# question about formulation of partial differential equations

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 $$x \in \partial \Omega$$?

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

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.

• How do I see that there is an infinite number of smooth function in the latter case?
– tgtt
Commented Oct 17, 2019 at 14:33
• 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 $u$. Commented Oct 18, 2019 at 16:56