MATH 3150: PDE FOR ENGINEERS
FINAL EXAM (VERSION A)
1. Draw the graph of
y = tan x
labelling all asymptotes and zeros. Include at least 3 periods in your graph.
What is the period of tan x?
Solution:
See figure 1. Asymptotes are at π/2 + πn for any integer n,
and zeros at πn for any integers n.
2.
(
x2
f (x) =
−x2
0≤x<π
−π ≤ x < 0
and f (x) has period 2π.
(a) Draw the graph of f (x).Include at least two periods in your graph.
Date: December 14, 2001.
2
y
–4
–2
1
0
2
4
x
–1
–2
Figure 1. Graph of the tangent function
1
2
MATH 3150: PDE FOR ENGINEERS
FINAL EXAM (VERSION A)
(b) Calculate the real Fourier amplitudes am and bm of f (x). Hint:
Z
x2 sin x dx = −x2 cos x + 2 cos x + 2x sin x .
(c) Find the energy of f (x).
(d) In the form of a fraction, find the total percentage of energy contained
among the amplitudes a0 , a1 , b1 , a2 , b2 . Do not use any decimal approximations.
(e) Show that about 60% of the energy is stored among the amplitudes
a0 , a1 , b1 , a2 , b2 . Hint:
π 2 ∼ 10
π 4 ∼ 100
π 6 ∼ 1000.
So collect all terms over a common denominator of π 6 .
Solution: All am = 0 because is it odd.
bm =
=
=
=
=
=
=
Z
πmx 2 L
f (x) sin
dx
L 0
L
Z π
2
x2 sin (mx) dx
π 0
Z
2 mπ
(u/m)2 sin u d(u/m)
π 0
Z mπ
2
u2 sin u du
πm3 0
u=mπ
2
−u2 cos u + 2 cos u + 2u cos u u=0
πm3
2 −(mπ)2 (−1)m + 2(−1)m − [2]
πm3
2(−1)m+1 π 4 1 + (−1)m+1
−
.
m
πm3
So
b1 = 2π −
8
π
b2 = −π.
The energy in these two is
T 2
64
2
2
b + b2 = π 5π − 32 + 2
2 1
π
The total energy in the function is
Z π
Z
f (x)2 dx = 2
−π
0
π
x4 dx =
2π 5
.
5
MATH 3150: PDE FOR ENGINEERS
FINAL EXAM (VERSION A)
3
1
0.8
0.6
0.4
0.2
–1
–0.5
y0
0.5
1
1
0.5
0
x
–0.5
–1
Figure 2. A round peak
The percentage of energy in these two amplitudes is
π 5π 2 − 32 + π642
5 5π 4 − 32π 2 − 64
=
5
2π /5
2
π6
5 500 − 320 + 64
∼
2
1000
61
=
.
100
So 61 per cent.
3. Suppose that u(x, y, t) satisfies
2 3 !
2 4 !
∂u
∂2u
∂ u
∂u
∂u
= −u
+
− exp
+
+1
∂t
∂x2
∂y 2
∂x
∂y
(which is so horrible an equation that no one can solve it). Suppose that
the function u at time t = 0 looks like a round peak as in figure 2. Does
the top of the peak move up or down? Explain.
Solution: It goes up, because at the top of the peak
u>0
∂u
∂x
∂u
∂y
∂2u
∂x2
∂2u
∂y 2
so that
=0
=0
<0
<0
∂u
= (−)(+) ((−) + (−)) − e0 + 1 = (+) .
∂t
4. Solve the heat equation
∂u
∂2u
= c2 2
∂t
∂x
4
MATH 3150: PDE FOR ENGINEERS
FINAL EXAM (VERSION A)
x
Figure 3. A circular ring, with x axis wound around it
in a circular ring of radius r (see figure 3) by finding a kernel K(x, t) so
that
Z 2πr
u(x, t) =
u(s, 0)K(x − s, t) ds .
0
Note that u(x, t) has period 2πr in x, so you can use complex Fourier series
(not Fourier transforms, since it isn’t infinitely long).
Solution: Writing
X
u(x, t) =
ak (t)eikx/r
k
we find that the heat equation becomes
2
dak
cik
=
ak
dt
r
so that
ak (t) = exp −
ck
r
2 !
t ak (0)
and
u(x, t) =
X
exp −
k
Z
ck
r
2 !
t ak (0)eikx/r
2πr
u(s, 0)K(x − s, t) ds
=
0
where
K(x, t) =
1 X −(ck/r)2 t+ikx/r
e
.
2πr
k
5. Find the Fourier transform of the function
(
1 if − 1 ≤ x ≤ 1
f (x) =
0 otherwise
MATH 3150: PDE FOR ENGINEERS
FINAL EXAM (VERSION A)
5
Solution:
Z 1
1
ˆ
e−iωx dx
f (ω) = √
2π −1
−iω
eiω
e
1
−
=√
iω
2π −iω
iω
−iω
1
e −e
.
=√
iω
2π
You might not recognize this: it is
r
fˆ(ω) =
2 sin(ω)
.
π ω
6. Show that according to the heat equation, heat propagates with infinite
speed. To do this, take initial temperature u(x, 0) which is positive near
x = 0, and not negative anywhere, and zero everywhere except in a little
region near x = 0. Now apply the heat kernel
2
G(x, t) =
2
e−x /4c t
√
2c πt
in the equation
Z
∞
u(s, 0)G(x − s, t) ds
u(x, t) =
−∞
giving final temperature at time t. How does this show you that a small
amount of heat spreads out to all points of space at as small a time as you
like?
Solution: Because the heat kernel is positive everywhere, and the function u(x, 0) is positive somewhere, negative nowhere, the function u(s, 0)G(x−
s, t) is positive somewhere, negative nowhere, so has positive integral over
s, for any value of x and t. Therefore u(x, t) is positive everywhere, no
matter how small t is. So the temperature is suddenly positive everywhere.
7. Take a disk of unit radius and heat the top half of the edge of the disk
to 1o and the bottom half of the edge to 0o . Let the disk sit with these
temperatures on its edges for a long time, until it reaches a steady state.
Center the disk at origin of coordinates. Try to estimate (without making
any decimal approximations—just use fractions) the temperature at the
point
π
(x, y) = 0,
.
10
6
MATH 3150: PDE FOR ENGINEERS
FINAL EXAM (VERSION A)
Hint: recall that if we write the temperature of the edge of the disk as f (θ),
and write the Fourier amplitudes of f (θ) as
Z 2π
1
a0 =
f (θ) dθ
2π 0
Z
1 2π
f (θ) cos(mθ) dθ
am =
π 0
Z
1 2π
f (θ) sin(mθ) dθ
bm =
π 0
then the steady state temperature inside the plate, in polar coordinates, is
∞
X
u(r, θ) = a0 +
rm (am cos (mθ) + bm sin (mθ))
m=1
Solution:
We calculate:
1
2
=0
a0 =
am
bm =
1 + (−1)m+1
πm
so that
u=
=
∞
1 X m 1 + (−1)m+1
+
r
sin (mθ)
2 m=1
πm
∞
1
2 X r2k+1
+
sin ((2k + 1) θ)
2 π
2k + 1
k=0
On the positive y axis, we find θ = π/2, so that
sin ((2k + 1) π) = (−1)k
and therefore
1
2
u= +
2 π
r3
r5
r−
+
+ ···
3
5
.
In particular, at r = π/10, we find
1
2 π
π2
u= +
−
+ ···
2 π 10 3000
so
1 1
7
u= + =
2 5
10
8. Suppose that I have a function
y = f (x)
which is periodic with period T . Suppose that it has energy E(f ) = kf k2 .
If I rescale the x variable by an amount α, making a new variable X = αx
and a new function F (X) = f (x),
(a) what is the period of F (X)? and
MATH 3150: PDE FOR ENGINEERS
FINAL EXAM (VERSION A)
7
(b) by what amount do I have to rescale the y variable to keep the same
energy as f ?
Solution:
(a) The
√ new period is αT .
(b) E( αF ) = E(f ).
9. Suppose that f (x) has period T and average value 0.
(a) Using complex Fourier series, show that the energy in the derivative
df /dx satisfies
2 2
df 2
≥ 2π
kf k .
dx T
(b) Which functions f (x) satisfy
2 2
df 2
= 2π
kf k ?
dx T
Solution:
(a) If
f (x) =
X
ak e2πikx/T
k
then
X
df
=
ak
dx
k
2πik
T
e2πikx/T
so that the energy of df /dx is
2
2
X
df 2πk
2
=T
|ak |
dx T
k
while the energy of f is
2
kf k =
X
2
|ak | .
k
Clearly each term in the first sum is at least (2π/T )2 larger than in
the second, except the a0 term. But the a0 term is the average, which
is zero. So we have the required inequality
2 2
df 2
≥ 2π
kf k .
dx T
(b) Equality can only occur in the a−1 and a1 terms, because all of the
others have a factor of
2 2
2πk
2π
>
.
T
T
So
f (x) = a−1 e−2πix/T + a1 e2πix/T .
10. Write down the general solution of the heat equation for temperature u(x, t)
in a wire of length L with left end at 0o and insulated right end. Explain
how you got your answer. (You don’t have to explain how to dig out the
amplitudes at time t = 0.)
8
MATH 3150: PDE FOR ENGINEERS
Solution:
FINAL EXAM (VERSION A)
The final result is
X
2 2
an e−c λn t sin (λn x)
u(x, t) =
n
where
λn =
π
L
n+
1
2