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Pembroke College, Oxford

MIKE FULLER’S

QUESTIONS FOR PRELIMS
Volume V: Paper V

Fourier Series and PDEs
Multivariable Calculus

April 8, 2017

Contents
I

Fourier Series and PDEs

2

State the Heat, Wave and Laplace equations
...

State the Fourier Convergence Theorem
...

State three orthogonality relations
...

State Parseval’s identity
...

State three advantages of a Fourier cosine series
...

List important symbols for the Heat equation and their SI units
...
[6]
...
[5]
...
[6]
...
[5]
...
[5]
...
[2]
...
[4]
...


...


...


...


...


...


...


Multivariable Calculus

3
4
5
6
7
8
9
10
11
12
13
14
15

16

Define Grad, Div and Curl
...

State the Divergence Theorem
...

Derive the 3D heat equation using Fourier’s Law
...

State, prove Green’s Theorem In The Plane (Divergence Form)
...
[6]
State Stokes’ Theorem
...

Derive Gauss’ Flux Theorem using Poisson’s equation
...


1


...


...


...


...


...


...


...


17
18
19
20
21
22
23

Part I
Fourier Series and PDEs

2

Fourier Series and PDEs

Mike Fuller’s Questions for Prelims

State the Heat, Wave and Laplace equations
...
e Tt = κTxx ,

where T (x, t) is a temperature at position x and time t, and κ > 0 is the thermal diffusivity
...
e ytt = c2 yxx ,

where y(x, t) is the transverse displacement of a stretched string at position x and time t, and
c > 0 is the wave speed
...
e Txx + Tyy = 0 or
∂x2
∂y 2

2

T = 0,

where, for example, T (x, y) may be a temperature and x, y are Cartesian coordinates in the
plane
...
[2]
[1] Let f be a function with period 2π, with f, f piecewise continuous on (−π, π)
...
e
∞

1
a0
+
[an cos(nx) + bn sin(nx)] = [f (x+ ) + f (x− )]
...

(2m + 1)2
8
m=0

4

Fourier Series and PDEs

Mike Fuller’s Questions for Prelims

State three orthogonality relations
...


and
−π

−π

Note:
•

SC is a trivial proof since it is an odd function,

•

SS, CC are proved by considering the cases m = n, m = n separately and using
CC(+) , SS(−) =

1
{cos((m − n)x) ± cos((m + n)x)}
...
[2]
[1] Let f (x) be a function of period 2L and let f, f be piecewise continuous on (−L, L)
...

{f (x)} dx = 0 +
n
n
2
n=1
2

−L

Note:
• To prove this, multiply the Fourier series by f (x) and integrate termwise from −L to L
...
[3]
• [1] Unlike a sine series, it converges to f everywhere,
• [1] Unlike a sine series, it has no Gibbs Phenomenon (overshooting at endpoints),
• [1] The rate of convergence is faster than a sine series
...
[5]
[5] There are 9 important symbols that come up:
Symbol

Quantity

SI Unit

x
t
T (x, t)
q(x, t)
A
ρ
c
k
k
κ or
ρc

Position (axial distance)
Time
Temperature
Heat flux (in the x direction)
Cross-sectional area of rod
Density of rod
Specific heat of rod
Thermal conductivity of rod

m
s
K
Jm−2 s−1
m2
kgm−3
Jkg−1 K−1
JK−1 m−1 s−1

Thermal diffusivity of rod

m2 s−1

Note:
• 1J = 1Nm

8

Fourier Series and PDEs

Mike Fuller’s Questions for Prelims

Derive the one-dimensional Heat equation
...

The rod has density ρ, specific heat c and thermal conductivity k, all positive and constant
...


[1] Consider any interval [a, a+h] of the rod
...


[1] By conservation of energy,
rate of change of internal energy + net heat flux out = 0
⇒

d
dt

a+h

ρcT (x, t) dx + [q(a + h, t) − q(a, t)] = 0
...

[1] Now substituting Fourier’s law, which states q(x, t) = −kTx , we arrive at the heat equation:
ρcTt = kTx ⇒ Tt = κTxx
...


9

Fourier Series and PDEs

Mike Fuller’s Questions for Prelims

Prove uniqueness (or non-uniqueness) of the Heat equation
...
Consider the difference D := T1 − T2
...


[2] Let
1
2

I(t) =

L

[D(x, t)]2 dx
...
By Leibniz’s rule, the derivative of I,
L

I (t) =

L

DDt dx = κ
0

L
2
(DDx )x − Dx dx,

DDxx dx = κ
0

0

the last equality coming from using the product rule: (DDx )x = Dx Dx + DDxx
...

[1] Hence
0 ≤ I(t) ≤ I(0) = 0 ⇒ I(t) = 0
L

[D(x, t)]2 dx = 0

⇒
0

for every t ≥ 0 and thus W = 0, i
...
T1 = T2 (uniqueness)
...


10

Fourier Series and PDEs

Mike Fuller’s Questions for Prelims

Derive the one-dimensional Wave equation
...


We start by assuming that |yx |

1 and ignoring gravity and air resistance
...
The vector t := rx = i + yx j is
a tangent vector to the string
...

[1] Hence T t is the force exerted by + on − (and the other way round for −T t)
...

[1] Consider an interval [a, a + h]
...

a

[2] By using Newton’s Second Law, then Leibniz’ rule and letting h → 0 (note vt = a),
net force = rate of change of momentum
⇒ T t(a + h, t) − T t(a, t) =
⇒T

t(a + h, t) − t(a, t)
h

d
dt

=

1
h

a+h

ρv(x, t) dx
a
a+h

ρa(x, t) dx
a

⇒ T tx (a, t) = ρa(a, t)
...


11

Fourier Series and PDEs

Mike Fuller’s Questions for Prelims

Prove uniqueness (or non-uniqueness) of the Wave equation
...
Consider the difference D := y1 − y2 and the associated energy
E(t) := Ekinetic + Estress =
remembering c =

L

1
2

2
2
ρDt + T Dx dx,
0

T /ρ
...
Then D satisfies
Dtt = c2 Dxx ,
D(x, 0) = 0,

0 < x < L, t > 0

Dt (x, 0) = 0,

0 ≤ x ≤ L,

and boundary conditions
D(0, t) = D(L, t) = 0,

t > 0
...

Hence E (t) = 0 from applying the condition on Dt
...

[1] So Dx , Dt = 0 ⇒ D = F (x) = G(t), that is, D is constant
...
e y1 = y2 (uniqueness)
...
[5]
Normal modes for a finite string: Solve the wave equation
ytt = c2 yxx
subject to boundary conditions y(0, t) = y(L, t) = 0 for t > 0, using separation of variables
...
Then by the wave equation
X(x)T (t) = c2 X (x)T (t)

⇒

X (x)
T (t)
= c2

...
We now have the ODEs
ω2
T (t) + ω T (t) = 0 = X (x) + 2 X(x),
c
2

where the ODE for X is subject to X(0) = X(L) = 0
...

L

[1] For a non-trivial solution we want A = 0, so
ω=

ωc
= nπ for n ∈ N+ , giving
L

nπc

...

Note:
• A normal mode is periodic in t: y(x, t + p) = y(x, t) with period p =

13

2L
2π
=

...
[2]
Stating your assumptions, write down the initial conditions y(x, 0) and yt (x, 0) as their Fourier
sine series expansions
...


Note:
• This comes directly from setting t = 0 in the equations for y and yt as superpositions
(sums) of normal modes ∞ yn , where
n=1
yn (x, t) = sin

nπx
L

an cos

14

nπct
L

+ bn

nπct
L


...
[4]
By making a change of variables of the form ξ = x − ct and η = x + ct, show that the general
solution of the wave equation is of the form
y(x, t) = F (x − ct) + G(x + ct)
...


[1] Seeking a solution for y(x, t) = Y (ξ, η), we have
yx = Yξ + Yη ,
yxx = Yξξ + 2Yξη + Yηη ,
yt = −cYξ + cYη ,
ytt = c2 Yξξ − 2c2 Yξη + c2 Yηη
...
e

∂
∂ξ

∂Y
∂η

= 0
...
But now Y − G(η) is a function of ξ only
...


15

Part II
Multivariable Calculus

16

Multivariable Calculus

Mike Fuller’s Questions for Prelims

Define Grad, Div and Curl
...
Then
grad φ =

∂φ ∂φ ∂φ
,
,
∂x ∂y ∂z

φ=


...
Then
div F =

·F=

∂F1 ∂F2 ∂F3
+
+
∂x
∂y
∂z

and
curl F =

i
j
k
∂/∂x ∂/∂y ∂/∂z
...
[4]
[1] Let R be a region of R3 with a piecewise smooth boundary ∂R
...

[1] Then
F · dS =
∂R

div F dV
R

[1] where dS is oriented in the direction of the outward pointing normal from R
...
[6]
Let T (x, t) denote the temperature at position x and at time t in an isotropic medium R with
thermal conductivity k, density ρ, specific heat c with heat flow determined by Fouriers Law,
which states that
q = −k T
where q is the heat flux
...


[1] Let S ⊆ R be an arbitrary subset of R
...

∂S

[1] So if the only heat loss/gain is via the boundary ∂S,
d
dt

ρcT dV = −

q · dS

S

∂S

[1] which by Fourier’s Law and the Divergence Theorem is
k T · dS =

=

· (k T ) dV =

∂S

S

k

2

T dV
...


[1] As this is true for an arbitrary subset S ⊆ R (we can prove that
by contradiction), it follows that
∂T
ρc
− k 2 T = 0,
∂t
which rearranges to the required result
...
[8]
[1] Let D be a closed bounded region in the (x, y) plane, whose boundary C is a piecewise
smooth simple closed curve,
[1] and let P (x, y), Q(x, y) have continuous first-order derivatives in D
...

[1] To prove this, let
F = (P (x, y), Q(x, y), 0)
and define the region
R := (x, y, z) ∈ R3 | (x, y) ∈ D, z ∈ [0, 1] ,
with boundary ∂R
...


=
∂R

[1] The contribution to the surface integral from the surfaces at z = 0, 1 are zero as the
integrand is zero
...

[1] Hence
1

1

F · n dS =
∂R

F · n ds dz =
0

C

(P, Q) · n ds dz
0

(P, Q) · n ds
...
[6]
[1] Let D be a closed bounded region in the (x, y) plane, whose boundary C is a piecewise
smooth simple closed curve,
[1] and let P (x, y), Q(x, y) have continuous first-order derivatives in D
...
e
...

[1] To prove this, let (P, Q) → (Q, −P ) in Greens Theorem In The Plane (Divergence Form)
...
T
...
T
...


21

Multivariable Calculus

Mike Fuller’s Questions for Prelims

State Stokes’ Theorem
...

[1] Let F be a smooth vector field on Σ ∪ ∂Σ
...


∂Σ

Σ

22

Multivariable Calculus

Mike Fuller’s Questions for Prelims

Derive Gauss’ Flux Theorem using Poisson’s equation
...


[1] We want to derive Gauss’ Flux Theorem, which states that for a smooth and bounded
region R, which contains matter of total mass M , we have
f · dS = −4πGM
...


23

φ, since φ is a



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