ODE and Control Systems

In this section, we will review some standard facts on ODE and set up notation and terminology of control theory.

Standard Facts on ODE

We consider a typical ordinary differential equation, $egin{cases} dot{x}(t)=f(t,x(t)), quad t in [0,T]\ x(0)=x_0 end{cases} ag{1}label{1}$ where, at time , the state $x(t)in mathbb{R}^n$ . We call this equation Cauchy Problem if it has such boundary conditions.

We first define the solution of this Cauchy Problem for given in $mathbb{R}^n$ .

Definition 1 The map $x: [0,T] ightarrow mathbb{R}^n?$ is a solution of （1） , if $x(cdot)in C([0,T];mathbb{R}^n)$ and satisfiesIt is of interest to know whether the solution exists, if it exists, is it unique. In order to answer this question, it is necessary to put some restrictions on .

We will make the following assumptions:

The map $f: [0,T] imesmathbb{R}^n ightarrowmathbb{R}^n$ is measurable,
The map $f(cdot,0)in L^1([0,T];mathbb{R}^n)$ ,
There exists a constant , such thatholds for all $tin [0,T], x,yinmathbb{R}^n$ .

Under the above assumptions, the existence and uniqueness of solution is established by our next theorem. This can be found in[1, ch.2].

Theorem 1 (Existence and uniqueness theorem) If satisfy the above assumptions, there exists a unique solution of equation (1).

Proof of the Theorem 1We begin by constructing Picard iterative sequences as follows, $egin{cases} x_0(t)=x(0), & t in [0,T]\ x_{k+1}(t)=x_0+int_0^t f(s,x_k(s))ds, & tin[0,T], kgeq 0 end{cases}$ In order to prove the existence of solution, it suffices to prove thatconverges uniformly to a limiting function on. Thus will be the solution of equation (1) . Consider the series $x_0(t)+ sum_{j=1}^{k}(x_{j}(t)-x_{j-1}(t))$ we only need to show that it converges uniformly on, which is clear from following inequality, $|x_{k+1}(t)-x_k(t)|leq frac{N(Mt)^k}{k!}, quad forall tin [0,T], kgeq0$ where $N=int_{0}^{T} |f(s,x_0)|ds$ . Note that the existence of N depended on the assumption 2. The above inequality can be proved by induction and the assumption 3.

We next prove the solution is unique. By definition, ifare solutions, thenThe uniqueness of solution is established by the following Gr?nwalls inequality.

Theorem 2 (Gr?nwalls inequality)Letand letbe continue functions on [0,T]. Assumeis non-negative. Ifthen $varphi(t)leq alpha e^{int_0^tpsi(s)ds} + int_0^t e^{int_s^tpsi(r)dr}eta(s)ds,quad tin[0,T].$

Proof of Theorem 2Letbe defined byThenMultiplying $e^{-int_s^tpsi(r)dr}$ to the both side, yields $frac{d}{dt}left(e^{-int_s^tpsi(r)dr} heta(t ight)leq e^{int_s^tpsi(r)dr} eta(t).$ Integrating both side, we thus obtain the conclusion.

Controllability of Linear Control System

We consider the time-varying linear control systemwhere the map $A(t)in L^{infty}([0,T];mathcal{L}(mathbb{R}^m,mathbb{R}^n))$ and $B(t)in L^{infty}([0,T];mathcal{L}(mathbb{R}^m,mathbb{R}^n)$ , the control $u(t)in L^{1}([0,T];mathbb{R}^n)$ .

By Theorem 1, there exists a unique solutionof equationfor given initial data $x(0)=x_0in mathbb{R}^n$ , which satisfiedMoreover, according to Gr?nwalls inequality, $|x(t)|leq C(|x_0|+|u|_{L^1}), quad t in [0,T],$ the constantis dependent ofand.

Note that we can get an explicit formula by using variation of constants method (see[2, ch.3]for more details).

Theorem 3 (variation of constant formula)The solution of equationiswheresatisfies $frac{dS(t)}{dt}=AS(t), S(0)=I$ . Heredenotes the identity matrix.

Proof of Theorem 3The solution set of equation $frac{dx(t)}{dt}=Ax(t)$ is a vector space of dimension. Indeed, given independent initial data, it follows from the existence and uniqueness of solution and linearity of equation. Hence, there exists a standard solution matrix, such that all solutions are given by, whereis in $mathbb{R}^n$ .

Therefore, all solutions of equation eqref{2} are given by, whereis any solution of the equation. Replacingby, setting, and substituting it into equation eqref{2} , we can get $S(t)frac{c(t)}{dt}=B(t)u(t)$ , and obtain $c(t)=int_0^t S^{-1}(s)B(s)u(s)ds$ . It follows that $ilde{x}(t)=int_0^t S(t-s)B(s)u(s)ds$ , which proves the theorem.

Now, we are interested in the controllability of the control system. We defined here two types of controllability (see[3]and references given there).

Definition 4 (exact controllability)Let. The control systemisexactly controllability in time Tif, for every $x_0, x_1in mathbb{R}^n$ , there exists $uin L^2([0,T];mathbb{R}^m)$ such that the solutionof the equationwith initial datasatisfies.

Definition 5 (null controllability)Let. The control systemisnull controllability in time Tif, for every $x_0, ilde x_0in mathbb{R}^n$ , there exists $uin L^2([0,T];mathbb{R}^m)$ such that the solutionof the equationwith initial datasatisfies.

According to the linearity, we get an equivalent definition of null controllability if, in the definition above, one assumes that $ilde x_{0}=0$ .

It is worth pointing out that the exact controllability is equivalent to null controllability in the case of equation, which we will prove later. The proof is similar ifis a strong continuous group of linear operators.

Theorem 6 (exact controllability/null controllability)The control systemis exactly controllable in timeif and only if it is null controllable in time.

Proof of Theorem 6Necessity is obvious.

Let $x_0 in mathbb{R}^n$ and $x_1in mathbb{R}^n$ . From null controllability applied to the initial data, there exists $uin L^2([0,T];mathbb{R}^m)$ such that the solutionof the equationwith above initial data satisfies. It follows that the solutionof the equationis given byIn particular,. This concludes the proof of the theorem.