# Solutions ofk-Hypergeometric Differential Equations

Authors: Shahid Mubeen; Mammona Naz; Abdur Rehman; Gauhar Rahman
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We solve the second-order linear differential equation called thek-hypergeometric differential equation by using Frobenius method around all its regular singularities. At each singularity, we find 8 solutions corresponding to the different cases for parameters and modified our solutions accordingly.


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## Body

## 1. Introduction

In 1769, Euler [1] formed the hypergeometric differential equation of the form
(1)z(1-z)w′′+[c-(a+b+1)z]w′-abw=0
which has three regular singular points: 0, 1, and ∞. The hypergeometric differential equation is a prototype: every ordinary differential equation of second-order with at most three regular singular points can be brought to the hypergeometric differential equation by means of a suitable change of variablesThe solutions of hypergeometric differential equation include many of the most interesting special functions of mathematical physics. Solutions to the hypergeometric differential equation are built out of the hypergeometric series. The solution of Euler’s hypergeometric differential equation is called hypergeometric function or Gaussian functionF12 introduced by Gauss [2].The equation has two linearly independent solutions at each of the three regular singular points0, 1, and ∞. Kummer [3] derived a set of 6 distinct solutions of hypergeometric differential equation. These include the hypergeometric function of Gauss and all of them could be expressed in terms of Gauss’s function. For more details on Kummer’s 24 solutions, see [4].Recently, Díaz et al. [5–7] have introduced Pochhammer k-symbol. They have introduced k-gamma and k-beta functions and proved a number of their properties. They have also studied k-zeta functions and k-hypergeometric functions based on Pochhammer k-symbols for factorial functions. In 2010, Kokologiannaki [8] and Krasniqi [9] followed the work of Diaz et al. and obtained some important results for the k-beta, k-gamma, k-hypergeometric, and k-zeta functions. Also, Krasniqi [10] gave a limit for the k-gamma and k-beta functions. In 2012, Mubeen and Habibullah [11, 12] introduced a variant of fractional integrals to be called k-fractional integral which was based on k-gamma function and gave its application. They also introduced an integral representation of some generalized confluent k-hypergeometric functions and k-hypergeometric functions by using the properties of Pochhammer k-symbols, k-gamma, and k-beta functions. Furthermore, in 2013, Mubeen [13] defined a second-order linear k-hypergeometric differential equation
(2)kz(1-kz)ω′′+[c-(a+b+k)kz]ω′-abω=0
having one solution in the form of k-hypergeometric function F1,k2.


## 2. Basic Concepts

Special functions are particular mathematical functions which have more or less established names and notations due to their importance in mathematical analysis, functional analysis, physics, or other applications. The solutions of hypergeometric differential equation include many of the most interesting special functions of mathematical physics. Solutions to the hypergeometric differential equation are built out of the hypergeometric series.Definition 1.
The Pochhammerk-symbol (a)n,k is defined as
(3)(a)n,k=a(a+k)(a+2k)⋯(a+(n-1)k)
and, for a≠0, (a)0,k=1, where k>0.Definition 2.
Thek-hypergeometric functions with three parameters a,b,c, two parameters a,b in the numerator and one parameter c in the denominator, are defined by
(4)2F1,k((a,k),(b,k);(c,k);z)=∑n=0∞‍(a)n,k(b)n,k(c)n,kznn!
for all a,b,c,c≠0,-1,-2,-3,…,|z|<1, where (a)n,k=(a)(a+k)(a+2k)⋯(a+(n-1)k);(a)0,k=1 and k>0.Definition 3.
In mathematics, the method of Frobenius [14], named after Ferdinand George Frobenius, is a method to find an infinite series solution for a second-order ordinary differential equation of the form
(5)P2(z)w′′+P1(z)w′+P0(z)w=0
about the regular singular point z0. After dividing this equation by P2(z), we obtain a differential equation of the form
(6)w′′+P1(z)P2(z)w′+P0(z)P2(z)w=0
which is not solvable with regular power series methods if either P1(z)/P2(z) or P0(z)/P2(z) is not analytic at z=z0. The Frobenius method enables us to obtain a power series solution to such a differential equation, provided that P1(z) and P0(z) are themselves analytic at z0 or, being analytic elsewhere, both their limits at z0 exist.


## 3. The Solutions of thek-Hypergeometric Differential Equation

In this section, we solve the following ordinary linear second-orderk-hypergeometric differential equation defined by Mubeen [13]
(7)kz(1-kz)w′′+[c-(a+b+k)kz]w′-abw=0
using Frobenius method. We usually use this method for complicated ordinary differential equations. This method is used to find an infinite series solution for a second-order ordinary differential equation about regular singular points of that equation. We prove that this equation has three regular singular points, namely, at z=0 and z=1/k and around ∞, and then we will be able to consider a solution in the form of a series.As this is a second-order differential equation, we must have two linearly independent solutions. The problem however will be that our assumed solutions may or may not be independent or worse may not be defined (depending on the values of the parameters of the equation). This is why we study the different cases for parameters and modify our assumed solutions accordingly.


### 3.1. Solution atz=0

Let(8)P0(z)=-ab,P1(z)=c-(a+b+k)kz,P2(z)=kz(1-kz).
Then P2(0)=0 and P2(1/k)=0.Hence,z=0 and z=1/k are singular points.Let us start withz=0.To see if it is regular, we study the following limits:(9)lim⁡z→z0(z-z0)P1(z)P2(z)=lim⁡z→0(z-0)(c-(a+b+k)kz)kz(1-kz)=ck,lim⁡z→z0(z-z0)2P0(z)P2(z)=lim⁡z→0(z-0)2(-ab)kz(1-kz)=0.
Hence, both limits exist and z=0 is a regular singular point.Therefore, we assume the solution of the form(10)w=∑n=0∞‍dnzn+β
with d0≠0.Hence(11)w′=∑n=0∞‍dn(n+β)zn+β-1,w′′=∑n=0∞‍dn(n+β)(n+β-1)zn+β-2.
By substituting these into the k-hypergeometric differential equation (7), we get
(12)k∑n=0∞‍dn(n+β)(n+β-1)zn+β-1-k2∑n=0∞‍dn(n+β)(n+β-1)zn+β+c∑n=0∞‍dn(n+β)zn+β-1-(a+b+k)k∑n=0∞‍dn(n+β)zn+β-ab∑n=0∞‍dnzn+β=0.
In order to simplify this equation, we need all powers of z to be the same, equal to n+β-1, the smallest power. Hence, we switch the indices as follows:
(13)k∑n=0∞‍dn(n+β)(n+β-1)zn+β-1-k2∑n=1∞‍dn-1(n+β-1)(n+β-2)zn+β-1+c∑n=0∞‍dn(n+β)zn+β-1-(a+b+k)k∑n=1∞‍dn-1(n+β-1)zn+β-1-ab∑n=1∞‍dn-1zn+β-1=0.
Thus isolating the first terms of the sums starting from 0, we get
(14)d0(kβ(β-1)+cβ)zβ-1+k∑n=1∞‍dn(n+β)(n+β-1)zn+β-1-k2∑n=1∞‍dn-1(n+β-1)(n+β-2)zn+β-1+c∑n=1∞‍dn(n+β)zn+β-1-(a+b+k)k∑n=1∞‍dn-1(n+β-1)zn+β-1-ab∑n=1∞‍dn-1zn+β-1=0.
Now, from the linear independence of all powers of z, that is, of the functions 1, z, z2, and so forth, the coefficients of zr vanish for all r. Hence, from the first term, we have
(15)d0(kβ(β-1)+cβ)=0
which is the indicial equation.Sinced0≠0, we have
(16)kβ(β-1)+cβ=0.
Hence, the solutions of the above indicial equation are given below:
(17)β1=0,β2=1-ck.
Also, from the rest of the terms, we have
(18)(n+β)[k(n+β-1)+c]dn=[k2(n+β-1)(n+β-2)+(a+b+k)k(n+β-1)+abk2]dn-1.
Hence, we get the following recurrence relation:
(19)dn=(a+k(n+β-1))(b+k(n+β-1))(n+β)(c+k(n+β-1))dn-1
for n≥1.Let us now simplify this relation by givingdn in terms of d0 instead of dn-1.From the recurrence relation, we have the following:(20)dn=(a+βk)n,k(b+βk)n,k(c+βk)n,k(β+1)nd0
for n≥1.Hence, our assumed solution takes the form(21)w=d0∑n=0∞‍(a+βk)n,k(b+βk)n,k(c+βk)n,kzn+β(β+1)n.


### 3.2. Analysis of the Solutions in terms of the Difference “(c/k)-1” of the Two Roots

Now, we study the solutions corresponding to the different cases for the expressionβ1-β2=(c/k)-1 (this reduces to studying the nature of the parameter c/k whether it is an integer or not).Case 1 (“c/k” not an integer).
Letc/k be not an integer. Then
(22)w1=w|β=0,w2=w|β=1-(c/k).
Since
(23)w=d0∑n=0∞‍(a+βk)n,k(b+βk)n,k(c+βk)n,kzn+β(β+1)n,
therefore, we have
(24)w1=d0∑n=0∞‍(a)n,k(b)n,k(c)n,kzn(1)nw1=d0F1,k2((a,k),(b,k);(c,k);z),w2=d0∑n=0∞‍(a+(1-(c/k))k)n,k(b+(1-(c/k))k)n,k(c+(1-(c/k))k)n,k×zn+1-(c/k)(2-(c/k))n=d0z1-(c/k)F1,k2×((a+k-c,k),(b+k-c,k);(2k-c,k);z).
Hence
(25)w=A′w1+B′w2.
Let
(26)A′d0=A,B′d0=B.
Then
(27)w=AF1,k2((a,k),(b,k);(c,k);z)+Bz1-(c/k)F21,k×((a+k-c,k),(b+k-c,k);(2k-c,k);z).Case 2 (“c/k=1” (i.e., c=k)).
Letc =k. Then
(28)w1=w|β=0.
Since we have
(29)w=d0∑n=0∞‍(a+βk)n,k(b+βk)n,k(c+βk)n,kzn+β(β+1)n,
using c=k, we get
(30)w=d0∑n=0∞‍(a+βk)n,k(b+βk)n,k((β+1)k)n,kzn+β(β+1)n.
Hence,
(31)w1=d0∑n=0∞‍(a)n,k(b)n,k(k)n,kzn(1)nw1=d0F1,k2((a,k),(b,k);(k,k);z),(32)w2=∂w∂β|β=0.
To calculate this derivative, let
(33)Mn=(a+βk)n,k(b+βk)n,k((β+1)k)n,k(β+1)n=(a+βk)n,k(b+βk)n,kkn(β+1)n2.
Then
(34)ln⁡(Mn)=ln⁡(a+βk)n,k(b+βk)n,kkn(β+1)n2=ln⁡(a+βk)n,k+ln⁡(b+βk)n,k-2knln⁡(β+1)n.
Since
(35)ln⁡(a+βk)n,k=ln⁡[(a+βk)(a+(β+1)k)⋯(a+(β+n-1)k)]=∑j=0n-1‍ln⁡(a+(β+j)k),
therefore
(36)ln⁡(Mn)=∑j=0n-1‍[ln⁡(a+(β+j)k)+ln⁡(b+(β+j)k)-2knln⁡(1+β+j)].
Differentiating both sides of the equation with respect to β, we get
(37)∂Mn∂β=(a+βk)n,k(b+βk)n,kkn(β+1)n2×∑j=0n-1‍[ka+(β+j)k+kb+(β+j)k-2kn1+β+j].
Since
(38)w=d0zβ∑n=0∞‍Mnzn,
therefore
(39)∂w∂β=d0zβ∑n=0∞‍(a+βk)n,k(b+βk)n,kkn(β+1)n2×[ln⁡z+∑j=0n-1(-2kn1+β+jka+(β+j)k+kb+(β+j)k‍∑j=0n-1222-2kn1+β+j)]zn.
For β=0, we get
(40)w2=d0∑n=0∞‍(a)n,k(b)n,k(k)n,k×[ln⁡z+∑j=0n-1‍(ka+jk+kb+jk∑j=0n-1-2kn1+j∑j=0n-1)]zn(1)n.
Hence,
(41)w=C′w1+D′w2.
Let
(42)C′d0=C,D′d0=D.
Then
(43)w=CF1,k2((a,k),(b,k);(k,k);z)+D∑n=0∞‍(a)n,k(b)n,k(k)n,k×[ln⁡z+∑j=0n-1‍(ka+jk+kb+jk-2kn1+j)]zn(1)n.Case 3 (“c/k” an integer and “c/k≠1”).
Here, we discuss the further two cases:
(i) “(c/k)<1.” Let (c/k)<1. Then from the recurrence relation
(44)dn=(a+(n+β-1)k)(b+(n+β-1)k)(n+β)(c+(n+β-1)k)dn-1,
we see that, when β=0 (the smaller root), d1-(c/k)→∞. Thus, we must make the substitution
(45)d0=g0(β-βi)k,
where βi is the root for which our solution is infinite.
Therefore, we take(46)d0=g0βk
and our assumed solution in equation (29) takes the new form
(47)wg=g0zβ∑n=0∞‍(βk)(a+βk)n,k(b+βk)n,k(c+βk)n,k(β+1)nzn.
Then
(48)w1=wg|β=0.
As we can see, all terms before
(49)(βk)(a+βk)1-(c/k),k(b+βk)1-(c/k),k(β+1)1-(c/k)(c+βk)1-(c/k),kz1-(c/k)
vanish because of the βk in the numerator.
Starting from this term, however, theβk in the numerator vanishes. To see this, note that
(50)(c+βk)1-(c/k),k=(c+βk)(c+(β+1)k)⋯(βk).
Hence, our assumed solution takes the form
(51)w1=g0(c)-(c/k),k∑n=1-(c/k)∞‍(a)n,k(b)n,k(1)n(k)n+(c/k)-1,kzj.
Now
(52)w2=∂wg∂β|β=1-(c/k).
To calculate this derivative, let
(53)Mn=(βk)(a+βk)n,k(b+βk)n,k(β+1)n(c+βk)n,k.
Then following the method in the previous case c/k=1, we get
(54)∂Mn∂β=Mn[1β+∑j=0n-1‍(ka+(β+j)k+kb+(β+j)k-1β+1+j-kc+(β+j)k)∑j=0n-1‍].
Since
(55)wg=g0zβ∑n=0∞‍Mnzn,
therefore
(56)∂wg∂β=g0zβ∑n=0∞‍(βk)(a+βk)n,k(b+βk)n,k(β+1)n(c+βk)n,k×[ln⁡z+1β+∑j=0j=n-1‍(ka+(β+j)k+kb+(β+j)k-1β+1+j-kc+(β+j)k)∑j=0j=n-1‍]zn.
At β=1-(c/k), we get
(57)w2=g0(k-c)z1-(c/k)×∑n=0∞‍(a+k-c)n,k(b+k-c)n,k(2k-c)n,k×[-1(1+j)k)ln⁡z+kk-c+k∑j=0n-1‍(1(a+k-c)+jk+1(b+k-c)+jk-1(2k-c)+jk-1(1+j)k)]znn!.
Hence
(58)w=E′w1+F′w2.
Let
(59)E′g0=E,F′g0=F.
Then
(60)w=E(c)-(c/k),k∑j=1-(c/k)∞‍(a)j,k(b)j,k(k)j+(c/k)-1,kzj(1)j+F(k-c)z1-(c/k)×∑n=0∞‍(a+k-c)n,k(b+k-c)n,k(2k-c)n,k×[ln⁡z+kk-c∑j=0n-1‍+k∑j=0n-1‍(1(a+k-c)+jk1(j)+1(b+k-c)+jk-1(2k-c)+jk-1(1+j)k)∑j=0n-1‍]znn!.
(ii) “(c/k)>1”. Let (c/k)>1. Then, from the recurrence relation
(61)dn=(a+(β+n-1)k)(b+(β+n-1)k)(n+β)(c+(β+n-1)k)dn-1,
we see that, when β=1-(c/k) (the smaller root), d(c/k)-1→∞. Thus, we must make the substitution
(62)d0=g0(β-βi)k,
where βi is the root for which our solution is infinite.
Hence we take(63)d0=g0(β+ck-1)k
and our assumed solution takes the new form
(64)wg=g0zβ∑n=0∞‍((β+(c/k)-1)k)(a+βk)n,k(b+βk)n,k(β+1)n(c+βk)n,kzn.
Then
(65)w1=wg|β=1-(c/k).
As we can see, all terms before
(66)((β+(c/k)-1)k)(a+βk)(c/k)-1,k(b+βk)(c/k)-1,k(β+1)(c/k)-1(c+βk)(c/k)-1,kz(c/k)-1
vanish because of the “β+(c/k)-1” in the numerator.
Starting from this term, however, the “β+(c/k)-1” in the numerator vanishes. To see this, note that
(67)(β+1)(c/k)-1=(β+1)(β+2)⋯(β+ck-1).
Hence, our solution takes the form
(68)w1=g0z1-(c/k)(2k-c)(c/k)-2,k∑n=(c/k)-1∞‍(a+k-c)n,k(b+k-c)n,k(k)n+1-(c/k),kzn(1)n.
Now
(69)w2=∂wg∂β|β=0.
To calculate this derivative, let
(70)Mn=((β+(c/k)-1)k)(a+βk)n,k(b+βk)n,k(β+1)n(c+βk)n,k.
Then following the method in the previous case (c/k)<1, we get
(71)∂Mn∂β=Mn[1β+(c/k)-1+∑j=0n-1‍(ka+(β+j)k+kb+(β+j)k-1β+1+j-kc+(β+j)k)],∂wg∂β=g0zβ∑n=0∞‍((β+(c/k)-1)k)(a+βk)n,k(b+βk)n,k(β+1)n(c+βk)n,k×[ln⁡z+1β+(c/k)-1∑j=0n-1‍+∑j=0n-1‍(ka+(β+j)k+kb+(β+j)k∑j=0n-1‍-1β+1+j-kc+(β+j)k)]zn.
At β=0, we get
(72)w2=g0(c-k)∑n=0∞‍(a)n,k(b)n,k(c)n,k×[ln⁡z+kc-kkj+∑j=0n-1‍(ka+jk+kb+jk-11+j-kc+jk)]znn!.
Hence
(73)w=G′w1+H′w2.
Let
(74)G′g0=G,H′g0=H.
Then
(75)w=G(2k-c)(c/k)-2,kz1-(c/k)×∑j=(c/k)-1∞‍(a+k-c)j,k(b+k-c)j,k(k)j+1-(c/k),k×zj(1)j+H(c-k)×∑n=0∞‍(a)n,k(b)n,k(c)n,k×∑n=0∞‍×[∑j=0n-1‍ln⁡z+kc-k+∑j=0n-1‍(ka+jk+kb+jk-11+j-kc+jk)∑j=0n-1‍]znn!.


### 3.3. Solution atz=1/k

Let us now study the singular pointz=1/k.To see if it is regular, we study the following limits:(76)lim⁡z→z0(z-z0)P1(z)P2(z)=lim⁡z→1/k(z-(1/k))(c-(a+b+k)kz)kz(1-kz)=c-(a+b+k)k,lim⁡z→z0(z-z0)2P0(z)P2(z)=lim⁡z→1/k(z-(1/k))2(-ab)kz(1-kz)=0.
As both limits exist, therefore z=1/k is a regular singular point.Now instead of assuming a solution in the form(77)w=∑n=0∞‍dn(z-1k)n+β,
we will try to express the solutions of this case in terms of the solutions for the point z=0. We proceed as follows.We have thek-hypergeometric differential equation of the form
(78)kz(1-kz)w′′+[c-(a+b+k)kz]w′-abw=0.
Let y=(1/k)-z. Then
(79)dwdz=dwdydydz=-dwdy=-w`,d2wdz2=ddz(dwdz)=ddz(-dwdy)=ddy(-dwdy)dydz=d2wdy2=w``.
Hence, k-hypergeometric differential equation (7) takes the form
(80)ky(1-ky)w``+[a+b-c+k-(a+b+k)ky]w`-abw=0.
Since y=(1/k)-z, the solution of the k-hypergeometric differential equation at z=1/k is the same as the solution for this equation at y=0. But the solution at y=0 is identical to the solution we obtained for the point z=0 if we replace c by a+b-c+k.Hence, to get the solutions, we just make the substitution in the previous results.Note also that, forz=0,
(81)β1=0,β2=1-ck.
Hence in our case,
(82)β1=0,β2=c-a-bk.


### 3.4. Analysis of the Solutions in terms of the Difference “(c-a-b)/k” of the Two Roots

Let us now find out the solutions. In the following we replace eachy by (1/k)-z.Case 1 (“(c-a-b)/k” not an integer).
Let(c-a-b)/k be not an integer. Then
(83)w=AF21,k((a,k),(b,k);(a+b-c+k,k);1k-z)+B(1k-z)(c-a-b)/k×F21,k((c-a,k),(c-b,k);(c-a-b+k,k);1k-z).Case 2 (“(c-a-b)/k=0”).
Let(c-a-b)/k=0. Then
(84)w=CF21,k((a,k),(b,k);(k,k);1k-z)+D∑n=0∞‍(a)n,k(b)n,k(k)n,k×[ln⁡(1k-z)∑j=0n-1‍+∑j=0n-1‍(ka+jk+kb+jk-2kn1+j)]×((1/k)-z)nn!.Case 3 (“(c-a-b)/k” an integer and “(c-a-b)/k≠0”).
Here, we discuss further two cases:
(i) “(c-a-b)/k>0.” Let (c-a-b)/k>0. Then
(85)w=E(a+b-c+k)(c-a-b)/k,k×∑n=(c-a-b)/k∞‍(a)n,k(b)n,k(k)n+((a+b-c)/k),k((1/k)-z)n(1)n+F(c-a-b)×(1k-z)(c-a-b)/k×∑n=0∞‍(c-a)n,k(c-b)n,k(c-a-b+k)n,k×[∑ln⁡(1k-z)+kc-a-b+k∑j=0n-1‍(1(1+j)k1(c-a)+jk+1(c-b)+jk-1(c-a-b+k)+jk-1(1+j)k)∑]((1/k)-z)nn!.
(ii) “(c-a-b)/k<0.” Let (c-a-b)/k<0. Then
(86)w=G(c-a-b+k)((a+b-c)/k)-1,k(1k-z)(c-a-b)/k×∑n=((a+b-c)/k)∞‍(c-a)n,k(c-b)n,k(k)n+((c-a-b)/k),k×((1/k)-z)n(1)n+H(a+b-c)×∑n=0∞‍(a)n,k(b)n,k(a+b-c+k)n,k×[ln⁡⁡(1k-z)+ka+b-c+∑j=0n-1‍(ka+jk+kb+jk-11+j-k(a+b-c+k)+jk)]×((1/k)-z)nn!.


### 3.5. Solution Around “∞”

Finally we study the singularity asz→∞. Since we cannot study this directly, therefore we let z=1/ks; then the solution of the equation as z→∞ is identical to the solution of the modified equation when s=0.We have thek-hypergeometric differential equation
(87)kz(1-kz)w′′+[c-(a+b+k)kz]w′-abw=0,dwdz=dwdsdsdz=-ks2dwds=-ks2w·,d2wdz2=k2(2s3w·+s4w··).
Hence the k-hypergeometric differential equation (7) takes the new form
(88)k2(s3-s2)w··+k[(2k-c)s2+(a+b-k)s]w·-abw=0.
Let
(89)P0(s)=-abP1(s)=k[(2k-c)s2+(a+b-k)s],P2(s)=k2(s3-s2).
Here, we only study the solutions when s=0. As we can see that this is a singular point because P2(0)=0.Let us now see thats=0 is a regular singular point, for this
(90)lim⁡s→s0(s-s0)P1(s)P2(s)=lim⁡s→0(s-0)k[(2k-c)s2+(a+b-k)s]k2(s3-s2)=k-a-bk,lim⁡s→s0(s-s0)2P0(s)P2(s)=lim⁡s→0(s-0)2(-ab)k2(s3-s2)=ab.
As both limits exist, therefore s=0 is a regular singular point.Thus, we assume the solution of the form(91)w=∑n=0∞‍dnsn+β
with d0≠0.Hence, we set the following:(92)w·=∑n=0∞‍dn(n+β)sn+β-1,w··=∑n=0∞‍dn(n+β)(n+β-1)sn+β-2.
By substituting these into the modified k-hypergeometric differential equation (88), we get
(93)k2∑n=0∞‍dn(n+β)(n+β-1)sn+β+1-k2∑n=0∞‍dn(n+β)(n+β-1)sn+β+k(2k-c)∑n=0∞‍dn(n+β)sn+β+1+k(a+b-k)∑n=0∞‍dn(n+β)sn+β-ab∑n=0∞‍dnsn+β=0.
In order to simplify this equation, we need all powers of s to be the same, equal to n+β, the smallest power. Hence, we switch the indices as follows:
(94)k2∑n=1∞‍dn-1(n+β-1)(n+β-2)sn+β-k2∑n=0∞‍dn(n+β)(n+β-1)sn+β+k(2k-c)∑n=1∞‍dn-1(n+β-1)sn+β+k(a+b-k)∑n=0∞‍dn(n+β)sn+β-ab∑n=0∞‍dnsn+β=0.
Thus, by isolating the first terms of the sums starting from 0, we get
(95)d0[-k2β(β-1)+k(a+b-k)β-ab]sβ+k2∑n=1∞‍dn-1(n+β-1)(n+β-2)sn+β-k2∑n=1∞‍dn(n+β)(n+β-1)sn+β+k(2k-c)∑n=1∞‍dn-1(n+β-1)sn+β+k(a+b-k)∑n=1∞‍dn(n+β)sn+β-ab∑n=1∞‍dnsn+β=0.
Now from the linear independence of all powers of s, that is, of the functions 1, s, s2, and so forth, the coefficients of sr vanish for all r. Hence, from the first term, we have
(96)d0[-k2β(β-1)+k(a+b-k)β-ab]=0
which is the indicial equation.Sinced0≠0, we have
(97)-k2β(β-1)+k(a+b-k)β-ab=0.
Hence, we get two solutions of this indicial equation:
(98)β1=ak,β2=bk.
Also, from the rest of the terms we have
(99)[k2(n+β-1)(n+β-2)+k2k(2k-c)(n+β-1)]dn-1+[-k2(n+β)(n+β-1)+k2k(a+b-k)(n+β)-ab]dn=0.
Hence, we get the recurrence relation of the following form:
(100)dn=(k(n+β)-c)k(n+β-1)(k(n+β)-a)(k(n+β)-b)dn-1
for n≥1.Let us now simplify this relation by givingdn in terms of d0 instead of dn-1.From the recurrence relation, we have(101)dn=(kβ)n,k(k(β+1)-c)n,k(k(β+1)-a)n,k(k(β+1)-b)n,kd0
for n≥1.Hence, our assumed solution in equation (91) takes the form
(102)w=d0∑n=0∞‍(kβ)n,k(k(β+1)-c)n,k(k(β+1)-a)n,k(k(β+1)-b)n,ksn+β.


### 3.6. Analysis of the Solutions in terms of the Difference “(a/k)-(b/k)” of the Two Roots

Now, we study the solutions corresponding to the different cases for the expressionβ1-β2=(a/k)-(b/k) (this reduces to studying the nature of the parameter (a/k)-(b/k) whether it is an integer or not).Case 1 (“(a/k)-(b/k)” not an integer).
Let(a/k)-(b/k) be not an integer. Then
(103)w1=w|β=a/k,w2=w|β=b/k.
Since
(104)w=d0∑n=0∞‍(kβ)n,k(k(β+1)-c)n,k(k(β+1)-a)n,k(k(β+1)-b)n,ksn+β,
therefore, we have
(105)w1=d0(kz)-(a/k)∑n=0∞‍(a)n,k(a+k-c)n,k(a+k-b)n,k1(k2z)n(1)n=d0(kz)-(a/k)×F21,k((a,k),(a+k-c,k);(a+k-b,k);1k2z),w2=d0(kz)-(b/k)×F21,k((b,k),(b+k-c,k);(b+k-a,k);1k2z).
Hence
(106)w=A′w1+B′w2.
Let
(107)A′d0=A,B′d0=B.
Then
(108)w=A(kz)-(a/k)×F21,k((a,k),(a+k-c,k);(a+k-b,k);1k2z)+B(kz)-(b/k)×F21,k((b,k),(b+k-c,k);(b+k-a,k);1k2z).Case 2 (“(a/k)-(b/k)=0”).
Let(a/k)-(b/k)=0. Then
(109)w1=w|β=a/k.
Also, we have
(110)w=d0∑n=0∞‍(kβ)n,k(k(β+1)-c)n,k(k(β+1)-a)n,k(k(β+1)-b)n,ksn+β
which can be written as
(111)w=d0∑n=0∞‍(β)n(β+1-(c/k))n(β+1-(a/k))n(β+1-(b/k))nsn+β.
Since we have (a/k)=(b/k), therefore
(112)w1=d0∑n=0∞‍(β)n(β+1-(c/k))n(β+1-(a/k))n2sn+β.
At β=a/k,
(113)w1=d0(kz)-(a/k)×F21,k((a,k),(a+k-c,k);(k,k);1k2z),w2=∂w∂β|β=a/k.
To calculate this derivative, let
(114)Mn=(β)n(β+1-(c/k))n(β+1-(a/k))n2,
and then we get
(115)∂Mn∂β=(β)n(β+1-(c/k))n(β+1-(a/k))n2×∑j=0n-1‍[1β+j+1β+1-(c/k)+j-2β+1-(a/k)+j],∂w∂β=d0sβ∑n=0∞‍(β)n(β+1-(c/k))n(β+1-(a/k))n2×[ln⁡s+∑j=0n-1‍(1β+j+1β+1-(c/k)+j-2β+1-(a/k)+j)∑j=0n-1]sn.
For β=a/k, we get
(116)w2=d0(kz)-(a/k)×∑n=0∞‍(a)n,k(a+k-c)n,kkn(k)n,k×[ln⁡(kz)-1∑j=0n-1‍+∑j=0n-1‍(ka+jk+ka+k-c+jk-21+j)]×(kz)-nn!.
Hence
(117)w=C′w1+D′w2.
Let
(118)C′d0=C,D′d0=D.
Then
(119)w=C(kz)-(a/k)∑n=0∞‍(a)n,k(a+k-c)n,kkn(k)n,k(kz)-nn!+D(kz)-(a/k)×∑n=0∞‍(a)n,k(a+k-c)n,kkn(k)n,k×[ln⁡⁡(kz)-1∑j=0n-1‍+∑j=0n-1‍(ka+jk+ka+k-c+jk-21+j)]×(kz)-nn!.Case 3 (“(a/k)-(b/k)” an integer and “(a/k)-(b/k)≠0”).
Here, we have further two cases:
(i) “(a/k)-(b/k)>0.” Let (a/k)-(b/k)>0.
Then, from the recurrence relation(120)dn=(k(n+β)-c)(k(n+β-1))(k(n+β)-a)(k(n+β)-b)dn-1,
we see that, when β=b/k (the smaller root), d(a/k)-(b/k)→∞. Thus, we must make the substitution
(121)d0=g0(β-βi)k,
where βi is the root for which our solution is infinite.
Hence we take(122)d0=g0(β-bk)k
and our assumed solution in equation (110) takes the new form
(123)wg=g0sβ∑n=0∞‍((β-(b/k))k)(k(β+1)-c)n,k(βk)n,k(k(β+1)-a)n,k(k(β+1)-b)n,ksn.
Then
(124)w1=wg|β=b/k.
As we can see, all terms before
(125)((β-(b/k))k)(βk)(a/k)-(b/k),k(k(β+1)-c)(a/k)-(b/k),k(k(β+1)-a)(a/k)-(b/k),k(k(β+1)-b)(a/k)-(b/k),k×s(a/k)-(b/k)
vanish because of the (β-(b/k)) in the numerator.
Starting from this term, however, the(β-(b/k)) in the numerator vanishes. To see this, note that
(126)(β+1-ak)(a/k)-(b/k)=(β+1-ak)(β+2-ak)⋯(β-bk).
Hence, our solution takes the form
(127)w1=g0(b+k-a)(a/k)-(b/k)-1,k×∑n=(a/k)-(b/k)∞‍(b)n,k(b+k-c)n,kkn+(b/k)-(a/k)(1)n+(b/k)-(a/k)sn(1)n=g0(b+k-a)(a/k)-(b/k)-1,k×∑n=(a/k)-(b/k)∞‍(b)n,k(b+k-c)n,k(k)n+(b/k)-(a/k),k(kz)-n(1)n.
Now
(128)w2=∂wg∂β|β=a/k.
To calculate this derivative, let
(129)Mn=((β-(b/k))k)(β)n(β+1-(c/k))n(β+1-(a/k))n(β+1-(b/k))n,
and then we get
(130)∂Mn∂β=Mn[1β-(b/k)+∑j=0n-1‍(1β+1-(c/k)+j+1β+j-1β+1-(a/k)+j-1β+1-(b/k)+j)],∂wg∂β=g0sβ∑n=0∞‍(β-(b/k))k(β)n(β+1-(c/k))n(β+1-(a/k))n(β+1-(b/k))n×[ln⁡s+1β-(b/k)+∑j=0n-1‍(1β+j+1β+1-(c/k)+j-1β+1-(a/k)+j-1β+1-(b/k)+j)]sn.
At β=a/k, we get
(131)w2=g0sa/k∑n=0∞‍(a-b)(a)n,k(a+k-c)n,k(a+k-b)n,k(k)n,k×[ln⁡s+ka-b+∑j=0n-1‍(ka+jk+ka+k-c+jk-ka+k-b+jk-11+j)]sn.
Replacing s by (kz)-1, we get w2 in terms of z:
(132)w2=g0(kz)-(a/k)∑n=0∞‍(a-b)(a)n,k(a+k-c)n,k(a+k-b)n,k×[ln⁡(kz)-1+ka-b+∑j=0n-1‍(ka+jk+ka+k-c+jk-11+j-ka+k-b+jk)]×(kz)-n(k)n,k.
Let
(133)w=E′w1+F′w2.
Let
(134)E′g0=E,F′g0=F.
Then
(135)w=E(b+k-a)(a/k)-(b/k)-1,k×∑n=(a/k)-(b/k)∞‍(b)n,k(b+k-c)n,k(k)n+(b/k)-(a/k),k(kz)-n(1)n+F(kz)-(a/k)∑n=0∞‍(a-b)(a)n,k(a+k-c)n,k(a+k-b)n,k×[ln⁡(kz)-1+ka-b+∑j=0n-1‍(ka+jk+ka+k-c+jk-11+j-ka+k-b+jk)]×(kz)-n(k)n,k.
(ii) “(a/k)-(b/k)<0.” Let (a/k)-(b/k)<0.
Then, from the symmetry of the situation here, we see that(136)w=G(a+k-b)(b/k)-(a/k)-1,k×∑n=(b/k)-(a/k)∞‍(a)n,k(a+k-c)n,k(k)n+(a/k)-(b/k),k(kz)-n(1)n+H(kz)-(b/k)∑n=0∞‍(b-a)(b)n,k(b+k-c)n,k(b+k-a)n,k×[ln⁡(kz)-1+kb-a+∑j=0n-1‍(kb+jk+kb+k-c+jk-11+j-kb+k-a+jk)]×(kz)-n(k)n,k.


## 3.1. Solution atz=0

Let(8)P0(z)=-ab,P1(z)=c-(a+b+k)kz,P2(z)=kz(1-kz).
Then P2(0)=0 and P2(1/k)=0.Hence,z=0 and z=1/k are singular points.Let us start withz=0.To see if it is regular, we study the following limits:(9)lim⁡z→z0(z-z0)P1(z)P2(z)=lim⁡z→0(z-0)(c-(a+b+k)kz)kz(1-kz)=ck,lim⁡z→z0(z-z0)2P0(z)P2(z)=lim⁡z→0(z-0)2(-ab)kz(1-kz)=0.
Hence, both limits exist and z=0 is a regular singular point.Therefore, we assume the solution of the form(10)w=∑n=0∞‍dnzn+β
with d0≠0.Hence(11)w′=∑n=0∞‍dn(n+β)zn+β-1,w′′=∑n=0∞‍dn(n+β)(n+β-1)zn+β-2.
By substituting these into the k-hypergeometric differential equation (7), we get
(12)k∑n=0∞‍dn(n+β)(n+β-1)zn+β-1-k2∑n=0∞‍dn(n+β)(n+β-1)zn+β+c∑n=0∞‍dn(n+β)zn+β-1-(a+b+k)k∑n=0∞‍dn(n+β)zn+β-ab∑n=0∞‍dnzn+β=0.
In order to simplify this equation, we need all powers of z to be the same, equal to n+β-1, the smallest power. Hence, we switch the indices as follows:
(13)k∑n=0∞‍dn(n+β)(n+β-1)zn+β-1-k2∑n=1∞‍dn-1(n+β-1)(n+β-2)zn+β-1+c∑n=0∞‍dn(n+β)zn+β-1-(a+b+k)k∑n=1∞‍dn-1(n+β-1)zn+β-1-ab∑n=1∞‍dn-1zn+β-1=0.
Thus isolating the first terms of the sums starting from 0, we get
(14)d0(kβ(β-1)+cβ)zβ-1+k∑n=1∞‍dn(n+β)(n+β-1)zn+β-1-k2∑n=1∞‍dn-1(n+β-1)(n+β-2)zn+β-1+c∑n=1∞‍dn(n+β)zn+β-1-(a+b+k)k∑n=1∞‍dn-1(n+β-1)zn+β-1-ab∑n=1∞‍dn-1zn+β-1=0.
Now, from the linear independence of all powers of z, that is, of the functions 1, z, z2, and so forth, the coefficients of zr vanish for all r. Hence, from the first term, we have
(15)d0(kβ(β-1)+cβ)=0
which is the indicial equation.Sinced0≠0, we have
(16)kβ(β-1)+cβ=0.
Hence, the solutions of the above indicial equation are given below:
(17)β1=0,β2=1-ck.
Also, from the rest of the terms, we have
(18)(n+β)[k(n+β-1)+c]dn=[k2(n+β-1)(n+β-2)+(a+b+k)k(n+β-1)+abk2]dn-1.
Hence, we get the following recurrence relation:
(19)dn=(a+k(n+β-1))(b+k(n+β-1))(n+β)(c+k(n+β-1))dn-1
for n≥1.Let us now simplify this relation by givingdn in terms of d0 instead of dn-1.From the recurrence relation, we have the following:(20)dn=(a+βk)n,k(b+βk)n,k(c+βk)n,k(β+1)nd0
for n≥1.Hence, our assumed solution takes the form(21)w=d0∑n=0∞‍(a+βk)n,k(b+βk)n,k(c+βk)n,kzn+β(β+1)n.


## 3.2. Analysis of the Solutions in terms of the Difference “(c/k)-1” of the Two Roots

Now, we study the solutions corresponding to the different cases for the expressionβ1-β2=(c/k)-1 (this reduces to studying the nature of the parameter c/k whether it is an integer or not).Case 1 (“c/k” not an integer).
Letc/k be not an integer. Then
(22)w1=w|β=0,w2=w|β=1-(c/k).
Since
(23)w=d0∑n=0∞‍(a+βk)n,k(b+βk)n,k(c+βk)n,kzn+β(β+1)n,
therefore, we have
(24)w1=d0∑n=0∞‍(a)n,k(b)n,k(c)n,kzn(1)nw1=d0F1,k2((a,k),(b,k);(c,k);z),w2=d0∑n=0∞‍(a+(1-(c/k))k)n,k(b+(1-(c/k))k)n,k(c+(1-(c/k))k)n,k×zn+1-(c/k)(2-(c/k))n=d0z1-(c/k)F1,k2×((a+k-c,k),(b+k-c,k);(2k-c,k);z).
Hence
(25)w=A′w1+B′w2.
Let
(26)A′d0=A,B′d0=B.
Then
(27)w=AF1,k2((a,k),(b,k);(c,k);z)+Bz1-(c/k)F21,k×((a+k-c,k),(b+k-c,k);(2k-c,k);z).Case 2 (“c/k=1” (i.e., c=k)).
Letc =k. Then
(28)w1=w|β=0.
Since we have
(29)w=d0∑n=0∞‍(a+βk)n,k(b+βk)n,k(c+βk)n,kzn+β(β+1)n,
using c=k, we get
(30)w=d0∑n=0∞‍(a+βk)n,k(b+βk)n,k((β+1)k)n,kzn+β(β+1)n.
Hence,
(31)w1=d0∑n=0∞‍(a)n,k(b)n,k(k)n,kzn(1)nw1=d0F1,k2((a,k),(b,k);(k,k);z),(32)w2=∂w∂β|β=0.
To calculate this derivative, let
(33)Mn=(a+βk)n,k(b+βk)n,k((β+1)k)n,k(β+1)n=(a+βk)n,k(b+βk)n,kkn(β+1)n2.
Then
(34)ln⁡(Mn)=ln⁡(a+βk)n,k(b+βk)n,kkn(β+1)n2=ln⁡(a+βk)n,k+ln⁡(b+βk)n,k-2knln⁡(β+1)n.
Since
(35)ln⁡(a+βk)n,k=ln⁡[(a+βk)(a+(β+1)k)⋯(a+(β+n-1)k)]=∑j=0n-1‍ln⁡(a+(β+j)k),
therefore
(36)ln⁡(Mn)=∑j=0n-1‍[ln⁡(a+(β+j)k)+ln⁡(b+(β+j)k)-2knln⁡(1+β+j)].
Differentiating both sides of the equation with respect to β, we get
(37)∂Mn∂β=(a+βk)n,k(b+βk)n,kkn(β+1)n2×∑j=0n-1‍[ka+(β+j)k+kb+(β+j)k-2kn1+β+j].
Since
(38)w=d0zβ∑n=0∞‍Mnzn,
therefore
(39)∂w∂β=d0zβ∑n=0∞‍(a+βk)n,k(b+βk)n,kkn(β+1)n2×[ln⁡z+∑j=0n-1(-2kn1+β+jka+(β+j)k+kb+(β+j)k‍∑j=0n-1222-2kn1+β+j)]zn.
For β=0, we get
(40)w2=d0∑n=0∞‍(a)n,k(b)n,k(k)n,k×[ln⁡z+∑j=0n-1‍(ka+jk+kb+jk∑j=0n-1-2kn1+j∑j=0n-1)]zn(1)n.
Hence,
(41)w=C′w1+D′w2.
Let
(42)C′d0=C,D′d0=D.
Then
(43)w=CF1,k2((a,k),(b,k);(k,k);z)+D∑n=0∞‍(a)n,k(b)n,k(k)n,k×[ln⁡z+∑j=0n-1‍(ka+jk+kb+jk-2kn1+j)]zn(1)n.Case 3 (“c/k” an integer and “c/k≠1”).
Here, we discuss the further two cases:
(i) “(c/k)<1.” Let (c/k)<1. Then from the recurrence relation
(44)dn=(a+(n+β-1)k)(b+(n+β-1)k)(n+β)(c+(n+β-1)k)dn-1,
we see that, when β=0 (the smaller root), d1-(c/k)→∞. Thus, we must make the substitution
(45)d0=g0(β-βi)k,
where βi is the root for which our solution is infinite.
Therefore, we take(46)d0=g0βk
and our assumed solution in equation (29) takes the new form
(47)wg=g0zβ∑n=0∞‍(βk)(a+βk)n,k(b+βk)n,k(c+βk)n,k(β+1)nzn.
Then
(48)w1=wg|β=0.
As we can see, all terms before
(49)(βk)(a+βk)1-(c/k),k(b+βk)1-(c/k),k(β+1)1-(c/k)(c+βk)1-(c/k),kz1-(c/k)
vanish because of the βk in the numerator.
Starting from this term, however, theβk in the numerator vanishes. To see this, note that
(50)(c+βk)1-(c/k),k=(c+βk)(c+(β+1)k)⋯(βk).
Hence, our assumed solution takes the form
(51)w1=g0(c)-(c/k),k∑n=1-(c/k)∞‍(a)n,k(b)n,k(1)n(k)n+(c/k)-1,kzj.
Now
(52)w2=∂wg∂β|β=1-(c/k).
To calculate this derivative, let
(53)Mn=(βk)(a+βk)n,k(b+βk)n,k(β+1)n(c+βk)n,k.
Then following the method in the previous case c/k=1, we get
(54)∂Mn∂β=Mn[1β+∑j=0n-1‍(ka+(β+j)k+kb+(β+j)k-1β+1+j-kc+(β+j)k)∑j=0n-1‍].
Since
(55)wg=g0zβ∑n=0∞‍Mnzn,
therefore
(56)∂wg∂β=g0zβ∑n=0∞‍(βk)(a+βk)n,k(b+βk)n,k(β+1)n(c+βk)n,k×[ln⁡z+1β+∑j=0j=n-1‍(ka+(β+j)k+kb+(β+j)k-1β+1+j-kc+(β+j)k)∑j=0j=n-1‍]zn.
At β=1-(c/k), we get
(57)w2=g0(k-c)z1-(c/k)×∑n=0∞‍(a+k-c)n,k(b+k-c)n,k(2k-c)n,k×[-1(1+j)k)ln⁡z+kk-c+k∑j=0n-1‍(1(a+k-c)+jk+1(b+k-c)+jk-1(2k-c)+jk-1(1+j)k)]znn!.
Hence
(58)w=E′w1+F′w2.
Let
(59)E′g0=E,F′g0=F.
Then
(60)w=E(c)-(c/k),k∑j=1-(c/k)∞‍(a)j,k(b)j,k(k)j+(c/k)-1,kzj(1)j+F(k-c)z1-(c/k)×∑n=0∞‍(a+k-c)n,k(b+k-c)n,k(2k-c)n,k×[ln⁡z+kk-c∑j=0n-1‍+k∑j=0n-1‍(1(a+k-c)+jk1(j)+1(b+k-c)+jk-1(2k-c)+jk-1(1+j)k)∑j=0n-1‍]znn!.
(ii) “(c/k)>1”. Let (c/k)>1. Then, from the recurrence relation
(61)dn=(a+(β+n-1)k)(b+(β+n-1)k)(n+β)(c+(β+n-1)k)dn-1,
we see that, when β=1-(c/k) (the smaller root), d(c/k)-1→∞. Thus, we must make the substitution
(62)d0=g0(β-βi)k,
where βi is the root for which our solution is infinite.
Hence we take(63)d0=g0(β+ck-1)k
and our assumed solution takes the new form
(64)wg=g0zβ∑n=0∞‍((β+(c/k)-1)k)(a+βk)n,k(b+βk)n,k(β+1)n(c+βk)n,kzn.
Then
(65)w1=wg|β=1-(c/k).
As we can see, all terms before
(66)((β+(c/k)-1)k)(a+βk)(c/k)-1,k(b+βk)(c/k)-1,k(β+1)(c/k)-1(c+βk)(c/k)-1,kz(c/k)-1
vanish because of the “β+(c/k)-1” in the numerator.
Starting from this term, however, the “β+(c/k)-1” in the numerator vanishes. To see this, note that
(67)(β+1)(c/k)-1=(β+1)(β+2)⋯(β+ck-1).
Hence, our solution takes the form
(68)w1=g0z1-(c/k)(2k-c)(c/k)-2,k∑n=(c/k)-1∞‍(a+k-c)n,k(b+k-c)n,k(k)n+1-(c/k),kzn(1)n.
Now
(69)w2=∂wg∂β|β=0.
To calculate this derivative, let
(70)Mn=((β+(c/k)-1)k)(a+βk)n,k(b+βk)n,k(β+1)n(c+βk)n,k.
Then following the method in the previous case (c/k)<1, we get
(71)∂Mn∂β=Mn[1β+(c/k)-1+∑j=0n-1‍(ka+(β+j)k+kb+(β+j)k-1β+1+j-kc+(β+j)k)],∂wg∂β=g0zβ∑n=0∞‍((β+(c/k)-1)k)(a+βk)n,k(b+βk)n,k(β+1)n(c+βk)n,k×[ln⁡z+1β+(c/k)-1∑j=0n-1‍+∑j=0n-1‍(ka+(β+j)k+kb+(β+j)k∑j=0n-1‍-1β+1+j-kc+(β+j)k)]zn.
At β=0, we get
(72)w2=g0(c-k)∑n=0∞‍(a)n,k(b)n,k(c)n,k×[ln⁡z+kc-kkj+∑j=0n-1‍(ka+jk+kb+jk-11+j-kc+jk)]znn!.
Hence
(73)w=G′w1+H′w2.
Let
(74)G′g0=G,H′g0=H.
Then
(75)w=G(2k-c)(c/k)-2,kz1-(c/k)×∑j=(c/k)-1∞‍(a+k-c)j,k(b+k-c)j,k(k)j+1-(c/k),k×zj(1)j+H(c-k)×∑n=0∞‍(a)n,k(b)n,k(c)n,k×∑n=0∞‍×[∑j=0n-1‍ln⁡z+kc-k+∑j=0n-1‍(ka+jk+kb+jk-11+j-kc+jk)∑j=0n-1‍]znn!.


## 3.3. Solution atz=1/k

Let us now study the singular pointz=1/k.To see if it is regular, we study the following limits:(76)lim⁡z→z0(z-z0)P1(z)P2(z)=lim⁡z→1/k(z-(1/k))(c-(a+b+k)kz)kz(1-kz)=c-(a+b+k)k,lim⁡z→z0(z-z0)2P0(z)P2(z)=lim⁡z→1/k(z-(1/k))2(-ab)kz(1-kz)=0.
As both limits exist, therefore z=1/k is a regular singular point.Now instead of assuming a solution in the form(77)w=∑n=0∞‍dn(z-1k)n+β,
we will try to express the solutions of this case in terms of the solutions for the point z=0. We proceed as follows.We have thek-hypergeometric differential equation of the form
(78)kz(1-kz)w′′+[c-(a+b+k)kz]w′-abw=0.
Let y=(1/k)-z. Then
(79)dwdz=dwdydydz=-dwdy=-w`,d2wdz2=ddz(dwdz)=ddz(-dwdy)=ddy(-dwdy)dydz=d2wdy2=w``.
Hence, k-hypergeometric differential equation (7) takes the form
(80)ky(1-ky)w``+[a+b-c+k-(a+b+k)ky]w`-abw=0.
Since y=(1/k)-z, the solution of the k-hypergeometric differential equation at z=1/k is the same as the solution for this equation at y=0. But the solution at y=0 is identical to the solution we obtained for the point z=0 if we replace c by a+b-c+k.Hence, to get the solutions, we just make the substitution in the previous results.Note also that, forz=0,
(81)β1=0,β2=1-ck.
Hence in our case,
(82)β1=0,β2=c-a-bk.


## 3.4. Analysis of the Solutions in terms of the Difference “(c-a-b)/k” of the Two Roots

Let us now find out the solutions. In the following we replace eachy by (1/k)-z.Case 1 (“(c-a-b)/k” not an integer).
Let(c-a-b)/k be not an integer. Then
(83)w=AF21,k((a,k),(b,k);(a+b-c+k,k);1k-z)+B(1k-z)(c-a-b)/k×F21,k((c-a,k),(c-b,k);(c-a-b+k,k);1k-z).Case 2 (“(c-a-b)/k=0”).
Let(c-a-b)/k=0. Then
(84)w=CF21,k((a,k),(b,k);(k,k);1k-z)+D∑n=0∞‍(a)n,k(b)n,k(k)n,k×[ln⁡(1k-z)∑j=0n-1‍+∑j=0n-1‍(ka+jk+kb+jk-2kn1+j)]×((1/k)-z)nn!.Case 3 (“(c-a-b)/k” an integer and “(c-a-b)/k≠0”).
Here, we discuss further two cases:
(i) “(c-a-b)/k>0.” Let (c-a-b)/k>0. Then
(85)w=E(a+b-c+k)(c-a-b)/k,k×∑n=(c-a-b)/k∞‍(a)n,k(b)n,k(k)n+((a+b-c)/k),k((1/k)-z)n(1)n+F(c-a-b)×(1k-z)(c-a-b)/k×∑n=0∞‍(c-a)n,k(c-b)n,k(c-a-b+k)n,k×[∑ln⁡(1k-z)+kc-a-b+k∑j=0n-1‍(1(1+j)k1(c-a)+jk+1(c-b)+jk-1(c-a-b+k)+jk-1(1+j)k)∑]((1/k)-z)nn!.
(ii) “(c-a-b)/k<0.” Let (c-a-b)/k<0. Then
(86)w=G(c-a-b+k)((a+b-c)/k)-1,k(1k-z)(c-a-b)/k×∑n=((a+b-c)/k)∞‍(c-a)n,k(c-b)n,k(k)n+((c-a-b)/k),k×((1/k)-z)n(1)n+H(a+b-c)×∑n=0∞‍(a)n,k(b)n,k(a+b-c+k)n,k×[ln⁡⁡(1k-z)+ka+b-c+∑j=0n-1‍(ka+jk+kb+jk-11+j-k(a+b-c+k)+jk)]×((1/k)-z)nn!.


## 3.5. Solution Around “∞”

Finally we study the singularity asz→∞. Since we cannot study this directly, therefore we let z=1/ks; then the solution of the equation as z→∞ is identical to the solution of the modified equation when s=0.We have thek-hypergeometric differential equation
(87)kz(1-kz)w′′+[c-(a+b+k)kz]w′-abw=0,dwdz=dwdsdsdz=-ks2dwds=-ks2w·,d2wdz2=k2(2s3w·+s4w··).
Hence the k-hypergeometric differential equation (7) takes the new form
(88)k2(s3-s2)w··+k[(2k-c)s2+(a+b-k)s]w·-abw=0.
Let
(89)P0(s)=-abP1(s)=k[(2k-c)s2+(a+b-k)s],P2(s)=k2(s3-s2).
Here, we only study the solutions when s=0. As we can see that this is a singular point because P2(0)=0.Let us now see thats=0 is a regular singular point, for this
(90)lim⁡s→s0(s-s0)P1(s)P2(s)=lim⁡s→0(s-0)k[(2k-c)s2+(a+b-k)s]k2(s3-s2)=k-a-bk,lim⁡s→s0(s-s0)2P0(s)P2(s)=lim⁡s→0(s-0)2(-ab)k2(s3-s2)=ab.
As both limits exist, therefore s=0 is a regular singular point.Thus, we assume the solution of the form(91)w=∑n=0∞‍dnsn+β
with d0≠0.Hence, we set the following:(92)w·=∑n=0∞‍dn(n+β)sn+β-1,w··=∑n=0∞‍dn(n+β)(n+β-1)sn+β-2.
By substituting these into the modified k-hypergeometric differential equation (88), we get
(93)k2∑n=0∞‍dn(n+β)(n+β-1)sn+β+1-k2∑n=0∞‍dn(n+β)(n+β-1)sn+β+k(2k-c)∑n=0∞‍dn(n+β)sn+β+1+k(a+b-k)∑n=0∞‍dn(n+β)sn+β-ab∑n=0∞‍dnsn+β=0.
In order to simplify this equation, we need all powers of s to be the same, equal to n+β, the smallest power. Hence, we switch the indices as follows:
(94)k2∑n=1∞‍dn-1(n+β-1)(n+β-2)sn+β-k2∑n=0∞‍dn(n+β)(n+β-1)sn+β+k(2k-c)∑n=1∞‍dn-1(n+β-1)sn+β+k(a+b-k)∑n=0∞‍dn(n+β)sn+β-ab∑n=0∞‍dnsn+β=0.
Thus, by isolating the first terms of the sums starting from 0, we get
(95)d0[-k2β(β-1)+k(a+b-k)β-ab]sβ+k2∑n=1∞‍dn-1(n+β-1)(n+β-2)sn+β-k2∑n=1∞‍dn(n+β)(n+β-1)sn+β+k(2k-c)∑n=1∞‍dn-1(n+β-1)sn+β+k(a+b-k)∑n=1∞‍dn(n+β)sn+β-ab∑n=1∞‍dnsn+β=0.
Now from the linear independence of all powers of s, that is, of the functions 1, s, s2, and so forth, the coefficients of sr vanish for all r. Hence, from the first term, we have
(96)d0[-k2β(β-1)+k(a+b-k)β-ab]=0
which is the indicial equation.Sinced0≠0, we have
(97)-k2β(β-1)+k(a+b-k)β-ab=0.
Hence, we get two solutions of this indicial equation:
(98)β1=ak,β2=bk.
Also, from the rest of the terms we have
(99)[k2(n+β-1)(n+β-2)+k2k(2k-c)(n+β-1)]dn-1+[-k2(n+β)(n+β-1)+k2k(a+b-k)(n+β)-ab]dn=0.
Hence, we get the recurrence relation of the following form:
(100)dn=(k(n+β)-c)k(n+β-1)(k(n+β)-a)(k(n+β)-b)dn-1
for n≥1.Let us now simplify this relation by givingdn in terms of d0 instead of dn-1.From the recurrence relation, we have(101)dn=(kβ)n,k(k(β+1)-c)n,k(k(β+1)-a)n,k(k(β+1)-b)n,kd0
for n≥1.Hence, our assumed solution in equation (91) takes the form
(102)w=d0∑n=0∞‍(kβ)n,k(k(β+1)-c)n,k(k(β+1)-a)n,k(k(β+1)-b)n,ksn+β.


## 3.6. Analysis of the Solutions in terms of the Difference “(a/k)-(b/k)” of the Two Roots

Now, we study the solutions corresponding to the different cases for the expressionβ1-β2=(a/k)-(b/k) (this reduces to studying the nature of the parameter (a/k)-(b/k) whether it is an integer or not).Case 1 (“(a/k)-(b/k)” not an integer).
Let(a/k)-(b/k) be not an integer. Then
(103)w1=w|β=a/k,w2=w|β=b/k.
Since
(104)w=d0∑n=0∞‍(kβ)n,k(k(β+1)-c)n,k(k(β+1)-a)n,k(k(β+1)-b)n,ksn+β,
therefore, we have
(105)w1=d0(kz)-(a/k)∑n=0∞‍(a)n,k(a+k-c)n,k(a+k-b)n,k1(k2z)n(1)n=d0(kz)-(a/k)×F21,k((a,k),(a+k-c,k);(a+k-b,k);1k2z),w2=d0(kz)-(b/k)×F21,k((b,k),(b+k-c,k);(b+k-a,k);1k2z).
Hence
(106)w=A′w1+B′w2.
Let
(107)A′d0=A,B′d0=B.
Then
(108)w=A(kz)-(a/k)×F21,k((a,k),(a+k-c,k);(a+k-b,k);1k2z)+B(kz)-(b/k)×F21,k((b,k),(b+k-c,k);(b+k-a,k);1k2z).Case 2 (“(a/k)-(b/k)=0”).
Let(a/k)-(b/k)=0. Then
(109)w1=w|β=a/k.
Also, we have
(110)w=d0∑n=0∞‍(kβ)n,k(k(β+1)-c)n,k(k(β+1)-a)n,k(k(β+1)-b)n,ksn+β
which can be written as
(111)w=d0∑n=0∞‍(β)n(β+1-(c/k))n(β+1-(a/k))n(β+1-(b/k))nsn+β.
Since we have (a/k)=(b/k), therefore
(112)w1=d0∑n=0∞‍(β)n(β+1-(c/k))n(β+1-(a/k))n2sn+β.
At β=a/k,
(113)w1=d0(kz)-(a/k)×F21,k((a,k),(a+k-c,k);(k,k);1k2z),w2=∂w∂β|β=a/k.
To calculate this derivative, let
(114)Mn=(β)n(β+1-(c/k))n(β+1-(a/k))n2,
and then we get
(115)∂Mn∂β=(β)n(β+1-(c/k))n(β+1-(a/k))n2×∑j=0n-1‍[1β+j+1β+1-(c/k)+j-2β+1-(a/k)+j],∂w∂β=d0sβ∑n=0∞‍(β)n(β+1-(c/k))n(β+1-(a/k))n2×[ln⁡s+∑j=0n-1‍(1β+j+1β+1-(c/k)+j-2β+1-(a/k)+j)∑j=0n-1]sn.
For β=a/k, we get
(116)w2=d0(kz)-(a/k)×∑n=0∞‍(a)n,k(a+k-c)n,kkn(k)n,k×[ln⁡(kz)-1∑j=0n-1‍+∑j=0n-1‍(ka+jk+ka+k-c+jk-21+j)]×(kz)-nn!.
Hence
(117)w=C′w1+D′w2.
Let
(118)C′d0=C,D′d0=D.
Then
(119)w=C(kz)-(a/k)∑n=0∞‍(a)n,k(a+k-c)n,kkn(k)n,k(kz)-nn!+D(kz)-(a/k)×∑n=0∞‍(a)n,k(a+k-c)n,kkn(k)n,k×[ln⁡⁡(kz)-1∑j=0n-1‍+∑j=0n-1‍(ka+jk+ka+k-c+jk-21+j)]×(kz)-nn!.Case 3 (“(a/k)-(b/k)” an integer and “(a/k)-(b/k)≠0”).
Here, we have further two cases:
(i) “(a/k)-(b/k)>0.” Let (a/k)-(b/k)>0.
Then, from the recurrence relation(120)dn=(k(n+β)-c)(k(n+β-1))(k(n+β)-a)(k(n+β)-b)dn-1,
we see that, when β=b/k (the smaller root), d(a/k)-(b/k)→∞. Thus, we must make the substitution
(121)d0=g0(β-βi)k,
where βi is the root for which our solution is infinite.
Hence we take(122)d0=g0(β-bk)k
and our assumed solution in equation (110) takes the new form
(123)wg=g0sβ∑n=0∞‍((β-(b/k))k)(k(β+1)-c)n,k(βk)n,k(k(β+1)-a)n,k(k(β+1)-b)n,ksn.
Then
(124)w1=wg|β=b/k.
As we can see, all terms before
(125)((β-(b/k))k)(βk)(a/k)-(b/k),k(k(β+1)-c)(a/k)-(b/k),k(k(β+1)-a)(a/k)-(b/k),k(k(β+1)-b)(a/k)-(b/k),k×s(a/k)-(b/k)
vanish because of the (β-(b/k)) in the numerator.
Starting from this term, however, the(β-(b/k)) in the numerator vanishes. To see this, note that
(126)(β+1-ak)(a/k)-(b/k)=(β+1-ak)(β+2-ak)⋯(β-bk).
Hence, our solution takes the form
(127)w1=g0(b+k-a)(a/k)-(b/k)-1,k×∑n=(a/k)-(b/k)∞‍(b)n,k(b+k-c)n,kkn+(b/k)-(a/k)(1)n+(b/k)-(a/k)sn(1)n=g0(b+k-a)(a/k)-(b/k)-1,k×∑n=(a/k)-(b/k)∞‍(b)n,k(b+k-c)n,k(k)n+(b/k)-(a/k),k(kz)-n(1)n.
Now
(128)w2=∂wg∂β|β=a/k.
To calculate this derivative, let
(129)Mn=((β-(b/k))k)(β)n(β+1-(c/k))n(β+1-(a/k))n(β+1-(b/k))n,
and then we get
(130)∂Mn∂β=Mn[1β-(b/k)+∑j=0n-1‍(1β+1-(c/k)+j+1β+j-1β+1-(a/k)+j-1β+1-(b/k)+j)],∂wg∂β=g0sβ∑n=0∞‍(β-(b/k))k(β)n(β+1-(c/k))n(β+1-(a/k))n(β+1-(b/k))n×[ln⁡s+1β-(b/k)+∑j=0n-1‍(1β+j+1β+1-(c/k)+j-1β+1-(a/k)+j-1β+1-(b/k)+j)]sn.
At β=a/k, we get
(131)w2=g0sa/k∑n=0∞‍(a-b)(a)n,k(a+k-c)n,k(a+k-b)n,k(k)n,k×[ln⁡s+ka-b+∑j=0n-1‍(ka+jk+ka+k-c+jk-ka+k-b+jk-11+j)]sn.
Replacing s by (kz)-1, we get w2 in terms of z:
(132)w2=g0(kz)-(a/k)∑n=0∞‍(a-b)(a)n,k(a+k-c)n,k(a+k-b)n,k×[ln⁡(kz)-1+ka-b+∑j=0n-1‍(ka+jk+ka+k-c+jk-11+j-ka+k-b+jk)]×(kz)-n(k)n,k.
Let
(133)w=E′w1+F′w2.
Let
(134)E′g0=E,F′g0=F.
Then
(135)w=E(b+k-a)(a/k)-(b/k)-1,k×∑n=(a/k)-(b/k)∞‍(b)n,k(b+k-c)n,k(k)n+(b/k)-(a/k),k(kz)-n(1)n+F(kz)-(a/k)∑n=0∞‍(a-b)(a)n,k(a+k-c)n,k(a+k-b)n,k×[ln⁡(kz)-1+ka-b+∑j=0n-1‍(ka+jk+ka+k-c+jk-11+j-ka+k-b+jk)]×(kz)-n(k)n,k.
(ii) “(a/k)-(b/k)<0.” Let (a/k)-(b/k)<0.
Then, from the symmetry of the situation here, we see that(136)w=G(a+k-b)(b/k)-(a/k)-1,k×∑n=(b/k)-(a/k)∞‍(a)n,k(a+k-c)n,k(k)n+(a/k)-(b/k),k(kz)-n(1)n+H(kz)-(b/k)∑n=0∞‍(b-a)(b)n,k(b+k-c)n,k(b+k-a)n,k×[ln⁡(kz)-1+kb-a+∑j=0n-1‍(kb+jk+kb+k-c+jk-11+j-kb+k-a+jk)]×(kz)-n(k)n,k.


## 4. Conclusion

In this research work, we derived thek-hypergeometric differential equation. Also, we obtained 24 solutions of k-hypergeometric differential equation around all its regular singular points by using Frobenius method. So, we conclude that, if k→1, we obtain Euler’s hypergeometric differential equation. Similarly, by letting k→1, we find 24 solutions of hypergeometric differential equation around all its regular singular points.



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*Source: 128787-2014-04-30.xml*