1

18 Series Solution and Special Functions

18.1 INTRODUCTION

Generally the solutions of ordinary differential equations are obtainable in explicit form called a

closed form of the solution. However, many differential equations arising in physical problems are

linear but have variable coefficients and do not permit a general solution in terms of known functions.

For such equations, it is easier to find a solution in the form of an infinite convergent series called

power series solution. The series solution of certain differential equations give rise to special

functions such as Bessel’s functions, Legendre’s polynomials, Lagurre’s polynomial, Hermite’s

polynomial, Chebyshev polynomials. Strum-Liovelle problem based on orthogonality of functions is

also included which shows that Bessel’s, Legendre’s and other equations can be determined from a

common point of view.

18.2 POWER SERIES SOLUTION OF DIFFERENTIAL EQUATIONS

Consider the differential equation

𝑃0 𝑥 𝑑2𝑦

𝑑𝑥2 + 𝑃1 𝑥 𝑑𝑦

𝑑𝑥+ 𝑃2 𝑥 𝑦 = 0 … (1)

where 𝑃𝑖 ′𝑠 are polynomials in 𝑥.

If 𝑃0 𝑎 ≠ 0, then 𝑥 = 𝑎 is called an ordinary point of (1), otherwise a singular point. Ordinary

point is also called a regular point of the equation.

A singular point 𝑥 = 𝑎 of (1) is called regular singular point if, (1) can be put in the form

𝑑2𝑦

𝑑𝑥2 +𝑄1 𝑥

𝑥−𝑎

𝑑𝑦

𝑑𝑥+

𝑄2 𝑥

𝑥−𝑎 2 𝑦 = 0 … (2)

provided 𝑄1 𝑥 and 𝑄2 𝑥 both possess derivatives of all orders in the neighborhood of 𝑎.

A singular point which is not regular is called an irregular singular point.

Note: The power series method sometimes fails to yield a solution

e.g. 𝑥2𝑦′′ + 𝑥𝑦′ + 𝑦 = 0 …(3)

dividing by 𝑥2 throughout, 𝑥2𝑦′′ + 𝑥𝑦′ + 𝑦 = 0 …(4)

Here neither of the terms 𝑃1 𝑥 =1

𝑥 and 𝑃2 𝑥 =

1

𝑥2 is defined at 𝑥 = 0, so we cannot find a

power series representation for 𝑃1 𝑥 or 𝑃2 𝑥 that converges in an open interval containing

𝑥 = 0.

Theorem I: If 𝑥 = 𝑎 is an ordinary point of the differential equation (1), i.e. 𝑃0 𝑎 ≠ 0, then

series solution of (1) can be found as:

𝑦 = 𝑎0 + 𝑎1(𝑥 − 𝑎) + 𝑎2(𝑥 − 𝑎)2 + ⋯ … (5)

2

Calculate the derivatives 𝑑𝑦

𝑑𝑥,𝑑2𝑦

𝑑𝑥2 from (5), and substitute the values of y and its derivatives in

differential equation (1).

The values of the constants 𝑎2 ,𝑎3 ,𝑎4 ,… are obtained by equating to zero the coefficients of

various powers of 𝑥.

Putting the values of these constants in the solution (5), the desired power series solution of (1) is

obtained with 𝑎0 ,𝑎1 as its arbitrary constants.

Theorem II: When 𝑥 = 𝑎 is a regular singularity of (1) at least one of the solutions can be expressed

as,

𝑦 = (𝑥 − 𝑎)𝑚 [𝑎0 + 𝑎1(𝑥 − 𝑎) + 𝑎2(𝑥 − 𝑎)2 + ⋯ ] …(6)

Theorem III: The series (5) and (6) are convergent at every point within the circle of convergence at

𝑎. A solution in series will be valid only if the series is convergent.

Example 1: Solve in series the equation 𝒅𝟐𝒚

𝒅𝒙𝟐− 𝒙𝒚 = 𝟎.

Solution: Given differential equation is

𝑑2𝑦

𝑑𝑥2 − 𝑥𝑦 = 0 … (1)

Here 𝑃0 𝑥 = 1, so 𝑃0 0 = 1, i.e. 𝑥 = 0 is the ordinary point of the differential equation (1).

Let the solution of differential equation (1) be

𝑦 = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + 𝑎3𝑥

3 + 𝑎4𝑥4 + 𝑎5𝑥

5 + ⋯ … (2)

Differentiating (2) w.r.t. 𝑥,

𝑑𝑦

𝑑𝑥= 𝑎1 + 2𝑎2𝑥 + 3𝑎3𝑥

2 + 4𝑎4𝑥3 + 5𝑎5𝑥

4 + ⋯ … (3)

Again differentiating w.r.t 𝑥

𝑑2𝑦

𝑑𝑥2 = 2𝑎2 + 6𝑎3𝑥 + 12𝑎4𝑥2 + 20𝑎5𝑥

3 + ⋯ … (4)

Substitute values of y from (2) and its derivative from (4) in the differential equation (1), we get

2𝑎2 + 6𝑎3𝑥 + 12𝑎4𝑥2 + 20𝑎5𝑥

3 + ⋯

−𝑥 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + 𝑎3𝑥

3 + 𝑎4𝑥4 + 𝑎5𝑥

5 + ⋯ = 0

=> 2𝑎2 + 6𝑎3 − 𝑎0 𝑥 + 12𝑎4 − 𝑎1 𝑥2 + 20𝑎5 − 𝑎2 𝑥

3 + ⋯ = 0

Equating each of the coefficients to zero, we obtain the identities,

2𝑎2 = 0, 6𝑎3 − 𝑎0 = 0, 12𝑎4 − 𝑎1 = 0, 20𝑎5 − 𝑎2 = 0

which further gives 𝑎2 = 0, 𝑎3 =1

6𝑎0 , 𝑎4 =

1

12𝑎1 , 𝑎5 =

1

20𝑎2 = 0

Generalizing the results, 𝑎𝑛+2 =𝑎𝑛−1

𝑛+2 (𝑛+1) … (5)

3

Putting 𝑛 = 4, 5, 6… in (5), we get

𝑎6 =1

6 (5)𝑎3 =

1

6 6 (5)𝑎0 =

1

180𝑎0 ,

𝑎7 =1

7 (6)𝑎4 =

1

12 7 (6)𝑎1 =

1

504𝑎1 ,

𝑎8 = 0.

Using the values of the constants in (2), the general solution of differential equation (1) becomes

𝑦 = 𝑎0 1 +1

6𝑥3 +

1

180𝑥6 + ⋯ + 𝑎1 𝑥 +

1

12𝑥4 +

1

504𝑥7 + ⋯ .

Example 2:

ASSIGNMENT 18.1

Solve the following differential equations in series

1. 𝑑2𝑦

𝑑𝑥2 + 𝑥𝑑𝑦

𝑑𝑥+ 𝑦 = 0.

2. 𝑑2𝑦

𝑑𝑥2 + 𝑥𝑦 = 0.

3. 1 − 𝑥2 𝑑2𝑦

𝑑𝑥2 − 𝑥𝑑𝑦

𝑑𝑥+ 4𝑦 = 0.

4. 𝑑2𝑦

𝑑𝑥2 + 𝑦 = 0, given 𝑦 0 = 0.

5. 1 − 𝑥2 𝑦′′ + 2𝑦 = 0,𝑔𝑖𝑣𝑒𝑛 𝑦 0 = 4, 𝑦′ 0 = 5.

ANSWERS

1. 𝑦 = 𝑎0 1 −𝑥2

2+

𝑥4

2.4−

𝑥6

2.4.6+ ⋯ + 𝑎1 𝑥 −

𝑥3

3+

𝑥5

3.5−

𝑥7

3.5.7+ ⋯

2. 𝑦 = 𝑎0 1 −1

3!𝑥3 +

1.4

6!𝑥6 −

1∙4.7

9!𝑥9 …

+𝑎1 𝑥 −1.2

4!𝑥4 +

2.7

7!𝑥7 + ⋯

3. 𝑦 = 𝑎0 1 − 2𝑥2 + 𝑎1𝑥 1 −𝑥2

2−

𝑥4

8−

3

6∙𝑥6

8−

5∙3

8∙6∙𝑥8

8−⋯

4. 𝑦 = 𝑎0 𝑥 −𝑥3

3!+

𝑥5

5!−⋯

5. 𝑦 = 4 + 5𝑥 − 4𝑥2 −5

3𝑥3 −

𝑥5

3−

𝑥7

7−⋯

4

18.3 FROBENIUS METHOD

This method is named after a German mathematician F.G. Frobenius (1849 – 1917) who is

known for his contributions to the theory of matrices and groups. This method is employed to find the

power series solution of the differential equation

𝑃0 𝑥 𝑑2𝑦

𝑑𝑥2 + 𝑃1 𝑥 𝑑𝑦

𝑑𝑥+ 𝑃2 𝑥 𝑦 = 0 … (1)

when 𝑥 = 0 is the regular singularity.

Working Procedure

(i) Let 𝑦 = 𝑥𝑚 (𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + 𝑎3𝑥

3 + ⋯+ 𝑎𝑛𝑥𝑛 + ⋯ ) … (2)

be the solution of the differential equation (1), where m is some real or complex number.

(ii) Substitute in (1) the values of 𝑦,𝑑𝑦

𝑑𝑥 ,

𝑑2𝑦

𝑑𝑥2 obtained by differentiating (2).

(iii) Find the indicial equation (a quadratic equation) by equating to zero the coefficient of the

lowest degree term in x.

(iv) Find the values of 𝑎1 , 𝑎2 , 𝑎3 , ⋯ in terms of 𝑎0 by equating to zero the coefficients of

other powers of x.

(v) Find the roots 𝑚1 , 𝑚2 (say) of the indicial equation. The complete solution depends on

the nature of roots of the indicial equation.

Case I: Roots 𝒎𝟏, 𝒎𝟐 are distinct and do not differ by an integer

In this case, the differential equation (1) has two linearly independent solutions of the following

forms:

𝑦1 = 𝑥𝑚1 (𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + 𝑎3𝑥

3 + ⋯ )

𝑦2 = 𝑥𝑚1 (𝑏0 + 𝑏1𝑥 + 𝑏2𝑥2 + 𝑏3𝑥

3 + ⋯ )

The complete solution of the differential equation is given by

𝑦 = 𝑐1𝑦1 + 𝑐2𝑦2.

Example 3: Solve 𝟒𝒙𝒅𝟐𝒚

𝒅𝒙𝟐+ 𝟐

𝒅𝒚

𝒅𝒙+ 𝒚 = 𝟎

Solution: Given 4𝑥𝑑2𝑦

𝑑𝑥2 + 2𝑑𝑦

𝑑𝑥+ 𝑦 = 0 … (1)

Here 𝑥 = 0 is a singular point, let its solution be

𝑦 = 𝑎0𝑥𝑚 + 𝑎1𝑥

𝑚+1 + 𝑎2𝑥𝑚+2 + 𝑎3𝑥

𝑚+3 + 𝑎4𝑥𝑚+4 + … … (2)

5

From equation (2)

𝑑𝑦

𝑑𝑥 = 𝑚𝑎0𝑥

𝑚−1 + 𝑚 + 1 𝑎1𝑥𝑚 + 𝑚 + 2 𝑎2𝑥

𝑚+1

+ 𝑚 + 3 𝑎3𝑥𝑚+2 + …… … (3)

𝑑2𝑦

𝑑𝑥2 = 𝑚 𝑚 − 1 𝑎0𝑥𝑚−2 + 𝑚 + 1 𝑚 𝑎1𝑥

𝑚−1

+ 𝑚 + 2 𝑚 + 1 𝑎2𝑥𝑚 + …… … (4)

Putting the above values in equation (1), we get

4𝑥 𝑚 𝑚 − 1 𝑎0𝑥𝑚−2 + 𝑚 + 1 𝑚 𝑎1𝑥

𝑚−1 + 𝑚 + 2 𝑚 + 1 𝑎2𝑥𝑚 + ……

+2 𝑚𝑎0𝑥𝑚−1 + 𝑚 + 1 𝑎1𝑥

𝑚 + 𝑚 + 2 𝑎2𝑥𝑚+1 + 𝑚 + 3 𝑎3𝑥

𝑚+2 + ……

+ 𝑎0𝑥𝑚 + 𝑎1𝑥

𝑚+1 + 𝑎2𝑥𝑚+2 + 𝑎3𝑥

𝑚+3 + 𝑎4𝑥𝑚+4 + …… = 0 …

(5)

Equating the coefficients of 𝑥𝑚−1 equal to zero

4𝑚 𝑚 − 1 𝑎0 + 2𝑚𝑎0 = 0 ⇒ 𝑎0 4𝑚2 − 4𝑚 + 2𝑚 = 0

Because 𝑎0 ≠ 0 ⇒ 4𝑚2 − 2𝑚 = 0 i.e. 𝑚 = 0,1

2

∴ The solution of the indicial equation is 𝑚1 = 0 and 𝑚2 =1

2.

Here, the roots are real, distinct and do not differ by an integer.

∴ Its solution is 𝑦 = 𝑐1𝑦1 + 𝑐2𝑦2 … (6)

On equating coefficients of 𝑥𝑚 , we get

4 𝑚 + 1 𝑚𝑎1 + 2 𝑚 + 1 𝑎1 + 𝑎0 = 0 or

2 𝑚 + 1 2𝑚 + 1 𝑎1 = −𝑎0

⇒ 𝑎1 =−𝑎0

2 𝑚+1 (2𝑚+1) … (7)

Likewise, 4 𝑚 + 2 𝑚 + 1 𝑎2 + 2 𝑚 + 2 𝑎2 + 𝑎1 = 0

𝑚 + 2 4𝑚 + 4 + 2 𝑎2 = −𝑎1 or

2 𝑚 + 2 2𝑚 + 3 𝑎2 = −𝑎1

⇒ 𝑎2 = −𝑎1

2 𝑚+2 (2𝑚+3)=

𝑎0

22 𝑚+2 𝑚+1 2𝑚+1 (2𝑚+3) … (8)

and 4 𝑚 + 3 𝑚 + 2 𝑎3 + 2 𝑚 + 3 𝑎3 + 𝑎2 = 0

𝑚 + 3 4𝑚 + 8 + 2 𝑎3 = −𝑎2

2 𝑚 + 3 2𝑚 + 5 𝑎3 = −𝑎0

22 𝑚+2 𝑚+1 2𝑚+1 (2𝑚+3)

⇒ 𝑎3 = −𝑎0

23 𝑚+3 𝑚+2 𝑚+1 2𝑚+1 2𝑚+3 (2𝑚+5) and so on. … (9)

6

Thus, for 𝑚 = 0, we get

𝑦(𝑚=0) = 𝑦1 = 𝑥𝑚 (𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + …… ) 𝑚=0

= 𝑎0 1 −1

2

𝑥

1.1+

1

22

𝑥2

2.1.1.3−

1

23

𝑥3

3.2.1.1.3.5+ ……

= 𝑎0 1 − 𝑥

2

2!+

𝑥 4

4!−

𝑥 6

6!+ …… = 𝑎0 cos 𝑥 … (10)

Likewise for 𝑚 =1

2, we get

𝑦(𝑚=

1

2)

= 𝑦2 = 𝑎0𝑥1

2 1 −1

21

𝑥3

2.2

+1

22

𝑥2

5

2.3

2.2.4

−1

23

𝑥3

7

2.5

2.3

2.2.4.6

+ ……

= 𝑎0 𝑥 − 𝑥

3

3!+

𝑥 5

5!−

𝑥 7

7!+ …… = 𝑎0 sin 𝑥 … (11)

Hence, on substituting the values of 𝑦1 and 𝑦2in equation (3), we get solution as:

𝑦 = 𝑐1𝑦1 + 𝑐2𝑦2 = 𝐶1 cos 𝑥 + 𝐶2 sin 𝑥 .

Example 4: Find the series solution of the equation

𝟐𝒙𝟐𝒅𝟐𝒚

𝒅𝒙𝟐− 𝒙

𝒅𝒚

𝒅𝒙+ 𝟏 − 𝒙𝟐 𝒚 = 𝟎

OR

Solve the equation 𝟐𝒙𝟐𝒅𝟐𝒚

𝒅𝒙𝟐− 𝒙

𝒅𝒚

𝒅𝒙+ 𝟏 − 𝒙𝟐 𝒚 = 𝟎 in power series.

Solution: Given 2𝑥2 𝑑2𝑦

𝑑𝑥2 − 𝑥𝑑𝑦

𝑑𝑥+ 1 − 𝑥2 𝑦 = 0 … (1)

Let its solution be

𝑦 = 𝑎0𝑥𝑚 + 𝑎1𝑥

𝑚+1 + 𝑎2𝑥𝑚+2 + 𝑎3𝑥

𝑚+3 + …… … (2)

So that 𝑑𝑦

𝑑𝑥= 𝑚𝑎0𝑥

𝑚−1 + 𝑚 + 1 𝑎1𝑥𝑚 + 𝑚 + 2 𝑎2𝑥

𝑚+1

+ 𝑚 + 3 𝑎3𝑥𝑚+2 + …… … (3)

And 𝑑2𝑦

𝑑𝑥2 = 𝑚 𝑚 − 1 𝑎0𝑥𝑚−2 + 𝑚 + 1 𝑚 𝑎1𝑥

𝑚−1

+ 𝑚 + 2 𝑚 + 1 𝑎2𝑥𝑚 + …… … (4)

On substituting the values of 𝑦, 𝑑𝑦

𝑑𝑥,

𝑑2𝑦

𝑑𝑥2 in the given equation, we get

2𝑥2 𝑚 𝑚 − 1 𝑎0𝑥𝑚−2 + 𝑚 + 1 𝑚 𝑎1𝑥

𝑚−1 + 𝑚 + 2 𝑚 + 1 𝑎2𝑥𝑚 + ……

−𝑥 𝑚𝑎0𝑥𝑚−1 + 𝑚 + 1 𝑎1𝑥

𝑚 + 𝑚 + 2 𝑎2𝑥𝑚+1 + 𝑚 + 3 𝑎3𝑥

𝑚+2 + ……

+ 1 − 𝑥2 𝑎0𝑥𝑚 + 𝑎1𝑥

𝑚+1 + 𝑎2𝑥𝑚+2 + 𝑎3𝑥

𝑚+3 + …… = 0

i.e. 2𝑚 𝑚 − 1 𝑎0𝑥𝑚 + 2 𝑚 + 1 𝑚𝑎1𝑥

𝑚+1 + 2 𝑚 + 2 𝑚 + 1 𝑎2𝑥𝑚+1 + ……

7

− 𝑚𝑎0𝑥𝑚 + 𝑚 + 1 𝑎1𝑥

𝑚+1 + 𝑚 + 2 𝑎2𝑥𝑚+2 + ……

+ 𝑎0𝑥𝑚 + 𝑎1𝑥

𝑚+1 + 𝑎2𝑥𝑚+2 + …… − 𝑎0𝑥

𝑚+2 + 𝑎1𝑥𝑚+3 + …… … (5)

On equating the coefficients of lowest power of 𝑥 (i.e. 𝑥𝑚 ) equal to zero on both sides,

2𝑚 𝑚 − 1 𝑎0 −𝑚𝑎0 + 𝑎0 = 0 … (6)

⇒ 𝑎0 2𝑚− 1 𝑚 − 1 = 0

⇒ Either 𝑎0 = 0 or 𝑚 = 1,1

2 .

Now equating the coefficients of 𝑥𝑚+1 equal to zero,

2 𝑚 + 1 𝑚 𝑎1 − 𝑚 + 1 𝑎1 + 𝑎1 = 0

⇒ 𝑎1𝑚 2𝑚− 1 = 0

Which implies either 𝑎1 = 0 or 𝑚 = 0, but 𝑚 ≠ 0,

∴ 𝑎1 = 0 … (7)

On comparing the coefficients of 𝑥𝑚+2,

2 𝑚 + 2 𝑚 + 1 𝑎2 − 𝑚 + 2 𝑎2 + 𝑎2 − 𝑎0 = 0

⇒ 2𝑚2 + 6𝑚 + 4 − 𝑚 + 1 𝑎2 = 𝑎0

⇒ 2𝑚2 + 5𝑚 + 3 𝑎2 = 𝑎0

⇒ 𝑎2 = 𝑎0

𝑚+1 2𝑚+3 . … (8)

Likewise, on comparing the coefficients of 𝑥𝑚+3,

2 𝑚 + 3 𝑚 + 2 𝑎3 − 𝑚 + 3 𝑎3 + 𝑎3 − 𝑎1 = 0

⇒ 2 𝑚 + 3 𝑚 + 2 𝑎3 − 𝑚 + 3 + 1 𝑎3 = 𝑎1

⇒ 𝑎3 = 0 (since 𝑎1 = 0) … (9)

Further, coefficients of 𝑥𝑚+4,

2 𝑚 + 4 𝑚 + 3 𝑎4 − 𝑚 + 4 𝑎4 + 𝑎4 − 𝑎2 = 0

⇒ 2 𝑚 + 4 𝑚 + 3 − 𝑚 + 4 + 𝑎2 = 𝑎2

2𝑚2 + 13𝑚 + 21 𝑎4 = 𝑎2

⇒ 𝑎4 = 𝑎2

𝑚+3 2𝑚+7 and so on … … (10)

Now for 𝑚 = 1, 𝑎2 = 𝑎0

1+1 2.1+3 =

𝑎0

2.5 from (8) … (11)

𝑎4 = 𝑎2

4.9=

𝑎0

2.5.4.9 from (10) … (12)

…… …… …… …… ……

For 𝑚 =1

2 , 𝑎2 =

𝑎0

1

2+1 2.

1

2+3

=𝑎0

3

2 . 4

=𝑎0

2.3 … (13)

𝑎4 =𝑎2

𝑚+1 2𝑚+7 =

𝑎0

2.3

1

1

2+3

2

2+7

=𝑎0

2.3.4.7 … (14)

…… …… …… …… ……

8

Thus 𝑦1 = 𝑦 𝑚=1 = 𝑎0𝑥𝑚 + 𝑎1𝑥

𝑚+1 + 𝑎2𝑥𝑚+2 + 𝑎3𝑥

𝑚+3 + ……

= 𝑎0𝑥 1 +𝑥2

2.5+

𝑥4

2.4.5.9+

𝑥6

2.4.5.9.6.13+ ……

𝑦2 = 𝑦 𝑚=

1

2

= 𝑎0𝑥1

2 1 +𝑥2

2.3+

𝑥4

2.3.4.7+

𝑥6

2.3.4.5.7.11+ ……

Hence 𝑦 = 𝐶1𝑦1 + 𝐶2𝑦2.

Case II: Roots 𝒎𝟏, 𝒎𝟐 are equal, i.e. 𝒎𝟏 = 𝒎𝟐.

In this case, one of the linearly independent solutions 𝑦1 is obtained by substituting 𝑚 = 𝑚1 and

the second solution is obtained as

𝑦2 = 𝜕𝑦

𝜕𝑚 𝑚=𝑚1

.

Thus the complete solution is given by

𝑦 = 𝑐1𝑦1 + 𝑐2 𝜕𝑦

𝜕𝑚 𝑚=𝑚1

.

Example 5: Solve 𝒙𝒅𝟐𝒚

𝒅𝒙𝟐+

𝒅𝒚

𝒅𝒙− 𝒚 = 𝟎.

Solution: Given 𝑥𝑑2𝑦

𝑑𝑥2 +𝑑𝑦

𝑑𝑥− 𝑦 = 0 … (1)

Let its solution be 𝑦 = 𝑎0𝑥𝑚 + 𝑎1𝑥

𝑚+1 + 𝑎2𝑥𝑚+2 + …… … (2)

∴ 𝑑𝑦

𝑑𝑥= 𝑚𝑎0𝑥

𝑚−1 + 𝑚 + 1 𝑎1𝑥𝑚 + 𝑚 + 2 𝑎2𝑥

𝑚+1 + …… … (3)

⇒ 𝑑2𝑦

𝑑𝑥2 = 𝑚 𝑚 − 1 𝑎0𝑥𝑚−2 + 𝑚 + 1 𝑚 𝑎1𝑥

𝑚−1

+ 𝑚 + 2 𝑚 + 1 𝑥𝑚 + …… … (4)

Putting the above values in equation (1), we have

𝑥 𝑚 𝑚 − 1 𝑎0𝑥𝑚−2 + 𝑚 + 1 𝑚 𝑎1𝑥

𝑚−1 + 𝑚 + 2 𝑚 + 1 𝑎2𝑥𝑚 + ……

+ 𝑚𝑎0𝑥𝑚−1 + 𝑚 + 1 𝑎1𝑥

𝑚 + 𝑚 + 2 𝑎2𝑥𝑚+1 + ……

− 𝑎0𝑥𝑚 + 𝑎1𝑥

𝑚+1 + 𝑎2𝑥𝑚+2 + …… … (5)

Equating the coefficients of 𝑥𝑚−1 to zero,

𝑚 𝑎0 + 𝑚(𝑚 − 1)𝑎0 = 0

9

⇒ Either 𝑎0 = 0 or 𝑚2 = 0

But 𝑎0 ≠ 0 ∴ 𝑚 = 0, 0.

Now equate the coefficients of 𝑥𝑚 on both sides,

𝑚 + 1 𝑎1 + 𝑚 𝑚 + 1 𝑎1 + 𝑎0 = 0 or (𝑚 + 1)2𝑎1 + 𝑎0 = 0

⇒ 𝑎1 = −𝑎0

(𝑚+1)2. … (6)

Next equate the coefficients of 𝑥𝑚+1 on both sides,

𝑚 + 2 𝑚 + 1 𝑎2 + 𝑚 + 2 𝑎2 − 𝑎1 = 0

⇒ 𝑚 + 2 𝑎2 𝑚 + 1 + 1 − 𝑎1 = 0 or 𝑚 + 2 2𝑎2 − 𝑎1 = 0

⇒ 𝑎2 = 𝑎1

𝑚+2 2 = 𝑎0

𝑚+1 2 𝑚+2 2 and so on. ... (7)

Putting the values of 𝑎1, 𝑎2, ….in the assumed series solution (2),

𝑦 = 𝑎0𝑥𝑚 1 +

𝑥

𝑚+1 2 +𝑥2

𝑚+1 2 𝑚+2 2 +𝑥3

𝑚+1 2 𝑚+2 2 𝑚+3 2 + …… … (8)

Differentiating (8) partially with respect to 𝑚

𝜕𝑦

𝜕𝑚 = 𝑎0𝑥

𝑚 log𝑥 1 +𝑥

(𝑚+1)2 +𝑥2

𝑚+1 2 𝑚+2 2 + ……

+𝑎0𝑥𝑚 0 −

2𝑥

𝑚+1 3 −2𝑥2

𝑚+1 2 𝑚+2 2 2𝑚+3

𝑚+1 (𝑚+2) + ……

= 𝑎0𝑥𝑚 log𝑥 1 +

𝑥

𝑚+1 2 +𝑥2

𝑚+1 2 𝑚+2 2 +𝑥3

𝑚+1 2 𝑚+2 2 𝑚+3 2 + ……

−2𝑎0𝑥𝑚

𝑥

𝑚+1 2 𝑚+1 +

𝑥2

𝑚+1 2 𝑚+2 2 1

𝑚+1 +

1

𝑚+2 +

𝑥3𝑚+12𝑚+22𝑚+32+ …… … (9)

Now 𝑦1 = 𝑦(𝑚=0) = 𝑎0𝑥 1 +𝑥

12 +𝑥2

12 .22 + …… … (10)

𝑦2 = 𝜕𝑦

𝜕𝑚 𝑚=0

= 𝑦1 log𝑥

−2𝑎0 𝑥

1!+

1

2!2 1 +1

2 𝑥2 +

1

3!2 1 +1

2+

1

3 𝑥3 + …… … (11)

Therefore, the complete solution is

10

𝑦 = 𝐶1 + 𝐶2 log𝑥 1 +𝑥

1!2 +𝑥2

2!2 +𝑥3

2!3 + ……

−2𝐶2 𝑥 +1

2!2 1 +1

2 𝑥2 +

1

3!2 1 +1

2+

1

3 𝑥3 + …… .

Case III: Roots 𝒎𝟏, 𝒎𝟐 are distinct and differ by an integer.

In this case, assume that 𝑚1 < 𝑚2 . If some of the coefficient of y series becomes infinite when

𝑚 = 𝑚1, we modify the form of y replacing 𝑎0 by 𝑏0 𝑚 −𝑚1 . Then the complete solution is given

by

𝑦 = 𝑐1(𝑦)𝑚2+ 𝑐1

𝜕𝑦

𝜕𝑚 𝑚1

Example 5: Solve the equation 𝒙 𝟏 − 𝒙 𝒅𝟐𝒚

𝒅𝒙𝟐− 𝟑

𝒅𝒚

𝒅𝒙+ 𝟐𝒚 = 𝟎

Solution: Given 𝑥 1 − 𝑥 𝑑𝑦

𝑑𝑥− 3

𝑑𝑦

𝑑𝑥+ 2𝑦 = 0 … (1)

Let its solution be 𝑦 = 𝑎0𝑥𝑚 + 𝑎1𝑥

𝑚+1 + 𝑎2𝑥𝑚+2 + …… … (2)

∴ 𝑑𝑦

𝑑𝑥= 𝑚𝑎0𝑥

𝑚−1 + 𝑚 + 1 𝑎1𝑥𝑚 + 𝑚 + 2 𝑎2𝑥

𝑚+1 + …… … (3)

and 𝑑2𝑦

𝑑𝑥2 = 𝑚 𝑚 − 1 𝑎0𝑥𝑚−2 + 𝑚 + 1 𝑚 𝑎1𝑥

𝑚−1

+ 𝑚 + 2 𝑚 + 1 𝑎2𝑥𝑚 + …… … (4)

On substituting these values of 𝑦, 𝑑𝑦

𝑑𝑥,

𝑑2𝑦

𝑑𝑥2 in the given differential equation,

𝑥 − 𝑥2 𝑚 𝑚 − 1 𝑎0𝑥𝑚−2 + 𝑚 + 1 𝑚 𝑎1𝑥

𝑚−1 + 𝑚 + 2 𝑚 + 1 𝑎2𝑥𝑚 + ……

−3 𝑚𝑎0𝑥𝑚−1 + 𝑚 + 1 𝑎1𝑥

𝑚 + 𝑚 + 2 𝑎2𝑥𝑚+1 + …… +2 𝑎0𝑥

𝑚 + 𝑎1𝑥𝑚+1 + 𝑎2𝑥

𝑚+2 +

……=0 … (5)

On equating the coefficients of lowest power of 𝑥 (i.e. 𝑥𝑚−1) on both sides,

𝑎0𝑚 𝑚 − 1 − 3𝑎0 = 0 or 𝑎0 𝑚 𝑚 − 4 = 0

⇒ Either 𝑎0 = 0 or 𝑚 𝑚 − 4 = 0

But as 𝑎0 ≠ 0 ∴ 𝑚 = 0, 4

Likewise, equate the coefficients of 𝑥𝑚 , 𝑥𝑚+1, 𝑥𝑚+2 equal to zero, and find out the values of

unknowns 𝑎0, 𝑎1, 𝑎2 etc.

For the coefficients of 𝑥𝑚 ,

11

−𝑚 𝑚 − 1 𝑎0 − 3 𝑚 + 1 𝑎1 + 𝑚 + 1 𝑚𝑎1 + 2𝑎0 = 0

⇒ 𝑚 − 3 𝑚 + 1 𝑎1 = 𝑚 − 2 𝑚 + 1 𝑎0

⇒ 𝑎1 = 𝑚−2

𝑚−3 𝑎0 … (6)

For the coefficient of 𝑥𝑚+1,

− 𝑚 + 1 𝑚𝑎1 + 𝑚 + 2 𝑚 + 1 𝑎2 − 3 𝑚 + 2 𝑎2 + 2𝑎1 = 0

⇒ 𝑚 + 2 𝑚 − 2 𝑎2 = 𝑚 − 1 𝑚 + 2 𝑎1

⇒ 𝑎2 = 𝑚−1

𝑚−2 𝑎1 =

𝑚−1

𝑚−3 𝑎0 … (7)

Similarly,

𝑎3 =𝑚

𝑚−1 𝑎2 =

𝑚

𝑚−1

𝑚−1

𝑚−3 𝑎0 =

𝑚

𝑚−3 𝑎0

𝑎4 = 𝑚+1

𝑚 𝑎3 =

𝑚+1

𝑚

𝑚

𝑚−3 𝑎0 =

𝑚+1

𝑚−3 𝑎0

𝑎5 = 𝑚+2

𝑚+1 𝑎4 =

𝑚+2

𝑚+1

𝑚+1

𝑚−3 𝑎0 =

𝑚+2

𝑚−3 𝑎0 … 𝑠𝑜 𝑜𝑛

… (8)

∴ 𝑦 = 𝑎0𝑥𝑚 1 +

𝑚−2

𝑚−3 𝑥 +

𝑚−1

𝑚−3 𝑥2 +

𝑚

𝑚−3 𝑥3 +

𝑚+1

𝑚−3 𝑥4 + …… … (9)

Now, 𝑦1 = 𝑦 𝑚=0 = 𝑎0 1 +2

3𝑥 +

1

3𝑥2 −

1

3𝑥4 − ……

and 𝑦2 = 𝑦 𝑚=4 = 𝑎0𝑥4 1 +

2

1𝑥 +

3

1𝑥2 +

4

1𝑥3 +

5

1𝑥4 + ……

Hence the complete solution, 𝑦 = 𝑐1𝑦1 + 𝑐2𝑦2.

ASSIGNMENT 18.2

Use Frobenius method to solve the following differential equations:

1. 9𝑥 1 − 𝑥 𝑑2𝑦

𝑑𝑥2 − 12𝑑𝑦

𝑑𝑥+ 4𝑦 = 0

2. 4𝑥𝑑2𝑦

𝑑𝑥2 + 2 1 − 𝑥 𝑑𝑦

𝑑𝑥− 𝑦 = 0

3. 𝑥𝑑2𝑦

𝑑𝑥2 +𝑑𝑦

𝑑𝑥+ 𝑥𝑦 = 0

4. 𝑥 1 − 𝑥 𝑑2𝑦

𝑑𝑥2 − 1 + 3𝑥 𝑑𝑦

𝑑𝑥− 𝑦 = 0

5. 𝑥𝑑2𝑦

𝑑𝑥2 + 2𝑑𝑦

𝑑𝑥+ 𝑥𝑦 = 0

6. 2𝑥2𝑦 ′′ + 𝑥𝑦 ′ − 𝑥 + 1 𝑦 = 0

7. 2𝑥 1 − 𝑥 𝑑2𝑦

𝑑𝑥2 + 1 − 𝑥 𝑑𝑦

𝑑𝑥+ 3𝑦 = 0

12

ANSWERS

1. 𝑦 = 𝐶1 1 +1

3𝑥 +

1.4

3.6𝑥2 +

1.4.7

3.6.9𝑥3 + ……

+𝐶2𝑥7/3 1 +

8

10𝑥 +

8.11

10.13𝑥2 +

8.11.14

10.13.16𝑥3 + ……

2. 𝑦 = 𝐶1 1 +1

2.1!𝑥 +

1

22 . 2!𝑥2 +

1

23 . 3!𝑥3 + ……

+𝐶2𝑥1

2 1 +1

1 .3𝑥 +

1

1.3.5𝑥2 +

1

1.3.5.7𝑥3 + ……

3. 𝑦 = (𝐶1 + 𝐶2 log𝑥) 1 −1

22 𝑥2 +

1

22 .42 𝑥4 −

1

22 .42 .62 𝑥6 + ……

+𝐶2 1

22 𝑥2 −

1

22 .42 1 +1

2 𝑥4 +

1

22 .42 .62 1 +1

2+

1

3 𝑥6 + ……

4. 𝑦 = (𝐶1 + 𝐶2 log𝑥) 1.2𝑥2 + 2.3𝑥3 + 3.4𝑥4 + ……

+𝐶2 −1 + 𝑥 + 5𝑥2 + 11𝑥3 + ……

5. 𝑦 = 𝑥−1(𝑎0 cos𝑥 + 𝑎1 sin𝑥)

6. 𝑦 = 𝑎0𝑥 1 +𝑥

5+

𝑥2

70+ ⋯ +

𝑎1

𝑥 1 − 𝑥 −

𝑥2

2+ ⋯

7. 𝑦 = 𝑎0 𝑥 1 − 𝑥 + 𝑎1 1 − 3𝑥 +3𝑥2

1.3+

3𝑥3

3.5+

3𝑥4

5.7+ ⋯

18.4 BESSEL’S EQUATION

In applied mathematics, many physical problems involving vibrations or heat conduction in

cylindrical regions give rise the differential equation

𝑥2 𝑑2𝑦

𝑑𝑥2 + 𝑥𝑑𝑦

𝑑𝑥+ 𝑥2 − 𝑛2 𝑦 = 0 … (1)

which is known as the Bessel’s differential equation of order n. The particular solutions of this

differential equation are called Bessel’s functions of order n.

Let 𝑦 = 𝑎0𝑥𝑚 + 𝑎1𝑥

𝑚+1 + 𝑎2𝑥𝑚+2 + ……

∴ 𝑑𝑦

𝑑𝑥= 𝑚𝑎0𝑥

𝑚−1 + 𝑚 + 1 𝑎1𝑥𝑚 + 𝑚 + 2 𝑎2𝑥

𝑚+1 + ……

𝑑2𝑦

𝑑𝑥2 = 𝑚 𝑚 − 1 𝑎0𝑥𝑚−1 + 𝑚 + 1 𝑚 𝑎1𝑥

𝑚

+ 𝑚 + 2 𝑚 + 1 𝑎2𝑥𝑚 + ……

Putting these in the given differential equation, we get

⇒𝑥2 𝑚 𝑚 − 1 𝑎0𝑥𝑚−1 + 𝑚 + 1 𝑚 𝑎1𝑥

𝑚 + 𝑚 + 2 𝑚 + 1 𝑎2𝑥𝑚 + …… + 𝑥 𝑚𝑎0𝑥

𝑚−1 +

𝑚+1𝑎1𝑥𝑚+𝑚+2𝑎2𝑥𝑚+1+ ……+[𝑥2−𝑛2𝑎0𝑥𝑚+𝑎1𝑥𝑚+1+𝑎2𝑥𝑚+2+ ……=0

Equating to zero, the coefficient of lowest degree term in 𝑥, i.e. 𝑥𝑚

𝑚 𝑚 − 1 𝑎0 + 𝑚𝑎0 − 𝑛2𝑎0 = 0, 𝑎0 ≠ 0

∴ Indicial equation 𝑚 𝑚 − 1 + 𝑚 − 𝑛2 = 0

13

⇒ 𝑚2 − 𝑛2 = 0 ∴ 𝑚 = ±𝑛

Now coefficients of 𝑥𝑚+1:

𝑚 + 1 𝑚𝑎1 + 𝑚 + 1 𝑎1 − 𝑛2𝑎1 = 0

𝑚 + 1 2 − 𝑛2 𝑎1 = 0

⇒ 𝑚 is fixed 𝑎1 = 0

Coefficient of 𝑥𝑚+2: 𝑚 + 2 𝑚 + 1 𝑎2 + 𝑚 + 2 𝑎2 − 𝑛2𝑎2 + 𝑎0 = 0

𝑚 + 2 2 − 𝑛2 𝑎2 + 𝑎0 = 0

∴ 𝑎2 = −𝑎0

𝑚+2 2−𝑛2

Similarly, 𝑚 + 3 2 − 𝑛2 𝑎3 + 𝑎1 = 0

𝑎3 = −𝑎1

𝑚+3 2−𝑛2 = 0, as 𝑎1 = 0

So 𝑎1 = 𝑎3 = 𝑎5 = …… = 0

𝑎4 = −𝑎2

𝑚+4 2−𝑛2 =𝑎0

𝑚+2 2−𝑛2 𝑚+4 2−𝑛2

So 𝑦 = 𝑎0𝑥𝑚 1 −

𝑥2

𝑚+2 2−𝑛2 +𝑥4

𝑚+2 2−𝑛2 𝑚+4 2−𝑛2 − … (2)

Case 1: For 𝑛 = 0, 𝑚 = 0 as 𝑚 = ±𝑛

𝑦𝐼 = 𝑎0 1 −𝑥2

22 +𝑥4

22 .42 − ……

𝜕𝑦

𝜕𝑚= 𝑦 log𝑥 + 𝑎0𝑥

𝑚 −𝑥2

𝑚+2 2−𝑛2 −2

𝑚+2 2−𝑛2 +

𝑥4

𝑚+2 2−𝑛2 𝑚+4 2−𝑛2

−2

𝑚+2 2−𝑛2 −−2

𝑚+4 2−𝑛2 + ……

𝑦𝐼𝐼 = 𝜕𝑦

𝜕𝑚 𝑚=0

= 𝑦𝐼 log𝑥 + 𝑎0 −𝑥2

2.−2

22 +𝑥4

22 .42 −2

22 −−2

42 + ……

So, the solution is 𝑦 = 𝐶1𝑦𝐼 + 𝐶2𝑦𝐼𝐼

𝑦 = 𝐶1𝑎0 + 𝐶2 log 𝑥 1 −𝑥2

22 +𝑥4

22 .42 − ……

+𝐶2𝑎0 𝑥2

2.

2

22 −2𝑥4

22 .42 1

22 +2

42 + ……

Case 2: For 𝑛 non integral and equal to 𝑛 (𝑛 = 𝑚) replace 𝑎0 in equation (2) by 1

2𝑛 𝑛+1

We get 𝑦0 =1

2𝑛 𝑛+1𝑥𝑛 1 −

𝑥2

22 𝑛+1 +

𝑥4

22 𝑛+1 4.2 𝑛+2 + ……

= 𝑥

2 𝑛

1

𝑛+1−

1

𝑛+2 𝑥

2

2+

1

2! 𝑛+3 𝑥

2

4+ ……

= −1 𝑟1

𝑟 ! 𝑛+𝑟+1 𝑥

2 𝑛+2𝑟

= 𝐽𝑛 𝑥 ∞𝑟=0

i.e. 𝐽𝑛 𝑥 = −1 𝑟1

𝑟 ! 𝑛+𝑟+1 𝑥

2 𝑛+2𝑟

∞𝑟=0 … (3)

Similarly by putting 𝑚 = −𝑛, we get the other solution

14

𝐽−𝑛 𝑥 = −1 𝑟1

𝑟! −𝑛+𝑟+1 𝑥

2 −𝑛+2𝑟

∞𝑟=0

The resulting solution is

𝑦 = 𝐶1 𝐽𝑛 𝑥 + 𝐶2 𝐽−𝑛 𝑥

𝐽𝑛 𝑥 & 𝐽−𝑛 𝑥 as defined as above.

Case 3: If 𝑛 is integral

Let 𝑦 = 𝑢 𝑥 𝐽𝑛 𝑥 𝑦′ = 𝑢′ 𝑥 𝐽𝑛 + 𝑢 𝐽𝑛′

𝑦′′ = 𝑢′′ 𝐽𝑛 + 2𝑢′ 𝐽𝑛′ + 𝑢 𝐽𝑛

′′

Putting these in 𝑥2 𝑑2𝑦

𝑑𝑥2 + 𝑥𝑑𝑦

𝑑𝑥+ 𝑥2 − 𝑛2 𝑦 = 0

⇒ 𝑥2 𝑢′′ 𝐽𝑛 + 2𝑢′ 𝐽𝑛′ + 𝑢 𝐽𝑛

′′ + 𝑥 𝑢′ 𝐽𝑛 + 𝑢 𝐽𝑛′ + 𝑥2 − 𝑛2 𝑢 𝐽𝑛 = 0

⇒ 𝑢 𝑥2𝐽𝑛′′ + 𝑥𝐽𝑛

′ + 𝑥2 − 𝑛2 𝐽𝑛 + 2𝑢′𝑥2𝐽𝑛′ + 𝑥2𝑢′′ 𝐽𝑛 + 𝑥𝑢′ 𝐽𝑛 = 0

Now 𝐽𝑛 𝑥 is a solution of 𝑥2 𝑑2𝑦

𝑑𝑥2 + 𝑥𝑑𝑦

𝑑𝑥+ 𝑥2 − 𝑛2 𝑦 = 0

∴ 𝑥2𝐽𝑛′′ + 𝑥𝐽𝑛

′ + 𝑥2 − 𝑛2 𝐽𝑛 = 0

We get,

2𝑢′𝑥2𝐽𝑛′ + 𝑥2𝑢′′ 𝐽𝑛 + 𝑥𝑢′ 𝐽𝑛 = 0

⇒ 2𝐽𝑛′

𝐽𝑛+

𝑢 ′′

𝑢 ′ +1

𝑥= 0 (divide by 𝐽𝑛𝑢

′𝑥2)

Integrating 2 log𝑒 𝐽𝑛 + log𝑒 𝑢′ log𝑒 𝑥 = log 𝑐

⇒ 𝑢′𝐽𝑛2𝑥 = 𝐵 Where 𝐵 is constant of integration.

⇒ 𝑢′ = 𝐵1

𝑥𝐽𝑛2

Integrating 𝑢 = 𝐴 + 𝐵 𝑑𝑥

𝑥 𝐽𝑛 𝑥 2

So the solution of 𝑦 in this case

𝑦 = 𝑢 𝑥 𝐽𝑛 𝑥

= 𝐴𝐽𝑛 𝑥 + 𝐵𝐽𝑛 𝑥 𝑑𝑥

𝑥 𝐽𝑛 𝑥 2

= 𝐴𝐽𝑛 𝑥 + 𝐵𝑦𝑛 𝑥

where 𝑦𝑛 𝑥 = 𝐽𝑛 𝑥 𝑑𝑥

𝑥 𝐽𝑛 𝑥 2 is the Bessel’s function of the second kind and 𝐽𝑛 𝑥 is Bessel’s

function of first kind.

18.5 RECURRENCE FORMULAE FOR 𝑱𝒏 𝒙

The following relations are the recurrence formulae for Bessel’s functions and are very useful in the

solution of Boundary value problems and in establishing various properties of Bessel’s functions:

1. 𝑑

𝑑𝑥 𝑥𝑛 𝐽𝑛(𝑥) = 𝑥𝑛 𝐽𝑛−1(𝑥)

2. 𝑑

𝑑𝑥 𝑥−𝑛 𝐽𝑛(𝑥) = −𝑥−𝑛 𝐽𝑛+1(𝑥)

15

3. 𝐽𝑛 𝑥 =𝑥

2𝑛 𝐽𝑛−1 𝑥 + 𝐽𝑛+1(𝑥)

4. 𝐽𝑛′ 𝑥 =

1

2 𝐽𝑛−1 𝑥 − 𝐽𝑛+1(𝑥)

5. 𝐽𝑛′ 𝑥 =

𝑛

𝑥 𝐽𝑛 𝑥 − 𝐽𝑛+1(𝑥)

6. 𝐽𝑛+1 𝑥 =2𝑛

𝑥 𝐽𝑛 𝑥 − 𝐽𝑛−1(𝑥)

18.6 EXPANSION FOR 𝑱𝟎 AND 𝑱𝟏

We know that 𝐽𝑛 𝑥 = −1 𝑟1

𝑟! 𝑛+𝑟+1 𝑥

2 𝑛+2𝑟

∞𝑟=0

Taking 𝑛 = 0 and 1 in above Bessel’s function, we get

𝐽0 𝑥 = 1 −1

1! 𝑥

2

2+

1

(2!)2 𝑥

2

4−

1

(3!)2 𝑥

2

6+ ⋯

and 𝐽1 𝑥 =𝑥

2 1 −

1

1! 2! 𝑥

2

2+

1

2! 3! 𝑥

2

4−

1

3! 4! 𝑥

2

6+ ⋯

18.7 VALUE OF 𝑱𝟏𝟐

(𝒙)

In Bessel’s functions, the function 𝐽1/2 is the simplest one, as it can be expressed in finite form.

Taking 𝑛 = 1/2 in the value of 𝐽𝑛 𝑥 , we get

𝐽1/2 𝑥 = 𝑥

2

1/2

1

Γ 3

2 −

1

1! Γ 5

2 𝑥

2

2+

1

2! Γ 7

2 𝑥

2

4−⋯

= 𝑥

2

1/2

11

2Γ

1

2 −

13

2∙1

2 Γ

1

2 𝑥

2

2+

1

2∙5

2∙3

2∙1

2 Γ

1

2 𝑥

2

4−⋯

= 𝑥

2Γ 1

2

2

1!−

2 𝑥2

3!+

2 𝑥4

5!−⋯

Now multiplying the series by 𝑥

2 and outside by

2

𝑥 , we get

𝐽1/2 𝑥 = 2

𝑥 𝜋 𝑥

1!−

𝑥3

3!+

𝑥5

5!−⋯ =

2

𝜋𝑥sin𝑥

Similarly, taking 𝑛 = 1/2 in the value of 𝐽−𝑛 𝑥 , we get

𝐽−1/2 𝑥 = 2

𝜋𝑥cos𝑥

16

18.8 GENERATING FUNCTION FOR 𝑱𝒏(𝒙)

To prove that 𝒆𝟏

𝟐𝒙(𝒕−

𝟏

𝒕) = 𝒕𝒏 𝑱𝒏(𝒙)∞

𝒏=−∞ .

We have 𝑒1

2𝑥(𝑡−

1

𝑡) = 𝑒

𝑥𝑡

2 × 𝑒−𝑥

2𝑡

= 1 +𝑥𝑡

2+

1

2 ! 𝑥𝑡

2

2+

1

3 ! 𝑥𝑡

2

3+ ⋯ + 1 −

𝑥

2𝑡+

1

2 ! 𝑥

2𝑡

2−

1

3 ! 𝑥

2𝑡

3+ ⋯

The coefficient of 𝑡𝑛 in this product is

1

𝑛 ! 𝑥

2 𝑛−

1

(𝑛+1) ! 𝑥

2 𝑛+2

+1

2 ! (𝑛+1) ! 𝑥

2 𝑛+4

−⋯ = 𝐽𝑛(𝑥)

As all the integral powers of t, both positive and negative occurs, we have

𝑒1

2𝑥 𝑡−

1

𝑡

= 𝐽0 𝑥 + 𝑡𝐽1 𝑥 + 𝑡2𝐽2 𝑥 + 𝑡3𝐽3 𝑥 + ⋯

+𝑡−1𝐽−1 𝑥 + 𝑡−2𝐽−2 𝑥 + 𝑡−3𝐽−3 𝑥 + ⋯

= 𝑡𝑛 𝐽𝑛(𝑥)∞𝑛=−∞

Thus the coefficients of different powers of t in the expansion of 𝑒1

2𝑥 𝑡−

1

𝑡 give Bessel’s functions of

various orders. Hence it is known as the generating function of Bessel’s functions.

Example 6: Evaluate 𝒆−𝒂𝒙∞

𝟎𝑱𝟎 𝒃𝒙 𝒅𝒙

Solution: We know that 𝐽𝑛 𝑥 =1

𝜋 𝑐𝑜𝑠 𝑛𝜃 − 𝑥 sin𝜃 𝜋

0𝑑𝜃

For 𝑛 = 0,

𝐽0 𝑥 =1

𝜋 cos 𝑥 cos𝜃 𝑑𝜃𝜋

0𝑜𝑟

𝐽0 𝑥 =1

𝜋 cos 𝑥 sin𝜃 𝑑𝜃𝜋

0

⇒ 𝐽0 𝑏𝑥 =1

𝜋 cos 𝑏𝑥 sin𝜃 𝑑𝜃 =

2

𝜋 cos 𝑏𝑥 sin𝜃 𝑑𝜃

𝜋

20

𝜋

0

So, 𝑒−𝑎𝑥∞

0𝐽0 𝑏𝑥 𝑑𝑥 = 𝑒−𝑎𝑥

2

𝜋 cos 𝑏𝑥 sin𝜃 𝑑𝜃

𝜋

20

∞

0𝑑𝑥

=2

𝜋

𝑒 −𝑎+𝑖 𝑏 sin 𝜃 𝑥+𝑒 −𝑎−𝑖 𝑏 sin 𝜃 𝑥

2

∞

0

𝜋

20

dx dy

=1

𝜋

𝑒 𝑖 𝑏 sin 𝜃−𝑎 𝑥

𝑖 𝑏 sin 𝜃−𝑎 +

𝑒− 𝑖 𝑏 sin 𝜃+𝑎 𝑥

− 𝑖 𝑏 sin 𝜃+𝑎

0

∞𝜋

20

𝑑𝜃

=1

𝜋

−1

𝑖 𝑏 sin 𝜃−𝑎 +

1

𝑖 𝑏 sin 𝜃+𝑎

𝜋

20

𝑑𝜃

=1

2𝜋

−2𝑎

𝑐2𝑏2𝑠𝑖𝑛 2𝜃−𝑎2 𝑑𝜃𝜋

20

17

=1

2𝜋

2𝑎

𝑎2+𝑏2𝑠𝑖𝑛 2𝜃𝑑𝜃

𝜋

20

=2𝑎

𝜋

𝑠𝑒𝑐 2𝜃

𝑎2𝑠𝑒𝑐 2𝜃+𝑏2𝑡𝑎𝑛 2𝜃𝑑𝜃

𝜋

20

=2𝑎

𝜋

𝑠𝑒𝑐 2𝜃

𝑎2+𝑏2 𝑡𝑎𝑛 2𝜃+𝑎2 𝑑𝜃𝜋

20

Now take 𝑎2 + 𝑏2 tan𝜃 = 𝑡

∴ 𝑎2 + 𝑏2 𝑠𝑒𝑐2𝜃 𝑑𝜃 = 𝑑𝑡

Further, if 𝜃 = 0 𝑖𝑡 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡 = 0

𝜃 =𝜋

2 𝑖𝑡 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡 = ∞

∴ 𝑒−𝑎𝑥 𝐽0 𝑏𝑥 𝑑𝑥 =2𝑎

𝜋

1

𝑡2+𝑎2

𝑑𝑡

𝑎2+𝑏2

∞

0

=2𝑎

𝜋 𝑎2+𝑏2

1

𝑎2+𝑡2

∞

0𝑑𝑡

=2𝑎

𝜋 𝑎2+𝑏2

1

𝑎 tan−1 𝑡

𝑎

0

∞

=2

𝜋 𝑎2+𝑏2 tan−1

∞

𝑎 − tan−1 0

=2

𝜋 𝑎2+𝑏2 𝜋

2− 0 =

1

𝑎2+𝑏2

Example 7: Show that 𝑱𝟒 𝒙 = 𝟒𝟖

𝒙𝟑−

𝟖

𝒙 𝑱𝟏 𝒙 + 𝟏 −

𝟐𝟒

𝒙𝟐 𝑱𝟎 𝒙

Solution: We know

𝐽𝑛+1 𝑥 =2𝑛

𝑥𝐽𝑛 𝑥 − 𝐽𝑛−1 𝑥

∴ for 𝑛 = 3

𝐽4 𝑥 =6

𝑥𝐽3 𝑥 − 𝐽2 𝑥 (1)

For 𝑛 = 2

𝐽3 𝑥 =4

𝑥𝐽2 𝑥 − 𝐽1 𝑥 (2)

For 𝑛 = 1

18

𝐽2 𝑥 =2

𝑥𝐽1 𝑥 − 𝐽0 𝑥 (3)

Substituting 𝐽2 𝑥 from (3) in (2), we get

𝐽3 𝑥 =4

𝑥

2

𝑥𝐽1 𝑥 − 𝐽0 𝑥 − 𝐽1 𝑥

= 8

𝑥2 − 1 𝐽1 𝑥 −4

𝑥𝐽0 𝑥 (4)

Now substituting for 𝐽2 𝑥 and 𝐽3 𝑥 in (1), we will have

𝐽4 𝑥 =6

𝑥

8

𝑥2 − 1 𝐽1 𝑥 −4

𝑥𝐽0 𝑥 −

2

𝑥𝐽1 𝑥 − 𝐽0 𝑥

= 48

𝑥3 −8

𝑥 𝐽1 𝑥 + 1 −

24

𝑥2 𝐽0 𝑥

Example 8: Show that

(i) 𝑱−𝟏

𝟐

𝒙 = 𝑱𝟏𝟐

𝒙 𝐜𝐨𝐭 𝒙

(ii) 𝑱−𝟑

𝟐

𝒙 = − 𝟐

𝝅𝒙 𝐬𝐢𝐧 𝒙 +

𝐜𝐨𝐬 𝒙

𝒙

(iii) 𝑱−𝟓

𝟐

𝒙 = 𝟐

𝝅𝒙 𝟑

𝒙𝐬𝐢𝐧𝒙 +

𝟑−𝒙𝟐

𝒙𝟐𝐜𝐨𝐬 𝒙

Solution: (i) We know

𝐽−

1

2

𝑥 = 2

𝜋𝑥 cos𝑥 and 𝐽1

2

𝑥 = 2

𝜋𝑥 sin𝑥

∴ 𝐽−

12

𝑥

𝐽12

𝑥 =

2

𝜋𝑥 cos 𝑥

2

𝜋𝑥 sin 𝑥

= cot𝑥

Hence 𝐽−

1

2

𝑥 = 𝐽1

2

𝑥 cot𝑥

(ii) We know

𝐽𝑛−1 𝑥 =2𝑛

𝑥𝐽𝑛 𝑥 − 𝐽𝑛+1 𝑥

∴ For 𝑛 = −1

2

𝐽−

3

2

𝑥 = −1

𝑥𝐽−

1

2

𝑥 − 𝐽1

2

𝑥

(iii) We know

𝐽𝑛−1 𝑥 =2𝑛

𝑥𝐽𝑛 𝑥 − 𝐽𝑛+1 𝑥

∴ For 𝑛 = −1

2

19

𝐽−

3

2

𝑥 = −1

𝑥𝐽−

1

2

𝑥 − 𝐽1

2

𝑥

= −1

𝑥

2

𝜋𝑥cos𝑥 −

2

𝜋𝑥sin𝑥

𝐽−

3

2

𝑥 = − 2

𝜋𝑥

cos 𝑥

𝑥+ sin𝑥

(iv) We know

𝐽𝑛−1 𝑥 =2𝑛

𝑥𝐽𝑛 𝑥 − 𝐽𝑛+1 𝑥 , 𝐽−1

2

𝑥 = 2

𝜋𝑥cos𝑥, 𝐽1

2

𝑥 =

2

𝜋𝑥sin𝑥

∴ for 𝑛 = −3

2

𝐽−

5

2

𝑥 = −3

𝑥𝐽−

3

2

𝑥 − 𝐽−

1

2

𝑥

(1)

and for 𝑛 = −1

2

𝐽−

3

2

𝑥 = −1

𝑥𝐽−

1

2

𝑥 − 𝐽1

2

𝑥 = − 2

𝜋𝑥

cos 𝑥

𝑥+ sin𝑥

(2)

∴ from (1) and (2), we will have

𝐽−

5

2

𝑥 = −3

𝑥× −

2

𝜋𝑥

cos 𝑥

𝑥+ sin𝑥 −

2

𝜋𝑥 cos𝑥

= 2

𝜋𝑥

3

𝑥2 − 1 cos𝑥 +3

𝑥sin𝑥

Example 9: Prove that

(i) 𝒅

𝒅𝒙 𝑱𝟎 𝒙 = −𝑱𝟏 𝒙 ,

(ii) 𝒅

𝒅𝒙 𝒙 𝑱𝟏 𝒙 = 𝒙𝑱𝟎 𝒙

(iii) 𝒅

𝒅𝒙 𝒙𝒏𝑱𝒏 𝒂𝒙 = 𝒂 𝒙𝒏𝑱𝒏−𝟏 𝒙

(iv) 𝒅

𝒅𝒙 𝒙−𝒏𝑱𝒏 𝒙 = −𝒙−𝒏𝑱𝒏+𝟏 𝒙

Solutions: (i) We know that 𝑑

𝑑𝑥 𝑥−𝑛𝐽𝑛 𝑥 = −𝑥−𝑛𝐽𝑛+1 𝑥

For 𝑛 = 0, we will have

𝑑

𝑑𝑥 𝑥0𝐽0 𝑥 = −𝑥0𝐽1 𝑥

20

𝑑

𝑑𝑥 𝐽0 𝑥 = −𝐽1 𝑥

(ii) We know

𝑑

𝑑𝑥 𝑥𝑛𝐽𝑛 𝑥 = 𝑥𝑛𝐽𝑛−1 𝑥

For 𝑛 = 1, it will give

𝑑

𝑑𝑥 𝑥 𝐽1 𝑥 = 𝑥𝐽0 𝑥

(iii) To prove 𝑑

𝑑𝑥 𝑥𝑛𝐽𝑛 𝑎𝑥 = 𝑎 𝑥𝑛𝐽𝑛−1 𝑥

Let 𝑎𝑥 = 𝑡 or 𝑥 =𝑡

𝑎

∴ 𝑥𝑛𝐽𝑛 𝑎𝑥 = 𝑡

𝑎 𝑛𝐽𝑛 𝑡

Differentiating with respect to ′𝑥′, we get

𝑑

𝑑𝑥 𝑥𝑛𝐽𝑛 𝑎𝑥 =

𝑑

𝑑𝑡

𝑡

𝑎 𝑛𝐽𝑛 𝑡 .

𝑑𝑡

𝑑𝑥

=1

𝑎𝑛 .𝑑

𝑑𝑡 𝑡𝑛𝐽𝑛 𝑡 .𝑎,

=1

𝑎𝑛−1 . 𝑡𝑛𝐽𝑛−1 𝑡 ,

=1

𝑎𝑛−1 . 𝑎𝑥 𝑛𝐽𝑛−1 𝑎𝑥 ,

= 𝑎𝑥𝑛𝐽𝑛−1 𝑎𝑥

(iv) To prove

𝑑

𝑑𝑥 𝑥−𝑛𝐽𝑛 𝑥 = −𝑥−𝑛𝐽𝑛+1 𝑥

We know

𝐽𝑛 𝑥 = −1 𝑟1

𝑟 ! 𝑛+𝑟+1 𝑥

2 𝑛+2𝑟

∞𝑟=0

𝑥−𝑛𝐽𝑛 𝑥 = −1 𝑟1

𝑟! 𝑛+𝑟+1 .

1

2𝑛+2𝑟 . 𝑥2𝑟∞𝑟=0

𝑑

𝑑𝑥 𝑥−𝑛𝐽𝑛 𝑥 = −1 𝑟

1

𝑟! 𝑛+𝑟+1

∞𝑟=1

1

2𝑛+2𝑟 . 2𝑟 𝑥2𝑟−1

= −𝑥−𝑛 −1 𝑟−1 1

𝑟−1 ! 𝑛+1+ 𝑟−1 +1

∞𝑟=1 .

𝑥𝑛+1+2𝑟

2𝑛−1+2𝑟

Taking 𝑟 − 1 = 𝑘

= −𝑥−𝑛 −1 𝑘 .1

𝑘! 𝑛+1+𝑘+1 .

𝑥

2 𝑛+1+2𝑟

∞𝑘=0

= −𝑥−𝑛𝐽𝑛+1 𝑥 .

21

Example 10: Show by the use of recurrence formula, that

(i) 𝑱𝟎′′ 𝒙 =

𝟏

𝟐 𝑱𝟐 𝒙 − 𝑱𝟎 𝒙

(ii) 𝑱𝟏′′ 𝒙 = −𝑱𝟏 𝒙 +

𝟏

𝒙𝑱𝟐 𝒙

Solutions: (i) We know

𝑑

𝑑𝑥 𝑥−𝑛𝐽𝑛 𝑥 = −𝑥−𝑛𝐽𝑛+1 𝑥

∴ for 𝑛 = 0

𝑑

𝑑𝑥 𝐽0 𝑥 = −𝐽1 𝑥

Differentiating with respect to ′𝑥′, we will have

𝑑2

𝑑𝑥2 𝐽0 𝑥 = −

𝑑

𝑑𝑥 𝐽1 𝑥

𝐽0′′ 𝑥 = −𝐽1

′ 𝑥

But 𝐽𝑛′ 𝑥 =

1

2 𝐽𝑛−1 𝑥 − 𝐽𝑛+1 𝑥

∴ for 𝑛 = 1

𝐽1′ 𝑥 =

1

2 𝐽0 𝑥 − 𝐽2 𝑥

∴ 𝐽0′′ 𝑥 = −

1

2 𝐽0 𝑥 − 𝐽2 𝑥

=1

2 𝐽2 𝑥 − 𝐽0 𝑥

(ii) We know

𝐽𝑛′ 𝑥 =

1

2 𝐽𝑛−1 𝑥 − 𝐽𝑛+1 𝑥

∴ 𝐽1′ 𝑥 =

1

2 𝐽0 𝑥 − 𝐽2 𝑥

Differentiating with respect to ′𝑥′, we get

𝐽1′′ 𝑥 =

1

2 𝐽0

′ 𝑥 − 𝐽2′ 𝑥

But 𝐽0′ 𝑥 = −𝐽1 𝑥 and also

𝐽𝑛′ 𝑥 = 𝐽𝑛−1 𝑥 −

𝑛

𝑥𝐽𝑛 𝑥

For 𝑛 = 2

𝐽2′ 𝑥 = 𝐽1 𝑥 −

2

𝑥𝐽2 𝑥

∴ 𝐽1′′ 𝑥 =

1

2 −𝐽1 𝑥 − 𝐽1 𝑥 +

2

𝑥𝐽2 𝑥

22

=1

𝑥𝐽2 𝑥 − 𝐽1 𝑥

Example 11: Show that

(i) 𝟒 𝑱𝟎′′′ 𝒙 + 𝟑 𝑱𝟎

′ 𝒙 + 𝑱𝟑 𝒙 = 𝟎

(ii) 𝟒 𝑱𝒏′′ 𝒙 = 𝑱𝒏−𝟐 𝒙 − 𝟐 𝑱𝒏 𝒙 + 𝑱𝒏+𝟐 𝒙 = 𝟎

Solution: (i) We know 𝑑

𝑑𝑥 𝑥−𝑛𝐽𝑛 𝑥 = −𝑥−𝑛𝐽𝑛+1 𝑥

for 𝑛 = 0

𝑑

𝑑𝑥 𝐽0 𝑥 = −𝐽1 𝑥

Differentiating with respect to ′𝑥′, we get

𝐽0′′ 𝑥 = −𝐽1

′ 𝑥

𝐽𝑛′ 𝑥 =

1

2 𝐽𝑛−1 𝑥 − 𝐽𝑛+1 𝑥 (1)

∴ for 𝑛 = 1

𝐽1′ 𝑥 =

1

2 𝐽0 𝑥 − 𝐽2 𝑥

Differentiating again, it will give

𝐽0′′′ 𝑥 =

1

2 −𝐽0

′ 𝑥 + 𝐽2′ 𝑥

=1

2 𝐽1 𝑥 + 𝐽2

′ 𝑥

From (1), for 𝒏 = 𝟐

𝐽2′ 𝑥 =

1

2 𝐽1 𝑥 − 𝐽3 𝑥

∴ 𝐽0′′′ 𝑥 =

1

2 𝐽1 𝑥 +

1

2 𝐽1 𝑥 − 𝐽3 𝑥

=1

4 3 𝐽1 𝑥 − 𝐽3 𝑥

=1

4 −3 𝐽0

′ 𝑥 − 𝐽3 𝑥

∴ 4 𝐽0′′′ 𝑥 + 3 𝐽0

′ 𝑥 + 𝐽3 𝑥 = 0

(ii) We know

𝐽𝑛′ 𝑥 =

1

2 𝐽𝑛−1 𝑥 − 𝐽𝑛+1 𝑥 (1)

Differentiating with respect to ′𝑥′, we get

𝐽𝑛′′ 𝑥 =

1

2 𝐽𝑛−1

′ 𝑥 − 𝐽𝑛+1′ 𝑥 (2)

23

From (1) 𝐽𝑛−1′ 𝑥 =

1

2 𝐽𝑛−2 𝑥 − 𝐽𝑛 𝑥

and 𝐽𝑛+1′ 𝑥 =

1

2 𝐽𝑛 𝑥 − 𝐽𝑛+2 𝑥

∴ From (2), we get

𝐽𝑛′′ 𝑥 =

1

2

1

2 𝐽𝑛−2 𝑥 − 𝐽𝑛 𝑥 −

1

2 𝐽𝑛 𝑥 − 𝐽𝑛+2 𝑥

=1

4 𝐽𝑛−2 𝑥 − 2 𝐽𝑛 𝑥 + 𝐽𝑛+2 𝑥

∴ 4 𝐽𝑛′′ 𝑥 = 𝐽𝑛−2 𝑥 − 2 𝐽𝑛 𝑥 + 𝐽𝑛+2 𝑥

Example 12: Prove that

(i) 𝒅

𝒅𝒙 𝑱𝒏

𝟐 𝒙 =𝒙

𝟐𝒏 𝑱𝒏−𝟏

𝟐 𝒙 − 𝑱𝒏+𝟏𝟐 𝒙

(ii) 𝒅

𝒅𝒙 𝑱𝒏

𝟐 𝒙 + 𝑱𝒏+𝟏𝟐 𝒙 = 𝟐

𝒏

𝒙𝑱𝒏𝟐 𝒙 −

𝒏+𝟏

𝟐𝑱𝒏+𝟏𝟐 𝒙

Solutions: (i) LHS = 2 𝐽𝑛 𝑥 𝐽𝑛′ 𝑥

But 𝐽𝑛+1 𝑥 =2𝑛

𝑥𝐽𝑛 𝑥 − 𝐽𝑛−1 𝑥

⇒ 𝐽𝑛 𝑥 =𝑥

2𝑛 𝐽𝑛+1 𝑥 + 𝐽𝑛−1 𝑥

and 𝐽𝑛′ 𝑥 =

1

2 𝐽𝑛−1 𝑥 − 𝐽𝑛+1 𝑥

LHS= 2𝐽𝑛 𝑥 𝐽𝑛′ 𝑥 = 2.

𝑥

2𝑛 𝐽𝑛+1 𝑥 + 𝐽𝑛−1 𝑥 ×

1

2 𝐽𝑛−1 𝑥 − 𝐽𝑛+1 𝑥

=𝑥

2𝑛 𝐽𝑛−1

2 𝑥 − 𝐽𝑛+12 𝑥 = RHS

Hence the result

(ii) LHS = 2𝐽𝑛 𝑥 𝐽𝑛′ 𝑥 + 2𝐽𝑛+1 𝑥 𝐽𝑛+1

′ 𝑥

But 𝐽𝑛′ 𝑥 =

𝑛

𝑥𝐽𝑛 𝑥 − 𝐽𝑛+1 𝑥

and 𝐽𝑛′ 𝑥 = 𝐽𝑛−1 𝑥 −

𝑛

𝑥𝐽𝑛 𝑥

⇒ 𝐽𝑛+1′ 𝑥 = 𝐽𝑛 𝑥 −

𝑛+1

𝑥𝐽𝑛+1 𝑥

∴ LHS = 2 𝐽𝑛 𝑥 . 𝑛

𝑥𝐽𝑛 𝑥 − 𝐽𝑛+1 𝑥 + 2 𝐽𝑛+1 𝑥 𝐽𝑛 𝑥 −

𝑛+1

𝑥𝐽𝑛+1 𝑥

= 2 𝑛

𝑥𝐽𝑛

2 𝑥 − 𝐽𝑛 𝑥 . 𝐽𝑛+1 𝑥 + 𝐽𝑛+1 𝑥 . 𝐽𝑛 𝑥 −𝑛+1

𝑥𝐽𝑛+1

2 𝑥

= 2 𝑛

𝑥𝐽𝑛

2 𝑥 −𝑛+1

𝑥𝐽𝑛+1

2 𝑥 = RHS

Example 13: Prove that

24

(i) 𝑱𝟎 𝒙 𝑱𝟏 𝒙 = −𝟏

𝟐 𝑱𝟎 𝒙

𝟐

(ii) 𝒙 𝑱𝟎 𝒂𝒙 𝒓

𝟎=

𝒓

𝒂𝑱𝟏 𝒂𝒓

(iii) 𝒆−𝒂𝒙𝑱𝟎 𝒃𝒙 ∞

𝟎=

𝟏

𝒂𝟐+𝒃𝟐

Solution: (i) We know 𝐽0′ 𝑥 = −𝐽1 𝑥

∴ 𝐽0 𝑥 𝐽1 𝑥 = − 𝐽0 𝑥 𝐽0′ 𝑥 𝑑𝑥

= −1

2 𝐽0 𝑥

2

(ii) Let 𝑎𝑥 = 𝑡, ∴ 𝑎𝑑𝑥 = 𝑑𝑡, 0 𝑡𝑜 𝑟 → 0 𝑡𝑜 𝑎𝑟

∴ 𝑥 𝐽0 𝑎𝑥 𝑟

0𝑑𝑥 =

𝑡

𝑎𝐽0 𝑡 .

𝑑𝑡

𝑎

𝑎𝑟

0

=1

𝑎2 𝑡𝐽0 𝑡 𝑑𝑡𝑎𝑟

0=

1

𝑎2 𝑑

𝑑𝑡

𝑎𝑟

0 𝑡 𝐽1 𝑡 𝑑𝑡

=1

𝑎2 𝑡 𝐽1 𝑡 0

𝑎𝑟 =1

𝑎2 𝑎𝑟 𝐽1 𝑎𝑟 − 0. 𝑥. 𝐽1 0

=1

𝑎𝑟 𝐽1 𝑎𝑟

(iii) 𝑒−𝑎𝑥 𝐽0 𝑏𝑥 𝑑𝑥∞

0

= 𝑒−𝑎𝑥 .1

𝜋 cos 𝑏𝑥 cos𝜑 𝑑𝜑 𝑑𝑥𝜋

0

∞

0

Integrating the order of integration, we get

=1

𝜋 𝑒−𝑎𝑥

∞

0cos 𝑏𝑥 cos𝜑 𝑑𝑥 𝑑𝜑

𝜋

0

= 1

𝜋

𝑒−𝑎𝑥

𝑎2+𝑏2𝑐𝑜𝑠2𝜑 −𝑎 cos 𝑏𝑥 cos𝜑 + 𝑏 cos𝜑 sin 𝑏𝑥 cos𝜑

0

∞

𝑑𝜑𝜋

0

=1

𝜋

𝑎

𝑎2+𝑏2𝑐𝑜𝑠2𝜑

𝜋

0𝑑𝜑 =

1

𝜋

𝑎 𝑠𝑒𝑐 2𝜑

𝑎2𝑠𝑒𝑐 2𝜑+𝑏2 𝑑 𝜑 =2

𝜋

𝜋

0 𝑎 𝑠𝑒𝑐 2𝜑

𝑎2+𝑏2 +𝑎2𝑡𝑎𝑛 2𝜑

𝜋

20

=2

𝜋𝑎 tan−1

𝑎 tan 𝜑

𝑎2+𝑏2

0

𝜋

2×

𝑎

𝑎2+𝑏2

=2

𝜋𝑎.

𝑎

𝑎2+𝑏2×

𝜋

2− 0 =

1

𝑎2+𝑏2

Example 14: Starting with series with generating functions, prove that

𝟐𝒏 𝑱𝒏 𝒙 = 𝒙 𝑱𝒏−𝟏 𝒙 = 𝒙 𝑱𝒏−𝟏 𝒙 + 𝑱𝒏+𝟏 𝒙 and

𝒙𝑱𝒏′ 𝒙 = 𝒏 𝑱𝒏 𝒙 − 𝒙 𝑱𝒏+𝟏 𝒙 (1)

Solutions: We know 𝑒1

2𝑥 𝑡−

1

𝑡

= 𝑡𝑛𝐽𝑛 𝑥 ∞−∞

Differentiating both sides with respect to ′𝑡′, we get

25

1

2𝑥 1 +

1

𝑡2 . 𝑒1

2𝑥 𝑡−

1

𝑡

= 𝑛𝑡𝑛−1𝐽𝑛 𝑥 ∞−∞

1

2𝑥 1 +

1

𝑡2 𝑡𝑛𝐽𝑛 𝑥 ∞−∞ = 𝑛 𝑡𝑛−1𝐽𝑛 𝑥

∞−∞

Equating the coefficients of 𝑡𝑛−1, we will have

1

2𝑥 𝐽𝑛−1 𝑥 +

1

2𝑥 𝐽𝑛+1 𝑥 = 𝑛𝐽𝑛 𝑥

⇒ 2𝑛 𝐽𝑛 𝑥 = 𝑥 𝐽𝑛−1 𝑥 + 𝐽𝑛+1 𝑥 (2)

Now differentiating with respect to ′𝑥′, we get

1

2 𝑡 −

1

𝑡 𝑒

1

2𝑥 𝑡−

1

𝑡

= 𝑡𝑛𝐽𝑛′ 𝑥 ∞

−∞

1

2 𝑡 −

1

𝑡 𝑡𝑛𝐽𝑛 𝑥

∞−∞ = 𝑡𝑛𝐽𝑛

′ 𝑥 ∞−∞

Equating the coefficients of ′𝑡𝑛 ′, we will have

1

2𝐽𝑛−1 𝑥 −

1

2𝐽𝑛+1 𝑥 = 𝐽𝑛

′ 𝑥

⇒ 𝐽𝑛′ 𝑥 =

1

2 𝐽𝑛−1 𝑥 − 𝐽𝑛+1 𝑥 (3)

From (2), substituting 𝐽𝑛−1 𝑥 in (3), we get

𝐽𝑛′ 𝑥 =

1

2

2𝑛

𝑥𝐽𝑛 𝑥 − 𝐽𝑛+1 𝑥 − 𝐽𝑛+1 𝑥

⇒ 𝐽𝑛′ 𝑥 =

𝑛

𝑥𝐽𝑛 𝑥 − 𝐽𝑛+1 𝑥

Example 15: Establish the Jacobi series

𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝜽 = 𝑱𝟎 − 𝟐𝑱𝟐 𝐜𝐨𝐬𝟐𝜽 + 𝟐𝑱𝟒 𝐜𝐨𝐬𝟒𝜽 −……

𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝜽 = 𝟐 𝑱𝟏 𝐜𝐨𝐬 𝜽 − 𝑱𝟑 𝐜𝐨𝐬𝟑𝜽 + 𝑱𝟓 𝐜𝐨𝐬 𝟓𝜽 −……

Solutions: We know 𝑒1

2𝑥 𝑡−

1

𝑡

= 𝑡𝑛𝐽𝑛 𝑥 ∞−∞

= 𝐽0 𝑥 + 𝐽𝑛 𝑥 𝑡𝑛 + −1 𝑛

1

𝑡𝑛 ∞

𝑛=1 (1)

Now, let 𝑡 = cos𝜃 + 𝑖 sin𝜃 and 1

𝑡= cos𝜃 − 𝑖 sin𝜃

To get 𝑡𝑝 = cos𝑝𝜃 + 𝑖 sin𝑝𝜃 and 𝑡−𝑝 = cos𝑝𝜃 − 𝑖 sin𝑝𝜃

and thus 𝑡𝑝 + 𝑡−𝑝 = 2 cos𝑝𝜃 and 𝑡𝑝 − 𝑡−𝑝 = 2 𝑖 sin𝑝𝜃

∴ From (1)

26

𝑒𝑖𝑥 sin 𝜃 = 𝐽0 𝑥 + 2𝑖 𝐽1 𝑥 sin𝜃 + 2𝐽2 𝑥 cos 2𝜃

+2𝑖𝐽3 𝑥 sin 3𝜃 + 2𝐽4 𝑥 cos 4𝜃 + ……

cos 𝑥 sin𝜃 + 𝑖 sin 𝑥 sin𝜃 = 𝐽0 𝑥 + 2𝐽2 𝑥 cos 2𝜃 + 2𝐽4 𝑥 cos 4𝜃 + ……

+𝑖 2𝐽1 𝑥 sin𝜃 + 2𝐽3 𝑥 sin𝜃 + ……

Equating the real and imaginary parts, we get

cos 𝑥 sin𝜃 = 𝐽0 𝑥 + 2 𝐽2 𝑥 cos 2𝜃 + 𝐽4 𝑥 cos 4𝜃 +……

and sin 𝑥 sin𝜃 = 2 𝐽 1 𝑥 sin𝜃 + 𝐽3 𝑥 sin 3𝜃 + ……

Replacing 𝜃 by 𝜋

2− 𝜃, we get

cos 𝑥 cos𝜃 = 𝐽0 𝑥 − 2 cos 2𝜃 𝐽2 𝑥 + 2𝐽4 𝑥 cos 4𝜃 + ……

and sin 𝑥 cos𝜃 = 2 𝐽1 𝑥 sin𝜃 − 𝐽3 𝑥 sin 3𝜃 + ……

Example 16: Prove that

(i) 𝐬𝐢𝐧 𝒙 = 𝟐 𝑱𝟏 𝒙 − 𝑱𝟑 𝒙 + 𝑱𝟓 𝒙 − ……

(ii) 𝐜𝐨𝐬 𝒙 = 𝑱𝟎 𝒙 − 𝟐𝑱𝟐 𝒙 + 𝟐𝑱𝟒 𝒙 − 𝟐𝑱𝟔 𝒙 + ……

(iii) 𝟏 = 𝑱𝟎 + 𝟐𝑱𝟐 + 𝟐𝑱𝟒 + 𝟐𝑱𝟔 + ……

Solution: We know

cos 𝑥 sin𝜃 = 𝐽0 𝑥 + 2 𝐽2 𝑥 cos 2𝜃 + 𝐽4 𝑥 cos 4𝜃 +……

and sin 𝑥 sin𝜃 = 2 𝐽1 𝑥 sin𝜃 + 𝐽3 𝑥 sin 3𝜃 + ……

On taking 𝜃 =𝜋

2 , we will have

(ii) cos𝑥 = 𝐽0 𝑥 + 2 𝐽2 𝑥 cos𝜋 + 𝐽4 𝑥 cos 2𝜋 + 𝐽6 𝑥 cos 3𝜋 + ……

= 𝐽0 𝑥 + 2 −𝐽2 𝑥 + 𝐽4 𝑥 − 𝐽6 𝑥 + ……

= 𝐽0 𝑥 − 2𝐽2 𝑥 + 2𝐽4 𝑥 − 2𝐽6 𝑥 + ……

(i) sin𝑥 = 2 𝐽1 𝑥 sin𝜋

2+ 𝐽3 𝑥 sin

3𝜋

2+ 𝐽5 𝑥 sin

5𝜋

2+ ……

= 2 𝐽1 𝑥 − 𝐽3 𝑥 + 𝐽5 𝑥 − ……

(iii) Taking 𝜃 = 0, we get

cos 0 = 1 = 𝐽 0 𝑥 + 2 𝐽2 𝑥 cos 2 × 0 + 𝐽4 𝑥 cos 4 × 0 + …… .

⇒ 1 = 𝐽0 𝑥 + 2𝐽2 𝑥 + 2𝐽4 𝑥 + ……

ASSIGNMENT 18.3

1. Compute 𝐽0 2 and 𝐽1 1 correct to three decimal places.

2. Express 𝐽5 𝑥 in terms of 𝐽0 𝑥 and 𝐽1 𝑥 .

3. Prove that

(a) 𝐽𝑛′′ 𝑥 =

1

4 𝐽𝑛−2 𝑥 − 2𝐽𝑛 𝑥 + 𝐽𝑛+2 𝑥

(b) 𝑑

𝑑𝑥 𝑥𝐽𝑛 𝑥 𝐽𝑛+1 𝑥 = 𝑥 𝐽𝑛

2 𝑥 − 𝐽𝑛+12 (𝑥) .

27

4. Prove that 𝐽5

2

𝑥 = 2

𝜋𝑥

3−𝑥2

𝑥2 sin𝑥 −3

𝑥cos𝑥 .

5. Prove that

(a) 𝐽3 𝑥 𝑑 𝑥 = 𝑐 − 𝐽2 𝑥 −2

𝑥 𝐽1(𝑥)

(b) 𝑥𝐽𝑛2 𝑥 𝑑𝑥 =

1

2𝑥2 𝐽0

2 𝑥 + 𝐽12(𝑥) .

6. Show that

a) 𝐽𝑛 𝑥 =1

𝜋 cos 𝑛𝜃 − 𝑥 sin𝜃 𝑑𝜃 , 𝑛𝜋

0 being an integer.

b) 𝐽0 𝑥 =1

𝜋 cos 𝑥 cos𝜃 𝑑𝜃𝜋

0

c) 𝐽02 + 2𝐽1

2 + 2𝐽22 + 2𝐽3

2 + ⋯ = 1.

ANSWERS

1. 0.224, 0.44

2. 𝐽5 𝑥 = 384

𝑥4 −72

𝑥2 − 1 𝐽1 𝑥 + 12

𝑥−

192

𝑥3 𝐽0 𝑥

18.9 EQUATIONS REDUCIBLE TO BESSEL’S EQUATION

In differential calculus, we come across such differential equations which can be easily reduced to

Bessel’s equation and thus can be solved by the means of Bessel’s functions. The following are some

examples of such differential equations:

1. Reduce the differential equation 𝒙𝟐𝒅𝟐𝒚

𝒅𝒙𝟐+ 𝒙

𝒅𝒚

𝒅𝒙+ 𝒌𝟐𝒙𝟐 − 𝒏𝟐 𝒚 = 𝟎 to the Bessel’s Equation.

Putting 𝑡 = 𝑘𝑥, so that 𝑑𝑦

𝑑𝑥= 𝑘

𝑑𝑦

𝑑𝑡 and

𝑑2𝑦

𝑑𝑥2 = 𝑘𝑑2𝑦

𝑑𝑡 2 in the above differential equation, we get

𝑡2 𝑑2𝑦

𝑑𝑡2 + 𝑡𝑑𝑦

𝑑𝑡+ 𝑡2 − 𝑛2 𝑦 = 0 , which is the Bessel’s Form of Equation.

∴ Its solution is 𝑦 = 𝑐1𝐽𝑛 𝑡 + 𝑐2𝐽−𝑛 𝑡 , 𝑛 is non-integral.

or 𝑦 = 𝑐1𝐽𝑛 𝑡 + 𝑐2𝑌𝑛 𝑡 , 𝑛 is integral.

Hence solution of the given differential equation is

𝑦 = 𝑐1𝐽𝑛 𝑘𝑥 + 𝑐2𝐽−𝑛 𝑘𝑥 , 𝑛 is non-integral.

or 𝑦 = 𝑐1𝐽𝑛 𝑘𝑥 + 𝑐2𝑌𝑛 𝑘𝑥 , 𝑛 is integral.

2. Reduce the differential equation 𝒙𝒅𝟐𝒚

𝒅𝒙𝟐+ 𝒂

𝒅𝒚

𝒅𝒙+ 𝒌𝟐𝒙𝒚 = 𝟎 to the Bessel’s Equation.

Putting 𝑦 = 𝑥𝑛𝑧, so that 𝑑𝑦

𝑑𝑥= 𝑥𝑛 𝑑𝑧

𝑑𝑥+ 𝑛𝑥𝑛−1𝑧

28

and 𝑑2𝑦

𝑑𝑥2 = 𝑥𝑛 𝑑2𝑧

𝑑𝑥 2 + 2𝑛𝑥𝑛−1 𝑑𝑧

𝑑𝑥+ 𝑛 𝑛 − 1 𝑥𝑛−2𝑧 in the above differential equation, we get

𝑥𝑛+1 𝑑2𝑧

𝑑𝑥2 + (2𝑛 + 𝑎)𝑥𝑛 𝑑𝑧

𝑑𝑥+ 𝑘2𝑥2 + 𝑛2 + 𝑎 − 1 𝑛 𝑥𝑛−1𝑧 = 0

Dividing throughout by 𝑥𝑛−1 and putting 2𝑛 + 𝑎 = 1, we get

𝑥2 𝑑2𝑧

𝑑𝑥2 + 𝑥𝑑𝑧

𝑑𝑥+ 𝑘2𝑥2 − 𝑛2 𝑧 = 0, which is the Bessel’s Form of Equation.

And its solution is 𝑧 = 𝑐1𝐽𝑛 𝑘𝑥 + 𝑐2𝐽−𝑛 𝑘𝑥 , 𝑛 is non-integral.

or 𝑦 = 𝑐1𝐽𝑛 𝑘𝑥 + 𝑐2𝑌𝑛 𝑘𝑥 , 𝑛 is integral.

Hence solution of the given differential equation is

𝑦 = 𝑥𝑛 𝑐1𝐽𝑛 𝑘𝑥 + 𝑐2𝐽−𝑛 𝑘𝑥 , 𝑛 is non-integral.

or 𝑦 = 𝑥𝑛 𝑐1𝐽𝑛 𝑘𝑥 + 𝑐2𝑌𝑛 𝑘𝑥 , 𝑛 is integral.

3. Reduce the differential equation 𝒙𝒅𝟐𝒚

𝒅𝒙𝟐+ 𝒄

𝒅𝒚

𝒅𝒙+ 𝒌𝟐𝒙𝒓𝒚 = 𝟎 to the Bessel’s Equation.

Putting 𝑦 = 𝑡𝑚 , so that 𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑡.𝑑𝑡

𝑑𝑥=

1

𝑚𝑡1−𝑚 𝑑𝑦

𝑑𝑡

and 𝑑2𝑦

𝑑𝑥2 =𝑑

𝑑𝑡

1

𝑚𝑡1−𝑚 𝑑𝑦

𝑑𝑡 .

1

𝑚𝑡1−𝑚 =

1

𝑚2 𝑡2−2𝑚 𝑑2𝑦

𝑑𝑡 2 +1−𝑚

𝑚2 𝑡1−2𝑚 𝑑𝑦

𝑑𝑡 in the above differential

equation, we get

1

𝑚2 𝑡2−𝑚 𝑑2𝑦

𝑑𝑡2 +1−𝑚+𝑐𝑚

𝑚2 𝑡1−𝑚 𝑑𝑦

𝑑𝑡+ 𝑘2𝑡𝑚𝑟 𝑦 = 0

Multiplying throughout by 𝑚2 𝑡1−𝑚 , we get

𝑡 𝑑2𝑦

𝑑𝑡2 + (1 −𝑚 + 𝑐𝑚)𝑑𝑦

𝑑𝑡+ (𝑘𝑚)2𝑡𝑚𝑟+𝑚−1𝑦 = 0

To reduce this equation to the equation at point 2. above, we set 𝑚𝑟 + 𝑚− 1 = 1

𝑖. 𝑒.𝑚 = 2/(𝑟 + 1) and 𝑎 = 1 −𝑚 + 𝑐𝑚 =𝑟+2𝑐−1

𝑟+1. Thus we get the equation as

𝑡 𝑑2𝑦

𝑑𝑡2 + 𝑎𝑑𝑦

𝑑𝑡+ (𝑘𝑚)2𝑡𝑦 = 0 which is similar to equation at point 2.

Hence its solution is

𝑦 = 𝑥𝑛/𝑚 𝑐1𝐽𝑛 𝑘𝑚 𝑥1/𝑚 + 𝑐2𝐽−𝑛 𝑘𝑚 𝑥1/𝑚 , 𝑛 is non-integral.

or 𝑦 = 𝑥𝑛/𝑚 𝑐1𝐽𝑛 𝑘𝑚 𝑥1/𝑚 + 𝑐2𝑌𝑛 𝑘𝑚 𝑥1/𝑚 , 𝑛 is integral.

18.10 ORTHOGONALITY OF BESSEL FUNCTIONS

29

Prove that 𝒙𝑱𝒏 𝜶𝒙 𝑱𝒏 𝜷𝒙 𝒅𝒙𝟏

𝟎=

𝟎, 𝜶 ≠ 𝜷𝟏

𝟐 𝑱𝒏+𝟏 𝜶

𝟐, 𝜶 = 𝜷

𝒘𝒉𝒆𝒓 𝒆 𝜶,𝜷 𝒂𝒓𝒆 𝒓𝒐𝒐𝒕𝒔 𝒐𝒇 𝑱𝒏 𝒙 = 𝟎.

Proof: Let 𝑢 = 𝐽𝑛(𝛼𝑥) and 𝑣 = 𝐽𝑛(𝛽𝑥) are the solutions of the following differential equations

𝑥2𝑢′′ + 𝑥𝑢′ + 𝛼2𝑥2 − 𝑛2 𝑢 = 0 (1)

𝑥2𝑣 ′′ + 𝑥𝑣 ′ + 𝛽2𝑥2 − 𝑛2 𝑣 = 0 (2)

Multiplying equation (1) by 𝑣/𝑥 and equation (2) by 𝑢/𝑥 and then on subtracting, we get

𝑥 𝑢′′ 𝑣 − 𝑢𝑣 ′′ + 𝑢′𝑣 − 𝑢𝑣 ′ + 𝛼2 − 𝛽2 𝑥𝑢𝑣 = 0

⇒ 𝑑

𝑑𝑥 𝑥(𝑢′𝑣 − 𝑢𝑣 ′) = (𝛽2 − 𝛼2)𝑥𝑢𝑣 (3)

Now, integrating both sides of equation (3) within the limits 0 to 1, we get

𝛽2 − 𝛼2 𝑥 𝑢𝑣 𝑑𝑥1

0= 𝑥(𝑢′𝑣 − 𝑢𝑣 ′) 0

1 = 𝑢′𝑣 − 𝑢𝑣 ′ (4)

Since 𝑢 = 𝐽𝑛(𝛼𝑥) and 𝑣 = 𝐽𝑛(𝛽𝑥)

∴ 𝑢′ = 𝛼 𝐽𝑛′(𝛼𝑥) and 𝑣 ′ = 𝛽 𝐽𝑛

′(𝛽𝑥)

Substituting these values in equation (4), we get

𝑥 𝐽𝑛 𝛼𝑥 𝐽𝑛(𝛽𝑥) 𝑑𝑥1

0=

𝛼 𝐽𝑛′ 𝛼 𝐽𝑛 𝛽 −𝛽 𝐽𝑛

′ (𝛽)𝐽𝑛 (𝛼)

𝛽2−𝛼2 (5)

Case I: 𝜶 ≠ 𝜷

Since 𝛼,𝛽 are roots of 𝐽𝑛 𝑥 = 0, so we have 𝐽𝑛 𝛼 = 𝐽𝑛 𝛽 = 0. Thus equation (5) results in

𝑥 𝐽𝑛 𝛼𝑥 𝐽𝑛(𝛽𝑥) 𝑑𝑥1

0= 0 (6)

Case II: 𝜶 = 𝜷

In this case RHS of (5) becomes 0/0 form. So to get its value, apply L’Hospital Rule, by taking 𝛼 as

constant and 𝛽 as variable approaching to 𝛼, we get

Lim𝛽→𝛼 𝑥 𝐽𝑛 𝛼𝑥 𝐽𝑛(𝛽𝑥) 𝑑𝑥1

0= Lim𝛽→𝛼

𝛼 𝐽𝑛′ 𝛼 𝐽𝑛 𝛽

𝛽2−𝛼2

0

0

or Lim𝛽→𝛼 𝑥 𝐽𝑛2 𝛼𝑥 𝑑𝑥

1

0= lim𝛽→𝛼

𝛼 𝐽𝑛′ 𝛼 𝐽𝑛

′ 𝛽

2𝛽

=1

2 𝐽𝑛

′(𝛼) 2

=1

2 𝐽𝑛+1(𝛼) 2 𝑢𝑠𝑖𝑛𝑔 𝐽𝑛

′ = −𝐽𝑛+1 (7)

The relations (6) and (7) are known as Orthogonality relations of Bessel functions.

30

18.11 FOURIER BESSEL EXPANSION

If 𝒇(𝒙) is a continuous function having finite number of oscillations in the interval 𝟎, 𝒂 , then we

can write

𝒇 𝒙 = 𝒄𝒋 𝑱𝒏(𝜶𝒋𝒙)∞𝒋=𝟏 = 𝒄𝟏 𝑱𝒏 𝜶𝟏𝒙 + 𝒄𝟐 𝑱𝒏 𝜶𝟐𝒙 + ⋯+ 𝒄𝒏 𝑱𝒏 𝜶𝒏𝒙 + ⋯(1)

where 𝜶𝟏,𝜶𝟐,… are the positive roots of 𝑱𝒏 𝒙 = 𝟎.

To determine the coefficients 𝒄𝒏, multiply both sides of (1) by 𝑥𝐽𝑛(𝛼𝑛𝑥) and integrating within the

limits 0 to a, we get

𝑥 𝑓 𝑥 𝐽𝑛 𝛼𝑛𝑥 𝑑𝑥𝑎

0= 𝑐𝑛 𝑥 𝐽𝑛

2 𝛼𝑛𝑥 𝑑𝑥𝑎

0= 𝑐𝑛

𝑎2

2 𝐽𝑛+1

2 𝑎 𝛼𝑛

⇒ 𝑐𝑛 =2

𝑎2𝐽𝑛+12 𝑎 𝛼𝑛

𝑥 𝑓 𝑥 𝐽𝑛 𝛼𝑛𝑥 𝑑𝑥𝑎

0

The relation (1) is called Fourier Bessel Expansion of f(x).

18.12 BER AND BEI FUNCTIONS

The differential equation generally encountered in the field of electrical engineering for finding the

distribution of alternating currents in wires of circular cross section is as follows:

𝑥𝑑2𝑦

𝑑𝑥2 +𝑑𝑦

𝑑𝑥− 𝑖 𝑥𝑦 = 0 (1)

which is the special case of first form of differential equation reducible to Bessel equation with 𝑛 = 0

and 𝑘2 = −𝑖, so that 𝑘 = −𝑖 = 𝑖 𝑖 = 𝑖3

2 (Refer Art 18.9).

Thus, the general solution of differential equation (1) is given by

𝑦 = 𝑐1𝐽0 𝑖3

2 𝑥 + 𝑐2𝑌0 𝑖3

2 𝑥

Now 𝐽0 𝑖3

2 𝑥 = 1 −𝑖3𝑥2

22 +𝑖6𝑥4

(2!)2 24 −𝑖9𝑥6

(3!)2 26 +𝑖12𝑥8

(4!)2 28 −⋯

= 1 −𝑥 4

22 .42 +𝑥8

22 .42 .62 .82 −⋯

+𝑖 𝑥2

22 −𝑥6

22 .42 .62 +𝑥10

22 .42 .62 .82 .102 −⋯ (2)

which is complex for x is real.

The series in the brackets of (2) is defined as

𝑏𝑒𝑟 𝑥 = 1 −𝑥4

22 .42 +𝑥8

22 .42 .62 .82 −⋯

= 1 + (−1)𝑚 .∞𝑚=1

𝑥4𝑚

22 .42 .62⋯(4𝑚)2

and 𝑏𝑒𝑖 𝑥 =𝑥2

22 −𝑥6

22 .42 .62 +𝑥10

22 .42 .62 .82 .102 −⋯

31

= − (−1)𝑚 .∞𝑚=1

𝑥4𝑚−2

22 .42 .62⋯(4𝑚−2)2

where ber stands for Bessel real and bei for Bessel imaginary.

Thus we have 𝐽0 𝑖3

2 𝑥 = 𝑏𝑒𝑟 𝑥 + 𝑖 𝑏𝑒𝑖(𝑥)

Similarly, decomposing 𝑌0 𝑖3

2 𝑥 into real and imaginary parts, we obtain another two functions

known as ker (x) and kei (x).

Properties of ber and bei functions

1. 𝑑

𝑑𝑥 𝑥.

𝑑

𝑑𝑥 𝑏𝑒𝑟 (𝑥) = −𝑥 𝑏𝑒𝑖 (𝑥)

2. 𝑑

𝑑𝑥 𝑥.

𝑑

𝑑𝑥 𝑏𝑒𝑖 (𝑥) = −𝑥 𝑏𝑒𝑟 (𝑥)

Example 17: Solve 𝒚′′ +𝒚′

𝒙+ 𝟏 −

𝟏

𝟗𝒙𝟐 𝒚 = 𝟎

Solution: 𝑦′′ +𝑦 ′

𝑥+ 1 −

1

9𝑥2 𝑦 = 0

⇒ 𝑥2𝑦′′ + 𝑥𝑦′ + 𝑥2 −1

9 𝑦 = 0

Comparing with Bessel’s equation

𝑥2𝑦′′ + 𝑥𝑦′ + 𝑥2 − 𝑛2 𝑦 = 0

We find 𝑛 =1

3

∴ The solution of the given equation is 𝑦 = 𝑐1𝐽1

3

𝑥 + 𝑐2𝑌1

3

𝑥

Example 18: Solve 𝒚′′ +𝒚′

𝒙+ 𝟏 −

𝟏

𝟔.𝟐𝟓 𝒙𝟐 𝒚 = 𝟎

Solution: 𝑦′′ +𝑦 ′

𝑥+ 1 −

1

6.25 𝑥 2 𝑦 = 0

⇒ 𝑦′′ +𝑦 ′

𝑥+ 1 −

100

625 𝑥2 𝑦 = 0

Comparing with the Bessel’s equation, we find 𝑛 =10

25=

2

5

∴ The solution of the given equation is 𝑦 = 𝑐1𝐽2

5

𝑥 + 𝑐2𝑌2

5

𝑥

Example 19: Solve 𝒙𝒚′′ + 𝒚′ +𝟏

𝟒𝒚 = 𝟎

Solution: Let 𝑡 = 𝑥1

𝑚 , so that

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑡.𝑑𝑡

𝑑𝑥=

1

𝑚𝑥

1

𝑚−1 .

𝑑𝑦

𝑑𝑡=

1

𝑚 𝑡 1−𝑚 .

𝑑𝑦

𝑑𝑡

𝑑2𝑦

𝑑𝑥2 =𝑑

𝑑𝑡

1

𝑚𝑡𝑚−1 .

𝑑𝑦

𝑑𝑡 𝑑𝑡

𝑑𝑥

= 1

𝑚. 1 −𝑚 𝑡−𝑚 .

𝑑𝑦

𝑑𝑡+

1

𝑚𝑡1−𝑚 𝑑2𝑦

𝑑𝑡2 ×1

𝑚𝑡1−𝑚

=1

𝑚2 1 −𝑚 𝑡1−2𝑚 𝑑𝑦

𝑑𝑡+

1

𝑚2 𝑡2−2𝑚 𝑑2𝑦

𝑑𝑡2

32

∴ 𝑥𝑑2𝑦

𝑑𝑥2 +𝑑𝑦

𝑑𝑥+

1

4𝑦

=𝑡𝑚

𝑚2 1 −𝑚 𝑡1−2𝑚 𝑑𝑦

𝑑𝑡+ 𝑡 2−2𝑚 𝑑2𝑦

𝑑𝑡2 +1

𝑚𝑡1−𝑚 𝑑𝑦

𝑑𝑡+

1

4𝑦 = 0

⇒ 1

𝑚2 𝑡2−𝑚 𝑑2𝑦

𝑑𝑡2 +1−𝑚

𝑚2 𝑡1−𝑚 𝑑𝑦

𝑑𝑡+

1

𝑚𝑡1−𝑚 𝑑𝑦

𝑑𝑡+

1

4𝑦 = 0

⇒ 𝑡2 𝑑2𝑦

𝑑𝑡2 + 1 −𝑚 + 𝑚 𝑡𝑑𝑦

𝑑𝑡+

1

4𝑚2𝑡𝑚𝑦 = 0

⇒ 𝑡2 𝑑2𝑦

𝑑𝑡2 + 𝑡𝑑𝑦

𝑑𝑡+

1

4𝑚2𝑡𝑚𝑦 = 0

Comparing with

𝑥𝑑2𝑦

𝑑𝑥2 + 𝑎𝑑𝑦

𝑑𝑥+ 𝑘2𝑛𝑦 = 0

We get 𝑎 = 1, 𝑘2 =𝑚2

4, 𝑚 − 1 = 1 it implies 𝑚 = 2

i.e. 𝑘2 = 1 and 𝑛 =1−𝑎

2= 0

∴ The solution of the given equation is

𝑦 = 𝑐1𝐽0 𝑡 + 𝑐2𝑌0 𝑡 = 𝑐1𝐽0 𝑥 + 𝑐2𝑌0 𝑥

Example 20: Solve 𝒙𝒚′′ + 𝟐𝒚′ +𝟏

𝟐𝒙𝒚 = 𝟎

Solution: Let 𝑦 = 𝑥𝑛𝑧 so that

𝑑𝑦

𝑑𝑥= 𝑥𝑛 𝑑𝑧

𝑑𝑥+ 𝑛𝑥𝑛−1𝑧

𝑑2𝑦

𝑑𝑥2 = 𝑥𝑛 𝑑2𝑧

𝑑𝑥2 + 2𝑛𝑥𝑛−1 𝑑𝑧

𝑑𝑥+ 𝑛 𝑛 − 1 𝑥𝑛−2𝑧

∴ 𝑥𝑦′′ + 2𝑦′ +1

2𝑥𝑦 = 0

⇒ 𝑥𝑛+1 𝑑2𝑧

𝑑𝑥2 + 2𝑛 + 2 𝑥𝑛 𝑑𝑧

𝑑𝑥+ 𝑛 𝑛 − 1 + 2𝑛 𝑥𝑛−1 +

1

2𝑥𝑛+1𝑧 = 0

⇒ 𝑥2 𝑑2𝑧

𝑑𝑥2 + 2 𝑛 + 1 𝑥𝑑𝑧

𝑑𝑥+ 𝑛 𝑛 + 1 +

1

2𝑥2 𝑧 = 0

Taking 2 𝑛 + 1 = 1 i.e. 𝑛 = −1

2

⇒ 𝑥2 𝑑2𝑧

𝑑𝑥2 + 𝑥𝑑𝑧

𝑑𝑥+

1

2𝑥2 −

1

4 𝑧 = 0

⇒ 𝑧 = 𝑐1𝐽1

2

1

2 𝑥 + 𝑐2𝑌1

2

1

2 𝑥

⇒ 𝑦 = 𝑥−1

2 𝑐1𝐽1

2

𝑥

2 + 𝑐2𝑌1

2

𝑥

2

33

Example 21: Solve 𝒙𝒚′′ + 𝒚 = 𝟎 (1)

Solution: Let 𝑡 = 𝑥1

𝑚 , so that

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑡

𝑑𝑡

𝑑𝑥=

1

𝑚𝑡1−𝑚 𝑑𝑦

𝑑𝑡

and 𝑑2𝑦

𝑑𝑥2 =𝑑

𝑑𝑡

1

𝑚𝑡1−𝑚 𝑑𝑦

𝑑𝑡 ×

𝑑𝑡

𝑑𝑥

= 1

𝑚𝑡1−𝑚 𝑑2𝑦

𝑑𝑡2 +1

𝑚 1 −𝑚 𝑡−𝑚 .

𝑑𝑦

𝑑𝑡

1

𝑚𝑡1−𝑚

∴ 𝑥𝑦′′ + 𝑦 = 0

⇒ 𝑡𝑚 1

𝑚2 𝑡2−2𝑚 𝑑2𝑦

𝑑𝑡2 +1

𝑚2 1 −𝑚 𝑡1−2𝑚 𝑑𝑦

𝑑𝑡 + 𝑦 = 0

⇒ 𝑡2−𝑚 𝑑2𝑦

𝑑𝑡2 + 1 −𝑚 𝑡1−𝑚 𝑑𝑦

𝑑𝑡+ 𝑚2𝑦 = 0

⇒ 𝑡𝑑2𝑦

𝑑𝑡2 + 1 −𝑚 𝑑𝑦

𝑑𝑡+ 𝑚2𝑡𝑚−1𝑦 = 0 (2)

Comparing both

𝑥𝑦′′ + 𝑎𝑦′ + 𝑘2𝑥𝑦 = 0

We will have

𝑎 = 1 −𝑚, 𝑘 = 𝑚 and 𝑚 − 1 = 1

i.e. 𝑚 = 2, 𝑘 = 2 and 𝑎 = 1 − 2 = −1

∴ 𝑛 =1−𝑎

2=

1+1

2= 1

Hence the solution of the equation (2) will be

𝑦 = 𝑡 𝑐1𝐽1 2𝑡 + 𝑐2𝑌1 2𝑡

⇒ 𝑦 = 𝑥1

2 𝑐1𝐽1 2 𝑥 + 𝑐2𝑌1 2 𝑥

Example 22: Solve 𝒚′′ + 𝟗𝒙 −𝟐𝟎

𝒙𝟐 𝒚 = 𝟎 (1)

Solution: Let 𝑡 = 𝑥1

𝑚 or 𝑥 = 𝑡𝑚 , so that

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑡

𝑑𝑡

𝑑𝑥=

1

𝑚𝑡1−𝑚 𝑑𝑦

𝑑𝑡

and 𝑑2𝑦

𝑑𝑥2 =𝑑

𝑑𝑡

1

𝑚𝑡1−𝑚 𝑑𝑦

𝑑𝑡 ×

𝑑𝑡

𝑑𝑥

= 1

𝑚𝑡1−𝑚 𝑑2𝑦

𝑑𝑡2 +1

𝑚 1 −𝑚 𝑡−𝑚 .

𝑑𝑦

𝑑𝑡

1

𝑚𝑡1−𝑚

⇒ 1

𝑚2 𝑡2−2𝑚 𝑑2𝑦

𝑑𝑡2 +1

𝑚2 1 −𝑚 𝑡1−2𝑚 𝑑𝑦

𝑑𝑡+ 9𝑡𝑚 −

20

𝑡2𝑚 𝑦 = 0

34

⇒ 𝑡2 𝑑2𝑦

𝑑𝑡2 + 1 −𝑚 𝑡𝑑𝑦

𝑑𝑡+ 𝑚2 9𝑡3𝑚 − 20 𝑦 = 0

⇒ 𝑡2 𝑑2𝑦

𝑑𝑡2 + 1 −𝑚 𝑡𝑑𝑦

𝑑𝑡+ 9𝑚2𝑡3𝑚 − 20𝑚2 𝑦 = 0

Taking 3𝑚 = 2 i.e. 𝑚 =2

3, we will have

𝑡2 𝑑2𝑦

𝑑𝑡2 +1

3𝑡𝑑𝑦

𝑑𝑡+ 4𝑡2 −

80

9 𝑦 = 0 (2)

Now let 𝑦 = 𝑡𝑛𝑧 𝑡 , so that

𝑑𝑦

𝑑𝑡= 𝑡𝑛

𝑑𝑧

𝑑𝑡+ 𝑛𝑡𝑛−1𝑧 ,

𝑑2𝑦

𝑑𝑡2 = 𝑡𝑛𝑑2𝑧

𝑑𝑡2 + 2𝑛𝑡𝑛−1 𝑑𝑧

𝑑𝑡+ 𝑛 𝑛 − 1 𝑡𝑛−2𝑧

Substituting these in (2), we get

𝑡𝑛+2 𝑑2𝑧

𝑑𝑡2 + 2𝑛 +1

3 𝑡𝑛+1 𝑑𝑧

𝑑𝑡+ 𝑛 𝑛 − 1 +

1

3𝑛 −

80

9 𝑡𝑛 + 4𝑡𝑛+2 𝑧 = 0 (3)

Now for 2𝑛 +1

3= 1, 𝑛 =

1

3

and 𝑛 𝑛 − 1 +1

3𝑛 −

80

9=

1

3× −

2

3+

1

3×

1

3−

80

9= −

81

9= −9

∴ Dividing (3) by 𝑡𝑛 and substituting for 𝑛, we will have

𝑡2 𝑑2𝑧

𝑑𝑡2 + 𝑡𝑑𝑧

𝑑𝑡+ 4𝑡2 − 9 𝑧 = 0 (4)

The solution of (4) is

𝑧 = 𝑐1𝐽3 2𝑡 + 𝑐2𝑌3 2𝑡

⇒ 𝑦 = 𝑡1

3 𝑐1𝐽3 2𝑡 + 𝑐2𝑌3 2𝑡

⇒ 𝑦 = 𝑥3

2

1

3 𝑐1𝐽3 2𝑥

3

2 + 𝑐2𝑌3 2𝑥3

2

= 𝑥1

2 𝑐1𝐽3 2𝑥3

2 + 𝑐2𝑌3 2𝑥3

2

Example 23: Show that

(i) 𝒙𝒏𝑱𝒏 𝒙 is a solution of the equation 𝒙𝒚′′ + 𝟏 − 𝟐𝒏 𝒚′ + 𝒙𝒚 = 𝟎 .

(ii) 𝒙−𝒏𝑱𝒏 𝒙 is the solution of the equation 𝒙𝒚′′ + 𝟏 + 𝟐𝒏 𝒚′ + 𝒙𝒚 = 𝟎

Solution: Let 𝑦 = 𝑥𝑛𝐽𝑛 𝑥

∴ 𝑑𝑦

𝑑𝑥= 𝑥𝑛𝐽𝑛

′ 𝑥 + 𝑛𝑥𝑛−1𝐽𝑛 𝑥

and 𝑑2𝑦

𝑑𝑥2 = 𝑥𝑛𝐽𝑛′′ 𝑥 + 2𝑛𝑥𝑛−1𝐽𝑛

′ 𝑥 + 𝑛 𝑛 − 1 𝑥𝑛−2𝐽𝑛 𝑥

35

∴ 𝑥𝑦′′ + 1 − 2𝑛 𝑦′ + 𝑥𝑦

= 𝑥𝑛+1𝐽𝑛′′ 𝑥 + 2𝑛𝑥𝑛𝐽𝑛

′ 𝑥 + 𝑛 𝑛 − 1 𝑥𝑛−1𝐽𝑛 𝑥

+ 1 − 2𝑛 𝑥𝑛𝐽𝑛′ 𝑥 + 𝑛𝑥𝑛−1𝐽𝑛 𝑥 + 𝑥𝑛+1𝐽𝑛 𝑥

= 𝑥𝑛+1𝐽𝑛′′ 𝑥 + 𝑥𝑛𝐽𝑛

′ 𝑥 2𝑛 + 1 − 2𝑛

+ 𝑛 𝑛 − 1 + 𝑛 1 − 2𝑛 𝑥𝑛−1 + 𝑥𝑛+1 𝐽𝑛 𝑥

= 𝑥𝑛−1 𝑥2𝐽𝑛′′ 𝑥 + 𝐽𝑛

′ 𝑥 + 𝑥2 − 𝑛2 𝐽𝑛 𝑥 = 0

As 𝐽𝑛 𝑥 is the Bessel function and is a solution of 𝑥2𝑦′′ + 𝑦′ + 𝑥2 − 𝑛2 𝑦 = 0

Hence, 𝑥𝑛𝐽𝑛 𝑥 satisfy the given equation and therefore is a solution of it.

Example 24: Show under the transformations 𝒚 =𝒖

𝒙 Bessel’s equation becomes 𝒖′′ +

𝟏 +𝟏−𝟒𝒏𝟐

𝟒𝒙𝟐 𝒖 = 𝟎; Hence find the solution of this equation.

Solution: We know that the Bessel’s equation is

𝑥2𝑦′′ + 𝑥𝑦′ + 𝑥2 − 𝑛2 𝑦 = 0 (1)

Taking 𝑦 =𝑢

𝑥

⇒ 𝑦′ =1

𝑥 𝑢′ + −

1

2 𝑥−

3

2𝑢,

and 𝑦′′ =1

𝑥 𝑢′′ + 2 −

1

2 𝑥−

3

2 𝑢′ + −1

2 −

3

2 𝑥−

5

2 𝑢

Substituting these into (1), we get

𝑥2 1

𝑥𝑢′′ − 𝑥−

3

2𝑢′ +3

4𝑥−

5

2 𝑢 + 𝑥 1

𝑥𝑢′ + −

1

2 𝑥−

3

2 𝑢 + 𝑥2 − 𝑛2 𝑢

𝑥= 0

⇒ 𝑥3

2 𝑢′′ + −𝑥1

2 + 𝑥1

2 𝑢′ + 3

4𝑥−

1

2 −1

2𝑥−

1

2 + 𝑥3

2 −𝑛2

𝑥 𝑢 = 0

⇒ 𝑥3

2 𝑢′′ + 𝑥3

2 +3−2−4𝑛2

4 𝑥 𝑢 = 0

⇒ 𝑢′′ + 1 +1−4𝑛2

4𝑥2 𝑢 = 0 (2)

Hence the Bessel’s equation (1) becomes (2) as desired.

Now the solution of (1) is 𝑦 = 𝑐1𝐽𝑛 𝑥 + 𝑐2𝑌𝑛 𝑥 (3)

⇒ 𝑢

𝑥= 𝑐1𝐽𝑛 𝑥 + 𝑐2𝑌𝑛 𝑥

⇒ 𝑢 = 𝑥 𝑐1𝐽𝑛 𝑥 + 𝑐2𝑌𝑛 𝑥

Example 25: By the use of the substitution 𝒚 =𝒖

𝒙 so that the solution of the equation 𝒙𝟐

𝒅𝟐𝒚

𝒅𝒙𝟐+

𝒙𝒅𝒚

𝒅𝒙+ 𝒙𝟐 −

𝟏

𝟒 𝒚 = 𝟎 can be written in the form 𝒚 = 𝒄𝟏

𝐬𝐢𝐧𝒙

𝒙+ 𝒄𝟐

𝐜𝐨𝐬 𝒙

𝒙.

36

Solution: Taking 𝑦 =𝑢

𝑥

⇒ 𝑑𝑦

𝑑𝑥=

1

𝑥

𝑑𝑢

𝑑𝑥−

1

2𝑥−

3

2 𝑢 and 𝑑2𝑦

𝑑𝑥2 = 𝑥−1

2 𝑑2𝑢

𝑑𝑥2 − 𝑥−3

2𝑑𝑦

𝑑𝑥+

3

4𝑥−

5

2 𝑢

Substituting these in the given equation, we get

𝑥2 𝑥−

1

2𝑑2𝑢

𝑑𝑥2 − 𝑥−3

2𝑑𝑢

𝑑𝑥+

3

4𝑥−

5

2 𝑢 + 𝑥 𝑥−

1

2𝑑𝑢

𝑑𝑥−

1

2𝑥−

3

2 𝑢 + 𝑥2 −1

4 𝑥−

1

2 𝑢 = 0

⇒ 𝑥3

2 𝑑2𝑢

𝑑𝑥2 − 𝑥1

2𝑑𝑢

𝑑𝑥+

3

4𝑥−

1

2𝑢 + 𝑥1

2𝑑𝑢

𝑑𝑥−

1

2𝑥−

1

2𝑢 + 𝑥3

2 −1

4𝑥−

1

2 𝑢 = 0

⇒ 𝑥3

2𝑑2𝑦

𝑑𝑥2 + 𝑥3

2 𝑢 = 0 It implies 𝑑2𝑦

𝑑𝑥2 + 𝑢 = 0

Its Auxiliary equation is 𝐷2 + 1 = 0 it implies 𝐷 = ±𝑖

∴ 𝑢 𝑥 = 𝑐1 cos 𝑥 + 𝑐2 sin𝑥

Hence 𝑦 =𝑢

𝑥= 𝑐1

cos 𝑥

𝑥+ 𝑐2

sin 𝑥

𝑥

Example 26: Show that

𝒙 𝒃𝒆𝒓𝟐𝒙 + 𝒃𝒆𝒊𝟐𝒙 𝒅𝒙 = 𝒑 𝒃𝒆𝒓 𝒑.𝒃𝒆𝒊′𝒑− 𝒃𝒆𝒊 𝒑.𝒃𝒆𝒓′ 𝒑 𝒑

𝟎

Solution: We know 𝑏𝑒𝑟 𝑥 = 1 + −1 𝑚𝑥4𝑚

22 .42 .62…. 4𝑚 2∞𝑚=1

and 𝑏𝑒𝑖 𝑥 = − −1 𝑚𝑥4𝑚−2

22 .42 .62…… 4𝑚−2 2∞𝑚=1

⇒ 𝑑

𝑑𝑥 𝑥 𝑏𝑒𝑖′ 𝑥 = 𝑥 𝑏𝑒𝑟 𝑥

∴ 𝑥 𝑏𝑒𝑟2 𝑥 + 𝑏𝑒𝑖2 𝑥 𝑑𝑥𝑝

0= 𝑥 𝑏𝑒𝑟 𝑥 .𝑏𝑒𝑟 𝑥 + 𝑥 𝑏𝑒𝑖 𝑥 𝑏𝑒𝑖 𝑥 𝑑𝑥

𝑝

0

= 𝑑

𝑑𝑥 𝑥 𝑏𝑒𝑖′ 𝑥 .𝑏𝑒𝑟 𝑥 −

𝑑

𝑑𝑥 𝑥 𝑏𝑒𝑟′ 𝑥 𝑏𝑒𝑖 𝑥 𝑑𝑥

𝑝

0

= 𝑏𝑒𝑟 𝑥. 𝑥 𝑏𝑒𝑖′ 𝑥 − 𝑏𝑒𝑟′ 𝑥 𝑥 𝑏𝑒𝑖′𝑥 𝑑𝑥 − 𝑏𝑒𝑖 𝑥 𝑥 𝑏𝑒𝑖′ 𝑥 +

𝑏𝑒𝑖′ 𝑥𝑥 𝑏𝑒𝑟′𝑥𝑑𝑥0𝑝

= 𝑝 𝑏𝑒𝑟 𝑝. 𝑏𝑒𝑖′𝑝 − 𝑏𝑒𝑖 𝑝. 𝑏𝑒𝑟′ 𝑝 hence proved

Example 27: If 𝒂𝟏,𝒂𝟐,𝒂𝟑,…… 𝒂𝒏 are the positive roots of 𝑱𝟎 𝒙 = 𝟎, prove that

(i) 𝟏

𝟐=

𝑱𝟎 𝜶𝒏 𝒙

𝜶𝒏𝑱𝟏 𝜶𝒏 ∞𝒏=𝟏 (ii) 𝒙𝟐 = 𝟐

𝜶𝒏𝟐−𝟒

𝜶𝒏𝟑𝑱𝟏 𝜶𝒏

𝑱𝟎 𝜶𝒏 𝒙 ∞𝒏=𝟏

Solutions: (i) Let the Fourier Bessel expression of 1

2 is

1

2= 𝑐𝑛𝐽0 𝛼𝑛 𝑥 ∞

𝑛=1 and integrating with

respect to ′𝑥′ from 0 to 1, we get

1

2𝑥𝐽0 𝛼𝑛 𝑥

1

0𝑑𝑥 = 𝑐𝑛 𝑥𝐽0

2 𝛼𝑛 𝑥 𝑑𝑥 = 𝑐𝑛1

2 𝐽1 𝛼𝑛

21

0

37

⇒ 𝑐𝑛1

2 𝐽1

2 𝛼𝑛 =1

2 𝑥 𝐽0 𝛼𝑛 𝑥

1

0

Let 𝛼𝑛𝑥 = 𝑡 it implies 𝑑𝑥 =𝑑𝑡

𝛼𝑛

𝑥 → 0, 1 It implies 𝑡 → 0 𝑡𝑜 𝛼𝑛

=1

2

𝑡

𝛼𝑛

𝛼𝑛

0𝐽0 𝑡

𝑑𝑡

𝛼𝑛

=1

2𝛼𝑛2 𝑡 𝐽0 𝑡 𝑑𝑡

𝛼𝑛

0=

1

2𝛼𝑛2

𝑑

𝑑𝑡 𝑡 𝐽1 𝑡 𝑑𝑡

𝛼𝑛

0

=1

2𝛼𝑛2 𝑡 𝐽1 𝑡 0

𝛼𝑛 =1

2𝛼𝑛2 𝛼𝑛𝐽1 𝛼𝑛

∴ 𝑐𝑛1

2 𝐽1

2 𝛼𝑛 =1

2𝛼𝑛𝐽1 𝛼𝑛

It implies 𝑐𝑛 =1

𝛼𝑛 𝐽1 𝛼𝑛 Hence

1

2=

1

𝛼𝑛 𝐽1 𝛼𝑛 𝐽0 𝛼𝑛 𝑥 ∞

𝑛=1

(ii) Let the Fourier-Bessel expansion of 𝑥2 is 𝑥2 = 𝑐𝑛𝐽0 𝛼𝑛 𝑥 ∞𝑛=1 and integrating from 0 to 1,

we get 𝑥3𝐽0 𝛼𝑛 𝑥 1

0= 𝑐𝑛 𝑥 𝐽0

2 𝛼𝑛 𝑥 𝑑𝑥

⇒ 𝑐𝑛1

2𝐽1

2 𝛼𝑛 = 𝑡3

𝛼𝑛3 𝐽0 𝑡

1

𝛼𝑛

𝛼𝑛

0𝑑𝑡, if 𝛼𝑛𝑥 = 𝑡 it implies 𝑑𝑥 =

𝑑𝑡

𝛼𝑛

=1

𝛼𝑛4 𝑡2 𝑑

𝑑𝑡 𝑡 𝐽1 𝑡

𝛼𝑛

0𝑑𝑡

=1

𝛼𝑛4 𝑡2 . 𝑡𝐽1 𝑡 − 2𝑡 . 𝑡𝐽1 𝑡 𝑑𝑡 0

𝛼𝑛

=1

𝛼𝑛4 𝑡

3𝐽1 𝑡 − 2 𝑑

𝑑𝑡 𝑡2𝐽2 𝑡 𝑑𝑡

0

𝛼𝑛

=1

𝛼𝑛4 𝑡

3𝐽1 𝑡 − 2 𝑑

𝑑𝑡 𝑡2𝐽2 𝑡 𝑑𝑡

0

𝛼𝑛

=1

𝛼𝑛4 𝑡3𝐽1 𝑡 − 2𝑡2𝐽2 𝑡 0

𝛼𝑛

=1

𝛼𝑛4 𝛼𝑛

3𝐽1 𝛼𝑛 − 2𝛼𝑛2𝐽 2 𝛼𝑛

=1

𝛼𝑛2 𝛼𝑛𝐽1 𝛼𝑛 − 2𝐽2 𝛼𝑛

=1

𝛼𝑛2 𝛼𝑛𝐽1 𝛼𝑛 − 2

2

𝛼𝑛 𝐽1 𝛼𝑛 − 𝐽0 𝛼𝑛

=1

𝛼𝑛2

𝛼𝑛2−4

𝛼𝑛 𝐽1 𝛼𝑛 − 𝐽0 𝛼𝑛

= 𝛼𝑛

2−4

𝛼𝑛3 𝐽1 𝛼𝑛 as 𝐽0 𝛼𝑛 = 0

∴ 𝑐𝑛 =2

𝐽1 𝛼𝑛

𝛼𝑛2−4

𝛼𝑛3

38

Hence 𝑥2 = 2 𝛼𝑛

2−4

𝛼𝑛3𝐽1 𝛼𝑛

𝐽0 𝛼𝑛 𝑥

Example 28: Expand 𝒇 𝒙 = 𝒙𝟐 in the interval 𝟎 < 𝑥 < 3 in terms of function 𝑱𝟏 𝜶𝒏 𝒙 where

𝜶𝒏 are determined by 𝑱𝟏 𝟑𝜶 = 𝟎.

Solution: Let the Fourier-Bessel expansion of 𝑓 𝑥 = 𝑥2 is

𝑥2 = 𝑐𝑛 𝐽1 𝛼𝑛 𝑥 ∞𝑛=1 , multiplying both sides by 𝑥𝐽 1 𝛼𝑛 𝑥 and integrating from 0

to 3,

we get

𝑥4 𝐽1 𝛼𝑛 𝑥 𝑑𝑥 = 𝑐𝑛 𝑥𝐽1 𝛼𝑛 𝑥 𝑑𝑥3

0

3

0

Let 𝑥 = 3𝑡 so that 𝑑𝑥 = 3𝑑𝑡

8 𝑡4𝐽1 3𝛼𝑛 𝑡 𝑡

0 3𝑑𝑡 = 𝑐𝑛 3𝑡 𝐽1

2 3𝛼𝑛 𝑡 3𝑑𝑡1

0

∴ 𝑐𝑛 𝑡𝐽121

0 3𝛼𝑛 𝑡 𝑑𝑡 = 27 𝑡4𝐽1 3𝛼𝑛 𝑡 𝑑𝑡

1

0

𝑐𝑛1

2𝐽2

2 3𝛼𝑛 = 27 𝑧4

81𝛼𝑛4 𝐽1 𝑧

𝑑𝑧

3𝛼𝑛

3𝛼𝑛

0 (where 3𝛼𝑛𝑡 = 𝑧 and 𝑑𝑡 =

𝑑𝑧

3𝛼𝑛)

=1

9𝛼𝑛5 𝑧4𝐽1 𝑧 𝑑𝑧

3𝛼𝑛

0

=1

9𝛼𝑛5 𝑧2 𝑑

𝑑𝑧 𝑧2𝐽2 𝑧 𝑑𝑧

3𝛼𝑛

0

=1

9𝛼𝑛5 𝑧2 . 𝑧2𝐽2 𝑧 − 2𝑧. 𝑧2𝐽2 𝑧 𝑑𝑧 0

3𝛼𝑛

=1

9𝛼𝑛5 𝑧

4𝐽2 𝑧 − 2 𝑑

𝑑𝑧 𝑧3𝐽 3 𝑧 𝑑𝑧

0

3𝛼𝑛

=1

9𝛼𝑛5 𝑧4𝐽2 𝑧 − 2𝑧3𝐽3 𝑧 0

3𝛼𝑛

=1

9𝛼𝑛5 81𝛼𝑛

4𝐽2 3𝛼𝑛 − 2 × 27𝛼𝑛3𝐽3 3𝛼𝑛

=1

𝛼𝑛2 9𝛼𝑛𝐽2 3𝛼𝑛 − 2𝐽3 3𝛼𝑛

∴ 𝑐𝑛 =6

𝛼𝑛2𝐽2

2 3𝛼𝑛 3𝛼𝑛𝐽2 3𝛼𝑛 − 2𝐽3 3𝛼𝑛

Hence 𝑥3 = 6 3𝛼𝑛 𝐽2 3𝛼𝑛 −2𝐽3 3𝛼𝑛

𝛼𝑛2𝐽2

2 3𝛼𝑛 𝐽1 𝛼𝑛 𝑥 ∞

𝑛=1

ASSIGNMENT 18.4

1. Solve the differential equations:

(i) 𝑦′′ +𝑦 ′

𝑥+ 8 −

1

𝑥2 𝑦 = 0

(ii) 4𝑦′′ + 9𝑥𝑦 = 0

39

(iii) 𝑥2𝑦′′ − 𝑥𝑦′ + 4𝑥2𝑦 = 0

2. If 𝛼1 , 𝛼2 ,… , 𝛼𝑛 are the positive roots of 𝐽0 𝑥 = 0, show that

1

2=

𝐽0(𝛼𝑛 𝑥)

𝛼𝑛 𝐽1(𝛼𝑛)

∞

𝑛=1

3. Expand 𝑓 𝑥 = 𝑥2 in the interval 0 < 𝑥 < 2 in terms of 𝐽2(𝛼𝑛 𝑥), where 𝛼𝑛 are determined

by 𝐽2 𝛼𝑛 = 0.

4. Prove that

(i) 𝑑

𝑑𝑥 𝑥.

𝑑

𝑑𝑥 𝑏𝑒𝑟 (𝑥) = −𝑥 𝑏𝑒𝑖 (𝑥)

(ii) 𝑑

𝑑𝑥 𝑥.

𝑑

𝑑𝑥 𝑏𝑒𝑖 (𝑥) = −𝑥 𝑏𝑒𝑟 (𝑥)

ANSWERS

1. (i) 𝑦 = 𝐶1𝐽1 2 2𝑥 + 𝐶2𝑌−1 2 2𝑥

(ii) 𝑦 = 𝑥 𝐶1𝐽1

3

𝑥3

2 + 𝐶2𝑌−1

3

𝑥3

2

(iii) 𝑦 = 𝑥 𝐶1𝐽1 2 𝑥 + 𝐶2𝑌1 2𝑥

3. 𝑥2 = 4 𝐽2(𝛼𝑛 𝑥)

𝛼𝑛 𝐽3(2 𝛼𝑛 )∞𝑛=1

18.13 LEGENDRE’S EQUATION

Legendre’s equation is one of the important differential equations occurring in applied mathematics,

particularly in boundary value problems for spheres. It is given as

1 − 𝑥2 𝑑2𝑦

𝑑𝑥2 − 2𝑥𝑑𝑦

𝑑𝑥+ 𝑛 𝑛 + 1 𝑦 = 0 (1)

where n is given real number. In most applications, n takes integral values.

The singularities of this equation are 𝑥 = ±1.

Substituting 𝑦 = 𝑎0𝑥𝑚 + 𝑎1𝑥

𝑚+1 + 𝑎2𝑥𝑚+2 + ⋯ (𝑎0 ≠ 0) in (1), we get

𝑎0 𝑚 𝑚 − 1 𝑥𝑚−2 + 𝑎1 𝑚 + 1 𝑚𝑥𝑚−1 + ⋯

+ 𝑎𝑟+2 𝑚 + 𝑟 + 2 𝑚 + 𝑟 + 1 − 𝑚 + 𝑟 𝑚 + 𝑟 + 1 − 𝑛 𝑛 + 1 𝑎𝑟 𝑥𝑚+𝑟

+ ⋯ = 0

Equating to zero the co-efficient lowest powers of x, i.e of 𝑥𝑚−2, we get

𝑎0 𝑚 𝑚 − 1 = 0 ⇒ 𝑚 = 0, 1 𝑎0 ≠ 0

Equating to zero the co-efficient of 𝑥𝑚−1 and 𝑥𝑚+𝑟 , we get

𝑎1 𝑚 + 1 𝑚 = 0 (2)

𝑎𝑟+2 𝑚 + 𝑟 + 2 𝑚 + 𝑟 + 1 − 𝑚 + 𝑟 𝑚 + 𝑟 + 1 − 𝑛 𝑛 + 1 𝑎𝑟 = 0 (3)

When 𝑚 = 0, (2) is satisfied and therefore 𝑎1 ≠ 0. Then (3) for 𝑟 = 0, 1,2, 3… gives

𝑎2 = −𝑛 𝑛+1

2!𝑎0; 𝑎3 = −

𝑛−1 𝑛+2

3!𝑎1;

40

𝑎4 = −(𝑛−2) 𝑛+3

4 .3𝑎2 =

𝑛 𝑛−2 𝑛+1 (𝑛+3)

4!𝑎0;

𝑎5 = −(𝑛−3) 𝑛+4

5 . 4𝑎3 =

(𝑛−1) 𝑛−3 𝑛+2 (𝑛+4)

5!𝑎1; 𝑒𝑡𝑐.

Therefore two independent solutions of (1) for 𝑚 = 0 are as follows:

𝑦1 = 𝑎0 1 −𝑛 𝑛+1

2!𝑥2 +

𝑛 𝑛−2 𝑛+1 (𝑛+3)

4!𝑥4 −⋯ (4)

𝑦2 = 𝑎1 𝑥 − 𝑛−1 𝑛+2

3!𝑥3 +

(𝑛−1) 𝑛−3 𝑛+2 (𝑛+4)

5!𝑥5 −⋯ (5)

When 𝑚 = 1, 2 gives that 𝑎1 = 0. Therefore (3) gives

𝑎3 = 𝑎5 = 𝑎7 = 0

𝑎2 = −𝑛 𝑛+1

2!𝑎0; 𝑎4 =

𝑛 𝑛−2 𝑛+1 (𝑛+3)

4!𝑎0; 𝑒𝑡𝑐

Thus for 𝑚 = 1, we get the solution (5) again. Hence the general solution of (1) is given by 𝑦 =

𝑦1 + 𝑦2.

Further, it is worth to note that if n is positive even integer, then (4) terminates at the term

containing 𝑥𝑛 and 𝑦1 becomes a polynomial of degree n. Similarly, if n is positive odd integer, then

𝑦2 becomes a polynomial of degree n. Thus, whenever n is a positive integer (even or odd), the

general solution of (1) always contains a polynomial of degree n and an infinite series.

These polynomial solutions, with 𝑎0 and 𝑎1 chosen properly so that the value of the polynomial

becomes one at 𝑥 = 1, are called Legendre’s Polynomials of degree n and is denoted by 𝑷𝒏(𝒙). The

infinite series with 𝑎0 and 𝑎1 chosen properly is called Legendre’s Function of second kind and is

denoted by 𝑸𝒏 𝒙 .

18.14 RODRIGUE’S FORMULA

Another presentation of Legendre’s Polynomials is given by

𝑷𝒏 𝒙 =𝟏

𝒏! 𝟐𝒏 𝒅𝒏

𝒅𝒙𝒏 𝒙𝟐 − 𝟏

𝒏 (1)

is known as Rodrigue’s Formula.

Proof: Let 𝑣 = 𝑥2 − 1 𝑛 , then 𝑣1 =𝑑𝑣

𝑑𝑥= 2𝑛𝑥 𝑥2 − 1 𝑛−1

i.e. 1 − 𝑥2 𝑣1 + 2𝑛𝑥 𝑣 = 0 (2)

Differentiating (2), n+1 times by Leibnitz’ theorem,

1 − 𝑥2 𝑣𝑛+2 + 𝑛 + 1 −2𝑥 𝑣𝑛+1 +1

2! 𝑛 + 1 𝑛 −2 𝑣𝑛

+2𝑛 𝑥𝑣𝑛+1 + (𝑛 + 1)𝑣𝑛 = 0

or 1 − 𝑥2 𝑑2(𝑣𝑛 )

𝑑𝑥2 − 2𝑥𝑑(𝑣𝑛 )

𝑑𝑥+ 𝑛 𝑛 + 1 (𝑣𝑛) = 0

which is Legendre’s Equation and 𝑐𝑣𝑛 is its solution. Also its finite series solution is 𝑃𝑛 𝑥 .

41

∴ 𝑃𝑛 𝑥 = 𝑐𝑣𝑛 = 𝑐 𝑑𝑛

𝑑𝑥 𝑛 𝑥2 − 1 𝑛 (3)

Putting 𝑥 = 1 in equation (3) for determining the value of the constant c, we get

1 = 𝑐 𝑑𝑛

𝑑𝑥 𝑛 𝑥 − 1 𝑛 𝑥 + 1 𝑛 𝑥=1

= 𝑐 𝑛! 𝑥 + 1 𝑛 + 𝑡𝑒𝑟𝑚𝑠 𝑤𝑖𝑡𝑕 𝑥 − 1 𝑎𝑛𝑑 𝑖𝑡𝑠 𝑝𝑜𝑤𝑒𝑟𝑠 𝑥=1

= 𝑐.𝑛! 2 𝑛 , 𝑖. 𝑒., 𝑐 =1

𝑛! 2𝑛

Substituting the value of c in (3), we get eqution (1) which is known as Rodrigue’s formula.

18.15 LEGENDRE’S POLYNOMIALS

By Rodrigue’s formula we have

𝑃0 𝑥 = 1, 𝑃1 𝑥 = 𝑥,

𝑃2 𝑥 =1

2 3𝑥2 − 1 , 𝑃3 𝑥 =

1

2 5𝑥3 − 3𝑥 ,

𝑃4 𝑥 =1

8 35𝑥4 − 30𝑥2 + 3 ,

𝑃5 𝑥 =1

8 63𝑥5 − 70𝑥3 + 15𝑥 .

In general, 𝑃𝑛 𝑥 = −1 𝑟 2𝑛−2𝑟 !

2𝑛 𝑟! 𝑛−𝑟 ! 𝑛−2𝑟 ! 𝑥𝑛−2𝑟𝑁

𝑟=0 where 𝑁 =1

2𝑛 𝑜𝑟

1

2 (𝑛 − 1) according as n is

even or odd.

This general expression for 𝑃𝑛 𝑥 in terms of sum of finite number of terms can be derived easily

from Rodrigue’s formula.

Example 29: Show that 𝑷𝒏 −𝒙 = −𝟏 𝒏𝑷𝒏 𝒙

Solution: 𝑃𝑛 𝑥 = −1 2 2𝑛−2𝑟 !

𝑟! 𝑛−𝑟 ! 𝑛−2𝑟 !𝑥𝑛±2𝑟𝑁

𝑟=0

Where 𝑁 =𝑛

2 or

𝑛−1

2

∴ Replacing 𝑥 by – 𝑥, we will get

𝑃𝑛 𝑥 = −1 𝑟 2𝑛−2𝑟 !

𝑟 ! 𝑛−2𝑟 ! 𝑛−𝑟 ! −1 𝑛−2𝑟𝑥𝑛−2𝑟𝑁

𝑟=0

= −1 𝑛 −1 𝑟 2𝑛−2𝑟 !

𝑟! 𝑛−𝑟 ! 𝑛−2𝑟 !𝑥𝑛−2𝑟∞

𝑟=0 , as −1 2𝑟 = 1

= −1 𝑛𝑃𝑛 𝑥

Example 30: Express the following in the Legendre Polynomials

(i) 𝟓𝒙𝟑 + 𝒙 (ii) 𝒙𝟑 + 𝟐𝒙𝟐 − 𝒙 − 𝟑 (iii) 𝟒𝒙𝟑 − 𝟐𝒙𝟐 − 𝟑𝒙 + 𝟖

42

Solution: We know 𝑃𝑛 𝑥 =1

𝑛! 2𝑛 𝐷𝑛 𝑥2 − 1 𝑛

∴ 𝑃0 𝑥 = 1 𝑃1 𝑥 = 𝑥 𝑃2 𝑥 =1

2 3x2 − 1 𝑃3 𝑥 =

1

2 5𝑥3 − 3𝑥

(i) 5𝑥3 + 𝑥 = 2.1

2 5𝑥3 − 3𝑥 + 4𝑥 = 2𝑃3 𝑥 + 4𝑃1 𝑥

𝑥3 =1

5 2𝑃3 𝑥 + 3𝑃1 𝑥 , 𝑥

2 =1

3 2𝑃2 𝑥 + 𝑃0 𝑥 , 𝑥 = 𝑃1 𝑥 , 1 = 𝑃0 𝑥

(ii) 𝑥3 + 2𝑥2 − 𝑥 − 3 =1

5 2𝑃3 𝑥 + 3𝑃1 𝑥 +

2

3 2𝑃2 𝑥 + 𝑃0 𝑥 − 𝑃1 𝑥 − 𝑃0 𝑥

=2

5𝑃3 𝑥 +

4

3𝑃2 𝑥 −

2

5𝑃1 𝑥 −

7

3𝑃0 𝑥

(iii) 4𝑥3 − 2𝑥2 − 3𝑥 + 8 =4

5 2𝑃3 𝑥 + 3𝑃1 𝑥 −

2

3 2𝑃2 𝑥 + 𝑃0 𝑥 − 3𝑃1 𝑥 + 8𝑃0 𝑥

=8

5𝑃3 𝑥 −

4

3𝑃2 𝑥 −

9

5𝑃1 𝑥 +

22

3𝑃0 𝑥

18.16 GENERATING FUCTION FOR 𝑷𝒏(𝒙)

To show that 𝟏 − 𝟐𝒙𝒕 + 𝒕𝟐 −𝟏

𝟐 = 𝒕𝒏𝑷𝒏 𝒙 ∞𝒏=𝟎

Proof: We know that

(1 − 𝑧)−1

2 = 1 +1

2𝑧 +

1

2.3

2

2! 𝑧2 +

1

2.3

2.5

2

3! 𝑧3 + ⋯

= 1 +2!

1! 2 22 𝑧 +4!

2! 2 24 𝑧2 +6!

3! 2 26 𝑧3 + ⋯

∴ 1 − 𝑡 2𝑥 − 𝑡 −

1

2 = 1 +2!

1! 2 22 𝑡 2𝑥 − 𝑡 +4!

2! 2 24 𝑡 2𝑥 − 𝑡 2

+ ⋯

+(2𝑛−2𝑟)!

(𝑛−𝑟)! 2 22𝑛−2𝑟 (𝑡 2𝑥 − 𝑡 )𝑛−𝑟 + ⋯+(2𝑛)!

𝑛 ! 2 22𝑛 (𝑡 2𝑥 − 𝑡 )𝑛 (1)

The term in 𝑡𝑛 from the term containing 𝑡𝑛−𝑟 2𝑥 − 𝑡 𝑛−𝑟

=(2𝑛−2𝑟)!

(𝑛−𝑟)! 2 22𝑛−2𝑟 𝑡𝑛−𝑟 .𝑛 − 𝑟𝐶𝑟 −𝑡 𝑟 2𝑥 𝑛−2𝑟

=(2𝑛−2𝑟)!

(𝑛−𝑟)! 2 22𝑛−2𝑟 × 𝑛−𝑟 !

𝑟 ! 𝑛−2𝑟 ! −1 𝑟𝑡𝑛 2𝑥 𝑛−2𝑟 =

−1 𝑟(2𝑛−2𝑟)!

2𝑛 𝑟! 𝑛−𝑟 ! 𝑛−2𝑟 !𝑡𝑛𝑥𝑛−2𝑟

Collecting all terms in 𝑡𝑛 which will occur in the term containing 𝑡𝑛 2𝑥 − 𝑡 𝑛 and the proceeding

terms, we see that terms in 𝑡𝑛

= −1 𝑟(2𝑛−2𝑟)!

2𝑛 𝑟! 𝑛−𝑟 ! 𝑛−2𝑟 !𝑡𝑛𝑥𝑛−2𝑟𝑁

𝑟=0 = 𝑃𝑛 𝑥 𝑡𝑛

where 𝑁 =1

2 𝑛 𝑜𝑟

1

2(𝑛 − 1) according as n is even or odd.

Hence (1) can be written as 1 − 2𝑥𝑡 + 𝑡2 −1

2 = 𝑡𝑛𝑃𝑛 𝑥 ∞𝑛=0 , which is known as generating

function of Legendre’s Polynomials.

18.17 RECURRENCE RELATION FOR 𝑷𝒏(𝒙)

I. 𝒏 + 𝟏 𝑷𝒏+𝟏 𝒙 = 𝟐𝒏 + 𝟏 𝑷𝒏 𝒙 − 𝒏𝑷𝒏−𝟏 𝒙 .

43

Proof: We have the generating functions

(1 − 2𝑥𝑡 + 𝑡2)−1/2 = 𝑃𝑛(𝑥)∞𝑛=0 𝑡𝑛 (1)

Differentiate partially w.r.t. t, we get

−1

2 1 − 2𝑥𝑡 + 𝑡2 −

3

2(−2𝑥 + 2𝑡) = 𝑃𝑛(𝑥)∞𝑛=0 𝑛𝑡𝑛−1

1 − 2𝑥𝑡 + 𝑡2 −3

2(𝑥 − 𝑡) = 𝑃𝑛(𝑥)∞𝑛=0 𝑛𝑡𝑛−1 (2)

1 − 2𝑥𝑡 + 𝑡2 −1

2(𝑥 − 𝑡) = 1 − 2𝑥𝑡 + 𝑡2 𝑃𝑛(𝑥)∞𝑛=0 𝑛𝑡𝑛−1

(𝑥 − 𝑡) 𝑃𝑛(𝑥)∞𝑛=0 𝑡𝑛−1 = 1 − 2𝑥𝑡 + 𝑡2 𝑃𝑛(𝑥)∞

𝑛=0 𝑛𝑡𝑛−1

Comparing the coefficients of 𝑡𝑛 from both sides, we get

𝑥 𝑃𝑛 𝑥 − 𝑃𝑛−1 𝑥 = 𝑛 + 1 𝑃𝑛+1 𝑥 − 2𝑥𝑛𝑃𝑛 𝑥 + (𝑛 − 1)𝑃𝑛−1(𝑥)

𝑛 + 1 𝑃𝑛+1 𝑥 = 2𝑛 + 1 𝑥 𝑃𝑛 𝑥 − 𝑛𝑃𝑛−1(𝑥)

II. 𝒏 𝑷𝒏 𝒙 = 𝒙𝑷𝒏′ 𝒙 − 𝑷′

𝒏−𝟏 𝒙 .

Proof: Differentiating (1) partially w.r.t x, we obtain

−1

2 1 − 2𝑥𝑡 + 𝑡2 −

3

2(−2𝑡) = 𝑃𝑛′(𝑥)∞

𝑛=0 𝑡𝑛

𝑡 1 − 2𝑥𝑡 + 𝑡2 −3

2 = 𝑃𝑛′(𝑥)∞

𝑛=0 𝑡𝑛 (3)

Dividing (2) by (3), we get

𝑥−𝑡

𝑡=

𝑛 𝑃𝑛′ (𝑥)∞

𝑛=0 𝑡𝑛−1

𝑃𝑛′ (𝑥)∞

𝑛=0 𝑡𝑛

𝑥 − 𝑡 𝑃𝑛′ 𝑥 ∞

𝑛=0 𝑡𝑛 = 𝑡. 𝑛 𝑃𝑛′(𝑥)∞

𝑛=0 𝑡𝑛−1 = 𝑃𝑛′(𝑥)∞

𝑛=0 𝑡𝑛

Comparing the coefficient of 𝑡𝑛 from both sides, we get

𝑥𝑃𝑛′ 𝑥 − 𝑃′

𝑛−1 𝑥 = 𝑛 𝑃𝑛 𝑥

III. (𝟐𝒏 + 𝟏) 𝑷𝒏 𝒙 = 𝑷′𝒏+𝟏 𝒙 − 𝑷′

𝒏−𝟏 𝒙 .

Proof: From relation I, we have

𝑛 + 1 𝑃𝑛+1 𝑥 = 2𝑛 + 1 𝑃𝑛 𝑥 − 𝑛𝑃𝑛−1 𝑥

Differentiating w.r.t x, we get

𝑛 + 1 𝑃′𝑛+1 𝑥 = 2𝑛 + 1 𝑃𝑛 𝑥 + 2𝑛 + 1 𝑥 𝑃𝑛

′ 𝑥 − 𝑛𝑃′𝑛−1 𝑥 (4)

Using 𝑥𝑃𝑛′ 𝑥 − 𝑃′

𝑛−1 𝑥 = 𝑛 𝑃𝑛 𝑥

Or 𝑥𝑃𝑛′ 𝑥 = 𝑛 𝑃𝑛 𝑥 + 𝑃′

𝑛−1 𝑥 (5)

Now eliminating the term 𝑥𝑃𝑛′ 𝑥 from (4) using (5), we get

𝑛 + 1 𝑃′𝑛+1 𝑥 = 2𝑛 + 1 𝑃𝑛 𝑥 + 2𝑛 + 1 𝑛 𝑃𝑛 𝑥 + 𝑃′

𝑛−1 𝑥 − 𝑛𝑃′𝑛−1 𝑥

𝑛 + 1 𝑃′𝑛+1 𝑥 = 𝑛 + 1 2𝑛 + 1 𝑃𝑛 𝑥 + 𝑛 + 1 𝑃′

𝑛−1 𝑥

𝑃′𝑛+1 𝑥 = 2𝑛 + 1 𝑃𝑛 𝑥 + 𝑃′

𝑛−1 𝑥

44

2𝑛 + 1 𝑃𝑛 𝑥 = 𝑃′𝑛+1 𝑥 − 𝑃′

𝑛−1 𝑥

IV. 𝑷𝒏′ 𝒙 = 𝒙 𝑷′

𝒏−𝟏 𝒙 − 𝒏 𝑷𝒏−𝟏 𝒙 .

Proof: Rewriting (4) as

𝑛 + 1 𝑃′𝑛+1 𝑥

= 2𝑛 + 1 𝑃𝑛 𝑥 + 𝑛 + 1 𝑥 𝑃𝑛′ 𝑥 + 𝑛 𝑥𝑃′

𝑛−1 𝑥 − 𝑃′𝑛−1 𝑥

= 2𝑛 + 1 𝑃𝑛 𝑥 + 𝑛 + 1 𝑥 𝑃𝑛′ 𝑥 + 𝑛2𝑃𝑛 𝑥

= 𝑛 + 1 𝑥 𝑃𝑛′ 𝑥 + 𝑛2 + 2𝑛 + 1 𝑃𝑛 𝑥

= 𝑛 + 1 𝑥 𝑃𝑛′ 𝑥 + (𝑛 + 1)2𝑃𝑛 𝑥

𝑃′𝑛+1 𝑥 = 𝑥 𝑃𝑛

′ 𝑥 + (𝑛 + 1)𝑃𝑛 𝑥

V. (𝟏 − 𝒙𝟐)𝑷𝒏′ 𝒙 = 𝒏 𝑷𝒏−𝟏 𝒙 − 𝒙 𝑷𝒏 𝒙 .

Proof: From Relation II, we have

𝑥𝑃𝑛′ 𝑥 − 𝑃′

𝑛−1 𝑥 = 𝑛 𝑃𝑛 𝑥 (6)

Also from relation IV, we have

𝑃′𝑛 𝑥 − 𝑥 𝑃𝑛−1

′ 𝑥 = 𝑛 𝑃𝑛−1 𝑥 (7)

Multiply equation (7) by x and subtracting form equation (6), we get

(1 − 𝑥2)𝑃𝑛′ 𝑥 = 𝑛 𝑃𝑛−1 𝑥 − 𝑥 𝑃𝑛 𝑥

18.18 ORTHOGONALITY OF LEGENDRE`S POLYNOMIALS

The Legendre Polynomial 𝑷𝒏 𝒙 satisfy the following orthogonality property

𝑷𝒎 𝒙 .𝑷𝒏 𝒙 𝒅𝒙 =

𝟎, 𝒎 ≠ 𝒏

𝟐

𝟐𝒏 + 𝟏, 𝒎 = 𝒏

𝟏

−𝟏

Proof: Both of the cases are discussed as follows:

Case I: 𝒎 ≠ 𝒏

Let the Legendre polynomials 𝑃𝑚 𝑥 and 𝑃𝑛 𝑥 satisfy the differential equations

1 − 𝑥2 𝑃′′𝑚 − 2𝑥 𝑃′

𝑚 + 𝑚 𝑚 + 1 𝑃𝑚 = 0 (1)

1 − 𝑥2 𝑃′′𝑛 − 2𝑥 𝑃′

𝑛 + 𝑛 𝑛 + 1 𝑃𝑛 = 0 (2)

Multiplying (1) by 𝑃𝑛 𝑥 and (2) 𝑃𝑚 𝑥 and then subtracting we get

1 − 𝑥2 𝑃′′𝑚 .𝑃𝑛 − 𝑃′′

𝑛 .𝑃𝑚 − 2𝑥 𝑃′𝑚 .𝑃𝑛 − 𝑃′

𝑛 .𝑃𝑚

+ 𝑚 𝑚 + 1 − 𝑛(𝑛 + 1) 𝑃𝑚 .𝑃𝑛 = 0

𝑑

𝑑𝑥 1 − 𝑥2 (𝑃′

𝑚 .𝑃𝑛 − 𝑃′𝑛 .𝑃𝑚 ) + 𝑚 − 𝑛 (𝑚 + 𝑛 + 1)𝑃𝑚𝑃𝑛 = 0

𝑚 − 𝑛 𝑚 + 𝑛 + 1 𝑃𝑚𝑃𝑛 = −𝑑

𝑑𝑥 1 − 𝑥2 (𝑃′

𝑚 .𝑃𝑛 − 𝑃′𝑛 .𝑃𝑚 )

Integrating from -1 to 1 both sides

𝑚 − 𝑛 𝑚 + 𝑛 + 1 𝑃𝑚 𝑥 .𝑃𝑛 𝑥 𝑑𝑥1

−1= − 1 − 𝑥2 (𝑃′

𝑚 .𝑃𝑛 − 𝑃′𝑛 .𝑃𝑚 ) −1

1 = 0

𝑃𝑚 𝑥 .𝑃𝑛 𝑥 𝑑𝑥 = 01

−1

45

Case II: 𝒎 = 𝒏

We know from generating functions that

1 − 2𝑥𝑡 + 𝑡2 −1

2 = 𝑡𝑛𝑃𝑛 𝑥 ∞𝑛=0 (3)

Squaring both sides and integrating w.r.t. x from -1 to 1, we get

1

1−2𝑥𝑡+𝑡2 𝑑𝑥1

−1= 𝑡𝑛𝑃𝑛 𝑥

∞𝑛=0 2𝑑𝑥

1

−1 (4)

Now 1

1−2𝑥𝑡+𝑡2 𝑑𝑥1

−1=

ln 1−2𝑥𝑡+𝑡2

−2𝑡 −1

1

= −1

2𝑡 ln 1 − 2𝑡 + 𝑡2 − ln 1 + 2𝑡 + 𝑡2

= −1

2𝑡 ln 1 − 𝑡 2 − ln 1 + 𝑡 2 = −

1

𝑡 ln(1 − 𝑡) − ln(1 + 𝑡 )

=1

𝑡 ln(1 + 𝑡) − ln(1 − 𝑡 )

=1

𝑡 𝑡 −

𝑡2

2+

𝑡3

3−⋯ − −𝑡 −

𝑡2

2−

𝑡3

3−⋯

= 2 1 +𝑡2

3+

𝑡4

5+ ⋯+

𝑡2𝑛

2𝑛+1+ ⋯ (5)

Also 𝑡𝑛𝑃𝑛 𝑥 ∞𝑛=0 2𝑑𝑥

1

−1= 𝑡𝑛𝑃𝑛 𝑥

∞𝑛=0 . 𝑡𝑛𝑃𝑛 𝑥

∞𝑛=0 𝑑𝑥

1

−1

= 𝑡2𝑛𝑃𝑛2 𝑥

1

−1∞𝑛=0 𝑑𝑥

= 𝑡2𝑛 𝑃𝑛2 𝑥

1

−1∞𝑛=0 𝑑𝑥 (6)

Using (5) and (6) in equation (4), we get

2 1 +𝑡2

3+

𝑡4

5+ ⋯+

𝑡2𝑛

2𝑛+1+ ⋯ = 𝑡2𝑛 𝑃𝑛

2 𝑥 1

−1∞𝑛=0 𝑑𝑥

Comparing the coefficient of 𝑡2𝑛 on both sides we get

𝑃𝑛2 𝑥

1

−1𝑑𝑥 =

2

2𝑛+1 .

18.19 FOURIER LEGENDRE EXPANSION

If 𝑓 𝑥 be a continuous function and having continuous derivatives over the interval [-1, 1], then we

can write

𝑓 𝑥 = 𝐶𝑛 𝑃𝑛(𝑥)∞𝑛=0 (1)

To determine the coefficient 𝐶𝑛 , multiply both sides by 𝑃 𝑛(𝑥) and integrate form -1 to 1, we get

𝑓 𝑥 .𝑃𝑛 𝑥 𝑑𝑥1

−1= 𝐶𝑛 𝑃𝑛

2(𝑥)1

−1𝑑𝑥

(Remaining terms vanishes by the orthogonal property)

= 𝐶𝑛 .2

2𝑛+1

𝐶𝑛 = 𝑛 +1

2 . 𝑓 𝑥 .𝑃𝑛 𝑥 𝑑𝑥

1

−1 (2)

The series in (1) converges uniformly in interval [-1, 1], and is known as Fourier-Legendre Expansion

of 𝑓 𝑥 .

Example 31: Prove that (i) 𝑷𝟐𝒏′ 𝟎 = 𝟎 and 𝑷𝟐𝒏+𝟏

′ 𝟎 = −𝟏 𝒏 𝟐𝒏+𝟏 !

𝟐𝟐𝒏 𝒏! 𝟐

46

Solution: We know 𝑡𝑛𝑃𝑛 𝑥 ∞𝑛=0 = 1 − 2𝑥𝑡 + 𝑡2 −

1

2

Differentiating with respect to ′𝑥′, we get

𝑡𝑛𝑃𝑛′ 𝑥 = −

1

2 1 − 2𝑥𝑡 + 𝑡2 −

3

2 −2𝑡

= 𝑡 1 − 2𝑥𝑡 + 𝑡2 −3

2

Putting 𝑥 = 0, 𝑃𝑛′ 0 ∞

𝑛=0 = 𝑡 1 + 𝑡2 −3

2

= 𝑡 1 −3

2𝑡2 +

−3

2×−

5

2

2!𝑡4 + ……+

−3

2×−

5

2×…… −

3

2−𝑛−1

𝑛 !𝑡2𝑛 + …

Equating the coefficients of 𝑡2𝑛 and 𝑡2𝑛+1, we get 𝑃2𝑛′ 0 = 0

𝑃2𝑛+1′ 0 = −1 𝑛

3×5×…… 2𝑛+1

2𝑛 𝑛!

= −1 𝑛 2𝑛+1 !

2𝑛𝑛!22𝑛

𝑃2𝑛+1′ 0 = −1 𝑛

2𝑛+1 !

22𝑛 𝑛!2

Example 32: Prove that

(i) 𝟏 − 𝒙𝟐 𝑷𝒏′ 𝒙 = 𝒏 + 𝟏 𝒙𝑷𝒏 𝒙 − 𝑷𝒏+𝟏 𝒙

(ii) 𝟐𝒏 + 𝟏 𝟏 − 𝒙𝟐 𝑷𝒏′ 𝒙 = 𝒏 𝒏 + 𝟏 𝑷𝒏−𝟏 𝒙 − 𝑷𝒏+𝟏 𝒙

(iii) 𝑷𝒏 𝒙 = 𝑷𝒏+𝟏′ 𝒙 − 𝟐𝒙𝑷𝒏

′ 𝒙 + 𝑷𝒏−𝟏′ 𝒙

Solution: We know 𝑡𝑛𝑃𝑛 𝑥 = 1 − 2𝑥𝑡 + 𝑡2 −1

2

(i) Differentiating with respect to ′𝑡′ and equating the coefficients of 𝑡𝑛 , we will get

𝑛 + 1 𝑃𝑛+1 𝑥 = 2𝑛 + 1 𝑥𝑃𝑛 𝑥 − 𝑛𝑃𝑛−1 𝑥 (1)

Now differentiating with respect to ′𝑥′ and using the derivative with respect ′𝑡′, we get

𝑛𝑃𝑛 𝑥 = 𝑥𝑃𝑛′ 𝑥 − 𝑃𝑛−1

′ 𝑥 (2)

From (1) & (2), we can derive

2𝑛 + 1 𝑃𝑛 𝑥 = 𝑃𝑛+1′ 𝑥 − 𝑃𝑛−1

′ 𝑥 (3)

𝑃𝑛′ 𝑥 = 𝑥𝑃𝑛−1

′ 𝑥 + 𝑛𝑃𝑛−1 𝑥 (4)

From (1) & (4) eliminate 𝑃𝑛−1 𝑥

𝑛 + 1 𝑃𝑛+1 𝑥 + 𝑃𝑛′ 𝑥 = 2𝑛 + 1 𝑥𝑃𝑛 𝑥 + 𝑥𝑃𝑛−1

′ 𝑥 (5)

= 2𝑛 + 1 𝑥𝑃𝑛 𝑥 + 𝑥 𝑥𝑃𝑛′ 𝑥 − 𝑛𝑃𝑛 𝑥 , From (4)

1 − 𝑥2 𝑃𝑛′ 𝑥 = 𝑛 + 1 𝑥𝑃𝑛 𝑥 − 𝑛 + 1 𝑃𝑛+1 𝑥

47

= 𝑛 + 1 𝑥𝑃𝑛 𝑥 − 𝑃𝑛+1 𝑥

(i) Eliminating 𝑃𝑛−1′ 𝑥 from (2) & (4), we get

1 − 𝑥2 𝑃𝑛′ 𝑥 = 𝑛 𝑃𝑛−1 𝑥 − 𝑥𝑃𝑛 𝑥

= 𝑛 𝑃𝑛−1 𝑥 −1

2𝑛+1 𝑛 + 1 𝑃𝑛+1 𝑥 + 𝑛𝑃𝑛−1 𝑥

=𝑛

2𝑛+1 2𝑛 + 1 − 𝑛 𝑃𝑛−1 𝑥 − 𝑛 + 1 𝑃𝑛+1 𝑥

2𝑛 + 1 1 − 𝑥2 𝑃𝑛′ 𝑥 = 𝑛 𝑛 + 1 𝑃𝑛−1 𝑥 − 𝑃𝑛+1 𝑥

(ii) (3)−2 ×(2) gives

𝑃𝑛 𝑥 = 𝑃𝑛+1′ 𝑥 − 2𝑥𝑃𝑛

′ 𝑥 + 𝑃𝑛−1′ 𝑥

Example 33: Using the Rodrigue’s formula, show that

𝒅

𝒅𝒙 𝟏 − 𝒙𝟐

𝒅

𝒅𝒙 𝑷𝒏 𝒙 + 𝒏 𝒏 + 𝟏 𝑷𝒏 𝒙 = 𝟎

Solution: We know that 𝑃𝑛 𝑥 =1

2𝑛 𝑛! 𝐷𝑛 𝑥2 − 1 𝑛 =

1

2𝑛𝑛!𝐷𝑛𝑉, 𝑉 = 𝑥2 − 1 𝑛

Now differentiating ′𝑉′ with respect to ′𝑥′ , we get

𝑉1 = 2𝑛𝑥 𝑥2 − 1 𝑛−1 or 𝑥2 − 1 𝑉1 = 2𝑛𝑥𝑉 or 1 − 𝑥2 𝑉1 + 2𝑛𝑥𝑉 = 0

Differentiating 𝑛 + 1 times, we get

1 − 𝑥2 𝑉𝑛+2 + 𝑛 + 1 𝑉𝑛+1 −2𝑥 + 𝑛 + 1 𝑛

2!𝑉𝑛 −2 + 2𝑛𝑥𝑉𝑛+1

+ 𝑛 + 1 2𝑛𝑉𝑛 = 0

1 − 𝑥2 𝑉𝑛+2 − 2𝑥 𝑛 + 1 − 𝑛 𝑉𝑛+1 + 𝑉𝑛 −𝑛 𝑛 + 1 + 2 𝑛 + 1 𝑛 = 0

1 − 𝑥2 𝑉𝑛+2 − 2𝑥𝑉𝑛+1 + 𝑛 𝑛 + 1 𝑉𝑛 = 0

1 − 𝑥2 𝑑2

𝑑𝑥2 𝑉𝑛 − 2𝑥𝑑

𝑑𝑥𝑉𝑛 + 𝑛 𝑛 + 1 𝑉𝑛 = 0

But 𝑉𝑛 = 𝐷𝑛𝑉 = 2𝑛𝑛!𝑃𝑛 𝑥

1 − 𝑥2 𝑑2

𝑑𝑥2 2𝑛𝑛!𝑃𝑛 𝑥 − 2𝑥

𝑑

𝑑𝑥 2𝑛𝑛!𝑃𝑛 𝑥 + 𝑛 𝑛 + 1 2𝑛𝑛!𝑃𝑛 𝑥 = 0

1 − 𝑥2 𝑑2

𝑑𝑥2 𝑃𝑛 𝑥 − 2𝑥𝑑

𝑑𝑥𝑃𝑛 𝑥 + 𝑛 𝑛 + 1 𝑃𝑛 𝑥 = 0

Example 34: Prove that

(i) 𝑷𝟐𝒏 𝒙 𝒅𝒙 = 𝟎𝟏

𝟎

(ii) 𝒙𝒎𝑷𝒏 𝒙 𝒅𝒙 = 𝟎𝟏

−𝟏 𝒎 < 𝑛

Solution: (i) we know 2𝑛 + 1 𝑃𝑛 𝑥 = 𝑃𝑛+1′ 𝑥 − 𝑃𝑛−1

′ 𝑥

48

∴ 4𝑛 + 1 𝑃2𝑛 𝑥 = 𝑃2𝑛+1′ 𝑥 − 𝑃2𝑛−1

′ 𝑥

Integrating both sides

4𝑛 + 1 𝑃2𝑛 𝑥 𝑑𝑥1

0= 𝑃2𝑛+1 𝑥 − 𝑃2𝑛−1 𝑥 0

1

= 𝑃2𝑛+1 1 − 𝑃2𝑛−1 1 − 𝑃2𝑛+1 0 − 𝑃2𝑛−1 0

= 1 − 1 − 0 − 0 = 0

(ii) 𝑥𝑚𝑃𝑛 𝑥 𝑑𝑥 = 𝑥𝑚 1

2𝑛𝑛!𝐷𝑛 𝑥2 − 1 𝑛

1

−1

1

−1

=1

2𝑛𝑛! 𝑥𝑚𝐷𝑛−1 𝑥2 − 1 𝑛 −1

1 − 𝑚𝑥𝑚−1𝐷𝑛−1 𝑥2 − 1 𝑛𝑑𝑥1

−1

=1

2𝑛𝑛! 0 −𝑚 𝑥𝑚−1𝐷𝑛−1 𝑥2 − 1 𝑛𝑑𝑥

1

−1

=1

2𝑛𝑛!× −1 𝑚 𝐷𝑛−𝑚 𝑥2 − 1 𝑛𝑑𝑥

1

−1

=1

2𝑛𝑛! −1 𝑚 𝐷𝑛−𝑚−1 𝑥2 − 1 𝑛 −1

1 = 0

As 𝐷𝑛−𝑚−1 𝑥 − 1 𝑛 𝑥 + 1 𝑛 = 0 will contain terms in 𝑥 − 1 and 𝑥 + 1 both and hence

when 𝑥 = ±, the value is zero.

Example 35: Prove that 𝑷𝒏 𝒙 𝟏 − 𝟐𝒙𝒉 + 𝒉𝟐 −𝟏

𝟐𝒅𝒙 =𝟐𝒉𝒏

𝟐𝒏+𝟏

𝟏

−𝟏.

Solution: We know 1 − 2𝑥𝑕 + 𝑕2 −1

2 = 𝑕𝑚𝑃𝑚 𝑥

∴ 𝑃𝑛 𝑥 1

−1 𝑕𝑚𝑃𝑚 𝑥 𝑑𝑥 = 𝑕𝑚 𝑃𝑛 𝑥 𝑃𝑚 𝑥 𝑑𝑥

1

−1

= 𝑕𝑚 0, 𝑛 ≠ 𝑚 2

2𝑛+1, 𝑛 = 𝑚

=2𝑕𝑛

2𝑛+1

Example 36: Show that

(i) 𝒙𝑷𝒏 𝒙 𝑷𝒏−𝟏 𝒙 𝒅𝒙 =𝟐𝒏

𝟒𝒏𝟐−𝟏

𝟏

−𝟏

(ii) 𝒙𝟐𝟏

−𝟏𝑷𝒏+𝟏 𝒙 𝑷𝒏−𝟏 𝒙 𝒅𝒙 =

𝟐𝒏 𝒏+𝟏

𝟐𝒏−𝟏 𝟐𝒏+𝟏 𝟐𝒏+𝟑

(iii) 𝟏 − 𝒙𝟐 𝟏

−𝟏 𝑷𝒏

′ 𝒙 𝟐𝒅𝒙 =𝟐𝒏 𝒏+𝟏

𝟐𝒏+𝟏

(iv) 𝟏 − 𝒙𝟐 𝑷𝒎′ 𝒙 𝑷𝒏

′ 𝒙 𝒅𝒙 = 𝟎𝟏

−𝟏

Solution: (i) We know 𝑛 + 1 𝑃𝑛+1 𝑥 = 2𝑛 + 1 𝑥𝑃𝑛 𝑥 − 𝑛𝑃𝑛−1 𝑥

𝑥𝑃𝑛 𝑥 =1

2𝑛+1 𝑛 + 1 𝑃𝑛+1 𝑥 + 𝑛𝑃𝑛−1 𝑥

∴ 𝑥𝑃𝑛 𝑥 𝑃𝑛−1 𝑥 𝑑𝑥1

−1

= 1

2𝑛+1 𝑛 + 1 𝑃𝑛+1 𝑥 + 𝑛𝑃𝑛−1 𝑥 𝑃𝑛−1 𝑥 𝑑𝑥

1

−1

49

=𝑛+1

2𝑛+1 𝑃𝑛+1 𝑥 𝑃𝑛−1 𝑥 𝑑𝑥 +

𝑛

2𝑛+1

1

−1 𝑃𝑛−12 𝑥

1

−1𝑑𝑥

=𝑛+1

2𝑛+1× 0 +

𝑛

2𝑛+1×

2

2𝑛−1=

2𝑛

4𝑛2−1

(ii) We know 𝑥𝑃𝑛 𝑥 =1

2𝑛+1 𝑛 + 1 𝑃𝑛+1 + 𝑛𝑃𝑛−1 𝑥

Changing 𝑛 → 𝑛 + 1

𝑥𝑃𝑛+1 𝑥 =1

2𝑛+1 𝑛 + 2 𝑃𝑛+2 𝑥 + 𝑛 + 1 𝑃𝑛 𝑥

and changing 𝑛 → 𝑛 − 1

𝑥𝑃𝑛−1 𝑥 =1

2𝑛−1 𝑛𝑃𝑛 𝑥 + 𝑛 − 1 𝑃𝑛−2 𝑥

∴ 𝑥2𝑃𝑛+1 𝑥 𝑃𝑛−1 𝑥 𝑑𝑥1

−1

= 1

2𝑛+3 𝑛 + 2 𝑃𝑛+2 𝑥 + 𝑛 + 1 𝑃𝑛 𝑥 × 𝑛𝑃𝑛 𝑥 + 𝑛 − 1 𝑃𝑛−2 𝑥 𝑑𝑥

1

−1

=1

2𝑛−1 2𝑛+3 0 + 0 + 𝑛 𝑛 + 1 ×

2

2𝑛+1+ 0

=2𝑛 𝑛+1

2𝑛−1 2𝑛+1 2𝑛+3

(iii) 1 − 𝑥2 𝑃𝑛′ 𝑥 2𝑑𝑥

1

−1= 1 − 𝑥2 𝑃𝑛

′ 𝑥 .𝑃𝑛′ 𝑥

1

−1

= 1 − 𝑥2 𝑃𝑛′ 𝑥 𝑃𝑛 𝑥 −1

1 − 𝑑

𝑑𝑥 1 − 𝑥2 𝑃𝑛

′ 𝑥 1

−1𝑃𝑛 𝑥 𝑑𝑥

= 0 − −𝑛 𝑛 + 1 𝑃𝑛 𝑥 𝑑𝑥1

−1= 𝑛 𝑛 + 1 𝑃𝑛

2 𝑥 1

−1𝑑𝑥 =

2𝑛 𝑛+1

2𝑛+1

(iv) 1 − 𝑥2 𝑃𝑚′ 𝑥 𝑃𝑛

′ 𝑥 𝑑𝑥1

−1

= 1 − 𝑥2 𝑃𝑚′ 𝑥 𝑃𝑛 𝑥 −1

1 − 𝑑

𝑑𝑥 1 − 𝑥2 𝑃𝑚

′ 𝑥 1

−1𝑃𝑛 𝑥 𝑑𝑥

= 0 − 0 + 𝑚 𝑚 + 1 𝑃𝑚 𝑥 𝑃𝑛 𝑥 1

−1𝑑𝑥 = 𝑚 𝑚 + 1 × 0 = 0

Example 37: Expand the following functions in terms of Legendre’s polynomials in the interval

𝟏,−𝟏

(i) 𝒇 𝒙 = 𝒙𝟑 + 𝟐𝒙𝟐 − 𝒙 − 𝟑 (ii) 𝒇 𝒙 = 𝒙𝟒 + 𝒙𝟑 + 𝟐𝒙𝟐 − 𝒙 − 𝟑

Solution: (i) We know 𝑓 𝑥 = 𝑐𝑛𝑃𝑛 𝑥 ∞𝑛=0

Where 𝑐𝑛 = 𝑛 +1

2 𝑓 𝑥 𝑃𝑛 𝑥 𝑑𝑥

1

−1

∴ 𝑐0 = 0 +1

2 𝑥3 + 2𝑥2 − 𝑥 − 3 × 1

1

−1𝑑𝑥

=1

2

𝑥4

4+ 2

𝑥3

3−

𝑥2

2− 3𝑥

−1

1

=1

2 0 +

4

3− 6 =

2

3− 3 = −

7

3

50

𝑐1 = 1 +1

2 𝑥3 + 2𝑥2 − 𝑥 − 3 𝑥 𝑑𝑥

1

−1=

3

2 𝑥4 + 2𝑥3 − 𝑥2 − 3𝑥

1

−1𝑑𝑥

= −3

2

2

5−

2

3 = 3

3−5

15 = −

6

15= −

2

5

𝑐2 = 2 +1

2 𝑥3 + 2𝑥2 − 𝑥 − 3

1

2 3𝑥2 − 1 𝑑𝑥

1

−1

=5

4 3𝑥5 − 𝑥3 + 6𝑥4 − 2𝑥2 − 3𝑥3 + 𝑥 − 9𝑥2 + 3 𝑑𝑥

1

−1

=5

4 6 ×

2

5−

4

3−

9×2

3+ 6 =

4

3

𝑓 𝑥 = −7

3𝑃0 𝑥 −

2

5𝑃1 𝑥 +

4

3𝑃2 𝑥 + ……

(i) 𝑓 𝑥 = 𝑥4 + 𝑥3 + 2𝑥2 − 𝑥 − 3 = 𝑐𝑛𝑃𝑛 𝑥

𝑐𝑛 = 𝑛 +1

2 𝑓 𝑥 𝑝𝑛 𝑥 𝑑𝑥

1

−1

𝑐0 = 0 +1

2 𝑥4 + 𝑥3 + 2𝑥2 − 𝑥 − 3 × 1 𝑑𝑥

1

−1

=1

2

2

5+

4

3− 6 =

1

2×

6+20−90

15= −

32

15

𝑐1 = 1 +1

2 𝑥4 + 𝑥3 + 2𝑥2 − 𝑥 − 3 × 𝑥 𝑑𝑥

1

−1

𝑐1 =3

2

2

5−

2

3 =

3

2×

6−10

15= −

2

5

𝑐2 = 2 +1

2 𝑥4 + 𝑥3 + 2𝑥2 − 𝑥 − 3 ×

1

2 3𝑥2 − 1

1

−1𝑑𝑥

=5

4 3𝑥6 − 𝑥4 + 3𝑥5 − 𝑥3 + 6𝑥4 − 2𝑥2 − 3𝑥3 + 𝑥 − 9𝑥2 + 3 𝑑𝑥

1

−1

=5

4

6

7−

2

5+

12

5−

4

3− 6 + 6 =

40

21

𝑓 𝑥 = −32

15𝑃0 𝑥 −

2

5𝑃1 𝑥 +

40

21𝑃2 𝑥 + ……

ASSIGNMENT 18.5

1. Show that 𝑃′𝑛 −𝑥 = −1 𝑛+1𝑃′

𝑛 𝑥 .

2. Evaluate the following:

(i) 𝑃23𝑛 𝑥 𝑑𝑥

1

0

(ii) 𝑥𝑚 .𝑃𝑛 𝑥 𝑑𝑥1

−1,𝑤𝑕𝑒𝑟𝑒 𝑚 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡𝑕𝑎𝑛 𝑛.

3. Express 8 𝑃5 𝑥 − 8 𝑃4 𝑥 − 2 𝑃2 𝑥 + 5𝑃0(𝑥) in terms of polynomial of x.

4. Use Rodrigues formulae to obtain 𝑃3(𝑥) and 𝑃4 𝑥 .

5. Find the value of cos 𝑡 .𝑃3(sin 𝑡) 𝑑𝑡𝜋/2

0.

6. Prove that

(i) 𝑃𝑛 (𝑥)

1−2𝑥𝑧+𝑧2 𝑑𝑥 =

2𝑧𝑛

2𝑛+1

1

−1

51

(ii) 𝑃𝑛(𝑥) 𝑑𝑥 = 01

−1 except when 𝑛 = 0 in which case the value of the integral is 2.

ANSWERS

2. (i) 1

6𝑛+1

(ii) 0

3. 63 𝑥5 − 35 𝑥4 − 70𝑥3 + 27𝑥2 + 15𝑥 + 3

4. 𝑃3 𝑥 =1

2 5𝑥3 − 3𝑥 ,𝑃4 𝑥 =

1

8 35𝑥4 − 30𝑥2 + 3

5. -1/8

18.20 STRUM – LIOUVILLE PROBLEMS

A differential equation of the form

𝑝 𝑥 𝑦′ ′ + 𝑞 𝑥 + 𝜆 𝑟(𝑥) 𝑦 = 0 (1)

is called Strum-Liouville Equation where 𝜆 is a real number.

Instead of initial conditions, this equation is usually subjected to the boundary conditions on the

interval 𝑎, 𝑏 as

𝛼1 𝑦 𝑎 + 𝛼2 𝑦′ 𝑎 = 0, 𝛽1 𝑦 𝑏 + 𝛽2 𝑦′ 𝑏 = 0 (2)

where 𝛼1 ,𝛼2 ,𝛽1 ,𝛽2 are real constants such that either 𝛼1 𝑜𝑟 𝛼2 are not zero and 𝛽1 𝑜𝑟 𝛽2 are not

zero.

The non trivial solutions of the differential equation (1) subjected to the conditions (2) exists only for

specific values of 𝜆, which values are termed as Eigen values or Characteristic values of the equation

(1). And the non trivial solution of (1) corresponding to these Eigen values are termed as Eigen

functions or Characteristic functions.

18.21 ORTHOGONALITY OF EIGEN FUNCTIONS

Two functions 𝑦𝑚 (𝑥) and 𝑦𝑛(𝑥) defined on some interval 𝑎, 𝑏 are said to be orthogonal on this

interval with respect to the weight function 𝑟 𝑥 > 0, if

𝑟 𝑥 𝑏

𝑎𝑦𝑚 𝑥 𝑦𝑛 𝑥 𝑑𝑥 = 0 𝑓𝑜𝑟 𝑚 ≠ 𝑛

Also the norm 𝑦𝑚 of the function 𝑦𝑚 (𝑥) is defined to be non negative square root of

𝑟 𝑥 𝑏

𝑎 𝑦𝑚 𝑥

2𝑑𝑥. Thus 𝑦𝑚 = 𝑟 𝑥

𝑏

𝑎 𝑦𝑚 𝑥

2𝑑𝑥.

The functions which are orthogonal and having the norm unity are said to be orthonormal functions.

52

Theorem: If 𝒚𝒎(𝒙) and 𝒚𝒏(𝒙) are two eigen functions of the Strum-Liouville problem

corresponding to eigen values 𝝀𝒎 and 𝝀𝒏 respectively (where 𝒎 ≠ 𝒏), then the eigen functions are

orthogonal w.r.t. the weight function 𝒓(𝒙) over the interval 𝒂,𝒃 .

Proof: Since distinct eigen values and their corresponding eigen functions are the solutions of the

Stum Liouville equation (1), so we can write it as

𝑝 𝑥 𝑦𝑚′ ′ + 𝑞 𝑥 + 𝜆𝑚 𝑟(𝑥) 𝑦𝑚 = 0

𝑝 𝑥 𝑦𝑛′ ′ + 𝑞 𝑥 + 𝜆𝑛 𝑟(𝑥) 𝑦𝑛 = 0

Multiplying first equation by 𝑦𝑛 and the second equation by 𝑦𝑚 , and then subtracting, we get

𝜆𝑚 − 𝜆𝑛 𝑟 𝑥 𝑦𝑚𝑦𝑛 = 𝑦𝑚 𝑟 𝑥 𝑦𝑛′ ′ − 𝑦𝑛 𝑟 𝑥 𝑦𝑚

′ ′

=𝑑

𝑑𝑥 𝑟 𝑥 𝑦𝑛

′ 𝑦𝑚 − 𝑟 𝑥 𝑦𝑚′ 𝑦𝑛

Now integrating both sides w.r.t. x from a to b, we get

𝜆𝑚 − 𝜆𝑛 𝑟 𝑦𝑚𝑦𝑛𝑑𝑥 = 𝑟 𝑥 𝑦𝑛′ 𝑦𝑚 − 𝑟 𝑥 𝑦𝑚

′ 𝑦𝑛 𝑎𝑏𝑏

𝑎

= 𝑟 𝑏 𝑦𝑛′ 𝑏 𝑦𝑚 𝑏 − 𝑦𝑚

′ 𝑏 𝑦𝑛 𝑏 − 𝑟 𝑎 𝑦𝑛′ 𝑎 𝑦𝑚 𝑎 − 𝑦𝑚

′ 𝑎 𝑦𝑛 𝑎

The R.H.S. will vanish if the boundary conditions are of one of the followings forms:

I. 𝑦 𝑎 = 𝑦 𝑏 = 0

II. 𝑦′ 𝑎 = 𝑦′ 𝑏 = 0

III. 𝛼1 𝑦 𝑎 + 𝛼2 𝑦′ 𝑎 = 0, 𝛽1 𝑦 𝑏 + 𝛽 2 𝑦′ 𝑏 = 0

where 𝛼1 ,𝛼2 ,𝛽1 ,𝛽2 are real constants such that either 𝛼1 𝑜𝑟 𝛼2 are not zero and 𝛽1 𝑜𝑟 𝛽2 are not

zero. Thus in each of the three cases we get

𝑟 𝑦𝑚𝑦𝑛𝑑𝑥 = 0, (𝑚 ≠ 𝑛)𝑏

𝑎

which shows that the eigen functions 𝑦𝑚 (𝑥) and 𝑦𝑛(𝑥) are orthogonal w.r.t. the weight function 𝑟(𝑥)

over the interval 𝑎, 𝑏 .

Example 38: For Strum-Liouville problem 𝒚′′ + 𝝀𝒚 = 𝟎, 𝒚 𝟎 = 𝟎, 𝒚 𝝅 = 𝟎 find the eigen

functions.

Solution: For 𝜆 = −𝛾2, the general solution of the equation is given by

𝑦 𝑥 = 𝐶1 𝑒𝛾𝑥 + 𝐶2 𝑒−𝛾𝑥

Using the above mentioned boundary conditions we get 𝐶1 = 𝐶2 = 0. Hence 𝑦 𝑥 = 0 is not an

eigen function.

Also for 𝜆 = 𝛾2, the general solution of the equation is given by

53

𝑦 𝑥 = 𝐶1 cos 𝛾𝑥 + 𝐶2 sin𝛾𝑥

Using 𝑦 0 = 0, we get 𝐶1 = 0

Using 𝑦 𝜋 = 0, we get 𝐶2 sin 𝛾𝜋 = 0 => sin𝛾𝜋 = 0

∴ 𝛾 𝜋 = 𝑛𝜋 => 𝛾 = 𝑛, 𝑛 = ±1, ±2, ±3,….

Thus the eigen values are 𝜆 = 0, 1, 4, 9,… and taking 𝐶2 = 1, we obtain the eigen functions as

𝑦𝑛 𝑥 = sin𝑛𝑥 , 𝑛 = 0, 1, 2,…

ASSIGNMENT 18.6

1. Find the eigen values of each of the following Stum Liouville problems and prove their

orthogonality:

i) 𝑦′′ + 𝜆𝑦 = 0, 𝑦 0 = 0, 𝑦 𝑙 = 0

ii) 𝑦′′ + 𝜆𝑦 = 0, 𝑦′ 0 = 0, 𝑦′ 𝑐 = 0

iii) 𝑦′′ + 𝜆𝑦 = 0, 𝑦 𝜋 = 𝑦(−𝜋), 𝑦′ 𝜋 = 𝑦′(−𝜋)

2. Show that the eigen values of the boundary value problem 𝑦′′ + 𝜆𝑦 = 0, 𝑦 0 = 0, 𝑦 𝜋 +

𝑦′ 𝜋 = 0 satisfies 𝜆 + tan 𝜆 𝜋 = 0.

ANSWERS

1. (i) sin𝑛𝜋𝑥

𝑙, 𝑛 = 0, 1, 2,…

(ii) cos𝑛𝜋𝑥

𝑐, 𝑛 = 0, 1, 2,…

(iii) 1, sin𝑥 , cos 𝑥 , sin 2𝑥 , cos 2𝑥 ,…