Fourier Series and PDES

Question 1

f : R→R is a periodic function if…

Answer 1

∃p >0 s.t. f(x+p) = f(x) ∀x∈R.

Question 2

The smallest p for which f is p-periodic is called…

Answer 2

The prime period.

Question 3

The periodic extension F : R→R of f: (α,α+p]→R is defined by…

Answer 3

F(x) = f(x−mp), where for each x∈R, m is the unique integer such that x−mp∈(α,α+p].

Question 4

5 Properties of periodic functions: If f and g are p–periodic, then:

Answer 4

(1) f,g are np–periodic ∀n∈N{0};
(2) αf+βg are p–periodic ∀α,β∈R;
(3) fg is p–periodic;
(4) f(λx) is p/λ–periodic ∀λ >0;
(5) ᵖ∫₀ f(x) dx = ᵃ⁺ᵖ∫ₐ f(x) dx ∀α∈R.

Question 5

f : R→R is odd if…

Answer 5

f(x) = −f(−x) ∀x∈R.

Question 6

f : R→R is even if…

Answer 6

f(x) = f(−x) ∀x∈R.

Question 7

4 Properties of odd/even functions: If f, f1 are odd and g, g1 are even, then:

Answer 7

(1) f(0) = 0;
(2) ᵃ∫₋ₐ f(x) dx = 0 ∀α∈R;
(3) ᵃ∫₋ₐ g(x) dx = 2ᵃ∫₀ g(x) dx ∀x∈R;
(4) fg is odd, ff1 is even, and gg1 is even.

Question 8

Let f : R→R be a periodic function of period 2π. We say a Fourier expansion for f is of the form

Answer 8

f(x) ∼ a₀/2 + ∞∑ₙ₌₁(aₙcos(nx) +bₙsin(nx)).

Question 9

What is a₀ of a 2π-periodic Fourier series for f(x)?

Answer 9

a₀ = 1/π ᵖᶦ∫₋ₚᵢ f(x) dx

Question 10

Let m,n∈N{0}. Then we have the orthogonality relations:
ᵖᶦ∫₋ₚᵢ cos(mx)cos(nx) dx =
ᵖᶦ∫₋ₚᵢ cos(mx)sin(nx) dx =
ᵖᶦ∫₋ₚᵢ sin(mx)sin(nx) dx =

Answer 10

ᵖᶦ∫₋ₚᵢ cos(mx)cos(nx) dx = πδₘₙ
ᵖᶦ∫₋ₚᵢ cos(mx)sin(nx) dx = 0
ᵖᶦ∫₋ₚᵢ sin(mx)sin(nx) dx = πδₘₙ

where δₘₙ is Kronecker’s delta.

Question 11

Prove what is aₘ of a 2π-periodic Fourier series for f(x), given the series. m =/= 0

Answer 11

ᵖᶦ∫₋ₚᵢ f(x)cos(mx) dx = 1/2 a₀ᵖᶦ∫₋ₚᵢ cos(mx) dx + ∞∑ₙ₌₁ (aₙᵖᶦ∫₋ₚᵢ cos(mx)cos(nx) dx + bₙᵖᶦ∫₋ₚᵢ cos(mx)sin(nx) dx)
giving ᵖᶦ∫₋ₚᵢ f(x)cos(mx) dx = 1/2 a₀·0 +∞∑ₙ₌₁(aₙπδₘₙ+bₙ·0) = πaₘ,
so that aₘ = 1/π ᵖᶦ∫₋ₚᵢ f(x)cos(mx) dx for m∈N{0}.

Question 12

What is bₘ of a 2π-periodic Fourier series for f(x)? m =/= 0

Answer 12

bₘ = 1/π ᵖᶦ∫₋ₚᵢ f(x)sin(mx) dx for m∈N{0}.

Question 13

What is the Fourier Sine series expansion for f(x)?

Answer 13

f(x) ∼ ∞∑ₙ₌₁ bₙsin(nx),

Question 14

What does bₙ equal in a 2π-periodic Fourier Sine series for f(x)?

Answer 14

bₙ = 2/π ᵖᶦ∫₀ f(x)sin(nx) dx

Question 15

What is the Fourier Cosine series expansion for f(x)?

Answer 15

f(x) ∼ 1/2 a₀ + ∞∑ₙ₌₁ aₙcos(nx),

Question 16

What does aₙ equal in a 2π-periodic Fourier Cosine series for f(x)?

Answer 16

aₙ = 2/π ᵖᶦ∫₀ f(x)cos(nx) dx

Question 17

The right-hand limit of f at c is…

Answer 17

f(c₊) = lim₍ₕ→₀, ₕ>₀₎ f(c+h) if it exists.

Question 18

The left-hand limit of f at c is…

Answer 18

f(c₋) = lim₍ₕ→₀, ₕ

Question 19

f is continuous at c iff…

Answer 19

f(c₋) = f(c) = f(c₊).

Question 20

f is piecewise continuous on (a,b)⊆R if…

Answer 20

There exists a finite number of points x₁,…,xₘ∈R with a = x₁< x₂< … < xₘ = b such that:

(i) f is defined and continuous on (xₖ, xₖ₊₁) for all k = 1,…,m−1;
(ii) f(xₖ₊) exists for k = 1,…,m−1;
(iii) f(xₖ₋) exists for k = 2,…,m.

Question 21

What is the Fourier Convergence Theorem?

Answer 21

Let f : R→R be 2π-periodic, with f and f′ piecewise continuous on (−π, π). Then, the Fourier coefficients
aₙ = 1/π ᵖᶦ∫₋ₚᵢ f(x)cos(nx) dx (n∈N),
bₙ = 1/π ᵖᶦ∫₋ₚᵢ f(x)sin(nx) dx for n∈N{0} exist,
and 1/2 (f(x₊) + f(x₋)) = a₀/2 + ∞∑ₙ₌₁(aₙcos(nx) + bₙsin(nx)) for x∈R.

[Proof not examinable, so fun times :-) ]

Question 22

Say the Fourier Convergence Theorem is true for 2π periodic f, then what is the formula for g, the even part of f, and h, the odd part of f. and their repective fourier series?

Answer 22

g(x) = 1/2 (f(x) + f(−x))
h(x) = 1/2 (f(x) − f(−x))

1/2 (g(x₊) + g(x₋)) = a₀/2 + ∞∑ₙ₌₁aₙcos(nx), for x∈R,
1/2 (h(x₊) + h(x₋)) = ∞∑ₙ₌₁bₙsin(nx) for x∈R,

Question 23

What conditions allow the Fourier Series to be integrated? And what does the integration equal?

Answer 23

If f is only 2π-periodic and piecewise continuous on (−π,π), then
ˣ∫₀ f(s) ds = 1/2 a₀x + ∞∑ₙ₌₁(aₙˣ∫₀ cos(ns) ds + bₙˣ∫₀ sin(ns) ds) for x∈R.

Question 24

What conditions allow the Fourier Series to be differentiated? And what does the differentiation equal?

Answer 24

If f is 2π-periodic and continuous on R with both f′ and f′′ piecewise continuous on (−π,π), then
1/2 (f′(x₊) +f′(x₋)) = ∞∑ₙ₌₁ (aₙ d/dx (cos(nx)) + bₙ d/dx (sin(nx))) for x∈R

Question 25

The smoother f, i.e.the more continuous derivatives it has…

Answer 25

the faster the convergence of the Fourier series for f.

Question 26

If the first jump discontinuity is in the pth derivative of f, with the convention that p = 0 if there is a jump discontinuity in f, then typically the slowest decaying aₙ and bₙ decay like…

Answer 26

1/(nᵖ⁺¹) as n→∞.

Question 27

Gibb’s Phenomenon is…

Answer 27

The persistent overshoot near a jump discontinuity. It happens whenever a jump discontinuity exists.

Question 28

(Gibb’s Phenomenon) As the number of terms in the partial sum tends to ∞, the width of the overshoot region…

while the total height of the overshoot region…

Answer 28

the width of the overshoot region tends to 0 (by the Fourier Convergence Theorem),

the total height of the overshoot region approaches γ|f(x₊) − f(x₋)|, where γ = 1/π ᵖᶦ∫₋ₚᵢ sin(x)/x dx ≈ 1.18,

[I think we just need to know the overshoot is 9% or something, 1.18 is 9% top and bottom overshoot]

Question 29

What is the Fourier series for a function of period 2L?

Answer 29

f(x) ~ a₀/2 + ∞∑ₙ₌₁(aₙcos(nπx/L) + bₙsin(nπx/L)),

Question 30

What are the Fourier coefficients of a Fourier series of period 2L?

Answer 30

aₙ = 1/L ᴸ∫₋ₗ f(x)cos(nπx/L) dx
bₙ = 1/L ᴸ∫₋ₗ f(x)sin(nπx/L) dx.

Question 31

What are the orthogonality relations equal to for 
ᴸ∫₋ₗ cos(mπx/L)cos(nπx/L) dx = 
ᴸ∫₋ₗ cos(mπx/L)sin(nπx/L) dx =
ᴸ∫₋ₗ sin(mπx/L)sin(nπx/L) dx = 
where n,m∈N\{0}.

Answer 31

ᴸ∫₋ₗ cos(mπx/L)cos(nπx/L) dx = Lδₘₙ
ᴸ∫₋ₗ cos(mπx/L)sin(nπx/L) dx = 0,
ᴸ∫₋ₗ sin(mπx/L)sin(nπx/L) dx = Lδₘₙ

Question 32

The even 2L-periodic extension fₑ : R→R of f : [0, L] → R is defined by…

Answer 32

fₑ(x) =
{ f(x) for 0≤x≤L,
{ f(−x) for −L

Question 33

What is the Fourier Cosine series for f : [0, L]→R,

and what are the coefficients equal to?

Answer 33

The Fourier Cosine series for f is equal to the Fourier series for fₑ,
fₑ(x) ~ a₀/2 + ∞∑ₙ₌₁(aₙcos(nπx/L))

aₙ = 2/L ᴸ∫₀ f(x)cos(nπx/L) dx, (n∈N)

Question 34

The odd 2L-periodic extension fₒ : R→R of f : [0, L] → R is defined by…

Answer 34

fₒ(x) =
{ f(x) for 0≤x≤L,
{ −f(−x) for −L

Question 35

What is the Fourier Sine series for f : [0, L]→R,

and what is the coefficients equal to?

Answer 35

bₙ = 2/L ᴸ∫₀ f(x)sin(nπx/L) dx, (n∈N{0})

Question 36

Note that the odd 2L-periodic extension fₒ : R→R of f : [0, L] → R is odd for x/L∈R\Z and odd (on R) iff…

Answer 36

f(0) = f(L) = 0.

Question 37

Consider the function f: [0,L]→R defined by f(x) = x for 0≤x≤L, why does the truncated cosine series give a better approximation to f on [0,L] than the truncated sine series?

[fₑ(x) ∼ L/2 − ∞∑ₘ₌₀ (4L)/((2m+1)²π²)) * cos((2m+1)πx/L).]
[fₒ(x) ∼∞∑ₙ₌₁ (2L(−1)ⁿ⁺¹)/(nπ) * sin(nπx/L).]

Answer 37

(1) it converges everywhere on [0,L];
(2) it converges more rapidly;
(3) it does not exhibit Gibb’s phenomenon.

Question 38

What is the Fundamental Theorem of Calculus?

Answer 38

If f(x) is continuous in a neighbourhood of a, then 1/h ᵃ⁺ʰ∫ₐ f(x) dx → f(a) as h→0.

Question 39

Leibniz’s Integral Rule: Let F(x,t) and ∂F/∂t be continuous in both x and t in some region R of the (x,t) plane containing the region S = {(x,t) : a(t)≤x≤b(t), t₀≤t≤t₁}, where the functions a(t) and b(t) and their derivatives are continuous for t∈[t₀,t₁]. Then d/dt ᵇ⁽ᵗ⁾∫ₐ₍ₜ₎ F(x,t) dx =

Answer 39

d/dt ᵇ⁽ᵗ⁾∫ₐ₍ₜ₎ F(x,t) dx = ᵇ⁽ᵗ⁾∫ₐ₍ₜ₎ ∂F/∂t(x,t) dx + db(t)/dt F(b(t),t)− da(t)/dt F(a(t),t).

Question 40

What is the Heat equation? And derive it in a rigid isotropic conducting rod of constant cross-sectional area A lying along the x-axis.

Answer 40

∂T/∂t = κ ∂²T/∂x²,

Derived using the whole entire pages of 2 and 3 lectures 5-6

Question 41

Notation: We denote by [p]…

Answer 41

the dimension of the quantity p in fundamental dimensions (M,L,T,Θ etc) or e.g. SI units (kg,m,s,K etc).

Question 42

Both sides of an equation modelling a physical process must have the same…

Answer 42

dimensions

Question 43

What do these symbols stand for and what are their SI units?

x, t, T, q, A, ρ, c, k, κ (kappa)

Answer 43

x Axial distance m 
t Time s
T Absolute temperature K
q Heat flux in positive x-direction Jm⁻²s⁻¹
A Cross-sectional area m²
ρ Rod density kgm⁻³
c Rod specific heat Jkg⁻¹K⁻¹
k Rod thermal conductivity JK⁻¹m⁻¹s⁻¹
κ Rod thermal diffusivity m²s⁻¹

Question 44

What is nondimensionalisation?

Answer 44

Method of scaling variables with typical values to derive dimension-less equations. These usually contain dimensionless parameters that characterise the relative importance of the physical mechanisms in the model.

Question 45

How would you typically nondimensionalise a problem?

Here the heat equation is used as an example

Answer 45

We nondimensionalise the variables by scaling x = Lxₕₐₜ, t = τtₕₐₜ, T(x,t) = T₂Tₕₐₜ(xₕₐₜ,tₕₐₜ),

By the chain rule, ∂T/∂t = T₂(∂Tₕₐₜ/∂tₕₐₜ)(dtₕₐₜ/dt) = (T₂/τ)(∂Tₕₐₜ/∂tₕₐₜ),
∂T/∂x = T₂(∂Tₕₐₜ/∂xₕₐₜ)(dxₕₐₜ/dx) = (T₂/L)(∂Tₕₐₜ/∂xₕₐₜ), etc.

The conditions can then be modified into their dimensionless counterparts.
Then heat problems can be compared on different scales, pg 5 lectures 5-6 for more info.

Question 46

What is Fourier’s Method for solving PDEs?

Answer 46

(I) Use the method of separation of variables to find the countably infinite set of nontrivial separable solutions satisfying the partial differential equation and boundary conditions, each containing an arbitrary constant

(II) Use the principle of superposition — that the sum of any number of solutions of a linear problem is also a solution (assuming convergence) — to form the general series solution that is the infinite sum of the separable solutions of the partial differential equation and boundary conditions.

(III) Use the theory of Fourier series to determine the constants in the general series solution for which it satisfies the initial condition

Question 47

Solve the heat equation with homogenous Dirichlet boundary conditions, where ∂T/∂t = κ(∂²T/∂x²) for 0< x < L, t >0,
with the boundary conditions T(0,t) = 0, T(L,t) = 0 for t >0,
and the initial condition T(x,0) = f(x) for 0< x < L,
where the initial temperature profile f(x) is given.

Answer 47

Lectures 5-6 bottom of page 6 to halfway down page 8.
(hence why its not on here)

Its the pretty standard method though, sep of variables, using boundary conditons here,making sure T is non-trivial, finding solutions that depend on n, superpositioning these solutions, then using the initial conditions to solve the constants Fourierly.

Question 48

What are some important implications of the general solution of the heat equation, where initial temperatures at the end of the rod are fixed to 0?
(Bad flashcard I know, but there might be some things worth remembering)

Answer 48

–the heat equation smoothes out instantaneously even irregular initial temperature profiles;
–as soon as t >0, most of the high frequency terms Tn(x,t) for n»1 will be extremely small, so that the solution may be well approximated by only a handful of terms;
–the temperature tends to zero exponentially quickly as κt/L² → ∞, i.e. on the timescale of heat conduction, with the thermal energy initially stored in the rod being conducted out of the ends of the rod on this timescale.

Question 49

How do you solve for uniqueness for the heat equation solutions?

Answer 49

We suppose that T₁(x,t) and T₂(x,t) are solutions to the heat equation: Let W(x,t) = T₁(x,t) − T₂(x,t) be their difference.
Work out the boundary/initial conditions for W.
Work out W = 0:
For example:

I(t) := 1/2 ᴸ∫₀ W(x,t)² dx.
Therefore I(t) ≥0 for t≥0 and I(0) = 0
[W(x,0) = T₁(x,0) − T₂(x,0) = f(x) − f(x) = 0]

dI/dt = ᴸ∫₀ W(∂W/∂t) dx
= ᴸ∫₀ Wκ(∂²W/∂x²) dx
= [κW(∂W/∂x)]ᴸ₀ 
− κ ᴸ∫₀ (∂W/∂x)(∂W/∂x) dx
= −κᴸ∫₀ (∂W/∂x)²dx ≤0

which means that I(t) cannot increase, so that I(t) ≤ I(0) = 0 for t≥0. So I(t) = 0 for all t≥0, and hence W(x,t) = 0 for 0≤x≤L, t ≥ 0

Question 50

Solve the heat equation with inhomogenous Dirichlet boundary conditions, where
∂T/∂t = κ(∂²T/∂x²) for 00,
with the inhomogeneous Dirichlet boundary conditions
T(0,t) = Tₗ, T(L,t) = Tᵣ for t >0,
and the initial condition T(x,0) = 0 for 0< x < L, where Tₗ and Tᵣ are prescribed constant temperatures, not both zero.

Answer 50

Lectures 7-8 bottom of page 2 to bottom of page 3

(On physical grounds, we conjecture that T(x,t) → S(x) as t→ ∞, where S(x) is the afore-mentioned steady-state solution, which satisfies 0 = κ(d²S/dx²) for 0< x < L, with S(0) = Tₗ and S(L) = Tᵣ. Thus, S(x) has the linear temperature profile given by S(x) = Tₗ(1−x/L) + Tᵣ(x/L); we note that in steady state thermal energy is conducted steadily along the rod with constantheat flux q = −k(∂T/∂x) = k(Tₗ − Tᵣ)/L,so that heat flows steadily in the positive x-direction for Tₗ> Tᵣ.

We now observe that if we let T(x,t) = S(x) + U(x,t), U has homogenous Dirichlet boundary conditions, and so can be solved normally.)

Question 51

Solve the heat equation with homogenous Neumann boundary conditions, where
∂T/∂t = κ(∂²T/∂x²) for 00,
with the homogeneous Neumann boundary conditions Tₓ(0,t) = 0, Tₓ(L,t) = 0 for t >0,
and the initial condition T(x,0) = f(x) for 0< x < L

Answer 51

Lectures 7-8 middle of page 4 to top of page 5

Solved pretty normally here, just note the ends must be thermally insulated due to the boundary conditions.

Question 52

Solve the heat equation with inhomogenous Neumann boundary conditions, where
ρc∂T/∂t = k(∂²T/∂x²) + Q(x,t) for 0< x < L, t >0,
with the inhomogeneous Neumann boundary conditions Tₓ(0,t) = φ(t), Tₓ(L,t) = ψ(t) for t >0,
and the initial condition T(x,0) = f(x) for 0< x < L
where the functions Q(x,t), φ(t), ψ(t) and f(x) are given.

Answer 52

Lectures 7-8 middle of page 5 to middle of page 7

(if we let T(x,t) = S(x,t) +U(x,t), where S(x,t) = −φ(t)(x−L)²/2L + ψ(t)x²/2L, then U(x,t) has homogenous boundary conditions and so can be solved normally.
If the new Q function for U is not identically 0, U should be solved as a cosine series however:
U(x,t) = U₀(t)/2 + ∞∑ₙ₌₁ Uₙ(t)cos(nπx/L), then by computing ρc(dUₙ/dt), and substituting in the integral to have Uₓₓ inside, integration by parts can be used to get ρc(dUₙ/dt) + kn²π²/L² Uₙ = newQₙ(t).
As the newQₙ(t) coeffiecient can be determined by its the fourier expansion of newQ(x,t), then Uₙ can be determined.

Question 53

Derive the wave equation in one dimension: Consider the small transverse vibrations of a homogeneous extensible elastic string stretched initially along the x-axis at time t = 0.
A point at xi at time t = 0 is displaced to r(x,t) = xi + y(x,t)j at time t >0, where the transverse displacement y(x,t) is to be determined.

(There’s a lot more conditions, look at lectures 7-8 bottom of page 7 to middle of page 8)

Answer 53

Lectures 7-8 middle of page 8 to bottom of page 9.

(Combine linear momentum, the fact that the tension is parallel to direction of string, the string has constant linear density, look at string between x = a + h and x = a, ignore gravity, air resistance, resisitence to bending by the string, use Newton’s second law, use the fundamental theorem of calculus, and assume |yₓ| &laquo_space;1)
c = √(T/ρ) is the wave speed

Question 54

On what timescale does a displacement travel a distance L when it follows the wave equation
yₜₜ = c²yₓₓ? (Non-dimensionalise)

Answer 54

The terms balance giving
(∂²yₕₐₜ/∂tₕₐₜ²) = (∂²yₕₐₜ/∂xₕₐₜ²) provided t₀ = L/c, which is therefore the timescale for a displacement to travel a distance L.

Question 55

Solve the wave equation yₜₜ = c²yₓₓ for 0 < x < L, with the Dirichlet boundary conditions y(0, t) = 0 and y(L, t) = 0, and the initial conditions y(x, 0) = f(x), (∂y/∂t)(x, 0) = g(x)

Answer 55

Lectures 9-10 bottom of page 1 to middle of page 2

(Use seperable solutions y = F(x)G(t),  then F''(x)/F(x) = G''(t)/c²G(t) = -λ ∈R.  
As F(0)G(t) = 0, and y non-trivial, F(0) = 0, and also F(L) = 0. Then F(x) can be solved, thus enabling the solving of G, and then all the solutions can be superposed together. Then the theory of Fourier series can be used with the initial conditions to find the constants aₙ and bₙ

Question 56

What are the normal modes for the wave equation yₜₜ = c²yₓₓ for 0 < x < L, with the boundary conditions y(0, t) = 0 and y(L, t) = 0?
And what is the prime period and frequency of the normal modes?
And what is the first normal mode, what is it called, and what’s special about it?

Answer 56

A normal mode is a solution for yₙ (a solution for a specific value of n)
Its prime period is p =
2π/(nπc/L) = 2L/nc
Its frequency (or pitch) is 1/p = nc/2L

The first normal mode y₁ is called the fundamental mode, with associated fundamental frequency c/2L. All of the other modes have a frequency that is an integer multiple of the fundamental frequency.

Question 57

What is the kinetic energy of the string whose motion is governed by the wave equation yₜₜ = c²yₓₓ for 0 < x < L, with the Dirichlet boundary conditions y(0, t) = 0 and y(L, t) = 0, and the initial conditions y(x, 0) = f(x), (∂y/∂t)(x, 0) = g(x), and also with line denisty ρ and tension T?

Also what is its elastic potential energy, to a first approximation?

And so what is the total energy of the string?

Prove that if y(x, t) satifies this wave equation, then E(t) (the energy) is constant.

Answer 57

pg 6 lectures 9-10