Math 381 Fall 2006 Final Exam Solutions
Instructor: Sabin Cautis
Take home: due back Friday, December 15, 2006
Instructions: This is a take home exam. You may use only your textbook and
notes for help. You have three hours. Do all the problems. You must show
your work to receive full credit on a problem. Use separate, stapled sheets to
do so.
Please print you name clearly here.
Print name:
Upon finishing please sign the pledge below:
On my honor I have neither given nor received any aid on this exam apart from
help from my textbook and notes.
1a.
/15
1b.
/5
1c.
/10
2.
/15
3a.
/5
3b.
/5
4a.
/15
4b.
/15
5a.
/10
5b.
/10
6a.
/15
6b.
/10
7.
/35
8.
/35
1. (a) [15 points] Give the Fourier series of e−|x| on the interval −L ≤ x ≤ L.
Note: you can leave the expressions for the Fourier coefficients as integrals
(ie you don’t need to evaluate them).
The general Fourier series is
e−|x| ∼ a0 +
∞
X
∞
an cos
n=1
where a0 =
1
2L
RL
−L e
−|x|
dx, an =
1
L
nπx
nπx X
+
bn sin
L
L
n=1
RL
e−|x| cos nπx
L dx and bn =
−L
1
L
RL
−L e
−|x|
(b) [5 points] Explain why the answer in part (a) is a cosine Fourier series
(ie there are only cosine terms).
Since e−|−x| = e−|x| the function e−|x| is even. This means the Fourier
series only contains the even terms, namely only the cosine terms (including the constant term). More precisely, using the expression for bn from
part (a) we can show bn = 0 for n ≥ 1 as follows
Z
L
e−|x| sin
−L
nπx
dx =
L
Z
0
e−|x| sin
−L
0
nπx
dx +
L
Z
L
e−|x| sin
0
nπx
dx
L
Z L
nπx
−nπx
(−dx) +
e−|x| sin
dx
L
L
0
L
Z 0
Z L
nπx
nπx
−|x|
=
e
sin
dx +
e−|x| sin
dx
L
L
L
0
= 0
=
Z
e−|−x| sin
where we used the fact that sin(−t) = −sin(t) to go from line two to
three.
(c) [10 points] Sketch the Fourier series of e−|x| from part (a) for −∞ <
x < ∞.
The Fourier series is the periodic extension of e−|x| from [−L, L].
2. [15 points] Let u(x, y, z) be a solution to the Laplace equation
∂2u ∂2u ∂2u
+ 2 + 2 =0
∂x2
∂y
∂z
on some connected domain D of R3 . Amazingly, such a solution always
satisfies the mean value theorem – namely, the mean value of u on any
sphere centered at a point p is equal to the value of u at the point p.
Explain why this means that if u is constant on the boundary of D then it is
constant everywhere. [Hint: if u is not constant can its maximal/minimal
value occur in the interior of D?]
If u(x, y, z) is not constant everywhere on D then there must exist some
point p ∈ D where u(x, y, z) achieves its maximal value and where u(x, y, z)
is not constant in a neighbourhood of p. We show this leads to a contradiction. Consider a small sphere centred at p. The average value of u on
this sphere equals u(p). But for any point q on this sphere u(q) ≤ u(p)
since u(p) is maximal. Both these things cannot happen unless u is constant and equal to u(p) on the sphere. This argument holds for any small
sphere centred at p. This means u is equal to u(p) in a neighbourhood of
p (contradiction).
3 −1
what is the
3. (a) [5 points] If L : R2 → R2 is the linear operator
0 4
adjoint operator L∗ of L (with respect to the standard dot product)?
sin nπx
L dx.
In general, if we use the standard inner (dot) product on Rn the adjoint
of a linear operator L represented by a matrix
by the
A is represented
3 0
3 −1
t
t
is A =
transpose A . In our case, the adjoint of A =
−1 4
0 4
(b) [5 points] Is L from part (a) self-adjoint? Why?
In general, L is self-adjoint if the matrix A representing A is symmetric
(ie A = At ). In part (a) this is not the case so L is not self-adjoint.
4. (a) [15 points] Let L be a self-adjoint linear operator on a vector space V
equipped with an inner product h·, ·i. If L(φ1 ) = λ1 φ1 and L(φ2 ) = λ2 φ2
where λ1 6= λ2 (ie we have two eigenvectors with distinct eigenvalues)
show that hφ1 , φ2 i = 0 (ie the eigenvectors are orthogonal).
We have
λ1 hφ1 , φ2 i = hλ1 φ1 , φ2 i = hL(φ1 ), φ2 i.
Using that L is self-adjoint we have
hL(φ1 ), φ2 i = hφ1 , L(φ2 )i = hφ1 , λ2 φ2 i = λ2 hφ1 , φ2 i.
Thus λ1 hφ1 , φ2 i = λ2 hφ1 , φ2 i. Since λ1 6= λ2 this means hφ1 , φ2 i = 0.
(b) [15 points] An example of a self-adjoint linear operator is the SturmRb
Liouville operator where the inner product is hf, gi = a f gσdx. Let
φ1 , φ2 , . . . denote the eigenfunctions of the Sturm-Liouville
P∞PDE which,
as weP
learned, have distinct eigenvalues.
Show
that
if
f
=
n=1 an φn and
P∞
∞
g = n=1 bn φn then hf, gi = n=1 an bn ||φn ||2 where ||φn ||2 = hφn , φn i
(you can assume the operations of summation and integration can be
interchanged).
We have the following calculation
hf, gi
= h
=
=
∞
X
i=1
∞
X
i,j=1
∞
X
a i φi ,
∞
X
b j φj i
j=1
hai φi , bj φj i
ai bj hφi , φj i
i,j=1
Since hφi , φj i = 0 unless i = j this sum simplifies to
P∞
n=1 an bn hφn , φn i.
P∞
i=j=1
ai bj hφi , φj i =
5. (a) [10 points] Explain why (for a fixed m) the associated Legendre funcR1
tions Pnm (x) are orthogonal with respect to hf, gi = −1 f gdx.
The Legendre functions Pnm (x) (for a fixed m) are eigenfunctions for the
(singular) Sturm-Liouville
dg
m2
d
(1 − x2 )
+ µ−
g=0
dx
dx
1 − x2
where −1 ≤ x ≤ 1 and the eigenvalue µ = n(n + 1). Here σ(x) = 1
so these Reigenfunctions are orthogonal with respect to the inner product
1
hf, gi = −1 f gdx.
(b) [10 points] Suppose f (x) is a Rsmooth function on [−1, 1] and f (x) =
1
P∞
f Pnm dx
m
R −1
n=1 an Pn (x) show that an = 1 P m P m dx .
−1
n
n
Since the Pnm (x) are orthogonal with respect to the inner product hf, gi =
R1
f gdx we have
−1
hf, Pnm (x)i = h
∞
X
n=1
an Pnm (x), Pnm (x)i = an hPnm (x), Pnm (x)i.
This gives
R1
f Pnm dx
hf, Pnm (x)i
−1
an =
=
.
R1
hPnm (x), Pnm (x)i
Pnm Pnm dx
−1
6. (a) [15 points] Show that the Fourier transform FT is a linear operator –
namely, show that
FT(c1 f (x) + c2 g(x)) = c1 FT(f (x)) + c2 FT(g(x)).
We have
FT(c1 f (x) + c2 g(x))
Z
∞
(c1 f + c2 g)eiωx dx
Z ∞
Z ∞
geiωx dx
f eiωx dx + c2
= c1
=
−∞
−∞
−∞
= c1 FT(f (x)) + c2 FT(g(x))
(b) [10 points] Is it true that FT(f (x)g(x)) = FT(f (x))FT(g(x))? Explain.
R∞
R∞
R∞
This is not true since, in general, −∞ f geiωx dx 6= −∞ f eiωx dx· −∞ geiωx dx.
7. [35 points] Compute (from the definition) the Fourier transform of e−|x| .
We have
FT(e−|x| ) =
Z
=
Z
∞
e−|x|eiωx dx
−∞
0
e
x+iωx
dx +
−∞
x+iωx
Z
∞
e−x+iωx dx
0
−x+iωx
e
e
]0−∞ + [
]∞
1 + iω
−1 + iω 0
1
1
−
1 + iω −1 + iω
2
(1 + iω)(1 − iω)
2
1 + ω2
= [
=
=
=
where to go from lines three to four we used that ex+iωx = ex · eiωx → 0 as
x → −∞ and e−x+iωx → 0 as x → ∞. This is because eiωx = cos(ωx) +
i sin(ωx) is bounded while ex → 0 as x → −∞ and e−x → 0 as x → ∞.
8. [35 points] Suppose u(x, t) is a solution to the 1-dimensional wave PDE
2
∂2u
2∂ u
=
c
∂t2
∂x2
where −∞ < x < ∞. Assuming u(x, t) → 0 as x → ±∞ find the general
solution for FT(u). [Hint: first apply the FT to the PDE]
∂FT(u)
We know that FT( ∂u
and if u(x, t) → 0 as x → ±∞ then
∂t ) =
∂t
2
2
∂u
FT( ∂x ) = (−iω)FT(u). Thus, if u satisfies ∂∂t2u = c2 ∂∂xu2 then, using the
fact that the FT is linear, we get that FT(u) satisfies:
∂ 2 FT(u)
= −c2 ω 2 FT(u).
∂t2
This is a differential equation of the form f 00 (t) = −f (t) which we know to
have general solutions cos(t) and sin(t). One can check that two solutions
are cos(cω) and sin(cω). Thus the general solution for FT(u) is
FT(u) = a(ω) cos(cω) + b(ω) sin(cω)
where a(ω) and b(ω) are functions of ω.
Note: the assumption that u(x, t) → 0 as x → ±∞ is not quite strong
enough. To conclude FT(uxx ) = (−iω)2 FT(u) one needs this as well as
ux (x, t) → 0 as x → ±∞. This is just something to keep in mind: you have
to be a little careful when you use that FT(ux ) = −iωFT(u) repeatedly.
The best thing is to go back to the proof of the result or to look at
a very precise statement of the result or (even better) to ask someone
knowledgeable in the field.