The Pressure Singularity for Compressible Stokes Flows in a Concave Polygon

Transcription

1 J. math. fluid mech. 11 (2009) /09/ c 2007 Birkhäuser Verlag, Basel DOI /s y Journal of Mathematical Fluid Mechanics The Pressure Singularity for Compressible Stokes Flows in a Concave Polygon Jae Ryong Kweon and R. Bruce Kellogg Communicated by H. Beirão da Veiga Abstract. A compressible Stokes system is studied in a polygon with one concave vertex. A corner singularity expansion is obtained up to second order. The expansion contains the usual corner singularity functions for the velocity plus an associated velocity singular function, and a pressure singular function. In particular the singularity of pressure is not local but occurs along the streamline emanating from the incoming concave vertex. It is observed that certain first derivatives of the pressure become infinite along the streamline of the ambient flow emanating from the concave vertex. Higher order regularity is shown for the remainder. Mathematics Subject Classification (2000). 76N10, 35Q30, 35J55. Keywords. Compressible flows, corner singularities, singular expansion. 1. Introduction In this paper we consider a compressible Stokes system in a polygon having only one concave vertex. A corner singular expansion for the solution up to second order is given. The expansion contains the usual corner singular functions for the velocity plus a new ingredient: an associated velocity singular function, and a pressure singular function. The singularity in the pressure singular function is not local but occurs along the streamline emanating from the incoming concave vertex. In particular, the results show that along this streamline, the first derivative of the pressure in the direction normal to the streamline becomes infinite. Higher order regularity is shown for the remainder. The expansion extends our earlier first order expansion, obtained in [8] for the compressible Stokes system and in [9, 10] for the (nonlinear) compressible Navier Stokes system. The non-local pressure singularity was shown in [11]; this paper isolates the singular function that represents this behavior. This work was supported by the Com2MaC-SRC/ERC program of MOST/KOSEF (grant R ), and by the U.S. National Science Foundation.

2 2 J. R. Kweon and R. B. Kellogg JMFM The system to be considered in this paper is µ u + p = f in Ω, divu + U p = g in Ω, u = 0 on Γ, p = 0 on Γ in, (1.1) where Ω is a concave polygon in the plane R 2 ; u = [u, v] is the fluid velocity and p is the fluid pressure; µ is a positive number representing the principal viscosity; U = [1, 0] is a specified convective vector; Γ in is the inflow boundary defined by Γ in = {(x, y) Γ : n 1 < 0}, where n = [n 1, n 2 ] denotes the outward pointing unit normal to Γ; f and g are given data. Let Ω be a polygon with only one concave vertex placed at the origin. Let ω 1 and ω 2 be numbers satisfying π < ω 1 < 0 < ω 2 < π, let a > 0, and let Ω a be the truncated sector of radius a defined by Ω a = {(r cosθ, r sinθ) : ω 1 < θ < ω 2, 0 < r < a}. We suppose that Ω 1 Ω = Ω 1 ; thus, Ω 1 coincides with Ω near the origin and the origin is a vertex of Ω. The 2 straight sides of Ω at the origin are denoted by Γ 1 and Γ 2. One sees that Γ 1 Γ 2 Γ in. We will also need the half-sectors Ω 1,+ and Ω 1, defined by the inequalities 0 < θ < ω 2 and ω 1 < θ < 0 respectively. Our goal is to develop a corner singularity expansion of the solution of (1.1) in a neighborhood of the origin. The angle of the polygon at the origin is ω = ω 2 ω 1 > π. We assume that ω is an irrational multiple of π to avoid further complications in an already complicated paper. Let α = π/ω. Let φ j = χ(r)r jα sin[jα(θ ω 1 )] be the j-th order corner singularity at the origin where χ is a smooth function which is 1 near r = 0 and which vanishes for r > 1. System (1.1) consists of 3 scalar equations. The first two equations in the system will be called the momentum equations, and the third equation will be called the continuity equation. The convection in the problem is represented by the derivative U p in the continuity equation. The system is obtained from a linearization of the compressible Navier Stokes equations [2] around a uniform ambient flow moving in the positive x direction. However we have stripped the equations of details that are not directly germane to the issue of corner singularities. Thus, we have not included a convective term in the momentum equations and we have removed the second viscosity from the system. The presence of these terms would change somewhat the details of the expansion, but would not alter its basic structure. System (1.1) is prevented from being elliptic by the presence of the term U p in the continuity equation. Thus, the well-developed theory of corner singularities for elliptic systems [4, 5, 6] cannot be immediately applied. On the positive side, the system satisfies a certain coercivity inequality, and one can solve the continuity equation for p and insert this solution into the momentum system, producing a system of 2 integro-differential equations. This device was used in our earlier works

3 Vol. 11 (2009) Compressible Flows 3 [7, 8, 9, 10] and is used here. For a good exposition of the corner singularities associated with the (incompressible) Stokes system, see [3]. A difference between the Stokes system and the compressible Stokes system is that in the former case, the singularities in the solution occur only at the vertices of the polygon. In the compressible Stokes system, the singularity in the pressure or density resulting from the corner may stretch into the domain along the streamline that emanates from the corner (see [11]). In this regard, it is also appropriate to cite [13], where it is shown that a jump discontinuity in the imposed boundary pressure may result in a jump discontinuity in the pressure along the streamline emanating from the boundary discontinuity. A difference in the behavior studied in [13] and the behavior studied here is that in [13] the pressure remains bounded as one approaches the critical streamline, whereas in the problem studied here, the first derivatives of pressure become unbounded as one approaches the critical streamline. This unboundedness is caused by the corner singularity of the problem. (See the discussion following the statement of Theorem 1.1.) We work in the context of Hilbert-Sobolev spaces [1, 4, 12]. For an open set O Ω, H s (O) and u s,o denote the usual Sobolev space of order s and the corresponding norm. If O = Ω, we write H s and u s. We will use vector valued versions of the above considered spaces, e.g., H s = H s H s, etc. For technical reasons Sobolev spaces of fractional order will be used. Let A : F u be the solution operator of the problem u = F in Ω, u = 0 on Γ. We also associate with this problem the singular functions φ j, j = 1, 2,... associated with the concave vertex at the origin. These singular functions are given, in polar coordinates, φ j (x, y) = χ(r)r jα sin jα(θ ω 1 ); see (2.2). Here χ is a smooth function on the real line which is 1 near the origin and has compact support. Let B : G q be the solution operator of the problem q x = G in Ω, q = 0 on Γ in. A formula for B is given in (2.7) In general the derivatives of BG have jumps along the horizontal lines emanating from the corners of Γ in. By imposing vanishing conditions on G at these vertices the jumps are eliminated. In the context of (1.1) it will be seen that these jumps are satisfied by divu (see Lemma 2.3), so it is reasonable to impose these vanishing conditions on G. For this a closed subspace H s on H s is defined as follows. If s 1, define H s = H s. If s = n > 1 is an integer, define H s = {v H n : m v(0, 0) = 0, m n 2}. (1.2) If s = n + σ > 1 with σ (0, 1), define H s = {v Hs : m v(0, 0) = 0, m n 1}. (1.3) The spaces H s can also be identified with certain weighted Sobolev spaces; we will not use this fact. We now state the main result of this paper, which is proved in Section 3.

4 4 J. R. Kweon and R. B. Kellogg JMFM Theorem 1.1 (Main Theorem). If [f, g] L 2 L 2, there is a unique solution [u, p] H 1 0 L 2 of problem (1.1), satisfying the inequality u 1 + p 0 C( f 0 + g ) with a constant C depending only on Ω. Let δ = min{1, 2α 1/2} and let s be any number with 2α + 1 < s < min{2 + δ, 1 + π/γ i } (1.4) where the γ i are the angles of the convex vertices of Ω. Suppose µ is large enough. Suppose [f, g] H s 2 H s 1 with g(0, 0) = 0. Then there are constant vectors C 1, C 2 and C 3 such that: setting ψ 1 = BC 1 φ 1 H 1 η 1 = µ 1 [χa( ψ 1 ) C 3 φ 1 ] H 2, the solution [u, p] of (1.1) may be decomposed as (1.5) u = C 1 φ 1 + C 2 φ 2 + η 1 + u R, p = ψ 1 + p R, (1.6) where the remainder [u R, p R ] H s H s 1. In addition the vectors C 1,C 2,C 3 depend linearly on [f, g] H s 2 H s 1 and one has the inequality 3 u R s + p R s 1 + C i C(Ω, µ)( f s 2 + g s 1 ). (1.7) 1 In particular, the pressure singular function ψ 1 has a singularity in the normal derivative along the streamline emanating from the concave corner (0, 0). In Theorem 1.1 the coefficient vectors C i are mechanically regarded the stress intensity factors of the corner singularities φ i of the velocity vector u and the function η 1 is an associate singularity function which is needed for extracting the second order singularity φ 2 from the velocity u (for details, see (2.27)). The function ψ 1 is a corner singularity of the pressure function p which shows propagation of the divergence of the first leading corner singularity C 1 φ 1 along the streamline. Theorem 1.1 has the following consequence, which may have physical significance: the derivatives of p in the direction normal to the streamlines coming from concave corners in general become infinite at these streamlines. This singular behavior was established in [11] for solutions to the nonlinear system that satisfy certain seemingly reasonable assumptions. In this paper, the singularity is isolated in terms of the pressure singular functions ψ 1. Indeed, the pressure singularity ψ 1 is given by ψ 1 = C 11 Bφ 1,x C 12 Bφ 1,y. The partial derivative of Bφ 1,y with respect to the y variable is given by y (Bφ 1,y)(x, y) = c 1 (s 2 + y 2 ) (α 3)/2 h 1 (s, y)ds c 2 (sgn y) y α 1 δ (y) where c 1 = α(α 1), c 2 = α(δ (y) 2 + 1) (α 1)/2 cos[(α 1)tan 1 (1/δ (y))] and h 1 (x, y) = y cos[(α 1)(θ ω 1 )] xsin[(α 1)(θ ω 1 )]. Since δ (y) < 0 for y > 0

5 Vol. 11 (2009) Compressible Flows 5 and δ (y) > 0 for y < 0, the function (sgn y) y α 1 δ (y) has a negative sign for y 0. So, if α < 1, then lim (sgn y 0 y) y α 1 δ (y) =. Also, lim y 0 (s2 +y 2 ) (α 3)/2 h 1 (s, y)ds =. So y (Bφ 1,y) becomes infinite at the line y = 0. A similar calculation pertains to y (Bφ 1,x). Consequently the normal derivatives of the pressure singular functions to the streamlines coming from the (concave) corners blow up at the streamlines. The paper is organized as follows. In Section 2 we give some facts concerning corner singularity expansions and the solution operator to the continuity equation. Furthermore we define some associate singular functions and investigate their properties. Section 3 contains the detailed construction of the high order singular functions and linear functionals. 2. Preliminary results In this section we give some facts concerning corner singularity expansions and the solution operator to the continuity equation. We start with a consideration of the problem u = F in Ω, u = 0 on Γ. (2.1) For F H 1 the problem (2.1) has a solution u H 1 0. Let A : F u denote the solution operator to (2.1). Thus, A is a bounded map from H 1 to H 1 0. The solution u has in general singularities at the origin, as well as the other vertices of Ω. If α is not an integer, the j-th corner singularity function at the origin is φ j (x, y) = χ(r)r jα sin[jα(θ ω 1 )], (2.2) where χ is a smooth function which is 1 near r = 0 and which vanishes for r > 1. If α is an integer, the formula for φ j involves a logarithm; we shall not consider this case. Note that φ j is smooth everywhere except at the origin. The singular functions associated with the other vertices of Ω are given by similar formulas. Next we give a corner singularity expansion of solution for problem (2.1) up to second order, without subtracting corner singularities corresponding to the convex vertices. Theorem 2.1. Let Ω be a concave polygon as specified in Section 1. Suppose α is irrational. (i) The function φ j given above is in H s for s < jα + 1 but not for s jα + 1. (ii) Let 2α + 1 < s < min{3α + 1, π/γ i + 1}, (2.3) where γ i s are the angles of the convex vertices of Ω. For each j there is a bounded linear functional Λ j on H s 2 for s > jα + 1 such that: if F H s 2 (Ω), and if

6 6 J. R. Kweon and R. B. Kellogg JMFM u = AF is the solution of (2.1), then u = 2 Λ j (F)φ j + u R (2.4) j=1 where u R H s and satisfies the inequality 2 u R H s + Λ j (F) C F H s 2. (2.5) j=1 Proof. Using basic results in [4, 5, 6] on corner singularities, one can derive the required results: For s < min{π/γ i +1}, near each vertex P i the solution u belongs to the space H s. For 2α+1 < s < 3α+1, near the concave vertex (0, 0) the singular functions φ j (j=1, 2) can be subtracted from the solution u and the remainder is in H s. Combining these and using a partition of unity, one can show that the decomposition (2.4) and the inequality (2.5) are valid for the regularity order s in (2.3). If F H s 2 with α + 1 < s < 2α + 1, we let A 1 : F u R := u Λ 1 (F)φ 1 denote the remainder. From Theorem 2.1, A 1 : H s 2 H s is a bounded operator. We also use a vector valued version of this operator. To deal with the continuity equation in (1.1), we consider the problem q x = G in Ω, q = 0 on Γ in. (2.6) The solution formula for (2.6) may be written in the following way. Let Γ k be a side of Γ in, and let D k denote the set of points in Ω which can be connected by a horizontal line lying in Ω to a point in Γ k. Thus, Ω = D k and the sets D k meet only along their boundaries. Let x = δ k (y) be the equation of the side Γ k. Then the solution q of (2.6) is q(x, y) = δ k (y) G(s, y)ds, (x, y) D k. (2.7) Let B : G q denote this solution operator. With the notation x = / x and y = / y we note the formulas BG(δ k (y), y) = 0, x BG = G, y BG(x, y) = BG y (x, y) δ k(y)g(δ k (y), y), y BG = BG y δ k (y)(g BG x). Lemma 2.1. For any s [0, 1], B : H s (Ω) H s (Ω) is a bounded map. (2.8) Proof. In the case s = 0 and s = 1 this result is proved in [8]. The result for s (0, 1) then follows by interpolation [12].

7 Vol. 11 (2009) Compressible Flows 7 From the Sobolev imbedding theorem, for each s 0, the space H s defined in (1.3) is a closed subspace of H s with finite codimension. Since smooth functions in Ω are dense in H s, it can be shown that smooth functions in Ω which satisfy (1.2) or (1.3) are dense in H s. It should be noted that the family of spaces {H s } s=0 is not an interpolating family; for example, since H 1 = H1, [H 2, L2 ] 1/2 H 1. On the other hand, for each integer n 0, the family {H s } with s (n, n + 1) is an interpolating family. This fact is used in the proof of the following lemma. Lemma 2.2. For s > 1, B : H s Hs is a bounded map. Proof. Let s > 1 and let G H s be a smooth function which satisfies (1.2) or (1.3). We first prove by induction that if s = n is an integer then BG H n for each k and satisfies (1.2). If n = 1 these assertions follow from Lemma 2.1 and from the solution formula (2.7). For n > 1 we use a decomposition of the solution and (2.7). Let r 1, r 2 be numbers with 0 < r 1 < r 2 < 1 and χ a smooth cutoff function such that χ = 1 in Ω r1 and χ = 0 outside Ω r2. Let G 1 = χg, G 2 = (1 χ)g. Since G 2 = 0 on Ω r1, BG 2 = 0 on Ω r1. Hence BG 2 H n. We now consider BG 1. Since G 1 H n 1, the inductive assumption implies that BG 1,x and BG 1,y are in H n 1 (Ω 1,± ) so (2.8) implies that BG 1 H n 1 (Ω 1,± ), so BG 1 H n (Ω 1,± ). Also, using (2.8) and induction it is seen that BG 1 satisfies (1.2). A similar inductive argument shows that if 0 < σ < 1 and if G H n+σ, then BG H n+σ (Ω 1,± ) and satisfies (1.3). We show using induction that m BG 1 is continuous across y = 0 for m = 0,...,n 2 in the case s = n is an integer, or m = 0,...,n 1 in the case s = n+σ with σ (0, 1). Inspection of (2.7) shows that the function BG 1 is continuous across y = 0, so the assertion is true in the case m = 0. If the derivative m contains an x-derivative, so m = x m 1, then m BG 1 = B m 1 G 1. Since m 1 G 1 (x, y) is continuous across y = 0, B m 1 G 1 is continuous across y = 0. Hence m BG 1 is continuous across y = 0. If the derivative m contains an y- derivative, so m = y m 1, then using (2.8) with y > 0, letting y y +0 and noting by the inductive assumption that m 1 G 1 (0, 0) = 0, we obtain The same argument shows that m BG 1 (x, y + 0) = B m 1 G 1 (x, +0). m BG 1 (x, y 0) = B m 1 G 1 (x, 0). Since by induction m 1 G 1 (x, y) is continuous across y = 0, m BG 1 is continuous across y = 0. Summarizing, it has been shown that if s = n is an integer, or if s = n+σ with σ (0, 1), then m BG 1 is continuous across y = 0 for all m n 1. Therefore, in the case σ 1 2, BG 1 H s and BG 1 s,ω1 C( BG 1 s,ω1,+ + BG 1 s,ω1, ) C G 1 s. (2.9) Since G 1 lies in a dense subset of H s, the inequality (2.9) holds for all G 1 H s,

8 8 J. R. Kweon and R. B. Kellogg JMFM so B is a bounded map from H s [H n+1/2+δ to itself provided σ 1 2. Since Hn+1/2 =, H s+1/2 δ ] 1/2 for δ (0, 1 2 ) it follows that B is a bounded map from H n+1/2 to itself. Note that the solution of the continuity equation in (1.1) is given by p = B(g divu). To get a higher regularity for p, from the view of Lemma 2.2 we need to show g divu H s (Ω), which means that divu should vanish at the origin. To prepare for this we need the following lemma. Lemma 2.3. (i) Let s = m + σ > 1 be a real number, where m = s and 0 < σ < 1. Let u H s H 1 0. Then u(0, 0) = 0. If s > 2, then u(0, 0) = 0. (ii) Suppose α is irrational. If s (α + 1, 2α + 1) and F H s 2, then A 1 F H s, and A 1 is a bounded map from H s 2 to H s. Proof. (i) Since s > 1, u H s is continuous and the boundary conditions imply that u(0, 0) = 0. If s > 2 then u is continuously differentiable. The tangential derivatives of the two boundary conditions at the origin yield (cosω l )u x (0, 0) + (sinω l )u y (0, 0) = 0 for l = 1, 2. Since ω π, the determinant of this linear system is not zero, so u x (0, 0) = u y (0, 0) = 0. (ii) Suppose s (α + 1, 2α + 1) and F H s 2. Set u = AF. Then A 1 F = u R is given by (2.4). From Theorem 2.1, A 1 F H s. From (i), k u R (0, 0) = 0 for k = 0, 1, so u R H s. In order to extract the second order singularity from the exact solution, we need to show that the derivatives of the function Bφ 2,y are in the appropriate Sobolev spaces. In this regard, note that φ 2 = 0 outside Ω 1, so the derivative estimations need to be made only in Ω 1. The partial derivative of Bφ 2,y with respect to the y variable is given by y (Bφ 2,y )(x, y) = c (s 2 + y 2 ) α 3/2 h 1 (s, y)ds + c y 2α 1, (2.10) where c = C(α) denotes a constant depending on α but will denote different constants in different places and h 1 (x, y) = ya(θ) xa (θ)/(2α 1), a(θ) = cos[(2α 1)(θ ω 1 )]. (2.11) Clearly h 1 (x, y) c(x 2 + y 2 ) 1/2 and h 1 (x, y) h 1 (x, y 1 ) c y y 1 where c = C(α). Lemma 2.4. Let δ = min{1, 2α 1/2}. For s < 2 + δ, B φ 2 H s 2 (Ω).

9 Vol. 11 (2009) Compressible Flows 9 Proof. For α < 1, recall that B φ 1 H s 2 (Ω) for s 1 < s 2 (see [8, 9]). Since 2α 1 < δ, the set {s : 2α + 1 < s < 2 + δ}. It will be enough to consider the derivatives for Bφ 2,y with respect to y variable because the other cases are simpler than this. The function in (2.10) can be estimated by y (Bφ 2,y (x, y)) c (s 2 + y 2 ) α 1 ds + c y 2α 1 /y = cy 2α 1 (t 2 + 1) α 1 dt + cy 2α 1 m c(x 2α 1 + y 2α 1 ). Since 2α 1 > 0, we see that y Bφ 2,y L 2 (Ω). On the other hand, from (2.10) we see that y Bφ 2,y (x, +0) = y Bφ 2,y (x, 0), so y Bφ 2,y (x, y) has a well-defined trace along the x-axis. Hence it is sufficient to show that y Bφ 2,y H σ (Ω ± ) for σ (2α 1, δ). We set q(x, y) = (s 2 + y 2 ) 2α 3 2 h 1 (s, y)ds, (2.12) so y (Bφ 2,y (x, y)) = c q(x, y) + c y 2α 1. Note that q 2 H σ (Ω) is equivalent to the following quantity: q q(x, y) q(x, y 1 ) 2 y y 1 1+2σ dxdy dy 1 q(x, y) q(x 1, y) 2 x x 1 1+2σ dxdx 1 dy. Since the above two integrands are symmetric in the pairs (y, y 1 ) and (x, x 1 ), it is sufficient to show that (I) = (II) = 1 y q(x, y) q(x, y 1 ) 2 y y 1 1+2σ dy 1 dy <, (2.13) q(x, y) q(x 1, y) 2 x x 1 1+2σ dx 1 dxdy <, (2.14) where 2α 1 < σ < δ. First, we show (2.13). Set γ = (3 2α)/2. Then where K 1 = K 2 = q(x, y) q(x, y 1 ) = K 1 + K 2 + K 3 (2.15) δ(y 1) δ(y 1) (s 2 + y 2 ) γ [h 1 (s, y) h 1 (s, y 1 )] ds, [(s 2 + y 2 ) γ (s 2 + y 2 1 ) γ ]h 1 (s, y 1 )ds,

10 10 J. R. Kweon and R. B. Kellogg JMFM K 3 = Setting s = yt, we have δ(y1) K 1 C (s 2 + y 2 ) γ h 1 (s, y)ds. δ(y 1) Cy 2α 2 /y (s 2 + y 2 ) 2α 3 2 ds y y 1 δ(y 1)/y Cy 2α 2 y y 1 Cy 2α 2 y y 1. To estimate K 2, we first note that (x 2 + y 2 ) γ (x 2 + y 2 1 )γ = 2γ (t 2 + 1) 2α 3 2 dt y y 1 (t 2 + 1) 2α 3 2 dt y and h 1 (x, y 1 ) C(x 2 + y1 2)1/2. We have [(x 2 + y 2 ) γ (x 2 + y1 2 ) γ ]h 1 (x, y 1 ) C(x 2 + y 2 ) γ (x 2 + y 2 1) γ+1/2 y Letting s = ξt, the term K 2 is estimated by K 2 C C where z(t) is given by y ξ y 1 δ(y 1) y /ξ ξ 2α 2 y 1 δ(y 1)/ξ y 1 ξ(x 2 + ξ 2 ) γ 1 dξ y 1 ξ(x 2 + ξ 2 ) γ 1 dξ. (s 2 + y 2 ) γ (s 2 + y 2 1) γ+1/2 (s 2 + ξ 2 ) γ 1 ds dξ z(t)dt dξ, z(t) = ( t 2 + (y/ξ) 2 ) γ( t 2 + (y 1 /ξ) 2) γ+1/2 (t 2 + 1) γ 1. Since 0 < y 1 < ξ < y, we have Using this, we have /ξ (t 2 + (y/ξ) 2 ) γ (t 2 + 1) γ 1 < (t 2 + 1) 1 < 1, δ(y 1)/ξ z(t)dt [t 2 + (y/ξ) 2 ] γ < [t 2 + (y 1 /ξ) 2 ] γ. 1 + δ(y 1)/ξ 1 1 δ ( t 2 + (y 1 /ξ) 2) γ+1/2 dt ( t 2 + (y 1 /ξ) 2) 2γ+1/2 dt t 2γ+1 dt + 1 t 4γ+1 dt, (2.16)

11 Vol. 11 (2009) Compressible Flows 11 which is finite because 1 2 < γ < 0, 3/2 < 2γ + 1/2 < 1/2, where = δy. Hence we have K 2 C y Finally the term K 3 is estimated by Combining (2.15) (2.18), y 1 ξ 2α 2 dξ C y 2α 1 y 2α 1 1. (2.17) K 3 Cy 2α 2 y y 1. (2.18) q(x, y) q(x, y 1 ) C ( y 2α 1 y 2α y 2α 2 y y 1 ). (2.19) Assuming that σ < δ and letting y 1 = ty, one can see that 1 y=0 y 0 y 2α 1 y1 2α 1 2 y y 1 1+2σ dy 1 dy y=0 y 2(2α 1 σ) dy 1 y=0 1 Thus (2.13) holds. Second, we show (2.14). Now q(x, y) q(x 1, y) 0 y 0 y 2(2α 2) y y 1 2 y y 1 1+2σ dy 1 dy 1 t 1 2σ dt <. x 1 (s 2 + y 2 ) 2α 3 2 h 1 (s, y) ds /y Cy 2α 1 (t + 1) 2α 2 dt x 1/y C (x + y) 2α 1 (x 1 + y) 2α 1. (2.20) Using (2.20) and letting x 1 + y = (x + y)t, we have x x 1 = (x + y)(1 t) and q(x, y) q(x 1, y) 2 x x 1 1+2σ dx 1 (x + y) 2(2α 1 σ) Using (2.21) we have (II) C Finally one easily sees that 1 y C(x + y) 4(α 1). 1 +y x+y (x + y) 4(α 1) dxdy <. y 2α 1 y 2α y y 1 1+2σ dy 1 dy <. 1 t 1 2σ dt (2.21) Thus y (Bφ 2,y ) H σ (Ω + ). Similarly, y (Bφ 2,y ) H σ (Ω ). The remaining derivatives: y (Bφ 2,x ), x (Bφ 2,y ) and x (Bφ 2,x ) can be computed in the same way. Thus the required result follows.

12 12 J. R. Kweon and R. B. Kellogg JMFM We now define some associate functions which will be used in subtracting the second order singular function from the solution of our problem. Define v 0 = 0, z 1 = B φ 1, v 1 = χa( z 1 ) Λ 1 ( z 1 )φ 1, z 2 = Bdivv 1 + B φ 2, (2.22) where χ is the cutoff function defined on page 1. The function v 1 H 1 0 is the solution of the problem v 1 = z 1 + Λ 1 ( z 1 ) φ 1 in Ω. (2.23) In the next lemma we establish the regularity properties of the functions z i. In the proof of this lemma, r 0,r 1 and r 2 are numbers satisfying 0 < r 2 < r 1 < r 0 < 1. Let χ C 0 (R 2 ) be a cutoff function satisfying χ = 1 for r r 2 and χ = 0 for r r 1. Lemma 2.5. Let s < 2 + δ with δ = min{1, 2α 1/2}. Then z 2 H s 2. Proof. From [8], we know z 1 H 0. We give an explicit form of the solution w for w = z 1 in Ω r2, w = 0 on Ω r2. (2.24) There exists a function Θ 1 such that z 1 = B φ 1 = r α 1 Θ 1 (θ) in Ω r2. This can be shown by expressing the solution BG of (2.7) in polar coordinates. Note that Θ 1 L 2 (ω 1, ω 2 ) because Θ 1 L 2 (ω 1,ω 2) = 2α/r z 1 0,Ωr2 <. We seek a function of the form w 1 = r α+1 Φ 1 (θ) satisfying w 1 = r α 1 Θ 1 (θ) in Ω r2. Since w 1 = r α 1 [Φ 1 + (α + 1) 2 Φ 1 ], the function Φ 1 satisfies Φ 1 (θ) + (α + 1)2 Φ 1 (θ) = Θ 1 (θ), Φ 1 (ω 1 ) = Φ 1 (ω 2 ) = 0. The eigenvalues of the corresponding homogeneous operator are (jα) 2, j = 1, 2,.... Since α + 1 jα, (α + 1) 2 is not an eigenvalue and the two point boundary value problem does have a solution Φ 1, which is in H 2 (ω 1, ω 2 ). We expand the function Φ 1 in a Fourier sine series as follows: Φ 1 (θ) = a 1,j sin[jα(θ ω 1 )], where a 1,j is the Fourier coefficient of Φ 1. Define w 2 = r2 α+1 a 1,j (r/r 2 ) jα sin[jα(θ ω 1 )]. j=1 j=1

13 Vol. 11 (2009) Compressible Flows 13 Note that w 2 = 0 in Ω r2 and w 2 = w 1 on r = r 2. Hence w and w 1 +w 2 solve the same boundary value problem (2.24), so w = r α+1 Φ 1 (θ) r α+1 2 a 1,j (r/r 2 ) jα sin[jα(θ ω 1 )] in Ω r2. (2.25) j=1 Note that Λ 1 is a bounded linear functional on the space H s 2 for s > α + 1 (see Theorem 2.1). Since z 1 H 0, Λ 1 ( z 1 ) is well-defined and we must have Using (2.25), we have, in Ω r2 Λ 1 ( z 1 ) = r 2 a 1,1. v 1 = A( z 1 ) Λ 1 ( z 1 )φ 1 = r α+1 Φ 1 (θ) r2 α+1 a 1,j (r/r 2 ) jα sin[jα(θ ω 1 )]. Since z 2 = B φ 2 + Bdivv 1, j=2 z 2 = B φ 2 + Bdiv(r α+1 Φ 1 ) a 1,j B [ (r/r 2 ) jα sin[jα(θ ω 1 )] ] in Ω r2. r α+1 2 j=2 Using Lemma 2.4, we have B φ 2 H s 2 (Ω r2 ) for s < 2 + δ. Next we show Bdiv(r α+1 Φ 1 ) H s 2 (Ω r2 ) for s < 2 + δ. Indeed, y ( Bdiv(r α+1 Φ 1 ) ) =α where θ := θ ω 1 = tan 1 (y/x), = m y and a( θ) = cos θ + sin θ, b( θ) = cos θ sin θ, ( s 2 α 2 +y 2) ( 2 y γ 1 (s, y) + s γ 2 (s, y) ) ds+c 1 y α, (2.26) γ 1 (x, y) = (α + 1)a( θ)φ 1 ( θ) + b( θ)φ 1 ( θ), γ 2 (x, y) = b( θ)φ 1( θ) + α a( θ)φ 1( θ) + (α + 1)b( θ)φ 1 ( θ), C 1 = m(1 + m 2 ) α 1 2 [(α + 1)(m + 1)Φ1 (tan 1 (1/m)) +(m 1)Φ 1 (tan 1 (1/m))]. Note that Φ 1 H 2 (ω 1, ω 2 ). Here we observe that (2.26) is not only in the form given in the right hand side of (2.10) but also smoother than that of (2.10) because α 2 > 2α 3 and α > 2α 1. Hence following the similar procedures as given in the proof of Lemma 2.4, one can show that y Bdiv(r α+1 Φ 1 ) H s 2 (Ω r2 ) for s < 2 + δ. Similarly x Bdiv(r α+1 Φ 1 ) H s 2 (Ω r2 ).

14 14 J. R. Kweon and R. B. Kellogg JMFM Note that φ j = r jα sin(jα θ) in Ω r2. Then ( a 1j y B y (r/r2 ) jα sin(jα θ) ) = a 1j r jα 2 y B y φ j. j=2 We want to show that the above term is in H s 2 (Ω r2 ) for s < 2+δ. Pick a number r 3 so that 0 < r 3 << r 2 and Ω r3 Ω r2. Now a 1j r jα 2 y B y φ j a 1j r s 2,Ωr3 jα 2 y B y φ j s 2,Ωr3 where j=2 S N = R N = N j=2 j=n+1 j=2 j=2 = S N + R N, a 1j r jα 2 y B y φ j s 2,Ωr3, a 1j r jα 2 y B y φ j s 2,Ωr3. From Lemma 2.4, we know that for j 2, y B y φ j H s 2 (Ω r1 ) for s < 2 + δ. If s < 2 + δ, then S N < for any finite number N. We show the finiteness of R N. If s < 2+δ, then s 2 1+τ with 0 < τ < 1. We use the following Sobolev imbedding result [4]: W 2 q(ω) H s 2 (Ω) where q = 2/(3 t) with t = s 2. Note that 0 < 2 t < 1 and 1 < q < 2. Note that n B y ( r jα sin(jα θ) ) C(jα) n+1( x jα n + y jα n) where C is a constant not depending on j. Hence for j 7, y B y φ j s 2,Ωr3 C y B y φ j 2,p,Ωr3 C(jα) 4( Ω r3 x p(jα 3) + y p(jα 3) dx C(jα) 4( r 3 r=0 = C(jα) 4 r jα 3+2/p 3 C (jα) 4 r jα 3 3 ) 1/p r p(jα 3)+1 dr / [(jα 3)p + 2] 1/p ) 1/p where C = C(ω, p). Using the Schwarz inequality and letting r = r 3 /r 2, R 6 Cr2 3 a 1j (jα) 4 (r 3 /r 2 ) jα Cr 3 2 j=7 ( a 1j 2 (jα) 4) 1/2( (jα) 4 r 2jα j=7 j=7 ) 1/2

15 Vol. 11 (2009) Compressible Flows 15 Cr 3 2 Φ 1 L 2 (ω 1,ω 2), where we used that the series j=1 (jα)4 r 2jα is convergent since r < 1, and C = C(ω, p). Thus z 2 H s 2 (Ω r3 ) for s < 2 + δ. Since z 2 is sufficiently smooth outside Ω r3, z 2 H s 2 (Ω r0 ). Let C 1 and C 2 be given constant vectors. Set = B. We define the following associate functions: η 0 = 0, ψ 1 = C 1 B φ 1, η 1 = µ 1 C 1 [χa( φ 1 ) Λ 1 ( ] φ 1 )φ 1, ψ 2 = B η 1 C 2 B φ 2. (2.27) Using the smoother part A 1 of A, the formula for η 1 can be written by η 1 = µ 1 C 1 A 1 ( φ 1 ), which is the solution of µ η 1 = ψ 1 Λ 1 ( ψ 1 ) φ 1 in Ω. (2.28) Using Lemma 2.5, we give some properties for η 1 and ψ 2. Basically Lemma 2.5 is equivalent to the following lemma but is rephrased into a convenient form, for later purpose. Lemma 2.6. Let s < 2 + δ with δ = min{1, 2α 1/2}. Then ψ 2 H s 2 and Bdiv η 1 s 2,Ω C C 1. (2.29) Proof. From Lemma 2.5, we know that ψ 1 L 2. The function η 1 is expressed by η 1 = µ 1 C 1 [χa( φ 1 ) Λ 1 ( φ 1 )φ 1 ] = µ 1 C 1 v 1. (2.30) Using Lemma 2.5, we have Bdivη 1 H s 2. Since ψ 2 = Bdiv(η 1 + C 2 φ 2 ) and using (2.30), ψ 2 = µ 1 C 1 v 1 C 2 φ 2. Using Lemmas 2.4 and 2.5, we have ψ 2 H s 2 and ψ 2 s 2,Ω C( C 1 + C 2 ) for a constant C. Using the operators A, B and A 1, we have

16 16 J. R. Kweon and R. B. Kellogg JMFM Lemma 2.7. There is a constant C such that if µ > C Bdiv : (a) for s < α+1, then A B div := A B divv sup Hs 1 0 v H v s 1 H s 1 <, (2.31) and A := ( I µ 1 A Bdiv ) 1 is bounded on H s 1 ; (b) for α + 1 < s < 2α + 1, A 1 B div := A 1 B divv H s sup 0 v H s v H s and A 1 := ( I µ 1 A 1 Bdiv ) 1 is bounded on H s. <, (2.32) Proof. Clearly (2.31) follows from [8]. Using Theorem 2.1 and Lemma 2.2, a sequence of the following mappings can be considered: for α + 1 < s < 2α + 1, H s div H s 1 B H s 1 H s 2 A 1 H s. Hence (2.32) follows. Let w = A 1 v for v Hs. So ( I µ 1 A 1 Bdiv ) w = v and µ(w v) = A 1 Bdivw. By the definition of A 1, µ(w v) = A( Bdivw) Λ 1 ( Bdivw)φ 1 where the linear functional Λ 1 is defined in Theorem 2.1. Hence, for F = Bdivw, µ w v s C( F s 2 + Λ 1 (F) φ 1 s 2 ) C Bdivw s 2 C Bdiv w s. (2.33) Using (2.33) and the triangle inequality, w s / v s µ/(µ C Bdiv ), which is finite if µ > C Bdiv. Thus A 1 is bounded on Hs. Before finishing this section, we give an existence result for (1.1). First, we define bilinear forms a(u,v) = µ( u, v) Ω for [u,v] H 1 H 1, b(q,v) = (q, divv) Ω for [q,v] L 2 H 1 and c(p, q) = (p x, q) Ω for [p, q] Q L 2, where (, ) Ω denotes the L 2 inner product. We define a weak solution of the problem (1.1) to be a pair [u, p] H 1 0 L 2 satisfying a(u,v) + b(p,v) = (f,v) Ω, v H 1 0, c(p, q) b(u, q) = (g, q) Ω, q L 2. (2.34) Lemma 2.8. (i) Suppose that [f, g] L 2 L 2. Then there is a unique weak solution [u, p] H 1 0 L2 of (1.1). The solution satisfies µ u 1 + p 0 C( f 0 + g 0 ). (ii) Suppose µ is large enough. If s < α+1, [f, g] H s 2 H s 1, and [u, p] is the solution of (1.1), then µ u s + p s 1 C( f s 2 + g s 1 ). Proof. (i) Let ã be the bilinear form defined by ã(v,w) = a(v,w)+(b v, w) Ω. One sees that ã is a bounded bilinear form on H 1 0. We show that ã is coercive on

17 Vol. 11 (2009) Compressible Flows 17 H 1 0 as follows: One has ã(v,v) = µ( v, v) Ω +(B v, v) Ω. Set q = B v, so q x = v. Then (B v, v) Ω = (q, q x ) Ω 0. Hence, also using the Poincaré inequality, ã(v,v) µ( v, v) Ω c v 1. By the Lax Milgram lemma, for f L 2 there is a unique u H 1 0 such that µ( u, v) Ω + (B u, v) Ω = (f,v) Ω for v H 1 0. (2.35) Set p = B u. Then (2.35) gives (2.33), so we have proved the existence of a solution to (1.1). The asserted inequality easily follows. (ii) Using A and B, the solutions are expressed by u = µ 1 A(f p) and p = B(g divu). Assuming that µ is large enough and applying Lemma 2.8, one obtains the solution formula u = µ 1 A A(f Bg), p = (I + µ 1 BdivA A )Bg µ 1 BdivA Af, which shows that [u, p] H s H s 1 provided [f, g] H s 2 H s 1 and s < α nd order expansion of corner singularities In this section, using the corner singularity functions φ i and the function [η 1, ψ 1 ] defined in (2.27) we obtain the second order corner singularity expansion for the solution [u, p] of (1.1) and establish a regularity for the remainder. On the lowest order level, to establish the required regularity of the remainder, it is sufficient to split only the velocity into singular and regular parts, without removing a singularity for the pressure (see [8]). However, in the higher order levels one must subtract the singularities, like ψ 1, in the pressure at the corner. Step 1. α + 1 < s 2. The first leading corner singularity decomposition is shown in [8]. If u = C 1 φ 1 + u 1 is written with the parameter C 1 chosen appropriately, the pair [u 1, p] solves µ u 1 + p = f + µc 1 φ 1 in Ω, p x + divu 1 = g C 1 φ 1 in Ω, u 1, p = 0 on Γ. (3.1) From [8, Theorem 2.2], a formula for C 1 is given with the properties: (i) C 1 C( f 0 + g 1 ) and (ii) u p 1 C( f 0 + g 1 ). Using interpolation between integer values, one obtains u 1 s + p s 1 + C 1 C( f s 2 + g s 1 ) for α + 1 < s 2. Step 2. Let s be any number in the interval 2α + 1 < s < s := min γ j<π {2 + δ, 1 + π/γ j} where δ = min{1, 2α 1/2} and γ j s are the angles of convex vertices.

18 18 J. R. Kweon and R. B. Kellogg JMFM In this level, we will extract the second leading singularity from the function u 1. For this, a difficulty is that the right hand side of (3.1b) is not sufficiently regular because of the term C 1 φ 1. To deal with this situation, we use the pair [η 1, ψ 1 ] in (2.27) to remove the trouble term C 1 φ 1 in the right hand side of (3.1b). Define ū 1 = u 1 η 1, p 1 = p ψ 1. (3.2) Inserting (3.2) into (3.1), µ ū 1 + p 1 = f + [µc 1 + Λ 1 ( ψ 1 )] φ 1 in Ω, p 1,x + divū 1 = g divη 1 in Ω, ū 1, p 1 = 0 on Γ. (3.3) If α + 1 < s 2 and [f, g] H s 2 H s 1, it follows from the property of η 1 that [ū 1, p 1 ] H s H s 1. We now want to remove the second singular function from ū 1. For this, let u R = ū 1 C 2 φ 2, p R = p 1, (3.4) where the vector C 2 will be constructed shortly. Inserting (3.4) into (3.3), the pair [u 2, p 2 ] solves µ u R + p R = f + f 2 + Λ 1 ( ψ 1 ) φ 1 in Ω, p R,x + divu R = g divη 1 C 2 φ 2 in Ω, u R = 0, p R = 0 on Γ, (3.5) where f 2 = µc 1 φ 1 + µc 2 φ 2. Applying Theorem 2.1 to the first equation in (3.5), we must require µ 1 Λ 2 ( f + f 2 + Λ 1 ( ψ 1 ) φ 1 p R ) = 0. (3.6) We will use this equation to express C 2 in terms of known data. From (3.5), u R = µ 1 [f + (µc 1 + Λ 1 ( ψ 1 )) φ 1 + µc 2 φ 2 p R ]. Since the first singular function has been removed from u 2, we can use the remainder operator A 1 associated with the Laplace operator to write Using (3.5b) and B, Inserting this into (3.7), we obtain u R = µ 1 A 1 [f + µc 2 φ 2 p R ]. (3.7) p R = B(g divu R divη 1 C 2 φ 2 ). (3.8) u R = µ 1 A 1 Bdivu R + µ 1 A 1 ( µc2 φ 2 + C 2 B φ 2 + f 2 ), where f 2 = Bdivη 1 + f Bg. So u R = µ 1 A 1 A ( 1 µc2 φ 2 + C 2 B φ 2 + f ) 2 (3.9)

19 Vol. 11 (2009) Compressible Flows 19 where A 1 = (I µ 1 A 1 Bdiv) 1. Note that the operators A 1 and A 1 depend on C 1 but not on C 2. From its definition, ψ 1 H 0 (Ω) so η 1 H 2. By Lemma 2.6 Bdivη 1 H s 2. By Lemma 2.4, B φ 2 H s 2. We observe that the requirement (3.6) enforces that u R H s for s < s. So, if C 2 can be chosen so that (3.6) holds, then A 1 (µc 2 φ 2 + C 2 B φ 2 + f 2 ) H s. (3.10) Inserting the function u R of (3.9) into (3.8), p R = C 2 BΨ 2 + B ( g divη 1 µ 1 K 1 f2 ) (3.11) where K 1 = div(a 1A 1 ), K 2 = A 1 A 1, Ψ 2 = φ 2 + K 2 φ 2 + µ 1 (K 1 B) φ 2. Using (3.11), where f + f 2 Λ 1 ( ψ 1 ) φ 1 p R = C 2 ( µ φ2 + BΨ 2 ) f 2 (3.12) f 2 = [f 21, f 22 ]T = (µc 1 Λ 1 ( ψ 1 )) φ 1 (I µ 1 BK 1 ) f 2. We set 1 = x, 2 = y and Φ ij = i B j (φ 2 + A 1 A 1 φ 2 ) + µ 1 i B(K 1 B) j φ 2. (3.13) Using (3.6) and (3.12), the coefficient C 2 = [C 21, C 22 ] is given by where d 2 = λ 2,11 λ 2,22 λ 2,21 λ 2,12 with C 21 = (M 21 λ 2,22 M 22 λ 2,12 )/d 2, C 22 = (M 22 λ 2,11 M 21 λ 2,21 )/d 2 (3.14) λ 2,11 = Λ 2 (µ φ 2 + Φ 11 ), λ 2,12 = Λ 2 (Φ 21 ), λ 2,21 = Λ 2 (Φ 12 ), λ 2,22 = Λ 2 (µ φ 2 + Φ 22 ), M 21 = Λ 2 (f 21 ), M 22 = Λ 2 (f 22 ). (3.15) We next show that the above numbers λ 2,ij and M 2i are well-defined and that the determinant d 2 0. Lemma 3.1. Suppose [f, g] H s 2 H s 1 for s < s. (i) The functions Φ ij, Bdivη 1 and BK 1 Bdivη 1 belong to H s 2. (ii) λ 2,ij is well-defined. (iii) If µ is large enough, then d 2 0. (iv) M 2 C( f s 2 + Bg s 1 ) and C 2 C( f s 2 + Bg s 1 ).

20 20 J. R. Kweon and R. B. Kellogg JMFM Proof. Since φ i = 0 near (0, 0), it is in H s 2. Using Lemmas 2.4 and 2.5, B φ 2 and Bdivη 1 are in H s 2 for s < s. So B(divA 1A 1 ) B φ 2 H s 2, which follows from A 1 B φ 2 H s (see (3.10)). A similar argument shows that BK 1 Bdivη 1 H s 2. If [f, g] H s 2 H s 1, then BK 1 f2 H s 2 by (3.10). Thus Φ ij, f 2 and BK 1 f 2 are in H s 2. So λ 2,ij is well-defined. Set a 2 = Λ 2 ( φ 2 ) and b 2 = a 2 [Λ 2 (h 11 ) + Λ 2 (h 22 )]. Hence d 2 = a 2 2 µ 2 + b 2 µ + o(1) 0, because a 2 0 and µ is large. Using Lemma 2.6, BK 1 divη 1 s 2 C C 1. Using Λ 1 ( ψ 1 ) C C 1 and using f 2, M 2 Λ 2 (f Bg) + Λ 2 ( Bdivη 1 ) + Λ 2 ( BK 1 divη 1 ) + C(µ C 1 + Λ 1 ( ψ 1 ) ) C( f s 2 + Bg s 1 + C 1 ) C( f s 2 + Bg s 1 ). Thus C 2 C( f s 2 + Bg s 1 ). We show the a priori estimate for [u R, p R ]. (3.16) Theorem 3.1. Let C 2 be given in (3.14). Suppose that µ > A 1 Bdiv, i.e., A 1 is bounded. If [f, g] H s 2 H s 1 for s < s, then [u R, p R ] satisfies the a priori estimate µ u R s + p R s 1 C ( f s 2 + Bg s 1 ). (3.17) Proof. We recall that v 1 = χa( φ 1 ) Λ 1 ( φ 1 )φ 1 = A 1 φ 1 with = B in (2.29). Using (3.9) and Lemmas 2.4, 2.6, µ u R s C ( f Bg s 2 + C 1 Bdivv 1 s 2 ) + C C 2 ( φ 2 s 2 + B φ 2 s 2 ) C ( f s 2 + Bg s 1 + C 1 + C 2 ) C ( f s 2 + Bg s 1 ), where C = C( A 1 A 1, µ). Using (3.11), p R s 1 C ( C 2 BΨ 2 s 1 + B(g divη 1 µ 1 K 1 f2 ) s 1 ) Thus (3.17) follows. (3.18) C ( f s 2 + Bg s 1 ). (3.19) We are now ready to show Theorem 1.1 in Section 1. It follows by Lemma 2.8, Lemma 3.1 and Theorem 3.1.

21 Vol. 11 (2009) Compressible Flows 21 References [1] R. A. Adams, Sobolev Spaces, Academic Press, New York, [2] J. D. Anderson, Jr., Fundamentals of Aerodynamics, 2nd edition, McGraw-Hill, New York, [3] M. Dauge, Stationary Stokes and Navier Stokes systems on two- or three-dimensional domains with corners. Part I. Linearized equations, SIAM J. Math. Anal. 20 (1989), [4] P. Grisvard, Elliptic problems in nonsmooth domains, Pitman Advanced Publishing Program, Boston London Melbourne, [5] R. B. Kellogg, Corner singularities and singular pertubations, Ann. Univ. Ferrara Sez. VII Sc. Mat. XLVII (2001), [6] V. A. Kozlov, V. G. Maz ya and J. Rossmann, Elliptic boundary value problems in domains with point singularities, AMS, [7] J. R. Kweon and R. B. Kellogg, Compressible Navier Stokes Equations in a bounded domain with Inflow Boundary Condition, SIAM J. Math. Anal. 28 (1997), [8] J. R. Kweon and R. B. Kellogg, Compressible Stokes problem on non-convex polygon, J. Differential Equations 176 (2001), [9] J. R. Kweon and R. B. Kellogg, Regularity of solutions to the Navier Stokes equations for compressible Barotropic flows on a polygon, Arch. Rational Mech. Anal. 163 (2002), [10] J. R. Kweon and R. B. Kellogg, Regularity of solutions to the Navier Stokes equations for compressible flows on a polygon, SIAM J. Math. Anal. 35 (2004), [11] J. R. Kweon and R. B. Kellogg, Singularities in the Density Gradient, J. Math. Fluid Mech. 8 (2006), [12] J. L. Lions and E. Magenes, Non-homogeneous boundary value problems and applications, Springer-Verlag, Berlin Heidelberg New York, [13] B. Liu and R. B. Kellogg, Discontinuous solutions of linearized, steady state, viscous, compressible flows, J. Math. Anal. Appl. 180 (1993), Jae Ryong Kweon Department of Mathematics Pohang University of Science and Technology Pohang , Kyoungbuk Republic of Korea kweon@postech.ac.kr R. Bruce Kellogg Department of Mathematics University of South Carolina Columbia SC USA kellogg@ipst.umd.edu (accepted: October 29, 2006; published Online First: June 12, 2007)