I'm trying to formulate a game so that at Nash equilibrium I achieve supply equales demand. Then I ran into this problem.
For all $i,$ $v_{i}\left(x_{i}\right)$ is concave in $x_{i}$. The value function of user $i.$ For all $k$ $c_{k}\left(y_{k}\right)$ is convex in $y_{k}$ and non negative. The cost function of producer $k.$ So that $\sum_{i}v_{i}\left(x_{i}\right)-\sum_{k}c_{k}\left(y_{k}\right)$ is concave.
Let $x=\left(x_{1},\,\cdots x_{N}\right)$ and $y=\left(y_{1},\cdots y_{M}\right).$
Let price per unit be
$$ p=a_{0}+a_{1}\sum_{i}x_{i}+a_{2}\sum_{k}y_{k}. $$
And the objective, which I have verified is an exact potential function of the game, is
$$ \Phi\left(x,\, y\right)=\sum_{i}v_{i}\left(x_{i}\right)-\sum_{k}c_{k}\left(y_{k}\right)-p\sum_{i}x_{i}+p\sum_{k}y_{k}. $$
Show that under some assumptions $\exists\, a_{0},\, a_{1},\, a_{2}$ such that if,
$$ (x^{'}, y^{'})=\underset{y_{k}\in Y_{k},\, x_{i}\in X_{i}}{\text{arg max }}\Phi\left(x, y\right) $$
then $\sum_{i\in I}x_{i}^{'}=\sum_{k\in K}y_{k}^{'}$.
I think this statement is equivalent to showing $\exists\, a_{0},\, a_{1},\, a_{2}$ such that if,
$$ \left(x^{"},y^{"}\right)=\underset{y_{k}\in Y_{k},\, x_{i}\in X_{i}}{\text{arg max }}\Phi\left(x,\, y\right). $$
and if $$ \left(x^{'}, y^{'}\right)=\underset{y_{k}\in Y_{k},\, x_{i}\in X_{i},\,\sum_{i\in I}x_{i}=\sum_{k\in K}y_{k}}{\text{arg max }}\Phi\left(x,\, y\right) $$
then
$\left(x^{'}, y^{'}\right)=\left(x^{"}, y^{"} \right)$
For the case when $a_{1}=a_{2}=0$ I have proved that $a_0$ is the Lagrange multiplier but I Can't solve for this general case.
