The Mean Field Games

Let us set $$\displaystyle{ a(x) = \frac{1} {2}\sigma {(x)\sigma }^{{\ast}}(x), }$$ (3.1) and introduce the second-order differential operator $$\displaystyle{ A\varphi (x) = -\text{tr }a(x){D}^{2}\varphi (x). }$$ (3.2) We define the dual operator $$\displaystyle{ {A}^{{\ast}}\varphi (x) = -\sum _{ k,l=1}^{n} \frac{{\partial }^{2}} {\partial _{x_{k}}\partial _{x_{l}}}(a_{kl}(x)\varphi (x)). }$$ (3.3) Since m(t) is the probability distribution of \(\hat{x}(t)\), it has a density with respect to the Lebesgue measure denoted by m(x, t), which is the solution of the Fokker–Planck equation $$\displaystyle\begin{array}{rcl} \frac{\partial m} {\partial t} + {A}^{{\ast}}m + \text{div }(g(x,m,\hat{v}(x))m)& =& 0, \\ m(x,0)& =& m_{0}(x).{}\end{array}$$ (3.4) We next want the feedback \(\hat{v}(x)\) to solve a standard control problem, in which m appears as a parameter. We can thus readily associate an HJB equation with this problem, parametrized by m.