Let $D\subset\mathbb R^n$ be open and bounded. Consider $\Delta u=f$ in $\Omega$ and $u=g$ on $\partial\Omega$. Let $g\in W^{1,2}(D)$ and $f\in L^\infty(D)$.

Then the minimizer of $$ I(u)=\int_\Omega |\nabla u|^2+2fu dx $$ over $U=\{u\in W^{1,2}(\Omega):u-g\in W_0^{1,2}(\Omega)\}$ solves the poisson equation given above.

My question is why do we have $u-g\in W_0^{1,2}(\Omega)$? Why do we get then $u=g$ on $\partial \Omega$? Are these conditions equivalent or does any implication hold?

  • $\begingroup$ You are minimizing over all $u$ that equal $g$ on the boundary. The purpose of $g$ is to make it easy to define the boundary condition, even though all you're really interested in is the trace of $g$ on $\partial\Omega$. $\endgroup$ – DisintegratingByParts Jun 7 '15 at 17:44
  • $\begingroup$ @T.A.E. why am I minimizing with $u=g$ on the boundary? I am just having $u-g\in W_0^{1,2}$? $\endgroup$ – user193603 Jun 7 '15 at 18:03
  • $\begingroup$ Because you have $u-g \in W^{1,2}_{0}$, which is how you have defined $U$. $\endgroup$ – DisintegratingByParts Jun 7 '15 at 18:05
  • $\begingroup$ @T.A.E. sorry, I've meant this. Why does follow $u=g$ on the boundary? $\endgroup$ – user193603 Jun 7 '15 at 18:08
  • $\begingroup$ Do you know of a better way to define $u=g$ on $\partial\Omega$ than requiring $u-g\in W_{0}^{1,2}$? You need some kind of trace theorem to define the boundary function; otherwise "boundary values" does not make sense. $\endgroup$ – DisintegratingByParts Jun 7 '15 at 19:05

The space $W^{1,2}_0(D)$ is appropriately differentiable functions with support in the interior of $D$. It is the natural space to consider when solving $\Delta u = f$ with $u=0$ on the boundary. To solve this problem, you develop machinery to work on $W^{1,2}_0$.

To tackle the general problem, you want to make use of the theory you've already developed. Therefore you translate the general problem into the homogeneous problem by considering functions with $u-g = 0$ on the boundary.

In other words, the only way you know to find the appropriate $u$ in $W^{1,2}$ is by using the existing machine to find $u-g$ in $W^{1,2}_0$.


It because $W^{1,2}_0(D)$ is the subset of functions $v$ of $W^{1,2}(D)$ such that $v$ is zero on $\partial D$ in the sense of trace. You can find it in the book "partial Differential Equations" of Evans. It is right after he define this space.

In this case, it means $u-g=0$ in the sense of trace.


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