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I would like to find a twice continuously differentiable function $f(x_1,x_2): \mathbb{R}^2 \rightarrow \mathbb{R}$. Denote $f_1, f_2, f_{11}, f_{12}, f_{22}$ its first and second order partial derivatives with respect to $x_1, x_2,\dots$. $f$ needs to fulfill the following equation \begin{equation} 1 - \exp(f) = \frac{f_1}{f_2}a \end{equation}

where $a$ is a given constant. There are no further constrains on the function. I tried to exploit the fact that $f_{12} = f_{21}$. Differentiating the equation above on both sides separately with respect to $x_1$ and $x_2$ and dividing yields:

\begin{align} 2f_{12} = f_{11}\frac{f_2}{f_1} + f_{22}\frac{f_1}{f_2} \end{align}

But I do not know how to proceed from here. Thank you for any help and suggestions.

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    $\begingroup$ In the original equation try the substitution $f = \log (1 + g)$. You end up with something very close to the inviscid Burgers’ equation for $g$. $\endgroup$ Mar 2, 2021 at 14:17

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As @A rural reader suggested, substituting $f=\ln(1+g)$ yields the inviscid Burgers' equation: $$ -g = \frac{g_1}{g_2} a \Leftrightarrow a g_1 + g~ g_2 = 0. $$ Solutions to this equation have to be classified through an initial condition, e.g. $g(0,x) = F(x)$ which has the implicit solution: $$ g(x_1,x_2) = F\left(x_2-\frac{x_1}{a}\cdot g \right) \Rightarrow g(x_1,x_2) = \cdots \Rightarrow f(x_1,x_2) = \ln(1+g). $$


A typical example for this would be $F(x) = x$, which leads to $$ g(x_1,x_2) = \frac{ax_2 }{a+ x_1 } \Rightarrow f(x_1,x_2) = \ln\left( 1 + \frac{ax_2 }{a+ x_1 } \right). $$

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