Synthesis of Optimal Bang–Bang Control for Cooperative Collision Avoidance for Aircraft (Ships) with Unequal Linear Speeds

Abstract

Close proximity encounters most often occur for situations in which participants have unequal linear speeds. Cooperative collision avoidance strategies for such situations are investigated. We show that, unlike the encounters of participants with equal linear speeds, bang–bang collision avoidance strategies are not always optimal when the linear speeds are unequal, and we establish the conditions for which no optimal bang–bang controls exist near the terminal time. Nevertheless, under certain conditions, we demonstrate that bang–bang collision avoidance strategies remain optimal for encounters of participants with unequal linear speeds. Such conditions are established, and it appears that they cover a wide range of important practical situations. The synthesis of bang–bang control is constructed, and its optimality is established.

Appendix

Lemma A.1

The value of ν which is consistent with the state equations and terminal constraint (9) is ν=0.

Proof

The transversality condition for free terminal time (20) can be rewritten as

(A.1)

First, consider the case u ₁=u ₂ (i.e., the RR and LL strategies). In this case, the last term in (A.1) disappears, and (A.1) can be rewritten, using Property 5.1, as

$$ \nu\bigl[\sin\phi_{T} + \gamma\sin(\theta_{0} -\phi_{T})\bigr] \bigl[ - u_{1} + \bigl(\sin\phi_{T}+ \gamma\sin(\theta_{0} - \phi_{T})\bigr)/r_{T}\bigr] = 0.$$

(A.2)

Assume that ν≠0. Then, condition (A.2) is satisfied if one of the conditions

(A.3)

(A.4)

holds. First, we assume that (A.3) is not satisfied. We also assume that ν is a continuous function of the initial value ρ ₀ and suppose that ν(ρ ₀)≠0. We fix θ ₀ and let $\hat{r}_{0}$ and $\hat{\phi}_{0}$ vary in the proximity of r ₀ and ϕ ₀, so that $\hat{\rho}_{0}: = (\hat{r}_{0},\hat{\phi }_{0},\theta_{0})^{*} = (r_{0} + \delta r,\phi_{0} +\delta\phi,\theta_{0})^{*}$. Then, there exists ε>0 such that ν(ρ) remains nonzero for all $\hat{\rho}_{0}$ such that |δr|≤ε,|δϕ|≤ε. The terminal constraint (9) and Property 5.1 imply that ϕ _T is a function of θ ₀ and γ and is independent of the initial values $\hat{r}_{0},\hat{\phi}_{0}$. Therefore, the left-hand side of (A.4) is a constant for all $\hat{\rho}_{0}$ such that |δr|≤ε,|δϕ|≤ε. At the same time, (60) implies that, for given θ ₀,γ,u ₁ and ϕ _T ,r _T is a function of the initial values $\hat{r}_{0},\hat{\phi}_{0}$ (through the identities $\hat{x}_{0}: = \hat{r}_{0}\sin\hat{\phi}_{0}$, $\hat{y}_{0}: = \hat{r}_{0}\cos\hat{\phi}_{0})$, and therefore r _T takes various values when $\hat{\rho}_{0}$ varies. Thus, (A.4) cannot be satisfied, and therefore ν=0.

We now assume that (A.3) is satisfied, which implies that (A.4) is not satisfied. The terminal constraint (9) implies

$$ \gamma^{2}\cos^{2}(\theta_{T} -\phi_{T}) = \cos^{2}\phi_{T},$$

(A.5)

while (A.3) implies

$$ \gamma^{2}\sin^{2}(\theta_{0} -\phi_{T}) = \sin^{2}\phi_{T}.$$

(A.6)

Adding conditions (A.5) and (A.6) yields γ=1, as the only case where both conditions can be satisfied. For γ=1, condition (A.3) is satisfied for θ ₀=0 and for ϕ _T =θ ₀/2−π/2−πk, while the terminal constraint (9) is satisfied for θ ₀=0 and for ϕ _T =θ ₀/2; ϕ _T =θ ₀/2−π. Therefore, for γ=1, the only case where both (9) and (A.3) are satisfied simultaneously is θ ₀=0, which is the case of no practical significance for γ=1. Thus, (A.3) cannot generally be satisfied, and therefore ν=0.

Now, consider the case u ₁=−u ₂ (i.e., the RL or LR strategies). In this case, condition (A.1) takes the form

(A.7)

or

$$ \nu\bigl\{ u_{1}\bigl[\gamma\sin(\theta_{T}- \phi_{T}) - \sin\phi_{T}\bigr] + \bigl[\sin \phi_{T} + \gamma\sin(\theta_{T} - \phi_{T})\bigr]^{2}/r_{T} \bigr\} = 0.$$

(A.8)

Assume that ν≠0. We start by assuming that sinϕ _T +γsin(θ _T −ϕ _T )≠0 and γsin(θ _T −ϕ _T )−sinϕ _T ≠0. Then, for (A.8) to be satisfied, the following condition

$$ r_{T} = - \frac{u_{1}[\gamma\sin(\theta_{T} - \phi _{T}) + \sin\phi_{T}]^{2}}{\gamma\sin(\theta_{T} - \phi_{T}) -\sin\phi_{T}}$$

(A.9)

must hold. Note that, for u ₁=−u ₂, the terminal condition (9) implies that θ _T is a function of ϕ _T and γ, and therefore the right-hand side of (A.9) is a constant for given ϕ _T ,u ₁ and γ. As before, we fix θ ₀ and let $\hat{r}_{0}$ and $\hat{\phi}_{0}$ vary in the proximity of r ₀ and ϕ ₀, so that $\hat{\rho}_{0} = (r_{0} + \delta r,\phi_{0} +\delta\phi,\theta_{0})$. If ν is a continuous function of the initial conditions and ν(ρ ₀)≠0, then there exists ε>0 such that ν(ρ) remains nonzero for all $\hat{\rho}_{0}$ satisfying |δr|≤ε,|δϕ|≤ε. Equations (62) implies that, for given γ,θ ₀ and ϕ _T , r _T  is a function of the initial values $\hat{r}_{0}$ and $\hat {\phi}_{0}$, and therefore r _T takes different values when $\tilde{\rho}_{0}$ varies. Thus, for given γ,θ ₀ and ϕ _T , and for all $\hat{\rho}_{0}$ such that |δr|≤ε, |δϕ|≤ε, (A.9) cannot be satisfied. Therefore, ν=0.

The remaining case where (A.8) can be satisfied is where γsin(θ _T −ϕ _T )+sinϕ _T =0 and γsin(θ _T −ϕ _T )−sinϕ _T =0 simultaneously. These two conditions are satisfied simultaneously if

$$ \sin\phi_{T} = 0 \quad\mbox{and} \quad\gamma\sin(\theta_{T} - \phi_{T}) = 0,$$

(A.10)

which implies ϕ _T =−π,0; θ _T =0,π. Substituting these expressions into the terminal condition (9) yields γ=1. In this case, the terminal constraint (9) is satisfied for θ _T =0 and for ϕ _T =θ _T /2 or ϕ _T =θ _T /2−π. Then, the only possible case where (A.10) and (9) are satisfied simultaneously is θ _T =0, ϕ _T =0,−π. As can be seen from [7], such a case is always suboptimal and therefore presents no practical interest. Thus, we come to contradiction, and therefore ν=0. □