# Fundamental solution of heat equation in a ball with homogeneous Dirichlet boundary condition

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...

• 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. Mar 15, 2021 at 21:42
• 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. Mar 15, 2021 at 21:57
• Not (-1,1), $(0,1)$!
– user145413
Mar 15, 2021 at 21:59
• 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. Mar 16, 2021 at 0:46
• Yes, it was my mistake latter, I mean $(0,1)$ instead of $(-1,1)$, thanks! Mar 16, 2021 at 0:47

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