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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 othere.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
Dont panic! Well 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
===
=
K
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
==
=
T
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
+=
+=
+=
rootscomplex :3 Case roots equal real, :2 Case roots unequal real, 1: Case
Second order* 0 0 222
=++=++ 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: 022
= kydx
ydwhere k can be +ve, -ve, 0
xFxEykkDCxyk
BeAeykk xx
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 +++++=
For example, if we are told: xBeyxy = 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
=
+
Laplaces 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 Laplaces Equation
( ) xeyxv y cos, =Consider the entirely different function: xe
xv y cos2
2=
We find
xeyv y cos2
2=
and
022
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
qLAqLA ==
===
: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
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(), with period
2, coefficients an and bn can be found such that
( ) 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
=Lx
T
T0
- T0
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
Well do both in the next lecture
A page scanned from HLT
( ) nbnaf nn
nn
sincos10
=
=
+= : that told are weSuppose
The orthogonality relationships massively simplify finding the coefficients. We first multiply the function f(), by cosm and integrate between 0 and 2
( ) =
dmcosf21 2
0
dmcosnsinb21dmcosncosa
21
n1n
2
0n
0n
2
0
=
=
+
Reversing the orders of summation and integration on the right hand side gives
( )
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
==
giving
( )2
admcosmcosa21dmcosf
21 m
m
2
0
2
0
==
giving
so
( )
dmcosf1a2
0m =
Similarly,
( )
dmsinfbm =2
0
1
These are the coefficients for the full-range series, ie those for which 0 < < 2. Orthogonality relationships also hold for half-range series (ie those for which 0 < < ) which are also useful. They are
( )
dncosf2a0
n = ( )
dnsinf2b0
n =
Three equations dominate
Diffusion (or heat) equation
Laplaces (or potential) equation
Wave Equation
022
2
2
=
+
yu
xu
2
21xu
tu
=
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 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, dAlemberts 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. Laplaces equation: first, separation of variables (again), Laplaces
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:
dmcosncos21 2
0
is periodic with period 2, and n and m are integer.
First take the case m n.
( ) ( )[ ]
( ) ( ) 0nmsinnm
1nmsinnm
141
dnmcosnmcos41dmcosncos
21
2
0
2
0
2
0
=
++
+=
++=
*Kreysig 8th Edn, page 530 & A3
Now take the case m = n.
( )212
21
4121
41
21 2
0
2
0
22
0
=
+=+=
nsinn
dncosdncos
We can do similar things for sinn sinm and sinn cosmand so obtain the orthogonality relationships:
( )nm0
0nm1nm5.0dmcosncos
21 2
0 =====
forwhenfor
( )nm
0nm0nm5.0dmsinnsin
21 2
0 =====
for 0whenfor
n,m0dmcosnsin21 2
0
allfor=