Comments on ‘Sweep algorithm for solving optimal control problem with multi-point boundary conditions’ by M Mutallimov, R Zulfuqarova, and L Amirova

Abstract

A counter example is given for the solution of the linear-quadratic optimization problem with three-point boundary conditions. The example shows that the solution obtained in (Mutallimov et al. in Adv. Differ. Equ. 2015:233, 2015) by using a sweep method is not optimal.

1 Introduction

In [1] the linear-quadratic optimization problem with multi-point boundary conditions, both in the continuous and the discrete cases, are considered. The sweep method [2, 3], which generalizes the results [4] for the two-point boundary conditions is given in [5]. However, the results obtained for the discrete case [1] are not optimal.

Not passing to the illustration of an example, we form the problem of discrete optimal control with multi-point boundary conditions [1, 4]. Let the motion of an object be described by the following linear system of finite-difference equations:

$$ x(i + 1) = \psi (i)x(i) + \Gamma (i)u(i)\quad (i = 0,1,\ldots, l - 1), $$

(1)

with nonseparate boundary conditions

$$ \Phi_{1}x(0) + \Phi_{2}x(s) +\Phi_{3}x(l) = q. $$

(2)

Here $x(l)$ is an n-dimensional phase vector, $u(i)$ an m-dimensional vector of control influences, $\psi (i)$, $\Gamma (i)$ ($i = 0,1,\ldots,l - 1$) matrices of the corresponding dimensions, being a controllability pair [4, 6], $\Phi_{1},\Phi_{2},\Phi_{3}$ are constant matrices, such that the system (2) satisfies the Kronecker-Capelli condition [3, 4], $0< s< l$.

It is required to find such a control $u(i)$ as minimizes the following quadratic functional:

$$ J = \sum_{i = 0}^{l - 1} \bigl( x'(i)Q(i)x(i) + u'(i)C(i)u(i) \bigr), $$

(3)

under the conditions (1), (2), where $Q(i) = Q'(i) \ge 0$, $C(i) = C'(i) \ge 0$ are the periodic matrices of the corresponding dimensions.

Let us illustrate this on the example from [4] in the one-dimensional case. Indeed, in the problem (23)-(25) from [1], let

$$\begin{aligned}& n = 1,\qquad m = 1,\qquad \psi (0) = \psi (1) = 1,\qquad \psi (2) = \psi (3) = 2, \\& \Gamma (0) = \Gamma (1) = \Gamma (2) = \Gamma (3) = 1,\qquad \Phi_{1} = \Phi_{2} = \Phi_{3} = 1, \qquad q = 1, \\& Q(0) = Q(1) = Q(2) = Q(3) = 1, \qquad C(0) = C(1) = C(2) = C(3) = 1. \end{aligned}$$

(4)

Using the algorithm given in [1] we can see that the ‘optimal’ phase trajectory and control, respectively, have the form

$$\begin{aligned}& x(0) = \frac{6}{19}, \qquad x(1) = \frac{5}{19},\qquad x(2) = \frac{4}{19},\qquad x(3) = \frac{7}{19}, \qquad x(4) = \frac{9}{19}, \\& u(0) = - \frac{1}{19}, \qquad u(1) = - \frac{1}{19},\qquad u(2) = - \frac{1}{19},\qquad u(3) = - \frac{5}{5}. \end{aligned}$$

Then it is easy to calculate [6, 7] that the ‘optimal’ value of the functional (25) of [1] will be $J \approx 0.8$.

However, the algorithm as given in [4, 6] gives other results, i.e.

$$\begin{aligned}& x(0) = \frac{5}{26}, \qquad x(1) = \frac{1}{26}, \qquad x(2) = \frac{1}{13},\qquad x(3) = \frac{7}{26}, \qquad x(4) = \frac{23}{26}, \\& u(0) = - \frac{2}{13},\qquad u(1) = \frac{3}{26}, \qquad u(2) = \frac{11}{26}, \qquad u(3) = \frac{9}{26}, \end{aligned}$$

and the functional (25) of [1] takes the value

$$J \approx 0.5. $$

Thus, the above solution in [1] is not optimal.