Page 82 - IJOCTA-15-1

E.M. Rabei et al. / IJOCTA, Vol.15, No.1, pp.71-81 (2025)
We find the eigenvalue λ N−2 = m N−3 (m N−3 +
D s − N − 1) where γ N−1 = D s−N−2 and the so- β 2β m N−3 β
2 D X β(N−2) = (1 − x ) 2 D u
x 1 (N−2)β
lution of eq.(18) is given by x 1 m N−3
2β
β
− m N−3 βx (1 − x ) 2 −1
γ N−1
X β(N−1) = T (x). (32) × u (N−2)β , (36)
βm N−3
Where T γ N−1 (x) is the Conformable Gegenbauer and
βm N−3
polynomials. 51
D β D β X β(N−2) (37)
x N−2 x N−2
Now, let us solve eq.(17) by using the change of 2β m N−3 β β
= (1 − x ) 2 D D u
variable: N−2 x N−2 x N−2 (N−2)β
m
β 2β N−3 −1 β
− 2m N−3 βx (1 − x ) 2 D u
N−2 N−2 x N−2 (N−2)β
X β(N−2) = Θ β(N−2) , − m N−3 β (1 − x 2β ) m N−3 −1 u (N−2)β
2
2
θ β N−2
β N−2 2 2β
x = cos , + m N−3 (m N−3 − 2)β x N−2
N−2 β m N−3
β × (1 − x 2β ) 2 −2 u (N−2)β ,
1 θ N−2
β N−2 β
∂ = − sin ∂ , 2β
θ N−2 x N−2 after multiplying this equation by (1 − x ), we
β β
β find
1 θ
2β 2 N−2 2β
∂ = sin ∂ ,
β β 2β β β
θ N−2 2 x N−2
2β (1 − x N−2 )D D X β(N−2) (38)
(1 − x ) x 1 x N−2
∂ 2β = N−2 ∂ 2β . 2β m N−3 +1 β β
u
θ N−2 β 2 x N−2 = (1 − x N−2 ) 2 D x N−2 D x N−2 (N−2)β
m
β 2β N−3 β
− 2m N−3 βx (1 − x ) 2 D x N−2 (N−2)β
u
Thus, eq.(17) becomes N−2 N−2
m N−3
2β
2
− m N−3 β (1 − x ) 2 u (N−2)β
N−2
h
2β 2β β 2 2β
(1 − x )∂ − β(D s − N + 1)∂ + m N−3 (m N−3 − 2)β x
N−2 x N−2 x N−2 N−2
m
# 2β N−3 −1
m N−3 (m N−3 + 2γ N−2 ) × (1 − x N−2 ) 2 u (N−2)β .
+ λ N−3 − 2β X β(N−2) = 0.
1 − x
N−2
(33) Substituting eqs.(35),(36), and (38) in (33) and
m
2β
multiplying by (1 − x ) 2 , we have
where γ N−2 = γ N−1 + 1 = D s−N−1 . To calculate
2 2
the eigenvalue λ N−3 let m N−3 = 0. Thus, this (1 − x 2β )D β D β u (39)
equation becomes N−2 x N−2 x N−2 (N−2)β
β β 2
− 2m N−3 βx D u − m N−3 β u
N−2 x N−2 (N−2)β (N−2)β
2β 2β 2 2β 2β −1
(1 − x )∂ X (34) + m N−3 (m N−3 − 2)β x (1 − x ) u
N−2 x N−2 β(N−2) N−2 N−2 (N−2)β
− β(D s − N + 1)∂ β X − 2βx β (γ N−2 + 1)D β u
x N−2 β(N−2) N−2 x N−2 (N−2)β
+ λ N−3 X β(N−2) = 0. + 2β(γ N−2 + 1)m N−3 βx 2β (1 − x 2β ) −1 u
N−2 N−2 (N−2)β
2
+ β [m N−4 (m N−4 + 2γ N−2 + 1)
Again comparing this equation with the Con- m N−3 (m N−3 + 2γ N−2 ) #
formable Gegenbauer equation, 51 we obtain − u (N−2)β .
1 − x 2β
N−2
λ N−3 = m N−4 (m N−4 + 2γ N−2 + 1) Thus, we arrived to
= m N−4 (m N−4 + D s − N).
2β β β
(1 − x )D D u
N−2 x N−2 x N−2 (N−2)β
β β
u
To solve eq.(33) we suppose − 2βx N−2 (m N−4 + γ N−2 + 1)D x N−2 (N−2)β
2
+ β [m N−4 (m N−4 + 2γ N−2 + 1) (40)
m
2β N−3
X β(N−2) = (1 − x N−2 ) 2 u (N−2)β , (35) − m N−3 (m N−3 + 2γ N−2 + 1)]u (N−2)β = 0.
where The Conformable Gegenbauer equation reads as 51
76