Partial Differential Equations

& waves

Professor Sir Michael Brady FRS FREng

Michaelmas 2005

Analysing physical systemsFormulate the most appropriate mathematical model for the

system of interest – this is very often a PDE

This is what a large part of Engineering science & practice is about.

• Diffusion of charge, flow of heat, absorption of a drug• Propagation of waves across water, electrical networks, with/without loss of energy• “steady state” – no further change – in stress analysis, heat or fluid flow, …

– We will recall from ODEs: a single equation can have lots of very different solutions, the boundary conditions determine which

Figure out the appropriate boundary conditions, apply them

In this course, solutions will be analytic = algebra & calculusReal life is not like that!! Numerical solutions include finite difference and finite element techniques

Solve the PDE

…but why partial differential equations

A physical system is characterised by its state at any point in space and time

now here,in re temperatu,),,,( tzyxu

tu

∂∂State varies over time:

yxu∂∂

∂2

like thingsState also varies over space:

Surely, we need to relate these variations to each other…e.g.

2

2

xuk

tu

∂∂

=∂∂

How do we relate spatial variations to temporal variations?

• Constituent equations which you met in vector calculus embody physical constraints such as– “conservation of mass”, – “conservation of enthalpy”

Don’t panic! We’ll work mostly in one spatial dimension

In the case of an insulated, diffusing distribution of heat, the equation (which we will derive later) is:

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∂∂

2

2

2

2

2

2

zu

yu

xuk

tu

That is, the spatial change is directly proportional k to the temporal change

An example of a PDE: the one-dimensional heat equation

2

22

xuc

tu

∂∂

=∂∂

material the ofdensity heatspecific

tyconductivi thermal

:case this In

===

=

ρσ

σρK

Kc2

Another example:the one-dimensional wave

equation

2

22

2

2

xuc

tu

∂∂

=∂∂

string the of length mass/unit string the in tension

:case this In

==

=

ρ

ρT

Tc2

Background to this course

Partial Differential Equations

Partial differentiation

Ordinary Differential Equations

Fourier series

Numerical methods

Vector calculus

Electrical engineering

Mechanical engineering

Civil engineering

Biomedical engineeringWe now give brief reminders of partial differentiation,

ODEs, and Fourier series. Please re-read the relevant parts of Kreysig if you are shaky on some particular part

Partial derivatives*

cyaxxux +=

∂∂ 2: to respect withderivative Partial

From which:

cuau

xy

xx

== 2

dcxyyxayxu ++−= )(),( 22 :function the Consider

* Please refer to Kreysig, 8th Edition, pages A57 – A60 for a refresher

xuxu

≡∂∂ :Notation

Partial differentiation with respect to y

cu

au

cxayu

yx

yy

y

=

−=

+−=

2

2

Evidently, changing the order of differentiation makes no difference:

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

yu

xxu

y

dcxyyxayxu ++−= )(),( 22

This is the case whenever u varies “smoothly” with respect to x and y. This is almost always so.

Chain rule*

),( vuC),( and ),( yxvvyxuu ==

Suppose that we are given a function

where

dvvCdu

uCdC

∂∂

+∂∂

=The total variation in C is

yvyuy

xvxux

vCuCCvCuCC

+=+=

From which we find

*Kreysig, 8th Edition, page 444

Ordinary differential equations

First order axAeyaydxdy −=⇒=+ 0

Kreysig, 8th Edn, pp 19-21

)sincos()(

, 2121

xBxAeyjaeBAxy

BeAey

ax

x

xx

ωωω

λ

λλλ

λλ

+=⇒±−

+=⇒

+=⇒

− rootscomplex :3 Case roots equal real, :2 Case roots unequal real, 1: Case

Second order* 0 0 22

2

=++⇒=++ cbmamcdxdyb

dxyda

Auxiliary equation

*Kreysig, 8th Edn, Chapter 2

Homogeneous equations: superposition of solutions

shomogeneou called then is ODE the , Ifs.derivative its of one or function unknown the contain

: of side left the on terms The

0)()(

)()(')(''

=

=++

xrxy

xryxqyxpy

Fundamental theorem* about homogeneous ODEs:

solution. a is

solutions of ionsuperposit linearany generally, More

constants. are wheresolution, a also is then ODE, given a to solutions are and if

∑

+

iii

i

xyc

cxycxycxyxy

)(

)()()()(

2211

21

*Kreysig, 8th Edn, p66

How we use superposition of solutions

Consider: 02

2

=− kydx

ydwhere k can be +ve, -ve, 0

xFxEykkDCxyk

BeAeykk xx

βββ

α αα

sincos say ,0 0 say ,0

2

2

+=⇒−=<

+=⇒=+=⇒=> −

We “superpose” these solutions, and leave it to analysis of the boundary conditions to help us figure out which bits are relevant in any given case

)sincos()()( xFxEDCxBeAey xx ββαα +++++= −

For example, if we are told: xBeyxy α−=∞→→ then , as ,0

Partial Differential Equations generally have many different solutions

axu 22

2

=∂∂

and ayu 22

2

−=∂∂

Evidently, the sum of these two is zero, and so the function u(x,y) is a solution of the partial differential equation:

0yu

xu

2

2

2

2

=∂∂

+∂∂

Laplace’s Equation

Recall the function we used in our reminder of partial derivatives:

dcxyyxayxu ++−= )(),( 22

This choice was not random! Recall that we showed:

A completely different solution to Laplace’s Equation

( ) xeyxv y cos, −=Consider the entirely different function:

xexv y cos2

2−−=

∂∂We find

xeyv y cos2

2−=

∂∂ and

02

2

2

2

=∂∂

+∂∂

yv

xv

So that the function v(x,y) also satisfies

Boundary conditions determine the solution in any particular case

solution a also is :ionsuperpositby Evidently, ),(),( yxvyxu +

Showing that particular functions satisfy particular PDEs is the subject of Q1, Q2 on the first tutorial sheet

An example of applying specific boundary conditions

Consider the superposition of the two solutions u(x,y)+v(x,y) suppressing constants, which would make no difference:

( ) (1) DCxy)yx(BxcosAey,xu 22y ++−+= −

And, suspending reality for a moment, suppose this represents the stress in an infinite plate with a circular hole:

F F

By considering x and y at infinity, it is clear that for (1) to be a physically plausible solution, then because the stress must remain finite, we conclude that B = C = 0.

x

y

Applying the problem-specific boundary condition that the end of the bar (x=L) is maintained at zero temperature, we have

LnqnqL

qLAqLAππ ==

===

:is that , so these, of first the in interested not are We

orthat so .0sin0,0sin

Every value of n corresponds to a solution, so we use superposition to find the general solution:

∑∞

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=0

sin),(2

22

n

Ltkn

n LxneAtxT π

π

How Fourier series enter the game

qxAetxT tkq sin),(2−=

Anticipating lecture 2, suppose we are solving a specific case of the Heat Equation, to find the temperature of a bar of length L. We will find that the solution is given (in that case) by the temperature

We then apply Fourier series to solve for the nA

Fourier series* in 3 steps1. Fourier theory asserts that for any periodic function, f(θ), with period

2π, coefficients an and bn can be found such that

( ) θθθ nsinbncosaf n1n

n0n

∑∑∞

=

∞

=

+=

*Kreysig, 8th Edn, Sections 10.1-10.4, p526

2. Many functions of interest are not specified as periodic; but they can be made so by judicious choices

Lx

T

To

πθ

=Lx

T

T0

- T0

θ

3. To find the constants an and bn, we proceed in one of two ways:

a. Look up the solution in HLT

b. Figure them out from first principles using “orthogonality relations”

We’ll do both in the next lecture

A page scanned from HLT

( ) θθθ nbnaf nn

nn

sincos10

∑∑∞

=

∞

=

+= : that told are weSuppose

The orthogonality relationships massively simplify finding the coefficients. We first multiply the function f(θ), by cosmθ and integrate between 0 and 2π

( ) =∫ θθθπ

π

dmcosf21 2

0

θθθπ

θθθπ

ππ

dmcosnsinb21dmcosncosa

21

n1n

2

0n

0n

2

0∑∫∑∫

∞

=

∞

=

+

Reversing the orders of summation and integration on the right hand side gives

( ) θθθπ

θθθπ

θθθπ

πππ

dmcosnsinb21dmcosncosa

21dmcosf

21

n

2

01nn

2

00n

2

0∫∑∫∑∫

∞

=

∞

=

+=

How to apply orthogonality relationships

This is always zeroThis is zero unless m=n

( )2

admcosmcosa21dmcosf

21 m

m

2

0

2

0

== ∫∫ θθθπ

θθθπ

ππ

giving

( )2

admcosmcosa21dmcosf

21 m

m

2

0

2

0

== ∫∫ θθθπ

θθθπ

ππ

giving

so

( ) θθθπ

π

dmcosf1a2

0m ∫=

Similarly,

( ) θθθπ

π

dmsinfbm ∫=2

0

1

These are the coefficients for the full-range series, ie those for which 0 < θ < 2π. Orthogonality relationships also hold for half-range series (ie those for which 0 < θ < π) which are also useful. They are

( ) θθθπ

π

dncosf2a0

n ∫= ( ) θθθπ

π

dnsinf2b0

n ∫=

Three equations dominate

• Diffusion (or heat) equation

• Laplace’s (or potential) equation

• Wave Equation

02

2

2

2

=∂∂

+∂∂

yu

xu

2

21xu

tu

∂∂

=∂∂

κ

2

22

2

2

xuc

tu

∂∂

=∂∂

Diffusion problems, transient heat transfer, concentration in fluids, transient electric potential

Steady state problems in stress analysis, heat transfer, electrostatics, fluid flow…..

Wave phenomena in mechanical systems (vibrations), fluids, electricity…..

The general second order PDE

),(),(),(),(

),(),(),(

yxGuyxFuyxEuyxD

uyxCuyxBuyxA

yx

yyxyxx

=++

+++

042 <− ACB042 =− ACB042 >− ACB

Elliptic, if

Parabolic, if

Hyperbolic, if

LaplaceDiffusionWave

The three PDEs arise most frequently in practice, and they cover the most interesting basic PDEs

Overview of the Course1. General introduction, revision of partial differentiation, ODEs, and

Fourier series2. Wave equation in 1D part 1: separation of variables, travelling

waves, d’Alembert’s solution3. Heat equation in 1D: separation of variables, applications4. limitation of separation of variables technique. Sometimes, one

way to proceed is to use the Laplace transform5. Laplace’s equation: first, separation of variables (again), Laplace’s

equation in polar coordinates, application to image analysis 6. Wave equation in 1D part 2: phase and transverse velocity,

characteristic impedance, wave number, circular fequency, standing waves; impedance boundaries, lossy (dispersive) waves, amplitude modulation

7. Water waves8. Another look at separation of variables: Sturm-Liouville Equations

and orthogonal functions. Legendre and Bessel functions.

Books

1. Kreyszig Advanced Engineering Mathematics 8th Edition. Very big, impresses fellow students, mostly unread but can support a stereo or three pints. Most of the course “follows” the treatment in this book.

2. James: Advanced Modern Engineering Mathematics. Again comprehensive, perhaps a bit easier than Kreyszig. Somewhat duller and less impressive for your tutor.

3. Main Vibrations and Waves in Physics. Used, with James for waves section. Unlikely that your tutor will believe that you bought it, or even read it.

4. Pain The Physics of Vibrations and Waves, ditto Main.

5. Pearson Partial Differential Equations. Wonderful book, if you are a mathematician at a US ivy league university. No pictures. Generally dull.

Moral of the tale: read the notes AND Kreysig. K has more detail, fewer jokes

Reminder of the orthogonality relations*

The orhogonality relations exploit values of integrals like:

θθθπ

π

dmcosncos21 2

0∫

θ is periodic with period 2π, and n and m are integer.

First take the case m ≠ n.

( ) ( )[ ]

( ) ( ) 0nmsinnm

1nmsinnm

141

dnmcosnmcos41dmcosncos

21

2

0

2

0

2

0

=⎥⎦

⎤⎢⎣

⎡−

−++

+=

−++= ∫∫π

ππ

θθπ

θθθπ

θθθπ

*Kreysig 8th Edn, page 530 & A3

Now take the case m = n.

( )212

21

4121

41

21 2

0

2

0

22

0

=⎥⎦⎤

⎢⎣⎡ +=+= ∫∫

πππ

θθπ

θθπ

θθπ

nsinn

dncosdncos

We can do similar things for sinnθ sinmθ and sinnθ cosmθand so obtain the orthogonality relationships:

( )nm0

0nm1nm5.0dmcosncos

21 2

0 ≠=====

∫ forwhenfor

θθθπ

π

( )nm

0nm0nm5.0dmsinnsin

21 2

0 ≠=====

∫ for 0whenfor

θθθπ

π

n,m0dmcosnsin21 2

0

allfor=∫ θθθπ

π