and Homsy model exists: the solutionof (10) is independent of σ . This property leads us to the non-existence ofthe eigenvalues. We recall (6) - (9) and we introduce the notations V L := kE ( a ) f ( a ) µ L kf ( a ) − µ f + x ( a ) , V R := kE ( b ) f ( b ) µ f − x ( b ) + kµ R f ( b ) , (11)Therefore the compatibility condition (8) is equivalent with V L = V R . Proposition 1.
The growth constants of the problem (6) - (10) can not exist.To this end, we prove that the condition (8) cannot be fulfilled.Proof. Suppose that there exist growth rates of the considered problem.The solution of (10) is f ( x ) = Ae kx + Be − kx , where A, B can depend on k and | f ( x ) | < ∞ , ∀ k . Thus the relations (7) become µ L ( Ae ka + Be − ka ) − µ ( Ae ka − Be − ka ) = E a σ ( Ae ka + Be − ka ) ,µ ( Ae kb − Be − kb ) + µ R ( Ae kb + Be − kb ) = E b σ ( Ae kb + Be − kb ) , and we get the system Ae ka ( µ L − µ − E a σ ) + Be − ka ( µ L + µ − E a σ ) = 0 ,Ae kb ( µ + µ R − E b σ ) − Be − kb ( µ − µ R − E b σ ) = 0 . (12)5 solution ( A, B ) = (0 , exists if the following condition is verified: e k ( a − b ) ( µ L − µ − E a σ )( µ − µ R − E b σ )+ e k ( b − a ) ( µ + µ R − E b σ )( µ L + µ − E a σ ) = 0 . (13) We study the values of the growth rates when k is large enough, such that e k ( a − b ) ≈ (recall a − b < , k > ). Thus ( µ + µ R − E b σ )( µ L + µ − E a σ ) = 0 . (14) The relation (14) is a second order equation for σ and we have two real roots: σ = E b µ + µ R , σ = E a µ L + µ . (15) We insert σ in (12) and get µ R Be − kb = 0 . We have µ R > , then itfollows B = 0 . So actually the solution to the equation (10) is f ( x ) = Ae kx .We use the notations (11) , we impose the condition V L = V R , then we get kU ( µ − µ L ) − k T ( a ) µ L − µ = kU ( µ R − µ ) − k T ( b ) µ + µ R . (16) We equate the coefficients of k, k and get µ R = 0 , µ [ T ( a ) + T ( b )] = µ L T ( b ) . (17) From the relation (16) with large k we obtain T ( a ) T ( b ) = µ L − µ µ + µ R ⇒ µ L > µ . (18) The relationship (17) and (18) contradict the hypothesis (9) .We insert σ in (12) and get µ Ae ka = 0 . The viscosity µ mustbe strictly positive, thus A = 0 and the solution of (10) becomes f ( x ) = Be − kx , x < , k > . The values of f ( x ) must be finite, thus we impose thecondition max k Be − kx < ∞ . We use the notations (11) with f ( x ) = Be − kx ,we impose the condition V L = V R , then it follows kU ( µ − µ L ) − k T ( a ) µ L + µ = kU ( µ R − µ ) − k T ( b ) − µ + µ R . (19)6 e equate the coefficients of k, k and get µ L = 0 , µ [ T ( a ) + T ( b )] = µ R T ( a ) . (20) The condition (19) for large enough k gives us T ( a ) T ( b ) = µ L + µ µ R − µ , (21) which is a restriction on the viscosities. The relations (20) , (21) contradictthe hypothesis (9) . The bottom line is: the condition (8) is not satisfied, thusthe problem (6) - (10) has no eigenvalues. (cid:3) Remark 1.
We consider now a constant intermediate viscosity µ = µ s.t.0 < µ L > µ , µ < µ R . We get the same relationship (17) - (18) and (20) - (21). This time, therelationship µ L > µ is not a contradiction, but µ R = 0 and µ L = 0 contra-dicts the hypothesis (9). For large k , the relations (18) and (21) representrestrictions on the viscosity µ , which also contradict the assumption (9). (cid:3) We use the above results for the case of N intermediate layers. Consider( N + 1) interfaces x i and N constant viscosities µ i in each layer ( x i − , x i ): x = a < x < x < ... < x N = b, < µ L ≡ µ < µ < µ < ... < µ i ... < µ N < µ R . (22)The surface tensions, amplitudes, viscosities and limit values in the points x i are denoted by T i = T ( x i ); f i = f ( x i ); f + , − x ( i ) = f + , − x ( x i ); µ + , − ( i ) = µ + , − ( x i ); µ ( x ) = µ i , ∀ x ∈ ( x i − , x i ); µ i = µ + ( i −
1) = µ − ( i ) . (23)The stability of the flow model with viscosities (22) is governed by the rela-tions (6) and the system (24)-(27) below: − f xx + k f = 0 , x ∈ ( a, b ) , x = x i ; (24)7 − ( i ) f + x ( i ) − µ + ( i ) f + x ( i ) = kE i f i σ ,E i = kU [ µ + ( i ) − µ − ( i )] − k T i ; (25) V = V = ... = V i = ... = V N ; (26) V i ≡ kE i f i µ − ( i ) f + x ( i ) − µ + ( i ) f + x ( i ) . (27) Proposition 2.
The above conditions (26) cannot be fulfilled.Proof. i) The solutions of the equations (24) in the intervals ( x i − , x ) could be f i ( x ) = A i e kx + B i e − kx , where A i , B i are absolute constants. However, for large k , the terms | B i e − kx | become very large. Thus | u ′ | also becomes very large and we exceed the frameof small perturbations. The linear stability analysis makes no sense. For thisreason, we consider only the solutions f i ( x ) = A i e kx , where A i are absoluteconstants. The amplitude f is continuous at the points x i . That means A i exp( kx i ) = A i +1 exp( kx i ) , ∀ k ⇒ A i = A i +1 , ∀ i. Therefore A i are the same for all x ∈ [ a, b ] and exists a constant, say, A suchthat A i = A, ∀ i .In the following, we show that the compatibility relation V = V N can notbe fulfilled. We will highlight some restrictions on the viscosities µ i whichcontradict the hypothesis (22) .We impose the condition V = V N with f ( x ) = Ae kx , thus kU ( µ − µ L ) − k T µ L − µ = kU ( µ R − µ N ) − k T N µ N + µ R . (28) We equate the coefficients of k, k and get µ R = 0 , µ N T + µ T N = µ L T N ,in contradiction with (22) . The equation (28) with large enough k givesus T /T N = ( µ L − µ ) / ( µ N + µ R ) , so µ L > µ , which also contradict thehypothesis (22) .ii) The most general solution of (24) is f ( x ) = A i e kx + Be − kxi , x ∈ ( x i − , x i ) , i = A i ( k ) , B i = B i ( k ) , ≤ i ≤ N. (29) In order to remain in the frame of small perturbations, the maximal valuesand the limits of A i e kx , B i e − kx (for large k ) must be finite. It seems naturalto impose the following conditions: max k { A i e kx } < ∞ , lim k →∞ A i e kx = AI < ∞ ;max k { B i e − kx } < ∞ , lim k →∞ B i e − kx = BI < ∞ . (30) We have to prove V = V N . To this end, we introduce the notations A = A , B = B , C = A N , D = B N ,f ( x ) = A ( k ) e kx + B ( k ) e − kx , x ∈ ( x , x ); f ( x ) = C ( k ) e kx + D ( k ) e − kx , x ∈ ( x N − , x N ) . Suppose V = V N , then E E N = µ L − µ Xµ N Y + µ R , (31) X = Ae ka − Be − ka Ae ka + Be − ka , Y = Ce kb − De − kb Ce kb + De − kb ,F := lim k →∞ X = A − B A + B , F N := lim k →∞ Y = A N − B N A N + B N . We consider large values of k in (31) , the above relations gives us T T N = µ L − µ F µ N F N + µ R . This is a restriction on the viscosities µ i which contradicts the hypothesis (22) . If B = B N = 0 , then we recover the formula (18) .Our conclusion is following: the conditions (26) cannot be fulfilled, thusthe eigenvalues of the problem (24) - (27) cannot exist. We use Remark 1 andget the same result for negative viscosity jumps in the flow direction (cid:3) Remark 2.
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