UDC 517.949

AN OPTIMAL CONTROL PROBLEM IN SYSTEMS WITH AFTEREFFECT

R. GABASOV, S. V. CHURAKOVA

For a system with aftereffect, we consider an optimal control problem [1] with retention of the trajectory at the origin on a finite time interval. Such an approach to the problem ensures preservation of the prescribed regime also subsequently, if at the end of the interval the control is switched off.

Let the motion of the object be described by a differential equation with deviating argument [2]:

\[ \dot{x}(t)=A(t)x(t)+B(t)x(t-h)+C(t)u(t). \tag{1} \]

Here \(x\) is an \(n\)-dimensional vector defined in the space \(X\); \(u(t)\) is an \(n\)-dimensional piecewise-continuous vector-valued control function belonging to the set of admissible controls \(U\); \(A(t)\), \(B(t)\), and \(C(t)\) are continuous matrix functions, with \(C\) a nonsingular matrix for all \(t\), \(0\le t\le T\); \(h\), \(h>0\), is a constant number (the delay). Initial conditions are given: a vector-function \(\varphi(t)\) with values in \(X\), \(x(t)\equiv \varphi(t)\) for \(-h\le t<0\), and the vector \(x^0=x(0)\).

Problem 1. In the class of admissible controls, choose a control \(u(t)\), \(0\le t\le T\), such that the trajectory of system (1), corresponding to the prescribed initial conditions and to the chosen control \(u(t)\), satisfies the condition

\[ x(t)\equiv 0,\qquad T-h\le t\le T. \tag{2} \]

In the space \(X\) consider the set \(G\) of vectors

\[ e(T-h)=\int_{-h}^{0} F(T-h,\tau+h)B(\tau+h)\varphi(\tau)\,d\tau +F(T-h,0)x^0. \]

Here \(F(t,\tau)\) is the fundamental matrix of solutions of the homogeneous differential equation corresponding to equation (1), under the condition

\[ F(t,\tau)= \begin{cases} E, & \tau=t,\\ 0, & \tau>t \end{cases} \]

(\(E\) is the identity matrix). The set \(G_1\), \(G_1\subset G\), of vectors \(e(T-h)\) formed with the aid of such functions \(\varphi(t)\) and vectors \(x^0\) for which Problem 1 has a solution, will be called the reachability domain. The set of controls \(u(t)\) which are solutions of Problem 1 for the reachability domain will be denoted by \(U_1\), \(U_1\subset U\). It can be shown that the set \(G_1\) is convex (see the appendix). Let \(G_1\) be closed.

From equation (1) and condition (2) it follows that every control \(u(t)\) solving problem 1 on the interval \(T-h \leq t \leq T\) has the form

\[ u(t)=C^{-1}(t)B(t)x(t-h). \]

Theorem 1. In order that, for given \(x^0\), \(\varphi(t)\), \(T\), and \(h\), problem 1 have a solution, it is necessary and sufficient that the inequality

\[ \max_{\|g\|=1}\left\{(g,e(T-h))+\min_{u\in U_1}\left(g,\int_0^{T-h}F(T-h,\tau)C(\tau)u(\tau)\,d\tau\right)\right\}\leq 0, \tag{3} \]

be satisfied, where \(g\) is an arbitrary vector of unit norm.

We shall carry out the proof according to the scheme of [3].

Necessity. There is a control \(u=u^*(t)\), \(u^*\in U\), which, for the given \(x^0\), \(\varphi(t)\), \(T\), and \(h\), ensures fulfillment of condition (2).

To prove: condition (3) is satisfied.

By the condition of the theorem, the trajectory generated by the control \(u^*(t)\), at the moment \(t=T-h\), arrives at the origin, i.e. \(x(T-h)=0\), or

\[ e(T-h)+\int_0^{T-h}F(T-h,\tau)C(\tau)u^*(\tau)\,d\tau=0. \]

Multiply all terms of the equality scalarly by an arbitrary vector \(g\), \(\|g\|=1\):

\[ (g,e(T-h))+\left(g,\int_0^{T-h}F(T-h,\tau)C(\tau)u^*(\tau)\,d\tau\right)=0. \]

Now consider the functional

\[ P(u)=(g,e(T-h))+\left(g,\int_0^{T-h}F(T-h,\tau)C(\tau)u(\tau)\,d\tau\right) \]

on the set of functions \(u(t)\in U_1\). Since the given control \(u^*(t)\in U_1\) and, moreover, \(P(u^*)=0\), we have

\[ \min P(u)=(g,e(T-h))+\min_{u\in U_1}\left(g,\int_0^{T-h}F(T-h,\tau)C(\tau)u(\tau)\,d\tau\right)\leq 0. \]

But the last inequality is satisfied for an arbitrary vector \(g\), \(\|g\|=1\); consequently,

\[ \max_{\|g\|=1}\left\{(g,e(T-h))+\min_{u\in U_1}\left(g,\int_0^{T-h}F(T-h,\tau)C(\tau)u(\tau)\,d\tau\right)\right\}\leq 0, \]

which was required to be proved.

Sufficiency. Given: for the specified \(x^0\), \(\varphi(t)\), \(T\), and \(h\), condition (3) is satisfied.

To prove: there exists at least one control \(u(t)\), \(u\in U\), ensuring fulfillment of condition (2).

Suppose the contrary: there exists no control \(u(t)\), \(u\in U\), such that, for the trajectory generated by this control,

the condition (2) be satisfied. This means that the vector \(e(T-h)\), for the given \(x^0\) and \(\varphi(t)\), does not belong to the reachability domain \(G_1\). Then, by virtue of the convexity of the closed set \(G_1\), there exists a supporting hyperplane and a vector \(g^*\), \(\|g^*\|=1\), normal to this hyperplane, such that the inequality

\[ (g^*, e(T-h)) > (g^*, e^*) \]

holds for all \(e^* \in G_1\).

But since \(e^*\) is a point of the reachability domain, there exists for it a control \(u^*(t)\), \(u^* \in U_1\), such that

\[ e^*+\int_0^{T-h} F(T-h,\tau)C(\tau)u^*(\tau)\,d\tau=0. \]

Hence

\[ e^*=-\int_0^{T-h} F(T-h,\tau)C(\tau)u^*(\tau)\,d\tau. \]

Then

\[ (g^*, e(T-h))+\left(g^*, \int_0^{T-h} F(T-h,\tau)C(\tau)u^*(\tau)\,d\tau\right)>0. \]

Further, since \(e^*\) is an arbitrary point of the set \(G_1\), the last inequality is satisfied for all \(u \in U_1\); therefore

\[ (g^*, e(T-h))+\min_{u\in U_1}\left(g^*, \int_0^{T-h} F(T-h,\tau)C(\tau)u(\tau)\,d\tau\right)>0. \]

By the condition of the theorem, for the given \(x^0\) and \(\varphi(t)\) inequality (3) is satisfied, which means that for any vector \(g\), \(\|g\|=1\), the inequality

\[ (g,e(T-h))+\min_{u\in U_1}\left(g,\int_0^{T-h} F(T-h,\tau)C(\tau)u(\tau)\,d\tau\right)\le 0 \]

holds.

However, we have managed to construct a vector \(g^*\), \(\|g^*\|=1\), for which the inequality of the opposite sense is satisfied. This contradiction proves the theorem.

In what follows we shall assume that the set \(U\) of admissible controls has the form

\[ U=\left\{u:\int_0^T (u,u)\,dt\le 1\right\}. \tag{4} \]

Condition (3) will be written in another form:

\[ \max_{\|g\|=1}\{(g,e(T-h))+\mu^0(g)\}\le 0, \]

\[ \mu^0(g)=\min_u \mu(g,u)=\min_u\left(g,\int_0^{T-h} F(T-h,\tau)C(\tau)u(\tau)\,d\tau\right). \]

under the condition

\[ \int_{0}^{T-h} (u(t),u(t))\,dt+ \int_{T-h}^{T} \left(C^{-1}(t)B(t)x(t-h),\, C^{-1}(t)B(t)x(t-h)\right)\,dt \leqslant 1 . \]

We transform the last inequality, replacing \(x(t-h)\) by Cauchy’s formula [4]:

\[ \int_{0}^{T-h} (u(t),u(t))\,dt+ \int_{T-h}^{T}\left(C^{-1}(t)B(t)\left[e(t-h)+ \right.\right. \]

\[ \left. +\int_{0}^{t-h} F(t-h,\tau)C(\tau)u(\tau)\,d\tau\right],\, C^{-1}(t)B(t)\left[e(t-h)+ \right. \]

\[ \left.\left. +\int_{0}^{t-h} F(t-h,\tau)C(\tau)u(\tau)\,d\tau\right]\right)\,dt \leqslant 1 . \]

Expanding the brackets in the second integral and changing the order of integration, we obtain

\[ \int_{0}^{T-h} (u(\tau),u(\tau))\,d\tau+N+ 2\int_{0}^{T-h} (f(\tau),u(\tau))\,d\tau+ \]

\[ +\int_{0}^{T-h}\left(\int_{0}^{T-h}\Phi(\tau,\theta)u(\theta)\,d\theta,\,u(\tau)\right)d\tau \leqslant 1 . \]

Here

\[ N=\int_{T-2h}^{T-h} \left(C^{-1}(t+h)B(t+h)e(t),\,C^{-1}(t+h)B(t+h)e(t)\right)\,dt, \]

\[ f(\tau)=\int_{T-2h}^{T-h} \left[C^{-1}(t+h)B(t+h)F(t,\tau)C(\tau)\right]^* C^{-1}(t+h)B(t+h)e(t)\,dt, \]

\[ \Phi(\tau,\theta)=\int_{T-2h}^{T-h} \left[C^{-1}(t+h)B(t+h)F(t,\tau)C(\tau)\right]^* \times \]

\[ \times C^{-1}(t+h)B(t+h)F(t,\theta)C(\theta)\,dt. \]

Next, introducing a Lagrange multiplier, we pass to the problem of an unconstrained extremum. We shall seek the minimum of the functional \(A(u)\):

\[ \min A(u)=\min_{u}\left\{\left(g,\int_{0}^{T-h}F(T-h,\tau)C(\tau)u(\tau)\,d\tau\right)+\right. \]

\[ +\lambda\left[\int_{0}^{T-h}(u(\tau),u(\tau))\,d\tau+N+ 2\int_{0}^{T-h}(f(\tau),u(\tau))\,d\tau+\right. \]

\[ \left.\left. +\int_{0}^{T-h}\left(\int_{0}^{T-h}\Phi(\tau,\theta)u(\theta)\,d\theta,\,u(\tau)\right)d\tau\right]\right\}. \]

We compute the variation of the functional

\[ \delta A(u)=\left.\frac{\partial A(u^*+\alpha v)}{\partial \alpha}\right|_{\alpha=0} = \int_0^{T-h} \left( L(\tau)g+\lambda\left[ 2u^*(\tau)+2f(\tau)+ \int_0^{T-h}K(\tau,\theta)u^*(\theta)\,d\theta \right],v(\tau) \right)d\tau, \]

where

\[ L(\tau)=[F(T-h,\tau)C(\tau)]^*,\qquad K(\tau,\theta)=\Phi(\tau,\theta)+\Phi^*(\theta,\tau). \]

Since the variation of the functional \(\delta A(u)\) must be equal to zero for any values of the variation of the function, we have

\[ L(\tau)g+\lambda\left[ 2u^*(\tau)+2f(\tau)+ \int_0^{T-h}K(\tau,\theta)u^*(\theta)\,d\theta \right]=0. \]

The solution of the obtained integral equation can be represented in the form of the sum of two solutions

\[ u^*(\tau)=u_1^*(\tau)+u_2^*(\tau) = u_1^*(\tau)+\frac{1}{\lambda}R(\tau)g, \]

where the vector \(u_1^*(\tau)\) and the matrix \(R(\tau)\) satisfy the following integral equations:

\[ 2u_1^*(\tau)+2f(\tau)+ \int_0^{T-h}K(\tau,\theta)u_1^*(\theta)\,d\theta=0 \tag{5} \]

and

\[ 2R(\tau)+L(\tau)+ \int_0^{T-h}K(\tau,\theta)R(\theta)\,d\theta=0. \tag{6} \]

The multiplier \(\lambda\) is found from the condition that the control thus found belong to the set \(U_1\):

\[ \int_0^{T-h} \left( u_1^*(\tau)+\frac{1}{\lambda}R(\tau)g,\, u_1^*(\tau)+\frac{1}{\lambda}R(\tau)g \right)d\tau+N+ \]

\[ +2\int_0^{T-h} \left( f(\tau),u_1^*(\tau)+\frac{1}{\lambda}R(\tau)g \right)d\tau+ \int_0^{T-h} \left( \int_0^{T-h} \Phi(\tau,\theta)\left[ u_1^*(\theta)+\frac{1}{\lambda}R(\theta)g \right]d\theta,\, u_1^*(\tau)+\frac{1}{\lambda}R(\tau)g \right)d\tau=1. \]

Hence, for \(\lambda\) we obtain the algebraic equation

\[ a\frac{1}{\lambda^2}+2b\frac{1}{\lambda}+c=0. \tag{7} \]

Here.

\[ a=\int_0^{T-h} (R(\tau)g,\ R(\tau)g)\,d\tau+ \int_0^{T-h}\left(\int_0^{T-h}\Phi(\tau,\theta)R(\theta)g\,d\theta,\ R(\tau)g\right)d\tau = \]

\[ =-\frac12\int_0^{T-h}(L(\tau),\ R(\tau)g)\,d\tau, \]

\[ b=\int_0^{T-h}(u_1^*(\tau),\ R(\tau)g)\,d\tau+ \int_0^{T-h}(f(\tau),\ R(\tau)g)\,d\tau+ \]

\[ +\frac12\int_0^{T-h}\left(\int_0^{T-h}K(\tau,\theta)u_1^*(\theta)\,d\theta,\ R(\tau)g\right)d\tau=0, \]

\[ c=\int_0^{T-h}(u_1^*(\tau),\ u_1^*(\tau))\,d\tau+N+ 2\int_0^{T-h}(f(\tau),\ u_1^*(\tau))\,d\tau+ \]

\[ +\int_0^{T-h}\left(\int_0^{T-h}\Phi(\tau,\theta)u_1^*(\theta)\,d\theta,\ u_1^*(\tau)\right)d\tau-1 = N-1+\int_0^{T-h}(f(\tau),\ u_1^*(\tau))\,d\tau. \]

Of the two roots of equation (7), the final minimum of the functional \(A(u)\) is always delivered by the positive root.

Thus, condition (3) for the set of admissible controls (4) has the form

\[ \max_{\|g\|=1} \left\{ (g,\ e(T-h))+ \left(g,\ \int_0^{T-h}F(T-h,\tau)C(\tau)u^*(\tau,g)\,d\tau\right) \right\} \le 0 \tag{8} \]

or, more briefly,

\[ \max_{\|g\|=1}\Psi(g,T)\le 0. \]

Here

\[ u^*(t,g)=u_1^*(t)+\frac1\lambda R(t)g, \]

where \(u_1^*(\tau)\) and \(R(\tau)\) are solutions of the integral equations (5) and (6), and \(\lambda\) is the positive root of equation (7).

Problem 2. In the class of admissible controls, find a control \(u^0(t)\) which, for given \(x^0\), \(\varphi(t)\), and \(h\), solves problem 1 in minimum time.

The control \(u^0(t)\) will be called optimal, and the time \(T^0\) of the transition process corresponding to this control will be called the time of fastest action.

From Theorem 1 it follows:

Theorem 2. The time of fastest action \(T^0\) is the smallest number for which inequality (8) is satisfied. The optimal control \(u^0(t)\) is determined as follows:

\[ u^0(t)= \begin{cases} u^*(t,g^0), & 0\le t\le T^0-h,\\ C^{-1}(t)B(t)x(t-h), & T^0-h\le t\le T^0, \end{cases} \]

where \(g^0\) is an element at which the maximum of the function \(\Psi(g,T^0)\) is attained.

Example:

\[ \dot{x}(t)=-x(t-1)+u(t), \tag{9} \]

\[ \varphi(t)\equiv 0,\quad -1\leq t<0,\quad x(0)=10,\quad T=3, \]

\[ \int_0^3 u^2(t)\,dt\leq M. \]

To simplify the calculations, instead of the time-optimal problem we consider another problem, directly related to the first.

Find an admissible control such that the trajectory corresponding to the prescribed initial conditions and the chosen control satisfies the condition \(x(t)\equiv 0,\ 2\leq t\leq 3\), while the functional \(\int_0^3 u^2(t)\,dt\) attains its minimum.

The solution of equation (9) is:

\[ x(t)=10+\int_0^t u(\tau)\,d\tau,\quad 0\leq t\leq 1; \]

\[ x(t)=-10t+20+\int_0^{t-1}(-t+\tau+2)u(\tau)\,d\tau+\int_{t-1}^t u(\tau)\,d\tau,\quad 1\leq t\leq 2; \]

\[ \begin{aligned} x(t)=&\,5[(t-3)^2-1]+\frac12\int_0^{t-2}[(t-\tau-3)^2-1]u(\tau)\,d\tau\\ &+\int_{t-2}^{t-1}(-t+\tau+2)u(\tau)\,d\tau+\int_{t-1}^t u(\tau)\,d\tau,\quad 2\leq t\leq 3. \end{aligned} \]

The integral equation for determining \(u(\tau,g)\) has the form

\[ L(\tau)g+\lambda\left[2u(\tau)+2f(\tau)+\int_0^2 K(\tau,\theta)u(\theta)\,d\theta\right]=0, \]

where

\[ L(\tau)= \begin{cases} \tau, & 0\leq \tau\leq 1,\\ 1, & 1\leq \tau\leq 2; \end{cases} \]

\[ f(\tau)= \begin{cases} \dfrac{10}{6}\tau^3-5\tau^2+5\tau+\dfrac{20}{6}, & 0\leq \tau\leq 1,\\[6pt] 5\tau^2-20\tau+20, & 1\leq \tau\leq 2; \end{cases} \]

\[ K(\tau,\theta)=\Phi(\tau,\theta)+\Phi(\theta,\tau),\quad \Phi(\tau,\theta)= \begin{cases} \Phi_1(\tau,\theta), & 0\leq \tau\leq 1,\\ \Phi_2(\tau,\theta), & 1\leq \tau\leq 2; \end{cases} \]

\[ \Phi_1(\tau,\theta)= \begin{cases} \left(-\dfrac12\tau^2+\dfrac32\right)\theta+\dfrac16\tau^3-\dfrac12\tau+\dfrac13, & 0\leq \theta\leq \tau,\\[6pt] \dfrac16\theta^3-\left(\dfrac12\tau+1\right)\theta^2+\left(2\tau+\dfrac32\right)\theta-\tau^2-\dfrac12\tau+\dfrac13, & \tau\leq \theta\leq 1. \end{cases} \]

\[ \Phi_1(\tau,\theta)= \begin{cases} -\theta-\tau^2+\tau+2, & 1\leq \theta\leq \tau+1,\\ \theta^2-(2\tau+4)\theta+4\tau+4, & \tau+1\leq \theta\leq 2; \end{cases} \]

\[ \Phi_2(\tau,\theta)= \begin{cases} 0, & 0\leq \theta\leq \tau-1,\\ \theta-\tau+1, & \tau-1\leq \theta\leq 1,\\ 2-\tau, & 1\leq \theta\leq \tau,\\ 2-\theta, & \tau\leq \theta\leq 2. \end{cases} \]

The solution of the integral equation can be represented in the form

\[ u(\tau)=\frac{g}{\lambda}u_1(\tau)+u_2(\tau). \]

After the corresponding computations we obtain

\[ u_1(\tau)= \begin{cases} \begin{aligned} &\frac{1}{2}e^{\frac{\sqrt3}{2}\tau} \left[(\sqrt3 A_1-A_2)\sin\frac{\tau}{2} -(A_1+\sqrt3 A_2)\cos\frac{\tau}{2}\right]\\ &\quad-\frac{1}{2}e^{-\frac{\sqrt3}{2}\tau} \left[(\sqrt3 A_3+A_4)\sin\frac{\tau}{2} +(A_3-\sqrt3 A_4)\cos\frac{\tau}{2}\right]+A_5, \quad 0\leq \tau\leq 1, \end{aligned}\\[1.2em] \begin{aligned} &e^{\frac{\sqrt3}{2}(\tau-1)} \left(A_1\sin\frac{\tau-1}{2}-A_2\cos\frac{\tau-1}{2}\right)\\ &\quad+e^{-\frac{\sqrt3}{2}(\tau-1)} \left(A_3\sin\frac{\tau-1}{2}-A_4\cos\frac{\tau-1}{2}\right), \quad 1\leq \tau\leq 2; \end{aligned} \end{cases} \]

\[ u_2(\tau)= \begin{cases} \begin{aligned} &\frac{1}{2}e^{\frac{\sqrt3}{2}\tau} \left[(\sqrt3 B_1-B_2)\sin\frac{\tau}{2} -(B_1+\sqrt3 B_2)\cos\frac{\tau}{2}\right]\\ &\quad-\frac{1}{2}e^{-\frac{\sqrt3}{2}\tau} \left[(\sqrt3 B_3+B_4)\sin\frac{\tau}{2} +(B_3-\sqrt3 B_4)\cos\frac{\tau}{2}\right]+B_5, \quad 0\leq \tau\leq 1, \end{aligned}\\[1.2em] \begin{aligned} &e^{\frac{\sqrt3}{2}(\tau-1)} \left(B_1\sin\frac{\tau-1}{2}-B_2\cos\frac{\tau-1}{2}\right)\\ &\quad+e^{-\frac{\sqrt3}{2}(\tau-1)} \left(B_3\sin\frac{\tau-1}{2}-B_4\cos\frac{\tau-1}{2}\right), \quad 1\leq \tau\leq 2. \end{aligned} \end{cases} \]

Here \(A_5=B_5=0\), and the remaining coefficients \(A_i\) and \(B_i\), \(i=1,2,3,4\), satisfy the following systems of algebraic equations:

\[ \begin{aligned} &e^{\frac{\sqrt{3}}{2}}\left(\sqrt{3}\sin\frac{1}{2}-\cos\frac{1}{2}\right)A_1 +\left[2-e^{\frac{\sqrt{3}}{2}}\left(\sin\frac{1}{2}-\sqrt{3}\cos\frac{1}{2}\right)\right]A_2\\ &\quad -e^{-\frac{\sqrt{3}}{2}}\left(\sqrt{3}\sin\frac{1}{2}+\cos\frac{1}{2}\right)A_3 +\left[2-e^{-\frac{\sqrt{3}}{2}}\left(\sin\frac{1}{2}-\sqrt{3}\cos\frac{1}{2}\right)\right]A_4=0; \end{aligned} \]

\[ \begin{aligned} &\left(\frac{\sqrt{3}}{2}+2e^{\frac{\sqrt{3}}{2}}\sin\frac{1}{2}\right)A_1 +\left(\frac{1}{2}-2e^{\frac{\sqrt{3}}{2}}\cos\frac{1}{2}\right)A_2 -\left(\frac{\sqrt{3}}{2}\right.\\ &\quad\left. -2e^{-\frac{\sqrt{3}}{2}}\sin\frac{1}{2}\right)A_3 +\left(\frac{1}{2}-2e^{-\frac{\sqrt{3}}{2}}\cos\frac{1}{2}\right)A_4=-1; \end{aligned} \]

\[ \begin{aligned} &\left[1-e^{\frac{\sqrt{3}}{2}}\left(\sin\frac{1}{2}-\sqrt{3}\cos\frac{1}{2}\right)\right]A_1 -\left[\sqrt{3}-e^{\frac{\sqrt{3}}{2}}\left(\sqrt{3}\sin\frac{1}{2}\right.\right.\\ &\quad\left.\left. +\cos\frac{1}{2}\right)\right]A_2 +\left[1-e^{-\frac{\sqrt{3}}{2}}\left(\sin\frac{1}{2}+\sqrt{3}\cos\frac{1}{2}\right)\right]A_3+\\ &\quad +\left[\sqrt{3}-e^{-\frac{\sqrt{3}}{2}}\left(\sqrt{3}\sin\frac{1}{2}-\cos\frac{1}{2}\right)\right]A_4=0; \end{aligned} \]

\[ \sqrt{3}A_1+A_2-\sqrt{3}A_3+A_4=0; \]

\[ \begin{aligned} &e^{\frac{\sqrt{3}}{2}}\left(\sqrt{3}\sin\frac{1}{2}-\cos\frac{1}{2}\right)B_1 +\left[2-e^{\frac{\sqrt{3}}{2}}\left(\sin\frac{1}{2}-\sqrt{3}\cos\frac{1}{2}\right)\right]B_2\\ &\quad -e^{-\frac{\sqrt{3}}{2}}\left(\sqrt{3}\sin\frac{1}{2}+\cos\frac{1}{2}\right)B_3 +\left[2-e^{-\frac{\sqrt{3}}{2}}\left(\sin\frac{1}{2}\right.\right.\\ &\quad\left.\left. -\sqrt{3}\cos\frac{1}{2}\right)\right]B_4=-\frac{20}{3}; \end{aligned} \]

\[ \begin{aligned} &\left(\frac{\sqrt{3}}{2}+2e^{\frac{\sqrt{3}}{2}}\sin\frac{1}{2}\right)B_1 +\left(\frac{1}{2}-2e^{\frac{\sqrt{3}}{2}}\cos\frac{1}{2}\right)B_2 -\left(\frac{\sqrt{3}}{2}\right.\\ &\quad\left. -2e^{-\frac{\sqrt{3}}{2}}\sin\frac{1}{2}\right)B_3 +\left(\frac{1}{2}-2e^{-\frac{\sqrt{3}}{2}}\cos\frac{1}{2}\right)B_4=-10; \end{aligned} \]

\[ \begin{aligned} &\left[1-e^{\frac{\sqrt{3}}{2}}\left(\sin\frac{1}{2}-\sqrt{3}\cos\frac{1}{2}\right)\right]B_1 -\left[\sqrt{3}-e^{-\frac{\sqrt{3}}{2}}\left(\sqrt{3}\sin\frac{1}{2}\right.\right.\\ &\quad\left.\left. +\cos\frac{1}{2}\right)\right]B_2 +\left[1-e^{-\frac{\sqrt{3}}{2}}\left(\sin\frac{1}{2}+\sqrt{3}\cos\frac{1}{2}\right)\right]B_3+ \end{aligned} \]

\[ +\left[\sqrt{3}-e^{-\frac{\sqrt{3}}{2}}\left(\sqrt{3}\sin\frac{1}{2}-\cos\frac{1}{2}\right)\right]B_4=10; \]

\[ \sqrt{3}B_1+B_2-\sqrt{3}B_3+B_4=-\frac{10}{3}. \]

The equation for the multiplier \(\lambda\) has the form

\[ a\left(\frac{g}{\lambda}\right)^2+2b\left(\frac{g}{\lambda}\right)+c=0, \]

\[ a\simeq 0.3273;\qquad b=0;\qquad c=-71.6011-M. \]

The optimal control has the form

\[ u^0(\tau)= \begin{cases} e^{\frac{\sqrt{3}}{2}\tau}\left(2.6521\sin\frac{\tau}{2}-2.5115\cos\frac{\tau}{2}\right) -e^{-\frac{\sqrt{3}}{2}\tau}\left(2.3118\sin\frac{\tau}{2} +4.6325\cos\frac{\tau}{2}\right), & 0\leq \tau\leq 1;\\[1.2em] e^{\frac{\sqrt{3}}{2}\tau}\left(1.1407\sin\frac{\tau}{2}-1.0301\cos\frac{\tau}{2}\right) +e^{-\frac{\sqrt{3}}{2}\tau}\left(12.2625\sin\frac{\tau}{2} +1.0359\cos\frac{\tau}{2}\right), & 1\leq \tau\leq 2;\\[1.2em] e^{\frac{\sqrt{3}}{2}\tau}\left(0.4901\sin\frac{\tau}{2}-0.4219\cos\frac{\tau}{2}\right) -e^{-\frac{\sqrt{3}}{2}\tau}\left(17.2647\sin\frac{\tau}{2} -23.6179\cos\frac{\tau}{2}\right)-8.3334\tau+19.9980, & 2\leq \tau\leq 3. \end{cases} \]

Appendix. Proof of the convexity of the set \(G_1\). By the definition of the set \(G_1\), if \(e(t)\in G_1\), then for it there exists such a \(u(t)\in U_1\) that \(x(t)\equiv 0\) for \(T-h\leq t\leq T\), where

\[ x(t)=e(t)+\int_0^t F(t,\tau)C(\tau)u(\tau)\,d\tau. \]

Consider two elements of the set \(G_1\): \(e_1(t)\) and \(e_2(t)\), and the corresponding controls \(u_1(t)\) and \(u_2(t)\). We shall show that the element \(e(t)=\alpha e_1(t)+(1-\alpha)e_2(t)\), \(0\leq\alpha\leq 1\), also belongs to the set \(G_1\), and moreover the control corresponding to it and ensuring fulfillment of condition (2) has the form

\[ u(t)=\alpha u_1(t)+(1-\alpha)u_2(t). \]

Indeed,

\[ x(t)=e(t)+\int_0^t F(t,\tau)C(\tau)u(\tau)\,d\tau =\alpha\left[e_1(t)+\int_0^t F(t,\tau)C(\tau)u_1(\tau)\,d\tau\right]+ \]

\[ +(1-\alpha)\left[e_2(t)+\int_0^t F(t,\tau)C(\tau)u_2(\tau)\,d\tau\right] =\alpha x_1(t)+(1-\alpha)x_2(t). \]

But since \(x_1(t)\equiv 0\) and \(x_2(t)\equiv 0\) for \(T-h\le t\le T\), it follows that \(x(t)\equiv 0\) for \(T-h\le t\le T\). The convexity of the set \(G_1\) is proved.

In conclusion, the authors express their gratitude to F. M. Kirillova for useful discussions.

References

1. Pontryagin L. S., Boltyanskii V. G., Gamkrelidze R. V., Mishchenko E. F. The Mathematical Theory of Optimal Processes. Moscow, Fizmatgiz, 1961.

2. Myshkis A. D. Linear Differential Equations with Retarded Argument. Gostekhizdat, 1951.

3. Gabasov R., Kirillova F. M. On the solution of certain problems in optimal-control theory. Automation and Remote Control, 25, No. 7, 1964.

4. Bellman R., Glicksberg I., Gross O. Some problems in the mathematical theory of control processes. IL, 1962.

Received by the editors
November 30, 1965

Ural Polytechnic Institute
named after S. M. Kirov