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Im trying to show that if we apply the Helmholtz decomposition on a smooth vector field $\mathbf{f} \in C^\infty_c(\Omega)^d$ with compact support, then the gradient potential is also smooth and has compact support: $\qquad \mathbf{f} = \nabla \phi + \mathbf{\psi},\qquad \phi \in C^\infty_c(\Omega), \psi \in C^\infty(\Omega)^d, \nabla \cdot \psi = 0$.

Im assuming $\Omega$ to be Lipschitz.

I looked through some proofs in the smooth case (e.g. The proof of the Helmholtz decomposition theorem through Neumann boundary value problem) where use was made of the Dirichlet fundamental solution $\Phi$:

$\phi(x) = -\int_\Omega \Phi(x-y)\nabla_y\cdot\mathbf{f}(y)dy + \int_{\partial \Omega}\Phi(x-y)\mathbf{f}(y)\cdot\mathbf{n}dS(y)$

$\psi(x) = \nabla\times(\int_\Omega \Phi(x-y)\nabla_y \times \mathbf{f}(y)dy-\int_{\partial \Omega}\Phi(x-y)\mathbf{n}\times\mathbf{f}(y)dS(y))$.

Initially I thought compact support on $f$ would induce compact support on $\phi$, but its not that easy. The Evans contains a theorem (Thm. 1, p. 23) stating that if $f$ is $C^2_c(R^n)$-smooth with compact support, then the singularity of the fundamental solution just renders $u(x) = \int_{R^n}\Phi(x-y)f(y)dy$ to be $C^2(R^n)$ WITHOUT compact support. Thus, currently I don't think I can show the existence of such a potential. What do you think?

Cheers, Uplink

Community
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Uplink
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  • If $f=0$ for $|x| \ge R$, then $\nabla\cdot f = 0$, gives $\nabla^2\phi=-\nabla\cdot\nabla\times F=0$ for $|x| > R$. Maybe it's possible to add something in the far field by sacrificing some smoothness at a chosen boundary? – Disintegrating By Parts Dec 30 '15 at 19:00

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I remember one lemma, hope it would give you some help. Let $B$ be a rectangle in $R^d$, $\phi\in C_c^\infty(B)$ with $\int_B\phi dx=0$. Then the divergence equation $$\nabla\cdot\psi=\phi,\ \ in\ \ B$$ has a solution $\psi\in C_c^\infty(B)^d$.

Ivy Hsu
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