Questions

Question 1

Show that φ=f(x-ct)+g(x+ct) satisfies the 1-D wave equation.

Answer 1

1-D wave equation:
∂²φ/∂x² = 1/c² ∂²φ/∂t²

Let φ = f(u) + g(v), with u = x − ct and v = x + ct

∂φ/∂x = df/du ∂u/∂x + dg/dv ∂v/∂x
= df/du + dg/dv

∂φ/∂t = df/du ∂u/∂t + dg/dv ∂v/∂t = −c df/du + c dg/dv

⇒ ∂²φ/∂x² = d²f/du² + d²g/dv²
∂²φ/∂t² = (−c)² d²f/du² + c² d²g/dv²

This substitutes into the 1-D wave equation

Question 2

General solution to y ∂f(x,y)/∂y = f(x, y)

Answer 2

ln f(x, y) = ln y + h(x)
⇒ f(x, y) = g(x)y, for any h(x) = ln g(x)

Question 3

Separation of variables for a stretched string

Answer 3

φ = X(x)T(t)

sub into the wave equation
T d²X/dx² = X 1/c² d²T/dt²

divide by φ = X(x)T(t)
1/X d²X/dx² = 1/c² 1/T d²T/dt² = −k²

now have two uncoupled ODEs
1/X d²X/dx² = −k²
1/T d²T/dt² = −c²k²

Question 4

Form of solutions for stretched string

Answer 4

d²X/dx² = −k²X
solution of the form X = Acos(kx) + Bsin(kx)
φ = 0 at x = 0, so A = 0
X ∝ sin(kx)
φ = 0 at x = L, so k = nπ/L = kₙ

d²T/dt² = −c²k²
solution of the form T = Aₙcos(ckₙt) + Bₙsin(ckₙt)

Combine all terms for superposition solution of weighted eigenmodes
φ(x, t) = Σₙ₌₁,₂ sin(kₙx) (Aₙcos(ωₙt) + Bₙsin(ωₙt))
where ωₙ = ckₙ

Question 5

Full solution for string plucked at center with an amplitude of L/2

Answer 5

φ(x, t) = Σₙ₌₁,₂ sin(kₙx) (Aₙcos(ωₙt) + Bₙsin(ωₙt))

The functional representation of plucking the string at t=0 is
φ(x, 0) = x if 0 ≤ x ≤ L/2
= L − x if L/2 ≤ x ≤ L

Sring starts from rest, initial speed is zero
∂φ/∂t |ₜ₌₀ = 0
⇒ Bₙ = 0

We are left with
φ(x, t) = Σₙ Aₙ sinkₙx cosωₙt

Evaluating at t = 0 results in a Fourier sine series
φ(x, 0) = Σₙ Aₙ sinkₙx

To evaluate Aₙ, use orthonality.
Multiply both sides by sin mπ/L and integrate over the length of the string.
∫₀ᴸ sin(mπx/L) φ(x,0) dx = Σₙ ∫₀ᴸ Aₙ sin(mπx/L) sin(nπx/L) dx

RHS
For m = n
Aₙ ∫₀ᴸ sin²(nπx/L) dx = Aₙ/2 ∫₀ᴸ (1 − cos(2nπx/L)) dx
= Aₙ/2 L

m /= n can be shown to equal 0

LHS (can also integrate by parts)
∫₀ᴸ sin(mπx/L) φ(x,0) dx = ∫₀ᴸ’² sin(mπx/L) x dx + ∫ₗ/₂ᴸ sin(mπx/L) (L − x) dx

First integral is given by
∫₀ᴸ’² x sin(nπx/L) dx = −L²/2nπ cos(nπ/2) + (L/nπ)² sin(nπ/2)

Second is given by
∫ₗ/₂ᴸ sin(mπx/L) (L − x) dx = ∫ₗ/₂ᴸ L sin(mπx/L) dx − ∫ₗ/₂ᴸ x sin(mπx/L) dx
∫ₗ/₂ᴸ x sin(mπx/L) dx = −L²/nπ cosnπ + L²/2nπ cos(nπ/2) − (L/nπ)² sin(nπ/2)
∫ₗ/₂ᴸ L sin(mπx/L) dx = −L²/nπ cosnπ + L²/2nπ cos(nπ/2)

Collecting terms
Aₙ L/2 = 2(L/nπ)² sin(nπ/2)
= 2L²/n²π² (−1)^((n−1)/2), ∀n = odd

Finally we have
φ(x, t) = 4L/π² [sin(πx/L) cos(πct/L) − 1/3² sin(3πx/L) cos(3πct/L) + 1/5² sin(5πx/L) cos(5πct/L) − …]
= 4L/π² Σ₁∞ (−1)^(n+1)/(2n−1)² sin((2n−1)πx/L) cos((2n−1)πct/L)

Question 6

Prove the orthogonality of
∫₀ᴸ sin(mπx/L) sin(nπx/L)dx

Answer 6

Using angle addition formulae
∫₀ᴸ sin(mπx/L) sin(nπx/L)dx = 1/2 ∫₀ᴸ cos((m−n)πx/L) − cos((n+m)πx/L)dx
= L/2π [sin((m−n)π)/m−n − sin((n+m)π)/m+n]
= 0 ∀m ≠ n

Question 7

Derive complex Fourier series from trigonometric

Answer 7

Trig form:
f(x) = a₀/2 + Σ(1to∞) aₙcos(nπx/L) + bₙsin(nπx/L)
where aₙ = 1/L ∫±L cos(nπx/L) f(x) dx
and bₙ = 1/L ∫±L sin(nπx/L) f(x) dx

Write cos and sin in exponential form:
cos(x) = (exp(ix) + exp(−ix)) / 2
sin(x) = (exp(ix) − exp(−ix)) / 2i

f(x) = a₀/2 + Σ(1to∞) aₙ((exp(inπx/L) + exp(−inπx/L)) / 2) + bₙ((exp(inπx/L) − exp(−inπx/L)) / 2i)
where aₙ = 1/2L ∫±L (exp(inπx/L) + exp(−inπx/L)/2) f(x) dx
and bₙ = 1/2L ∫±L (exp(inπx/L) − exp(−inπx/L)/2i) f(x) dx

Using 1/i = −i, this can be rearranged into:
f(x) = a₀/2 + Σ(1to∞) ((aₙ−ibₙ)/2 exp(inπx/L) + (aₙ+ibₙ)/2 exp(−inπx/L))

Define
cₙ = (aₙ−ibₙ)/2 = 1/2L ∫±L (cos(nπx/L) − isin(nπx/L)) f(x) dx
= 1/2L ∫±L exp(−inπx/L) dx
Also
(aₙ+ibₙ)/2 = 1/2L ∫±L (cos(nπx/L) + isin(nπx/L)) f(x) dx
= 1/2L ∫±L exp(inπx/L) dx
= c₋ₙ

Substituting these in:
f(x) = a₀/2 + Σ(1to∞) cₙexp(inπx/L) + Σ(1to∞) c₋ₙexp(−inπx/L)

Change limits:
f(x) = a₀/2 + Σ(1to∞) cₙexp(inπx/L) + Σ(-∞to-1) cₙexp(inπx/L)

Constant term is given by:
c₀ = 1/2L ∫±L f(x) exp(i0πx/L) dx
= 1/2L ∫±L f(x) dx = a₀/2

Combine for complex form:
f(x) = Σ±∞ cₙexp(−ikₙx)
where cₙ = 1/2L ∫±L f(x) exp(−ikₙx) dx
and kₙ = nπ/L

Question 8

Derive the Fourier transform from the Fourier series

Answer 8

Exponential form of the Fourier series
f(x) = Σ±∞ cₙ exp(ikₙx)
cₙ = 1/2L ∫±L f(x) exp(−ikₙx)

Using F(k) = 2Lacₙ, and writing kₙ = nπ/L, ∆k = π/L
F(kₙ) = a ∫±L f(x) exp(−ikₙx) dx
f(x) = 1/2πa Σ±∞ F(kₙ) exp(ikₙx)∆k

In the limit of L → ∞
F(k) = a ∫±∞ f(x) exp(−ikx) dx
f(x) = 1/2πa ∫±∞ F(k) exp(ikx) dk

F(k) is the Fourier transform of f(x)
f(x) is the inverse Fourier transform of F(k)

The constant a is unity for the physics definition, and 1/√2π’ for the maths definition.

Question 9

Fourier series of a square wave

Answer 9

Square function of period 2L, defined as
f(x) = 1 for 0 < x < L
= −1 for 0 > x > −L

a0 = −1/L ∫(-Lto0) dx + ∫(0toL) dx = 0
an = −1/L ∫(-Lto0) cos(nπx/L) dx + ∫(0to1) cos(nπx/L) dx = 0

Components of an do not contribute. Can also use symmetry of the problem to determine this.

bn = −1/L ∫(-Lto0) sin(nπx/L) dx + ∫(0to1) sin(nπx/L) dx
Replace x with -x in one of the integrals:
bn = 2/L∫(0to1) sin(nπx/L) dx
= 2/nπ (1−cosnπ) = 2L/nπ (1−(−1)ⁿ)

Expansion becomes
f(x) = 2Σ(1to∞) (1−(−1)ⁿ)/nπ sin(nπx/L)
= 4/π Σ(1to∞) sin[(2n−1)πx/L]/(2n−1)
where n is replaced with 2n−1 as only odd numbers contribute to the sum.

Question 10

Fourier series of a repeating parabola.
f(x) = x²
Function is confined to ±L and outside this region it repeats.

Answer 10

Function is even, so bₙ = 0
Use integration by parts twice to get
aₙ = 4L²/π²n² (−1)ⁿ
Also calculate
a₀ = 1/L ∫±L x² dx = 2/3 L²

Substitute into Fourier series equation to get
f(x) = L²/3 + 4L²/π² Σ(1to∞) (−1)ⁿ/n² cos(nπx/L)

Question 11

Fourier series of a sawtooth function
y = x/L
Function is confined to ±L and outside this region it repeats.

Answer 11

Odd function, so aₙ = a₀ = 0
bₙ = 1/L ∫±L sin(nπx/L) x/L dx
= 1/L² [x(-L/nπ cos(nπx/L)) + ∫L/nπ cos(nπx/L) dx ]
= 1/L² [-Lx/nπ cos(nπx/L)) + L²/n²π² sin(nπx/L)]
sin(nπ) ≡ 0, so ignore sine term
= -1/L² (L²/nπ cos(nπ) + L²/nπ cos(-nπ))
cos(nπ) ≡ cos(-nπ)
= -2 (-1)ⁿ/nπ

f(x) = Σ(1to∞) -2 (-1)ⁿ/nπ sin(nπx/L)
= 2/π Σ(1to∞) (-1)ⁿ⁺¹/n sin(nπx/L)

Question 12

Consider a 2D box of sides L such that 3 of the sides are grounded and held at potential φ = 0, and the remaining (right-hand) side is held at a potential φ = V.
Find the potential φ(x, y)

Answer 12

Need to solve Laplace’s equation
∂²φ/∂x² + ∂²φ/∂y² = 0

Boundary conditions:
φ(x, 0) = φ(x, L) = φ(0, y) = 0
φ(L, y) = V

Use separation of variables:
φ(x, y) = X(x)Y(y)
Laplace’s equation becomes
Y ∂²X/∂x² + X ∂²Y/∂y² = 0
Dividing by φ = XY
1/X d²X/dx² + 1/Y d²Y/dy² = 0

Now have two separable equations
1/X d²X/dx² = k²
1/Y d²Y/dy² = −k²

Y equation solved by
Y(y) = Acos(ky) + Bsin(ky)

Y boundary conditions
Y(0) = 0
⇒ Acos(0) + Bsin(0)
A = 0

Y(L) = 0
Bsin(kL) = 0
k = nπ/L = kₙ
k is constrained so B is irrelevant

X equation solved by
X(x) = Aₙexp(kₙx) + Bₙexp(−kₙx)
with boundary condition X(0) = 0
⇒ Aₙ = −Bₙ
⇒ Xₙ = 2Aₙ sinh(nπx/L)

Sum over all possible excitations
φ(x, y) = X(x)Y(y) = Σ(1to∞) 2Aₙ sinh(nπx/L) sin(nπy/L)

Final boundary condition, φ(L,y) = V
φ(L, y) = Σₙ 2Aₙ sinh(nπ) sin(nπy/L) = V

Determine Aₙ by orthogonality, multiplying by sin(mπy/L) and integrating over all x (0 to L)
∫₀ᴸ sin(mπy/L) sin(nπy/L) dx = L/2 if m = n

2Aₙ sinh(nπ) L/2 = ∫₀ᴸ V sin(nπy/L) dx
= VL/nπ [−cos(nπy/L)]₀ᴸ
= VL/nπ (1−cos(nπ))

Aₙ = V/π 1/nsinh(nπ) (1+(−1)ⁿ⁺¹)

Only non-zero for odd values of n.
Final solution is
φ(x, y) = 4V/π Σ(odd n) 1/nsinh(nπ) sinh(nπx/L) sin(nπy/L)

Question 13

Derive the diffusion equation for heat flowing in one dimension

Answer 13

Diffusion equation:
∇²T = 1/α² ∂T/∂t

Heat energy in general:
Q = cmT
where c is specific heat capacity, m is body mass and T is temperature

Fourier’s law of heat transfer temperature gradient: rate of heat transfer per unit area is proportional to negative temperature gradient
Q̇ 1/A = −K ∂T/∂x
where Q̇ is time derivative of heat, K is thermal conductivity, and A is cross-sectional area

Consider a uniform rod of length L with non-uniform temperature lying on the x-axis from x = 0 to x = L.
Uniform rod, so density ρ, specific heat C, thermal conductivity K, cross-sectional area A, ALL constant. Assume sides of the rod are insulated and only the ends are exposed.

Arbitrary thin slice of the rod of from x to x + ∆x. Temperature throughout the slice is T(x, t).

Heat energy of segment = Q = cρA∆xT(x,t)

Change in heat energy of segment in ∆x = Heat in from left boundary − Heat out from right boundary:
Q(x+∆x, t) − Q(x,t) = −cρA∆xT

Differentiate with respect to time t
(Q̇(x+∆x, t) − Q̇(x, t))/∆x = −cρA ∂T/∂t

Use Fourier’s law Q̇ = −AK ∂T/∂x
∂Q̇/∂x = −AK ∂²T/∂x² = −cρA ∂T/∂t

A terms cancel, now have the Heat Equation in one dimension
∂²T/∂x² = 1/α² ∂T/∂t
where α² = K/cρ is the thermal diffusivity.

Note:
* Steady state solution has ∂T/∂t = 0 and reduces
to the Laplace equation.
* There are two associated boundary conditions
– (a) fix the temperature on some particular surface
– (b) fix the rate of heat flow across a surface

Question 14

Boundary condition for no heat flowing across a plane perpendicular to the x-axis at x = L.

Answer 14

i.∇T |ₓ₌ₗ = 0
i.e.
∂T/∂x |ₓ₌ₗ = 0

Question 15

Boundary condition for no heat flowing out of a sphere of radius a

Answer 15

r.∇T|ᵣ₌ₐ = 0
i.e.
∂T/∂r |ᵣ₌ₐ = 0

Question 16

Show that
Σ±∞ δ(x−2nπ) = 1/2π Σ±∞ exp(inx)

Answer 16

Recalling exponential Fourier series
f(x) = Σ±∞ cₙexp(inπx/L)
Replaced πx/L with x (succinct normalised form)
= Σ±∞ cₙexp(inx)
where cₙ = 1/2π ∫±π f(x) exp(−inx) dx

Evaluate the coefficients of the function
f(x) = Σ±∞ δ(x−2nπ)
by integrating over one period
cₙ = 1/2π ∫(0to2π) f(x) exp(−inx) dx = 1/2π

hence
Σ±∞ δ(x−2nπ) = 1/2π Σ±∞ exp(inx)

Question 17

Fourier transform of the rectangular (“top hat”) function, bound by ±a

Answer 17

f(x) = 1 when −a ≤ x ≤ a, 0 elsewhere

Using mathematical Fourier transform:
F(k) = 1/√2π’ ∫±∞ f(x) exp(−ikx) dx = 1/√2π’ ∫±a exp(−ikx) dx
= 1/√2π’ (1/ik) [− exp(−ika) + exp(ika)]
[ ] part is 2i sin(ka)
= √2/π’ sin(ka)/k = √2/π’ a sinc(ka)

Question 18

Fourier transform of the Gaussian function

Answer 18

Gaussian
f(x) = 1/√2π’σ exp(−x²/2σ²)

Physics Fourier tranform
F(k) = ∫±∞ f(x) exp(−ikx) dx = 1/√2π’σ ∫±∞ exp(−x²/2σ²) exp(−ikx) dx

Complete the square of the exponent
−x²/2σ² −ikx = −( x/√2’σ + ikσ/√2’ )² − k²σ²/2
Defining ξ = x/√2’σ + ikσ/√2’

Fourier integral becomes
F(k) = 1/√2π’σ ∫±∞ exp(−ξ²) exp(−k²σ²/2) dξ (dξ/dx)⁻¹
= exp(−k²σ²/2)/√π’ ∫±∞ exp(−ξ²)
where we have used dξ/dx = 1/√2’σ

Final result
F(k) = exp(−σ²k²/2)

Question 19

Prove Parseval’s theorem

Answer 19

Consider the inverse Fourier symmetric transform
∫±∞|f(x)|² dx = 1/2π ∫±∞∫±∞ F(k) exp(ikx) dk ∫±∞ F∗(k′) exp(−ik′x) dk′dx
= ∫±∞ F(k) dk ∫±∞ F∗(k′) dk′ 1/2π ∫±∞ exp(i(k−k′)x) dx
Note that 1/2π ∫±∞ exp(i(k−k′)x) dx = δ(k−k′)
= ∫±∞|F(k)|² dk

Question 20

Prove the relation
H(k) = F(k)G(k)

Answer 20

h(x) = ∫±∞ f(x−x′) g(x′) dx′
H(k) = ∫±∞ ∫±∞ dx dx′ f(x−x′) g(x′) exp(−ikx)

Define ζ = x − x′
H(k) = ∫±∞ ∫±∞ dx′ dζ f(ζ) g(x′) exp(−ik(ζ+x′))
= ∫±∞ g(x′) exp(−ikx′) dx′ ∫±∞ f(ζ) exp(−ikζ) dζ
= G(k)F(k)

Question 21

Convolution of a rectangle with itself (with a=1/2)

Answer 21

Π(x) = 1 for −a ≤ x ≤ a, 0 elsewhere
h(x) = Π(x) ∗ Π(x) = ∫±∞ Π(x′) Π(x−x′) dx′ = Λ(x)

Fourier transform H(k) is equal to the square of the transform of the rectangle function.
H(k) = sinc²(k/2) = ( sin(k/2) / k/2 )²

Question 22

Taylor expansion of eˣ about x=0, prove series converges

Answer 22

f⁽ⁿ⁾(x) = dⁿ/dxⁿ eˣ = eˣ
f⁽ⁿ⁾(0) = 1

Taylor expansion
eˣ = Σ(0to∞) 1/n! xⁿ

To find if the series converges, compute the
ratio of successive terms
Analyse the function
f(x) = Σ(0to∞) aₙxⁿ

If the ratio of successive terms is less than unity then the series converges:
lim(n→∞) [aₙ₊₁ xⁿ⁺¹ / aₙxⁿ] = lim(n→∞) [aₙ₊₁ x / aₙ] < 1

For f(x) = eˣ
aₙ₊₁/aₙ x = xⁿ⁺¹/(n+1)! n!/xⁿ
= x/n+1 → 0 as n → ∞ ∀x

Therefore the Taylor expansion of eˣ about x = 0 converges for all values of x.

Question 23

Taylor series of f(x) = (1 − x)⁻¹, and when does it converge

Answer 23

1/1−x = Σ(0to∞) xⁿ

Coefficients of succesive terms are identical (aₙ₊₁ = aₙ)
Ratio of successive terms
aₙ₊₁/aₙ x = x

Series converges provided |x| < 1, i.e. −1 < x < 1
Outside this range it diverges.

Question 24

Derive the Hermite polynomials

Answer 24

Hermite’s equation
d²y/dx² − 2x dy/dx + 2ny = 0
for n > 0

Consider a summation of terms as a general solution
y = Σ(k=0to∞) aₖxᵏ

Substitute in
Σ(k=2to∞) aₖk(k−1)xᵏ⁻² − 2x Σ(k=1to∞) aₖkxᵏ⁻¹ + 2n Σ(k=0to∞) aₖxᵏ = 0

Rearrange
Σ(k=0to∞) [aₖ₊₂(k+2)(k+1)xᵏ − 2aₖkxᵏ + 2naₖkxᵏ] = 0

Equate coefficients of xᵏ
aₖ₊₂(k+2)(k+1) − 2kaₖ + 2naₖ = 0

now have recurrence relation
aₖ₊₂ = 2(k−n)/(k+2)(k+1) aₖ

Final solution
y = a₀ [1 − 2n x²/2! − 2n(4−2n) x⁴/4! − …] + a₁ [x + (2−2n) x³/3! + (2−2n)(6−2n) x⁵/5! + …]

a₀ = 1 and a₁ = 0 for even solutions, a₀ = 0 and a₁ = 1
recurrence relation gives aₖ₊₂ = 0 at k = n

normalise by multiplying by 2ⁿ