**f:**R^{n}→R^{n} is defined by **f**(**x**) =(f_{1}(**x**), f_{2}(**x**), ... , f_{n}(**x**)), where **x** = (x_{1}, x_{2}, ... , x_{n}) and the n functions f_{i}**:**R^{n}→R have continuous 2nd order partial derivatives and satisfy ∂f_{i}/∂x_{j} - ∂f_{j}/∂x_{i} = c_{ij} (for all 1 ≤ i, j ≤ n) for some constants c_{ij}. Prove that there is a function g**:**R^{n}→R such that f_{i} + ∂g/∂x_{i} is linear (for all 1 ≤ i ≤ n).

**Solution**

*Easy.*

Well, it *is* easy, provided one can assume the standard result of the vector calculus that if ∂f_{i}/∂x_{j} = ∂f_{j}/∂x_{i} for all i, j, then there is a function g**:**R^{n}→R such that **f** = **∇**g.

But can one assume this result? Well, it is pure bookwork, and relatively time-consuming to prove, so presumably yes. Of course, it is a *bad* question, because it is unclear that you can assume it. The question does not *tell* you that you can assume it, because that would make it trivial - the only difficulty of the question is that it is slightly inverted, apparently asking about some linear expression rather than about g.

Let h_{i}(**x**) = (c_{i1}x_{1} + c_{i2}x_{2} + ... + c_{in}x_{n})/2 - **f**(**x**). Then ∂h_{i}/∂x_{j} - ∂h_{j}/∂x_{i} = c_{ij}/2 - c_{ji}/2 - c_{ij} = 0, since it follows directly from the definition that c_{ij} = - c_{ji}.

© John Scholes

jscholes@kalva.demon.co.uk

30 Sep 1999