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I have a function $f(x,y)$, where both $x$ and $y$ are $n$-dimensional vectors, $n\ge 2$. I know that this function has the following property: $$ \frac{\partial}{\partial x_j} \frac{\partial}{\partial y_k} f = a_j(x,y) b_k(x,y) $$ This can be expressed saying that the $n\times n$ block of the Hessian, out of diagonal, is a diadyc matrix. Examples of such a $f$ are: $$ f(x,y) = A(x)B(y) + C(x) + D(y) $$ Actually, this example is similar to the form used in the "separation of variables" method for solving partial differential equations.

The second equation implies the first. My question is: is it also true that the first equation implies the second? Or a counter-example can be found?

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The answer is 'no, the first equation does not imply the second when $n=2$'.

The reason is that when $n=2$, the equation is essentially equivalent to requiring that the off-diagonal $n$-by-$n$ block of the Hessian of $f$ have determinant equal to zero.

This is one (non-linear) second-order equation for $f$ as a function on $\mathbb{R}^4$, and it is easy to show that the local solutions depend on 3 functions of 4 variables in Cartan's sense. (Basically, one can prescribe $f$ and its first normal derivative along a generic hypersurface in $\mathbb{R}^4$ subject to some generic conditions.)

However, the second form depends only on 4 functions of 2 variables, so not every local solution can be put in the second form, even locally.

In higher dimensions, the condition is equivalent to requiring that that $n$-by-$n$ matrix have rank at most equal to $1$. This is $(n{-}1)^2$ second order equations for $f$, so it's overdetermined when $n>2$. One would need to do the Cartan-Kähler analysis of this system to determine the generality of its space of solutions, but, probably, the answer is still 'no'.

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  • $\begingroup$ Could you please write an explicit example of $f$ which meets the first equation but not the second, for the 2D case? $\endgroup$
    – Doriano Brogioli
    Commented Sep 7, 2020 at 11:56
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    $\begingroup$ @DorianoBrogioli: Yes, for example, let $f = x_1^2y_1+x_1y_1^2$. $\endgroup$
    – Robert Bryant
    Commented Sep 7, 2020 at 12:11
  • $\begingroup$ It seems that this very simple example works for every $n$. $\endgroup$
    – Doriano Brogioli
    Commented Sep 7, 2020 at 12:16
  • $\begingroup$ @DorianoBrogioli: Yes, that's true. In fact, any $f(x_1,y_1)$ satisfies the equation, but the generic one is not of the desired form. $\endgroup$
    – Robert Bryant
    Commented Sep 7, 2020 at 12:17

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