How Do the Two Dirichlet Laplacian Operators Differ?

The symbol $-\Delta$ can describe two related operators, but the spaces in the arrow matter. One is the weak Dirichlet Laplacian, which takes a function in $H_0^1(\Omega)$ to a continuous linear functional on $H_0^1(\Omega)$. The other takes a more regular function in $H^2(\Omega)\cap H_0^1(\Omega)$ to its Laplacian in $L^2(\Omega)$. They express the same differential equation at different levels of regularity.

Assume for the basic discussion that $\Omega$ is a bounded open set. Boundary regularity becomes important when asking whether every $L^2$ right-hand side has an $H^2$ solution.

The weak operator $H_0^1\to H^{-1}$

Here $H^{-1}(\Omega)$ means the continuous dual of $H_0^1(\Omega)$ (often written $(H_0^1(\Omega))^*$). Define $A:H_0^1(\Omega)\to H^{-1}(\Omega)$ by

$$\langle Au,v\rangle_{H^{-1},H_0^1}=\int_\Omega \nabla u\cdot\nabla v\,dx,\qquad v\in H_0^1(\Omega).$$

This is the weak meaning of $Au=-\Delta u$. It is well-defined even if $u$ has only one weak derivative in $L^2$: the right-hand side is bounded as a functional of $v$, by the Cauchy–Schwarz inequality. Thus $A$ is a bounded linear map between these Sobolev spaces. For bounded $\Omega$, Poincaré’s inequality makes the gradient norm equivalent to the usual $H_0^1$ norm.

This formulation is useful for weak solutions. Given $f\in H^{-1}(\Omega)$, the equation $Au=f$ means

$$\int_\Omega \nabla u\cdot\nabla v\,dx=\langle f,v\rangle\quad\text{for every }v\in H_0^1(\Omega).$$

The bilinear form is coercive on $H_0^1(\Omega)$, so the Lax–Milgram theorem gives a unique weak solution for each such $f$. Equivalently, this weak operator is an isomorphism from $H_0^1$ onto $H^{-1}$, with the usual bounded-domain assumptions.

The stronger operator $H^2\cap H_0^1\to L^2$

If $u\in H^2(\Omega)$, its second weak derivatives belong to $L^2(\Omega)$. Therefore $-\Delta u=-\sum_{j=1}^n\partial_j^2u$ is an $L^2$ function. Restricting to functions in $H_0^1(\Omega)$ gives the map

$$B:H^2(\Omega)\cap H_0^1(\Omega)\to L^2(\Omega),\qquad Bu=-\Delta u.$$

This definition demands more from $u$ than the weak one: $H_0^1$ controls first derivatives, while $H^2$ also controls second derivatives. Equipped with the $H^2$ norm, the displayed map is bounded. But if one regards $B$ as an operator on $L^2(\Omega)$, its domain is the proper subspace $H^2\cap H_0^1$; it is then an unbounded operator on $L^2$, not a bounded map defined on all of $L^2$.

Whether this domain gives an onto map to $L^2$ depends on $\Omega$. For example, on a bounded domain with sufficiently smooth boundary (such as a $C^{1,1}$ domain), Dirichlet elliptic regularity gives an $H^2$ solution for every $f\in L^2$. On rougher domains, a weak solution may fail to belong to $H^2$, so the displayed operator need not be onto. More generally, the natural $L^2$ realization of the Dirichlet Laplacian can be defined on $\{u\in H_0^1(\Omega):\Delta u\in L^2(\Omega)\}$; this domain need not equal $H^2\cap H_0^1$ on an irregular domain.

How the two definitions agree

For every $u\in H^2(\Omega)\cap H_0^1(\Omega)$ and $v\in H_0^1(\Omega)$, integration by parts in the weak sense gives

$$\int_\Omega \nabla u\cdot\nabla v\,dx=\int_\Omega (-\Delta u)v\,dx.$$

The right side treats $-\Delta u\in L^2$ as a functional on $H_0^1$. This is valid because the inclusion $L^2(\Omega)\hookrightarrow H^{-1}(\Omega)$ is continuous on a bounded domain: an $L^2$ function acts on $v$ by $\int_\Omega fv\,dx$. Thus $B$ is the restriction of the weak operator after identifying its output in $L^2$ with the corresponding element of $H^{-1}$.

For a concrete example, let $\Omega=(0,1)$ and $u(x)=x(1-x)$. Then $u\in H^2(0,1)\cap H_0^1(0,1)$ and $-u''=2$. The strong operator returns the function $2\in L^2(0,1)$. The weak operator returns the functional

$$v\longmapsto\int_0^1 2v(x)\,dx,$$

which is the same output viewed as an element of $H^{-1}(0,1)$.

Common mistakes

• Treating the arrows as interchangeable. The codomains differ: $H^{-1}$ contains functionals, while $L^2$ contains functions. An $L^2$ function can be embedded in $H^{-1}$, but not every $H^{-1}$ functional is an $L^2$ function.
• Assuming $u\in H_0^1$ has an $L^2$ Laplacian. The weak operator is defined for such $u$, but its output is generally only in $H^{-1}$.
• Assuming the second map is always onto. That conclusion requires elliptic regularity assumptions on the domain; the weak solution exists under much weaker assumptions.
• Confusing boundedness with the unbounded-operator viewpoint. The map is bounded when its domain has the $H^2$ norm. As an operator in $L^2$ with the $L^2$ norm on its domain, it is unbounded.