Can you have solution for the equation $|\nabla f|^2=f\,\Delta f$, for a homogeneous polynomial $f$ with $\deg(f)>2$?

Question

Consider the following equation \begin{equation} |\nabla f|^2=f\,\Delta f, \end{equation} where $\nabla f$ is the gradient of $f$ and $\Delta f$ is the Laplacian of $f$. Does this equation have a solution $f\colon \mathbb{R}^n\to \mathbb{R}$ such that $f$ is a homogenous polynomial with $\deg(f)>2$ and $n>1$?

Answer 1

Answer: Yes. Moreover the equation has homogeneous polynomial solutions of any even degree.

Proof. It is easy to check that the function $u(x_1,\ldots,x_n)=\left(x_1^2+x_2^2\right)^N$ satisfies the equation $$u_{x_1}^2+u_{x_2}^2=u\left(u_{x_1x_1}+u_{x_2x_2}\right)$$ for any $N\in\mathbb N$ (and, in fact, for any $N\in\mathbb R$). This may also be understood without calculation using the first comment and that $\ln(x^2+y^2)$ is essentially the Green's function of the 2D Laplacian. But then $u$ also satisfies $|\nabla u|^2=u\Delta u$.

One may also construct examples of symmetric homogeneous polynomial solutions in the form of products of the above elementary solutions. Specifically, $$ v(x_1,\ldots,x_n)=\prod_{1\leq i<j\leq n}\left(x_i^2+x_j^2\right)^N$$ gives a homogeneous polynomial solution of degree $n(n-1)N$ for any $N\in\mathbb N$. For $n>2$, one may set $N=1$ and still have a solution with required properties.