Monotone Meshfree Methods for Linear Elliptic Equations in Non-divergence Form via Nonlocal Relaxation

Abstract

We design a monotone meshfree finite difference method for linear elliptic equations in the non-divergence form on point clouds via a nonlocal relaxation method. The key idea is a novel combination of a nonlocal integral relaxation of the PDE problem with a robust meshfree discretization on point clouds. Minimal positive stencils are obtained through a local $l_1$-type optimization procedure that automatically guarantees the stability and, therefore, the convergence of the meshfree discretization for linear elliptic equations. A major theoretical contribution is the existence of consistent and positive stencils for a given point cloud geometry. We provide sufficient conditions for the existence of positive stencils by finding neighbors within an ellipse (2d) or ellipsoid (3d) surrounding each interior point, generalizing the study for Poisson’s equation by Seibold (Comput Methods Appl Mech Eng 198(3–4):592–601, 2008). It is well-known that wide stencils are in general needed for constructing consistent and monotone finite difference schemes for linear elliptic equations. Our result represents a significant improvement in the stencil width estimate for positive-type finite difference methods for linear elliptic equations in the near-degenerate regime (when the ellipticity constant becomes small), compared to previously known works in this area. Numerical algorithms and practical guidance are provided with an eye on the case of small ellipticity constant. At the end, we present numerical results for the performance of our method in both 2d and 3d, examining a range of ellipticity constants including the near-degenerate regime.

Appendices

Proof of Theorem 3.2

We show the proof of Theorem 3.2 in Sect. 1. We begin with some useful lemmas before proving the theorem.

Lemma A.1

Let $r({\varvec{v}},{\varvec{x}}_i)$ denote the radius of the inscribed ball in $T_i ({\mathcal {C}}_\delta ^{{\varvec{v}}}({\varvec{x}}_i))$ and h be the fill distance asscociated with $X = \{ {\varvec{x}}_i\}_{i=1}^M\subset \varOmega _\delta$. If

$$\begin{aligned} h < \min _{{\varvec{v}}\in {\mathbb {R}}^d, |{\varvec{v}}|=1} r({\varvec{v}},{\varvec{x}}_i), \end{aligned}$$

then $S_{\delta , h, 2}({\varvec{x}}_i)$ and ${\overline{S}}_{\delta , h, 2}({\varvec{x}}_i)$ are not empty.

Proof

Notice that by the definition of the fill distance in Eq. (16), there are no holes with a radius larger than h. Suppose $S_{\delta , h, 2}({\varvec{x}}_i)$ or ${\overline{S}}_{\delta , h, 2}({\varvec{x}}_i)$ is empty, then by Corollary 3.1, there exists ${\varvec{v}}$ such that $T_i({\mathcal {C}}^{{\varvec{v}}}_\delta ({\varvec{x}}_i))$ contains no point in $X\backslash \{ {\varvec{x}}_i\}$. Therefore the inscribed ball in $T_i({\mathcal {C}}^{{\varvec{v}}}_\delta ({\varvec{x}}_i))$ is a hole with radius larger than h by the assumption, which gives a contradiction. $\square$

From the lemma above, our goal is then to get a lower bound for $\min _{{\varvec{v}}\in {\mathbb {R}}^d, |{\varvec{v}}|=1} r({\varvec{v}},{\varvec{x}}_i)$ for each ${\varvec{x}}_i\in \varOmega$. We first present a result in 2d which will also be useful for the 3d estimates. In 2d, we assume that ${\mathcal {C}}^{{\varvec{v}}}_\delta ({\varvec{x}}_i)$ is a cone with total opening angle $2\phi$. In addition, without loss of generality, we fix ${\varvec{x}}_i\in \varOmega$ and assume that

$$\begin{aligned} A({\varvec{x}}_i) = \begin{pmatrix} \varrho &{} 0 \\ 0 &{} 1 \\ \end{pmatrix}. \end{aligned}$$

(A.1)

From symmetry, we only need to consider ${\varvec{v}}(\theta ) = (\cos (\theta ), \sin (\theta ))$ for $\theta \in [0,\frac{\pi }{2}]$.

Lemma A.2

Consider $d=2$ and $A({\varvec{x}}_i)$ given by Eq. (A.1). Assume that ${\mathcal {C}}^{{\varvec{v}}}_\delta ({\varvec{x}}_i)$ is a cone with total opening angle $2\phi$ for $\phi \in (0,\frac{\pi }{8}]$, and $r({\varvec{v}},{\varvec{x}}_i)$ denote the radius of the inscribed circle in $T_i ({\mathcal {C}}_\delta ^{{\varvec{v}}}({\varvec{x}}_i))$. In addition, let ${\varvec{v}}(\theta ) = (\cos (\theta ), \sin (\theta ))$ for $\theta \in [0,\frac{\pi }{2}]$. Then, there exists a constant $K =K (\phi )>0$ such that

$$\begin{aligned} \min _{\theta \in [0, \frac{\pi }{2}]} r({\varvec{v}}(\theta ),{\varvec{x}}_i) \ge K \delta \sqrt{\varrho }. \end{aligned}$$

Proof

We try to fit a cone in $T_i({\mathcal {C}}_\delta ^{{\varvec{v}}(\theta )}({\varvec{x}}_i))$ and then find the inscribed circle in the cone. First, notice that for a cone with a radius R and total opening angle $\alpha \in (0,\pi )$, the radius of the inscribed circle is given by the formula

$$\begin{aligned} \frac{\sin (\alpha /2)}{1+\sin (\alpha /2)} R \ge \frac{1}{2} \sin (\alpha /2) R. \end{aligned}$$

Notice that $\frac{1}{2} \sin (\alpha /2) R$ increases with $\alpha \in (0,\pi )$ and R. Now for $\theta \in [0, \frac{\pi }{2}]$, we let $\varGamma (\theta )$ denote the opening angle of $T_i({\mathcal {C}}_\delta ^{{\varvec{v}}(\theta )}({\varvec{x}}_i))$ and define

$$\begin{aligned} R(\theta ):= \min _{\varphi \in [\theta -\phi , \theta +\phi ]}\delta \sqrt{\varrho \cos ^2(\varphi ) + \sin ^2(\varphi )} = \min _{\varphi \in [\theta -\phi , \theta +\phi ]}\delta \sqrt{\varrho + (1-\varrho ) \sin ^2(\varphi )}, \end{aligned}$$

then it is easy to see that a cone with radius $R(\theta )$ and total opening angle $\varGamma (\theta )$ is contained in $T_i({\mathcal {C}}_\delta ^{{\varvec{v}}(\theta )}({\varvec{x}}_i))$. Therefore we have

$$\begin{aligned} \min _{\theta \in [0, \frac{\pi }{2}]} r({\varvec{v}}(\theta ),{\varvec{x}}_i) \ge \min _{\theta \in [0, \frac{\pi }{2}]} \frac{1}{2} \sin (\varGamma (\theta )/2) R(\theta ). \end{aligned}$$

By calculation we have

$$\begin{aligned} \varGamma (\theta ) = \left\{ \begin{aligned} \arctan (\sqrt{\varrho ^{-1}} \tan (\theta + \phi )) - \arctan (\sqrt{\varrho ^{-1}} \tan (\theta - \phi )), \quad&\theta \in [0, \pi /2-\phi ), \\ \pi {+} \arctan (\sqrt{\varrho ^{-1}} \tan (\theta {+} \phi )) {-} \arctan (\sqrt{\varrho ^{-1}} \tan (\theta {-} \phi )), \quad&\theta \in ( \pi /2-\phi , \pi /2], \end{aligned} \right. \end{aligned}$$

and

$$\begin{aligned} R(\theta ) = \left\{ \begin{aligned}&\delta \sqrt{\varrho }, \quad&\theta \in [0, \phi ], \\&\delta \sqrt{\varrho + (1-\varrho ) \sin ^2(\theta - \phi )}, \quad&\theta \in [\phi , \pi /2]. \end{aligned} \right. \end{aligned}$$

For $\theta \in [0, \phi ]$, $\varGamma (\theta )$ decreases and $R(\theta ) = \delta \sqrt{\varrho }$, so

$$\begin{aligned} \min _{\theta \in [0,\phi ]} \frac{1}{2} \sin (\varGamma (\theta )/2) R(\theta ) = \frac{1}{2} \sin (\varGamma (\phi )/2) \delta \sqrt{\varrho } \ge \frac{\sin (\phi )}{2}\delta \sqrt{\varrho } \end{aligned}$$

where we have used $2\phi \le \varGamma (\phi )\le \pi /2$.

For $\theta \in [\pi /2 -2\phi , \pi /2]$, $\varGamma (\theta )$ decreases and $R(\theta )$ increases, so

$$\begin{aligned} \begin{aligned}&\min _{\theta \in [\pi /2 -2\phi , \pi /2] } \frac{1}{2} \sin (\varGamma (\theta )/2) R(\theta )\\&\quad \ \ge \frac{1}{2} \sin (\varGamma (\pi /2)/2) R(\pi /2 -2\phi ) \ge \frac{\delta }{2}\sin (\varGamma (\pi /2)/2) \sin (\pi /8) \\&\quad \ = \frac{\sin (\pi /8) \delta }{2} \frac{\sqrt{\varrho }}{\sqrt{\varrho + \tan ^2(\pi /2-\phi )}} \ge \frac{\sin (\pi /8)}{2\sqrt{1 + \tan ^2(3\pi /8)}} \delta \sqrt{\rho }. \end{aligned} \end{aligned}$$

where we have used $\phi \le \pi /8$ and $\varGamma (\pi /2) = 2 \arctan (\sqrt{\varrho } \cot (\pi /2-\phi ))$.

Now for $\theta \in [\phi , \pi /2-2\phi ]$, we use the formulas for $\varGamma (\theta )$ and $R(\theta )$ to compute $\frac{1}{2} \sin (\varGamma (\theta )/2) R(\theta )$. Denote $\alpha = \arctan (\sqrt{\varrho ^{-1}} \tan (\theta + \phi ))$ and $\beta = \arctan (\sqrt{\varrho ^{-1}} \tan (\theta - \phi ))$. Use the formula

$$\begin{aligned} \sin \left( \frac{\alpha -\beta }{2}\right) = \sqrt{\frac{1-\cos (\alpha -\beta )}{2}} = \sqrt{\frac{1-\cos (\alpha ) \cos (\beta ) - \sin (\alpha )\sin (\beta )}{2}}, \end{aligned}$$

and the fact that

$$\begin{aligned} \begin{aligned}&\sin (\alpha ) = \frac{\sqrt{\varrho ^{-1}} \tan (\theta + \phi )}{ \sqrt{1+\varrho ^{-1}\tan ^2(\theta +\phi )}},\; \cos (\alpha ) = \frac{1}{ \sqrt{1+\varrho ^{-1}\tan ^2(\theta +\phi )}}, \\&\sin (\beta ) = \frac{\sqrt{\varrho ^{-1}} \tan (\theta - \phi )}{ \sqrt{1+\varrho ^{-1}\tan ^2(\theta -\phi )}},\; \cos (\beta ) = \frac{1}{ \sqrt{1+\varrho ^{-1}\tan ^2(\theta -\phi )}}, \end{aligned} \end{aligned}$$

we can obtain the formula for $\frac{1}{2} \sin (\varGamma (\theta )/2) R(\theta )$ where $\theta \in [\phi , \pi /2-2\phi ]$. In particular, denote $g_1(\theta ) = \tan ^2(\theta - \phi )$, $g_2(\theta ) = \tan ^2(\theta + \phi )$, and $g_3(\theta )= \sin ^2(\theta -\phi )$, we fine

$$\begin{aligned} \begin{aligned}&\frac{1}{2} \sin (\varGamma (\theta )/2) R(\theta ) \\&\quad =\frac{\delta }{2\sqrt{2}} \sqrt{\frac{\sqrt{(\varrho +g_1(\theta )) (\varrho +g_2(\theta )) } -\varrho - \sqrt{g_1(\theta )g_2(\theta )} }{\sqrt{(\varrho +g_1(\theta )) (\varrho +g_2(\theta )) }} \left( \varrho + (1-\varrho ) g_3(\theta ))\right) } \\&\quad = \frac{\delta \sqrt{\rho }}{2} \sqrt{\frac{ \left( (g_1(\theta ))^{1/2}-(g_2(\theta ))^{1/2}\right) ^2 \left( \varrho + (1-\varrho ) g_3(\theta ))\right) }{\sqrt{(\varrho +g_1(\theta )) (\varrho +g_2(\theta )) } \left( \sqrt{(\varrho +g_1(\theta )) (\varrho +g_2(\theta )) } +\varrho + \sqrt{g_1(\theta )g_2(\theta )}\right) }} \\&\quad =: \frac{\delta \sqrt{\rho }}{2} G(\varrho , \theta ). \end{aligned} \end{aligned}$$

Notice that $G(\varrho , \theta )$ defined in the above is a continuous function on $(\varrho , \theta ) \in [0,1]\times [\phi , \pi /2-2\phi ]$. Therefore it attains a minimum value at some $(\varrho ^*, \theta ^*) \in [0,1]\times [\phi , \pi /2-2\phi ]$. Next we show that we must have $G(\varrho ^*, \theta ^*)>0$. Indeed, if $\varrho ^*>0$, then it is easy to see that $G(\varrho ^*, \theta ^*)>0$. Now if $\varrho ^*=0$, then

$$\begin{aligned} \begin{aligned} G(0, \theta )&= \sqrt{\frac{ \left( (g_1(\theta ))^{1/2}-(g_2(\theta ))^{1/2}\right) ^2 g_3(\theta ) }{2g_1(\theta ) g_2(\theta )}} \\&= \left( \tan (\theta + \phi ) - \tan (\theta - \phi )\right) \frac{\sin (\theta -\phi )}{2 \tan (\theta - \phi ) \tan (\theta + \phi ) } \\&= \left( \tan (\theta + \phi ) - \tan (\theta - \phi )\right) \frac{\cos (\theta -\phi )}{2 \tan (\theta + \phi ) }>0 \end{aligned} \end{aligned}$$

for any $\theta \in [\phi , \pi /2-2\phi ]$. Therefore, we can take

$$\begin{aligned} K(\phi ) = \min \left\{ \frac{\sin (\phi )}{2},\frac{\sin (\pi /8)}{2\sqrt{1 + \tan ^2(3\pi /8)}}, G(\varrho ^*, \theta ^*)/2 \right\} >0 \end{aligned}$$

for the claim to be true. $\square$

Proof of Theorem 3.2

Let $\lambda _j=\lambda _j({\varvec{x}}_i)$ denotes the jth smallest eigenvalue of $A({\varvec{x}}_i)$. For $d=2$, we apply Lemma A.2 with $\phi = \pi /8$ on a rescaled ellipse of $T_i(B_\delta ({\varvec{x}}_i))$, we get

$$\begin{aligned} \min _{\theta \in [0, \frac{\pi }{2}]} r({\varvec{v}}(\theta ),{\varvec{x}}_i) \ge C_1 \delta \sqrt{\lambda _2}\sqrt{\frac{\lambda _1}{\lambda _2}} =C_1 \delta \sqrt{\lambda _1}, \end{aligned}$$

where $C_1 = K(\pi /8)$.

Now consider $d=3$. First of all, we can assume without loss of generality that $\lambda _3=1$ by the method of rescaling. By the discussions at the beginning of Sect. 3.4, ${\mathcal {C}}^{{\varvec{v}}}_\delta ({\varvec{x}}_i)$ is a 3d cone with total opening angle $33.7^\circ$ for a given unit vector ${\varvec{v}}\in {\mathbb {R}}^3$. Let $P_{{\varvec{v}}} \subset {\mathbb {R}}^3$ be a 2d plane that contains the vector ${\varvec{v}}$, then we see that $P_{{\varvec{v}}} \cap B_\delta ({\varvec{x}}_i)$ is a circular domain and $P_{{\varvec{v}}} \cap {\mathcal {C}}^{{\varvec{v}}}_\delta ({\varvec{x}}_i)$ is a 2d cone with total opening angle $33.7^\circ$. With the transform $T_i$, we see that $T_i (P_{{\varvec{v}}} \cap B_\delta ({\varvec{x}}_i))$ is a 2d ellipse and $T_i( P_{{\varvec{v}}} \cap {\mathcal {C}}^{{\varvec{v}}}_\delta ({\varvec{x}}_i))$ is a section of the ellipse. Therefore, the 2d calculations in Lemma A.2 can be applied. Notice that for each $P_{{\varvec{v}}}$, there exists $\rho _1$ and $\rho _2$ with $\lambda _1 \le \rho _1 \le \rho _2\le 1$ such that the lengths of the semi-axes of the ellipse $T_i (P_{{\varvec{v}}} \cap B_\delta ({\varvec{x}}_i))$ are given by $\sqrt{\rho _1}$ and $\sqrt{\rho _2}$. We can then rescale the the ellipse $T_i (P_{{\varvec{v}}} \cap B_\delta ({\varvec{x}}_i))$ and use Lemma A.2 with $\phi =33.7^\circ /2$ we see that the radius of the inscribed circle in $T_i( P_{{\varvec{v}}} \cap {\mathcal {C}}^{{\varvec{v}}}_\delta ({\varvec{x}}_i))$ has a lower bound

$$\begin{aligned} {\tilde{C}} (\delta \sqrt{\rho _2}) \sqrt{\frac{\rho _1}{\rho _2}} \ge {\tilde{C}} \delta \sqrt{\lambda _1}, \end{aligned}$$

where ${\tilde{C}} =K(33.7^\circ /2)>0$. Notice that $P_{{\varvec{v}}}$ is an arbitrary plane that contains ${\varvec{v}}$, and in addition, the average length of the line segments that connect $T_i({\varvec{x}}_i)$ and the edge of $T_i( P_{{\varvec{v}}} \cap {\mathcal {C}}^{{\varvec{v}}}_\delta ({\varvec{x}}_i))$ is at the same scale for different plane $P_{{\varvec{v}}}$. Therefore, it can be shown that there exists $C_2>0$ such that

$$\begin{aligned} \min _{\theta \in [0, \frac{\pi }{2}]} r({\varvec{v}}(\theta ),{\varvec{x}}_i) \ge C_2 \delta \sqrt{\lambda _1}. \end{aligned}$$

At last, by Lemma A.1 and the above discussions, there exists $C=C(d)>0$ such that whenever $h\le C \delta \sqrt{\lambda _1({\varvec{x}}_i)}$, $S_{\delta , h, 2}({\varvec{x}}_i)$ and ${\overline{S}}_{\delta , h, 2}({\varvec{x}}_i)$ are not empty. Therefore taking $c(d)=1/C(d)$ the conclusion is true. By our assumption, $\varrho \le \lambda _1({\varvec{x}}_i)$ for all ${\varvec{x}}_i$, and therefore $\delta \ge c h (\varrho )^{-1/2}$ implies the existence of positive stencils. $\square$

Point Cloud Generation and Adjustment

In this appendix, we discuss the generation of proper point clouds that satisfy Definition 4.1. We first initialize a random point cloud using the Quasi-Monte Carlo method [53] (see Fig 16a), then adjust this point cloud to make it proper. Adjustment contains three steps in each loop:

Step 1. add points until h satisfies condition (i) (see Fig. 16b);

Step 2. map points until $\kappa$ satisfies condition (iii) (see Fig. 16c);

Step 3. merge points until $\zeta$ satisfies condition (ii) (see Fig. 16d).

Adjustment stops when the conditions in Definition 4.1 are satisfied. In practice, it usually takes a few loops to make the point cloud proper.

Fig. 16

The process of proper point cloud generation. The grey circular domain is $\varOmega$. The square domain is a bounding box of $\varOmega _{\delta _0}$ for some $\delta _0>0$. A Initialize a point cloud by the Quasi-Monte Carlo method. B Use the Voronoi diagram [66, 67] for the calculation of the fill distance, and then add the green points to the point cloud so that h satisfies condition (i). C Map points near the boundary of $\varOmega$ to the interior so that $\kappa$ satisfies condition (iii). D Merge points whose distances are less than $2 c_\zeta h$ so that $\zeta$ satisfies condition (ii). Notice that after merging of points, the fill distance may increase, as a result, the adjustment loop may be needed again

Remark B.1

Since the domain $\varOmega _\delta$ may be irregular, in practice, we always generate point clouds on a larger bounding box of $\varOmega _{\delta _0}$ for $\delta \in (0,\delta _0]$, as indicated by Fig. 16a–d. The formulas for the fill distance in Eq. (16) and condition (i) Definition 4.1 are then modified accordingly. In practice, $\delta _0$ is determined by the largest discretization parameter as well as the ratio $\varrho$ as indicated by neighborhood estimate in Sects. 3.4 and 4.3.

Remark B.2

One can also use the Quasi-Monte Carlo method for generating the initial point cloud and perform only step 2 without the adding and merging steps. Our adjustment algorithm provides explicit control over the fill distance and the separation distance and it leads to smaller fill distances for the same number of interior points compared with point clouds without adjustment. This is a trade-off situation, namely, one can save memory by using extra time adjusting the initialized point cloud, or vice versa.

Inscribed Circle Search Algorithm

We show the details of finding the radius of the inscribed circle in $T_{i}({\mathcal {C}}_{1}^{{\varvec{v}}}({\varvec{x}}_{i}))$ that contained in an ellipse given by $x^{2} / \varrho + y^{2} = 1$. Let $(x_{0}, y_{0})$ be the center of the inscribed circle with radius r, then it can only sit on the angle bisector of $T_{i}({\mathcal {C}}_{1}^{{\varvec{v}}}({\varvec{x}}_{i}))$. Let $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ represent the two corner points of $T_{i}({\mathcal {C}}_{1}^{{\varvec{v}}}({\varvec{x}}_{i}))$. Then by some elementary calculations, there exists some $t>0$ such that

$$\begin{aligned} (x_{0}, y_{0}) = t \left( \sqrt{x_{2}^{2} + y_{2}^{2}} x_{1} + \sqrt{x_{1}^{2} + y_{1}^{2}} x_{2}, \sqrt{x_{2}^{2} + y_{2}^{2}} y_{1} + \sqrt{x_{1}^{2} + y_{1}^{2}} y_{2} \right) \end{aligned}$$

and

$$\begin{aligned} r = t |x_{1} y_{2} - y_{1} x_{2}|. \end{aligned}$$

To determine $t>0$, we find the closest point to the circle center on the ellipse and choose $t>0$ such that the point is also on the circle. The closest point to the circle center on the ellipse can be found by the minimization problem

$$\begin{aligned} \min _{\theta } \Vert (\sqrt{\varrho } \cos (\theta ), \sin (\theta )) - (x_{0}, y_{0}) \Vert _{2}^{2}, \end{aligned}$$

which can be solved by, e.g., Newton’s method. At last, one may use a numerical method, e.g., the bisection method, to determine $t>0$ such that the point is also on the circle.