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By imposing periodic boundary conditions you end up with the 3D version of the particle on a ring (particle on a 3-torus?). If we drop down to 1D to keep things simple, then the difference between the ring and the 1D box is that on a ring we have two waves propagating round the ring in opposite directions e.g. one clockwise and one anti-clockwise. The two ...

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The open boundary condition means, as stated in the question, that at the boundary no force acts on the end of the string in the direction of elongation. As the tip of the string has infinitesimal mass, we can argue as if we were considering conditions for static equilibrium (if the forces caused by the string would differ from the forces caused by the wall ...

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First Part. No. The vortex is caused by some internal forces of the fluid. It can't end inside the fluid. (Helmholtz's theorem) Some rotation can, and will be transferred like in any surface contact, but this will never create any vortex. Theoretically it might be possible if you have some thin layer of less dense fluid above the vortex fluid. But in case ...

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$B$ actually behaves as you explain, however there's a problem with $H$. You say: "However, the field lines must also curl around and meet the top surface of the magnet, where $H$ will therefore need to point in the negative z direction." Why the field lines need to meet the top surface of the magnet? Outside the magnet $B$ and $H$ are proportional so ...

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This is, unfortunately, not a simple task in general. My experience on non-reflecting boundary conditions is for the Navier-Stokes equations, but you should be able to do a similar approach for your system. As you noted, a fixed boundary ($u=0$) will lead to one type of wave while a free boundary ($\partial u/\partial x = 0$) leads to another wave. What you ...

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