FACULTY OF NATURAL SCIENCES
CONSTANTINE THE PHILOSOPHER UNIVERSITY NITRA
ACTA MATHEMATICA 12
DIFFERENTIAL EQUATIONS WITH SYMMETRICAL SOLUTIONS
KLEMENT HRKOTA, ML.
ABSTRACT. In this article, the existence of certain pairs of types of solutions of linear
differential equations on symmetrical interval is being discussed. The existence of these
solutions is being examined in relation to the parity of coefficients of these differential
equations.
1. Introduction
While building a theory of linear differential equations, differential equations with
constant coefficients serve as a model which shows many properties that can be applied on
differential equations with variable coefficients. We can see it on differential equations
with property A, B or binomial type of differential equation. In cases like this, the identity
of signs of coefficients seems like one of the basic and natural conditions on interval of
solutions. In this article, we are examining the existence of certain pairs of types of
solutions of differential equations in relation to the parity of coefficients of these
differential equations. Specifically, we focus on searching conditions of existence of
solutions we call symmetric for coefficients of differential equations.
2. Symmetrical solutions
Consider the linear differential equation:
Lx :  x ( n )  p1 (t ) x ( n1)  ...  pn (t ) x  0
in interval I   a, a  , where 0  a   and coefficients pi, i  1, ..., n are continuous
functions in I.
DEFFINITION 1. Differential equation Lx  0 is said to have symmetrical solutions, if for
each solution ut  , t  I is function u t  solution of differential equation Lx  0 , t  I
too.
Here are some examples of differential equations with constant coefficients class.
EXAMPLE 1.
Differential equations
x  x  0
with solution x  c1e t  c2 e t
Supported by grant no. 1/0519/08 of the scientific grant agency VEGA of Slovak Republic
KLEMENT HRKOTA, ML.
x ( 4)  x  0
with solution x  c1e t  c2 e t  c3 cos t  c4 sin t
have symmetrical solutions.
Differential equation
x  x  0
with solution x  c1e  c2 e
t
1
 t
2
1
 t
3
3
cos
t  c3 e 2 sin
t
2
2
has not symmetrical solutions.
Now, for all t  I , next assumptions for coefficients of differential equation
Lx  0 will be used :
H1)
  0, if i is odd
pi t  
are even, if i is even
H2)
 0, if i is even
pi t 
 are odd , if i is odd
THEOREM 1. Let assumption H1) is satisfied. Then differential equation Lx  0 has
symmetrical solutions
Proof: For simplicity, we will divide the proof in two parts.
Consider at first, that n is even. Let w is arbitrary solution of differential equation
Lx  0 . It means that following equation holds for all t  I :
w( n ) t   p2 (t ) w( n2) t   ...  pn2 (t ) wt   pn (t ) wt   0
(1)
Calculate now. Let
n 
n  2 

Lw t   w t   p2 (t )w t 
 ...  pn2 (t )w t   pn (t )w t 
 d
where   .
dt
By functions differentiation in components we obtain:




Lw t   w( n) t  t t 1  p 2 (t ) w( n2) t  t t 1
n
n2
 ...  p n2 (t )wt t t 1  p n (t )w t 
2
(2)
(3)
DIFFERENTIAL EQUATIONS WITH SYMMETRICAL SOLUTIONS
Because of the fact that n is even and coefficients of differential equation Lx  0 are even
functions (3) we can write in the form:




Lw t   w ( n ) t  t  t  p 2 (t ) w ( n  2) t  t  t  ...  p n  2 (t )wt t  t  p n (t ) w t 
(4)
Because equation (1) holds for all t  I can be rewritten in next form:
w
(n)
t t  t  p 2 (t )w ( n  2) t t  t  ...  p n  2 (t )wt t  t  p n (t )w t   0
(1*)
Hence Lw t   0 for all t  I and w t  is solution of differential equation Lx  0 in
I.
In the n odd case is situation similarly, but equation (1) has the form:
w( n ) t   p1 (t ) w( n1) t   ...  pn3 (t ) wt   pn1 (t ) wt   0 .
Next by similarly way as in the n even case it is possible to show that Lw t   0 for all
t  I , which completes the proof of this theorem.
THEOREM 2. Let assumption H2) is satisfied. Then differential equation Lx  0 has
symmetrical solutions
The proof of this theorem is similar to the theorem 1 proof and will be omitted.
Referring to the formulation of conditions H1) and H2) it is good to realize that if
we say n is odd and z  x in the theorem 1, obtained differential equation Lz  0 also
satisfies assumptions of theorem 1 for n being even. On the contrary, if we say n is even
and z  x in theorem 2, obtained differential equation Lz  0 satisfies assumptions of
theorem 2 for n being odd.
3. Symmetrical equations
Now consider next differential equation:
Kx :  x ( n )  r1 (t ) x ( n1)  ...  rn (t ) x  0
on interval I   a, a  , where 0  a   and coefficients ri, i  1, ..., n are continuous
functions on I.
DEFFINITION 2. We say that differential equations Lx  0 a Kx  0 are symmetrical, if
for each solution ut  of differential equation Lx  0 , function u t  is solution of
differential equation Kx  0 .
KLEMENT HRKOTA, ML.
EXAMPLE 2.
Differential equations
1
 t
2
1
 t
3
3
t  c3 e 2 sin
t a
2
2
1
1
t
t
3
3
t
2
t  c3 e 2 sin
t
with solution x  c1e  c2 e cos
2
2
x  x  0
with solution x  c1e t  c2 e
x  x  0
cos
are symmetrical.
In this part of the document we will assume that for coefficients of considered differential
equations is:
pi t   ri t  , i  1,2,..., n for every t  I .
(5)
Let next assumptions for coefficients of differential equation Lx  0 for all t  I
are satisfying:
H3)
 0, if i is odd
pi t 
 are odd , if i is even
H4)
 0, if i is even
pi t 
 are even, if i is odd
THEOREM 3. If assumption H3 (H4) is satisfied for coefficients of differential equation
Lx  0 , then differential equations Lx  0 and Kx  0 are symmetrical.
It is clear, that if for coefficients of differential equation Lx  0 H3 (H4) is true,
then by (5) assumption H3 (H4) holds for coefficients of differential equation Kx  0 too.
Proof: We give just one part of proof of this theorem for n even and assumption H3 here.
The proofs of other parts of this theorem are similar and will be omitted.
Let w is arbitrary solution of differential equation Lx  0 . It means, that for all
holds
equations (1) a (1*).
t I
Same way as in the proof of the theorem 1, we obtain, for differential equation
Kx  0 , that




Kw t   w ( n ) t  t  t  r2 (t ) w ( n  2 ) t  t  t  ...  rn  2 (t )wt t  t  rn (t ) w t 
(6)
Because of the fact that coefficients ri , i  2,4,..., n of differential equations Lx  0 are
odd it is possible to rewrite (6) to the form:
DIFFERENTIAL EQUATIONS WITH SYMMETRICAL SOLUTIONS




Kw t   w ( n ) t  t  t  r2 (t ) w ( n  2) t  t  t  ...  rn  2 (t )wt t  t  rn (t ) w t  . (7)
By (5) and from (1*)




Kw t   w ( n ) t  t t  p 2 (t ) w ( n 2) t  t  t  ...  pn2 (t )wt t  t  pn (t ) w t   0 . (8)
Last equation implies that w t  is solution of differential equation Kx  0 which
completes this part of proof.
RNDr. Klement Hrkota, ml.
Department of informatics
Faculty of Mechatronics
Alexander Dubček University in Trenčín
Študentská 2
Sk- 911 01 Trenčín
e-mail: hrkota@tnuni.sk