Journal of Computational Mathematics Vol.xx, No.x, 200x, 1–25. http://www.global-sci.org/jcm doi:10.4208/jcm.1703-m2015-0340 OPTIMAL QUADRATIC NITSCHE EXTENDED FINITE ELEMENT METHOD FOR INTERFACE PROBLEM OF DIFFUSION EQUATION * Fei Wang School of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an, Shaanxi 710049, China. Department of Mathematics, Pennsylvania State University, University Park, PA 16802, USA. Email: [email protected]Shuo Zhang LSEC, ICMSEC, NCMIS, Academy of Mathematics and System Sciences, Chinese Academy of Sciences, Beijing 100190, Peoples Republic of China. Email: [email protected]Abstract In this paper, we study Nitsche extended finite element method (XFEM) for the inter- face problem of a two dimensional diffusion equation. Specifically, we study the quadratic XFEM scheme on some shape-regular family of grids and prove the optimal convergence rate of the scheme with respect to the mesh size. Main efforts are devoted onto classifying the cases of intersection between the elements and the interface and prove a weighted trace inequality for the extended finite element functions needed, and the general framework of analysing XFEM can be implemented then. Mathematics subject classification: 65N30, 65N12, 65N15. Key words: Interface problems, Extended finite element methods, Error estimates, Nitsche’s scheme, Quadratic element. 1. Introduction Many problems in physics, engineering, and other fields contain a certain level of coupling between different physical systems, such as the coupling between fluid and structure in fiuid- structure interaction problems, and the coupling among different flows in multi-phase flows problems. An interface where the coupling takes place is generally encountered in such kind of problems, and, consequently, the numerical discretization of the interface problem is important in applied sciences and mathematics. In this paper, we take the diffusion equation −∇ · (α(x)∇u)= f, (1.1) as a model problem, and study its interface problem. Namely, the underlying domain is assumed to be divided to two subdomains by an interface, such that α is smooth on each subdomain, but not smooth on the whole domain. The model problem is a fundamental one in numerical analysis, and very incomplete literature review can be found in, e.g., [4,5,14,30,31] and below. Different from problem with smooth coefficient, the existence of an interface for diffusion equation can invalidate the global smoothness of the solution of the system, and the accuracy of the standard finite element method is limited when used for such problems. As the loss of * Received July 22, 2015 / Revised version received November 14, 2016 / Accepted March 23, 2017 / Published online xxxxxx xx, 20xx /
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Journal of Computational Mathematics
Vol.xx, No.x, 200x, 1–25.
http://www.global-sci.org/jcm
doi:10.4208/jcm.1703-m2015-0340
OPTIMAL QUADRATIC NITSCHE EXTENDED FINITEELEMENT METHOD FOR INTERFACE PROBLEM OF
DIFFUSION EQUATION*
Fei Wang
School of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an, Shaanxi 710049, China.
Department of Mathematics, Pennsylvania State University, University Park, PA 16802, USA.
a finite number of curves with ends, and ~r ′(t)·~e is a piecewise smooth function defined on Γ′, and
there exists a constant 0 < G′~e < 1, such that |~r ′(t)·~e| ≤ 1−G′
~e. Then, supP∈Γ′
|~r ′(tP )·~e| ≤ 1−G′~e.
Now, given P ∈ Γ, if P ∈ Γ′, then the condition 2 holds; otherwise, there exists a j ∈ [1, C~e],
such that P ∈ U(Qj , rj), and thus condition 1 holds. This finishes the proof.
3.2.2. Classifications for all the possible cases of sub-elements
First, let us make a classification for all the possible cases as follow. Denote ∂Ki = (∂Ki)1 ∪(∂Ki)2, with (∂Ki)1 ⊂ ∂K, (∂Ki)2 = eK ; moreover, both (∂Ki)1 and (∂Ki)2 are continuous.
Denote the ends of (∂Ki)1 by L and R. Now we make the classification for Ki with respect to
(∂Ki)2 and (∂Ki)1.
1. There is no turning point in the interior of (∂Ki)2, and Ki is convex
(a) L and R are on the same edge (including the ends), then there can be 0 or 3 vertices
located in the interior of (∂Ki)1;
i. there is no vertex in the interior of (∂Ki)1; for this case, we refer to K1 in Case
A in Figure 3.1 for illustration;
Optimal Quadratic Nitsche Extended FEM for Interface Problem 13
ii. there are three vertices in the interior of (∂Ki)1, with Ki being convex if and
only if Ki = K; we omit this trivial case.
(b) L and R are located on two different edges, then there can be 1 or 2 vertices located
in the interior of (∂Ki)1
i. there is only one vertex in the interior of (∂Ki)1; in other words, there is one
corner of the original triangle contained in Ki; the two extreme cases are eK is
close to the vertex and eK is tangential to the opposite edge of the vertex, see
K1 in Case B and K2 in Case E in Figure 3.1 for illustration, respectively;
ii. there are two vertices in the interior of (∂Ki)1; in other words, there are two
corners of the original triangle contained in Ki; refer to K2 in Case C and K2
in Case D in Figure 3.1 for illustration;
e
A
K2
e
B
K2
e
C
K2
e
D
K2
e
E
K2
e
F
K2
Fig. 3.1. Different cases of how the interface intersects with elements. K1 = K \K2.
2. There is no turning point on the interior of (∂Ki)2, and K \Ki is convex
(a) L and R are on a same edge (including the ends), then there can be 0 or 3 vertices
located in the interior of (∂Ki)1;
i. there is no vertex in the interior of (∂Ki)1; for this case, K\Ki can be convex if
and only if K\Ki = K. We omit this trivial case.
ii. there are three vertices in the interior of (∂Ki)1; see K2 in Case A in Figure 3.1
for illustration;
(b) L and R are located on two different edges, then there can be 1 or 2 vertices located
in the interior of (∂Ki)1;
i. there is only one vertex in the interior of (∂Ki)1; in other words, there is one
angle of the original triangle contained in Ki; two extreme cases are
14 F. WANG AND S. ZHANG
A. eK is tangential to one of the edges where the ends of (∂Ki)1 lie; K1 in Case
C in Figure 3.1 for illustration.
B. eK is away from the edges where the ends of (∂Ki)1 lie; refer to K1 in Case
D in Figure 3.1 for illustration;
ii. there are two vertices in the interior of (∂Ki)1; in other words, there are two
angles of the original triangle contained in Ki; two extreme cases are
A. eK is tangential to the edge of the two vertices; see K1 in Case E in Figure
3.1 for illustration;
B. eK is away from the edge of the two vertices, ; refer to K2 in Case B in
Figure 3.1 for illustration;
3. There is one turning point on the interior of (∂Ki)2
By Lemma 3.6, we know that a tangent point and a turning point can be avoided to
be present at same element if mesh size is small enough, so obviously, there are two
possibilities, namely Ki is a curved triangle and Ki is a curved quadrilateral, respectively;
see K1 and K2 in Case F in Fig 3.1 for an illustration.
By Lemma 3.1, there is no other possibility. We obtain for anyK ∈ T Γh such that eK = Γ∩K
is continuous and any subelement Ki = K ∩ Ωi, there exist triangles Ki, and K0i ⊂ Ki ⊂ K∗
i ,
such that (3.8) holds when Ki is convex, K \Ki is convex, or there is a turning point on eK ,
by Lemmas 3.7, 3.8 and 3.9, respectively. Then, the proof of Lemma 3.2 is finished by Lemma
3.5.
C
F
A BD EO
G
K2
∗
∗
∗∗
∗
∗∗
Case A
A
B
C
D
E
K2
F
∗∗∗
∗
∗
Case B
e∗ ∗∗
∗ ∗C A
B
D E
K2
Case C
Fig. 3.2. Cases that eK does not contain a turning point.
3.2.3. Proof of Lemma 3.2 for convex sub-element Ki
Lemma 3.7. Fix K ∈ T Γh and let eK = Γ ∩ K with a convex sub-element Ki = K ∩ Ωi.
Then, there exists a (curved) triangle Ki ⊂ Ki with eK as one of its edges, and there exist two
triangles, K0i and K∗
i , such that K0i ⊂ Ki ⊂ K∗
i and (3.8) holds true with Ci and Di defined
in (3.5) and (3.6).
Proof. Now, we verify the three possible cases one by one.
Optimal Quadratic Nitsche Extended FEM for Interface Problem 15
1. Possibility 1-(a)-i, namely K1 in Case A (see Fig 3.2). In this case, the interface Γ
only intersects with one edge of element K at two points A and B. Let C be the extremal
point, that is,
C = arg maxx∈arcAB
dist(x,AB).
Let us consider the sector ACB, denote it as K1 and choose K1 := K1. Connect AC and
BC, and draw two lines DF and EF parallel to AC and BC respectively, such that DF
and EF tangent with e and intersect at point F . Denote O as the intersection point of
line FC and AB. We choose triangles ABC and DEF as K01 and K∗
Fig. 3.3. Cases that eK does not contain a turning point.
2. Possibility 1-(b)-i , namely K1 in Case B (Fig 3.2) and K2 in Case E (Fig 3.3).
Let us prove (3.8) for K1 in Case B here and same argument applies to K2 in Case E.
In this case, we choose K1 = K1. Connect B and C to get triangle ABC denoted by
K01 . Then, draw a line DE parallel to line BC and tangential to e. Let K∗
1 denote the
triangle ADE, then K01 ⊂ K1 = K1 ⊂ K∗
1 . If R = ADAB = AE
AC ≤ 2, we know that
C1 ≤ 4R2 and D1 ≤ R2 are bounded, which finishes the proof of (3.8). If R > 2, that
is, |DB| > |AB|, then 2|K1| > |K1|. So we only need to apply the same argument of
K1 in Case A for K1 = K1\K01 to prove that, there exist K0
1 ⊂ K1 ⊂ K∗1 , such that
C1 = supv∈P1(K∗
1 )
‖v‖20,∞,K∗
1
‖v‖2
0,∞,K01
and D1 = |K1|
|K01 |
are bounded. Then, by Lemma 3.5, we have
1
|e|
∫
e
v2 ds ≤ CC1D11
|K1|
∫
K1
v2 dx ≤ 2CC1D11
|K1|
∫
K1
v2 dx,
which completes the proof of (3.8).
3. Possibility 1-(b)-ii, namely K2 in Case C (see Fig 3.2) and K2 in Case D (see Fig
3.3).
Here, let us take K2 in Case D to show the proof. Connect the non-adjacent vertices
to get diagonals CG and BH , and they intersect at point O. Denote ∠BOG, ∠OCB
and ∠CBO by α, β, and γ, respectively. Let K2 ⊂ K2 be the (curved) triangle with e
as one of its edge and make choice so that the area of K2 is larger. Namely, we choose
(curved) triangle BCG as K2 if β ≥ γ, or choose (curved) triangle BCH as K2 otherwise.
Without loss of generality, we assume that β ≥ γ. Then we know that β < α ≤ 2β due to
α = β+γ. If β ≤ π/4, we have sinαsin β ≤ sin 2β
sin β = 2 cosβ < 2. If β > π/4, then sinβ > 1/√2,
and sinαsin β <
√2. So, for both cases,
|K2||K2|
≤12 |BH ||CG| sinα12 |BC||CG| sin β
<2|BH ||BC| . (3.10)
Optimal Quadratic Nitsche Extended FEM for Interface Problem 17
From the previous argument, we can prove there exists K02 ⊂ K2 ⊂ K∗
2 , such that
C2 = supv∈P1(K∗
2 )
‖v‖20,∞,K∗
2
‖v‖20,∞,K0
2
, D2 =|K2||K0
2 |
are bounded and (3.7) hold true. Therefore, we get
C2D2hK |eK ||K2|2|K|−2|K2|−1 < 2C2D2hK |eK ||K2||K|−2|BH ||BC|−1 ≤ C.
Summing all above finishes the proof.
3.2.4. Proof of Lemma 3.2 for sub-element Ki, with K \Ki being convex
Lemma 3.8. Let K ∈ T Γh and Ki = K ∩Ωi be a sub-element with its complement K\Ki being
convex. There is a (curved) triangle Ki ⊂ Ki with eK being one of its edges and two triangles,
K0i and K∗
i with K0i ⊂ Ki ⊂ K∗
i such that (3.8) holds true with Ci and Di defined in (3.5) and
(3.6).
Proof. Now we verify the possible cases one by one.
1. Possibility 2-(b)-i-A, namely K1 in Case C ( see Fig 3.2). Interface Γ is tangent
with AC ⊂ ∂K at point C, and intersects ∂K at another point B. The curved triangle
ABC is K1, and let K1 = K1. Connect points B and C, and draw a line, BD, tangent
to e at point B, and intersecting AC at point D. Let BE be perpendicular to AC and
intersect with AC at point E. Denote the triangles ABC and DAB by K∗1 and K0
1 ,
respectively. Then we have K01 ⊂ K1 = K1 ⊂ K∗
1 . Similarly, let R = |AC||AD| , we have
C1 ≤ 4R2. Furthermore, vD =|CD||AC| vA +
(1− |CD|
|AC|
)vC . Then, we have
vA =
(1− |CD|
|AC|
)−1(vD − |CD|
|AC| vB)
= R(vD − |CD|
|AC| vB).
So, C1 ≤ 4R2 and in addition,
D1 =|K1||K0
1 |≤ |K∗
1 ||K0
1 |=
0.5|AC||BE|0.5|AD||BE| = R.
In local coordinates, set point C ((f(t0), g(t0)) as the origin, and CA as ξ-coordinate.
Then, Γ can be expressed as η = ψ(ξ) in these local coordinates with ψ(0) = ψ′(0) = 0.
By definition 3.2, we know that there exists a number, n ≤ N , such that ψ(n)(0) 6= 0
([25]), so we have
ψ(ξ) =1
n!ψ(n)(0)ξn +
1
(n+ 1)!ψ(n+1)(ξ)ξn+1.
Then, there is a small number ǫ depending on Γ and h such that
1− ǫ
n!ψ(n)(0)ξn ≤ ψ(ξ) ≤ 1 + ǫ
n!ψ(n)(0)ξn,
1− ǫ
(n− 1)!ψ(n)(0)ξn−1 ≤ ψ′(ξ) ≤ 1 + ǫ
(n− 1)!ψ(n)(0)ξn−1
18 F. WANG AND S. ZHANG
where
ǫ = sup0≤ξ≤l
ξ
n+ 1
∣∣∣∣∣ψ(n+1)(ξ)
ψ(n)(0)
∣∣∣∣∣ .
Here, l = |CE|. As the number of possible tangential points is uniformly finite with
respect to Γ and T0, for all these points tiPi=1, ~r(n)(ti) ·~ν(ti) has a uniform lower bound,
meaning |ψ(n)(0)| has a uniform lower bound, C0. Therefore, ǫ→ 0 as h→ 0.
Elementary calculus leads to
1− ǫ
(1 + ǫ)nl ≤ ψ(l)
ψ′(l)≤ 1 + ǫ
(1− ǫ)nl,
so (1− 1 + ǫ
(1− ǫ)n
)l ≤ l − ψ(l)
ψ′(l)≤(1− 1− ǫ
(1 + ǫ)n
)l.
Hence, we have
R =
(1− |CD|
|AC|
)−1
≤(1− |CD|
|CE|
)−1
≤ (1 + ǫ)n
1− ǫ,
which is bounded, because n is uniformly bounded for non-degenerate smooth curves and
ǫ→ 0 when h→ 0. Therefore,
C1D1hK |eK ||K1||K|−2 ≤ 4R3h−2K |K1|
is bounded. The proof of (3.8) is finished for this case.
2. Possibility 2-(a)-ii, namely K1 in Case D (see Fig 3.3). Here, curved triangle ABC
is K1 and choose K1 = K1. Without loss of generality, we assume that |AB| ≤ |AC|.We draw a line CE tangent to e at point C and intersecting AB at point E. Denote the
triangles ABC and ACE by K∗1 and K0
1 , respectively. Then, we have K01 ⊂ K1 = K1 ⊂
K∗1 . As in Case C, we have C1 ≤ 4R2 and D1 ≤ R, with R = |AB|
|AE| . If R is bounded, then
the proof is complete. If not, we see that |AC||AE| does not have a lower bound. Then by
Lemma 3.6, there is a pointM on Γ near C, and its the tangential direction of Γ is parallel
to AH . Draw a line BD tangent with e at point B, and intersect with AC at point D.
Draw a tangential line of Γ from M , which intersects with the extension of GA at Q.
Connect BM which intersects AH at S. Note that S is not necessarily between A and H .
Extend BD to intersect MQ at T . The inner triangle K01 is ABD, the outer triangle K∗
1
is ABS. Note |AS|/|AD| = |QM |/|QT | and |∆ABS|/|∆ABD| = |∆QBM |/|∆QBT |.Then, we use the conclusion of Case C to finish the proof of (3.8) for this case.
3. Possibility 2-(b)-ii-A, namely K1 in Case E (see Fig 3.3). Next, let us consider the
case that interface Γ intersects with ∂K at 3 points, denoted by C, O and D. Note that
O is a tangent point, which separates the eK = Γ ∩K into two parts denoted by e1 and
e2. Let us consider the inequality for K1. The tangent point O separates the K1 as two
curved triangles, K(1)1 (curved triangle AOC) and K
(2)1 (curved triangle BOD). Again, set
O as the origin and OB as the ξ-coordinate. Then, locally the interface eK is expressed
as
ψ(ξ) =1
n!ψ(n)(0)ξn +
1
(n+ 1)!ψ(n+1)(ξ)ξn+1,
Optimal Quadratic Nitsche Extended FEM for Interface Problem 19
and there is a small number ǫ depending on Γ and h such that
1− ǫ
n!ψ(n)(0)ξn ≤ ψ(ξ) ≤ 1 + ǫ
n!ψ(n)(0)ξn. (3.11)
Note that n should be an even number due to the shape of e in this case. Without loss of
generality, we assume that |AO| ≥ |BO|, by the relation (3.11) and direct calculation, we
know that there exist two constants C1 and C2, only depending on ǫ such that |BD| ≤C1|AC| and |K(2)
1 | ≤ C2|K(1)1 |. So |K1| = |K(2)
1 |+ |K(1)1 | ≤ (1 + C2)|K(1)
1 |.Choose K1 = K1 and draw a line CE that is tangent to e1 at point C. Let us denote
the triangle AEC by K01 . Connect points C and D, and denote the quadrilateral ABDC
by K∗1 . From previous argument, we know that R = |AO|
|AE| ≤ (1+ǫ)n1−ǫ and
|K(1)1 |
|K01 |
≤ R.
Therefore, we have|AB||AE| ≤
2|AO||AE| ≤ 2(1 + ǫ)n
1− ǫ.
We know that
vB =|AB||AE|
(vE − |BE|
|AB|vA),
vF =|AF ||AC|
(vC − |FC|
|AF |vA),
vF =|BF ||BD|
(vD − |FD|
|BF |vB).
By some manipulations, we get
vD =|BD||AF ||AC||BF | vC − |BD||FC|
|AC||BF | vA +|FD||BF |vB .
Because we assume that each element K is shape regular, there exists a constant C3 only
depending on the shape-regularity of the grid such that
|AF | ≤ C3|BF |.
By simple calculation, we have
|vB | ≤2(1 + ǫ)n
1− ǫ(|vE |+ |vA|),
|vD| ≤(C1C3 +
2(1 + ǫ)n
1− ǫ
)(|vA|+ |vC |+ |vE |),
which implies that C1 ≤ 9(C1C3 +
2(1+ǫ)n1−ǫ
)2. Furthermore, we have
D1 =|K1||K0
1 |=
|K1||K(1)1 |
|K(1)1 ||K0
1 |≤ (1 + C2)
(1 + ǫ)n
1− ǫ.
We see that both C1 and D1 are bounded, so (3.8) holds true for this case.
Note, if point O does not touch AB, but the distance to AB is very small, then we
draw a tangent line of eK on point O, which intersects with AF and BF at A′ and B′,
respectively. Let K1 denote K1\ABB′A′. Then, we make similar argument for K1 to
prove (3.8).
20 F. WANG AND S. ZHANG
4. Possibility 2-(a)-ii and 2-(b)-ii-B, namely K2 in Case A and B (see Fig 3.2).
K2 in Case A. Let us connect points GA and GB and choose the curved triangle GAB
as K2. Then, from the previous argument, we prove that there exists K02 ⊂ K2 ⊂ K∗
2 ,
and C2 and D2 are two bounded constants such that (3.7) holds true. Furthermore, it is
easy to see that|K2||K2|
<2hK|eK | .
Therefore, we get
C2D2hK |eK ||K2|2|K|−2|K2|−1 < 2C2D2h2K |K2||K|−2 < C.
K2 in Case B. Let us connect points F and C and choose the curved triangle FCB as
K2. Then, from the previous argument, we again prove that there exists K02 ⊂ K2 ⊂ K∗
2 ,
and C2 and D2 are two bounded constants such that (3.7) holds true. Furthermore, by
the same argument for (3.10), we get
|K2|+ ǫ
|K2|+ ǫ≤ 2hK
|eK | ,
where ǫ = |CDFE| − |K2| ≤ C0h3K . So, if |DF | > h2K , equivalently, ǫ ≤ C0|K2|, then
|K2||K2|
≤ C0hK|eK | .
Therefore, we get
C2D2hK |eK ||K2|2|K|−2|K2|−1 < 2C2D2h2K |K2||K|−2 < C.
If |DF | ≤ h2K , then it is reduced to K1 in Case E.
Summing all above finishes the proof.
3.2.5. Proof of Lemma 3.2 when eK contains a turning point
Lemma 3.9. Let eK = Γ ∩K contain a turning point. Then, we can find a (curved) triangle
Ki ⊂ Ki with eK as one of its edges, and two triangles, K0i and K∗
i , such that K0i ⊂ Ki ⊂ K∗
i
and (3.8) holds true with Ci and Di defined in (3.5) and (3.6).
Proof. Obviously, there are two possibilities, namely Ki is a curved triangle or Ki is a
curved quadrilateral, respectively; see K1 and K2 in Case F in Fig 3.4 for an illustration. We
analyze the two cases below.
1. K1 in Case F (see Fig 3.4). Here, K1 is the curved triangle ABC. We choose K1 = K1.
Then, draw a line CD going through point C, tangent to eK and intersect AB at D to
get triangle ADC denoted by K01 , and then draw a line EF parallel to CD, and tangent
with eK to get triangle AEF denoted by K∗1 . We still have K0
1 ⊂ K1 = K1 ⊂ K∗1 . Let
R = AEAD = AF
AC , then C1 ≤ 4R2 and D1 ≤ R2. We know that if R is bounded, the proof
is finished. Otherwise, let H be the turning point on eK , separating eK as e1 and e2,
Optimal Quadratic Nitsche Extended FEM for Interface Problem 21
∗
∗
∗
∗∗∗∗
A
D
EB
F C
K1
K2
G
H
I
Case F
Fig. 3.4. Case that eK contains a turning point.
which do not contain turning point. Let HI be perpendicular to AC and intersect with
AC at point I, and K1 is divided into K(1)1 and K
(2)1 by HI.
Applying the similar argument of K1 in Case B for K(1)1 , and similar argument of K2 in
Case B for K(2)1 , we prove that there exist K
(i)01 ⊂ K
(i)1 ⊂ K
(i)∗1 , such that C
(i)1 and D
(i)1
are bounded and
∫
ei
v2 ds ≤ CC(i)1 D
(i)1
|ei||K(i)
1 |
∫
K(i)1
v2 dx, i = 1, 2.
Without loss of generality, we assume that |e2|
|K(2)1 |
≤ |e1|
|K(1)1 |
. Then there exists a constant C1
such that δ1 ≤ C1δ2, where δ1 = supx∈arcCH dist(x,AC) and δ2 = supx∈arcHB dist(x,AC).
Let l1 = |CI| and l2 = |IA|. If l1 ≤ l2, then |K(1)1 | ≤ C1|K(2)
1 |, and we choose K(2)01
as K01 , and the convex hull of K1 as K∗
1 to apply the same argument of K1 in Case D
to prove (3.8). Otherwise, l2 = lα1 for some α > 1. Then, C1 ≤ l1+l2l2
= 1 + l1−α1 and
D1 ≤ D(2)1
l1+l2l2
= D(2)1 (1 + l1−α
1 ).
Because |e2|
|K(2)1 |
≤ |e1|
|K(1)1 |
, we have
hKβ22‖v‖20,eK ≤ ChK
|K1|2|K|2
∑
i=1,2
|ei||K(i)
1 |
∫
K(i)2
v2 dx
≤ 2ChK(|K(1)
1 |+ |K(2)1 |)2
|K|2|e1|
|K(1)1 |
∫
K1
v2 dx
= 2ChK|K(1)
1 |2 + 2|K(1)1 ||K(2)
1 |+ |K(2)1 |2
|K|2|e1|
|K(1)1 |
∫
K1
v2 dx.
It is easy to see thathk|e1||K
(1)1 |
|K|2 andhk|e1||K
(2)1 |
|K|2 are bounded, so we only need to consider
22 F. WANG AND S. ZHANG
whether the termhk|e1||K
(2)1 |2
|K|2|K(1)1 |
is bounded. We know that if δ1 ≥ l22δ22
h3K
,
hk|e1||K(2)1 |2
|K|2|K(1)1 |
≤ Cl22δ
22
h3Kδ1≤ C.
Otherwise,
C1D1hK |eK ||K1|2|K|−2|K1|−1 ≤ Cl2−2α1 |K1||K|−1 ≤ C
l21δ1l1l22h
2K
≤ Cl31δ
22
h5K≤ C.
This completes the proof for this case.
2. K2 = K\K1 in Case F (see Fig 3.4). Here, K2 is a curved quadrilateral. Let us connect
points C and G and choose the curved triangle CBG as K2. Then, by the previous
argument, it is proved that there exists K01 ⊂ K1 = K1 ⊂ K∗
1 and two bounded constants
C2 and D2 such that (3.7) holds true. Furthermore, by the same argument for (3.10), we
get|K2|+ ǫ
|K2|+ ǫ≤ 2hK
|eK | ,
where ǫ ≤ C0h3K . Then, if |BG| > h2K , namely ǫ ≤ C0|K2|, we have
|K2||K2|
≤ C02hK|eK | .
Therefore, we get
C2D2hK |eK ||K2|2|K|−2|K2|−1 ≤ 2C2D2h2K |K2||K|−2 ≤ C.
If |DF | ≤ h2K , then it is reduced to K1 in the Case F, which have been proved before.
Summing up all above finishes the proof.
4. Numerical Example
In this section, we present numerical verification of the convergence rate of the quadratic
Nitsche-XFEM for interface problem.
Denote the domain Ω := (0, 1)2 and let the interface Γ be the zero level set of the function
φ(x) = (x1 − 0.5)2 + (x2 − 0.5)2 − 1/8. We consider the model problem (2.1). Specifically, the
right hand side is chosen so that the exact solution is
u(x) =
1/α1 exp(x1x2), x ∈ Ω1 = x ∈ Ω, φ(x) < 0,1/α2 sin(πx1) sin(πx2), x ∈ Ω2 = x ∈ Ω, φ(x) > 0.
We apply the quadratic Nitsche-XFEM with the penalty parameter ηeK = 10.
Figure 4.1 shows the numerical results of the quadratic Nitsche-XFEM method for solving
the interface problem. The blue cross (×) and red star (*) denote the errors in H1 norm and
L2 norm, respectively. The black and blue lines are the reference lines of slopes -2 and -3. It
can be observed that the convergence orders match well with the theoretical predictions.
Optimal Quadratic Nitsche Extended FEM for Interface Problem 23
100
101
102
log10
(1/h)
10-6
10-5
10-4
10-3
10-2
10-1
log
10(e
rror)
Error in H1 norm
Error in L2 norm
h2
h3
Fig. 4.1. Numerical errors of the quadratic Nitsche-XFEM method.
5. Concluding Remarks
In this paper, we study the quadratic Nitsche-XFE scheme for the elliptic interface problem,
and prove its optimal convergence rate. A key step in the proof of the stability is to show a
trace inequality, for which we study in detail the cases how the interface may intersect the edges
of the grid, and finally verify the result for each of them. The convergence result then follows
under the general framework.
As the interface problem of the diffusion equation is encountered more and more frequently
from both theoretical research and practical application, and the accuracy and the flexibility
would be among issues with most interests, it is natural to generalize the construction and
analysis in the present paper to the schemes of higher degrees and of other DG types. These
will be discussed in the future.
Acknowledgements. The first author is partially supported by the U.S. Department of
Energy, Office of Science, Office of Advanced Scientific Computing Research as part of the
Collaboratory on Mathematics for Mesoscopic Modeling of Materials under Award Number
DE-SC-0009249, and the Key Program of National Natural Science Foundation of China with
Grant No. 91430215. The second author is supported by State Key Laboratory of Scientific
and Engineering Computing (LSEC), National Center for Mathematics and Interdisciplinary
Sciences of Chinese Academy of Sciences (NCMIS), and National Natural Science Foundation of
China with Grant No. 11471026; he is thankful to the Center for Computational Mathematics
and Applications, the Pennsylvania State University, where he worked on this manuscript as
a visiting scholar. The authors are grateful to Professor Jinchao Xu, Dr. Yuanming Xiao and
Dr. Maximilian Metti for their valuable suggestions and discussions, to Professor Haijun Wu
for his valuable help on preparing the numerical example, and to the anonymous referee for the
valuable comments and suggestion which lead to improvements of the paper.
24 F. WANG AND S. ZHANG
References
[1] D.N. Arnold, F. Brezzi, B. Cockburn and L.D. Marini, Unified analysis of discontinuous Galerkin
methods for elliptic problems, SIAM J. Numer. Anal., 39 (2002), 1749–1779.
[2] Ivo Babuska, The finite element method for elliptic equations with discontinuous coefficients, Com-
puting, 5(1970): 207-213.
[3] T. Belytschko and T. Black, Elastic crack growth in finite elements with minimal remeshing,
International Journal for Numerical Methods in Engineering, 45 (1999), 601–620.
[4] J.H. Bramble, and J.T. King, A finite element method for interface problems in domains with
smooth boundaries and interfaces, Adv. Comput. Math., 6 (1996), 109-138.
[5] Z. Chen and J. Zou, Finite element methods and their convergence for elliptic and parabolic