An Application of Petersen’s Theorem To Finite
Element Triangulations
Alistair Bentley,∗
August 28, 2015
Abstract
This letter shows that for every finite element triangulation in R2 , there
exists a coloring where each triangle has exactly one red and two black
edges. This result is shown using Petersen’s Theorem which states that
all bridgeless 3-regular graphs have a perfect matching1 . The existence of
such a coloring has useful applications when the finite element method is
applied to problems with solutions in the Hdiv vector space.
Key Words RTk basis, BDMk basis, Hdiv , Graph Coloring
AMS Mathematics subject classifications. 05C15, 65N30, 65M60
1
Introduction
The finite element method (FEM) is a popular method used for approximating
solutions to partial differential equations. As a description of the FEM is beyond
the scope of this letter, interested readers are referred to [4] for details. This
article focuses on the FEM in R2 for which the solution domain is partioned
into triangles.
Of interest is whether an arbitrary FEM triangulation can be colored so that
each triangle has exactly one red and two black edges. This question arises when
the FEM is applied to problems with solutions in the Hdiv vector space, where
specialized basis elements like Raviart Thomas (RT) and Brezzi-Douglas-Marini
(BDM) are needed (for details on the Hdiv vector space see [8], [2]). Indeed,
one method of implementing RT and BDM elements requires a red and black
coloring of the triangulation described above [6], [1], [9].
This letter shows that for every finite element triangulation in R2 there
exists a coloring such that each triangle has exactly one red and two black
edges. Moreover, such a coloring can be found in O(n) time. We establish this
∗ Department of Mathematical Sciences, Clemson University, Clemson, SC, 29634, [email protected].
1 After writing this letter, it was discovered a similar result was presented in [7].
1
2
result with the of Petersen’s Theorem, which states that all bridgeless 3-regular
graphs have a perfect matching.
To apply Petersen’s Theorem, we show that the dual graph of a finite element
triangulation is isomorphic to a bridgeless 3-regular graph. It follows that this
graph has a perfect matching, which is equivalent to the existence of the desired
red and black coloring. Furthermore, since the resulting graph is planar, an
O(n) algorithm exists to find the coloring [3].
2
Notation
Let Th be a finite element triangulation and let T = {t1 , · · · , tn } be the set of
triangles in the triangulation. This letter assumes that each triangle has three
vertices and three edges where every triangle edge connects two of its vertices.
In the finite element context, this means the triangulation has no hanging nodes.
Thus, one can represent Th as an undirected graph Th = (V, E), where
V = {v1 , · · · , vn } is the collection of triangle vertices and E = {e1 , · · · em } is
the collection of triangle edges.
An edge is a boundary edge if it corresponds to a single triangle, and we
denote the set of boundary edges in Th as E∂Ω . Edges that are not boundary
edges are interior edges. Two triangles are adjacent if they share a common
edge. If all of a triangle’s edges belong to two triangles, the triangle is an interior
triangle. Triangles that are not interior triangles are exterior triangles.
Definition 1. Let Th be a finite element triangulation with no hanging nodes,
of a bounded domain Ω ⊂ R2 . A conforming coloring assigns every edge in Th
the color red or black so that every triangle in Th has exactly one red edge.
Some useful definitions related to a graph G = (V, E) are:
Definition 2. A perfect matching is a subset of E such that each vertex corresponds to exactly one edge.
Definition 3. A planar graph is graph that can be drawn in the plane in a way
that no edges cross each other.
Definition 4. A cycle is path for which the first and last vertices are the same.
Definition 5. A graph is k-regular if every vertex is a vertex for exactly k
edges.
In this letter, we are especially interested in 3-regular graphs.
3
The Dual Graph
Next we consider the dual graph GTh of Th . The dual graph is helpful, because
a perfect matching of GTh gives a conforming coloring of Th .
3
Figure 1: Example of a Dual Graph
Given Th = (V, E), let GTh = (V, E) be the undirected graph where
V = {i : ti ∈ T }
E = {(i, j) : if triangles i, j share a common edge in Th }.
Figure 1 provides an illustration of the dual graph for a 2D triangulation.
The dots are elements of V and the double lines are elements of E. Notice that
the boundary edges of Th are not included in the dual graph. If a vertex in V
has fewer than 3 edges, we call it an exterior vertex. Thus, exterior triangles in
Th correspond to exterior vertices in V.
Next, define the functions κ : V → T and φ : E → E \ E∂Ω where
κ(i) = ti , and
φ((i, j)) = em , where em is the edge shared by triangles ti and tj .
Lemma 1. κ : V → T and φ : E → E \ E∂Ω are bijections.
Proof. First consider κ. Suppose κ(i) = κ(j), then ti = tj → i = j, so κ is
one-to-one. Let ti ∈ T , then by the definition of VTh , there exists an i ∈ V such
that κ(i) = ti . Therefore, κ is onto.
Next consider φ. If em ∈ E \ E∂Ω , then there exist ti , tj ∈ T such that em is
an edge of ti and tj . Since ti , tj ∈ T share a common edge, there exists i, j ∈ V
and (i, j) = (j, i) ∈ E such that φ((i, j)) = em . Therefore, φ is onto.
Next, suppose φ((i, j)) = φ((s, t)). If φ((i, j)) = em , then the edge em connects
triangles i and j. Since every edge in E \ ∂E∂E belongs to exactly two triangles,
it follows that (i, j) = (s, t). Therefore, φ is one-to-one.
Lemma 2. A perfect matching of GTh corresponds to a Th conforming coloring.
Proof. Let Pe = {e1 , · · · , en } ⊂ E be a perfect matching of GTh . This implies
that every v ∈ V, is uniquely associated with an element of Pe .
4
Figure 2: Example of Extending a Dual Graph of Th
To find a conforming coloring, color every element in {φ(e) : e ∈ Pe } red and
the remaining elements in E black. Because κ(V) = T , it follows that every
ti ∈ T now has exactly one red edge.
Lemma 3. In 2D, GTh is a planar graph.
Proof. This follows from the fact Th is a valid triangulation and can be drawn
in the plane. See Figure 2.
Hence our problem has now become one of finding a perfect matching for
GTh . For this, consider Petersen’s Theorem.
Theorem 1 (Petersen’s Theorem). Every bridgeless 3-regular graph has a perfect matching.
Proof. See [5].
Unfortunately, GTh ’s exterior vertices do not have three edges, so GTh is not
3-regular (see Figure 2). However, as the following Lemma shows, a 3-regular
extension of GTh exists.
Lemma 4. The dual graph, GTh , of any finite element triangulation can be
extened to a 3-regular graph G3 .
Proof. The proof is constructive in nature. That is, we can add a set of artifical
nodes Vs and edges Es to GTh , that results in the 3-regular graph G3 = (V3 , E3 ).
Recall E∂Ω ⊂ E is the set of boundary edges in Th = (T, E), with |E| = m
and |E∂Ω | = p. Since Th is finite, |E∂Ω | = p < ∞. Next we re-index Th so that
E∂Ω = {m − p, m − p + 1, · · · , m} forms a clockwise path circumscribing Th .
5
Assume |T | = n and define Vs = {n + 1, · · · , n + p} where every Vs [i]
corresponds to E∂Ω [i] for i = m − p, · · · m. Organizing Vs in this way serves
two purposes. First, every vi ∈ Vs corresponds to an exterior vertex in VTh .
Second, this ensures that vi ∈ Vs is adjacent to vi−1 and vi+1 (if i = n + 1, then
vi is adjacent to vn+2 and vn+p ). Finally, let V3 = VTh ∪ Vs = {1, · · · , n, n +
1, · · · , n + p}.
To complete the extension of GTh to G3 , a set of artificial edges Es is
created. First, let Es1 = {m+1 , · · · , m+p } denote a set of edges connecting each node in Vs with its corresponding exterior vertex in VTh . Second,
Es2 = {m+p+1 , · · · , m+2p } connects each artificial vertex with its adjacent artificial vertices. Then Es = Es1 ∪ Es2 and E3 = ETh ∪ Es . Thus every vertex in
G3 corresponds to exactly three 3 edges.
Lemma 5. G3 has no bridges.
Proof. From [5], an edge of G3 is a bridge if it does not lie on a cycle of G3 .
Thus, if is a bridge, then 6∈ Es because the edges in Es circumvent the graph,
and form a cycle.
Instead, suppose ij = (i, j) ∈ ETh , the edge connecting nodes i and j, is a
bridge. If we can construct paths from i and j to Es , then (i, j) would be part
of a cycle and not a bridge.
To find such a path, let κ(V) and φ(E \ ij ) represent a modified form of the
triangulation Th . With φ(ij ) removed, both ti and tj have two remaining edges.
Let L be a line that passes through ti , tj and one of each triangles’ remaining
edges. Such a line exists because there are only a finite number of vertices in
the triangulation that can obstruct L from reaching the boundary.
Let {ei1 , ei2 , · · · , ein1 } ⊂ E and {ej1 , ej2 , · · · , ejn2 } ⊂ E denote the ordered
set of edges intersecting L when tracing over L and moving towards the boundary starting from ti and tj respectively. See Figure 3 for an illustration.
Therefore {φ−1 (ei1 ), φ−1 (ei2 ), · · · , φ−1 (ein1 )} ⊂ E and {φ−1 (ej1 ), φ−1 (ej2 ), · · · ,
φ−1 (ejn2 )} ⊂ E form paths from node i and j respectively to Es . Thus, ij lies
on a cycle and cannot be a bridge.
This leads to the main result of this letter.
Theorem 2. Every 2D finite element triangulation Th with no hanging nodes
has a conforming coloring.
Proof. Lemmas 1 and 2, show that a perfect matching of the dual graph GTh
provides a conforming coloring of the finite element triangulation Th . Next,
Lemmas 4 and 5 show that the dual graph of GTh can be extended to a bridgeless
3-regular graph G3 . Finally, applying Petersen’s Theorem to G3 guarantees the
existence of a perfect matching PE3 .
It remains to show that PE3 gives a conforming coloring for Th . Following
the method used in Lemma 2, we first color every element in {φ(e) : ∈ PE3 ∩E}
red. For the remaining edges in PE3 ∩ Es , if ∈ PE3 ∩ Es connects two elements
in Vs it is ignored. Otherwise the edge e corresponds to a boundary edge of Th
6
Figure 3: Construction of a Cycle Containing Edge ij
which is colored red. The remaining edges of Th are then colored black, yielding
the desired conforming coloring.
4
Conclusion
As a result of Petersen’s Theorem, it is always possible to find a coloring of a
finite element triangulation in R2 such every triangle has exactly one red and two
black edges. Further, since GTh is planar, [3] provide an O(n)-time algorithm
for finding a perfect matching and red-black coloring.
An open question remains whether a red and black coloring exists in which
every simplex has three black and one red face for an aribitrary finite element
tetrahedra in R3 .
References
[1] A. Bentley Computational Bases for Hdiv . MSc Thesis, 2014. Clemson
University, USA.
[2] F. Brezzi, and M Fortin, Mixed and Hybrid Finite Element Methods,
Springer-Verlag, 1991.
[3] T. Biedl, P. Bose, E. Demaine and A. Lubiw Efficient Algorithms for
Petersen’s Matching Theorem, Journal of Algorithms 38: 110-134, Jan 2001.
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[4] A. Ern and J.L. Guermond, Theory and Practice of Finite Elements,
Springer-Verlag, New York, 2004.
[5] G. Chartrand and P. Zhang Introduction to Graph Theory, McGrawHill Companies, 2004.
[6] V.J. Ervin, Computational Bases for RT and BDM on Triangles, Computers and Mathematics with Applications, 69 (2012), pp. 2765-2774.
[7] W.F. Mitchell, Unified Multilevel Adaptive Finite Element Methods for
Elliptic Problems, PhD thesis; Technical Report UIUCDCS-R-88-1436, Department of Computer Science, University of Illinois, Urbana, IL, 1988.
[8] P.A. Raviart, and J.M. Thomas A mixed finite element method for 2-nd
order elliptical problems, Mathematical Aspects of Finite Element Methods.
Lecture Notes in Mathematics 606, Springer, Berlin-Heidelberg-New York,
1977.
[9] M.E. Rognes, R.C. Kirby, and A. Logg Efficient Assembly of H(div)
and H(curl) Conforming Finite Elements, SIAM J. Sci. Comput., Vol. 31,
No. 6, pp 4130-4151.