# Solve this homogeneous linear PDE

$$\prod_{j When $$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 where $$f_{j,k}$$ are arbitrary functions. How can we approach this question?

• 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 $L_{1}u_{1} = 0, L_{2}u_{2} = 0, \dots, L_{n}u_{n} = 0$. – mattos Mar 25 at 15:59
• @mattos That's true! My question is more on the side of uniqueness---how do we know this solution is a general solution? – Jiyuan Zhang Mar 26 at 2:53