# Compensation Control Strategy for Orbital Pursuit-Evasion Problem with Imperfect Information
## 3 Construction of Guaranteed Cost Strategy with Perfect Information
### Time-To-Go Calculation
#### 3.2.1 Time-To-Go Calculation Based on the Homotopy Method  
For the sake of brevity, ignore the superscripts and subscripts, and use \(F\) for the functions \(F_{P}\) and \(F_{E}\), and \(\theta\) for the variables \(\theta^{p}\) and \(\theta^{E}\). To solve the nonlinear equation \(F(\theta)=0\), we construct the fixed-point homotopy equation as follows:  
\[H(\theta,p)=pF(\theta)+(1-p)(\theta-\theta_{0})\ =\ 0 \tag{20}\]  
where \(p\ \in\ [0,\ 1]\). The solution of \(H(\theta,1)\ =\ 0\) is the desired solution, whereas the solution of \(H(\theta,0)\ =\ 0\) is \(\theta_{0}\), which can be any specified value. As shown in Figure 3, when the variable \(p\) increases from \(0\) to \(1\), the solution of the homotopy equation changes continuously from \(\theta_{0}\) to the expected solution along the homotopy path. The detailed procedure is shown as follows. \begin{table}
\begin{tabular}{c c c c c c c} \hline
**Case** & \(X\) **(m)** & \(Y\) **(m)** & \(Z\) **(m)** & \(V_{X}\) **(m \(\times\) s\({}^{-1}\))** & \(V_{Y}\) **(m \(\times\) s\({}^{-1}\))** & \(V_{Z}\) **(m \(\times\) s\({}^{-1}\))** \\ \hline Case 1 & 3000 & 200 & 1000 & \(-3\) & \(-8\) & \(-5\) \\ Case 2 & 413 & \(-32\) & 331 & \(-5\) & \(-0.5\) & \(-12.5\) \\ Case 3 & 413 & \(-32\) & 331 & \(5\) & \(0.5\) & \(12.5\) \\ Case 4 & 306 & \(-16\) & 166 & \(-10\) & \(0\) & \(-5\) \\ \hline \end{tabular}
\end{table} 
Table 1: Current relative states of different cases. Figure 2: The function \(F\) change with range \(\theta\) in (**a**) case 1, (**b**) case 2, (**c**) case 3 and (**d**) case 4. To find the solution along the homotopy path, first, we need to get the unit tangent vector of the homotopy path. Let \(\theta~{}=~{}\theta(s)\) and \(p~{}=~{}p(s)\), where \(s\) is the arc length along the homotopy path. The derivative of \(H(\theta,p)\) with respect to \(s\) is  
\[\frac{\partial H}{\partial\theta}\frac{d\theta}{ds}+\frac{\partial H}{\partial p }\frac{dp}{ds}=0 \tag{21}\]  
where  
\[\left\{ \begin{array}{l}\frac{\partial H}{\partial\theta}=p\frac{\mathrm{d}F} {\mathrm{d}\theta}+(1-p)\\ \frac{\mathrm{d}H}{\partial p}=F-\theta+\theta_{0}\\ \frac{\mathrm{d}F}{\mathrm{d}\theta^{p}}=\frac{Z^{\left(\theta^{p}\right)}DA \phi\left(\theta^{p}\right)x_{PE}}{\omega\left\|Z\left(\theta^{p}\right)\right\| }+\frac{1}{\omega}\left(-\rho_{P}\lambda_{\mathrm{min}}^{1/2}\left(\theta^{p }\right)+\rho_{E}\lambda_{\mathrm{max}}^{1/2}\left(\theta^{p}\right)\right)\\ \frac{\mathrm{d}F_{E}}{\mathrm{d}\theta^{p}}=\frac{Z^{\left(\theta^{p}\right)} DA\phi\left(\theta^{p}\right)x_{PE}}{\omega\left\|Z\left(\theta^{2}\right)\right\| }+\frac{1}{\omega}\left(-\rho_{P}\lambda_{\mathrm{max}}^{1/2}\left(\theta^{E }\right)+\rho_{E}\lambda_{\mathrm{min}}^{1/2}\left(\theta^{E}\right)\right) \end{array} \right. \tag{22}\]  
Considering the arc length constraint, the relationship between \(\theta,p\), and \(s\) is as follows:  
\[\left(\frac{d\theta}{ds}\right)^{2}+\left(\frac{dp}{ds}\right)^{2}=1. \tag{23}\]  
The unit tangent vector of the homotopy path \(\boldsymbol{V}~{}=~{}\left[\frac{d\theta}{ds},\frac{dp}{ds}\right]^{T}\) can be obtained by combining Equations (21) and (23):  
\[\left\{ \begin{array}{l}\frac{dp}{ds}=\frac{\|H_{\theta}\|}{\sqrt{H_{\theta} ^{2}+H_{\theta}^{2}}}\\ \frac{d\theta}{ds}=-\frac{H_{p}}{H_{\theta}}\frac{\|H_{\theta}\|}{\sqrt{H_{ \theta}^{2}+H_{p}}^{2}}\end{array} \right., \tag{24}\]  
where \(H_{\theta}=\frac{\partial H}{\partial\theta}\), \(H_{p}=\frac{\partial H}{\partial p}\). By integrating the unit tangent vector until \(p=1\) is satisfied, the desired solution can be obtained. The iterative equation is as follows:  
\[\left\{ \begin{array}{l}\left[\begin{array}{l}\theta\\ p\end{array}\right]^{(k+1)}=\left[\begin{array}{l}\theta\\ p\end{array}\right]^{(k)}+h\boldsymbol{V}^{(k)},\quad k=0,1,\cdots\\ \left[\begin{array}{l}\theta\\ p\end{array}\right]^{(0)}=\left[\begin{array}{l}\theta_{0}\\ 0\end{array}\right]\end{array} \right., \tag{25}\]  
where \(h\) is the step length.