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On the unmapped tent pitching for the heterogeneous wave equation

Marcella Bonazzoli, Gabriele Ciaramella, Ilario Mazzieri

TL;DR

This paper extends Unmapped Tent Pitching (UTP) to a heterogeneous 1D wave equation with piecewise-constant speed, and analyzes how to optimally decompose the space–time domain for parallel solution. By comparing configurations across regions with speeds c1 and c2, it demonstrates that using identical subdomain dimensions in both regions, governed by the maximum speed c_max, minimizes time to solution. The key result is that the optimal choice is L2 = L1 and H2 = H1 (i.e., m2 = m1), leading to a total computational cost T_tot = L^2/(2 m1 c1) and allowing the heterogeneous case to proceed with the same footprint as the homogeneous case by solving heterogeneous local problems with c(x). This provides a principled guideline for applying UTP to heterogeneous media and informs extension to higher spatial dimensions.

Abstract

The Unmapped Tent Pitching (UTP) algorithm is a space--time domain decomposition method for the parallel solution of hyperbolic problems. It was originally introduced for the homogeneous one-dimensional wave equation in [Ciaramella, Gander, Mazzieri, 2024]. UTP is inspired by the Mapped Tent Pitching (MTP) algorithm [Gopalakrishnan, Sch{ö}berl, Wintersteiger, 2017], which constructs the solution by iteratively building polytopal space--time subdomains, referred to as tents. In MTP, each physical tent is mapped onto a space--time rectangle, where local problems are solved before being mapped back to the original domain. In contrast, UTP avoids the nonlinear and potentially singular mapping step by computing the solution directly on a physical space--time rectangle that contains the tent, at the expense of redundant computations in the region outside the tent. In this work, we investigate several strategies to extend UTP to heterogeneous media, where the wave propagation speed is piecewise constant over two subregions of the domain. Among the considered approaches, the most efficient in terms of computational time is the one employing space--time subdomains with identical spatial and temporal dimensions in both regions, determined by the maximum propagation speed.

On the unmapped tent pitching for the heterogeneous wave equation

TL;DR

This paper extends Unmapped Tent Pitching (UTP) to a heterogeneous 1D wave equation with piecewise-constant speed, and analyzes how to optimally decompose the space–time domain for parallel solution. By comparing configurations across regions with speeds c1 and c2, it demonstrates that using identical subdomain dimensions in both regions, governed by the maximum speed c_max, minimizes time to solution. The key result is that the optimal choice is L2 = L1 and H2 = H1 (i.e., m2 = m1), leading to a total computational cost T_tot = L^2/(2 m1 c1) and allowing the heterogeneous case to proceed with the same footprint as the homogeneous case by solving heterogeneous local problems with c(x). This provides a principled guideline for applying UTP to heterogeneous media and informs extension to higher spatial dimensions.

Abstract

The Unmapped Tent Pitching (UTP) algorithm is a space--time domain decomposition method for the parallel solution of hyperbolic problems. It was originally introduced for the homogeneous one-dimensional wave equation in [Ciaramella, Gander, Mazzieri, 2024]. UTP is inspired by the Mapped Tent Pitching (MTP) algorithm [Gopalakrishnan, Sch{ö}berl, Wintersteiger, 2017], which constructs the solution by iteratively building polytopal space--time subdomains, referred to as tents. In MTP, each physical tent is mapped onto a space--time rectangle, where local problems are solved before being mapped back to the original domain. In contrast, UTP avoids the nonlinear and potentially singular mapping step by computing the solution directly on a physical space--time rectangle that contains the tent, at the expense of redundant computations in the region outside the tent. In this work, we investigate several strategies to extend UTP to heterogeneous media, where the wave propagation speed is piecewise constant over two subregions of the domain. Among the considered approaches, the most efficient in terms of computational time is the one employing space--time subdomains with identical spatial and temporal dimensions in both regions, determined by the maximum propagation speed.
Paper Structure (4 sections, 7 equations, 7 figures, 1 algorithm)

This paper contains 4 sections, 7 equations, 7 figures, 1 algorithm.

Figures (7)

  • Figure 1: Iterations of UTP with $N=9$ in the homogeneous case (left), and in the heterogeneous case with the choice $H_2 = H_1$, $m_2=m_1=2$ (right). Here $c_1=2c_2$. Red and black rectangles are the space-time subdomains constructed by UTP at odd and even iterations. We display iterations $k=1, k=2$ at the top, iterations $k=3, k=4$ in the middle, iteration $k=5$ at the bottom of the figure. The hatched regions are the portions of the domain where UTP computes the exact solution.
  • Figure 2: The possible choices for the length $L_2 = L/(2m_2)$, with $m_2 \in \mathbb{N}\backslash\{0\}$, and the height $H_2$ of the space-time rectangles in the right region $(L/2,L)$, with respect to the reference rectangle of length $L_1 = L/(2m_1)$ and height $H_1 = L_1/(2c_1)$ in the left region $(0,L/2]$. We define $m_2^*$ such that $m_2^*/m_1 = c_1/c_2$, so $m_2^* > m_1$ for the present case $c_1>c_2$. The hatched region is the portion where the exact solution can be computed following the characteristic lines.
  • Figure 3: Iterations of UTP with the choice $H_2 = H_1$, $1=m_2 < m_1=2$ (left), and the choice $H_2 = H_1$, $4=m_2=m_2^* \,(>m_1=2)$ (right).
  • Figure 4: First eight iterations of UTP with the choice $H_2 = H_1$, $8=m_2>m_2^*=4 \,(>m_1=2)$. We display iterations $k=1, k=2$ at the top left, iterations $k=3, k=4$ at the top right, iterations $k=5, k=6$ at the bottom left, and iterations $k=7, k=8$ at the bottom right of the figure.
  • Figure 5: Iterations of UTP with the choice $H_2 > H_1$, $m_2=m_1=2$, case $(a)$ (left), and the choice $H_2 > H_1$, $m_2=m_1=2$, case $(b)$ (right).
  • ...and 2 more figures