A note on the optimal degree of the weak gradient of the stabilizer free weak Galerkin finite element method

1 Introduction

In this note, similar to [1], we will consider the following Poisson equation

(1)		$- Δ u$	$=$	$f in Ω,$
(2)		$u$	$=$	$0 on \partial Ω,$

as the model problem, where $Ω$ is a polygonal domain in $R^{2}$ . The variational formulation of the Poisson problem (1)-(2) is to seek $u \in H_{0}^{1} (Ω)$ such that

(3)

(\nabla u, \nabla v)

=

(f, v), \forall v \in H_{0}^{1} (Ω) .

A stabilizer free weak Galerkin finite element method is proposed by Ye and Zhang in [1] as a new method for the solution of Poisson equation on polytopal meshes in 2D or 3D. Let $j$ be the degree of polynomials for the calculation of weak gradient. It has been proved in [1] that there is a $j_{0} > 0$ so that the SFWG method converges with optimal order of convergence for any $j \geq j_{0}$ . However, when $j$ is too large, the magnitude of the weak gradient can be extremely large, causing numerical instability. In this note, we provide the optimal $j_{0}$ to improve the efficiency and avoid unnecessary numerical difficulties, which has mathematical and practical interests.

2 Weak Galerkin Finite Element Schemes

Let $T_{h}$ be a partition of the domain $Ω$ consisting of polygons in 2D. Suppose that $T_{h}$ is shape regular in the sense defined by (2.7)-(2.8) and each $T$ is convex. Let $E_{h}$ be the set of all edges in $T_{h}$ , and let $E_{h}^{0} = E_{h} ∖ \partial Ω$ be the set of all interior edges. For each element $T \in T_{h}$ , denote by $h_{T}$ the diameter of $T$ , and $h = m a x_{T \in T_{h}} h_{T}$ the mesh size of $T_{h}$ .

On $T$ , let $P_{k} (T)$ be the space of all polynomials with degree no greater than $k$ . Let $V_{h}$ be the weak Galerkin finite element space associated with $T \in T_{h}$ defined as follows:

(4)

V_{h}

=

{v = {v_{0}, v_{b}} : v_{0} |_{T} \in P_{k} (T), v_{b} |_{e} \in P_{k} (e), e \in \partial T, T \in T_{h}},

where $k \geq 1$ is a given integer. In this instance, the component $v_{0}$ symbolizes the interior value of $v$ , and the component $v_{b}$ symbolizes the edge value of $v$ on each $T$ and $e$ , respectively. Let $V_{h}^{0}$ be the subspace of $V_{h}$ defined as:

(5)

V_{h}^{0}

=

{v : v \in V_{h}, v_{b} = 0 on \partial Ω} .

For any $v = {v_{0}, v_{b}} \in V_{h} + H^{1} (Ω),$ the weak gradient $\nabla_{w} v \in [P_{j} (T)]^{2}$ is defined on $T$ as the unique polynomial satisfying

(6)

(\nabla_{w} v, q)_{T}

=

- (v_{0}, \nabla \cdot q)_{T} + ⟨ v_{b}, q \cdot n ⟩_{\partial T}, \forall q \in [P_{j} (T)]^{2}, j \geq k .

where $n$

is the unit outward normal vector of

\partial T

For simplicity, we adopt the following notations,

(v, w)_{T_{h}} = \sum T \in T_{h} (v, w)_{T} = \sum T \in T_{h} \int_{T} v w d x,

{⟨ v, w ⟩}_{\partial T_{h}} = \sum T \in T_{h} {⟨ v, w ⟩}_{\partial T} = \sum T \in T_{h} \int_{\partial T} v w d x .

For any $v = {v_{0}, v_{b}}$ and $w = {w_{0}, w_{b}}$ in $V_{h}$ , we define the bilinear forms as follows:

(7)

A (v, w)

= \sum T \in T_{h} (\nabla_{w} v, \nabla_{w} w)_{T} .

Stabilizer free Weak Galerkin Algorithm 1

A numerical solution for (1)-(2) can be obtain by finding $u_{h} = {v_{0}, v_{b}} \in V_{h}^{0}$ , such that the following equation holds

(8)

A (u_{h}, v) = (f, v_{0}), \forall v = {v_{0}, v_{b}} \in V_{h}^{0} .

where $A (\cdot, \cdot)$ is defined in (7).

Accordingly, for any $v \in V_{h}^{0}$ , we define an energy norm $| | | \cdot | | |$ on $V_{h}^{0}$ as:

(9)

| | | v {| | |}^{2} = \sum T \in T_{h} (\nabla_{w} v, \nabla_{w} v)_{T} = \sum T \in T_{h} ∥ \nabla_{w} v ∥_{T}^{2} .

An $H^{1}$ norm on $V_{h}^{0}$ is defined as:

∥ v ∥_{1, h}^{2} = \sum T \in T_{h} (∥ \nabla v_{0} ∥_{T}^{2} + h_{T}^{- 1} ∥ v_{0} - v_{b} ∥_{\partial T}^{2}) .

The following well known lemma and corollary will be needed in proving our main result. Suppose that $D_{1} \subset D_{2}$ are convex regions in $R^{d}$ such that $D_{2}$ can be obtain by scaling $D_{1}$ with a factor $r > 1$ . Then for any $p \in P_{n} (D_{2})$ ,

∥ p ∥_{D_{1}} \leq ∥ p ∥_{D_{2}} \leq C r^{n} ∥ p ∥_{D_{1}},

where $C$ depends only on $d and n$ .

Corollary 2.1

Suppose that $D_{1} and D_{2}$ satisfy the conditions of Lemma 2 and $D_{1} \subseteq D \subseteq D_{2}$ . Then

∥ p ∥_{D_{1}} \leq ∥ p ∥_{D} \leq C r^{n} ∥ p ∥_{D_{1}}, \forall p \in P_{n} (D),

where $C$ depends only on $d and n$ .

The following Theorem shows that for certain $j_{0} > 0$ , whenever $j \geq j_{0}$ , $∥ \cdot ∥_{1, h}$ is equivalent to the $| | | \cdot | | |$ defined in (9), which is crucial in establishing the feasibility of SFWG. Later on, we will show that $j_{0}$ is optimal. (cf.[1], Lemma 3.1) Suppose that $\forall T \in T_{h}, T$ is convex with at most $m$ edges and satisfies the following regularity conditions: for all edges $e_{t}$ and $e_{s}$ of $T$

(10)

| e_{s} | < α_{0} | e_{t} |;

for any two adjacent edges $e_{t} and e_{s}$ the angle $θ$ between them satisfies

(11)

θ_{0} < θ < π - θ_{0},

where $1 \leq α_{0} and θ_{0} > 0$ are independent of $T$ and $h$ . Let $j_{0} = k + m - 2 or j_{0} = k + m - 3$ when each edge of $T$ is parallel to another edge of $T$ . When $d e g \nabla_{w} v = j \geq j_{0}$ , then there exist two constants $C_{1}, C_{2} > 0$ , such that for each $v = {v_{0}, v_{b}} \in V_{h}$ , the following hold true

where $C_{1} and C_{2}$ depend only on $α_{0} and θ_{0}$ . For simplicity, from now on, all constant are independent of $T$ and $h$ , unless otherwise mentioned. They may depend on $k, α_{0}, and θ_{0}$ . Suppose $\nabla_{w} v \in [P_{j} (T)]^{2}, v \in P_{k} (T)$ . We know that

(12)

(\nabla_{w} v, \to q)_{T} = (\nabla v, \to q)_{T} + {⟨ v_{b} - v_{0}, \to q \cdot \to n ⟩}_{\partial T} \forall \to q \in [P_{j} (T)]^{2} .

Suppose $\partial T = (\cup_{i = 0}^{m - 1} e_{i})$ , and we want to construct a $\to q \in [P_{j_{0}} (T)]^{2} \subseteq [P_{j} (T)]^{2}$ , where $j_{0} \leq j$ , so that

(13)	$(\to q, p)_{T}$	$=$	$0, \forall p \in [P_{k - 1} (T)]^{2},$
(14)	$\to q \cdot {\to n}_{i}$	$=$	$0 on e_{i}, i = 1, \dots, m - 1,$
(15)	${⟨ v_{b} - v_{0} - \to q \cdot n_{0}, p ⟩}_{e_{0}}$	$=$	$0, p \in P_{k} (e_{0}),$

where ${\to n}_{i}$ is the unit outer normal to $e_{i}$ . Without loss of generality, we may assume that $e_{0} = {(x, 0) | x \in (0, 1)}, \forall (x, y) \in T, y > 0$ , and one of the vertices of $e_{1}$ is $(0, 0)$ . Let $ℓ_{i} \in P_{1} (T), i = 2, \dots, m - 1$ be such that $ℓ_{i} (e_{i}) = 0, ℓ_{i} (p) \geq 0, \forall p \in T, and {max}_{p \in T} ℓ_{i} (p) = 1$ . Let

(16)

\to q = ℓ_{2} \dots ℓ_{m - 1} Q_{1} \to t = L Q_{1} \to t = Q \to t,

where $\to t$ is a unit tangent vector to $e_{1}$ and $Q_{1} \in P_{j_{0} - m + 2} (T)$ . It is easy to see that $\to q \cdot {\to n}_{i} |_{e_{i}} = 0, i = 1, \dots, m - 1$ . We want to find $Q_{1}$ so that

(17)		${⟨ v_{b} - v_{0} - ℓ_{2} \dots ℓ_{m} Q_{1} ({\to t}_{1} \cdot {\to n}_{0}), p ⟩}_{e_{0}}$	$=$	$0, \forall p \in P_{j_{0} - m + 2} (e_{0}),$
(18)		$(ℓ_{2} \dots ℓ_{m} Q_{1}, p)_{T}$	$=$	$0, \forall p \in P_{k - 1} (T) .$

Let $j_{0} - m + 2 = k$ or $j_{0} = k + m - 2$ . It is easy to see that if $Q_{1}$ satisfies (17)-(18), then $\to q$ satisfies (13)-(15). In addition, (17)-(18) has $(k + 1) (k + 2) / 2$ equations and the same number of unknowns. To show (17)-(18) has a unique solution $Q_{1} \in P_{k} (T)$ we set $v_{b} - v_{0} = 0$ . Setting $p = Q_{1}$ in (17) yields

{⟨ ℓ_{2} \dots ℓ_{m - 1} Q_{1}, Q_{1} ⟩}_{e_{0}} = 0.

Since $L = ℓ_{2} \dots ℓ_{m - 1} > 0$ inside $e_{0}$ , $Q_{1} (x, 0) \equiv 0$ . Thus $Q_{1} = y Q_{2}$ , where $Q_{2} \in P_{k - 1} (T)$ . Similarly

(y ℓ_{2} \dots ℓ_{m - 1} Q_{2}, p)_{T} = 0 \forall p \in P_{k - 1} (T),

implies $Q_{2} = 0$ and thus $Q_{1} = 0$ . Note that for any $j \geq j_{0}$ , since $\to q \in [P_{j_{0}} (T)]^{2} \subset [P_{j} (T)]^{2}$ , there exists at least one such $\to q$ in $[P_{j} (T)]^{2}$ . Note also that when every edge of $\partial T$ is parallel to another one, we can lower the number of $ℓ_{i}$ in (16) by one. Thus for such a $T$ , we have $j_{0} = m + k - 3$ . Next, we want to show that the unique solution of (13)-(15) satisfies

(19)

∥ \to q ∥_{T} \leq γ_{0} ∥ \to q \cdot_{0} ∥_{e_{0}},

for some $γ_{0} > 0$ . Note that

(\to q \cdot_{0})^{2} = Q^{2} {sin}^{2} θ = \to q \cdot \to q {sin}^{2} θ \geq \to q \cdot \to q {sin}^{2} θ_{0},

where $θ$ is the angle between $e_{0}$ and $e_{1}$ . Thus, (19) is equivalent to

(20)

∥ Q ∥_{T} \leq γ_{0} ∥ Q ∥_{e_{0}},

for some $γ_{0} > 0$ and we may assume

e_{1} = {(0, y) | 0 \leq y \leq 1},

because of the assumption (10). Now let $T_{r}$ be the triangle spanned by $e_{0}$ and $e_{1}$ and label the third edge of $T_{r}$ as $e_{t}$ . Let ${^T}_{r} \subset T_{r}$ be the triangle with vertices $(0, 0.25), (0, 0.75)$ and $(0.5, 0.25)$ . Label the edge of ${^T}_{r}$ connecting $(0, 0.75)$ and $(0.5, 0.25)$ as ${^e}_{t}$ . For $i = 2, \dots, m - 1$ , let

	$d_{i}$	$=$	$d i s t ({^T}_{r}, L_{i}),$
	$D_{i}$	$=$	$max p \in T d i s t (p, L_{i}),$

where $L_{i}$ is the line containing $e_{i}$ . Then

	$d_{i}$	$\geq$	$d i s t ({^e}_{t}, e_{t}) = \sqrt{2} / 4,$
(21)		$D_{i}$	$\leq$	$(\| e_{0} \| + \| e_{1} \| + \dots + \| e_{m - 1} \|) / 2$
(21)			$\leq$	$\frac{m α_{0}}{2} .$

Thus $\forall p \in {^T}_{r}$ ,

(22)

ℓ_{i} (p) \geq \frac{\sqrt{2}}{2 m α_{0}} = ε, i = 2, \dots, m - 1.

Now we will prove (20) in $3$ steps:

$I : β_{1} ∥ Q_{1} ∥_{e_{0}}$	$\leq$	$∥ Q ∥_{e_{0}},$
$I I : β_{2} ∥ Q ∥_{T_{r}}$	$\leq$	$∥ Q_{1} ∥_{e_{0}},$
$I I I : β_{3} ∥ Q ∥_{T}$	$\leq$	$∥ Q ∥_{T_{r}} .$

Let $^e={(x,0)|0≤x≤0.75}$ . Then similar to (22), $\forall p \in^e, ℓ_{i} (p) \geq ε, i = 2. \dots, m - 1$ . Thus

$ε^{m - 2} ∥ Q_{1} ∥_{^e} \leq ∥ Q ∥_{^e} \leq ∥ Q ∥_{e_{0}} .$

It follows from Corollary 2.1 that

(23) $∥ Q ∥_{e_{0}} \geq ε^{m - 2} ∥ Q_{1} ∥_{^e} \geq C ε^{m - 2} ∥ Q_{1} ∥_{e_{0}} = β_{1} ∥ Q_{1} ∥_{e_{0}} .$
It is easy to see that

$∥ Q ∥_{T_{r}}^{2}$ $=$ $\int_{0}^{1} \int_{0}^{1 - y} L^{2} Q_{1}^{2} d x d y = \int_{0}^{1} \int_{0}^{1 - y} L^{2} (Q_{1} - Q_{1} (x, 0) + Q_{1} (x, 0))^{2} d x d y$

$\leq$ $2 \int_{0}^{1} \int_{0}^{1 - y} [L^{2} (Q_{1} - Q_{1} (x, 0))^{2} + L^{2} {(Q_{1} (x, 0))}_{1}^{2}] d x d y .$

It follows from (22) and Corollary 2.1 that

(24) $∥ L Q_{1} (x, 0) ∥_{T_{r}}^{2}$ $=$ $\int_{0}^{1} \int_{0}^{1 - y} L^{2} {(Q_{1} (x, 0))}_{1}^{2} d x d y$

$\leq$ $\int_{0}^{1} \int_{0}^{1} {(Q_{1} (x, 0))}_{1}^{2} d x d y$

$=$ $∥ Q_{1} ∥_{e_{0}}^{2}$

Note that

$Q_{1} - Q_{1} (x, 0) = y ¯ Q, ¯ Q \in P_{k - 1},$

and

(25) $\int_{T} [L Q_{1} (x, 0) + L y ¯ Q] P d A = 0, \forall p \in P_{k - 1} .$

Since $L y \geq \frac{1}{4} ε for (x, y) \in {^T}_{r} and L y \geq 0, \forall (x, y) \in T$ ,

(26) $\int_{T} L y {¯ Q}^{2} d A \geq \frac{ε}{4} \int_{{^T}_{r}} {¯ Q}^{2} d A = δ_{3} ∥ ¯ Q ∥_{{^T}_{r}}^{2} .$

Since $∥ \cdot ∥_{{^T}_{r}} and ∥ \cdot ∥_{T_{r}}$ are equivalent norms on $P_{k - 1} (T_{r})$ ,

(27) $δ_{3} ∥ ¯ Q ∥_{{^T}_{r}} \geq β_{3} ∥ ¯ Q ∥_{T_{r}} .$

Let $T_{e}^{c}$ be the circumscribed isosceles-right triangle obtained by scaling $T_{r}$ . It is easy to see that

(28) $d i a m (T_{e_{0}}^{c}) \leq 2 m α_{0} .$

Since $T_{r} \subseteq T \subseteq T_{e}^{c}$ , it follows from (26), (27), and Corollary 2.1 that

$\int_{T} L y {¯ Q}^{2} d A \geq β_{3} ∥ ¯ Q - 1 ∥_{T_{r}}^{2} \geq δ_{4} ∥ ¯ Q ∥_{T}^{2},$

where $δ_{4}$ depend on (28) and $k$ . Letting $p = ¯ Q$ in (25) yields

$\int_{T} L y {¯ Q}^{2} d A = - \int_{T} L Q_{1} (x, 0) ¯ Q d A \leq ∥ Q_{1} (x, 0) ∥_{T} ∥ ¯ Q ∥_{T},$

and thus

$∥ ¯ Q ∥_{T} \leq \frac{1}{δ_{4}} ∥ Q_{1} (x, 0) ∥_{T} .$

Then

$∥ ¯ Q ∥_{T}^{2} \leq δ_{5} \int_{0}^{N} \int_{0}^{M} Q_{1}^{2} (x, 0) d x d y = C N \int_{0}^{M} Q_{1}^{2} (x, 0) d x = C ∥ Q_{1} ∥_{[0, M]}^{2},$

where $N = {max}_{(x, y) \in T} y \leq m α_{0} and M = {max}_{(x, y) \in T} x ε \leq m α_{0}$ . It follows from Corollary 2.1 and (26)

(29) $∥ L y ¯ Q ∥_{_{r}} \leq ∥ ¯ Q ∥_{T} \leq C ∥ Q_{1} ∥_{e_{0}} .$

Combining (29) and (24) yields

(30) $∥ Q ∥_{T_{r}} \leq C ∥ Q_{1} ∥_{e_{0}} .$
Applying Corollary 2.1 again yields

(31) $C ∥ Q ∥_{T} \leq ∥ Q ∥_{T_{r}} .$

Now we have completed the $3$ -step argument.
By a scaling argument, we have

(32) $∥ Q ∥_{T} \leq γ_{0} h^{1 / 2} ∥ Q ∥ e_{0} .$

Plugging $\to q$ into (12) yields

(33) $∥ Q ∥_{T} ∥ \nabla_{w} v ∥_{T}$ $\geq$ $(\nabla_{w} v, \to q \to n)_{^T} = {⟨ v_{b} - v_{0}, \to q \to n \cdot \to n ⟩}_{e_{0}}$

$=$ $∥ v_{b} - v_{0} ∥_{e_{0}}^{2}$

$=$ $∥ v_{b} - v_{0} ∥_{e_{0}} ∥ Q ∥_{e_{0}} .$

It follows from (32) that

$∥ \nabla_{w} v ∥_{T} \geq C h^{1 / 2} ∥ v_{b} - v_{0} ∥_{e_{0}} .$

Similarly,

∥ v_{b} - v_{0} ∥_{e_{i}} \leq C h_{T}^{1 / 2} ∥ \nabla_{w} v ∥_{T}, i = 1, \dots, m - 1.

Thus

∥ v_{b} - v_{0} ∥_{\partial T} \leq C h_{T}^{1 / 2} ∥ \nabla_{w} v ∥_{T} .

By letting $q = \nabla_{w} v$ in (12), we arrive at

(\nabla_{w} v, \nabla_{w} v)_{T} = (\nabla v_{0}, \nabla_{w} v)_{T} + {⟨ v_{b} - v_{0}, \nabla_{w} v \cdot n ⟩}_{\partial T} .

It follows from the trace inequality and the inverse inequality that

	$∥ \nabla_{w} v ∥_{T}^{2}$	$\leq$	$∥ \nabla v_{0} ∥_{T} ∥ \nabla_{w} v ∥_{T} + ∥ v_{0} - v_{b} ∥_{\partial T} ∥ \nabla_{w} v ∥_{\partial T}$
		$\leq$	$∥ \nabla v_{0} ∥_{T} ∥ \nabla_{w} v ∥_{T} + C h_{T}^{1 / 2} ∥ v_{0} - v_{b} ∥_{\partial T} ∥ \nabla_{w} v ∥_{T} .$

Thus

∥ \nabla_{w} v ∥_{T}

\leq

C (∥ \nabla v_{0} ∥_{T} + h_{T}^{1 / 2} ∥ v_{0} - v_{b} ∥_{\partial T}) .

To show that $j_{0} = k + m - 2 or j_{0} = k + m - 3$ is the lowest possible degree for the weak gradient, we give the following examples.

Example 2.1

Suppose $Ω = (0, 1) \times (0, 1)$ and is partitioned as shown in Figure 1.

Let $j = k = 1 = j_{0} - 1$ . To see if (8) has a unique solution in $V_{h}^{0}$ , we set $f = 0$ and $v = u_{h}$ . Then $\nabla_{w} u_{h} = 0$ . By definition,

0 = (\nabla_{w} u_{h}, \to q)_{T_{h}} = - (u_{h}, \nabla \to q)_{T_{h}} + {⟨ u_{b}, \to q \cdot \to n ⟩}_{\partial T_{h}}, \forall \to q \in [P_{1} (T_{h})]^{2} .

Note that the degree of freedom of

[P_{1} (T_{h})]^{2}

36

while the degree of freedom of

V_{h}^{0}

39

. Thus, there exist

u_{h} \neq 0

so that

\nabla_{w} u_{h} = 0

. Thus Algorithm

1

will not work.

Example 2.2

Let $Ω = (0, 1) \times (0, 1)$ and is partitioned as shown in Figure 2.

Let $j = k = 1 = j_{0} - 1$ . Note that the degree of freedom of $[P_{1} (T_{h})]^{2}$ is $72$ while the degree of freedom of $V_{h}^{0}$ is $75$ . A similar argument as in example 2.1 shows that Algorithm $1$ will not work.

For the completeness, we list the following from [1]. Let $u_{h} \in V_{h}$ be the weak Galerkin finite element solution of (8). Assume that the exact solution $u \in H^{k + 1} (Ω)$ and $H^{2}$ -regularity holds true. Then, there exists a constant $C$ such that

(34)		$\| \| \| u - u_{h} \| \| \|$	$\leq$	$C h^{k} \| u \|_{k + 1},$
(35)		$∥ u - u_{0} ∥$	$\leq$	$C h^{k + 1} \| u \|_{k + 1} .$

3 Numerical Experiments

We are devoting this section to verify our theoretical results in previous sections by two numerical examples. The domain in all examples is $Ω = (0, 1) \times (0, 1)$ . We will implement the SFWG finite element method (8) on triangular meshes. A uniform triangulation of the domain $Ω$ is used. By refining each triangle for $N = 2, 4, 8, 16, 32, 64, 128$ , we obtain a sequence of partitions. The first two levels of grids are shown in Figure 3. We apply the SF-WG finite element method with $(P_{k} (T), P_{k} (e), [P_{k + 1} (T)]^{2}), k = 1, 2$ finite element space to find SFWG solution $u_{h} = {u_{0}, u_{b}}$ in the computation.

Figure 3: The first two triangle grids in computation

Example 3.1

In this example, we solve the Poisson problem (1)-(2) posed on the unit square $Ω$ , and the analytic solution is

(36)

u (x, y) = sin (π x) sin (π y) .

The boundary conditions and the source term $f (x, y)$ are computed accordingly. Table 1 lists errors and convergence rates in $H^{1}$ -norm and $L^{2}$ -norm.

$k$	$h$		Rate	$∥ u_{0} - Q_{0} u ∥$	Rate
	1/2	8.8205E-01	-	8.0081E-02	-
	1/4	4.3566E-01	1.02	2.5055E-02	1.68
	1/8	2.1565E-01	1.01	6.7091E-03	1.90
1	1/16	1.0747E-01	1.00	1.7092E-03	1.97
	1/32	5.3684E-02	1.00	4.2940E-04	1.99
	1/64	2.6836E-02	1.00	1.0749E-04	2.00
	1/128	1.3417E-02	1.00	2.6871E-05	2.00
	1/2	2.5893E-01	-	1.0511E-02	-
	1/4	6.5338E-02	1.99	1.2889E-03	3.03
	1/8	1.6252E-02	2.00	1.5596E-04	3.05
2	1/16	4.0538E-03	2.09	1.9193E-05	3.02
	1/32	1.0129E-03	2.00	2.3837E-06	3.00
	1/64	2.5321E-04	2.00	2.9713E-07	3.00
	1/128	6.3304E-05	2.00	3.7094E-08	3.00

Table 1: Error profiles and convergence rates for

P_{k}

elements with

P_{k + 1}^{2}

weak gradient,

(k = 1, 2)

Example 3.2

In this example, we consider the problem $- Δ u = f$ with boundary condition and the exact solution is

(37)

u (x, y) = x (1 - x) y (1 - y) .

The results obtained in Table 2 show the errors of SFWG scheme and the convergence rates in the $| | | \cdot | | |$ norm and $∥ \cdot ∥$ norm. The simulations are conducted on triangular meshes and polynomials of order $k = 1, 2$ . The SFWG scheme with $P_{k}$ element has optimal convergence rate of $O (h^{k})$ in $H^{1}$ -norm and $O (h^{k + 1})$ in $L^{2}$ -norm.

$k$	$N$	$\| \| \| e_{h} \| \| \|$	Rate	$∥ e_{0} ∥$	Rate
	1/2	5.6636E-02	-	5.3064E-03	-
	1/4	2.9670E-02	0.93	1.6876E-03	1.65
	1/8	1.4948E-02	0.99	4.5066E-04	1.90
1	1/16	7.4848E-03	1.00	1.1461E-04	1.98
	1/32	3.7435E-03	1.00	2.8778E-05	1.99
	1/64	1.8719E-03	1.00	7.2024E-06	2.00
	1/128	9.3596E-04	1.00	1.8011E-06	2.00
	1/2	1.3353E-02	-	5.6066E-04	-
	1/4	3.4831E-03	1.94	6.8552E-05	3.03
	1/8	8.7712E-04	1.99	8.2466E-06	3.06
2	1/16	2.1961E-04	2.00	1.0094E-06	3.03
	1/32	5.4967E-05	2.00	1.2482E-07	3.02
	1/64	1.3747E-05	2.00	1.5518E-08	3.00
	1/128	3.4375E-06	2.00	1.9344E-09	3.00

Table 2: Error profiles and convergence rates for

P_{k}

elements with

P_{k + 1}^{2}

weak gradient,

(k = 1, 2)

Overall, the numerical experiment’s results shown in the formentioned examples match the theoretical study part in the previous sections.

References

[1] X. Ye, S. Zhang, On stabilizer-free weak Galerkin finite element method on polytopal meshes, arXiv:1906.06634v2, 2019.

A note on the optimal degree of the weak gradient of the stabilizer free weak Galerkin finite element method

Authors

The lowest-order stabilizer free Weak Galerkin Finite Element Method

A new weak gradient for the stabilizer free weak Galerkin method with polynomial reduction

A stabilizer-free C^0 weak Galerkin method for the biharmonic equations

A Weak Galerkin Method for Elasticity Interface Problems

Numerical investigation on weak Galerkin finite elements

Weak Galerkin Method for Electrical Impedance Tomography

A comparison of matrix-free isogeometric Galerkin and collocation methods for Karhunen–Loève expansion

1 Introduction

2 Weak Galerkin Finite Element Schemes

Stabilizer free Weak Galerkin Algorithm 1

Corollary 2.1

Example 2.1

Example 2.2

3 Numerical Experiments

Example 3.1

Example 3.2

References

	$∥ Q ∥_{T_{r}}^{2}$	$=$	$\int_{0}^{1} \int_{0}^{1 - y} L^{2} Q_{1}^{2} d x d y = \int_{0}^{1} \int_{0}^{1 - y} L^{2} (Q_{1} - Q_{1} (x, 0) + Q_{1} (x, 0))^{2} d x d y$
		$\leq$	$2 \int_{0}^{1} \int_{0}^{1 - y} [L^{2} (Q_{1} - Q_{1} (x, 0))^{2} + L^{2} {(Q_{1} (x, 0))}_{1}^{2}] d x d y .$

(24)	$∥ L Q_{1} (x, 0) ∥_{T_{r}}^{2}$	$=$	$\int_{0}^{1} \int_{0}^{1 - y} L^{2} {(Q_{1} (x, 0))}_{1}^{2} d x d y$
		$\leq$	$\int_{0}^{1} \int_{0}^{1} {(Q_{1} (x, 0))}_{1}^{2} d x d y$
		$=$	$∥ Q_{1} ∥_{e_{0}}^{2}$

(33)	$∥ Q ∥_{T} ∥ \nabla_{w} v ∥_{T}$	$\geq$	$(\nabla_{w} v, \to q \to n)_{^T} = {⟨ v_{b} - v_{0}, \to q \to n \cdot \to n ⟩}_{e_{0}}$
		$=$	$∥ v_{b} - v_{0} ∥_{e_{0}}^{2}$
		$=$	$∥ v_{b} - v_{0} ∥_{e_{0}} ∥ Q ∥_{e_{0}} .$

A note on the optimal degree of the weak gradient of the stabilizer free weak Galerkin finite element method

Authors

Related Research

The lowest-order stabilizer free Weak Galerkin Finite Element Method

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A Weak Galerkin Method for Elasticity Interface Problems

Numerical investigation on weak Galerkin finite elements

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This week in AI

1 Introduction

2 Weak Galerkin Finite Element Schemes

Stabilizer free Weak Galerkin Algorithm 1

Corollary 2.1

Example 2.1

Example 2.2

3 Numerical Experiments

Example 3.1

Example 3.2

References