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I have a question on the use of moving mesh to solve the inviscid euler equations. I have solved the following equations: $$\frac{\partial}{\partial t}\left[\begin{array}{c} \rho\\ \rho u \end{array}\right]=\frac{\partial}{\partial x}\left[\begin{array}{c} \rho(u-w)\\ \rho u(u-w)+p \end{array}\right],$$ where p is a linear function of $p=f(\rho)$ and $w$ is the mesh velocity.

I have used the integral form of FVM, treated the right hand side in its conservative form and have evaluated the results at the cell centres. At the left BC I have wall conditions, so $u_L = w_L = 0$ and at the right hand BC, $u_R = w_R = f(t)$. I have used linear interpolation for the other boundary terms. I am moving the mesh using $dx/dt = w$.

I know that moving the mesh should not effect the solution, however, my results seem to be effected since for instance for the $\rho u (u-w)$ term of the momentum equation, on the right hand side, when $w=0$, I have $\rho_R u_R u_R$ but when $w\neq 0$ I will have $\rho_R u_R (0-0) = 0$. Can someone please let me know if I am doing the right thing here, and if it is possible to avoid this issue somehow? Thank you.

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  • $\begingroup$ Does anyone have any idea???? $\endgroup$ – Hooman Sep 13 '15 at 0:38

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