0
$\begingroup$

I have this particular instance of the Heat Equation: Find u such that

$$\label{eq:Neumann_parabolico} \left\lbrace \begin{array}{rl} u_t + \Delta u = 0 & \mbox{in } B(0,1)\times (0,T], \\ u = 0 & \mbox{in } \partial B(0,1) \times (0,T],\\ u(\cdot,0) = \delta_0 & \mbox{in } B(0,1), \end{array} \right. $$ for some final $T>0$. Here $B(0,1) \subset \mathbb{R}^2$ is the unitary ball centered at 0, and $\delta_0$ is the Dirac delta function.

My question: Is there a closed-form expression for the solution $u(x,y,t)$?

I tried to use the fact that $u$ should have radial symetry, but I didn't succeed...

$\endgroup$
7
  • $\begingroup$ Only the equation in shperical coortinated... not much. I used the radial laplacian, $\Delta_r u = \frac{1}{r} \partial r (r u_r)$ and replaced B(0,1) by (-1,1). Then I got stuck there, I don't know how to deal with the delta type initial data. $\endgroup$
    – JorgeNN
    Mar 15, 2021 at 21:42
  • $\begingroup$ I started to give a hint but realized it was for Laplace's equation. My mistake. I'll check back later and if you still need a hint I'll lend a hand. $\endgroup$ Mar 15, 2021 at 21:57
  • $\begingroup$ Not (-1,1), $(0,1)$! $\endgroup$
    – user145413
    Mar 15, 2021 at 21:59
  • $\begingroup$ Using Bessel functions seems to be a good idea. But I think there is a problem if we separate variables. We should have $T(t)R(r) \to \delta_0$ when $t \to 0$... and this doesn't seem to be possible. $\endgroup$
    – JorgeNN
    Mar 16, 2021 at 0:46
  • $\begingroup$ Yes, it was my mistake latter, I mean $(0,1)$ instead of $(-1,1)$, thanks! $\endgroup$
    – JorgeNN
    Mar 16, 2021 at 0:47

1 Answer 1

2
$\begingroup$

For the standard heat equation ($\partial_t u-\Delta u=0$), there is an explicit solution formulated in Gegenbauer polynomial when dimension is not less than 3. See Theorem 2 in

Hsu, P. (1986). Brownian Exit Distribution of a Ball. In: Çinlar, E., Chung, K.L., Getoor, R.K., Glover, J. (eds) Seminar on Stochastic Processes, 1985. Progress in Probability and Statistics, vol 12. Birkhäuser Boston. https://doi.org/10.1007/978-1-4684-6748-2_8

But as the author said, the same method applies but the final formula looks different. This fundamental solution is the heat kernel for Brownian motion killed upon leaving the unit ball.

For the backward heat equation ($\partial_t u+\Delta u=0$), I think the same method still applies.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .