The weak formulation of the Poisson equation

17. The weak formulation of the Poisson equation#

We are now able to give a precise definition of the weak formulation of the Poisson problem as introduced in the first unit, and analyze the existence and uniqueness of a weak solution.

Let \(\Omega\) be a bounded domain. Its boundary \(\partial \Omega\) is decomposed as \(\partial \Omega = \Gamma_D \cup \Gamma_N \cup \Gamma_R\) according to Dirichlet, Neumann and Robin boundary conditions.


  • \(u_D \in H^{1/2}(\Gamma_D)\),

  • \(f \in L_2(\Omega)\),

  • \(g \in L_2(\Gamma_N \cup \Gamma_R)\),

  • \(\alpha \in L_\infty(\Gamma_D), \alpha \geq 0\).

Assume that there holds

  • The Dirichlet part has positive measure \(| \Gamma_D | > 0\),

  • or the Robin term has positive contribution \(\int_{\Gamma_R} \alpha \, dx > 0\).

Define the Hilbert space

\[ V := H^1(\Omega), \]

the closed sub-space

\[ V_0 = \{ v : \operatorname{tr}_{\Gamma_D} \, v = 0 \}, \]

and the linear manifold

\[ V_D = \{ u \in V : \operatorname{tr}_{\Gamma_D} \, u = u_D \}. \]

Define the bilinear-form \(A(.,.) : V \times V \rightarrow {\mathbb R}\)

\[ A(u,v) = \int_\Omega \nabla u \, \nabla v \, dx + \int_{\Gamma_R} \alpha u v \, ds \]

and the linear form

\[ f(v) = \int_\Omega f v \, dx + \int_{\Gamma_N \cup \Gamma_R} g v \, dx. \]

Theorem: The weak formulation of the Poisson problem: find \(u \in V_D\) such that

\[ A(u,v) = f(v) \qquad \forall \, v \in V_0 \]

has a unique solution \(u \in V\).

Proof: The bilinear-form \(A(.,.)\) and the linear-form \(f(.)\) are continuous on \(V\). Tartar’s theorem of equivalent norms proves that \(A(.,.)\) is coercive on \(V_0\).

Since \(u_D\) is in the closed range of \(\operatorname{tr}_{\Gamma_D}\), there exists an \(\tilde u_D \in V_D\) such that

\[ \operatorname{tr} \, \tilde u_D = u_D \qquad \mbox{and} \qquad \| \tilde u_D \|_V \preceq \| u_D \|_{H^{1/2}(\Gamma_D)} \]

Now, pose the problem: Find \(z \in V_0\) such that

\[ A(z,v) = f(v) - A(\widetilde u_D, v) \qquad \forall \, v \in V_0. \]

The right hand side is the evaluation of the continuous linear form \(f(.) - A(\widetilde u_D, .)\) on \(V_0\). Due to Lax-Milgram, there exists a unique solution \(z\). Then, \(u := \widetilde u_D + z\) solves the weak formulation of the problem. The choice of \(\widetilde u_D\) is not unique, but, the constructed \(u\) is unique. \(\Box\)

17.1. Shift theorems#

Let us restrict to Dirichlet boundary conditions \(u_D = 0\) on the whole boundary. The variational problem: Find \(u \in V_0\) such that

\[ A(u,v) = f(v) \qquad \forall \, v \in V_0 \]

is well defined for all \(f \in V_0^\ast\), and, due to Lax-Milgram there holds

\[ \| u \|_{V_0} \leq c \| f \|_{V_0^\ast}. \]

Vice versa, the bilinear-form defines the linear functional \(A(u,.)\) with norm

\[ \| A(u,.) \|_{V_0^\ast} \leq c \| u \|_{V_0} \]

This dual space is called \(H^{-1}\):

\[ H^{-1} := [H_0^1(\Omega)]^\ast \]

Since \(H_0^1 \subset L_2\), there is \(L_2 \subset H^{-1}(\Omega)\). All negative spaces are defined as \(H^{-s}(\Omega) := [H_0^s]^\ast(\Omega)\), for \(s \in {\mathbb R}^+\). There holds

\[ \ldots H_0^{2} \subset H_0^{1} \subset L_2 \subset H^{-1} \subset H^{-2} \ldots \]

The solution operator of the weak formulation is smoothing twice. The statements of shift theorem are that for \(s > 0\), the solution operator maps also

\[ f \in H^{-1+s} \rightarrow u \in H^{1+s} \]

with norm bounds

\[ \| u \|_{H^{1+s}} \preceq \| f \|_{H^{-1+s}}. \]

In this case, we call the problem \(H^{1+s}\) - regular.

Shift theorems

  1. Assume that \(\Omega\) is convex. Then, the Dirichlet problem is \(H^2\) regular.

  2. Let \(s \geq 2\). Assume that \(\partial \Omega \in C^s\). Then, the Dirichlet problem is \(H^s\)-regular.

We give a proof of 1. for the square \((0,\pi)^2\) by Fourier series. Let

\[ V_N = \mbox{span} \{ \sin (k x) \sin (ly) : 1 \leq k,l \leq N \} \]

For an \(u = \sum_{k,l = 1}^N u_{kl} \sin (kx) \sin (ly) \in V_N\), there holds

\[\begin{align*} \| u \|_{H^2}^2 & = \| u \|_{L_2}^2 + \| \partial_x u \|_{L_2}^2 + \| \partial_y u \|_{L_2}^2 + \| \partial _x^2 u \|_{L_2}^2 + \| \partial_x \partial_y u \|_{L_2}^2 + \| \partial_y^2 u \|^2 \\ & \simeq \sum_{k,l = 1}^N (1 + k^2 +l^2 + k^4 + k^2 l^2 + l^4) u_{kl}^2 \\ & \simeq \sum_{k,l = 1}^N (k^4 + l^4) u_{kl}^2, \end{align*}\]

and, for \(f = -\Delta u\),

\[ \| - \Delta u \|_{L_2}^2 = \sum_{k,l=1}^N (k^2+l^2)^2 u_{kl}^2 \simeq \sum_{k,l=1}^N (k^4+l^4) u_{kl}^2. \]

Thus we have \(\| u \|_{H^2} \simeq \| \Delta u \|_{L_2} = \| f \|_{L_2}\) for \(u \in V_N\). The rest requires a closure argument: There is \(\{ -\Delta v : v \in V_N \} = V_N\), and \(V_N\) is dense in \(L_2\). \(\Box\)

Indeed, on non-smooth non-convex domains, the \(H^2\)-regularity is not true. Take the sector of the unit-disc

\[ \Omega = \{ (r \cos \phi, r \sin \phi) : 0 < r < 1, \; 0 < \phi < \omega \} \]

with \(\omega \in (\pi, 2 \pi)\). Set \(\beta = \pi/\omega < 1\). The function

\[ u = (1-r^2) r^\beta sin (\phi \beta) \]

is in \(H_0^1\), and fulfills \(\Delta u = -(4 \beta + 4) r^{\beta} sin (\phi \beta) \in L_2\). Thus \(u\) is the solution of a Dirichlet problem. But \(u \not\in H^2\).

On non-convex domains one can specify the regularity in terms of weighted Sobolev spaces. Let \(\Omega\) be a polygonal domain containing \(M\) vertices \(V_i\). Let \(\omega_i\) be the interior angle at \(V_i\). If the vertex belongs to a non-convex corner (\(\omega_i > \pi\)), then choose some

\[ \beta_i \in (1 - \frac{\pi}{\omega}, 1) \]


\[ w(x) = \prod_{\mbox{non-convex} \atop \mbox{Vertices V}_i} | x - V_i |^{\beta_i} \]

Theorem: If \(f\) is such that \(w f \in L_2\). Then \(f \in H^{-1}\), and the solution \(u\) of the Dirichlet problem fulfills

\[ \| w D^2 u \|_{L_2} \preceq \| w f \|_{L_2}. \]