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I am kind of confused by the following problem, and wondering if someone could give me some hints. Many thanks!

In optimal control theory, the target set is a description of restrictions on the endpoints. For example, a fixed-time, free-endpoint problem could be formulated as $$\min_{u(t)}\int_0^TL(t,x(t),u(t))\,dt$$ $$\text{s.t.}\,\,\dot{x}(t)=f(t,x(t),u(t)),\,\,x(0)=x_0>0, \,\,u(t)\in U\subseteq \mathbb R.$$ That is, there are no constraints on the value of $x(T)$ and hence it can take any achievable value. But it is possible that due to the setup of the problem, $x(T)$ can only reach a proper subset of $\mathbb R$ rather than the whole $\mathbb R$. For example, if $U=[0,1]$ and $$x(t)=x_0\exp\left\{-\int_0^tu(s)\,ds\right\},$$ then $x(T)$ can only reach a point in $(x_0e^{-T},x_0]$ for any admissible control. In this case, should the target set be $\{T\}\times \mathbb R$ or $\{T\}\times (x_0e^{-T},x_0]$? The answer to this question is important to correctly using the Maximum Principle, since the necessary conditions for optimality depend on the form of the target set.

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    The *target set* (or *target manifold*) is the set of states one wants to reach (we are happy if we reach *any* state in the target set). It should be whatever corresponds to the goal of the control system. By contrast, the set of all states that can be reached with the given control dynamics ($\dot{x} = \ldots$) is called the *reachable set* (or *attainable set*).2017-01-30
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    @ avs So it means that target set is basically one part of the setup rather than a derivative of other parts of the setup, right?2017-01-30
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    Yes. Furthermore, one can, in principle, pick a target set that has no reachable states, thereby obtaining an impossible control problem. (E.g., hypothetically, consider a self-guiding missile with enough fuel to go at most 200 miles, while the actual physical target is 350 miles away.)2017-01-30
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    @ avs So if there are $n$ real-valued state variables and we are "happy" with any terminal state, then we can always set the target set to be $\mathbb R^n$, although actually not all states in $\mathbb R^n$ are reachable?2017-01-30
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    Yes, that is correct.2017-01-30

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