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Jun 28, 2022 at 0:42 comment added Willie Wong I don't know the most general form of the solution. But, observe that if $k\in \mathbb{C}^n$, then $$ g(X) = \sum_{j = 1}^n k_j x_j $$ solves the PDE iff (I almost got caught out by the double conjugate) $$ \sum_{j = 1}^n k_j^2 = - k_1 $$ This defines a variety in $\mathbb{C}^n$ and shows that there are infinitely solutions just under this ansatz. When $n \geq 3$ there are also multiple solutions with $k_1 = 0$.
Jun 27, 2022 at 22:38 comment added Schemer1 Thank you @WillieWong. I corrected the statement of the problem to state that $ g(X): \mathbb{C}^{n} \to \mathbb{C} $ is a polynomial in $ n $-indeterminates $ x_{1},\dots,x_{n} $ and that I consider a solution $ g(X) $ to be equivalent to one of the form $ g(X)+c $ where $ c \in \mathbb{C} $.
Jun 27, 2022 at 22:34 history edited Schemer1 CC BY-SA 4.0
The question about uniqueness was flawed. I corrected it. I added constraints on the domain of $ g(X) $
Jun 27, 2022 at 19:18 comment added Willie Wong But the solution shouldn't be unique, since if $g$ is a solution to the PDE, then so is $g + C$ for any constant. Are you prescribing some sort of boundary condition?
Jun 27, 2022 at 19:17 comment added Willie Wong What are the range and domain of the function $g$?
Jun 27, 2022 at 14:15 history asked Schemer1 CC BY-SA 4.0