Fourier Analysis Workshop 1: Fourier Series

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Fourier Analysis
Workshop 1: Fourier Series
Professor John A. Peacock
School of Physics and Astronomy
jap@roe.ac.uk
Session: 2013/14
24th & 27th September 2013
1. By writing sin A and cos B in terms of exponentials, prove that
2 sin A cos B = sin(A + B) + sin(A
B).
2. If f (x) and g(x) are periodic with fundamental period X, show that the following are also
periodic with the same period:
(a) h(x) = a f (x) + b g(x)
(b) j(x) = c f (x) g(x)
where a, b, c are constants.
3. Find the fundamental periods for the following functions:
(a) cos 2x
(b) 3 cos 3x + 2 cos 2x
(c) cos2 x
(d) | cos x|
(e) sin3 x.
4. Show that
Z
Printed: September 22, 2013
L
dx sin
L
⇣ m⇡x ⌘
L
sin
⇣ n⇡x ⌘
1
L
=
⇢
0 m 6= n
L m=n
5. (a) Sketch f (x) = (1 + sin x)2 and determine its fundamental period.
(b) Using a trigonometric identity for sin2 x in terms of cos 2x, write down the Fourier Series
for f (x) (don’t do any integrals to obtain the coefficients).
6. Show that the Fourier Series expansion of the periodic function
f (x) =
is
⇢
1
+1
⇡<x<0
0<x<⇡
1
4 X sin[(2k + 1)x]
f (x) =
.
⇡ k=0
2k + 1
7. (a) Show that the Fourier Series for f (x) = x in the range ⇡ < x < ⇡ is
1
X
( 1)m+1
f (x) = 2
sin mx.
m
m=1
(b) Hence, by carefully choosing a value of x, show that
1
1 1
+
3 5
1
⇡
... = .
7
4
8. Using a trigonometric identity, or otherwise, compute the Fourier Series for f (x) = x sin x
for
⇡ < x < ⇡, and hence show that
⇡
1
1
= +
4
2 1⇥3
Printed: September 22, 2013
1
1
+
3⇥5 5⇥7
2
...
Fourier Analysis
Workshop 2: More on Fourier Series
Professor John A. Peacock
School of Physics and Astronomy
jap@roe.ac.uk
Session: 2013/14
1st & 4th October 2013
Handin Deadline: 4 p.m. Friday 11th October 2013
1. Consider the function f (x) = | cos x|.
(a) What is its fundamental period?
(b) Sketch the function for 2⇡ < x < 2⇡
(c) Show that the Fourier Series expansion for f (x) is
f (x) =
1
2
4 X ( 1)m+1
+
cos(2mx).
⇡ ⇡ m=1 4m2 1
2. Let f (x) = 1 + cos2 (⇡x).
(a) Sketch f (x) and determine its fundamental period.
(b) Using a trigonometric identity, and without doing any calculations, write down a Fourier
Series for f (x).
3. If f (x) = sin x for 0  x  ⇡,
(a) compute the fundamental period for a sine series expansion
(b) compute its Fourier sine series
(c) sketch the function in (b) for
2⇡ < x < 2⇡.
(d) compute the fundamental period for a cosine series expansion
(e) compute its cosine series and show it is
2
sin x =
⇡
(f) sketch the function in (e) for
1
4X
1
cos(2mx)
2
⇡ m=1 4m
1
2⇡ < x < 2⇡.
4. Consider f (x) = e x , defined in 0 < x < 1. Expand it as a Fourier sine series, and sketch
Printed: September 22, 2013
3
the function for
2 < x < 2.
5. If f (x) = x for 1  x  1,
(a) show that its Fourier Series is
f (x) =
1
X
( 1)n+1
n=1
(b) Hence show that
2
sin(n⇡x).
n⇡
1
X
( 1)k
⇡
= .
2k + 1
4
k=0
6. Compute the complex Fourier Series for f (x) = x, ⇡ < x < ⇡.
7. If f (x) = |x| for ⇡  x  ⇡,
(a) show that its Fourier Series is
f (x) =
(b) Hence show that
⇡
2
1
X
n=0
Printed: September 22, 2013
1
4 X cos[(2n + 1)x]
.
⇡ n=0 (2n + 1)2
1
⇡2
=
.
(2n + 1)2
8
4
Fourier Analysis
Workshop 3: Parseval, and ODEs by Fourier Series
Professor John A. Peacock
School of Physics and Astronomy
jap@roe.ac.uk
Session: 2013/14
8th & 11th October 2013
1. Compute the complex Fourier Series for
f (x) =
and show it is cn = (cos(kn )
⇢
1
+1
1<x<0
0<x<1
1)i/kn .
2. Prove Parseval’s theorem for the (sine and cosine) Fourier Series.
3. You are given that the Fourier Series of f (x) = |x| (defined for ⇡  x  ⇡) is
⇡
f (x) =
2
State Parseval’s theorem, and prove that
1
X
n=0
1
4 X cos[(2n + 1)x]
.
⇡ n=0 (2n + 1)2
1
⇡4
=
.
(2n + 1)4
96
4. You are given that the Fourier Series of f (x) = x (defined for 1  x  1) is
f (x) =
1
X
( 1)n+1
n=1
Using Parseval’s theorem, show that
2
sin(n⇡x).
n⇡
1
X
1
⇡2
=
.
n2
6
n=1
5. (a) By expanding both sides as Fourier Sin Series, show that the solution to the equation
d2 y
+ y = 2x
dx2
Printed: September 22, 2013
5
with boundary conditions y(x = 0) = 0, y(x = 1) = 0 is
1
4 X ( 1)n+1
y(x) =
sin(n⇡x).
⇡ n=1 n(1 n2 ⇡ 2 )
(b) Show that the r.m.s. value of y(x) is
v
u 1
p
4u
1X
1
2
hy (x)i = t
2
⇡ 2 n=1 n (1 n2 ⇡ 2 )2
6. If the function f (x) is periodic with period 2⇡ and has a complex Fourier Series representation
f (x) =
1
X
fn einx
n= 1
then show that the solution of the di↵erential equation
dy
+ ay = f (x)
dx
is
y(x) =
1
X
n=
fn inx
e .
a
+
in
1
7. An RLC series circuit has a sinusoidal voltage V0 sin !t imposed, so the current I obeys:
L
d2 I
dI
+ R + CI = !V0 cos !t.
dt2
dt
(a) What is the fundamental period of the voltage?
(b) Write I(t) as a Fourier Series,
1
a0 X
I(t) =
+
[an cos(n!t) + bn sin(n!t)]
2
n=1
and show that an and bn satisfy
C
1
⇥
⇤
a0 X
+
an Ln2 ! 2 cos(n!t) Rn! sin(n!t) + C cos(n!t) +
2
n=1
⇥
⇤
bn Ln2 ! 2 sin(n!t) + Rn! cos(n!t) + C sin(n!t)
= !V0 cos !t.
(c) Hence show that only a1 and b1 survive, with amplitudes
!V0 ( L! 2 + C)
(C L! 2 )2 + R2 ! 2
! 2 V0 R
=
.
(C L! 2 )2 + R2 ! 2
a1 =
b1
Printed: September 22, 2013
6
8. A simple harmonic oscillator with natural frequency !0 and no damping is driven by a driving
acceleration term f (t) = sin t + sin 2t.
(a) Write down the di↵erential equation which the displacement y(t) obeys.
(b) Compute the fundamental period of the driving terms on the right hand side, and hence
T (where the solution is assumed periodic on T < t < T ).
(c) Assuming the solution is periodic with the same fundamental period as the driving term,
find the resultant motion.
(d) Calculate the r.m.s. displacement of the oscillator.
Printed: September 22, 2013
7
Fourier Analysis
Workshop 4: Fourier Transforms
Professor John A. Peacock
School of Physics and Astronomy
jap@roe.ac.uk
Session: 2013/14
15th & 18th October 2013
Handin Deadline: 4 p.m. Friday 25th October 2013
1. Prove that, for a real function f (x), its Fourier Transform satisfies f˜( k) = f˜⇤ (k).
2. In terms of f˜(k), the Fourier Transform of f (x), what are the Fourier Transforms of the
following?
(a) g(x) = f ( x)
(b) g(x) = f (2x)
(c) g(x) = f (x + a)
(d) g(x) = df /dx.
(e) g(x) = xf (x)
3. Consider a Gaussian quantum mechanical wavefunction
(x) = A exp
✓
x2
2 2
◆
,
where A is a normalisation constant, and the width of the Gaussian is .
(a) Compute the Fourier Transform ˜(k) and show that it is also a Gaussian.
(b) Noting that the probability density function is | (x)2 |, show by inspection
that the
p
uncertainty in x (by which we mean the width of the Gaussian) is x = / 2.
(c) Then, using de Broglie’s relation between the wavenumber k and the momentum, p = h̄k,
compute the uncertainty in p, and demonstrate Heisenberg’s Uncertainty Principle,
p x=
h̄
.
2
4. Express the Fourier Transform of g(x) ⌘ eiax f (x) in terms of the FT of f (x).
5. The function f (x) is defined by
f (x) =
Printed: September 22, 2013
⇢
x
e
0
8
x>0
x<0
(a) Calculate the FT of f (x), and, using Q4, of eix f (x) and of e
ix
f (x).
(b) Hence show that the FT of g(x) = f (x) sin x is
1
(1 + ik)2 + 1
(c) Finally, calculate the FT of h(x) = f (x) cos x.
6. (a) Show that the FT of f (x) = e
a|x|
is f˜(k) = 2a/(a2 + k 2 ), if a > 0.
(b) Sketch the FT of the cases a = 1 and a = 3 on the same graph, and comment on the
widths.
(c) Using the result of question 4, show that the FT of g(x) = e
ha (x) ⌘
⇢
sin x is
4ik
.
4 + k4
g̃(k) =
7. Let
|x|
ax
e
0
(a) Show that the FT of ha (x) is
h̃a (k) =
x 0
x < 0.
1
.
a + ik
(b) Take the FT of the equation
df
(x) + 2f (x) = h1 (x)
dx
and show that
f˜(k) =
(c) Hence show that f (x) = e
x
e
2x
1
1 + ik
1
.
2 + ik
is a solution to the equation (for x > 0).
(d) Verify your answer by solving the equation using an integrating factor.
(e) Comment on any di↵erence in the solutions.
8. Compute the Fourier Transform of a top-hat function of height h and width 2a, which is
centred at x = d = a. Sketch the real and imaginary parts of the FT.
Printed: September 22, 2013
9
Fourier Analysis
Workshop 5: Dirac Delta Functions
Professor John A. Peacock
School of Physics and Astronomy
jap@roe.ac.uk
Session: 2013/14
22nd & 25th October 2013
1. By using the result that if, for all functions f (x),
Z
1
f (x)g(x)dx =
1
Z
1
f (x)h(x)dx
1
then g(x) = h(x), show that
(a) ( x) = (x)
Hint: show that
(b) (ax) =
(x)
|a|
(c) (x2 a2 ) =
(d) x (x) = 0.
Z
1
f (x) ( x)dx = f (0) =
1
Z
1
f (x) (x)dx
1
(x a)+ (x+a)
.
2|a|
2. Evaluate
R
1
(a) 1 f (x) (2x 3) dx
R2
(b) 1 f (x) (x 3) dx
3. Evaluate
Z
⇡+0.1
dx
0.1
Z
4
dy (sin x) (x2
y2)
1
4. Show that the derivative of the Dirac delta function has the property that
Z
1
1
d (t)
f (t) dt =
dt
5. What are the Fourier Transforms of:
(a) (x)
(b) (x d)
Printed: September 22, 2013
10
df
dt
t=0
(c) (2x)?
By writing (x) as an integral (i.e. as an Inverse Fourier Transform) show that
(d) ⇤ (x) = (x)
6. Evaluate
Z
where a is a constant.
7. Compute
where b is a constant.
Z
1
1
Z
1
1
Z
1
1
1
1
Z
Z
1
1
t2 e
iax itx
e
dxdt
1
eixy e
ixz ibz iyt
e e e
t3
dxdydzdt
1
8. A one-dimensional harmonic oscillator with natural frequency !0 is driven with a driving
acceleration a(t), so obeys
d2 z
+ !02 z = a(t).
dt2
(a) Take the Fourier Transform of this equation (from t to !) and show that
z̃(!) =
(b) Hence show that
1
z(t) =
2⇡
Z
1
1
ã(!)
!02 ! 2
ã(!) i!t
e d!.
!02 ! 2
(c) If a(t) = sin2 ⌦t, find ã(!).
(d) Hence find a solution for z(t) (ignore solutions to the homogeneous equation).
Printed: September 22, 2013
11
Fourier Analysis
Workshop 6: Convolutions, Correlations
Professor John A. Peacock
School of Physics and Astronomy
jap@roe.ac.uk
Session: 2013/14
29th October & 1st November 2013
Handin Deadline: 4 p.m. Friday 8th November 2013
1. Show that convolving a signal f (t) with a Gaussian smoothing function
1
g(t) = p
exp
2⇡
✓
t2
2 2
◆
results in the Fourier Transform being ‘low-pass filtered’ with a weight exp(
2
! 2 /2).
2. Show that the FT of a product h(x) = f (x)g(x) is a convolution in k-space:
1 ˜
h̃(k) =
f (k) ⇤ g̃(k) =
2⇡
Z
1
1
3. Show that the convolution of a Gaussian of width
dk 0 ˜ 0
f (k )g̃(k
2⇡
k 0 ).
with a Gaussian of width 2 gives another Gaussian, and calculate its width. (A Gaussian of width has the form N exp[ x2 /(2 2 )]).
1
4. Show that the FT of the cross-correlation h(x) of f (x) and g(x),
h(x) =
5. A signal f (x) = e
Z
1
f ⇤ (x0 )g(x0 + x)dx0
is h̃(k) = f˜⇤ (k)g̃(k).
1
x
for x > 0 and zero otherwise.
(a) Show that the Fourier Transform is f˜(k) = (1 + ik) 1 .
(b) Using Parseval’s theorem, relate the integral of the power |f˜(k)|2 to an integral of |f (x)|2 .
(c) Hence show that
Z
1
1
dk
= ⇡.
1 + k2
(d) If the signal is passed through a low-pass filter, which sets the Fourier transform coefficients to zero above |k| = k0 , calculate k0 such that the filtered signal has 90% of the original
power.
Printed: September 22, 2013
12
6. (a) Compute the Fourier Transform of
h(t) =
⇢
e
bt
t 0
t < 0.
0
(b) A system obeys the di↵erential equation
dz
+ !0 z = f (t)
dt
By using Fourier transforms, show that a solution of the equation is a convolution of f (t)
with
⇢ !t
e 0 t 0
g(t) =
0
t<0
i.e.
z(t) =
Z
1
f (t0 )g(t
t0 ) dt0 ,
1
and write down the full expression for z(t).
7. (a) Show that the FT of h(t) = e
a|t|
, for a > 0 is
h̃(!) =
a2
2a
.
+ !2
(b) A system obeys the di↵erential equation
d2 z
dt2
!02 z = f (t).
Calculate z̃(!) in terms of f˜(!).
(c) By considering the form of z̃(!), show using the convolution theorem that a solution of
the equation is the convolution of f (t) with some function g(t).
(d) Using your answer to (a), find the function g(t) and write down explicitly a solution to
the equation.
8. Compute the Fourier Transform of
h(x) =
⇢
1 |x|  1
0 otherwise
Show that the convolution H(x) ⌘ h(x) ⇤ (x a) is 1 if a 1 < x < a + 1 and zero otherwise,
and compute its Fourier transform directly, and via the convolution theorem.
9. A triple slit experiment consists of slits which each have a Gaussian transmission with Gaussian width , and they are separated by a distance d
far from the slits, and sketch it.
Printed: September 22, 2013
13
. Compute the intensity distribution
Fourier Analysis
Workshop 7: Sampling, and Green’s Functions
Professor John A. Peacock
School of Physics and Astronomy
jap@roe.ac.uk
Session: 2013/14
5th & 8th November 2013
1. (a) Expand
1
1 z
as a Taylor series about z = 0, or as a power series.
(b) Hence show that
1
X
eijk
x
1
=
1
j=0
z
where z = exp(ik x).
(c) Similarly, show that
0
X
eijk
x
=
j= 1
(d) Finally, show that if z 6= 1,
1
X
e
1
.
1 1/z
ijk x
= 0.
j= 1
2. Letting p = dy/dt, and then using an integrating factor, show that the general solution to
d2 y dy
+
=0
dt2
dt
is y(t) = A + Be t , where A and B are constants.
3. Show that the Green’s function for the range x
0, satisfying
@ 2 G(x, z)
+ G(x, z) = (x
@x2
z)
with boundary conditions G(x, z) = @G(x, z)/@x = 0 at x = 0 is
⇢
cos z sin x sin z cos x x > z
G(x, z) =
0
x<z
Printed: September 22, 2013
14
4. Consider the equation, valid for t
0
d2 f
df
+ 5 + 6f = e t ,
2
dt
dt
subject to boundary conditions f = 0, df /dt = 0 at t = 0. Find the Green’s function G(t, z),
showing is is zero for t < z, and for t > z it is
G(t, z) = e2z
2t
e3z
3t
.
(You may find the complementary function (homogeneous solution) by using a suitable trial
function).
Hence show that the solution to the equation is
1
f (t) = e
2
t
e
2t
1
+ e
2
3t
.
5. (a) Show that the Green’s function for the equation, valid for t
0
2
d y dy
+
= f (t),
dt2
dt
with y = 0 and dy/dt = 0 at t = 0, is
G(t, T ) =
(b) Hence show that if f (t) = Ae
2t
⇢
1
0
eT
t
t<T
.
t>T
, the solution is
y(t) =
A
1
2
2e
t
+e
2t
.
6. The equation for a driven, damped harmonic oscillator is
d2 y
dy
+ 2 + (1 + k 2 )y = f (t)
2
dt
dt
(a) If the initial conditions are y = 0 and dy/dt = 0 at t = 0, show that the Green’s function,
valid for t 0, is
⇢
A(T )e t cos kt + B(T )e t sin kt 0 < t < T
G(t, T ) =
C(T )e t cos kt + D(T )e t sin kt
t>T
(b) Show that A = B = 0 and so G(t, T ) = 0 for t < T .
(c) By matching G(t, T ) at t = T , and requiring dG/dt to have a discontinuity of 1 there,
show that, for t > T
G(t, T ) =
eT t
( sin kT cos kt + cos kT sin kt) .
k
(d) Hence if f (t) = e t , find the solution for y(t).
Printed: September 22, 2013
15
Fourier Analysis
Workshop 8: Partial Di↵erential Equations
Professor John A. Peacock
School of Physics and Astronomy
jap@roe.ac.uk
Session: 2013/14
12th & 15th November 2013
Handin Deadline: 4 p.m. Friday 22nd November 2013
1. Find solutions u(x, y) by separation of variables to
(a)
x
@u
@x
y
@u
=0
@y
xy
@u
=0
@y
(b)
@u
@x
2. Consider a particle of mass m which is confined within a square well 0 < x < ⇡, 0 < y < ⇡.
The steady-state 2D Schrödinger equation inside the well (where the potential is zero) is
h̄2 2
r =E .
2m
The walls have infinite potential, so = 0 on the boundaries.
(a) Find separable solutions (x, y) = X(x)Y (y) and show that they are
(x, y) = A sin(rx) sin(ny)
for integers r, n.
(b) The wavefunction is normalised so that
R
| (x, y)|2 dx dy = 1. For given r, n, find A.
(c) Show that the energy levels corresponding to the quantum numbers m, n are
E = (r2 + n2 )
h̄2
.
2m
3. Show by direct substitution into the equation that
u(x, t) = f (x
ct) + g(x + ct)
where f and g are arbitrary functions, is a solution of the 1D wave equation,
@ 2u
1 @ 2u
=
,
@x2
c2 @t2
Printed: September 22, 2013
16
where the sound speed c is a constant. You may recall that the partial derivative of f (y)
with respect to x (where y may be a function of several variables y(x, t, . . .)) is
@f
@y df
=
.
@x
@x dy
4. Consider the wave equation (with sound speed unity) for t > 0
@ 2u
@ 2u
=
@x2
@t2
with initial conditions u(x, 0) = h(x) and @u/@t|t=0 = v(x).
(a) Write down d’Alembert’s solution for u(x, t).
(b)If h(x) and v(x) are known only for 0 < x < 1, then find the regions in the x, t plane for
which the solution for u can be determined, and sketch it.
5. (a) Consider separable solutions for the temperature u(x, t) = X(x)T (t) of the 1D heat
equation
@ 2u
@u
=
2
@x
@t
and find the di↵erential equations which X and T must satisfy, giving your reasoning.
(b) Solving for T , show that the separable solutions which are finite as t ! 1 are of the
form
[A cos(kx) + B sin(kx)] exp( k 2 t).
where k 2 > 0.
(c) There is one more (rather simple) permitted solution. What is it?
(d) Following on from the last question, find all solutions for which u(0, t) = u(⇡, t) = 0 at
all times. Hint: the answer is not a single term, but rather a sum.
(e) If the initial temperature (at t = 0) is u(x, 0) = sin x cos x, what is the full solution
u(x, t)?
6. This is a question which looks hard, because it uses polar coordinates, but you can solve it
in exactly the same way as the cartesian equations.
Laplace’s equation in polar coordinates r, ✓ is
@ 2 u 1 @u
1 @ 2u
+
+
=0
@r2
r @r r2 @✓2
(a) Show that for solutions which are separable, u(r, ✓) = R(r)⇥(✓),
⇥00 (✓) =
k 2 ⇥;
r2 R00 (r) + rR0 (r)
k 2 R(r) = 0
for some constant k 2 .
(b) Argue that the solution must be periodic in ✓, and say what the period must be.
Printed: September 22, 2013
17
(c) As a consequence, what values of k are permitted?
(d) By trying power-law solutions R(r) / r↵ , find the general solution which is finite at the
origin.
(e) Find the solution for a situation where u is fixed on a circular ring at r = 1 to be
u(r = 1, ✓) = sin2 ✓ + 2 sin ✓ cos ✓.
Printed: September 22, 2013
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Fourier Analysis
Workshop 9: Fourier PDEs
Professor John A. Peacock
School of Physics and Astronomy
jap@roe.ac.uk
Session: 2013/14
19th & 22nd November 2013
1. If we have a function u(x, t), we may do a partial Fourier Transform, changing x to k but
leaving t in the equations. We have used the result that

@u(x, t)
@ ũ(k, t)
FT
=
@t
@t
Show this (you can probably fit it on one line).
2. Consider the 1D wave equation
@ 2u
1 @ 2u
=
,
@x2
c2 @t2
with boundary conditions at t = 0 that u(x, t) = e a|x| for some a > 0, and @u(x, t)/@t = 0.
(a) By applying a Fourier Transform with respect to x, show that the FT of the general
solution is of the form
ũ(k, t) = A(k)e ikct + B(k)eikct .
(b) Show that at t = 0,
2a
.
+ k2
(c) Hence, applying the boundary conditions, show that
ũ(k, 0) =
ũ(k, t) =
a2
a2
a
e
+ k2
ickt
+ eickt .
Note that you will need to argue that the boundary condition on @u/@t also applies to each
Fourier component individually.
(d) Finally deduce that
u(x, t) =
1
e
2
a|x ct|
+e
a|x+ct|
3. Consider the 1D heat equation for the temperature u(x, t),
@ 2u
@u
=
@x2
@t
where the initial condition is that u(x, t = 0) = (x).
Printed: September 22, 2013
19
(a) Take the Fourier Transform with respect to x, i.e.
Z 1
ũ(k, t) =
u(x, t)e
ikx
dx
1
Note that the transform is still a function of t. Show that it obeys
@ ũ(k, t)
=
@t
k2
ũ(k, t).

(b) Now fix the value of k for now, and use an integrating factor to find
ũ(k, t) = f˜(k)e
k2 t/
for some (arbitrary) function f˜(k).
(c) From the initial condition u(x, 0) = (x) show that
f˜(k) = ˜(k) ) ũ(k, t) = ˜(k)e
k2 t/
.
p
2
2
(d) Using the result that the FT of e x /(4t) is 4⇡t/e k t/ , show using the convolution
theorem that the general solution for u(x, t) in terms of (x) is
p Z 1

0 2
p
u(x, t) =
e (x x ) /(4t) (x0 ) dx0
4⇡t 1
(e) If (x) = (x
1), what is u(x, t)?
4. Using d’Alembert’s method, show that the solution to the wave equation
2
@ 2u
2@ u
=
c
@t2
@x2
with the boundary conditions
@u
(x, t = 0) = v(x) = x
@t
u(x, t = 0) = h(x) = 0
is
⇤
1 ⇥
(x + ct)2 (x ct)2 .
4c
If v(x) = x only for 0 < x < 1 (and is zero otherwise), what is the solution? Note - you will
have to consider many di↵erent combinations depending on the values of x ct and x + ct be guided by the spacetime diagram which is in the notes.
u(x, t) =
5. The charge density ⇢ and the electrostatic potential
r2 (x) =
Printed: September 22, 2013
20
⇢(x)
✏0
are related by Poisson’s equation
where we assume that there is no time-dependence. Treating this as a one-dimension problem
(so r2 ! d2 /dx2 ), show using a Fourier Transform that a Gaussian potential
(x) =
x2 /(2
e
2)
is sourced by a charge density field
⇢(x) =
✏0
2
e
x2 /(2
2)
In doing this, you will demonstrate that
Z 1
dk 2 k2 2 /2 ikx
1
k e
e =p
2⇡
2⇡
1
✓
3
x2
1
2
x2 /(2
e
◆
.
2)
✓
1
x2
2
◆
.
You can use this result without proof in handin question 6.
Verify the solution by direct di↵erentiation of
(x).
p
R1
2
2
2 2
2
You may assume that the Fourier Transform of e x /(2 ) is 2⇡ e k /2 , and that 1 e u /2 du =
p
2⇡. (This method is of more practical use if ⇢ is known and you want , when the direct
method here cannot be employed).
6. (Hint: do all of this in cartesian coordinates - do not be tempted to use spherical polars,
despite the symmetry of the problem).
The charge density ⇢ and the electrostatic potential
r2 (r) =
are related by Poisson’s equation
⇢(r)
✏0
where we assume that there is no time-dependence. Treating this now as a 3D problem (so
r2 ! @ 2 /@x2 + @ 2 /@y 2 + @ 2 /@z 2 ), show using a Fourier Transform that a Gaussian potential
(r) =
e
r2 /(2
2)
(where r2 = x2 + y 2 + z 2 ) has a charge density FT given by
⇢˜(k) = (2⇡)3/2
3
✏0 k 2 e
k2
2 /2
where k 2 = kx2 + ky2 + kz2 .
and so the potential is sourced by a charge density field
✓
◆
✏0
r2
2
⇢(r) = 2 3
e r /(2
2
2)
.
p
2
2
2 2
You may Rassume that thepFourier Transform (w.r.t. x, ! kx ) of e x /(2 ) is 2⇡ e kx /2 ,
2
2
2
1
and that 1 e u /2 du = 2⇡. You can also assume the inverse FT of k 2 e k /2 which you
proved in question 5. Hint: you will be faced with an integral with 3 terms in it (involving
kx2 + ky2 + kz2 ). Do one of them only, and engage brain to write down the answer for the other
two without doing more algebra.
Printed: September 22, 2013
21
Fourier Analysis
Workshop 10: Revision: Green’s functions and convolutions
Professor John A. Peacock
School of Physics and Astronomy
jap@roe.ac.uk
Session: 2013/14
26th & 29th November 2013
Marks out of 25 (like exam paper). There are no hand-ins from this revision workshop.
1. On the mysterious and enigmatic planet Pendleton, a planetary explorer vehicle falls o↵ a
cli↵ at t = 0. The acceleration due to gravity is a constant g, and the vehicle attempts
to slow down its motion by applying an upward acceleration f (t). Unfortunately the fuel
in the vehicle rapidly runs out, so the upward acceleration decays with time according to
f (t) = ae t , for a constant a.
The equation of motion for the height z(t) is evidently
d2 z(t)
= f (t) g ⌘ F (t).
dt2
(a) If the top of the cli↵ is at z = 0, then evidently z(t = 0) = 0. What is dz/dt at t = 0?
(b) Write down the equation for the Green’s function G(t, T ).
(c) Hence show that G(t, T ) = 0 for t < T .
(d) Show that the solution for G(t, T ) for t > T is
G(t, T ) = t
T
t > T.
(e) Hence show that the general solution for the vertical height as a function of time is
z(t) = a(e
t
1 2
gt .
2
1 + t)
(f) Verify your answer by directly integrating the equation, with the appropriate boundary
conditions.
(g) Show that the early time behaviour is z(t) = (a
g)t2 /2 + O(t3 ).
2. The e↵ects of magnet errors in a synchrotron require the solution of the equation
d2 y(✓)
+ ! 2 y(✓) = g(✓)
d✓2
where ✓ is an angle which lies in the range 0  ✓  2⇡ and ! is fixed. The solution is
periodic, so the boundary conditions are
y(0) = y(2⇡);
Printed: September 22, 2013
dy
d✓
22
=
✓=0
dy
d✓
.
✓=2⇡
(a) Write down the equation which the Green’s function G(✓, z) satisfies.
(b) Write down the solutions for ✓ < z and ✓ > z in terms of complex exponentials.
(c) Applying the supplied boundary conditions and continuity of G(✓, z) at ✓ = z, simplify
the Green’s function to
G(✓, z) = A(z) ei!✓
e
i!✓+2i!z 2⇡i!
for some arbitrary function A(z).
(d) Similarly show that
G(✓, z) = A(z)(e
2⇡i!+i!✓
+ e2i!z
i!✓
)
✓>z
(e) Applying the boundary condition for the derivative of G at ✓ = z, show that
A(z) =
1
2i!(e
2⇡i!
1)
Hence show that the Green’s function is
e i!z
2i!(e 2⇡i!
e i!z
G(✓, z) =
2i!(e 2⇡i!
G(✓, z) =
1)
1)
ei!✓ + e
e
i!✓+2i!z 2⇡i!
2⇡i!+i!✓
+ e2i!z
i!✓
✓<z
✓>z
3. Find the Fourier transform of the ‘aperture function’ of a double slit. This consists of a
function which is two top hats, centred on x = a and x = a. Each has a width 2b and
height 1. i.e. the function h(x) is equal to zero, except for a b < x < a + b and
a b < x < a + b, where it is equal to unity.
Do this using convolutions. i.e. note that h(x) is the convolution of the sum of two delta
functions, f (x) = (x a) + (x + a) with a top hat g(x) = 1 if |x| < b, and g = 0 otherwise.
(a) Show that the FT of f (x) is f˜(k) = 2 cos(ka).
(b) Show that the FT of g(x) is
g̃(k) =
2 sin(kb)
.
k
(c) Hence write down (assuming the convolution theorem) the Fourier transform of h(x).
4. In this question, the logic is the important part, not the algebra so much, so write plenty of
words so it is clear that you understand what is going on.
Consider the equation
dy
+ y = f (t)
dt
for some as yet unspecified function f (t), for t > 0. The function f (t) = 0 for t < 0.
Printed: September 22, 2013
23
(a) Solve this first by using an integrating factor to show that
Z t
0
y(t) =
e (t t ) f (t0 )dt0 + Ae t
0
where A is a constant.
(b) This is a convolution of f with what function (answer carefully)?
(c) Now solve the equation using a Fourier Transform w.r.t. t:
Z 1
ỹ(!) =
y(t)e i!t dt
1
to show that
ỹ(!) =
f˜(!)
i! + 1
where f˜(!) is the FT of f (t).
(d) We see that this is a multiplication in Fourier space. What does this mean for the
solution in real (t) space?
(e) By inspecting the solution in (a) obtained with an integrating factor, can you guess what
function has the Fourier Transform (1+i!) 1 ? Support your answer by explicitly calculating
the FT of the function.
(f) Hence write down the general solution obtained by the Fourier Transform method. It
is not quite the same as in part (a). Pay careful attention to the limits, which you should
justify.
(g) Why are we allowed to add an extra term Ae
(h) If f (t) = e
2t
t
to the FT solution?
, and y(0) = 1, find the full solution.
Printed: September 22, 2013
24

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