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
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
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