Chow T.L. Mathematical Methods for Physicists: A Concise Introduction
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Ordinary and singular points of a diÿerential equation
We shall concentrate on the linear second-order diÿerential equation of the form
d
2
y
dx
2
Px
dy
dx
Qxy 0 2:24
which plays a very important part in physical problems, and introduce certain
de®nitions and state (without proofs) some important results applicable to equa-
tions of this type. With some small modi®cations, these are applicable to linear
equation of any order. If both the functions P and Q can be expanded in Taylor
series in the neighborhood of x , then Eq. (2.24) is said to possess an ordinary
point at x . But when either of the functions P or Q does not possess a Taylor
series in the neighbo rhood of x , Eq. (2.24) is said to have a singular point at
x .If
P x=x ÿ and Q x=x ÿ
2
and x and x can be expanded in Taylor series near x . In such cases,
x is a singular point but the singularity is said to be regular.
Frobenius and Fuchs theorem
Frobenius and Fuchs showed that:
(1) If Px and Q x are regular at x , then the diÿerential equation (2.24)
possesses two distinct solutions of the form
y
X
1
0
a
x ÿ
a
0
6 0: 2:25
(2) If Px and Qx are singul ar at x , but x ÿ Px and x ÿ
2
Qx
are regular at x , then there is at least one solution of the diÿerential
equation (2.24) of the form
y
X
1
0
a
x ÿ
a
0
6 0; 2:26
where is some constant, which is valid for jx ÿ j <ÿwhenever the Taylor
series for x and x are valid for these values of x.
(3) If Px and Qx are irregular singular at x (that is, x and x are
singular at x , then regular solutions of the diÿerential equation (2.24)
may not exist.
86
ORDINARY DIFFERENTIAL EQUATIONS
The proofs of these results are beyond the scope of the book, but they can be
found, for example, in E. L. Ince's Ordinary Diÿerential Equations, Dover
Publications Inc., New York, 1944.
The ®rst step in ®nding a solution of a second-order diÿerential equation
relative to a regular singular point x is to determine possible values for the
index in the solution (2.26). This is done by substituting series (2.26) and its
appropriate diÿerential coeci ents into the diÿerential equation and equating to
zero the resulting coecient of the lowest power of x ÿ .This leads to a quadratic
equation, called the indicial equation, from which suitable values of can be
found. In the simplest case, these values of will give two diÿerent series solutions
and the general solution of the diÿerential equation is then given by a linear
combination of the separate solutions. The complete procedure is shown in
Example 2.14 below.
Example 2.14
Find the general solution of the equation
4x
d
2
y
dx
2
2
dy
dx
y 0:
Solution: The origin is a regular singul ar point and, writing
y
1
0
a
x
a
0
6 0 we have
dy=dx
X
1
0
a
x
ÿ1
; d
2
y=dx
2
X
1
0
a
ÿ 1x
ÿ2
:
Before substituting in the diÿerential equation, it is convenient to rewrite it in the
form
4x
d
2
y
dx
2
2
dy
dx
()
y
fg
0:
When a
x
is substituted for y, each term in the ®rst bracket yiel ds a multiple of
x
ÿ1
, while the second bracket gives a multiple of x
and, in this form, the
diÿerential equation is said to be arranged according to weight, the weights of the
bracketed terms diÿering by unity. When the assumed series and its diÿerential
coecients are substituted in the diÿerential equation, the term containing the
lowest power of x is obtained by writing y a
0
x
in the ®rst bracket. Since the
coecient of the lowest power of x must be zero and, since a
0
6 0, this gives the
indicial equation
4 ÿ 12 22 ÿ 10;
its roots are 0, 1=2.
87
SOLUTIONS IN POWER SERIES
The term in x
is obtained by writing y a
1
x
1
in ®rst bracket and
y a
x
in the second. Equating to zero the coecient of the term obtained in
this way we have
f4 1 2 1ga
1
a
0;
giving, with replaced by n,
a
n1
ÿ
1
2 n 12 2n 1
a
n
:
This relation is true for n 1; 2; 3; ... and is called the recurrence relation for the
coecients. Using the ®rst root 0 of the indicial equation, the recurrence
relation gives
a
n1
1
2n 12n 1
a
n
and hence
a
1
ÿ
a
0
2
; a
2
ÿ
a
1
12
a
0
4!
; a
3
ÿ
a
2
30
ÿ
a
0
6!
; ...:
Thus one solution of the diÿerential equation is the series
a
0
1 ÿ
x
2!
x
2
4!
ÿ
x
3
6!
ÿ
ý!
:
With the second root 1=2, the recurrence relation becomes
a
n1
ÿ
1
2n 32n 2
a
n
:
Replacing a
0
(which is arbitrary) by b
0
, this gives
a
1
ÿ
b
0
3 2
ÿ
b
0
3!
; a
2
ÿ
a
1
5 4
b
0
5!
; a
3
ÿ
a
2
7 6
ÿ
b
0
7!
; ...:
and a second solution is
b
0
x
1=2
1 ÿ
x
3!
x
2
5!
ÿ
x
3
7!
ÿ
ý!
:
The general solution of the equation is a linear combination of these two solu-
tions.
Many physical problems require solutions which are valid for large values of
the independent variable x. By using the transformation x 1=t, the diÿerential
equation can be transformed into a linear equation in the new variable t and the
solutions required will be those valid for small t.
In Example 2.14 the indicial equation has two distinct roots. But there are two
other possibilities: (a) the indicial equation has a double root; (b ) the roots of the
88
ORDINARY DIFFERENTIAL EQUATIONS
indicial equation diÿer by an integer. We now take a gen eral look at these cases.
For this purpose, let us consider the following diÿerential equation which is highly
important in mathematical physics:
x
2
y
00
xgxy
0
hxy 0; 2:27
where the functions gx and hx are analytic at x 0. Since the coecients are
not analyic at x 0, the solution is of the form
yxx
r
X
1
m0
a
m
x
m
a
0
6 0: 2:28
We ®rst expand gx and hx in power series,
gxg
0
g
1
x g
2
x
2
hxh
0
h
1
x h
2
x
2
:
Then diÿerentiating Eq. (2.28) term by term, we ®nd
y
0
x
X
1
m0
m ra
m
x
mrÿ1
; y
00
x
X
1
m0
m rm r ÿ 1a
m
x
mrÿ2
:
By inserting all these into Eq. (2.27) we obtain
x
r
rr ÿ 1a
0
g
0
g
1
x x
r
ra
0
h
0
h
1
x x
r
a
0
a
1
x 0:
Equating the sum of the coecients of each power of x to zero, as before, yields a
system of equations involving the unknown coecients a
m
. The smallest power is
x
r
, and the corresponding equation is
rr ÿ 1g
0
r h
0
a
0
0:
Since by assumption a
0
6 0, we obtain
rr ÿ 1g
0
r h
0
0orr
2
g
0
ÿ 1 r h
0
0: 2:29
This is the indicial equation of the diÿerential equation (2.27). We shall see that
our series method will yield a fundamental system of solutions; one of the solu-
tions will always be of the form (2.28), but for the form of other solution there will
be three diÿerent possibilities corresponding to the follo wing cases.
Case 1 The roots of the indicial equation are distinct and do not diÿer by an
integer.
Case 2 The indicial equation has a double root.
Case 3 The roots of the indicial equati on diÿer by an integer.
We now discuss these cases separately.
89
SOLUTIONS IN POWER SERIES
Case 1 Distinct roots not diÿering by an integer
This is the simplest case. Let r
1
and r
2
be the roots of the indicial equation (2.29).
If we insert r r
1
into the recurrence relation and determine the coecients
a
1
, a
2
; ... successively, as before, then we obtain a solution
y
1
xx
r
1
a
0
a
1
x a
2
x
2
:
Similarly, by inserting the second root r r
2
into the recurrence relation, we will
obtain a second solution
y
2
xx
r
2
a
0
*
a
1
*
x a
2
*
x
2
:
Linear independence of y
1
and y
2
follows from the fact that y
1
=y
2
is not constant
because r
1
ÿ r
2
is not an integer.
Case 2 Double roots
The indicial equation (2.29) has a double root r if, and only if,
g
0
ÿ 1
2
ÿ 4h
0
0, and then r 1 ÿ g
0
=2. We may determine a ®rst solution
y
1
xx
r
a
0
a
1
x a
2
x
2
r
1ÿg
0
2
2:30
as before. To ®nd another solution we may apply the method of variation of
parameters, that is, we replace constant c in the solution cy
1
x by a function
ux to be determined, such that
y
2
xuxy
1
x2:31
is a solution of Eq. (2.27). Inserting y
2
and the derivatives
y
0
2
u
0
y
1
uy
0
1
y
00
2
u
00
y
1
2u
0
y
0
1
uy
00
1
into the diÿerential equation (2.27) we obtain
x
2
u
00
y
1
2u
0
y
0
1
uy
00
1
xgu
0
y
1
uy
0
1
huy
1
0
or
x
2
y
1
u
00
2x
2
y
0
1
u
0
xgy
1
u
0
x
2
y
00
1
xgy
0
1
hy
1
u 0:
Since y
1
is a solution of Eq. (2.27), the quantity inside the bracket vanishes; and
the last equation reduces to
x
2
y
1
u
00
2x
2
y
0
1
u
0
xgy
1
u
0
0:
Dividing by x
2
y
1
and inserting the power series for g we obtain
u
00
2
y
0
1
y
1
g
0
x
u
0
0:
Here and in the following the dots designate terms which are constant s or involve
positive powers of x. Now from Eq. (2.30) it follows that
y
0
1
y
1
x
rÿ1
ra
0
r 1a
1
x
x
r
a
0
a
1
x
1
x
ra
0
r 1a
1
x
a
0
a
1
x
r
x
:
90
ORDINARY DIFFERENTIAL EQUATIONS
Hence the last equation can be written
u
00
2r g
0
x
u
0
0: 2:32
Since r 1 ÿ g
0
=2 the term 2r g
0
=x equals 1=x, and by dividing by u
0
we
thus have
u
00
u
0
ÿ
1
x
:
By integration we obtain
ln u
0
ÿln x or u
0
1
x
e
...
:
Expanding the exponential function in powers of x and integrating once more, we
see that the expression for u will be of the form
u ln x k
1
x k
2
x
2
:
By inserting this into Eq. (2.31) we ®nd that the second solution is of the form
y
2
xy
1
xln x x
r
X
1
m1
A
m
x
m
: 2:33
Case 3 Roots diÿering by an integer
If the roots r
1
and r
2
of the indicial equation (2.29) diÿer by an integer, say, r
1
r
and r
2
r ÿ p, where p is a positive integer, then we may always determine one
solution as before, namely, the solution corresponding to r
1
:
y
1
xx
r
1
a
0
a
1
x a
2
x
2
:
To determine a second solution y
2
, we may proceed as in Case 2. The ®rst steps
are literally the same and yield Eq. (2.32). We determine 2r g
0
in Eq. (2.32).
Then from the indicial equation (2.29), we ®nd ÿr
1
r
2
g
0
ÿ 1. In our case,
r
1
r and r
2
r ÿ p, therefo re, g
0
ÿ 1 p ÿ 2r. Hence in Eq. (2.32) we have
2r g
0
p 1, and we thus obtain
u
00
u
0
ÿ
p 1
x
:
Integrating, we ®nd
ln u
0
ÿp 1ln x or u
0
x
ÿp1
e
...
;
where the dots stand for some series of positive powers of x. By expanding the
exponential function as before we obtain a series of the form
u
0
1
x
p1
k
1
x
p
k
p
x
k
p1
k
p2
x :
91
SOLUTIONS IN POWER SERIES
Integrating, we have
u ÿ
1
px
p
ÿk
p
ln x k
p1
x : 2:34
Multiplying this expression by the series
y
1
xx
r
1
a
0
a
1
x a
2
x
2
and remem bering that r
1
ÿ p r
2
we see that y
2
uy
1
is of the form
y
2
xk
p
y
1
xln x x
r
2
X
1
m0
a
m
x
m
: 2:35
While for a double root of Eq. (2.29) the second solution always contains a
logarithmic term, the coecient k
p
may be zero and so the logarithmic term may
be missing, as shown by the following example.
Example 2.15
Solve the diÿerential equation
x
2
y
00
xy
0
x
2
ÿ
1
4
y 0:
Solution: Substituting Eq. (2.28) and its derivatives into this equation, we obtain
X
1
m0
m rm r ÿ 1m rÿ
1
4
a
m
x
mr
P
1
m0
a
m
x
mr2
0:
By equating the coecient of x
r
to zero we get the indicial equation
rr ÿ 1r ÿ
1
4
0orr
2
1
4
:
The roots r
1
1
2
and r
2
ÿ
1
2
diÿer by an integer. By equating the sum of the
coecients of x
sr
to zero we ®nd
r 1r r ÿ 1ÿ
1
4
a
1
0 s 1: 2:36a
s rs r ÿ 1s r ÿ
1
4
a
s
a
sÿ2
0 s 2; 3 ; ...: 2:36b
For r r
1
1
2
, Eq. (2.36a) yields a
1
0, and the indicial equation (2.36b)
becomes
s 1sa
s
a
sÿ2
0:
From this and a
1
0 we obtain a
3
0, a
5
0, etc. Solving the indicial equation
for a
s
and setting s 2p, we get
a
2p
ÿ
a
2pÿ2
2p2p 1
p 1; 2; ...:
92
ORDINARY DIFFERENTIAL EQUATIONS
Hence the non-zero coecients are
a
2
ÿ
a
0
3!
; a
4
ÿ
a
2
4 5
a
0
5!
; a
6
ÿ
a
0
7!
; etc:;
and the solution y
1
is
y
1
xa
0
x
p
X
1
m0
ÿ1
m
x
2m
2m 1!
a
0
x
ÿ1=2
X
1
m0
ÿ1
m
x
2m1
2m 1!
a
0
sin x
x
p
: 2:37
From Eq. (2.35) we see that a secon d independent solution is of the form
y
2
xky
1
xln x x
ÿ1=2
X
1
m0
a
m
x
m
:
Substituting this and the derivatives into the diÿerential equation, we see that the
three expressions involving ln x and the expressions ky
1
and ÿky
1
drop out.
Simplifying the remaining equation, we thus obtain
2kxy
0
1
X
1
m0
mm ÿ 1a
m
x
mÿ1=2
X
1
m0
a
m
x
m3=2
0:
From Eq. (2.37) we ®nd 2kxy
0
ÿka
0
x
1=2
. Since there is no further term
involving x
1=2
and a
0
6 0, we must have k 0. The sum of the coecients of the
power x
sÿ1=2
is
ss ÿ 1a
s
a
sÿ2
s 2; 3; ...:
Equating this to zero and solving for a
s
, we have
a
s
ÿa
sÿ2
=ss ÿ 1 s 2; 3; ... ;
from which we obtain
a
2
ÿ
a
0
2!
; a
4
ÿ
a
2
4 3
a
0
4!
; a
6
ÿ
a
0
6!
; etc:;
a
3
ÿ
a
1
3!
; a
5
ÿ
a
3
5 4
a
1
5!
; a
7
ÿ
a
1
7!
; etc:
We may take a
1
0, because the odd powers would yield a
1
y
1
=a
0
. Then
y
2
xa
0
x
ÿ1=2
X
1
m0
ÿ1
m
x
2m
2m!
a
0
cos x
x
p
:
Simultaneous equations
In some physics and engineering problems we may face simultaneous dif-
ferential equations in two or more dependent variables. The general solution
93
SIMULTANEOUS EQUATIONS
of simultaneous equations may be found by solvi ng for each dependent variable
separately, as shown by the following example
Dx 2y 3x 0
3x Dy ÿ 2y 0
)
D d=dt
which can be rewritten as
D 3x 2y 0;
3x D ÿ 2y 0:
)
We then operate on the ®rst equation with (D ÿ 2) and multiply the second by a
factor 2:
D ÿ 2D 3x 2D ÿ 2y 0;
6x 2D ÿ 2y 0:
)
Subtracting the ®rst from the second leads to
D
2
D ÿ 6x ÿ 6x D
2
D ÿ 12x 0;
which can easily be solved an d its solut ion is of the form
xtAe
3t
Be
ÿ4t
:
Now inserting xt back into the original equation to ®nd y gives:
ytÿ3Ae
3t
1
2
Be
ÿ4t
:
The gamma and beta functions
The factorial notation n! nn ÿ 1 n ÿ 23 2 1 has proved useful in
writing down the coecien ts in some of the series solutions of the diÿerential
equations. However, this notation is meaningless when n is not a positive integer.
A useful extension is provided by the gamma (or Euler) function, which is de®ned
by the integral
ÿ
Z
1
0
e
ÿx
x
ÿ1
dx >02:38
and it follows immediately that
ÿ1
Z
1
0
e
ÿx
dx ÿe
ÿx
1
0
1: 2:39
Integration by parts gives
ÿ 1
Z
1
0
e
ÿx
x
dx ÿe
ÿx
x
1
0
Z
1
0
e
ÿx
x
ÿ1
dx ÿ: 2:40
94
ORDINARY DIFFERENTIAL EQUATIONS
When n, a positive integer, repeated application of Eq. (2.40) and use of Eq.
(2.39) gives
ÿn 1nÿnnn ÿ 1ÿn ÿ 1... nn ÿ 13 2 ÿ1
nn ÿ 1 3 2 1 n!:
Thus the gamma function is a generalization of the factorial function. Eq. (2.40)
enables the values of the gamma function for any positive value of to be
calculated: thus
ÿ
7
2
5
2
ÿ
5
2
5
2
3
2
ÿ
3
2
5
2
3
2
1
2
ÿ
1
2
:
Write u
x
p
in Eq. (2.38) and we then obtain
ÿ2
Z
1
0
u
2ÿ1
e
ÿu
2
du;
so that
ÿ
1
2
2
Z
1
0
e
ÿu
2
du
p
:
The function ÿ has been tabulated for values of between 0 and 1.
When <0 we can de®ne ÿ with the help of Eq. (2.40) and write
ÿÿ 1=:
Thus
ÿÿ
3
2
ÿ
2
3
ÿÿ
1
2
ÿ
2
3
ÿ
2
1
ÿ
1
2
4
3
p
:
When ! 0;
R
1
0
e
ÿx
x
ÿ1
dx diverges so that ÿ0 is not de®ned.
Another function which will be useful later is the beta function which is de®ned
by
Bp; q
Z
1
0
t
pÿ1
1 ÿ t
qÿ1
dt p; q > 0: 2:41
Substituting t v = 1 v , this can be written in the alternati ve form
Bp; q
Z
1
0
v
pÿ1
1 v
ÿpÿq
dv: 2:42
By writing t
0
1 ÿ t we deduce that Bp; qBq; p.
The beta function can be expressed in terms of gamma functions as follows:
Bp; q
ÿpÿq
ÿp q
: 2:43
To prove this, write x at (a > 0) in the integral (2.38) de®ning ÿ, and it is
straightforward to show that
ÿ
a
Z
1
0
e
ÿat
t
ÿ1
dt 2:44
95
THE GAMMA AND BETA FUNCTIONS