3
$\begingroup$

Given an open bounded set $D\subset \mathbb R^N$, let $f\in W^{-1,q}(D)$ and let $u$ be a Sobolev function $u\in W_0^{1,p}(D)$ such that $u$ solves the PDE $$ \begin{cases} -\Delta_p u=f\;\text{in $D$}\\ u\in W_0^{1,p}(D) \end{cases} $$

Given $\Omega\subset D$ (which can be assumed open, or quasi open) we can define $P_{\Omega}:W_0^{1,p}(D)\to W_0^{1,p}(\Omega)$, $P_{\Omega}=\mathrm{Proj}_{W_0^{1,p}(\Omega)}$ (the projection of the function onto the subspace $W_0^{1,p}(\Omega)$).

I wish to prove that the function $u_{\Omega}=P_{\Omega}u$ solves the PDE

$$ \begin{cases} -\Delta_p u_{\Omega}=f\;\text{in $\Omega$}\\ u_{\Omega}\in W_0^{1,p}(\Omega) \end{cases} $$

It would suffice if I could find a proof for $f\equiv 1$.

$\endgroup$
2
  • 2
    $\begingroup$ Look at the case $p=2$, $D=[-2,2]$, $\Omega=(-1,1)$, $u=-\frac{1}{2}x^2+2$. $\endgroup$ Commented Jul 31, 2018 at 11:26
  • 1
    $\begingroup$ Your projection is not well defined, see the example of Liviu Nicolaescu. $\endgroup$ Commented Jul 31, 2018 at 15:04

1 Answer 1

1
$\begingroup$

This works (only?) for $p = 2$. Let us denote the solution of the PDE on $\Omega$ by $v$.

Then, the variational formulations of the PDEs are $$\int_D \nabla u \cdot \nabla z - fz \,\mathrm{d}x = 0 \quad\forall z \in H_0^1(D)$$ and $$\int_\Omega \nabla v \cdot \nabla z - fz \,\mathrm{d}x = 0 \quad\forall z \in H_0^1(\Omega).$$ Thus, $$\int_\Omega (\nabla v - \nabla u) \cdot\nabla z \, \mathrm{d}x=0\quad\forall z \in H_0^1(\Omega).$$ This means that $v$ is the projection of $u$ onto $H_0^1(\Omega)$.

For $p \ne 2$, this does no longer work. Instead, one can check that $v$ is the minimizer (in $H_0^1(\Omega)$) of the functional $$J(z) = \int_\Omega \frac1p|\nabla z|^p - |\nabla u|^{p-2}\nabla u\cdot\nabla z \, \mathrm{d}x,$$ but this is not really a projection problem. Note that for $p = 2$ this is a projection due to $$J(z) = \frac12 \,\int_\Omega |\nabla u - \nabla z|^2 - |\nabla u|^2\,\mathrm{d}x.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.