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 Aug28 comment Solving the 2D Poisson equation with variable boundary location Wow, Awesome, thanks! Minor correction, you missed a minus sign in $A_1$ i.e. $A_1=\frac{C}{4}\frac{b^2-a^2}{a^2+b^2}$ Aug28 comment Solving the 2D Poisson equation with variable boundary location I do see that they indeed use the elliptic coordinates that @rajb245 mentioned so that seems to be the right route Aug28 comment Solving the 2D Poisson equation with variable boundary location @Semiclassical I can access the paper behind the paywall, but to be honest I don't exactly get what I am looking at. Aug26 comment Solving the 2D Poisson equation with variable boundary location @rajb245 Ah, ok. Didn't know that coordinate system. I'm curious how it goes! Aug25 comment Solving the 2D Poisson equation with variable boundary location @rajb245 Terrific thanks. Note that the bottom part of my question contains a transformation of (1) to (what I would call) elliptical coordinates. Might save you some work?! Aug25 comment Solving the 2D Poisson equation with variable boundary location @Semiclassical Yes, that is indeed what I am looking for. Aug20 comment Transforming the Laplace operator from Polar to Cartesian coordinates @BeniBogosel Isn't the mixed derivative supposed to have a factor 2? Aug19 comment Solving the 2D Poisson equation with variable boundary location @Dmoreno ok, I think I get it now. Thanks! I actually know roughly what the solution to this problem should look like (because of the physical shape it represents) and I know that there isn't a singularity at $r=0$, but I will work with the mathematically correct formulation you propose! Aug19 comment Solving the 2D Poisson equation with variable boundary location @Dmoreno I will certainly try the other choice of coordinates, indeed that would turn the boundary condition simply into $r=r_0=1$. I'm not sure I completely understand your second point, I can see that $r=0$ is indeed a singular point in the differential equation, but would the boundary condition be something like $\lim_{r\to0}$ $z_r\to0$ (sorry about the notation, don't quite know how to write that) ? Aug14 comment Are there any surfaces that contain both positive and negative Gaussian curvature? Could you clarify what you mean with k1 and k2?! --- And just to clarify my own wording: with inside I mean at the side of the torus inside the hole, I do not mean actually inside the 'tube' of the torus Aug7 comment What is this semicircle-like shape called? @Lucian Smiley?! Aug6 comment What is this semicircle-like shape called? @Semiclassical cutting the major axis of the stadium would result in what is often called an extended or elongated semicircle, so perhaps semi-stadium is not that bad?! Aug2 comment Area of the polygon formed by cutting a cube with a plane Great, thank you! Aug2 comment Area of the polygon formed by cutting a cube with a plane Terrific explanation! Quick question: is there an easy way to deal with the case for which $m_i=0$ occurs? Or should I just compute the limit using l'hopital's rule? Aug2 comment Area of the polygon formed by cutting a cube with a plane @CalvinLin I will certainly try that, thanks! I do wonder whether that easily generalizes to polygons with a different number of vertices (due to a different cut of the box) Aug2 comment Area of the polygon formed by cutting a cube with a plane @StephenNand-Lal Good point, completely forgot the word for it. Yes I do, edited it. May8 comment Integration and differentiation of an approximation to a function - order of approximation So indeed the order of the approximation changes with integration/differentation?! Interesting, I wasn't expecting that! Mar13 comment Scaling a function with two 'asymptotes' of which one is non-constant The slope would be fixed if that's what you mean. Essentially it will be the same as the graph in my question but with all curves having (all the same) slope $dy/dx\neq0$ at $x=0$ instead of the current case with $dy/dx=0$ at $x=0$ Mar13 comment Scaling a function with two 'asymptotes' of which one is non-constant Just a quick follow-up, if the lefthand asymptote is also a linear function instead of a constant, could I still apply a similar transformation?! Mar12 comment Scaling a function with two 'asymptotes' of which one is non-constant I was indeed looking for a linear transformation. Sorry, forgot to mention that. Thanks for the answer! - Just for future reference: the choice of $t$ to map everything on the red curve in my case would be $t=(1-y_{as})/(y_0-y_{as})$ where $y_{as}$ is the value of the horizontal asymptote of the given function