# Euler equations & expansion shocks

Posted April 27, 2009 at 19:23 by technophobe

I'm specifically talking about shocks occuring on an airfoil although I guess the theory would apply generally. I suppose I'm really asking about the reversibility of the Euler equations. I understand that expansion shocks are non-physical but I'm reading that they are a possible solution of the Euler equations. Essentially the Euler equations don't 'know' that the expansion shocks are unphysical and so in some applications an entropy boundary condition is required (which basically rules out expansion shocks)

I have been told that the Euler equations are reversible whereas the Navier stokes equations are not. The idea seems to be that if we take the x-momentum equation (assuming u,v and w all positive) and reverse the sign of all the velocity components (i.e. the flow is running backwards) then the temporal term becomes negative - because time is running in reverse. The velocity components all appear in pairs so the u*du/dx etc have the same sign as before.

According to my colleague, if we solve this new equation we'd get the same solution as if we'd not reversed the flow velocity. The only difference being that the flow is going in the opposite direction. The same doesn't apply to the NS equations as they contain a lone u-term that does change sign under this reversal.

Apparently this is why the Euler equations can admit expansion shocks (because we can reverse time and still get the solution). The Navier Stokes equations can't admit such shocks.

My problem with this idea is: what happens to the dp/dx term? under time reversal this would surely remain unchanged and so solving the reversed equations would surely give a

Any info on this matter would be greatly appreciated.

T

I have been told that the Euler equations are reversible whereas the Navier stokes equations are not. The idea seems to be that if we take the x-momentum equation (assuming u,v and w all positive) and reverse the sign of all the velocity components (i.e. the flow is running backwards) then the temporal term becomes negative - because time is running in reverse. The velocity components all appear in pairs so the u*du/dx etc have the same sign as before.

According to my colleague, if we solve this new equation we'd get the same solution as if we'd not reversed the flow velocity. The only difference being that the flow is going in the opposite direction. The same doesn't apply to the NS equations as they contain a lone u-term that does change sign under this reversal.

Apparently this is why the Euler equations can admit expansion shocks (because we can reverse time and still get the solution). The Navier Stokes equations can't admit such shocks.

My problem with this idea is: what happens to the dp/dx term? under time reversal this would surely remain unchanged and so solving the reversed equations would surely give a

*different*solution to the standard form?Any info on this matter would be greatly appreciated.

T

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