PDE and Boundary-Value Problems Winter Term 2014/2015 PDF

FOURIER TRANSFORMS

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Fourier sine transform and Fourier cosine transform one can solve many important problems of physics with very simple way. • Thus we will learn from this

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Example 4 Show that Fourier sine and cosine transforms of are and respectively. 2.5 Applications of Fourier Transforms to boundary value problems.

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PDE and Boundary-Value Problems

Winter Term 2014/2015

Lecture 7

Saarland University

24. November 2014

c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 1 / 26

Purpose of Lesson

To introduce the idea of

integ raltr ansforms and sho who wthe y transform PDEs innvariables into differential equations inn1 variables.To introduce thesine and cosine tr ansformsand use them to solv e an infinite-diffusion problem.To define theF ourierand in verseF ouriertr ansforms,to illustr ate several useful properties of the Fourier transform and to show how these properties can be used to solve PDEs. c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 2 / 26

Integral Transforms (Sine and Cosine Transforms)

General philosophy of transformsObstacleHardproblemEasyproblemSolu4onofeasyproblemSolu4onofhardproblemIntegraltransform(Step1)Inversetransform(Step3)Solvetheeasyproblem(Step2)c

Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 3 / 26

Integral Transforms (Sine and Cosine Transforms)

Integral transformation

An integral transformation is a transformation that assigns to one functionf(t)a new functionF(s)by means of a formula like f(t)!F(s) =B Z A

K(s;t)f(t)dt

Note that we

st art

with a function of tandend with a function of s.The functionK(s;t)is called thek ernelof tr ansformation. It is the

major ingredient that distinguishes one transform from another; it

is chosen so that the transform has certain properties.The limitsAandBalso change from transformation to

transformation. c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 4 / 26

Integral Transforms (Sine and Cosine Transforms)

Integral transformation

The general philosophy behind integral transformations is that they eliminate par tialder ivatives with respect to one of the variables; hence, the new equation has one less variable.Example

If we apply a transform to the PDE

u t=uxx for the purpose of eliminating the time derivative, then we would arrive at an ODE inx.The transform and its inverse together form what is called a transform pair c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 5 / 26 Integral Transforms (Sine and Cosine Transforms)Sine and Cosine Transforms

Sine and Cosine transforms

8 >>>>>:F s[f] =F(!) =2 1 Z 0 f(t)sin(!t)dt(Fourier sine transform) F

1s[F] =f(t) =1

Z 0

F(!)sin(!t)d!(inverse sine transform)

8 >>>>>:F c[f] =F(!) =2 1 Z 0 f(t)cos(!t)dt(Fourier cosine transform) F

1c[F] =f(t) =1

Z 0

F(!)cos(!t)d!(inverse cosine transform)c

Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 6 / 26 Integral Transforms (Sine and Cosine Transforms)Sine and Cosine Transforms

Sine and Cosine transforms of derivatives

1F s[f0] =!Fc[f](proved by integration by parts)2F s[f00] =2 !f(0)!2Fs[f]3F c[f0] =2 f(0) +!Fs[f]4F c[f00] =2 !f(0)!2Fc[f]c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 7 / 26 Integral Transforms (Sine and Cosine Transforms)Sine and Cosine Transforms

Fourier Sine Transform

f(x) =1R

0F(!)sin(!x)dxF(!) =2

1 R

0f(x)sin(!x)d!0 a F!a 2:e ax2!(a2+!2)3:x 1=2r2
4:H(ax)2
[1cos(!a)]5:x11 c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 8 / 26 Integral Transforms (Sine and Cosine Transforms)Sine and Cosine Transforms
Fourier Sine Transform (cont.)
f(x) =1R
0F(!)sin(!x)dxF(!) =2
1 R
0f(x)sin(!x)d!0 x 2+a2e a!7:x x 4+41 2 e!sin(!)8:tan 1ax 1ea!!
9:x2f(x)2

F00(!)c
Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 9 / 26 Integral Transforms (Sine and Cosine Transforms)Sine and Cosine Transforms
Fourier Sine Transform (cont.)
f(x) =1R
0F(!)sin(!x)dxF(!) =2
1 R
0f(x)sin(!x)d!0 x2 pa 2
1ea!2!
;a>0Here
H(ax) =(1;x6a

0;x>a(Reflected Heaviside function);
erfc(x) =2p1 Z x e t2dt(complimentary-error function):c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 10 / 26 Integral Transforms (Sine and Cosine Transforms)Sine and Cosine Transforms
Fourier Cosine Transform
f(x) =1R
0F(!)cos(!x)dxF(!) =2
1 R
0f(x)cos(!x)d!0 a F!a 2:e ax2a(a2+!2)3:x 1=2r2
4:H(ax)2
sin(a!)5:(x)2=c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 11 / 26 Integral Transforms (Sine and Cosine Transforms)Sine and Cosine Transforms
Fourier Cosine Transform (cont.)
f(x) =1R
0F(!)cos(!x)dxF(!) =2
1 R
0f(x)cos(!x)d!0 ax21pae!2=(4a)7:sin(ax)xH(a!)8:a x 2+a2e a!9:x2f(x)2
F00(!)c
Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 12 / 26
Integral Transforms (Sine and Cosine Transforms)Solution of an Infinite-Diffusion Problem via the Sine Transform
Solution of an Infinite-Diffusion Problem via the Sine
Transform
We now show how the sine transform can solve an important IBVP (the infinite diffusion problem).Problem 7-1
To find the functionu(x;t)that satisfies

PDE:ut=2uxx;0
BC:u(0;t) =A;0
IC:u(x;0) =0;06x61

To solve this, we carry out the following steps.
c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 13 / 26
Integral Transforms (Sine and Cosine Transforms)Solution of an Infinite-Diffusion Problem via the Sine Transform
Step 1. (Transformation)We transform thex-variablevia the F ouriersine tr ansformso that we get an ODE in time. F s[u] =2 1 Z 0 u(x;t)sin(!;x)dx=:U(!;t) =U(t); F s[ut] =2 1 Z 0 u t(x;t)sin(!;x)dx @@t2 4 2 1 Z 0 u(x;t)sin(!;x)dx3 5 =ddt
Fs[u] =ddt
U(t); F s[uxx] =2 !u(0;t)!2Fs[u] =2A! !2U(t):c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 14 / 26
Integral Transforms (Sine and Cosine Transforms)Solution of an Infinite-Diffusion Problem via the Sine Transform
Step 1. (Transformation)Transformation of the IC provides F s[u(x;0)] =U(0) =0:Substituting all these expressions into our IBVP, we change the original problem into an initial-value problem
ODE:U0(t) =2
!2U(t) +2A!
IC:U(0) =0:(7:1)c
Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 15 / 26
Integral Transforms (Sine and Cosine Transforms)Solution of an Infinite-Diffusion Problem via the Sine Transform
Step 2. (Solving the IVP for ODE)To solve (7.1), we could use a variety of elementary techniques from ODEs (integrating factor, homogeneous and particular solution); in any case, the solution is
U(t) =2A!

1e!22t
Step 3. (Inverse transform)The last step is to find the inverse transform ofU(t); that is, u(x;t) =F1s[U]: We can either evaluate the inverse transform directly from the integral or else resort to the tables. Using the tables we see that u(x;t) =Aerfcx2pt: c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 16 / 26
Integral Transforms (Sine and Cosine Transforms)Solution of an Infinite-Diffusion Problem via the Sine Transform

Remark

The exact values of the
complimentar y-errorfunction erfc(x) =2p 1 Z x e t2dt can be found in special tables.There is also the so-callederror function erf(x) =2p x Z 0 e t2dterf(x) +erfc(x) =1.The integrals in erf(x)and erfc(x)cannot be integrated by the usual elementary tricks of calculus. c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 17 / 26
The Fourier Transform and Its Application to PDEs

Exponential Fourier transforms:

F[f]F() =1p21
Z 1h f(x)eixdxi (Fourier transform - FT) F
1[F]f(x) =1p21
Z 1h
F()eixdi
(inverse FT) c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 18 / 26
The Fourier Transform and Its Application to PDEs

Exponential Fourier transforms:

Remarks
The Fourier transformF()can be acomple xfunction ; for example, the Fourier transform of f(x) =(0;x60 e x;x>0 isF() =1p21i1+2.Not all functions have Fourier transforms; in fact,f(x) =c, sin(x), e x,x2, donot ha veF ouriertr ansforms.Only functions that go to zero sufficiently fast asjxj ! 1have transforms.c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 19 / 26 The Fourier Transform and Its Application to PDEsProperties of the Fourier Transform
Properties of the Fourier Transform:

Transformation of partial derivatives:

F[ux] =1p21
Z 1 u x(x;t)eixdx=iF[u]
F[uxx] =1p21
Z 1 u xx(x;t)eixdx=2F[u]
F[ut] =1p21
Z 1 u t(x;t)eixdx=@@tF[u]
F[utt] =1p21
Z 1 u tt(x;t)eixdx=@2@t2F[u]c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 20 / 26 The Fourier Transform and Its Application to PDEsProperties of the Fourier Transform
Properties of the Fourier Transform:
The Fourier transform is a linear transformation; that is F[af+bg] =aF[f] +bF[g]The transform of a product of two functionsf(x)g(x)isnot the product of the individual transforms; that is,
F[f(x)g(x)]6=F[f] F[g]Convolution property:

F[fg] =F[f] F[g];
where (fg)(x) =1p21 Z 1 f(x)g()d:c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 21 / 26 The Fourier Transform and Its Application to PDEsProperties of the Fourier Transform
Properties of the Fourier Transform:

From the convolution property it follows that
fg=F1(F[f] F[g]):Remark
Hence, to find
F
1(F[f] F[g]);
all we have to do is find the inverse transform of each f actorto get f andgandthen compute their con volution.c Daria Apushkinskaya (UdS)PDE and BVP lecture 724. November 2014 22 / 26 The Fourier Transform and Its Application to PDEsProperties of the Fourier Transform
Exponential Fourier transform
f(x) =F1[F]F(!) =F[f]1:f (n)(x) (nthderivative)(i!)nF(!)2:f(ax);a>01 a F!a
3:f(xa)e
ia!F(!)4:equotesdbs_dbs12.pdfusesText_18

[PDF] PDE and Boundary-Value Problems Winter Term 2014/2015

PDE and Boundary-Value Problems

Winter Term 2014/2015

Lecture 7

Saarland University

24. November 2014

Purpose of Lesson

To introduce the idea of

Integral Transforms (Sine and Cosine Transforms)

Integral Transforms (Sine and Cosine Transforms)

Integral transformation

K(s;t)f(t)dt

Note that we

Integral Transforms (Sine and Cosine Transforms)

Integral transformation

If we apply a transform to the PDE

Sine and Cosine transforms

1s[F] =f(t) =1

F(!)sin(!t)d!(inverse sine transform)

1c[F] =f(t) =1

F(!)cos(!t)d!(inverse cosine transform)c

Sine and Cosine transforms of derivatives

Fourier Sine Transform

0F(!)sin(!x)dxF(!) =2

4:H(ax)2

Fourier Sine Transform (cont.)

0F(!)sin(!x)dxF(!) =2

9:x2f(x)2

F00(!)c

Fourier Sine Transform (cont.)

0F(!)sin(!x)dxF(!) =2

1ea!2!

H(ax) =(1;x6a

0;x>a(Reflected Heaviside function);

Fourier Cosine Transform

0F(!)cos(!x)dxF(!) =2

4:H(ax)2

Fourier Cosine Transform (cont.)

0F(!)cos(!x)dxF(!) =2

F00(!)c

Transform

To find the functionu(x;t)that satisfies

IC:u(x;0) =0;06x61

To solve this, we carry out the following steps.

Fs[u] =ddt

ODE:U0(t) =2

IC:U(0) =0:(7:1)c

U(t) =2A!

1e!22t

Remark

The exact values of the

The Fourier Transform and Its Application to PDEs

Exponential Fourier transforms:

F[f]F() =1p21

1[F]f(x) =1p21

F()eixdi

The Fourier Transform and Its Application to PDEs

Exponential Fourier transforms:

Remarks

Properties of the Fourier Transform:

Transformation of partial derivatives:

F[ux] =1p21

F[uxx] =1p21

F[ut] =1p21

F[utt] =1p21

Properties of the Fourier Transform:

F[f(x)g(x)]6=F[f] F[g]Convolution property:

F[fg] =F[f] F[g];

Properties of the Fourier Transform:

From the convolution property it follows that

Hence, to find

1(F[f] F[g]);

Exponential Fourier transform

3:f(xa)e