Weyl's lemma (Laplace equation)

Last updated May 30, 2024

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

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

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

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

Idea of the proof

To prove Weyl's lemma, one convolves the function $u$ with an appropriate mollifier $\varphi _{\varepsilon }$ and shows that the mollification $u_{\varepsilon }=\varphi _{\varepsilon }\ast u$ satisfies Laplace's equation, which implies that $u_{\varepsilon }$ has the mean value property. Taking the limit as $\varepsilon \to 0$ and using the properties of mollifiers, one finds that $u$ 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 $\left(\rho _{\varepsilon }\right)_{\varepsilon >0}$ be the standard mollifier.

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

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

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

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

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

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

Recall that

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

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

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

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

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

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

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

and if we set

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

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

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

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

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

\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

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

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

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

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

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

.

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

Generalization to distributions

More generally, the same result holds for every distributional solution of Laplace's equation: If $T\in D'(\Omega )$ satisfies $\langle T,\Delta \varphi \rangle =0$ for every $\varphi \in C_{c}^{\infty }(\Omega )$ , then $T=T_{u}$ is a regular distribution associated with a smooth solution $u\in C^{\infty }(\Omega )$ 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 $P$ with smooth coefficients is hypoelliptic if the singular support of $Pu$ is equal to the singular support of $u$ for every distribution $u$ . The Laplace operator is hypoelliptic, so if $\Delta u=0$ , then the singular support of $u$ is empty since the singular support of $0$ is empty, meaning that $u\in C^{\infty }(\Omega )$ . In fact, since the Laplacian is elliptic, a stronger result is true, and solutions of $\Delta u=0$ are real-analytic.

Notes

↑ Hermann Weyl, The method of orthogonal projections in potential theory, Duke Math. J., 7, 411–444 (1940). See Lemma 2, p. 415
↑ The mean value property is known to characterize harmonic functions in the following sense. Let $h\in \mathrm {C} (\Omega )$ . Then $h$ is harmonic in the usual sense (so $h\in \mathrm {C} ^{2}(\Omega )$ and $\left.\Delta h=0\right)$ if and only if for all balls $B_{r}\left(x_{0}\right)\Subset \Omega$ we have
$h\left(x_{0}\right)={\frac {1}{\omega _{n-1}r^{n-1}}}\int _{\partial B_{r}\left(x_{0}\right)}h(x)\mathrm {d} S_{x}.$
where $\omega _{n-1}r^{n-1}$ is the (n− 1)-dimensional area of the hypersphere $\partial B_{r}\left(x_{0}\right)$ . Using polar coordinates about $x_{0}$ we see that when $h$ is harmonic, then for $B_{r}\left(x_{0}\right)\Subset \Omega$ ,
$h\left(x_{0}\right)=\left(\rho _{r}*h\right)\left(x_{0}\right).$
↑ Bernard Dacorogna, Introduction to the Calculus of Variations, 2nd ed., Imperial College Press (2009), p. 148.
↑ Stroock, Daniel W. "Weyl's lemma, one of many" (PDF).
↑ Lars Gårding, Some Points of Analysis and their History, AMS (1997), p. 66.
↑ Lars Hörmander, The Analysis of Linear Partial Differential Operators I, 2nd ed., Springer-Verlag (1990), p.110

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[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 $h\in \mathrm {C} (\Omega )$ . Then $h$ is harmonic in the usual sense (so $h\in \mathrm {C} ^{2}(\Omega )$ and $\left.\Delta h=0\right)$ if and only if for all balls $B_{r}\left(x_{0}\right)\Subset \Omega$ we have
$h\left(x_{0}\right)={\frac {1}{\omega _{n-1}r^{n-1}}}\int _{\partial B_{r}\left(x_{0}\right)}h(x)\mathrm {d} S_{x}.$
where $\omega _{n-1}r^{n-1}$ is the (n− 1)-dimensional area of the hypersphere $\partial B_{r}\left(x_{0}\right)$ . Using polar coordinates about $x_{0}$ we see that when $h$ is harmonic, then for $B_{r}\left(x_{0}\right)\Subset \Omega$ ,
$h\left(x_{0}\right)=\left(\rho _{r}*h\right)\left(x_{0}\right).$

[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

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