1
$\begingroup$

Let

  • $d\in\mathbb N$
  • $\Omega\subseteq\mathbb R^d$ be open
  • $\mathcal D:=C_c^\infty(\Omega)$ and $$H=\overline{\mathcal D}^{\langle\;\cdot\;,\;\cdot\;\rangle_H}\tag 1$$ with $$\langle\phi,\psi\rangle_H:=\langle\phi,\psi\rangle_{L^2(\Omega)}+\langle\nabla\phi,\nabla\psi\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for }\phi,\psi\in\mathcal D$$

Let $u\in L^1_{\text{loc}}(\Omega)$ be weakly differentiable (with $\nabla u\in L^1_{\text{loc}}(\Omega,\mathbb R^d)$). Then, $v\in L^1_{\text{loc}}(\Omega)$ is called weak Laplacian of $u$ $:\Leftrightarrow$ $$\langle\phi,v\rangle_{L^2(\Omega)}=-\langle\nabla u,\nabla\phi\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for all }\phi\in\mathcal D\tag 2\;.$$ In that case, we write $\Delta u:=v$.

Let $f\in L^2(\Omega)$. Can we show that $\Delta f$ with $$(\Delta f)(u):=\langle\Delta u,f\rangle_{L^2(\Omega)}\;\;\;\text{for }u\in H_0^1(\Omega)$$ is an element of $H_0^1(\Omega)'$?

$\endgroup$
6
  • 1
    $\begingroup$ how is it not a duplicate of math.stackexchange.com/questions/1855228/… ? you think replacing $\partial_{x_i}^2 f$ by $\Delta f$ changes what ? $\endgroup$
    – reuns
    Commented Jul 14, 2016 at 10:31
  • $\begingroup$ Your notation is quite confusing. $\Delta u$ is not really a function in $L^2$, so $\langle \Delta u, f\rangle$ does not really mean the usual $L^2$ product. However, if you use $-\langle \nabla u, \nabla f\rangle_{L^2(\Omega, \mathbb R^d)}$, it is not clear why $\nabla f $ exists. Which texts are you following? $\endgroup$
    – user99914
    Commented Jul 14, 2016 at 11:15
  • $\begingroup$ @ArcticChar I'm not sure about the definition of the weak Laplacian. Obviously, in order for $(2)$ to make sense, we need that $u\in L_{\text{loc}}^1(\Omega)$ and that $u$ has a weak gradient $\nabla u\in L^1_{\text{loc}}(\Omega,\mathbb R^d)$. A solution $v$ needs to be in $L^1_{\text{loc}}(\Omega)$. However, it's clear that (see my comment below the answer of user1952009) any $u\in H_0^2(\Omega)$ has a weak Laplacian in the sense of $(2)$. $\endgroup$
    – 0xbadf00d
    Commented Jul 14, 2016 at 11:46
  • $\begingroup$ @ArcticChar : he follows mat.tuhh.de/veranstaltungen/isem18/images/0/02/Lecture13.pdf $\endgroup$
    – reuns
    Commented Jul 14, 2016 at 12:01
  • $\begingroup$ @0xbadf00d : Note that you have asked several question recently and it seems that you are quite messed up with the definitions. There are at least two different definitions of weak Laplacian, it will be easier for you and me to if we follow a fixed reference. $\endgroup$
    – user99914
    Commented Jul 14, 2016 at 13:23

1 Answer 1

2
$\begingroup$

let's take $\Omega = [0,1]$, $\varphi,f \in C^\infty_c((0,1))$.

Then $$\langle f'',\varphi \rangle = \int_0^1 f''(x) \varphi (x) dx = f'(1)\varphi (1)-f'(0)\varphi (0) - \int_0^{1} f'(x) \varphi '(x) dx$$

$\varphi,f$ have their support strictly inside $(0,1)$ so $f'(0)\varphi (0) = f'(1)\varphi (1)= 0$ and

$$\langle f'',\varphi \rangle =-\langle f',\varphi' \rangle$$

Then with $\|u\|_{H^1_0} = \|u\|_{L^2}+\|u'\|_{L^2} \ge \|u'\|_{L^2}$ and the Cauchy inequality you have $$|\langle f',\varphi' \rangle| \le \|f'\|_{L^2}\|\varphi'\|_{L^2}\le \|f'\|_{L^2}\|\varphi\|_{H^1_0} \le \|f\|_{H^1_0}\|\varphi\|_{H^1_0}$$

The conclusion is that since $C^\infty_c(\Omega)$ is dense in $H^1_0(\Omega)$, all this stays true for any $\varphi,f \in H^1_0([0,1])$ (this is where $H^1_0$ makes a huge difference to $H^1$)

Replacing $'$ by $\nabla$ it extends to any domain $\Omega \subset \mathbb{R}^d$ and

  • $u \mapsto \langle \Delta f,u \rangle$ is a bounded operator $H^1_0(\Omega) \to \mathbb{C}$ whenever $f \in H^1_0(\Omega)$.

  • $f \mapsto \langle \Delta f,. \rangle$ is bounded when seen as an operator $H^1_0(\Omega) \to H^1_0(\Omega)'$

$\endgroup$
16
  • $\begingroup$ If $v\in L^1_{\text{loc}}(\Omega)$ such that $v_i$ is weakly partially differentiable wrt the $i$-th variable, then $v$ has a weak divergence $\nabla\cdot u\in L^1_{\text{loc}}(\Omega)$. Since the weak Laplacian of $u$ is the weak divergence of $\nabla u$, we should need that $\partial_iu$ is weakly partially differentiable wrt the $i$-th variable, i.e. we should need something like $u\in H_0^2(\Omega)$. Or is there something wrong in my argumentation? $\endgroup$
    – 0xbadf00d
    Commented Jul 14, 2016 at 11:41
  • $\begingroup$ I just proved that $\langle f'',\varphi \rangle =-\langle f',\varphi' \rangle$ whenever $f,\varphi \in H^1_0$ ( $f''$ is the distributional derivative of $f'$). And please take $\Omega = [0,1]$ @0xbadf00d $\endgroup$
    – reuns
    Commented Jul 14, 2016 at 11:45
  • $\begingroup$ What do you denote by $\langle\;\cdot\;,\;\cdot\;\rangle$? The duality bracket? $\endgroup$
    – 0xbadf00d
    Commented Jul 14, 2016 at 11:48
  • $\begingroup$ I wrote it.. 1st line (I make things simple, you make them complicated) @0xbadf00d $\endgroup$
    – reuns
    Commented Jul 14, 2016 at 11:49
  • $\begingroup$ So, the $L^2$-inner product?! But then, how do you make sense of $\langle f'',\varphi \rangle =-\langle f',\varphi' \rangle$, if you consider $f''$ as a distribution? $\endgroup$
    – 0xbadf00d
    Commented Jul 14, 2016 at 11:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .