Pragabhava
Reputation
3,816
Next privilege 5,000 Rep.
Approve tag wiki edits
 Apr 6 comment Separation of Variables PDE on Klein Gordon Equ The coordinates you choose depend on the geometry of the system. Mar 2 comment Directional Derivative @guanglei Mathematica Feb 19 comment Confusion over Bifurcation Diagram @bifurcationLost One looks for bifurcation on families of solutions. The question is, solutions of what? Feb 12 comment Trigonometric identities from differential equation Check out this answer. Feb 11 comment How do i prove $\Gamma (m) \Gamma (1-m)= \frac{\pi}{\sin(m\pi)}$ @TaylorTed This is the simplest and more transparent proof I know. The other one involves the infinite product representation of $\sin \pi z$. Jan 28 comment Centre of mass of bounded region conformation of numerical answer The object is of constant density and symmetric with respect to the $y$-axis, which intuitively justify the $y=0$ component. What would be the cause of it going to the right? Regarding the $z$ calculation, I did it both in euclidean and cylindrical coordinates, and ended up with the same result. Of course, I could be wrong. Why don't you post your calculations so we can compare? Jan 27 comment Solve the screened Poisson equation $(\Delta-\lambda^2) u(\vec r)=-c$ David, what are the boundary conditions and the domain? Jan 25 comment Centre of mass of bounded region conformation of numerical answer See edit for answer. Jan 25 comment Centre of mass of bounded region conformation of numerical answer See my edited answer for the details. Note that I did have a small error, as I forgot to multiply the integral by two. Jan 25 comment Centre of mass of bounded region conformation of numerical answer Let me put it in another way: what is the volume under the surface $z = \sqrt{x^2 + y^2}$, where $x^2 + y^2 = 2 x$? Jan 24 comment Centre of mass of bounded region conformation of numerical answer Why should they be different? The only thing that changes is the weight of the integral. You should be careful when a different order of integration has to be done, but I don't think is the case here. Jan 23 comment Centre of mass of bounded region conformation of numerical answer Well, $0 < z < r$ is straight forward. As can be seen from the second graph, $0 < r < 2$, and the limitation then means that $\theta$ has to go from $0$ to $\pi \over 2$. You have to be careful with the signs, though. Jan 22 comment Proving uniqueness of a steady state Why don't you study the expression $(1+\alpha)\bar{p}^\alpha -\alpha y \bar{p}^{\alpha-1}$? Jan 20 comment Partial derivatives of a function with conditions dependent on parameters Why don't you use a smooth characteristic function? Jan 20 comment Literature on the differential operator $-\frac{d^2}{dx^2}-\frac{2\nu+1}{x}\frac{d}{dx} + x^2$ The operator $\frac{1}{r^\beta}\frac{\partial}{\partial r}\left(r^\beta \frac{\partial}{\partial r}\right)$ has something to do with anomalous diffusion. I can't remember now, but I'll see what I can find. Jan 20 comment Literature on the differential operator $-\frac{d^2}{dx^2}-\frac{2\nu+1}{x}\frac{d}{dx} + x^2$ You said "any possible literature on the differential operator[...]". What if I had referred you to a Bessel functions book? Jan 19 comment the solution of a system of nonlinear differential equations What does the crossoperator stands for? Jan 11 comment Particular solution to nonhomogeneneous 2nd order ODE @Ian I agree with all your concerns. In my experience, students tend to omit BCs because they don't understand their role and importance. That's why I'm so emphatic about them, specially on this site, where lack of context is almost a given. If this problem had arisen form a PDE on a cylinder, two out of three students wouldn't know what to do with the missing boundary condition, and the implications on the eigenvalues, eigenfunctions, etc. All the fun starts when BCs are imposed :P Jan 11 comment Particular solution to nonhomogeneneous 2nd order ODE @usumdelphini Sorry about that. It has been corrected. Jan 11 comment What is this ODE called and how to solve it It can be easily rewritten as an Euler's equation.