I'm trying to solve the heat equation (e.g a ball at room temperature is dropped into a lake which is at freezing temperature) $$ \dot{u}=\Delta u,\hspace{.3 in} x\in\mathbb{D}^2:=\big\{x\in\mathbb{R}^2\,:\,|x|\leq 1\big\},\hspace{.3 in} u\big|_{t=0}20^{\circ}C,\hspace{.3 in}u\big|_{\mathbb{S}^1}=0^{\circ}C $$ in two dimensions. I'm ultimately trying to do this for four dimensions but I first want to get a handle on this in two dimensions, so I'd like to avoid any methods that take advantage of the smaller dimension.
Using the trial function $u(r,\theta)=1-r^2$, I've computed that the lowest eigenvalue $\lambda^2$ of the laplacian (i.e. $-\Delta\varphi=\lambda^2\varphi$ for some eigenfunction $\varphi$) is approximately $$ \lambda^2\approx\frac{\int_{\mathbb{D}^2}\big|\nabla u\big|^2dV}{\int_{\mathbb{D}^2}\big|u\big|^2dV}=\frac{1}{1/6}=6. $$ However I'm not quite sure how to leverage this information to actually approximate the solution.
Thoughts
If I assume a solution $u$ is of the form $u=Z(x,y)T(t)$ then we get that $$ 0=u_t-\Delta u\hspace{.5 in}\Leftrightarrow\hspace{.5 in}\frac{\Delta Z}{Z}(x,y)=\frac{T'}{T}(t); $$ therefore $\Delta Z(x,y)=\sigma Z(x,y)$ and $T'(t)=\sigma T(t)$ for some constant $\sigma$. From the first equation $\sigma$ must be an eigenvalue, so we assume $\sigma=-6$. It's easy to see that $T(t)=Ce^{-6t}$, however I don't know what to do with the eigenvalue equation $\Delta Z(x,y)=-6 Z(x,y)$. This might be (relatively) easily solvable in $\mathbb{R}^2$ but I don't see how to even approximate a solution in $\mathbb{R}^4$. Thoughts and advice are appreciated.