Partial Differential Equations & wavesjmb/lectures/pdelecture1.pdf · Partial Differential Equations Partial ... of the partial differential equation: 0 y u x u 2 ... A completely

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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=