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Exercises

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Exercises
SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
coefficient of the highest power z N ; such a power now exists because of our
assumed form of solution.
By assuming a polynomial solution find the values of λ in (16.34) for which such a solution
exists.
We assume a polynomial solution to (16.34) of the form y =
form into (16.34) we find
N
N
n=0
an z n . Substituting this
n(n − 1)an z n−2 − 2znan z n−1 + λan z n = 0.
n=0
Now, instead of starting with the lowest power of z, we start with the highest. Thus,
demanding that the coefficient of z N vanishes, we require −2N + λ = 0, i.e. λ = 2N, as we
found in the previous example. By demanding that the coefficient of a general power of z
is zero, the same recurrence relation as above may be derived and the solutions found. 16.6 Exercises
16.1
Find two power series solutions about z = 0 of the differential equation
(1 − z 2 )y − 3zy + λy = 0.
16.2
Deduce that the value of λ for which the corresponding power series becomes an
Nth-degree polynomial UN (z) is N(N + 2). Construct U2 (z) and U3 (z).
Find solutions, as power series in z, of the equation
4zy + 2(1 − z)y − y = 0.
16.3
Identify one of the solutions and verify it by direct substitution.
Find power series solutions in z of the differential equation
zy − 2y + 9z 5 y = 0.
16.4
Identify closed forms for the two series, calculate their Wronskian, and verify
that they are linearly independent. Compare the Wronskian with that calculated
from the differential equation.
Change the independent variable in the equation
d2 f
df
+ 4f = 0
(∗)
+ 2(z − a)
dz 2
dz
from z to x = z − α, and find two independent series solutions, expanded about
x = 0, of the resulting equation. Deduce that the general solution of (∗) is
f(z, α) = A(z − α)e−(z−α) + B
2
16.5
∞
(−4)m m!
(z − α)2m ,
(2m)!
m=0
with A and B arbitrary constants.
Investigate solutions of Legendre’s equation at one of its singular points as
follows.
(a) Verify that z = 1 is a regular singular point of Legendre’s equation and that
the indicial equation for a series solution in powers of (z − 1) has roots 0
and 3.
(b) Obtain the corresponding recurrence relation and show that σ = 0 does not
give a valid series solution.
550
16.6 EXERCISES
(c) Determine the radius of convergence R of the σ = 3 series and relate it to
the positions of the singularities of Legendre’s equation.
16.6
Verify that z = 0 is a regular singular point of the equation
z 2 y − 32 zy + (1 + z)y = 0,
and that the indicial equation has roots 2 and 1/2. Show that the general solution
is given by
∞
(−1)n (n + 1)22n z n
(2n + 3)!
n=0
∞
(−1)n 22n z n
z 1/2 1/2
3/2
+ b0 z + 2z −
.
4 n=2 n(n − 1)(2n − 3)!
y(z) = 6a0 z 2
16.7
16.8
Use the derivative method to obtain, as a second solution of Bessel’s equation
for the case when ν = 0, the following expression:
n
∞
(−1)n 1 z 2n
,
J0 (z) ln z −
(n!)2
r
2
n=1
r=1
given that the first solution is J0 (z), as specified by (18.79).
Consider a series solution of the equation
zy − 2y + yz = 0
(∗)
about its regular singular point.
(a) Show that its indicial equation has roots that differ by an integer but that
the two roots nevertheless generate linearly independent solutions
y1 (z) = 3a0
y2 (z) = a0
∞
(−1)n+1 2nz 2n+1
,
(2n + 1)!
n=1
∞
(−1)n+1 (2n − 1)z 2n
.
(2n)!
n=0
(b) Show that y1 (z) is equal to 3a0 (sin z − z cos z) by expanding the sinusoidal
functions. Then, using the Wronskian method, find an expression for y2 (z) in
terms of sinusoids. You will need to write z 2 as (z/ sin z)(z sin z) and integrate
by parts to evaluate the integral involved.
(c) Confirm that the two solutions are linearly independent by showing that
their Wronskian is equal to −z 2 , as would be expected from the form of (∗).
16.9
Find series solutions of the equation y − 2zy − 2y = 0. Identify one of the series
as y1 (z) = exp z 2 and verify this by direct substitution. By setting y2 (z) = u(z)y1 (z)
and solving the resulting equation for u(z), find an explicit form for y2 (z) and
deduce that
x
∞
n!
2
2
e−v dv = e−x
(2x)2n+1 .
2(2n + 1)!
0
n=0
16.10
Solve the equation
z(1 − z)
d2 y
dy
+ (1 − z)
+ λy = 0
dz 2
dz
as follows.
(a) Identify and classify its singular points and determine their indices.
551
SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
(b) Find one series solution in powers of z. Give a formal expression for a
second linearly independent solution.
(c) Deduce the values of λ for which there is a polynomial solution PN (z) of
degree N. Evaluate the first four polynomials, normalised in such a way that
PN (0) = 1.
16.11
Find the general power series solution about z = 0 of the equation
z
16.12
dy
d2 y
4
+ (2z − 3)
+ y = 0.
dz 2
dz
z
Find the radius of convergence of a series solution about the origin for the
equation (z 2 + az + b)y + 2y = 0 in the following cases:
(a) a = 5, b = 6;
16.13
16.14
(b) a = 5, b = 7.
Show that if a and b are real and 4b > a2 , then the radius of convergence is
always given by b1/2 .
For the equation y + z −3 y = 0, show that the origin becomes a regular singular
point if the independent variable is changed from z to x = 1/z. Hence find a
−n
series solution of the form y1 (z) = ∞
0 an z . By setting y2 (z) = u(z)y1 (z) and
expanding the resulting expression for du/dz in powers of z −1 , show that y2 (z)
has the asymptotic form
ln z
,
y2 (z) = c z + ln z − 12 + O
z
where c is an arbitrary constant.
Prove that the Laguerre equation,
z
d2 y
dy
+ (1 − z)
+ λy = 0,
dz 2
dz
has polynomial solutions LN (z) if λ is a non-negative integer N, and determine
the recurrence relationship for the polynomial coefficients. Hence show that an
expression for LN (z), normalised in such a way that LN (0) = N!, is
LN (z) =
16.15
N
(−1)n (N!)2 n
z .
(N − n)!(n!)2
n=0
Evaluate L3 (z) explicitly.
The origin is an ordinary point of the Chebyshev equation,
(1 − z 2 )y − zy + m2 y = 0,
n
which therefore has series solutions of the form z σ ∞
0 an z for σ = 0 and σ = 1.
(a) Find the recurrence relationships for the an in the two cases and show that
there exist polynomial solutions Tm (z):
(i) for σ = 0, when m is an even integer, the polynomial having 12 (m + 2)
terms;
(ii) for σ = 1, when m is an odd integer, the polynomial having 12 (m + 1)
terms.
(b) Tm (z) is normalised so as to have Tm (1) = 1. Find explicit forms for Tm (z)
for m = 0, 1, 2, 3.
552
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