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How can we characterize polynomials $p(x,y)$ in $\mathbb{R}^2$ (in two variables) that are harmonic (that is $\Delta p(x,y) = 0$)?

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This is a much-studied question.

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    I see, but where can I find a proof of the results in dimension $2$? – 2017-01-17
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(Since the Laplacian in two variables can be written as $\partial\circ\overline{\partial}$, for example), all harmonic polynomials in $x,y$ are linear combinations of $(x+iy)^n$ and $(x-iy)^n$, for non-negative integers $n$.

EDIT: one way to prove this is to use $z=x+iy$ and $\overline{z}=x-iy$, and observe that every (complex) polynomial in $x,y$ can be rewritten as such in $z$ and $\overline{z}$. Let $\partial=\partial/\partial z$ and $\overline{\partial}=\partial/\partial \overline{z}$. Now, of course, it's not clear what these are, except as expressing them in terms of $x,y$. Indeed. Ok, but it is not hard to prove the key lemma, that $\overline{\partial}z=0$ and $\partial \overline{z}=0$. Also, check that the Laplacian is $\partial \circ \overline{\partial}=\overline{\partial}\circ \partial$. Also, check that the obviously-different monomials in $z$ and $\overline{z}$ are linearly independent. Then show that $\partial(P(z,\overline{z})=0$ for a polynomial in these two occurs if and only if $z$ does not appear. And similarly for $\overline{z}$.

Then annihilation by the Laplacian is $\partial(\overline{\partial}P)=0$, so $\overline{\partial}P$ must have no $z$'s in it. Thus, $P$ itself must have only monomials that purely involve $z$, or purely involve $\overline{z}$.

(Yes, this argument is an algebraic emulation of some things about holomorphic functions...)

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    I don't quite understand why. Can you add some details? – 2017-01-17
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    Let me see if I've understood your reasoning. We have that $\partial(P) = 0$ iff $P$ does not depend on $z$ and $\overline{\partial}(P) = 0$ iff $P$ does not depend on $\overline{z}$. In addition, $z$ and $\overline{z}$ are linearly independent. Right? How does it follow that $P(x,y)$ is harmonic if it is given by linear combinations of $z^n$ and $\overline{z}^n$? Also, it is not true that all linear combinations of $z$ and $\overline{z}$ are harmonic polynomials of two real variables, right? – 2017-01-17
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    Not only $z$ and $\overline{z}$, but also monomials. That is, what "look like" polynomials _are_ polynomials, and behave as expected. And, yes, all linear combinations of _pure_ monomials in $z$ and pure monomials in $\overline{z}$ are harmonic polys in two variables. – 2017-01-17
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    I don't quite understand your comment. Could you elaborate? The points still unclear to me are: **1.** Are $z^n$ and $\overline{z}^n$ linearly independent? **2.** How do we conclude that $P(x,y)$ is harmonic if it is given by linear combinations of $z^n$ and $\overline{z}^n$? **3.** Is it true that all linear combinations of $z^n$ and $\overline{z}^n$ are harmonic polynomials of two real variables? – 2017-01-17
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    1. Yes, $z^n$ and $\overline{z}^n$ are linearly independent. 2. The pure monomials in $z^n$ are annihilated by $\overline{\partial}$ and the pure monomials in $\overline{z}^n$ are annihilated by $\partial$. 3. Yes, the latter proves that all linear combinations of those pure monomials are harmonic. The only slightly less obvious point is that these are the _only_ harmonic polys. – 2017-01-17
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    Then why are the linear combinations of $z$ and $\overline{z}$ the only harmonic polynomials in two real variables? – 2017-01-17