The Fourier transform is often useful for finding Green’s functions. Consider th

Question

The Fourier transform is often useful for finding Green’s functions. Consider the following partial differential equation for the function u(x, t):

(a) Apply a Fourier transform with respect to x to obtain a 2nd-order ordinary differential equation for the function ˆu(?, t).

(b) Find ˆu(?, t) by solving the differential equation from part (a).

(c) Invert the Fourier transform to obtain u(x, t). If you cannot invert the transform analytically, you can at least express u as an integral.

Explanation / Answer

Ans-

a function f(x) : [L, L] C, we have the orthogonal expansion f(x) = X n= cne inx/L, cn = 1 2L Z L L f(y)e iny/Ldy. (1) We think of cn as representing the “amount” of a particular eigenfunction with wavenumber kn = n/L present in the function f(x). So what if L goes to ? Notice that the allowed wavenumbers become more and more dense. Therefore, when L = , we expect f(x) is a superposition of an uncountable number of waves corresponding to every wavenumber k R, which can be accomplished by writing f(x) as a integral over k instead of a sum over n. Now let’s take the limit L formally. Setting kn = n/L and k = /L and using (1), one can write f(x) = 1 2 X n= Z L L f(y)e ikny dy e iknxk. Notice this is a Riemann sum for an integral on the interval k (, ). Taking L is the same as taking k 0, which gives f(x) = 1 2 Z F(k)e ikxdk, (2) where F(k) = Z f(x)e ikxdx. (3) The function F(k) is the Fourier transform of f(x). The inverse transform of F(k) is given by the formula (2). (Note that there are other conventions used to define the Fourier transform). Instead of capital letters, we often use the notation ˆf(k) for the Fourier transform, and F(x) for the inverse transform. 1.1 Practical use of the Fourier transform The Fourier transform is beneficial in differential equations because it can transform them into equations which are easier to solve. In addition, many transformations can be made simply by applying predefined formulas to the problems of interest. A small table of transforms and some properties is given below. Essentially all of these result from using elementary calculus techniques on definitions (3) and (2). 1 A Brief table of Fourier transforms Description Function Transform Delta function in x (x) 1 Delta function in k 1 2(k) Exponential in x ea|x| 2a a 2+k 2 (a > 0) Exponential in k 2a a 2+x2 2ea|k| (a > 0) Gaussian e x 2/2 2ek 2/2 Derivative in x f0 (x) ikF(k) Derivative in k xf(x) iF0 (k) Integral in x R x f(x 0 )dx0 F(k)/(ik) Translation in x f(x a) e iakF(k) Translation in k eiaxf(x) F(k a) Dilation in x f(ax) F(k/a)/a Convolution f(x)*g(x) F(k)G(k) Typically these formulas are used in combination. Preparatory steps are often required (just like using a table of integrals) to obtain exactly one of these forms. Here are a few examples. Example 1. The transform of f 00(x) is (using the derivative table formula) f 00(x) = ik f 0 (x) = (ik) 2 ˆf(k) = k 2 ˆf(k). Notice what this implies for differential equations: differential operators can be turned into “multiplication” operators. Example 2. The transform of the Gaussian exp(Ax2 ) is, using both the dilation and Gaussian formulas, exp(Ax2 ) = h exp([ 2Ax] 2 /2)i = 1 2A exp(x 2 /2) (k/ 2A) = r A exp([k/ 2A] 2 /2) = r A exp(k 2 /(4A)). Example 3. The inverse transform of e 2ik/(k 2 + 1) is, using the translation in x property and then the exponential formula, e 2ik k 2 + 1 = 1 k 2 + 1 (x + 2) = 1 2 e |x+2| . Example 4. The inverse transform of kek 2/2 uses the Gaussian and derivative in x formulas: h kek 2/2 i = i h ikek 2/2 i = i d dx h e k 2/2 i = = i 2 d dx h 2ek 2/2 i = i 2 d dx e x 2/2 = ix 2 e x 2/2 . 1.2 Convolutions Unfortunately, the inverse transform of a product of functions is not the product of inverse transforms. Rather, it is a binary operation called convolution, defined as (f g)(x) = Z f(x y)g(y)dy. (4) 2 Using the definition, the Fourier transform of this is (f g) = Z Z f(x y)g(y)e ikxdy dx Using the change of variables z = x y, this becomes Z Z f(z)g(y)e ik(y+z) dydz = Z f(z)e ikzdz Z g(y)e ikydy = ˆf(k)ˆg(k), which is just the last formula in the table. 1.3 Divergent Fourier integrals as distributions The formulas (3) and (2) assume that f(x) and F(k) decay at infinity so that the integrals converge. If this is not the case, then the integrals must be interpreted in a generalized sense. In addition, some of the table formulas must be adjusted to take this into account. Notice that the transform of (x) is one, so at least formally, the inverse transform of one should be a delta function: (x) = 1 2 Z e ikxdk. (5) This is very troublesome: the integral does not even converge, so what could such a statement mean? Of course the issue is that the integral represents a distribution, not a regular function. We can define the inverse transform of F(k) more generally as a distribution which is the limit of the regular functions fL(x) = 1 2 Z L L exp(ikx)F(k)dk as L (recall the fact that distributions can always be approximated by regular functions). This means that the inverse transform f(x) is a distribution which acts on smooth functions like f[] = lim L Z fL(x)(x)dx. (6) Let’s look at the integral in (5) and see what distribution it represents. For this case, fL(x) = 1 2 Z L L exp(ikx)dk = 1 x sin(Lx). Then if one asks how the limit of fL acts as a distribution, one computes f[] = lim L 1 Z sin(Lx) x (x)dx = lim L 1 Z sin(y) y (y/L)dy. The limit L can be taken inside the integral (this can be justified with a little effort), and this results in f[] = (0) Z sin(y) y dy = (0). So the inverse transform really is the delta function!

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