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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 $n>1$?

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    $\begingroup$ Essentially you are asking for $\Delta\ln|f|=0$? $\endgroup$ Jan 7 '19 at 21:55
  • $\begingroup$ Maybe you are right also, but I think the equation above is equivalent to $\Delta\ln f^2=0$. $\endgroup$
    – user73454
    Jan 8 '19 at 13:48
  • $\begingroup$ As $\ln f^2=2\ln|f|$, this constant factor does not change the equation, as the Laplace operator is linear. $\endgroup$ Jan 8 '19 at 14:02
  • $\begingroup$ @LutzL, yes, you are right, thanks! Do you have any idea if my question has a positive or negative answer? $\endgroup$
    – user73454
    Jan 8 '19 at 14:34
  • $\begingroup$ @user73454 In case the solutions suggested previously look pathological for you for some reason, I updated my answer to include solutions given by symmetric polynomials. $\endgroup$ Jan 9 '19 at 16:15
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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.

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  • $\begingroup$ (+1) This is all so clear once we realize that $f\Delta f=|\nabla f|^2$ is equivalent to $\Delta\log(f)=0$. Then the product of two solutions is also a solution. $\endgroup$
    – robjohn
    Jan 9 '19 at 16:42
  • $\begingroup$ @Start ..., thanks for the answer. Do you think that there exist solutions that are differents of these kind of solutions presented by you? $\endgroup$
    – user73454
    Jan 23 '19 at 19:56
  • $\begingroup$ @user73454 I do not have the general answer, but for $n=2$ it should be possible to prove that there are no other solutions: in that case we can write the general solution as $f(x_1,x_2)=F(x_1+ix_2) \tilde{F}(x_1-ix_2)$, so there seems to be no room for different homogeneous polynomial solutions. $\endgroup$ Jan 31 '19 at 15:54
  • $\begingroup$ @Start..., could you give me a proof that for n=2 those solutions are the unique solutions? $\endgroup$
    – user73454
    Feb 1 '19 at 16:43
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    $\begingroup$ @user73454 This is just because the 2D Laplacian can be written as $\Delta=\partial_{z\bar{z}}$, therefore the solutions of $\Delta g=0$ have the form $g(z,\bar{z})=G(z)+\tilde{G}(\bar{z})$ and $e^g$ has the form $F(z)\tilde{F}(\bar{z})$. $\endgroup$ Feb 5 '19 at 17:01

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