Recent content by kmot

Yes. It's a density depending on pressure.

g is a generic function of which we know just that it depends only on F(X) grad(F(x)) is gradient of F(x)

I've found yet another way to prove this. Convert line integral to surface integral using Stokes theorem: ##\oint{\frac{1}{\rho} \cdot \nabla p \, dl } = \iint_{A}{\nabla \times \left( \frac{1}{\rho} \cdot \nabla p \right) \cdot \vec{n}\, dA } = \iint_{A}{\left(\nabla\frac{1}{\rho}\right) \times...

That's wonderful. Thank you. OK, Kelvin theorem doesn't say that per se, but it is implied. It says: For barotropic fluid, i.e. fluid where ##\rho## is a function of only ##p## ##\frac{D\Gamma}{Dt} = \frac{D}{Dt}(\oint \vec{v} \,dl ) = 0## After some manipulations, mainly inserting Newton's...

This problem comes from fluid dynamics where Kelvin circulation theorem states, that if density "rho" is a function of only pressure "p", then closed line integral of grad(p) / rho(p) equals zero. It seems so trivial, so that no one ever gives reason for this claim. When trying to solve it...

I found the solution. Here are the intermediate steps

Hi, In a text describing solution to linearized shallow water equations, I am not able to move forward. It's a 1 dimensional shallow water setup. There is a steady state (velocity) and (height of free surface). On top of this steady state there are u' and h' as disturbances. The goal is to...