The multiscale hybrid mixed method in general polygonal meshes

Abstract

This work extends the general form of the multiscale hybrid-mixed (MHM) method for the second-order Laplace (Darcy) equation to general non-conforming polygonal meshes. The main properties of the MHM method, i.e., stability, optimal convergence, and local conservation, are proven independently of the geometry of the elements used for the first level mesh. More precisely, it is proven that piecewise polynomials of degree k and $k+1$, $k \ge 0$, for the Lagrange multipliers (flux), along with continuous piecewise polynomial interpolations of degree $k+1$ posed on second-level sub-meshes are stable if the latter is fine enough with respect to the mesh for the Lagrange multiplier. We provide an explicit sufficient condition for this restriction. Also, we prove that the error converges with order $k+1$ and $k+2$ in the broken $H^1$ and $L^2$ norms, respectively, under usual regularity assumptions, and that such estimates also hold for non-convex; or even non-simply connected elements. Numerical results confirm the theoretical findings and illustrate the gain that the use of multiscale functions provides.

Appendix

Lemma 4

Let $k\ge 0$, and let us define the following space

$$\begin{aligned} \mathscr {X}=\left\{ q\in H^1_0(0,1): q|_{(0,0.5)}^{}\in \mathbb {P}_{k+1}^{}\left( 0,0.5\right) \;\text {and}\; q|_{(0.5,1)}^{}\in \mathbb {P}_{k+1}^{}\left( 0.5,1\right) \right\} . \end{aligned}$$

Then, if $\mu$ satisfies

$$\begin{aligned} \mu \in \mathbb {P}_{k}^{}(0,1) \;\text {and}\;\int _0^1\mu (x)q(x)dx=0\quad \text {for all }q\in \mathscr {X}, \end{aligned}$$

(101)

then $\mu =0$ in [0, 1].

Proof

The case $k=0$ will be treated first. For $k=0$ let

$$\begin{aligned} q(x)=\left\{ \begin{array}{ll} x &{} \quad \text {if}\; x\le 0.5 \\ 1-x &{} \quad \text {if}\; x > 0.5\end{array}\right. . \end{aligned}$$

Then, $q\in \mathscr {X}$. Thus, if $\mu \in \mathbb {R}{\setminus }\{0\}$, then

$$\begin{aligned} \int _0^1\mu q(x)dx = \mu \int _0^1q(x)dx =\frac{\mu }{4}\not = 0, \end{aligned}$$

which contradicts the hypothesis.

Now, let $k\ge 1$, and let $\mu \in \mathbb {P}_{k}^{}(0,1)$ satisfying (101). Let $q_1^{},q_2^{}\in \mathscr {X}$ be defined by

$$\begin{aligned} q_1^{}(x)=\left\{ \begin{array}{ll} x(0.5-x) &{} \quad \text {if}\; x\le 0.5 \\ 0 &{} \quad \text {if}\; x> 0.5\end{array}\right. ,\quad q_2^{}(x)=\left\{ \begin{array}{ll} 0 &{} \quad \text {if}\; x\le 0.5 \\ (x-0.5)(1-x) &{} \quad \text {if}\; x > 0.5\end{array}\right. . \end{aligned}$$

(102)

Since

$$\begin{aligned} \int _0^1\mu (x)q_1(x)dx=\int _0^{0.5} \mu (x)x(0.5-x)\,dx=0, \end{aligned}$$

(103)

and $q_1^{}(x)\ge 0$ in (0, 0.5), then $\mu$ changes sign in (0, 0.5), which implies that $\mu$ has at least one root $x_1^{}\in (0,0.5)$. Analogously, since

$$\begin{aligned} \int _0^1\mu (x)q_2(x)dx=\int _{0.5}^{1} \mu (x)(x-0.5)(1-x)\,dx=0, \end{aligned}$$

(104)

then $\mu$ has also a root $x_2^{}\in (0.5,1)$. So, if $k=1$, then $\mu \in \mathbb {P}_{1}^{}(0,1)$ has two roots in (0, 1), and thus it needs to be identically zero. If $k\ge 2$, we define the following functions in $\mathscr {X}$:

$$\begin{aligned} q_3^{}(x)=q_1^{}(x)(x-x_1^{})\quad \text {and}\quad q_4^{}(x)=q_2^{}(x)(x-x_2^{}). \end{aligned}$$

(105)

If $\mu$ does not have any other root in (0, 0.5), then, since $\mu$ and $q_3^{}$ change signs at the exact same points in (0, 0.5), then

$$\begin{aligned} \int _0^1\mu (x)q_3^{}(x)dx=\int _0^{0.5}\mu (x)q_3^{}(x)\,dx\not =0, \end{aligned}$$

(106)

which contradicts (101). In an analogous way, using the function $q_4^{}$, we deduce the existence of a fourth root $x_4^{}\in (0.5,1)$. If $k=3$, then $\mu$ is a quadratic polynomial with four distinct roots in (0, 1), and then it needs to be identically zero.

For $k\ge 3$, following completely analogous steps to the ones just described, we deduce that $\mu$ needs to vanish at, at least, $2(k+1)-2$ different points in (0, 1), which, for a polynomial $\mu \in \mathbb {P}_{k}^{}(0,1)$, is impossible unless $\mu =0$. This finishes the proof. $\square$

The following result extends the previous result to the $\mathbb {P}_{k+1}^{}(F)\times \mathbb {P}_{k+1}^{}(\mathfrak {T})$ case.

Lemma 5

Let $k\ge 0$. Then, if $\mu$ satisfies

$$\begin{aligned} \mu \in \mathbb {P}_{k+1}^{}(0,1)\; \text {and}\;\int _0^1\mu (x)q(x)\,dx=0\quad \text {for all }q\in \mathscr {X}, \end{aligned}$$

(107)

then $\mu =0$ in [0, 1], where the space $\mathscr {X}$ is defined as:

• For $k=0$:

  $$\begin{aligned}&\mathscr {X}=\left\{ q\in H^1_0(0,1): \text {the restriction of }\;q\;\text {to}\;(0,1/4),(1/4,1/2),(1/2,3/4), \right. \\&\quad \,\,\quad \qquad \left. \text {and} \;(3/4,1)\;\text {belongs to}\; \mathbb {P}_1^{}\right\} . \end{aligned}$$

• For $k=1$:

  $$\begin{aligned}&\mathscr {X} = \left\{ q\in H^1_0(0,1): \text {the restriction of }\;q\;\text {to}\;(0,1/3),(1/3,2/3), \right. \\&\quad \,\,\quad \qquad \left. \text {and} \;(2/3,1)\;\text {belongs to}\; \mathbb {P}_2^{}\right\} . \end{aligned}$$

• For $k\ge 2$:

  $$\begin{aligned} \mathscr {X}=\left\{ q\in H^1_0(0,1): q|_{(0,0.5)}^{}\in \mathbb {P}_{k+1}^{}\left( 0,0.5\right) \;\text {and}\; q|_{(0.5,1)}^{}\in \mathbb {P}_{k+1}^{}\left( 0.5,1\right) \right\} . \end{aligned}$$

Proof

We split the proof in the three cases described above:

${k=0}$: Let $\mu \in \mathbb {P}_1^{}(0,1)$ satisfying (101), and let $\ell _1^{},\ell _2^{}\in \mathscr {X}$ be defined as follows:

$$\begin{aligned} \ell _1^{}(x)=\left\{ \begin{array}{ll} x &{} \quad \text {if}\; x\le 0.25 \\ 0.25-x &{} \quad \text {if}\; 0.25 \le x\le 0.5 \\ 0 &{} \quad \text {if}\; 0.5\le x \end{array}\right. ,\quad \ell _2^{}(x)=\left\{ \begin{array}{ll} 0 &{} \quad \text {if}\; x\le 0.5 \\ x-0.5 &{} \quad \text {if}\; 0.5 \le x\le 0.75 \\ 1-x &{} \quad \text {if}\; 0.75\le x \end{array}\right. . \end{aligned}$$

(108)

Since $\mu$ satisfies (101) then

$$\begin{aligned} \int _0^1\mu (x)\ell _1^{}(x)\,dx=\int _0^{0.5} \mu (x)\ell _1^{}(x)\,dx=0, \end{aligned}$$

(109)

which implies that $\mu$ has a root in (0, 0.5) since $\ell _1^{}$ is positive in that interval. Analogously, since

$$\begin{aligned} \int _0^1\mu (x)\ell _2^{}(x)\,dx=\int _{0.5}^{1} \mu (x)\ell _2^{}(x)\,dx=0, \end{aligned}$$

(110)

and $\ell _2^{}$ is positive in (0.5, 1), we deduce that $\mu$ has a root also in (0.5, 1). Since $\mu \in \mathbb {P}_1^{}(0,1)$ and has two different roots in (0, 1), then it is identically zero.

${k=1}:$ Let $\mu \in \mathbb {P}_2^{}(0,1)$ satisfying (101), and let $\ell _3^{},\ell _4^{}, \ell _5^{}\in \mathscr {X}$ be defined as follows:

$$\begin{aligned}&\ell _3^{}(x)=\left\{ \begin{array}{ll} x(1/3-x) &{} \quad \text {if}\; x\le 1/3 \\ 0 &{} \quad \text {if}\; 1/3\le x \end{array}\right. ,\quad \ell _4^{}(x)=\left\{ \begin{array}{ll} (x-1/3)(2/3-x) &{} \quad \text {if}\; 1/3 \le x\le 2/3 \\ 0 &{} \quad \text {else} \end{array}\right. \\&\ell _5^{}(x)=\left\{ \begin{array}{ll} 0 &{} \quad \text {if}\; x\le 2/3 \\ (x-2/3)(1-x) &{} \quad \text {if}\; 2/3\le x \end{array}\right. . \end{aligned}$$

Since $\mu$ satisfies (101) then

$$\begin{aligned} \int _0^1\mu (x)\ell _3^{}(x)\,dx=\int _0^{1/3} \mu (x)x(1/3-x)\,dx=0, \end{aligned}$$

(111)

which, due to the positivity of $\ell _3^{}$ implies the existence of $x_1^{}\in (0,1/3)$ such that $\mu (x_1^{})=0$. In a completely analogous way, we deduce there exist $x_2^{}\in (1/3,2/3)$ and $x_3^{}\in (2/3,1)$ such that $\mu (x_2^{})=\mu (x_3^{})=0$. Thus, $\mu$ must necessarily vanish.

${k\ge 2}$: From the proof of Lemma 4, any function $\mu$ that satisfies (107) has at least 2k different roots in (0, 1). Now, for $k\ge 2$, $2 k \ge k+2$, and thus, if $\mu \in \mathbb {P}_{k+1}^{}(0,1)$ satisfies (107) it will vanish at at least $k+2$ different points in (0, 1), which implies that $\mu$ is identically zero. This finishes the proof. $\square$