Given a bounded open set $\Omega$ and some differential operator, for example the laplacian operator $\Delta$, one PDE formulation is to find $u$

$$ \Delta u = f $$ in $\Omega$, and $u = g$ on $\partial \Omega$. In this formulation, do we mean the PDE should be satisfied only in $\Omega$? or do we also want the PDE to hold on the boundary, that is

$$\Delta u(x) = f(x)$$ for $x \in \partial \Omega$?

  • $\begingroup$ differentiation is what's called an "open condition" $\endgroup$ Commented Oct 15, 2019 at 17:20

1 Answer 1


We want $\Delta u = f$ to be satisfied only in $U$ because derivative operators are not well-defined in the boundary $\partial U$.

Derivative operators in PDEs are defined using limit processes; for example, the usual Frechét derivative is the (unique) linear operator $A$ such that

$$\lim_{h \to 0} \frac{\|u(x+h)-u(x)-Ah\|}{\|h\|} = 0$$

The problem with taking this limit at the boundary is that we don't have any information whatsoever about $u(x+h)$ for $x+h \not\in \overline{U}$, thus the limit from that direction is not defined. Similar problems arise with the laplacian.

One could ask (assuming that $f$ is actually defined outside $U$):

Could we extend $u(x,t)$ by forcing it to satisfy $\Delta u = f$ outside $U$ , say, for another open set $V \supset \overline{U}$, while still having $u=g$ in $\partial U$?

Then you lose uniqueness; there is an infinite number of smooth functions that satisfy the equation in $V$ while still satisfying the "boundary" condition in $\partial U$. And, if you impose conditions over $\partial V$, well, it's no different than considering the problem in $V$ in the first place.

  • $\begingroup$ How do I see that there is an infinite number of smooth function in the latter case? $\endgroup$
    – tgtt
    Commented Oct 17, 2019 at 14:33
  • $\begingroup$ Take a harmonic function $v(x)$ on $V\backslash\overline{U}$ such that $v(x) = 0$ in $\partial U$ and satisfy any boundary condition in $\partial V$. You can then extend $v(x)$ inside $U$ by setting $v(x) = 0$ in $U$. Now you can add any multiple of this function to $u$. $\endgroup$ Commented Oct 18, 2019 at 16:56

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