# Relationship between the distributional Laplacian and the weak Laplacian

Let

• $d\in\mathbb N$
• $\Omega\subseteq\mathbb R^d$ be open
• $\langle\;\cdot\;,\;\cdot\;\rangle$ denote the $L^2(\Omega)$- or $L^2(\Omega,\mathbb R^d)$-inner product (depending on the context)
• $\mathcal D:=C_c^\infty(\Omega)$ and $$H:=\overline{\mathcal D}^{\langle\;\cdot\;,\;\cdot\;\rangle_H}$$ with $$\langle\phi,\psi\rangle_H:=\langle\phi,\psi\rangle+\langle\nabla\phi,\nabla\psi\rangle\;\;\;\text{for }\phi,\psi\in\mathcal D$$

Let $$\frac{\partial p}{\partial x_i}(\phi):=-p\left(\frac{\partial\phi}{\partial x_i}\right)\;\;\;\text{for }\phi\in\mathcal D$$ and $\nabla p:=\left(\frac{\partial p}{\partial x_1},\ldots,\frac{\partial p}{\partial x_d}\right)^T$ for $p\in\mathcal D'$ and $$(\nabla\cdot p)(\phi):=\sum_{i=1}^d\frac{\partial p_i}{\partial x_i}(\phi)\;\;\;\text{for }\phi\in\mathcal D$$ for $p\in(\mathcal D')^d$. Then, $$\Delta p:=\nabla\cdot\nabla p$$ is the distributional Laplacian of $p\in\mathcal D$ and it's easy to see that $$(\Delta p)(\phi)=p(\Delta\phi)\;\;\;\text{for all }\phi\in\mathcal D\;.\tag 1$$

Now, each $f\in L_{\text{loc}}^1(\Omega)$ can be identified with some unique $\langle f\rangle\in\mathcal D'$ via $$\langle f\rangle:=\left.\langle\;\cdot\;,f\rangle\right|_{\mathcal D}\;.$$ Let $\phi\in\mathcal D$. We can show that $\Delta\langle\phi\rangle$ has a unique extension $F\in H'$ with $$F(v)=-\langle\nabla v,\nabla\phi\rangle\;\;\;\text{for all }v\in H\;.\tag 2$$ We continue to denote $F$ by $\Delta\langle\phi\rangle$. In the same way $\Delta\langle\;\cdot\;\rangle:\mathcal D\to H'$ has a unique extension $L\in\mathfrak L(H,H')$ with $$(Lu)v=-\langle\nabla v,\nabla u\rangle\;\;\;\text{for all }u,v\in H\;.\tag 3$$

If $u\in L_{\text{loc}}^1(\Omega)$ is weakly differentiable, then $v\in L_{\text{loc}}^1(\Omega)$ is called weak Laplacian of $u$, if $$\langle\phi,v\rangle=-\langle\nabla u,\nabla\phi\rangle\;\;\;\text{for all }\phi\in\mathcal D\tag 4\;.$$ In that case, we write $\Delta u:=v$ and $$(\Delta\langle u\rangle)\psi=\langle\Delta u\rangle(\psi)=\langle\psi,\Delta u\rangle=-\langle\nabla\psi,\nabla u\rangle=(Lu)\psi\;\;\;\text{for all }\psi\in\mathcal D\;.\tag 5$$

If each $u\in H$ would admit a weak Laplacian $\Delta u$, it would make sense to continue to denote $L$ by $\Delta\langle\;\cdot\;\rangle$ and the relationship to the weak Laplacian would be clear. So, does each $u$ admit a weak Laplacian in the sense of $(4)$?

• For $u \in H$, we know that $Lu \in H'$. Moreover, we have that $\langle Lu, \varphi\rangle_{H', H} := \langle \nabla u, \nabla \varphi \rangle_{L^2(\Omega)}$ for all $\varphi \in H$. For $u$ to admit a weak Laplacian, we need a function $\phi \in L_{\text{loc}}^1(\Omega)$ such that $\langle Lu, \varphi\rangle_{H', H} = \langle \phi, \varphi\rangle_{L^2(\Omega)} =: \langle \phi, \varphi\rangle_{H', H}$ for all $\varphi \in H$. Thus, $u$ admits a weak Laplacian iff $Lu \in L_{\text{loc}}^1(\Omega)$. Does this help you to answer your question? – Mike Jul 15 '16 at 10:36
• they are the same on $C^\infty_c$ and $H^1_0$ a dense subset, that's all we need to know. When the distributional Laplacian is really a distribution (say with Dirac deltas) then the weak-Laplacian too (or it isn't defined). I think you really need to take some examples, for example $u(x) = 1-|x|$ on $H^1_0([-1,1])$. – reuns Jul 15 '16 at 10:45
• And did you define the tempered distributions (the dual of Schwartz functions) ? They are really simple, and let you define almost every distributions, in particular all the derivatives of any compactly supported continuous function. And you didn't explain your problem about the Stokes operator yet. Why do you care about ? What do you want to show ? – reuns Jul 15 '16 at 10:50
• @Mike I can't follow your reasoning. Let $u\in L_{\text{loc}}^1(\Omega)$ be weakly differentiable and $v\in L_{\text{loc}}^1(\Omega)$. Then, $v$ is the weak Laplacian of u in the sense of $(4)$ if and only if $$\langle\phi,v\rangle=(Lu)\phi\;\;\;\text{for all }\phi\in\mathcal D\;.$$ How do you deduce $Lu\in L_{\text{loc}}^1(\Omega)$ from that? And what exactly do you understand by the claim $\Phi\in L_{\text{loc}}^1(\Omega)$ for $\Phi\in H'$? – 0xbadf00d Jul 15 '16 at 19:11
• @0xbadf00d:You already mentioned that a function $v \in L_{\text{loc}}^1(\Omega)$ can be identified with a distribution $\langle v \rangle \in \mathcal{D}'$. In fact, $\langle v \rangle \in H'$. So every weak Laplacian $v \in L_{\text{loc}}^1(\Omega)$ induces a distribution $\langle v \rangle \in H'$. Let $v$ be the weak Laplacian of $u$. I claim that $\langle v \rangle$ is the distributional Laplacian of $u$. To see this, note that $(4)$ holds for all $\phi \in H$ (density of $\mathcal{D}$). Now your question boils down to: – Mike Jul 15 '16 at 20:01