1
3. Differential Equations
3.1 Introduction
Almost all the elementary and numerous advanced parts of theoretical physics are
formulated in terms of differential equations(DE). Sometimes these are ordinary DE in
one variable (ODE). More often the equations are partial DE(PDE) in two or more
variables.
Since the dynamics of many physical systems involve just two derivatives, e.g.,
acceleration in classical mechanics and the kinetic energy operator (-2) in quantum
mechanics, DE of second order occur most frequently in physics.
Examples of PDEs
1. Laplace's eq., 2
This very common and important eq. occurs in studies of
a. electromagnetic phenomena,
b. hydrodynamics,
c. heat flow,
d. gravitation.
2. Poisson's eq. 2 -0.
In contrast to the homogeneous Laplace eq., Poisson's eq. is nonhomogeneous with a
source term -0.
3. The wave (Helmholtz) and time-independent diffusion eqs., 2  k2 = 0.
These eqs. appear in such diverse phenomena as
a. elastic waves in solids,
b. sound or acoustics,
c. electromagnetic waves,
d. nuclear reactors.
4. The time-dependent diffusion eq.
2 
1 

a 2 t
5. The time-dependent wave eq.,
 1 2

         2  2  2 
 c t

is a four-dimensional analog of the Laplacian.
2
where  2

2
6.The scalar potential eq.,  2     0
7.The Klein-Gordon eq.,  2     , and the corresponding vector eqs. in which 
is replaced by a vector function.
8.The Schrodinger wave eq.

2 2

   V  i
2m
t
and

2 2
   V  E
2m
for the time-independent case.
Some general techniques for solving second-order PDEs:
1. Separation of variables, where the PDE is split into ODEs that are related by common
constants which appear as eigenvalues of linear operators, L = l, usually in
variable.
2. Conversion of a PDE into an integral eq. using Green's functions applies to
inhomogeneous PDEs.
3. Other analytical methods such as the use of integral transforms.
4. Numerical calculations
 Nonlinear PDEs
Notice that the above mentioned PDEs are linear (in ). Nonlinear ODEs an PDEs are a
rapidly growing and important field. The simplest nonlinear wave eq


 C ( )
0
t
t
results if the local speed of propagation, c, depends on the wave . When a nonlinear eq.
has a solution of the form (x,t) = Acos(kx-t) where (k) varies with k so that ''(k) 0,
3
then it is called dispersive. Perhaps the best known nonlinear eq. of second is the
Korteweg-deVries (KdV) eq.

  3


0
t
x x3
which models the lossless propagation of shallow water waves and other phenomena. It is
widely known for its soliton solutions. A solition is a traveling wave with the properties
of persisting through an interaction with another soliton: After they pass through each
other, they emerge in the some shape and with the same velocity and acquire no more
than a phase shift.
On the other aspect, another important and interesting phenomenon, chaos, can be
numerically found in some nonlinear PDEs in certain parameter regime.

Classes of PDEs and boundary conditions* (Optional Reading)
For a linear PDE (two variables)
 2

2
2


a

2
b

c
d
e
 f    F ( x1 , x2 )

2
2
x1x2
x2
x1
x2
 x1

we say it is
(i) elliptic PDE, if D = ac - b2 > 0; (Laplace, Possion in x, y)
(ii) parabolic PDE, if D = 0; ( Diffusion eq. in (x,t))
(iii) hyperbolic PDE, if D < 0. (wave eq. in (x,t))
For a physical system, we want to find solutions that match given points, curves, or
surfaces (boundary value problems). Eigenfunctions usually are required to satisfy certain
imposed boundary conditions. These boundary conditions may take these forms:
1. Cauchy boundary conditions. The value of a function and normal derivative specified
on the boundary.
2. Dirichlet boundary conditions. The value of a function specified on the boundary.
3. Neumann boundary conditions. The normal derivative (normal gradient) of a function
specified on the boundary.
4
A summary of the relation of these three types of boundary conditions to the three types
of 2D PDEs is given in the table for reference.
The tern boundary conditions includes as a special case the concept of initial conditions.
Boundary
conditions
Neumann
Dirichlet
Cauchy
Open surface
Closed
surface
Open surface
Type of partial differential equation
Elliptic
Hyperbolic
Parabolic
Laplace Poission Wave function in Diffusion
in (x,y)
(x,t)
equation in (x,t)
Unphysical
Unique, stable
Too restrictive
result(instability) solution
Too restrictive
Too restrictive
Too restrictive
Insufficient
Insufficient
Closed
surface
Open surface
Unique, stable
solution
Insufficient
Solution not
unique
Insufficient
Closed
surface
Unique, stable
solution
Solution not
unique
Unique, stable
solution in one
direction
Too restrictive
Unique, stable
solution in one
direction
Too restrictive
3.2 First-order Differential Equations
Physics includes some first order DEs. As well, they are sometimes intermediate steps
for second order DEs.
We consider here DEs of the general form
dy
P ( x, y )
 f ( x, y )  
dx
Q ( x, y )
(3.1)
The eq. is clearly a first-order ODE. It may or may not be linear, although we shall treat
the linear case explicitly later.
5

Separable variables
Frequently, the above eq. will have the special form
dy
P( x)
 f ( x, y )  
dx
Q( y )
or
P(x)dx + Q(y) dy = 0
Integrating from (x0, y0) to (x, y) yields,

x
x0
y
P( x)dx   Q ( y )dy  0
y0
Since the lower limits x0 and y0 contribute constants, we may ignore the lower limits of
integration and simply add a constant of integration.
Example Boyle's Law
In differential form Boyle's gas law is
dV
V

dP
P
V-volume, P - Pressure.
or
ln V + ln P = C.
If we set C = ln k, PV = k.

Exact Differential Equations
Consider
P(x,y) dx + Q(x,y) dy = 0 .
This eq. is said to be exact if we match the LHS of it to a differential dφ,
d 


dx 
dy
x
y
.
Since the RHS is zero, we look for an unknown function (x,y) = const. and d= 0.
6
We have
P( x, y )dx  Q( x, y )dy 


dx 
dy
x
y
and


 P( x, y ),
 Q ( x, y )
x
y
The necessary and sufficient for our eq. to be exact is that the second, mixed partial
derivatives of assumed continuous) are independent of the order of differential:
.
 2
P( x, y ) Q( x, y )  2



yx
y
x
xy
If (x,y) exists then (x,y) = C.
It may well turn out the eq. is not exact. However, there always exists at least one and
perhaps an infinity of integrating facts, (x,y), such that
(x,y) P(x,y) dx + (x,y)Q(x,y) dy = 0
is exact. Unfortunately, an integrating factor is not always obvious or easy to find. Unlike
the case of the linear first-order DE to be considered next, there is no systematic way to
develop an integrating factor for a general eq.

Linear First-order ODE
If f (x,y) has the form -p(x)y + q(x), then
dy
 p ( x) y  q( x)
(3.2)
dx
It is the most general linear first-order ODE. If q(x) = 0, Eq.(3.2) is homogeneous (in y).
A nonzero q(x) may represent a source or deriving term. The equation is linear; each term
is linear in y or dy/dx. There are no higher powers; That is, y2, and no products, ydy/dx.
This eq. may be solved exactly.
.
Let us look for an integrating factor α(x) so that
 ( x)
dy
  ( x) p( x) y   ( x)q( x)
dx
may be rewritten as
(3.3)
7
d
 ( x) y    ( x)q( x).
dx
(3.4)
The purpose of this is to make the left –hand side of Eq.(3.2) a derivative so that it can be
integrated—by inspection. It also, incidentally, makes Eq. (3.2) exact. Expanding Eq.
(3.4), we obtain
 ( x)
dy d

y   ( x)q ( x).
dx dx
Comparison with Eq.(3.3) shows that we must require
d ( x)
  ( x) p ( x).
dx
Here is a differential equation for α(x) , with the variables α and x separable. We separate
variables, integrate, and obtain
 ( x)  exp  p( x)dx
x


(3.5)
as our integrating factor.
With α(x) known we proceed to integrate Eq.(3.4). This, of course, was the point of
introducing α(x) in the first place. We have

x
x
d
 ( x) y ( x)dx   p( x)dx.
dx
Now integrating by inspection, we have
x
 ( x) y ( x)    ( x)q ( x)dx  C.
The constants from a constant lower limit of integration are lumped into the constant C.
Dividing by α(x) , we obtain
y ( x)   ( x)
1
  ( x)q( x)dx  C.
x
Finally, substituting in Eq.(3.5) for α yields
x
x
 s p(t )dt q( s)ds  C .
y( x)  exp   p(t )dt  
exp



 



(3.6)
8
Here the(dummy)variables of
integration have been rewritten to make them
unambiguous. Equation (3.6) is the complete general solution of the linear, first-order
differential equation, Eq.(3.2). The portion
x
y1 ( x)  C exp   p(t )dt 


corresponds to the case q(x) =0 and is a general solution of the homogeneous differential
equation. The other term in Eq.(3.6),
x
x
s
y 2 ( x)  exp   p(t )dt  
exp  p(t )dt q(s)ds 

.


 



is a particular solution corresponding to the specific source term q(x) .
The reader might note that if our linear first-order differential equation is
homogeneous(q =0),then it is separable. Otherwise, apart from special cases such as
p=constant, q=constant, or q(x) =ap(x) , Eq.(3.2) is not separable.
Example
RL Circuit
For a resistance-inductance circuit Kirchhoff’s law leads to
L
dI (t )
 RI (t )  V (t )
dt
for the current I(t) , where L is the inductance and R the resistance, both constant. V(t)
is the time-dependent impressed voltage.
From Eq.(3.5) our integrating factor α(t) is
x R
 (t )  exp  dt
L
Rt L
e .
Then by Eq.(3.6)
V (t )
 t

I (t )  e  Rt L   e Rt L
dt  C ,
L


with the constant C to be determined by an initial condition (a boundary condition).
For the special case V(t)=V0 , a constant,
V L

I (t )  e  Rt L  0  e Rt L  C 
L R


V0
 Ce  Rt L .
R
9
If the initial condition is I(0) =0, then c  V0 R
I (t ) 

V0
1  e  Rt
R
L
and
.
3.3 SEPARATION OF VARIABLES
The equations of mathematical physics listed in Section 3.1 are all partial differential
equations. Our first technique for their solution splits the partial differential equation of n
variables into ordinary differential equations. Each separation introduces an arbitrary
constant of separation . If we have n variables, we have to introduce n-1 constants,
determined by the conditions imposed in the problem being solved.
Cartesian Coordinates
In Cartesian coordinates the Helmholtz equation becomes
 2  2  2
 2  2  k 2  0
x 2
y
z
(3.7)
Using  ( x, y, z ) for the Laplacian. For the present let k2 be a constant. Perhaps the
simplest way of treating a partial differential equation such as (3.7) is to split it into a set
of ordinary differential equations. This may be done as follows. Let
2
 ( x, y, z )  X ( x)Y ( y ) Z ( z )
YZ
(3.7a)
d2X
d 2Y
d 2Z

XZ

XY
 k 2 XYZ  0
2
2
2
dx
dy
dz
Dividing by   XYZ and rearranging terms, we obtain
1 d2X
1 d 2Y 1 d 2 Z
2


k


.
X dx 2
Y dy 2 Z dz 2
(3.8)
Equation (3.8) exhibits one separation of variables. The left-hand side is a function of x
alone, whereas the right-hand side depends only on y and z. So Eq.(3.8) is a sort of
paradox. A function of x is equated to a function of y and z , but x , y , and z are all
independent coordinates. This independence means that the behavior of x as an
independent variable is not determined by y and z . The paradox is resolved by setting
each side equal to a constant, a constant of separation. We choose
10
1 d2X
 l 2 ,
X dx
 k2 
(3.9)
1 d 2Y 1 d 2 Z

 l 2 .
Y dy 2 Z dz 2
(3.10)
Now, turning our attention to Eq.(3.10), we obtain
1 d 2Y
1 d 2Z
2
2
 k  l 
,
Y dy 2
Z dz 2
and a second separation has been achieved. Here we have a function of z equated to a
function of and the same paradox appears. We resolve it as before by equating each side
to another constant separation, -m2,
1 d 2Y
(3.11)
 m 2 ,
2
Y dy
1d Z
 k 2  l 2  m 2  n 2 ,
2
Z dz
(3.12)
introducing a constant n2 by k2=l2+m2+n2
to produce a symmetric set of
equations. Now we have three ordinary differential equations((3.9),(3.11),and (3.12)) to
replace Eq.(3.7). Our assumption (Eq.(3.7a))has succeeded and is thereby justified.
Our solution should be labeled according to the choice of our constants l, m ,and
n ,that is ,
lmn ( x, y, z )  X l ( x)Ym ( y)Z n ( z ).
(3.13)
Subject to the conditions of the problem being solved and to the condition
k2=l2+m2+n2 ,we may choose l , m ,and n as we like, and Eq.(3.13)will still be a solution
of Eq.(3.7), provided Xl(x) is a solution of Eq.(3.9), and so on . We may develop the
most general solution of Eq.(3.7)by taking a linear combination of solutions ψlmn ,

a
lmn
lmn
(3.14)
l , m, n
The constant coefficients almn
conditions of the problem.
are finally chosen to permit ψ to satisfy the boundary
Circular Cylindrical Coordinates
With our unknown function ψ dependent on ρ, φ , and z
becomes
, the Helmholtz equation
11
2(  , , z )  k 2(  , , z )  0.
(3.15)
or
1 

1  2  2
(
) 2
 2  k 2   0,
2
  
 
z
(3.16)
As before, we assume a factored form for ψ ,
 (  , , z )  P(  ) ( ) Z ( z ).
(3.17)
Substituting into Eq.(3.16), we have
`
Z d
dP
PZ d 2 
d 2Z
(
) 2

P

 k 2 PZ  0.
2
2
 d d
 d
dz
(3.18)
All the partial derivatives have become ordinary derivatives. Dividing by PΦZ
moving the z derivative to the right-hand side yields
1 d
dP
1 d 2
1 d 2Z
2
( )  2

k


.
P d d
Z dz 2
  d 2
(3.19)
Then
d 2Z
 l 2Z ,
2
dz
(3.20)
And
1 d
dP
1 d 2
(
) 2
 k 2  l 2 .
2
P d d
  d
(3.21)
Setting k2+l2=n2, multiplying by ρ2, and rearranging terms, we obtain
 d
dP
1 d 2
(  )  n2  2  
.
P d d
 d 2
(3.22)
We may set the right-hand side to m2 and
d 2
 m 2.
2
d
(3.23)
and
12
Finally, for the ρ dependence we have

d
dP
(
)  (n 2  2  m 2 ) P  0.
d
d
(3.24)
This is Bessel’s differential equation . The solution and their properties are presented in
Chapter 6. The separation of variables of Laplace’s equation in parabolic coordinates also
gives rise to Bessel’s equation. It may be noted that the Bessel equation is notorious for
the variety of disguises it may assume. For an extensive tabulation of possible forms the
reader is refered to Tables of Functions by Jahnke and Emde.
The original Helmholtz equation , a three-dimension partial differential equation, has
been replaced by three ordinary differential equations, Eqs.(3.20),(3.23),and (3.24). A
solution of the Helmholtz equation is
 (  ,  , z )  P(  )( ) Z ( z ).
(3.25)
Identifying the specific P, Φ, Z solutions by subscripts, we see that the most general
solution of the Helmholtz equation is a linear combination of the product solutions:
 (  ,  , z )   a mn Pmn (  ) m ( ) Z n ( z ).
(3.26)
m,n
Spherical Polar Coordinates
Let us try to separate the Helmholtz equation, again with k2 constant , in spherical
polar coordinates:
1
2
r sin 

 2 


1  2 
sin

(
r
)

(sin

)

 k 2. (3.27)

2 

r

r




sin





Now, in analogy with Eq.(3.7a) we try
 (r , , )  R(r )( )( ).
(3.28)
By substituting back into Eq.(3.27) and dividing by RΘΦ , we have
1 d 2 dR
1
d
d
1
d 2
(r
) 2
(sin 
)
 k 2 . (3.29)
2
2
2
2
Rr dr
dr
r sin  d
d
r sin  d
Note that all derivatives are now ordinary derivatives rather than partials. By multiplying
by r2sin2θ , we can isolate (1 )( d 2 d 2 ) to obtain
13
1 d 2
1 d
dR
1
d
d 

 r 2 sin 2   k 2  2 (r 2 )  2
(sin 
) . (3.30)
2
 d
dr
d 
Rr dr
r sin  d

Equation(3.30) relates a function of φ alone to a function of r and θ alone. Since r , θ,
and φ are independent variables, we equate each side of Eq.(3.30) to a constant. Here a
little consideration can simplify the later analysis. In almost all physical problems φ will
appear as an azimuth angle. This suggests a periodic solution rather than an exponential.
With this in mind, let us use -m2 as the separation constant. Any constant will do, but this
one will make life a little easier. Then
1 d 2 ( )
 m 2
2
 d
(3.31)
and
1 d 2 dR
1
d
d
m2
(r
)
(sin 
) 2
 k 2 . (3.32)
2
2
2
dr
d
Rr dr
r sin  d
r sin 
Multiplying Eq.(3.32) by r2 and rearranging terms, we obtain
1 d 2 dR
1
d
d
m2
2 2
(r
)r k 
(sin 
)
.
R dr
dr
 sin  d
d
sin 2 
(3.33)
Again, the variables are separated. We equate each side to a constant
obtain
1 d
d
m2
(sin 
)
  Q  0.
sin  d
d
sin 2 
Q
and finally
(3.34)
QR
1 d 2 dR
(r
)  k 2 R  2  0.
2
dr
r dr
r
(3.35)
Once more we have replaced a partial differential equation of three variables by three
ordinary differential equations. Eq.(3.34) is identified as the associated Legendre equation
in which the constant Q becomes l(l+1); l is an integer. If k2 is a (positive) constant, Eq.
(3.35) becomes the spherical Bessel equation.
Again, our most general solution may be written
Qm (r , , )   RQ (r )Qm ( ) m ( ).
(3.36)
Q,m
The restriction that k2 be constant is unnecessarily severe. The separation process will be
possible for k2 as general as
14
k 2  f (r ) 
1
1
g ( )  2 2 h( )  k 2 .
2
r
r sin 
(3.37)
In the hydrogen atom problem, one of the most important examples of the Schrodinger
wave equation with a closed form solution is k2=f(r). Equation (3.35) for the hydrogen
atom becomes the associated Legendre equation.
The great importance of this separation of variables in spherical polar coordinates
stems from the fact that the case k2= k2(r) covers a tremendous amount of physics: a
great deal of the theories of gravitation, electrostatics, atomic physics, and nuclear
physics. And with k2= k2(r), the angular dependence is isolated in Eq. (3.31) and (3.34),
which can be solved exactly.
Finally, as an illustration of how the constant m in Eq.(3.31) is restricted, we note that
φ in cylindrical and spherical polar coordinates is an azimuth angle. If this is a classical
problem, we shall certainly require that the azimuthal solution Φ(φ) be singled valued,
that is,
 (  2 )   ( ).
(3.38)
This is equivalent to requiring the azimuthal solution to have a period of 2π or some
integral multiple of it. Therefore m must be an integer. Which integer it is depends on the
details of the problem. Whenever a coordinate corresponds to an axis of translation or to
an azimuth angle the separated equation always has the form
d 2 ( )
 m 2 ( )
2
d
For φ, the azimuth angle, and
d 2Z ( z)
 a 2 Z ( z)
(3.39)
dz 2
For z, an axis of translation in one of the cylindrical coordinate systems. The solutions, of
course, are sinaz and cosaz for –a2 and the corresponding hyperbolic function (or
exponentials) sinhaz and coshaz for +a2 .
Table Solution in Circular Cylindrical Coordinates *
   am m ,
2  0
m ,
a.
 J () cos m ez 
m   m

 z 
 N m () sin m  e 
b.
 I () cos m cosz 
m   m



K m () sin m  sin z 
15
c. If α=0 (no z-dependence)
  m cos m 
m    m 

  sin m 
* Reference for the radial functions are Jm(αρ), Nm(αρ), Im(αρ) and Km(αρ) in Chapter 6.
For the Helmholtz and the diffusion equation the constant  k is added to the
2
separation constant  a to define a new parameter  or
2
2
 2 >0) we get Jm(γρ) and Nm(γρ). For the choice   2
  2 . For the choice  2 (with
(with  >0) we get Im(γρ) and
2
Km(γρ) as previously.
3.4 Singular Points
Let us consider a general second order homogeneous DE (in y) as
y'' + P(x) y' + Q(x) y = 0
(3.40)
where y' = dy/dx. Now, if P(x) and Q(x) remain finite at x = x0, point x = x0 is an
ordinary point. However, if either P(x) or Q(x) ( or both) diverges as x  x, x0 is a
singular point.
Using Eq. (3.40), we are able to distinguish between two kinds of singular points.
(1) If either P(x) or Q(x) diverges as x x0 but (x - x0)P(x) and (x-x0)2Q(x) remain
finite, then x0 is called a regular or non-essential singular point.
(2) If either (x - x0)P(x) or (x - x0)2Q(x) remain still diverges ax x  x0, then x0 is labeled
an irregular or essential singularity.
These definitions hold for all finite values of x0. The analysis of x   is similar to the
treatment of functions of a complex variable. We set x = 1/z, substitute into the DE, and
then let z  0. By changing variables in the derivative, we have
dy ( x) dy ( z 1 ) dz
1 dy ( z 1 )
dy

 2
 z 2
dx
dz dx
dz
dz
x
2
1
d 2 y ( x) d  dy ( x)  dz
dy ( z 1 )
2 
2 d y( z ) 


(

z
)

2
z

z


dz  dx  dx
dz
dx 2
dz 2 

dy ( z 1 )
d 2 y ( z 1 )
 2z 3
 z4
dz
dz 2
(3.41)
(3.42)
16
Using these results, we transform Eq.(3.40) into
z4


d2y
dy
 2 z 3  z 2 P( z 1 )
 Q( z 1 ) y  0
2
dz
dz
(3.43)
The behavior at x =  (z = 0) then depends on the behavior of the new coefficients
2 z  P( z 1 )
Q( z 1 )
~
~
P ( z) 
and Q( z ) 
z4
z2
as z  0. If these two expressions remain finite, point x =  is an ordinary point. If they
diverge no more rapidly than that 1/z and 1/z2, respectively, x =  is a regular singular
point, otherwise an irregular singular point.
Example
Bessel's eq. is
x2y'' + x y' + (x2 - n2) y = 0.
Comparing it with Eq. (3.40) we have
P(x) = 1/x, Q(x) = 1 - n2/x2,
which shows that point x = 0 is a regular singularity. As x   (z  0), from Eq. (3.43),
we have the coefficients
1  n2 z 2
1
P( z ) 
and Q( z ) 
z
z4
4
Since the Q(z) diverges as z , point x =  is an irregular or essential singularity.
We list , in Table 3.4, several typical ODEs and their singular points.
Table 3.4
Equation
1. Hypergeometric
Regular
singularity
0,1, 
Irregular
singularity
___
2. Legendre
-1,1, 
___
-1,1, 
___
x( x  1) y  (1  a  b) x  cy  aby  0
(1  x ) y  2 xy  l (l  1) y  0
2
3. Chebyshev
(1  x ) y  xy  n y  0
2
2
17
4. Confluent hypergeometric
0

5. Bessel
0

6. Laguerre
0

7. Simple harmonic oscillator
___

___

xy  (c  x) y  ay  0
x y  xy  ( x  n ) y  0
2
2
2
xy  (1  x) y  ay  0
y   2 y  0
8. Hermite
y  2 xy  2y  0
3.5 Series Solutions
In this section, we develop a method of a series expansion for obtaining one solution of
the linear, second-order, homogeneous DE. This method will always work, provided the
point of expansion is no worse than a regular singular point. In physics this very gentle
condition is almost always satisfied.
A linear, second-order, homogeneous ODE may be written in the form
y'' + P(x) y' + Q(x) y = 0.
In this section we develop (at least) one solution. In the next section we develop a second,
independent solution and prove that no third, independent solution exists. Therefore, the
most general solution may be written as
y = c1y1(x) + c2y2(x).
Our physical problem may lead to a nonhomogeneous, linear, second-order DE
y'' + P(x) y' + Q(x) y = F(x).
F(x) represents a source or driving force. Specific solution of this eq., yp, could be
obtained by some special techniques. Obviously, we may add to yp any solution of the
corresponding homogeneous eq.
Hence,
y(x) = c1y1(x) + c2y2(x) + yp(x).
18
The constants c1 and c2 will eventually be fixed by boundary conditions.
For the present, we assume F(x) = 0, and we attempt to develop a solution of the eq. by
substituting in a power series with undetermined coefficients. Also available as a
parameter is the power of the lowest nonvanishing term of the series. To illustrate, we
apply the method to two important DEs. First, the linear oscillator eq.
y'' + 2y = 0,
(3.44)
with known solutions y = sinx, cos t.
We try

y ( x)  x k (a0  a1 x  a2 x 2  ...)   a x k  
 0
with k and a still undetermined. Note that k need not be an integer. By differentiating
twice, we obtain

y   a (k   )(k    1) x k    2
 0
By substituting into Eq. (3.44), we have


 0
x 0
 a (k   )(k    1) x k    2   2  a  x k     0
(3.45)
From the uniqueness of power series the coefficients of each power of x on the LHS must
vanish individually.
The lowest power is xk-2, for  = 0 in the first summation. The requirement that the
coefficient vanishes yields
a0 k(k-1) = 0.
Since, by definition, a0  0, we have
k(k-1) = 0
This eq., coming from the coefficient of the lowest power of x, is called the indicial
equation. The indicial eq. and its roots are of critical importance to our analysis. If k = 1,
the coefficient a1(k+1)k of x k-1 must vanish so that a1 = 0. Clearly, in this example, we
must require either that k = 0 or k = 1.
19
Now, we return to Eq. (3.45) and demand that the remaining net coefficients vanish. We
set  = j+2 in the first summation and ' = j in the second. This results in
aj+2 (k+j+2)(k+j+1) + 2aj = 0
or
a j  2  a j
2
(k  j  2)( k  j  1)
This is a two-term recurrence relation. Given aj, we may compute aj+2 and then aj+4, aj+6,
and so on as far as desired. We note that, for this example, if we start with a0, the above
eq. leads to a2, a4 and so on, and ignores a1, a3, a5 and so on. Since a1 is arbitrary if k = 0
and necessarily zero if k =1, let us set it equal to zero and then a3=a5=... = 0.
We first try the solution k = 0. The recurrence relation becomes
a j  2  a j
2
( j  2)( j  1)
,
which leads to
2
a 2  a 0
a4  a2
2 1
2
43


2
2!
4
4!
a0
a0
……
a2 n  (1) n
 2n
(2n)!
a0
……
So our solution is
y ( x) k  0
2n

 (x) 2 (x) 4

n (x)
 a 0 1 

 ...  a 0  (1)
 a 0 cos(x)
2!
4!
(2n)!
n 0


If we choose the indicial eq. root k = 1, the recurrence relation becomes
a j  2  a j
2
( j  3)( j  2)
20
Again, we have
a2 n  (1) n
 2n
(2n  1)!
For this choice, k = 1, we obtain
y( x) k 1
 (x) 2 (x) 4
 a
 a 0 x 1 

 ...  0
3!
5!

 
(x) 2 n 1 a 0
(1)
 sin( x)

(2n  1)! 
n 0

n
.
This series substitution, known as Frobenius' method, has given us two series solution
of the linear oscillation eq. However, two points must be strongly emphasized:
(1) The series solution should be substituted back into the DE, to see if it works, as a
precaution against algebraic and logical errors.
(2) The acceptability of a series solution depends on its convergence (including
asymptotic convergence).

Expansion above x0
It is perfectly possible to write

y ( x)   a ( x  x0 ) k   , a0  0
 0
Indeed, for the Legendre, Chebyshev and hypergeometric eqs the choice x = 1 has some
advantages. The point x should not be chosen at an essential singularity -or our
Frobenius method will probably fail.

Symmetry of solutions
It is interesting to note that we obtained one solution of even symmetry, y1(x) = y1(-x),
and one of odd symmetry y2(x) = -y2(-x). This is not just an accident but a direct
consequence of the form of the DE. Writing a general DE as
L(x) y(x) = 0,
in which L(x) is the differential operator, we see that for the linear oscillator eq, L(x) is
even; namely L(x) = L(-x). Often this is described as ever parity.
Whenever the differential operator has a specific parity or symmetry, either even or odd,
we may interchange +x and -x, and have
21
 L(x) y (-x) = 0.
Clearly, if y(x) is a solution, y(-x) is also a solution. Then any solution may be resolved
into even and odd parts,
y(x) = [y(x)+y(-x)]/2 + [y(x)-y(-x)]/2,
the first bracket giving an even solution, the second an odd solution.
If we refer back to section 3.4, we can see that Legendre, Chebyshev, Bessel, simple
harmonic oscillator, and Hermite eqs. all exhibit this even parity. i.e., their P(x) is odd
and Q(x) even. Our emphasis on parity stems primarily from the importance of it in
quantum mechanics.
We find that wave functions usually are either even or odd, meaning that they have a
definite parity. Most interactions (beta decay is the big exception) are also even or odd
and the result is that parity is conserved.

Limitations of Series Approach
This attack on the linear oscillator eq. was perhaps a bit too easy.
To get some idea of what can happen we try to solve Bessel's eq.
x2 y'' + x y' + (x2-n2) y = 0.
(3.46)
Again, assuming a solution of the form

y ( x)   a  x k  
 0
we differentiate and substitute into Eq. (3.46). The result is




 0
 0
 0
 0
 a  (k   )(k    1) x k  1   a  (k   ) x k    a x k   2   a  n 2 x k   0 .
By setting= 0, we get the coefficient of xk,
a[k(k-1) + k-n2] = 0.
The indicial equation
k2 - n2 = 0
22
with solution k =  n. For the coefficients of xk+1, we obtain
a1[(k+1)k + k+1 -n2] = 0.
For k =  n (k  1/2), [ ] does not vanish and we must require a1 = 0.
Proceeding to the coefficient of xk+j for k = n, we set  = j in the 1st, 2nd, and 4th terms
and  = j-2 in the 3rd term. By requiring the resultant coefficient of xk+j to vanish, we
obtain
aj[(n+j)(n+j-1)+(n+j)-n2] + aj-2=0.
When j j+2, this can be written for j≥0 as
a j  2  a j
1
( j  2)( 2n  j  2)
,
(3.47)
which is the desired recurrence relation. Repeated application of this recurrence relation
leads to
1
a n!
a2  a0
 2 0
2(2n  2)
2 1!(n  1)!
a 4  a 2
a n!
1
 4 0
4(2n  4)
2 2!(n  2)!
a 6  a 4
a n!
1
, and so on, and in general
 6 0
2(2n  2)
2 3!(n  3)!
a 2 p  (1) p
a 0 n!
2
2p
p!(n  p)!
Inserting these coefficients in our assumed series solution, we have


n! x 2
n! x 4
y( x)  a0 x n 1  2
 4
 ...
 2 1!(n  1)! 2 2!(n  2)!

In summation form

y ( x)  a0  (1) j
j 0

n! x n  2 j
1
x
n

a
2
n
!
(1) j
( )n  2 j

0
2j
2 j!(n  j )!
j!(n  j )! 2
j 0
(3.48)
23
The final summation is identified as the Bessel function Jn(x).
When k = -n and n is not integer, we may generate a second distinct series to be labeled
J-n(x). However, when -n is a negative integer, trouble develops. The recurrence relation
is still given by Eq (3.47), but with 2n replaced by -2n. Then when j+2 = 2n, aj+2 blows
up and we have no series solution. This catastrophe can be remedied in Eq. (3.48). Since
m! =   for a negative integer m (note that (Z-1)!=Z!/Z ),

J n ( x)   (1) j
j 0

1
x
1
x
( ) n2 j   (1) jn
( ) 2 jn  (1) n J n ( x)
j!( j  n)! 2
j !( j   n)! 2
j 0
The second solution simply reproduces the first. We have failed to construct a second
independent solutions for Bessel's eq. by this series technique when n is an integer.
By substituting in an infinite series, we have obtained two solutions for the linear
oscillator eq. and one for Bessel's eq. (two if n is not integer) Can we always do this? Will
this method always work? The answer is no!

Regular and Irregular Singularities
The success of the series substitution method depends on the roots of the indicial eq. and
the degree of singularity. To have clear understanding on this point, consider four simple
eqs.
6
y0
x2
6
y   3 y  0
x
1
a2



y  y  2 y0
x
x
1
a2
y   2 y   2 y  0
x
x
y 
(3.49a)
(3.49b)
(3.49c)
(3.49d)
For the 1st eq., the incicial eq. is
k(k-1) - 6 =0,
giving k = 3, -2. Since the eq. is homogeneous in x ( counting d2/dx2 as x-2), here is no
recurrence relation. However, we are left with two perfectly good solution, x3 and x-2.
For the 2nd eq., we have -6a0 = 0, with no solution at all, for we have agreed that a0  0.
The series substitution broke down at Eq. (3.49b) which has an irregular singular point at
the origin.
24
Continuing with the Eq. (3.49c), we have added a term y'/x. The indicial eq. is k2-a2=0,
but again, there is no recurrence relation. The solutions are y = xa, x-a, both perfectly
acceptable one term series.
For Eq. (3.49d), (y'/x  y'/x2), the indicial eq. becomes k = 0. There is a recurrence
relation
a j 1  a j
a 2  j ( j  1)
j 1
Unless a is selected to make the series terminate we have
lim
j 
a j 1
aj
 lim
j 
j ( j  1)

j 1
Hence our series solution diverges for all x 0.

Fuchs's Theorem
We can always obtain at least one power-series solution, provided we are expanding
about a point that is an ordinary point or at worst a regular singular point. If we attempt
an expansion about irregular or essential singularity, our method may fail. Fortunately,
the more important eqs. of mathematical physics listed in Section 3.4 have no irregular
singularities in the finite plane.
SUMMARY
If we are expanding about an ordinary point or at worst about a regular singularity,
the series substitution approach will yield at least one solution (Fuchs’s theorem).
Whether we get one or two distinct solutions depends on the roots of the indicial
equation.
1.
2.
3.
If the two roots of the indicial equation are equal, we can obtain only one
solution by this series substitution method.
If the two roots differ by a nonintegeral number, two independent solutionsmay
be obtained.
If the two roots differ by an integer, the larger of the two will yield a solution.
The smaller may or may not give a solution,depending on the behavior of the coefficients.
In the linear oscillator equation we obtain two solutions; for Bessel’s equation, only one
solution.
25
The usefulness of the series solution in the terms of what is the solution(i.e.,
numbers) depends on the rapidity of the convergence of the series and the availability of
the coefficients. Many, probably most, differential equations will not yield nice simple
recurrence relations for the coefficients. In general, the available series will probably be
useful for |x| (or |x-x0|) very small. Computers can be used to determine additional series
coefficients using a symbolic language such as Mathematica, Maple or Reduce. Often,
however, for numerical work a direct numerical integration will be preferred.
3.6 A Second Solution
In this section we develop two methods of obtaining a second independent solution: an
integral method and a power series containing a logarithmic term. First, however we
consider the question of independence of a set of function.

Linear Independence of Solutions
Given a set of functions, , the criterion for linear dependence is the existence of
a relation of the form
 k  0 ,
(3.50)
in which not all the coefficients k are zero. On the other hand, if the only solution of Eq.
(3.50) is k=0 for all , the set of functions  is said to be linearly independent.
Let us assume that the functions  are differentiable. Then, differentiating Eq.
(3.50) repeatedly, we generate a set of eq.
n
k   0


1
n
k     0


1
…
n
k 


1
( n 1)
0
.
This gives us a set of homogeneous linear eqs. in which k are the unknown quantities.
There is a solution k 0 only if the determinant of the coefficients of the k's vanishes,
26
1
 1
2
 2
...
...
...
...
...
...
( n 1)
...  n
 1( n 1)  2( n 1)
n
 n
0
This determinant is called the Wronskian.
1. If the wronskian is not equal to zero, then Eq.(3.50) has no solution other than k = 0.
The set of functions  is therefore independent.
2. If the Wronskian vanishes at isolated values of the argument, this does not
necessarily prove linear dependence (unless the set of functions has only two functions).
However, if the Wronskian is zero over the entire range of the variable. the functions 
are linearly dependent over this range.
Example: Linear Independence
The solution of the linear oscillator eq. are  = sinx,  = cos x. The
Wronskian becomes
sin x
cosx
   0
 cosx   sin x
 and  are therefore linearly independent. For just two functions this means that one is
not a multiple of the other, which is obviously true in this case.
You know that
sin x  (1  cos2 x)1 2
but this is not a linear relation.
Example Linear Dependence
Consider the solutions of the one-dimensional diffusion eq.: y   y  0 . We have  = ex
and  = e-x, and we add  = cosh x, also a solution. Then
ex
ex
ex
ex
 e x
ex
cosh x
sinh x  0
cosh x
,
27
because the first and third rows are identical. Here, they are linearly dependent, and
indeed, we have
ex + e-x - 2 coshx = 0

with k  0.
A Second Solution
Returning to our 2nd order ODE
y'' + P(x) y' + Q(x) y = 0
Let y1 and y2 be two independent solutions. Then Wroskian is
W= y1y2' - y1'y2.
By differentiating the Wronskian, we obtain
W   y1 y2  y1 y2  y1y2  y1 y2
 y1  P( x) y2  Q( x) y2   y2  P( x) y1  Q( x) y1 
(3.51)
  P( x)( y1 y2  y1 y2 ).
In the special case that P(x) = 0, i.e.
y'' + Q(x) y = 0,
(3.52)
W = y1y2'-y1'y2 = constant.
Since our original eq. is homogeneous, we may multiply solutions y1 and y2 by whatever
constants we wish and arrange to have W =1 ( or -1). This case P(x) = 0, appears more
frequently than might be expected. ( 2 in Cartesian coordinates, the radical dependence
of 2(r) in spherical polar coordinates lack a first derivative). Finally, every linear 2ndorder ODE can be transformed into an eq. of the form of Eq.(3.52).
For the general case, let us now assume that we have one solution by a series
substitution ( or by guessing). We now proceed to develop a 2nd, independent solution
for which W  0.
dW
  P ( x1 )dx1
W
We integrate from x1 =a to x1 = x to obtain
.
28
ln
x
W ( x)
   P( x1 )dx1 ,
a
W (a)
or
x
W ( x)  W (a) exp   P( x1 )dx1 .
 a

But
(3.53)
W ( x)  y1 y2  y1 y2
 y12
(3.54)
d y2
( ).
dx y1
By combining Eqs. (3.53) and (3.54), we have
x
exp   P( x1 )dx1 
d y2
 a

( )  W (a)
2
dx y1
y1
.
(3.55)
Finally, by integrating Eq. (3.55) from x2=b to x2=x we get
x2
exp   P( x1 )dx1 
 a

y2 ( x)  y1 ( x)W (a) 
dx2 .
2
b
y1 ( x2 )
x
Here a and b are arbitrary constants and a term y1(x)y2(b)/y1(b) has been dropped, for it
leads to nothing new. As mentioned before, we can set W(a) =1 and write
y 2 ( x)  y1 ( x) 
x
x2
exp   P( x1 )dx1 


dx 2 .
2
y1 ( x 2 )
(3.56)
If we have the important special case of P(x) = 0. The above eq. reduces to
y 2 ( x)  y1 ( x) 
x
dx 2
y1 ( x 2 )2
.
Now, we can take one known solution and by integrating can generate a second
independent solution.
Example
A Second Solution for the Linear Oscillator eq.
From d2y/dx2 + y = 0 with P(x) = 0, let one solution be y1 = sin x.
29
y 2 ( x)  sin( x)


x
dx 2
sin 2 x 2
 sin x( cot x)   cos x.
Series Form of the Second Solution* (Optional Reading)
Further insight may be obtained by the following sequence of operations:
1. Express P(x) and Q(x) as
P( x) 

 pi xi ,
and Q( x) 
i  1

q
j  2
j
x j.
2. Develop the first few terms of a power-series solution, as in section 3.5.
3. Using this solution as y1, obtain y2, with Eq. (3.56), integrating term by term.
Proceeding with Step 1, we have
y  ( p1 x 1  p0  p1 x  ...) y  (q 2 x 2  q1 x 1  ...) y  0 (3.57)
in which x = 0 is at worst a regular singular point. If p-1 = q-1 = q2 = 0, it reduces to an
ordinary point. Substituting

y   a x k  
 0
(Step 2), we obtain

(k   )(k    1)a  x


k  2
0


  pi x i  (k   )a  x k   1 
i  1
 0


a x
q x 

j
j  2
j
k 
0 .
0
Assuming that p-10, q2 0, our indicial eq. is
k(k-1) + p-1k + q-2 = 0,
which sets the net coefficient of xk-2 equal to zero. We denote the two roots by k =  and
k = -n, where n is zero or positive integer. (If n is not an integer, we expect two
independent series solutions by the method of Section 3.5 and there is no problem). Then
(k-) (k-+ n) = 0,
or
k2 + (n-2)k + (-n) = 0.
30
We have
p-1 -1 = n-2.
(3.58)
The known series solution corresponding to the larger root k =  may be written as

y1  x  a x  .
 0
Substituting this series solution into Eq. (3.56) (Step 3), yields,
y2 ( x)  y1 ( x) 
x

x  ( 
x2

a
i  1
exp(  
pi x1i dx1 )

a x )2
0  2
2
2
dx2 ,
(3.59)
where the solutions y1 and y2 have been normalized so that the Wronskian, W(a)=1.
Tracking the exponential factor first, we have

  p x dx
x2
a
i
i 1
i  1
1

pk k 1
x2  f (a ).
k 0 k  1
 p1 ln x2  
Hence
exp(  
x2
a
 p x dx )
i
i
1
1
i

 exp  f (a )x 2 p1 exp( 
p k k 1
x2 )
k 0 k  1

p
1  p


 exp  f (a )x 2 p1 1   k x 2k 1  ( k x 2k 1 ) 2  ....
2! k 0 k  1
 k 0 k  1

This final series expansion of the exponential is certainly convergent if the original
expansion of the coefficient P(x) was convergent.
The denominatorin Eq. (3.59) may be handled by writing
 2 

 x 2 ( a x 2 ) 2 


 0

1

x 22 (

 a x2 ) 2
 0
 x 22

 b x2 .
 0
(3.60)
31
Neglecting constant factor that will be picked up anyway by the requirement that W(a)=1,
we obtain

y2 ( x)  y1 ( x)  x2 p 1  2 ( c x2 )dx2 .
x
(3.61)
 0
By Eq. (3.58),
x2 p1  2  x2 n 1
and we have assumed here that n is an integer. Substituting this result into Eq. (3.61), we
obtain
x
y2 ( x)  y1 ( x)  (c0 x2 n 1  c1 x2 n  c2 x2n 1  ...  cn x21  ...)dx2 . (3.62)
The integration indicated in Eq. (3.62) leads to a coefficient of y1(x) consisting of two
parts:
1. A power series starting with x-n.
2. A logarithm term from the integration of x-1 (when λ=n). This term always
appears when n is an integer unless cn fortuitously happens to vanish.