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4.2.6 Key topological lemma

Up until now, we have not yet used the fact that $ u^*$ is an optimal control and $ y^*$ is an optimal trajectory. As discussed in Section 4.2.1 and demonstrated in Figure 4.2, optimality means that no other trajectory $ y$ corresponding to another control $ u$ can reach the line $ S'$ (the vertical line through $ \Big({\textstyle{0}\atop
\textstyle{x_1}}\Big)$ in the $ y$ -space) at a point below $ y^*(t^*)$ . Since the terminal cone $ C_{t^*}$ is a linear approximation of the set of points that we can reach by applying perturbed controls, we expect that the terminal cone should face ``upward."

To formalize this observation, consider the vector

$\displaystyle \mu:=\begin{pmatrix}-1&0&\cdots&0\end{pmatrix}^T\in\mathbb{R}^{n+1}$ (4.26)

and let $ \vec\mu$ be the ray generated by this vector (which points downward) originating at $ y^*(t^*)$ . Optimality suggests that $ \vec\mu$ should be directed outside of $ C_{t^*}$ , a situation illustrated in Figure 4.9. Since $ C_{t^*}$ is only an approximation, the correct claim is actually slightly weaker.

Figure: Illustrating the statement of Lemma 4.1

$\vec\mu$\ does not intersect the interior of the
cone $C_{t^*}$. \end{Lemma}

In other words, $ \vec\mu$ can in principle touch $ C_{t^*}$ along the boundary, but it cannot lie inside it. We note that since $ C_{t^*}$ is a cone, $ \vec\mu$ intersects its interior if and only if all points of $ \vec\mu$ except $ y^*(t^*)$ are interior points of $ C_{t^*}$ .

Let us see what would happen if the statement of the lemma were false and $ \vec\mu$ were inside $ C_{t^*}$ . By construction of the terminal cone, as explained at the end of Section 4.2.5, there would exist a (spatial plus temporal) perturbation of $ u^*$ such that the terminal point of the perturbed trajectory $ y$ would be given by

$\displaystyle y(t_f)=y^*(t^*)+\varepsilon \beta\mu+o(\varepsilon )

for some (arbitrary) $ \beta>0$ . Writing this out in terms of the components $ (x^0,x)$ of $ y$ and recalling the definition (4.26) of $ \mu$ and the relation (4.6) between $ x^0$ and the cost, we obtain

$\displaystyle J( u)$ $\displaystyle =J(u^*)-\varepsilon \beta+o(\varepsilon )$    
$\displaystyle x(t_f)$ $\displaystyle =x_1+o(\varepsilon )$    

where $ u$ is the perturbed control that generates $ y$ . Presently there is no direct contradiction with optimality of $ u^*$ yet, because the terminal point $ x(t_f)$ of the perturbed trajectory $ x$ is different from the prescribed terminal point $ x_1$ , i.e., $ x$ need not hit the target set. Thus we see that although Lemma 4.1 certainly seems plausible, it is not obvious.

Let us try to build a more convincing argument in support of Lemma 4.1. If the statement of the lemma is false, then we can pick a point $ \hat y$ on the ray $ \vec\mu$ below $ y^*(t^*)$ such that $ \hat y$ is contained in $ C_{t^*}$ together with a ball of some positive radius $ \varepsilon $ around it; let us denote this ball by $ B_\varepsilon $ . For a suitable value of $ \beta>0$ , we have $ \hat
y=y^*(t^*)+\varepsilon \beta\mu$ . Since the points in $ B_\varepsilon $ belong to $ C_{t^*}$ , they are of the form (4.25) and can be written as $ y^*(t^*)+\varepsilon \nu$ where the vectors $ \varepsilon \nu$ are first-order perturbations of the terminal point arising from control perturbations constructed earlier. We know that the actual terminal points of trajectories corresponding to these control perturbations are given by

$\displaystyle y^*(t^*)+\varepsilon \nu+o(\varepsilon ).$ (4.27)

We denote the set of these terminal points by $ \widetilde B_\varepsilon $ ; we can think of it as a ``warped" version of $ B_\varepsilon $ , since it is $ o(\varepsilon )$ away from $ B_\varepsilon $ .

In the above discussion, $ \varepsilon >0$ was fixed; we now make it tend to 0. The point $ y^*(t^*)+\varepsilon \beta\mu$ , which we relabel as $ \hat
y_\varepsilon $ to emphasize its dependence on $ \varepsilon $ , will approach $ y^*(t^*)$ along the ray $ \vec\mu$ as $ \varepsilon \to 0$ (here $ \beta$ is the same fixed positive number as in the original expression for $ \hat y$ ). The ball $ B_\varepsilon $ , which now stands for the ball of radius $ \varepsilon $ around $ \hat
y_\varepsilon $ , will still belong to $ C_{t^*}$ and consist of the points $ y^*(t^*)+\varepsilon \nu$ for each value of $ \varepsilon $ . Terminal points of perturbed state trajectories (the perturbations being parameterized by $ \varepsilon $ ) will still generate a ``warped ball" $ \widetilde B_\varepsilon $ consisting of points of the form (4.27). Figure 4.10 should help visualize this construction.

Figure: Proving Lemma 4.1

Since the center of $ B_\varepsilon $ is on $ \vec\mu$ below $ y^*(t^*)$ , the radius of $ B_\varepsilon $ is $ \varepsilon $ , and the ``warping" is of order $ o(\varepsilon )$ , for sufficiently small $ \varepsilon $ the set $ \widetilde B_\varepsilon $ will still intersect the ray $ \vec\mu$ below $ y^*(t^*)$ . But this means that there exists a perturbed trajectory $ x$ which hits the desired terminal point $ x_1$ with a lower value of the cost. The resulting contradiction proves the lemma.

The above claim about a nonempty intersection between $ \widetilde B_\varepsilon $ and $ \vec\mu$ seems intuitively obvious. The original proof of the maximum principle in [PBGM62] states that this fact is obvious, but then adds a lengthy footnote explaining that a rigorous proof can be given using topological arguments. A conceivable scenario that must be ruled out is one in which the set $ \widetilde B_\varepsilon $ has a hole (or dent) in it and the ray $ \vec\mu$ goes through this hole. It turns out that this is indeed impossible, thanks to continuity of the ``warping" map that transforms $ B_\varepsilon $ to $ \widetilde B_\varepsilon $ . In fact, it can be shown that $ \widetilde B_\varepsilon $ contains, for $ \varepsilon $ small enough, a ball centered at $ \hat
y_\varepsilon $ whose radius is of order $ \varepsilon -o(\varepsilon )$ . One quick way to prove this is by applying Brouwer's fixed point theorem (which states that a continuous map from a ball to itself must have a fixed point).

rigorously that $\widetilde B_\varepsilon $\ and $\vec\mu...
...e a
nonempty intersection for sufficiently small~$\varepsilon $.

next up previous contents index
Next: 4.2.7 Separating hyperplane Up: 4.2 Proof of the Previous: 4.2.5 Terminal cone   Contents   Index
Daniel 2010-12-20