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4.2.5 Terminal cone

We now want to describe geometrically the combined effect of the temporal and spatial control perturbations on the terminal state. The vector $ \varepsilon \delta(w,I)$ describes the infinitesimal (first-order in $ \varepsilon $ ) perturbation of the terminal state caused by the needle perturbation with parameters $ w$ and $ I$ . (It corresponds to the vector labeled as $ \varepsilon \psi(t^*)$ in Figure 4.6.) From the definition (4.24) of $ \delta(w,I)$ , it is clear that its direction depends only on $ w$ and $ b$ , but not on $ a$ . We let $ \vec\rho(w,b)$ denote the ray in this direction originating at $ y^*(t^*)$ . If we keep $ w$ , $ b$ , and $ \varepsilon $ fixed, $ \vec\rho(w,b)$ consists of the points $ y^*(t^*)+\varepsilon \delta(w,I)$ for various values of $ a$ . The construction of $ \vec\rho(w,b)$ is analogous to that of $ \vec\rho$ in Section 4.2.2, except that $ \vec\rho(w,b)$ is unidirectional (because both $ a$ and $ \varepsilon $ are positive) whereas $ \vec\rho$ was bidirectional. We also let $ \vec P$ denote the union of the rays $ \vec\rho(w,b)$ for all possible values of $ w$ and $ b$ . Then $ \vec P$ is a cone with vertex at $ y^*(t^*)$ . Note that this cone is not convex; for example, if the control set $ U$ (in which $ w$ takes values) is finite, then $ \vec P$ will in general be a union of isolated rays starting at $ y^*(t^*)$ .

Let us now ask ourselves the following question: is there a spatial control perturbation such that the corresponding first-order perturbation of the terminal point is, say, $ \varepsilon \delta(w_1,I_1)+\varepsilon \delta(w_2,I_2)$ for some control values $ w_1,w_2$ and intervals $ I_1=(b_1-\varepsilon a_1, b_1]$ and $ I_2=(b_2-\varepsilon a_2, b_2]$ ? We will now see that the right way to ``add" two needle perturbations is to concatenate them, i.e., to perturb $ u^*$ both on $ I_1$ (by setting it equal to $ w_1$ there) and on $ I_2$ (by setting it equal to $ w_2$ ). Here we are assuming that $ b_1< b_2$ , so that for $ \varepsilon $ small enough $ I_1$ and $ I_2$ do not overlap. Such a spatial perturbation is shown in Figure 4.7.

Figure: ``Adding" spatial perturbations

The resulting first-order perturbation of the terminal point will then be the sum $ \varepsilon \delta(w_1,I_1)+\varepsilon \delta(w_2,I_2)$ . This is true simply because the variational equation, which propagates first-order state perturbations up to the terminal time, is linear. Indeed, according to the formulas derived in the two previous subsections, we will have

$\displaystyle y(b_1)=y^*(b_1)+\nu_{b_1}(w_1)\varepsilon a_1+o(\varepsilon )

at the end of the first perturbation interval, then

$\displaystyle y(b_2)=y^*(b_2)+\varepsilon \big(
\Phi_*(b_2,b_1)\nu_{b_1}(w_1)a_1+\nu_{b_2}(w_2)a_2\big)+o(\varepsilon )

at the end of the second perturbation interval, and finally, since $ \Phi_*(t^*,b_2)\Phi_*(b_2,b_1)=\Phi_*(t^*,b_1)$ by the semigroup property of the transition matrix,

$\displaystyle y(t^*)$ $\displaystyle =y^*(t^*)+\varepsilon \Phi_*(t^*,b_2)\big( \Phi_*(b_2,b_1)\nu_{b_1}(w_1)a_1+\nu_{b_2}(w_2)a_2\big)+o(\varepsilon )$    
  $\displaystyle = y^*(t^*)+\varepsilon \Phi_*(t^*,b_1)\nu_{b_1}(w_1)a_1+\varepsilon \Phi_*(t^*,b_2)\nu_{b_2}(w_2)a_2 +o(\varepsilon )$    
  $\displaystyle = y^*(t^*)+\varepsilon \delta(w_1,I_1)+\varepsilon \delta(w_2,I_2)+o(\varepsilon ).$    

Explain how to \lq\lq add'' two
needle perturbations with $b_1=b_2$.... mind that $w_1+w_2$
might not be an admissible control value.)

More generally, if we want to generate the infinitesimal perturbation $ \varepsilon \beta_1\delta(w_1,I_1)+\varepsilon \beta_2\delta(w_2,I_2)$ for some $ \beta_1,\beta_2>0$ , then it is easy to see that we need to adjust the lengths of the intervals on which the two needle perturbations are acting: we need to set $ u(t)=w_1$ on $ \bar
I_1:=(b_1-\varepsilon \beta_1a_1,b_1]$ and $ u(t)=w_2$ on $ \bar
I_2:=(b_2-\varepsilon \beta_2a_2,b_2]$ . It is also clear that this construction immediately extends to linear combinations (with positive coefficients) of more than two terms.

Recall that $ \vec P$ is the cone with vertex at $ y^*(t^*)$ formed by the rays $ \vec\rho(w,b)$ corresponding to all simple (individual) needle perturbations. The preceding discussion demonstrates that by concatenating different needle perturbations, we generate a larger cone (with the same vertex) which consists exactly of convex combinations of points in $ \vec P$ . We denote this convex cone by co$ (\vec P)$ .

In Section 4.2.2 we also constructed the line $ \vec\rho$ of perturbation directions arising from the temporal perturbations of $ u^*$ . We now add this line to the convex cone co$ (\vec P)$ , in the sense of adding vectors attached to the point $ y^*(t^*)$ . More precisely, we consider the set of points of the form

$\displaystyle y=y^*(t^*)+\varepsilon \Big(\beta_0\delta(\tau)+\sum_{i=1}^m \beta_i\delta(w_i,I_i)\Big)$ (4.25)

where $ \varepsilon >0$ , $ \delta(\tau)$ comes from (4.9) for some $ \tau$ , $ \delta(w_i,I_i)$ comes from (4.24) for some $ w_i$ and $ I_i$ , and $ \beta_0,\beta_1,\dots,\beta_m$ are arbitrary nonnegative numbers. We denote this set by $ C_{t^*}$ and call it the terminal cone. It is easy to check that $ C_{t^*}$ is again a convex cone, with vertex at $ y^*(t^*)$ . This construction is illustrated in Figure 4.8, where $ C_{t^*}$ is the infinite ``wedge" between the two half-planes.

Figure: The terminal cone

By the same reasoning as before, we can show that for every point $ y\in C_{t^*}$ given by (4.25) there exists a perturbation of $ u^*$ such that the terminal point of the perturbed trajectory satisfies

$\displaystyle y(t_f)=y+o(\varepsilon ).

To obtain the desired control perturbation, we need to apply a concatenated spatial perturbation as explained above, followed by the temporal perturbation that adjusts the terminal time by $ \beta_0\varepsilon \tau$ . Since the intervals $ I_i$ are strictly inside $ [t_0,t^*]$ , they do not interfere with the temporal perturbation (for small enough $ \varepsilon $ ). The fact that the resulting total first-order perturbation of the terminal point is indeed the correct one hinges on the linearity of the variational equation and on the linear dependence of $ \delta(\tau)$ on $ \tau$ .

next up previous contents index
Next: 4.2.6 Key topological lemma Up: 4.2 Proof of the Previous: 4.2.4 Variational equation   Contents   Index
Daniel 2010-12-20