*Nihan GÜNGÖR; [email protected], Tel: (0456) 213 10 00, orcid.org/0000 -0003-1235-2700 ISSN: 2146-538X http://dergipark.gov.tr/gumusfenbil GÜFBED/GUSTIJ (2020) 10 (3): 814-829 Gümüşhane Üniversitesi Fen Bilimleri Enstitüsü Dergisi DOI: 10.17714/gumusfenbil.709376 Araştırma Makalesi / Research Article BG-Volterra Integral Equations and Relationship with BG-Differential Equations BG-Volterra İntegral Denklemleri ve BG-Diferansiyel Denklemlerle İlişkisi Nihan GÜNGÖR* Gümüşhane Üniversitesi, Mühendislik ve Doğa Bilimleri Fakültesi, Matematik Mühendisliği Bölümü, 29100, Gümüşhane • Geliş tarihi / Received: 25.03.2020 • Düzeltilerek geliş tarihi / Received in revised form: 20.06.2020 • Kabul tarihi / Accepted: 23.06.2020 Abstract In this study, the Volterra integral equations are defined in the sense of bigeometric calculus by the aid of bigeometric integral. The main aim of the study is to research the relationship between bigeometric Volterra integral equations and bigeometric differential equations. Keywords: Bigeometric Calculus, Bigeometric Differential Equations, Bigeometric Volterra Integral Equations Öz Bu çalışmada, bigeometrik integral yardımıyla bigeometrik Volterra integral denklemleri tanımlanmıştır. Çalışmanın asıl amacı bigeometrik manada Volterra integral denklemleri ile bigeometrik manada diferansiyel denklemler arasındaki ilişkiyi araştırmaktır. Anahtar kelimeler: Bigeometrik Hesap, Bigeometrik Diferansiyel Denklemler, Bigeometrik Volterra İntegral Denklemleri
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GÜFBED/GUSTIJ (2020) 10 (3): 814-829 Gümüşhane Üniversitesi Fen Bilimleri Enstitüsü Dergisi
DOI: 10.17714/gumusfenbil.709376 Araştırma Makalesi / Research Article
BG-Volterra Integral Equations and Relationship with BG-Differential
Equations
BG-Volterra İntegral Denklemleri ve BG-Diferansiyel Denklemlerle İlişkisi
Nihan GÜNGÖR* Gümüşhane Üniversitesi, Mühendislik ve Doğa Bilimleri Fakültesi, Matematik Mühendisliği Bölümü, 29100, Gümüşhane
• Geliş tarihi / Received: 25.03.2020 • Düzeltilerek geliş tarihi / Received in revised form: 20.06.2020 • Kabul tarihi / Accepted: 23.06.2020
Abstract
In this study, the Volterra integral equations are defined in the sense of bigeometric calculus by the aid of bigeometric
integral. The main aim of the study is to research the relationship between bigeometric Volterra integral equations and bigeometric differential equations.
Keywords: Bigeometric Calculus, Bigeometric Differential Equations, Bigeometric Volterra Integral Equations
Öz
Bu çalışmada, bigeometrik integral yardımıyla bigeometrik Volterra integral denklemleri tanımlanmıştır. Çalışmanın
asıl amacı bigeometrik manada Volterra integral denklemleri ile bigeometrik manada diferansiyel denklemler
Theorem 8. ( BG -Leibniz Formula) Let ,A I be exp -open set and f be a BG -continuous function on
A I into exp . If BG
xf exists and is BG -continuous on A I , ( ) ( ),u x v x are BG -continuously
differentiable functions of A into I , then
( )
( )
( ) ( )( )( )
( )
( )( ) ( )( ), , , ,
v xv x
BG BG BG BG BGBG BG x x xB
B
xx
G
G
uu
df x t dt f x t dt f x v x v f x u x u
dx
=
.
Proof. Take ( ) ( ) ( ), , ,BG
BG
tBGf x t F x t F x t
t
= =
. Hence we find
( )
( )
( )( )
( )
( ) ( )( ) ( )( ), , , ,
v x v x BGBG BG
BG BGBG
u x u x
f x t dt F x t dt F x v x F x u xt
= =
.
Therefore, we obtain
( )
( )
( ) ( )( ) ( )( )( ) ( )( ) ( )( ), , , = , ,
v x
BGBG
BG BG B
BG BG B
G B
x
G BG
G
u
d d d df x t dt F x v x F x u x F x v x F x u x
dx dx dx dx
=
(1)
by using the properties of BG - derivative. We can write as
( )( ) ( )( ) ( ) ( )( ), , ,BG BG BG
x xu xB
B
G
GdF x u x F x u x F x u x u
dx= (2)
Güngör / GUFBED 10(3) (2020) 814-829
824
and
( )( ) ( )( ) ( ) ( )( ), , ,BG BG BG
x xv xB
B
G
GdF x v x F x v x F x v x v
dx= (3)
from BG -chain rule. Hence by using Theorem 7, we get
( )
( )
( ) ( )( ) ( )( ) ( ) ( )( ) ( ) ( )( )
( )( )
( )
( )( ) ( )( )
, , , , ,
, , ,
v x
BG BG BG BG BG BG BBG
GBG x x x xv x u xBG
u x
v x BGBG BG BG BG
BG x x xBG
u x
df x t dt F x v x F x u x F x v x v F x u x u
dx
F x t dt f x v x v f x u x ut
=
=
( )( )
( )
( )( ) ( )( )
( )( )( )
( )
( )( ) ( )( )
, , ,
, , ,
v x BGBG BG BG BG
BG t x xBG
u x
v x
BG BG BG BGBG x x x
u x
F x t dt f x v x v f x u x ux
f x t dt f x v x v f x u x u
=
=
from the expressions (1), (2) and (3).
Example 2. Show that the BG -Volterra integral equation ( ) ( )1
sin
x
x BGBGu x x e u t dt= can be
transformed to BG -differential equation.
Solution. If we consider the equation and differentiate it by using BG -Leibniz formula, we obtain
( ) ( )1
sin
xBG BG BGx BG
BGBG BG BG
u x x u td d d
dx dx de dt
x
=
( ) ( )( ) ( )( ) ( )( )
( ) ( )( )
( ) ( )( )
cot
1
cot
1
cot
1
.
1
x BGBG x x x BG x x
BG
x
BG x x BG xB
BG BG
x x
G
BG x x x
BGu x e u t e dt u x e u e
u x e edt u x e
u x
xx
e x u x e
=
=
=
Thus the BG -Volterra integral equation is equivalent to the BG -differential equaiton
( ) ( )( ) cotBG x x xu x u x e e x= .
3.1.2. The Conversion of the BG -Linear Diferential Equations to BG -Volterra Integral Equations
In this part, we prove that it is converted to BG -Volterra integral equations by defining BG - linear
differential equation with constant coefficients and variable.
Definition 12. The equation of the form
( ) ( ) ( ) ( ) ( )1
1 1BG BG
nn BG
n ny a x y a x y a x y f x−
− =
where f is a bipositive function, is called thn order BG -linear differential equation. If the coefficients
( )na x are constants, then the equation is called as BG - linear differential equation with constant
coefficients; if not it is called BG -linear differential equation with variable coefficients.
Theorem 9. If n is a positive integer and expa with expx a , then we have
( ) ( )( )
( )( )
( )1
exp
exp1 !exp
x x xnBG BG BG
BG BG BG
a a a
en u t dt dt x t u t dt
n
−=
−
Proof. Take
Güngör / GUFBED 10(3) (2020) 814-829
825
( )( )
( )exp1
xn BG
BGn
a
I x t u t dt−
= . (4)
If it is taken ( ) ( )( )
( )exp1
,n
F x t x t u t−
= , we write
( )
( )( ) ( )( ) ( )( )
,
, , ,
xBG BGBGn
BGBG BG
a
x BGBG BG BG
BG x xBG
a
d I dF x t dt
dx dx
F x t dt F x x x F x a ax
=
=
( )
( )
,
,
xF x tx xF x t BG
BG
a
e dt
= (5)
by using BG -Leibniz rule. Since ( ) ( )( )
( )
( ) ( ) ( )
( )2 2
exp
ln
ln ln ln1
,
n nu tx x
u tt tn x x
F x t x t u tt t
− −
−
= = =
,
( ) ( ) ( ) ( )( )2
1, 1 , ln ln
n
xF x t n F x t u t
x x t
−
= −
. Therefore we find
( )
( )( ) ( )
( ) ( ) ( )( )( )
( )( )
1
2 2 2ln,
1 ln ln ln ln1,
n
n n nx
u tF x t x x xx x x xBG x n u t
nF x t t t tBG BG BG BGnBG BG BG BG
BG
a a a a
d Ie dt e dt e dt e u t dt
dx
−
− − − − −
= = = =
( )
( ) ( )
( )
( ) ( )
( ) ( )( )
( ) ( )( )
( )
( )
( )( )
3
2
1
exp exp
1
lnln ln
1 1
3 31
1
1 1
n
n
n
n
xx xx x t
n nt BG BGtBG BG
a a
x xn nn BG BG
BG BG
a a
n
n n
e e u t dt e e u t dt
e x t u t dt x t u t dt
I e I
−
−
−
−
− −
− −−
−
− −
= =
= =
= =
from the equation (5). Hence we get
( )( )
( )
( )1
exp3 1
1
n
xBGn BG nn
BG nBG
a
d Ix t u t dt e I
dx
−
− −
−
= = (6)
for 1n . Since ( )1
x
BGBG
a
I u t dt= for 1n = , then we write
( ) ( )1
xBG BGBG
BGBG BG
a
d I du t dt u x
dx dx
= =
. (7)
If it is taken BG -derivative of the equation (6) by using BG -Leibniz formula, then
( ) ( )( )
( ) ( ) ( )( )
( )( )
( )( )( )
( )( ) ( ) ( )
( )( ) ( ) ( )
exp exp
3
exp
23 31 1
2
1 2 ln ln31 2 1 2
2 .
BG
BG
n
x xBG BGn nn nBG BGn
BG BGBG BG
a a
xx xn n u t
nn n n nBG BGtBG BG n
a a
d I de x t u t dt e x t u t dt
dx xdx
e dt e e x t u t dt e e I
−
− −− −
− − −− − − −
−
= =
= = =
By proceeding similarly, we get ( )
( )
( ) ( ) ( ) ( )1
1 2 1 !1
1 1 exp 111 !
BG
BG
n
n n nn
n
d Ie e e I e I n I
dx
−
− − −
−= = = − .
Hence we write
( ) ( )exp1 !BG
BG
n
n
n
d In u x
dx= −
Güngör / GUFBED 10(3) (2020) 814-829
826
from the equation (7). Now, we will take BG -integral by considering the above relations. From the equation
(7), ( ) ( )1 1 1
x
BGBG
a
I x u x dx= . Also, we have
( ) ( ) ( )1
1
2 1 1 1 1 1 2
xx x
BG BG BGBG BG BG
a a a
I x e I x dx e u x dx dx= =
where 1x and
2x are parameters. By proceeding likewise, we get
( ) ( ) ( )
( ) ( )
3 2
3 2
1 !
1 1 2 1
exp 1 1 2 1 1 !
n
n
x x xxn BG BG BG BG
BG BG BG BGn n n
a a a a
x x xx
BG BG BG BGBG BG BG BG n n
a a a a
I x e u x dx dx dx dx
n u x dx dx dx dx
−
−
−
=
= −
where1x ,
2x , ,nx are parameters. If we write the equation (4) instead of the statement
nI , then we find
( )( )
( ) ( ) ( )3 2
exp1
exp 1 1 2 11 !nx x xx x
n BG BG BG BG BGBG BG BG BG BG n n
a a a a a
x t u t dt n u x dx dx dx dx−
−= − .
If it is taken 1 2 nx x x x= = = = , then we obtain
( )( )
( ) ( ) ( ) ( )exp1
exp1 !
x x xn BG BG BG
BG BG BG
a a a
x t u t dt n n u t dt dt−
= − .
Therefore, we get
( ) ( )( )
( )( )
( )exp1
exp
exp1 !
x x xnBG BG BG
BG BG BG
a a a
en u t dt dt x t u t dt
n
−=
−
and this completes the proof.
Let thn order BG -linear differential equation
( ) ( ) ( ) ( ) ( )1
1 1BG BG
nn BG
n ny a x y a x y a x y f x−
− = (8)
given with the initial conditions
( ) ( ) ( ) ( )1
0 1 11 , 1 , , 1BGnBG
ny c y c y c−
−= = = (9)
This thn order BG -linear differential equation can be reduced to a BG -Volterra integral equation. Hence the
solution of (8)-(9) may be reduced to a solution of some BG -Volterra integral equation.
Taking ( )BGny u x= , we can write
( ) ( ) ( )1
BG
BGn
BG
dy x u x
dx
−= . By BG -integrating both sides of this equality,
( )( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
1
1 1
1 1
1
1
1
1
1
BG
BG BG
BG
x xnBG BG
BG BG
xn n BG
BG
xn BG
BGn
d y u t dt
y x y u t dt
y x c u t dt
−
− −
−
−
=
=
=
By proceeding similarly, we find
Güngör / GUFBED 10(3) (2020) 814-829
827
( )( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( )( ) ( )exp
2
1
1 1 1
2 2
1
1 1 1
2
2 1
1 1
2
1 2 1
1 1
1
1 .
BG
BG BG
BG
x x xnBG BG BG
BG BG BGn
x x xn n BG BG BG
BG BG BGn
x xn BG BG
BG BGn n
x xnBG BG BG
BG BGn
d y c u t dt dt
y x y c dt u t dt dt
y x c c x u t dt dt
y c c x c x n u t dt dt
−
−
− −
−
−
− −
−
−
=
=
=
= −
Therefore, we obtain,
( )( ) ( )exp exp
12
0 1 2 1
1 1
x xn BG BG
BG BGny c c x c x c x n u t dt dt−
−= .
If we take into account the above expressions, the BG -linear differential equation is written as follows
( ) ( ) ( ) ( ) ( )
( )( )
( ) ( ) ( )exp exp
1 1 2 2 1
1 1 1
12
0 1 2 1
1 1
x x x
BG BG BGBG BG BGn n n
x xn BG BG
BG BGn n
u x a x c u t dt a x c c x u t dt dt
a x c c x c x c x n u t dt dt f x
− − −
−
−
=
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )
( )( )exp exp
1 2
1 1 1 1 1
1 2
1 2 1 2 3 2
x x x x x
BG BG BG BG BGBG BG BG BG BGn
n n
n n n n
u x a x u t dt a x u t dt dt a x n u t dt dt
f x a x a x x a x x c a x a x x a x x c− −
− −
=
( ) 0na x c (10)
If we set
( ) ( ) ( )( )
( )
( ) ( ) ( )( )
( )
( ) ( )
exp
exp
1
1 2 1
2
2 3 2
0
n
n n
n
n n
n
a x a x x a x x f x
a x a x x a x x f x
a x f x
−
−
−
−
=
=
=
and
( ) ( ) ( ) ( ) ( )1 1 2 2 0 0n n n nF x f x f x c f x c f x c− − − −=
Then, one can see that the equation (10) is in the following form:
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( )
1 2
1 1 1 1 1
.
x x x x x
BG BG BG BG BGBG BG BG BG BGnu x a x u t dt a x u t dt dt a x n u t dt dt
F x
=
By using Theorem 9, we get
( ) ( ) ( ) ( ) ( ) ( ) ( )( )
( )( )
( )
( )
1 1exp exp1 2
exp exp11! 1 !
.
exp exp
x x xnBG BG BG
BG BG BGn
a a
e eu x a x u t dt a x x t u t dt a x x t u t dt
n
F x
−
−
=
If we edit this equation as
( ) ( ) ( ) ( ) ( )( )
( )( )
( ) ( )1 1
exp exp1 2
exp exp11! 1 !
exp exp
xn BG
BG n
e eu x u t a x a x x t x t a x dt F x
n
− =
−
and set
( ) ( ) ( ) ( )( )
( )( )
( )1 1
exp exp1 2
exp exp
,1! 1 !
exp expn
n
e eK x t a x a x x t x t a x
n
−=
−
as the kernel function, then the equation (8) is turned into
Güngör / GUFBED 10(3) (2020) 814-829
828
( ) ( ) ( ) ( )1
,
x
BGBGu x u t K x t dt F x =
which is a BG -Volterra integral equation of the second kind.
Example.3. Find BG -Volterra integral equation corresponding to the BG -differential 2 2BG BGy x y y e = with the initial conditions ( )1y e= , ( )1 1BGy = .
Solution. Let ( )2
2
2
BG
BG
BG
dy u x
dx= = . Since ( )
2
2
BG
BG
BGBG
BG
d dy u x
dxdx= = , then we write
( ) ( )
( ) ( ) ( )
( ) ( )
1 1
1
1
1
x x
BG BG BGBG BG
x
BG BG BGBG
x
BG BGBG
d y u t dt
y x y u t dt
y x u t dt
=
=
=
Therefore, we find
( )
( ) ( ) ( ) ( )
( ) ( ) ( )
1 1 1
exp
exp 1
1
11!
x x x
BG BG BGBG BG BG
x
BGBG
x
BGBG
d y u t dt dt
ey x y x t u t dt
y x e e x t u t dt
=
=
=
If we replace the findings above into the given BG -differential equation, we obtain
( ) ( ) ( ) ( ) 2
1 1
x x
BG BGBG BGu x x u t dt e e x t u t dt e
=
From this, we get BG -Volterra integral equation as ( ) ( )2
1
( )
x
BGBGu x e x t u t dt= .
4. Conclusion
In this paper, the Volterra integral equations are
defined in the sense of bigeometric calculus by using the concept of bigeometric integral. The
Leibniz formula is proved in bigeometric calculus
and aid of this the bigeometric Volterra integral equations are converted to bigeometric differential
equations. By defining the bigeometric linear
differential equations with constant coefficients and variable coefficients, they are converted to
bigeometric Volterra equations is proved.
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