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Peter Kravchuk has already given a good answer. Here we will follow the programming hint given in the Exercise 1.6. How would one program this minimization problem? By discretization. So the positions ${\bf r}_n$ live on discrete times $$t_n~=~n\Delta t,\qquad\Delta t ~:=~\frac{T}{N},\qquad n\in\{0,\ldots,N\};$$ and velocities are discretized as e.g. ...

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If your action only depends on first time derivatives, it is then not required for the trajectory to have second time derivative -- i.e. an abrupt change in velocity does not by itself give a contribution to the action. In other words, there is no penalty for changing your speed instantaneously. It then means that you can ignore the boundary values for ...

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Ok, it was obviously not apparent to people reading this question - there is a perfectly conducting, infinite cylinder (with a circular cross-section) at some distance from the plane of separation. After some research, I am convinced there is not analytic solution to the problem, and instead used numeric methods to receive an approximate solution.

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The way I would proceed is first to exchange the spherical electrode by a point charge at the centre : $$Q = \frac{VR}{K}$$ With : $$K = \frac{1}{4 \pi \epsilon_0}$$ Then that charge would have an image across the dielectric surface, the method has been described quite well in introduction to electrodynamics or here The image charge will itself ...

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On the one hand, you can't solve for the magnetic field without appropriate boundary conditions (e.g. there could always be an incoming electromagnetic wave that hasn't yet impinged on your cylinder). On the other hand if you have a fixed charge and current distribution you can always use Jefimenko's equations to find a solution to Maxwell's equations, and ...

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I) Well, $r=0$ is the boundary of a $d$-dimensional spherical coordinate patch, i.e. an artifact of our choice of coordinate system, but $r=0$ does not correspond to a physical boundary per se, other than what we can deduce from the TISE. The mantra is that a boundary condition (for finite $r$) should only be imposed if it is a consequence of the TISE. See ...

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