Fourier series
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Fourier Series. Dr. K.W. Chow Mechanical Engineering. Introduction. Conceptual question: While one can readily see that two vectors can be ‘perpendicular’ or ‘orthogonal’, how can we extend this concept to a sequence of functions?. Introduction.

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Fourier series

Fourier Series

Dr. K.W. Chow

Mechanical Engineering


Introduction

Introduction

  • Conceptual question: While one can readily see that two vectors can be ‘perpendicular’ or ‘orthogonal’, how can we extend this concept to a sequence of functions?


Introduction1

Introduction

  • A general formulation: For a sequence of functions {φn} and

    f(x) = Σcn φn

    What is cn?

    IF ∫ φm φn dx = 0 for m, n different, then

    cn can be found fromthis ‘orthogonal’ property.


Introduction2

Introduction

  • A general theory has been developed for linear, second order differential equations regarding these issues:

    (a) Orthogonal ‘eigenfunctions’?

    (b) Completeness in terms of expansion? (i.e. is it possible for any arbitrary function f(x) to be represented as a sum of φn(x)?)


Introduction3

Introduction

  • Sin(x) and Cos(x) are the solutions of the simplest second order ordinary differential equations (ODEs)

  • d2y/dx2 + y = 0,

  • subject to certain boundary conditions.


Introduction4

Introduction

  • Fourier series is an infinite series of sine and cosine

  • Dirichlet’s theorem (Sufficient, but not necessary):

    If a function is 2L-periodic and piecewise continuous, then its Fourier series converges

  • At a point of discontinuity, the series converges to the mean value.


Fourier series1

Fourier series

  • an and bn are calculated by the orthogonal properties of sines and cosines.

  • If one uses a0 as the constant term, two schemes for defining an, n = 0and n > 0.

  • If one uses a0/2, then one definition for all an .


Introduction5

Introduction

  • Consider:


Fourier series

f(x)


Fourier series

f(x)

n = 3


Fourier series

f(x)

n = 3

n = 7


Fourier series

f(x)

n = 3

n = 7

n = 15


Introduction6

Introduction

  • The n = 15 series gives a very good approximation to the original function


Introduction7

Introduction

Even function

Odd function


Introduction8

Introduction

  • Even extension of a function results in Fourier cosine series

  • Odd extension of a function results in Fourier sine series

    (Assuming function is given only in half the interval.)


Introduction9

Introduction

Even extension

Odd extension


Introduction10

Introduction

Even extension of f : Fourier series of f1will have cosine terms only

Odd extension of f : Fourier series of f2will have sine terms only


Fourier series

  • Consider

    using odd extension


Fourier series

f(x)


Fourier series

f(x)

n = 3


Fourier series

f(x)

n = 3

n = 7


Fourier series

f(x)

n = 3

n = 7

n = 50


Introduction11

Introduction

The Fourier series converges to this value at the discontinuity.


Introduction12

Introduction


Introduction13

Introduction

Approximation using n = 1 series


Introduction14

Introduction

Approximation using n = 2 series


Introduction15

Introduction

Approximation using n = 3 series


Introduction16

Introduction

  • Gibbs phenomenon – large oscillations of the series near a discontinuity.


Introduction17

Introduction

  • Consider the square wave again:

5 terms


Introduction18

Introduction

25 terms


Introduction19

Introduction

25 terms


Introduction20

Introduction

  • Animation to show the Gibbs phenomenon using different partial sums.


Differentiation of series

Differentiation of series

  • Fourier coefficients are determined uniquely using orthogonality of trigonometric functions.

    only when the series on the right is uniformly convergent. (d/dx = a local operator)


Integration of series

Integration of series

  • Always permitted, but the resulting series is not a Fourier series, unless the constant term a0 is zero.

  • (Integration = a global operator).


1d heat conduction in finite domain

1D heat conduction in finite domain

  • Configuration:

Density of conductor:

Specific heat capacity: c

Heat conduction coefficient: k

Cross-sectional area: A

Temperature:

Temperature:

L


1d heat conduction in finite domain1

1D heat conduction in finite domain

Thermal conductivity defined by

  • Heat flux = - k A∂u/∂x

    where A = cross sectional area,

    u = temperature

    (i.e. k is heat flux per unit area per unit temperature gradient).


1d heat conduction in finite domain2

1D heat conduction in finite domain

Assumptions:

  • Heat flow in the xdirection only

  • No external heat source

  • No heat loss


1d heat conduction in finite domain3

1D heat conduction in finite domain

  • Consider heat conduction across an infinitesimal element of the conductor:

Heat out

Heat in


1d heat conduction in finite domain4

1D heat conduction in finite domain

  • Heat flux at the left surface : - k A∂u(x,t)/∂x

  • Heat flux at the right surface:

    - k A ∂u(x,t)/∂x - ∂[k A ∂u(x,t)/∂x]/∂x dx + …

  • Net heat flux INTO the element:

  • ∂[k A ∂u(x,t)/∂x]/∂x dx


1d heat conduction in finite domain5

1D heat conduction in finite domain

  • Net heat must be used to heat up the element (c = specific heat capacity):


1d heat conduction in finite domain6

1D heat conduction in finite domain

  • Therefore (if k and A not functions of x):


1d heat conduction in finite domain7

1D heat conduction in finite domain

  • When , the above equation becomes exact:

    This is the 1D heat conduction equation in finite domain.


1d heat conduction in finite domain8

1D heat conduction in finite domain

Solution procedure

  • Separation of variables:

  • Substitute back into the heat equation:


1d heat conduction in finite domain9

1D heat conduction in finite domain


1d heat conduction in finite domain10

1D heat conduction in finite domain

  • Assume both ends are kept at :

  • For non-trivial solution, choose:


1d heat conduction in finite domain11

1D heat conduction in finite domain

  • F satisfies the differential equation:

  • For non-trivial solutions:

  • The temporal part:


1d heat conduction in finite domain12

1D heat conduction in finite domain

  • Overall solution:

  • Using superposition principle, we obtain general solution:


1d heat conduction in finite domain13

1D heat conduction in finite domain

  • is the Fourier sine coefficient:


1d heat conduction in finite domain14

1D heat conduction in finite domain

  • is called the eigen-value

  • is called the

    corresponding eigen-function


1d heat conduction in finite domain15

1D heat conduction in finite domain

  • Consider:


Fourier series

t = 0


Fourier series

t = 0

t = 0.5


Fourier series

t = 0

t = 0.5

t = 1.2


Fourier series

t = 0

t = 0.5

t = 1.2

t = 5


Fourier series

t = 0

t = 0.5

t = 1.2

t = 5

t = 15


1d heat conduction in finite domain16

1D heat conduction in finite domain

  • The above procedure cannot be applied directly when the end points are not at

  • First find steady-state temperature distribution v:

Note that the steady-state temperature depends on x only.


1d heat conduction in finite domain17

1D heat conduction in finite domain

  • Introduce a new function (transient):

  • w satisfies the heat equation with homogeneous boundary conditions.

  • Solve for w with separation of variables and hence u can be found.


1d heat conduction in finite domain18

1D heat conduction in finite domain

  • Consider:


Fourier series

t = 0


Fourier series

t = 0

t = 0.5


Fourier series

t = 0

t = 0.5

t = 1.2


Fourier series

t = 0

t = 0.5

t = 1.2

t = 5


Fourier series

t = 0

t = 0.5

t = 1.2

t = 5

t = 12


Laplace s equation

Laplace’s equation

  • For heat conduction in higher dimensions,

    where is the Laplacian.


Laplace s equation1

Laplace’s equation

  • The steady-state solution in 2D satisfies:

    which is the Laplace’s equation

  • In the presence of heat sources, u satisfies the Poisson’s equation:


Laplace s equation2

Laplace’s equation

3 types of boundary conditions:

  • Dirichlet boundary condition

  • Neumann boundary condition

  • Robin boundary condition


Laplace s equation3

Laplace’s equation

  • Dirichlet boundary condition


Laplace s equation4

Laplace’s equation

  • Neumann boundary condition


Laplace s equation5

Laplace’s equation

  • Robin boundary condition


Laplace s equation6

Laplace’s equation

Consider (Dirichlet b.c.)


Laplace s equation7

Laplace’s equation

u(x, y) = F(x) G(y)

F’’/F = – G’’/G = constant

F(0) = F(a) = 0 and thus the constant is

– n2π2 / L2 .

Hence

F ~ sin (n πx/L)

G ~ cosh


Laplace s equation8

Laplace’s equation

  • Solution is obtained using separation of variables:


Laplace s equation9

Laplace’s equation

  • Consider


Fourier series

y

x

Isotherms (curves joining points with the same temperature) of the problem


1d wave equation in finite domain

1D wave equation in finite domain

  • Configuration:

  • Assumptions:

    - CONSTANT TENSION and density,

    - the slope of the vibration is small,

    - gravity much smaller than tension.

L


1d wave equation in finite domain1

1D wave equation in finite domain


1d wave equation in finite domain2

1D wave equation in finite domain

  • Vertical force at the left end:

    Vertical force at the right end:

  • u = u(x, t) = displacement of the string


1d wave equation in finite domain3

1D wave equation in finite domain

  • Net vertical upward force on the element:


1d wave equation in finite domain4

1D wave equation in finite domain

  • Newton’s second law (ρ = linear density):


1d wave equation in finite domain5

1D wave equation in finite domain

  • When , the equation becomes exact.

  • If no external force is present:


1d wave equation in finite domain6

1D wave equation in finite domain

  • c = speed of the wave

  • Check the dimensions: Square root of (force/mass per unit length).

  • Mathematically, signs of two second derivatives same (contrast with Laplace equation).


1d wave equation in finite domain7

1D wave equation in finite domain

  • Consider a vibrating string with two ends fixed and initial position and velocity given;

  • i.e. initial and boundary conditions:


Separation of variables

Separation of Variables

u(x, y) = F(x) G(t)

F’’/F = G’’/(c2 G) = constant

F(0) = F(a) = 0 and thus the constant is

– n2π2 / L2 .

Hence

F ~ sin (n πx/L)

G ~ C1sin (c n πt/L)+C2 cos (c n πt/L)


1d wave equation in finite domain8

1D wave equation in finite domain

  • Solution obtained by separation of variables:


1d wave equation in finite domain9

1D wave equation in finite domain

  • Modes of vibration are the profiles of the envelopes corresponding to different n

First mode

Second mode

Third mode


1d wave equation in finite domain10

1D wave equation in finite domain

  • The mode shapes will oscillate with time


1d wave equation in finite domain11

1D wave equation in finite domain

  • Another example – Vibration of a stretched string with a triangular initial profile.

  • Qualitatively, the shapes of the string at subsequent times are similar to the n = 1 mode previously. However, the shapes remain piecewise linear, and some ‘corners’ persist.


1d wave equation in finite domain12

1D wave equation in finite domain

  • Consider:


1d wave equation in finite domain13

1D wave equation in finite domain


Fourier series

t = 0

Forward cycle


Fourier series

t = 0

Forward cycle

t = L/6c


Fourier series

t = 0

Forward cycle

t = L/6c

t = L/3c


Fourier series

t = 0

Forward cycle

t = L/6c

t = L/3c

t = L/2c


Fourier series

t = 0

Forward cycle

t = L/6c

t = L/3c

t = L/2c

t = 2L/3c


Fourier series

t = 0

Forward cycle

t = L/6c

t = L/3c

t = L/2c

t = 2L/3c

t = 5L/6c


Fourier series

Backward cycle

t = L/c


Fourier series

Backward cycle

t = 7L/6c

t = L/c


Fourier series

Backward cycle

t = 4L/3c

t = 7L/6c

t = L/c


Fourier series

Backward cycle

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c


Fourier series

Backward cycle

t = 5L/3c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c


Fourier series

Backward cycle

t = 11L/6c

t = 5L/3c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c


Fourier series

Backward cycle

t = 2L/c

t = 11L/6c

t = 5L/3c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c


1d wave equation in finite domain14

1D wave equation in finite domain

  • General solution (d’Alembert) to the wave equation:

  • Using trigonometric identities, the Fourier series solution can also be rewritten in the above form.


1d wave equation in finite domain15

1D wave equation in finite domain

  • x- ctrepresents a wave traveling to the right with speed c


1d wave equation in finite domain16

1D wave equation in finite domain

  • x+ ct represents a wave traveling to the left with speed c


1d wave equation in finite domain17

1D wave equation in finite domain

  • The lines are known as characteristic lines (or just characteristics).

  • The forms of functions Fand G depend on the initial conditions.

  • The initial profile splits into two waves of same amplitude traveling in opposite directions


1d wave equation in finite domain18

1D wave equation in finite domain


1d wave equation in finite domain19

1D wave equation in finite domain

  • Method of characteristics is of tremendous theoretical significance but less practical interest.

  • Usually we just use separation of variables for finite domains and integral transform for infinite ones.


Pdes in infinite domain

PDEs in infinite domain

  • Fourier series gives information in the interval

  • Fourier series will give the periodic extension outside this domain.

  • Fourier ‘fails’ if the given function is already defined along the whole real-axis.


From fourier series to fourier integrals

From Fourier series to Fourier integrals

  • To represent a function from minus infinity to plus infinity, we use a Fourier series over the interval (- L, L) and let L go to infinity.

  • Result (Fourier integral):

  • f = integral (integral f dξ) dx


From fourier series to fourier integrals1

From Fourier series to Fourier integrals

  • f(x) = sum over An cos (nπx/L)

  • An = 2/L integral f(ξ) cos (nπξ/L)dξ

  • Hence

  • f(x) = sum over n [ 2/L (integral of

    f(ξ) cos (nπ (x – ξ)/L) dξ)]

    Now convert ‘sum over n and (1/L)’ into another integral.


Pdes in infinite domain1

PDEs in infinite domain

  • Separation of variables is usually not feasible or will fail.

  • Use integral transforms:

    - Fourier transform

    - Laplace transform

    - …


Pdes in infinite domain2

PDEs in infinite domain

  • Fourier integrals are analogous to Fourier series:


Pdes in infinite domain3

PDEs in infinite domain

  • For functions defined in semi-infinite domain,

    even extension Fourier cosine integral

    odd extension Fourier sine integral


Pdes in infinite domain4

PDEs in infinite domain

  • Fourier transform pair:


Pdes in infinite domain5

PDEs in infinite domain

  • In some alternative versions, the + and - signs in the exponentials are interchanged.

  • There is no universally accepted format.

  • The constants in front of the integral signs are arbitrary, as long as their product is


Applications of fourier transform

Applications of Fourier transform

Idea:

  • A PDE defined in an infinite domain is given.

  • Apply transform on each term in the equation, with respect to a certain independent variable, e.g. x

  • Derivatives in x become algebraic in ω.


Applications of fourier transform1

Applications of Fourier transform

  • The transformed equation becomes an ODE in t (ω is a parameter, no derivatives in ω), rather than PDE in x and t.

  • Solve for the transformed function.

  • Apply inverse transform to obtain solution in the original coordinates.


Applications of fourier transform2

Applications of Fourier transform

Common techniques:

  • Integration by parts

  • Exchange order of integrations

  • Contour integrals

  • Gaussian integrals


Applications of fourier transform3

Applications of Fourier transform

  • Example : heat equation

    Spatial conditions: u(x, t) decaying in far field.

    This represents an initially concentrated source of unit intensity at the origin


Applications of fourier transform4

Applications of Fourier transform


Fourier series

t = 0.05


Fourier series

t = 0.05

t = 0.1


Fourier series

t = 0.05

t = 0.1

t = 0.5


Fourier series

t = 0.05

t = 0.1

t = 0.5

t = 5


Fourier series

t = 0.05

t = 0.1

t = 0.5

t = 5

t = 50


Fourier series

  • This represents a uniform initial temperature distribution


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