Global Journal of Pure and Applied Mathematics.
ISSN 0973-1768 Volume 12, Number 3 (2016), pp. 1921-1945
© Research India Publications
http://www.ripublication.com/gjpam.htm
The K-Exponential Matrix to solve systems of
differential equations deformed
Ruthber Rodríguez Serrezuela
Faculty of Biomedical,
Electronic and Mechatronics Engineering,
Universidad Antonio Nariño,
Osmin Ferrer Villar
Faculty of Education,
Program of Math, Universidad de Sucre,
Sincelejo, Republic of Colombia.
Jorge Ramirez Zarta
Faculty of Engineering,
Corporación Universitaria del Huila, CORHUILA,
Yefrén Hernández Cuenca
Neiva, Republic of Colombia.
Abstract
With this work we introduce exponential matrix notions and deformed logarithms,
wherof we realize a complete study of their properties with a deformation generator
introduced by G. Kaniadakis 1. Furthermore, we establish deformed differential
equations and some of their solution techniques, which permit to solve common
physical problems which are modeled with the mentioned equations. We show an
application of K-differential equations to a physical application of basic engineering demonstrating its effectiveness for engineering calculations.
AMS subject classification:
Keywords:
1922
1.
Ruthber Rodríguez Serrezuela et al.
Introduction
In this article we define the k-exponential matrix and the k-logarithmic matrix of a square
matrix A, which we will denote expk (A), and lnk (A) respectively. Additionally, we will
present some convergence properties and criteria for the same matrices [1], [2]. In the first
part we will discuss the deformation generator according to G. Kaniadakis, some aspects
of deformed k-algebra and properties of the k-exponential and k-logarithmic functions
[3], [4], [5]. In the second part we will define the k-exponential and k-logarithmic matrices of an A-square matrix by using their performances in a series of convergent matrices.
Finally, in the last part, we define the k-differential equations, understood as differential
equations with kaniadakis deviations and k-deformation parameters. Also, we present
some techniques for solving k-differential equations and k-differential equation systems,
where the k-exponential matrix forms part of the solutions for some of these systems.
2.
Deformed mathematics after Kaniadakis
Deformation Generator
In [1] a real g function is defined, which depends on k ∈ R parameter. A deformation
generator has to show the following properties:
i) g(x) ∈ C ∞ (R)
ii) g(−x) = −g(x)
d
g(x) > 0
dx
iv) g(±∞) = ±∞
iii)
v) g(x) ≈ x when x → 0.
1
G. Kaniadiakis [1] defines the real function as xk = arcsenhg(kx) and its inverse
k
1 −1
k
x = g (sinh(kx)) which complies with the following:
k
i) xk ∈ C ∞ (R)
ii) (−x)k = −xk
d
xk > 0
dx
iv) (±∞)k = ±∞
iii)
v) x−k = xk
vi) xk ≈ x when k → 0(x0 = x)
vii) xk ≈ x when k → 0(0k = 0).
The K-Exponential Matrix to solve systems of differential equations
1923
Deformed K-Algebra
In [15] the construction of a deformed algebra and calculus to solve problems outlined
thermostatic not extended. Similarly, in [1] the k-sum is defined which generalizes the
sum of real numbers by means of
k
1
x ⊕ y = (xk + yk )k = g −1 [senh[arcsenh(g(kx)) + arcsenh(g(ky))]
k
(2.1)
The following formulas are k− sum properties as demonstrated by [1] and [3].
k
i) (x ⊕ y) = xk + yk
0
ii) x ⊕ y = x + y
k
k
iii) x ⊕ y = y ⊕ x
k
k
k
k
iv) (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)
k
v) x ⊕ 0 = x
k
k
vi) x ⊕ (−k) = (−x) ⊕ x = 0
k
k
Remark 2.1. The x ⊕ (−y) is written as x y and it is called k-difference.
The k−product is also introduced in the following way:
k
1
x ⊗ y = (xk · yk )k = g −1 (senh[arcsenh(g(kx)) · arcsenh(g(ky))])
k
The following properties relate k-sum and k-product, [1]:
0
i) x ⊗ y = xy
k
k
k
k
ii) (x ⊗ y) ⊗ z = x ⊗ (y ⊗ z)
k
k
k
k
k
iii) z ⊗ (x ⊕ y) = (z ⊗ x) ⊕ (z ⊗ y)
k
iv) (R − {0}, ⊗) is an abelian group [4].
k
v) I = 1k is such as x ⊗ I = x.
k
k
k
1
appears like in x ⊗ x = x ⊗ x = I
vi) x =
xk
(2.2)
1924
Ruthber Rodríguez Serrezuela et al.
1
vii) x y = x ⊗ y = x ⊗
yk
k
k
k
k
.
In [6] the exponential deformed k-function or k-exponential is defined as
1
arcsenh(kx) = ( 1 + k 2 x 2 + kx)1/k
expk (x) = exp
k
(2.3)
with 0 < k < 1 [13].
The expk (x) function consists of real numbers, is part of the C ∞ (R) classification,
is strictly growing and obeys the exponential rules when k → 0. Furthermore, the kderivation and the usual derivation show the following properties:
i)
lim (expk (x)) = 0
x→−∞
ii) lim (expk (x)) = ∞
x→∞
1
d
expk (x) [14],
(expx (x)) = √
dx
1 + k2x 2
√
k2x − 1 + x 2k2
iv)
expk (x)dx =
expk (x) + C
k2 − 1
iii)
v) exp−k (x) = expk (x)
vi) expk (x)expk (−x) = 1
vii) (expk (x))r = expk/r (rx), (r = 0)
viii) expk (0) = 1
k
ix) expk (x)expk (y) = expk (x ⊕ y)
k
x) expk (x)/expk (y) = expk (x y).
The inverse k-exponential function equals the k-logarithmic function, denoted as lnk (x),
defined in real positive numbers and for k = 0 it behaves as ln0 (x) = lnx. When k = 0
it is defined as,
x k − x −k
1
(2.4)
lnk (x) = senh(klnx) =
k
2k
The lnk (x) function [2] belongs to the C∞ (R+ ) class is strictly growing, concave and
adheres to:
i) lim (lnk (x)) = −∞
x→0+
The K-Exponential Matrix to solve systems of differential equations
1925
ii) lim (lnk (x)) = ∞
x→∞
iii) lnk (1) = 0
iv) ln−k (x) = lnk (x)
1
v) lnk ( ) = −lnk (x) [14],
x
vi) lnk (x r ) = rlnrk (x); with r ∈ R
k
vii) lnk (xy) = lnk (x) ⊕ lnk (y)
k
viii) lnk (x/y) = lnk (x) lnk (y)
1 ln2k (tb)
d
lnk (tb) =
ix)
t lnk (tb)
dt
t
ln2k (tb)
.
lnk (tb) −
x)
lnk (tb)dt =
1 − k2
lnk (tb)
Returning to the k-exponential [7], one can use the Taylor expansion series for exp(·)
and proceed with x0 = 0, which is given by:
expk (x) =
∞
an (k)
n=0
xn
, k2x 2 < 1
n!
(2.5)
where the coefficients an are defined by a0 (k) = a1 (k) = 1 and
a2m (k) =
m−1
[1 − (2j ) k ]; a2m+1 (k) =
2 2
j =0
m
[1 − (2j − 1)2 k 2 ]
(2.6)
j =1
It is important to note that an (0) = 1 and an (−k) = an (k). According to [5], the Taylor
expansion of lnk (1 + x) [7] converges if −1 < x ≤ 1 and if it has the following form:
lnk (1 + x) =
∞
n=1
n−1 x
bn (k)(−1)
n
n
(2.7)
with bn (k) = 1, bn (0) = 1 bn (−k) = bn (k). for n > 1
1
k
k
1
k
k
bn (k) = (1 − k) 1 −
··· 1 −
+ (1 + k) 1 +
··· 1 +
2
2
n−1
2
2
n−1
(2.8)
1926
3.
Ruthber Rodríguez Serrezuela et al.
K-Exponential matrix
In this part we explain how the definition of the k-exponential matrix behaves when
joined with an A-square matrix, denoted as expk (A), as an extension of the execution
of a series of k-exponential matrix, in such a way that when k −→ 0 then the usual
exponential function of a matrix is obtained [10]. Moreover, some important properties
of an k-exponential matrix are verified.
Theorem 3.1. The matrix series
∞
ap (k)
p=0
p!
Ap converges if A <
1
with ap (K) as
|k|
in (2.6).
∞ ∞ ∞ ap (k) p ap (k) p
ap (k) p
Proof. Note that
p! A =
p! A ≤
p! A .
p=0
p=0
p=0
Effectivley,
∞ ∞ ∞ a2n+1 (k) a2m (k) ap (k) 2n+1
p
2m
(2n + 1)! A
2m! A +
p! A =
n=1
m=0
p=0
2m
2
a
(k)A A 2m+2
2
bm+1 (2m + 2)(2m + 1)(2m)! (k)A
a
2m+2
=
=
b 2m
(2m
+
1)(2m
+
2)a
(k)
a2m (k)A
m
2m
(2m)!
[1 − (2m)2 k 2 ]A2 =
(2m + 2)(2m + 1) Then,
bm+1 [1 − (2m)2 k 2 ]
2
A
= lim
lim
m→∞ bm m→∞ (2m + 2)(2m + 1)
 
1
2
−
k
(2m)2
= lim  A2 
2
1
m→∞
1 + 2m 1 + 2m
= | − k 2 |A2
1
Therefore, the series is convergent, when k 2 A2 < 1, which is the same as A <
.
|k|
∞ a2n+1 (k) 2n+1
Similarly, it shows that the series
is convergent [11].
(2n + 1)! A
n=1
The K-Exponential Matrix to solve systems of differential equations
1927
1
,
|k|
we will define the k-exponential expk (A) as the matrix n × n given by the convergent
series:
∞
ap (k) p
(3.9)
A with ap (k) as in (2.6)
expk (A) :=
p!
Definition 3.2. [k-Exponential of a matrix] Given a matrix A ∈ C n×n , with A <
p=0
Note that A is a nilpotent matrix, therefore, expk (A) is a finite series and, consequently,
convergent.
Proposition 3.3. If O = [0] ∈ Cn×n is the zero matrix, than expk (O) = In .
Proof.
O a2 (k) 2 a3 (k) 3
+
O +
O + · · · = In where k 2 O = 0 < 1
1!
2!
3!
expk (O) = In +
Proposition 3.4. When D ∈ Cn×n is a diagonal matrix, D = diag{d1 , . . . , dn } with
1
|di | <
, for i = 1, 2, 3, . . . , n so that, expk (D) converges to diag{expk (d1 ), . . . ,
|k|
expk (dn )}.
Proof.
a2 (k) 2 a3 (k) 3
expk (D) = In + D +
D +
D + ···
2!
3!


1 + d1 + a2 (k)d12 /2! + · · ·
0


...
=

2
0
1 + dn + a2 (k)dn /2! + · · ·


0
expk (d1 )
=
..
0
.
expk (dn )
 = diag {expk (d1 ), . . . , expk (dn )} with |di | < 1 ,
|k|
f or i = 1, 2, 3, . . . , n
Proposition 3.5.

d
[expk (tB)] = B 
dt
∞
ap+1 (k)
p=0
p!

(tB)p  .
1928
Ruthber Rodríguez Serrezuela et al.
(m)
Proof. This time bij
shall be the element i, j of (tB)m , therefore the element i, j of
(tB)m t m (m)
is
b , consequently,
m!
m! ij


∞
ap+1 (k) p p+1
a2 (k) 2 (2) a3 (k) 3 (3)
d
(1)
= 
t bij + · · ·
+
1 + tbij
t bij +
t bij  then,
3!
dt
2!
p!
p=0
∞
∞
ap+1 (k) p p+1
ap+1 (k)
d
(tB)]
=
B
=
B
t
(tB)p
[expk
dt
p!
p!
p=0
p=0
Proposition 3.6. For D = diag{d1 , d2 , . . . , dn } the derivative of the k-exponential
matrix expk (tD) is given by:


1
0
 1 + (ktd )2

1


d


...
(3.10)
[expk (tD)] = D 
 expk (tD)


dt
1


0
1 + (ktdn )2
Proof.

∞
ap+1 (k)
d
(expk (tD)) = D 
dt
p=0
∞
p!
ap+1 (k)

 p=0

= D




p!

(tD)p 

p
0
(td1 )
..
.
∞
ap+1 (k)
0
p=0
p!

1
 1 + (ktd )2 expk (td1 )
1


...
= D


0
= D · diag
(tdn )
1
1 + (ktdi )2






p
0
1
1 + (ktdn )2
· expk (tD)
expk (tdn )






The K-Exponential Matrix to solve systems of differential equations
1929
When k → 0, one obtains,
d
(exp(tD)) = D exp(tD).
dt
Proposition 3.7. For a diagonalizable matrix B = SDS −1 we obtain
expk = S [expk (D)] S −1 .
Proof.
expk (B) =
∞
ap (k)
p!
p=0

=S
Bp =
∞
ap (k) p=0
∞
ap (k)D
p!
p=0
p!
SDS −1

 S −1
= S expk (D) S −1
And also
d
d
[expk (tB)] = [S (expk (tD)) S −1 ]
dt
dt
1
= SD · diag · expk (tD)
1 + (ktdi )2
Proposition 3.8.
∞
ap−1
expk (tB)dt = B −1
p=1
p!
(tB)p + C
where B and C are matrices with the size of n × n.
Proof.
expk (tB)dt =


∞
p=0

ap (k)
(tB)p  dt
p!
∞
ap (k) p p
(t) B dt
=
p!
p=0
(3.11)
1930
Ruthber Rodríguez Serrezuela et al.
one element (i,j) out of (3.11) has the following form,
(1)
1 + tbij
a2 (k) 2 (2) a3 (k) 3 (3)
+
t bij +
t bij + · · · dt
2!
3!
t 2 (1) a2 (k) t 3 (2) a3 (k) t 4 (3)
b +
b + ... + Cij
+
= t + bij
2! 3 ij
3! 4 ij
2
=
∞
ap−1 t p
p!
p=1
(p−1)
bij
Consequently,
expk (tB)dt =
∞
ap−1 (k)t p
p!
p=1
= B −1
∞
ap−1
p=1
p!
B p−1 + C
(tB)p + C
Then, for a matrix D = diag{d1 , . . . , dn } one obtains,

2 2 2
 k t d1 −



expk (tD)dt = D −1 




k2
4.

1 + k 2 t 2 d12
−1




 expk (tD) + C



2 2 2
2
2
2
k t d n − 1 + k t dn 
0
..
.
0
When k → 0 one obtains
k2 − 1
exp(tD)dt = D −1 exp(tD) + C.
K-Logarithmic matrix
Through the series performance of the deformed k-logarithms we define a square matrix
A as the logarithmic matrix and we call it lnk (A). Also, we define some properties of
the logarithmic k-matrix which reduce the usual properties of logarithmic performance
of a square matrix.
Proposition 4.1. The matrix series
∞
bp (k)(−1)p−1
p=1
p
(A − In )p
is convergent when A − In < 1. where bp (K) as in (2.8).
The K-Exponential Matrix to solve systems of differential equations
Proof.
∞ ∞
bp (k)(−1)p−1
bp (k)
n
(A − In ) ≤
A − In p
p
p
p=1
p=1
Given
bp (k)
A − In p ,
p
ap =
then:
ap+1 pbp+1 (k)A − In p+1 a = (p + 1)b (k)A − I p p
p
n
pbp+1 (k) A − In = (p + 1)bp (k) k
When p −→ ∞, 1 ±
p
−
+
(k) + bp+1
(k), with,
−→ 1, if bp+1 (k) = bp+1
k
1
= (1 − k) · · · 1 −
2
p
−
(k)
bp+1
and
+
(k)
bp+1
k
1
.
= (1 + k) · · · 1 +
2
p
Furthermore,
lim
p→∞
y lim
−
bp+1
= lim (1 − k/p) = 1
p→∞
bp−
+ bp+1
bp+
p→∞
= lim (1 + k/p) = 1,
p→∞
therefore
lim
p→∞
=
−
+
bp+1
+ bp+1
bp− + bp+
bp− + bp+
bp− + bp+
= 1,
1931
1932
Ruthber Rodríguez Serrezuela et al.
then
ap+1 lim p→∞ ap = A − In Consequently, the series
∞
bp (k)
p
p=1
(A − In )p
converges when A − In < 1.
Definition 4.2. [k-logarithms of a matrix] Given a matrix A ∈ Mn×n (C), with A −
In < 1. We define the k-logarithms of a lnk (A) matrix as the n × n matrix given by the
convergent series:
lnk (A) :=
∞
bp (k)(−1)p−1
p=1
p
(A − In )p , for n − 1 < A < n + 1
Proposition 4.3. Proves that lnk (In ) = 0, where 0 is the zero matrix with the size of
n × n.
Proof.
lnk (In ) =
=
∞
bp (k)(−1)p−1
p=1
∞
p=1
p
(In − In )p
bp (k)(−1)p−1 p
(0) = 0
p
Proposition 4.4. When D is a diagonal matrix D = diag{d1 , d2, . . . , dn } then lnk (D) =
diag{lnk (d1 ), . . . , lnk (dn )}.
Proof.
lnk (D) =
∞
bn (k)(−1)p−1
p
p=1
with

(d1 − 1)p
Ap = 

0
..
0
Ap

.
(dn − 1)
p
The K-Exponential Matrix to solve systems of differential equations
one element (i, j ) of
∞
bn (k)(−1)p−1
p
p=1
has the form
∞
bp (k)(−1)p−1
p
p=1
1933
Ap
(di − 1)p for i = j
and “0” for i = j .
Then
∞
bp (k)(−1)p−1
p
p=1
(di − 1)p = lnk (1 + (di − 1)) = lnk (di ).
Proposition 4.5. For a diagonal matrix D = diag(d1 , . . . , dn ) one obtains
ln(expk (tD)) = tD.
Proof.
lnk (expk (tD)) = lnk (diag(expk (td1 ), . . . , expk (tdn )))
= diag(lnk (expk (td1 )), . . . , lnk (expk (tdn ))) = tD.
Proposition 4.6. For a diagonizable matrix A = RDR −1 with D = diag{λ1 ,
λ2 , . . . , λn }, we conclude that lnk (A) is convergent when 0 < λi ≤ 2, where λi are
values belonging to A.
Proof.
A − In = RDR −1 − In = RDR −1 − RR −1 = R(D − In )R −1
= R diag{λ1 − 1, . . . , λn − 1}R −1
Then
lnk (A) =
∞
bp (k)(−1)p−1
p
p=1

= R
∞
bp (k)(−1)p−1
p
p=1
The element (i, i) of
∞
bp (k)(−1)p−1
p=1
[R(D − In )R −1 ]p
p

(D − In )p R −1 
(D − In )p
1934
Ruthber Rodríguez Serrezuela et al.
is
∞
bp (k)(−1)p−1
p=1
Then
p
(λi − 1)p = lnk (λi )

lnk (A) = R 

0
lnk (λ1 )
..
 R −1 .
.
0
(4.12)
lnk (λn )
It is converged when 0 < λi ≤ 2, through (2.7).
Proposition 4.7.
1
d
lnk (tD) = lnk (lD) (lnk (tD))−1 .
dt
t
Proof.

∞
bp (k)(−1)p−1

d 
d
lnk (tD) =
(tD − In )p 
dt
dt
p
p=1


d
0

 dt lnk (tλ1 )


.

.
= 
.




d
0
lnk (tλn )
dt
1

−1
ln (tλ )(lnk (tλ1 ))
0
 t 2k 1



...
=



1
−1
0
ln2k (tλn )(lnk (tλn ))
t
1
= ln2k (tD)(lnk (tD))−1 .
t
Remark 4.8. For a matrix B = RDR −1 applies
1
d
lnk (tB) = R ln2k (tD) (lnk (tD))−1 R −1 .
dt
t
It is important to note that when k → 0, we obtain
1
1
d
ln(tB) = R ln(tD)(ln(tD))−1 R −1 = In .
t
t
dt
The K-Exponential Matrix to solve systems of differential equations
1935
Proposition 4.9. To obtain a diagonalizable matrix
B = RDR −1 = R diag{λ1 , . . . , λn }R −1 , we have to
1
t
−1
lnk (tB)dt = R
lnk (tD) −
ln2k (tD)(lnk (tD)) + C R −1
1 − k2
1 − k2
Proof.
lnk (tD)dt R −1
1
−1
= Rdiag
(lnk (tλi ) − ln2k (tλi )(lnk (tλi )) ) + Ci R −1
1 − k2
t
1
−1
=R
lnk (tD) −
ln2k (tD)(lnk (tD)) + C R −1
1 − k2
1 − k2
lnk (tB)dt = R
Then k → 0
ln(tB)dt = R (tln(tD) − tIn + C) R −1
= tRln(tD)R −1 − tIn + CR
when CR = RCR −1 .
5.
K-Differential equations
k
We consider two algebraic structures (X, ⊕, ·) and (Y, +, ·) and the complex of the F
functions: F : {f : X → Y } with f ⊆ C ∞ (R). Kanadianis [8] defines the differential
k
dk x as: dk x = lim x z where dk x = dxk . Moreover, he defines the k-deviation of the
z→x
f -function as:
df (x)
f (x) − f (z)
= lim
k
z→x
dk x
xz
df (x)
df (x)
=
=
dx
dx
k
k
df (x)
1
=
dxk /dx
dx
d
= 1 + k 2 x 2 f (x)
dx
fk =
with x, z ∈ R y f (x), f (z) ∈ R. When k → 0, the k-deviation is reduced to the usual
derivation.
1936
Ruthber Rodríguez Serrezuela et al.
Figure 1: K-derivation concept
According to figure 1, one can observe that the k−derivation as defined by G Kaniadakis is interpreted as the reason for change between the image variations of a function.
Recognizing the image variations of the xk deformator we see that when x tends to a x0
f/x
df (x0 )/dx
AN
AN
AD
=
=
=
xk /x
d(x0 )k /dx
BF
BF
AD
1
df (x0 )
df (x0 )
=
=
d(x0 )k /dx
dx
dk x
Proposition 5.1. The k-derivation in point x0 is the inclination of a hyperbola tangent
in point x0 given by the equation
k
y = fk (x0 )(x x0 ) + f (x0 ).
Proof.
k
y = fk (x0 )(x x0 ) + f (x0 ) = fk (x0 )(x 1 + k 2 x02 − x0 1 + k 2 x 2 ) + f (x0 )
(5.13)
If we express (5.13) through the formula Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. We
have to
2
A = (fk (x0 )) ; B = −2fk (x0 ) 1 + k 2 x02 ;
C = 1; D = 2fk (x0 )f (x0 ) 1 + k 2 x02 ;
E = −2f (x0 ); F = (f (x0 ))2 − x02 (fk (x0 ))2 .
The K-Exponential Matrix to solve systems of differential equations
1937
Figure 2: Hyperbola tangent of the function f (x) = x 2 in point x0 = 2.
With
2
B 2 − 4AC = 4k 2 x02 fk (x0 ) ≥ 0.
Consequently, it is verified that y = g(x) is the straight tangent to the curve y = f (x)
when k = 0 and it is the hyperbola for para k = 0. Moreover,
i.)
g(x0 ) = fk (x0 ) x0 1 + k 2 x02 − x0 1 + k 2 x 2 +f (x0 ) = f (x0 )(0)+f (x0 ) = f (x0 )
ii)
x0 xk 2
2
2
1 + k x0 − √
g (x) =
1 + k2x 2


2
2
2
2
2
 1 + k x0 1 + k x0 − x0 x0 k 
= fk (x0 ) 

1 + k 2 x02
1
1
= fk (x0 ) =
· 1 + k 2 x02 f (x0 ) = f (x0 ).
1 + k 2 x02
1 + k 2 x02
fk (x0 )
In figure 2 the function f (x) = x 2 is visualized, which is the tangent to the hyperbola
g(x) and the point x0 = 2 for the k-values between (−1, 1) .
Proposition 5.2. When f, g ∈ F, B ∈ R then apply
1938
Ruthber Rodríguez Serrezuela et al.
d
d
d
(f + g) =
f+
g
dk x
dk x
dk x
d
d
d
ii)
(f · g) = g
f +f
g
dk x
dk x
dk x
1
d
d
d
f (g(x)) ·
(g(x))
iii)
[f (g(x))] =
dk x
dk x/dx dx
dx
i)
Definition 5.3. When f ∈ F where F : {f : X → Y } with f ⊆ C ∞ (R), then
1
dxk
f (x)dx =
f (x)dk x =
f (x)dx,
√
dx
1 + k2x 2
k
[13] with f (x)dk x = f (x)dx.
0
Definition
5.4. A k-differential equation is an expression of the form
E k, x, y, yk , yk , . . . , yk(n) = 0 a represents deformations of a usual differential equation, in such a way that when k → 0, both, the usual differential equation and its solution,
reduce.
For example in a k-differential equation yk − 2yk + (1 − k 2 )y = 0 it can be verified
that one solution is y = xekx and when k → 0 yk − 2yk + (1 − k 2 )y = 0 which is
reduced to yk − 2yk + y = 0 and y = xekx which is reduced to y = xex . Where y = xex
is solution of y − 2y + y = 0.
Proposition 5.5. A separable k-differential equation has the following form
f (x) · g(y) and a solution is given by
1
1
dy =
f (x)dx.
√
g(y)
1 + k2x 2
dy
=
dk x
Remark 5.6. The k-exponential is invariant to the k-derivation, which is effective for
dy
=y
x
d
k
1
1
dy =
dx
√
2x 2
y
1
+
k
1
=
dx
√
2x 2
1
+
k
1
d(kx)
=
k
1 + (kx)2
1
= ln |secθ + tanθ | + C
k
The K-Exponential Matrix to solve systems of differential equations
when kx = tanθ, then
and accordingly
when k → 0
1939
1/k 2
2
ln(y) = ln ( 1 + k x + kx) + C1
y = ec1 ( 1 + k 2 x 2 + kx)1/k = C2 expk (x)
dy
= y it results in y = Cex .
dx
Proposition 5.7. A linear K-diferential equation with constant coefficients has the form
dy
= ay + f (x) and its general solution is
dk x
y = Cexpk/a (ax) + expk/a (ax) f (s) exp−k/a (−as) dk s.
k
dy
= ay with a general solution y = C2 expk/a (ax).
dk x
When C2 = v(x) then y = v(x) expk/a (ax) then
dy
dv(x) d
=
expk/a (ax) + v(x)
expk/a (ax).
dk x
dk x
dk x
Proof. For f (x) = 0 we obtain
but
d
d
(expk/a (ax)) = 1 + k 2 x 2 (expk (x))a = a(expk/a (ax)).
dk x
dx
Therefore,
dy
= 1 + k 2 x 2 v (x) expk/a (ax) + av(x) expk/a (ax)
dk (x)
(5.14)
Replacing (5.14) in the linear k-diferential equation with constant coeficients we obtain,
1 + k 2 x 2 v (x)expk/a (ax) + av(x)(expk/a (ax)) = av(x)expk/a (ax) + f (x),
wherefrom
v(x) =
k
f (x) exp−k/a (−ax) dk x.
dy
= ay + f (x) is
dk x
y = Cexpk/a (ax) + expk/a (ax) f (s) exp−k/a (−as) dk s
Consequently, the general solution of
(5.15)
k
dy
It is observed that when k → 0, the k-differential equation is reduced to
= ay +f (x)
dx
and its respective solution is y = Cexp(ax) + exp(ax) f (y)(exp(−ay))dy.
1940
6.
Ruthber Rodríguez Serrezuela et al.
K-Differential equation systems
Definition 6.1. Given A ∈ C n×n as a constant matrix e Y as a matrix function with in
dY
order to n × n the k-differential matrix equation
= AY denominates a homogeneous
dk t
linear with constant coefficients.
Proposition 6.2. When given D = diag{λ1 , λ2 , . . . , λn }, the matrix function is

Y =

0
C1 expk/λ1 (λ1 t)
..

.
0
Cn expk/λn (λn t)
which is the solution of the k-differential system
dY
= DY.
dk t
Proof.


C1 expk/λ1 (λ1 t)
0
dY
d 
..

=
.
dk t
dk t
0
Cn expk/λn (λn t)

 d
 dk t

=

..
0

=
λ1
..
0
0
C1 expk/λ1 (λ1 t)
.
.




d
Cn expk/λn (λn t)
dk t

0
C1 expk/λ1 (λ1 t)
0
..

.
λn )
0

 = DY
Cn expk/λn (λn t)
Proposition 6.3.

 


Y11 · · · Y1n
d1 · · · 0
Y11 · · · Y1n
dY
d  .
.  .
..  =  .. . .
..  = DY
...
...
=
 ..
. ..   ..
.  .
. 
dk t
dk t
0 · · · dn
Yn1 · · · Ynn
Yn1 · · · Ynn
1941
The K-Exponential Matrix to solve systems of differential equations
obtains as solution

..
C11 expk/d1 (d1 t) .
···
..
.
Cij expk/di (di t)

Y =

Cn1 expk/dn (dn t)

0
···
0

···

C1n expk/d1 (d1 t)

..

.
..

. Cnn expk/dn (dn t)

 .
n  ..



=
Cii · · · Cin  expk/di (di t)


i=1  ...

0
···
(6.16)
0
Remark 6.4.
i) If D = In , then Y (t) = Cexpk (tIn ).
ii) When k → 0, then
dY
= DY obtains as solution Y = Cexp(tD).
dt
Proposition 6.5. Given A = P DP −1 as a diagonalizable matrix, YA = P Y (t)P −1 is
dY (t)
the solution of the k-differential equation
= AY (t).
dk t
Proof.
d
d
YA (t) = P
(Y (t))P −1 = P DY (t)p−1 ,
dk t
dk t
d
YA (t) = AP Y (t)P −1 = AYA (t).
but, A = P DP −1 implies AP = P D, then
dk t
Proposition 6.6. When A is a diagonalizable matrix where A = P DP −1 with D =
diag(λ1 , · · · , λn ) then YA = P Y (t)P −1 with Yk (t) = DY is solution of Yk (t) = AY ,
with Y (t) as in (6.16).
Proof.
d
d
YA =
(P Y (t)P −1 ) = P Yk (t)P −1 = P DY (t)P −1
dk t
dk t
= AP Y (t)P −1 = AYA (t)
Remark 6.7.
i) When D = In , YA = P expk (tIn )P −1 = expk (tA)
ii) When k → 0, YA = P exp(tD)P −1 = exp(tA) is solution of
dY (t)
= AY
dk t
1942
Ruthber Rodríguez Serrezuela et al.
Proposition 6.8. The k-differential linear system
dY
= AY + F (t), where F (t) =
dk t
diag (fi (t)) obtains as a general solution
Y (t) = YA (t)C + YA (t)
k
Remark 6.9.
i) when D = In then Y =
ektA C
+ ektA
ii) when k → 0; Y = etA C + etA
ek−SA F (S)dS
e−SA F (S)dS is solution of
Proposition 6.10. The system


a11 0 · · ·
a21 0 · · ·
Yk (t) = 
 ... ... . . .
am 0 · · ·
k
YA−1 (s)F (s)dk s.
0 
Y11 · · ·
0
 ...
.. 

.
Yn1 · · ·
0
dY
= AY + F (t).
dt


a11 Y11 a11 Y12 · · ·
a Y
a Y
···
 =  22. 11 21. 12 .
 ..
..
..
Ynn
an1 Y11 an1 Y12 · · ·
Y1n

a11 Y1n
a21 Y1n 
.. 
. 
an1 Ynn
obtains as solution

C11
···
 a21 C11 · · ·
 a11
Y (t) = 
..
 ...
.

an1
C11 · · ·
a11
C1n

a21
C1n 

a11
expk/a(11) (a11 t).
.. 
. 

an1
C1n
a11
Physical applications
→
A relativistic linear movement of a mass particle m0 at rest is moving with a speed of −
v
towards a reference system S
→
v
m0 −
−
→
→
= m−
v
P =
v2
1 − c2
m0
where m = .
1 − v 2 /c2
1
a particular mass of, γ = defines the Lorentz factor [9], [14].
1 − v 2 /c2
The relativistic moment is
−
→
→
P = m−
v γ.
The K-Exponential Matrix to solve systems of differential equations
1943
It is known that the relativistic force F exerted on a particle with the moment P is defined
as:
−
→
→
d(γ m0 −
v)
−
→ dP
=
F =
dt
dt d
m0 v
=
dt
1 − v 2 /c2
d =
m0 v(1 − v 2 /c2 )−1/2
dt
It is to be proven that the relativistic force according to Kaniadakis is:
dv
m0
−
→
2
2
F k = 1+k t
(1 − k 2 v 2 )3/2 dt
Effectively,
−
→
d dP
=
m0 v(1 − v 2 /c2 )−1/2
dt
dt
dv
−1
2 2 −1/2
2 2 −3/2 −2v dv
= m0 (1 − v /c )
(1 − v /c )
+ m0 v
dt
2
c2
dt
m0 dv/dt
v
m0 dv/dt
=
+
1 − v 2 /c2 (1 − v 2 /c2 )3/2 c2
m0 dv
dt
=
[1+
1 − v 2 /c2
1
v 2 /c2
if k =
2
2
1 − v /c
c
m0 dv
k2v2
dt
=√
1+
1 − k2v2
1 − k2v2
m0 dv
1
dt
=√
1 − k2v2 1 − k2v2
m0 dv
dt
=
(1 − k 2 v 2 )3/2
then
−
→
dP
−
→
2
2
F k = 1+k t
dt
dv
m
0
dt
= 1 + k2t 2
(1 − k 2 v 2 )3/2
dv
m0
−
→
2
2
F k = 1+k t
(1 − k 2 v 2 )3/2 dt
1944
Ruthber Rodríguez Serrezuela et al.
like
k
P (v1 ) k P (v2 )
P (v1 v2 )
=
m1
m2
m1
moreover, the kaniadrik hyperbolization is supposed to be expressed through the following form
k
h(x) = fk (x0 )(x x0 ) + f (x0 ).
Resolving the k-differential through this expression, the following result is obtained:
k
x x0 =
h(x) − f (x0 )
.
fk (x0 )
For the case of the relativistic linear moment P1 and P2 we obtain
h(P (v1 )/m1 ) − f (P2 (v2 )/m2 )
P (v1 ) k P (v2 )
=
m1
m2
fk (P2 (v2 )/m2 )
c
P (v1 v2 )
=
m1
for the identic function f (x) = x we have to:
2
2
fk (x0 ) = 1 + k x0 f (x0 ) = 1 + k 2 x02 .
Then
h(P (v1 )/m1 ) − P2 (v2 )/m2
h(P (v1 )/m1 ) − P2 (v2 )/m2
=
1 + k 2 (P2 (v2 )/m2 )2
1 + k 2 (v2 /1 − k 2 v22
h(P (v1 )/m1 ) − P2 (v2 )/m2
1/1 − k 2 v22
p(v1 )
p(v2 )
2
2
= 1 − k v2 h
−
m1
m
2
p(v2 )
p(v1 )
2
2
2
2
= 1 − k v2 h
− 1 − k v2
m1
m2


c
P (v1 v1 ) 
p(v1 )
× 1 − k 2 v22 h
− v2 =
m1
m1
=
The author(s) declare(s) that there is no conflict of interest regarding the publication of
this article.
The K-Exponential Matrix to solve systems of differential equations
1945
References
[1] G. Kaniadakis, Statistical mechanics in the context of special relativity II. Physical
Review E 72(3), 036108, 2005.
[2] Santos, A. P., Silva, R., Alcaniz, J. S., and Anselmo, D. H., Kaniadakis statistics
and the quantum H-theorem, Physics Letters A, 375(3), 352–355, 2011.
[3] Dora Esther Dcossa Casas, About exponential and logarithmic functions are deformed according Kaniadakis, EAFIT University master’s thesis. Pág. 8–9, 2011.
[4] Scarfone, A. M., Entropic forms and related algebras, Entropy, 15(2), 624–649,
2015.
[5] I.S. Gradshteyn and I.M. Ryzhil, Table of Integrals, Series and Product, Academic
Press, London, 2000.
[6] Kaniadakis, G., Non-linear kinetics underlying generalized statistics, Physica A:
Statistical Mechanics and its Applications, 296(3), 405–425, 2001.
[7] Clementi, F., Gallegati, M., and Kaniadakis, G., A I-generalized statistical mechanics approach to income analysis, Journal of Statistical Mechanics: Theory
and Experiment, 2009(02), P02037, 2009.
[8] Kaniadakis, G., and Scarfone, A. M., A new one-parameter deformation of the
exponential function, Physica A: Statistical Mechanics and its Applications, 305(1),
69–75, 2002.
[9] Abe, S., and Okamoto, Y., Nonextensive statistical mechanics and its applications,
Springer Science and Business Media., vol. 2001.
[10] Call, G. S., and Velleman, D. J., Pascal’s matrices, American Mathematical
Monthly, 372–376, 1993.
[11] Eliasson, L. H., Absolutely convergent series expansions for quasi periodic motions,
Stockholms Universitet. Matematiska Institutionen, 1998.
[12] Borges, E. P., A possible deformed algebra and calculus inspired in nonextensive
thermostatistics, Physica A: Statistical Mechanics and its Applications, 340(1),
95–101, 2004.
[13] Kaniadakis, G., Theoretical foundations and mathematical formalism of the powerlaw tailed statistical distributions, Entropy, 15(10), 3983–4010, 2013.
[14] Naudts, J., Deformed exponentials and logarithms in generalized thermostatistics,
Physica A: Statistical Mechanics and its Applications, 316(1), 323–334, 2002.
[15] Abe, S., A note on the q-deformation-theoretic aspect of the generalized entropies
in nonextensive physics, Physics Letters A, 224(6), 326–330, 1997.
[16] Borges, E. P., A possible deformed algebra and calculus inspired in nonextensive
thermostatistics, Physica A: Statistical Mechanics and its Applications, 340(1),
95–101, 2004.
[17] Umarov, S., Tsallis, C., and Steinberg, S., On a q-central limit theorem consistent
with nonextensive statistical mechanics, Milan journal of mathematics, 76(1), 307–
328, 2008.