Michailov criterion for exponential polynomial

Krall, A. M.

English

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N A S A  C O N T R A C T O R  
R E P O R T  - 
co 
ON THE  MICHAILOV  CRITERION 
FOR EXPONENTIAL  POLYNOMIALS 
Prepared by r 
PENNSYLVANIA STATE UNIVERSITY 
University Park,  Pa. 
. $  
NATIONAL  ERONAUTICS  AND PACE  ADMINISTRATION WASHINGTON,  D. C .  MAY 1966  
https://ntrs.nasa.gov/search.jsp?R=19660016021 2020-03-16T19:57:58+00:00Z
TECH LIBRARY KAFB. NY 
ON THE MICHAILOV CRITERION FOR EXPONENTIAL POLYNOMIALS 
By Allan M. Kral l  
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information exchange. Responsibility for the contents 
resides  in  the  author or  organization  that  prepared it. 
Prepared  under  Contract No. NGR-39-009-041  by 
PENNSYLVANIA STATE UNIVERSITY 
University  Park, Pa. 
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NATIONAL AERONAUTICS AND SPACE ADMINISTRATION 
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On the Michailov Criterion for Exponential  Polynomials 
Allan M. K r a l l  1 
We consider an equation of the form 
where 0 5 w1 < w2 ... < wm a r e   r e a l  numbers  and 
i = 1, . . . m, j = 1, . . . n a r e  complex numbers. N “ i j t  
We assume tha t  the,re i s  a t  l e a s t  one c o e f f i c i e n t  a 
d i f f e r e n t  from zero when i 1 - 1. 
i j  
I n  many app l i ca t ions  i t  i s  necessary to know whether 
or n o t   F ( z )  has any   zeros   in   the  r i g h t  ha l f   p lane .  The 
b e s t  known technique for answering this  ques t ion  i s  due t o  
Pont.rjagin [l] and [2]. Pont r jag in’s   c r i te r ion ,   howeverg  
i s  v e r y  d i f f i c u l t  t o  a p p l y .  I n  f a c t  i t  has  been  found  unsat- 
i s f ac to ry  excep t  as a t h e o r e t i c a l  r e s u l t  ( s e e  [ 3 ,  page 4201 ) 
Popov, [ 3 ,  page 4201 ), s t a t e s  tha t  t o  provide a more u s e f u l  
c r i t e r i o n ,  A.  A. Sokolov  [4]  and N. N. Miasnikov  extended 
the  Michai lov  c r i te r ion  to cover exponential polynomials 
such as F ( z ) .  Unfortunately no re ference  i s  g.iven f o r  
1 
McAll is ter  Bui lding,  The Pennsylvania  State  Universi ty ,  
University Park,  Pennsylvania.  
" 
Miasnikov Is work and Sokolov's i s  i n a c c e s s i b l e  i n  t h i s  
country.   Since  the  Michai lov  cr i ter ion i s  e a s i l y  stated, 
easi ly  proved and easy to apply,  i t  seems advantageous 
even a t  t h e  r i sk  o f  d u p l i c a t i o n  to presen t  i t  he re .  
Dividing by eWmZ and l e t t i n g  xij = am-i+l, we 
transform (1) i n t o  
Here ri = wm-w > 0 f o r  i = 2, 3, . . . m and m - i + l  
r1 = 0. The zeros  of F ( z )  and G ( z )  a r e  t h e  same,  and i t  
i s  G ( z )  we w i l l  consider  f rom now on. 
I n  g e n e r a l  G ( z )  w i l l  c o n t a i n  a n  i n f i n i t e  number 
of zeros which occur in  cha ins  having  the  fo l lowing  
p rope r t i e s .  
1. The imaginary  par ts  o f  the   zeros   a re  
O(k)  f o r  k = + - 1, + - 2, ... 2 n, . O .  
2. The r e a l  p a r t s  o f  t he   ze ros   a r e  
O(1og k ) ,  (advanced  type) ,   constant ,   (neutral  
type)  or - O(1og k )  ( r e t a r d e d   t y p e ) ,  
Any combination o f  advanced, neutral  or re ta rded  types  
may occur.  
The term aln z i s  s a i d  to be the   p r inc ip le   t e rm n 
o f  G( z ) .  If the   p r inc ip l e   t e rm i s  p re sen t ,   t he re   a r e  
no zeros  of  advanced  type. However, t h e r e  s t i l l  may be 
2 
a n  i n f i n i t e  number of z e r o s  i n  the r i g h t  half plane 
because of the presence of zeros  of n e u t r a l  type .  
We w i l l  need the following .hypotheses i n  what 
follows. 
H 2 .  a i n = O ,  i = 2 J  3, ... m. 
H 3. A l l  t h e   c o e f f i c i e n t s  a a r e  r e a l .  
i j  
Theorem 1. Let H 1 h o l d .  Let 
Then i n  t h e  r igh t  half plane G ( z )  has a t  most a f i n i t e  
number o f  zeros  a l l  of  which l i e  in s ide  a semic i rc le  o f  
r ad ius  M + 1 centered a t  t h e   o r i g i n .  
Proof.  Let  IzI > M + 1. Then 
I f  G(z) had a n  i n f i n i t e  number o f   z e r o s   i n   t h e   r i g h t  
half plane,  they would approach  in f in i ty  g iv ing  a con- 
tra,dic t ion.  
We write 
where 
and 
Note t ha t  i n  the r i g h t  half  plane including the imaginary 
a x i s  I @ ( z ) I  < 1 and  Y(z) = O ( 1 )  as 1.1 4 03 whenever H 1 
holds 
Theorem 2. (Michai lov  Cri ter ion)   Let  H 1 hold.  A s  z 
va r i e s   a long  the imaginary axis f rom - i CQ t o  i 00 then 
G(z) passes through  the  origin  each  t ime G ( z )  has  an 
irnagqinary zero.  I f  G ( z )  has no imaginary  zeros,  then 
4 
n 1 N =  I " 2 2.n P ( -  i y ,   i y )  arg G ( Z )  + ( 8 )  
+ w - i Y > > l s  
where N is the number of zeros of G( z )  i n  the r igh t  
half plane,  y i s  any number g rea t e r  t han  M + 1 of theorem 1, 
l a r g e  enough so tha t  I @ ( z )  + Y ( z ) l  < 1 when l z l  2 y and 
A ( iy, - iy )  a rg  G( z )  i s  the n e t  change i n  arg G( z ) as z 
varies  f rom - iy  t o  i y .  
Proof. We choose a semicircular   contour  C varying 
from - i y  to i y  along a c i r c l e  c e n t e r e d  a t  the 0r igl .n  with 
r ad ius  y and then along the imaginary axis from i y  t o  - iy .  
If G ( z )  has no zeros  on  t.he imaginary  axis  then it  i s  
well. known that  
where A c  denotes  the  net   change i n  a r g  G( z )  along the 
s e n i c i r c l e .  We see  from ( 7 )  t h a t  
a r g  ~ ( z )  = n A~ a r g  z + .A C a r g  (1 + @ ( z )  + Y ( z ) ) .  (IO) 
A, a r g  G ( Z )  = n T + a r g  (1. + a( i y )  + Y (  i y ) )  - 
s i n c e  1 + a(.) + Y ( z )  cannot wind around the o r ig in .  In -  
s e r t i n g  (11) i n  (9) achieves the r e s u l t .  
5 
Corol.lary 1.. If H 1 holds,  and G( z )  has. ,no  imaginary 
zeros ,   then N = 0 i f  and  only i f  
Corol la ry  2 .  If H 1 and H 3 hold,  and G ( z )  has no i m -  
ag inary  zeros ,  then  
Proof. When H 3 holds the argument f o r  y a r e   t h e  - 
negat ive  of  those  for  y. 
Coro l la ry  3. I f  H 1 and H 3 hold ,  and G ( z )  has no imaginary 
zeros ,  then N = 0 i f  and only i f  
Corol lary 4. If H 1 and H 2 hold ,  and G ( z )  has no 
imaginary zeros then  
Proof. When H 2 holds,  @( z )  0. 
Since   y (z)  = 0(1), 1 + Y(z) + 1 a s  I z l  -c 03, Thus t h e  l a s t  
two terms of (8)  approach 0. 
Coroll.ary 5 If H 1 and H 2 hold , and G( z) has no i m -  
ag inary  zeros ,  then  N = 0 i f  and only i f  
6 
Corol lary -~ " . .  6 -  If H - .  1, H .. 2 and . B .3 hold,  and G ( z )  has 
no imaginary zeros,  then 
Corol lary ".  , . 7. . . - . If . H - . 1, H. .  2 . . . and . . H - . . 3 hold,  and G ( z )  has no 
. imaginary - - . . . . zeros,   then N = 0 i f  and  only i f  . ~. 
A ( 0 ,  i m )  a r g  G ( Z )  = n7r 
If these s ta tements  are compared to those o f  
Pontr jagin,  i t  i s  easy to s e e  t h a t  they imply each other .  
The main advantage to the  Michai lov  c r i te r ion  i s  the  re- 
moval o f  the perpe tua l  osc i l1a t . ion  present  in  Pont r jag in ' s  
c r i t e r i o n .  
It is a f a i r l y  simple mat.ter to see i f  a r g  G ( Z )  
varies t.hrough the appropr ia te  number o f  quadrants or 
not, This procedure  should  prove  quite  useful En 
a p p l i c a t i o n .  
7 
1. Richard  Bellman  and Kemeth L. Cooke, " D i f f e r e n t i a l  - 
Difference Equations",  Academic Press, New York, 1963, 
2. Le S Pontr   jagin,  "On the  Zeros  of Some Elementary 
Trancendental   Functions",  A m e r .  Math. SOC. Trans., 
3. E .  P. Popov, "The Dynamics of  Automatic  Control 
Systems",  Addison-Wesley,  Reading,  Mass., 1962. 
4. A .  A.  Sokolov, "Kriteri i  u s to l ch ivos t i   l i ne inykh  
sistem r e g u l i r o v a n i i a  s raspredelennymi parametrami 
i ego primeneniia.  (The s t a b i l i t y  c r i t e r i o n  i n  
. l b e a r   r e g u l a t i o n  sys tems with d i s t r i b u t e d  parameters 
and i t s  appl ica t ion) , "  Inzhenernyi  Sborn ik  2,2 (1946) 
8 NASA-Langley, 1966 CR-489 


On the Michailov criterion for exponential polynomials

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