Journal of Indian Acad. Math.  
ISSN: 0970-5120  
Vol. 48, No. 1 (2026) pp. 112–120.  
GENERATING FUNCTIONS FOR TWO  
VARIABLE LEGENDRE POLYNOMIALS  
B. S. Desale1 and G. S. Kadu2  
Abstract. In this paper we have used Weisner’s group theoretic method to derive  
generating functions involving two variable Legendre polynomials Pn(x, y) by suitable  
interpretation to the index variable n.  
Keywords: Two variable Legendre polynomials, group theoretic method, generating func-  
tions.  
2010 AMS Subject Classification: Primary 33C50; Secondary 33C70.  
1. Introduction  
It was Weisner [1, 2, 3] who introduced the group theoretic method to obtain generating  
functions for hypergeometric, Hermite, and Bessel functions. The Weisner’s method was  
the subject of Miller’s [4] research, and he presented it in a systematic manner to establish  
a solid foundation for us. During the same period, McBride ([5]) used this method to  
obtain generating functions for Laguerre, modified Laguerre, Bessel, and Gegenbauer  
polynomials. Recently, M. G. Bin-saad and M. J. S. Shahwan [6], H. M. Srivastava et al.  
[7], G. Yasmin and A. Muhyi [8], and S. Khan and M. Ali [9] have used the method adopted  
by Miller to obtain relations for multivariable special functions, namely 2D-Laguerre,  
2I4V1P Hermite, 3V2P Hermite, and 3V Laguerre-Hermite polynomials, respectively. In  
this paper, we have used a method adopted by McBride to obtain generating functions  
for two-variable Legendre polynomials.  
The two-variable extension of Legendre polynomials (TVLP) introduced by G. Dattoli  
et al. [10] is defined as  
ꢀ
ꢃ
ꢁ ꢂ  
(2n)!  
22n(n!)2  
n
1 − n∗ 1 − 2n 4y  
Pn(x, y) =  
(−x)n F1  
−
,
;
;
(1)  
2
2
2
2
x2  
where  
ꢄ
n,  
if n is even  
n∗ =  
n + 1, if n is odd  
and its generating function is given by,  
∞
ꢅ
1
Pn(x, y)tn =  
.
(2)  
ꢆ
1 + xt + yt2  
n=0  
In the following section we will use group theoretic method to find generating functions  
of TVLP.  
112  
GENERATING FUNCTIONS FOR TWO VARIABLE LEGENDRE POLYNOMIALS  
113  
2. Group-Theoretic Method  
The first and important step in the group theoretic method is to obtain two indepen-  
dent differential recurrence relations; using (??eq. 2), we have determined the following  
differential recurrence relations for TVLP.  
∂Pn(x, y)  
(x2 − 4y)  
= n [xPn(x, y) + 2yPn−1(x, y)] ,  
(3)  
∂x  
∂Pn(x, y)  
(4y − x2)  
= (n + 1) [2Pn+1(x, y) + xPn(x, y)] ,  
(4)  
(5)  
∂x  
∂Pn(x, y)  
(4y − x2)  
= n[2Pn(x, y) + xPn−1(x, y)].  
∂y  
By using the above recurrence relations (3) and (4), we verify that the following partial  
differential equation is satisfied by TVLP:  
∂
(4y − x2)D2Pn(x, y) − 2xDPn(x, y) + n(n + 1)Pn(x, y) = 0, where D =  
.
(6)  
∂x  
We can recast the differential equation (6) in the form of an operator as follows:  
ꢀ
ꢃ
ꢇ
ꢈ
∂
L x, y,  
, n Pn(x, y) = (4y − x2)D2 − 2xD + n(n + 1) Pn(x, y) = 0.  
(7)  
∂x  
∂
Now we replace n by z∂z and Pn(x, y) by f(x, y, z) in (7), to get the following partial  
differential equation  
ꢉ
ꢊ
2
2
∂
∂
∂
∂
(4y − x2)  
− 2x  
+ 2z  
+ z2  
f(x, y, z) = 0.  
(8)  
∂x2  
∂x  
∂z  
∂z2  
We can recast the partial differential equation (8) in the form of operator as  
ꢀ
ꢃ
∂
∂
L x, y,  
, z  
f(x, y, z) = 0.  
∂x  
∂z  
Now we proceed to find power raising and lowering operators, which is one of the key  
steps in the group theoretic method. By simple calculations we have determined that the  
following linear partial differential operators  
∂
A = z  
,
(9)  
∂z  
(4y − x2) ∂  
x ∂  
B =  
+
,
(10)  
yz  
∂x  
y ∂z  
and  
∂
∂
C = z(4y − x2)  
− xz2  
− xz  
(11)  
∂x  
∂z  
are such that their action on Pn(x, y)zn is as follows:  
A[Pn(x, y)zn] = nPn(x, y)zn,  
(12)  
B[Pn(x, y)zn] = −2nPn−1(x, y)zn−1  
,
(13)  
(14)  
C[Pn(x, y)zn] = 2(n + 1)Pn+1zn+1  
.
114  
B. S. Desale and G. S. Kadu  
Here operator (10) is a power-lowering operator, whereas operator (11) is a power-raising  
operator. Both these operators play an important role in determining the generating func-  
tions of TVLP. Now we proceed to determine commutator relations satisfied by operators  
A, B, and C, which are presented in the form of the following proposition.  
Proposition 2.1. The operators A, B and C satisfy the following commutator relations  
(i) [A, B] = −B  
(ii) [B, C] = −8A − 4  
(iii) [A, C] = C  
2
Proof. Let f = f(x, y, z) ∈ C be a arbitrary function.  
Firstly we will prove the commutator relation satisfied by operators A and B. Now  
ꢉ
ꢊ
∂
(4y − x2) ∂f  
x ∂f  
ABf = z  
+
.
∂z  
yz  
∂x  
y ∂z  
Further simplification gives leads to  
−(4y − x2) ∂f  
(4y − x2) ∂2f  
zx ∂2f  
y ∂z2  
ABf =  
+
+
.
(15)  
(16)  
yz  
∂x  
y
∂z∂x  
Similary, we get  
x df  
(4y − x2) ∂2f  
xz ∂2f  
y ∂z2  
BAf =  
+
+
.
y dz  
Using equations (15) and (16), we get  
y
∂x∂z  
[A, B]f = ABf − BAf = −Bf.  
Therefore,  
[A, B] = −B.  
Now we will prove the commutator relation for operators B and C; by doing elementary  
calculations, we get  
−(4y − x2)z ∂f  
(4y − x2)  
2x2z ∂f  
x2f  
y
z(4y − x2) ∂f  
[B, C]f =  
−
f −  
−
−
y
∂z  
y
y ∂z  
y
∂z  
∂f  
= −8z  
− 4f  
∂z  
= (−8A − 4)f.  
Therefore,  
∂f  
[B, C]f = BCf − CBf = −8z  
− 4f = (−8A − 4)f.  
(17)  
∂z  
Hence [B, C] = −8A − 4.  
Finally, we will prove the commutator relation satisfied by operators A and C. Doing  
some simplification, we get  
∂f  
∂f  
[A, C]f = z(4y − x2)  
− xz2  
− xzf = Cf.  
∂x  
∂z  
Therefore  
[A, C] = C.  
(18)  
□
GENERATING FUNCTIONS FOR TWO VARIABLE LEGENDRE POLYNOMIALS  
115  
Now by using the commutator relations proved in proposition 2.1, we prove the following  
proposition.  
Proposition 2.2. The operators given by (9), (10), and (11) commute with the operator  
(4y − x2)  
L
y
where L is the operator given by (7).  
2
Proof. Let f = f(x, y, z) ∈ C be a arbitraray function. Here,  
∂2f  
∂f  
∂f  
∂2f  
∂z2  
Lf = (4y − x2)  
− 2x  
+ 2z  
+ z2  
(19)  
∂x2  
∂x  
∂z  
and  
CBf =  
(4y − x2)2 ∂2f  
2x(4y − x2) ∂f  
z(4y − x2) ∂f  
x2z2 ∂2f  
y ∂z2  
x2z ∂f  
−
+
−
−
. (20)  
y
∂x2  
y
∂x  
y
∂z  
y ∂z  
Using (19) and (20), we have  
ꢉ
ꢊ
ꢉ
ꢊ
(4y − x2)  
∂
∂
2
L − CB f = 4 z2  
+ z  
f = 4A2f.  
y
∂z2  
∂z  
Since f is arbitrary, we conclude that  
(4y − x2)  
L = CB + 4A2.  
(21)  
y
Using (17) and (18), we have following assertions  
BC − CB = −8A − 4  
A2C − CA2 = 2CA + C.  
(22)  
(23)  
Now using (21), (22) and (23), we have  
ꢉ
ꢊ
ꢉ
ꢊ
(4y − x2)  
(4y − x2)  
(4y − x2)  
L, C f =  
LC − C  
L f  
y
y
y
= [C(BC − CB) + 4(A2C − CA2)]f  
= 0.  
2
2
Hence (4y−x ) L commutes with C. Similary, we can verify that the operator (4y−x ) L  
y
y
commutes with the operators A and B respectively.  
□
Now we proceed to find the extended form of group of operators; the operators exp(aA),  
exp(bB) and exp(cC) are called the extended forms of operators A, B and C, where a, b, c  
are arbitrary constants. In order to find the extended form of the group of operator  
corresponding to C, we firstly will change the form of the differential operator C =  
∂
∂
∂
∂
∂
∂
∂
C1 ∂x + C2 ∂y + C3 ∂z + C4 to E = C1 ∂x + C2 ∂y + C3 ∂z and finally to D =  
by using  
change of variables. If a function φ(x, y, z) is such that C[φ] = 0, then φ−1∂CXφ = E =  
∂
∂
∂
C1 ∂x +C2 ∂y +C3 ∂z , refer McBride [5] for the proof. Hence, firstly, we will find a function  
φ such that C[φ] = 0.  
116  
Now C[φ] = 0, implies that  
B. S. Desale and G. S. Kadu  
∂φ  
∂φ  
z(4y − x2)  
− xz2  
− xzφ = 0.  
∂x  
∂z  
For z = 0, we get  
∂φ  
∂φ  
(4y − x2)  
− xz  
− xφ = 0.  
∂x  
∂z  
To get φ, we solve the following system  
dx  
(4y − x2)  
dz  
dφ  
=
=
.
−xz  
xφ  
The general solution of the above system is, σ(zφ, y) = 0, where σ is an arbitrary function.  
We choose a special case,  
z
φ−1 = .  
(24)  
y
Now for any function u(x, y, z), we get  
∂u  
∂u  
φ−1Cφu = z(4y − x2)  
− xz2  
.
∂x  
∂z  
Thus  
∂
∂
E = φ−1Cφ = z(4y − x2)  
− xz2  
.
(25)  
∂x  
∂z  
∂
Now we have to choose variables X, Y & Z, so that E will be transformed into k∂X where  
k2 = 1. We put  
∂u  
∂u ∂X  
∂u ∂Y  
∂u ∂Z  
=
=
+
+
+
+
∂x  
∂u  
∂X ∂x  
∂u ∂X  
∂Y ∂x  
∂u ∂Y  
∂Z ∂x  
∂u ∂Z  
∂y  
∂X ∂y  
∂Y ∂y  
∂Z ∂y  
∂u  
∂u ∂X  
∂u ∂Y  
∂u ∂Z  
=
+
+
∂z  
∂X ∂z  
∂Y ∂z  
∂Z ∂z  
in  
∂u  
∂u  
Eu = z(4y − x2)  
− xz2  
∂x  
∂z  
to get  
ꢉ
ꢊ
ꢉ
ꢊ
∂X  
∂X  
∂u  
∂Y  
∂Y  
∂u  
Eu = z(4y − x2)  
− xz2  
+ z(4y − x2)  
− xz2  
∂x  
∂z  
∂X  
∂x  
∂z  
∂Y  
ꢉ
ꢊ
∂Z  
∂Z  
∂u  
+ z(4y − x2)  
− xz2  
.
∂x  
∂z  
∂Z  
We set  
∂Y  
∂Y  
∂Z  
∂Z  
∂X  
∂X  
z(4y − x2)  
− xz2  
= 0, z(4y − x2)  
− xz2  
= 0, z(4y − x2)  
− xz2  
= k.  
∂x  
∂z  
∂x  
∂z  
∂x  
∂z  
For z = 0, we have  
∂Y  
∂Y  
(4y − x2)  
− xz  
= 0,  
(26)  
(27)  
(28)  
∂x  
∂Z  
∂z  
∂Z  
(4y − x2)  
− xz  
= 0,  
∂x  
∂z  
∂X  
∂X  
(4y − x2)  
− xz  
= kz−1.  
∂x  
∂z  
GENERATING FUNCTIONS FOR TWO VARIABLE LEGENDRE POLYNOMIALS  
117  
General solutions of (26), (27) and (28) are  
ꢀ
ꢃ
ꢀ
ꢃ
4y − x2  
x
Xy  
k
σ1  
, Y  
= 0, τ (y, Z) = 0 and η  
−
, y  
= 0  
(29)  
z2  
4z  
respectively, where σ1, τ & and η are arbitrary functions. We choose following particular  
solutions  
4y − x2  
Y =  
,
(30)  
(31)  
z2  
Z = y,  
and  
x
X =  
.
(32)  
4yz  
Solving (30), (31) and (32), gives us  
3
1
2
2
8Z X  
2Z  
Y + 16Z2X2  
√
√
x =  
, y = Z, z =  
,
(33)  
Y + 16Z2X2  
∂
which transforms E to ∂X . Now we are in position to derive extended form of operators.  
Firstly we will find extended form of operator C.  
−1  
ecCf(x, y, z) = ecφEφ f(x, y, z)  
ꢇ
ꢈ
= φ(x, y, z)ecE φ−1(x, y, z)f(x, y, z) .  
Using (24) the above equation can be simplified as  
ꢇ
ꢈ
ecCf(x, y, z) = yz−1ecE y−1zf(x, y, z) .  
(34)  
∂
Now use of (33) in (34) will transform E to D, where D = ∂X . Making this substitution  
(34) can be recasted as  
ꢋ
ꢌ
ꢍꢎ  
3
1
2
2
2
8Z X  
2Z  
Y + 16Z2X2  
ecCf(x, y, z) = yz−1ecD  
√
f
, Z,  
.
√
√
√
Y + 16Z2X2  
Z Y + 16Z2X2  
(35)  
Now using Taylor’s theorem in the right hand side of the above equation gives  
ecCf(x, y, z)  
ꢋ
ꢌ
ꢍꢎ  
3
1
2
2
2
8Z (X + c)  
2Z  
Y + 16Z2(X + c)2  
= yz−1  
f
, Z,  
.
√ ꢆ  
ꢆ
ꢆ
Z
Y + 16Z2(X + c)2  
Y + 16Z2(X + c)2  
Finally using reverse substitution, we get the following extended form of the operator C.  
ecCf(x, y, z)  
(36)  
ꢌ
ꢍ
1
x + 4cyz  
z
ꢆ
ꢆ
ꢆ
=
× f  
, y,  
.
1 + 4c2yz2 + 2xzc  
1 + 4c2yz2 + 2xzc  
1 + 4c2yz2 + 2xzc  
Similarly we can derive extended form of the operators B and A, which are as follows:  
ꢌ
ꢍ
ꢆ
√
yz2 + 4b2 + 2xzb  
(xz + 4b) y  
ebBf(x, y, z) = f  
, y,  
(37)  
(38)  
ꢆ
√
yz2 + 4b2 + 2xzb  
y
ꢏ
ꢐ
x
eaAf(x, y, z) = f  
e−a, y, z .  
z
118  
B. S. Desale and G. S. Kadu  
Using (36) and (37) we get,  
1
ecCebBf(x, y, z) =  
× f(ζ, y, ξ)  
(39)  
ꢆ
1 + 4c2yz2 + 2xzc  
where,  
√
y[zx(1 + 8bc) + 4cyz2(1 + 4bc) + 4b]  
ꢆ
ꢆ
ζ =  
,
1 + 4c2yz2 + 2xzc yz2(1 + 4bc)2 + 2xzb(1 + 4bc) + 4b2  
ꢆ
yz2(1 + 4bc)2 + 2xzb(1 + 4bc) + 4b2  
ꢆ
ξ =  
.
√
y 1 + 4c2yz2 + 2xzc  
In the next section, we will use transformation operator (39) to derive some generating  
functions for TVLP.  
3. Generating functions  
Now to obtain generating functions of TVLP we will transform f(x, y, z) using (39).  
We discuss the following particular cases.  
Case (i): b = 1, c = 0  
For b = 1, c = 0 the equation (39) takes the following form:  
ꢌ
ꢍ
ꢆ
√
yz2 + 4 + 2xz  
y(xz + 4)  
eBf(x, y, z) = f  
, y,  
.
ꢆ
√
yz2 + 4 + 2xz  
y
Now replace f(x, y, z) by Pn(x, y)zn in the above equation, to get  
ꢌ
ꢍ
n
2
ꢀ
ꢃ
√
yz2 + 4 + 2xz  
(xz + 4) y  
eB[Pn(x, y)zn] =  
Pn  
, y .  
ꢆ
yz2 + 4 + 2xz  
y
Using (13) in the above equation gives us  
ꢌ
ꢍ
n
2
ꢀ
ꢃ
√
∞
q
2 (−n)qPn−q(x, y)zn−q  
yz2 + 4 + 2xz  
ꢅ
(xz + 4) y  
ꢆ
=
Pn  
, y .  
yz2 + 4 + 2xz  
q!  
y
q=0  
Replace z−1 by t in the above equation to obtain the following generating function  
ꢌ
ꢍ
n
2
ꢀ
ꢃ
√
∞
q
2 (−n)qPn−q(x, y)tq  
y + 4t2 + 2xt  
ꢅ
(x + 4t) y  
ꢆ
=
Pn  
, y .  
(40)  
y + 4t2 + 2xt  
q!  
y
q=0  
Case (ii) b = 0, c = 1  
For b = 0, c = 1 the equation (39) takes the following form  
ꢌ
ꢍ
1
x + 4yz  
z
eCf(x, y, z) =  
× f  
, y,  
.
ꢆ
ꢆ
ꢆ
1 + 4yz2 + 2xz  
1 + 4yz2 + 2xz  
1 + 4yz2 + 2xz  
Now replace f(x, y, z) by Pn(x, y)zn in the above equation, to get  
ꢌ
ꢍ
zn  
x + 4yz  
eC[Pn(x, y)zn] =  
Pn  
, y .  
ꢆ
(1 + 4yz2 + 2xz)1+n  
1 + 4yz2 + 2xz  
2
GENERATING FUNCTIONS FOR TWO VARIABLE LEGENDRE POLYNOMIALS  
119  
Use of (14) in above equation leads to  
ꢌ
ꢍ
∞
q
2 (n + 1)q  
zn  
x + 4yz  
ꢅ
Pn+q(x, y)zn+q  
=
Pn  
, y .  
ꢆ
(1 + 4yz2 + 2xz)1+n  
1 + 4yz2 + 2xz  
q!  
2
q=0  
Now we divide by zn to get the following generating function  
ꢌ
ꢍ
∞
q
ꢅ
2 (n + 1)q  
1
x + 4yz  
Pn+q(x, y)zq =  
Pn  
, y .  
(41)  
ꢆ
(1 + 4yz2 + 2xz)1+n  
1 + 4yz2 + 2xz  
q!  
2
q=0  
1
1
Case (iii) b = , c = −  
2
2
1
1
For b = , c = − the equation (39) takes the following form  
2
2
C
B
1
e− e [f(x, y, z)] =  
× f(ξ, y, τ),  
(42)  
(43)  
2
2
ꢆ
1 + yz2 − xz  
where  
√
y[2 − xz]  
1
ꢆ
ꢆ
ξ =  
, τ =  
.
√
1 + yz2 − xz  
y 1 + yz2 − xz  
Replace f(x, y, z) by Pn(x, y)zn in the above equation to get,  
ꢌ
ꢍ
√
n
2
ꢑ
ꢒ−  
y[2 − xz]  
C
B
1
n
2
e− e [Pn(x, y)z ] =  
× Pn  
, y  
1 + yz2 − xz  
y−  
.
n
2
2
ꢆ
ꢆ
1 + yz2 − xz  
1 + yz2 − xz  
Now using (13) and (14), we get  
∞
(−n)p(−1)q(n − p + 1)q  
ꢅ
Pn−p+q(x, y)zn−p+q  
p!q!  
p,q=0  
ꢌ
ꢍ
√
n
2
ꢑ
ꢒ−  
y[2 − xz]  
1
n
2
=
× Pn  
, y  
1 + yz2 − xz  
y− . (44)  
ꢆ
ꢆ
1 + yz2 − xz  
1 + yz2 − xz  
Now we divide the above equation by zn to get the following generating function  
∞
(−n)p(−1)q(n − p + 1)q  
Pn−p+q(x, y)z−p+q  
p!q!  
ꢅ
p,q=0  
ꢌ
ꢍ
n
2
√
n
2
(1 + yz2 − xz)− y−  
y[2 − xz]  
1
ꢆ
ꢆ
=
×
× Pn  
, y . (45)  
zn  
1 + yz2 − xz  
1 + yz2 − xz  
4. Conclusion remarks  
In this paper we have used one of the powerful tools, i.e., Weisner’s group theoretic  
method to derive generating functions of two variable Legendre polynomials Pn(x, y). In  
the beginning, we have determined the power maintaining, lowering, and raising group of  
operators A, B, and C. Subsequently, we have derived their extended forms and utilized  
them to find the generating functions, which are represented by the equations (40), (41),  
and (45).  
120  
B. S. Desale and G. S. Kadu  
Conflicts of Interest  
The authors declare no conflict of interest.  
REFERENCES  
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(Received, July 26, 2025)  
(Revised, September 07, 2025)  
1Mathematics Department,  
University of Mumbai, Mumbai 400 098, India.  
E-mail: bhausaheb.desale@mathematics.mu.ac.in  
2Department of Mathematics,  
Arts, Commerce and Science College,  
Lanja 416701, India.  
E-mail: 4ganeshkadu@gmail.com