Here is the system of equations: \begin{eqnarray*} &&{\frac {\theta_1\,y}{ \left( x+y \right) ^{2}}}-q \left( z \right) -x{ \frac {d }{d z}}q \left( z \right)\frac{\partial z}{\partial x} =0\\ && {\frac {\theta_2\,x}{ \left( x+y \right) ^{2}}}-q \left( z \right) -y{ \frac {d }{d z}}q \left( z \right)\frac{\partial z}{\partial y} =0, \end{eqnarray*}
where
- $x\geq 0$ and $y\geq 0$;
- $z:=x+y$ (thus $\frac{\partial z}{\partial x}=1$ and $\frac{\partial z}{\partial y}=1$);
- $\theta_1>0$ and $\theta_2>0$;
- $q:=\mathbb{R}_{++}\longrightarrow \mathbb{R}_{++}$, differentiable with $\frac{d }{d z}q \left( z \right)<0$, and;
- $\lim_{z\rightarrow 0}q(z)=\infty$ and $\lim_{z\rightarrow \infty}q(z)=0$.
$(x,y)$ is a pair that solves the system and I want to retrieve $q$. The problem that I face comes from the fact that $\theta_1\not=\theta_2$. If $\theta_1=\theta_2$, then by symmetry $x=y$ and $q$ is easily found. So what is $q$ when $\theta_1\not=\theta_2$.
Thanks for your help!