# Heat equation problem with initial condition in a disk

I was solving this problem

Find the solution $$u(x,y,t)$$ of the problem $$\begin{cases} u_t=D\nabla^2u \\[5pt] u(x,y=0,t) = 0\\[5pt] u(x,y,0) = \phi(x,y) \end{cases}$$ where $$\phi(x,y) = \begin{cases} c &(x-x_0)^2+(y-y_0)^2\le R^2 \\[5pt] 0 & (x-x_0)^2+(y-y_0)^2\gt R^2 \end{cases}$$ then find an approximate solution in the region $$(x-x_0)^2+(y-y_0)^2\gg R^2$$ $$\underline{\textbf{Hint:}}$$ Note that the Dirac delta function is given by the limit when $$R\rightarrow \infty$$ of the function $$\frac{1}{\pi R^2}\chi_R (x,y)$$ where $$\chi_R(x,y) = \begin{cases} 1& x^2+y^2\le R \\ 0&x^2+y^2\gt R\end{cases}$$

and I got stuck evaluating the following integral: taking the convolution between the heat kernel and the initial condition $$u(x,y,t) = \frac{1}{4\pi Dt}\iint_{\mathbb R^2} e^{-\frac{|\underline{x}-\underline{x}'|^2}{4Dt}}u(\underline{x}',0)\,\mathrm d^2\underline x' = \frac{c}{4\pi Dt}\iint\limits_{\Omega} e^{-\frac{|\underline{x}-\underline{x}'|^2}{4Dt}}\,\mathrm d^2\underline x'$$ where $$\Omega$$ is the region limited by the disk centered in $$(x_0, y_0)$$ of radius $$R$$.

I've tried going to polar coordinates, which seems to me the right choice, but I haven't got very far: $$\begin{cases} x' = x_0+r\cos\theta \\ y'=y_0+r\sin\theta\end{cases}$$ which gives me $$u(x,y,t) = \frac{c}{4\pi Dt}e^{(x^2+y^2)+(x_0^2+y_0^2)-2(x_0+y_0)}\int\limits_0^R\int\limits_0^{2\pi}r^2\sin(\theta) e^{-2r[\cos\theta(1-x_0)+\sin\theta(1-y_0)]+r^2}\,\mathrm dr\,\mathrm d\theta$$ here's where I'm stuck.

How do I proceed? Are my calculations correct?

Sorry for the late correction, I forgot the Jacobian when changing the variables.

• Are you sure you don't miss some square somewhere? cos and sin would be inside norm squared, wouldnt they? – mathreadler Sep 23 '18 at 12:29
• @mathreadler Yes, in fact there's the $r^2$ term in the exponential which comes from $r^2\sin^2\theta+r^2\cos^2\theta$! – Davide Morgante Sep 23 '18 at 12:31
• Ah, ok. Yes, that is what I suspected. In general a good idea to take smaller steps when simplifying. It's very easy to miss something otherwise. – mathreadler Sep 23 '18 at 12:32
• @mathreadler Yes, my bad. I wrote down less steps that I should had :( – Davide Morgante Sep 23 '18 at 12:33
• I am also pretty sure there is some way, but it was such a long time ago I did this on paper. I will try bumping it for you and maybe some integral master comes along. Hmm, seems there is some time limit for bumping. Maybe after work tomorrow. – mathreadler Sep 23 '18 at 13:31

WLOG, suppose $$x_0=0,\ y_0>0,\ D=\frac12,\ c\pi R^2=1$$.

1. $$y_0\ge R$$.

By Green's function of the heat equation $$u(x,y\le0,t)=0$$.

For $$y>0$$, the $$u$$ in the upper half plane is equivalent to the upper half plane solution of whole plane solution of from the original source and its negative image $$-\Omega$$ reflected across the line $$y=0$$, since the latter satisfies the PDE and its zero boundary condition on $$y=0$$. \begin{align} u(x,y>0,t) &= \frac c{\sqrt{2\pi t}}\Big(\int_{\Omega}dudv\, e^{-\frac{(x-u)^2+(y-v)^2}{2t}}-\int_{-\Omega}dudv\, e^{-\frac{(x-u)^2+(y-v)^2}{2t}}\Big) \\ &\underset{x^2+y^2\to\infty}\sim\ \frac1{\sqrt{2\pi t}}\Big( e^{-\frac{x^2+(y-y_0)^2}{2t}}-e^{-\frac{x^2+(y+y_0)^2}{2t}}\Big) \\ &=\sqrt{\frac2{\pi t}}\,e^{-\frac{x^2+y^2+y_0^2}{2t}}\sinh\Big(\frac{yy_0}t\Big). \end{align} We have treated the source as one point, as at large distance the exact shape $$\Omega$$ of the source impacts little on the final approximation.

1. $$y_0.

The disk is cut into two parts by $$y=0$$. We still treat the upper and lower half of the plane separately. This time the total "energy" of positive source is a fraction proportional to its area above $$y=0$$. $$u$$ no longer vanishes in the lower half plane but is evaluated similarly to that in the upper half plane.

• I couldn't have asked for more! Thank you so much, the bounty is yours :) – Davide Morgante Oct 1 '18 at 12:43
• @DavideMorgante: You are welcome. :-) – Hans Oct 1 '18 at 16:08