# Gluing of two solutions to the same parabolic equation

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 \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.

Suppose I also have $$u_2(x,t) \in C^\infty([0,1] \times [0,T])$$ solving \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.

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 $$u(1,t) = g_1(t)$$.

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.
• well, for example $u_3$ is certainly not a solution of the PDE: think of what happens when the spatial derivatives in $L$ "hit" the indicator functions $\chi_{[0,1/2]}$ and $\chi_{[1/2,1]}$... Nov 9, 2021 at 16:33