Title: Chapter 3 Differential Equations
1Chapter 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).  Newtons Laws
 Maxwell equations
 Schrodinger and Dirac equations
etc.
Since the dynamics of many physical systems
involve just two derivatives, DE of second order
occur most frequently in physics. e.g.,
acceleration in classical mechanics
the kinetic energy operator in quantum mechanics
Ordinary differential equation (ODE) Partial
differential equation (PDE)
2Examples of PDEs
 1. Laplace's eq.
 This very common and important eq. occurs
in studies of  a. electromagnetic phenomena, b.
hydrodynamics,  c. heat flow,
d. gravitation.  2. Poisson's eq.,
 In contrast to the homogeneous Laplace eq.,
Poisson's eq. is nonhomogeneous with a
source term
3. The wave (Helmholtz) and timeindependent
diffusion eqs., . 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
timedependent diffusion eq.
3 5. The timedependent wave eq.,
 where is a fourdimensional analog of
the Laplacian.  6.The scalar potential eq.,
 7.The KleinGordon eq., ,
and the corresponding vector eqs. in which ? is
replaced by a vector function.  8.The Schrodinger wave eq.
 and

 for the timeindependent case.


4Some general techniques for solving secondorder
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,
LY lY, 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
Perhaps the best known nonlinear eq. of second is
the KortewegdeVries (KdV) eq.
53.2 Firstorder Differential Equations
We consider here the general form of firstorder
DE
(3.1)
The eq. is clearly a firstorder ODE. It may or
may not be linear, although we shall treat the
linear case explicitly later.
Separable variables
Frequently, the above eq. will have the special
form
6or P(x)dx Q(y) dy
0 Integrating from (x0, y0) to (x, y) yields,
Since the lower limits and 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
Vvolume, P  Pressure. or
ln V ln P C. If we set C ln k, PV k.
7Exact 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 dj,
Since the RHS is zero, we look for an unknown
function ? j(x,y) const. and dj? 0.
We have
and
The necessary and sufficient for our eq. to be
exact is that the second, mixed partial
derivatives of j (????assumed continuous) are
independent of the order of differential
If such j(x,y) exists then the solution is
j(x,y)C.
8There always exists at least one integrating
facts, a(x,y), such that
Linear Firstorder ODE
If f (x,y) has the form p(x)y q(x), then
(3.2)
It is the most general linear firstorder 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.
9Let us look for an integrating factor a(x) so
that
(3.3)
may be rewritten as
(3.4)
The purpose of this is to make the left hand
side of Eq.(3.2) a derivative so that it can be
integratedby inspection. Expanding Eq. (3.4),
we obtain
Comparison with Eq.(3.3) shows that we must
require
10Here is a differential equation for a(x) , with
the variables a and x separable. We separate
variables, integrate, and obtain
(3.5)
as our integrating factor.
With a(x) known we proceed to integrate
Eq.(3.4). This, of course, was the point of
introducing a(x) in the first place. We have
Now integrating by inspection, we have
The constants from a constant lower limit of
integration are lumped into the constant C.
Dividing
by a(x) , we obtain
11Finally, substituting in Eq.(3.5) for a yields
(3.6)
Equation (3.6) is the complete general solution
of the linear, firstorder differential
equation, Eq.(3.2). The portion
corresponds to the case q(x) 0 and
is a general solution of the Homogeneous
differential equation. The other term in
Eq.(3.6),
is a particular solution corresponding to the
specific source term q(x) .
12V
Example RL Circuit For a
resistanceinductance circuit Kirchhoffs law
leads to
for the current I(t) , where L is the
inductance and R the resistance, both constant.
V(t) is the timedependent impressed voltage.
From Eq.(3.5) our integrating factor a(t) is
13Then by Eq.(3.6)
with the constant C to be determined by an
initial condition (a boundary condition).
For the special case V(t)V0 , a constant,
If the initial condition is I(0) 0, then
and
143.3 SEPARATION OF VARIABLES
A very important PDE in physics
Electromagnetic field, wave transition etc.
Cartesian coordinates Cylindrical
coordinates Spherical coordinates
m mass of an electron hbar Plank constant
15 Certain partial differential equations can be
solved by separation of variables. The method
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 n1 constants, determined by the
conditions imposed in the problem being solved. 
 Using for the Laplacian. For the
present let be a constant.
Cartesian Coordinates In Cartesian
coordinates the Helmholtz equation becomes
(3.7)
16Let
(3.7a)
Dividing by
and rearranging terms, we obtain
(3.8)
The lefthand side is a function of x alone,
whereas the righthand side depends only on y
and z, but x , y , and z are all independent
coordinates. The only possibility is setting each
side equal to a constant, a constant of
separation. We choose
(3.10)
(3.9)
Rearrange Eq.3.10
where a second separation constant has been
introduced.
17Similarly
(3.11)
(3.12)
introducing a constant by
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 solution should be
labeled according to the choice of our constants
l, m ,and n ,that is ,
(3.13)

 Subject to the conditions of the problem being
solved.  We may develop the most general solution of
Eq.(3.7)by taking a linear combination of
solutions ,
(3.14)
Where the constant coefficients are
determined by the boundary conditions
18Example Laplace equation in rectangular
coordinates
z
Consider a rectangle box with dimensions (a,b,c)
in the (x,y,z) directions. All surfaces of the
box are kept at zero potential, except the
surface zc, which is at a potential V(x,y). It
is required to find the potential everywhere
inside the box.
zc
yc
y
xa
x
19Example Laplace equation in rectangular
coordinates
Where
Then the solutions of the three ordinary
differential equations are
20To have F0 at xa and yb, we must have aanp,
and bbmp. Then
Since FV(x,y) at zc
We have the coefficients
Here the features of Fourier series have been
used.
21Circular Cylindrical Coordinates
 With our unknown function ? dependent on ?, f ,
and z , the Helmholtz equation becomes
(3.15)
or
(3.16)
As before, we assume a factored form for ? ,
(3.17)
22Substituting into Eq.(3.16), we have
(3.18)
All the partial derivatives have become ordinary
derivatives. Dividing by PFZ and moving the z
derivative to the righthand side yields
(3.19)
Then
(3.20)
And
(3.21)
23Setting , multiplying by ,
and rearranging terms, we obtain
(3.22)
We may set the righthand side to m2 and
(3.23)
Finally, for the ? dependence we have
(3.24)
This is Bessels differential equation . The
solution and their properties are presented in
Chapter 6. 
The original Helmholtz equation has been replaced
by three ordinary differential equations. A
solution of the Helmholtz equation is
A general Sol.
(3.26)
24 Spherical Polar Coordinates
Let us try to separate the Helmholtz equation in
spherical polar coordinates
(3.27)
(3.29)
(3.30)
25 we use as the separation constant. Any
constant will do, but this one will make life a
little easier. Then
(3.31)
and
(3.32)
Multiplying Eq.(3.32) by and rearranging
terms, we obtain
(3.33)
26Again, the variables are separated. We equate
each side to a constant Q and finally obtain
 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(l1) l is an
integer. If is a (positive) constant, Eq.
(3.35) becomes the spherical Bessel equation.  Again, our most general solution may be
written
(3.34)
(3.35)
(3.36)
27The restriction that k2 be a constant is
necessarily. The separation process will Still be
possible for k2 as general as
In the hydrogen atom problem, one of the most
important examples of the Schrodinger Wave
equation with a closed form solution is k2f(r)
Finally, as an illustration of how the constant m
in Eq.(3.31) is restricted, we note that f in
cylindrical and spherical polar coordinates is an
azimuth angle. If this is a classical problem, we
shall certainly require that the azimuthal
solution F(f) be singled valued, that is,
28 This is equivalent to requiring the azimuthal
solution to have a period of 2p 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
293.4 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 , point x is an ordinary
point. However, if either P(x) or Q(x) ( or both)
diverges as x ?approaches to , is a
singular point.
Using Eq.(3.40), we are able to distinguish
between two kinds of singular points
30These definitions hold for all finite values of
x0. The analysis of is similar to
the treatment of functions of a complex variable.
We set x 1/z, substitute into the DE, and then
let . By changing variables in the
derivative, we have
(3.41)
(3.42)
Using these results, we transform Eq.(3.40) into
(3.43)
The behavior at x ? (z 0) then depends on the
behavior of the new coefficients
31 and
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/ ,
respectively, x is a regular singular point,
otherwise an irregular singular point.
Example
Bessel's eq. is
Comparing it with Eq. (3.40) we have
P(x) 1/x, Q(x) 1  ,
which shows that point x 0 is a regular
singularity. As x ?? ? (z ?? 0), from Eq.
(3.43), we have the coefficients
and
Since the Q(z) diverges as , point x ? is
an irregular or essential singularity.
32 We list , in Table 3.4, several typical ODEs and
their singular points.
Table 3.4
Equation
Irregular singularity
___
0,1,
2. Legendre
___
1,1,
3. Chebyshev
___
1,1,
4. Confluent hypergeometric
5. Bessel
6. Laguerre
7. Simple harmonic oscillator
___
___
8. Hermite
333.5 Series Solutions
(A)
(B)
The Bessels function
A linear secondorder homogeneous ODE
y'' P(x) y' Q(x) y 0.
34Linear secondorder ODE
 In this section, we develop a method of a
series expansion for obtaining one solution of
the linear, secondorder, homogeneous DE.  A linear, secondorder, homogeneous ODE may
be written in the form  y'' P(x) y' Q(x) y 0.
 the most general solution may be written
as 
 Our physical problem may lead to a
nonhomogeneous, linear, secondorder DE  y'' P(x) y' Q(x) y F(x).
 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,
The constants c1 and c2 will eventually be
fixed by boundary conditions
35To seek a solution with the form
Fuchss Theorem We can always get at least one
powerseries solution, provided we are expanding
about a point that is an ordinary point or at
worst a regular singular point.
36 .
 To illustrate the series solution, we apply
the method to two important DEs.  First, the linear oscillator eq.
 , (3.44)

 with known solutions y sin wx, cos wx.

 We try
with k and al? still undetermined. Note that k
need not be an integer. By differentiating twice,
we obtain
37By substituting into Eq. (3.44), we have
(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 , for l? 0 in
the first summation. The requirement that the
coefficient vanishes yields  k(k1) 0.
 Since, by definition, a0 ? 0, we have
 k(k1) 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. k0 or
k1  If k 1, the coefficient (k1)k of
must vanish so that  0.
 We set ?l j2 in the first summation
and ?l' j in the second. This results in
382 (kj2)(kj1) 0
or
 This is a twoterm recurrence relation.
We first try the solution k 0. The recurrence
relation becomes
which leads to
39 So our solution is
If we choose the indicial eq. root k 1, the
recurrence relation becomes
Again, we have
For this choice, k 1, we obtain
40 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.  (2) The acceptability of a series
solution depends on its convergence (including
asymptotic convergence).  Expansion above
 It is perfectly possible to write
Indeed, for the Legendre eq the choice x0 1
has some advantages. The point x0 should not
be chosen at an essential singularity or the
method will probably fail.
41 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. 

(3.46) 
 Again, assuming a solution of the form
42 we differentiate and substitute into Eq.
(3.46). The result is 
 By setting? l? 0, we get the coefficient of
,  a0k(k1) k 0.
 The indicial equation

 with solution k n or n. For the
coefficients of x(k1), we obtain 

 For k n or n (k is not equal 1/2),
does not vanish and we must require 0.  Proceeding to the coefficient of for k
n, we set ?l j in the 1st, 2nd, and 4th
terms and ?l j2 in the 3rd term. By requiring
the resultant
coefficient of to vanish, we obtain
(nj)(nj1)(nj) 0.
43When j? j2, this can be written for j0 as
(3.47)
which is the desired recurrence relation.
Repeated application of this recurrence relation
leads to
, and so on, and in general
Inserting these coefficients in our assumed
series solution, we have
44In summation form
(3.48)
 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
Jn(x). However, when n is a negative integer,
trouble develops. 
 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.
45Will this method always work? The answer is
no! 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 (Fuchss theorem).  Whether we get one or two distinct
solutions depends on the roots of the indicial
equation.  1. If the two roots of the indicial
equation are equal, we can obtain only one
solution by this series substitution method.  2. If the two roots differ by a
noninteger number, two independent solutions
may be obtained.  3. 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 Bessels equation, only one
solution.
46
 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.
(3.49a)
(3.49b)
(3.49c)
(3.49d)
47For the 1st eq., the indicial eq. is
k(k1)  6 0,
 giving k 3, 2. Since the eq. is
homogeneous in x ( counting as
), here is no recurrence relation. However, we
are left with two perfectly good solution,
and .  For the 2nd eq., we have 6 0, with no
solution at all, for we have agreed that ? 0.
The series substitution broke down at Eq. (3.49b)
which has an irregular singular point at the
origin.  Continuing with the Eq. (3.49c), we have
added a term y'/x. The indicial eq. is
, but again, there is no recurrence
relation. The solutions are  y , 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 
48Unless a is selected to make the series terminate
we have
Hence our series solution diverges for all x ?0.
493.6 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 

(3.50) 
 in which not all the coefficients k are
zero. On the other hand, if the only solution of
Eq. (3.50) is k0 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.
50 This gives us a set of homogeneous linear eqs. in
which k? are the unknown quantities. There is a
solution kl? ?0 only if the determinant of the
coefficients of the k?'s vanishes,
.
51 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
. The Wronskian becomes 
 and f1 and f2 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.
52 but this is not a linear relation.
Example Linear Dependence
Consider the solutions of the onedimensional
diffusion eq. .
We have ?f1 ex and f2 ex, and we add ?3
cosh x, also a solution. Then
because the first and third rows are identical.
Here, they are linearly dependent, and indeed,
we have  2 coshx 0 with kl? ? 0.
53  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 .
 By differentiating the Wronskian, we obtain
In the special case that P(x) 0,
i.e. y'' Q(x) y 0,
(3.52)
W constant.
54 Since our original eq. is homogeneous, we
may multiply solutions and 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. ( in Cartesian
coordinates, the radical dependence of (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.
We integrate from x1 a to x1 x to obtain
(3.53)
55 (3.54)
By combining Eqs. (3.53) and (3.54), we have
(3.55)
Finally, by integrating Eq. (3.55) from x2b to
x2x we get
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
56 (3.56)
If we have the important special case of P(x)
0. The above eq. reduces to
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 y 0 with P(x) 0, let
one solution be y1 sin x.
57Chapter 4. Orthogonal Functions (Optional
Reading)
 4.1 Hermitian Operators (HO)
 HO in quantum mechanics (QM)
 As we know, is an HO
operator. As is customary in QM, we simply assume
that the wave functions satisfy appropriate
boundary conditions vanishing sufficiently
strongly at infinity or having periodic behavior.
The operator L is called Hermitian if  The adjoint A of an operator A is defined
by  Clearly if A A (selfadjoint) and
satisfies the above mentioned boundary
conditions, then A is Hermitian. The expectation
value of an operator L is defined as