Condition number of the BEM matrix arising from the Stokes equations in 2D

Condition numer of the BEM matrix arising from the Stokes equations in D W. Dijkstra a, R.M.M. Mattheij Department of Mathematics and Computing Science, Eindhoven University of Technology, P.O. Box 53, 56 MB Eindhoven, The Netherlands a w.dijkstra@tue.nl Astract We study the condition numer of the system matrices that appear in the Boundary Element Method when solving the Stokes equations at a two-dimensional domain. At the oundary of the domain we impose Dirichlet conditions or mixed conditions. We show that for certain critical oundary contours the underlying oundary integral equation is not solvale. As a consequence, the condition numer of the system matrix of the discrete equations is infinitely large. Hence, for these critical contours the Stokes equations cannot e solved with the Boundary Element Method. To overcome this prolem the domain can e rescaled. Several numerical examples are provided to illustrate the solvaility prolems at the critical contours. Introduction Nowadays the Boundary Element Method (BEM) is widely used y mathematicians and engineers in a variety of fields, such as heat and fluid flow, structural mechanics, acoustics, etc. Despite the succes of the BEM, little attention has een paid to a fundamental issue as solvaility of the equations that appear in the BEM. In general the BEM amounts to solving a system of linear algeraic equations. The solvaility of this system depends to a large extent on the condition numer of the system matrix. Hence knowledge aout this condition numer is essential when investigating the solvaility of the system of equations. It was shown that the condition numer of the BEM-matrix for the Laplace equation in D with Dirichlet conditions may ecome infinitely large when the logarithmic capacity of the domain of interest is equal to one [, 3, 4, 6, 9]. This is also true for the Laplace equation with mixed oundary conditions []. The condition numer of the BEM matrix for the Helmholtz equation may also ecome infinitely large [, 5, 6]. This happens at a countale set of critical wave numers. In all these cases where the condition numer gets infinitely large, solving the linear system is very difficult. In general, to investigate the condition numer of the BEM-matrix, the underlying oundary integral equations need to e studied. These equations and the operators that appear in them have een the topic of several papers. Again it is noted that the oundary integral equation for the Laplace equation with Dirichlet conditions is not uniquely solvale when the logarithmic capacity of the domain is equal to one [3, 4, 9, ]. A similar result is true for the iharmonic equation with Dirichlet conditions [5, 8, ]. In this paper we concentrate on the Stokes equations supplemented with Dirichlet or mixed oundary conditions. We investigate the solvaility of the oundary integral formulation of these equations. For this goal we study the single-layer operator that appears in the formulation and see under what conditions it admits zero eigenvalues. In Section we riefly descrie the oundary

integral equation for the Stokes prolem, introducing the single and doule layer operator. For the case of a circular domain the eigenvalues and eigenfunctions of the single-layer operator can e calculated analytically. These results are presented in Section 3. In Section 4 we show that the single-layer operator admits zero eigenvalues under certain conditions. As a consequence the oundary integral equation for Stokes flow with Dirichlet conditions is not always solvale. The same holds for the oundary integral equation for Stokes flow with mixed oundary conditions. This is proved in Section 5. Section 6 illustrates the findings y computing the condition numer of the BEM-matrix for several examples. Boundary integral equations for D Stokes flow Let Ω e a two-dimensional simply-connected domain Ω with smooth oundary contour. The Stokes equations for a flow in Ω read v p =,.v =, () where v is the velocity field of the fluid and p its pressure. Let e divided into a part at which the velocity v is prescried, and a part at which the pressure p is prescried, =. The Stokes equations are suject to the oundary conditions v = ṽ, x, p = p, x. Either or can e empty, leading to a pure Neuman or Dirichlet prolem respectively. The Stokes equations in differential form can e transformed to the oundary integral equation (for details aout the derivation we refer to [7, 9, ]) v i(x) + q ij (x, y)v j (y)d y = () u ij (x, y) j (y)d y, x, i =,. (3) Here a repeated index means summation over all possile values of that index. The vector function is the normal stress at the fluid oundary, := T (p, v)n, with n the outward unit normal at the oundary and the stress tensor T defined as ( vi T ij (p, v) := pδ ij + + v ) j. (5) x j x i Hence the oundary integral formulation involves two variales, the velocity v and the normal stress. At each point of the oundary either v or is prescried, v = ṽ, x, = pn, x. (6) The kernels u ij and q ij in the integral equations are defined as q ij (x, y) = u ij (x, y) = 4π (x i y i )(x j y j )(x k y k )n k π x y 4, { δ ij log x y + (x i y i )(x j y j ) x y (4) }. (7) We introduce oundary integral operators, (Gϕ) i (x) = u ij (x, y)ϕ j (y)d y, (Hψ) i (x) = q ij (x, y)ψ j (y)d y, (8)

which enales us to write (3) as ( I + H)v = G. (9) The operators G and H are often called the single and doule layer operator for the Stokes equations. For the Dirichlet prolem the velocity v at the oundary is given ( = ) and we aim to reconstruct the normal stress at the oundary y solving G = f(ṽ) := ( I + H)ṽ. () To solve this equation we need to invert the operator G. This can only e done when all eigenvalues of G are unequal to zero. In this paper we investigate under what conditions G admits a zero eigenvalue. For the mixed prolem the velocity at is prescried and the normal stress at is prescried. By rearranging known and unknown terms (see Section 5) the system (9) can e written as Ax = g. () Again, to solve this equation we need to invert the operator A. This can only e done when all eigenvalues of A are unequal to zero. We will show that zero eigenvalues occur under the same conditions as for the Dirichlet prolem. 3 Eigensystem of the single layer operator G at a circular domain In this section we compute eigenfunctions and eigenvalues of the operator G for a circular oundary. Here we make use of the fact that the first term in the kernel u ij is equal (apart from a constant) to the fundamental solution for the Laplace operator in D. Hence the first term from the single-layer operator G for the Stokes equations is related to the single-layer operator (K s ) for the Laplace equation y 4π ( log x y )(y))d y = (Ks )(x). () For a circle with radius R, the single-layer potential K s admits the following eigensystem ([9]), K s cos(kt) = R k cos(kt), K s sin(kt) = R k sin(kt), K s = R log R. (3) We introduce the following polar coordinates, x = R[cost, sint], y = R[coss, sins]. (4) Then the distance squared etween x and y is computed as x y = R ( cos(t s)) and we find u (t, s) = { δ ij log [ R ( cos(t s)) ] } (cost coss) +, 8π cos(t s) u (t, s) = (cost coss)(sin t sins), 8π cos(t s) u (t, s) = 8π u (t, s) = 8π (sin t sins)(cos t coss) = u (t, s) cos(t s) { δ ij log [ R ( cos(t s)) ] + } (sin t sin s). (5) cos(t s) 3

eigenvalue R R 4 [ R 4k, (k =, 3,...) eigenfunctions [ ] cost sin t [ ] sint cost [ ] [ ] cost sin t, sint cost ] [ cos(kt), sin(kt) [ ] [ ] R log R + 4 R, cos(kt) sin(kt) ], [ sin(kt) cos(kt) ], [ sin(kt) cos(kt) ] Tale : Eigenvalues and eigenfunctions of G. Using the integral kernels in polar coordinates given in (5) it is straightforward to compute the eigenvalues and eigenvectors of the operator G. The results are given in Tale. We oserve that the function (x) = n = [cost, sin t] T is an eigenfunction with eigenvalue zero. This can e explained as follows: when we apply a stress in the direction of the normal at each point of the oundary, each stress having the same magnitude, the net contriution will e equal to zero. This phenomenon does not restrict to the circle, ut applies to any shape of the oundary, as will e proven in Section 4. For the case R = R := exp(/), we see that there is another zero eigenvalue which has an eigenspace with dimension two. The eigenfunctions correspond to a uniform stress distriution; at each point at the oundary an equal stress is applied in the same direction. This particular stress distriution will cause a translation of the ody. The circular oundary contour with R = R is often called a critical contour (or -contour [4]). The radius or scale R is often called the critical scale or degenerate scale. Equation () at the critical contour cannot e solved as the operator G admits a zero eigenvalue. The situation in which the oundary integral equation is not solvale for a critical size of the circle also occurs for the Dirichlet Laplace equation in D [, 3, 4, 6, 9]. It is shown that the single-layer operator for the unit circle admits a zero eigenvalue, and consequently it cannot e inverted. Hence, the same phenomenon appears for oth the Laplace and Stokes equations. The critical scale however is not the same for the Laplace and Stokes equations. 4 Invertiility of single layer operator on general domain In this section we study the solvaility of the oundary integral equation () on a smooth closed oundary contour. We search for eigenfunctions of the oundary integral operator G with zero eigenvalue, hence G =. If such eigenfunctions exist the integral equation () is not uniquely solvale. Theorem For any smooth oundary the outward unit normal n is an eigenfunction of the oundary integral operator G with eigenvalue zero. 4

Proof. The i-th component of Gn equals (Gn) i = u ij (x, y)n j (y)d y = [ δ ij log 4π x y + (x ] i y i )(x j y j ) x y n j (y)d y = [ δ ij log 4π Ω x j x y + (x ] i y i )(x j y j ) x y dω y = u i j x dω y = div u i dω =. (6) j Ω Ω Here u i is the flow velocity due to a Stokeslet [3]. This velocity satisfies the incompressiility condition div u i =. In Section 3 we already saw that for a circular oundary the normal vector n is an eigenfunction of G with eigenvalue zero. The current theorem generalizes this result to aritrary smooth closed oundary contours. In the sequel of this section we assume that the solutions of the Dirichlet prolem () are sought in a function space that excludes the unit normal. Hence the eigenfunctions of G we are looking for cannot e equal to n. We now show that for each oundary there exist (atmost) two critical scalings of the oundary such that the operator G in the Dirichlet prolem () is not invertile. This phenomenon has een oserved and proven y [] and we will partly present their analysis here. Theorem For all given functions f and constant vectors d the system of equations { G + c = f d = d, (7) has a unique solution pair (, c), where is a function and c a constant vector. Proof. The main idea is to show that the operator that maps the pair (, c) to the left-hand side of (7) is an isomorphism. For details we refer to []. We proceed y introducing the two unit vectors e = [, ] T and e = [, ] T. Theorem quarantees that two pairs (, c ) and (, c ) exist that are the unique solutions to the two systems { G + c = d = e, We define the matrix C as C := [c c ]. { G + c = d = e. Theorem 3 If det(c ) =, then the operator G is not invertile. Proof. Suppose that det(c ) =, then the columns c and c are dependent, say c = αc for some α R, α. In that case = ( G + c ) α ( G + c ) = G( α ) + αc αc = G( α ). (9) The function α cannot e equal to zero, since this would require ( α )d to e equal to zero, while we have ( α )d = e αe. () So α is an eigenfunction of G with zero eigenvalue. (Note that this eigenfunction cannot e equal to the normal n at the oundary, since n satisfies nd =.) (8) Corollary There are (atmost) two critical scalings of the domain for which the operator G is not invertile. 5

Proof. We rescale the domain y a factor a, i.e. a. With the definition of the operator G it can e shown that G G a := 4π a log a d + ag. () Then the two systems in (8) change into { ag j + c j 4π a log a j d = a j d = e j, j =,. () Define j a := a j for j =,, then we otain { G j a + c j 4π log a j ad = j a d = e j, j =,. Sustituting the second equation into the first equation, we get (3) { G j a + c j 4π log a e j = j a d = e j, j =,. These systems have the same form as the original systems in (8), with c j c j 4π log a e j for j =,. Define the new matrix C a y C a := C 4π log a I, (5) then G a is not invertile when det(c a ) =. Hence when 4π log a is an eigenvalue of C the operator G a is not invertile. This implies that, when C has two distinct eigenvalues, there are two critical sizes a for which G a is not invertile. If C has one eigenvalue with doule multiplicity these critical scalings coincide. This result shows that the solvaility of the oundary integral equation for the Dirichlet Stokes prolem depends on the scale of the domain. There exist (atmost) two critical scales for which the equations are not solvale. The solvaility prolem is an artifact of the oundary integral formulation; the Stokes equations inn differential form are well-posed. For the Laplace equation with Dirichlet conditions a similar phenomenon is oserved [3, 4, 9, ]. In its differential form the prolem is well-posed, while the corresponding oundary integral equation is not solvale at critical oundary contours. Remark: The oundary integral equations for the Stokes flow in D are similar to the equations for plane elasticity. Hence the BIE for the latter equations suffer from the same solvaility prolems as the Stokes equations. A proof of this phenomenon for plane elasticity has een presented in [5] and is similar to the proof sketched aove. 5 Invertiility of operator on general domain with mixed oundary conditions In the previous section we showed that the oundary integral operator G for the Dirichlet Stokes equations is not invertile for all oundaries. In this section we show that this phenomenon extends to the Stokes equations with mixed oundary conditions. The starting point is again the BIE for the Stokes equations, v + Hv = G, at. (6) Suppose that the oundary is split into two parts, =. At we prescrie the velocity v while the normal stress is unknown. At we prescrie the normal stress while the (4) 6

velocity v is unknown. The oundary integral operators G and H are split accordingly, [G] i = u ij j d = u ij j d + u ij j d =: [G ] i + [G ] i, [Hv] i = q ij v j d = q ij vj d + q ij vj d =: [H v ] i + [H v ] i. (7) With these notations the BIE is written in the following way, vk + H v + H v = G + G, at k, k =,. (8) We arrange the terms in such a way that all unknowns are at the left-hand side and all knowns are at the right-hand side, H v G = G H v v, at, v + H v G = G H v, at. (9) Now we can define an operator A that assigns to the pair (, v ) the two functions at the left-hand side of (9), [ ] [ A H v v G ] v + H v G (3) To study the invertiility of this operator we need to study the homogeneous version of the equations in (9), H v G =, at, v + H v G =, at. (3) Theorem 4 There are (atmost) two scalings of such that the homogeneous equations (3) have a non-trivial solution, i.e. A is not invertile. Proof. From the Dirichlet prolem we know that there are (atmost) two scalings of for which G is not invertile. So there are a I and a II R and vector functions q I and q II such that Gq k =, at a k, k = I, II. (3) The scalings a k and functions q k may coincide ut this does not affect the proof of the theorem. Denote the nullspace of G y N(G). The normal vector n is always in N(G). If is a critical contour then also q I and q II are in N(G). Assume that is such a critical contour. Let q N(G), that is q = α q I + α q II + α 3 n, for some α, α and α 3 in R, and Gq =. Consider the homogeneous equations (3). We will show that there is a non-trivial pair (, v ) that satisfies these equations. We choose v = and = q + h. Here q i is the restriction of q to i, i =,, and h is still unknown. Sustituting these functions in the left-hand sides of the homogeneous equations yields for oth equations G = G q G h = +G q G h, (33) where we made use of Gq = G q + G q =. If we can find a function h such that G h = G q, then the left-hand sides of the homogeneous equations yield zero and we have found a non-trivial solution. Our task is then to prove that a solution h of G h = G q exists. A function is in N(G) if the inner product of q and is equal to zero, i.e. q i (x) i (x)d x =. (35) (34) 7

First we note that the right-hand side of (34) is contained in N(G), since q i [G q ] i d x = q i (x) u ij (x, y)qj(y)d y d x G = qj(y) u ij (x, y)q i (x)d x d y = qj(y)[gq] j d y =. (36) So we may generalize our task: prove that a solution h of G h = φ exists, for any φ N(G). If so, then φ = G q completes the proof. It is known that the operator G is a Fredholm operator with index zero [7]. Hence the Fredholm alternative can e applied [8, p. 37]. This states that for the homogeneous equation two mutually exclusive possiilities exist:. Gh = only has the trivial solution;. Gh = has exactly p linearly independent solutions. We are in the second siuation, as the nullspace of G is non-empty and is spanned y n, q I and q II. The Fredholm alternative then states that Gh = φ is solvale if and only if φ is perpendicular to the p independent solutions of G f =, where G denotes the adjoined operator of G. However, the single layer operator for the Stokes flow is self-adjoined, so G = G. Consequently the solvaility condition says that φ has to e perpendicular to the solutions of Gh =. This is the case since we defined φ N(G). Consequently there exists an h with Gh = φ, for φ N(G). We split this h into two parts, { h, at h =, h (38), at. Recall that we search for h such that G h = φ. We add G h to oth sides of this equation, otaining or shorter G h + G h = φ + G h, (39) Gh = φ + G h. The right-hand side of this equation is in N(G), since φ N(G) and q i [G h ] i d x = q i (x) u ij (x, y)h j (y)d yd x G = h j (y) u ij (x, y)q i (x)d x d y = h j(y)[gq] j d y =, (4) so G h is also in N(G). Recall that Gh = φ+g h is solvale if the right-hand side is in N(G), so there exists an h satisfying (4). Then we may construct h y simply restricting h to, i.e. h = h. This results shows that also the oundary integral equation for the Stokes equations with mixed oundary conditions is not always solvale. This happens for the same critical oundary contours as for the Stokes equations with Dirichlet oundary conditions. Hence the mixed prolem inherits the solvaility prolems from the Dirichlet prolem. The division of the oundary into a Dirichlet and a Neuman part does not play a role in this. Note that the Laplace equation exhiits the same ehavior. The oundary integral equation for the Laplace equation with mixed oundary conditions also inherits the solvaility prolems from the oundary integral equation for the Dirichlet case []. (37) (4) 8

6 Numerical examples To solve the oundary integral equation (9), the oundary is discretised into a set of linear elements. We choose a set of N nodes x p (p =,...,N) at the oundary where two consecutive nodes x k and x k are connected y a straight line segment k. A point y at the element k is descried y y = φ (s)x k + φ (s)x k, where φ and φ are linear shape functions, given y φ (s) = ( s) and φ (s) = (+s), s. At each element k the velocity v and normal stress are approximated with the same shape functions φ i, according to v = φ (s)v k + φ (s)v k, = φ (s) k + φ (s) k. (4) The vectors v p and p are the velocity and normal stress at the node x p. We sustitute x = x p at the oundary integral equation (3) and replace the integral over the oundary y a sum of integrals over the oundary elements k. This results in v i(x p ) + = N ( q ij (x p, y(s)) k= k N ( u ij (x p, y(s)) k k= φ (s)v k j φ (s) k j ) + φ (s)vj k ds y(s) ) + φ (s) k j ds y(s), i =,. (43) Varying p from to N, equation (43) yields N equations for 4N coefficients v p j and p j. However at each point at the oundary we either prescrie v or, thus reducing the numer of coefficients y N. Hence we otain a system of N linear algeraic equations for N unknown coefficients. We introduce vectors v and of length N containing the coefficients of v and in the nodes. Then the system of equations can e written in short-hand notation as ( I + H)v = G. (44) To otain the matrices H and G integrals of the form q ij (x p, y(s))φ r (s)ds y(s), k u ij (x p, y(s))φ r (s)ds y(s), k (45) have to e evaluated, respectively. These integrals can e approximated numerically y applying a Gaussian quadrature formula. When x p is in the element k over which integration is performed, the integrands will have a singularity in the integration domain. However, in the case of linear elements such as we use here, these integrals can e calculated analytically. 6. Dirichlet oundary conditions For the Dirichlet prolem, the coefficients of the velocity vector v are given, say v = ṽ, and we need to solve G = f(ṽ) := ( I + H)ṽ (46) to find the unknown coefficients of the normal stress vector. If the oundary integral operator G is not invertile then also its discrete counterpart, the matrix G is not invertile. To visualize this we compute the condition numer of G: if the condition numer is infinitely large, then the matrix is not invertile. As a consequence, (46) cannot e inverted to otain. In the following example we choose an ellipsoidal oundary contour. We construct the matrix G and compute its condition numer. Then we rescale the oundary y a factor a, i.e. a. Again we compute the condition numer of the matrix G. We do this for several values of a. According to the theory in the previous section there are two critical values for a for which the condition numer of G goes to infinity. 9

In Figure we show the condition numer as a function of the scale a. We do this for an ellipse with aspect ratio.4 (ellipse ) and for an ellipse with aspect ratio.7 (ellipse ). We oserve that for oth cases two critical scaling exist for which the condition numer goes to infinity. Moreover, these critical values differ significantly for the two ellipses. 6 aspect ratio =.4 aspect ratio =.7 5 cond(g) 4 3.6.8..4.6.8 3 3. a Figure : Condition numer of G for an ellipsoidal domain with aspect ratios.4 and.7 as a function of scaling parameter a. 4.5 4 3.5 3.5.5 a a.5 =a/d.5.5.5 3 3.5 4 4.5 5 5.5 6 Figure : The critical sizes of an ellipse for which the condition numer of G is very large. a Figure visualises all ellipses for which the condition numer of G is very large. At the horizontal axis we put the length a of the horizontal semi-axis of the ellipse, at the vertical axis the length of the vertical semi-axis of the ellipse. We compute the condition numer of G for several values of a and. We call the values of a and for which the condition numer goes to infinity the critical values and the corresponding ellipse the critical ellipse. At the critical values we plot a dot in the (a, )-plane of Figure. We see that the critical values lie on two curved lines,

which are symmetric around the line a =. The lack dashed line with slope /d represents all ellipses with aspect ratio d := a/. It can e concluded that for a fixed d, i.e. an ellipse with a fixed aspect ratio, two critical sizes exist. For a circle, where d =, only one critical size exists. The values corresponding to this critical size are approximately a = =.65, which agrees with the eigenvalue of exp(/).649 that we found in Section 3. 4 4.5 3.5 4 3.5 3.5 a a 3.5.5 a a.5.5.5.5.5.5 3 3.5 4 4.5 5 5.5 6 a.5.5.5 3 3.5 4 4.5 a Figure 3: The critical sizes of two fluid domains. In Figure 3 we have shown the critical sizes for two other oundary contours. For oth cases the results look similar to those of the ellipse. Again there are always two critical sizes, except when the aspect ratio of the domain is equal to one. 6. Mixed oundary conditions In the case of mixed oundary conditions we may rearrange terms in (44) and put all known coefficients at the right-hand side and all unknown coefficients at the left-hand side. Then we otain a linear system of the form Ax = g. (47) The matrix A consists of a lock from the matrix G and a lock from the matrix H. We compute the condition numer of the matrix A for the case of an ellipsoidal oundary contour. The ellipse is approximated y 6 linear elements. At the first eight elements we impose Dirichlet oundary conditions and at the last eight elements we impose Neuman oundary conditions. Then we rescale the oundary y a factor a and again compute the condition numer. Figure 4 shows the condition numer of the matrix for an ellipse with aspect ratio.4 and an ellipse with aspect ratio.7. We see that there exist two critical scalings for each ellipse. For these critical ellipses the condition numer of A ecomes infinitely large. The critical sizes of the ellipse in the case of mixed conditions are close to the critical sizes for the case of Dirichlet conditions. In Figure 5 we show all critical sizes of the ellipse for the case of mixed conditions. One can oserve that this plot is almost similar to the plot in Figure for Dirichlet conditions. This illustrates the idea that the oundary integral equation for the mixed case inherits the solvaility prolem from the oundary integral equation for the Dirichlet case. Still some small differences can e distinguished in the critical scalings for the Dirichlet and mixed case. These differences are caused y errors in the numerical computations. When the numer of oundary elements increases, the differences etween the critical scalings descreases.

7 6 aspect ratio =.4 aspect ratio =.7 5 cond(a) 4 3.4.6.8..4.6.8 3 3. a Figure 4: Condition numer of A for an ellipsoidal domain with aspect ratios.4 and.7 as a function of scale a. 5.5 5 4.5 4 3.5 3.5.5.5.5.5.5 3 3.5 4 4.5 5 5.5 Figure 5: The critical sizes of an ellipse for which the condition numer of A is very large. a 6.3 Deformation of viscous drop due to surface tension We now turn our attention to a time-dependent prolem. We study a viscous drop of fluid of ellipsoidal shape that deforms to a circle due to surface tension. The evolution of the oundary of the drop is governed y the Stokes equations and can e solved using the Boundary Element Method [, 4]. The velocity v of the oundary and the normal stress at the oundary are related y the oundary integral equation (9). The normal stress is proportional to the mean curvature κ of the oundary, = γκn, (48)

with γ the surface tension and n the outward normal at the oundary. At time level t = t we compute the velocity v at all discretisation nodes x. Then we update the oundary of the drop with an Euler forward step, x x + tv(x), (49) otaining a new oundary contour. For this new oundary we again compute the velocity, and perform another Euler forward step. In this way we can study the shape evolution of the oundary. In Figure 6 we see the evolution of the ellipse to a circle. We choose 4 points to discretise the oundary, the size of the time step is t =.375 and we compute time steps. The initial shape is an ellipse with aspect ratio.5 and the longest semi-axis has a length.. Figure 6: An ellipse deforming to a circle. In the prolem descried aove the normal stress is prescried and the velocity has to e reconstructed with equation (). For the inverse prolem we try to reconstruct the normal stress given the oundary velocity, and therefore we need to invert the matrix G in each time step. We know that there are certain critical oundary contours for which the matrix G will not e invertile. In the prolem of the deforming ellipse we go through a whole range of ellispes with different shapes, and we risk to encounter one or more of those critical contours. 9 9 8 8 7 7 cond(g) 6 cond(g) 6 5 5 4 4 3...3.4.5.6 time (a) Without scaling 3...3.4.5.6 time () With scaling close to critical size Figure 7: The condition numer of G at each time step. In Figure 7(a) we show the condition numer of the matrix G at each time step. An increase in the condition numer indicates that the oundary contour under consideration is close to a 3

without scaling with scaling difference area.5979837.59785.35e 5 length.975356.9757.4e 4 a.7595789.7594955 8.674e 5.48664.4968 3.544e 5 Tale : Difference in the BEM with and without temporarily rescaling. Listed are the surface area, the length of the oundary and the length of the semi-axes at time t =.6. critical contour. We see that there is one time level in which such a critical contour is reached. In this case we choose discretisation points, a time step size t =. and 3 time steps. There is a simple way to avoid a critical contour. When the size of the contour gets close to the critical contour, we temporarily scale the domain. In Figure 7() we show the outcome of this strategy. At each time step etween t =.38 and t =.46 we scale the domain y a factor.. Then we solve the BEM prolem in each of these time steps and the solutions are rescaled y a factor /.. As a consequence the condition numer of the BEM-matrices in these time steps does not ecome very large. The scaling of the domain during some time steps does hardly affect the outcome of the test. In Tale we give the surface area, the length of the oundary contour and the sizes of the semi-axes at the final time step. Their values hardly change when we perform temporary scaling. 4.5 4 3.5 3.5.5.5.5.5.5 3 3.5 4 4.5 5 5.5 6 a Figure 8: Each solid line represents the size of an ellipse as it deforms to a circle. The initial size, i.e. the lengths a and of the semi-axes, is denoted with a large dot. The dashed-dotted lines represent the critical sizes. The thick lack dashed line represents all circles. Of course, for this strategy certain knowledge is needed aout the critical contours. However, we do not need to know the exact critical contour. We only have to make sure we do not get too close to it. In the current test we only scaled at a restricted numer of time steps. In general one 4

could scale at every time step, thus excluding the possiility that a critical contour is encountered. 8 7 6 cond(g) 5 4 3 3 4 5 6 time Figure 9: The condition numer for the ellipse with initial values a = 4 and =.8. During the deformation to a circle this particular ellipse gets a critical size twice. The size of the initial ellipse strongly affects the condition numer of the matrix at each time level as the ellipse deforms. In Figure 8 we show the size (a and ) of an ellipse as it is deforming. The size of the initial ellipse is shown with a large dot, whereas the sizes it takes as it deforms to a circle is denoted y the trajectory starting in the dot. All ellipses deform to a circle, which is seen from the fact that all trajectories converge to the straight thick dashed line where a =. The dashed-dotted lines represent all critical ellispes, c.f. Fifure. When the trajectory of an ellipse crosses one of the two dashed-dotted lines, it means that the ellipse at that time level is a critical ellipse. For this particular ellipse the condition numer will e large. The trajectories of some ellipses never cross such a critical line. This means that they never get a critical size as they deform to a circle. For other ellipses there is one point along their trajectory where the ellipse is a critical ellipse. It is also possile that an ellipse ecomes critical twice during the deformation. An example is the ellipse with initial size a = 4 and =.8. The trajectory of this ellipse crosses the critical line twice. Figure 9 shows the condition numer for this particular case. It can e seen that the condition numer gets very large at the eginning and at the end of the time interval. 7 Discussion The oundary integral equation for the Stokes equations with Dirichlet conditions at a twodimensional domain is not always uniquely solvale. The existence of a unique solution depends on the size of the domain. For each domain there is at least one size, or etter one scaling, fow which the oundary integral does not have a unique solution. For most domains there are even two of such scalings. In order to find a unique solution the domain can e rescaled y a suitale factor. Similar solvaility prolems appear in the oundary integral equation for the Laplace prolem with Dirichlet conditions at a two-dimensional domain. In this case the critical size of the domain 5

for which the equation is not uniquely solvale relates to the logarithmic capacity of the domain. We have proven that, in analogy to the Laplace prolem with mixed oundary conditions, also the Stokes prolem with mixed oundary conditions does not always have a unique solution. This happens at the same critical domains as for the Dirichlet prolem. Hence, the division of the oundary into a Dirichlet part and a Neuman part does not play a role in this. By using a suitale rescaling of the domain the solvaility prolems can e avoided adequately. However, to choose the right scaling some knowledge aout the critical scalings is useful. References [] S. Amini and S.M. Kirkup. Solution of Helmholtz equation in the exterior domain y elementary oundary integral methods. J. Comput. Phys., 8:8, 995. [] S. Christiansen. Condition numer of matrices derived from two classes of integral equations. Mat. Meth. Appl. Sci, 3:364 39, 98. [3] S. Christiansen. On two methods for elimination of non-unique solutions of an integral equation with logarithmic kernel. Appl. Anal., 3: 8, 98. [4] S. Christiansen and P.C. Hansen. The effective condition numer applied to error analysis of certain oundary colocation methods. J. Comp. Appl. Math., 54:5 35, 994. [5] S. Christiansen and P. Hougaard. An investigation of a pair of integral equations for the iharmonic prolem. J. Inst. Maths. Applics, :5 7, 978. [6] S. Christiansen and J. Saranen. The conditioning of some numerical methods for the first kind oundary integral equations. J. Comp. Appl. Math., 67:43 58, 996. [7] M. Costael. Boundary integral operators on Lipschitz domains: elementary results. SIAM J. Math. Anal., 9(3):63 66, 988. [8] M. Costael and M. Dauge. On invertiility of the iharmonic single-layer potential operator. Integr. Equ. Oper. Theory, 4:46 67, 996. [9] W. Dijkstra and R.M.M. Mattheij. The condition numer of the BEM-matrix arising from Laplace s equation. Electron. J. Bound. Elem., 4():67 8, 6. [] W. Dijkstra and R.M.M. Mattheij. A relation etween the the logarithmic capacity and the condition numer of the BEM-matrices. Comm. Numer. Methods Engrg., ():, 7. [] V. Domínguez and F.J. Sayas. A BEM-FEM overlapping algorithm for the Stokes equation. Appl. Math. Comput., pages 69 7, 6. [] B. Fuglede. On a direct method of integral equations for solving the iharmonic dirichlet prolem. ZAMM, 6:449 459, 98. [3] G.C. Hsiao. On the staility of integral equations of the first kind with logarithmic kernels. Arch. Ration. Mech. Anal., 94:79 9, 986. [4] M.A. Jaswon and G. Symm. Integral Equation Methods in Potential Theory and Elastostatics. Academic Press, London, 977. [5] S.M. Kirkup. The influence of the weighting parameter on the improved oundary element solution of the exterior Helmholtz equation. Wave Motion, 93:93, 99. [6] R. Kress and W.T. Spassov. On the condition numer of oundary integral operators for the exterior Dirichlet prolem for the Helmholtz equation. Numer. Math., pages 77 95, 983. 6

[7] O.A. Ladyzhenskaya. The Mathematical Theory of Viscous Incompressile Flow. Gordon and Beach, New York-London, 963. [8] W. McLean and T. Tran. A preconditioning strategy for oundary element galerkin methods. Numer. Methods Partial Differential Eq., 3:83 3, 997. [9] H. Power and L.C. Wroel. Boundary integral methods in fluid mechanics. Computational Mechanics Pulications, Southampton, 995. [] C. Pozrikidis. Boundary integral and singularity methods for linearized viscous flows. Camridge University Press, Camridge, 99. [] A.R.M. Primo, L.C. Wroel, and H. Power. An indirect oundary-element method for slow viscous flow in a ounded region containig air ules. J. Engrg. Math., 37:35 36,. [] I.H. Sloan and A. Spence. The Galerkin method for integral equations of the first kind with logarithmic kernel: theory. IMA J. Numer. Anal., 8:5, 988. [3] G.A.L. van der Vorst. Modelling and numerical simulation of viscous sintering. PhD thesis, Eindhoven University of Technology, 994. [4] G.A.L. van der Vorst, R.M.M. Mattheij, and H.K. Kuiken. Boundary element solution for two-dimensional viscous sintering. J. Comput. Phys., :5 63, 99. [5] R. Vodicka and V. Mantic. On invertiility of elastic single-layer potential operator. J. Elasticity, 74:47 73, 4. 7