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

Consider the following equation $$$$|\nabla f|^2=f\,\Delta f,$$$$ where $$\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$$?

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

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, $$v(x_1,\ldots,x_n)=\prod_{1\leq i gives a homogeneous polynomial solution of degree $$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.

• (+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.
– robjohn
Jan 9 '19 at 16:42
• @Start ..., thanks for the answer. Do you think that there exist solutions that are differents of these kind of solutions presented by you? Jan 23 '19 at 19:56
• @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. Jan 31 '19 at 15:54
• @Start..., could you give me a proof that for n=2 those solutions are the unique solutions? Feb 1 '19 at 16:43
• @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 $F(z)\tilde{F}(\bar{z})$. Feb 5 '19 at 17:01