\qquad\qquad\qquad-\displaystyle\frac{1}{2}(DU(\xi,m,t))^{*}BR^{-1}B^{*}DU(\xi,m,t)+(DU(\xi,m,t))^{*}(A\xi+\bar{A}y)\bigg]m(\xi)d\xi=0\\
    V(m,T)=\displaystyle\int_{\mathbb{R}^{n}}\frac{1}{2}\xi^{*}(Q_{T}+\bar{Q}_{T})\xi m(\xi)d\xi-\frac{1}{2}y^{*}(S_{T}^{*}\bar{Q}_{T}+\bar{Q}_{T}S_{T})y+\frac{1}{2}y^{*}S_{T}^{*}\bar{Q}_{T}S_{T}y\, m_{1}.
\end{array}
\right.$$

We look for a solution $V$ in (\[eq:5.11\]) of the form in which $P(t)$ and $\Sigma(t,m_1)$ are symmetric matrices $$V(m,t)=\frac{1}{2}\int_{\mathbb{R}^{n}}\xi^{*}P(t)\xi m(\xi)d\xi+\frac{1}{2}y^{*}\Sigma(t,m_{1})y+\lambda(t,m_{1}).\label{eq:5.13}$$

Clearly, the terminal condition of (\[eq:5.11\]) yields that $$\begin{aligned}
P(T) & =Q_{T}+\bar{Q}_{T},\;\Sigma(T,m_{1})=S_{T}^{*}\bar{Q}_{T}S_{T}\, m_{1}-(S_{T}^{*}\bar{Q}_{T}+\bar{Q}_{T}S_{T}),\:\lambda(T,m_{1})=0.\label{eq:5.14}\end{aligned}$$

Next $$U(\xi,m,t)=\frac{1}{2}\xi^{*}P(t)\xi+y^{*}\Sigma(t,m_{1})\xi+\frac{1}{2}y^{*}\frac{\partial\Sigma(t,m_{1})}{\partial m_{1}}y+\frac{\partial\lambda(t,m_{1})}{\partial m_{1}},\label{eq:5.15}$$ and hence $$\begin{array}{c}
    D_{\xi}U(\xi,m,t)=P(t)\xi+\Sigma(t,m_{1})y,\\ D_{\xi}^{2}U(\xi,m,t)=P(t),\\
    \displaystyle\frac{\partial U(\xi,m,t)}{\partial m}(\eta)=\eta^{*}\Sigma(t,m_{1})\xi+y^{*}\frac{\partial\Sigma(t,m_{1})}{\partial m_{1}}(\xi+\eta)+\frac{1}{2}y^{*}\frac{\partial^{2}\Sigma(t,m_{1})}{\partial m_{1}^{2}}y+\frac{\partial^{2}\lambda(t,m_{1})}{\partial m_{1}^{2}}.
\end{array}$$

We can see that the property (\[eq:4.8\]) is satisfied, thanks to the symmetry of the matrix $\Sigma(t,m_{1}).$ We need $$\sum_{i=1}^{n}\frac{\partial}{\partial\xi_{i}}\frac{\partial}{\partial\eta_{i}}\frac{\partial U(\xi,m,t)}{\partial m}(\eta)=\text{tr }\Sigma(t,m_{1}).$$

With these calculations, we can proceed on equation (\[eq:5.11\]) and obtain $$\begin{array}{l}
    \displaystyle\frac{1}{2}\int_{\mathbb{R}^{n}}\xi^{*}\frac{d}{dt}P(t)\xi m(\xi)d\xi+\frac{1}{2}y^{*}\frac{\partial}{\partial t}\Sigma(t,m_{1})y+\frac{\partial\lambda(t,m_{1})}{\partial t}\\
    \qquad+\displaystyle(\frac{1}{2}\text{tr}\, aP(t)+\frac{\beta^{2}}{2}\text{tr}P(t))m_{1}+\frac{1}{2}\beta^{2}\text{tr}\Sigma(t,m_{1})(m_{1})^{2}+\displaystyle\int_{\mathbb{R}^{n}}\bigg[\frac{1}{2}\xi^{*}(Q+\bar{Q})\xi-\xi^{*}\bar{Q}Sy+\frac{1}{2}y^{*}S^{*}\bar{Q}Sy\\
    \qquad-\displaystyle\frac{1}{2}(P(t)\xi+\Sigma(t,m_{1})y)^{*}BR^{-1}B^{*}(P(t)\xi+\Sigma(t,m_{1})y)+(P(t)\xi+\Sigma(t,m_{1})y)^{*}(A\xi+\bar{A}y)\bigg]m(\xi)d\xi=0.
    
\end{array}$$

We can identify terms and obtain $$\frac{\partial\lambda(t,m_{1})}{\partial t}+(\frac{1}{2}\text{tr}\, aP(t)+\frac{\beta^{2}}{2}\text{tr}P(t))m_{1}+\frac{1}{2}\beta^{2}\text{tr}\Sigma(t,m_{1})(m_{1})^{2}=0.$$

Therefore, from the final condition $\lambda(T,m_{1})=0,$ it follows that $$\lambda(t,m_{1})=\int_{t}^{T}(\frac{1}{2}\text{tr}\, aP(s)+\frac{\beta^{2}}{2}\text{tr}P(s))ds\, m_{1}+\frac{1}{2}\beta^{2}\int_{t}^{T}\text{tr}\Sigma(s,m_{1})ds\,(m_{1})^{2}.\label{eq:5.16}$$

Recalling that $y=\int_{\mathbb{R}^{n}}\xi m(\xi)d\xi$ and identifying quadratic terms in $\xi$ (within the integral) and in $y$ respectively, it follows easily that $$\left\{
\begin{array}{l}
\displaystyle\frac{dP}{dt}+PA+A^{*}P-PBR^{-1}B^{*}P+Q+\bar{Q}=0,\\
P(T)=Q_{T}+\bar{Q}_{T},\label{eq:5.17}\\
\end{array}
\right.$$ and $$\left\{
\begin{array}{l}
\displaystyle\frac{d\Sigma}{dt}+\Sigma(A+\bar{A}m_{1}-BR^{-1}B^{*}P)+(A+\bar{A}m_{1}-BR^{-1}B^{*}P)^{*}\Sigma\label{eq:5.18}\\
\qquad\qquad -\displaystyle\Sigma BR^{-1}B^{*}\Sigma m_{1}+S^{*}\bar{Q}Sm_{1}-\bar{Q}S-S^{*}\bar{Q}+P\bar{A}+\bar{A}\,^{*}P=0;\\
\Sigma(T,m_{1})=S_{T}^{*}\bar{Q}_{T}S_{T}\, m_{1}-(S_{T}^{*}\bar{Q}_{T}+\bar{Q}_{T}S_{T}).
\end{array}
\right.$$

We obtain formula (\[eq:5.13\]) with the values of $P(t),\,\Sigma(t,m_{1}),\,\lambda(t,m_{1})$ given by equations (\[eq:5.17\]), (\[eq:5.18\]), (\[eq:5.16\]). We next turn to the Master equation. The function $U(x,m,t)$ is given by (\[eq:5.15\]). Let us set $\Gamma(t,m_{1})=\frac{\partial\Sigma(t,m_{1})}{\partial m_{1}}$. From (\[eq:5.18\]) we obtain easily

$$\left\{
\begin{array}{l}
