Variational formulations: the Poisson equation
Problem statement ¶ { − Δ u = f on Ω , u = 0 on Γ D , ∂ n u = 0 on Γ N . \left\{
\begin{aligned}
-\Delta u &= f \ &\text{on }\Omega, \\
u &= 0 \ &\text{on }\Gamma_D, \\
\partial_n u &= 0 \ &\text{on }\Gamma_N.
\end{aligned}
\right. ⎩ ⎨ ⎧ − Δ u u ∂ n u = f = 0 = 0 on Ω , on Γ D , on Γ N . The weak formulation is obtained by multiplying by a test function v ∈ H 0 1 ( Ω ) v \in H_0^1(\Omega) v ∈ H 0 1 ( Ω ) Ω \Omega Ω
− ∫ Ω Δ u v d V = ∫ Ω f v d V . -\int_\Omega \Delta u\: v \:\mathrm{d}V = \int_\Omega f v\:\mathrm{d}V. − ∫ Ω Δ u v d V = ∫ Ω f v d V . Using Green’s theorem, we can express the left-hand term as:
− ∫ Ω Δ u v d V = ∫ Ω ∇ u ‾ ⋅ ∇ v ‾ d V − ∫ ∂ Ω ∂ n u v d S , -\int_\Omega \Delta u\: v \:\mathrm{d}V = \int_\Omega \underline{\nabla u}\cdot\underline{\nabla v}\:\mathrm{d}V - \int_{\partial\Omega} \partial_n u\:v\:\mathrm{d}S, − ∫ Ω Δ u v d V = ∫ Ω ∇ u ⋅ ∇ v d V − ∫ ∂ Ω ∂ n u v d S , Resulting in the following weak formulation:
∫ Ω ∇ u ‾ ⋅ ∇ v ‾ d V = ∫ Ω f v d V + ∫ ∂ Ω ∂ n u v d S . \int_\Omega \underline{\nabla u}\cdot\underline{\nabla v}\:\mathrm{d}V = \int_\Omega f v\:\mathrm{d}V + \int_{\partial\Omega} \partial_n u\:v\:\mathrm{d}S. ∫ Ω ∇ u ⋅ ∇ v d V = ∫ Ω f v d V + ∫ ∂ Ω ∂ n u v d S . However, in the case of our problem, ∂ n u = 0 \partial_n u =0 ∂ n u = 0 Γ N \Gamma_N Γ N v = 0 v=0 v = 0 Γ D \Gamma_D Γ D ∂ Ω = Γ D ∪ Γ N \partial\Omega = \Gamma_D \cup \Gamma_N ∂ Ω = Γ D ∪ Γ N
Boundary conditions ¶ We distinguish two types of boundardy conditions: ones which are applied directly to the definition space of the function (Dirichlet boudnary conditions, also called essential boundary conditions), and conditions enforced in the weak formulation which appear when using Green’s theorem (called Neumann or natural boundary conditions). The latter is enforced through the weak form.