I have the given problem to solve:
A very long beam with quadratic intersection ($0-L$ for x and y) lies on a surface with a temperature of $0$ degrees. The rest of the beam's surface is in contact with air which holds $10$ degrees. Determine the stationary temperature distribution in the beam under the assumption that it is infinitely long.
So, since the beam is clearly two-dimensional I considered the PDE:
$\alpha\Delta u=u_t$
where the solution is:
\begin{equation} u(x,y,t)=\begin{cases} \bigg(\frac{1}{2\sqrt{\pi \alpha t}}\bigg)^2\int_0^Le^{-(x-\xi_1)/4\alpha t}\phi(\xi_1,\xi_2)d\xi_1\int_0^Le^{-(y-\xi_2)/4\alpha t}d\xi_2,\ \ \ \ \ \ t>0 \\ \phi(x,y) , \ \ \ \ t=0 \end{cases} \end{equation}
However, since the rod is defined by $-\infty<z<\infty$ we use the reflection method and define:
\begin{equation} u(z)=\begin{cases} \phi(z) ,\ \ \ \ \ \ \ \ \ z>0 \\ \phi(-z,), \ \ \ \ \ \ \ \ \ z<0 \end{cases} \end{equation}
Then we evaluate
$u(x,y,t)=\phi(\xi_1,\xi_2)\mid_{0}^{L}$
So we get, with $\phi(\xi_1,\xi_2)=10$:
\begin{equation} u(x,y,t)=\frac{10}{4\pi\alpha t}\int_0^Le^{-(x-\xi_1/4\alpha t)}d\xi_1\int_0^Le^{-(y-\xi_2/4\alpha t)}d\xi_2 \\ \end{equation}
with substitution $p=\frac{x-\xi_1}{\sqrt{4t}}$ and $q=\frac{y-\xi_2}{\sqrt{4t}}$ we get:
\begin{equation} u(x,y,t)=\frac{4\alpha t}{4\pi \alpha t}\int_{\frac{x}{\sqrt{4\alpha t}}}^{\frac{x-L}{\sqrt{4\alpha t}}}e^{-p^2}dp\int_{\frac{y}{\sqrt{4\alpha t}}}^{\frac{y-L}{\sqrt{4\alpha t}}}e^{-q^2}dq \\ \end{equation}
which then gives:
\begin{equation} u(x,y,t)=\frac{1}{4}\bigg[\textrm{erf}\bigg(\frac{x-L}{\sqrt{4\alpha t}}\bigg)-\textrm{erf}\bigg(\frac{x}{\sqrt{4\alpha t}}\bigg)\bigg]\bigg[\textrm{erf}\bigg(\frac{y-L}{\sqrt{4\alpha t}}\bigg)-\textrm{erf}\bigg(\frac{y}{\sqrt{4\alpha t}}\bigg)\bigg] \end{equation}
But I am not sure this is correct, since the zero temperature is not considered. Of course, I could go over to Kelvin, and use 273 and 283 K. But that is not rectifying the problem eventually. Also, the question is mentioning "stationary temperature distribution", and that would mean $u_t=0$, so I am not sure if I have solved the right problem at all.
Any hints?
Thanks