## 3 Modified SDRE Method
To solve the problem, consider the Hamiltonian  
\[H=\frac{1}{2}(\mathbf{x}^{T}\mathbf{Q}\mathbf{x}+\mathbf{u}^{T}\mathbf{R} \mathbf{u})+\mathbf{p}^{T}((\mathbf{A}_{1}+\mathbf{A}_{2}(\mathbf{x}))\mathbf{ x}+\mathbf{B}\mathbf{u}), \tag{25}\]  
where \(\mathbf{p}\in\mathbb{R}^{n}\) is the co-state vector. If \(\mathbf{u}\in C[t_{0},\infty]\) is a control that minimizes Equation (19) subject to the state Equation (20), and if \(\mathbf{x}\) denotes the corresponding state, then there exists a \(\mathbf{p}\in C[t_{0},\infty]\) such that  
\[J(\mathbf{u})=\frac{1}{2}\int\limits_{0}^{\infty}(\mathbf{x}^{T}\mathbf{Q} \mathbf{x}+\mathbf{u}^{T}\mathbf{R}\mathbf{u})dt, \tag{19}\]0  
and  
\[J(\mathbf{u})=\frac{1}{2}\int\limits_{0}^{\infty}(\mathbf{x}^{T}\mathbf{Q} \mathbf{x}+\mathbf{u}^{T}\mathbf{R}\mathbf{u})dt, \tag{19}\]1  
where \(\mathbf{p}\) satisfies the condition for optimal control that minimizes Equation (19). It follows that any optimal input \(\mathbf{u}\in\mathbb{R}^{m}\) and the corresponding state \(\mathbf{x}\in\mathbb{R}^{m}\) satisfy Equation (26); that is,  
\[J(\mathbf{u})=\frac{1}{2}\int\limits_{0}^{\infty}(\mathbf{x}^{T}\mathbf{Q} \mathbf{x}+\mathbf{u}^{T}\mathbf{R}\mathbf{u})dt, \tag{19}\]2  
Thus, the optimal control is  
\[J(\mathbf{u})=\frac{1}{2}\int\limits_{0}^{\infty}(\mathbf{x}^{T}\mathbf{Q} \mathbf{x}+\mathbf{u}^{T}\mathbf{R}\mathbf{u})dt, \tag{19}\]3  
and subsequently, the adjoint differential equation is  
\[J(\mathbf{u})=\frac{1}{2}\int\limits_{0}^{\infty}(\mathbf{x}^{T}\mathbf{Q} \mathbf{x}+\mathbf{u}^{T}\mathbf{R}\mathbf{u})dt, \tag{19}\]4  
for \(t\in[t_{0},\infty],\mathbf{x}(t_{0})=\mathbf{x}_{0},\mathbf{A}_{2}(\mathbf{x} _{0})=\mathbf{A}_{2_{0}}\), and \(\mathbf{p}(\infty)=\mathbf{0}\), where  
\[J(\mathbf{u})=\frac{1}{2}\int\limits_{0}^{\infty}(\mathbf{x}^{T}\mathbf{Q} \mathbf{x}+\mathbf{u}^{T}\mathbf{R}\mathbf{u})dt, \tag{19}\]5  
Consider \(\mathbf{p}\) as a linear and nonlinear combination of the state of the system Equation (18), which is the second modification of the classic approach  
\[J(\mathbf{u})=\frac{1}{2}\int\limits_{0}^{\infty}(\mathbf{x}^{T}\mathbf{Q} \mathbf{x}+\mathbf{u}^{T}\mathbf{R}\mathbf{u})dt, \tag{19}\]6  
where \(\mathbf{K}(\mathbf{x}),\mathbf{K}_{1},\mathbf{K}_{2}(\mathbf{x})\in\mathbb{R }^{n\times n}\). Consider \(\mathbf{x}\) as the solution of the nonlinear state equation  
\[J(\mathbf{u})=\frac{1}{2}\int\limits_{0}^{\infty}(\mathbf{x}^{T}\mathbf{Q} \mathbf{x}+\mathbf{u}^{T}\mathbf{R}\mathbf{u})dt, \tag{19}\]7  
for \(t\in[t_{0},\infty],\mathbf{x}(t_{0})=\mathbf{x}_{0}\), and \(\mathbf{A}_{2}(\mathbf{x}_{0})=\mathbf{A}_{20}\). Note that, in this approach, two feedback compensators are defined. Rearranging Equation (33), it is possible to obtain  
\[J(\mathbf{u})=\frac{1}{2}\int\limits_{0}^{\infty}(\mathbf{x}^{T}\mathbf{Q} \mathbf{x}+\mathbf{u}^{T}\mathbf{R}\mathbf{u})dt, \tag{19}\]8  
The first bracket of Equation (34) is state-independent, and the second one is state-dependent; thus, it is possible to linearize it and solve the state-dependent gain matrix-\(\mathbf{K}_{2}(\mathbf{x})\) from:  
\[J(\mathbf{u})=\frac{1}{2}\int\limits_{0}^{\infty}(\mathbf{x}^{T}\mathbf{Q} \mathbf{x}+\mathbf{u}^{T}\mathbf{R}\mathbf{u})dt, \tag{19}\]9  
as follows:  
\[\dot{\mathbf{x}}=\mathbf{A}(\mathbf{x})\mathbf{x}+\mathbf{B}\mathbf{u}= \mathbf{A}_{1}\mathbf{x}+\mathbf{A}_{2}(\mathbf{x})\mathbf{x}+\mathbf{B} \mathbf{u}, \tag{20}\]0  
Matrix \(\mathbf{B}\mathbf{R}^{-1}\mathbf{B}^{T}\) is singular; thus, the state-dependent matrix gain \(\mathbf{K}_{2}(\mathbf{x})\) may be computed only by the pseudoinverse operation \([\dots]^{+}\). In order to perform this procedure, a Moore-Penrose pseudoinverse is used, and the matrix is unique [27; 35; 36].