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For $k\leq{}n$ the elementary symmetric polynomials are defined by: $$e_k(x_1,\ldots,x_n)=\sum_{1\leq{}i_1<...<i_k\leq{}n}x_{i_1}\cdots x_{i_k}$$ I believe I can prove (by a complex brute force method) that the function: $$f(x_1,\ldots,x_n)=e_k(x_1,\ldots,x_n)/(x_1+\cdots+x_n)^k$$ satisfies the following PDE: $$\left(\sum_{i=1}^nx_i\right)\left(\sum_{i=1}^nx_i\frac{\partial^2f}{\partial{}{x_i}^2}\right)+k(k-1)f=0$$ I have some questions, since both symmetric polynomials and PDEs are out of my area:

  1. Is the above (assuming it is correct) known or easily proven, e.g. is there a reference I could use or a relatively simple proof?

  2. Does PDE theory tell us anything about uniqueness of solutions of the given PDE(s)?

Thanks

Helmut

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    $\begingroup$ I think ultimately many symmetric polynomials can be obtained as eigenfunctions of Laplace-Beltrami operators; see for instance page 9 of www-math.mit.edu/~rstan/pubs/pubfiles/73.pdf discusses this for Jack polynomials; for elementary symmetric polynomials, this may simplify / specialize. $\endgroup$ – Suvrit Jan 21 '16 at 11:34
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For your first question, here's a simple proof.

  1. Observe that $\partial_{ii}^2 e_k = 0$ for any $i\in \{1,\ldots,n\}$. This implies $$ \partial^2_{ii} (\sigma^k f) = 0 \tag{1}$$ for any $i$, where $\sigma = \sum x_i$.
  2. Observe that $$ \partial_i \sigma = 1 $$ for any $i$, and hence $$ \partial_i \sigma^k = k \sigma^{k-1} \tag{2}$$ by the chain rule.
  3. The product rule applied to (1) using (2) gives $$ 0 = k(k-1)\sigma^{k-2} f + 2 k \sigma^{k-1} \partial_i f + \sigma^k \partial^2_{ii} f$$ from which we divide by $\sigma^k$ to obtain $$ \partial^2_{ii} f + k(k-1)\sigma^{-2} f + 2 k \sigma^{-1} \partial_i f = 0 \tag{3}$$ Recall that this holds individually for any $i$.
  4. Multiply now the expression (3) by $x_i$. And sum over $i$, we get finally $$ \sum_{i = 1}^n x_i \partial^2_{ii} f + k(k-1) \sigma^{-2} f \underbrace{\sum x_i}_{= \sigma} + 2 k \sigma^{-1} \sum_{i = 1}^{n} x_i \partial_i f = 0 \tag{4} $$
  5. Now observe that $f$ is, by definition, a homogeneous function of degree 0. So the homogeneous derivative $\sum x_i \partial_i f = 0$. This kills the last term in (4). The remainder is exactly what you wanted to show.

Concerning uniqueness, observe that only two ingredients were used in the derivation of the PDE above:

  1. That the original function $e_k$ satisfies $\partial^2_{ii} e_k = 0$ for any $i$.
  2. That the original function $e_k$ is homogeneous of degree $k$.

There are a lot more functions that satisfy the same property. For example, when $n = 3$ and $k = 2$, we know that it works for $e_k = x_1 x_2 + x_2 x_3 + x_3 x_1$. But the same also works for $x_1 x_2$ or $x_2 x_3$ or $x_3 x_1$ or any linear combination thereof.

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