Delta Shock as Free Piston in Pressureless Euler Flows

TL;DR

This paper demonstrates that delta shocks in pressureless Euler flows can be understood as free pistons, simplifying the analysis of fluid-solid interactions by reducing complex boundary problems to a single ODE.

Contribution

It establishes the equivalence between delta shocks and free pistons in pressureless Euler equations using three different approaches, providing new insights into their physics and simplifying related problems.

Findings

Delta shocks are equivalent to free pistons in pressureless Euler flows.

The ODE of the piston trajectory can be derived from three different perspectives.

The equivalence simplifies the analysis of fluid-solid interaction problems.

Abstract

We establish the equivalence of free piston and delta shock, for the one-space-dimensional pressureless compressible Euler equations. The delta shock appearing in the singular Riemann problem is exactly the piston that may move freely forward or backward in a straight tube, driven by the pressureless Euler flows on two sides of it in the tube. This result not only helps to understand the physics of the somewhat mysterious delta shocks, but also provides a way to reduce the fluid-solid interaction problem, which consists of several initial-boundary value problems coupled with moving boundaries, to a simpler Cauchy problem. We show the equivalence from three different perspectives. The first one is from the sticky particles, and derives the ordinary differential equation (ODE) of the trajectory of the piston by a straightforward application of conservation law of momentum, which is physically simple and clear. The second one is to study a coupled initial-boundary value problem of pressureless Euler equations, with the piston as a moving boundary following the Newton's second law. It depends on a concept of Radon measure solutions of initial-boundary value problems of the compressible Euler equations which enables us to calculate the force on the piston given by the flow. The last one is to solve directly the singular Riemann problem and obtain the ODE of delta shock by the generalized Rankine-Hugoniot conditions. All the three methods lead to the same ODE.