What boundary conditions do I need to uniquely determine a vector field in 3D space, given that the field has zero curl and zero divergence?

I tried to start with an $xy$-plane that was curl-free in the $z$ component, but then I realized that this overdetermines the solution, since the evolution of the field in the $z$-direction can give the field a non-zero curl.


1 Answer 1


Let $G(\vec r,\vec r')=\frac{1}{4\pi |\vec r-\vec r'|}$. Helmholtz's Theorem states that a twice continuously differentiable vector field $\vec A(\vec r)$ can be written

$$\begin{align} \vec A(\vec r)&=-\nabla \left(\int_D G(\vec r, \vec r')\,\nabla'\cdot\vec A(\vec r')\,dV' -\oint_{\partial D}G(\vec r;\vec r')\,\hat n'\cdot \vec A(\vec r')\,dS'\right)\\\\ &+\nabla \times\left(\int_D G(\vec r, \vec r')\,\nabla'\times\vec A(\vec r')\,dV' -\oint_{\partial D}G(\vec r;\vec r')\,\hat n'\times \vec A(\vec r')\,dS'\right) \tag1 \end{align}$$

Therefore, if both $\nabla \cdot \vec A(\vec r)$ and $\nabla \times \vec A(\vec r)$, $\vec r\in D$ are given , then $\vec A(\vec r)$ is uniquely determined by specification of its normal and tangential components on the boundary $\partial D$.

In the special case for which $\nabla \cdot \vec A(\vec r)=0$ and $\nabla \times \vec A(\vec r)=0$, we see that

$$\begin{align} \vec A(\vec r)&=\nabla \oint_{\partial D}G(\vec r;\vec r')\,\hat n'\cdot \vec A(\vec r')\,dS'-\nabla \times\oint_{\partial D}G(\vec r;\vec r')\,\hat n'\times \vec A(\vec r')\,dS' \\\\ &= \oint_{\partial D}(\hat n'\cdot \vec A(\vec r'))\nabla G(\vec r;\vec r')\,dS'+ \oint_{\partial D}(\hat n'\times \vec A(\vec r'))\times \nabla G(\vec r;\vec r')\,dS' \end{align}$$

  • $\begingroup$ why on earth people are so stingy with upvotes? Voting is essential for making this page a good place to do math... (+1) $\endgroup$
    – tired
    Sep 23, 2016 at 19:10
  • $\begingroup$ @tired I cannot explain the curious (even bizarre) behavior of some on this site. From the anonymous down voters to the rude and contentious users, the integrity of this site is compromised. Of course, that is my opinion ... but mine is the only one I have. ;-)) $\endgroup$
    – Mark Viola
    Sep 23, 2016 at 19:37
  • $\begingroup$ @Dr.MV Thank you very much for the answer, Mark. Your answer is very helpful! I'm currently reading into Helmholtz's Theorem to have a better understanding of it. I tried to give you an upvote, but unfortunately, I don't have enough reputation to do so. Sorry about that. $\endgroup$
    – PeaBrane
    Sep 23, 2016 at 20:16
  • $\begingroup$ +1. I agree with your opinion about this site. There are users that behaves as 'MSE police officer' or/and MSE-owners. I have one that check my answers, check my $\LaTeX$, check editions I did for some OP questions and, usually, tries to delete my answers and down-vote. Never go back and delete a comment whenever we recheck our answer. Appears everywhere trying to close questions, on hold, etc... Moderators don't do anything about it. $\endgroup$ Sep 25, 2016 at 3:02
  • $\begingroup$ @FelixMarin First, thank you for the +1! Much appreciated my friend. I am curious; have you reported the "stalker" to MSE? If this culprit is doing anything that you feel exhibits inappropriate behavior, then I would report this to the MSE moderators. They will hopefully admonish the user to refrain from continuing those activities. Does this person have a user name? -Mark $\endgroup$
    – Mark Viola
    Sep 25, 2016 at 3:12

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