I am trying to go through the papers by Gueant, Lions and Lasry on Mean field games. One of their examples is the Mexican wave (which happens in football stadia). Straight to the point the Lagrangian of the problem is
\begin{equation} \mathscr{L}=\sup_{z(.,x)} \liminf_{T \to +\infty}{1\over T} \int_0^T \left( \left[ -\frac{1}{\epsilon^2} \int_\mathbb{R} \bigl( z(t,x) - z(t,x-y) \bigr)^2 \frac{1}{\epsilon} g \left( \frac{y}{\epsilon} \right) dy \right] + u \bigl( z(t,x) \bigr) - \frac{\dot{z}(t,x)^2}{2} \right) dt \end{equation} g is a standard normal kernel. [Where the first term is a 'local' interaction term which models how people like to mimic their neighbours, the seconf term is the cost associated to being in a certain position (standing/seated) and the third term if the cost of changing position.]
Then the authors claim that this ergodic control problem can be formally transformed in a differential way and we get:
\begin{equation} -\frac{2}{\epsilon^2} \int_\mathbb{R} \bigl( z(t,x) - z(t,x-y) \bigr) \frac{1}{\epsilon} g \left( \frac{y}{\epsilon} \right) dy + u' \bigl( z(t,x) \bigr) = -\partial_{tt}^2 z(t,x) \end{equation}
I think this is an Euler-Lagrange equation ($d\mathscr{L}/dz=d/dt(d\mathscr{L}/d\dot{z})$), the right hand side gives the third term and but I am not sure how to differentiate the integral with respect to z. Could anyone explain that?
Then the authors continue that if $\epsilon \rightarrow 0$ then we get in the distribution sense: \begin{equation} \frac{\partial^2 z}{\partial t^2} (t,x) + \frac{\partial^2 z}{\partial x^2} (t,x) = - u' \bigl( z(t,x) \bigr). \end{equation}
I am not too sure what they mean, but I tried using Taylor expansion $$z(t,x-y)=z(t,x)-y\partial_xz(t,x)+{y^2 \over 2}\partial_{xx}^2z(t,x) ...$$ to get $$-\frac{2}{\epsilon^2} \int_\mathbb{R} \bigl( y\partial_xz(t,x)-{y^2 \over 2}\partial_{xx}^2z(t,x) \bigr) \frac{1}{\epsilon} g \left( \frac{y}{\epsilon} \right) dy$$ the first term in the integrand is zero because it is the mean of the standard normal. However I get $$\frac{2}{\epsilon^2} \int_\mathbb{R} \bigl( {y^2 \over 2}\partial_{xx}^2z(t,x) \bigr) \frac{1}{\epsilon} g \left( \frac{y}{\epsilon} \right) dy= \frac{1}{\epsilon} \partial_{xx}^2z(t,x)\int_\mathbb{R} \bigl({y \over \epsilon}\bigr)^2 g \left( \frac{y}{\epsilon} \right) dy=\frac{1}{\epsilon} \partial_{xx}^2z(t,x)$$
which is a $1/\epsilon$ factor different. Can someone explain how to do this properly? Also the integral term is supposed to represent the local interaction between agents. I understand that the $\epsilon$ in the g represents the fast decay of interaction but could someone explain the role of the other $\epsilon$? Thanks!