Let $f \in L^2(\Omega)$ for some smooth domain $\Omega \subset \mathbb{R}^n$, and let $u \in H^1_0(\Omega)$ such that $$ \sum_{i=1}^{n} \int_\Omega \partial_{x_i}u \partial_{x_i}v = \int_\Omega fv $$ for all $v \in H^1_0(\Omega)$.

Then I have questions about the following statements.

1) $-\Delta u = f$ holds in distributional sense.

Does this mean that if $T_u$ denotes the distribution "generated" by $u$ via

$$ \langle T_u, \phi \rangle = \int_\Omega u \phi $$ for smooth test function $\phi$, then the (negative) distributional Laplacian of $u$, $-\Delta T_u$ is the same distribution as the distribution $T_f$ generated by $f$ as

$$ \langle T_f,\phi\rangle = \int_{\Omega}f\phi? $$ $\qquad$

2) Since $f\in L^2(\Omega)$, the equation is $-\Delta u = f$ is true in $L^2(\Omega)$, hence almost everywhere in $\Omega$.

What does this statement mean?

All I know from 1) is that $-\Delta T_u$ and $T_f$ are the same mappings on the set of test functions $\phi$, but this statement seems to say that the pointwise negative Laplacian of $u$ is defined almost everywhere and square integrable, and equal to $f$. How is $-\Delta u$ defined for almost every $x \in \Omega$?


1 Answer 1


If $u$ has ordinary second order derivatives and $\phi$ is a test function, then $$\langle T_{-\Delta u}, \phi \rangle = \int_\Omega (-\Delta u) \phi = \int_\Omega Du \cdot D\phi.$$ The last integral makes sense even if $u$ belongs only to $H^1(\Omega)$, and the distribution $T$ defined by $$\langle T,\phi \rangle = \int_\Omega Du \cdot D\phi$$ is called the distributional Laplacian of $u$. If it happens that $$\int_\Omega Du \cdot D\phi = \int_\Omega f\phi$$ for all test functions $\phi$ then you have $T = T_f$. This is what is meant by $f = -\Delta u$ in the distributional sense.

Your second question requires some regularity theory to ensure that $u$ has second order weak derivatives.

  • $\begingroup$ Am I right that your saying $u$ having ordinary second order derivatives can be replaced by $u \in H^2(\Omega)$? $\endgroup$
    – tgtt
    Oct 12, 2018 at 21:13
  • $\begingroup$ Yes. That's part of the regularity theory which is a bit sophisticated. $\endgroup$
    – Umberto P.
    Oct 13, 2018 at 1:56

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