Skip to main content
6 votes

Basic Solution to the Heat Equation

As of your final answer, it is indeed a correct one. In more general terms, you would need to find a so-called fundamental solution $\Phi(x,t)$, which is the solution to the problem: $$ \left\{ \begin{...
Egor Larionov's user avatar
4 votes
Accepted

Why are kernels often singular on the diagonal?

A partial answer is that many kernels of interest are Green's functions of some PDE, and if that PDE is translation invariant, then the Green's function will be of the form $G(x,y) = \Phi(x-y)$ and ...
whpowell96's user avatar
  • 6,119
3 votes
Accepted

Basic Solution to the Heat Equation

There is a conceptual bridge between stochastic models of motion of particles by random steps in time and the smooth Gaussian as the probability density of the final distribution the the position ...
Roland F's user avatar
  • 3,234
2 votes

On max-min representation for the principal eigenvalue of nonsymmetric elliptic operators

This solution is due to a brilliant friend of mine. I promissed him the bounty if he wrote his solution, but I guess he is too busy to write a MSE post. If he does end up writing it I will naturally ...
Kadmos's user avatar
  • 2,234
1 vote

Why is the term $H(x,y)u_{yx}$ omitted in every definition of a linear 2nd order PDE in two independent variables?

Clairaut's Theorem and its variants makes it so you generally are in a situation where $u_{xy}$ and $u_{yx}$ are equal, so that you only need one term to handle both of them.
Mark S.'s user avatar
  • 24.3k
1 vote

Confusion in Partial Derivation of an Equation containing Quaternion

$ \newcommand\dd{\mathbf d} $It's safest to assume you can't take the derivative with respect to a constrained variable. You're better off using an unnormalized quaternion. Then the rotation is $g \...
Nicholas Todoroff's user avatar
1 vote

Saying solution of a PDE is continuous to the boundary

No. Let $\Phi$ be the fundamental solution to Laplace's equation (aka the Newton potential). It has a singularity at $0$. Now if $y \in \partial \Omega$ then the map $u(x) = \Phi(x-y)$ satisfies $\...
Glitch's user avatar
  • 8,496

Only top scored, non community-wiki answers of a minimum length are eligible