# Numerical Computation of the Helmholtz-Leray Projection

Let

• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$ be bounded and open
• $u\in L^2(\Lambda,\mathbb R^d)$

It's well-known that there is a weakly differentiable $p\in L_{\text{loc}}^2(\Lambda)$ with $\nabla p\in L^2(\Lambda)$, $$\Delta p=\nabla\cdot u\;,\tag 1$$ i.e. $$\langle\nabla p,\nabla\phi\rangle_{L^2(\Lambda,\:\mathbb R^d)}=\langle u,\nabla\phi\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for all }\phi\in C_c^\infty(\Lambda)\;,\tag 2$$ and $$\left.\frac{\partial p}{\partial\nu}\right|_{\partial\Lambda}={\rm n}\cdot u\tag 3\;,$$ where $\nu$ denotes the outer unit normal field of $\partial\Lambda$.

Now, I want to compute $p$ numerically in the case $\Lambda=(0,a)\times(0,b)$ with $a,b>0$.

I'm not searching for a state-of-the-art algorithm; just a simple solver. However, I would be happy about any book/paper where this problem is considered; no matter which method is used.

• Your weak formulation $(2)$ is somewhat different from $$\int_\Lambda \nabla p \nabla \phi d\lambda = \int_\Lambda (\mathbf f\nabla)\phi d\lambda$$ used, for eample, in FEM methods. Also, note, you're dot producting $\nabla$ with a scalar $\phi$ – uranix Dec 16 '16 at 23:36
• @uranix You're right, the right-hand side of $(2)$ was wrong. I've fixed that. But the left-hand side is equal to the left-hand side of your equation. Do you know a reference for my specific problem? I can't imagine that there is no literature related to the numerical computation of the Helmholtz-Leray decomposition, but I couldn't find a suitable reference. – 0xbadf00d Dec 17 '16 at 1:12
• Your problem is just a Poisson's equation with Neumann boundary conditions. FEM allows to discretize a problem based on a weak formulation, just search for "FEM for Poisson equation" – uranix Dec 17 '16 at 6:38
• @uranix What do I do with the divergence on the right-hand side of $(1)$? Usually, there is only a function $f\in L^2$ on the right-hand side of the Poisson equation. – 0xbadf00d Dec 17 '16 at 11:08
• There's no divergence of $\mathbf f$ in the weak formulation, only $\mathbf f \cdot \nabla \phi$ – uranix Dec 17 '16 at 11:37