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


1 Answer 1


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.

  • $\begingroup$ Thank you so much! I would like to know what's wrong with my reason in the above. $\endgroup$
    – mnmn1993
    Nov 9, 2021 at 16:30
  • 2
    $\begingroup$ 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 $\chi_{[1/2,1]}$... $\endgroup$ Nov 9, 2021 at 16:33

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