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18.034 SOLUTIONS TO PROBLEM SET 8
Due date: Friday, April 23 in lecture. Late work will be accepted only with a medical note or for
another Instituteapproved reason. You are strongly encouraged to work with others, but the final
writeup should be entirely your own and based on your own understanding.
This problem set is essentially a reading assignment. I had originally intended to present the material
in the problem set in lecture. However, this material is less relevant to other topics in this course,
and there is no time to present it in lecture. Each student will receive 10 points simply for reading
through this problem set. At several places, you are asked to workthrough and writeup details in
the derivation. You will turn in this writeup to be graded.
Problem 1(30 points) A collection of N identical particles of mass m 0 = · · · = mN − 1 = m
are allowed to oscillate about their equilibrium positions. Denote by x 0 (t), x 1 (t),... , xN − 1 (t) the
displacement of the masses from equilibrium. The mass m 0 is connected to a motionless base by a
spring. It is also connected to mass m 1 by a spring. For i = 1,... , N − 2, mass mi is connected
to mass mi− 1 and mass mi+1 by springs. Finally, mass mN − 1 is connected to mass mN − 2 and to a
motionless base by a spring. Each spring is identical and has spring constant κ. Time is measured
in units so that the frequency κ/m equals 1. The equations of motion for this system are,
⎡ ⎤
x
�� = AN x, x =
x 0
x 1
. . .
xN − 1
where AN is the N × N matrix,
⎡ ⎤ − 2 1 0 0... 0 0
1 − 2 1 0... 0 0
0 1 − 2 1... 0 0
0 0 1 − 2... 0 0
..
AN =
. . ..^ ..
In other words,
(AN )i,j =
1 , j = i + 1,
− 2 , j = i,
1 , j = i − 1 ,
0 , otherwise
In this problem, you will determine the general solution of this system of linear differential equa
tions. Because of the special nature of this problem (namely, every eigenvalue of A has multiplicity
1), it is not necessary to reduce to a system of firstorder linear differential equations.
(a), Step 1(10 points) In this step, you will find, for each integer n, all the roots of a polynomial
pn that is defined inductively below. Please read the whole derivation. In your writeup, only fill in
the missing steps in the third paragraph below. You do not need to fill in the missing steps in the
other paragraphs (but you are encouraged to work them out for yourself).
1
Let u be an indeterminate. Let p 0 , p 1 ,... , pn,... be a sequence of real polynomials in u that satisfy
the linear difference equation, ⎧ ⎨
pn+2 − 2 upn+1 + pn = 0,
p 0 = 1, (1)
p 1 = 2u
3 = 4u
2 where n varies among all nonnegative integers. In particular, p 2 − 1, and p 3 = 8u
is a polynomial with integer coefficients.) In this step, you will prove that a closed
− 4 u. (In
fact, each pn
formula for pn is, n
pn = 2
n (u − cos(kπ/(n + 1))).
k=
As a reality check, observe this gives,
p 1 = 2(u − cos(π/2)) = 2u 2 p 2 = 4(u − cos(π/3))(u − cos(2π/3)) = 4(u − 1 /2)(u + 1/2) = 4u − 1 , 3 p 3 = 8(u − cos(π/4))(u − cos(2π/4))(u − cos(3π/4)) = 8(u − 1 /
2)(u)(u + 1/
- = 8u − 4 u
Write up a careful proof of the missing steps in the next paragraph. To prove the formula,
observe that it suffices to check when |u| > 1; two polynomials in u that agree for infinitely many
n n
values of u are equal (you don’t have to write a proof of this). Assume that p (^) n =C (^) +r (^) + +C−r (^) −,
where C (^) + , C (^) − , r (^) +, r − are continuous functions in u that do not depend on n. Forr = r (^) +, r =r−,
prove that r satisfies the characteristic polynomial,
2 r − 2 ur + 1 = 0.
Conclude that,
r = a ± b, a = u, b = u^2 − 1. ±
Plugging this in to the equations for p 0 and p 1 , conclude that,
C+ + C = 1 ,
(C+ + C−)a + (C+ − C−)
−
b = 2 a,
Solve this system of linear equations to get,
C+ = (a + b)/ 2 b,
C (^) − = −(a − b)/ 2 b
Therefore one solution of Equation 1 for |u| > 1 is,
(a + b)
n+ − (a − b)
n+1 (^2) pn = [ ]/ 2 b, a = u, b = u − 1.
But, of course, there is a unique solution: for each n ≥ 2, pn can be determined recursively in terms
of pn− 1 and pn− 2. Therefore, this is the solution of Equation 1.
n n Solution: Substituting in pn = C+r+ + C−r− to the finite difference equation yields,
2 n 2 n C+(r+ − 2 ur+ + 1)r+ + C−(r− − 2 ur− + 1)r− = 0,
2 for every n ≥ 0. Clearly, if r± − 2 ur (^) ±+ 1 = 0, this equation is satisfied. This is all that is required
for the rest of the argument: the strategy is to find one solution of the finite difference equation,
not every solution (in the last step, we observe there is a unique solution). However, it is reasonable
n to check that every solution of the finite difference equation of the form pn = C+r+ + C−r
n does 2 2
−
satisfy r± − 2 ur + 1 = 0. Suppose that r+ − 2 ur+ + 1 = 0. Plugging in n = 0, 1 above gives,
2 2 C−(r− − 2 ur− + 1) = −C+(r+ − 2 ur+1), 2 C+(r+ − 2 ur+ + 1)(r+ − r−) = 0
2
Also, 2 + 2 cos(θ) = 4 cos
2 (θ/2). Therefore, u
2 = cos
2 (πk/(n + 1)). So u = ± cos(πk/(n + 1)). But
of course − cos(πk/(n + 1)) = cos(π(n + 1 − k)/(n + 1)). Therefore, every root of pn(u) = 0 with
|u| < 1 is of the form cos(πk/(n + 1)) for some integer k = 1,... , n.
Reversing the steps above, cos(πk/(n + 1)) is a root of pn for every k = 1,... , n. Also, for 0 < θ < π,
the function cos(θ) is strictly decreasing. Therefore the real numbers cos(πk/(n+1)) are all distinct.
This gives n distinct real roots of the degree n polynomial pn. Since a polynomial of degree n has
at most n real roots, counted with multiplicity, every root of pn is of the form cos(πk/(n + 1)), and
each of these roots has multiplicity 1. It is straightforward to compute that the leading coefficient
of pn is 2
n
. Therefore,
n
pn = 2
n (u − cos(πk/(n + 1))).
k=
(b), Step 2(10 points) For each integer N ≥ 1, define PN (λ) to be the characteristic polynomial
det(λIN ×N − AN ). Define P 0 (λ) = 1. Using cofactor expansion along the first row, prove
λ that the sequence of polynomials P 0 , P 1 , P 2 ,... satisfies Equation 1 where u = + 1. 2
Solution: By direct computation, P 0 = 1 and P 1 = λ + 2. Let N ≥ 0 and consider PN +2. By
cofactor expansion along the first row,
λ + 2 − 1 0 0... 0 0
− 1 λ + 2 − 1 0... 0 0
0 − 1 λ + 2 − 1... 0 0
0 0 − 1 λ + 2... 0 0
..
PN +2 = det
0 0 0 0... λ + 2 − 1
0 0 0 0... − 1 λ + 2
equals,
⎡ ⎤ ∗ ∗ ∗ ∗... ∗ ∗ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ∗ λ + 2 − 1 0... 0 0
∗ − 1 λ + 2 − 1... 0 0
∗ 0 − 1 λ + 2... 0 0
..
(λ + 2)det
.. (^)... .... ....
∗ 0 0 0... λ + 2 − 1
∗ 0 0 0... − 1 λ + 2 ⎡ ∗ ∗ ∗ ∗... ∗ ∗ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ − 1 ∗ − 1 0... 0 0
0 ∗ λ + 2 − 1... 0 0
0 ∗ − 1 λ + 2... 0 0
..
+1det
.. (^)... .... ....
0 ∗ 0 0... λ + 2 − 1
0 ∗ 0 0... − 1 λ + 2
where the asterisks are just placeholders for then entries to be deleted in computing the determi
nant. The first determinant above, is simply PN +1. For the second determinant, perform cofactor
4
expansion along the first column (which has only 1 nonzero entry),
⎡ ⎤ ∗ ∗ ∗ ∗... ∗ ∗
det
0 ∗ λ + 2 − 1... 0 0
0 ∗ − 1 λ + 2... 0 0
..
0 ∗ 0 0... λ + 2 − 1
0 ∗ 0 0... − 1 λ + 2 ⎡ ⎤ ∗ ∗ ∗ ∗... ∗ ∗
∗ ∗ ∗ ∗... ∗ ∗
∗ ∗ λ + 2 − 1... 0 0
∗ ∗ − 1 λ + 2... 0 0
..
−1det
.. (^)... .... ....
∗ ∗ 0 0... λ + 2 − 1
∗ ∗ 0 0... − 1 λ + 2
This is −PN. Therefore, PN +2 = (λ + 2)PN +1 − PN. Defining u =
⎧ ⎨
λ 2
, this is,
PN +2 − 2 uPN +1 + PN = 0,
P 0 = 1,
P 1 = 2u
End of the solution
It then follows that the characteristic polynomial of AN is,
�^ N
PN (λ) = (λ + 2(1 − ck)), ck = cos(πk/(N + 1)).
k=
Therefore the eigenvalues of AN are λk = −2 + 2ck = −(2 sin(πk/(N + 1)))
2 , for k = 1,... , N.
(c), Step 3(10 points) Let the integer N ≥ 1 be fixed. For each integer k = 1,... , N , consider the
eigenvalue λk. Denote, ⎡ ⎤
x(k), 0
x(k), 1
. . .
x(k),N − 1
x(k) =
the eigenvector of AN with eigenvalue λk that is normalized by the condition x(k), 0 = 1.
Prove that the sequence x(k), 0 , x(k), 1 ,... satisfies Equation 1 where u = ck. It then follows
that for each n = 0,... , N − 1, the entry x(k),n is given by,
n
x(k),n = 2
n (cos(πk/(N + 1)) − cos(πl/(n + 1))).
l=
Solution: Because x(k) is an eigenvector with eigenvalue λk, Ax(k) = λkx(k). Expanding this out,
⎧ ⎨
(−2)x(k), 0 + x(k), 1 = λkx(k), 0 ,
x(k),n + (−2)x(k),n+1 + x(k),n+2 = λkx(k),n+1, n = 0,... , N − 3 ,
x(k),N − 2 + (−2)x(k),N − 1 = λkx(k),N − 1
5
Let v be an nvector and consider g(S, t)v. Write up the proof of the following fact. The
function of t, g(S, t)v is C
∞ and for every nonnegative integer r,
r ∂
r d (g(S, t)v) = g (S, t)v. dtr^ ∂tr
Solution: It suffices to prove that for some choice of ordered basis B, every entry of the matrix
A = [g(S, t)]B,B is a C
∞ function, and
dr dtr^ (Ai,j^ ) is the (i, j)entry of [(^ ∂tr^ g)(S, t)]B,B^.
∂r
Because S is diagonalizable, there exists an ordered basis B = (v 1 ,... , vn) such that each vj is an
eigenvector for S. For j = 1,... , n, define μj by Svj = μj vj. By definition, [g(S, t)]B,B is a diagonal
matrix whose (j, j)entry is g(μj , t). Therefore every entry is a C
∞ function, and
r d ∂
r g g(μj , t) = r
(μj , t), dtr^ ∂t
which is the (j, j)entry of the diagonal matrix [(
∂r ∂tr^
g)(S, t)]B,B.
End of the solution
(b), Step 2(5 points) Next, let A be an n × n real matrix whose eigenvalues are all real, but such
that the eigenspaces of A might be deficient. Denote by λ 1 ,... , λk the eigenvalues of A. For each
i = 1,... , k, denote by V
gen λ the generalized eigenspace of^ A. Define^ S^ to be the unique matrix such i
that for each i = 1,... , k, S preserves V
gen and the restriction of S to V
gen is λi times the identity λi λi
matrix. Define N = A − S. Then S commutes with N , the matrix S is diagonalizable, and the
matrix N is nilpotent, i.e. N
e+ = 0 for some integer e.
Define g(A, t) to be the matrix,
�^ e^ �^ � 1 ∂
m
g(A, t) = N
m m
g (S, t). m! ∂λ m=
Let v be an nvector and consider g(A, t)v. Write up the proof of the following fact. The
function of t, g(A, t)v is C
∞ and for every nonnegative integer r,
r ∂
r d (g(A, t)v) = g (A, t)v. dtr^ ∂tr
(Hint: Use the equality of mixed partial derivatives, and reduce to part (a) ).
∂m Solution: First of all, by part (a), each of the functions wm(t) = ( ∂λm^
g)(S, t)v is a C
∞ function.
Therefore each (1/m!)N
m wm(t) is a C
∞ function (after all, N
m is a constant matrix). Thus the
finite sum g(A, t)v is a C
∞ function.
The equation, � � r ∂
r d (g(A, t)v) = g (A, t)v, dtr^ ∂tr
is proved by induction on the positive integer r. Let r = 1. By part (a), each of the functions
∂m ( ∂λm^
g)(S, t)v is C
∞ and,
� � � � �� ∂
m ∂
m
g (S, t)v = m
g (S, t)v.
d ∂
dt ∂λm^ ∂t ∂λ
Because g is C
∞ , mixed partial derivatives are equal, i.e. � � � � �� ∂
m
g (S, t)v =
m ∂g (S, t)v.
d
dt ∂λm^ ∂λm^ ∂t
7
Therefore,
e �^ �^ ��^ �^ � d N
m (g(A, t)v) =
� (^1) ∂m^ ∂g (S, t)v =
∂g (A, t)v. dt m! ∂λm^ ∂t ∂t m=
This is the case r = 1 in the induction above.
∂g By way of induction, suppose that r > 1 and that the result is proved for r − 1. Define h(λ, t) =. ∂t
Then, by the induction hypothesis,
d
r− 1 ∂
r− 1 h h(A, t)v = (A, t)v. dtr−^1 ∂tr−^1
Therefore,
d
r d
r− 1 d d
r− 1 ∂g g(A, t)v = g(A, t)v = (A, t)v = dtr^ dtr−^1 dt dtr−^1 ∂t
d
r− 1
h(A, t)v =
r− 1 h (A, t)v =
r g (A, t)v. dtr−^1 ∂tr−^1 ∂tr
The second equality follows from the case r = 1 and the fourth equality follows from the induction
hypothesis. Therefore the result is proved for r. So the result is proved by induction on r.
End of the solution
(c), Step 3( 0 points) Let g(λ, t) and h(λ, t) be C
∞ functions on the (λ, t)plane. Let A be an n × n
matrix all of whose eigenvalues are real. Then g(S, t), g(A, t), h(S, t) and h(A, t) all commute with
each other. Moreover,
g(S, t)h(S, t) = (g · h)(S, t), and g(A, t)h(A, t) = (g ·h)(A, t).
To prove that g(S, t) and h(S, t) commute with each other, that they commute with g(A, t) and
h(A, t), and that g(S, t)h(S, t) = (g h)(S, t), it suffices to prove this after restricting to V
gen λ for i
every i = 1,... , k. But for any C
∞ function f (λ, t), the restriction of f (S, t) to V
gen is simply f (λi, t) λi
times the identity matrix. Denote this by f (S, t)i. Therefore f (S, t)i commutes with every linear
operator on V
gen λ. In particular,^ g(S, t)i^ and^ h(S, t)i^ commute with each other and with^ g(A, t)i^ and i
h(A, t)i. Also, g(S, t)i · h(S, t)i is simply g(λi, t) ·h(λi, t) times the identity matrix. This is the same
as (g · h)(S, t)i. Therefore g(S, t) · · · h(S, t) = (g ·h)(S, t).
To prove that g(A, t) and h(A, t) commute with each other and that g(A, t)h(A, t) = (g ·h)(A, t),
it suffices to prove this after restricting to V
gen λ for every^ i^ = 1,... , k.^ Denote by^ Si^ and^ Ni^ the i
restrictions of S and N respectively to V
gen
. By definition, λi
�^ e^ �^ �^ e^ �^ � 1 ∂
l (^) � 1 ∂
m
g(A, t)ih(A, t)i = N
l
∂λl^
g (S, t)i N
m h (S, t)i. l!
i ·^ m!
i ∂λm l=0 m=
All of the matrices in this last sum commute. Therefore we may rearrange the (finite) sum in
whatever order we wish,
e e ∂
m h g(A, t)ih(A, t)i =
N
l+m (
l g )(S, t)i. l! m!
i ∂λl^
∂λm l=0 m=
Define p = l + m. For p > e, N p^ = 0. Therefore the sum equals,
e
p �^ � � 1 � (^) p ∂p−mg ∂mh
g(A, t)ih(A, t)i = N
p (S, t)i. p!
i m ∂λp−m^
∂λm p=0 m= 8
(f ), Step 6(0 points) This is somewhat beside the point, but if g(λ, t) is an analytic function in λ
whose expansion about the origin is,
∞
g(λ, t) = cr(t)λ
r ,
r=
and if all the eigenvalues of A are within the radius of convergence of this power series, then the
following series converges to g(A, t), ∞
cr(t)A
r .
r= λt In particular, this holds if g(λ, t) is a polynomial in λ or if g(λ, t) = e.
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