Weyl's lemma (Laplace equation)

In mathematics, Weyl's lemma, named after Hermann Weyl, states that every weak solution of Laplace's equation is a smooth solution. This contrasts with the wave equation, for example, which has weak solutions that are not smooth solutions. Weyl's lemma is a special case of elliptic or hypoelliptic regularity.

Statement of the lemma

Let ${\displaystyle \Omega }$ be an open subset of ${\displaystyle n}$-dimensional Euclidean space ${\displaystyle \mathbb {R} ^{n}}$, and let ${\displaystyle \Delta }$ denote the usual Laplace operator. Weyl's lemma^[1] states that if a locally integrable function ${\displaystyle u\in L_{\mathrm {loc} }^{1}(\Omega )}$ is a weak solution of Laplace's equation, in the sense that

${\displaystyle \int _{\Omega }u(x)\,\Delta \varphi (x)\,dx=0}$

for every test function (smooth function with compact support) ${\displaystyle \varphi \in C_{c}^{\infty }(\Omega )}$, then (up to redefinition on a set of measure zero) ${\displaystyle u\in C^{\infty }(\Omega )}$ is smooth and satisfies ${\displaystyle \Delta u=0}$ pointwise in ${\displaystyle \Omega }$.

This result implies the interior regularity of harmonic functions in ${\displaystyle \Omega }$, but it does not say anything about their regularity on the boundary ${\displaystyle \partial \Omega }$.

Idea of the proof

To prove Weyl's lemma, one convolves the function ${\displaystyle u}$ with an appropriate mollifier ${\displaystyle \varphi _{\varepsilon }}$ and shows that the mollification ${\displaystyle u_{\varepsilon }=\varphi _{\varepsilon }\ast u}$ satisfies Laplace's equation, which implies that ${\displaystyle u_{\varepsilon }}$ has the mean value property. Taking the limit as ${\displaystyle \varepsilon \to 0}$ and using the properties of mollifiers, one finds that ${\displaystyle u}$ also has the mean value property,^[2] which implies that it is a smooth solution of Laplace's equation.^[3][4] Alternative proofs use the smoothness of the fundamental solution of the Laplacian or suitable a priori elliptic estimates.

Proof

Let ${\displaystyle \left(\rho _{\varepsilon }\right)_{\varepsilon >0}}$ be the standard mollifier.

Fix a compact set ${\displaystyle \Omega ^{\prime }\Subset \Omega }$ and put ${\displaystyle \varepsilon _{0}=\operatorname {dist} \left(\Omega ^{\prime },\partial \Omega \right)}$ be the distance between ${\displaystyle \Omega ^{\prime }}$ and the boundary of ${\displaystyle \Omega }$.

For each ${\displaystyle x\in \Omega ^{\prime }}$ and ${\displaystyle \varepsilon \in \left(0,\varepsilon _{0}\right)}$ the function

${\displaystyle y\longmapsto \rho _{\varepsilon }(x-y)}$

belongs to test functions ${\displaystyle {\mathcal {D}}(\Omega )}$ and so we may consider

${\displaystyle \left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle }$

We assert that it is independent of ${\displaystyle \varepsilon \in \left(0,\varepsilon _{0}\right)}$. To prove it we calculate ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\rho _{\varepsilon }(x-y)}$ for ${\displaystyle x,y\in \mathbb {R} ^{n}}$.

Recall that

${\displaystyle \rho _{\varepsilon }(x-y)=\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)}$

where the standard mollifier kernel ${\displaystyle \rho }$ on ${\displaystyle \mathbb {R} ^{n}}$ was defined at Mollifier#Concrete_example. If we put

${\displaystyle \theta (t)={\begin{cases}{\frac {1}{c_{n}}}\mathrm {e} ^{\frac {1}{t-1}}&{\text{ if }}t<1,\\0&{\text{ if }}t\geqslant 1,\end{cases}}}$

then ${\displaystyle \rho (x)=\theta \left(|x|^{2}\right)}$.

Clearly ${\displaystyle \theta \in \mathrm {C} ^{\infty }(\mathbb {R} )}$ satisfies ${\displaystyle \theta (t)=0}$ for ${\displaystyle t\geqslant 1}$. Now calculate

${\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left(\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)\right)&=-n\varepsilon ^{-n-1}\rho \left({\frac {x-y}{\varepsilon }}\right)-\varepsilon ^{-n}\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon ^{2}}}\\&=-{\frac {1}{\varepsilon ^{n+1}}}\left(n\rho \left({\frac {x-y}{\varepsilon }}\right)+\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon }}\right)\end{aligned}}}$

Put ${\displaystyle K(x)=-n\rho (x)-\nabla \rho (x)\cdot x}$ so that

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left(\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)\right)={\frac {1}{\varepsilon ^{n+1}}}K\left({\frac {x-y}{\varepsilon }}\right).}$

In terms of ${\displaystyle \rho (x)=\theta \left(|x|^{2}\right)}$ we get

${\displaystyle K(x)=-\operatorname {div} (\rho (x)x)=-\operatorname {div} \left(\theta \left(|x|^{2}\right)x\right)}$

and if we set

${\displaystyle \Theta (t)={\frac {1}{2}}\int _{t}^{\infty }\theta (s)\mathrm {d} s}$

then ${\displaystyle \Theta \in \mathrm {C} ^{\infty }(\mathbb {R} )}$ with ${\displaystyle \Theta (t)=0}$ for ${\displaystyle t\geqslant 1}$, and ${\displaystyle \Theta ^{\prime }(t)=-{\frac {1}{2}}\theta (t)}$. Consequently

${\displaystyle -\theta \left(|x|^{2}\right)x=\nabla \left(\Theta \left(|x|^{2}\right)\right)}$

and so ${\displaystyle K(x)=\operatorname {div} \nabla \left(\Theta \left(|x|^{2}\right)\right)=(\Delta \Phi )(x)}$, where ${\displaystyle \Phi (x)=\Theta \left(|x|^{2}\right)}$. Observe that ${\displaystyle \Phi \in {\mathcal {D}}\left({\overline {B_{1}(0)}}\right)}$, and

${\displaystyle {\begin{aligned}-{\frac {1}{\varepsilon ^{n+1}}}\left(n\rho \left({\frac {x-y}{\varepsilon }}\right)+\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon }}\right)&={\frac {1}{\varepsilon ^{n+1}}}\Delta _{y}\left(\Phi \left({\frac {x-y}{\varepsilon }}\right)\right)\\&=\Delta _{y}\left(\varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)\right).\end{aligned}}}$

Here ${\displaystyle y\mapsto \varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)}$ is supported in ${\displaystyle {\overline {B_{\varepsilon }(x)}}\subset \Omega }$, and so by assumption

${\displaystyle \left\langle u,\Delta _{y}\left(\varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)\right)\right\rangle =0}$.

Now by considering difference quotients we see that

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =\left\langle u,{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\rho _{\varepsilon }(x-\cdot )\right\rangle }$.

Indeed, for ${\displaystyle \varepsilon ,\varepsilon ^{\prime }>0}$ we have

${\displaystyle {\begin{aligned}{\frac {\rho _{\varepsilon +\varepsilon ^{\prime }}(x-y)-\rho _{\varepsilon }(x-y)}{\varepsilon ^{\prime }}}&{\stackrel {\mathrm {FTC} }{=}}\int _{0}^{1}{\frac {\mathrm {d} }{\mathrm {d} t}}\rho _{\varepsilon +t\varepsilon ^{\prime }}(x-y)\mathrm {d} t\\&\left.{\underset {\varepsilon ^{\prime }\searrow 0}{\longrightarrow }}{\frac {\mathrm {d} }{\mathrm {d} s}}\right|_{s=\varepsilon }\rho _{s}(x-y)\end{aligned}}}$

in ${\displaystyle {\mathcal {D}}^{\prime }(\Omega )}$ with respect to ${\displaystyle y}$, provided ${\displaystyle x\in \Omega ^{\prime }}$ and ${\displaystyle 0<\varepsilon <\varepsilon _{0}}$ (since we may differentiate both sides with respect to ${\displaystyle y)}$. But then ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =0}$, and so ${\displaystyle \left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle }$ for all ${\displaystyle \varepsilon \in \left(0,\varepsilon _{0}\right)}$, where ${\displaystyle \varepsilon _{1}\in \left(0,\varepsilon _{0}\right)}$. Now let ${\displaystyle \varphi \in {\mathcal {D}}\left(\Omega ^{\prime }\right)}$. Then, by the usual trick when convolving distributions with test functions,

${\displaystyle {\begin{aligned}\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x&=\left\langle u,\int _{\Omega ^{\prime }}\rho _{\varepsilon }(x-\cdot )\varphi (x)\mathrm {d} x\right\rangle \\&=\left\langle u,\rho _{\varepsilon }*\varphi \right\rangle \end{aligned}}}$

and so for ${\displaystyle \varepsilon \in \left(0,\varepsilon _{1}\right)}$ we have

${\displaystyle \left\langle u,\rho _{\varepsilon }*\varphi \right\rangle =\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x}$.

Hence, as ${\displaystyle \rho _{\varepsilon }*\varphi \rightarrow \varphi }$ in ${\displaystyle {\mathcal {D}}(\Omega )}$ as ${\displaystyle \varepsilon \searrow 0}$, we get

${\displaystyle \langle u,\varphi \rangle =\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x}$.

Consequently ${\displaystyle \left.u\right|_{\Omega ^{\prime }}\in \mathrm {C} ^{\infty }\left(\Omega ^{\prime }\right)}$, and since ${\displaystyle \Omega ^{\prime }}$ was arbitrary, we are done.

Generalization to distributions

More generally, the same result holds for every distributional solution of Laplace's equation: If ${\displaystyle T\in D'(\Omega )}$ satisfies ${\displaystyle \langle T,\Delta \varphi \rangle =0}$ for every ${\displaystyle \varphi \in C_{c}^{\infty }(\Omega )}$, then ${\displaystyle T}$ is a regular distribution associated with a smooth solution ${\displaystyle u\in C^{\infty }(\Omega )}$ of Laplace's equation.^[5]

Connection with hypoellipticity

Weyl's lemma follows from more general results concerning the regularity properties of elliptic or hypoelliptic operators.^[6] A linear partial differential operator ${\displaystyle P}$ with smooth coefficients is hypoelliptic if the singular support of ${\displaystyle Pu}$ is equal to the singular support of ${\displaystyle u}$ for every distribution ${\displaystyle u}$. The Laplace operator is hypoelliptic, so if ${\displaystyle \Delta u=0}$, then the singular support of ${\displaystyle u}$ is empty since the singular support of ${\displaystyle 0}$ is empty, meaning that ${\displaystyle u\in C^{\infty }(\Omega )}$. In fact, since the Laplacian is elliptic, a stronger result is true, and solutions of ${\displaystyle \Delta u=0}$ are real-analytic.

Notes

1. Hermann Weyl, The method of orthogonal projections in potential theory, Duke Math. J., 7, 411–444 (1940). See Lemma 2, p. 415
2. The mean value property is known to characterize harmonic functions in the following sense. Let ${\displaystyle h\in \mathrm {C} (\Omega )}$. Then ${\displaystyle h}$ is harmonic in the usual sense (so ${\displaystyle h\in \mathrm {C} ^{2}(\Omega )}$ and ${\displaystyle \left.\Delta h=0\right)}$ if and only if for all balls ${\displaystyle B_{r}\left(x_{0}\right)\Subset \Omega }$ we have

   ${\displaystyle h\left(x_{0}\right)={\frac {1}{\omega _{n-1}r^{n-1}}}\int _{\partial B_{r}\left(x_{0}\right)}h(x)\mathrm {d} S_{x}.}$

   where ${\displaystyle \omega _{n-1}r^{n-1}}$ is the (n − 1)-dimensional area of the hypersphere ${\displaystyle \partial B_{r}\left(x_{0}\right)}$. Using polar coordinates about ${\displaystyle x_{0}}$ we see that when ${\displaystyle h}$ is harmonic, then for ${\displaystyle B_{r}\left(x_{0}\right)\Subset \Omega }$,

   ${\displaystyle h\left(x_{0}\right)=\left(\rho _{r}*h\right)\left(x_{0}\right).}$

3. Bernard Dacorogna, Introduction to the Calculus of Variations, 2nd ed., Imperial College Press (2009), p. 148.
4. Stroock, Daniel W. "Weyl's lemma, one of many" (PDF).
5. Lars Gårding, Some Points of Analysis and their History, AMS (1997), p. 66.
6. Lars Hörmander, The Analysis of Linear Partial Differential Operators I, 2nd ed., Springer-Verlag (1990), p.110

References

• Gilbarg, David; Neil S. Trudinger (1988). Elliptic Partial Differential Equations of Second Order. Springer. ISBN 3-540-41160-7.
• Stein, Elias (2005). Real Analysis: Measure Theory, Integration, and Hilbert Spaces. Princeton University Press. ISBN 0-691-11386-6.