POINTWISE CONVERGENCE OF THE COMBINATION TECHNIQUEFOR LAPLACE'S EQUATION 1H.-J. BUNGARTZ, M. GRIEBEL, D. R�OSCHKE, AND C. ZENGER 2Abstract. For a simple model problem| the Laplace equation on the unit square with a Dirichletboundary function vanishing for x = 0, x = 1, and y = 1, and equaling some suitable g(x) for y = 0| we present a proof of convergence for the combination technique, a modern, e�cient, and easilyparallelizable sparse grid solver for elliptic partial di�erential equations that recently gained importancein �elds of applications like computational uid dynamics. For full square grids with meshwidth h andO(h�2) grid points, the order O(h2) of the discretization error using �nite di�erences was shownin [12], if g(x) 2 C2[0; 1]. In this paper, we show that the �nite di�erence discretization error ofthe solution produced by the combination technique on a sparse grid with only O �(h�1log2(h�1)� gridpoints is of the order O �h2log2(h�1)�, if the Fourier coe�cients bk of ~g, the 2-periodic and 0-symmetricextension of g, ful�ll jbkj � cg � k�3�" for some arbitrary small positive ". If 0 < " � 1, this is vaildfor g 2 C4[0; 1] and g(0) = g(1) = g00(0) = g00(1) = 0, e.g.. A simple transformation even shows thatg 2 C4[0; 1] is su�cient. Furthermore, we present results of numerical experiments with functions g ofvarying smoothness.Key words. Combination technique, sparse grids, elliptic partial di�erential equations, order ofdiscretization error.AMS(MOS) subject classi�cations. 35J05, 65N06, 65N15, 65N22, 65N55.1. Introduction. Since their presentation in 1990 [17], sparse grids have turnedout to be a very interesting tool for the e�cient solution of elliptic boundary value prob-lems. Besides the implementation of a hierarchical �nite element algorithm on sparsegrids [1], it has been primarily the combination technique [9] that attained attention inthe sparse grid context. Here, �rst, the problem is solved on several rectangular gridswith di�erent meshwidths hx and hy in x- and y-direction, respectively. Afterwards, thecomputed solutions are combined linearly in order to get the desired sparse grid solution.Consequently, one of the main advantages of the combination technique stems from theproperties of sparse grids [1]: In comparison to the standard full grid approach, thenumber of grid points can be reduced signi�cantly from O(Nd) to O(N(log2(N))d�1)in the d-dimensional case, whereas the accuracy of the calculated approximation to thesolution is only slightly deteriorated from O(N�2) (see [12]) to O(N�2(log2(N))d�1).Another advantage has to be seen in the simplicity of the combination concept, its in-herent parallel structure, and its framework property allowing the integration of existingsolvers for partial di�erential equations.Up to now, the combination technique has been applied to several types of ellipticpartial di�erential equations including problems originating from computational uiddynamics. Its excellent parallelization properties have been veri�ed on di�erent parallelarchitectures and on workstation networks [5, 6, 7, 8, 10].1 This work is supported by the Bayerische Forschungsstiftung via FORTWIHR | The BavarianConsortium for High Performance Scienti�c Computing.2 Institut f�ur Informatik der Technischen Universit�at M�unchen, D-80290 M�unchen, Germany.1

For a proof of the approximation properties mentioned above, in the two-dimensionalcase the existence of an error splittinguhx;hy(x; y)� u(x; y) = C1(x; y; hx)h2x + C2(x; y; hy)h2y + C(x; y; hx; hy)h2xh2y(1)with jC1(x; y; hx)j, jC2(x; y; hy)j, and jC(x; y; hx; hy)j bounded by some positive B(x; y)for all meshwidths hx and hy has to be shown [9]. Here, u(x; y) is the exact solu-tion of the given boundary value problem, and uhx;hy(x; y) denotes the solution on therectangular full grid with meshwidths hx and hy resulting from a �nite di�erence dis-cretization. For a simple model problem | the Laplace equation on the unit squarewith a Dirichlet boundary function vanishing for x = 0, x = 1, and y = 1, and equalingsome suitable g(x) for y = 0 | we present a proof for the existence of such an errorsplitting, if g satis�es a certain smoothness requirement: The Fourier coe�cients bk ofthe 2-periodic and 0-symmetric extension of g have to be of the order O(k�3�") for somearbitrary small positive ".In section 2, we give a short survey of the combination technique and its mostimportant properties. Then, the following sections deal with the proof of (1). Since, forthe error splitting presented in section 6, we need explicit representations of u, uhx;hy ,and of the solutions uhx;0 and u0;hy of the Laplacian discretized in only one direction,a Fourier-based representation of those functions is given in section 3. After showingsome properties of the components of u, uhx;hy , uhx;0, and u0;hy in sections 4 and 5, weare able to de�ne an error splitting according to (1). Finally, in section 7, we presentnumerical examples with Dirichlet boundary functions of varying smoothness.2. The Combination Technique. Let L be an elliptic operator of second order.Consider the partial di�erential equation Lu = f on the unit square := ]0; 1[2 withappropriate boundary conditions. Furthermore, let Gi;j be the rectangular grid on �with meshwidths hx := hi := 2�i in x- and hy := hj := 2�j in y-direction (i; j 2 IN). In[9], the combination technique was introduced, which combines the solutions uhx;hy ofthe discrete problems Lhx;hyuhx;hy = fhx;hy associated to Lu = f on di�erent rectangulargrids Gi;j : ucn := Xi+j=n+1 uhi;hj � Xi+j=n uhi;hj :(2)The resulting solution ucn is given on the sparse grid ~Gn;n with O(2nn) grid pointsinstead of the usual full grid Gn;n with O(4n) points. Figure 1 shows the sparse grids~G3;3 and ~G4;4. For a detailed introduction to sparse grids, see [17] and [1].Note that we have to solve n di�erent problems with O(2n+1) grid points (i+ j =n + 1) and n � 1 di�erent problems with O(2n) grid points (i+ j = n). These 2n � 1problems are independent from one another and can be solved in parallel. Finally, theirbilinearly interpolated solutions are combined according to (2).Concerning the error ecn(x; y) of the combined solution ucn(x; y), we cite a resultfrom [9]. To complete the picture, we give the proof, too.2

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* * * * * * * * *n=3 Fig. 1. Sparse grids ~G3;3 and ~G4;4:Theorem 1. Assume that, for each inner point (x; y) 2 , the pointwise errorof the solution uhx;hy = u2�i;2�j obtained on the rectangular grid Gi;j with meshwidthshx = hi = 2�i and hy = hj = 2�j is of the form (1) with jC1(x; y; hx)j, jC2(x; y; hy)j,and jC(x; y; hx; hy)j bounded by some positive B(x; y) for all hx; hy 2 ]0; 12]. Then, theerror ecn(x; y) of the combined solution ucn(x; y) on the sparse grid ~Gn;n ful�llsjecn(x; y)j := jucn(x; y)� u(x; y)j � B(x; y) � h2n � �1 + 54 � log2(h�1n )�;(3)where hn = 2�n.Proof. Looking at de�nition (2), we haveecn(x; y) = ucn(x; y)� u(x; y)= Xi+j=n+1 uhi;hj(x; y) � Xi+j=n uhi;hj(x; y) � u(x; y)= Xi+j=n+1�uhi;hj (x; y)� u(x; y)� � Xi+j=n�uhi;hj (x; y)� u(x; y)�;and, assuming (1), we getecn(x; y) = Xi+j=n+1�C1(x; y; hi) � h2i + C2(x; y; hj) � h2j + C(x; y; hi; hj) � h2ih2j�� Xi+j=n�C1(x; y; hi) � h2i + C2(x; y; hj) � h2j + C(x; y; hi; hj) � h2ih2j�= nXi=1C1(x; y; hi) � h2i � n�1Xi=1 C1(x; y; hi) � h2i+ nXj=1C2(x; y; hj) � h2j � n�1Xj=1 C2(x; y; hj) � h2j+ Xi+j=n+1C(x; y; hi; hj) � h2ih2j � Xi+j=nC(x; y; hi; hj) � h2ih2j3

= C1(x; y; hn) � h2n + C2(x; y; hn) � h2n+ 14 � Xi+j=n+1C(x; y; hi; hj) � Xi+j=nC(x; y; hi; hj)! � h2n:This leads to the estimationjecn(x; y)j � jC1(x; y; hn)j � h2n + jC2(x; y; hn)j � h2n+ 14 � Xi+j=n+1 jC(x; y; hi; hj)j + Xi+j=n jC(x; y; hi; hj)j! � h2n� B(x; y) � h2n � �2 + n4 + n� 1�= B(x; y) � h2n � �1 + 54 � log2(h�1n )�:Thus, if (1) holds, the pointwise error of the combined solution is of the orderO(h2nlog2(h�1n )), but the computation of ucn only involves theO(h�1n log2(h�1n )) grid pointsof the sparse grid ~Gn;n. In other words, on a sparse grid with signi�cantly less gridpoints than on a standard full grid, the combination technique produces a solution thatis nearly as good as a full grid solution.In this paper, we deal with the question which smoothness requirements have tobe ful�lled by the Dirichlet boundary function in order to get a behaviour of the erroruhx;hy(x; y)� u(x; y) on Gi;j as indicated in (1).For the remainder, we concentrate on the Laplace equation�u(x; y) = 0; (x; y) 2 ;(4)with the Dirichlet boundary conditionu(x; y) = ( g(x) : (x; y) 2 ��; y = 00 : (x; y) 2 ��; y > 0:(5)Throughout the paper, we assume that the Fourier coe�cients bk of ~g, the 2-periodicand 0-symmetric extension of g, ful�lljbkj := ����Z 1�1 ~g(x) sin(k�x)dx���� = ����2 Z 10 g(x) sin(k�x)dx���� � cgk3+"(6)for some arbitrary small positive " and a constant cg depending only on g. It is awell-known fact from Fourier theory that this is valid with 0 < " � 1 if ~g 2 C4(IR) orif g 2 C4[0; 1] and g(0) = g(1) = g00(0) = g00(1) = 0.3. Explicit Solutions. As a �rst step on the way to a proof of (1), we introducea Fourier-based representation of u, uhx;hy , uhx;0, and u0;hy. Here, as mentioned before,uhx;0 and u0;hy denote the solutions of the Laplacian discretized only in x- or y-direction,respectively. These Fourier series will enable us to �nd an appropriate splitting ofuhx;hy � u in section 6. 4

3.1. Continuous Problem. For an explicit representation of the solution u(x; y)of (4) and (5), we cite a result from [16]:Theorem 2. u(x; y) := 1Xk=1 bk sin(k�x)T (k; y)(7)with bk de�ned in (6) and T (k; y) := sinh(k�(1� y))sinh(k�)(8)solves (4) and satis�es the boundary condition (5).3.2. Discrete Problem. In contrast to the usual square grid approach, we wantto allow nets for the discretization with di�erent meshwidths hx and hy in x- and iny-direction. Therefore, the well-known �ve-point-stencil normally used in connectionwith �nite di�erences has to be modi�ed. The discrete counterpart to the continuousproblem then reads(9) �hx;hyuhx;hy(x; y) = uhx;hy(x+ hx; y) + uhx;hy(x� hx; y)� 2uhx;hy(x; y)h2x+ uhx;hy(x; y + hy) + uhx;hy(x; y � hy)� 2uhx;hy(x; y)h2y= 0;where x = ihx; 1 � i � Nx � 1; hx = 1=Nx; Nx 2 IN n f1g ;y = jhy; 1 � j � Ny � 1; hy = 1=Ny ; Ny 2 IN n f1g :In the grid points on the boundary ��, we postulateuhx;hy(ihx; jhy) = ( g(ihx) : (ihx; jhy) 2 ��; j = 00 : (ihx; jhy) 2 ��; j > 0:(10)On the analogy of the continuous case, we call functions satisfying (9) discrete harmonic.Again, we present a Fourier-type representation of a function uhx;hy(x; y) that solves (9)and ful�lls (10):Theorem 3. uhx;hy(x; y) := 1Xk=1 bk sin(k�x)T (�k; y)(11)with 1� cos(k�hx)h2x = cosh(�k�hy)� 1h2y ;(12) 5

de�ning �k implicitly as a function of k, hx, and hy, is discrete harmonic and evensatis�es the continuous boundary condition (5). For the de�nitions of bk and T (�k; y),see (6) and (8), respectively.Proof. The proof follows [12]. In (9), we use the separation approachuhx;hy(x; y) = X(x) � Y (y);and getY (y)X(x+ hx) +X(x � hx)� 2X(x)h2x + X(x)Y (y + hy) + Y (y � hy)� 2Y (y)h2y = 0:Note that the following relations hold:sin�k�(x+ hx)� + sin�k�(x� hx)� = 2 sin(k�x) � cos(k�hx)andsinh���(1� y + hy)� + sinh���(1� y � hy)� = 2 sinh���(1 � y)� � cosh(��hy):These considerations suggest the choiceuhx;hy(x; y) = sin(k�x) � sinh (��(1� y)) ;(13)and it can be shown, indeed, that for each k 2 IN uhx;hy(x; y) given in (13) is solutionof (9) on condition that 1 � cos(k�hx)h2x = cosh(��hy)� 1h2y :So, relation (12) has been shown. To indicate the dependence on k, we are going towrite �k instead of � from now on.Since the discrete operator �hx;hy is linear,1Xk=1�k sin(k�x) sinh��k�(1� y)�(14)with arbitrary �k 2 IR solves (9), too, if we assure convergence. Note that (14) alreadysatis�es the continuous boundary condition given in (5) for y > 0. For y = 0, we requestthat our solution also satis�es the according boundary condition from (5):1Xk=1�k sinh(�k�) sin(k�x) = g(x); 0 � x � 1:(15)Fourier theory and comparison of the coe�cients provide�k sinh(�k�) = bk:The resulting series given in (11) converges due to (6) and because jT (t; y)j � 1. Thiscompletes our proof. 6

3.3. Discrete Problem in x-, Continuous Problem in y-Direction. Dis-cretizing the Laplacian only in x-direction leads to(16) �hx;0uhx;0(x; y) = uhx;0(x+ hx; y) + uhx;0(x� hx; y)� 2uhx;0(x; y)h2x+ @2uhx;0(x; y)@y2= 0;where x = ihx; 1 � i � Nx � 1; hx = 1=Nx; Nx 2 IN n f1g ;y 2 ]0; 1[:In the grid points on the boundary ��, we postulateuhx;0(ihx; y) = ( g(ihx) : (ihx; y) 2 ��; y = 00 : (ihx; y) 2 ��; y > 0:(17)The limits for hy ! 0 from equations (11) and (12) suggestTheorem 4. uhx;0(x; y) := 1Xk=1 bk sin(k�x)T (�k; y)(18)with �2k�22 = 1 � cos(k�hx)h2x ;(19)de�ning �k as a function of k and hx, solves equation (16) and even satis�es the con-tinuous boundary condition (5).Proof. The convergence of the series in (18) is, again, assured by (6). Using theseparation approach uhx;0(x; y) = X(x) � Y (y)in equation (16), we get(20) �hx;0uhx;0(x; y) = X(x + hx) +X(x� hx)� 2X(x)h2x Y (y) +X(x)Y 00(y) = 0:With our proposed solutions X(x) = sin(k�x) and Y (y) = T (�k; y), we obtain a directveri�cation of equation (20):�hx;0uhx;0(x; y) = 2 sin(k�x) cos(k�hx)� 2 sin(k�x)h2x � sinh(�k�(1� y))sinh(�k�)7

+ sin(k�x) � �2k�2 sinh(�k�(1� y))sinh(�k�)= 2 sin(k�x) � T (�k; y) � cos(k�hx)� 1h2x + �2k�22 != 0;if condition (19) holds.Our solution (18) satis�es the boundary condition (5) for y > 0. For y = 0, we getuhx;0(x; 0) = 1Xk=1 bk sin(k�x);which is exactly the Fourier series of g(x). Thus, again even the continuous boundarycondition (5) are ful�lled, which completes our proof.3.4. Continuous Problem in x-, Discrete Problem in y-Direction. Thistime, we discretize the Laplacian only in y-direction:(21) �0;hyu0;hy(x; y) = @2u0;hy(x; y)@x2+ u0;hy(x; y + hy) + u0;hy(x; y � hy)� 2u0;hy(x; y)h2y= 0;where x 2 ]0; 1[;y = jhy; 1 � j � Ny � 1; hy = 1=Ny ; Ny 2 IN n f1g :In the grid points on the boundary ��, we postulateu0;hy(x; jhy) = ( g(x) : (x; jhy) 2 ��; j = 00 : (x; jhy) 2 ��; j > 0:(22)The limits for hx ! 0 from equations (11) and (12) suggestTheorem 5. u0;hy(x; y) := 1Xk=1 bk sin(k�x)T (�k; y)(23)with k2�22 = cosh(�k�hy)� 1h2y ;(24)de�ning �k implicitly as a function of k and hy, solves equation (21) and even satis�esthe continuous boundary condition (5).We omit the proof, because it exactly follows the argumentation in the proof oftheorem 4. 8

4. Properties of �k; �k; and �k. When we compare the explicit representa-tions (7), (11), (18), and (23) of u, uhx;hy , uhx;0, and u0;hy, respectively, we see thatthere is only one di�erence, namely the �rst argument of the function T . Therefore, toprepare the discussion of u, uhx;hy, uhx;0, and u0;hy, it is important to study both thosedi�erent �rst arguments �k(hx; hy), �k(hx), and �k(hy) and the function T itself.4.1. Properties of �k. Since T (t; y) is an even function of t for every y, we canchoose the following de�nition for �k instead of condition (19):�k�hx2 = sin k�hx2 ! :(25)This leads toLemma 1. For all k 2 IN with k < Nx := 1hx , �k satis�es the estimates2�k < �k < k:(26)Here, the upper bound k for �k is valid for all k 2 IN. Furthermore, limhx!0 �k = k forall k 2 IN.Proof. From (25), we get�k � k = k � sin �k�hx2 �� k�hx2k�hx2 = k � sin(w) �ww ;where w := k�hx2 . For arbitrary k 2 IN, we have w 2 ]0;1[, sin(w)�ww < 0, and, thus,�k < k. If k < Nx = 1hx , we get w 2 ]0; �2 [, sin(w)�ww 2 ]2��� ; 0[, and, therefore, 2�k < �k.Finally, limhx!0 �k = k follows directly from (25).Next, we make a statement concerning the di�erence �k � k:Lemma 2. For all k 2 IN, we have�k � k = k3h2xc1(k; hx);(27)where jc1(k; hx)j < �224 and limhx!0 c1(k; hx) = ��224 .Proof. According to (25) and (27), c1(k; hx) can be written asc1(k; hx) := �k � kk3h2x = 1k3h2x � 0@sin �k�hx2 ��hx2 � k1A= sin �k�hx2 �� k�hx2k3�h3x2= sin(w)� ww3 � �24with w := k�hx2 , showing both jc1(k; hx)j < �224 and limhx!0 c1(k; hx) = ��224 .9

4.2. Properties of �k. Since T (t; y) is an even function in its �rst argument, weassume that �k > 0 without loss of generality. Then, relation (24) is equivalent tok�hy2 = scosh(�k�hy)� 12 = sinh �k�hy2 ! :(28)This leads toLemma 3. For all k 2 IN, �k satis�es the estimates4� ln k�2 ! < �k < k;(29)and limhy!0 �k = k.Proof. �k < k immediately follows from (28), because sinh(x) > x for all x > 0.On the other hand, let us assume that �k � 4� ln �k�2 �:�k � 4� ln k�2 ! ) sinh �k�4 ! � sinh ln k�2 !! < 12 exp ln k�2 !! = k�4 ;which contradicts (28) for hy = 12 . Finally, limhy!0 �k = k follows directly from (28).Now, we have a closer look on the di�erence �k � k:Lemma 4. For all k 2 IN, we have�k � k = k3h2yc2(k; hy);(30)where jc2(k; hy)j < �224 and limhy!0 c2(k; hy) = ��224 .Proof. According to (28) and (30), c2(k; hy) can be written asc2(k; hy) := �k � kk3h2y = 1k3h2y � 0@arsinh �k�hy2 ��hy2 � k1A= arsinh �k�hy2 �� k�hy2k3�h3y2= arsinh(w)� ww3 � �24with w := k�hy2 , showing both jc2(k; hy)j < �224 and limhy!0 c2(k; hy) = ��224 .4.3. Properties of �k. Again, we pro�t from the fact that T (t; y) is an evenfunction of t. Then, relations (12) and (19) lead to�k�hy2 = sinh �k�hy2 ! :(31)Thus, we get 10

Lemma 5. For all k 2 IN with k < Nx, �k satis�es the estimates4� ln k�4 ! < �k < k:(32)Here, the upper bound k for �k is valid for all k 2 IN. Furthermore, limhy!0 �k = �kand limhx!0 �k = �k for all k 2 IN.Proof. Applying the argumentation of lemma 3 to (31), we immediately get4� ln��k�2 � < �k < �k:Together with the bounds for �k from lemma 1, we obtain (32). Finally, limhy!0 �k = �kfollows directly from (31), and limhx!0 �k = �k results from (28), (31), and the factthat limhx!0 �k = k.5. Properties of T (t; y). In this section, we �nish the discussion of u, uhx;hy ,uhx;0, and u0;hy with some remarks on important properties of T (t; y).Lemma 6. Let y 2 ]0; 1[ be arbitrary, but �xed. With T (0; y) := 1 � y, T (t; y)becomes an even function of t continuous and di�erentiable for all t 2 IR, andT (t; y) � minne�jtj�y; 1 � yo :(33)Furthermore,@T (t; y)@t = T (t; y) �P (t; y) =: T (t; y) � � � �(1� y) coth(t�(1� y))� coth(t�)�;(34)where jP (t; y)j � �y:(35)Proof. We learn from de�nition (8) that T (t; y) is an even function in its �rstargument. Sincelimt!0 T (t; y) = limt!0 sinh(t�(1� y))sinh(t�) = (1 � y) limt!0 cosh(t�(1� y))cosh(t�) = 1� y;T (t; y) can be extended continuously in t = 0. Furthermore, T (t; y) is di�erentiable int = 0:@T (t; y)@t �����t=0 = limt!0 T (t; y)� T (0; y)t = limt!0 sinh(t�(1� y))� (1� y) sinh(t�)t sinh(t�) = 0;if we apply L'Hospital's rule twice.Next, we turn to the bounds for T (t; y). As T (t; y) is positive and even in t, weonly have to show that T (t; y) decreases monotonously for t > 0. Consider@T (t; y)@t = �(1� y) sinh(t�) cosh(t�(1� y)) � � sinh(t�(1� y)) cosh(t�)(sinh(t�))2 :11

Consequently,@T (t; y)@t < 0 , (1 � y) cosh(t�(1� y)) sinh(t�) < sinh(t�(1� y)) cosh(t�), (1 � y) tanh(t�) < tanh(t�(1� y)), y tanh(0) + (1 � y) tanh(t�) < tanh(y0 + (1� y)t�);which is ful�lled due to the fact that tanh is strictly concave in (0;1). Thus, T (t; y) �T (0; y) = 1� y for all t 2 IR. Furthermore, we haveT (t; y) = T (jtj; y) = sinh(jtj�(1� y)sinh(jtj�) = e�jtj�y 1 � e�2jtj�(1�y)1 � e�2jtj� < e�jtj�y;which completes the proof of (33).Since (34) can be easily veri�ed, we �nally show (35). First of all, note that P (t; y)is an odd function in t with limt!1 P (t; y) = ��y due to limt!1 coth(t) = 1 andlimt!0 P (t; y) = 0 because @T (t;y)@t vanishes for t = 0. Furthermore, P (t; y) decreasesmonotonously for t > 0:@P (t; y)@t < 0 , ����(1� y)2 1sinh2(t�(1� y)) + � 1sinh2(t�)� < 0, sinh(t�(1� y)) < (1� y) sinh(t�), 1Xl=0 (t�)2l+1(1� y)2l+1(2l + 1)! < 1Xl=0 (t�)2l+1(1 � y)(2l + 1)! ;which is valid for positive t. This completes the proof of (35).6. Error Splitting. Now, we de�ne a splitting of the discretization error uhx;hy�uinto three terms �(1)hx , �(2)hy , and �hx;hy . Here, �(1)hx and �(2)hy are discretization errors ofone-dimensional problems, and �hx;hy involves all solutions u, uhx;hy , uhx;0, and u0;hy.The results from the previous sections concerning these four solutions and the functions�k, �k, �k, and T will enable us to show under which circumstances this splitting is ofthe type (1). Thus, writinguhx;hy � u = �(1)hx + �(2)hy + �hx;hy ;(36)where(37) �(1)hx := uhx;0 � u;�(2)hy := u0;hy � u;�hx;hy := uhx;hy � uhx;0 � u0;hy + u;we have to study the requirements on the Dirichlet boundary function g, i.e. on bk, sothat �(1)hx = h2x � C1(x; y; hx);(38) �(2)hy = h2y � C2(x; y; hy);(39) �hx;hy = h2xh2y � C(x; y; hx; hy)(40) 12

hold, where jC1j, jC2j, and jCj are bounded from above by some B(x; y).6.1. �(1)hx . We start with �(1)hx . From (18) and (7), we know that�(1)hx = 1Xk=1 bk sin(k�x)�T (�k; y)� T (k; y)� = Nx�1Xk=1 bk sin(k�x)�T (�k; y)� T (k; y)�+ 1Xk=Nx bk sin(k�x)�T (�k; y)� T (k; y)�:First, we deal with the �nite sum. Looking at lemmas 6 and 2, we can verifyimmediately that���T (�k; y)� T (k; y)��� = ���T (�k; y)P (�k; y) (�k � k)��� � e��k�y � �y � k3h2xjc1(k; hx)jfor a �k 2 (�k; k). Since 0 < k < Nx, (26) shows that �k > 2�k and, consequently, that�����Nx�1Xk=1 bk sin(k�x)�T (�k; y)� T (k; y)������ � h2x � �3y24 � Nx�1Xk=1 jbkjk3e�2ky� h2x � cg�3y24 � 1Xk=1 k�"e�2ky;where the series converges for all 0 < y < 1.Next, we turn to the in�nite sum in our representation of �(1)hx . Because of (6) andlemma 6, we get������ 1Xk=Nx bk sin(k�x)�T (�k; y)� T (k; y)������� � 2 � 1Xk=Nx jbkj � 2cg � 1Xk=Nx 1k3+"� h2x � 2cg � 1Xk=Nx 1k1+" :Thus, (38) has been proved valid.6.2. �(2)hy . Next, we look at �(2)hy . u0;hy and u were de�ned in (23) and (7), respec-tively: �(2)hy = u0;hy � u = 1Xk=1 bk sin(k�x)�T (�k; y)� T (k; y)�:Therefore,����(2)hy ��� � h2y � 1Xk=1 jbkj T (�k; y)� T (k; y)h2y � h2y � cg � 12y2 � 1Xk=1 1k3+" ;if we take into account the followingLemma 7. For all k 2 IN, we have the relationT (�k; y)� T (k; y)h2y � e��k�y � e�k�yh2y � 12y2 :13

Proof. First, the results of lemma 6 show that T (t; y)�e�t�y increases monotonouslyin t > 0: @@t�T (t; y)� e�t�y� = T (t; y)P (t; y) + �ye�t�y � 0;because T (t; y) 2 [0; e�t�y] and P (t; y) 2 [��y; 0]. Since �k < k for all k 2 IN, we getT (�k; y)� T (k; y)h2y � e��k�y � e�k�yh2y :With the de�nition F (t; h) := arsinh �th�2�h�2 ;(41)we get �k = F (k; hy) and limhy!0 F (k; hy) = k. Thus, we can writee��k�y � e�k�yh2y = e�F (k;hy)�y � e�F (k;0)�yh2y= 1hy � @@hy e�F (k;hy)�y������hy= 1hy � �y � e�F (k;�hy)�y � ����������� k(�2 )2�hyq1+(k �2 �hy)2 � �2 arsinh �k �2�hy���2�2 �2h2y �����������for some � 2 (0; 1). De�ning w := �2�hyk and using e�x � 3!x3 for x > 0, we �nallyobtain e��k�y � e�k�yh2y � e�F (k;�hy)�y � �y ��2�2 � � ������� wp1+w2 � arsinh(w)��2�hy�3 �������� 6�3y3 � �y�24 � ������� wp1+w2 � arsinh(w)F 3(k; �hy) � ��2�hy�3 �������= 32y2 � ����� wp1+w2 � arsinh(w)arsinh3(w) ������ 12y2 ;which completes the proof of lemma 7 and (39).14

6.3. �hx;hy . Finally, we turn to �hx;hy . With the de�nitions from section 3, we get�hx;hy := uhx;hy � uhx;0 � u0;hy + u= Nx�1Xk=1 bk sin(k�x)�T (�k; y)� T (�k; y)� T (�k; y) + T (k; y)�+ 1Xk=Nx bk sin(k�x)�T (�k; y)� T (�k; y)� T (�k; y) + T (k; y)�:Again, let us start with the �nite sum. With the function F (t; h) de�ned in (41)and with S(t; h; y) := T�F (t; h); y�� T (t; y);(42)the relations �k = F (k; hy) and �k = F (�k; hy) (cf. (28) and (31)) lead toT (�k; y)� T (�k; y)� T (�k; y) + T (k; y)= S(�k; hy; y)� S(k; hy; y)= St(�k; hy; y) � (�k � k)= hTt�F (�k; hy); y� � Ft(�k; hy)� Tt(�k; y)i � (�k � k)for a �k 2 (�k; k) � � 2�k; k�, where St, Tt, and Ft denote the partial derivatives of S, F ,and T with respect to their �rst argument. Furthermore, since limh!0 F (t; h) = t andlimh!0 Ft(t; h) = 1, we obtainTt�F (�k; hy); y� � Ft(�k; hy)� Tt(�k; y)= Tt�F (�k; hy); y� � Ft(�k; hy)� Tt�F (�k; 0); y� � Ft(�k; 0)= @@hy Tt�F (�k; hy); y� � Ft(�k; hy)!������hy � hyfor some � 2 (0; 1). Together with (34), we receive@@hy Tt�F (�k; hy); y� � Ft(�k; hy)!������hy= Ttt�F (�k; �hy); y� � Fhy(�k; �hy) � Ft(�k; �hy) + Tt�F (�k; �hy); y� � Ft;hy(�k; �hy)= T � �P 2 + Pt� � Fhy(�k; �hy) � Ft(�k; �hy) + P � Ft;hy(�k; �hy)!;if missing arguments are expanded to �F (�k; hy); y�. Looking at the partial derivativesFhy(�k; �hy) = �2��2�hy�2 � wp1 + w2 � arsinh(w)! ;Ft(�k; �hy) = 1p1 + w2 ; andFt;hy(�k; �hy) = ��k �2 � wp1 + w23 ;15

where w := �k �2�hy 2 (0;1), we obtainT � ������P 2 + Pt� � Fhy(�k; �hy) � Ft(�k; �hy) + P � Ft;hy(�k; �hy)������ T � ���P 2 + Pt��� � �2p1 + w2 � ������� wp1+w2 � arsinh(w)��2�hy�2 �������+ jP j � �k �2 � wp1 + w23!= hy�k � T � ���P 2 + Pt��� � ��2�2 � � w��2�hy� � p1 + w2 � ������� wp1+w2 � arsinh(w)��2�hy�3 ������� ++ jP j � ��2�2 � � w3��2�hy�3 � p1 + w23!� hy�k � ��2�2 � e�F (�k;�hy)��y � ���P 2 + Pt��� � w��2�hy� � p1 + w2 � ������� wp1+w2 � arsinh(w)��2�hy�3 �������+ jP j � w3��2�hy�3 � p1 + w23!;according to (33). Since e�x � m!xm for each x > 0 and an arbitrary integer m > 0, weget T � ������P 2 + Pt� � Fhy(�k; �hy) � Ft(�k; �hy) + P � Ft;hy(�k; �hy)����� �� hyk � ��2�3 � 4!�4y4 ���P 2 + Pt��� � w�F (�k; �hy)�hy �2� � p1 + w2 � ������� wp1+w2 � arsinh(w)�F (�k; �hy)�2�hy�3 �������+ 3!�3y3 jP j � w3�F (�k; �hy)�2�hy�3p1 + w23!= hyk � �38 � 24�4y4 ���P 2 + Pt��� � warsinh(w) � p1 + w2 � ����� wp1+w2 � arsinh(w)arsinh3(w) ����� ++ 6�3y3 jP j � w3arsinh3(w) � p1 + w23!� hyk � �38 � 8�4y4 ���P 2 + Pt��� + 6�3y3 jP j!;because 0 � warsinh(w)p1 + w2 � 1; 0 � ����� wp1+w2 � arsinh(w)arsinh3(w) ����� � 13 :From (35), we know that jP (t; y)j � �y. Analogously, we can show that jPt(t; y)j ��23 (2y � y2). These relations, �nally, lead us toT � ������P 2 + Pt� � Fhy(�k; �hy) � Ft(�k; �hy) + P � Ft;hy(�k; �hy)����� � hy � 3�k � y316

and, consequently, to�����Nx�1Xk=1 bk sin(k�x)�T (�k; y)� T (�k; y)� T (�k; y) + T (k; y)������� Nx�1Xk=1 jbkj � ���Tt�F (�k; hy); y� � Ft(�k; hy)� Tt(�k; y)��� � j�k � kj� Nx�1Xk=1 jbkj � hy � hy � 3�k � y3 � h2xk3jc1(k; hx)j� h2xh2y � �38y3 � cg � 1Xk=1 1k1+" ;because jbkj � cgk3+" (see (6)).Finally, we turn to the in�nite sum in our representation of �hx;hy in order tocomplete the proof of (40):������ 1Xk=Nx bk sin(k�x)�T (�k; y)� T (�k; y)� T (�k; y) + T (k; y)�������� h2y � 1Xk=Nx jbkj � �����T (�k; y)� T (�k; y)h2y � T (�k; y)� T (k; y)h2y ������ h2xh2y � cg � 1Xk=Nx 1k1+" � �����T (�k; y)� T (�k; y)h2y �����+ �����T (�k; y)� T (k; y)h2y �����! :In section 6.2, we have already shown that�����T (�k; y)� T (k; y)h2y ����� = T (�k; y)� T (k; y)h2y � 12y2for all k and hy. Now, we turn to �T (�k; y)� T (�k; y)�=h2y. From (25) and (31),�k = sin � k�2Nx��2Nx = sinh ��k�hy2 ��hy2 ;we can see the 4Nx-periodic behaviour of �k and, hence, of �k, and the periodic be-haviour of the relationship between �k and �k (see the proof of lemma 5):k 6= l � 2Nx; l 2 ZZ ) 0 < j�kj < j�kj :Only for k = l � 2Nx, l 2 ZZ, we get �k = �k = 0 and, consequently, T (�k; y) = T (�k; y).Thus, without loss of generality, let �k 6= 0. Then, due to the proof of lemma 7,�����T (�k; y)� T (�k; y)h2y ����� = T (j�kj ; y)� T (j�kj ; y)h2y � e�j�k j�y � e�j�kj�yh2y :17

The rest of the proof follows exactly the lines of section 6.2, just with j�kj instead of k,leading to T (�k; y)� T (�k; y)h2y � 12y2 :This completes the proof of (40) and, hence, of the error splitting (36) and (37).Consequently, our smoothness requirement (6) on the Dirichlet boundary function gis su�cient for the existence of a splitting of uhx;hy(x; y) � u(x; y) according to (1).Summing up our results of this section, we get the followingTheorem 6. If the Fourier coe�cients bk of ~g, the 2-periodic and 0-symmetricextension of the Dirichlet boundary function g, satisfy jbkj � cgk3+" , the error uhx;hy(x; y)�u(x; y) on the grid Gi;j with meshwidths hx = 2�i and hy = 2�j can be written asuhx;hy(x; y)� u(x; y) = C1(x; y; hx)h2x + C2(x; y; hy)h2y + C(x; y; hx; hy)h2xh2y;(43)where jC1j, jC2j, and jCj are bounded from above by some B(x; y) independent of hxand hy.Note that, actually, the condition concerning the Fourier coe�cients bk is a smooth-ness requirement on the Dirichlet boundary function g. If, e.g., g 2 C4([0; 1]) withg(0) = g(1) = g00(0) = g00(1) = 0, it is a well-known fact from Fourier theory thatbk � cgk4 . In the more general case, an arbitrary continuous boundary function that isfour times partially di�erentiable on each edge of � can be reduced to the situationof our model problem (4), (5). For that, we have to add (harmonic) functions of thetype cos(�2x) sinh(�(1 � y)) and cos(�2 (1 � x)) sinh(�(1 � y)) on each edge in order toforce the second partial derivatives to zero at the corners. Afterwards, the (harmonic)bilinear interpolant of the values at the four corners of � has to be subtracted, and,�nally, superposition enables us to consider only functions vanishing on three edges of�. Here, it is important that all those transformations do not in uence the convergencebehaviour. Therefore, it is su�cient to require g 2 C4([0; 1]) in order to obtain bk � cgk4 .Finally, combining theorems 1 and 6, we get the crucial result of this paper: If theDirichlet boundary function g ful�lls the smoothness requirement of theorem 6, then,according to theorem 1, the error ecn(x; y) of the combined solution ucn(x; y) is of theorder O(h2n log2(h�1n )), i.e., apart from the logarithmic factor, the sparse grid solutionucn(x; y) is of the same accuracy as the full grid solution uhn;hn .7. Numerical Experiments. In the following, we study the convergence be-haviour of the combination technique in the case of Laplace's equation on = ]0; 1[2with Dirichlet boundary conditions and the solutionsu(�)(x; y) := Im�(x� 12 + iy)�� = r�(x; y) � sin�� � '(x; y)�; � > 0;where r(x; y) = q(x� 12)2 + y2 and '(x; y) = arg(x � 12 + iy). Here, the parameter �indicates the smoothness of u(�) in the critical point (12 ; 0). If, e.g., � � 4, then we haveg(x) := u(�)(x; 0) 2 C4([0; 1]). Figure 2 shows u( 12) with its singularity in (12; 0).18

0

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0.2

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0.60.4

0.20

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0.60.4

0.20 Fig. 2. u( 12 ).All of the following computations were performed on a HP 9000/755 workstationwith 128 MBmain storage. The code has been written in FORTRAN. For the discretiza-tion on the full grids Gi;j, we used �nite di�erences as indicated in (9). The resultingsystems were solved with the help of the NAG library routine D03EDF written by P.Wesseling.As introduced before, ecn(x; y) := ucn(x; y)�u(x; y) denotes the error of the combinedsolution. Concerning the pointwise error, we consider the two points P1 := (12 ; 12) andP2 := (14 ; 14). The quotients �Pi := ���ecn�1(Pi)=ecn(Pi)��� show the decrease in the pointwiseerror when proceeding from level n � 1 to n. For n = 2l � 1 and n = 2l (l 2 IN), thediscrete maximum- and L2-norm are both computed on the full grid Gl;l, the �nest fullgrid contained in ~Gn;n. Therefore, we always compare the norms on levels n� 2 and nand de�ne �1 := kecn�2k1=kecnk1 and �2 := kecn�2k2=kecnk2.In table 1, we present the results for � = 0:5. From [12], we know that for fullgrids Gn;n, a pointwise convergence rate of O(h3=2n ) (hn = 2�n) can be expected in asu�cient distance from the singularity. Now, for the combined solutions on sparse grids,we observe values of �P1 and and �P2 close to 23=2 = p8, too. Looking at P2, we cansee that the combination technique seems to cause an additional logarithmic factor likein the smooth case. Thus, the combined solution achieves nearly full grid accuracy, butinvolves signi�cantly less grid points. Concerning the norms, we can observe a behaviourcomparable to full grids. Proceeding from n � 2 to n leads to the same convergencerates �2 of 23=2 = p8 for the L2-norm and �1 of 21=2 = p2 for the maximum-norm aswe get for full grids when proceeding from n� 1 to n (see [12] and [9]). Since, in both19

Table 1� = 0:5n ecn(P1) �P1 ecn(P2) �P2 jjecnjj1 �1 jjecnjj2 �21 �4:8552 � 10�02 | 4:8552 � 10�02 4:8552 � 10�022 �1:9259 � 10�02 2:521 | | 1:9259 � 10�02 | 1:9259 � 10�02 |3 �4:7039 � 10�03 4:094 �4:6221 � 10�03 | 3:5937 � 10�02 1:351 1:2571 � 10�02 3:8624 �1:7749 � 10�04 26:502 �2:4920 � 10�03 1:855 1:3863 � 10�02 1:389 4:9091 � 10�03 3:9235 4:0667 � 10�04 0:436 �1:1839 � 10�03 2:105 2:5689 � 10�02 1:399 3:9633 � 10�03 3:1726 2:2504 � 10�04 1:807 �5:1783 � 10�04 2:286 9:9078 � 10�03 1:399 1:5510 � 10�03 3:1657 8:9342 � 10�05 2:519 �2:1420 � 10�04 2:418 1:8199 � 10�02 1:412 1:3215 � 10�03 2:9998 3:3083 � 10�05 2:701 �8:5143 � 10�05 2:516 7:0177 � 10�03 1:412 5:1653 � 10�04 3:0039 1:1989 � 10�05 2:759 �3:2867 � 10�05 2:591 1:2871 � 10�02 1:414 4:5301 � 10�04 2:91710 4:2925 � 10�06 2:793 �1:2414 � 10�05 2:648 4:9633 � 10�03 1:414 1:7701 � 10�04 2:91811 1:5260 � 10�06 2:813 �4:6134 � 10�06 2:691 9:1015 � 10�03 1:414 1:5767 � 10�04 2:87312 5:4037 � 10�07 2:824 �1:6937 � 10�06 2:724 3:5097 � 10�03 1:414 6:1605 � 10�05 2:87313 1:9095 � 10�07 2:830 �6:1608 � 10�07 2:749 6:4357 � 10�03 1:414 5:5307 � 10�05 2:85114 6:7406 � 10�08 2:833 �2:2255 � 10�07 2:768 2:4817 � 10�03 1:414 2:1610 � 10�05 2:85115 2:3787 � 10�08 2:834 �7:9972 � 10�08 2:783 4:5508 � 10�03 1:414 1:9478 � 10�05 2:83916 8:3938 � 10�09 2:834 �2:8623 � 10�08 2:794 1:7548 � 10�03 1:414 7:6103 � 10�06 2:840cases, the number of grid points increases approximately by a factor of 4 (the in uenceof the logarithmic term being neglected), the full grid solution uhn;hn and the combinedsolution uc2n are of the same quality. Table 2� = 0:9n ecn(P1) �P1 ecn(P2) �P2 jjecnjj1 �1 jjecnjj2 �21 �7:4688 � 10�03 | 7:4688 � 10�03 7:4688 � 10�032 �2:5289 � 10�03 2:953 | | 2:5289 � 10�03 | 2:5289 � 10�03 |3 �5:3009 � 10�04 4:771 �3:7809 � 10�04 | 3:9451 � 10�03 1:893 1:3566 � 10�03 5:5064 �1:4235 � 10�05 37:238 �1:4551 � 10�04 2:598 1:2745 � 10�03 1:984 4:3797 � 10�04 5:7745 3:7429 � 10�05 0:380 �5:2373 � 10�05 2:778 2:0998 � 10�03 1:879 3:1188 � 10�04 4:3506 1:8191 � 10�05 2:058 �1:7967 � 10�05 2:915 6:7916 � 10�04 1:877 1:0103 � 10�04 4:3357 6:3389 � 10�06 2:870 �5:9319 � 10�06 3:029 1:1252 � 10�03 1:866 7:8141 � 10�05 3:9918 2:0267 � 10�06 3:128 �1:8981 � 10�06 3:125 3:6396 � 10�04 1:866 2:5313 � 10�05 3:9919 6:2366 � 10�07 3:250 �5:9230 � 10�07 3:205 6:0299 � 10�04 1:866 2:0270 � 10�05 3:85510 1:8689 � 10�07 3:337 �1:8116 � 10�07 3:269 1:9505 � 10�04 1:866 6:5657 � 10�06 3:85511 5:4910 � 10�08 3:404 �5:4526 � 10�08 3:322 3:2314 � 10�04 1:866 5:3451 � 10�06 3:79212 1:5892 � 10�08 3:455 �1:6198 � 10�08 3:366 1:0452 � 10�04 1:866 1:7314 � 10�06 3:79213 4:5456 � 10�09 3:496 �4:7602 � 10�09 3:403 1:7317 � 10�04 1:866 1:4209 � 10�06 3:76214 1:2879 � 10�09 3:529 �1:3862 � 10�09 3:434 5:6013 � 10�05 1:866 4:6026 � 10�07 3:76215 3:6213 � 10�10 3:556 �4:0057 � 10�10 3:461 9:2797 � 10�05 1:866 3:7924 � 10�07 3:74716 1:0117 � 10�10 3:579 �1:1497 � 10�10 3:484 3:0017 � 10�05 1:866 1:2284 � 10�07 3:747The tables 2{5 show some results for increasing values of �. In the case of � = 0:9,we again can observe convergence rates �P1 , �P2, �1, and �2 that are only slightly worsethan the corresponding full grid values 20:9 � 1:866 and 21:9 � 3:732, respectively, whichis probably due to the logarithmic factor. Like for � = 0:5, this means a signi�cantgain in comparison with full grids regarding the pointwise rates, but no improvement ifnorms are of interest.This situation changes if we look at � 2 f1:5; 2:5; 3:5g. Here, the pointwise er-rors seem to decrease like O(h2n log2(h�1n )). The convergence rates �1 and �2 show a20

Table 3� = 1:5n ecn(P1) �P1 ecn(P2) �P2 jjecnjj1 �1 jjecnjj2 �21 7:2260 � 10�03 | 7:2260 � 10�03 7:2260 � 10�032 2:2000 � 10�03 3:285 | | 2:2000 � 10�03 | 2:2000 � 10�03 |3 5:1219 � 10�04 4:295 1:1748 � 10�03 | 2:4573 � 10�03 2:941 9:7681 � 10�04 7:3984 8:7919 � 10�05 5:826 3:6735 � 10�04 3:198 6:9889 � 10�04 3:148 2:8435 � 10�04 7:7375 1:0937 � 10�05 8:039 1:0996 � 10�04 3:341 8:4134 � 10�04 2:921 1:5243 � 10�04 6:4086 9:9056 � 10�07 11:041 3:2374 � 10�05 3:397 2:3941 � 10�04 2:919 4:4250 � 10�05 6:4267 2:2457 � 10�10 4410:919 9:3957 � 10�06 3:446 2:9634 � 10�04 2:839 2:5350 � 10�05 6:0138 �4:8418 � 10�08 0:005 2:6848 � 10�06 3:500 8:4386 � 10�05 2:837 7:3529 � 10�06 6:0189 �2:3117 � 10�08 2:094 7:5565 � 10�07 3:553 1:0474 � 10�04 2:829 4:3444 � 10�06 5:83510 �8:3321 � 10�09 2:774 2:0986 � 10�07 3:601 2:9828 � 10�05 2:829 1:2597 � 10�06 5:83711 �2:6829 � 10�09 3:106 5:7643 � 10�08 3:641 3:7030 � 10�05 2:829 7:5605 � 10�07 5:74612 �8:1369 � 10�10 3:297 1:5691 � 10�08 3:674 1:0546 � 10�05 2:828 2:1920 � 10�07 5:74713 �2:3790 � 10�10 3:420 4:2402 � 10�09 3:701 1:3092 � 10�05 2:828 1:3261 � 10�07 5:70114 �6:7868 � 10�11 3:505 1:1389 � 10�09 3:723 3:7284 � 10�06 2:829 3:8446 � 10�08 5:70215 �1:9025 � 10�11 3:567 3:0432 � 10�10 3:742 4:6288 � 10�06 2:828 2:3350 � 10�08 5:67916 �5:2638 � 10�12 3:614 8:0966 � 10�11 3:759 1:3182 � 10�06 2:828 6:7698 � 10�09 5:679behaviour that is quite comparable to the results one gets using Richardson extrapola-tion. In fact, numerical experiments in [14] indicate the same rates for the combinedsolution uc2n and the extrapolated solution uRn := 13 �4uhn=2;hn=2 � uhn;hn�. Note that,except for a logarithmic factor, computing uc2n costs the same as the computation of uRndoes. Table 4� = 2:5n ecn(P1) �P1 ecn(P2) �P2 jjecnjj1 �1 jjecnjj2 �21 5:1095 � 10�03 | 5:1095 � 10�03 5:1095 � 10�032 1:5775 � 10�03 3:239 | | 1:5775 � 10�03 | 1:5775 � 10�03 |3 4:5941 � 10�04 3:434 �1:2014 � 10�05 | 1:0051 � 10�03 5:084 5:6306 � 10�04 9:0754 1:2927 � 10�04 3:554 �2:0791 � 10�05 0:578 2:9221 � 10�04 5:399 1:6407 � 10�04 9:6155 3:5944 � 10�05 3:596 �1:0301 � 10�05 2:018 1:8969 � 10�04 5:299 5:9037 � 10�05 9:5376 9:9555 � 10�06 3:610 �4:0176 � 10�06 2:564 5:4353 � 10�05 5:376 1:6558 � 10�05 9:9097 2:7393 � 10�06 3:634 �1:3937 � 10�06 2:883 3:3093 � 10�05 5:732 5:8185 � 10�06 10:1468 7:4777 � 10�07 3:663 �4:4966 � 10�07 3:099 9:4489 � 10�06 5:752 1:5981 � 10�06 10:3619 2:0268 � 10�07 3:689 �1:3818 � 10�07 3:254 5:8241 � 10�06 5:682 5:5103 � 10�07 10:55910 5:4603 � 10�08 3:712 �4:1033 � 10�08 3:368 1:6627 � 10�06 5:683 1:4963 � 10�07 10:68011 1:4624 � 10�08 3:734 �1:1885 � 10�08 3:453 1:0286 � 10�06 5:662 5:0887 � 10�08 10:82912 3:9043 � 10�09 3:746 �3:3781 � 10�09 3:518 2:9365 � 10�07 5:662 1:3730 � 10�08 10:89813 1:0375 � 10�09 3:763 �9:4630 � 10�10 3:570 1:8180 � 10�07 5:658 4:6239 � 10�09 11:00514 2:7475 � 10�10 3:776 �2:6202 � 10�10 3:612 5:1902 � 10�08 5:658 1:2429 � 10�09 11:04715 7:2529 � 10�11 3:788 �7:1866 � 10�11 3:646 3:2138 � 10�08 5:657 4:1578 � 10�10 11:12116 1:9092 � 10�11 3:799 �1:9557 � 10�11 3:675 9:1748 � 10�09 5:657 1:1151 � 10�10 10:798As a last example, we want to look at the Dirichlet problem with the solutionu(x; y) := sin(�y) � sinh(�(1� x))sinh(�) :Note that u vanishes on three edges of the unit square. For x = 0, we have g(y) :=u(0; y) = sin(�y) and, especially, b1 = 1 and bk = 0 for all k > 1. Therefore, g perfectly21

Table 5� = 3:5n ecn(P1) �P1 ecn(P2) �P2 jjecnjj1 �1 jjecnjj2 �21 �1:7166 � 10�02 | 1:7166 � 10�02 1:7166 � 10�022 �5:5140 � 10�03 3:113 | | 5:5140 � 10�03 | 5:5140 � 10�03 |3 �1:7080 � 10�03 3:228 �2:4374 � 10�03 | 2:4374 � 10�03 7:043 1:7699 � 10�03 9:6994 �5:1273 � 10�04 3:331 �7:1068 � 10�04 3:430 7:1068 � 10�04 7:759 5:1468 � 10�04 10:7135 �1:5001 � 10�04 3:418 �2:0325 � 10�04 3:497 2:7319 � 10�04 8:922 1:6012 � 10�04 11:0546 �4:3006 � 10�05 3:488 �5:7314 � 10�05 3:546 7:4642 � 10�05 9:521 4:4061 � 10�05 11:6817 �1:2132 � 10�05 3:545 �1:5971 � 10�05 3:589 2:5915 � 10�05 10:542 1:3333 � 10�05 12:0098 �3:3783 � 10�06 3:591 �4:4052 � 10�06 3:625 6:9060 � 10�06 10:808 3:5685 � 10�06 12:3479 �9:3097 � 10�07 3:629 �1:2046 � 10�06 3:657 2:2259 � 10�06 11:642 1:0535 � 10�06 12:65610 �2:5434 � 10�07 3:660 �3:2701 � 10�07 3:684 5:8216 � 10�07 11:863 2:7759 � 10�07 12:85511 �6:8985 � 10�08 3:687 �8:8219 � 10�08 3:707 1:8835 � 10�07 11:818 8:0189 � 10�08 13:13812 �1:8596 � 10�08 3:710 �2:3671 � 10�08 3:727 4:8946 � 10�08 11:894 2:0921 � 10�08 13:26813 �4:9865 � 10�09 3:729 �6:3219 � 10�09 3:744 1:6303 � 10�08 11:553 5:9337 � 10�09 13:51414 �1:3310 � 10�09 3:746 �1:6815 � 10�09 3:760 4:2279 � 10�09 11:577 1:5376 � 10�09 13:60615 �3:5384 � 10�10 3:762 �4:4563 � 10�10 3:773 1:4288 � 10�09 11:410 4:2951 � 10�10 13:81516 �9:3734 � 10�11 3:775 �1:1772 � 10�10 3:786 3:7029 � 10�10 11:418 1:1075 � 10�10 13:884ful�lls the smoothness requirement (6). In fact, we learn from table 6 that both thepointwise error and the error norms decrease like O(h2n log2(h�1n )). Remember that �1and �2 result from comparing levels n � 2 and n.Table 6sin(�y) sinh(�(1� x))= sinh(�)n ecn(P1) �P1 ecn(P2) �P2 jjecnjj1 �1 jjecnjj2 �21 5:0732 � 10�02 | 5:0732 � 10�02 5:0732 � 10�022 1:6274 � 10�02 3:117 | | 1:6274 � 10�02 | 1:6274 � 10�02 |3 5:0475 � 10�03 3:224 7:6309 � 10�03 | 9:3968 � 10�03 5:399 5:6525 � 10�03 8:9754 1:5164 � 10�03 3:329 2:1985 � 10�03 3:471 2:7439 � 10�03 5:931 1:6439 � 10�03 9:9005 4:4371 � 10�04 3:418 6:2177 � 10�04 3:536 1:0765 � 10�03 8:729 5:2966 � 10�04 10:6726 1:2717 � 10�04 3:489 1:7349 � 10�04 3:584 2:8984 � 10�04 9:467 1:4527 � 10�04 11:3167 3:5858 � 10�05 3:546 4:7887 � 10�05 3:623 1:0661 � 10�04 10:098 4:4719 � 10�05 11:8448 9:9818 � 10�06 3:592 1:3101 � 10�05 3:655 2:8205 � 10�05 10:276 1:1920 � 10�05 12:1879 2:7498 � 10�06 3:630 3:5576 � 10�06 3:683 8:8161 � 10�06 12:093 3:5591 � 10�06 12:56510 7:5105 � 10�07 3:661 9:6000 � 10�07 3:706 2:2943 � 10�06 12:294 9:3415 � 10�07 12:76011 2:0366 � 10�07 3:688 2:5765 � 10�07 3:726 6:9502 � 10�07 12:685 2:7211 � 10�07 13:08012 5:4890 � 10�08 3:710 6:8824 � 10�08 3:744 1:7972 � 10�07 12:766 7:0742 � 10�08 13:20513 1:4716 � 10�08 3:730 1:8309 � 10�08 3:759 5:3551 � 10�08 12:979 2:0194 � 10�08 13:47514 3:9274 � 10�09 3:747 4:8530 � 10�09 3:773 1:3657 � 10�08 13:160 5:2160 � 10�09 13:56315 1:0440 � 10�09 3:762 1:2822 � 10�09 3:785 3:9338 � 10�09 13:613 1:4647 � 10�09 13:78716 2:7652 � 10�10 3:775 3:3778 � 10�10 3:796 9:9975 � 10�10 13:660 3:7657 � 10�10 13:85117 7:3010 � 10�11 3:787 8:8755 � 10�11 3:806 2:8500 � 10�10 13:803 1:0434 � 10�10 14:038Figure 3 shows the ratios (n = log2(h�1n ))ecn�1(P2)=(n � 1)ecn(P2)=n ;i.e. the decrease in ecn(P2)=n proceeding from level n � 1 to level n for our di�erentmodel problems. We see the O(h1+�n log2(h�1n ))-behaviour for � 2 f0:5; 0:9g and theO(h2n log2(h�1n ))-convergence for � 2 f1:5; 2:5; 3:5g and for our last example with g(y) =sin(�y). 22

Fig. 3. Decrease in ecn(P2)=n proceeding from level n� 1 to level n.In �gure 4, we see the plots of the pointwise error ecn(P2) divided by h2n. For� 2 f1:5; 2:5; 3:5g and g(y) = sin(�y), we observe a linear dependence on n, whichshows again the O(h2n log2(h�1n ))-behaviour of ecn(P2). However, for � 2 f0:5; 0:9g, theconvergence is worse.8. Concluding Remarks. In this paper, we have dealt with questions of conver-gence of the combination technique for the Laplacian. We have seen that for su�cientlysmooth Dirichlet boundary functions g, i.e., if the Fourier coe�cients bk of the 2-periodicand 0-symmetric extension of g are of the order O(k�3�"), the pointwise error of thesolution resulting from a �nite di�erence discretization on a rectangular grid with mesh-widths hx and hy allows a splitting of the typeuhx;hy(x; y)� u(x; y) = C1(x; y; hx)h2x + C2(x; y; hy)h2y + C(x; y; hx; hy)h2xh2ywith jC1(x; y; hx)j, jC2(x; y; hy)j, and jC(x; y; hx; hy)j bounded by some positive B(x; y)for all hx; hy 2 ]0; 12]. If g 2 C4([0; 1]) and g(0) = g(1) = g00(0) = g00(1) = 0, the sameresult recently has been shown by the authors in a di�erent approach based on theinteractive use of tools of symbolic computation (see [2]). From [9], we know that theabove splitting is a su�cient condition for the O(h2n log2(h�1n ))-behaviour of the errorecn(x; y) of the combined solution ucn(x; y), which is typical for the sparse grid context.In several numerical experiments, we have studied the convergence of the pointwiseerror and of error norms for both smooth and non-smooth boundary functions. Evenin the latter situation, the pointwise rates of convergence seem to be only slightlydeteriorated by a logarithmic factor that stems from the use of sparse grids.23

Fig. 4. Pointwise error ecn(P2) divided by h2n.Furthermore, if we compare our results with those in [14] for Richardson extrapola-tion, we observe the same rates of convergence of the norms. However, the combinationtechnique shows a better behaviour of the pointwise error in the non-smooth case, i.e,if the Dirichlet boundary function does not ful�ll the smoothness requirement bk < 1k3+"of theorem 6. REFERENCES[1] H.-J. Bungartz, D�unne Gitter und deren Anwendung bei der adaptiven L�osung der dreidimen-sionalen Poisson-Gleichung, Dissertation, Technische Universit�at M�unchen, (1992).[2] H.-J. Bungartz, M. Griebel, D. R�oschke, and C. Zenger, A proof of convergence forthe combination technique for the Laplace equation using tools of symbolic computation, inProceedings of the International IMACS Symposium on Symbolic Computation, Lille, June1993, G. Jacob, N. Oussous, and S. Steinberg, eds., IMACS / Universit�e des Sciences etTechnologies de Lille, Villeneuve d'Ascq, 1993, pp. 56{61.[3] W. Gordon, Distributive lattices and the approximation of multivariate functions, in Approxi-mation with Special Emphasis on Spline Functions, I. Schoenberg, ed., Academic Press, NewYork, 1969, pp. 223{277.[4] W. Gordon and C. Hall, Trans�nite element methods: Blending-function interpolation overarbitrary curved element domains, Numerische Mathematik, 21 (1973), pp. 109{129.[5] M. Griebel, The combination technique for the sparse grid solution of PDEs on multiprocessormachines, Parallel Processing Letters, 2 (1992), pp. 61{70.[6] M. Griebel, W. Huber, and T. St�ortkuhl, The combination technique for parallel sparse-grid-preconditioning or -solution of PDEs on workstation networks, in Lecture Notes in Com-puter Science, Parallel Processing: CONPAR92-VAPP V, L. Boug�e, M. Cosnard, Y. Robert,and D. Trystram, eds., Springer, Heidelberg, 1992, pp. 217{228.24

[7] M. Griebel, W. Huber, T. St�ortkuhl, and C. Zenger, On the parallel solution of 3DPDEs on a network of workstations and on vectorcomputers, in Lecture Notes in ComputerScience, Computer Architecture: Theory, Hardware, Software, Applications, A. Bode andM. DalCin, eds., Springer, Heidelberg, 1993.[8] M. Griebel, W. Huber, and C. Zenger, A fast Poisson solver for turbulence simulationon parallel computers using sparse grids, in Notes on Numerical Fluid Mechanics (vol. 38),E. Hirschel, ed., Vieweg, Braunschweig, 1993, pp. 101{113.[9] M. Griebel, M. Schneider, and C. Zenger, A combination technique for the solution ofsparse grid problems, in Proceedings of the IMACS International Symposium on IterativeMethods in Linear Algebra, Brussels, April 1991, P. de Groen and R. Beauwens, eds., Elsevier,Amsterdam, 1992, pp. 263{281.[10] M. Griebel and V. Thurner, The e�cient solution of uid dynamics problems by the combi-nation technique, Institut f�ur Informatik der Technischen Universit�at M�unchen, SFB-Report342/1/93 A, (1993).[11] J. Hennart and E. Mund, On the h- and p-versions of the extrapolated Gordon's projectorwith applications to elliptic equations, SIAM J. Sci. Stat. Comput., 9 (1988), pp. 773{791.[12] P. Hofmann, Asymptotic expansions of the discretization error of boundary value problems ofthe Laplace equation in rectangular domains, Numerische Mathematik, 9 (1967), pp. 302{322.[13] T. L�u, S. Tsi-Minh, and L. Chin-Bo, An analysis of splitting extrapolation for multidimen-sional problems, Systems Science and Mathematical Sciences, 3 (1990), pp. 261{272.[14] C. Pflaum, Convergence of the combination technique for the �nite element solution of Pois-son's equation, Institut f�ur Informatik der Technischen Universit�at M�unchen , SFB-Report342/14/93 A, (1993).[15] D. R�oschke, �Uber eine Kombinationstechnik zur L�osung partieller Di�erentialgleichungen,Diplomarbeit, Institut f�ur Informatik der Technischen Universit�at M�unchen, (1991).[16] J. Walsh and D. Young, On the degree of convergence of solutions of di�erence equations tothe solution of the Dirichlet problem, J. Math. & Phys., 33 (1954), pp. 80{83.[17] C. Zenger, Sparse grids, in Parallel Algorithms for Partial Di�erential Equations: Proceedingsof the Sixth GAMM-Seminar, Kiel, January 1990, Notes on Numerical Fluid Mechanics (vol.31), W. Hackbusch, ed., Vieweg, Braunschweig, 1991.

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