error analysis of lagrange interpolation on tetrahedrons · 2019-09-06 · the standard position...

SCAN2016 Kobayashi-Tsuchiya Lagrange Interpolation on Tetrahedrons 1 / 24

Error Analysis ofLagrange Interpolation on Tetrahedrons

Kenta KOBAYSHI (Hitotsubashi University)Takuya TSUCHIYA (Ehime University)

SCAN2016

Uppsala University September 29th, 2016

Contents


• The kth-order Lagrange interpolation on tetrahedrons

• The known error estimations

• The projected circumradius of tetrahedrons and the main theorem

• The Squeezing Theorem.

• The standard position for tetrahedrons and affine maps.

The kth-order Lagrange interpolation on tetrahedrons


k : a positive interger,

Pk : the set of polynomials whose order are at most k,

K ⊂ R3 : any tetrahedron in R

3,

(λ1, λ2, λ3, λ4) : the barycentric coordinate on K,

ai : integers,

Σk(K) :=

(a1k, · · · ,

a4k

)∈ K

∣∣∣ 0 ≤ ai ≤ k,4∑

i=1

ai = k

.


K and Σk(K), k = 1, k = 2, k = 3.

For v ∈ C0(K), define IkKv ∈ Pk by

(IkKv)(x) = v(x), ∀x ∈ Σk(K).

An important thing is to obtain an error estimation such as

|v − IkKv|m,p,K ≤ Chk+1−m

K |v|k+1,p,K .

The piecewise Pk finite element method


Ω ⊂ Rd,d = 1, 2, 3 : a bounded polygonal domain

τ : a proper (face-to-face) triangulation of Ω

Sh :=vh ∈ C0(Ω) ∩H1

0 (Ω)∣∣ v|K ∈ Pk, ∀K ∈ τ

Set of piecewise linear functions on τ

Model problem Find u ∈ H10 (Ω) such that

−∆u = f for a given f ∈ L2(Ω).

Weak form Find u ∈ H10 (Ω) such that

∫

Ω∇u · ∇vdx =

∫

Ωfvdx for ∀v ∈ H1

0 (Ω).

Pk FEM Find uh ∈ Sh such that∫

Ω∇uh · ∇vhdx =

∫

Ωfvhdx for ∀vh ∈ Sh.


Let u ∈ H10 (Ω) ∩H2(Ω) and uh ∈ Sh be the exact and finite element

solutions, respectively. Then, by Cea’s Lemma, we have

‖u− uh‖1,2,Ω ≤ C infvh∈Sh

‖u− vh‖1,2,Ω

≤ C‖u− Ikτ u‖1,2,Ω

= C

(∑

K∈τ

‖u− IkKu‖21,2,K

)1/2

,

where C is a positive constant.

Therefore, estimating ‖u− IkKu‖1,2,K is very important for the error analysis

of the finite element methods.

The reference tetrahedrons


Let K and K be the tetrahedrons that have the following vertices:

K has the vertices (0, 0, 0)⊤, (1, 0, 0)⊤, (0, 1, 0)⊤, (0, 0, 1)⊤,

K has the vertices (0, 0, 0)⊤, (1, 0, 0)⊤, (1, 1, 0)⊤, (0, 0, 1)⊤.

Figure 1: K and K.

We denote the reference tetrahedron by K, that is, K is either K or K .

The Basic Idea


An arbitrary triangle K ⊂ R3 can be obtained by a affine transformation

ϕK(x) := Ax+ b as K = ϕK(K).

The important factors are ‖A‖ and ‖A−1‖. It seems that

if K becomes very “flat”, then the estimation would become very “poor.”

It seems that we need a geometric condition on K to obtain an error estimation.

The Standard Error Estimation


Let hK := diamK and ρK be the diameter of the inscribed sphere of K.

Theorem 1 Let γ > 0 be a constant. If hK/ρK ≤ γ, there exists a constant

C = C(γ) independent of hK such that

|v − IkKv|m,p,K ≤ Chk+1−m

K |v|k+1,p,K , ∀v ∈ W k+1,p(K).

Ciarlet, The Finite Element Methods for Elliptic Problems,

North Holland, 1978, reprint by SIAM 2008.

Brenner-Scott, The Mathematical Theory of Finite Element Methods, 3rd ed.Springer, 2008.

If a triangulation τ satisfies maxK∈τ hK/ρK ≤ C , τ is called regular. The

value maxK hK/ρK is called the chunkiness parameter .

Krızek’s maximum angle condition


Theorem 2 Let θ2 (π/3 ≤ θ2 < π) be a constant. Let γK be the maximum

angle of faces of a tetrahedron K and ϕK be the maximum angle between

faces of K. If γK ≤ θ2, ϕK ≤ θ2, and hK ≤ 1, then there exists a constantC = C(θ2) that is independent of hK such that

‖v − I1Kv‖1,p,K ≤ ChK |v|2,p,K , ∀v ∈ W 2,p(K), 2 < p ≤ ∞.

M. Krızek, On the maximum angle condition for linear tetrahedral elements,

SIAM J. Numer. Anal., 29 (1992), 513–520.

R.G. Duran, Error estimates for 3-d narrow finite elements,

Math. Comp., 68 (1999), 187–199.

The Circumradius estimation


Theorem 3 (Kobayashi-Tsuchiya) Let K ⊂ R2 be an arbitrary triangle. Let

RK be the circumradius of K and hK := diamK. For any positive integer kand p, 1 ≤ p ≤ ∞, there exists a constant Ck,p independent of K such that,

for m = 0, 1, · · · , k and ∀v ∈ W k+1,p(K),

|v − Ikv|m,p,K ≤ Ck,pRmKhk+1−2m

K |v|k+1,p,K

= Ck,p

(RK

hK

)m

hk+1−mK |v|k+1,p,K .

Note that no geometric condition is imposed on K.

Kobayashi, Tsuchiya, A priori error estimates for Lagrange interpolations on

triangles, Applications of Mathematics, 60 (2015) 485–499.

Kobayashi, Tsuchiya, Extending Babuska-Aziz’s theorem to higher-order

Lagrange interpolation, Applications of Math., 61 (2016) 121–133.

The error estimation of Lagrange interpolation


Let T kp (K) and Bm,k

p (K) be defined by

T kp (K) :=

v ∈ W k+1,p(K)

∣∣∣ v(x) = 0, ∀x ∈ Σk(K),

Bm,kp (K) := sup

v∈T kp (K)

|v|m,p,K

|v|k+1,p,K.

From the definitions, we have v−IkKv ∈ T k

p (K) for any v ∈ W k+1,p(K) and

|v − IkKv|m,p,K ≤ Bm,k

p (K)|v|k+1,p,K .

Note that

Bm,kp (K) = inf

C; |v − Ik

Kv|m,p,K ≤ C|v|k+1,p,K , ∀v ∈ W k+1,p(K).

That is, Bm,kp (K) is the best constant for the error estimation

|v − IkKv|m,p,K ≤ C|v|k+1,p,K .

The projected circumradius of tetrahedrons


Let K be an arbitrary tetrahedron and B be any facet of K. We regard B as

the base of K.

Consider any plane H perpendicular to B and the orthogonal projection δH on

H . The image δH(K) is a triangle on P . Let

RP := maxH

circumradius of δH(K).

The projected circumradius RK of a tetrahedron K is defined by

RK := minB

RBRP

hB, (1)

where the minimum is taken over all the facets of K.

The main theorem


Theorem 4 Let K be an arbitrary tetrahedron. Let hK := diamK and RK be

the projected circumradius of K. Assume that k, m are integers with k ≥ 1,0 ≤ m ≤ k, and p is taken as

k −m = 0 =⇒ 2 < p ≤ ∞,

k = 1, m = 0 =⇒3

2< p ≤ ∞,

k ≥ 2, k −m ≥ 1 =⇒ 1 ≤ p ≤ ∞.

Then, for arbitrary v ∈ W k+1,p(K), there exists a constant C = C(k,m, p)independent of K such that

|v − IkKv|m,p,K ≤ CRm

Khk+1−2mK |v|k+1,p,K

= C

(RK

hK

)m

hk+1−mK |v|k+1,p,K .

Note that no geometric condition is imposed on K.

The reference tetrahedrons


Let K and K be the tetrahedrons that have the following vertices:

K has the vertices (0, 0, 0)⊤, (1, 0, 0)⊤, (0, 1, 0)⊤, (0, 0, 1)⊤,

K has the vertices (0, 0, 0)⊤, (1, 0, 0)⊤, (1, 1, 0)⊤, (0, 0, 1)⊤.

Figure 2: K and K.

We denote the reference tetrahedron by K, that is, K is either K or K .

The Squeezing Theorem


We now generalize the squeezing map. Let α, β, and γ ∈ R be positive. We

then define the squeezing map sqαβγ2 : R3 → R3 by

sqαβγ2 (x, y, z) := (αx, βy, γz)⊤, (x, y, z)⊤ ∈ R3.

Theorem 5 Let Kαβγ := sqαβγ2 (K). Assume that k ≥ 1, 0 ≤ m ≤ k, and pis taken as

k −m = 0 =⇒ 2 < p ≤ ∞,

k = 1, m = 0 =⇒3

2< p ≤ ∞,

k ≥ 2, k −m ≥ 1 =⇒ 1 ≤ p ≤ ∞.

We then have

Bm,kp (Kαβγ) := sup

v∈T kp (Kαβγ)

|v|m,p,Kαβγ

|v|k+1,p,Kαβγ

≤ (maxα, β, γ)k+1−mCk,m,p.

The Standard Position for Tetrahedrons


Let K be a tetrahedron with vertices x1,x2,x3,x4.

Let B be the facet of K with vertices x1, x2, x3 ⇐= the base of K.Let α, β, 0 < β ≤ α, be the longest and shortest lengths of the edges of B.

We assume that x1x2 is the longest edge of B; |x1 − x2| = α.

Consider cutting R3 with the plane that contains the midpoint of the edge x1x2

and is perpendicular to the vector x1 − x2. Then, there exist two cases:

(i) x3 and x4 belong to the same half-space,

(ii) x3 and x4 belong to different half-spaces.

Let γ := |x1 − x4|.

The Standard Positions for Tetrahedrons


x1 x2

x3

x4

α

β

γ

Case 1

Kαβγ

x1 x2

x3

x4

α

βγ

Case 2

Kαβγ


Under appropriate rotation, translation, and reflection operations, these

situations can be written using the parameters

0 < β ≤ α, 0 < γ, s21 + t21 = 1, s1 > 0, t1 > 0, βs1 ≤

α2 ,

s221 + s222 + t22 = 1, t2 > 0, γs21 ≤α2 ,

asx1 = (0, 0, 0)⊤, x2 = (α, 0, 0)⊤, x4 = (γs21, γs22, γt2)

⊤,x3 = (βs1, βt1, 0)

⊤ for the case (i)

x3 = (α− βs1, βt1, 0)⊤ for the case (ii)

.

We refer to the above coordinates as the standard position of a tetrahedron.

In the following, we sometimes write hB := α.Let RB be the circumradius of B.

General tetrahedrons


Define the matrices A, A ∈ GL(3,R) by

A :=

1 s1 s210 t1 s220 0 t2

, A :=

1 −s1 s210 t1 s220 0 t2

,

s21 + t21 = 1, t1 > 0s221 + s222 + t22 = 1, t2 > 0

We then have

K = A(Kαβγ) for case (i) or K = A(Kαβγ) for case (ii).

Note that A and A can be decomposed as A = XY and A = XY with

X :=

1 0 s210 1 s220 0 t2

, Y :=

1 s1 00 t1 00 0 1

, Y :=

1 −s1 00 t1 00 0 1

,

respectively.


The eigenvalues of Y ⊤Y and Y ⊤Y are 1, 1± s1 with s1 := |s1|The eigenvalues of X⊤X are 1, 1± s2 with s2 :=

√s221 + s222.

Therefore, for a ∈ R3, we have

(1− s2)|a|2 ≤ |Xa|2 ≤ (1 + s2)

2|a|2,

(1− s1)|a|2 ≤ |Za|2 ≤ (1 + s1)|a|

2, Z = Y or Z = Y ,

2∏

i=1

(1− si)|a|2 ≤ |V a|2 ≤

2∏

i=1

(1 + si)|a|2, V = A or V = A.

Theorem 6 Let K be an arbitrary tetrahedron at the standard position. Let k,

m be integers with k ≥ 1 and 0 ≤ m ≤ k. Let p be taken as in the SqueezingTheorem according to k and m. Then, we have

Bm,kp (K) := sup

v∈T kp

|v|m,p,K

|v|k+1,p,K≤ C

(maxα, β, γ)k+1−m

∏2i=1(1− si)m/2

,

where C = C(k,m, p) is a constant independent of K.

A geometric interpretation


Lemma 7 Let K be a tetrahedron at the standard position. We then have

2∏

i=1

(1− si)−1/2 ≤ C

RBRP

hB maxα, β, γ,

where C is a constant independent of K.

Corollary 8 The following estimation holds;

Bm,kp (K) := sup

v∈T kp

|v|m,p,K

|v|k+1,p,K≤ C

(RBRP

hB

)m

(maxα, β, γ)k+1−2m .

Conclusions


• A new error estimation of Lagrange interpolation on tetrahedrons is

obtained.

• In the new error estimation, the error of Lagrange interpolation isestimated in terms of the projected circumradius and the diameter of

tetrahedrons. Geometric conditions are not imposed.

• The authors are not so sure if the projected circumradius is the bestinterpretation of the singular values of the linear transformation.

• Further consideration on the geometry of tetrahedrons is required.

• Theoretically interesting problem: extend the estimation to the case of

d-simplex, d ≥ 4.

References


• Babuska, Aziz; On the angle condition in the finite element method, SIAM

J. Numer. Anal., 13 (1976) 214–226.

• Kobayashi, Tsuchiya; A Babuska-Aziz type proof of the circumradius

condition, Japan J. Indust. Appl. Math., 31 (2014) 193–210.

• Kobayashi, Tsuchiya; On the circumradius condition for piecewise linear

triangular elements, Japan J. Indust. Appl. Math., 32 (2015) 391–396.

• Kobayashi, Tsuchiya; A priori error estimates for Lagrange interpolations

on triangles, Applications of Math., 60 (2015) 485–499.

• Kobayashi, Tsuchiya; Extending Babuska-Aziz’s theorem to higher-order

Lagrange interpolation, Applications of Math., 61 (2016) 121–133.

• Kobayashi, Tsuchiya; Error analysis of Lagrange interpolation on

tetrahedrons, submitted. arXiv:1606.03918.

error analysis of lagrange interpolation on tetrahedrons · 2019-09-06 · the standard position...

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