Korotov, Sergey

Stando, Jacek

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Miskolc Mathematical Notes HU e-ISSN 1787-2413Vol. 3 (2002), No 1, pp. 39-46 DOI: 10.18514/MMN.2002.49Yellow-red and nonobtuse reﬁnements ofplanar triangulationsSergey Korotov and Jacek Sta«doMathematical Notes, Miskolc, Vol. 3., No. 1., (2002), pp. 39–46YELLOW-RED AND NONOBTUSE REFINEMENTS OFPLANAR TRIANGULATIONSSergey KorotovDepartment of Mathematical Information TechnologyUniversity of Jyva¨skyla¨, FIN–40351 Jyva¨skyla¨, Finlandkorotov@mit.jyu.fiJacek Stan´doInstitute of Mathematics, Technical University of ÃLo´dz´Al. Politechniki 11, 90-924 ÃLo´dz´, Polandstandoj@ck-sg.p.lodz.pl[Received: January 15, 2002]Abstract. A new yellow-red refinement of nonobtuse triangles, combining standard redand yellow refinements, is proposed. This approach is also adapted for the construction ofnonobtuse global refinement of an initial triangulation in the case when some obtuse trianglesin it are created by a grid generator.Mathematical Subject Classification: 65N30, 65N50Keywords: nonobtuse triangulation, polygonal domain, finite element method, discrete maxi-mum principle, maximum angle condition, red and yellow refinements, grid generation1. IntroductionIn this paper, we consider conforming (planar) triangulations of a polygon into tri-angles, i.e., the union of all triangles of any particular triangulation is always equalto our polygon, and any two different triangles in any (particular) triangulation mayonly have a common edge, or a common vertex, or no common points (cf. [11]). Inmost cases namely such conforming triangulations are used in finite element modellingand finite element analysis.A planar triangulation is called nonobtuse if all interior angles in all triangles ofthe triangulation are less than or equal to pi2 . This condition was first introduced in[4] and [15], where it was particularly used to prove the discrete maximum principlefor the piecewise linear finite element approximations to the Poisson equation withthe Dirichlet boundary condition. Later this condition was imposed on the interiorangles between faces of tetrahedra in tetrahedral partitions, and again used to provethe discrete maximum principle for the finite element solutions to a class of nonlinearelliptic problems for which the continuous maximum principle holds, see [12]. Ob-viously, a similar question, under which conditions on the triangulations used the40 S. Korotov and J. Stan´dob) yellow refinementa) red refinementFigure 1discrete maximum principle holds, naturally appears in the case of parabolic prob-lems. For these problems there are also some other important qualitative propertiesclosely related to the preservation of the maximum principle, we refer to [5], [6], and[14] in this respect.Note that in nonobtuse triangulations, all triangles satisfy the maximum anglecondition with the same constant, pi2 . This condition is very important for the finiteelement analysis, since the optimum interpolation properties of the finite elements arepreserved [11, Chapt. 4].The construction of a conforming triangulation of a polygon into (arbitrary) tri-angles is, obviously, simple; moreover, there are proofs of the existence of nonobtusetriangulations of polygons [1], [7]. Thus, having a nonobtuse initial triangulation, andthen applying the red refinement [2] and/or the yellow refinement [9] (see Figure 1),we can easily build a family of globally refined triangulations where no obtuse anglesoccur.Nevertheless, in practice, the initial (coarse) triangulation of the polygonal domainis performed by a grid generator, and it is most probable that this initial triangulationis not nonobtuse. In this paper, we propose, first, a new way of refinement of a sin-gle nonobtuse triangle, combining the standard yellow and red refinement techniques(which we will call the yellow-red refinement). Thus, a nonobtuse triangulation can,in fact, be globally refined using red and/or yellow and/or yellow-red refinements forits triangles (in any desired combination). Further, evolving the approach used foryellow-red refinement, we also analyse a possibility of refining globally the given trian-gulation, initially containing several isolated obtuse triangles (cf. Figure 2), into theconforming nonobtuse triangulation. In particular, we present the algebraic condi-tions on the coordinates of vertices of a quadrilateral formed by obtuse and nonobtusetriangles in the triangulation (where those two triangles are adjacent along the longestside of the obtuse triangle), under which it is possible to refine the quadrilateral intononobtuse subtriangles (cf. Figure 4a). In the partition proposed in this paper, themidpoints of non-common sides of the initial triangles and the center of the circum-scribed circle around the obtuse triangle, which is inside of the quadrilateral, are used.Nonobtuse triangular refinements 41Thus, if the conditions presented are satisfied for all the obtuse triangles and nonob-tuse triangles adjacent to them (in the above–described manner), occurring in theinitial triangulation, we can conformly (globally) refine the given initial triangulationso that the next (refined) triangulation (after one refinement step) already becomesnonobtuse.a) obtuse triangle is inside of solution domain b) obtuse triangle is adjacent to the boundary Figure 2It is also worth mentioning here, that the nonobtusity condition is only sufficientto prove the discrete maximum principle: it is possible to observe the validity of thisprinciple also in situations when some of the interior angles in the triangulation areslightly obtuse. This issue, leading to so-called weakened acute type conditions, isaddressed in [13] and [10] in the two- and three-dimensional cases, respectively.Other known refinements of triangles are green [2], and blue [8], they are used forlocal refinement procedures. For a comprehensive survey of the results on the gridgeneration issues, we refer to monograph [3].2. Nonobtuse triangle and yellow-red refinement2.1. Criterion for a nonobtuse triangle. In this section, we give a simple criterionfor a triangle to be nonobtuse, and introduce the idea of yellow-red refinement.We present a simple criterion, which employs the coordinates of vertices of a trian-gle, to decide whether this triangle is nonobtuse. From now on, ABC always standsfor a triangle with vertices A = (xA, yA), B = (xB , yB), and C = (xC , yC) (seeFigure 3).Lemma 1. Triangle ABC is nonobtuse if and only if the following three inequalitieshold(xA − 2xB + xC)2 + (yA − 2yB + yC)2 ≥ (xC − xA)2 + (yC − yA)2, (2.1a)(xC − 2xA + xB)2 + (yC − 2yA + yB)2 ≥ (xB − xC)2 + (yB − yC)2, (2.1b)(xB − 2xC + xA)2 + (yB − 2yC + yA)2 ≥ (xA − xB)2 + (yA − yB)2. (2.1c)Proof. Conditions (2.1a)–(2.1c) imply that |BK| ≥ 12 |AC|, |AL| ≥ 12 |CB|, |CM | ≥12 |AB|, where K, L, M are midpoints of AC, CB, AB, respectively. Geometrically,42 S. Korotov and J. Stan´dob) yellow-red refinementBK LMA A BSK LZX YMC Ca) acute triangle ABCFigure 3it means that (2.1a) is equivalent to the case when vertex B does not lie inside ofa circle having AC as its diameter (and K as its center), i.e., the angle at B isnonobtuse; the same holds for vertices A and C, from (2.1b) and (2.1c), respectively(cf. Figure 3a). ¤2.2. Yellow-red refinement. The idea of red refinement from [2] (cf. Figure 1a)consists of the usage of the midpoints of sides of the triangle: connecting them, weget four subtriangles similar to the original one; red refinement can be applied to anytriangle.The idea of yellow refinement was recently introduced in [9] for tetrahedra andtheir faces. It essentially uses the assumption that the center of the circumscribedcircle of the given triangle lies in the closure of the triangle. Using this center and themidpoints of sides of the triangle, we get partition into four or six right subtriangles,depending on the value of the maximum angle in the given triangle (cf. Figure 1b).Thus, yellow refinement can only be applied to a nonobtuse triangle.Now, we show that it is also possible to combine both techniques inside of a nonob-tuse triangle. The idea of such a construction is explained (in terms of Figure 3b)as follows: let K, L, M be defined as before, the dashed line via point M be themid-perpendicular to side AB, points X, Y , Z be the midpoints of AK, BL, KL,respectively (these three points are also centers of three dotted circles having AK,BL, and KL as their diameters). If we can take point S from trapezoid AKLB(⊂ ABC), lying on the mid-perpendicular via M , so that it does not lie inside of allthree (dotted) circles, then triangle ABC is, obviously, decomposed into six nonob-tuse subtriangles – KLC, KSA, KLS, LSB, SMA, and SMB (the last two are,moreover, right ones at the vertex M).Remark 1. In general, there exist many choices for the position of point S. Never-theless, there may occur a situation when the only possible choice of S is to be atNonobtuse triangular refinements 43point M , and it only happens if ABC is the right triangle. In this case, the yellow-redrefinement coincides with the standard red refinement of triangle ABC.2.3. Nonobtuse refinements. In this section, we discuss the issue of the construc-tion of (refined) nonobtuse triangulations from the initially generated one, containingobtuse triangles; it is, particularly, shown how the yellow-red refinement techniquecan be adapted for this purpose.2.4. Nonobtuse refinement of two adjacent triangles. The approach, used inSection 2.2, can be adapted to the problem of partitioning the triangulations withobtuse triangles. For this purpose, we make the following natural observation – weneed not always make the refinement only in a single triangle, we can also performthe refinement, for example, for a pair of adjacent triangles.To show how this observation and the yellow-red refinement can be employed, weassume (in denotations of Figure 4a) that triangle ADB is obtuse at D = (xD, yD)and triangle ABC is nonobtuse as before. Also, let X = (xX , yX), Y = (xY , yY ), andZ = (xZ , yZ) be midpoints of AK, BL, and KL, respectively.Lemma 2. The coordinates of X, Y , and Z are given by the formulaexX =34xA +14xC , yX =34yA +14yC ; (2.2a)xY =34xB +14xC , yY =34yB +14yC ; (2.2b)xZ =12xC +14xA +14xB , yZ =12yC +14yA +14yB . (2.2c)Further, let S = (xS , yS) be the center of the circumscribed circle of ADB, itstands at the intersection of three mid-perpendiculars to sides AD, DB and AB, andlies outside of ADB, since it is obtuse.Lemma 3. The coordinates of S are given by the formulaexS =(yB − yD)(x2A + y2A − x2D − y2D) + (yD − yA)(x2B + y2B − x2D − y2D)2 · [(xA − xD)(yB − yD)− (xB − xD)(yA − yD)] , (2.3a)yS =(xD − xB)(x2A + y2A − x2D − y2D) + (xA − xD)(x2B + y2B − x2D − y2D)2 · [(xA − xD)(yB − yD)− (xB − xD)(yA − yD)] . (2.3b)The calculation in both lemmas is straightforward, therefore omitted.Further, the following theorem holds, where | · | denotes the standard distancebetween two points in plane.Theorem 1. Let a triangle ABC, adjacent to an obtuse triangle ADB, be nonobtuse.If[(xC − xA)(yA − yB) + (yC − yA)(xB − xA)]×× [(xS − 12(xC + xA))(yA − yB) + (yS −12(yC + yA))(xB − xA)] ≤ 0 (2.4)44 S. Korotov and J. Stan´dob) refinement of quadrilateral is not possibleBCK LZMYSAXDSA BDCP QP QK La) refinement of quadrilateral is possibleFigure 4.and|SX| ≥ 14|AC| , |SZ| ≥ 14|AB| , |SY | ≥ 14|BC| , (2.5)where coordinates of points S, X, Y , and Z are given by (3.1)–(3.2), then subtrianglesKLC, KSA, KLS, LSB, SAP , SPD, SDQ, and SQB are nonobtuse (the last fourare, moreover, right triangles). Above, P and Q are the midpoints of sides AD andDB, respectively (see Figure 4a).Proof. Condition (3.3) means that points C and S are in different half-planes withrespect to straight-line KL, i.e., S lies in the trapezoid AKLB. Further, condition(3.4) means that point S is outside of three circles defined as in Figure 3b. Thus,all the above mentioned subtriangles make a nonobtuse partition of a quadrilateralADBC. ¤Remark 2. Obviously, the conditions of Theorem 5 cannot be fulfilled in a generalcase. For example, in Figure 4b, two adjacent (obtuse and nonobtuse) triangles aresuch that the point S is completely outside of the quadrilateral made by the triangles.Therefore, a partition of the quadrilateral in the manner, proposed in Theorem 5, isnot possible.Remark 3. One of the other possibilities to get nonobtuse triangles, particularly, forthe case of Figure 4b, would be to use a swapping of diagonals in the quadrilateral.Thus, in the decomposition of ADBC we can employ two triangles ACD and BCD(which could be both nonobtuse) instead of ABC and ABD. However, the analysis ofsuch an approach, is beyond the scope of the present paper.Nonobtuse triangular refinements 452.5. On nonobtuse refinements of coarse triangulations. In this subsection,we shortly discuss how the ideas from the previous subsection can be used to makethe nonobtuse triangulation from the coarse triangulation, with obtuse triangles in it,built by the grid generator.So far, various refinement techniques proposed – red, yellow, and yellow-red –are only applied to the single triangle. As we mentioned above, the red refinement,applied to the obtuse triangle, gives obtuse subtriangles; the other two techniques canonly be applied to nonobtuse triangles by their definitions. However, the yellow-redtechnique can be used for the union of adjacent triangles, one of which is obtuse andthe other is nonobtuse, as it was shown in Section 3.1.Thus, we can make a refinement of the initial (generated) triangulation in thefollowing manner. Using the criterion of Section 2.1, we check whether a triangle takenfrom the mesh is nonobtuse; and, thus, mark all obtuse triangles. If the (marked)obtuse triangle is adjacent to the boundary along its longest edge (see Figure 2b), welet the perpendicular from the interior vertex of this triangle onto the boundary, andadd a new node and two right-angled triangles to the triangulation.Further, if we have an obtuse triangle inside of the solution domain (see Figure 2a),we consider the quadrilateral formed by it and adjacent to it and try to apply thetechnique of Section 3.1.Remark 4. A similar question – how to build nonobtuse tetrahedral triangulationsfrom triangulations containing some tetrahedra with obtuse interior angles betweenfaces, naturally appears in the three-dimensional case. To the authors’ knowledgethis is an open problem. Another related open question is how to make local nonob-tuse refinements of planar and tetrahedral triangulations. It is also interesting tomake research on which geometrical properties of triangulations in the two- and three-dimensional cases provide the validity of the discrete maximum principle for parabolicproblems.3. Numerical experimentsNumerical experiments on the effectivity of the proposed technique have been per-formed using Matlab in Intel 700 MgHz/128 RAM. We (pseudo)randomly generated100 000 times the coordinates of vertices A,B,C,D (the time needed for this was1131 seconds): the situation when one of triangles ABC or ABD had an obtuse an-gle at the vertex C or D, and the other triangle had a nonobtuse angle at D or C,respectively, occured 17 235 times. For those 17 235 “undesirable” cases, we couldsuccessfully apply 2 353 times the result of Theorem 5 (the time needed for that was56 seconds). Also, the swapping procedure of Remark 7 was “successful” for 2 488cases from those 17 235 “undesirable” cases (time costs for this checking was 24 sec-onds) , and there was no intersection between “successful” cases of Theorem 5 andRemark 7 (the time needed for this was 22 seconds).Another interesting result was that the situation, when both angles C and D inthe triangles ABC and ABD were obtuse, occurred 11 563 times, and we could46 S. Korotov and J. Stan´do“successfully” apply the swapping procedure from Remark 7 to 2 856 cases fromthose 11 563 (time costs in this experiment was 119 seconds).Acknowledgement. The authors are grateful to Prof. Michal Krˇ´ızˇek for valuable commentson the paper. The first author was supported by Grant no. 49051 of the Academy of Finland,the second author was supported by the Academy of Poland.REFERENCES[1] Baker, B. S., Grosse, E. and Rafferty, C. 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Yellow-red and nonobtuse refinements of planar triangulations

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