# Parabolic equation with discontinuous boundary condition

Consider a parabolic initial-boundary value problem in $\Omega\times (0,T]$

$$\frac{\partial u(x,t)}{\partial t}=\mathcal{L}u(x,t),$$

with $$u(x,0)=0, x\in \bar{\Omega} \text{ and } u(x,t)=\chi_{[t_0-1/n,t_0+1/n]}(t), (x,t)\in \partial\Omega\times (0,T],$$

where $\Omega\subset \mathbb{R}^d$ is a bounded domain with smooth boundary, $\mathcal{L}=\sum a_{ij}(x,t)\frac{\partial^2}{\partial x_i\partial x_j}+\sum b_i(x,t)\frac{\partial}{\partial x_i}-c$ is a uniformly elliptic operator with nice coefficients and $\chi_A$ denotes the characteristic function of set $A$, $t_0\in (0,T)$.

This equation has a unique solution $u_n$ that belongs to $C^{2,1}(\Omega\times (0,T])$, and is continuous in $\bar{\Omega}\times[0,T]$ except for $\partial\Omega\times\{t_0-1/n,t_0+1/n\}$.

My question: is it true that, as $n\to\infty$, $u_n$ converges uniformly to $0$ on any compact domain in $\bar{\Omega}\times ([0,T]\backslash\{t_0-1/n,t_0+1/n\})$ (bounded away from the discontinuity on the boundary)? If true, how to prove this (or is there any literature on this)?

I feel that some kind of $L^2$ estimates may do this: the $L^2$-norm of the boundary data converges to 0, so I expect that the $L^2$-norm of the solution should also converge to 0. However, all the $L^2$ estimate I found are for zero boundary condition. I am aware that I can transform the present problem to one with zero boundary condition, however this would put the time-derivative of the boundary data into the right hand side term of the PDE, whose $L^2$ norm is exploding as $n\to\infty$. I am currently stuck with this.