MATHEMATICAL PHYSICS UNIT – I

LEGENDRE’S EQUATION

DR. RAJESH MATHPAL ACADEMIC CONSULTANT

SCHOOL OF SCIENCESU.O.U. HALDWANI

UTTRAKHANDMOB:9758417736,7983713112

CONTENTS

1. LEGENDRE’S EQUATION

1.1. LEGENDRE’S POLYNOMIAL Pn(x)

1.2. LEGENDRE’S FUNCTION OF THE SECOND KIND i.e. Qn(x)

1.3 GENERAL SOLUTION OF LEGENDRE’S EQUATION

2. RODRIGUE’S FORMULA

3. A GENERATING FUNCTION OF LEGENDRE’S POLYNOMIAL

4. ORTHOGONALITY OF LEGENDRE POLYNOMIALS

5. RECURRENCE FORMULAE

6. NUMERICAL PROBLEMS

7. LEGENDRE’S POLYNOMIALS APPLICATIONS

8. NUMERICAL PROBLEMS

1. LEGENDRE’S EQUATION

Now,

1 2

0 1 2

0

0

2

2

( ...) ...(2)

dyso that ( )

dx

d y (

dx

m

m r

r

r

m r l

r

r

r

y x a a x a x

y a x

a m r

and a m

2

0

)( ) m r

r

r m r l x

Substituting these values in (1), we have

2 2

0 0 0

2

0

(1- ) ( )( ) 2 ( ) ( 1) 0

( )( ) ( 1) 2( ) ( )( ) 0

( )(

m r m r l m r

r r r

r r r

m r m r

r

r

x a m r m r l x x a m r x n n a x

m r m r l x n n m r m r m r l x a

m r

2

0

) ( 1) ( ) ( ) 0 ...(3)m r m r

r

r

m r l x n n m r m r l x a

The equation (3) is an identity and therefore coefficients of various powers of x

must variable. Now equating to zero the coefficient of xm i.e. by substituting r = 0

in the second summation we get,

a0 {n(n + 1) – m (m + 1)} = 0

But a0 ≠ 0, as it is the coefficient of the very first term in the series

Here, n(n + 1) – m (m + 1) = 0 …(4)

n2 + n – m2 – m = 0 ⇒ (n2 – m2) + (n – m) = 0

i.e., (n – m)(n + m + 1) = 0, This is the indicial equation.

⇒ which gives m = n or m = - n – 1 …(5)

Next equating to zero the coefficient of xm – 1 by putting r = 1, in the second

summation a1[n(n + 1) – (m – 1)m] = 0

⇒ a1(n2 + n – m2 + m) = 0 ⇒ a1[(n2 – m2) + n + m] = 0

⇒ a1((m + n) (m – n – l)] = 0

which gives a1 = 0 …(6)

Since (m + n)(m – n – l) ≠ 0, by (5)

Again to find a relation in successive coefficients ar, etc., equating the

coefficient of xm – r - 2 to zero, we get

(m – r)(m – r – l) ar + [n(n + 1) – (m – r – 2)(m – r – l)] ar + 2 = 0 …(7)

Now, n(n + 1) – (m – r – 2)(m – r – l) = n2 + n – (m – r – l – l)(m – r – l)

= − 𝑚 − 𝑟 − 𝑙 2 − 𝑚 − 𝑟 − 𝑙 − 𝑛2 − 𝑛

= − 𝑚 − 𝑟 − 𝑙 + 𝑛 𝑚 − 𝑟 − ! − 𝑛 − (𝑚 − 𝑟 − 𝑙 + 𝑛)

= − 𝑚 − 𝑟 − 𝑙 + 𝑛 𝑚 − 𝑟 − 𝑙 − 𝑛 − 𝑙

= − 𝑚 − 𝑟 + 𝑛 − 𝑙 𝑚 − 𝑟 − 𝑛 − 2

On simplification (7) becomes

⇒ (m – r)(m – r – l) ar – (m – r + n – l)(m – r – n – 2) ar + 2 = 0

⇒ 𝑎𝑟+2 = 𝑚−𝑟 𝑚−𝑟−𝑙

𝑚−𝑟+𝑛−𝑙 𝑚−𝑟−𝑛−2 𝑎𝑟 …(8)

Now since a1 = a3 = a5 = a1 = … = 0

For the two values given by (5) there arises following two cases.

Case I: When m = n

𝑎𝑟+2 = − 𝑛−𝑟 𝑛−𝑟−𝑙

2𝑛−𝑟−𝑙 𝑟+2 𝑎𝑟 [From(8)]

If r = 0 𝑎2 = −𝑛 𝑛−𝑙

2𝑛−𝑙 2𝑎0

If r = 2, 𝑎4 = − 𝑛−2 𝑛−3

2𝑛−3 ×4𝑎2 =

𝑛 𝑛−1 𝑛−2 𝑛−3

2𝑛−1 2𝑛−3 2.4𝑎0

and so on and a1 = a3 = a5 = … = 0

Hence the series (2) becomes

𝑦 = 𝑎0 𝑥𝑛 𝑛 𝑛−1

2𝑛−1 .2𝑥𝑛−2 +

𝑛 𝑛−1 𝑛−2 (𝑛−3)

2𝑛−1 2𝑛−3 2.4. 𝑥𝑛−2 − ⋯

Which is a solution of (1).

Case II: When m = − (n + 1), we have

𝑎𝑟+2 = − 𝑛+𝑟+1 𝑛+𝑟+2

𝑟+2 2𝑛+𝑟+3 𝑎𝑟 [From(8)]

If r = 0, 𝑎2 = − 𝑛+1 𝑛+2

2 2𝑛+3 𝑎0;

If r = 2, 𝑎4 = − 𝑛+3 𝑛+4

4. 2𝑛+5 𝑎2 =

𝑛+1 𝑛+2 𝑛+3 𝑛+4

2.4 2𝑛+3 2𝑛+5 𝑎0 and so on.

Hence the series (2) in this case becomes

𝑦 = 𝑎0 𝑥−𝑛−1 + 𝑛+1 𝑛+2

2. 2𝑛+3 𝑥−𝑛−3 +

𝑛+1 𝑛+2 𝑛+3 𝑛+4

2.4 2𝑛+3 2𝑛+5 . 𝑥−𝑛−5 + + ⋯ …(9)

This gives another solution of (1) in a series of descending powers of x.

Note. If we want to integrate the Legendre’s equation in a series of ascending

powers of x, we any proceed by taking

4 1 1 2

0 1 2

0

...k k r

r

r

y a x a x a x a x

But integration in descending powers of x is more important than that in ascending

powers of x.

1.1. LEGENDRE’S POLYNOMIAL Pn(x)

1.2. LEGENDRE’S FUNCTION OF THE SECOND KIND i.e. Qn(x)

1.3 GENERAL SOLUTION OF LEGENDRE’S EQUATION

Since Pn(x) and Qn(x) are two independent solution of Legendre’s equation,

Therefore the most general solution of Legendre’s equation is

y = APn(x) + Qn(x)

Where A and B are two arbitrary constants

2. RODRIGUE’S FORMULA

𝑃𝑛 𝑥 = 1

2𝑛 .𝑛 !

𝑑𝑛

𝑑𝑥 𝑛 𝑥2 − 1 𝜋

Proof. Let v = (x2 – 1)n …(1)

Then 𝑑𝑣

𝑑𝑥= 𝑛 𝑥2 − 1 𝑛−1 2𝑥

Multiplying both sides by (x2 – 1), we get

𝑥2 − 1 𝑑𝑣

𝑑𝑥= 2𝑛 𝑥2 − 1 𝑛𝑥.

⇒ 𝑥2 − 1 𝑑𝑣

𝑑𝑥= 2𝑛𝑣𝑥 [Using (1)] …(2)

Now differentiating (2), (n + 1) times Leibnitz’s theorem, we have

𝑥2 − 1 𝑑𝑛+2𝑣

𝑑𝑥 𝑛+2+ 𝑛+1 𝐶1 2𝑥

𝑑𝑛+1𝑣

𝑑𝑥 𝑛+1+ 𝑛+1 𝐶2 2

𝑑𝑛𝑣

𝑑𝑥 𝑛= 2𝑛 𝑥

𝑑𝑛+1𝑣

𝑑𝑥 𝑛+1+ 𝑛+1 𝐶1 𝑙

𝑑𝑛𝑣

𝑑𝑥 𝑛

⇒ 𝑥2 − 1 𝑑𝑛+2𝑣

𝑑𝑥 𝑛+2+ 2𝑥 𝐶1 − 𝑛

𝑛+1

𝑑𝑛+1𝑣

𝑑𝑥 𝑛+1+ 2 𝐶2 − 𝑛. 𝑛+1 𝑛+1

𝐶1 𝑑𝑛𝑣

𝑑𝑥 𝑛= 0

⇒ 𝑥2 − 1 𝑑𝑛+2𝑣

𝑑𝑥 𝑛+2+ 2𝑥

𝑑𝑛+1𝑣

𝑑𝑥 𝑛+1− 𝑛 𝑛 + 1

𝑑𝑛𝑣

𝑑𝑥 𝑛= 0 …(3)

If we put 𝑑𝑛𝑣

𝑑𝑥 𝑛= 𝑦, (3) becomes

𝑥2 − 1 𝑑2𝑦

𝑑𝑥 2+ 2𝑥

𝑑𝑦

𝑑𝑥− 𝑛 𝑛 + 1 𝑦 = 0

⇒ 1 − 𝑥2 𝑑2𝑦

𝑑𝑥 2− 2𝑥

𝑑𝑦

𝑑𝑥+ 𝑛 𝑛 + 1 𝑦 = 0

This shows that 𝑦 =𝑑𝑛𝑣

𝑑𝑥 𝑛 is a solution of Legendre’s equation.

∴ 𝐶𝑑𝑛𝑣

𝑑𝑥 𝑛= 𝑃𝑛 𝑥 …(4)

Where C is a constant.

But v = (x2 – 1)n = (x + 1)n (x – 1)n

so that 𝑑𝑛𝑣

𝑑𝑥 𝑛= 𝑥 + 1 𝑛 𝑑𝑛

𝑑𝑥 𝑛 𝑥 − 1 𝑛 + 𝐶1. 𝑛 𝑥 + 1 𝑛 𝑛−1 𝑑𝑛−1

𝑑𝑥 𝑛−1 𝑥 − 1 𝑛 + ⋯ +

𝑥 − 1 𝑛 𝑑𝑛

𝑑𝑥 𝑛 𝑥 + 1 𝑛 = 0

when x = 1, then 𝑑𝑛𝑣

𝑑𝑥 𝑛= 2𝑛 . 𝑛!

All the other terms disappear as (x – 1) is a factor in every term except first.

Therefore when x = 1, (4) gives

C.2n.n! = Pn(1) = 1 [Pn(1) = 1]

𝐶 =1

2𝑛 .𝑛 ! …(5)

Substituting the value of C from (5) in (4), we have

𝑃𝑛 𝑥 =1

2𝑛 .𝑛 ! 𝑑𝑛𝑣

𝑑𝑥 𝑛

𝑃𝑛 𝑥 =1

2𝑛 .𝑛 !

𝑑𝑛

𝑑𝑥 𝑛 𝑥2 − 1 𝑛 ∵ 𝑣 = 𝑥2 − 1 𝑛

3. A GENERATING FUNCTION OF LEGENDRE’S POLYNOMIAL

Thus coefficient of zn in the expansion of (1) is sum of (2), (3) and (4) etc.

=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)

Thus coefficient of z, z2, z3 … etc. in (1) are P1(x), P2(x), P3(x) …

Hence

(1 – 2xz + z2)-1/2 = P0(x) + zP1(x) + z2P2(x) + z3P3(x) + … + zn Pn(x) + …

i.e.,

1/ 22

0

1 2 ( ). Proved.n

n

n

n

xz z P x z

5. RECURRENCE FORMULAE

6. NUMERICAL PROBLEMS

7. LEGENDRE’S POLYNOMIALS APPLICATIONS

Similarly 𝑃3 𝑥 =1

2 5𝑥3 − 3𝑥

𝑃4 𝑥 =1

8 35𝑥4 − 30𝑥4 + 3

𝑃5 𝑥 =1

8 63𝑥5 − 70𝑥3 + 15𝑥

……………………………………………….

2

0

( 1) (2 2 )!( )

2 . !( )!( 2 )!

rnn r

n nr

n rP x x

r n r n r

where 𝑁 =𝑛

2 if n is even

𝑁 =1

2 𝑛 − 1 if n is odd

Note. We can evaluate Pn(x) by differentiating (x2 – 1)n, n times.

2 2 2 2

0 0

2 2 2

0

2

0

!( 1) ( ) ( 1) ( 1)

!( )!

1 1 !( ) ( 1) ( 1)

2 . ! 2 . ! !( )!

( 1) (2 2 )!

!( )!( 2 )!

n r n r nn n n r r r n r

rnr r

n r nn n n r

n n n nr

rNn r

r

d nx C x x

dx r n r

d nP x x x

n dx n r n r

n rx

r n r n r

Either x0 or x1 is in the last term.

∴ n – 2r = 0 or 𝑟 =𝑛

2 (n is even)

N – 2r = 1 or 𝑟 =1

2 𝑛 − 1 (n is odd)

8. NUMERICAL PROBLEMS

Example 6. Prove that Pn(1) = 1.

Solution. We know that

(1 – 2xz + z2)-1/2 = 1 + zP1(x) + z2 P2(x) + z3 P3(x) + … + zn Pn(x) + …

Substituting 1 for x in the above equation, we get

(1 – 2 z + z2)-1/2 = 1 + zP1(1) + z2 P2(1) + z3 P3(1) + … + zn Pn(1) + …

1/ 2 12

0

1 2 3

(1 ) (1) 1 (1)

(1) 1 1 ... ...

n n

n n

n

n n

n

z z P z z P

z P z z z z z

Equating the coefficient of zn on both sides, we get

Pn(1) = 1

Example 7. Prove that

0

1( ) .

2 2n

n

P xx

Solution. We know that

1

2 2

0

(1 2 ) ( ) ...(1)n

n

n

xz z z P x

Putting z = 1 in (1), we get

1

2

0

0

1 2 1 ( )

1( ) Proved.

2 2

n

n

n

n

x P x

P xx

Example 8. Prove that:

1

0

1 1( ) log

1 2 1

n

n

n

x xP x

n x

Solution. We know that

1/ 22

0

( ) 1 2n

n

n

h P x xh h

Integrating both sides w.r.t. h from 0 to h, we get

1

2 2 20 0 0

( ) ; 1 Here x is constant h is variable.1 1 2 1

h hn

n

n

h dh dhP x if x

n hx h h x x

= log ℎ−𝑥 + ℎ2−2ℎ𝑥+1

1−𝑥

𝑑ℎ

ℎ2+𝑎2= log

ℎ+ ℎ2+𝑎2

𝑎

Putting h = x in the expression, we get

1 2

0

1 1 1( ) log log Proved.

1 1 2 1

n

n

n

x x xP x

n x x

Example 9. Show that

Pn(-x) = (-1)n Pn (x) and Pn(-1) = (-1)n.

Solution. We know that

12 2

0

1 2 ( ) ...(1)n

n

n

xz z z P x

Putting –x for x in both sides of (1), we get

12 2

0

1 2 ( ) ...(2)n

n

n

xz z z P x

Again putting –z for z in (1), we obtain

12 2

0

1 2 ( 1) ( ) ...(3)n n

n

n

xz z z P x

From (2) and (3), we have

0 0

( ) ( 1) ( )n n n

n n

n n

z P x z P x

Comparing the coefficients of zn from both sides of (4), we obtain …(4)

Pn(-x) = (-1)n Pn(x)

Pn(-1) = (-1)n Pn(1) = (1)(-1)n …(5)

Putting x = 1 in (5), we get [Pn(1)n1]

(i) If n is even, then from (5)

Pn(-x) = Pn (x),

So, Pn(x), is even function of x.

(ii) If n is odd, then from (5)

Pn(-x) = -Pn(x), so Pn(x) is odd function.

Example 12. Show that

3/ 22

0

1(2 1) .

1 2

in

n

n

zn P z

xz z

Solution. We know that

12 2

0

1 2 ( ) ...(1)n

n

n

xz z z P x

Differentiating both sides of (1) with respect to z, we get

3/ 2

2 1

0

1

2 3/ 20

11 2 ( 2 2 ) . ( )

2

( ) ...(2)(1 2 )

n

n

n

n

n

n

xz z x z nz P x

x znz P x

xz z

Multiplying both sides of (2) by 2z, we get

21

2 3/ 20

2 22 ( ) ...(3)

(1 2 )

n

n

n

xz znz P x

xz z

2 2

20

2

2 3/ 20

1 2 2 2 (2 1) ( )

(1 2 )

1 (2 1) ' Proved.

(1 2 )

n

n

n

n

n

n

xz z xz zn z P x

xz z

zn z P

xz z

Example 13. Prove that

12

0

1 1( + )

1 2

n

n n

n

zP P z

zz xz z

Solution.

1 1

0 0 0

. . . ( + ) +n n n

n n n n

n n n

R H S P P z z P z P

1

1

0 0

1 ...(1)n n

n n

n n

z P z Pz

2 3

0 1 2 3

0

1 2 3

1 1 2 3

0

But ...

And ...

n

n

n

n

n

n

z P P zP z P z P

z P zP z P z P

= -P0 + P0 + zP1 + z2P2 + z3P3 + … = -P0 + 𝑧𝑛𝑃𝑛

⇒ log1+sin

𝜃

2

sin𝜃

2

= 1 +1

2𝑃1 cos 𝜃 +

1

3𝑃2 cos 𝜃 +

1

4𝑃3 cos 𝜃 + …

Example 17. Assuming that a polynomial f(x) of degree n can be written as

0

( ) ( ).x

m m

m

f x C P x

show that 𝐶𝑚 =2𝑚+1

2 𝑓 𝑥 𝑃𝑚 𝑥 𝑑𝑥

1

1

Solution. We have,

0

( ) ( )m m

m

f x C P x

= C0P0(x) + C1P1(x) + C2P2(x) + C3P3(x) + C4P4(x) + … + CmPn(x)+…+

Multiplying both sides by Pm(x), we get

Pm(x)f(x)=C0P0(x) Pm(x) + C1P1(x) Pm(x) + C2P2(x) Pm(x) + … + Cm𝑃𝑚2 (x) + …

𝑓(𝑥)+1

−1 Pm(x) dx = [

+1

−1C0P0(x) Pm(x) + C1P1(x) Pm(x) + C2P2(x) Pm(x) + …

Cm𝑃𝑚2 (x) + …]

= 0 + 0 + ⋯ + 𝐶𝑚2

2𝑚+1+ ⋯ =

2𝐶𝑚

2𝑚+1

𝐶𝑚2𝑚+1

2 𝑓 𝑥 𝑃𝑚 𝑥 𝑑𝑥

+1

−1

THANKS