Heat equation - Step function initial condition Suppose you want to solve the usual heat equation on the real line $[-\infty ,+\infty ]$
\begin{equation}
\begin{cases}
 \partial_t u(x,t)= \partial_{xx} u(x,t)\\
 u(x,0)=f(x)\\
\end{cases},
\end{equation}
where the initial condition is given by a step function
\begin{equation}
f(x)= 
\begin{cases}
0 \ \ \ \ \text{if} \ \ x<0\\
1 \ \ \ \ \text{if} \ \ x>0\\
\end{cases}.
\end{equation}
By solving the equation using Green's function method it's possible to find a solution of the form
\begin{equation}
u(x,t)=\frac{1}{2}\left( 1+\mathrm{erf} \left( \frac{x}{\sqrt{4t}} \right) \right)
\end{equation}
A short derivation of this result can be found at this link.
It seems to me that the limit $(x,t)\to (0,0)$ of this function does not exist (in the sense of limit of functions); this is of course in conflict with the IC imposed.
Why is this happening? Am I forgetting something while calculating the solution? Or am I misinterpreting the word "limit" for this particular case?
 A: I will use the problem setup in your link. Convergence to the initial condition can be understood in the following sense. Fix any x > 0. Then, rewrite the solution as 
\begin{equation*}
u(x,t) = \frac{1}{2} + \frac{1}{\sqrt{\pi}} \int_0^{x/\sqrt{4kt}} e^{-p^2} \, dp = \frac{1}{2} + \frac{1}{\sqrt{\pi}} \int_0^\infty e^{-p^2} \chi_{[0, \frac{x}{\sqrt{4kt}}]} (p) \, dp, 
\end{equation*}
where $\chi_{[a,b]}$ is the characteristic function of the interval $[a,b]$. $\chi_{[0, \frac{x}{\sqrt{4kt}}]} (p)$ converges pointwise to $1$ as $t \to 0^+$ for $p > 0$, and the integrand is dominated by the Gaussian, which is integrable, so by the dominated convergence theorem, we obtain 
\begin{equation*}
\lim_{t \to 0^+} \int_0^{x/\sqrt{4kt}} e^{-p^2} \, dp = \int_0^\infty e^{-p^2} \, dp = \frac{\sqrt{\pi}}{2}
\end{equation*}
for each $x > 0$. For each $x < 0$, the limit is $-\frac{\sqrt{\pi}}{2}$ instead. Therefore, we have 
\begin{equation*}
\lim_{t \to 0^+} u(x,t) = \begin{cases}
0 &\text{ if } x < 0, \\
1 &\text{ if } x > 0,
\end{cases}
\end{equation*}
which matches the initial condition. We have convergence to the initial condition pointwise almost everywhere. We do not have, for instance, uniform convergence, since the solutions for $t>0$ are continuous, and a uniform limit of continuous functions is continuous, but the initial condition is not continuous. 
