Parallel analysis system for free-surface flow using MPS method with explicitly represented polygon wall boundary model

Abstract

This study develops a parallel solver of free-surface flow based on a mesh-free particle method, the explicit MPS method, with polygon boundary representation. We adopt the explicitly represented polygon (ERP) wall boundary model, which expresses wall boundaries as arbitrarily shaped triangular polygons. A bucket-based domain decomposition algorithm for dynamic load balancing is expanded to the polygon-based computation of the ERP model. The validity and parallel efficiency of the developed solver are tested by analyzing a dam break problem, and the numerical results are compared with experimental results. Our developed solver can simplify the evaluation of the integrity of coastal structures since it can be easily connected to finite element analyses of the structures.

Appendix: Computation of the closest point on polygons

We take a triangle ABC that consists of points A, B, and C. The closest point on ABC corresponding to polygon k and particle i is computed as follows.

We define the constants $v_A$, $v_B$, and $v_C$ as

$$\begin{aligned} v_A&= d_3 d_6 - d_5 d_4, \\ v_B&= d_5 d_2 - d_1 d_6, \\ v_C&= d_1 d_4 - d_3 d_2, \end{aligned}$$

where $d_i(i = 1, 2, \cdots , 6)$ are the following:

$$\begin{aligned} d_1 = \varvec{x}_{AB} \cdot \varvec{x}_{Ai},&\qquad d_2 = \varvec{x}_{AC} \cdot \varvec{x}_{Ai}, \qquad \\ d_3 = \varvec{x}_{AB} \cdot \varvec{x}_{Bi},&\qquad d_4 = \varvec{x}_{AC} \cdot \varvec{x}_{Bi}, \qquad \\ d_5 = \varvec{x}_{AB} \cdot \varvec{x}_{Ci},&\qquad d_6 = \varvec{x}_{AC} \cdot \varvec{x}_{Ci}. \end{aligned}$$

• 1. If

  $$\begin{aligned} d_1 \le 0 \ \ \wedge \ \ d_2 \le 0, \end{aligned}$$

  point i orthogonally projects outside ABC within the Voronoi region of A. Then, the closest point is given by A, that is,

  $$\begin{aligned} \varvec{x}^{\mathrm {near}}_{i,k} = \varvec{x}_{A}, \end{aligned}$$

  and the point in barycentric coordinates is given by

  $$\begin{aligned} \{ \xi , \eta \} = \{ 1, 0 \}. \end{aligned}$$

• 2. If

  $$\begin{aligned} d_3 \ge 0 \ \ \wedge \ \ d_4 \le d_3, \end{aligned}$$

  point i projects outside ABC within the Voronoi region of B. Then, the closest point is given by B, that is,

  $$\begin{aligned} \varvec{x}^{\mathrm {near}}_{i,k} = \varvec{x}_{B}, \end{aligned}$$

  and the point in barycentric coordinates is given by

  $$\begin{aligned} \{ \xi , \eta \} = \{ 0, 1 \}. \end{aligned}$$

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• 3. If

  $$\begin{aligned} v_C \le 0 \ \ \wedge \ \ d_1 \ge 0 \ \ \wedge \ \ d_3 \le 0, \end{aligned}$$

  point i projects outside ABC within the edge region of AB. Then, the closest point is on the edge AB and is given by

  $$\begin{aligned} \varvec{x}^{\mathrm {near}}_{i,k} = \varvec{x}_{A} + \frac{d_1}{d_1 - d_3} \varvec{x}_{AB} \end{aligned}$$

  and the point in barycentric coordinates is given by

  $$\begin{aligned} \{ \xi , \eta \} = \left\{ 1- \frac{d_1}{d_1 - d_3}, \frac{d_1}{d_1 - d_3} \right\} . \end{aligned}$$

• 4. If

  $$\begin{aligned} d_6 \ge 0 \ \ \wedge \ \ d_5 \le d_6, \end{aligned}$$

  point i projects outside ABC within the Voronoi region of C. Then, the closest point is given by C, that is,

  $$\begin{aligned} \varvec{x}^{\mathrm {near}}_{i,k} = \varvec{x}_{C} \end{aligned}$$

  and the point in barycentric coordinates is given by

  $$\begin{aligned} \{ \xi , \eta \} = \{0, 0 \}. \end{aligned}$$

• 5. If

  $$\begin{aligned} v_B \le 0 \ \ \wedge \ \ d_2 \ge 0 \ \ \wedge \ \ d_6 \le 0, \end{aligned}$$

  point i projects outside ABC within the edge region of AC. Then, the closest point is on the edge AC and is given by

  $$\begin{aligned} \varvec{x}^{\mathrm {near}}_{i,k} = \varvec{x}_{A} + \frac{d_2}{d_2 - d_6} \varvec{x}_{AC} \end{aligned}$$

  and the point in barycentric coordinates is given by

  $$\begin{aligned} \{ \xi , \eta \} = \left\{ 1- \frac{d_2}{d_2 - d_6}, 0 \right\} . \end{aligned}$$

• 6. If

  $$\begin{aligned} v_A \le 0 \ \ \wedge \ \ d_4 \ge d_3 \ \ \wedge \ \ d_5 \ge d_6, \end{aligned}$$

  point i projects outside ABC within the edge region of BC. Then, the closest point is on the edge BC and is given by

  $$\begin{aligned}&\varvec{x}^{\mathrm {near}}_{i,k} = \varvec{x}_{B} + \frac{(d_4-d_3)}{(d_4-d_3) + (d_5-d_6)} \varvec{x}_{BC} \\&\{ \xi , \eta \} = \left\{ 0 \ , 1- \frac{(d_4-d_3)}{(d_4-d_3) + (d_5-d_6)} \right\} . \end{aligned}$$

• 7. If none of the above conditions hold, i orthogonally projects inside ABC. Then the closest point is in the triangle ABC and is given by

  $$\begin{aligned} \begin{aligned} \varvec{x}^{\mathrm {near}}_{i,k}&= \frac{v_A}{v_A+v_B+v_C} \varvec{x}_A + \frac{v_B}{v_A+v_B+v_C} \varvec{x}_B \\&\qquad + \frac{v_C}{v_A+v_B+v_C} \varvec{x}_C \end{aligned} \end{aligned}$$

  and the point in barycentric coordinates is given by

  $$\begin{aligned} \{ \xi , \eta \} = \left\{ \frac{v_A}{v_A+v_B+v_C}, \ \frac{v_B}{v_A+v_B+v_C} \right\} . \end{aligned}$$