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All right, so my Professor claims that it is impossible to find a vector field that has both div and curl zero and yet vanishes at infinity (that is, becomes the zero vector).

I don't know if it is correct. I decide to trust my professor and check for myself.
So I conclude that,
if the vector field is $\vec{F} =\vec{F_1} \hat{i}+\vec{F_2} \hat{j}+\vec{F_3} \hat{k} $, then, $$\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}=0$$ $$\text{and}$$ $$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x},\frac{\partial F_1}{\partial z}=\frac{\partial F_3}{\partial x},\frac{\partial F_2}{\partial z}=\frac{\partial F_3}{\partial y}$$

I have no idea how to combine all of these to get the required result. We haven't been taught differential equations yet.

Any ideas and hints are much appreciated.

SBF
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Math Monk
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    I guess curl being zero implies that $F$ is a gradient of some function, whereas divergence of gradient being zero implies that this function is harmonic. Those are unbounded unless they are constant – SBF Apr 15 '22 at 11:59
  • You may be interested in the Hemholtz decomposition https://en.m.wikipedia.org/wiki/Helmholtz_decomposition – JackT Apr 15 '22 at 12:20

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First of all, if $\mathrm{curl}\;F = 0$ then there exists a function $U$ such that $F = \nabla U$. For some reason I can't find a proper reference, but see e.g. here. Then $\mathrm{div}F = 0$ means $\Delta U = 0$ where $$ \Delta = \mathrm{div}(\mathrm{curl) = }\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} $$ is the Laplace operator. Solutions of $\Delta U = 0$ are called harmonic functions. In particular, the Liouville theorem states that a harmonic function that is bounded below (or above) is necessarily constant, the latter link provides the statement and intuition as well.

If that's not enough by itself, you also get the harmonic functions are real analytic and hence infinitely continuously differentiable. As a result, the order of taking derivatives does not matter, hence any partial derivative of a harmonic function is harmonic itself. In particular, any component of its gradient is a harmonic function. Applying Liouville's theorem to the latter you get non-vanishing gradient, unless it's zero of course.

Note that as far as I recall nothing of the facts above requires general theory of ODEs or PDEs, I think one mostly uses mean-value property of harmonic functions. And integrating zero curls should be also pretty straight forward, essentially you can define $U$ as a path integral of $F$ and show that due to curl being zero the integral only depends on the end points, not on the path itself. I guess, it was some time.

SBF
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