# PDE satisfied by projection of a function on Subspace

Partially Solved

Posted online: 2018-07-31 10:13:04Z by Harish Shrivastava117

Cite as: P-180731.1

• Analysis of PDEs

### Problem's Description

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 $u$ solve the PDE $$\begin{cases} -\Delta_p u=f\;\text{in D}\\ u\in W_0^{1,p}(D) \end{cases}$$

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

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

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

It will also sufficient to me if I could find a proof just for $f\equiv 1$.

• ## Partial SolutionExistence of a (non-linear) map with desired property that is idempotent and is the identity operator on $W^{1,p}_0(\Omega)$.

• Jacob Ralph Mirra2018-07-31 12:11:04Z

How do you define this "projection" onto a subspace of $W^{1,p}(D)$ if $p \neq 2$?

• New solution is added on 2018-09-27 20:30:34Z View the solution

• Created at: 2018-07-31 10:13:04Z