Applications of Fourier transform PDF

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Applications of Fourier transform

Applications of Fourier transform. A. Eremenko. April 16 2021. 1. Heat equation on the line. ut = kuxx

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Applications of Fourier transform

A. Eremenko

April 16, 2021

1. Heat equation on the line.

u t=kuxx; u(x;0) =f(x):(1) Let us assume thatfandx7!u(x;t) tend to 0 forx! 1suciently fast so that we can take Fourier transforms in the variablex. Then we obtain ^ut=ks2^u;^u(s;0) =^f(s): (Dierentiation with respect totcan be performed under the integral sign). This is the initial value problem for a rst order linear ODE whose solution is u(s;t) =^f(s)eks2t: Since the inverse Fourier transform of a product is a convolution, we obtain the solution in the form u(x;t) =K(x;t)? f(x); whereK(x;t) is the inverse Fourier transform ofeks2t. Using Example

2 (formula (5)) from the previous lecture \Fourier Transform" witha=

1=(2kt), we obtain

K(x;t) =12

pkt ex24kt:(2) This is called theheat kernel. So the solution of our problem is u(x;t) =12 pkt Z 1 1 e(xy)24ktf(y)dy=12 pkt Z 1 1 ey24ktf(xy)dy: 1 Notice that this solution makes sense under a very mild restrictions onf: because of the factorey2, the integral will be convergent for functions of suciently slow growth, even if Fourier transform is not dened for them. On the other hand, once this formula is written, it is easy to verify that it indeed solves our problem. Indeed, the heat kernel itself does satisfy the heat equation, which is veried by dierentiation (do this!), so its convolution with any function also satises heat equation.

Now the heat kernel has these three properties:

(i)K(x;t)>0 (ii) R1

1K(x;t)dx= 1 for allt >0, (check this!)

(iii) For every >0 we haveK(x;t)!0 ast!0,uniformlyforjxj :

Check this!

Any functionK(x;t) with these properties is called apositive kernel, and it is easy to see that for every continuousf2L1we have (K(:;t)? f)(x)!f(x); t!0: So we obtained a solution of the initial value problem for the heat equation on the line for a large class of initial conditions. Actually this solution is unique under a mild restriction of the growth at innity, namely thatu(x;t) as a function ofxhas slower growth thanex2.

2. Heat equation on half-line with zero boundary condition.

Problem.Solve the heat equation (1) on the half-linex >0 with the boundary condition u(0;t) = 0; t >0;(3) and constant initial condition u(x;0) =f(x):(4)

Thenf(x) = const, this is problem occurs as the

at Earth approximation of the problem of cooling the Earth, solved by Kelvin:xis the depth, and the temperature is supposed to depend on depth and time only. The surface tem- perature is constant and we may take it as zero. (The constant temperature is a crude approximation, of course). 2 The question is how to reduce this problem to a problem on the whole line which we just solved. The answer isconsider the odd extension of the initial condition!. The heat kernelK(x;:) is even (as a function ofx), and a convolution of an even function and odd function is odd: if we suppose thatKis even and fis odd, then the convolution u(x) :=Z

K(xy)f(y)dy

will be odd. Indeed u(x) =Z

K(xy)f(y)dy=Z

K(x+y)f(y)dy

K(xu)f(u)du=Z

K(xu)f(u)du=u(x):

Since every odd function is zero at zero, we can replace the boundary con- dition byu(x;0) =~f(x) where~fis the odd extension off, and apply the solution for the whole real line obtained in the previous section. For example, whenfis constant, sayf(x) =T, thenu+Twill have boundary values 0 for negativexand 2Tfor positivex, and the formula from section 1 gives: u(x;t) =Tpkt Z 1 0 e(xy)24ktdyT:(5)

So we obtained a solution.

To determine the age of Earth, we compute the gradientux(0;t), and since this gradient can be measured, we can determine the aget. Dierentiating and substitutingx= 0, we obtain u x(0;t) =Tpkt See the handout \Age of the Earth" for a further discussion.

3. Heat equation on a half-line with arbitrary boundary condition.

In the previous section we solved heat equation on the half-linex >0 with initial conditionu(x;0) = 1; x >0 and the boundary conditionu(0;t) = 3

0; t >0:It is given by formula (5) which we write as

u(x;t) = 2Z 1 0

K(xy;t)dy1;

whereK(x;t) is the heat kernel given by formula (2). Now it is easy to obtain a solution of the heat equation with initial and boundary conditionsv(x;0) = 0;x >0;v(0;t) = 1:it is simply v(x;t) = 1u(x;t) = 22Z 1 0

K(xy;t)dt:

Now we make an important remark: for everyx6= 0, the functiont7!K(x;t) has limit 0 ast!0, and moreover, all partial derivatives with respect tot have limit 0, whent!0. See the discussion of such functions in \Fourier transform", in the Example in section 3. This means that we can extend the functionK(x;t) tot <0 by deningK(x;t) = 0 fort <0, and the extended function will satisfy the heat equation for allx >0 and allt. So our function v(x;t) also satises heat equation forx >0 and allt, and it equals 0 for t0.

Let us denote byHs(t) theHeaviside function,

H(t) =1; t0;

0; t <0:

We can solve now the heat equation forx >0,1< t <1with the boundary conditionu(0;t) =H(ts) : v s(x;t) = 22Z 1 0

K(xy;ts)dy; vs(0;t) =H(ts);(6)

where we use the convention thatK(x;t) = 0 fort <0. Once we can solve the problem whose boundary condition is Heaviside function, we can also solve it with any linear combination of Heaviside functions. And every reasonable function oftcan be approximated by a linear combination of Heaviside functions: suppose for example thatf(t) is continuous and has bounded support which is contained in the positive ray. Then the linear combinations nX j=0(f(sj+1)f(sj))H(tsj) 4 approximatefuniformly when the partitions0< s1< ::: < sn+1of the support offis ne enough, andf(s0) =f(sn+1) = 0. We can write this as n X j=0f(sj+1)f(sj)s j+1sjH(tsj)(sj+1sj); which is the integral sum of an integral, and we obtain the representation f(s) =Z 1 0

H(ts)f0(s)ds;

in the form of convolution with the Heaviside function. Since we solved the heat equation with boundary condition in the form of a Heaviside function, we can now use the superposition principle, and solve it with arbitrary boundary function: u(x;t) =Z 1 0 v s(x;t)f0(s)ds; wherevsis the solution (6). Integrating this by parts, we obtain a convolution formula representing the solution in terms of boundary conditionf(t): u(x;t) =Z 1 0 f(s)dds vs(x;t)ds: It remains to compute the derivative of the explicit functionvs. The com- putation is simplied by the fact that it satises the heat equation. By dierentiating (6) and using the heat equationKt=kKxxwe obtain: dds vs(x;t) =2dds Z 1 0

K(xy;ts)dy= 2Z

1 0 K t(xy;ts)dy = 2kZ 1 0 K xx(xy;ts)dy= 2kKx(x;ts); and simple explicit dierentiation of the heat kernel shows that

L(x;t) := 2kKx(x;t) =x2

pk t3=2ex24kt; and we obtain the nal result u(x;t) = (L ? f)(x;t) =x2 pk Z t 0 (ts)3=2exp x24k(ts) f(s)ds:(7) 5 Notice that integral from 0 to1is actually from 0 totsinceK(x;ts) = 0 fors > t. FunctionL(x;t) is called thelateral heat kernel. Once the formula for a solution is found, one can prove it without any mentioning of its derivation:

Exercise 1.Show that:

a) The kernel

L(x;t) =x2

pk t3=2ex22kt; as a function oftis a positive kernel in the sense dened in Section 1, and b) It satises the heat equationLt=kLxxforx >0; t >0 and is innitely dierentiable when extended by 0 fort <0. c) Conclude that (7) is a solution of the heat equation with the boundary functionu(x;t) =f(t) and initial conditionu(x;0) = 0. See the handout \Transatlantic Cable" for a discussion of this solution.

4. Laplace equation in a half-plane.

Consider the Dirichlet problem

u xx+uyy= 0;1< x <1; y >0; with the boundary condition u(x;0) =f(x); and some condition at1, for example thatutends to zero suciently fast so that Fourier transform can be applied.

Then Fourier transform with respect toxgives

s2^u(s;y) + ^uyy= 0;^u(s;0) =^f(s): This is an second order linear ODE whose general solution is C

1(s)esy+C2(s)esy:

One boundary conditionC1(s) +C2(s) =^f(s) is not sucient, but the sim- plest way to satisfy it

1is to takeC1(s) = 0 fors >0 andC2(s) = 0 fors <0,1

Of course, this is not rigorous. But once this correct formula is guessed, it can be proved without any reference to Fourier transform. Uniqueness of solution is a separate problem. In fact it has uniqueboundedsolution, but we do not prove this result here. 6

This gives

^u(s;y) =^f(s)eyjsj: Since this is a product, the originalumust be a convolution offwith the inverse transform ofeyjsj. The last transform we know from Example 3 of the previous lecture. It is

P(x;y) =y(x2+y2):

This is called thePoisson kernelfor the upper half-plane. So we obtain u(x;y) = (P ? f)(x;y) =y Z 1

1f(t)(xt)2+y2dt;

which is called thePoisson formulafor the upper half-plane. This formula has an appealing geometric interpretation. Let us set rst f(x) = 1; x <0 andf(x) = 0;x >0. Then the Poisson integral can be evaluated:y Z 0

1dt(xt)2+y2=1

arctanyx in other words,u(x;y) in this case is just the polar angle of the point (x;y), divided by. Check that the boundary values fory= 0 are correct! More generally, iff(x) = 1; x2(a;b) and 0 otherwise, we conclude that u(x;y) is the angle under which the interval (a;b) is seen from the point (x;y), divided by.

5. Wave equation on the whole line.

u tt=c2uxx; and let us take the initial conditions u(x;0) =f(x); ut(x;0) =g(x);1< x <1:

Applying Fourier's transform, we obtain

^utt=c2s2^u;^u(s;0) =^f(s);^ut(s;0) = ^g(s):

The general solution is

1(s)cos(cst) +C2(s)sin(cst);

7 and the rst initial conditions implyC1(s) =^fs; C2(s) = ^g(s)=(cs):So we solved the problem in the sense that we found the Fourier transform of the solution ^u(s;t) =^f(s)cos(cst) + ^g(s)sin(cst)cs :(8) To obtain an explicit formula of convolution type, we would like to know inverse transforms of cos(cst) and sin(cst)=(cs):There is no problem with the second one, this is essentially Example 1 in \Fourier Transform", and it isquotesdbs_dbs20.pdfusesText_26

[PDF] Applications of Fourier transform

Applications of Fourier transform

A. Eremenko

April 16, 2021

1. Heat equation on the line.

2 (formula (5)) from the previous lecture \Fourier Transform" witha=

1=(2kt), we obtain

K(x;t) =12

Now the heat kernel has these three properties:

1K(x;t)dx= 1 for allt >0, (check this!)

Check this!

2. Heat equation on half-line with zero boundary condition.

Thenf(x) = const, this is problem occurs as the

K(xy)f(y)dy

K(xy)f(y)dy=Z

K(x+y)f(y)dy

K(xu)f(u)du=Z

K(xu)f(u)du=u(x):

So we obtained a solution.

3. Heat equation on a half-line with arbitrary boundary condition.

0; t >0:It is given by formula (5) which we write as

K(xy;t)dy1;

K(xy;t)dt:

Let us denote byHs(t) theHeaviside function,

H(t) =1; t0;

0; t <0:

K(xy;ts)dy; vs(0;t) =H(ts);(6)

H(ts)f0(s)ds;

K(xy;ts)dy= 2Z

L(x;t) := 2kKx(x;t) =x2

Exercise 1.Show that:

L(x;t) =x2

4. Laplace equation in a half-plane.

Consider the Dirichlet problem

Then Fourier transform with respect toxgives

1(s)esy+C2(s)esy:

1is to takeC1(s) = 0 fors >0 andC2(s) = 0 fors <0,1

This gives

P(x;y) =y(x2+y2):

1f(t)(xt)2+y2dt;

1dt(xt)2+y2=1

5. Wave equation on the whole line.

Applying Fourier's transform, we obtain

The general solution is

1(s)cos(cst) +C2(s)sin(cst);