Consider the domain $[0,1] \times [0,T]$ and the uniformly parabolic operator $L -\partial_t$ with smooth coefficient. Suppose I have $u_1(x,t) \in C^\infty([0,1] \times [0,T])$ solving \begin{equation} \left\{\begin{aligned} &L u_1 -\partial_t u_1= 0& \hspace{10pt} &\text{for $(x,t) \in (0,1) \times (0,T]$} ;\\ & u_1(0,t) =f_1(t) & \hspace{10pt} &\text{for $t \in \big[0,T\big]$;}\\ & u_1(1,t) =g_1(t) & \hspace{10pt} &\text{for $t \in \big[0,T\big]$;}\\ & u_1(x,0) =h(x) & \hspace{10pt} &\text{for $x \in \big(0,1\big)$.}\\ \end{aligned}\right. \end{equation}

Suppose I also have $u_2(x,t) \in C^\infty([0,1] \times [0,T])$ solving \begin{equation} \left\{\begin{aligned} &L u_2 -\partial_t u_2 = 0& \hspace{10pt} &\text{for $(x,t) \in (0,1/2) \times (0,T]$} ;\\ & u_2(0,t) =f_2(t) & \hspace{10pt} &\text{for $t \in \big[0,T\big]$;}\\ & u_2(1/2,t) =u_1(1/2,t) & \hspace{10pt} &\text{for $t \in \big[0,T\big]$;}\\ & u_2(x,0) =h(x) & \hspace{10pt} &\text{for $x \in \big(0,1\big)$.}\\ \end{aligned}\right. \end{equation}

Question: Is $\partial_x u_1(1/2,t) = \partial_x u_2(1/2,t)$?

Reason: I think it is true since $u_1$ and $u_2$ solve the same equation. Also, their values at $x=1/2$ are the same. Therefore $u_3=u_1 \chi_{[1/2,1]\times [0,T]} +u_2 \chi_{[0,1/2)\times [0,T]}$ is at least Lipschitz in the variable $x$. So $u_3$ is a smooth unique solution solving the equation with boundary conditions $u(0,t) = f_2(t)$ and $u(1,t) = g_1(t)$.