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\title{Ship Handling when the Environmental Parameters Varied as the Function of the Way}
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\begin{document}

             \author[1]{Nguyen Xuan  Phuong}

             \affil[1]{  Ho Chi Minh City University of Transport}

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\date{\small \em Received: 10 February 2015 Accepted: 4 March 2015 Published: 15 March 2015}

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\begin{abstract}
        


The paper devotes the algorithm of ship handling when the environmental parameters varied as the function of the way. In nautical practice, when the ships sail in the channel, they often arrange as the convoy with the leader ship. In order to ensure the maritime safety, the mariner should establish the algorithm for control of ship engine system and steering gear complex. In this research, the author uses the maximum principle of Pontryagin L.S to establish the similar control. However, in order to obtain these ranges of numerical solutions like this, sometimes it?s difficult to use the maximum principle. Because, there is not enough the initial condition for using of the auxiliary vector that is the quantity to define the time of control variation. These obstacles shall be cleared by the selection of the transversal conditions. The problems are solved under Maier?s and G. Kelly?s condition as well as the Hamiltonian operator and Cardano?s formula.

\end{abstract}


\keywords{function of way, maier?s condition, G. kelly?s condition, hamiltonian operator.}

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\let\tabcellsep& 	 	 		 
\section[{I. Introduction}]{I. Introduction}\par
n order to control the navigation of ship in following the object in the sub-system of higher order, example in the coastal system, there should be the unique complex of the programs (or algorithm) for controlling of power system and steering complex in normal and emergency situation. Basing on these programs, the mariner evaluates the situation and the context around the ship and obtains the objective information that he can make the good decision and give the proper solution. In necessary case, it can be transferred the program control to the diesel system. These programs like this can help the mariner estimate the controlled movement of ship as the time.\par
Those algorithms should be considered as the evaluation and auxiliary. Their objective is to help obtaining the proper controls that are not compulsory to use directly on board the vessel. Also, they can be used as the initial information for maneuvering of the specified vessel to escape from the emergency and critical situation.\par
These algorithm creations are carried out by the way how the classes of limited condition are used for imposing on the control action and phase co-ordinate. Also, it may be easily extended for the limitations that applied for the speed of variable control or acceleration, where the general form of obtained algorithm is fully preserved. That property of them is to help the mariner using the given algorithms to synthesize the systems of engine complex control and ship steering gear that is for purpose of safety and economical navigation. 
\section[{II. Literature Review}]{II. Literature Review}\par
In this subject, there are researches of authors such as Krasovsky A.A.  {\ref (1999)}   {\ref 2010}). Their works are based on the classical methods of construction of automatic control systems and in particular the ship's course allows classifying the type of techniques used by the mathematical model of the vessel, processed information, methods of adaptation, design features. In some cases, the sufficient condition purely is the evaluation of the proposal algorithm that is to create the exactly controls. But, it is often required the more detail solutions that means the numerical ones.\par
In this research, the author uses the maximum principle of Pontryagin L.S to establish the similar control. However, in order to obtain these ranges of numerical solutions like this, sometimes it's difficult to use the maximum principle. Because, there is not enough the initial condition for using of the auxiliary vector that is the quantity to define the time of control variation. These obstacles shall be cleared by the selection of the transversal conditions. 
\section[{III. Method of Research}]{III. Method of Research}\par
It's assumed that the external environment is changing its characteristics as the function of the way. This happens when the parameters are considered as characteristic of the depth, width, and tortuosity of the channel[1, 2, 3 and 4], then (s)? = ? (1)\par
Equations of the ship complex in this case will be:c c c c c h v g g g g g h m g g K K dv 1 v (s), dt T T T K K d 1 h v, dt T T T dG ds K K h, v. dt dt ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? + ? ? ? ? = ? ? + + = ? =\textbf{(2)}\par
And the restricted conditions that applied on the control action hare ? ? max 0 h h (3) It's necessary to find the control law for the given complex in the dynamic condition:= h h(t)\par
Which the function can be minimized in the sailing time T:? = m m G G (v,s)\textbf{(4)}\par
Basing on the principle of maximum, it's developed and solved the problem of optimal control in the form of Mayer  {\ref [6, 7, 8 and 11]}. The problem relates to the problem of fixed right and left ends. The boundary conditions are written:= = ? = = = ? ? = ? ? = ? 0 0 0 T 0 T T T\par
at the left end t 0, v G s 0; at the right end t T, v , free quantities, s s\par
The Hamiltonian of equation ( \hyperref[formula_1]{2}) is:c c 1 c c c h h g g h g g 4 2 3 g g g K K 1 V (S) T T T K K 1 h V K K h V T T T H ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? + + ? ?+ + ? + ? ? + ? =\textbf{(5)}\par
The function for finding the vector ? will be:v g 1 1 2 4 c g h 2 c 1 2 g g 3 c g 3 4 c 1 c K d 1 , dt T T d K 1 K K h , dt T T d d K d (s) 0, . dt dt T ds ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? =? ? + ? ? ? ? ? ? ? ? ? = = ? ? ?\textbf{(6)}\par
The transversal conditions are:m 3 1 T 4 2 0 [(1 ) G H t v s] 0 + ? ? ? ? + ? ? + +? ??+ ? ? = (7)\par
Then the above problem has the 1 st order integral form:c c 1 c c c h v g g 2 g g g h g g 4 3 K K 1 v (s) T T T K K 1 h v T T T K K h v c ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? + + ? ? + + ? + + ? ? + ? =\textbf{(8)}\par
The equality \hyperref[b7]{(8)} is only relied on the contingent selection of variations m G , v, ? ? ?? when:\} 3T 1T 2T 4T\par
1, H c 0, 0, 0, 0? = ? = = ? = ? = ? = (9)\par
The structure of resulted control action is investigated:h h g 2 g g 3 g K H K K T h ? ? = ? + ?? ? (10)\par
Thence, there is the variable law of the control action as:h h g max 2 g g 3 g h h g 2 g g 3 g K h h at K K 0, T K h 0 at K K 0 T ? ? ? = ? + ?? > ? ? ? ? = ? + ?? < ? ?\textbf{(11)}\par
From the equation ( \hyperref[formula_6]{6}), it is obtained the solution of component vector: Basing on this, the control will be varied as following law:h h g max 2 g g g h h g 2 g g g K h h at K K 0, T K h 0 at K K 0 T ? ? = ? ? ? > = ? ? ? < ? ? ? ? ?\textbf{(12)}\par
Now, it is going to integrate the equations of the problem. From the equation: Therefore the equations of 4 ? (equation 6) can be rewritten as following:T 0 4 c 1 c d vdt d K 1 , dt T v dt ? ? ? ? ? ? ? ? ? ? ? = ? ? Or 4 c 1 c d K 1 d (t) dt T v dt ? ? ? = ?\textbf{(13)}\par
So the given task can be converted to the problem of variable external conditions that change as the function of time  {\ref [14, 15, 16 and 19]}.\par
On the basis of the equation \hyperref[b12]{(13)}, it should be integrated the following differential equations with the control action h = h max on the interval time 0÷t*, it means the time where:h h g 2 g g 3 g c c c c c h v g g max g g g h m g g max v g 1 1 2 4 c g h 2 c 1 2 g g max 3 c g 3 4 c 1 c K K K : T K K dv 1 v (v,t), dt T T T K K d 1 h v, dt T T T dG ds K K h , v, dt dt K d 1 , dt T T d K 1 K K h , dt T T d d K 1 d (t) 0, . dt dt T v dt ? ? ? ? ? ? ? ? ? ? + ?? ? ? ? = ? + ? ? ? ? ? ? ? = ? ? + + ? ? = ? = ? ? = ? ? ? ? ? ? =? ? + ? ? ? ? ? ? = = ? ? ? ? ? ? ? ? ? ? ? ?\textbf{(14)}\par
And, it's continuously integrated that equation in range of t*÷T until:h h g 2 g g 3 g K K K 0 T ? ? + ?? < .\par
When the actions control h = 0:c c c c c v g g g v g 1 1 2 4 c g 2 c 3 1 2 c g 4 c 1 c K K dv 1 v (v,t), dt T T T K d 1 ds v, v, dt T T dt K d 1 , dt T T d K d 1 , 0, dt T T dt d K 1 d (t) . dt T v dt ? ? ? ? ? = ? + ? ? ? ? ? ? ? = ? ? + = ? ? ? ? = ? ? ? ? ? ? ? ? ? ? =? ? + ? = ? ? ? ? ? = ? ? ?\textbf{(15)}\par
The initial conditions about the solution of the equations (15) will be the values ofthe phase coordinate calculated by the solution of the equation ( \hyperref[formula_15]{14}) at time t*.\par
The solutions of the equations ( \hyperref[formula_15]{14}) and ( \hyperref[formula_17]{15})can be numerical. In order to consider the structure of the control action and the switching functions ( \hyperref[formula_12]{11}) and ( \hyperref[formula_13]{12}), it should be rewritten the equations ( \hyperref[formula_6]{6}): ? ? = ? ? ? ? ? ? ? ? ? =? ? + ? ? ? ? ? ? ? ? = = ? ? ?\textbf{(16)}\par
Where:v g c 11 12 21 14 c g c h c 22 23 g g 41 g c K K 1 a , a , a , a 1, T T T K d (s) 1 a , a K K h, a . T T ds ? ? ? ? = = = = ? ? ? ? ? ? = = = ? ?\textbf{(17)}\par
The typical determinant of the given equations is:  p q p q p q p 0? + ? =\textbf{(18)}\par
That typical equation is invariable for the action control, because of: q a a ,q a a a a a a , q a a a\} 1 22= + = + ? = (19)\par
Now, it is continuously found the solutions of equation ( \hyperref[formula_18]{16}) in the form: 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 p t p t p t (1) (2) (3) 1 1 2 3 1 1 1 p t (4) 4 1 p t p t p t (1)\textbf{(2) (3)}c l c l c l c l , c l c l c l c l , c l c l c l c l , c l c l c l c ? = ? + ? + ? + + ? ? = ? + ? + ? + + ? ? = ? + ? + ? + + ? ? = ? + ? + ? + + ? 4 p t (4) l . ? ? ? ? ? ? ? ? ? ? ? (20)\par
Wherein k i ? is constant factor that specified for each solution k of the typical equation \hyperref[b17]{(18)} in the following systems: 1. P 1 = 0 (1)\textbf{(1)}? ? ? ? ? ? = ? ? ? + ? ? ? = ? = ?\textbf{(21)}\par
Basing on ( \hyperref[formula_26]{21}), it's obtained (1) 1 0 ? =and (2) 2 1 ? = ,? ? ? ? ? ? ? = ? ? ? + ? ? ? ? = ? ? ? ? = ? ? ? = ?\textbf{(22)}\par
Basing on the equation ( \hyperref[formula_28]{22}), it's obtained:\par
(\par
( ? ? ? = ? = ? ? ? ? ? ? = ? = ? ?\textbf{(23)}\par
3. P = P 3(3) (3) (3) 11 3 1 12 2 14 4 (3) (3) (2) 21 1 22 3 2 23 3 (3) (3) (3) 3 3 41 1 3 4 (a p ) a a 0, a (a p ) a 0, p 0 , a p 0. ? ? ? ? ? ? ? = ? ? ? + ? ? ? ? = ? ? ? ? = ? ? ? = ?\textbf{(24)}\par
From that, there'll be:\par
( ? ? ? ? ? ? ? = ? ? ? + ? ? ? ? = ? ? ? ? = ? ? ? = ?\textbf{(26)}\par
Basing on equation ( \hyperref[formula_32]{26}), it's obtained: ? ? ? = ? = ? ? ? ? ? = ? = ?\textbf{(27)}\par
Analyzing of solutions of typical equation ( \hyperref[formula_20]{18}), there is:3 2 1 2 3\par
p(p q p q p q ) 0 ? + ? = Substituting the below -mentioned into the given equation: \par
Where:2 3 1 2 1 1 2 3 q 9q q 2q b q , c q 3 27 ? =? = ?\par
The equation ( 28) will be solved by Cardano formula: = ? + + + + ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?\textbf{(29)}\par
Understanding that each of three roots:2 3 c c b 2 2 3 ? ? ? ? ? = ? + + ? ? ? ? ? ? ? ?\par
It should be chosen one value of solution:2 3 c c b 2 2 3 ? ? ? ? ? = ? ? + ? ? ? ? ? ? ? ?\par
In which, the following condition should be carried out: b 3 ?? = ? On the basis of that condition, it can be written the root of the typical equation as following:1 2 1 1 1 3 1 1 1 1 1 4 1 1 1 1 q p ( ) , 3 q 1 3 p ( ) i ( ) , 2 2 3 q 1 3 p ( ) i ( ) 2 2 3 ? = ? + ? ? ? ? ? =? ? + ? + ? ? ? ? ? ? ? =? ? + ? ? ? ? ? ? ? ?\textbf{(30)}\par
It is clarified a matter of the obtained structure of control.h g h 2 g g 3 g K H K K h T ? ? = ? + ?? ?\par
From there, there will be:h g h max 2 g g 3 g h g h 2 g g 3 g K h h khi K K 0, T K h 0 khi K K 0. T ? ? ? = ? + ?? > ? ? ? ? = ? + ?? < ? ?\textbf{(31)}\par
On the basis of symbol \hyperref[b26]{(27)}, roots of equations (  {\ref 20}), ( \hyperref[formula_26]{21}), ( \hyperref[formula_28]{22}), ( \hyperref[formula_30]{23}), ( \hyperref[formula_31]{24}), (  {\ref 25}), \hyperref[b25]{(26)}, \hyperref[b26]{(27)} and the expression solutions of the typical equation, it can be affirmed that: 2 2 3 3 (2) (2) 4 4 2 2 (2) (2) (3) (3) 4 4 2 2 (3) (3) (4)\textbf{(4}? ? ? = = ? ? ? ? ? = ? = ? ? ? ? ?? ? = ? ? = ? ? ? ? ?? ? = ? ? = ? ? ? ? ? ? = ? ? ?\textbf{(32)}\par
In the function (32), A is vector of parameters of equation \hyperref[b15]{(16)}. The given expressions will be right for the constant case of value A and d / ds ? (method of freezing factor)  {\ref [16, 17, 18 and 21]}.\par
It is necessary to find On the basis of (20÷27) and (32),it can be written:\par
( 1 c T ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ?? ? ? ? ? ? ? ? = + ? + ? ? + ? + ? ? + ? ? = ? ? ? ? ? ? ?\textbf{(33)}\par
The integral constants of the given case are defined in correspondence with the obtained edge conditions \hyperref[b8]{(9)}.\par
The above problem of the optimal control when the external environment is function of way is required for the practical implementation of the algorithm found by measuring the magnitude d ds ? and hence the value (s) ? = ? . As a rule, this information is available especially on canals and rivers. The main difficulty is to find the ways of formalizing this information. Such methods must be simple in structure, and at the same time provide a minimum amount of information loss in finding the controls. For these purposes, it may be proposed a method described in previously.\par
It's analyzed the possibility of existing of special control in the systems (2) and ( \hyperref[formula_6]{6}) and transformed the Hamiltonian (5):h g h 2 g g 3 g c c 1 1 1 c c c v g 4 2 g K K H K K K K h T 1 v (s) T T T v T ? ? ? ? = ? ? ? ? ? ? ? + ? + ?? ? ? ? ? ? ? ? ? ? ? + ?? ? ? ? + + ? + ?\textbf{(34)}\par
It's marked:c 0 1 1 c c v g c 4 1 2 c g h g h g g 1 2 3 g K K K K K 1 H v T T (s) v , T T H K . T ? ? ? ? ? ? ? ? ? ? ? ? ? ? =? ? + ?? ? ? ? ? + ? + ? = ? + ?? (35)\par
In correspondence with the result in \hyperref[b4]{[5,}\hyperref[b19]{20]}, it is found:h g h 2 1 g g 3 g d d d H K . dt T dt dt K K ? ? ? = + ? (36)\par
The given expression is obtained from the condition 3 const. 
\section[{? =}]{? =}\par
The special control may be existed in (35) only if the derivative H 1 is even order, so it can be found:h 2 2 2 g h 2 1 g g 3 2 2 2 g d d d H K 0. T dt dt dt K K ? ? ? = + ? = (37)\par
In which:2 h 2 c 1 2 g g 3 2 c g h v 2 g g 2 g g g d d d 1 dh K , T dt T dt dt dt d 1 d dh v. T dt T dt T dt K K K K ? ? ? ? ? = ? + ? ? ? ? = ? + + ? ? ? ? ? ? ?\textbf{(38)}\par
Substituting (38), respectively, ( \hyperref[formula_1]{2}) and ( \hyperref[formula_6]{6}), it's obtained:v 2 g 2 c 1 2 4 2 c c g h c 1 2 g g 3 g c g h g g 3 h v 2 g g 2 g g g g h v g g g g d 1 T T T dt 1 1 K h T T T dh K , dt d 1 1 h v T T T T dt dh v. T dt T K K K K K K K K K ? ? ? ? ? = ? ? ? ? ? ? + + ? + ? ? ? ? ? ? ? = ? ? ? + + + + + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (39)\par
On the basis of (39), it can be rewritten (37) in the developed form, as following:2 H H H 1 1 1 2 2 3 4 2 H H H H 4 1 2 3 d H b b b dt b h c v c c h 0. = ? + ? + ? ? ? + + ? ? =\textbf{(40)}\par
In which:h h h c g c g g H H 1 2 2 2 3 g c g c g v h h g c g c g H 3 2 g c g c h g H h 4 g g 3 2 g h v g g H h h g g g g 1 3 3 2 g g h g H h H h g g g g 2 3 3 3 2 2 g g K K K K K K K K K K K K K K K K K K K b , b T T T T T , b , T T T T b K , T c K K , T T 1 c K , c K . T T ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? = + + = = ? = ? ? + ? = ? = ? (41)\par
From the expression (40), it is found the special control: J H H H H H 1 1 2 2 3 4 1 2 H H 4 3 b b b c v c h b c ? + ? + ? + + = +\textbf{(42)}\par
It's rechecked the optimum of the special control (42) under the G. Kelly's condition \hyperref[b4]{[5,}\hyperref[b19]{20]} as following:2 H H 1 4 3 2 d H (b c ) h dt ? = ? + ?\par
In correspondence with the conditions of transversal action \hyperref[b6]{(7)}, \hyperref[b8]{(9)}, and equation ( \hyperref[formula_6]{6}), as well as the signs inserted into (41), it's obtained:3 1 ? =? and H H 4 3 b 0, c 0 < < , therefore: 2 1 2 d H 0 h dt ? > ?\textbf{(43)}\par
The G. Kelly's condition is satisfied and the special controls are optimal. Now, it's re-examined the answer of the given problem with the less dimension of the model of the mobile system. This less dimension is carried out by excluding of the diesel equation from the equations (32). The problem setting is done as same as  {\ref [9, 10, 12 and 13]}, the differential equations are following:c c c c c 2 m g K K dv 1 v (s), dt T T T dG K , dt ds v. dt ? ? ? ? = ? + ? ? ? ? ? ? = ? ? ? ? = ? ?\textbf{(44)}\par
The limitation that is necessarily imposed for the control action (Frequency of diesel rotation) will be: max 0 ? ? ? ? It's necessarily found the control law in dynamics (t) ? = ? to ensure that at the interval T, the given movement time is reached to minimum for the function:m m G G (v,s) ? =\textbf{(45)}\par
The problem is defined under Maier's condition and solved by the maximum principle  {\ref [18, 22 and 23]}. The edge condition is rewritten as following: At the left end:0 0 0 T 0 t 0, v G s 0 = =? = = =\par
At the rights end:T T t T, v , = ? at free T s s =\par
The Hamiltonian of the equations ( \hyperref[formula_58]{44}) is:c c 1 c c c 2 g 2 3 1 H v (s) T T T v . K K K ? ? ? = ? + ? ? ? ? + + ? ? + ? ? ? ? ? ? ? ? ? ? ?\textbf{(46)}\par
The transversal conditions are:T 2 m 1 3 0 [(1 ) G c t v s ] 0 + ? ? ? ? + ? ? + ? ? =\textbf{(47)}\par
The considered problem is 1 st order integral:c c 1 c c c 2 g 2 3 K K 1 v (s) T T T K v K 0. ? ? ? ? ? ? + ? ? ? ? + ? ? ? ? + ? ? + ? = =\par
The equality (47) can be only existed at the arbitrary selection of variation ofm v G , ? ? ,when 1T 2T 3T 0, 1, 0 ? = ? = ? ? =.\par
The structure of control is obtained as following:c 1 g 2 c K H 2K 0 T ? ? ? = ? + ?? = ?? (48) c max 1 g 2 c c 1 g 2 c K when 2K 0, T K 0 when 2K 0. T ? ? ? ? ? ? = ? ? + ?? > ? ? ? ? ? = ? + ?? < ? ?\textbf{(49)}\par
The equations for finding the vector ? are:1 1 3 c 2 3 c 1 c d 1 , dt T d 0, dt d K d (s) dt T ds ? ? ? = ? ? ? ? ? ? ? = ? ? ? ? ? = ? ? ?\textbf{(50)}\par
Understanding that 2 c c ? ? = , in correspondence with the transversal condition, there's 2 c 1 ? = ? .The 2 nd equation doesn't relate to the remained tasks, thence it's found the solution of equation (50). Excluding 3 ? from equation (50), it's obtained:2 1 1 1 2 1 2 d d a a 0 dt dt ? ? ? ? ? + ? =\textbf{(51)}\par
In which:c 1 2 c c K 1 d (s) a , a . T T ds ? ? ? ? = =\par
The typical equation will be: For the ship complex, the below-expression is always right  {\ref [24, 25 and 26]}:( ) 2 1 2 a a ? ? 
\section[{<<}]{<<}\par
Therefore the solutions of the equation (52) will be synchronization with the real positive part. On that basis, the quantity 1 1 (t) ? =? will be changed the sign for one more time. In order to find the analytic expression for the commutative function (49): \par
It's defined 1 c ? and 2 c ? from the transversal condition 1T 0 ? = and from the 1 st order integral of the problem. The 1 st order integral at all the control interval 0 ÷ T when t = T is defined that is equal 0.Applying the edge condition at the left end and max ? = ? , it can be written the integral as following:2 c c 10 10 max g max 20 c c K K (s) K 0 T T ? ? ? ? ? ? + ? ? + ? ? = (54)\par
Or with the condition at the interval 0 ÷ T, ? 2 = constant = -1, it's obtained:2 c c 10 max g max c c K K (s) K T T ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? (55) 2 c g max 10 max c T K K K (s) ? ? ? ? ? = ? ? ?\textbf{(56)}\par
At time t = 0, there is the algebraic equation as: ? = ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ?\textbf{(58)}\par
The given problem has the analytic solution. In order to analyze the particular navigational condition, it should be known the function (s) ? = ? . Now, it will be examined the appearance possibility of the special control in the set of equation (44), it is shown the Hamiltonian as following  {\ref [27, 28 and 29]}:c 1 1 3 c c c 1 g 2 c K 1 H v (s) v T T K K T ? ? ? ? ? = ? ? ? ? ? + ? + ? ? ? ? ? ? +? ? + ?? ? ? ? ?\textbf{(59)}\par
It's marked:c 0 1 1 3 c c c 1 1 g 2 c K 1 H v (s) v , T T K H K . T ? ? ? ? =? ? ? ? ? + ? ? ? ? ? = ? + ?? ? ?\textbf{(60)}\par
Because of 2 1 ? = ? at the interval 0 T ÷ , so there is:c 1 1 g c K H K . T ? ? = ? ? ?\textbf{(61)}\par
It's found:c 1 1 g c K d d d H K dt T dt dt ? ? ? ? = ? (62) And 2 2 2 c 1 1 g 2 2 2 c K d d d H K 0 T dt dt dt ? ? ? ? = ? =\textbf{(63)}\par
Applying the equation ( 53), it's found: From the equation (63), there is:1 2 2 2 2 p t p t c 1 1 c 2 2 2 c g c g K c p K c p d e e dt T K T K ? ? ? ? ? ? ? = +\textbf{(64)}\par
It's integrated respectively the equations (64), it's obtained the special controls:    
\section[{IV. Discussion}]{IV. Discussion}\par
The above problem of the optimal control when the external environment is function of way is required for the practical implementation of the algorithm found by measuring the magnitude d ds ? and hence the value (s) ? = ? . As a rule, this information is available especially on canals and rivers. The main difficulty is to find the ways of formalizing this information. Such methods must be simple in structure, and at the same time provide a minimum amount of information loss in finding the controls. For these purposes, it may be proposed a method described in previously. 
\section[{V. Conclusion}]{V. Conclusion}\par
The research is obtained the results:\par
It's proposed the establishing method of extremum principle control on the basis of the selection of the transversal condition that helps us to obtain not only the quality solutions but also the quantitative solution.\par
It's obtained the control algorithms of engine system that allow the following vessel approaching to the leader ship.\par
It's researched the programs of control for the steering complex that ensures the meeting movement of ships.\par
It's obtained the programs of control for engine system and steering complex that is solved the problem of head-on navigation in the confined water.\par
It's established the programs of control for engine system when the parameters of external environment are varied as function of time, way, and the parameters of ship's sailing are nonlinear variation.\begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-2.png}
\caption{\label{fig_0}}\end{figure}
        \begin{figure}[htbp]
\noindent\textbf{} \par 
\begin{longtable}{P{0.5274096385542169\textwidth}P{0.0051204819277108436\textwidth}P{0.010240963855421687\textwidth}P{0.0051204819277108436\textwidth}P{0.010240963855421687\textwidth}P{0.0051204819277108436\textwidth}P{0.05632530120481928\textwidth}P{0.15873493975903613\textwidth}P{0.010240963855421687\textwidth}P{0.0051204819277108436\textwidth}P{0.010240963855421687\textwidth}P{0.04096385542168675\textwidth}P{0.0051204819277108436\textwidth}}
(1) 4 ? = ?\tabcellsep a\tabcellsep 12\tabcellsep a\tabcellsep 14\tabcellsep ,\tabcellsep \multicolumn{2}{l}{(1) 3 ? = a}\tabcellsep 22\tabcellsep a\tabcellsep \multicolumn{2}{l}{23}\tabcellsep .\\
2. P = P 2\tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \\
\multicolumn{7}{l}{(2) 2 1 p ) (2) 21 1 11 (a 22 a (a (2) 2 3 p 0 , a a p ) 12 2 (2) (2) 2 2 (2) 41 1}\tabcellsep \multicolumn{4}{l}{(2) 14 4 (2) a a 23 3 (2) 2 4 p}\tabcellsep 0, 0, 0.\end{longtable} \par
 
\caption{\label{tab_4}}\end{figure}
 			\footnote{© 20 15 Global Journals Inc. (US) Global Journal of Researches in Engineering ( ) Volume XV Issue VI Version I} 		 		\backmatter  			  				\begin{bibitemlist}{1}
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