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Can you point me at an existence theorem for an optimal control problem with binary control set? In particular,

$$ \max_u \int_0^T u(t) e^{-f(A(t))}(v(A(t))-c) \, dt $$ $$\text{s.t. } \dot{A}(t)=u $$

where $u \in \{0,1\}$. The existence results that I found typically required the control set to be convex.

Michael Hardy
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ctmntlds
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  • Depending on the $f$ and $v$ functions, the maximum may be $\infty$. For example, if $v(a) = c+1$, $f(a)=a$ for all $a \in \mathbb{R}$, $u(t)=1$ for all $t$, and $A(t) = t-d$ for any constant $d \in \mathbb{R}$, then the integral is $e^d\int_0^T e^{-t}dt$, which can be arbitrarily large by choosing $d\rightarrow\infty$. – Michael Sep 11 '16 at 05:59
  • I can show that the value is always bounded. Also, the functions are all continuous and differentiable. – ctmntlds Sep 11 '16 at 15:18
  • Well then what about my example above that shows the value can be arbitrarily large? In other words, fix $T>0$, $c \in \mathbb{R}$, and consider $v(a)=c+1$ (a constant function) and $f(a)=a$ for all $a \in \mathbb{R}$. You cannot bound the value of solutions in this case. – Michael Sep 11 '16 at 17:30
  • Sorry, but in my case these are particular functions in my example. And for those, the value cannot grow arbitrarily large. I am wondering about existence theorems that give conditions on when an optimal control exists. In particular, for control sets with finite elements. – ctmntlds Sep 11 '16 at 18:21

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Let us temporarily remove the binary constraint and solve the problem without that. Then, we shall see if we can achieve the same value under the binary constraint.

Fix $T>0$, $c \in \mathbb{R}$, and let $f(x)$ and $v(x)$ be given differentiable functions of $x \in \mathbb{R}$. For $t \in \mathbb{R}$ define: $$ H(t) = \int_0^t e^{-f(x)}(v(x)-c)dx $$ Let $A(t)$ be a given differentiable function of $t \in \mathbb{R}$. Suppose that $A(t)$ is nondecreasing and has $A'(t) \in [0,1]$ for all $t$. So $A(t)$ increases by at most $T$ over any interval of duration $T$. Then the objective function value for this given $A(t)$ is: \begin{align} val &= \int_0^T A'(t) e^{-f(A(t))}(v(A(t))-c)dt \\ &= \int_0^T \frac{d}{dt}[H(A(t))]dt\\ &= H(A(T))-H(A(0)) \\ &\leq \sup_{z \in [0,T]} [H(A(0)+z)-H(A(0))]\\ &\leq \sup_{a \in \mathbb{R}, z \in [0,T]} [H(a+z)-H(a)] \end{align}

Suppose the supremum is finite and is achieved at particular values $a^* \in \mathbb{R}, z^* \in [0,T]$. Then the optimal value is: $$ val^* = H(a^*+z^*) - H(a^*) $$ and we can achieve this value using the function $A(t) = a^* + (z^*/T)t$. So linear functions are optimal for this problem. Of course, the objective value only depends on the $A(t)$ function at its endpoints, so many other kinds of functions are also optimal.

We can also achieve $val^*$ using binary control: Define: $$ u(t) = \left\{ \begin{array}{ll} 0 &\mbox{ if $t <T-z^*$} \\ 1 & \mbox{ if $t \geq T-z^*$} \end{array} \right.$$ Define $A(t) = a^* + \int_0^t u(\tau)d\tau$. Then $A(T-z^*)=a^*$ and $A(T)=a^*+z^*$. We get: \begin{align} val &= \int_0^T u(t)e^{-f(A(t))}(v(A(t))-c)dt \\ &= \int_0^{T-z^*} u(t)e^{-f(A(t))}(v(A(t))-c)dt + \int_{T-z^*}^Tu(t)e^{-f(A(t))}(v(A(t))-c)dt \\ &= \int_{T-z^*}^T A'(t) e^{-f(A(t))}(v(A(t))-c)dt\\ &=H(A(T))-H(A(T-z^*))\\ &=H(a^*+z^*)-H(a^*)\\ &=val^* \end{align}

Michael
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  • Note that $H(t)=1-e^{-t}$ for the example case I gave in my comments, that is, $v(a)=c+1$ [constant function] and $f(a)=a$ [identity function]. In that case $val^*=\sup_{a \in \mathbb{R}, z \in [0,T]} [H(a+z)-H(a)]=\infty$. – Michael Sep 12 '16 at 16:01