given by formula (\[eq:5.13\]) with $P(t),$$\Sigma(t,m_{1}),\lambda(t,m_{1})$ given respectively by equations (\[eq:5.17\]), (\[eq:5.18\]), (\[eq:5.16\]). The solution of the Master equation (\[eq:2.8-1\]) is given by formula (\[eq:5.21\]), in which $\Gamma(t,m_{1})$ is the solution of (\[eq:5.19\]).

\[sub:Mean Field Type\]SYSTEM OF STOCHASTIC HJB-FP EQUATIONS FOR MEAN FIELD TYPE CONTROL 
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We now consider the solution of the stochastic Fokker-Planck equation $$\left\{
\begin{array}{l}
\partial_{t}m+(\mathcal{A}^{*}m-\frac{1}{2}\beta^{2}\Delta m+\text{div}(G(x,m,DU)m))dt+\beta Dm db(t)=0,\label{eq:5.25}\\
m(x,0)=m_{0}(x),
\end{array}
\right.$$ for the linear quadratic Mean Field type control problem, assuming the initial condition $m_{0}$ is Gaussian with mean $\bar{x}_{0}$ and covariance matrix $\Pi_{0}.$ If we call $m(t)=m(x,t)$ the solution, then $$m_{1}(t)=\int_{\mathbb{R}^{n}}m(\xi,t)d\xi=1,\: y(t)=\int_{\mathbb{R}^{n}}\xi m(\xi,t)d\xi.$$

Next $$DU(x,m(t),t)=P(t)x+\Sigma(t)y(t),\;\Sigma(t)=\Sigma(t,1),$$ $$G(x,m,DU(x,m(t),t))=(A-BR^{-1}B^{*}P(t))x+(\bar{A}-BR^{-1}B^{*}\Sigma(t))y(t).$$

From (\[eq:5.25\]) by using the test function $x$, and then integrating in $x,$ we get easily that $$\label{eq:5.26}
\left\{
\begin{array}{l}
    dy=(A+\bar{A}-BR^{-1}B^{*}P(t))y(t)dt-BR^{-1}B^{*}\Sigma(t)y(t)dt+\beta db(t),\\
    y(0)=\bar{x}_{0}.\\
\end{array}
\right.$$

We define $$u(x,t)=U(x,m(t),t)=\frac{1}{2}x^{*}P(t)x+x^{*}r(t)+s(t),$$ with $$\label{add_1}
r(t)=\Sigma(t)y(t),\: s(t)=\frac{1}{2}y(t)^{*}\Gamma(t)y(t)+\mu(t),$$ in which $\Gamma(t)=\Gamma(t,1)$ and $\mu(t)=\mu(t,1)$. An easy calculation, by taking account of (\[eq:5.18\]) and (\[eq:5.26\]) yields $$\label{eq:5.27}
\left\{
\begin{array}{l}
    -dr=(A^{*}+\bar{A}^{*}-PBR^{-1}B^{*})rdt+(S^{*}\bar{Q}S-\bar{Q}S-S^{*}\bar{Q}+P\bar{A}+\bar{A}\,^{*}P)ydt-\beta\Sigma db(t),\\
    r(T)=(S_{T}^{*}\bar{Q}_{T}S_{T}\,{\bf 1}-S_{T}^{*}\bar{Q}_{T}-\bar{Q}_{T}S_{T})y(T),
\end{array}
\right.$$ and we can rewrite (\[eq:5.26\]) as

$$\label{eq:5.28}
\left\{
\begin{split}
dy&=(A+\bar{A}-BR^{-1}B^{*}P(t))y(t)dt-BR^{-1}B^{*}r(t)dt+\beta db(t),\\
y(0)&=\bar{x}_{0},\\
\end{split}
\right.$$

and the pair $(y(t),r(t))$ becomes the solution of a system of forward-backward SDE. Considering the fundamental matrix $\Phi_{P}(t,s)$ associated to the matrix $A+\bar{A}-BR^{-1}B^{*}P(t)$: $$\left\{
\begin{array}{l}
    \displaystyle\frac{\partial\Phi_{P}(t,s)}{\partial t}=(A+\bar{A}-BR^{-1}B^{*}P(t))\Phi_{P}(t,s),\,\forall t>s,\\
    