If $-\Delta u=f$ in the distributional sense, can we conclude that $\int_Λ(-\Delta u)v\:{\rm d}λ=\int_Λfv\:{\rm d}λ$ for all $v\in H^1(\Lambda)$?

Question

Let

• $d\in\mathbb N$
• $\lambda$ denote the Lebesgue measure on $\mathbb R^d$
• $\Lambda\subseteq\mathbb R^d$ be open
• $u\in H^2(\Lambda)$
• $f\in L^2(\Lambda)$

Assume $$-\Delta u=f\tag 1$$ in the sense of distributions. Why can we conclude that $$\int_\Lambda(-\Delta u)v\:{\rm d}\lambda=\int_\Lambda fv\:{\rm d}\lambda\tag 2$$ for all $v\in H^1(\Lambda)$.

I've read that in a lecture note (see page 78) and don't understand this. Clearly, the claim would be true for $v\in H_0^1(\Lambda)$. Maybe they use an approximation argument that I'm not aware of (with that in mind, please feel free to add any regularity assumption on $\Lambda$).

Answer 1

This comes from the regularity of $u$. For $u\in H^2$, $$-\Delta u=f\quad\text{in}\quad\mathcal{D}'(\Lambda)\tag{#}$$ means $$\int_\Lambda(-\Delta u)\varphi\;\mathrm{d}\lambda=\int_\Lambda f\varphi\;\mathrm{d}\lambda\quad\forall\ \varphi\in\mathcal{D}(\Lambda)$$ which implies $$-\Delta u=f\quad\text{a.e.}\tag{$*$}$$ by the du Bois-Reymond Lemma. Multiplying $(*)$ by $v\in H^1(\Lambda)$ and integrating over $\Lambda$ you get the result (without any further regularity on $\Lambda$).

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Edit

Let $\Lambda\subset\mathbb{R}^d$ be open and $f,u\in L^1 _{\text{loc}}(\Lambda)$. The distributions $f$, $u$ and $\Delta u$ are defined by $$\langle f,\varphi\rangle=\int_\Lambda f\varphi\;\mathrm{d}\lambda,\quad \langle u,\varphi\rangle=\int_\Lambda u\varphi\;\mathrm{d}\lambda,\quad \langle\Delta u,\varphi\rangle=\langle u,\Delta\varphi\rangle=\int_\Lambda u\Delta \varphi\;\mathrm{d}\lambda,\quad\varphi\in\mathcal{D}(\Lambda).$$ Assume that $u,f\in L^1_{\text{loc}}(\Lambda)$ and $-\Delta u=f$ in $\mathcal{D}'(\Lambda)$. Then $$-\int_\Lambda u\Delta \varphi\;\mathrm{d}\lambda=\int_\Lambda f \varphi\;\mathrm{d}\lambda.$$

If, additionally, $u\in H^2(\Lambda)$ and $\Lambda$ is bounded with Lipschitz boundary, then (Corollary 3.46 here) $$\int_\Lambda (-\Delta u) \varphi\;\mathrm{d}\lambda=-\int_\Lambda u\Delta \varphi\;\mathrm{d}\lambda=\int_\Lambda f \varphi\;\mathrm{d}\lambda.$$ Therefore by the Du Bois Reymond Lemma (Lema 1.16 here) $$-\Delta u=f\quad\text{a.e.}$$

Conclusion: $(\#)$ implies that $$\int_\Lambda(-\Delta u)v\;\mathrm{d}\lambda=\int_\Lambda fv$$ for all $v\in H^1(\Lambda)$ provided that $\Lambda$ is a bounded open set with Lipschitz boundary and $u\in H^2(\Lambda)$.

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Edit 2

Print of the third link: