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I'm considering the compressible euler equations without any heat addition, body forces, energy addition, etc. In other words, i'm using the following equations with the assumption of an ideal gas.

$$ \begin{split} \frac{\partial{\rho}}{\partial t} + \nabla \cdot (\rho\mathbf{v}) &= 0,\\ \frac{\partial{(\rho\mathbf{v})}}{\partial t} + \nabla(p+\rho|\mathbf{v}|^2) &= 0,\\ \frac{\partial{(\rho E)}}{\partial t} + \nabla \cdot ((p + \rho E) \mathbf{v}) &=0. \end{split} $$

If I have a two-dimensional domain with inflow from the left and properly defined boundary conditions that are independent of time, will the solution of these equations always converge to a steady solution? Or is it possible that the solution exhibits some kind of periodicity similar to the Navier-Stokes equations?

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  • $\begingroup$ Are you using them with an equation of state or assumptions on pressure or energy? $\endgroup$
    – kricheli
    Jun 20, 2023 at 10:15
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    $\begingroup$ @kricheli i'm assuming an ideal gas, i've added it to the question now $\endgroup$ Jun 20, 2023 at 10:29

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"Do the Euler equations always converge to a steady solution if the boundary conditions are steady?"

I would say no. To see why, consider a situation where the system is not subject to external driving forces and/or moving boundaries: the home experiment would be stirring your cup of tea to prepare the initial condition and then letting the system evolve. We first consider Navier-Stokes, then the Euler equations.

Navier-Stokes - Navier-Stokes equations contain the idea that "things diffuse" (this is implemented by the Laplacian, see this answer). Therefore, the initial condition (possibly complex and turbulent, i.e. with gradients on many scales) will tend to become more and more homogeneous. The final homogeneous time-independent state is the equilibrium state (the tea in your cup finally settles and no macroscopic motion is visible). This macroscopic equilibrium is achieved through the so-called "energy cascade": kinetic energy in the macroscopic flow cascades down to smaller and smaller scales till it is dissipated at the "viscous scale". Structures (i.e. eddies, velocity and density gradients) smaller than the viscous timescale are "dissipated" in the sense that they are smoothed.

Therefore: 1) pump energy in the macroscopic flow with your spoon. 2) energy cascades down to smaller scales. 3) at the viscous scale, energy is "lost" in the microscopic motion of particles (that is not treated within Navier Stokes, this gives rise to the feeling that "energy is lost", even though it is obviously still there: to see it you have to model also the internal energy of a fluid element, i.e. account for frictional heating). 4) this gives rise to the "homogenization" of the initial state and, eventually, the viscous fluid relaxes to a time-independent stable equilibrium configuration.

Euler - Now consider the situation where you neglect viscosity: if you remove the diffusion terms in Navier Stokes, then you end up with the Euler equation. Now, the "viscous scale" is zero: instead of homogenization, you will end up with a sort of fractal solution, where the flow structures pumped at large scales in the initial condition give rise to finer and finer structures but are never dissipated.

Note: There are several integrals of motion associated with the Euler equation, like helicity. Therefore, if you start with an initial condition with a certain non-zero helicity, this automatically tells you that you will never achieve a static equilibrium. Still, it may be possible to think of complex flows that are "independent of time" or "periodic flows" that are reached at large enough times. This means that there would be periodic attractors for the Euler dynamics. This seems to be in contrast with the idea of a cascade.

Practically: When you solve the Euler equations, you do that on a grid. The grid provides you with an ultimate length scale, so in the end, you will never see the fractal structure in a simulation (or, better, you can see it evolving up to a certain scale defined by the grid).

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