Preamble

DIFFERENTIAL EQUATIONS 1967, Vol. III, No. 3

Let $H$ be a separable Hilbert space, and let $A$ be a linear bounded operator mapping $H$ into $H$. Consider the equation:

$$\frac{dx}{dt} = Ax + bu, \quad (1)$$

where $u$ is a scalar function of time ($t < \infty$) called the control. Let us initially assume that $u(t) \in L_2$. We shall define the function $x = x(t)$ (where $t$ is a real argument and $x(t) \in H$) as a solution to equation (1) given $u = u(t)$ on an interval if the equality holds almost everywhere within that interval.

^ ± =Ax(t) + bu(t). dt

It is easily verified that the solution to equation (1) with the initial condition $x(0) = x_0$ is given by the Cauchy formula:

$$x(t) = [\exp At] x_0 + \int_0^t [\exp A(t - \tau)] b u(\tau) d\tau.$$

By definition, we shall refer to the generalized solution of the equation as the function $x(t) = \exp(At)x_0 + \int_0^t [\exp A(t - \tau)] b d\sigma(\tau)$, where $d\sigma(\tau)$ is the Stieltjes differential of a function of bounded variation $\sigma$. In particular, if $\sigma(t)$ is a differentiable function, then $d\sigma(t) = u(t)dt$. We define the dynamical system described by equation (1) as controllable if, for any point in the space $H$, there exists a function of bounded variation (an admissible control) and a finite time $T > 0$ such that the state reaches that point. Here, the state is determined by formula (2), and $x_0$ denotes the initial point in the space $H$. It is well known that in the case where the space $H$ is $n$-dimensional, the linear independence of the vectors $b, Ab, A^2b, \dots, A^{n-1}b$ ensures the controllability of the system. Naturally, the question arises as to what role the sequence $b, Ab, A^2b, \dots, A^nb, \dots$ plays when the space $H$ is infinite-dimensional. We shall call a system a basis of the space $H$ if every element $x \in H$ can be represented (not necessarily uniquely) in the form of a series with coefficients $c_n$.

F. A. SHOLOKHOVICH
Theorem: System (1) is controllable if the set...

E = {b,Ab,A 2 b,..., A m b,...}

is a basis in the space.
Proof. Let $x$ be an arbitrary point in the space $R$. For this point, there exists an admissible control $u$ and a number $T$ such that

$$x(T) = [\exp AT] x_0 + \int_{0}^{T} [\exp A(T - \tau)] B u(\tau) d\tau = q.$$

From this, we obtain $[\exp(-AT)] x(T) - x_0 = \int_{0}^{T} [\exp(-A\tau)] B u(\tau) d\tau$. The integrand can be represented as a series. We shall now demonstrate the validity of term-by-term integration for this series. By applying Theorem 3.7.5 and Corollary 1 of Section IV (see \cite{1}, pp. 95 and 78), we obtain:
$$\int_{0}^{T} \left\| \sum_{n=0}^{\infty} \frac{(-A\tau)^n}{n!} B u(\tau) \right\| d\tau \leq \sum_{n=0}^{\infty} \frac{\|A\|^n T^n}{n!} \|B\| \int_{0}^{T} \|u(\tau)\| d\tau < \infty.$$

< { sup

$Var.[a(x)]$. As $n \to \infty$, the latter expression tends to zero, which proves the validity of term-by-term integration. Performing term-by-term integration on the left side of equality (3), we obtain:
$$\int_0^T t^k da(x) = A b^k \int_0^T d\sigma(x) + (-1)^k b^k \int_0^T d\sigma(x)$$
The theorem is proved. It is easy to establish that the infinite dimensionality of the linear span of a set implies the linear independence of any finite number of elements taken from this set. Now, let the set serve as a basis for the space $H$ and, furthermore, let the following requirement be satisfied (the condition for an arbitrary $H$): the expansion is uniquely determined.

The condition will be satisfied, for example, if in the coordinate Hilbert space $l_2$ we take the vector $\{1, 0, 0, \dots, 0, \dots\}$ as $x$ and define the operator $A$ using a matrix. Obviously, $Ax_1 = \{0, 1, 0, \dots\}$, $Ax_2 = \{0, 0, 1, 0, \dots\}$, and so on. This condition is quite restrictive. When it is satisfied, the operator $A$, for instance, cannot have eigenvalues other than zero. Indeed, let $Ax = \lambda x$; for simplicity, assume that $\lambda$ is a real number. Represent the eigenvector of $\lambda$ in the form $x = \sum \alpha_i e_i$. By the continuity of the operator, we have $Ax = \sum \alpha_i A e_i$. Using the equality $\lambda x = Ax$, we obtain $\lambda \sum \alpha_i e_i = \sum \alpha_i e_{i+1}$. From this, it follows that $\alpha_0 = 0$, and consequently, $\alpha_1 = 0$. From this equality, it further follows that $\alpha_2 = 0$, and in general, $\alpha_n = 0$ for $n = 0, 1, 2, \dots$. Thus, $x = 0$, contrary to the assumption. We shall now show that in an infinite-dimensional space, Theorem 1 is not reversible if condition $C$ is satisfied. Let some point be transferred by the control $u(t)$ to the initial point $\theta$ in time $T$. Then equality (3) holds, and consequently, equality (4) holds as well. Suppose that $x(T) = 0$. Using equality (4), we obtain:
$$\int_0^T t^m da(t) = (-1)^m \frac{m!}{b^m} x_m \quad (m=0, 1, 2, \dots)$$
Conversely, if there exist a number $T$ and a function of bounded variation $a(t)$ such that equalities (5) hold, then the point is transferred to point $\theta$ by the control $u(t)$ in time $T$.

SHOLOKHOVICH. Thus, the question of the existence of a control for a given point that transfers the point to the equilibrium position in finite time is equivalent to the question of the solvability of the power moment problem (5). We perform a change of variables by setting $t = \tau/T$ and denote $g(t) = a(T t)$. From equality (5), we obtain:
$$\int_0^1 t^m dg(t) = (-1)^m \frac{m!}{b^m T^m} x_m$$
We utilize the following theorem by Hausdorff (\cite{2}, p. 209): In order for there to exist a function of bounded variation $g(t)$ such that...

1 \t m dg(t):-ll m (til = • 0, 1, 2, . . . ) ,

Theorem and Proof of Reachability

It is necessary and sufficient that for $n = 1, 2, \dots$, the binomial coefficients and the $n$-th differences of the sequence satisfy the specified conditions. Let $C$ be a certain positive constant. We denote the $n$-th differences of the sequence as $\Delta^n a_k$.

Theorem

If the condition is satisfied starting from a certain index $n$, and the numbers in the expansion are non-negative (or non-positive), then at these points, the system cannot be transferred to the origin from the initial state by any admissible control $u(t)$ in finite time according to the dynamics of equation (2). In other words, such a point does not belong to the reachability set of the initial point.

Proof

Indeed, let us assume that $t > 0$ and consider the properties of the control trajectory. Suppose there exists an admissible control that brings the system to the target state. By analyzing the sequence of differences and the signs of the coefficients in the expansion, we can demonstrate a contradiction regarding the finite-time reachability. Specifically, if the coefficients maintain a consistent sign (non-negative or non-positive) beyond a certain threshold, the integral representation of the trajectory under equation (2) precludes the state from reaching the origin within a finite interval.

Let

$\hat{T}_m = (-1)^m \hat{T}^{(m)}$ (8)

For an arbitrary, fixed number $k$, the formula $C_m^k = \frac{m(m-1)\dots(m-k+1)}{k!}$ holds. Taking into account the conditions of the theorem and formula (8), we obtain

Consequently, as $m \to \infty$, we have

m 2c |A M -|i*|->«>.

Condition (7), which is necessary for the solvability of the moment problem, is satisfied. The theorem is proved.

In particular, points whose expansions contain only a finite number of non-zero coefficients (where at least one of these coefficients has a positive index) belong to the reachable set of the origin. We define the dynamical system described by the equation as stabilizable if, for any point $x_0$, there exists an admissible control $u(t)$ such that the corresponding solution to the equation satisfies the property:

$$\lim_{t \to \infty} x(t, x_0, u) = 0.$$

Theorem: If the real-valued function $\| \exp(At) \|$ is bounded and the set $\{b, Ab, A^2b, \dots, A^n b, \dots\}$ forms a basis for the space $H$, then the dynamical system corresponding to the equation is stabilizable.

Proof: By transforming the Cauchy formula, we obtain:
$$x(t) = \exp(At) \left( x_0 + \int_0^t \exp(-A\tau) b u(\tau) d\tau \right)$$
We must establish the existence of a control $u(t)$ that ensures stabilization. Applying term-by-term integration here, we obtain:
$$u_0(x) - \int_0^t \tau A b u(\tau) d\tau + \dots + \frac{(-1)^m}{m!} \int_0^t \tau^m A^m b u(\tau) d\tau + \dots \quad (11)$$
Consider one of the expansions of the element $x_0$ in the basis $\{b, Ab, \dots, A^m b, \dots\}$ (where uniqueness of the expansion is assumed):
$$x_0 = \sum_{m=0}^{\infty} x_m^0 A^m b$$
To ensure that condition (11) is satisfied, it is sufficient to select a function $u(\tau)$ that solves the moment problem:

$$\int_0^\infty \tau^m u(\tau) d\tau = (-1)^m x_m^0 m! \quad (m = 0, 1, 2, \dots).$$

As is well known (\cite{3}, p. 103, Theorem 3.11), this problem has an infinite set of solutions for arbitrary right-hand sides. Let $u(t)$ be one such solution. Then

SHOLOKHOVICH

From the formula $\left\| x(t) \right\| \leq \left\| x(0) \right\| \left\| \exp(At) \right\|$, by virtue of the given condition and the boundedness of $\left\| \exp(At) \right\|$, we obtain

$$\lim_{t \to \infty} \| x(t) \| = 0.$$

The theorem is proved. The author expresses his gratitude to N. N. Krasovskii for discussing this work.

References

Hille, E., Phillips, R. S. Functional Analysis and Semi-groups. Moscow, IL, 1962. Kantorovich, L. V., Akilov, G. P. Functional Analysis in Normed Spaces. Fizmatgiz, 1959. Shohat, J. A., Tamarkin, J. D. The Problem of Moments. Math. Surveys, Amer. Math.

Soc., New York, 1943. Received by the editorial board on June 27, 1966. Ural State University named after A. M. Gorky.