$$\prod_{j<k}(\partial_{x_k}-\partial_{x_j})u(x_1,\ldots,x_n)=0.$$ When $n=2$ this is a typical transport equation $u_x-u_y=0$ with solution $u(x,y)=u(x+y)$. Mathematica tells me that for general $n$ its general solution must be $$\sum_{j<k} f_{j,k}(x_j+x_k,x_{other})$$ where $f_{j,k}$ are arbitrary functions. How can we approach this question?

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    $\begingroup$ If you have operators $L_{i}$ such that $$\prod_{i}^{n} L_{i} u = 0$$ then a solution to just one of the operators, $L_{j}u = 0$ say, is a solution to the full problem as $$\prod_{i}^{n} L_{i} u_{j} = \left( \prod_{i, i \ne j}^{n} L_{i} \right) L_{j} u = \left( \prod_{i, i \ne j}^{n} L_{i} \right)(0) = 0$$ As we can do this for each $j$, we see that the full solution is a sum of each of the individual solutions $L_{1}u_{1} = 0, L_{2}u_{2} = 0, \dots, L_{n}u_{n} = 0$. $\endgroup$ – mattos Mar 25 at 15:59
  • $\begingroup$ @mattos That's true! My question is more on the side of uniqueness---how do we know this solution is a general solution? $\endgroup$ – Jiyuan Zhang Mar 26 at 2:53

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