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Symmetry/slip wall boundary condition in potential flow around a cylinder

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Old   January 23, 2020, 16:02
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ilia Nikiforov
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Quote:
Originally Posted by FMDenaro View Post
That means


Div v =0


dv/dt + div (vv) + Grad phi = 0

That is a first order hyperbolic system of equation in the velocity field, the "pressure" phi being only a lagrangian multiplier (just apply the curl to the momentum). Again, only one condition is required on the boundary. Since the fluid is ideal, you cannot set the physical non-slip condition and only the normal velocity is prescribed to fulfill the continuity equation.
To have two BCs you need a second order equation, that is the momentum equation with the diffusive term in the RHS.
Thank you very much for your help, looks like I need to review systems of PDEs in general, thank you for pointing me in the right direction.
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