ANALYTIC SOLUTION OF TRANSCENDENTAL EQUATIONS

Transcription

1 Int J Appl Math Comput Sci 00 ol 0 No DOI: 0478/v ANALYTIC SOLUTION OF TRANSCENDENTAL EQUATIONS HENRYK GÓRECKI Faculty of Informatics Higher School of Informatics ul Rzgowska 7a Łódź Poland head@novaiaaghedupl A decomposition technique of the solution of an n-th order linear differential equation into a set of solutions of -nd order linear differential equations is presented Keywords: transcendental equations zeros etrema Introduction Let us consider the differential equation determining the transient error in a linear control system of the n-th order with lumped and constant parameters a i i =n: d n (t dt n + a d n (t d(t dt n + + a n + a n (t =0 dt ( with the initial conditions which in general are different from zero: (i (0 = c i 0 for i = n The solution of Eqn ( takes the following form: (t = A k e skt ( s k are the simple roots of the characteristic equation s n + a s n + + a n s + a n =0 ( In order to obtain an eplicit form for A k we need higher derivatives of (t: d p (t dt p = s p k A ke skt p = n (4 The formulae ( and (4 represent a system of n linear equations with respect to unknown terms A k e s kt Itsmatri of coefficients is the andermonde matri s s s n (5 s n s n s n n According to the assumption that s i s j for i j the matri (5 has an inverse and the system ( and (4 can be solved We denote by the andermonde determinant of the matri (5 and by k the andermonde determinant of order (n of the variables s s k s k+ s n We also denote by Φ (k r the fundamental symmetric function of the r-th order of (n variables s s k s k+ s n for r =0 n : Φ (k 0 = Φ (k = s + s + + s k + s k+ + s n = a s k Φ (k = s s + s s + + s s k + s s k+ + s s + + s s k + s s k+ + s s n + = a s s k s s k s n s k Φ (k n = n i=i k s i =( n a n s k (6 It can be shown that the elements of the matri inverse to the matri (5 have the form α ik = ( i+k Φ (i n k k (7

2 67 H Górecki or The solution of the system ( and (4 is as follows: A k e s kt = = α kj (j (t ( k+j A k e s kt = ( k k ( j Φ (k Φ (k n j k (j (t n j (j (t (8 k = n (9 For t =0 we know (j (0 = c j and the substitution t =0into (9 gives A k = ( k k ( j Φ (k n j c j (0 Using the relation (6 we can formulate the following theorem Theorem The eplicit form of the coefficient A is as follows: A = ( n ( c n s j c n n + + ij i s j s i c n (s n s (s n s (s s ( n s i s j s k c n + ijk (s n s (s n s (s s +( n n i s i c (s n s (s n s (s s ( Then the coefficients A A A n can be obtained by the sequential change of the indices of s i according to the scheme s s s s n s n s Eample The solution of the -rd order equation is as follows: The coefficient (t =A e st + A e st + A e st A = ( ( j Φ ( j c j = = s s s s s s s s = s s =(s s (s s (s s Φ ( 0 = Φ ( = s + s Φ ( = s s ( (s s A = (s s (s s (s s ] ( Φ ( c +Φ ( c Φ ( 0 c = ( ] s s c +(s + s c c (s s (s s = c (s + s c + s s c (s s (s s A = c (s + s c + s s c (s s (s s A = c (s + s c + s s c (s s (s s After the substitution of (0 into (9 we obtain e s kt ( k k ( j Φ (k = ( k k n j (j (0 ( j Φ (k and finally for k = nwehave e s kt ( j Φ (k n j c j = ( j Φ (k n j (j (t n j (j (t ( Premultiplying both sides of ( and using iete s relations between the roots s i and the coefficient a of the characteristic equation s k = a ( we obtain the main result formulated as Theorem Theorem (Górecki and Turowicz 968 The relation between cofficients a i i = n the initial values c j j = nand solutions (j (t is as follows: e at n ( j Φ (k = n n j c j ( j Φ (k n j (j (t (4

3 Analytic solution of transcendental equations Both the sides of Eqn ( are composed of symmetric polynomials of variables s s n Accordingly it is possible to epress these terms as polynomials of the coefficiets a a n Using ieta s relations it is possible to replace the roots s k by the coefficients a i and to avoid calculating the roots by solving algebraic equations Eample For n =wehave e at {a c +a a c c +(a a + a c c +(a a a c +(a a +a c c c + a a c c } +a c c +(a + a c c +a c c + c ] = a (t +a a ( (t (t ] +(a a + a ( (t ] (t+(a a a ( (t ] +(a a +a (t ( (t ( (t ] + a a (t ( (t+a (t ( (t ] +(a + a ( (t ] ( (t+a ( (t ( (t ] + ( (t ] (5 a = (s + s + s a = s s + s s + s s a = s s s Eample For n =4 similarly e at a 4c 4 +a a 4c c +a a 4c c + a a 4c c 4 +(a a a 4 a 4 c c +(4a a +a a 4 a 4 c c c +(a a +a 4 a 4 c c c 4 + ( a + a a +a 4 a4 c c + ( a a +a a4 c c c 4 + a a 4 c c 4 +(a +a a a 4 a a 4 c c +(a a + a a 4 +a a a 4 a 4 c c c +(a a + a a a 4 +5a a 4 c c c 4 +(a a + a a +5a a a 4 a a 4 c c c +(a a a +a a 4 +a +4a a 4 c c c c 4 +(a a +a a 4 c c c 4 + ( a a a + a a 4 a +a a 4 c c + ( a a + a a +5a a 4 c c c 4 + ( a a +a 4 c c c 4 + a c c 4 +(a a a a a 4 + a 4c 4 +(a a + a a a a a 4 a a 4 c c +(a a a a a 4 + a a a 4 c c 4 +a (a +a a a 4 c c +(a a + a a +5a a a a 4 c c c 4 +(a + a a +a 4 c c 4 +(a a + a a a a a a 4 c c +(a a + a +a a a 4 c c c 4 +(4a a +a c c c 4 +a c c 4 ( a a a a + a 4 c 4 + ( a +a a a c c 4 + ( ] a + a c c 4 +a c c 4 + c4 4 = a 4 (t4 +a a 4 (t ( (t +a a 4 (t ( (t+a a 4 (t ( (t + ( a + a a 4 a4 (t ( (t ] + ( 4a a +a a 4 a4 (t ( (t ( (t + ( a a +a 4 a4 (t ( (t ( (t + ( a + a a +a 4 a4 (t ( (t ] + ( a a +a a4 (t ( (t ( (t +a a 4 (t ( (t ] + ( a +a a a 4 a a 4 (t ( (t ] + ( a a + a a 4 +a a a 4 a4 (t ( (t ] ( (t+ ( a a + a a a 4 +5a a 4 (t ( (t ] ( (t + ( a a + a a +5a a a 4 a a 4 (t ( (t ( (t ] + ( a a a +a a 4 +a +4a a 4 (t ( (t ( (t ( (t + ( a a +a a 4 (t ( (t ( (t ] + ( a a a + a a 4 a +a a 4 (t ( (t ] + ( a a + a a +5a a 4 (t ( (t ] ( (t + ( a a +a 4 (t ( (t ( (t ] 67 +a (t ( (t ] + ( a a a a a 4 + a 4 ( (t ] 4 ( a a + a a a a a 4 a a 4 ( (t ] ( (t + ( a a a a a 4 + a +a a 4 ( (t ] ( (t ( +a a +a a a 4 ( (t ] ( (t ] + + ( a a + a a +5a a a a 4 ( (t ] ( (t ( (t + ( a + a a +a 4 ( (t ] ( (t ] + ( a a + a a a a a a 4 ( (t ( (t ] ( a a + a +a a a 4 ( (t ( (t ] ( (t

4 674 H Górecki + ( 4a a +a ( (t ( (t ( (t ] +a c ( (t ] + ( a a a a + a 4 ( (t ] 4 + ( a +a a a ( (t ] ( (t + ( a + a ( (t ] ( (t ] +a ( (t ( (t ] + ( (t ] 4 (6 Analytical method of determining zeroes and etremal values of the solution (t described by the relation ( Basic results The general relation analogous to the formulae (5 or (6 for the equation of the n-th order is very complicated For that reason we illustrate the method on eamples of equations of the -rd and 4-th orders We assume that at the etremal point t e or at the zero t 0 of the solution ( the second derivative d /dt 0 We can write the relation (5 in the following form: ] { ( ] a ( + (a ( ] a + a a ( ( ( + (a a + a ( ] +(a a ( +a ( + a ( ( ( ( ( + (a a a +(a ( ( + a ( ( ]} +a + ( { ( = e at c a c ( c ( c + a a + a a c c c ( + (a a + a c +(a a +a c ] c + a c c c ( c ( + (a a a +(a c c + a c +a c c + ] } (7 Setting = c = u ( c (8 ( = c = v ( c (9 we can write the relations (7 in the following form: { ] ( e a t c} { a u +(a a v + a a u + (a a + a v + a ] u + (a a a v +(a + a v +a v + ]} =0 (0 If we assume that c = 0 then from (9 we have ( (t e =0and v =0 It is a necessary condition for etremum In this case the equation (0 has a simple form: { ( ] } e a t e a c u + a a u + a u + ] =0 ( If we assume c =0 then from (8 we obtain that (t 0 =0and u =0 It is a necessary condition for (t to be zero In this case Eqn (0 has the following form: { ( ] e a t 0 c} a a a v +(a + a v +a v + ] =0 ( It is possible to find the relations between the roots of the equation a u + a a u + a u +=0 ( and the roots s s and s of the characteristic equation s + a s + a s + a =0 (4 Setting u = y (5 a in Eqn ( we obtain the following: y + a y + a y +=0 (6 a a Similarly setting s = a z (7 in Eqn (4 we obtain that z + a z + a z +=0 (8 a a Equations (6 and (8 are identical As a result we have that y = z or a u = s (9 a Finally from (9 we conclude that u = s a (0 Returning to (9 we find that if ( = c =0at the etremum point t e the following relations hold: (t e ( (t e = c = s i i = ( c a Taking into account in ( that a = s s s we finally obtain that (t e ( (t e = c = or c s s (t e ( (t e = c = or ( c s s (t e ( (t e = c = c s s

5 Analytic solution of transcendental equations Theorem From the relations ( it is possible to determine etrema (if they eist using the relations s s (t e + ( (t e =0 s s (t e + ( (t e =0 ( s s (t e + ( (t e =0 under the constraints that c and c fullfil the same relations Following a similar procedure with the equation (a a a v +(a + a v +a v +=0 (4 we can find that in the article by Górecki and Szymkat (98 it is proved that the roots of the equation 8r +8a r +(a + a r + a a a =0 (5 are as follows: r = s + s r = s + s r = s + s Setting r = p in (5 we obtain the equation p +a p +(a + a p + a a a =0 (6 whose roots are p = s + s p = s + s p = s + s Let p = q (7 Then Eqn (6 has the following form: (a a a q +(a + a q +a q +=0 (8 and its roots are q = s + s q = s + s q = Finally from (9 and (9 we obtain that ( (t 0 ( (t 0 = c = c s + s ( (t 0 ( (t 0 = c = c s + s ( (t 0 ( (t 0 = c = c s + s s + s (9 (40 Theorem 4 From the relation (40 it is possible to determine the zeros of (t 0 (if they eist using the relations ( (t 0 (s + s ( (t 0 =0 ( (t 0 (s + s ( (t 0 =0 (4 ( (t 0 (s + s ( (t 0 =0 under the constraints that c and c fullfil the same relations 675 A generalization of these result relations ( and (4 to higher order equations may be obtained directly due to the following remark Remark The relations ( and (4 may be obtained directly from the following propositions Let the coefficients A i of the solution (t fullfil the relations A =0 A 0 A 0 A =0 A 0 A 0 (4 A =0 A 0 A 0 In this way we obtain equations which contain only two eponential terms and such equations can be solved in analytical form The relations (4 are more general than ( and (4 because they are also valid when c 0or c 0 Moreover they also hold for higher order equations For such equations to obtain only two eponential terms it is necessary to assume more than one coefficient A i equal to zero Basic result Theorem 5 The equation or (t = ( (t = A i e sit (4 i= s i A i e sit (44 i= can be decomposed into a system of equations containing a set of equations composed of only two terms The set contains ( n = n(n n equations with two eponential terms Eample 4 For n = we have the following equations: (t =A e st + A e st + A e st (45 ( (t =A s e st + A s e st + A s e st (46 A = c (s + s c + s s c (47 (s s (s s A = c (s + s c + s s c (48 (s s (s s A = c (s + s c + s s c (49 (s s (s s Looking for an etremum we use Eqn (46 the necessary condition is ( (t =0 Assuming that ( (t =0 A =0 (50

6 676 H Górecki after eliminating the initial condition c from (50 we obtain that e (s ste = c s c c s c (5 e (s ste = c s c c s c (5 e (s ste = c s c (5 c s c c = (s + s c c ] s s Similarly asumming A =0 we obtain after eliminating c that e (s ste = c s c c s c e (s ste = c s c c s c s s (54 s s (55 e (s ste = c s c s c s c (56 s Finally asumming A =0 after eliminating c we obtain e (s ste = c s c s (57 c s c s e (s ste = c s c c s c s s (58 e (s ste = c s c s (59 c s c s Similarly the equation (t =A e st + A e st + A e st + A 4 e s4t =0 can be decomposed into the following set of equations: A e st + A e st =0 A =0 and A 4 =0 A e st + A e st =0 A =0 and A 4 =0 A e st + A 4 e s4t =0 A =0 and A =0 A e st + A e st =0 A =0 and A 4 =0 A e st + A 4 e s4t =0 A =0 and A =0 A e st + A 4 e s4t =0 A =0 and A =0 It is a set of ( 4 = 4 =6 equations with only two eponential terms Remark It is evident that looking for (t 0 =0instead of ( (t e =0 we must multiply the relations (5 (59 a propriately by s i /s j For eample andsoon e (sj sit0 = c s j c c s i c s i s j Remark If Eqn ( has repeated roots then the relations ( and ( must be transformed by properly passing to the limit In the particular case when s = s = = s n = s we obtain n (t =e st A k t k k (k (0( i s i A k = k = n i!(k i! i=0 The necessary condition for the eistence of the local etremum of the solution ( is ( (t =0andtheproblem is reduced to an algebraic one 4 Conclusion A k st (k e +(k t k ] e =0 It was shown that every differential equation of the n-th order can be decomposed into a set of n(n equations of the -nd order which can be solved in analytical form References Górecki H (004 A new method for analytic determination of etremum of the transients in linear systems Control and Cybernetics (: Górecki H and Szymkat M (98 Application of an elimination method to the study of the geometry of zeros of real polynomials International Journal of Control 8(: 6 Górecki H and Turowicz A (968 Determination of the dependence of the maimal deviation of the controlled quantity and the time of deviation on the parameters in linear control systems Avtomatika i Telemekhanika (6: 79 8 Górecki H (009 A new method for analytic determination of etremum of the transients in linear systems Bulletin of the Polish Academy of Sciences: Technical Sciences 57(: 5 55 Henryk Górecki was born in Zakopane in 97 He received the MSc and PhD degrees in technical sciences from the AGH University of Science and Technology in Cracow in 950 and 956 respectively Since the beginning of his academic activity he has been attached to the Faculty of Electrical Engineering Automatics and Electronics of the AGH University of Science and Technology In 97 he became a full professor and up to 997 he was the director of the Institute of Automatics He has lectured etensively in automatics control theory optimization and technical cybernetics He is a pioneer of automatics in Poland as the author of the first book on this topic in the country published in 958 For many years he was the head of doctoral studies and the supervisor of 78 PhD students He is the author or co-author of 0 books and among them a monograph on control systems with delays in 97 and about 00 scientific articles in international journals His current research interests include optimal control of systems with time

7 Analytic solution of transcendental equations 677 delay distributed parameter systems and multicriteria optimization Professor Górecki is an active member of the Polish Mathematical Society the American Mathematical Society and the Committee on Automatic Control and Robotics of the Polish Academy of Sciences a Life Senior Member of the IEEE a member of technical committees of the IFAC as well as many Polish and foreign scientific societies He was chosen a member of the Polish Academy of Arts and Sciences (PAU in 000 He was granted an honorary doctorate of the AGH University of Science and Technology in Cracow in 997 He has obtained many scientific awards from the Ministry of Science and Higher Education as well as the award of the Prime Minister (008 the Academy of Sciences and the Mathematical Society He was honored with the Commander s Cross of the Order of Polonia Restituta in 99 Received: January 00 Revised: 7 April 00