On the choice of a Stabilizing Subgrid for Convection-Diffusion Problems

Transcription

1 On the choice of a tabilizing ubgrid for Convection-Diffusion Problems F. Brezzi,2, L. D. Marini,2, and A. Russo,2 Abstract UPG and Residual-Free Bubbles are closely related methods that have been used with success to stabilize a certain number of problems, including advection dominated flows. In recent times, a slightly different idea has been proposed: to choose a suitable subgrid in each element, and then solving tandard Galerkin on the Augmented Grid. For this, however, the correct location of the subgrid node(s) plays a crucial role. Here, for the model problem of linear advection-diffusion equations, we propose a simple criterion to choose a single internal node such that the corresponding plain-galerkin scheme on the augmented grid provides the same a priori error estimates that are typically obtained with UPG or RFB methods. EYWORD: Finite elements, advection-diffusion, stabilizations, augmented grids Introduction We consider, for the sake of simplicity, the model problem of a linear elliptic convectiondiffusion equation in a polygonal domain Ω: { Lu = f in Ω () u = on Ω, where Lu = ε u + u. (2) Let T h = {} be a family of regular discretizations of Ω into triangles, and let h = diam(), h = max Th h. We assume that the diffusion ε is a positive constant, and both the convection field and the right-hand side f are piecewise constant with respect to the triangulation T h. If the operator L is convection-dominated, it is well known that the exact solution of () can exhibit boundary and s, i.e., very narrow regions where the solution and its derivatives change abruptly. As a consequence, if we employ a classical finite element method with a discretization scale which is too big Dipartimento di Matematica, Università di Pavia, Via Ferrata, 27 Italy 2 IMATI del CNR, Via Ferrata, 27 Pavia, Italy Dipartimento di Matematica e Applicazioni, Università di Milano Bicocca, Italy

2 to resolve the layers, the solution that we get has in general large numerical oscillations spreading all over the domain, and can be completely unrelated to the true solution. To properly resolve the layers, the mesh size (at least in the layer regions) must be of the same size as the ratio between diffusion and convection. In many problems, this choice would lead to a huge number of degrees of freedom, making the discretization intractable. In recent years, many stabilization methods have been proposed to cope with this kind of problems. Among them, the most popular is the UPG method (treamline- Upwind Petrov/Galerkin), first described in [2], which has been successfully applied to many different situations (see e.g. [4] and the references therein). As well known, the method corresponds to adding a consistent term providing an additional diffusion in the streamline direction (see () below). The amount of such additional diffusion is tuned by a parameter τ that must be chosen in a suitable way. According to thumb-rule arguments and a lot of numerical tests, several recipes have been proposed for the choice of τ (one of them being recalled in () and (2) below). The method has been proved to have a solid mathematical basis in several cases of practical interest (see e.g. [9], or [22]). Nevertheless, the need for a suitable convincing argument to guide the choice of τ is still considered as a major drawback of the method by several users. Later on, UPG has been related to the process of addition and elimination of suitable bubble functions (see [, 2]), that aroused considerable interest, although the problem of the optimal choice of τ was simply translated into the problem of the optimal choice of the bubble space. This was partly solved by the Residual Free Bubble approach, started in [] and further developed in [6]. An alternative viewpoint, the Multiscale Method, was proposed in [8], but the two approaches were shown to be essentially equivalent in []. Roughly speaking this approach proposes to find the optimal τ through the solution of a suitable boundary value problem (obviously, strictly related to the original one in Ω) in each element. For the case described above, and if we employ continuous, piecewiselinear elements, this corresponds to solving, in each, the following boundary-value problem: find b H () such that Lb = in () (which is, in a sense, as difficult as ()), and then setting τ = (/ ) b. (4) For the limit case ε, one can compute the limit solution in some special cases (included the present one, as shown in []), but a general approach is still lacking. The RFB method for advection dominated problems has also been analyzed from the theoretical point of view, and a priori error bounds were proved, similar to the ones for UPG, for the case of piecewise linear elements in [4], and in [] in the general case. Additional results, including local error analysis were proved in [24], [25]. In more recent times several authors tried to deal with problems of the type () by providing an approximate solution with the use of suitable subgrid problems. This is the case of [9], where the subgrid consisted of a single internal node per triangle, but the rationale for choosing the location of such node was not truly elementary. This is also 2

3 the case, among others, of [5], where a subgrid consisting of a few hishkin elements (cfr e.g. []) was used. In general, however, a satisfactory error analysis was lacking for these methods, apart from the case of the limit behavior for ε, where they all reproduce the same behaviour as UPG. Then a slightly different point of view, based on all these previous attempts, was proposed in [7], [8]. The basic idea is to consider both the original grid and the subgrid at the same time as an augmented grid, and to solve with tandard Galerkin method on such Augmented Grid (GAG). In practice, the internal nodes added with the subgrid can still be eliminated by static condensation, so that the method could still be regarded as a variant of the RFB approach. GAG point of view, however, looks philosophically more appealing. Indeed, once a convenient subgrid has been decided, the idea of using a plain Galerkin code with no smart tricks is surely of interst. In [8] abstract conditions on the choice of the subgrid were given (for advection dominated problems) that ensured the same a priori error estimates of the original RFB method in all regimes. The idea was further developed in [5] for one-dimensional advection-reaction-diffusion problems, where a simple recipe was proposed for the choice of the subgrid (that, there, consisted of two nodes per element). Essentially, the idea is to compute the coefficients of the subgrid stiffness matrix as functions of the distance between each internal node and its closest node at the boundary of the element. Then the distance is chosen in such a way that the coefficient (in the row of the internal node) corresponding to the adjacent boundary node becomes zero. The recipe provides an unsuitable location for the downwind internal node when the problem is diffusion dominated, and hence if the required distance is bigger than h / we set it equal to h /. imilarly, the recipe provides an unsuitable location of the upwind internal node unless the problem is reaction dominated. Hence, again, if the required distance is bigger than h / we set it equal to h /. The recipe is actually more easy to implement than to describe. Here we adapt the same idea of [5] to the case of advection dominated two dimensional problems. Having no reaction terms we can get away with just one internal node. If the triangle has only one inflow edge, the location of the internal node is set, a priori, on the mediane connecting the (upwind) midpoint of the inflow edge with the opposite (downwind) vertex, and the precise position along the mediane is chosen by requiring the coefficient (of the row of the internal node) corresponding to the downwind vertex to be zero. If, on the contrary, the triangle has two inflow edges, the location of the internal node is set, a priori, on the mediane connecting the (downwind) midpoint of the outflow edge with the opposite (upwind) vertex, and the precise position along the mediane is chosen by requiring the sum of the coefficients (of the row of the internal node) corresponding to the two vertices of the outflow edge to be zero. imilarly to the onedimensional case, the position is stopped at the barycenter when it would be required (by the recipe) to be too much far away from the downwind vertex or the downwind midpoint, in the two cases. Again, the recipe is actually more easy to implement than to describe. With the abovementioned choice for the position of the internal node, and hence of the subgrid, we are then able to prove that the abstract assumptions of [8] are satisfied,

4 and hence our choice provides the same error bounds of the RFB methods in all regimes. The layout of the paper is as follows. In ection 2 we briefly recall the basic ideas of the UPG method, of RFB method and the tandard Galerkin on the Augmented Grid method. In ection we describe our choice of the subgrid, and in ection 4 we prove the corresponding a priori error estimates. 2 UPG, RFB, and GAG We consider the model convection-diffusion problem ()-(2), and we recall its variational formulation: { find u H (Ω) such that a (u, v) = F (v) for all v H (Ω) (5) where a (u, v) = ε u v + ( u) v (6) Ω Ω is a continuous and coercive bilinear form on the Hilbert space H (Ω) and v F (v) = fv (7) is in H (Ω). A Galerkin approximation of problem () consists in taking a finitedimensional subspace V h of H (Ω), and then solving the variational problem (5) in V h. For the sake of simplicity, from now on we will restrict ourselves to the case of continuous, piecewise linear elements, i.e., we will consider the finite element space V L = { v H (Ω), v } linear for all T h (8) Ω so that the approximation of (5) reads { find ul V L such that a (u L, v L ) = F (v L ) for all v L V L. (9) As already pointed out, if the problem is convection-dominated, then, unless the mesh size h is of the same size of ε/, the solution of (9) will exhibit strong oscillations spreading all over the domain. The UPG method consists in adding to the original bilinear form a (, ) a term which introduces a suitable amount of artificial diffusion in the direction of streamlines, but without upsetting consistency. In the case of problem () (and with linear elements) the UPG method reads find u L V L such that for all v L V L a (u L, v L ) + τ ( u L f) ( v L ) = F (v L ), () T h where τ is a stabilization parameter depending on the local character of the discretization: in elements whose diameter is not small enough to resolve all scales, τ h / 4

5 and elsewhere τ. More precisely, we can introduce a mesh Péclet number in the following way: for each T h, P e = h, () 6ε and then define τ element by element according to the size of P e : τ = h 2 if P e, τ = h2 2ε if P e <. (2) cheme () leads to a reasonable numerical solution, where of course layers are not resolved, but they are very well localized, and away from the layers the accuracy is very good. A priori error estimates for the UPG method were proved in [2] to be of the type ε u u L 2,Ω + h (u u L) 2,Ω C (εh 2s 2 u 2 s, + h 2s u 2 s,) () T h T h (where u L is the UPG discrete solution) whenever the solution belongs to H s (Ω) for some s with < s 2. We refer to [4, 9 2, 26, 27] for further details. ee also [22] for a more complete presentation. A possible drawback of the UPG method is the sensitivity of the solution to the stabilization parameter τ, whose value is not determined precisely by the available theory. A way to recover intrinsically the value of τ is to use the residual-free bubbles approach (see [,, 6]). The idea is to enlarge the finite element space V L in the following way. For each element, we define the space of bubbles in as B = H (), the enlarging space V B as V B = Th B, and set V h = V RF B h = V L V B. (4) By (4) we have that any v h V h can be split into a linear part v L V L and into a bubble part v b V B in a unique way: v h = v L + v b V L V B, and the bubble part itself can be uniquely split element by element: v b = T h v b, v b B. (5) Then, the variational problem (5) is approximated as follows: find u h = u L + u b V L V B such that for all v L V L, T h, and vb B a (u L + u b, v L ) = F (v L ) a (u L + u b, v b ) = F (v b ), (6) where the subscript ( ) indicates that the integrals involved are restricted to the element. Of course, we cannot expect to solve exactly problem (6), because V h is infinitedimensional. But, for the moment, just assume that we can do it (say, with paper and pencil). 5

6 We introduce now, in each element, the operator M that to every right-hand side g, say, in L 2 () associates the unique solution ϕ := M (g) of Lϕ = g in, ϕ = on. (7) We now see that the second equation in (6) determines u b in terms of u L and f as ubstituting into the first equation of (6) we easily have u b = M (f Lu L ). (8) a (u L, v L ) + T h a (M (f Lu L ), v L ) = (f, v L ),Ω v L V L. (9) Introducing L as the formal adjoint of L on (with zero boundary conditions on ), satisfying a (v b, v L ) = (v b, L v L), for all v b V B and v L V L, (9) can also be written as a (u L, v L ) + (M (f Lu L ), L v L), = (f, v L ),Ω v L V L. (2) T h }{{} effect of residual-free bubbles onto linears ince the coefficients of the operator are piecewise constant, and the elements of V L are piecewise linear, we have in each (f Lu L ) = (f u L ) = constant, (L v L ) = ( v L ) = constant. (2) In particular we have that M (f Lu L ) = (f u L ) M (). Using this, and some simple manipulations, the resulting scheme (2) becomes find u L V L such that for all v L V L a (u L, v L ) + (22) τ ( u L f)( v L ) = F (v L ) where τ = M (). (2) We see that (22) and the UPG scheme () have an identical structure; we need only to compare the two constants τ and τ. etting b = M () we see by (7) that b solves the following boundary value problem on : Lb = in, b = on. (24) We are left with the problem of evaluating, possibly in some approximate way, the integral of b. For strongly convection-dominated cases (the most interesting ones) we can argue as in []: If ε h, then b will be very close (in L ()) to the solution of the purely convective problem b = with boundary conditions b = on the inflow 6

7 part of. This is a pyramid whose volume can be computed by hand. If we define h as the length of longest segment parallel to and contained in, we have so that b Volume of the pyramid = τ = b h, (25) h. (26) Using a scaling argument (see [2]), we can also show that when ε is large with respect to h, we have τ C h 2 /ε, where C still depends on and h but can be uniformly bounded from above and from below if we have a regular family of triangulations. We then see that the values of τ and τ are very close in both limits. A priori error estimates for the RFB method were proved in [4] for linear elements: if the solution u belongs to H s (Ω) for some s with < s 2, then, as in UPG, ε u u R L 2,Ω + h (u u R L) 2,Ω C (εh 2s 2 u 2 s, + h 2s u 2 s,) (27) T h T h where u R L is the linear component of the RFB solution. The same type of estimates (in the more general case of piecewise polynomials of degree k ) were proved in [] also for the error u u h, where u h = u L + u B. ee also [24], [25] for additional results. The method of tandard Galerkin on the Augmented Grid (GAG) starts as a method to compute an approximate solution of the local equation (24). As we have just seen, originally this was done only for the case of ε h, by taking the solution of the limit hyperbolic problem. ubsequently, several researchers tried to construct finite dimensional subspaces Bh B in such a way that the solution of the discrete local problem find b h B h such that a (b h, b h) = (, b h ) b h Bh (28) could produce a solution b h such that b h b (29) where b is again the solution of (24). This was the case, for instance, of the Pseudo Residual Free Bubbles in [9], where a suitable one-node subgrid was constructed in order to satisfy (29). This was also done by the Two-Level FEM in [5], where a suitable subgrid of hishkin type was used to solve (28), and also in [6], where the use of a subgrid with one node in the barycenter was suggested with a suitable subgrid viscosity (as in [7] but with a finely tuned viscosity parameter). No error estimates however were available for all these variants, unless in the limit for ε. As pointed out in [8], most of these methods could be regarded from a slightly different point of view. Indeed, we can consider that we augmented the original space V L with the subgrid space Bh, forming Vh A = V L Th Bh, () 7

8 and then solve with tandard Galerkin in the space Vh A. Indeed in [8] it was proved that if the Bh satisfies certain sufficient conditions (that will be reported later on), then the solution of the tandard Galerkin { find uh V A h such that a (u h, v) = F (v) for all v V A h satisfies the same a priori error bounds as UPG or RFB methods. In the next section we are going to choose a convenient subgrid, consisting of a single node per element, and in the following section we shall use the general results of [8] to show that our choice satisfies the sufficient conditions therein, and therefore the same error estimates as in the UPG and RFB methods hold true for the new method. The choice of the subgrid As we have seen in the previous section, the basic idea is to construct a subgrid in each element, and then solve the problem () on the augmented space, essentialy made of piecewise linear functions on the augmented grid (that is the union of the original grid and of the subgrid). As announced, we are going to take a subgrid that contains just one additional node P = P in each element. The node is then joined to the three vertices, thus splitting the triangle in three subgrid triangles. To further specify our strategy of choice, we prescribe that the location of P should be chosen along one of the three medianes of. The choice of the mediane, and the position of P on it will depend on the direction of, and will be made precise in the sequel. As a first step, suppose the P s are given. In V A h () we choose a basis made, as usual, of functions having value at one node and at the other nodes. The basis function attached to the each point P will have support contained in. The other three basis functions that are different from zero in will have value one at one vertex, and at P and at the other vertices. As we are going to discuss each element separately, we drop most of the indices, and take a local numbering for the vertices, that will be denoted by V i (i =, 2, ) using, as usual, the counterclockwise ordering. The basis functions that are different from on will then be denoted by b P, ϕ, ϕ 2, ϕ, where and b P (P ) =, b P (V i ) = (i =, 2, ), (2) ϕ i (V i ) =, ϕ i (V j ) = j i, ϕ i (P ) =. () In the final stiffness matrix corresponding to (), the row corresponding to the point P will have the form a (b P, b P )u h (P ) + a (ϕ i, b P )u h (V i ) = (f, b P ). (4) i= 8

9 V V 2 δ δ reg P M Σ N E W V e 2 z 2 e P z z 2 Figure Case : two inflow edges e M V 2 δ δ reg P M Σ N E W V M e z z 2 P 2 e 2 z e Figure 2 Case 2: one inflow edge V In order to specify our choice for P we need some additional notation. As in Fig. and Fig. 2 we denote by e i (i =, 2, ) the edges of, with e i opposite to V i ; e i will denote the length of e i, n i the outward unit normal to e i, and ν i = e i n i. The actual numbering of the vertices will be chosen according to the direction of. Then, as announced, P will be a point on the mediane m from V to the midpoint M of edge e, that is, m = (e e 2 )/2. Finally, we denote by i, (i =, 2, ) the three subtriangles obtained by connecting P with the vertices V i, by i the area of i, and by z i (i =, 2, ) the vectors from V i pointing to P. (see Fig. or Fig. 2). In order to choose the position of P, we have to distinguish among three cases. Case : The inflow boundary is made of two edges of. Referring to Fig., let e 2, e be the two inflow edges. The position of P along the mediane from V will be determined by annihilating the sum of the contributions of V 2 and V to P. More precisely we look for P such that Writing we have then a (ϕ 2, b P ) + a (ϕ, b P ) =. (5) P = ( t)v + tm, < t <, (6) z = mt, 2 = = z H/2 = m Ht/2, = m H, = 2 = m H( t), H being the height from V 2 (or V ) to m. Then we have: ( (e, z ) a (ϕ 2, b P ) = ε (e, z ) ) 4 4 ( (e2, z ) a (ϕ, b P ) = ε (e, z 2 ) ) (, ν2 ), 6 (, ν ). 6 (7) (8)

10 By summing and using (7) and the geometrical properties e +e 2 +e = ν +ν 2 +ν =, e + z z 2 =, we obtain ( (e, z z 2 ) a (ϕ 2, b P ) + a (ϕ, b P ) = ε + (e 2 e, z ) ) + (, ν ) ( (e, e ) = ε 4 m H( t) + (e 2 e, m) ) + (, ν ) (9) 2 m H 6 ( e 2 = ε 4 ( t) + e 2 e 2 ) + (, ν ) =. 4 6 olving equation (9) for t gives t = + ε e 2 ε e 2 e 2 2 (, ν )/. (4) As actual value for t, however, we do not always take that given by (4). Indeed, for ε not too small (that is, for diffusion dominated problems) this type of stabilization would be unnecessary, and actually the value provided by (4) could be meaningless. Hence we take t = t, if ε ε 2 (, ν )/ e 2 + e 2 e 2 (4) t = 2/ otherwise. Notice that for ε = ε we have exactly t = t = 2/, so that (4) actually gives a continuous dependence of t upon ε. Moreover for < ε < ε we have > t > 2/ so that, for every ε >,we have 2 t <. (42) We also point out explicitly that there exist two constants γ, and γ, depending only on and the minimum angle of such that γ, h ε γ h, (4) where, here and in all the sequel, h denotes the diameter of. Case 2: The inflow boundary is made of one edge of. Referring to Fig. 2, let e be the inflow edge. In this case we determine the position of P along the mediane from V by annihilating the contribution of V to P, that is, by imposing a (ϕ, b P ) =, (44) where ϕ is still given in (). With the notation of the previous case, since e +z 2 z =, e 2 + z z =, equation (44) gives ( (e, z 2 ) a (ϕ, b P ) = ε (e 2, z ) ) (, ν ) ( (e, z e ) = ε (e 2, z + e 2 ) ) (, ν ) (45) ( = ε e e 2 + (e e 2, m) ) (, ν ) =. 2 t 2 6

11 olving equation (45) for t gives: t 2 = ε( e e 2 ) ε e 2 e 2 /2 (, ν )/. (46) As we did in Case, however, we do not take t = t 2 for every value of ε, but only for convection dominated problems. In particular we take here t = t 2, if ε ε 2 (, ν )/ 2 ( e e 2 ) e 2 e 2 (47) t = 2/ otherwise. Notice that for ε = ε 2 we have exactly t = t 2 = 2/, so that (47) actually gives a continuous dependence of t upon ε. Moreover for < ε < ε 2 we have < t 2 < 2/ so that, for every ε >, we have < t 2. (48) We also point out explicitly that there exist two constants γ,2 and γ2, depending only on and the minimum angle of such that γ,2 h ε 2 γ 2 h. (49) In the next section we are going to show that our choice of the subgrid provides optimal error bounds (that is, the same of UPG and RFB) for all values of ε >. As we have seen, however, the choice of the subgrid can also be interpreted as a mean to solve (24) in an approximate way, in order to compute a reasonable approximation of the stabilizing parameter τ. We approximate the solution b of (24) with the function b P = αb P (x), unique solution of a (b P, b P ) = b P b P. (5) ince is constant, an easy computation gives α(p ) = b P ε b P 2, (5) and notice that α does not depend on the convection coefficient. Recalling that b P (x) = /, (52) and b P 2 = b P 2 dx = e i i i i i 2 4 i = e i i i, (5) the corresponding stabilization parameter τ approximating τ given by (26), becomes τ = b P = ( b P ) 2 ε b P = 4 2 9ε i e i 2 / i. (54)

12 The dependence of τ on P is in the denominator, and it is worth performing an asymptotic analysis for ε. From formulae (7) we have ε = ε ( t), ε 2 = ε = 2ε t, with t = t given by (4) or (46). It is then easy to see that ε for t given by (4) lim ε t = ν for t given by (46) Hence, lim ε ε t = ( e e 2 ) 2 ν for t given by (4) e 2 for t given by (46) (55) lim τ = 2 ε ν = h = lim τ, (56) ε where τ is the stabilization coefficient given by the residual-free bubble see (26). If instead diffusion dominates, that is, ε > ε for Case, or ε > ε 2 for Case 2, then t = 2/, and i = / (i =, 2, ), so that from (54) we easily have τ = 4 2 9ε i e i 2 Ch2 /ε, (57) where C depends on but can be uniformly bounded from above and from below if we have a regular family of triangulations. For instance, if θ =minimum angle of, we have sin 2 (θ) h2 4ε 8 τ h2 ε. (58) Case : One edge of the triangle is parallel to. As it will be clear in the next ection, from the error estimates point of view in this case we can define the point P either by following the recipe of Case, or the recipe of Case 2. The solution of course will change, but not very much, as we will see with some numerical experiments. 4 Error Estimates In this section we shall prove that the present choice of the subgrid satisfies the abstract Assumptions made in [8] in order to keep the same error estimates that we have for the exact Residual Free Bubble, as given for instance in []. Before recalling the results of [] we need to introduce some further notation. For this, to every function ϕ we associate the function v L (ϕ) defined as the unique function of the form v L (ϕ) = ϕ + µb P that satisfies a (v L (ϕ), b P ) = (59) 2

13 (which actually determines µ in a unique way). We are now ready to recall the following result, that can easily be deduced from the more general results in [8] as a particular case. Theorem Assume that, in each element, the subgrid is made by a single internal node P = P (), and let b P be the bubble defined in (2). Assume further that the bubble space satisfies the following two assumptions and C : T h, b P, h /2 ε/2 b P, (6) C 2 : T h, ϕ P, ϕ, C 2 h /2 ε/2 v L (ϕ),, (6) where v L (ϕ) has been defined in (59). Let u and u h be the solutions of (5) and () respectively, and assume that u H s (Ω) for some s with < s 2. Then there exists a constant C, independent of h, such that ( ) /2 ε /2 u u h,ω + h (u u h ) 2, C (h s ε /2 u s,ω +h s /2 u s,ω ). (62) T h The proof follows immediately combining Theorem 2 and Theorem of [8], plus standard approximation results. Indeed, (6) is the sufficient condition [[8]: (4.28)] that ensures Assumption of [8] in the case of a subgrid consisting only of one bubble, and (6) is precisely Assumption 2 of [8], upon abserving that the function v used in [8] coincides with ϕ whenever ϕ is a polynomial of degree. Before proving that our choice of P guarantees that conditions (6) and (6) are satisfied, we are going to prove two lemmata that give us the values of b P (P ) in the cases when ε ε or ε ε 2 (respectively in Case and Case 2). Lemma Assume that, in Case, ε ε. Then there exist two constants C and C,, depending only on and on the minimum angle in, such that C, h b P (P ) C h. (6) Proof. Remember that in Case we have two inflow boundary edges, V is their common vertex and the edge opposite to V has midpoint M and outward normal proportional to ν. Consider the function ψ := ν (x V ). (64) ν It is clear that that easily implies ε ψ + ψ =, (65) a (ψ, b) = (, b) b H (). (66)

14 Being linear everywhere, ψ is, in particular, piecewise linear on the subgrid of. Hence ψ can be seen as the unique function, piecewise linear on the subgrid of, that verifies a (ψ, b P ) = (, b P ) ψ (V ) =, ψ (V 2 ) = ψ (V ) = ν (V 2 V ), (67) ν having used the fact that ν (V 2 V ) = since ν is orthogonal to the edge e, opposite to V. We consider now, for every real number ξ, the problem of finding a function ϕ ξ such that a (ϕ ξ, b P ) = (, b P ) ϕ ξ (V ) =, ϕ ξ (V 2 ) = ϕ ξ (V ) = ξ. (68) Writing ϕ ξ in terms of the basis ()-(2) we have ϕ ξ = ϕ + ξ (ϕ 2 + ϕ ) + ϕ ξ (P ) b P, and the equation in (68) gives easily that, surprisingly enough, gives ϕ ξ (P )a (b P, b P ) + ξ(a (ϕ 2, b P ) + a (ϕ, b P )) = (, b P ) (69) ϕ ξ (P ) = (, b P ) a (b P, b P ) (7) for every real number ξ, since the coefficient of ξ vanishes due to (5). Hence, the value ϕ ξ (P ) does not depend on ξ. Comparing (67) and (68) for ξ = ν (V 2 V )/ν, we see that ϕ ξ coincides with ψ. Hence ϕ ξ (P ) = ν (P V ) ν (7) for all ξ. However, if we take ξ = in (68) we get that its solution is exactly b P. We conclude that b P (P ) = ψ (P ) = ν (P V ), (72) ν and (6) follows immediately since, as we have seen, P V = t(m V ) and t 2/. The next lemma is the counterpart of the previous one for Case 2. Lemma 2 Assume that, in Case 2, ε ε 2. Then there exist two constants C 2 and C,2, depending only on and on the minimum angle in, such that C,2 h b P (P ) C 2 h. (7) Proof. Remember that in Case 2 we have one inflow boundary edge e, with midpoint M and outward normal proportional to ν, and that V is the vertex opposite to it. Consider the function ψ 2 := ν (x M). (74) ν 4

15 Notice that ψ 2 as the numerator is and the denominator negative. It is clear that ε ψ 2 + ψ 2 =, ψ 2 (V 2 ) = ψ 2 (V ) =, ψ 2 (V ) = ν (V M). (75) ν Hence ψ 2 is the unique function, piecewise linear on the subgrid of, that verifies a (ψ 2, b P ) = (, b P ) ψ 2 (V 2 ) = ψ 2 (V ) =, ψ 2 (V ) = ν (V M). (76) ν However the condition (44), valid for ε ε 2, implies that the value in P of the solution of (67) will coincide with the value in P of any other piecewise linear function ϕ ξ that verifies a (ϕ ξ, b P ) = (, b P ) ϕ ξ (V 2 ) = ϕ ξ (V ) =, ϕ ξ (V ) = ξ (77) for every real number ξ (by the same argument used in the previous lemma). In particular, if we take ξ = in (77) we get that its solution is exactly b P. We conclude that b P (P ) = ψ 2 (P ) = ν (P M), (78) ν and (7) follows immediately since, as we have seen, V P = t(v M), so that P M = P V + V M = ( t)(v M) and t 2/. The next lemma ensures that, both in Case and in Case 2, the upper bound in (6) and (7) are satisfied for every value of ε >. Lemma With the choice of P made in the previous section, we always have C : T h b P (P ) C h, (79) with C depending only on the minimum angle in T h and the maximum value of. Proof. The result follows immediately, in both Case and Case 2, if ε ε and ε ε 2 respectively. Otherwise, from equation (5) we deduce ε b P 2 = b P. (8) We consider Case, as the analysis of Case 2 is practically identical. For ε ε we have t = 2/, and P is the barycenter of. Hence, from (5) we see that b P 2, = b P (P )2 b P 2, C b P (P )2, where C depends only on the minimum angle of. On the other hand, the integral of b P over is b P (P ) /, so that (8) yields b P (P ) = / ε b P / 2 ε C, (8) whenever ε ε. ince estimate (4) implies that ε γ, h, (79) follows from (8). For Case 2 we just change ε into ε 2 and use (49). We can now prove that conditions (6) and (6) are satisfied with our choice for P. We start with (6). 5

16 Proposition Assume that, for every element T h, the position of the internal node P is such that the function b P, solution of (5), satisfies (79), as in the thesis of Lemma. Then the condition (6) holds true, with constant C independent of h and. Proof. We start by noticing that, since b P is linear in each triangle i, we can use the midpoints of the edges integration formula for computing the integral of its square: b 2 P = i= so that, comparing with (52), we deduce b P 2, = 2 On the other hand, equation (5) implies i ( ) 2 2 = 2 6, (82) b P (P )ε b P 2, = and (6) follows immediately from (79) and (8)-(84). We deal now with (6). b P. (8) b P, (84) Proposition 2 Assume that, for every T h the position of the internal node P is chosen according with (4) (in Case ) or (47) (in Case 2). Then (6) is satisfied. Proof. As we did before, we deal first in detail with Case, as the proof for Case 2 is almost identical. From the definition (59) of v L (ϕ) we obtain ε v L (ϕ) b P = v L (ϕ) b P = ( ϕ) b P. (85) Consider first the subcase ε ε, and observe that (85) holds for every b P, so that we can take it for b P = b P. As ϕ is constant in we easily deduce that ϕ, = /2 ε v L(ϕ) b P b P Using Cauchy-chwarz inequality and then (84) in (86) we deduce. (86) ϕ, /2 ε /2 ε /2 b P v L (ϕ),, b P = /2 ε/2 v L (ϕ), ( b P )/2 = ε/2 v L (ϕ), (b P (P ))/2. (87) 6

17 ince we are in the subcase ε ε we can apply inequality (6) of Lemma to obtain ε /2 v L (ϕ), (b P (P ) ) /2 ε /2 (C, h ) /2 v L (ϕ),, (88) and the result follows (for ε ε ) by inserting (88) in (87). For ε ε, instead, we restart from (85) with b P = b P and use, this time, equality ϕ = v L(ϕ) b P, (89) thus obtaining ϕ, = /2 v L(ϕ) b P b P b P /2 v L (ϕ), b P, b P after using the Cauchy-chwarz inequality. We go back now to (8) to recover that b P, / b P = b P, / b P = /2 /2. By inserting this into (9), and using (4), for ε ε we finally get: (9) ϕ, /2 v L (ϕ), = /2 ε /2 ε /2 v L (ϕ), (γ,2 h ) /2 /2 ε /2 v L (ϕ),, (9) and the result follows. The proof for Case 2, as we said, is almost identical, just replacing ε with ε 2. 5 Numerical results In this ection we present some numerical experiments to validate our method. We consider the test case shown in Figure for which the exact solution exhibits an internal and a boundary layer. We compare our method with the classical UPG method; for completeness, we also show the solutions obtained with the plain Galerkin method. The basic mesh is made of 762 triangles, and finer meshes are obtained by iterative refinements, dividing each triagle into four triangles. As exact solution we take the solution of the plain Galerkin method on the grid generated by four refinements, thus made of = 9572 triangles. Figures 4, 5, and 6 plot the solutions obtained with the plain Galerkin Method, UPG, and our method, respectively. The left column corresponds to the initial coarse grid (762 elements) and the right column corresponds to a grid of = 48 elements obtained with the first refinement. We report the L 2 errors in each case. We see that the L 2 error is slightly smaller for our method despite the fact that oscillations are a little bigger. We remark that the error estimates are the same for both methods. We also computed the error in L and the results are qualitatively the same. We point out that in the presence of a boundary layer only, the results are similar. 7

18 P 2 M e eσ 2 en z ze 2 Wz M P δ reg δ P M Σ N E W u = u = u = u = u = = (, ) boundary layer boundary layer u = u = u = = (, ) u =.4 L 2 error =.7e- 2 e e 2 e z z 2 z M P δ reg δ P M Σ Figure The test Ncase E W u = u = u = u = u = = (, ) boundary.2 layer Figure 4 Plain Galerkin Method Equation: ε u + u = f with ε =., = (, ), f =..4 L 2 error = 5.7e-2.2 The situation is different if we consider a case with an only; indeed, for the value of ε used so far no stabilization is necessary, and with a smaller ε our method coincides with the RFB method. In this last case, there is no difference with the UPG method. In Fig. 7 (left) we show the complete solution computed with the tandard Galerkin on the Augmented Grid, i.e., the linear part (which lives on the original mesh) plus the bubble part (which lives on the subgrid). We see that the complete solution is very close to the linear part only shown in Fig. 6, despite the fact that in the layers the bubble part shown in Fig. 7 (right) is not negligeable. Finally, we want to check that in the case of a grid aligned with, there is essentially no difference in defining the point P as in Case (i.e. by considering the parallel edge as inflow) or in Case 2 (parallel edge is considered outflow). We study a very particular case where all the triangles are equal and have an edge parallel to. The domain is the 8

19 eσ en 2 e ze Wz 2 u = z u = M P δ reg δ = (, P) M Σ N E W u = u = u = u = boundary layer u = u = = (, ) boundary layer L 2 error =.2e-.4 L 2 error = 8.2e-2 eσ en 2 e ze Wz 2 u = z u = M.4 P.2 δ reg δ.2 = (, P) M Σ Figure 5 UPG Method N E W u = u = u = u = boundary.2 layer u = u = = (, ) boundary.2 layer Figure 6 Our Method L 2 error = 6.54e-2.4 L 2 error = 4.7e unit square and the grid is uniform, with each side divided in 2 parts. For the sake of simplicity, we identify the sides of the square with the four cardinal points (N,, E, W); the oblique edge of the triangles goes from N-W to -E. Experiment. In this case we take = (, ), f =, homogenous Neumann condition at N and, u = at W, u = at E. The continuos problem is one-dimensional, while the discrete one is not because of the different orientations of triangles. Experiment 2. We take = (, ), f =, u = at N and E, u = at and W. We start by plotting in Figures 8 and 9 the two values of the stabilization parameter τ given by (54): the solid line represent the recipe of Case (parallel = inflow), while the dotted line is Case 2 (parallel = outflow). The value is of course the same for all triangles. For Experiment, the worst value for ε (i.e. the value for which the difference is the largest) is about.58e-, while for Experiment 2 the worst value for ε is.26e-. Figures and show the solutions corresponding to these two values of ε obtained with τ given by the Case recipe (parallel = inflow). In Figures 2 and we plot 9

20 M 2 eσ en 2 e ze Wz 2 u = z u = M P u = u = δ reg δ = (, ) P boundary layer lineare+bolla M 2 eσ en 2 e ze Wz 2 u = z u = M P u = u = = (, ).2 M boundary layer.4 M tandard Σ Galerkin on the Augmented Grid Σ Figure 7 N Our Method N.8.8 E E.6.6 W W.4.4 u = u =.2.2 u = u = u = u = = (, ) boundary layer Figure 8 Experiment : τ vs ε δ reg δ P u = u = = (, ) boundary layer bolla Bubble part Figure 9 Experiment 2: τ vs ε.2 the difference between the two solutions obtained in correspondence of the two values of the stabilization parameter τ. We see that in the presence of an only, the difference is negligible, while if there is a boundary layer the choice of P makes a difference. However, we point out that we have chosen the worst possible case: all triangles are aligned with, and the value of ε maximizes the difference between the stabilization parameters. References [] C. Baiocchi, F. Brezzi, and L.P. Franca. Virtual bubbles and the GaL. Comput. Methods Appl. Mech. Engrg., 5:25 4, 99. [2] F. Brezzi, M.O. Bristeau, L.P. Franca, M. Mallet, and G. Rogé. A relationship between stabilized finite element methods and the Galerkin method with bubble 2

21 δ reg δ P M Σ ag replacements N V VE 2 WV u = 2 u = u = δ reg δ P M Σ Pfrag replacements N V VE 2 WV u =.2 u = u = 2 e e.4.4 e 2 e e e.2 = (, ) z = (, ).2 z boundary layer z 2.4 boundary layer z 2.4 z.2.2 z M Figure M Figure Experiment : ε =.58e-, τ given by Case Experiment 2: ε =.26e-, τ given by Case P δ δ reg P M Σ N E W u = u = u = u = = (, ) boundary layer.2.4 u = u = P δ δ reg P M Σ N E W u = u = u = u = = (, ) boundary layer Figure 2 Experiment : Difference of the solutions x.2.4 Figure Experiment 2: Difference of the solutions.2.4 2

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