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Lecture notes on mathematics
Dr. Sunil KumarAssociate professor
Department of Mathematics,National Institute of Technology, Jamshedpur.
April 1, 2020
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Outlines
1 Series solution2 Frobenius method3 Legendre differential equation4 Bessel’s differential equation5 Recurrence formula6 Generating functions7 Orthogonality
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Series solution
Consider the differential equation with variable coefficients
P0(x)d2ydx2 + P1(x)
dydx
+ P2(x)y = 0 (1)
where P0(x), P1(x) and P2(x) are polynomial in x.(a). Ordinary point-: A point x = a is called ordinary point of the above differentialequation (DE) if P0(x) 6= 0 at x = a.(b). Singular point-: A point x = a is called Singular point of the above differentialequation (DE) if P0(x) = 0 at x = a.(i). Regular Singular point-: If P1(x)
P0(x) and P2(x)P0(x) are differentiable in the neighborhood
of x = a.(ii). Irregular Singular point-: If not regular then irregular singular point.
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Series solution
Working Rule:-I. In the given DE (1), now P0(x) 6= 0 at x = 0 is the ordinary point.II. Consider the series solution of DE (1)
y = a0 + a1x + a2x2 + a3x3 + ...+ anxn + ... (2)
find dydx and d2y
dx2 .
III. Put y, dydx and d2y
dx2 into the DE (1) and simplifying it. Now, equating the coefficientsof x0, x1, x2, ..., xn to zero. It gives the values of a2,a3,a4,...,an in terms of a0 and a1.IV. Put the values of a2, a3, a4,...,an into (2) we get the solution.
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Frobenius Method
When x = 0 is the regular singular point of the DE (Frobenius method).Working Rule:-I. Given DE
P0(x)d2ydx2 + P1(x)
dydx
+ P2(x)y = 0 (3)
x = 0 is the regular singular point.II. Let y = xm(a0 + a1x + a2x + ...+ anxn + ...) be the series solution of (3). Find dy
dx
and d2ydx2 .
II. Put y, dydx and d2y
dx2 into (3) and simplify it. Equating the coefficient of lowest degreeof x to zero, we obtain a quadratic equation in m. This equation is called indicalequation. Solve indical equation we get two roots m1 and m2 (say). Moreover, thecomplete solution depends on the nature of roots.Case-I. If m1 6= m2 and m1 − m2 6= an integer then the complete solution is
y = C1(y)m=m1 + C2(y)m=m2
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Frobenius Method
Case-II. If m1 = m2 then the complete solution is
y = C1(y)m1 + C2(∂y∂m
)m1 .
Case-III. If m1 6= m2 and m1 − m2 = an integer, then the complete solution is
y = C1(y)m=m1 + C2(∂y∂m
)m2 , (m1 > m2).
AssignmentQ.1 Solve 2x2 d2y
dx2 + x dydx − (x + 1)y = 0.
Q.2 Solve x d2ydx2 + dy
dx + xy = 0.
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Legendre differential equation
Legendre differential equation is defined as
(1− x2)d2ydx2 − 2x
dydx
+ α(α+ 1)y = 0 (4)
where n is positive integer.
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Legendre differential equation
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Legendre differential equation
On substituting it into the equation (4) we obtain
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Legendre differential equation
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Bessel’s differential equation
Bessel’s differential equation is defined as
x2 d2ydx2 + x
dydx
+ (x2 − n2)y = 0. (5)
Or,d2ydx2 +
1x
dydx
+ (1− n2
x2 )y = 0. (6)
Since, x = 0 is regular singular point. Let the series solution is y =∑∞
r=0 arxk+r
find dydx and d2y
dx2 and put it into equation (5)
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Recurrence formula
Recurrence Relations of Bessel’s Function
1. ddx [xnJn(x)] = xnJn−1(x)
Proof
Jn(x) =∞∑
r=0
(−1)r( x2)n+2r 1
r!Γ(n + r + 1)
⇒ xnJn(x) =∞∑
r=0
(−1)r x2n+2r
2n+2r
1r!Γ(n + r + 1)
⇒ ddx
[xnJn(x)] =∞∑
r=0
(−1)r 2(n + r)x2n+2r−1
2n+2r
1r!(n + r)Γ(n + r)
∵ Γ(n + r + 1) = (n + r)Γ(n + r)
⇒ ddx
[xnJn(x)] = xn∞∑
r=0
(−1)r( x2)(n−1)+2r 1
r!Γ((n− 1) + r + 1)
⇒ ddx
[xnJn(x)] = xnJn(x). �
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2. ddx [x−nJn(x)] = −x−nJn+1(x)
Proof
Jn(x) =∞∑
r=0
(−1)r( x2)n+2r 1
r!Γ(n + r + 1)
⇒ x−nJn(x) =
∞∑r=0
(−1)r x2r
2n+2r
1r!Γ(n + r + 1)
⇒ ddx
[x−nJn(x)] =∞∑
r=1
(−1)r 2rx2r−1
2n+2r
1(r − 1)!rΓ(n + r)
= x−n∞∑
r=1
(−1)r( x2)n+2r−1 1
(r − 1)!Γ(n + r + 1)
= x−n∞∑
k=0
(−1)k( x2)(n+1)+2k 1
k!Γ((n + 1) + k + 1)
Putting r = k + 1
⇒ ddx
[x−nJn(x)] = x−nJn+1(x). �
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3. J′n(x) = Jn−1(x)− nx Jn(x)
Proof
From recurrence relation (1)ddx [xnJn(x)] = xnJn−1(x)
⇒ xnJ′n(x) + nxn−1Jn(x) = xnJn−1(x)Dividing by xn, we getJ′n(x) + n
x Jn(x) = Jn−1(x)⇒ J′n(x) = Jn−1(x)− n
x Jn(x).
4. J′n(x) = −Jn−1(x) + nx Jn(x)
Proof
From recurrence relation (2)ddx [x−nJn(x)] = −x−nJn+1(x)
⇒ x−nJ′n(x)− nx−n−1Jn(x) = −xnJn+1(x)Dividing by x−n, we get⇒ J′n(x) = −Jn+1(x) + n
x Jn(x).
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5. J′n(x) = 12 [Jn−1(x)− Jn+1(x)]
Proof
Adding recurrence relation (3) and (4), we getJ′n(x) = 1
2 [Jn−1(x)− Jn+1(x)].
6. 2nJn(x) = x[Jn−1(x) + Jn+1(x)]
Proof
Subtracting recurrence relation (3) and (4), we get2 n
x Jn(x) = Jn−1(x) + Jn+1
⇒ 2nJn(x) = x[Jn−1(x) + Jn+1(x)].
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Generating function for Pn(x)
The function (1−2xz + z2)−12 is called the generating function of Legendre’s ployno-
mials as (1− 2xz + z2)−12 =
∑∞n=0 Pn(x)zn
Proof: (1− 2xz + z2)−12 = [1− (2xz− z2)]−
12
= 1 + 12 (2xz− z2) + 1
234 (2xz− z2)2 + . . .+ 1
234 . . .
2k−12 (2xz− z2)k + . . .
∵ (1− t)−12 = 1 + 1
2 t + 12
32
t2
2! + . . .= 1 +
∑∞k=0
12
34 . . .
2k−12k (2xz− z2)k, ....(1)
Again (2xz− z2)k = zk[2x− z]k
= zk[(2x)k − k(2x)k−1z + k(k−1)2! (2x)k−2z2 − . . .+ (−1)kzk], ....(2)
Using (2) in (1) we get,(1− 2xz + z2)−
12 =
1+∑∞
k=012
34 . . .
2k−12k [(2x)kzk−k(2x)k−1zk+1+ k(k−1)
2! (2x)k−2zk+2−. . .+(−1)kz2k], ....(3)Coefficient of zn in expression (3) is given by12
34 . . .
2n−12n (2x)n− 1
234 . . .
(2n−3)(2n−2) (n−1)(2x)n−2+ 1
234 . . .
(2n−5)(2n−4)
(n−2)(n−3)2! (2x)n−4−. . .
= 1.3.5...(2n−1)n!
[xn − n(n−1)
2(2n−1) xn−2 + n(n−1)(n−2)(n−3)2.4(2n−1)(2n−3) xn−4 − . . .
]= Pn(x)
∴ (1− 2xz + z2)−12 =
∑∞n=0 Pn(x)zn. �
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Recurrence formula
Recurrence Relations of Legendre’s Function
1. (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)
Proof: From generating function
(1− 2xz + z2)−12 =
∞∑n=0
znPn(x) (7)
Differentiating both side of (7) partially with respect to z, we get− 1
2 (1− 2xz + z2)−32 (−2x + 2z) =
∑∞n=0 nzn−1Pn(x)
⇒ (x− z)(1− 2xz + z2)−32 =
∑∞n=0 nzn−1Pn(x)
⇒ (x− z)(1− 2xz + z2)−12 = (1− 2xz + z2)
∑∞n=0 nzn−1Pn(x)
⇒ (x− z)∑∞
n=0 znPn(x) = (1− 2xz + z2)∑∞
n=0 nzn−1Pn(x), by using (7)Equating coefficient of zn on both sidexPn(x)− Pn−1(x) = (n + 1)Pn+1(x)− 2xnPn(x) + (n− 1)Pn−1(x)(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x).
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2. Pn(x) = P′n+1(x)− 2xP′n(x) + P′n−1(x)
Proof: Differentiating recurrence relation (7) partially with respect to x, we get− 1
2 (1− 2xz + z2)−32 (−2z) =
∑∞n=0 znP′n(x)
⇒ z(1− 2xz + z2)−12 = (1− 2xz + z2)
∑∞n=0 znP′n(x)
⇒ z∑∞
n=0 znPn(x) = (1− 2xz + z2)∑∞
n=0 znP′n(x), by using (7)Equating coefficient of zn+1 on both sidePn(x) = P′n+1(x)− 2xP′n(x) + P′n−1(x).
3. nPn(x) = xP′n(x)− P′n−1(x)
Proof: Differentiating recurrence relation (7) partially with respect to x, we get(n + 1)P′n+1(x) = (2n + 1)xP′n(x) + (2n + 1)Pn(x)− nP′n−1(x), . . . (a)Also from recurrence relation (2)P′n+1(x) = Pn(x) + 2xP′n(x)− P′n−1(x), . . . (b)Using (a) and (b) , we get(n + 1)[Pn(x) + 2xP′n(x)−P′n−1(x)] = (2n + 1)xP′n(x) + (2n + 1)Pn(x)− nP′n−1(x)⇒ nPn(x) = xP′n(x)− P′n−1(x)
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4. (n + 1)Pn(x) = P′n+1(x)− xP′n(x)
Proof: Adding recurrence relation (2) and (3), we get(n + 1)Pn(x) = P′n+1(x)− xP′n(x).
5. (2n + 1)Pn(x) = P′n+1(x)− P′n−1(x)
Proof: Adding recurrence relation (3) and (4), we get(2n + 1)Pn(x) = P′n+1(x)− P′n−1(x)
6. (1− x2)P′n(x) = n[Pn−1(x)− xPn(x)]
Proof: Replacing n by (n-1) in recurrence relation (4)nPn−1(x) = P′n(x)− xP′n−1(x) . . . (c)Also multiplying recurrence relation (3) by xnxPn(x) = x2P′n(x)− xP′n−1(x) . . . (d)Subtracting (d) from (c)(1− x2)P′n(x) = n[Pn−1(x)− xPn(x)]
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AssignmentProve following
(1) Pn(1) = 1. (2) Pn(−1) = (−1)n.
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Orthogonality
�Orthogonality of Legendre’s ploynomialOrthogonality property of Legendre’s polynomials is given by the relations∫ 1−1 Pm(x)Pn(x)dx = 0 , when m 6= n
and∫ 1−1 P2
n(x)dx = 22n+1 , when m = n.
where m and n are positive integers.
�Orthogonality of Bessel’s functionIf α and β be the roots of Jn(x) = 0, then∫ 1
0 xJn(αx)Jn(βx)dx = 0 , if α 6= βand∫ 1
0 xJn(αx)Jn(βx)dx = 12 J2
n+1(α), if α = β.where m and n are positive integers.