Partial Differential Equations

Reference Books
E. Kreyszig, Advanced Engineering Mathematics,
Ch. 11. (CH12 9th Ed)
Jain and Iyengar: Sec 9.5
INTRODUCTION
Let u be a variable which depends on independent variables x, y
and related to them by a relation of the form
u  f ( x, y)

If the function on right is differentiable partially with respect to

x and y respectively, we write partial derivatives of u as
First Order Partial Derivatives
u u
ux  , uy 
x y
Second Order Partial Derivatives
2 u 2 u 2 u
u xx  , u xy  , u yy 
2
x x y y 2
An equation containing x, y, u, ux , uy , that is, F ( x , y , u, u x , u y )  0
is a First Order partial differential equation.
The equation is linear, if it is linear in ux, uy.
An equation containing x, y, u, ux, uy,, uxx, uxy, uyy that is,
F ( x , y , u, u x , u y , u xx , u xy , u yy )  0

is a Second Order partial differential equation.

The second order PDE is linear, if it is linear in derivative
terms.
If each term of a PDE contains dependent variable or its partial
derivative, the equation is said to be Homogeneous; otherwise
it is said to be Non-Homogeneous.
In most of applications the quantities involved (dependent
variable and coefficients) are functions of position defined by
the coordinates x, y, z and time t.
 CLASSIFICATION OF LINEAR SECOND
ORDER PDES
CONSIDER THE SECOND ORDER, LINEAR HOMOGENEOUS PARTIAL DIFFERENTIAL EQUATION

 2u  2u  2u u u
A B C D E  Fu 0
2 xy 2 x y
x y
WHERE A, B, C, D, E AND F ARE FUNCTIONS OF X, Y OR CONSTANTS.

THE PDE IS SAID TO BE

B 2  4 AC  0
B 2  4 AC 0
ELLIPTIC IF

B 2  4 AC  0
PARABOLIC IF

HYPERBOLIC IF

NOTE THAT FIRST ORDER TERMS HAVE NO ROLE IN CLASSIFICATION.

CLASSIFICATION IS HELPFUL IN DETERMINING THE NUMBER AND TYPE OF CONDITIONS TO BE PRESCRIBED SO

THAT PROBLEM IS WELL FORMULATED AND HAS A UNIQUE SOLUTION.
 Some Important linear Partial Differential Equations
2 u 2  2
u One dimensional Wave equation.
c
2
t x 2 Hyperbolic
2 One dimensional diffusion (or heat) equation.
u 2 u
c
t x 2 Parabolic

2 u 2 u Two dimensional Laplace equation.

 0
x 2 y 2 Elliptic

2 u  2 u 2 u 
2
c    Two dimensional Wave equation.
t 2  2 2 
 x y 

2 u 2 u 2 u
  0 Three dimensional Laplace equation.
2 2 2
x y z
All these equations are Homogeneous.
 2 u 2 u
  f ( x, y) Two dimensional Poisson equation.
2 2
x y
This is a Non-homogeneous equation.
A solution of a PDE in some region R of the space of
independent variables is a function, which has all the derivatives
appearing in PDE and satisfies the equation everywhere in R.
In general, the totality of solutions of a PDE is very large.
For example 2-D Laplace equation
2 u 2 u has the following solutions:
 0
2 2
x y

u  x2  y2 u e x cos y u ln( x 2  y 2 )
These solutions are entirely different from each other.
However we can find unique solutions by using the additional
conditions defined in the problem, such as
Boundary Conditions: The dependent variable u and its
derivatives are known on the boundary of the region.
Initial Conditions: When time t is one of the variables, the
dependent u and its derivatives are known at some fixed time,
usually at t = 0.
As in the case of homogeneous linear ordinary differential
equations, we can construct more solutions of a homogeneous
linear partial differential equation by using
Linearity or Superposition Principle.

Theorem: If u1 and u2 are two solutions of a linear homogeneous

PDE in some region R, then u = c1 u1 + c2 u2 with any constants c1
and c2 is also a solution of that equation in R.
SOLUTION OF SECOND ORDER PDE
WITH CONSTANT COEFFICIENTS
METHOD OF SEPARATION OF VARIABLES OR FOURIER METHOD
LET IN A PDE WITH CONSTANT COEFFICIENTS U BE THE DEPENDENT VARIABLE AND X, Y BE
THE INDEPENDENT VARIABLES.
WE PROPOSE A SOLUTION OF PDE IN THE FORM OF PRODUCT OF A FUNCTION OF X AND A
FUNCTION Y.
WE TAKE u ( x, y )  X ( x)Y ( y )
THEN WE HAVE

u  u   2u  2u
 XY   X Y  XY   X Y   X Y  X Y 
x x y y x 2 y 2

du du d 2u d 2u
where X  Y X  Y  
dx dy dx 2 dy 2

THESE EXPRESSIONS ARE SUBSTITUTED IN PDE, VARIABLES ARE SEPARATED RESULTING IN

ODES WHICH ARE SOLVED FOR X AND Y.
EXAMPLE
USING THE METHOD OF SEPARATION OF VARIABLES SOLVE THE
HYPERBOLIC EQUATION
 2u  2u
16
2
t x 2

TAKING
u ( x, t )  X ( x) T (t )
WE GET
T  X 
X T  16 X T OR 
16T X

NOW LEFT HAND SIDE IS A FUNCTION OF T ALONE AND RIGHT HAND SIDE
IS FUNCTION X ALONE.
THEREFORE BOTH LEFT AND RIGHT HAND SIDES MUST BE EQUAL TO A
CONSTANT .
IT MAYBE 0 OR K2 OR - K2 T  0 X  0
at
CASE I : CONSTANTT IS b
ZERO X cx  d
IN THIS CASE WE HAVE u ( x, y )  Axt  Bx  Ct  D
INTEGRATING WE GET
CASE 2 WHEN CONSTANT IS K2, WE OBTAIN
X  k 2 X T  16k 2T

SOLUTIONS OF THESE EQUATIONS ARE

X ae kx  be  kx T ce 4kt  de  4t
u ( x, t )  Ae k ( x  4t )  Be  k ( x  4t )  Ce k ( x  4t )  De  k ( x  4t )

HENCE
X   k 2 X T   16k 2T

CASE 3 WHEN CONSTANT IS –K2, WE OBTAIN

X a cos(kx)  b sin(kx) T c cos(4kt )  d sin(4kt )

or X  A cos(kx   ) T B cos(4kt   )
SOLUTIONS OF THESE EQUATIONS ARE
u  AB cos(kx   ) cos(4kt   )
Vibrating string and Wave equation
Mechanical disturbance propagates through a material medium
(solid, liquid, or gas) in the form of a wave at a wave speed
which depends on the elastic and inertial properties of that
medium. There are two basic types of wave motion for
mechanical waves: longitudinal waves and transverse waves.
In a Longitudinal wave the particle displacement is parallel to the
direction of wave propagation.
In a Transverse wave the particle displacement is perpendicular
to the direction of wave propagation.

First we obtain an equation governing small transverse vibrations

of an elastic string.
 We place the string along x-axis, stretch it to length L, and fix it
at the ends x = 0 and x = L.
We then distort it vertically, and at some instant, say, t = 0, we
release it and allow it to vibrate.
u

O x L X

Problem:
To determine the vibrations of the string, or
To find the deflection u(x, t) at any point x and at any time t > 0.
We first establish a differential equation which will be a
mathematical model of our physical system.
As two independent variables are involved, it will be a Partial
differential equation.
To obtain a simple equation, we make simplifying assumptions:
1. The mass of the string per unit length is constant (i.e. The
String is homogenous).
2. The string is perfectly elastic and does not offer any resistance
to bending.
3. The tension caused by stretching the string before fixing it is
so large that the action of gravitational force on the string can
be neglected.
4. When released, the string performs small motion in a vertical
plane; that is, every particle of the string moves strictly
vertically and the deflection and the slope at every point of the
string always remain small in absolute value.
 u

T2 
Q
P

T1

O x x+x L X

We consider a small portion PQ of the string. Since string does

not offer any resistance to bending, the tension is tangential to the
curve of string at each point.
Let T1 and T2 be the tension at the endpoints P and Q of that portion
making angles  and  with horizontal.
Since the particles of the string move vertically only, hence the
horizontal component of the force due to tension must vanish.
T2 cos   T1 cos   0
 or T2 cos  T1 cos  T (1)
For the motion in the vertical direction, we apply Newton’s
second law of motion
The mass of the portion of the string times the acceleration must
be equal to the vertical component of the force.
2
 x  u
 T2 sin   T1 sin 
t 2
Using (1), we divide left side by T and the terms on right by
T2 cos  and T1 cos  respectively, and obtain
 x  2 u T2 sin  T1 sin 
   tan   tan  ( 2)
T t 2 T2 cos  T1 cos 
 u 
Now tan  is the slope of the string at x , therefore tan   
 x x
 u 
Similarly tan   
 x  x  x
From (2) therefore
  2 u 1   u   u  
       
T t 2 x   x  x  x  x  x 
Letting x approach zero, we obtain
2 u  2
u T
c 2 , c2 
t 2 x 2 
This is the equation which governs our problem and is known
as One dimensional wave equation.
The nature of motion of the string is given by the solution of
this equation which satisfies the conditions imposed by the
physical system.
In the present case, string is fixed at the ends. Therefore
u(0, t )  0, u( L, t )  0 for all t  0. Boundary Conditions
The form of the motion will depend on the initial deflection
and initial velocity of the string
Denoting the initial deflection by f(x) and initial velocity by
g(x), we obtain the two Initial Conditions
u
u( x , 0)  f ( x ),  g( x ).
t t 0

Thus to find out how a vibrating string moves, we solve One

dimensional wave equation or the PDE
2 u 2
2 u
c (1)
2 2
t x
where u(x, t) satisfies
Boundary Conditions:
u(0, t )  0, u( L, t )  0 for all t  0. ( 2)
Initial Conditions:
u
u( x , 0)  f ( x ),  g( x ). (3)
t t 0
Solution :
We apply method of Separation of Variables.
Let the solution of the wave equation (1) be
u( x , t )  F ( x ) G ( t ) ( 4)
By differentiating (4), we obtain
2 u d 2G  2 u d2F
F F G and  G  F G
t 2 dt 2 x 2 dx 2
Here the ‘dots’ indicate derivative with respect to time t, and
‘primes’ indicate derivatives with respect to space variable x.
Using these expressions for derivatives in (1), we get

G F 
FG   c F G
2 or 
2 F
c G
Now left side depends on t only and right depends on x only,
therefore
 
G F 
 k
2 F
c G
This gives two ordinary linear homogeneous equations
F   k F  0 (5)
and
  k c 2G  0
G ( 6)
The constant k is arbitrary.
Boundary Conditions:
Using (4) in boundary conditions (2), we get
u(0, t )  F (0)G ( t )  0, u( L, t )  F ( L)G ( t )  0 for all t  0.
If G ( t )  0, then u( x , t )  0 and this solution is of no interest. Thus
G ( t )  0 and
F (0)  0, F ( L)  0 (7)
These are boundary conditions for (5).
Now we solve
F   k F  0 (5)
where F(x) satisfies the boundary conditions
F (0)  0, F ( L)  0 (7)

Case I : If k = 0, then general solution of (5) is

F  ax  b
Applying (7), we get a = 0, b = 0 and therefore F = 0.
This is of no interest because then u ≡ 0.
Case II: If k is positive, we take k = µ2. Then the general
solution is
F  ae  x  be   x
Applying (7), we get a = 0, b = 0 and therefore F = 0.
As before we are not interested in this solution.
Only alternative left is that of negative k.
 Case III: k is negative, and we take k = -p 2. Then (5)
becomes F   p 2 F  0
Its general solution is F ( x )  a cos px  b sin px
Applying boundary conditions (7), we get
F (0)  a  0, and F ( L)  b sin pL  0
We must have b ≠ 0, otherwise F ≡ 0. Hence sin pL  0
n
Therefore pL  n so that p  .
L
Thus we obtain infinitely many independent solutions
n
Fn ( x ) sin x, n 1, 2, 3, ..........
L
We consider only positive integers n, because negative integers
give the same solutions since sin(  x )  sin x .
Solution of Equation (6)
2
 n  2
Now with k taking the values k    ,
thepequation
n (6)
becomes  L 
   2G  0, cn
G n where n 
L
A general solution is
Gn ( t )  An cos n t  Bn sin n t

Hence a solution of (1) satisfying the boundary conditions (2),

can be written as
n
un ( x , t )  Fn ( x )Gn ( t ) ( An cos n t  Bn sin n t ) sin x
L
n 1, 2, 3, ..................
Satisfying the Initial Conditions (3)
A single solution un(x, t) in general does not satisfy the initial
conditions. Applying the principal of superposition, we construct
the complete solution in the form of infinite series
 
n
u( x , t )   un ( x , t )   ( An cos n t  Bn sin n t ) sin x
n 1 n 1
L
Applying Initial Conditions (3)

n
u( x , 0)   An sin x  f ( x) (8)
n 1
L

u n
  Bn n sin x g ( x ). ( 9)
t t 0 n 1 L
m
Multiplying (8) by sin andx integrating with respect to x from
L
0 to L, we get
  L L
n m m
 An sin L x sin L xdx f ( x )sin L x dx
n 1 0 0
L
n m
Now for m  n sin L
x sin
L
xdx  0
0

L L 2
n m  n  L
and for m  n, sin x sin xdx   sin x  dx 
L L  L  2
0 0

L
2 m
Therefore Am  f ( x )sin x dx
L L
0
Similarly
L L
2 m 2 m
Bm 
Lm  g ( x )sin
L
x dx 
m c  g ( x )sin
L
x dx
0 0
Thus the complete solution of One dimensional Wave equation

satisfying the boundary conditions and initial conditions is

2 u 2  2
u
c (1)
2 2
t x
u(0, t )  0, u( L, t )  0 for all t  0. ( 2)
u
u( x , 0)  f ( x ),  g( x ). (3)
t t 0

n c n c n
u( x , t )   ( An cos t  Bn sin t ) sin x
n 1
L L L
where
L L
2 m 2 m
Am  f ( x )sin x dx , Bm  g( x )sin L x dx
L L m c
0 0
Observe that from the equations (8) and (9) we have

n
 An sin L x  f ( x ) (8.1)
n 1

n
 Bn n sin
L
x g ( x ). (9.1)
n 1
cn
where n 
L
L L
2 m 2 m
Am  f ( x )sin x dx , Bm  g( x )sin L x dx
L L m c
0 0

The infinite series in (8.1) and (9.1) are the Fourier Sine
Series for f(x) and g(x) respectively.
Definitions:
The solutions
n
un ( x , t ) ( An cos n t  Bn sin n t ) sin x of the
L
PDE cn
with n  , n 1, 2, 3, ..................
L
2 u 2 u
2
c
satisfying the boundary
t 2 conditions
x 2
are called Eigenfunctions (or characteristic
functions) t ) the
u(0, and 0, values  0called
u( x ,Ln )are all t  0
foreigenvalues (or
characteristic values).

The set {1, 2, 3, …., n , …….} is called Spectrum.

 Example: A string stretched to length L between two
points on x-axis is initially deflected to form a triangle with
base on x axis as shown in the figure.
k
If the initial velocity is zero,
determine the deflection of the string
at point x at time t > 0. O L
X
L
2

Solution: We solve One dimensional wave equation or the PDE

2 u 2  2
u
c (1)
2 2
t x
where u(x, t) satisfies
Boundary Conditions: u(0, t ) 0, u( L, t ) 0 for all t 0. ( 2)
u
Initial Conditions: u( x , 0)  f ( x ),  g( x ). (3)
t t 0
We know that the complete solution of the system of equation
(1) to (3) is given 
by
n c n c n
u( x , t )   ( An cos t  Bn sin t ) sin x
n 1
L L L
where
L L
2 m 2 m
L g( x )sin
Am  f ( x )sin x dx , Bm  x dx
L m c L
0 0
In the present problem
2k
f ( x)  x when 0 x L
L 2
2k L
 ( L  x ) when xL and g ( x )  0
L 2
Therefore
 L2 L 
2 m m
Am    2k x sin x dx   2k ( L  x )sin x dx 
L L L L
L L 
 0 2 
  L2 L 
4k  m m 
2   
Am  x sin x dx  ( L  x )sin x dx
L  0 L L 
L
 2 

4k  L2 m L2 m
Am    cos( ) sin( )
L  2m
2 2 2
m  2 2

L2 m L2 m L2 
 cos( ) sin( ) sin( m ) 
2m 2 m 2 2 2 2
m  2

4k m
Am 
2 2
 2 sin( 2 )  sin( m )
m 
8k
A2m  1 (  1)m  1 , A2m  0, m 1, 2, 3, ........
2 2
( 2m  1) 
L
2 m
and Bm 
m c g( x )sin L
x dx  0
0
 
( 2n  1) c ( 2n  1)
u( x , t )   A2n  1 cos t sin x
n 1
L L

8k   ct x 1 3 ct 3 x
u( x , t )  cos sin  cos sin
2  L L 3 2 L L

1 5 ct 5 x 
 cos sin  ............ 
2 L L
5 
 Diffusion Equation
2
u 2 u
c (1)
t x 2
Let the solution of the diffusion equation (1) be
u( x , t )  F ( x ) G ( t ) ( 2)
By differentiating (2), we obtain
u dG 2u d 2F
F FG and  G  F G
t dt x 2 dx 2
Using these expressions for derivatives in (1), we get
G F 
FG c 2 F G 2

cG F
Now left side depends on t only and right depends on x only,
therefore
 G F 
 k
2 F
c G
This gives two ordinary linear homogeneous equations
F   k F  0 (3)
and
G  kc 2G 0 ( 4)
The constant k is arbitrary.
We first solve the equation (4) for different values of k.
Case I For k = 0, the equation (4) reduces to
G 0 Therefore G  constant.
Hence u does not vary with time. This solution is of little
practical interest.
 kc 2 t
Case II For k ≠ 0, the equation (4) has the basic solution e .
Since c2 and t are positive, this solution becomes infinite if k > 0,
and remains finite and tends to zero with the passage of time if k is
negative.
Therefore the natural choice is that k should be negative and we
take k = - p2. Then the basic solution of (4) is
 p2 c 2 t
e .
The equation (3) now becomes F   p 2 F  0

It has the general solution F ( x )  a cos px  b sin px

Therefore from (2), we obtain the general solution of (1) as
 p2 c 2 t
u( x , t )  F ( x ) G ( t ) [a cos px  b sin px ] e
The constants a, b and p are determined using Initial and
Boundary conditions.
 Example
A thin rod of length L is first immersed in boiling water so
that its temperature is 1000C throughout, then it is removed
from the water at time t = 0 and its ends are immediately put in
ice so that they are kept at temperature 0 0C. Find the
temperature at a point of the rod at a subsequent time.

Let u(x, t) be the temperature at a point distance x from one end at

time t.
Then u(x, t) satisfies
2
u 2 u
Heat equation c (1)
t 2
x
Initial Condition u( x , 0)  f ( x ) 100 0 x L ( 2)

Boundary Conditions u(0, t )  0  u( L, t ) t 0 (3)

To solve (1) by separation of variables we take
u( x , t )  F ( x ) G ( t )
Then from (1) we get
G F 
F G  c 2 F G or 
2 F
c G
Now left side depends on t only and right depends on x only,
therefore G F 
  p 2
c 2G F
This gives two ordinary linear homogeneous equations
F   p 2 F  0 ( 4)
and
G  p 2 c 2G  0 (5)
The constant p is arbitrary.
The general solutions of (4) and (5) respectively are.
 F ( x )  a cos px  b sin px
 p2 c 2 t
G(t )  A e
Therefore the general solution of (1) is
 p2 c 2 t
u( x , t )  F ( x ) G ( t ) [a cos px  b sin px ]e
Applying the boundary conditions
 p2 c 2 t
u(0, t )  a e 0  a 0
 p2 c 2 t  sin pL  0
u( L, t )  b sin pL e 0
n
pL  n or p , n 1, 2, 3, ..............
L
Thus for each n we get a solution
n x  p2 c 2 t
un ( x , t )  bn sin e
L
Combining all these solutions, we get the solution
  
n x  p2 c 2 t
u( x , t )   un ( x , t )   bn sin e ( 6)
n 1 n 1
L

Applying initial condition (2), we get


n x
u( x , 0)   bn sin 100
n 1
L
m x
Multiplying the above equation by sin and integrating with
L
respect to x from 0 to L, we get
 L L
n x m x m x
 bn sin
L
sin
L
dx  100 sin
L
dx
n 1 0 0

L 100 L
bm  [1  cos m ]
2 m
 Laplace Equation

Three dimensional Laplace equation is

2 u 2 u 2 u
  0
2 2 2
x y z
This equation is a special case of 3D heat equation
 2 2 2 
u  u  u  u
c 2    
t  x 2 y 2 z 2 
 
when c2 → ∞.
It then gives steady state temperature distribution in a body.
Laplace equation also arises in many problems of Engineering
and Physics which deal with Potential Theory.
The solution of Laplace equation is a Harmonic Function and
defines Potential related to the problem.
Problem
A rectangular plate with insulated surface is 8cm wide and is so long
as compared to its width that it may be considered of infinite length
without introducing an appreciable error. If temperature along one
short edge y = 0 at time t = 0 is given by
x
u( x , 0) 10 sin   , 0  x  8.
 8 
while the long edges x = 0 and x = 8 as well as the other short edge
are kept at 00C. Determine the steady state temperature u(x, y) at the
point (x, y) of the plate.
Y

X
o X=8
The steady state temperature u(x, y) is the solution of 2D Laplace
equation 2 u 2 u
 0
x 2 y 2
satisfying the boundary conditions
x
u( x , 0) 10 sin   , 0  x  8.
 8 
u(0, y )  0  u(8, y ) y  0 Lim u( x , y )  0
y 
We take u( x , y )  F ( x ) G ( y )
d 2F d 2G
Then Laplace equation takes the form GF 0
2 2
dx dy
1 d 2F 1 d 2G
Separating the variables we get 
F dx 2 G dy 2
Since x and y are independent variables, this result holds only if
each side is equal to a constant, say k.
Then we have two ordinary differential equations
d 2F d 2G
 kF  0 and  kG  0
2 2
dx dy
We solve these equations for
(a) k = 0 (b) k = p2 > 0 (c) k = - p2 < 0
For k = 0 we get
F a x  b and G c y  d
For k = p2 > 0, we get
F  a e px  b e  px and G  c cos py  d sin py
For k = - p2 < 0, we get
F  a cos px  b sin px and G  c e py  d e  py
Therefore the possible solutions are
 u( x , y )  FG (a x  b )(c y  d ) (1)
u( x , y )  FG (a e px  b e  px )(c cos py  d sin py ) ( 2)
u( x , y )  FG (a cos px  b sin px )(c e py  d e  py ) (3)
We will select one which is consistent with given boundary
conditions
x
u( x , 0) 10 sin   , 0  x  8. ( 4)
 8 
u(0, y )  0 y0 (5)
u(8, y )  0 y0 ( 6)
Lim u( x , y )  0 (7)
y 
Solution (2) cannot satisfy BC (7). Solution (1) cannot satisfy BC
(5) and (6) for all values of y. Therefore the only possible choice is
solution (3).
Applying the boundary conditions, we get
x
u( x , 0) (a cos px  b sin px )(c  d ) 10 sin   (8)
 8 
u(0, y )  a (c e py  d e  py )  0 (9)

u(8, y ) (a cos 8 p  b sin 8 p)(c e py  d e  py )  0 (10)

Lim u( x , y )  Lim (a cos px  b sin px )(c e py  d e  py )  0 (11)

y  y 
From (9) a = 0, from (11) c = 0. Then from (10) we get sin 8 p  0
n
 p , n 1, 2, 3.............
8
From (8) we get x
b d sin px 10 sin  
 8 
To satisfy this we must take n =1, then bd = 10
y
Hence u( x , y ) (b sin px )(d e  py ) 10 sin
x 
e 8
8
 Vibrations of Circular Membrane

Circular membranes occur in drums, microphones, telephones

and many other appliances. Hence their great importance.
Whenever a circular membrane is plane and its material is elastic,
but offers no resistance to bending its vibrations are governed by
2D wave equation
2 u  2
2  u  u
2 
c   
t 2  2 2 
 x y 
Since the membrane is Circular, we take the wave equation in
terms of polar coordinates
2 u   2 u 1 u 1  2 u 
2
c     (1)
t 2  r 2 r r r 2
 2 
 
We solve the problem by separating the variables.
First Separation: Substituting u( r ,  , t )  F (we
r , get
) G ( t ),

1 d 2G 1   2 F 1 F 1 2 F 
    
2
c G dt 2 F  r 2 r r r 2  2 
 
As r, , t are independent, this holds only if each side is a
constant. We take
1 d 2G 1   2 F 1 F 1 2 F 
      p 2
c 2G dt 2 F  r 2 r r r 2  2 
so that
d 2G
  2G  0 where   cp ( 2)
2
dt
2 F 1 F 1 2 F
   p2 F 0 (3)
r 2 r r r 2  2
 Second Separation: Substituting F ( r ,  )  Ein((3) ( ) separating
r ) Hand
the variables, we get
r 2  d 2 E 1 dE 2
 1 d 2
H
   p E  

E  dr 2 r dr  H d 2

As r,  are independent variables, the above equation holds if
each side is equal to a constant, say k2, then
d2H
 k 2 H 0 ( 4)
2
d
d2E
r 2
dr 2
r
dE
dr
 
 r 2 p2  k 2 E 0 (5)

it
Two independent solutions of (2) are e
The general solution of (4) is
H  a cos k  b sin k
Since r,  are polar coordinates, and
( r ,  ) and ( r ,   2n ) n 1, 2, 3, ...........
are coordinates of the same point, therefore solution of (4) must
be periodic with period 2π. Hence k = n = 1, 2, 3, 4………. and
H a cos n  b sin n
The equation (5) now becomes
2
2 d E dE 2 2 2
r 2
 r  ( r p  n ) E 0
dr dr
This is a Bessel equation of order n and argument pr.
As we interested in a solution which remains finite for all values
of r, we take
H  J n ( pr )
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