If $ \vec{u} $ (respectively $ \vec{v} $) is an element of $ \mathbb{C}^{n} $ with components $ u_{i} $ (respectively $ v_{i} $) and $ \langle \vec{u},\vec{v} \rangle $ is the Hermitian inner product $ \sum_{i=1}^{n} u_{i} \overline{v_{i}} $, then does anyone know a form for solutions of the PDE $ \langle \nabla g(X), \overline{\nabla g(X)} \rangle = -\frac{\partial g(X)}{\partial x_{1}} $ for a polynomial function $ g(X): \mathbb{C}^{n} \to \mathbb{C} $? Is such a solution unique up to addition of a constant in $ \mathbb{C} $? For any such solution does $ \frac{\partial g(X)}{\partial x_{1}} $ equal zero?
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$\begingroup$ What are the range and domain of the function $g$? $\endgroup$– Willie WongCommented Jun 27, 2022 at 19:17
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$\begingroup$ 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? $\endgroup$– Willie WongCommented Jun 27, 2022 at 19:18
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$\begingroup$ 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} $. $\endgroup$– Schemer1Commented Jun 27, 2022 at 22:38
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$\begingroup$ 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 $k_1 = 0$. $\endgroup$– Willie WongCommented Jun 28, 2022 at 0:42
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