A Type of the Cauchy-Euler Equations: A Unique Real Root

Vol. 18, No. 1, 2018, 27-35

ISSN: 2279-087X (P), 2279-0888(online) Published on 17 July 2018

www.researchmathsci.org

DOI: http://dx.doi.org/10.22457/apam.v18n1a4

27

Annals of

A Type of the Cauchy-Euler Equations:

A Unique Real Root

Wichai Jisabuy and Gumpon Sritanratana1

Department of Mathematics, Rajabhat Mahasarakham University

Mahasarakham 44000, Thailand. E-mail: [email protected]

1

Corresponding author. E-mail: [email protected]

Received 2 June 2018; accepted 14 July 2018

Abstract. In this research, we give the family of all homogeneous Cauchy-Euler

equations such that each equation has general solution depending only on a unique real number.

Keywords: Linear Differential Equations, Cauchy-Euler equations. AMS Mathematics Subject Classification (2010): 34A30, 34B30 1. Introduction

Consider a homogeneous Cauchy-Euler equation of order n of the form

1 ... 1 0 0.

1 1

1 + + + =

+ − −₋

− a y

dx dy x a dy

y d x a dx

y d x

a _n

n n n n n n

n (1.1)

where a0,a1,...,an are real numbers with an ≠0. Details for methods to find solutions of

the equation (1.1) was explained in [2, 4, 5, 8]. Moreover, Sabuwala and Leon [6] studied the particular solution for the most general n-th order Euler differential equation when the non-homogeneity is a polynomial. They found a formula which can be used to compute the unknown coefficients in the form of the particular solution. It is well known that the general solution of (1.1) can be found from the characteristic equation

∑

∏

= =

= + + −

n

j j

i

j m i a

a

1 1

0 0

) 1

( (1.2)

of the linear ordinary differential equation with constant coefficients

, 0 1

1 1

0 =

  

  

+

   

 

+ −

∑

_∏

= =

y a i

dt d a

n

j j

i j

where t=ln x. In general, the general solution of any homogeneous Cauchy-Euler

equations depends on zeros of the polynomial

∑

∏

= =

+ + −

n

j j

i

j m i a

a

1 1

0.

) 1

28

The aim of this paper is to give the family of all Cauchy-Euler equations (1.1)

such that

∑

= −

= n

i i i x

c x y

1 1

ln

α _{is the general solution of (1.1) on}

) , 0

( ∞ for some real

number

α

.

2. Preliminary

In this section, we shall give the related basic notions that can be found in [1, 5, 7].

Let n∈ℕ_, ∈

n

a a

a0, 1,..., ℝ with an ≠0. An ordinary differential equation of the

form

1 ... 1 0 0

1

1 + + + =

+ − −₋ a y

dx dy a dy

y d a dx

y d

a _n

n n n n

n (2.1)

is said to be a homogeneous linear ordinary differential equation with constant

coefficients. By a transformation mx,

e

y= where mis a suitable number, the equation

(2.1) is transformed into the polynomial equation

0

... _{1 0}

1

1 + + + =

+ −

−m am a

a m

a n

n n

n ,

which is said to be the characteristic equation of (2.1).

Theorem 2.1. [7] Let n∈ℕ_, ∈

n

a a

a₀, ₁,..., ℝ _with ≠_0.

n

a Then the real numbers

α

is

the zero of multiplicity n of the polynomial a_nmn+a_n−1mn−1+_⋯+a₁m+a₀=0, if and

on only if

∑

= −

= n

i i i x

c x y

1 1

ln α

is the general solution of homogeneous linear ordinary

differential equation (2.1) on (0,∞), where c₁,...,c_n are arbitrary constants.

A linear ordinary differential equation form

0

0 1

1 1 1

1 + + + =

+ − −₋

− a y

dx dy x a dy

y d x a dx

y d x

a _n

n n n n n n

n ⋯ (2.2)

is called a homogeneous Cauchy-Euler equation.

The following theorem tells us that each Cauchy-Euler equation (2.2) can be transformed into linear ordinary differential equation with constant coefficients by the transformation x=et.

Theorem 2.2. [5] Let n∈ℕ_, ∈

n

a a

a0, 1,..., ℝ with an ≠0. Then the transformation t

e

x= transforms equation (2.2) into the equation

1 0

1 1

0 =

  

  

+

   

 

+ −

∑

∏

= =

y a i

dt d a

n

j j

i

j , (2.3)

29

Corollary 2.1. [7] Let n∈ℕ _and ∈

n

a a

a₀, ₁,..., ℝ _with ≠_0.

n

a Then the transformation

mt

e

x= transforms equation (2.4) into the equation

. 0 )

1 (

1 1

0

∑

_∏

= =

= + + −

n

j j

i

j m i a

a (2.4)

Theorem 2.3. [5] Let n∈ℕ_, ∈

n

a a

a₀, ₁,..., ℝ _with ≠_0.

n

a Then the real numbers

α

is

the zero of multiplicity n of the polynomial

∑

∏

= =

+ + −

n

j j

i

j m i a

a

1 1

0

) 1

( if and only if

∑

=

−

= n

i i i x

c x y

1 1

ln α

is the general solution of Cauchy-Euler equation (2.2) on (0,∞),

where c ,...,1 cn are arbitrary constants.

3. Main theorems

Definition 3.1. For eachj,k∈ℕ _{with j}≤_{k we define}

{

k

}

N_k:= 1,2,..., ,

k j j

k

j aa a a a a N

P, :={ 1 2⋯ : 1, 2,..., ∈ and a1<a2<⋯<aj},

, :

,

,

∑

∈ =

k j

P p k

j p

N N_0,0:=1 and N_0,k:=1,

and for every integers j,k with k> j we define Nk,j:=0.

For example; N_2,3 =1⋅2+1⋅3+2⋅3=11.

Form above definition, it is important to note that

k k k k N

kN −1, −1= ,

for every positive integer k. In addition, we have the following applicable lemma.

Lemma 3.1. Let n be a positive integer. Then

N_i,n+(n+1)N_i−1,n =N_i,n+1 (3.1)

for all i=1,2,3,...,n.

Proof: Let all i=1,2,3,...,n and

{

1 2 1 1 2 1 1 1

}

1

,n+ = i− ⋅( +1): , ,..., i− ∈ n, < < i− i aa a n a a a N a a

Q ⋯ ⋯

Before we proof (3.1), We shall prove that Pi,n and Qi,n+1 form a partition of Pi,n+1, that

is P_i,n ∩Q_i,n+1is empty and P_i,n∪Q_i,n+1=P_i,n+1. For P_i,n∩Q_i,n+1 is empty we suppose, to the contrary, that P_i,n∩Q_i,n+1is not empty. We can let b∈P_i,n∩Q_i,n+1. Since

,

,n i

Q

30

) 1 (

1 2

1 ⋅ +

=aa a− n

b ⋯ _i but b∈Pi,n it follows that a1,a2,...,ai−1,n+1∈Nn and

therefore n+1∈Nn which is a contradiction.

Now we shall prove that P_i,n∪Q_i,n+1=P_i,n+1. We see that P_i,n∪Q_i,n+1⊂ P_i,n+1.

since both Pi,n and Qi,n+1 are subsets of Pi,n+1. Now we shall show that

.

1 , , 1

,n+ ⊂ in∪ in+ i P Q

P Let b∈Pi,n+1. Then there exist a1,a2,...,ai∈Nn+1 such that

i

a a a

b= 1 2⋯ and a1<a2 <⋯<ai. Therefore ai≤n or ai =n+1 since ai∈Nn+1.

Case 1. Let ai≤n. Then a1,a2,...,ai∈Nn since a1<a2<⋯<ai. Therefore

n i i P

a a a

b= 1 2⋯ ∈ , and thus b∈Pi,n∪Qi,n+1. Consequently, Pi,n+1⊂Pi,n∪Qi,n+1.

Case 2. Let ai =n+1 . Then b=a1a2⋯ai∈Qi,n+1 since a1,a2,...,ai−1∈Nn.

Therefore b∈P_i,n∪Q_i,n+1. Consequently, P_i,n+1⊂P_i,n∪Q_i,n+1. Now P_i,n and Q_i,n+1 form a partition of P_i,n+1.

Next, we consider

. )

1 ( )

1 (

1 , ,

1 ,

1 ,

, 1

,

∑

∑

∑

∑

+

− ∈ ∈

∈ ∈

− = + + = +

+ +

n i n n

i n

i p P p P p Q

P p n i n

i n N p n p p p

N

Because P_i,n and Q_i,n+1, form a partition of P_i,n+1, we obtain

.

1 ,

1 ,

, , 1 , , 1

+ ∈

∈ ∈ ∈ ∪

= =

=

+

∑

∑

∑

∑

+

+ +

n i P

p P

p p Q p P Q

N p p

p

n i n

i in in in

It follows that Ni,n+(n+1)Ni−1,n =Ni,n+1. □

Lemma 3.2. Let m∈ℂ_{. For every n}∈ℕ_,

∏

∑

( )

=

−

=

− − −

= + −

n

i

n

i

i n n i i

m N i

m

1

1

0

1 ,

1 )

1

( (3.2)

Proof: We shall proof by mathematical induction on n. Since

( )

1 ,

) 1 ( )

1 (

1

1

1 1

0

1 1 1 , 0

1 1 1 , 0 0

∏

∑

=

−

=

− − −

− = − −

= = + −

i i

i i i

m N m

N m

i m

we obtain (3.2) is true for n=1.

Let k∈ℕ _{be arbitrary. Suppose that}

( )

∏

∑

=

−

=

− − −

= + −

k

i

k

i

i k k i i

m N i

m

1

1

0

1 ,

1 )

1 (

is true. Since

∏

∏

+

= =

+ − −

= + −

1

1 1

) 1 (

) ( ) 1 (

k

i

k

i

i m k m i

m , by the inductive hypothesis we

have

∏

+

=

+ −

1

1

) 1 (

k

i

i

m

∑

−

=

− − −

−

= 1

0

1 ,

) 1 ( ) (

k

i

i k k i i

m N k

31

∑

−

∑

= − = − − + + − − + − − = 1 0 1 0 1 , 1 1 1

, ( 1)

) 1 ( k i k i i k k i i i k k i i m kN m N

∑

−

∑

= = + − − − + − − + − − = 1 0 1 1 1 , 1 1 1

, ( 1)

) 1 ( k i k j j k k j j i k k i i m kN m N

∑

−

∑

= = + − − − + − − + − − = 1 0 1 1 1 , 1 1 1

, ( 1)

) 1 ( k i k i i k k i i i k k i i m kN m N

∑

−

∑

= − = + − − − + − − − + − + − − = + 1 1 1 1 1 1 , 1 1 1 , 1 , 0 0 ) 1 ( ) 1 ( ) 1 ( 1 k i k i i k k i i i k k i i

k m N m kN m

N k

+(−1)kk [(k−1)!]m

(

)

∑

− = + − − − − − + − + + − − = + 1 1 1 1 , 1 1 , 1 , 0 0 ) 1 ( ) 1 ( ) 1 ( 1 k i k i k k i k i i

k m N kN m k

N k _!m.

Therefore, by Lemma 3.1 and k!=Nk,k, we obtain

∏

+ = + − 1 1 ) 1 ( k i i m

∑

− = + − − + − + − − = + 1 1 , 1 , 1 , 0 0 ) 1 ( ) 1 ( ) 1 ( 1 k i k k k i k k i i

k m N m N m

N k . ) 1 ( 0 1 ,

∑

= + − − = k i i k k i i m N

Thus (3.2) is true for n=k+1. The proof is complete by mathematical induction

□

Lemma 3.3 Let m∈ℂ _and ∈

n

a

a ,...,1 ℝ with an ≠0. Then

, ) 1 ( ) 1 ( 0 1 , 1 1 1 1

∑

∑

∏

∑

= +− − +− − = − = = − + = + − j i j i n j i n i i n j j n n n j i n j

j m i am m N a

a (3.3)

for every n∈ℕ _{with n}≥_2.

Prove: We prove by induction. For n=2, consider

∑

∏

= = + − j i j

j m i

a 1 2 1 ) 1 (

∏

∏

= = + − + + − = 2 1 2 1 1

1 ( 1) ( 1) i i i m a i m a ) 1 ( 2 1 + −

=am am m m a a m

a 2 ( _{1 2})

2 + −

=

(

N a N a

)

m

m

a 1,1 2

1 1 0 , 0 0 2

2 + (−1) +(−1)

= i i i i i a N m m a ₊ =

∑

− +

= , 1

1

0 2

2 ( 1)

( 1) .

0 2 1 , 1 1 2 2

2

∑

∑

= − + +− = − ₋ + = j i j i j i i i j j a N m m a

It follows that (3.3) is true for n=2.

32 . ) 1 ( ) 1 ( 0 1 , 1 1 1 1

∑

∑

∏

∑

= +− − +− − = − = = − + = + − j i j i k j i k i i k j j k k k j i k j

j m i a m m N a

a

We must prove that (3.2) is true for n=k+1. Consider

. ) 1 ( ) 1 ( ) 1 ( 1 1 1 1 1 1 1

1

∏

∏

∑

∏

∑

+ = + = = = + = + − + + − = + − k i k j i k j j j i k j

j m i a m i a m i

a

Therefore, by inductive hypothesis and Lemma 3.2, we obtain

1 , 0 1 0 1 , 1 1 1 1 1 ) 1 ( ) 1 ( ) 1 ( −+ = + − + = +− − − = − = + =

∏

∑

∑

∑

∑

− + = + − + − k i

k i k i i k j i k j i j i k i i k j j k k k j i k j

j m i a m m N a a N m

a

, 1 1

0 1 , 0 1 1 ) 1 ( ) 1 ( ₊ −+ = − + − − + = − = −

_∑

_∑

∑

− + − +

= k i

k k i k i i j i k j i k i j i i k j j k k

km m N a N a m

a k k k k k j i k j i k i j i i k j j k k

km m N a a m N a m

a 1 1 _{1, 1}

1 1 , 0 1 1 ) 1 ( [ ) 1 ( _{+− − +− +} + ₊ = − = − _{− + + −} + =

∑

∑

( 1) ( 1) 2 ( 1) _{, 1} ]

1 , 1 1 1 1 , 2 2 m a N m a N m a

N k _{kk k}

k k k k k k

k − + +

− − + + + − + − − + ⋯

∑

= − − + − + − + + + − + − = 1 0 1 1 2 , 1 , 0 0 1

1 ( 1) ( 1) i k i k i k i i k k k k

k m N am N a m

a

∑

∑

− = + − − + − + = − + + − + 1 0 1 , 2 2 3 , 2 0 ) 1 ( ) 1 ( k i i i i i k i k i k i i i m a N m a N _⋯

1, 1 2

1 1 1 , 2 2 1 , 1 1 ) 1 ( ) 1 ( ) 1

( N a m N a m Nk kak m

k k k k k k

k − +

− − + + + − + + − − + ⋯

+(−1)kN_k,ka_k+1m

] ) 1 ( ) 1

[( 1, 1

1 1 , 0 0 1

1 − +

+

+ + − + −

= k k k k

k k

k m m N a N a

a _      − + − + +− + = +− −

_∑

1 , 2 2 1 1 0 2 , 1 ) 1 ( ) 1

( _{k i k k}

i i k i i k a N a N m _      − + − + + + + − =

∑

, 1 , 1

1 0 ) 1 ( ) 1

( kk k

k i i i k i i a N a N m ⋯

∑

∑

= +− +− − + − + − = + + + − + − = 2 0 1 2 , 1 1 , 1 0 1

1 ( 1) ( 1) i i k i k i i k i k i k i k i i k k

k m m N a m N a

a _{, 1} 0 2 1 0 1 , 2 ) 1 ( ) 1 ( ₊ = + − = +

∑

∑

− + − +

+ ii i

k i i i k i i i i a N m a N m ⋯

∑

∑

= +− +− + = − + + + + − = j i j i k j i k i i k j j k k

k m m N a

a 0 1 , 1 1 1

1 ( 1) .

Thus (3.3) is true for n=k+1. The proof is complete. □

33 , ) 1 ( ) 1 ( ₀ 0 1 , 1 1 0 1 1 a a N m m a a i m a j i j i n j i n i i n j j n n n j i n j

j

∏

− + + = +

∑

∑

− +

∑

= +− − +− − = − = =

and thus by Theorem 2.3, we obtain the following corollary.

Corollary 3.1. Let n∈ℕ _and ∈

n

a a a

m,

α

, 0, 1,..., ℝ with an ≠0. Then

α

is the zero of

multiplicity n of the polynomial

0 0 1 , 1 1 ) 1

( N a a

m m a j i j i n j i n i i n j j n n

n +

∑

∑

− +

= +− − +− −

= −

(3.4)

if on only if

∑

= − = n i i i x c x y 1 1 ln

α _{is the general solution of the Cauchy-Euler equation}

0 0 1 1 1 1

1 + + + =

+ − −₋

− a y

dx dy x a dy y d x a dx y d x a _n n n n n n n

n ⋯ (3.5)

on the open interval (0,∞), where c1,...,cn are arbitrary constants.

Lemma 3.4. Let n∈ℕ _and ∈

n

a a a

m,

α

, 0, 1,..., ℝ with an ≠0. Then

j n j n n

n j j n j i j i n j i n i i n j j n n n m j n m a a N m m

a ( 1) ( 1)

α

( 1)

α

1 1 0 0 1 , 1 1 − +       − + = + − + − − = = +− − +− − = −

_∑

_∑

∑

(3.6) if and only if

a0 =(−1)n

α

n, an =1and 1, 1

1 0 ) 1 ( ) 1 ( _{+ + + +} − − = − − ₊

_∑

₋       −

= i i k i k

k n i i k n k n

k N a

k n

a

α

(3.7)

for every k=1,2,...,n−1.

Proof: Because the set of 1,m,m2,...,mn are linearly independent on ℝ_{, we have (3.6) is}

true if and only if a₀ =(−1)n

α

n, a_n =1 and for every k=1,2,...,n−1,

( 1) ( 1) .

0 1 , k n k n k i k n i k i i i k n n a

N ₊ − −

− = + −      − − = −

∑

α

(3.8)

Since N_0,k−1=1, _

     =       − k n k n n and

∑

∑

− = + − + − + − = + − − + =

− n k

i k i k i i i k k k i k n i k i i i a N a N a N 1 1 , 1 , 0 0 1

, ( 1)

) 1

( ( 1) ,

1 1 ,

∑

− = + − + − +

= n k

i k i k i i i

k N a

a

from (3.8), we have

. ) 1 ( ) 1

( , 1

1 k n k n k i k i i k n i i k k n a N

a _{+ − +} − −

− =      − = − +

∑

α

Since − − _{+ − +} =

−

=

∑

ii k i k

k n

i i

a N, 1 1 ) 1 ( − _{+ − +} = − = +

∑

ii k i k

k n

i i

a N, 1 1

1

) 1

( ( 1) 1, 1,

1 0 + + + + − − =

∑

− i i k i k k

n

i i

a

34

( 1) ( 1) _{1, 1}.

1

0

+ + + + −

−

= −

− ₊

_∑

₋

     

−

= i i k i k

k n

i i k

n k n

k N a

k n

a

α

(3.9)

Therefore (3.8) and (3.9) are equivalent. This proof is complete. □

By binomial theorem,

, ) 1 ( )

1 ( )

(

1

1

n n j

n j n

j j n

n

m j n m

m

α

_

α

+ −

α

    

− + =

− − −

=

∑

we obtain

α

is the zero of multiplicity n of the polynomial , ) 1 ( )

1 (

1

1

n n j

n j n

j j n

m j n

m _

α

+ −

α

    

−

+ − −

=

∑

the next corollary is a direct consequence of Lemma 3.4. and Corollary 3.1.

Corollary 3.2. Let n∈ℕ _and ∈

n

a a a

m,

α

, ₀, ₁,..., ℝ _with ≠_0.

n

a Then

α

is the zero of

multiplicity n of polynomial (3.4) if and only if (3.7) is true for every k=1,2,...,n−1. The next main theorem is an immediately consequence of Corollary 3.1 and

Corollary 3.2.

Theorem 3.1. Let n∈ℕ _and ∈

n

a a a

m,

α

, 0, 1,..., ℝ with an ≠0. Then

∑

=

−

= n

i i i x

c x y

1 1

ln α

is the general solution of the homogeneous Cauchy-Euler equation

(3.5) on (0,∞), where c ,...,1 cn are arbitrary constants if and only if (3.7) is true for

every k=1,2,...,n−1.

The application of this theorem is to give a Cauchy-Euler equation of order

nfrom a given general solution on (0,∞) in the form ln ,

1 1

∑

=

−

= n

i i i x

c x

y α where

α

is

a given real number.

4. Conclusion

We give every Cauchy-Euler differential equation from its general solution that depends only on a given real numbers. In the future, we will devote our attention to the family of all Cauchy-Euler differential equations that have general solutions depending on several real numbers.

Acknowledgements. The authors are thankful to the reviewers for valuable comments and

suggestions on the manuscript and thank the Faculty of Science and Technology,

35

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