Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 3 pdf

Differentiating this relation yields
f

(x) ∼ −
1
x
2
as x → ∞.
However, this is not true since
f

(x) = −
1
x
2
−
2
x
3
sin(x
3
) + 2 cos(x
3
)
∼ −
1
x
2
as x → ∞.
The Controlling Factor. The controlling factor is the most rapidly varying factor in an asymptotic relation.

Consider a function f(x) that is asymptotic to x
2
e
x
as x goes to infinity. The controlling factor is
e
x
. For a few
examples of this,
• x log x has the controlling factor x as x → ∞.
• x
−2
e
1/x
has the controlling factor
e
1/x
as x → 0.
• x
−1
sin x has the controlling factor sin x as x → ∞.
The Leading Behavior. Consider a function that is asymptotic to a sum of terms.
f(x) ∼ a
0
(x) + a
1
(x) + a
2
(x) + ··· , as x → x
0

.
where
a
0
(x)  a
1
(x)  a
2
(x)  ··· , as x → x
0
.
The first term in the sum is the leading order behavior. For a few examples,
• For sin x ∼ x −x
3
/6 + x
5
/120 − ··· as x → 0, the leading order behavior is x.
• For f(x) ∼
e
x
(1 − 1/x + 1/x
2
− ···) as x → ∞, the leading order behavior is
e
x
.
1254
24.2 Leading Order Behavior of Differential Equations
It is often useful to know the leading order behavior of the solutions to a differential equation. If we are considering
a regular point or a regular singular point, the approach is straight forward. We simply use a Taylor expansion or the

Frobenius method. However, if we are considering an irregular singular point, we will have to be a little more creative.
Instead of an all encompassing theory like the Frobe niu s method which always gives us the solution, we will use a
heuristic approach that usually gives us the solution.
Example 24.2.1 Consider the Airy equation
y

= xy.
We
1
would like to know how the solutions of this equation b ehave as x → +∞. First we need to classify the point at
infinity. The change of variables
x =
1
t
, y(x) = u(t),
d
dx
= −t
2
d
dt
,
d
2
dx
2
= t
4
d
2

dt
2
+ 2t
3
d
dt
yields
t
4
u

+ 2t
3
u

=
1
t
u
u

+
2
t
u

−
1
t
5

u = 0.
Since the equation for u has an irregular singular point at zero, the equation for y has an irregular singular point at
infinity.
1
Using ”We” may be a bit presumptuous on my part. Even if you don’t particularly want to know how the solutions behave, I
urge you to just play along. This is an interesting section, I promise.
1255
The Controlling Factor. Since the solutions at irregular singular points often have exponential behavior, we make
the substitution y =
e
s(x)
into the differential equation for y.
d
2
dx
2

e
s

= x
e
s

s

+ (s

)
2


e
s
= x
e
s
s

+ (s

)
2
= x
The Dominant Balance. Now we have a differential equation for s that appears harder to solve than our equation
for y. However, we did not introduce the substitution in order to obtain an equation that we could solve exactly. We
are looking for an equation that we can solve approximately in the limit as x → ∞. If one of the terms in the equation
for s is much smaller that the other two as x → ∞, then dropping that term and solving the simpler equation may give
us an approximate solution. If one of the terms in the equation for s is m uch smaller than the others then we say that
the remaining terms form a dominant balance in the limit as x → ∞.
Assume that the s

term is much smaller that the others, s

 (s

)
2
, x as x → ∞. This gives us
(s


)
2
∼ x
s

∼ ±
√
x
s ∼ ±
2
3
x
3/2
as x → ∞.
Now let’s check our assumption that the s

term is small. Assuming that we can differentiate the asymptotic relation
s

∼ ±
√
x, we obtain s

∼ ±
1
2
x
−1/2
as x → ∞.
s


 (s

)
2
, x → x
−1/2
 x as x → ∞
Thus we see that the behavior we found for s is consistent with our assumption. The controlling factors for solutions
to the Airy equation are exp(±
2
3
x
3/2
) as x → ∞.
1256
The Leading Order Behavior of the Decaying Solution. Let’s find the leading order behavior as x goes to
infinity of the solution with the controlling factor exp(−
2
3
x
3/2
). We substitute
s(x) = −
2
3
x
3/2
+ t(x), where t(x)  x
3/2

as x → ∞
into the differential equation for s.
s

+ (s

)
2
= x
−
1
2
x
−1/2
+ t

+ (−x
1/2
+ t

)
2
= x
t

+ (t

)
2
− 2x

1/2
t

−
1
2
x
−1/2
= 0
Assume that we can differentiate t  x
3/2
to obtain
t

 x
1/2
, t

 x
−1/2
as x → ∞.
Since t

 −
1
2
x
−1/2
we drop the t


term. Also, t

 x
1/2
implies that (t

)
2
 −2x
1/2
t

, so we drop the (t

)
2
term.
This gives us
−2x
1/2
t

−
1
2
x
−1/2
∼ 0
t


∼ −
1
4
x
−1
t ∼ −
1
4
log x + c
t ∼ −
1
4
log x as x → ∞.
Checking our assumptions about t,
t

 x
1/2
→ x
−1
 x
1/2
t

 x
−1/2
→ x
−2
 x
−1/2

1257
we see that the behavior of t is consistent with our assumptions.
So far we have
y(x) ∼ exp

−
2
3
x
3/2
−
1
4
log x + u(x)

as x → ∞,
where u(x)  log x as x → ∞. To continue, we s ubs titute t(x) = −
1
4
log x + u(x) into the differential equation for
t(x).
t

+ (t

)
2
− 2x
1/2
t


−
1
2
x
−1/2
= 0
1
4
x
−2
+ u

+

−
1
4
x
−1
+ u


2
− 2x
1/2

−
1
4

x
−1
+ u


−
1
2
x
−1/2
= 0
u

+ (u

)
2
+

−
1
2
x
−1
− 2x
1/2

u

+

5
16
x
−2
= 0
Assume that we can differentiate the asymptotic relation for u to obtain
u

 x
−1
, u

 x
−2
as x → ∞.
We know that −
1
2
x
−1
u

 −2x
1/2
u

. Using our assumptions,
u

 x

−2
→ u


5
16
x
−2
u

 x
−1
→ (u

)
2

5
16
x
−2
.
Thus we obtain
−2x
1/2
u

+
5
16

x
−2
∼ 0
u

∼
5
32
x
−5/2
u ∼ −
5
48
x
−3/2
+ c
u ∼ c as x → ∞.
1258
Since u = c + o(1),
e
u
=
e
c
+o(1). The behavior of y is
y ∼ x
−1/4
exp

−

2
3
x
3/2

(
e
c
+o(1)) as x → ∞.
Thus the full leading order behavior of the decaying solution is
y ∼ (const)x
−1/4
exp

−
2
3
x
3/2

as x → ∞.
You can show that the leading behavior of the exponentially growing solution is
y ∼ (const)x
−1/4
exp

2
3
x
3/2


as x → ∞.
Example 24.2.2 The Modified Bessel Equation. Consider the modified Bessel equation
x
2
y

+ xy

− (x
2
+ ν
2
)y = 0.
We would like to know how the solutions of this equation behave as x → +∞. First we need to classify the point at
infinity. The change of variables x =
1
t
, y(x) = u(t) yields
1
t
2
(t
4
u

+ 2t
3
u


) +
1
t
(−t
2
u

) −

1
t
2
+ ν
2

u = 0
u

+
1
t
u

−

1
t
4
+
ν

2
t
2

u = 0
Since u(t) has an irregular singular point at t = 0, y(x) has an irregular singular point at infinity.
The Controlling Factor. Since the solutions at irregular singular points often have exponential behavior, we make
the substitution y =
e
s(x)
into the differential equation for y.
x
2
(s

+ (s

)
2
)
e
s
+xs

e
s
−(x
2
+ ν
2

)
e
s
= 0
s

+ (s

)
2
+
1
x
s

− (1 +
ν
2
x
2
) = 0
1259
We make the assumption that s

 (s

)
2
as x → ∞ and we know that ν
2

/x
2
 1 as x → ∞. Thus we drop these
two terms from the equation to obtain an approximate equation for s.
(s

)
2
+
1
x
s

− 1 ∼ 0
This is a quadratic equation for s

, so we can solve it exactly. However, let us try to simplify the equation even further.
Assume that as x goes to infinity one of the three terms is much smaller that the other two. If this is the case, there
will be a balance between the two dominant terms and we can neglect the third. Let’s check the three possib ili ties.
1.
1 is small. → (s

)
2
+
1
x
s

∼ 0 → s


∼ −
1
x
, 0
1 
1
x
2
, 0 as x → ∞ so this balance is inconsistent.
2.
1
x
s

is small. → (s

)
2
− 1 ∼ 0 → s

∼ ±1
This balance is consistent as
1
x
 1 as x → ∞.
3.
(s

)

2
is small. →
1
x
s

− 1 ∼ 0 → s

∼ x
This balance is not consistent as x
2
 1 as x → ∞.
The only dominant balance that makes sense leads to s

∼ ±1 as x → ∞. Integrating this relationship,
s ∼ ±x + c
∼ ±x as x → ∞.
Now let’s see if our assumption that we made to get the simplified equation for s is valid. Assuming that we can
differentiate s

∼ ±1, s

 (s

)
2
becomes
d
dx


± 1 + o(1)



± 1 + o(1)

2
0 + o(1/x)  1
1260
Thus we see that the behavior we obtained for s is consistent with our initial assumption.
We have found two controlling factors,
e
x
and
e
−x
. This is a good sign as we know that there must be two line arly
independent solutions to the equation.
Leading Order Behavior. Now let’s find the full leading behavior of the solution with the controlling factor
e
−x
.
In order to find a better approximation for s, we substitute s(x) = −x + t(x), where t(x)  x as x → ∞, into the
differential equation for s.
s

+ (s

)
2

+
1
x
s

−

1 +
ν
2
x
2

= 0
t

+ (−1 + t

)
2
+
1
x
(−1 + t

) −

1 +
ν
2

x
2

= 0
t

+ (t

)
2
+

1
x
− 2

t

−

1
x
+
ν
2
x
2

= 0
We know that

1
x
 2 and
ν
2
x
2

1
x
as x → ∞. Dropping these terms from the equation yields
t

+ (t

)
2
− 2t

−
1
x
∼ 0.
Assuming that we can differentiate the asymptotic relation for t, we obtain t

 1 and t


1
x

as x → ∞. We can
drop t

. Since t

vanishes as x goes to infinity, (t

)
2
 t

. Thus we are left with
−2t

−
1
x
∼ 0, as x → ∞.
Integrating this relationship,
t ∼ −
1
2
log x + c
∼ −
1
2
log x as x → ∞.
1261
Checking our assumptions about the behavior of t,
t


 1 → −
1
2x
 1
t


1
x
→
1
2x
2

1
x
we see that the solution is consistent with our assumptions.
The leading order behavior to the solution with controlling factor
e
−x
is
y(x) ∼ exp

−x −
1
2
log x + u(x)

= x

−1/2
e
−x+u(x)
as x → ∞,
where u(x)  log x. We substitute t = −
1
2
log x + u(x) into the differential equation for t in order to find the
asymptotic behavior of u.
t

+ (t

)
2
+

1
x
− 2

t

−

1
x
+
ν
2

x
2

= 0
1
2x
2
+ u

+

−
1
2x
+ u


2
+

1
x
− 2

−
1
2x
+ u



−

1
x
+
ν
2
x
2

= 0
u

+ (u

)
2
− 2u

+
1
4x
2
−
ν
2
x
2
= 0
Assuming that we can differentiate the asymptotic relation for u, u



1
x
and u


1
x
2
as x → ∞. Thus we see that
we can neglect the u

and (u

)
2
terms.
−2u

+

1
4
− ν
2

1
x
2

∼ 0
u

∼
1
2

1
4
− ν
2

1
x
2
u ∼
1
2

ν
2
−
1
4

1
x
+ c
u ∼ c as x → ∞
1262

Since u = c + o(1), we can expand
e
u
as
e
c
+o(1). Thus we can write the leading order behavior as
y ∼ x
−1/2
e
−x
(
e
c
+o(1)).
Thus the full leading order behavior is
y ∼ (const)x
−1/2
e
−x
as x → ∞.
You can verify that the solution with the controlling factor
e
x
has the leading order behavior
y ∼ (const)x
−1/2
e
x
as x → ∞.

Two linearly independent solutions to the mod ified Bessel equation are the modified Bessel functions, I
ν
(x) and
K
ν
(x). These functions have the asymptotic behavior
I
ν
(x) ∼
1
√
2πx
e
x
, K
ν
(x) ∼
√
π
√
2x
e
−x
as x → ∞.
In Figure 24.1 K
0
(x) is plotted in a solid line and
√
π
√

2x
e
−x
is plotted in a dashed line. We se e that the leading order
behavior of the solution as x goes to infinity gives a good approximation to the behavior even for fairly small values of
x.
24.3 Integration by Parts
Example 24.3.1 The complementary error function
erfc(x) =
2
√
π

∞
x
e
−t
2
dt
1263
0
1
2
3
4
5
0.25
0.5
0.75
1

1.25
1.5
1.75
2
Figure 24.1: Plot of K
0
(x) and it’s leading order behavior.
is used in statistics for its relation to the normal probability distribution. We would like to find an approximation to
erfc(x) for large x. Using integration by parts,
erfc(x) =
2
√
π

∞
x

−1
2t


−2t
e
−t
2

dt
=
2
√

π

−1
2t
e
−t
2

∞
x
−
2
√
π

∞
x
1
2
t
−2
e
−t
2
dt
=
1
√
π
x

−1
e
−x
2
−
1
√
π

∞
x
t
−2
e
−t
2
dt.
1264
We examine the residual integral in this expression.
1
√
π

∞
x
t
−2
e
−t
2

dt ≤
−1
2
√
π
x
−3

∞
x
−2t
e
−t
2
dt
=
1
2
√
π
x
−3
e
−x
2
.
Thus we see that
1
√
π

x
−1
e
−x
2

1
√
π

∞
x
t
−2
e
−t
2
dt as x → ∞.
Therefore,
erfc(x) ∼
1
√
π
x
−1
e
−x
2
as x → ∞,
and we expect that

1
√
π
x
−1
e
−x
2
would be a good approximation to erfc(x) for large x. In Figure 24.2 log(erfc(x)) is
graphed i n a solid line and log

1
√
π
x
−1
e
−x
2

is graphed in a dashed line. We see that this first approximation to the
error function gives very good results even for moderate values of x. Table 24.1 gives the error in this first approximation
for various values of x.
If we continue integrating by parts, we might get a better approximation to the complementary error function.
erfc(x) =
1
√
π
x
−1

e
−x
2
−
1
√
π

∞
x
t
−2
e
−t
2
dt
=
1
√
π
x
−1
e
−x
2
−
1
√
π


−
1
2
t
−3
e
−t
2

∞
x
+
1
√
π

∞
x
3
2
t
−4
e
−t
2
dt
=
1
√
π

e
−x
2

x
−1
−
1
2
x
−3

+
1
√
π

∞
x
3
2
t
−4
e
−t
2
dt
=
1
√

π
e
−x
2

x
−1
−
1
2
x
−3

+
1
√
π

−
3
4
t
−5
e
−t
2

∞
x
−

1
√
π

∞
x
15
4
t
−6
e
−t
2
dt
=
1
√
π
e
−x
2

x
−1
−
1
2
x
−3
+

3
4
x
−5

−
1
√
π

∞
x
15
4
t
−6
e
−t
2
dt
1265
0.5
1
1.5
2
2.5 3
-10
-8
-6
-4

-2
2
Figure 24.2: Logarithm of the Approximation to the Complementary Error Function.
The error in approximating erfc(x) with the first three terms is given in Table 24.1. We see that for x ≥ 2 the three
terms give a much better approximation to erfc(x) than just the first term.
At this p oint you might guess that you could continue this process indefinitely. By repeated application of integration
by parts, you can obtain the series expansion
erfc(x) =
2
√
π
e
−x
2
∞

n=0
(−1)
n
(2n)!
n!(2x)
2n+1
.
1266
x erfc(x) One Term Relative Error Three Term Relative Error
1 0.157 0.3203 0.6497
2 0.00468 0.1044 0.0182
3 2.21 × 10
−5
0.0507 0.0020

4 1.54 × 10
−8
0.0296 3.9 · 10
−4
5 1.54 × 10
−12
0.0192 1.1 · 10
−4
6 2.15 × 10
−17
0.0135 3.7 · 10
−5
7 4.18 × 10
−23
0.0100 1.5 · 10
−5
8 1.12 × 10
−29
0.0077 6.9 · 10
−6
9 4.14 × 10
−37
0.0061 3.4 · 10
−6
10 2.09 × 10
−45
0.0049 1.8 · 10
−6
Table 24.1:
This is a Taylor expansion about infinity. Let’s fin d the radius of convergence.

lim
n→∞




a
n+1
(x)
a
n
(x)




< 1 → lim
n→∞




(−1)
n+1
(2(n + 1))!
(n + 1)!(2x)
2(n+1)+1
n!(2x)
2n+1
(−1)

n
(2n)!




< 1
→ lim
n→∞




(2n + 2)(2n + 1)
(n + 1)(2x)
2




< 1
→ lim
n→∞




2(2n + 1)
(2x)
2





< 1
→




1
x




= 0
Thus we see that our series diverges for all x. Our conventional mathematical sense would tell us that this series is
useless, however we will see that this series is very useful as an asymptotic expansion of erfc(x).
1267
Say we are working with a convergent series expansion of some function f(x).
f(x) =
∞

n=0
a
n
(x)
For fixed x = x
0

,
f(x
0
) −
N

n=0
a
n
(x
0
) → 0 as N → ∞.
For an asymptotic series we have a quite different behavior. If g(x) is asymptotic to

∞
n=0
b
n
(x) as x → x
0
then for
fixed N,
g(x) −
N

0
b
n
(x)  b
N

(x) as x → x
0
.
For the complementary error function,
For fixed N, erfc(x) −
2
√
π
e
−x
2
N

n=0
(−1)
n
(2n)!
n!(2x)
2n+1
 x
−2N−1
as x → ∞.
We say that the error function is asymptotic to the series as x goes to infinity.
erfc(x) ∼
2
√
π
e
−x
2

∞

n=0
(−1)
n
(2n)!
n!(2x)
2n+1
as x → ∞
In Figure 24.3 the logarithm of the difference between the one term, ten term and twenty term approximations and
the complementary error function are graphed in coarse, medium, and fine dashed lines, respectively.
*Optimal Asymptotic Series. Of the three approximations, the one term is best for x  2, the ten term is
best for 2  x  4, and the twenty term is best for 4  x. This leads us to the concept of an optimal asymptotic
approximation. An optimal asymptotic approximation contains the number of terms in the series that best approximates
the true behavior.
1268
1
2
3
4
5
6
-60
-40
-20
Figure 24.3: log(error in approximation)
In Figure 24.4 we see a plot of the number of terms in the approximation versus the logarithm of the error for x = 3.
Thus we see that the optimal asymptotic approximation is the first nine terms. After nine terms the error gets larger.
It was inevitable that the error would start to grow after some point as the series diverges for all x.
A good rule of thumb for finding the optimal series is to find the smallest term in the series and take all of the

terms up to but not including the smallest term as the optimal approximation. This makes sen se, because the n
th
term
is an approximation of the error incurred by using the first n − 1 terms. In Figure 24.5 there is a plot of n versus the
logarithm of the n
th
term in the asymptotic expansion of erfc(3). We see that the tenth term is the smallest. Thus, in
this case, our rule of thumb predicts the actual optimal series.
1269
5
10
15
20 25
-18
-16
-14
-12
Figure 24.4: The logarithm of the error in using n terms.
24.4 Asymptotic Series
A function f(x) has an asymptotic series expansion about x = x
0
,

∞
n=0
a
n
(x), if
f(x) −
N


n=0
a
n
(x)  a
N
(x) as x → x
0
for all N.
An asymptotic series may be convergent or divergent. Most of the asymptotic series you encounter will be divergent.
If the series is convergent, then we have that
f(x) −
N

n=0
a
n
(x) → 0 as N → ∞ for fixed x.
1270
5
10
15
20 25
-16
-14
-12
Figure 24.5: The logarithm of the n
th
term in the expansion for x = 3.
Let 

n
(x) be some set of gauge functions. The example that we are most familiar with is 
n
(x) = x
n
. If we say that
∞

n=0
a
n

n
(x) ∼
∞

n=0
b
n

n
(x),
then this means that a
n
= b
n
.
1271
24.5 Asymptotic Expansions of Differential Equations
24.5.1 The Parabolic Cylinder Equation.

Controlling Factor. Let us examine the behavior of the bounded solution of the parabolic cylinder equation as
x → +∞.
y

+

ν +
1
2
−
1
4
x
2

y = 0
This equation has an irregular singular point at infinity. With the substitution y =
e
s
, the equation becomes
s

+ (s

)
2
+ ν +
1
2
−

1
4
x
2
= 0.
We know that
ν +
1
2

1
4
x
2
as x → +∞
so we drop this term from the equation. Let us make the assumption that
s

 (s

)
2
as x → +∞.
Thus we are left with the equation
(s

)
2
∼
1

4
x
2
s

∼ ±
1
2
x
s ∼ ±
1
4
x
2
+ c
s ∼ ±
1
4
x
2
as x → +∞
Now let’s check if our assumption is consistent. Substituting into s

 (s

)
2
yields 1/2  x
2
/4 as x → +∞ which

is true. Since the equation for y is second order, we would expect that there are two different behaviors as x → +∞.
This is confirmed by the fact that we found two behaviors for s. s ∼ −x
2
/4 corresponds to the solution that is bounded
at +∞. Thus the controlling factor of the leading behavior is
e
−x
2
/4
.
1272
Leading Order Behavior. Now we attempt to get a better approximation to s. We make the substitution
s = −
1
4
x
2
+ t(x) into the equation for s where t  x
2
as x → +∞.
−
1
2
+ t

+
1
4
x
2

− xt

+ (t

)
2
+ ν +
1
2
−
1
4
x
2
= 0
t

− xt

+ (t

)
2
+ ν = 0
Since t  x
2
, we assume that t

 x and t


 1 as x → +∞. Note that this in only an assumption since it is not
always valid to differentiate an asymptotic relation. Thus (t

)
2
 xt

and t

 xt

as x → +∞; we drop these terms
from the equation.
t

∼
ν
x
t ∼ ν log x + c
t ∼ ν log x as x → +∞
Checking our assumptions for the derivatives of t,
t

 x →
1
x
 x t

 1 →
1

x
2
 1,
we see that they were consistent. Now we wish to refine our approximation for t with the substitution t(x) =
ν log x + u(x). So far we have that
y ∼ exp

−
x
2
4
+ ν log x + u(x)

= x
ν
exp

−
x
2
4
+ u(x)

as x → +∞.
We can try and determine u(x) by substituting the expression t(x) = ν log x + u(x) into the equation for t.
−
ν
x
2
+ u


− (ν + xu

) +
ν
2
x
2
+
2ν
x
u

+ (u

)
2
+ ν = 0
After suitable simplification, this equation becomes
u

∼
ν
2
− ν
x
3
as x → +∞
1273
Integrating this asymptotic relation,

u ∼
ν − ν
2
2x
2
+ c as x → +∞.
Notice that
ν−ν
2
2x
2
 c as x → +∞; thus this procedure fails to give us the behavior of u(x). Further refinements to
our approximation for s go to a constant value as x → +∞. Thus we have that the leading behavior is
y ∼ cx
ν
exp

−
x
2
4

as x → +∞
Asymptotic Expansion Since we have factored off the singular behavior of y, we might expect that what is left
over is well behaved enough to be expanded in a Taylor series about infinity. Let us assume that we can expand the
solution for y in the form
y(x) ∼ x
ν
exp


−
x
2
4

σ(x) = x
ν
exp

−
x
2
4

∞

n=0
a
n
x
−n
as x → +∞
where a
0
= 1. Di fferentiating y = x
ν
exp

−
x

2
4

σ(x),
y

=

νx
ν−1
−
1
2
x
ν+1

e
−x
2
/4
σ(x) + x
ν
e
−x
2
/4
σ

(x)
y


=

ν(ν − 1)x
ν−2
−
1
2
νx
ν
−
1
2
(ν + 1)x
ν
+
1
4
x
ν+2

e
−x
2
/4
σ(x) + 2

νx
ν−1
−

1
2
x
ν+1

e
−x
2
/4
σ

(x)
+ x
ν
e
−x
2
/4
σ

(x).
Substituting this into the differential equation for y,

ν(ν − 1)x
−2
− (ν +
1
2
) +
1

4
x
2

σ(x) + 2

νx
−1
−
1
2
x

σ

(x) + σ

(x) +

ν +
1
2
−
1
4
x
2

σ(x) = 0
σ


(x) + (2νx
−1
− x)σ

(x) + ν(ν − 1)x
−2
σ = 0
x
2
σ

(x) + (2νx − x
3
)σ

(x) + ν(ν − 1)σ(x) = 0.
1274
Differentiating the expression for σ(x),
σ(x) =
∞

n=0
a
n
x
−n
σ

(x) =

∞

n=1
−na
n
x
−n−1
=
∞

n=−1
−(n + 2)a
n+2
x
−n−3
σ

(x) =
∞

n=1
n(n + 1)a
n
x
−n−2
.
Substituting this into the differential equation for σ(x),
∞

n=1

n(n + 1)a
n
x
−n
+ 2ν
∞

n=1
−na
n
x
−n
−
∞

n=−1
−(n + 2)a
n+2
x
−n
+ ν(ν − 1)
∞

n=0
a
n
x
−n
= 0.
Equating the coefficient of x

1
to zero yields
a
1
x = 0 → a
1
= 0.
Equating the coefficient of x
0
,
2a
2
+ ν(ν − 1)a
0
= 0 → a
2
= −
1
2
ν(ν − 1).
From the coefficient of x
−n
for n > 0,
n(n + 1)a
n
− 2νna
n
+ (n + 2)a
n+2
+ ν(ν − 1)a

n
= 0
(n + 2)a
n+2
= −[n(n + 1) − 2νn + ν(ν −1)]a
n
(n + 2)a
n+2
= −[n
2
+ n − 2νn + ν(ν −1)]a
n
(n + 2)a
n+2
= −(n − ν)(n − ν + 1)a
n
.
1275
Thus the recursion formula for the a
n
’s is
a
n+2
= −
(n − ν)(n − ν + 1)
n + 2
a
n
, a
0

= 1, a
1
= 0.
The first few terms in σ(x) are
σ(x) ∼ 1 −
ν(ν − 1)
2
1
1!
x
−2
+
ν(ν − 1)(ν −2)(ν − 3)
2
2
2!
x
−4
− ··· as x → +∞
If we check the radius of convergence of this series
lim
n→∞




a
n+2
x
−n−2

a
n
x
−n




< 1 → lim
n→∞




−
(n − ν)(n − ν + 1)
n + 2
x
−2




< 1
→
1
x
= 0
we see that the radius of convergence is zero. Thus if ν = 0, 1, 2, . . . our asymptotic expansion for y
y ∼ x

ν
e
−x
2
/4

1 −
ν(ν − 1)
2
1
1!
x
−2
+
ν(ν − 1)(ν −2)(ν − 3)
2
2
2!
x
−4
− ···

diverges for all x. However this solution is still very useful. If we only use a finite number of terms, we will get a very
good numerical approximation for large x.
In Figure 24.6 the one term, two term, and three term asymptotic approximations are shown in rough, medium, and
fine dashing, respectively. Th e numerical solu tion is plotted in a solid line.
1276
1
2
3

4
5
6
-2
2
4
6
Figure 24.6: Asymptotic Approximations to the Parabolic Cylinder Function.
1277