Obtaining a second solution

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
is required, i.e. z = 0, is in fact an ordinary point of the ODE rather than a
regular singular point, then substitution of the Frobenius series (16.12) leads to
an indicial equation with roots σ = 0 and σ = 1. Although these roots differ by
an integer (unity), the recurrence relations corresponding to the two roots yield
two linearly independent power series solutions (one for each root), as expected
from section 16.2.
16.4 Obtaining a second solution
Whilst attempting to construct solutions to an ODE in the form of Frobenius
series about a regular singular point, we found in the previous section that when
the indicial equation has a repeated root, or roots differing by an integer, we can
(in general) find only one solution of this form. In order to construct the general
solution to the ODE, however, we require two linearly independent solutions y1
and y2 . We now consider several methods for obtaining a second solution in this
case.
16.4.1 The Wronskian method
If y1 and y2 are two linearly independent solutions of the standard equation
y + p(z)y + q(z)y = 0
then the Wronskian of these two solutions is given by W (z) = y1 y2 − y2 y1 .
Dividing the Wronskian by y12 we obtain
d
W
1
y2
y1
y2
d y2
=
−
y
=
+
=
y
,
2
2
y1
y1
dz y1
dz y1
y12
y12
which integrates to give
y2 (z) = y1 (z)
z
W (u)
du.
y12 (u)
Now using the alternative expression for W (z) given in (16.4) with C = 1 (since
we are not concerned with this normalising factor), we find
u
z
1
y2 (z) = y1 (z)
exp
−
p(v)
dv
du.
(16.25)
y12 (u)
Hence, given y1 , we can in principle compute y2 . Note that the lower limits of
integration have been omitted. If constant lower limits are included then they
merely lead to a constant times the first solution.
Find a second solution to (16.21) using the Wronskian method.
For the ODE (16.21) we have p(z) = 3/(z − 1), and from (16.24) we see that one solution
544
16.4 OBTAINING A SECOND SOLUTION
to (16.21) is y1 = z/(1 − z)2 . Substituting for p and y1 in (16.25) we have
u
z
z
(1 − u)4
3
exp
−
y2 (z) =
dv
du
(1 − z)2
u2
v−1
z
(1 − u)4
z
exp [−3 ln(u − 1)] du
=
(1 − z)2
u2
z
z
u−1
=
du
(1 − z)2
u2
1
z
.
ln z +
=
(1 − z)2
z
By calculating the Wronskian of y1 and y2 it is easily shown that, as expected, the two
solutions are linearly independent. In fact, as the Wronskian has already been evaluated
as W (u) = exp[−3 ln(u − 1)], i.e. W (z) = (z − 1)−3 , no calculation is needed. An alternative (but equivalent) method of finding a second solution is simply to
assume that the second solution has the form y2 (z) = u(z)y1 (z) for some function
u(z) to be determined (this method was discussed more fully in subsection 15.2.3).
From (16.25), we see that the second solution derived from the Wronskian is
indeed of this form. Substituting y2 (z) = u(z)y1 (z) into the ODE leads to a
first-order ODE in which u is the dependent variable; this may then be solved.
16.4.2 The derivative method
The derivative method of finding a second solution begins with the derivation of
a recurrence relation for the coefficients an in a Frobenius series solution, as in the
previous section. However, rather than putting σ = σ1 in this recurrence relation
to evaluate the first series solution, we now keep σ as a variable parameter. This
means that the computed an are functions of σ and the computed solution is now
a function of z and σ:
y(z, σ) = z σ
∞
an (σ)z n .
(16.26)
n=0
Of course, if we put σ = σ1 in this, we obtain immediately the first series solution,
but for the moment we leave σ as a parameter.
For brevity let us denote the differential operator on the LHS of our standard
ODE (16.7) by L, so that
L=
d2
d
+ p(z) + q(z),
dz 2
dz
and examine the effect of L on the series y(z, σ) in (16.26). It is clear that the
series Ly(z, σ) will contain only a term in z σ , since the recurrence relation defining
the an (σ) is such that these coefficients vanish for higher powers of z. But the
coefficient of z σ is simply the LHS of the indicial equation. Therefore, if the roots
545
SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
of the indicial equation are σ = σ1 and σ = σ2 then it follows that
Ly(z, σ) = a0 (σ − σ1 )(σ − σ2 )z σ .
(16.27)
Therefore, as in the previous section, we see that for y(z, σ) to be a solution of
the ODE Ly = 0, σ must equal σ1 or σ2 . For simplicity we shall set a0 = 1 in the
following discussion.
Let us first consider the case in which the two roots of the indicial equation
are equal, i.e. σ2 = σ1 . From (16.27) we then have
Ly(z, σ) = (σ − σ1 )2 z σ .
Differentiating this equation with respect to σ we obtain
∂
[Ly(z, σ)] = (σ − σ1 )2 z σ ln z + 2(σ − σ1 )z σ ,
∂σ
which equals zero if σ = σ1 . But since ∂/∂σ and L are operators that differentiate
with respect to different variables, we can reverse their order, implying that
∂
y(z, σ) = 0
at σ = σ1 .
L
∂σ
Hence, the function in square brackets, evaluated at σ = σ1 and denoted by
∂
y(z, σ)
,
(16.28)
∂σ
σ=σ1
is also a solution of the original ODE Ly = 0, and is in fact the second linearly
independent solution that we were looking for.
The case in which the roots of the indicial equation differ by an integer is
slightly more complicated but can be treated in a similar way. In (16.27), since L
differentiates with respect to z we may multiply (16.27) by any function of σ, say
σ − σ2 , and take this function inside the operator L on the LHS to obtain
L [(σ − σ2 )y(z, σ)] = (σ − σ1 )(σ − σ2 )2 z σ .
(16.29)
Therefore the function
[(σ − σ2 )y(z, σ)]σ=σ2
is also a solution of the ODE Ly = 0. However, it can be proved§ that this
function is a simple multiple of the first solution y(z, σ1 ), showing that it is not
linearly independent and that we must find another solution. To do this we
differentiate (16.29) with respect to σ and find
∂
{L [(σ − σ2 )y(z, σ)]} = (σ − σ2 )2 z σ + 2(σ − σ1 )(σ − σ2 )z σ
∂σ
+ (σ − σ1 )(σ − σ2 )2 z σ ln z,
§
For a fuller discussion see, for example, K. F. Riley, Mathematical Methods for the Physical Sciences
(Cambridge: Cambridge University Press, 1974), pp. 158–9.
546
16.4 OBTAINING A SECOND SOLUTION
which is equal to zero if σ = σ2 . As previously, since ∂/∂σ and L are operators
that differentiate with respect to different variables, we can reverse their order to
obtain
∂
[(σ − σ2 )y(z, σ)] = 0
L
at σ = σ2 ,
∂σ
and so the function
∂
[(σ − σ2 )y(z, σ)]
∂σ
(16.30)
σ=σ2
is also a solution of the original ODE Ly = 0, and is in fact the second linearly
independent solution.
Find a second solution to (16.21) using the derivative method.
From (16.23) the recurrence relation (with σ as a parameter) is given by
(n + σ − 1)an = (n + σ)an−1 .
Setting a0 = 1 we find that the coefficients have the particularly simple form an (σ) =
(σ + n)/σ. We therefore consider the function
y(z, σ) = z σ
∞
an (σ)z n = z σ
n=0
∞
σ+n n
z .
σ
n=0
The smaller root of the indicial equation for (16.21) is σ2 = 0, and so from (16.30) a
second, linearly independent, solution to the ODE is given by
#
.
∞
∂
∂
[σy(z, σ)]
=
(σ + n)z n
.
zσ
∂σ
∂σ
σ=0
n=0
σ=0
The derivative with respect to σ is given by
∞
∞
∞
∂
(σ + n)z n = z σ ln z
(σ + n)z n + z σ
zn,
zσ
∂σ
n=0
n=0
n=0
which on setting σ = 0 gives the second solution
y2 (z) = ln z
∞
nz n +
n=0
∞
zn
n=0
z
1
ln z +
(1 − z)2
1−z
1
z
ln z + − 1 .
=
2
(1 − z)
z
=
This second solution is the same as that obtained by the Wronskian method in the previous
subsection except for the addition of some of the first solution. 16.4.3 Series form of the second solution
Using any of the methods discussed above, we can find the general form of the
second solution to the ODE. This form is most easily found, however, using the
547