A function $u$ on the square $\Omega=[0,L]\times[0,L]$ is said to satisfy periodic boundary conditions (PBCs) if $u(x,0)=u(x,L)$ and $u(0,y)=u(L,y)$ for all $x,y\in[0,L]$. Now consider the Poisson equation $$\Delta u=f \;\;\text{ on }\;\;\Omega,$$ where $f\in L^2(\Omega)$. Suppose that we seek a solution $u$ such that $u$, $\partial_xu$ and $\partial_yu$ satisfy periodic boundary conditions.

Question. Which theorem or method of solving should we use to tackle such a problem? If there is a solution, is it unique?

For instance, in variational form we have $$\langle \nabla u,\nabla v\rangle=\langle f,v\rangle\;\;\;\;\text{ for all }\;\;\;\;\;v\in H:=\{\phi\in H^1(\Omega):Trace(\phi) \text{ satisfies PBCs}\}$$ (the boundary integral will vanish after applying Green's formula). So it almost seems we can apply Lax-Milgram on $H$ provided we know that $a(u,v)=\langle \nabla u,\nabla v\rangle$ is coercive on $H$. However that does not seem to be the case!

I further noticed that the divergence theorem implies that $$\int_\Omega f=\int_\Omega \Delta u=\int_{\partial\Omega} \nabla u\cdot\vec{n}=0, $$ where the last equality holds because of the periodicity of $\partial_xu$ and $\partial_yu$.


1 Answer 1


$H^1(\Omega)$ is not the correct space to analyze this problem. The correct space to analyze is to use $H^1(\mathbb{T}^2)$, where $\mathbb{T}^2$ is the 2D torus. The $L^2$-norm can be most easily understood using Parseval's identity for Fourier series: let $k\in \mathbb{Z}^2$ and $$\hat{v}_k := \int_{[0,L]^2} e^{-2\pi i k\cdot x} v(x)\, dx \quad \text{ and } \quad \|v\|_{L^2(\mathbb{T}^2)}^2 = c\sum_{k\in \mathbb{Z}^2} |k|^2, $$ where $c$ is a constant determined by the Fourier basis you choose but fixed once the basis are chosen. $H^1$ follows naturally. $$ H^1(\mathbb{T}^2) := \{v\in L^2(\mathbb{T}^2): \|v\|_{H^1(\mathbb{T}^2)}<\infty\}, $$ where $$ \|v\|_{H^1(\mathbb{T}^2)}^2 = \sum_{k \in\mathbb{Z}^2}(1 +|k|^2) |\hat{v}_k|^2. $$ Poincare inequality is naturally true for this space (which is needed to prove Lax-Milgram), and you are good to go.


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