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

Partial Solution

Posted online: 2018-09-27 20:30:34Z by Shane Cooper 55

Cite as: S-180927.1

PDE satisfied by projection of a function on Subspace

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 $f\equiv 1$.

### Solution Description

Here we will show there exists a map $P_\Omega: W^{1,p}_0(D) \rightarrow W^{1,p}_0(\Omega)$ that satisfies the desired properties of the question and 'behaves' like a projector in the sense that $P^2_\Omega = P_\Omega$ and $P = I$ on $W^{1,p}_0(\Omega)$. Yet $P_\Omega$ is only a linear map if $p=2$.

Set $\Omega \subset D$, $1< p< \infty$. For fixed $u\in W^{1,p}_0(D)$ let $w\in W^{1,p}(\Omega)$ satisfy $$F(w) = \min_{v \in W^{1,p}_0(\Omega)} F_u(v), \qquad \text{where} \qquad F_u(v) : = \frac{1}{p} \int_{\Omega} | \nabla v|^p - \int_{\Omega} |\nabla u|^{p-2}\nabla u \cdot \nabla v.$$ Notice, the minimiser exists and is unique by usual methods (i.e. direct method for existence and uniqueness since $|\cdot|^p$ is strictly mid-point convex.)

Now $w$ weakly solves the following Euler-Lagrange equation for $F_u$: $$\int_\Omega |\nabla w|^{p-2}\nabla w \cdot \nabla v = \int_{\Omega} |\nabla u|^{p-2}\nabla u \cdot \nabla v\qquad \forall v \in W^{1,p}_0(\Omega).$$ That is $w\in W^{1,p}_0(\Omega)$ weakly solves $-\Delta_p w = -\Delta_p u$ in $\Omega$.

Let us define $P_{\Omega} u : = w$.

1) Suppose that $u\in W^{1,p}_0(D)$ solves $$\int_D |\nabla u|^{p-2}\nabla u \cdot \nabla \phi = \langle F, \phi \rangle \qquad \forall \phi \in W^{1,p}_0(D),$$ (i.e. $u$ weakly solves $-\Delta_p u = F$ in $D$). Then, by taking the Euler-Lagrange equation for $w$ with test function $v\in W^{1,p}_0(\Omega)$ and setting $\phi = v$ above, we deduce that $w = P_{\Omega} u$ solves $$\int_\Omega |\nabla w|^{p-2}\nabla w \cdot \nabla v = \int_{\Omega} |\nabla u|^{p-2}\nabla u \cdot \nabla v = \langle F, v \rangle \qquad \forall v \in W^{1,p}_0(\Omega).$$ That is $w \in W^{1,p}_0(\Omega)$ weakly solves $-\Delta_p w = F$ in $\Omega$.

Now $P_{\Omega}$ is not a linear map (unless $p=2$) yet we will that $P_\Omega$ is idempotent and equal to identity on $W^{1,p}_0(\Omega)$.

2) Claim: $P_\Omega^2 = P_\Omega$. Proof: let $w = P_\Omega u$ and let $\tilde{w} = P_\Omega w$. Then for $v \in W^{1,p}_0(\Omega)$ it follows that $$\int_\Omega |\nabla \tilde{w}|^{p-2}\nabla \tilde{w} \cdot \nabla v = \int_\Omega |\nabla w|^{p-2}\nabla w \cdot \nabla v = \int_\Omega |\nabla u|^{p-2}\nabla u \cdot \nabla v.$$ It follows that $\tilde{w}$ minimises $F_u$ (as well as $w$) and so by uniqueness of the minimiser we deduce that $\tilde{w} = {w}$.

3) Claim: $P_\Omega u = u$ for $u \in W^{1,p}_0(\Omega)$. Proof: Notice that $$F_u(u) = (\tfrac{1}{p}-1) \int_\Omega |\nabla u|^p = - \tfrac{1}{q} \int_\Omega |\nabla u|^p,$$ where $q$ is the conjugate exponent of $p$. Let us now show that $u$ minimises $F_u$. Indeed, for $v\in W^{1,p}_0(\Omega)$ it follows by Holder's inequality and Young's inequality that $$\left| \int_{\Omega} |\nabla u|^{p-2}\nabla u \cdot \nabla v \right| \le \| |\nabla u|^{p-2}\nabla u \|_q \| \nabla v\|_p = \| \nabla u \|_p^{(p-1)} \| \nabla v\|_p \le \tfrac{1}{q} \| \nabla u \|_p^p + \tfrac{1}{p} \| \nabla v\|_p^p.$$ Therefore \begin{align} F_u(v) & = \frac{1}{p} \int_{\Omega} | \nabla v|^p - \int_{\Omega} |\nabla u|^{p-2}\nabla u \cdot \nabla v \\ & \ge \frac{1}{p} \int_{\Omega} | \nabla v|^p - \left( \tfrac{1}{q} \| \nabla u \|_p^p + \tfrac{1}{p} \| \nabla v\|_p^p\right) \\ & = - \tfrac{1}{q} \| \nabla u \|_p^p = F_u(u). \end{align} That is $u$ minimises $F_u$ and so $P_\Omega u = u$.

1. ## Other own calculation

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• Created at: 2018-09-27 20:30:34Z