IX.2 THE FOURIER TRANSFORMvps/ME505/IEM/09 02.pdfIX.2.5 Solution of PDEs in the Infinite Region 728 2.5.1 The Heat Equation – Gauss’s Kernel – Green’s Function 728 2.5.2 The

Chapter IX The Integral Transform Methods IX.2 The Fourier Transform November 8, 2020

721

IX.2 THE FOURIER TRANSFORM IX.2.1 The Fourier Transform – Definition 722 IX.2.2 Properties 723

IX.2.3 Examples 724

IX.2.4 Solution of ODE 726

IX.2.5 Solution of PDEs in the Infinite Region 728

2.5.1 The Heat Equation – Gauss’s Kernel – Green’s Function 728

2.5.2 The Laplace Equation – Poisson’s Integral Formula 732

IX.2.6 Fourier Integrals (Fourier Integral Representations) 736

IX.2.7 Solution of PDE’s in the Semi-Infinite Regions 738 IX.2.8 Review Questions and Exercises 750 IX.2.9 Derivation of D’Alambert’s solution 757

Chapter IX The Integral Transform Methods IX.2 The Fourier Transform November 8, 2020 722

IX.2.1 DEFINITION The Fourier Integral Transform Pair The complex Fourier transform of a function ( )f t is defined as

( ){ } ( ) i tˆF f t f ( ) f t e dtωω∞

−

−∞

= = ∫

and the inverse Fourier Transform is defined as

( ) ( ){ } ( )1 i t1ˆ ˆf t F f f e d2

ωω ω ωπ

∞−

−∞

= = ∫

The function ( )f t from the “time domain” is translated to its

spectrum – the function ( )F ω in the “frequency domain”.

The inverse Fourier transform reconstructs the function ( )f t from its spectrum:

Fourier integral formula ( ) ( )i t i i t1 1ˆf ( t ) f e d f e d e d2 2

ω ωξ ωω ω ξ ξ ωπ π

∞ ∞ ∞−

−∞ −∞ −∞

= =

∫ ∫ ∫

The question arises if the function reconstructed from its spectrum coincides with the original function. The answer to this question is given for functions satisfying certain conditions.

Theorem 9.2 The Fourier Integral Theorem (Fourier, 1822)

Let the function ( )f t satisfy the conditions (Dirichlet’s conditions):

i) ( )f t has only a finite number of finite discontinuities (jumps) in

the interval ( ),−∞ ∞ and has no infinite discontinuities.

ii) ( )f t has only the finite number of extrema (maxima and minima)

in the interval ( ),−∞ ∞ .

And let the function ( )f t be absolutely integrable on ( ),−∞ ∞ :

( )f t dt∞

−∞

< ∞∫ .

Then the Fourier integral converges to the function ( )f t at the points

( )t ,∈ −∞ ∞ where ( )f t is continuous, and it converges to

( ) ( )1 f t 0 f t 02

− + + (average value of the jump) at the points where

the function ( )f t is discontinuous:

( )( )

( ) ( )i i t

f t f is cont. at t1 f e d e d f t f t2 f is discont. at t

2

ωξ ωξ ξ ωπ

∞ ∞− − +

−∞ −∞

→ +

∫ ∫

The Theorem 9.2 provides only the sufficient condition of the existence of the Fourier and the inverse Fourier transforms – there are many other functions for which the transform also exists.

( )Jean-Baptiste Joseph Fourier 1768 - 1830

( )f t

( ) ( )i i t1 f e d e d f t2

ωξ ωξ ξ ωπ

∞ ∞−

−∞ −∞

=

∫ ∫

( ) ( )1 f t 0 f t 02

− + +

point of discontinuity

( )f t is continuous

1This definition w/o factor 2

is consistent with Mapleπ


723

IX.2.2 PROPERTIES Let ( )f t satisfy the conditions of the Fourier integral theorem on ( ),−∞ ∞ . Denote the Fourier transform and the inverse Fourier transform by

( ) ( ){ } ( ) i tf F f t f t e dtωω∞

−

−∞

≡ = ∫ ( ){ } ( )1 i t1ˆ ˆF f f e d2

ωω ω ωπ

∞−

−∞

≡ ∫

Consider some important properties of the Fourier transform:

1) Linearity: ( ) ( ){ } ( ){ } ( ){ }F af t bg t aF f t bF g t+ = +

( ) ( ){ } ( ){ } ( ){ }1 1 1ˆ ˆˆ ˆF f g F f F gα ω β ω α ω β ω− − −+ = +

2) Shifting in ω : ( ){ } ( )0i t0

ˆF e f t fω ω ω= − 0ω ∈

3) Shifting in t: ( ){ } ( ){ } ( )0 0i t i t0

ˆF f t t e F f t e fω ω ω− −− = = 0t ∈

4) Scaling: ( ){ } 1 ˆF f at fa a

ω =

a ∈

5) Duality: ( ){ } ( )ˆF f t f ω= − (not a misprint)

6) Convolution: ( ) ( )f g f t x g x dx∞

−∞

∗ = −∫

{ } { } { } ( ) ( )ˆ ˆF f g F f F g f gω ω∗ = =

( ) ( ){ }1 ˆ ˆF f g f gω ω− = ∗

Operational properties 7) Derivatives: Assume ( )tlim f t 0→∞

= and ( ) ( )k

tlim f t 0→∞

= , k 1,2,...=

( ){ } ( )ˆF f t i fω ω′ =

( ){ } ( )2 ˆF f t fω ω′′ = −

( ) ( ){ } ( ) ( )nn ˆF f t i fω ω=

Denote the Fourier transform in the x variable of the function ( )t,xu as

{ } ( ) ( ) i xˆF u( x,t ) u ,t u x,t e dxωω∞

−

−∞

≡ = ∫

Assume that ( )xlim u x,t 0→±∞

= and ( )k

kx

u x,tlim 0

x→±∞

∂=

∂, k 1,2,...= , then

8) Derivative in t: ( ) ( )i x ˆF u( x,t ) u x,t e dx u ,tt t t

ω ω∞

−

−∞

∂ ∂ ∂ = = ∂ ∂ ∂ ∫

9) Derivatives in x: ( )ˆF u( x,t ) i u ,tx

ω ω∂ = ∂

( )2

22

ˆF u( x,t ) u ,tx

ω ω ∂

= − ∂

( ) ( )n

nn

ˆF u( x,t ) i u ,tx

ω ω ∂

= ∂

( )

Fourier Transform in x variable of the function of two variables u x,t


IX.2.3 EXAMPLES 1. Unit Pulse Function – gate function – filter function

The unit pulse function can be defined with the help of the Heaviside unit step function

( ) ( ) ( ) 0 x a

f t H t a H t a 1 x a0 x a

< −= + − − = < >

a 0>

The Fourier transform of this function can be determined as

( ) ( ) i tf f t e dtωω∞

−

−∞

= ∫ a

i t

a

e dtω−

−

= ∫

ai t

a

1 ei

ω

ω−

−

− =

( )i a i a1 e ei

ω ω

ω−−

= − use Euler’s formula

cos a2 ω

ω=

i sin a cos aω ω+ − i sin a2i

ω +

( )2 sin aωω

=

( )sin a

2aa

ωω

= see p.484 for sinc

2. Two-sided exponential function

Consider the even two-sided exponential function:

( ) a t f t e−= a 0>

Then the Fourier transform of this function can be evaluated as

( ) ( ) i tf f t e dtωω∞

−

−∞

= ∫ a t i te e dtω∞

− −

−∞

= ∫

( ) ( )0

a i t a i t

0

e dt e dtω ω∞

− − +

−∞

= +∫ ∫

( ) ( )0a i t a i t

0

1 1e ea i a i

ω ω

ω ω

∞− + − +

−∞ = − − +

1 1a i a iω ω

= +− +

2 2

2aa ω

=+

3. Dirac delta ( ) ( )f t tδ=

( ){ } ( ) i t 0F t t e dt e 1ωδ δ∞

−

−∞

= = =∫

( ) ( )a

i i t i t i t

a aa

1 not confitmed

1 1 1 sin at e d e d e d lim e d lim2 2 2

ωξ ω ω ω ωδ δ ξ ξ ω ω ωπ π π ω

∞ ∞ ∞−

→∞ →∞−∞ −∞ −∞ −

= = = =

∫ ∫ ∫ ∫

( ) ( ) ( )f t H t a H t a= + − −

a− a

( ) a t f t e−=

( ) 2 2

2afa

ωω

=+

( )tδ

0

( )f ω2a


725

4. Gaussian distribution function

( ) 2atf t e−= , a 0>

The Fourier transform of ( ) 2atf t e−= can be evaluated as

{ }2atF e− 2at i te e dtω

∞− −

−∞

= ∫

2i t ate dtω∞

− −

−∞

= ∫

2 2ia t2a 4ae dtω ω ∞ − + −

−∞

= ∫

22 ia t

2a4ae e dtωω ∞ − +−

−∞

= ∫

22 ia t2a4a1 ie e d a t

2aa

ωω

π

ω ∞ − +−

−∞

= + ∫

2

4aea

ωπ −

=

The inverse Fourier transform of

2ae ω− can be evaluated as

{ }21 aF e ω− − 2a i t1 e e d

2ω ω ω

π

∞−

−∞

= ∫

2 2

2it ta2a 4a1 e d

2

ω

ωπ

∞ − − +

−∞

= ∫

22 itt a

2a4a1 e e d2

ωω

π

− ∞ − −

−∞

= ∫

2

2u4a1 e e du2 a

ω

π

π

∞− −

−∞

= ∫

2t4a1 e

4 aπ

−

=

5. Derive the Fourier transform of the derivative (property 7):

( )dF f tdt

( ) i td f t e dtdt

ω∞

−

−∞

= ∫

( )i te d f tω∞

−

−∞

= ∫

( ) ( )i t i te f t f t deω ω∞∞− −

−∞−∞

= − ∫ integration by parts

( ) i t0 i f t e dtωω∞

−

−∞

= + ∫

( )î fω ω=

( ) 2atf t e−=

( )2

4af ea

ωπω−

=2ue du π

∞−

−∞

=∫

u-substitutioniu a t2a

1 du dta

ω = +

=

2ue du π∞

−

−∞

=∫

u-substitutionitu a2a

1 du da

ω

ω

= −

=


Application of integral transform for the solution of PDE

According to properties 7) and 9), application of the Fourier Transform eliminates the derivatives with respect to time or to space variables. This fact will be used for the solution of the differential equations. First, we transform a differential equation to eliminate the derivative of unknown function, then we solve (algebraic) transformed equation in the frequency space, and finally, the solution of the original problem will be obtained by the inverse transform:

IX.2.4 SOLUTION OF THE ORDINARY DIFFERENTIAL EQUATIONS Example 4 (Steady-State Conduction) Solve the 2nd order ordinary differential equation

( )− + =2

22

d y a y f x 0dx

( )∈ −∞ ∞x ,

with the help of the Fourier transform.

Solution: Apply the Fourier transform ( ) ( )∞

−

−∞

= ∫ i xy y x e dxωω to the

given equation, using Property (7) for the transform of the 2nd

derivative, assuming ( )→±∞

=xlim u x 0 , ( )

→±∞

∂=

∂

k

kx

u xlim 0

x, k 1,2= :

( ) ( ) ( )2 2 ˆˆ ˆy a y f 0ω ω ω ω− − + = transformed equation

Solve the transformed equation for ( )y ω

( )y ω ( )2 2

fa

ωω

=+

transformed solution

Note that the function ( ) 2 2

1ga

ωω

=+

is the Fourier

transform of the two-sided exponential function (Example 2):

( ) ( ){ }− = = = + a x

2 21 1g F e F g x

a 2aω

ω

Then the transformed solution can be written as the product of two transformed functions:

( )y ω ( ) ( )ˆg fω ω=

DE

TE solutionof TE

solutionof DE

F 1F −


727

The solution of the differential equation can be found by inverse transform with application of the convolution theorem:

( )y x ( ){ }1 ˆF y ω−= ( ) ( ){ }1 ˆˆF g fω ω−=

g f= ∗ convolution

( ) ( )∞

−∞

= −∫ g x s f s ds

( )a x s1 e f s ds2a

∞− −

−∞

= ∫

Case of a point source ( ) ( )= − 0f x x xδ Consider the case when the source function ( )f x is the Dirac delta function

( ) ( )= − 0f x x xδ

defining the constant point source at = 0x x .

Then integration using the property of the delta function

( ) ( ) ( )0 0u s s x ds u xδ∞

−∞

− =∫

yields the solution of the differential equation:

( )y x ( )a x s0

1 e s x ds2a

δ∞

− −

−∞

= −∫

( )y x − −= 0a x x1 e2a

solution

The graph of the solution is the graph of the double-sided exponential function (Example 2) centered at = 0x x .

2

c

The problem can be interpreted as a steady state conduction in the thin rod infinite in both directions (lumped capacitance model) exposed to convective enviroment at 0 temperature,

hPa kA

with

=

0 a point heat source at x x .=

0xx

( )− 0x xδ

( ) − −= 0a x x1y x e2a

0x

x

( ) ( )= − 0f x x xδ∞ =T 0, h


IX.2.5 SOLUTION OF PDEs IN THE INFINITE REGION IX.2.5.1 Heat Equation – Gauss’s Kernel – Green’s Function

Example 5 (1-D heat conduction with heat generation

in the infinite region - Gauss’s Kernel) Consider the initial-boundary value problem for non-homogeneous heat

equation in the infinite slab (see Chapter 4):

( ) ( ) ( )t

t,xuat,xSx

t,xu 22

2

∂∂

=+∂

∂ 0t > , ( )x ,∈ −∞ ∞

with the initial condition: ( ) ( )xu0,xu 0= ( )x ,∈ −∞ ∞ The source function defines the heat generation in the slab:

( ) ( )k

t,xgt,xS = , where ( ) 3

Wg x,t ,m

volumetric heat source

There are no boundary conditions for the infinite region; but from physical considerations, we assume that both the unknown function and its derivative vanish when approaching ±∞ :

( )x

u x,t 0→±∞

= ( )x

u x,t 0

x→±∞

∂=

∂

This assumption will allow application of property (7) of the Fourier transform which will be used for solution of the given IBVP. Transformed equation Take the Fourier integral transform of the equation (applying properties (1) and (7) for the transform of the derivatives)

( ) ( ) ( )t

t,uat,St,u 22

∂∂

=+−ωωωω

with the transformed initial condition: ( ) ( )ωω 0u0,u = where the following notations for the transformed functions are used

[ ] ( ) ( )i x ˆF u( x,t ) u x,t e dx u ,tω ω∞

−

−∞

= =∫

[ ] ( ) ( )i x0 0 0ˆF u ( x ) u x e dx uω ω

∞−

−∞

= =∫

[ ] ( ) ( )i x ˆF S( x,t ) S x,t e dx S ,tω ω∞

−

−∞

= =∫

Solution of transformed equation. The obtained transformed equation is an ordinary linear differential equation in variable t with constant coefficients:

( ) ( ) ( )22

2

at,St,u

att,u ωωωω

=+∂

∂ ( ) ( )ωω 0u0,u =

the general solution of which can be obtained by the variation of parameter formula (see Table ODE):

transformed solution ( ) ( ) ( )2 2 2

2 2 2 tt t t

a a a0 2

0

1 ˆˆ û ,t u e e e S ,t dta

ω ω ω

ω ω ω− − ′

′ ′= + ∫ (◊)

[ ] ( ) ( )i x

transformation w.r.t variable x

ˆF u( x,t ) u x,t e dx u ,tω ω∞

−

−∞

= =∫


729

Inverse transform Solution of the given IBVP now can be obtained

by the inverse Fourier integral transform of the expression (◊):

( ) ( ){ }1 û x,t F u ,tω−= ( )( )

( )2 2

2 2tt t t

1 1a a0 2

0

1 ˆˆF u e F e S ,t dta

ω ω

ω ω− ′− −

− − ′ ′= +

∫

Functions t

a2

2

eω−

and ( )

2

2 t taeω− ′−

which appear in the integrand are the Fourier transforms of the Gaussian distribution functions (p.725).

Gauss’s Kernel ( )2

2x

t4a

et2

at,xG−

=π

( )( )

( )2

2a x4 t taG x,t t e

2 t tπ

−′−′− =

′−

( )2

2 taG ,t eω

ω−

= ( )( )

2

2 t taG ,t t eω

ω− ′−

′− =

( ) ( ){ }1 û x,t F u ,tω−= ( ) ( ){ } ( ) ( )t

1 10 2

0

1ˆ ˆ ˆˆF u G ,t F S ,t G ,t t dta

ω ω ω ω− − ′ ′ ′= + − ∫

( ) ( ){ } ( ) ( ){ }t

1 10 2

0

1ˆ ˆ ˆˆF u G ,t F S ,t G ,t t dta

ω ω ω ω− − ′ ′ ′= + −∫

Then the first integral in the solution is the inverse Fourier transform of the product of two transformed functions Gu0 which according to property (3) is a convolution of these two functions

{ }10

ˆˆF u G−0u G= ∗ ( ) ( )0u s G x s,t ds

∞

−∞

= −∫ ( ) ( )

2 2a x s4t

0a e u s ds

2 tπ

−∞ −

−∞

= ∫

Consider the second term in the solution. It leads to the convolution:

{ }1 ˆ ˆF GS− G S= ∗ ( ) ( )G x s,t t S s ds∞

−∞

′= − −∫ ( )

( )( ) ( )

22a x st4 t t

0

a e S s,t dsdt2 t tπ

− −∞′−

−∞

′ ′=′− ∫ ∫

Then a formal solution of the given IBVP is:

Solution: ( ) ( ) ( )∫∞

∞−

−−

= dssuet2

at,xu 0

sxt4

a 22

π

( )( )

( )( )

22a x st 4 t t

0

1 e S s,t dsdt2ak t tπ

− −′∞ −

−∞

′ ′+′−∫ ∫


Particular cases: Consider two particular cases: a) Homogeneous equation (no heat generation, ( ) 0t,xS = ) Let the initial temperature distribution have a step-wise variation:

Then solution is given by the integral over the finite region:

( ) ( )∫−

−−

=1

1

sxt4

a

dset2

at,xu2

2

π

Use change of variable: ( )sxt2

av −= dst2

adv −=

( ) ( )∫−

−−=

1

1

sxt4

a

dset2

at,xu2

2

π

( )

( )

∫

−

+

−−=t21xa

t21xa

v dvea

t2t2

a 2

π

( ) ( )

−= ∫∫

−

−

+

−t21xa

0

vt21xa

0

v dve2dve221 22

ππ

( ) ( )

−−

+=

t21xaerf

21

t21xaerf

21

Therefore, the solution of the Heat Equation in the infinite region with no heat source is given by

( )t,xu ( ) ( )

−−

+=

t21xaerf

21

t21xaerf

21

The graph shows the temperature profiles for 2a =

( ) ( ) ( )0u x H x 1 H x 1= + − −

( ) ( ) ( )0u x H x 1 H x 1= + − −


731

b) Non-homogeneous equation - Green’s function

Consider a heat conduction problem with a zero initial condition

( ) 0xu0 =

and a point heat source located at 0xx = which instantaneously released energy at time 0tt = of strength 0S (impulse point source):

impulse point source ( ) ( ) ( )0 0 0,S x t S x x t tδ δ= − −

Then the first term in the solution disappears because of the zero initial condition, and the solution becomes

( )( )

( )( ) ( ) ( )

22a x st4 t t0

0 00

S 1u x,t e s x t t dsdt2ak t t

δ δπ

− −∞′−

−∞

′ ′= − −′−

∫ ∫

Then integration of the expression with the Dirac delta functions yields the solution

Green’s Function ( )

( )( )

( )

220

0

a x x4 t t

00

0

0

S e t tu x,t 2ak t t

0 t t

π

− −

− >= − ≤

This solution of the problem with the impulse point source is called the Green function for the infinite region in the Cartesian coordinate system. It is used for solution of non-homogeneous partial differential equations. Example The solution curves at different moments of time for

2a = , 0S 1= , 2k = , 1t0 = , 2x0 =

( ) ( ) ( )0 0g x,t g x x t tδ δ= − −

( )0x xδ −

0x

( )u x,t


IX.2.5.2 Wave Equation Example 6 Wave equation for an infinite string – D’Alambert solution

( ) ( )2

2

22

2

tt,xu

v1

xt,xu

∂∂

=∂

∂ 0t > , ∞<<−∞ x

with initial conditions: ( ) ( )xu0,xu 0= ∞<<−∞ x initial displacement

( ) ( )xut

0,xu1=

∂∂ initial velocity

The coefficient in the Wave Equation v is a physical property of the

string and represents a speed of wave propagation along the string. It is determined through the equation

2

22

Tg mv w s

=

where T is a tension, g is the acceleration of gravity, w is a weight of the string per unit length. 1) Transformed equation Apply the Fourier transform to the wave

equation and initial conditions:

( ) ( )2

2

22

tt,u

v1t,u

∂∂

=−ωωω

( ) ( )ωω 0u0,u =

( ) ( )ωω1u

t0,u

=∂

∂

This is the initial value problem for a 2nd order linear ODE in t with constant coefficients

( ) ( ) 0t,uvt

t,u 222

2

=+∂

∂ ωωω

The general solution is

( ) vtsincvtcosct,u 21 ωωω +=

where coefficients can be found from initial conditions. ( )ω01 uc =

( )1

2

uc

vω

ω=

then the solution of the ODE becomes

( ) ( ) ( )10

uˆ û ,t u cos vt sin vt

vω

ω ω ω ωω

= +

2) Inverse transform The solution of the IBVP can now be obtained using the inverse Fourier transform

( ) ( ) ( )1 i x0

u1 û x,t u cos vt sin vt e d2 v

ωωω ω ω ω

π ω

∞

−∞

= +

∫

This is the general form of the solution, which includes Fourier transform of i.c.’s. For some functions, it can be integrated explicitly.


733

x

0

t

It can be shown (V.S.Vladimirov Equation of Mathematical Physics, p.200, Debnath, p.75) that this solution can be written in the form of D’Alambert solution with arbitrary functions ( ) ( )2

0u x C∈ and

( ) ( )11u x C∈ (see derivation on p.757):

D’Alambert solution ( ) ( ) ( ) ( )x vt

0 0 1x vt

1 1u x,t u x vt u x vt u s ds2 2v

+

−

= − + + + ∫

The solution of the IVP for the Wave Equation is unique.

Particular cases: 1. Consider the IBVP for an infinite string with initial conditions

( ) ( ) ( )1xH1xHxu0 −−+=

( ) 0xu1 =

Fourier transform of initial conditions yields

( ) ( ) ( )ω

ωω

ωπδω

ωπδω ωω sin21e1eu ii0 =

−−

−= −

( )1u 0ω = Then the solution of the IBVP becomes (compare to D’Alambert soln):

( ) ( ) ( )( )

( ) ( )( )0 0u x vt u x vt

1u x,t H x 1 vt H x 1 vt H x 1 vt H x 1 vt2

− +

= + − − − − + + + − − +

Solution curves for v 2= are shown in the figure (F-1.mws)

2. Consider the IBVP for an infinite string with initial conditions

( ) ( ) ( )20 1u x exp x , u x 0= − =

3. Consider the IBVP for an infinite string with initial conditions

[Debnath, p.80] ( )0u x 0= , ( ) ( )1u x xδ=

( ) ( ) ( )0

initial condition: (gate function)u x H x 1 H x 1= + − −

( )u x,t


( ) ( ) ( )0u x H x 1 H x 1= + − −

( )u x, y

IX.2.5.3 The Laplace Equation Example 7 (Laplace’s Equation in a semi-infinite plane – Dirichlet problem – Poisson integral formula)

Consider a 2-dimensional Laplace’s equation in the semi-infinite plane

y 0> , but with ( )∞∞−∈ ,x :

( ) ( )2 2

2 2

u x, y u x, y0

x y∂ ∂

+ =∂ ∂

( )∞∞−∈ ,x , ( )y 0,∈ ∞

with the boundary condition at y 0= :

( ) ( )xu0,xu 0= Dirichlet

Transformed equation We will apply the Fourier transform in the x variable

[ ] ( ) ( )i x ˆF u( x, y ) u x, y e dx u , yω ω∞

−

−∞

= =∫

( ) ( )i x0 0 0ˆF f ( x ) u x e dx uω ω

∞−

−∞

= = ∫

Transformed equation

0yuu 2

22 =

∂∂

+− ω

has the general solution: y

2y

1 ececu ωω += − The solution is bounded ⇒ 0c2 = for 0>ω 0c1 = for 0<ω Then two terms of the solution can be combined

yceu ω−= Application of the boundary condition yields

y0euu ω−=

Inverse transform Function ye ω− is a Fourier transform of

Poisson’s kernel 22 yxy1+π

then the solution of the transformed equation can be written as

+= 220 yx

y1Fuuπ

which is a product of two Fourier transforms. Then the inverse transform of u can be written as a convolution of these two functions:

Poisson’s integral formula ( ) ( )( )∫

∞

∞− +−= ds

ysxsuyy,xu

220

π

It is a solution of the Dirichlet problem for Laplace’s equation in the semi-plane which is called Poisson’s integral formula for the upper half-plane.

For the initial condition ( ) ( ) ( )1xH1xHxu0 −−+=

the solution is given by: ( )y

1xarctan1y

1xarctan1y,xu −−

+=

ππ


735

F-1 Poisson integral.mws Solution of Laplace's Equation in the upper half-plane - Poisson's integral formula > restart; > u0:=Heaviside(s+1)-Heaviside(s-1);

:= u0 − ( )Heaviside + s 1 ( )Heaviside − s 1

Poisson's Integral Formula: > u(x,y):=simplify(int(u0/((x-s)^2+y^2),s=-infinity..infinity)*y/Pi);

:= ( )u ,x y −− +

arctan + 1 x

y

arctan − + 1 x

yπ

> plot3d(u(x,y),x=-4..4,y=0..2,grid=[50,50]);

> with(plots): > densityplot(u(x,y),x=-4..4,y=0..2,grid=[150,150],style=patchnogrid,axes=boxed);

> contourplot(u(x,y),x=-4..4,y=0..2,axes=boxed,coloring=[blue,yellow],filled=true);

( )u x, y


IX.2.6 FOURIER INTEGRALS a) Complex Fourier integral representation Suppose that the function f : → satisfies the Dirichlet

condition in every finite interval of (see p.696) and suppose that there exists an improper integral

( )f t dt∞

−∞

< ∞∫

( ) ( ) i tf t F e dωω ω∞

−∞

= ∫

where the coefficient function is given by

( ) i t1F( ) f t e dt2

ωωπ

∞−

−∞

= ∫

b) Standard Fourier integral representation Suppose that the function ( )f : 0,∞ → satisfies the Dirichlet

condition in every finite interval of and

( )0

f t dt∞

< ∞∫

( ) ( ) ( )0

f t A cos t B sin t dω ω ω ω ω∞

= + ∫

where the real coefficient functions are given by

( )1A( ) f t cos tdtω ωπ

∞

−∞

= ∫

( )1B( ) f t sin tdtω ωπ

∞

−∞

= ∫

c) Fourier cosine integral representation

( ) ( )0

f t A cos tdω ω ω∞

= ∫


( )0

2A( ) f t cos tdtω ωπ

∞

= ∫

d) Fourier sine integral representation

( ) ( )0

f t B sin tdω ω ω∞

= ∫


( )0

2B( ) f t sin tdtω ωπ

∞

= ∫

( )f t


737

Convergence All given Fourier integral representations converge to:

( )( )

( ) ( ) f t if f is continuous at t

f t f t f t if f is discontinuous at t

2

− +

→ +

FI-01.mws Fourier Integral Representation W = the upper limit in the Fourier Integral (defines the accuracy of approximation)

> W:=20; := W 20

> f(t):=Heaviside(t)-Heaviside(t-1); := ( )f t − ( )Heaviside t ( )Heaviside − t 1

Standard Fourier Integral: > A(w):=int(f(t)*cos(w*t),t=-infinity...infinity)/Pi;

:= ( )A w ( )sin w

w π

> B(w):=int(f(t)*sin(w*t),t=-infinity...infinity)/Pi;

:= ( )B w −− + 1 ( )cos w

w π

> u(t):=int(A(w)*cos(w*t)+B(w)*sin(w*t),w=0..W);

:= ( )u t − − ( )Si − 20 t 20 ( )Si 20 t

π

> plot({f(t),u(t)},t=-2..2,color=[red,black]);

Fourier Cosine Integral: > A(w):=2*int(f(t)*cos(w*t),t=-infinity...infinity)/Pi;

:= ( )A w 2 ( )sin ww π

> u(t):=int(A(w)*cos(w*t),w=0..W);

:= ( )u t − ( )Si + 20 20 t ( )Si − 20 t 20π


Fourier Sine Integral: > B(w):=2*int(f(t)*sin(w*t),t=-infinity...infinity)/Pi;

:= ( )B w −2 ( )− + 1 ( )cos w

w π

> u(t):=int(B(w)*sin(w*t),w=0..W);

:= ( )u t − − 2 ( )Si 20 t ( )Si + 20 20 t ( )Si − 20 t 20

π



IX.2.7 INTEGRAL FOURIER TRANSFORM IN THE SEMI-INFINITE REGIONS Consider a semi-infinite region ∞<≤ x0 .

Define the integral transform pair of the function ( )xu as

{ }F u = ( ) ( ) ( )∫∞

=0

dxx,Kxuu ωω

{ }1 ˆF u− = ( ) ( ) ( )∫∞

=0

dx,Kuxu ωωω

1. Fourier Integral Transform kernel We are looking for a specific form of the kernel ( )ω,xK which can be

applied for particular form of the boundary condition at 0x = . In the following Table, we specify the kernel ( )ω,xK for three types of classical homogeneous boundary conditions:

Boundary condition Kernel ( )ω,xK Integral transform pair

I [ ]x 0u 0

== xsin2 ω

π { }IF u ( ) ( )I

0

2u u x sin xdxω ωπ

∞

= = ∫

{ }1I IˆF u− ( ) ( )I

0

2 û x u sin xdω ω ωπ

∞

= = ∫

II 0dxdu

0x

=

=

xcos2 ωπ

{ }IIF u ( ) ( )II0

2u u x cos xdxω ωπ

∞

= = ∫

{ }1II IIˆF u− ( ) ( )II

0

2 û x u cos xdω ω ωπ

∞

= = ∫

III 0Hudxdu

0x

=

+−

=

22 H

xsinHxcos2

+

+

ω

ωωωπ

{ }IIIF u ( ) ( )III 2 20

2 cos x H sin xu u x dxH

ω ω ωωπ ω

∞ += =

+∫

{ }1III ˆF u− ( ) ( )III III 2 2

0

2 cos x H sin xû x u dH

ω ω ωω ωπ ω

∞ += =

+∫

The kernel corresponding to the Dirichlet boundary condition is based on the Fourier sine integral transform pair (note, that the coefficient 2 π is split into two factors 2 π ):

( ) ( )∫∞

=0

xdxsinxu2u ωπ

ω ( ) ( )∫∞

=0

xdsinu2xu ωωωπ

The kernel corresponding to the Neumann boundary condition is based on the Fourier cosine integral transform pair:

( ) ( )∫∞

=0

xdxcosxu2u ωπ

ω ( ) ( )∫∞

=0

xdcosu2xu ωωωπ

The integral transform pair with the kernel corresponding to Robin boundary condition:

( ) ( )∫∞

+

+=

022

dxH

xsinHxcosxu2uω

ωωωπ

ω ( ) ( )∫∞

+

+=

022

dH

xsinHxcosu2xu ωω

ωωωωπ

II

Fourier cosinetransform F

I

Fourier sinetransform F

x0

III

Fouriertransform F


739

The kernel of integral transform for the case of a Robin boundary condition is obtained from the solution of the auxiliary boundary value problem for the semi-infinite region with parameter [ )∞∈ ,0ω :

( ) ( ) 0xXxX 2 =+′′ ω [ )∞∈ ,0x

( ) ( ) 00HX0X =+′−

It can be verified that the function xsinHxcosX ωωω +=

is a solution of the auxiliary BVP. Convolution Formulas: [See also Debnath, p.94]

Let { }IF u ( ) ( )I0

2u u x sin xdxω ωπ

∞

= = ∫

{ }IIF v ( ) ( )II0

2v v x cos xdxω ωπ

∞

= = ∫

( ) ( ){ }1I I IIˆ ˆF u vω ω− ⋅ ( ) ( ) ( )

0

1 u s v s x v s x ds2π

∞

= − − + ∫ (I-I-II)

( ) ( ){ }1II II IIˆ ˆF u vω ω− ⋅ ( ) ( ) ( )

0

1 u s v x s v x s ds2π

∞

= + + − ∫ (II-II-II)

( ) ( ){ }1II I Iˆ ˆF u vω ω− ⋅ ( ) ( ) ( )

0


∞

= + + − ∫ (II-I-I)

( ) ( ){ }1

I I Iˆ ˆF u vω ω− ⋅ (I-I-I)

( ) ( ){ }1I II IIˆ ˆF u vω ω− ⋅ (I-II-II)

( ) ( ){ }1I I IIˆ ˆF u vω ω− ⋅ ( ) ( )I II

0

2 ˆ û v sin xdω ω ω ωπ

∞

= ∫ (I-I-II)

( ) ( )II0 0

2 2 û s sin sds v sin xdω ω ω ωπ π

∞ ∞ =

∫ ∫

( ) ( )II0 0

2 û s v sin s sin xd dsω ω ω ωπ

∞ ∞ =

∫ ∫

( ) ( ) ( ) ( )II0 0

1 û s v cos s x cos s x d dsω ω ω ωπ

∞ ∞ = − − +

∫ ∫

( ) ( ) ( ) ( ) ( )II II0 0 0

1 ˆ û s v cos s x d v cos s x d dsω ω ω ω ω ωπ

∞ ∞ ∞ = − − +

∫ ∫ ∫

( ) ( ) ( ) ( ) ( )II II0 0 0

1 2 2ˆ û s v cos s x d v cos s x d ds2π ω ω ω ω ω ω

π π π

∞ ∞ ∞ = − − +

∫ ∫ ∫

( ) ( ) ( )0


∞

= − − + ∫


( ) ( ){ }1I II IIˆ ˆF u vω ω− ⋅ ( ) ( )II II

0

2 ˆ û v sin xdω ω ω ωπ

∞

= ∫ (I-II-II)

( ) ( )II0 0

2 2 û s cos sds v sin xdω ω ω ωπ π

∞ ∞ =

∫ ∫

( ) ( )II0 0

2 û s v sin x cos sd dsω ω ω ωπ

∞ ∞ =

∫ ∫

( ) ( ) ( ) ( )II0 0

1 û s v sin x s sin x s d dsω ω ω ωπ

∞ ∞ = − + +

∫ ∫

( ) ( ) ( ) ( )0 0 0

1 2u s v p cos pdp sin x s sin x s d dsω ω ω ωπ π

∞ ∞ ∞ = − + +

∫ ∫ ∫

( ) ( ) ( ) ( )0 0 0

1 2 u s v p cos p sin x s sin x s d dp dsω ω ω ωπ π

∞ ∞ ∞ = − + +

∫ ∫ ∫

( ) ( ) ( ) ( )0 0 0

undefined

1 2 u s v p cos p sin x s sin x s d dp dsω ω ω ωπ π

∞ ∞ ∞ = − + +

∫ ∫ ∫

( ) ( ){ }1II II IIˆ ˆF u vω ω− ⋅ ( ) ( ) ( )

0

1 u s v x s v x s ds2π

∞

= + + − ∫ (II-II-II)

[Debnath, 2.14.7, p.94]


741

( ) ( ){ }1II I Iˆ ˆF u vω ω− ⋅ ( ) ( ) ( )

0


∞

= + + − ∫ (II-I-I)

[Debnath, 2.14.10-11, p.95]

( ) ( ){ }1

I I Iˆ ˆF u vω ω− ⋅ (I-I-I)


OPERATIONAL PROPERTIES OF THE INTEGRAL FOURIER TRANSFORMS

0 x≤ < ∞ We consider the function ( )xu as a function describing some physical quantity in the semi-infinite region [ )∞∈ ,0x . At 0x = , it is specified by one of the boundary conditions. When x approaches infinity, we assume that both the function ( )xu and its derivative are equal to zero:

( )x

u x 0→∞

= ( )x

u x 0x →∞

∂=

∂

Integral transform of 2

2

xu

∂∂ Apply the integral transform to the second partial derivative of u

( )2

2

u xF

x ∂

∂

( ) ( )2

20

u xK ,x dx

xω

∞ ∂=

∂∫

Boundary Condition at 0x = Kernel ( )ω,xK ( ) ( )

2

20

u xK x, dx

xω

∞ ∂

∂∫

I Dirichlet [ ] ( )0x 0u f t=

= xsin2 ωπ

( ) ( )20

2u f tω ω ωπ

− +

II Neumann ( )0x 0

u f tx =

∂ = ∂ xcos2 ω

π ( ) ( )2

02u f tω ωπ

− −

III Robin ( )0

x 0

f tu Hux k=

∂ − + = ∂

22 H

xsinHxcos2

+

+

ω

ωωωπ

( ) ( )02

2 2

f t2u + kH

ωω ωπ ω

−+

1) Dirichlet boundary condition, [ ] 0u 0x == , ( ) xsin2x,K ωπ

ω = :

( )2

I 2

u xF

x ∂

∂

( ) ( )2

20

u xK ,x dx

xω

∞ ∂=

∂∫ ( )∫∞

∂∂

=0

2

2

xdxsinx

xu2 ωπ

( )∫∞

∂∂

=0 x

xuxdsin2 ωπ

( ) ( ) [ ]∫∞∞

∂∂

−

∂∂

=00

xsindxxu2

xxuxsin2 ω

πω

π

( )∫∞

∂∂

−=0

dxxxuxcos2 ω

πω

( )[ ]∫∞

−=0

xuxdcos2 ωπ

ω

( )[ ] ( ) [ ]∫∞

∞ +−=0

0 xcosdxu2xxucos2 ωπ

ωωπ

ω

( ) ( ) ( ) ( ) ( ) [ ]0x0

2 2cos x u x cos 0 f t u x d cos xω ω ω ωπ π

∞

→∞ = − − + ∫

[ ] ( )0x 0

in the case of non-homogeneous conditionu f t

==

x0


743

( ) ( )20

0

2 2f t u x sin xdxω ωπ π

∞

= − ∫

( ) ( )20

2 ˆf t uω ω ωπ

= −

( ) ( )∫∞

∂∂

02

2

dxx,Kx

xu ω ( ) ( )20

2 ˆf t uω ω ωπ

= −

( ) ( )∫

∞

∂∂

02

2

dxx,Kx

xu ω ( )2uω ω= − in a case of ( )0f t 0=

The integral transform eliminates the derivative. Similar results can be obtained for the remaining two boundary conditions:

2) Neumann b.c., 0dxdu

0x

=

=

: ( ) xcos2x,K ωπ

ω =

( )2

II 2

u xF

x ∂

∂

( ) ( )2

20

u xK ,x dx

xω

∞ ∂=

∂∫( )2

20

u x2 cos xdxx

ωπ

∞ ∂=

∂∫

( )

0

u x2 cos xdx

ωπ

∞ ∂=

∂ ∫

( ) ( ) [ ]

00

u x u x2 2cos x d cos xx x

ω ωπ π

∞∞ ∂ ∂

= − ∂ ∂

∫

( ) ( ) ( ) [ ]

0x 0

u x u 0 u x2 2cos x cos0 d cos xx x x

ω ωπ π

∞∞

→∞

∂ ∂ ∂ = − −

∂ ∂ ∂ ∫

( ) ( )

0

u 0 u x2 2 sin xdxx x

ω ωπ π

∞∂ ∂= − +

∂ ∂∫

( ) ( )

0

u 02 2 sin xd u xx

ω ωπ π

∞∂ = − + ∂ ∫

( ) ( ) ( ) ( ) [ ]

x0

u 02 2 2sin xu x sin0u 0 u x d sin xx

ω ω ω ωπ π π

∞

→∞

∂ = − + − − ∂ ∫

( ) ( )2

0

u 02 2 u x cos xdxx

ω ωπ π

∞∂= − −

∂ ∫

( ) ( )2u 02 ux

ω ωπ

∂= − −

∂

( )2uω ω= − (for homogeneous boundary condition, ( )u 0

0x

∂=

∂)

If condition is non-homogeneous, ( )0x 0

du f tdx =

= , then

( ) ( )∫∞

∂∂

02

2

dxx,Kx

xu ω ( ) ( )20

2 ˆf t uω ωπ

= − −


3) Robin b.c., ( )0

x 0

f tu Hux k=

∂ − + = ∂ , ( )

22 H

xsinHxcos2x,K+

+=

ω

ωωωπ

ω

( )0x 0

du Hu f tdx =

− + = , hH

k= [see Ozicik (2-124), p.74]

( )22 H

xsinHxcos2x,K+

+=

ω

ωωωπ

ω ( ) x 0 2 2

2K ,xH

ωωπ ω=

=+

( )2 2

2 sin x H cos xK ,xx H

ω ω ωω ωπ ω

∂ − +=

∂ + ( )

2 2x 0

2 HK ,xx H

ω ωπ ω=

∂=

∂ +

( ) ( )2

2 22 2 2

2 cos x H sin xK ,x K ,xx H

ω ω ωω ω ω ωπ ω

∂ − −= = −

∂ +

2

II 2

uFx

∂

∂ ( )

2

20

u K ,x dxx

ω∞ ∂

=∂∫

( )0

uK ,x dx

ω∞ ∂ = ∂ ∫

( ) ( )0 0

u uK ,x d K ,xx x

ω ω∞ ∞∂ ∂ = − ∂ ∂ ∫

( ) ( )0 0

u uK ,x K ,x dxx x x

ω ω∞∂ ∂ ∂ = − − ∂ ∂ ∂ ∫

( ) ( )x 0 0

u K ,0 K ,x dux x

ω ω∞

=

∂ ∂= − −

∂ ∂∫

( ) ( ) ( )x 0 0 0

u K ,0 u K ,x ud K ,xx x x

ω ω ω∞ ∞

=

∂ ∂ ∂ = − − + ∂ ∂ ∂ ∫

( ) ( ) ( )2

2x 0 0 0

u K ,0 u K ,x u K ,x dxx x x

ω ω ω∞

=

∂ ∂ ∂ = − + + ∂ ∂ ∂ ∫

( ) ( ) ( )2x 0

x 0 0

u K ,0 u K ,0 uK ,x dxx x

ω ω ω ω∞

==

∂ ∂= − + −

∂ ∂ ∫

( )2x 02 2 2 2

x 0 0

u 2 2 Hu uK ,x dxx H H

ω ω ω ωπ πω ω

∞

==

∂= − + −

∂ + +∫

( )2

2 2x 0 0

2 u Hu uK ,x dxxH

ω ω ωπ ω

∞

=

∂ = − + − ∂ +∫

( ) ( )0 2

2 2

f t2 ukH

ω ω ωπ ω

= −+


745


IX.2.7.1 Heat equation in the semi-infinite region Consider the homogeneous one-dimensional heat equation

( ) ( )t

t,xuax

t,xu 22

2

∂∂

=∂

∂ , [ )∞∈ ,0x , 0t >

initial condition: ( ) ( )xu0,xu 0= , 0t = boundary condition: ( ) ( )tft,0u 0= , 0t > (II) 1. Case ( ) 0tf0 = 1) Transformed equation. Apply the integral transform

corresponding to the Dirichlet problem (Fourier sine transform) to the differential equation:

( ) ( )t

t,uat,u 22

∂∂

=−ωωω

This is an ODE for the transformed function ( )u ,tω as a function of t with the variable ω treated as a parameter:

( ) ( ) 0t,uat

t,u2

2

=+∂

∂ ωωω

with an initial condition which can be obtained by integral transform of the original initial condition:

( ) ( )ωω 0u0,u = ( ) ( )∫∞

=0

0 dxx,Kxu ω

Solution of this linear 1st order homogeneus ODE with constant coefficients is given by

( ) ( )2

2 ta

0ˆ û ,t u eω

ω ω−

= transformed solution

Application of the convolution formula (I-I-II): 2) Inverse transform – solution of IBVP.

( )u x,t ( )t,xu ( ) ( )2

2 ta

00

u e K ,x dω

ω ω ω−∞

= ∫

{ }1I ˆF u−= ( )

2

2 ta

00

2 u e sin xdω

ω ω ωπ

−∞

= ∫

( )2

2 t1 aI 0ˆF u e

ω

ω−

− =

This solution can be reduced to the traditional form [Ozisik]:

( ){ }2

2 t1 aI I 0ˆF F u e

ω

ω−

− = ⋅

( )t,xu ( )∫∞ −

=0

ta

0 xdsineu2 2

2

ωωωπ

ω

( ){ }

( )2

2

G x,t

a x1 4tI I 0 II

aˆF F u F e2t

ω−

−

= ⋅

( )∫ ∫∞ −∞

′′′=

0

ta

00 xdsinexdxsinxu22 2

2

ωωωππ

ω

( ) ( ) ( )00

1 u s G s x,t G s x,t ds2π

∞

= − − + ∫ ( )∫ ∫∞ ∞ −

′

′′=

0 0

ta

0 xdxdsinxsine2xu2 2

2

ωωωππ

ω

( )( ) ( )

2 22 2a ax s x s

4t 4t0

0

1 a au s e e ds2 2t 2tπ

− −∞ − + = −

∫ ( ) ( ) ( )

∫∞ ′+

−′−−

′

−′=

0

xxt4

axxt4

a

0 xdeet8

axu2 22

22

π

( )( ) ( )

2 22 2a ax s x s

4t 4t0

0

a u s e e ds2 tπ

− −∞ − + = −

∫ ( )

( ) ( )2 2

2 2a ax x x x4t 4t

00

a u x e e dx2 tπ

− −∞ ′ ′− + ′ ′= −

∫


747

FS-1-1h.mws HE in the semi-infinite space - integral transform solution (Fourier sine transform) B.C.: f0=0 ( ) ( )1xH1xu0 −−=

( )ω

ωπ

ω cos12u0−

=

> restart;

> a:=2;

:= a 2

Kernel of integral transform

> K(omega,x):=sqrt(2/Pi)*sin(omega*x);

:= ( )K ,ω x 2 ( )sin ω xπ

Initial condition:

> u0(x):=1-Heaviside(x-1);

:= ( )u0 x − 1 ( )Heaviside − x 1

> u(omega):=factor(int(u0(x)*K(omega,x),x=0..infinity));

:= ( )u ω −2 ( ) − ( )cos ω 1

π ω

Inverse transform - solution of IBVP: the upper limit defines the accuarcy of approximation (similar to the number of terms in the truncated Fourier series) > u(x,t):=int(u(omega)*exp(-omega^2/a^2*t)*K(omega,x), omega=0..40);

:= ( )u ,x t d⌠

⌡

0

40

−2 ( ) − ( )cos ω 1 e( )− /1 4 ω 2 t

( )sin ω xπ ω

ω

> u00:=subs(t=0.0,u(x,t)):u1:=subs(t=0.1,u(x,t)): u2:=subs(t=0.5,u(x,t)):u3:=subs(t=1.0,u(x,t)): > plot({u0(x),u00,u1,u2,u3},x=0..4,color=black);

0

boundary conditionf 0=

( ) ( )0

initialconditionu x 1 H x 1= − −

t 0=

t 0.1=

t 0.5=

t 1.0=

( )u x,t


2. Case ( ) 0tf0 ≠ Non-homogeneous time-dependent boundary condition.

The transform of the derivative:

( )2

I 2

u xF

x ∂

∂

( )2

20

u x2 sin xdxx

ωπ

∞ ∂=

∂∫ ( ) ( )20

2 ˆf t uω ω ωπ

= −

Then the transformed equation is

( ) ( ) ( )t

t,uat,utf2 220 ∂

∂=−

ωωωωπ

( ) ( ) ( )tfa

2t,uat

t,u022

2 ωπ

ωωω=+

∂∂

The initial condition is the same: ( ) ( )ωω 0u0,u = Then the transformed solution is obtained by variation of

parameter

( ) ( ) ( )2 2 2

2 2 2tt t

a a a0 02

0

2ˆ û ,t u e e e f da

ω ω ω τωω ω τ τπ

− −

= + ∫

The solution of the IBVP can be found by the inverse transform:

( )t,xu ( ){ }1I ˆF u ,tω−=

( )( )

( )2 2

2 2 tt t

1 1a aI 0 I 02

0

2ˆF u e F e f da

ω ω τωω τ τπ

−− −

− − = +

∫

( )( ) ( )

2 22 2a ax s x s

4t 4t0

0

a u s e e ds2 tπ

− −∞ − + = −

∫

( )( )

2

2 t t

1 aI 02

0

2F e f da

ω τω τ τπ

− −−

+

∫

( )( ) ( )

2 22 2a ax s x s

4t 4t0

0

a u s e e ds2 tπ

− −∞ − + = −

∫

( )( )

2

2 t t

1 aI 02

0

2 1 F e f da

ω τω τ τ

π

− −−

+

∫

( )( ) ( )

2 22 2a ax s x s

4t 4t0

0

a u s e e ds2 tπ

− −∞ − + = −

∫

( )

( )( )

2 2a x4 t t

302 3

0

2 1 2 xea f d4a t

τ

τ τπ τ

−−

+−

∫ integration with Maple

( )t,xu ( )( ) ( )

2 22 2a ax s x s

4t 4t0

0

a u s e e ds2 tπ

− −∞ − + = −

∫

( )

( )( )

2 2a x4 t t

030

ax e f d2 t

τ

τ τπ τ

−−

+−

∫ see p.706

Case 0f const=

( )t,xu ( )( ) ( )

2 22 2a ax s x s

4t 4t0

0

a u s e e ds2 tπ

− −∞ − + = −

∫ 0

ax 2 axf 1 erfax2 2 t

ππ

+ −

integration with Maple

( )t,xu ( )( ) ( )

2 22 2a ax s x s

4t 4t0

0

a u s e e ds2 tπ

− −∞ − + = −

∫ 0

axf erfc2 t

+

see p.707

Attempt of application of convolution formula (failed so far):

( )( ) ( )

2 22 2a ax s x s

4t 4t0

0

a u s e e ds2 tπ

− −∞ − + = −

∫ ( )

( )

( )2

2

G x,t

a t x4 t1

I II II 02 20

2 1 2 1 aF F F e f da x 2t

τ

τ τ τπ π

−

−−−

+ −

∫


749

x

IX.2.7.2 Heat conduction with heat generation in a semi-infinite region See Ozisik, p.72.

I. Conduction with heat generation ( )2

2 2

g x,tu 1 uk tx a

∂ ∂+ =

∂∂, 0t > , x 0>

initial condition: ( ) ( )xu0,xu 0= , 0t = boundary condition: [ ] ( )0x 0u f t

== , 0t > (I)

II. Conduction with heat generation ( )2

2 2

g x,tu 1 uk tx a

∂ ∂+ =

∂∂, 0t > , x 0>

initial condition: ( ) ( )xu0,xu 0= , 0t =

boundary condition: ( )0x 0

u f tx =

∂ − = ∂ , 0t > (II)

III. Conduction with heat generation ( )2

2 2

g x,tu 1 uk tx a

∂ ∂+ =

∂∂, 0t > , x 0>

initial condition: ( ) ( )xu0,xu 0= , 0t =

boundary condition: ( )0

x 0

f tdu Hudx k=

− + = , 0t > (III)

0


Laplace’s Equation in the 1st quadrant Consider 2-D Laplace’s Equation in semi-infinite strip:

( ) ( ) ( ) 0y

y,xux

y,xuy,xu 2

2

2

22 =

∂∂

+∂

∂≡∇ [ )∞∈ ,0x , [ ]∞∈ ,0y

with boundary conditions: 0u

0x=

= 0t > Dirichlet

( ) ( )xf0,xu 0= 0t > Dirichlet

Solution: We will apply the Fourier sine transform in the x variable corresponding to the Dirichlet problem in the semi-infinite region:

( ) ( ) ( ) ( )∫∫∞∞

==00

xdxsiny,xu2dxx,Ky,xuy,u ωπ

ωω

The transformed equation:

0yuu 2

22 =

∂∂

+− ω

The integral transform of the second boundary condition:

( ) ( )∫∞

=0

00 xdxsinxf2f ωπ

ω

The general solution of the ODE is y

2y

1 ececu ωω += − The solution is bounded ⇒ 0c2 =

Apply the first boundary condition

0y = ⇒ ( )ω01 fc = The solution of the transformed equation:

( ) y0

û f e ωω −=

The solution of the BVP for Laplace’s Equation

( ) ( )∫∞

−=0

y0 xdsinef2y,xu ωωω

πω


751

FS-1-2h.mws LE in the 1st quadrant - Dirichlet problem - Fourier sine transform > restart;

> f(x):=Heaviside(x-1)-Heaviside(x-2);

:= ( )f x − ( )Heaviside − x 1 ( )Heaviside − x 2

> plot(f(x),x=0..5);

> ft:=simplify(int(f(x)*sin(omega*x),x=0..infinity));

:= ft −− + − ( )cos ω 2 ( )cos ω 2 1

ω

Transformed solution: > ut:=ft*exp(-omega*y);

:= ut −( )− + − ( )cos ω 2 ( )cos ω 2 1 e

( )−ω y

ω

Inverse transform -solution of BVP: the upper limit of integral defines the accuracy of approximation > u(x,y):=int(ut*sin(omega*x),omega=0..30)*2/Pi: > plot3d(u(x,y),x=0..4,y=0..4,axes=boxed);

( )u x, y

( ) ( ) ( )f x H x 1 H x 2= − − −


Laplace’s Equation in the semi-infinite strip Consider a 2-D Laplace’s Equation in the semi-infinite strip:

( ) ( ) ( ) 0y

y,xux

y,xuy,xu 2

2

2

22 =

∂∂

+∂

∂≡∇ ( )x 0,∈ ∞ , ( )y 0,M∈

with boundary conditions:

0xu

0x

=∂∂

=

0t > Neumann (insulation)

( ) ( )xf0,xu 0= 0t > Dirichlet ( ) ( )xfM,xu M= 0t > Dirichlet

We will apply the Fourier cosine transform in the x variable which corresponds to the Neumann problem in the semi-infinite region:

( ) ( ) ( ) ( )∫∫∞∞

==00

xdxcosy,xu2dxx,Ky,xuy,u ωπ

ωω

Transformed equation:

0yuu 2

22 =

∂∂

+− ω (2nd order ODE)

The integral transform of the boundary conditions:

( ) ( )∫∞

=0

00 xdxcosxf2f ωπ

ω

( ) ( )∫∞

=0

MM xdxcosxf2f ωπ

ω

Then the 2nd order ODE has to be solved with boundary conditions. The general solution of the ODE is ysinhcycoshcu 21 ωω +=

Find coefficients 1c and 2c from boundary conditions, then the solution of the IBVP will be given by the inverse integral transform:

( ) ( ) ( ) ( )∫∫∞∞

==00

xdcosy,u2dx,Ky,uy,xu ωωωπ

ωωω

Example Let ( ) 0xf0 = ( ) 0f0 =ω

( ) ( ) ( )2xH1xHxfM −−−= ( ) =ωMf

Apply first boundary condition 0y = ⇒ 0c1 =

My = ⇒ ( )ωω M2 fMsinhc = ⇒ ( )Msinh

fc M2 ω

ω=

Then the transformed solution is ( )Mfu sinh y

sinh Mω

ωω

=

Solution of IBVP (inverse transform):

( )y,xu [ ]0

2 u cos xdω ωπ

∞

= ∫


753


IX.2.8 REVIEW QUESTIONS 1) How is the Fourier transform defined?

2) What conditions guarantee the existence of the Fourier transform?

3) How is the inverse Fourier transform defined?

4) What are the main properties of the Fourier transform and the inverse

Fourier transform?

5) How can the convolution theorem be applied for evaluation of the inverse

Fourier transform?

6) What property allows application of the Fourier transform for solution of

differential equations?

7) What are the main steps in the procedure of application of the Fourier

transform for solution of the differential equations?

EXERCISES 1. a) Derive the scaling property

( ){ } 1 ˆF f at fa a

ω =

b) Using integration by parts derive the Fourier transform of the second order derivative (property 7b):

( ) ( )2

22

d ˆF f t fdt

ω ω

= −

2. Using the definition of the Fourier transform derive the transform of the one-sided exponential function:

( )ate t 0

f t0 t 0

− ≥=

< ( ) atH t e , a 0−= ⋅ >

and sketch the graph of the real part ( )ˆRe f ω and the imaginary part

( )Îm f ω of the transformed function for a 1= .

3. Using the result of the previous problem and the convolution theorem, evaluate the inverse Fourier transform:

( )

12

1Fa iω

−

+

and sketch the graph of the solution for a 1= .

4. The current ( )y t in the electrical circuit is modeled with the help of the 1st order differential equation

( )dy ay g tdt

+ = a 0> ( )t ,∈ −∞ ∞

where ( )f t represents the applied electromagnetic force.

Use the Fourier transform for solution. Write the solution in the form of a convolution. Then evaluate the solution for the case of instantaneous force defined by the Dirac delta function ( ) ( )g t tδ= . Sketch the graph of the solution curve.

5. Use the Poisson integral formula (p. 734) to derive the solution of the

Laplace equation in the semi-infinite region ( y 0> ) with the boundary condition defined by the Dirac delta function:

( ) ( ) ( )0u x,0 u x xδ= = and sketch the graph of the solution. x

y

( )xδ

0


755

6. Consider IX.2.7.1, p.744. Find the solution of the IBVP for :

2a = , ( ) ( ) ( )0u x H x 2 H x 4= − − − , ( ) ( ) ( )[ ]1tHtH2tf0 −−=

and visualize.

7. Verify operational property of Integral Fourier transform in a case of non-homogeneous Robin boundary condition (p.744).

8. Following Section IX.2.6, find the standard Fourier integral representation,

the Fourier cosine integral representation, and the Fourier sine integral representation of the function

( )0 t 0

f t cos t 0 t 20 t 2

ππ

<= < < >

( ) ( )H t H t 2 cos tπ= − − ⋅

9. Neutron transfer in infinite medium. ( )u x,t is the number of neutrons per

unit volume per unit time. ( ) ( )x tδ δ is the source function.

( ) ( )2

2 2

u 1 ux ttx a

δ δ∂ ∂+ =

∂∂, x−∞ < < ∞ , 0t > ,

( ) ( )u x,0 xδ= , x−∞ < < ∞ , 0t =

( )u x,t 0→ , x → ± ∞ , 0t >

Sketch the solution curves for a 1= and time t 1= , 4.

10. Solve the initial-boundary problem for one-dimensional heat equation.

2

2 2

u 1 utx a

∂ ∂=

∂∂, x−∞ < < ∞ , 0t > ,

( ) ( )u x,0 H x= , x−∞ < < ∞ , 0t =

Sketch the solution curves for a 1= and for time t 1= , t 2= .

11. Solve the initial-boundary problem for one-dimensional wave equation. 2 2

2 2 2

u 1 ux v t

∂ ∂=

∂ ∂, x−∞ < < ∞ , 0t > ,

( ) 2xu x,0 e−= , x−∞ < < ∞ , 0t =

( )u x,0 0t

∂=

∂, x−∞ < < ∞ , 0t =

Sketch the solution curves for v 1= and several moments of time.

12. Solve the initial-boundary problem for one-dimensional wave equation. 2 2

2 2 2

u 1 ux v t

∂ ∂=

∂ ∂, x−∞ < < ∞ , 0t > ,

( )u x,0 0= , x−∞ < < ∞ , 0t =

( ) ( )u x,0 xt

δ∂=

∂, x−∞ < < ∞ , 0t =

Sketch the solution curves for v 1= and several moments of time.



757

IX.2.9 Derivation of D’Alambert’s solution of the wave equation of an infinite string (p.733)

1) Transformed equation and initial conditions:

( ) ( )2

2

22

tt,u

v1t,u

∂∂

=−ωωω

( ) ( )ωω 0u0,u =

( ) ( )ωω1u

t0,u

=∂

∂

This is the initial value problem for a 2nd order linear ODE in t with constant coefficients

( ) ( ) 0t,uvt

t,u 222

2

=+∂

∂ ωωω

The roots of characteristic equation are ,1 2r i vω= ±

The general solution is

( )ˆ , i v t i v t1 2u t c e c eω ωω − += +

( )ˆ , i v t i v t1 2u t i v c e i vc e

tω ωω ω ω− +∂

= − +∂

where coefficients can be found from initial conditions.

( )ˆ0 1 2u c cω = + ⇒ ( ) ( )ˆ ˆ1 0 11 1c u w u w2 2i vω

= −

( )ˆ1 1 2u i v c i v cω ω ω= − + ⇒ ( ) ( )ˆ ˆ2 0 11 1c u w u w2 2i vω

= − +

Then the solution of the ODE becomes

( ) ( ) ( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ, i v t i v t0 1 0 1

1 1 1 1u t u w u w e u w u w e2 2i v 2 2i v

ω ωωω ω

− + = − + − +

( ) ( ) ( ) ( ) ( )ˆ ˆ ˆ, i v t i v t i v t i v t0 1

1 1u t e e u w e e u w2 2i v

ω ω ω ωωω

− + + −= − + −

( ) ( ) ( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ, i v t i v t i v t i v t0 0 1 1

1 1 1 1u t e u w e u w e u w e u w2 2 2i v 2i v

ω ω ω ωωω ω

− + + −= − + −

Definition of the inverse Fourier transform

( ) ( ){ } ( )1 i x1ˆ ˆf x F f f e d2

ωω ω ωπ

∞−

−∞

= = ∫


Use the inverse Fourier transform to find the solution

( ) ( ) ( ) ( ) ( )ˆ ˆˆ ˆ, 1 1i v t i x i v t i x i v t i x i v t i x

0 0

u w u w1 1 1 1 1 1 1 1u x t e u w e d e u w e d e e d e e d2 2 2 2 2 2i v 2 2 2i v

ω ω ω ω ω ω ω ωω ω ω ωπ π π ω π ω

− + + −= − + −∫ ∫ ∫ ∫

( ) ( ) ( )

( )

( ) ( )

( )

( ) ( ) ( ) ( )ˆ ˆˆ ˆ,

0 0

i x vt i x vt i x vt i x vt1 10 0

u x vt u x vt

u w u w1 1 1 1 1 1 1 1u x t u w e d u w e d e d e d2 2 2 2 2 2 i v 2 2 i v

ω ω ω ωω ω ω ωπ π π ω π ω

− + + −

− +

= − − + −∫ ∫ ∫ ∫

( ) ( ) ( ) ( ) ( ) ( )( )ˆ, i x vt i x vt1

0 0

u w1 1 1 1u x t u x vt u x vt e e d2 2 2 2 i v

ω ω ωπ ω

+ −= − − + + −∫

( ) ( ) ( ) ( ) ( )ˆ

,x vt

1 i s0 0

x vt

u w1 1 1 1u x t u x vt u x vt e d i s d2 2 2 2 i v

ω ω ωπ ω

∞ +

−∞ +

= − − + +

∫ ∫

( ) ( ) ( ) ( )ˆ,s x vt

i s0 0 1

s x vt

1 1 1 1u x t u x vt u x vt u w e ds d2 2 2v 2

ω ωπ

∞ = +

−∞ = +

= − − + +

∫ ∫

( ) ( ) ( ) ( )

( )

ˆ,

1

s x vti s

0 0 1s x vt

u s

1 1 1 1u x t u x vt u x vt u e d ds2 2 2v 2

ωω ωπ

= + ∞

= + −∞

= − − + +

∫ ∫

D’Alambert solution with arbitrary functions ( ) ( )20u x C∈ and ( ) ( )1

1u x C∈ :

( ) ( ) ( ) ( )x vt

0 0 1x vt

1 1 1u x,t u x vt u x vt u s ds2 2 2v

+

−

= − + − + ∫

Documents

IX.2 THE FOURIER TRANSFORMvps/ME505/IEM/09 02.pdfIX.2.5 Solution of PDEs in the Infinite Region 728 2.5.1 The Heat Equation – Gauss’s Kernel – Green’s Function 728 2.5.2 The