SCIENCE ONE: MATHEMATICS ASSIGNMENT 3 (Due Oct. 12, 2011)
There are two parts to this assignment. The first part is online and is due at 6 pm on Tue. Oct. 11; you will
need to login at www.mathxl.com. The second part consists of the questions on this page and is due at 10
am on Wed. Oct. 12. You are expected to provide full solutions with complete arguments and justifications.
You will be graded primarily on the correctness, clarity and elegance of your solutions. Your answers must
be typeset or very neatly written. They must be stapled, with your name and student number at the top of
each page.
1. Let m > 0 and ~ > 0 be given constants (m is the mass of a particle/wave in a one-dimensional box of
length 1, ~ is Planck’s constant divided by 2π), and consider the problem of finding a wave function Ψ(x)
that is continuously differentiable on the closed interval [0, 1] (i.e. the derivative Ψ0 (x) is defined and is
a continuous function for 0 ≤ x ≤ 1, where we use the right-hand derivative Ψ0+ (0) at x = 0 and the
left-hand derivative Ψ0− (1) at x = 1, p. 134 in the textbook), is twice continuously differentiable on the
open interval (0, 1) (i.e. the second derivative Ψ00 (x) is defined and is a continuous function for 0 < x < 1),
satisfies the second order ordinary differential equation
Ψ00 +
2mE
Ψ = 0,
~2
0 < x < 1,
(1)
where E is a positive constant (energy), satisfies the boundary condition at x = 0
Ψ(0) = 0,
(2)
Ψ0 (1) + Ψ(1) = 0.
(3)
and also the boundary condition at x = 1
(a) Notice that the trivial function, Ψ(x) = 0 for all 0 ≤ x ≤ 1, is a solution of the problem. Show that
there is a nontrivial solution Ψ(x) satisfying (1), (2) and (3) only if E satisfies
!
√
√
2mE
2mE
= − tan
(4)
~
~
(b) Show graphically that there are infinitely many
discrete positive values of E that satisfy (4) (where
√
2mE
m and ~ are fixed constants). Hint: Put θ = ~ > 0, plot the graphs of two functions of θ, look
for intersections of the graphs.
(c) You could use the Intermediate Value Theorem, applied to some suitable function, to prove that (4)
has infinitely many strictly positive solutions E. Recall that the Intermediate Value Theorem applies
to continuous functions on closed and bounded (i.e. not infinite) intervals, so you would have to avoid
any points of discontinuity. Think about it, but you don’t have to write down a rigourous proof as it
could be quite tedious. Instead, use the method used in the proof of the Intermediate Value Theorem
(called the bisection method), together with a calculator, to approximate the smallest positive value
of E, in terms of m and ~, that satifies (4). Use the bisection method on a suitable continuous
function on a closed and bounded interval. Express your answer as
E=X
~2
,
2m
where X is an explicit numerical approximation that you have carefully determined to be guaranteed
within 0.4 of the exact value.
2. The Chain Rule can be used to make a differential equation “look nicer” so that it is easier to solve.
For example, it can be shown that the wave functions for the hydrogen atom have the form Ψ(r, θ, φ) =
R(r)H(θ)G(φ), where R, H and G are functions of one variable and satisfy ordinary differential equations. The solutions of these ordinary differential equations give the orbital shapes you have seen in the
Chemistry lectures. It can be shown that the ordinary differential equation that H must satisfy is the
rather nasty-looking
d2 H
dH
m2
+
cot
θ
+
`(`
+
1)
−
H = 0,
(5)
dθ2
dθ
sin2 θ
where ` and m are the quantum numbers you learned about in the Chemistry lectures. Now let
H(θ) = P (x),
where
x = cos θ.
(a) Use the Chain Rule to show that
dH
dP
= − sin θ
dθ
dx
and
d2 H
d2 P
dP
= sin2 θ
− cos θ
.
2
dθ
dx2
dx
(b) Use part (a) to help show that if H(θ) is a solution of (5), then P (x) must be a solution of
2
dP
m2
2 d P
+ `(` + 1) −
P = 0.
(1 − x ) 2 − 2x
dx
dx
1 − x2
(6)
(c) It can be shown that (6) has polynomial solutions if m = 0. Verify that i) if m = 0 and ` = 1 then
P (x) = x is a solution of (6) (therefore H(θ) = cos θ is a solution of (5)); ii) if m = 0 and ` = 2 then
P (x) = 3x2 − 1 is a solution of (6) (therefore H(θ) = 3 cos2 θ − 1 is a solution of (5)).
3. Let m > 0 and k > 0 be given constants (m is the mass of a particle attached to a spring that has a Hooke’s
Law constant k), and consider the problem of finding the position function x(t) that is continuously
differentiable on the interval [0, ∞), is twice continuously differentiable on the interval (0, ∞), satisfies
the ordinary differential equation
mx00 + kx = 0, 0 < t < ∞,
and satisfies at t = 0 the two initial conditions
x(0) = x0 ,
x0+ (0) = v0 .
The solution can be expressed as
x(t) = A cos(ωt − φ0 ),
0 ≤ t < ∞.
Suppose m = 2 kg and k = 18 N · m−1 . Find A, ω and φ0 (including their units) if the initial position
and velocity of the mass is:
(a)
(b)
(c)
(d)
x0
x0
x0
x0
= 3 m, and v0 = 1 m · s−1 .
= 3 m, and v0 = −1 m · s−1 .
= −3 m, and v0 = 1 m · s−1 .
= −3 m, and v0 = −1 m · s−1 .