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 $f\equiv 1$.

  • 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


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 $$J(z) = \frac12 \,\int_\Omega |\nabla u - \nabla z|^2 - |\nabla u|^2\,\mathrm{d}x.$$


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