Physics 360/Math 360
MΘF 12:00 - 12:50 PM, Room: Dennis 214
Instructor: Michael Lerner, Dennis 221, Phone: 727-LERNERM
Assignment 4, Due Tuesday Feb 14
Covering the method of Frobenius and our introduction to Bessel functions. I’m trying a new
model for this homework assignment, and I think I’ll likely continue it in future assignments if it’s
successful. You are required to do the first few problems.
After that, I’ve assigned two “challenge problems”. These are longer, harder, and require a bit
more work on your own. They also focus on one of the main skills that we’ll be developing in this
class: figuring out how to model new situations. Although the right answer certainly counts, the
main thing that I’ll be grading on the challenge problems is your method of attacking the problem.
If you cannot solve them on your own, you are encouraged to come talk to me. These problems are
also meant to take up some of the wiggle-room in how long a particular assignment takes you. If
you really only have 6 hours to spend on a problem set and it’s clearly going to take you longer than
that to solve all of the problems, solve the non-challenge problems completely and then write down
the general method you’d use to solve the challenge problems, along with as much of a solution as
you can come up with in time.
1
1.1
Method of Frobenius
MoF 1
a) Find all of the regular singular points of
xy 00 + 2xy 0 + 6ex y = 0
(1)
Answer. In standard notation, we have
P (x) = x Q(x) = 2x R(x) = 6ex
6ex
p(x) = 2 q(x) =
x
So we see that x = 0 is a singular point. We now notice that xp(x) = 2x and x2 q(x) = 6xex are
analytic at zero, so 0 is a regular singular point.
b) Determine the indicial equation and the eponents at the singularity for each regular singular
point.
Answer. Taking the limits as we approach zero, we see that p0 = q0 = 0 so the indicial equation is
just F (r) = r(r − 1) = 0. Then the exponents at the singularity are r = 0, 1.
c) Find the first three nonzero terms in each of the two solutions (not multiples of each other)
about x = 0. This is Boyce and DiPrima §5.6 #14, and their solution is well written.
Answer.
1.2
MoF 2
a) Find all of the regular singular points of
x(x − 1)y 00 + 6x2 y 0 + 3y = 0
(2)
Answer. In standard notation, we have
P (x) = x(x − 1) Q(x) = 6x2 R(x) = 3
3
6x
q(x) =
p(x) =
x−1
x(x − 1)
So we see that x = 0, 1 are singular points.
3x
6x2
and x2 q(x) = x−1
are both zero, and the point is clearly a
For x = 0, the limits of xp(x) = x−1
regular singular point.
3(x−1)2
For x = 1, we consider (x − 1)p(x) = 6x(x−1)
=
6x
→
6
and
= 3(x−1)
→ 0, so x = 1 is
x−1
x(x−1)
x
also a regular singular point.
b) Determine the indicial equation and the eponents at the singularity for each regular singular
point.
Answer. For x = 0, we have the same indicial equation as before, r(r − 1) = 0 so r1 = 1, r2 = 0.
For x = 1, we have F (r) = r(r − 1) + p0 r + q0 = 0 = r(r − 1) + 6r = 0 = r2 − r + 6r = r2 − 5r =
r(r − 5) so r1 = 5, r2 = 0
c) Find the first three nonzero terms in each of the two solutions (not multiples of each other)
about x = 0.
Answer. This is Boyce and DiPrima §5.6 #15, and their solution is well written.
2
Particle in a box
This problem modified slightly from notes graciously provided by Nathan A. Baker The
“particle in a box” is a classic quantum mechanics problems and illustrates how quantum energy
levels change as the size of the confining box changes. All of these examples use the time-independent
Schrödinger equation
~2 2
−
∇ ψn (~x) + V (~x)ψn (~x) = En ψn (~x),
(3)
2m
where ψn is the particle wavefunction, ~ is a fundamental constant, m is the mass of the particle
under consideration, V (~x) is the potential energy of the particle at the location ~x, and En is the
current quantum energy level of the system.
2.1
Particle in a linear box
A particle in a linear box is a reasonable simplified model for electron delocalization along a linear
chain of double bonds or along a symmetric nanotube. This model involves Eq. 3 with V (x) = 0
for 0 ≤ x ≤ L with zero Dirichlet boundary conditions
ψn (0) = ψn (L) = 0.
(4)
Solve Eq. 3 for values of E and ψ which satisfy this problem.
To get you started, especially since we haven’t officially covered ∇2 , Our equation is
−
~2 2
∇ ψn (x) = En ψn (x)
2m
(5)
or
2mEn
ψn (x).
(6)
~2
You will need to write down the solutions to this equation that satisfy the above boundary
conditions. This problem may be review from last semester.
−ψn00 (x) =
Answer. Solutions to this equation have the form
!
!
r
r
2mEn
2mEn
x + c2 sin
x .
ψn (x) = c1 cos
~2
~2
Clearly the boundary conditions can only be satisfied if c1 = 0 and
r
2mEn
= nπ
~2
2mEn
= n2 π 2
~2
~2 n2 π 2
,
En =
2m
where n is a positive integer.
2.2
(7)
(8)
Particle in a disk
A particle in a linear box is a simplified model for electron delocalization in a planar disk-like
molecule; e.g. porphyrins, hemes, nanobowls, etc. This model involves Eq. 3 with V (~x) = 0 for
0 ≤ r ≤ a in cylindrical coordinates with zero Dirichlet boundary conditions
ψn (a) = 0.
(9)
Solve Eq. 3 for values of E and ψ which satisfy this problem.
Again, to get you started, our equation is
~2 2
−
∇ ψn (x) = En ψn (x)
2m
(10)
or
1
2mEn
ψn00 (r) + ψn0 (r) +
ψn (r) = 0.
r
~2
bearing in mind that, in polar coordinates,
∇2 =
∂2
1 ∂
1 ∂2
+
+
.
∂r2 r ∂r r2 ∂θ2
(11)
(12)
Answer. This is a form of Bessel’s equation with solutions
!
!
r
r
2mEn
2mEn
ψn (r) = c1 J0
r + c2 Y 0
r
2
~
~2
(13)
where J0 is the zero-order Bessel function of the first kind and Y0 is the zero-order Bessel function
of the second kind. In order to avoid discontinuous solutions, we set c2 = 0. We can satisfy the
boundary conditions if
r
2mEn
= ξn
~2
2mEn
= ξn2
~2
~2 ξn2
,
(14)
En =
2m
where ξn is the n-th root of the Bessel function J0 .
3
Challenge problem 1
Find two solutions (not multiples of each other) of the Bessel equation of order 32 :
9
x2 y 00 + xy 0 + (x2 − )y = 0,
4
4
x>0
(15)
Challenge problem 2
Boas uses slightly different notation for Bessel functions. Having covered this in Boyce and DiPrima,
you should be able to read through the relevant parts of Boas Chapter 12 §12-18 to do the following:
Read through Boas Ch. 12 §18: The Lengthening Pendulum. Do Boas Ch. 12 §18 Problem
9, the “shortening pendulum”. You do not need to do the very last part (restating the results of
problem 8), but it’s physically informative to do so.