# Vector fields whose divergence is Gaussian

Let f be the pdf of a $$n$$ dimensional $$N(0,C)$$ distribution i.e up to a multiplicative constant, $$f(x) = \exp(-\frac{1}{2} x'C^{-1}x)$$.

Which vector fields $$F$$ are so that $${\rm div} (F)= f$$ ?

• first transform the coordinates so that $C=1$, and then choose $F_i={\rm erf}\,(x_i)\exp(-\sum_{j\neq i}x_j^2)$. – Carlo Beenakker Nov 27 '18 at 7:28
• @Carlo Beenakker "div" is a metric dependent operator. By choosing a convenient orthonormal basis the best one can hope is to make $C$ diagonal. Then a suitable modification of your argument above will do the trick. – Liviu Nicolaescu Nov 27 '18 at 9:03
• After the suggested tricks, take $F=\nabla\phi$ and you will get a Poisson equation that has a known solution. – Jon Nov 27 '18 at 16:00