## 2 Problem Statement
### Transformation of the Game (1)-(3)  
In what follows, we assume:  
**(A4)**\(\det\Big{(}\mathcal{B}_{\mathrm{u}}^{T}(t)\mathcal{D}_{\mathrm{u}}(t)\mathcal{B }_{\mathrm{u}}(t)\Big{)}\
eq 0\;\;\forall t\in[0,t_{f}]\). **(A5)** The matrix-valued functions \(\mathcal{A}(t)\), \(\mathcal{B}_{\mathrm{v}}(t)\), \(\mathcal{D}_{\mathrm{v}}(t)\), \(G_{\mathrm{u,v}}(t)\) and \(G_{\mathrm{v,u}}(t)\) are continuously differentiable in the interval \([0,t_{f}]\). **(A6)** The matrix-valued functions \(\mathcal{B}_{\mathrm{u}}(t)\) and \(\mathcal{D}_{\mathrm{u}}(t)\) are twice continuously differentiable in the interval \([0,t_{f}]\). Let \(\mathcal{B}_{\mathrm{c}}(t)\) be a complement matrix to the matrix \(\mathcal{B}_{\mathrm{u}}(t)\) for any \(t\in[0,t_{f}]\). Thus, the dimension of the matrix \(\mathcal{B}_{\mathrm{c}}(t)\) is \(n\times(n-r)\), and the block-form matrix \(\big{(}\mathcal{B}_{\mathrm{c}}(t),\mathcal{B}_{\mathrm{u}}(t)\big{)}\) is invertible for all \(t\in[0,t_{f}]\). Due to Assumption (A6) and the results from [14] (Section 3.3), the matrix-valued function \(\mathcal{B}_{\mathrm{c}}(t)\) can be chosen as twice continuously differentiable in the interval \([0,t_{f}]\). Consider the following matrix-valued functions of \(t\in[0,t_{f}]\):  
\[\mathcal{L}_{\mathrm{u}}(t)=\mathcal{B}_{\mathrm{c}}(t)-\mathcal{B}_{\mathrm{u }}(t)\big{(}\mathcal{B}_{\mathrm{u}}^{T}(t)\mathcal{D}_{\mathrm{u}}(t) \mathcal{B}_{\mathrm{u}}(t)\big{)}^{-1}\mathcal{B}_{\mathrm{u}}^{T}(t)\mathcal{ D}_{\mathrm{u}}(t)\mathcal{B}_{\mathrm{c}}(t),\;\;\;\mathcal{R}_{\mathrm{u}}(t)= \big{(}\mathcal{L}_{\mathrm{u}}(t),\mathcal{B}_{\mathrm{u}}(t)\big{)}. \tag{7}\]  
**Remark 2**. : _Due to the results of [14] (Section 3.3), the matrix \(\mathcal{R}_{\mathrm{u}}(t)\) is invertible for all \(t\in[0,t_{f}]\). Moreover, the matrix-valued function \(\mathcal{R}_{\mathrm{u}}(t)\) is twice continuously differentiable with respect to \(t\in[0,t_{f}]\)._  
Using the matrix \(\mathcal{R}_{\mathrm{u}}(t)\), we transform the state in the game (1)-(3) as  
\[Z(t)=\mathcal{R}_{\mathrm{u}}(t)z(t),\quad t\in[0,t_{f}], \tag{8}\]  
where \(z(t)\in E^{n}\) is a new state variable. Using Remark 1, we obtain (quite similarly to the results of [14] (Section 3.3)) the following assertion. **Proposition 1**. : _Let Assumptions (A1)-(A6) be valid. Then, Transformation (8) converts System (1) and Functionals (2) and (3) into the following new system and functionals:_  
\[\frac{dz(t)}{dt} =A(t)z(t)+B_{\rm u}(t)u(t)+B_{\rm v}(t)v(t),\ \ t\in[0,t_{f}],\ \ z(0)=z_{0}, \tag{9}\] \[J_{\rm u}(u,v) =\frac{1}{2}\int_{0}^{t_{f}}\big{[}z^{T}(t)D_{\rm u}(t)z(t)+ \varepsilon^{2}u^{T}(t)u(t)+v^{T}(t)G_{\rm u,v}(t)v(t)\big{]}dt,\] (10) \[J_{\rm v}(u,v) =\frac{1}{2}\int_{0}^{t_{f}}\big{[}z^{T}(t)D_{\rm v}(t)z(t)+v^{T} (t)v(t)+\varepsilon^{2}u^{T}(t)G_{\rm v,u}(t)u(t)\big{]}dt,\] (11) _where_ \[A(t)=\mathcal{R}_{\rm u}^{-1}(t)\big{[}\mathcal{A}(t)\mathcal{R}_{\rm u}(t )-d\mathcal{R}_{\rm u}(t)/dt\big{]},\] (12) \[B_{\rm u}(t)=\mathcal{R}_{\rm u}^{-1}(t)\mathcal{B}_{\rm u}(t)=\left( \begin{array} []{c}0\\ I_{r}\end{array} \right),\ \ B_{\rm v}(t)=\mathcal{R}_{\rm u}^{-1}(t)\mathcal{B}_{\rm v}(t),\] (13) \[D_{\rm u}(t)=\mathcal{R}_{\rm u}^{T}(t)\mathcal{D}_{\rm u}(t)\mathcal{R}_{\rm u }(t)=\left( \begin{array}{cc}D_{\rm u,1}(t)&0\\ 0&D_{\rm u,2}(t)\end{array} \right),\] (14) \[D_{\rm u,1}(t)=\mathcal{L}_{\rm u}^{T}(t)\mathcal{D}_{\rm u}(t)\mathcal{L}_{ \rm u}(t),\ \ D_{\rm u,2}(t)=\mathcal{B}_{\rm u}^{T}(t)\mathcal{D}_{\rm u}(t) \mathcal{B}_{\rm u}(t),\] \[D_{\rm v}(t)=\mathcal{R}_{\rm u}^{T}(t)\mathcal{D}_{\rm v}(t)\mathcal{R}_{ \rm u}(t),\ \ z_{0}=\mathcal{R}_{\rm u}^{-1}(0)Z_{0}. \tag{15}\]  
_The matrix \(D_{\rm u,1}(t)\) is symmetric positive semi-definite, while the matrix \(D_{\rm u,2}(t)\) is symmetric positive definite for all \(t\in[0,t_{f}]\).