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[Edited on 29-March-2020 to make the question clearer]

Let $f, g : [0,1]^2 \to \mathbb{R}$ be two smooth functions, which are strictly increasing and concave in each coordinate. That is, for every $0 < x,y < 1$ we have $\frac{\partial f}{\partial x}(x,y) > 0$, $\frac{\partial f}{\partial y}(x,y) > 0$, $\frac{\partial^2 f}{\partial x^2}(x,y) < 0$, $\frac{\partial^2 f}{\partial y^2}(x,y) < 0$, and the same holds for $g$.

A point $(x_0,y_0) \in [0,1]^2$ is a solution if $$\frac{\partial f}{\partial x}(x_0,y_0) = \frac{\partial g}{\partial x}(1-x_0,1-y_0),$$ $$\frac{\partial f}{\partial y}(x_0,y_0) = \frac{\partial g}{\partial y}(1-x_0,1-y_0).$$

I would like to know under which conditions on $f$ and $g$ this system has a unique solution.

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  • $\begingroup$ Isn't this impossible, since $g(1-x,1-y)$ is now strictly decreasing in each variable rather than increasing like $f(x,y)$? Or am I missing something? $\endgroup$
    – Igor Khavkine
    Commented Mar 28, 2020 at 13:28
  • $\begingroup$ The function $g(1-x,1-y)$ is indeed decreasing in each variable, but the directional derivative of $g$ in each variable at the point $(1-x,1-y)$ is still positive, or do I miss your question? $\endgroup$
    – Eilon
    Commented Mar 28, 2020 at 20:54
  • $\begingroup$ Sorry, I still see a contradiction. For simplicity freeze $y$ and let $h(x) = g(1-x,1-y)$. According to what you are saying, $\frac{\partial}{\partial x} h(x) > 0$ but $h(x)$ is monotone decreasing. Is that not a contradiction? $\endgroup$
    – Igor Khavkine
    Commented Mar 28, 2020 at 23:05
  • $\begingroup$ I guess the confusion is because of the way we interpret the expression $\frac{\partial g}{\partial x}(1-x)$: for me, $\frac{\partial g}{\partial x}(1-x)$ is the derivative of $g$ evaluated at the point $1-x$. For you, this is the derivative of the function $h(x) = g(1-x)$ evaluated at $x$. And these two derivatives are indeed of different signs. I apologize for the confusion. $\endgroup$
    – Eilon
    Commented Mar 29, 2020 at 5:35
  • $\begingroup$ OK, now I understand! Thanks for explaining. But then the solution is pretty straightforward. In one variable, using the above notation, you want $f'(x) = g'(1-x) = -h'(x)$. The only possible solution is $h(x)=-f(x)$ or $g(x) = -f(1-x) + C$. Or with two variables, $g(x,y) = -f(1-x,1-y) + C$, for a constant $C$. $\endgroup$
    – Igor Khavkine
    Commented Mar 29, 2020 at 10:46

1 Answer 1

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Let $z=z_0$ with $F(z)$, $G(z)$ solve the 1-dimensional version of your problem, as you described in the comments. Then $f(x,y) = F(x+y)$ and $g(x,y) = G(x+y)$ satisfy your hypotheses, while any $(x_0,y_0)$ on the line $x_0+y_0=z_0$ is a solution to your equation, meaning that the solution is not unique.

If you strengthen your hypotheses such that the full Hessian of $f$ is negative definite as a matrix, same for $g$, you can exclude such counterexamples. Then uniqueness does hold: if there are two distinct solutions, draw a straight line between them, restrict $f$ and $g$ to that line, and repeat the 1-dimensional uniqueness argument.

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