PA214 Waves and Fields Fourier Methods

Blue book New

chapter 12

• Fourier sine series• Application to the wave equation

• Fourier cosine series• Fourier full range series

• Complex form of Fourier series• Introduction to Fourier transforms

and the convolution theorem

Fourier Methods

Dr Mervyn Roy (S6)www2.le.ac.uk/departments/physics/people/academic-staff/mr6

PA214 Waves and Fields Fourier Methods

Lecture notes www2.le.ac.uk/departments/physics/people/academic-staff/mr6

214 course texts • Blue book, new chapter 12 available on Blackboard

Notes on Blackboard• Notes on symmetry and on trigonometric identities• Computing exercises• Exam tips

• mock papersBooks

• Mathematical Methods in the Physical Sciences (Mary L. Boas)• Library!

Resources

PA214 Waves and Fields Fourier Methods

The wave equation

for a string fixed at and has harmonic solutions

Introduction

𝜕2 𝑦𝜕 𝑥2

= 1𝑐2𝜕2 𝑦𝜕𝑡 2

,

𝑦 (𝑥 , 𝑡 )=sin 𝑛𝜋 x𝐿 (𝑏𝑛cos

𝑛𝜋𝑐𝑡𝐿

+𝑎𝑛sin𝑛𝜋𝑐𝑡𝐿 ) .

Superposition tells us that sums of such terms must also be solutions,

𝑦 (𝑥 , 𝑡 )=∑𝑛

sin𝑛𝜋 x𝐿 (𝑏𝑛 cos

𝑛𝜋 𝑐𝑡𝐿

+𝑎𝑛sin𝑛𝜋𝑐𝑡𝐿 ) .

PA214 Waves and Fields Fourier Methods

𝑦 (𝑥 , 𝑡 )=∑𝑛

sin𝑛𝜋 x𝐿 (𝑏𝑛 cos

𝑛𝜋 𝑐𝑡𝐿

+𝑎𝑛sin𝑛𝜋𝑐𝑡𝐿 )

Set coefficients from initial conditions, e.g. string released from rest with then

PA214 Waves and Fields Fourier Methods

What happens if the initial shape of the string is something more complex?

In general can be any function

The implication is that we can represent any function as a sum of sines

… and/or cosines or complex exponentials.

This is Fourier’s theorem

PA214 Waves and Fields Fourier Methods

We will find that a function in the range can be represented by the Fourier sine series

where

Fourier sine series (half-range)

𝑏𝑛=2𝐿∫0

𝐿

𝑓 (𝑥 ) sin 𝑛𝜋 𝑥𝐿

𝑑𝑥 .

PA214 Waves and Fields Fourier Methods

How does this work ?Need a standard integral (new chapter 12 – A.2)…

Fourier sine series

∫0

𝐿

sin𝑚𝜋 𝑥𝐿

sin𝑛𝜋 𝑥𝐿

𝑑𝑥= 𝐿2𝛿𝑛𝑚

PA214 Waves and Fields Fourier Methods

Square wave,

𝑓 (𝑥 )=∑𝑛=1

∞

𝑏𝑛sin𝑛𝜋 𝑥𝐿

,

𝑏𝑛=2𝐿∫0

𝐿

𝑓 (𝑥 ) sin 𝑛𝜋 𝑥𝐿

𝑑𝑥 .

PA214 Waves and Fields Fourier Methods

Square wave,

𝑓 (𝑥 )=∑𝑛=1

∞

𝑏𝑛sin𝑛𝜋 𝑥𝐿

,

𝑏𝑛=2𝐿∫0

𝐿

𝑓 (𝑥 ) sin 𝑛𝜋 𝑥𝐿

𝑑𝑥 .

𝑓 (𝑥 )≈ 4𝜋sin

𝜋 𝑥𝐿

PA214 Waves and Fields Fourier Methods

Square wave,

𝑓 (𝑥 )=∑𝑛=1

∞

𝑏𝑛sin𝑛𝜋 𝑥𝐿

,

𝑏𝑛=2𝐿∫0

𝐿

𝑓 (𝑥 ) sin 𝑛𝜋 𝑥𝐿

𝑑𝑥 .

𝑓 (𝑥 )≈ 4𝜋sin

𝜋 𝑥𝐿

+43𝜋

sin3𝜋 x𝐿

PA214 Waves and Fields Fourier Methods

Square wave,

𝑓 (𝑥 )=∑𝑛=1

∞

𝑏𝑛sin𝑛𝜋 𝑥𝐿

,

𝑏𝑛=2𝐿∫0

𝐿

𝑓 (𝑥 ) sin 𝑛𝜋 𝑥𝐿

𝑑𝑥 .

𝑓 (𝑥 )≈ 4𝜋sin

𝜋 𝑥𝐿

+43𝜋

sin3𝜋 x𝐿

+45𝜋

sin5𝜋 𝑥𝐿

PA214 Waves and Fields Fourier Methods

Square wave,

𝑓 (𝑥 )=∑𝑛=1

∞

𝑏𝑛sin𝑛𝜋 𝑥𝐿

,

𝑏𝑛=2𝐿∫0

𝐿

𝑓 (𝑥 ) sin 𝑛𝜋 𝑥𝐿

𝑑𝑥 .

𝑓 (𝑥 )≈ 4𝜋sin

𝜋 𝑥𝐿

+43𝜋

sin3𝜋 x𝐿

+45𝜋

sin5𝜋 𝑥𝐿

+47𝜋

sin7𝜋 𝑥𝐿

PA214 Waves and Fields Fourier Methods

Square wave,

𝑓 (𝑥 )=∑𝑛=1

∞

𝑏𝑛sin𝑛𝜋 𝑥𝐿

,

𝑏𝑛=2𝐿∫0

𝐿

𝑓 (𝑥 ) sin 𝑛𝜋 𝑥𝐿

𝑑𝑥 .

𝑓 (𝑥 )≈ 4𝜋sin

𝜋 𝑥𝐿

+43𝜋

sin3𝜋 x𝐿

+…+419𝜋

sin19𝜋 𝑥𝐿

PA214 Waves and Fields Fourier Methods

Periodic extension of Fourier sine series

𝑓 (𝑥 )=∑𝑛=1

∞

𝑏𝑛sin𝑛𝜋 𝑥𝐿

sin𝑛𝜋 𝑥𝐿

=− sin−𝑛𝜋 𝑥

𝐿,

sin𝑛𝜋 𝑥𝐿

=sin𝑛𝜋 (𝑥+2𝐿)

𝐿.

and that

We know that sine waves have odd symmetry,

PA214 Waves and Fields Fourier Methods

Within can expand any function as a sum of sine waves,

𝑓 (𝑥 )=∑𝑛=1

∞

𝑏𝑛sin𝑛𝜋 𝑥𝐿

.

How does this expansion behave outside of the range ?

PA214 Waves and Fields Fourier Methods

square wave

sawtooth wave

exp wave (odd)

PA214 Waves and Fields Fourier Methods

String fixed at and

The wave equation

Initial conditions and

𝑦 (𝑥 , 𝑡 )=∑𝑛

sin𝑛𝜋 x𝐿 (𝐵𝑛cos

𝑛𝜋𝑐𝑡𝐿

+𝐴𝑛sin𝑛𝜋𝑐𝑡𝐿 ).

𝐵𝑛=2𝐿∫0

𝐿

𝑝 (𝑥 ) sin 𝑛𝜋 𝑥𝐿

𝑑𝑥

𝑛𝜋 𝑐𝐴𝑛

𝐿= 2𝐿∫0

𝐿

𝑞 (𝑥 ) sin 𝑛𝜋 𝑥𝐿

𝑑𝑥

PA214 Waves and Fields Fourier Methods

e.g. and , then (see workshop 1, exercises 1 & 2)

PA214 Waves and Fields Fourier Methods

e.g. and , then (see workshop 1, exercises 1 & 2)

PA214 Waves and Fields Fourier Methods

e.g. and (see new chapter 12, exercise 12.5)

PA214 Waves and Fields Fourier Methods

e.g. and (see new chapter 12, exercise 12.5)

PA214 Waves and Fields Fourier Methods

Can go through the same procedure with the solutions to other PDEse.g. Laplace equation (see workshop 1 exercise 3),

𝜕2𝜙𝜕𝑥2

+ 𝜕2𝜙𝜕 𝑦2

=0.

Imagine the boundary conditions are and then

𝜙 (𝑥 , 𝑦 )=∑𝑛=1

∞

𝐵𝑛sin𝑛𝜋 x𝐿

𝑒−𝑛𝜋 𝑦 /𝐿

and can find coefficients from boundary condition for