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?
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.