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I'm wondering, for no particular reason: are there differentiable vector-valued functions $\vec{f}(\vec{x})$ in three dimensions, other than the constant function $\vec{f}(\vec{x}) = \vec{C}$, that have zero divergence and zero curl? If not, how would I prove that one doesn't exist?

I had a vague memory of learning some reason that such a function doesn't exist, but there's a pretty good chance my mind is just making things up to trick me ;-) But I thought about it for a little while and couldn't think of another divergence-free curl-free function off the top of my head, so I'm curious whether I was thinking of real math or not.

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    $\begingroup$ You might've been thinking of en.wikipedia.org/wiki/Helmholtz_decomposition . See, however, en.wikipedia.org/wiki/Laplacian_vector_field . $\endgroup$ Jan 28 '11 at 23:50
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    $\begingroup$ Indeed, the key phrase in the Helmholtz decomposition theorem is "rapidly decaying". There are infinitely many non-constant Laplacian vector fields, as Sivaram's answer implies, but none of them decay sufficiently rapidly at infinity. This is why they (almost?) never arise in physical problems involving vector fields in free space: the corresponding system would have infinite energy. $\endgroup$
    – user856
    Jan 29 '11 at 1:49
  • $\begingroup$ @DavidZ So, there are no other differentiable vector field, other than the constant vector field $\vec{f}(\vec{x}) = \vec{C}$, that have zero divergence and zero curl?....In other words any vector fields with no divergence or curl are always constant and hence can be represented by a potential function satisfying Laplace equation? $\endgroup$
    – GRANZER
    Mar 29 '19 at 5:33
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How about Gradient of a harmonic function?

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For example a constant field is harmonic, but doesn't satisfy the normal boundary conditions at infinity. In any topological space, there is one harmonic field per winding number satisfying given boundary conditions (Hodge decomposition theorem).

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