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In linear control theory, we consider a system of the form $$ \dot x = A\,x(t) + B\, u(t) \qquad \star $$ where $x(t)$ is the state function, $u(t)$ is the control function and $A,B$ are matrices.

We denote by $x(t,x_0,u)$ the solution at time $t$ starting from $x_0$ and associated to the control $u(t)$, and we have $$ x(t,x_0,u) = \exp(t\,A)\left(x_0+\int_0^T \exp(-s\,A)Bu(s) \; ds \right) $$ and the attainable set from $x_0$ at time $T$ is the set $$ A(x_0,T) = \cup_{u(\cdot)}\; x(t,x_0,u). $$

We can show that: $$ A(x_0,T)=\exp(At)x_0+A(0,T). $$

My question is the following.

Previously, I didn't specify the set where $u$ belongs. Typically we consider $u\in L^1([0,T],\mathbb{R}^m)$ for some $T>0$.

What if we take $u\in L^\infty([0,T],\mathbb{R}^m)$ ? $u\in L^2([0,T],\mathbb{R}^m)$ ?

What is the influence of the set where $u$ belongs ? Is that affect the attainable set ?

prolea
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1 Answers1

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The standard assumption is that $u$ is piecewise continuous, which guarantees the existence and uniqueness of the solution to ($\star$) on any finite time interval. For a finite-horizon problem, $u$ doesn't need to be $L_p$, and in most cases it won't be. Consider for instance a step command, ramp command, or sinusoid. All of these are common inputs and none of them are $L_p.$ Most likely you are looking at infinite-horizon problems, in which you need integrals like

$$ J := \int_0^\infty (x^TQx + u^TRu)\ dt $$

to converge.

Concerning your question: if you restrict the input space, then in general you will change the reachable (attainable) set. This follows directly from the definition you gave plus the fact that if $A,B$ are sets, $x:A\rightarrow B$ is function, and $C\subset A$, then

$$ \bigcup_{c\in C}\{x(c)\} \subset\bigcup_{a\in A}\{x(a)\}. $$

Note: for $L_p-L_q$ inclusion theorems, see $L^p$ and $L^q$ space inclusion.

JMJ
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  • Thanks for your good answer. Why does a sinusoid is not in $L_p$? the sinusoid defined on $[0,T]$, $T<\infty$ is integrable on $[0,T]$, isn't it? – prolea Oct 23 '18 at 20:50
  • @Smilia most of the time $L_p$ implies functions with unbounded support. Isn't $L_p$ on $[0,T]$ essentially the same as saying piecewise continuous? I'm not sure but I'll give it some thought. – JMJ Oct 24 '18 at 14:36
  • ok, thanks. Is it not obvious that on a finite interval, a piecewise continuous function may be not $L_p$?

    Take on $[0,2]$ the following function:

    $f(x)=-1/(x-1)$ if x<1, $f(1)=0$ and $f(x)=1/(x-1)$ if x>1. This function is piecewise continuous but not in $L_1$.

     – prolea Oct 31 '18 at 15:49
  • @Smilia It may be, but the function you give is not piecewise continuous on $[0,2]$, since it is not defined at $x = 1$. Perhaps we are using different definitions of piecewise continuous? – JMJ Oct 31 '18 at 15:59
  • Ok my definition is the following: "A function or curve is piecewise continuous if it is continuous on all but a finite number of points at which certain matching conditions are sometimes required."

    So my function is defined on $[0,2]$ and $f(1)=0$ and it is piecewise continuous because there is a finite number of points of discontinuity (here one).

     – prolea Oct 31 '18 at 16:45
  • @Smilia OK. I think I see the problem. The definition of piecewise continuous used in existence and uniqueness theorems for ODEs is: a function is piecewise continuous on a domain $\Omega$ if (1) it is defined everywhere on $\Omega$, (2) there exists a partition of $\Omega = \cup \Omega_i$ such that $f|{int(\Omega_i)}$ is continuous and (3) the limit of $f|{int(\Omega_i)}(x_j)$, where $x_j$ is any sequence which approaches the boundary of $\Omega_i$ is finite. – JMJ Oct 31 '18 at 17:01
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    For example, consider $\dot{x} = \frac{1}{1-t}$. By your definition of piecewise continuous, this function is pc on $[0,2]$. The solution is $\ln\frac{1}{1-t}$, which has no solution beyond $[0,1)$, so the existence fails. – JMJ Oct 31 '18 at 17:05
  • ok I see thanks. – prolea Nov 01 '18 at 07:19