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 $u(1,t) = g\_1(t)$.



Gluing of two solutions to the same parabolic equation
Absolutely not! Taking the difference $v=u\_1-u\_2$, you see that $v(x,t)$ solves
$$
\begin{cases}
(L-\partial\_t) v=0 & \mbox{for }(x,t)\in(0,1/2)\times(0,T];
\\
v(0,t)=f(t) & \mbox{for }x=0,\,t\in(0,T]\\
v(1/2,t)=0 & \mbox{for }x=1/2,\,t\in(0,T]\\
v(x,0) =0 & \mbox{for }x\in (0,1/2),\,t=0
\end{cases}
$$
for some left boundary data $f=f\_1-f\_2$ which is a priori non-zero, and your question amounts to asking whether the overdetermined Neumann boundary condition $\partial\_x v(1/2,t)=0$ is satisfied as well at the rightmost endpoint $x=1/2$. Of course there is no reason why this should hold. For a counterexample, just take $f(t)=t$, so that by the maximum principle $v(t,x)\geq 0$ and in fact $v(t,x)>0$ for all $(x,t)\in (0,1/2)\times(0,T)$. Hopf's boundary lemma then guarantees that, since $v$ attains its minimum value $v(1/2,t)=0$ on the right boundary, necessarily $\partial\_x v(1/2,t)<0$ there.


