Lecture notes on mathematics

Dr. Sunil KumarAssociate professor

Department of Mathematics,National Institute of Technology, Jamshedpur.

April 1, 2020

Outlines

1 Series solution2 Frobenius method3 Legendre differential equation4 Bessel’s differential equation5 Recurrence formula6 Generating functions7 Orthogonality

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.

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.

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

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.

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.

Legendre differential equation

Legendre differential equation

On substituting it into the equation (4) we obtain

Legendre differential equation

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)

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). �

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). �

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).

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)].

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. �

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).

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)

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)]

AssignmentProve following

(1) Pn(1) = 1. (2) Pn(−1) = (−1)n.

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.