PPT HERMITE BY DR. RAJESH MATHPAL-converted.pdf

Hermite polynomial

1. MATHEMATICAL PHYSICS
UNIT –3
HERMITE FUNCTION
DR. RAJESH MATHPAL
ACADEMIC CONSULTANT
SCHOOL OF SCIENCES
UTTARAKHAND OPEN UNIVERSITY
TEENPANI, HALDWANI
UTTRAKHAND
MOB:[phone]
Email: [email protected]

2. STRUCTURE OF UNIT
•3.1 INTRODUCTION
•3.2 HERMITE’S EQUATION
•3.3. SOLUTION OF HERMITE’S EQUATION
•3.4. GENERATING FUNCTION OF HERMITE POLYNOMIALS
(RODRIGUE FORMULA )
•3.5. ORTHOGONAL PROPERTY
•3.6. RECURRENCE FORMULAE FOR H
n(X) OF HERMITE
EQUATION

3. 3.1 INTRODUCTION
Hermitepolynomialswere defined by Pierre-Simon Laplace in 1810, though in
scarcely recognizable form, and studied in detail by PafnutyChebyshevin 1859.
They were consequently not new, althoughHermitewas the first to define the
multidimensionalpolynomialsin his later 1865 publications.

4. 3.2 HERMITE’S EQUATION
The differential equation of the form
??????
2
�
??????�
2
−2�
??????�
??????�
+2��=0 …(1)
Is called Hermiteequation.
The solution of (1) is known as Hermite’spolynomial.

5. 3.3. SOLUTION OF HERMITE’S
EQUATION
•Here, we have
??????
2
�
??????�
2
−2�
??????�
??????�
+2��=0 …(1)
•Suppose its series solution is
y = a
0x
m
+ a
1x
m+1
+ a
2x
m+2
+ a
3x
m+3
+ …… + a
kx
m+k 0
or ...(2)
mk
k
k
y a x

+
=
=

6. ��
��
= �
�
�+� �
�+�−1
2
2
2
0
( )( 1)
mk
k
k
dy
a m k m k x
dx

+−
=
= + + −

Putting the values of y,
��
��
and
�
2
�
��
2
in (1), we get
⇒ �
�
�+� �+�−1 �
�+�−2
−2� �
�
�+� �
�+�−1
+2� �
��
�+�
=0
⇒ �
�
�+� �+�−1 �
�+�−2
−2� �
�
�+� �
�+�
+2� �
��
�+�
=0
⇒ �
�
�+� �+�−1 �
�+�−2
−2� �
�
�+� −� �
�+�
=0 …(3).
This equation holds good for k = 0 and all positive integer. By our assumption k cannot
be negative.

7. To get the lowest degree term x
m-2
, we put k = 0 in the first summation of (3)
and we cannot have x
m-2
from the second summation. Since k ≠−2.
The coefficient of x
m-2
is
a
0m (m –1) = 0 ⇒m = 0, m = 1, since a
0≠0 …(4)
This is the indicial equation.
Now equating the coefficient of next lowest degree term x
m-1
, zero in (3), we
get (by putting k = 1 in the first summation and we cannot have x
m-1
from the
second summation since (k ≠−1).

8. �
1��+1=0
⇒ቈ
�
1������������������ℎ���=0
�
1=0,�ℎ���=1
�+1≠0���??????�
������������������
Again equating the coefficient of the general term x
m-k
to zero, we get
a
k+2(m + k + 2) (m + k + 1) –2a
k(m + k –n) = 0
�
??????+2=
2�+??????−�
�+??????+2�+??????+1
�
?????? …(5)
If m = 0, then,�
??????+2=
2??????−�
??????+2??????+1
�
?????? …(6)
If m = 1, then,�
??????+2=
2??????−1−�
??????+3??????+2
�
?????? …(7)

9. Case I.When m = 0, a
k+2=
�??????−??????
??????+�??????+�
??????
??????
If k = 0, then,�
2=
−2�
2
�
0=−��
0
If k = 1, then,�
3=
21−�
6
�
1=−2
�−1
3!
�
1
If k = 2, then,�
4=
22−�
12
�
2=2
2−�
12
−��
0=2
2
��−2
4!
�
0
If k = 3, then,�
5=
23−�
20
�
3=2
23−�
20
−
2�−1
3!
�
1=2
2
�−1�−3
5!
�
1
�
2??????=
−2
??????
��−2�−4………�−2??????+2
2??????!
�
0
�
2??????+1=
−2
??????
�−1�−3………�−2??????+1
2??????+1!
�
1=0
When m = 0, then there are two possibilities

10. Possibility I. When, a1 = 0, then a3 = a5 = a7 = a2r+1 = ….. = 0.
Possibility II. When a1 ≠ 0
0
k
k
k
y a x

=
=
i.e. y = a0 + a1x + a2x
2 + a3x
3 + a4x
4 + a5x
5 + …………………
= a0 + a2x
2 + a4x
4 +………………… + a 1x + a3x
3 + a5x
5. …(8)
Putting the values of a0, a1, a2, a3, a4 and a5 in (8), we get
=�
0 1−
2�
2!
�
2
+
2
2
� �−2
4!
�
4
−⋯+ −1
�
2
2� !
� �−2 … �−2�+2 �
2�
+⋯
+�
1� 1−
2 �−1
3!
�
2
+
2
2
�−1 �−3
5!
+ −1
�
2
�
2�+1 !
�−1 �−3 …(�−2�+1)�
2�
+
⋯ ….(9)
2
0
1
21
0 1 0
1
( 1) 2
1 ( 2)...( 2 2)
(2 )!
( 1) 2
( 1)( 3)...( 2 2) (If a =a ) ...(10)
(2 1)
rr
r
r
rr
r
r
a n n n r x
r
a x n n n r x
r

=

+
=
 −
= + − − +


 −
= + − − − +

+



11. Case II.When m = 1, then a
1= 0 and so by putting k = 0, 1, 2, 3, ………. In (7), we get
�
??????+2=
2 ??????+1−�
??????+3??????+2
�
??????
�
2=−
2 �−1
3!
�
0
�
4=
2
2
�−1�−3
5!
�
0
…………………………………….
�
2??????−1
??????
=
2
??????
�−1�−3…�−2??????+1
2??????+1!
�
0
Hence, the solution is
=�
0� 1−
2�−1
3!
�
2
+
2
2
�−1�−3
5!
�
4
…+
−1
??????
2
??????
�−1�−3…�−2??????+1
2??????+1!
�
2??????
+⋯…(11)

12. It is clear that the solution (11) is included in the second part of (9) except that a
0is
replaced by a
1and hence in order that the Hermiteequation may have two
independent solutions, a
1must be zero, even if m = 0 and then (9) reduce to
=�
0� 1−
2�−1
3!
�
2
+
2
2
�−1�−3
5!
�
4
…+
−1
??????
2
??????
�−1�−3…�−2??????+1
2??????+1!
�
2??????
+⋯
…(12)
The complete integral of (1) is then given by
�=�1−
2�
2!
�
2
+
2
2
��−2
4!
�
4
−⋯+�1−
2�−1
3!
�
2
+
2
2
�−1�−3
5!
�
4
…
…(13)
where A and B are arbitrary constants.

13. 3.4. GENERATING FUNCTION OF
HERMITE POLYNOMIALS
(RODRIGUE FORMULA )We know that
�
�
2??????
�
??????
�
�
− 1−�
2

=??????
�
� +??????
�+1
� �+??????
�+2
� .�
2
+ ……… …(1)
Now differentiating �
− 1−�
2

�.�.�.,�, we get

??????
??????�
�
− 1−�
2

=−2 �−� �
− �−�
2


Taking limit when �→0, we get
 
2
2()
0
lim 2 ...(2)
tx
x
t
e xe
t
−−
−
→

=


14. Again differentiating �
− 1−�
2

w.r.t. ‘x’, we get

??????
??????�
�
− 1−�
2

= −1
2
�−� �
− �−�
2


Taking limit when t → 0, we get
 
2
2()
0
lim 2 ...(3)
tx
x
t
e xe
x
−−
−
→

=

From (2) and (3), we have
   
22
( ) ( )
1
00
lim ( 1) lim
t x t x
tt
ee
tx
− − − −
→→

=−


15. Similarly,
   
22
22
( ) ( )
2
22
00
lim ( 1) lim
t x t x
tt
ee
tx
− − − −
→→

=−

……………………………………………………………………………..
………………………………………………………………………………
   
22
2( ) ( )
00
lim ( 1) lim ( 1)
n n n
t x t x
n n x
n n n
tt
d
e e e
t x dx
− − − −
−
→→

= − = −

[differentiating n times] …(4)
Putting t = 0 in (1), we get
 
2
2 ()
0
lim ( ) ...(5)
n
tx
x
nn
t
e e H x
t
−−
→

=


16. Putting the value of
 
2
()
0
lim
n
tx
n
t
e
t
−−
→

 from (4) in (5), we get
(-1)
n �
�
2�
�
��
�
�
−�
2
=Hn (x)
Hn (x) = (-1)
n �
�
2�
�
��
�
�
−�
2
…(6)
n = 0
On putting n = 0 in (6), we get
H0(x) = (-1)
0 �
�
2
�
−�
2
=1
H0(x) = 1
n = 1
On putting n = 1 in (6), we get
H1(x) = (-1)
1 �
�
2�
��
�
−�
2
=−�
�
2
−2� �
−�
2
=2�
H1(x) = 2x

17. n = 2
On putting n = 2 in (6), we get
H
2(x) = (-1)
2
�
�
2??????
2
??????�
2
�
−�
2
=�
�
2??????
??????�
−2��
−�
2
=�
�
2
−2�
�
2
−2� −2� �
−�
2
=−2+4x
2
H
2(x) = 4x
2
–2
n = 3
On putting n = 3 in (6), we get
H
3(x) = (-1)
3
�
�
2??????
2
??????�
3
�
−�
2
=−�
�
2??????
2
??????�
3
−2��
−�
2
=−�
�
2??????
??????�
−2��
−�
2
+−2�−2��
−�
2
=−�
�
2??????
??????�
−2+4�
2
�
−�
2
=�
�
2
8��
−�
2
+4�
2
−2−2��
−�
2
=−8�+4�
2
−2−2�=−8�+8�
3
−4�=8�
3
−12�
H
3(x) = 8x
3
–12x

18. SimilarlyH
4(x) = 16x
4
–48 x
2
+ 12
H
5(x) = 32x
5
–160 x
3
+ 120x
H
6(x) = 64x
6
–480 x
4
+ 720 x
2
–120
H
7(x) = 128x
7
–1344 x
5
+ 3360 x
3
–1680x

19. Example 1.Convert Hermitepolynomial
2H
4(x) + 3 H
3(x) –H
2(x) + 5 H
1(x) + 6 H
0
into ordinary polynomial.
Solution.Here, we have
2H
4(x) + 3 H
3(x) –H
2(x) + 5 H
1(x) + 6 H
0
=
2
16
�
4
−
48
�
2
+
12+
3
8�
3
−
12�−4
�
2
−
2+
5
2�+
6
1
= 32x
4
–96x
2
+ 24 + 24x
3
–36x –4x
2
+ 2 + 10x + 6
= 32x
4
+ 24x
3
–100x
2
–26x + 32

20. Example 2.Convert ordinary polynomial
64x
4
+ 8x
3
–32x
2
+ 40x + 10
into Hermitepolynomial.
Solution.Here, we have
Let 64x
4
+ 8x
3
–32x
2
+ 40x + 10 AH
4(x) + BH
3(x) + CH
2(x) + DH
1(x) EH
0(x)
= A (16x
4
–48x
2
+ 12) + B (8x
3
–12x) + C (4x
2
–2) + D (2x) + E (1)
= 16Ax
4
+ 8Bx
3
(-48A + 4C)x
2
+ (-12B + 2D) x + 12A –2C + E
Equating the coefficients of like powers of x, we get
16A = 64⇒A = 4
8B = 8 ⇒B = 1
−48A + 4C = −32⇒4C = −32 + 192 ⇒C = 40
−12B + 2D = 40⇒−12 + 2D = 40 ⇒2D = 52⇒D = 26
12A –2C + E = 10⇒12 ×4 –2 (40) + E = 10 ⇒E = 42
The required Hermitepolynomial is
4 H
4(x) + H
3(x) + 40 H
2(x) + 26 H
1(x) + 42 H
0(x)

21. 3.5. ORTHOGONAL PROPERTYThe orthogonal property of Hermites polynomials is
2
2
0,
( ) ( )
!,
x
mn
mn
e H x H x dx
n n m n

−
−

=
=


22. Solution. We know that
�
�
2
− �1−�
2

=
??????�
�
�!
�
1
�
(generating function) …(1)
and �
�
2
− �2−�
2

=
??????�
�
�!
�
1
�
…(2)
Multiplying (1) and (2), we get
() 
() 
22
22
2
1
11
00
( ) ( )
!!
x t x
x t x
nmnm
nm
H x H x
e e t t
nm
−− 
−−
==
   
=
   
   

12
0
( ) ( )
!!
nm
mm
m
tt
H x H x
nm

=
=  


23. Multiplying both the sides of this equation by �
−�
2
and then integrating with
the limits from −∞ to ∞, we have ()  () 
2222
22
12
12
( ) ( ) .
!!
nm
x t x x t x
xx
nm
nm
tt
e H x H x dx e e e dx
nm
 − − − −
−−
− −

=



( )  () 
22 2
12 12
2
...(3)
tt x x t t
e e dx
−+ − + +
−
= 
We have already learnt that
 
2
2
2
2
b
ax bx
a
e dx e
−+
−
= [standard formula] … (4)
Replacing 2b by (t1 + t2) and a by 1 in (4), we get
()  ()
2
12
12
2
2
...(5)
x x t t
tt
e dx e 
− + +
+
−
=

24. Putting the value of
() 
2
12
2x x t t
e dx
− + +
−
from (5) in R.H.S. of (3), we get
�
− �1+�2

2

. ??????�
�1+�2

2
= ??????�
−�
1
2
−�
2
2
+�
1
2
+�
2
2
+2�1+�2
= ?????? 1+2�
1+�
2+
2�1�2

2
2!
+
2�1�2

3
3!
+⋯ = ??????
2�1�2

�
�!

= ??????
2
�
�
1
�
�
2
�

�!
= ?????? 2
�
�
1
�
�
2
�
??????
�.�
∞
�=0
�=0
�
2
�
=�
2
�
??????
�.�

From (3), we have
2
12
1 2 .
2
( ) ( )
! ! !
nm n
x n m
n m n m
nm nm
tt
e H x H x dx t t
n m n


−
−

=



25. On equating the coefficients of �
1
�
, �
2
�
on both sides, we get
2
.
( ) ( ) 2
! ! !
n
x nm
nm
H x H x
e dx
n m n



−
−
= 2
.
( ) ( ) 2 !
xn
n m n m
e H x H x dx m

−
−
= 
2
,
.
0 0,
( ) ( ) 2 !
1, 2 ! ,
nmxn
m n n m
n
mn if m n
e H x H x dx m
if m nn m n




−
−
 =
=  
=== 


26. Example 3. Find the value of
2
23
( ) ( ) .
x
e H x H x dx

−
−
Solution. We known that
2
( ) ( ) 0
x
mn
e H x H x if m n

−
−
=
Here m = 2 and n = 3, �≠�
Hence,
2
23
( ) ( ) 0
x
e H x H x

−
−
=

27. Example 4. Find the value of
 
2 2
2
()
x
e H x dx

−
−
Solution. We know that
 
2 2
( ) 2 ( )!
xn
n
e H x dx n 

−
−
=
 
2 2
2
2
( ) 2 (2!) 8
x
e H x dx 

−
−
==

28. 3.6. RECURRENCE FORMULAE FOR
H
n(X) OF HERMITE EQUATION
Four recurrence Relations
•2n H
n-1(x)= H
’
n(x)
•2x H
n(x) = 2n H
n-1(x) + H
n+1(x)
•H
’
n(x) = 2x H
n(x) –H
n+1(x)
•H
’
n(x) = x H
’
n(x) + 2n H
n(x) = 0

29. THANKS