 Hello, so we continue with the general discussion on Fourier transforms. The topic that we shall take up now is Fourier transforms of functions of several variables. So far our functions have been one variable functions like the characteristic function of the interval we completed the Fourier transform of the Gaussian e to the power minus a x squared etcetera. Now we take a functions defined from R n to R f of x 1, x 2, x n. So, before we take it up we must introduce the Schwarz's class of functions that are infinitely differentiable and decay very rapidly namely we take the function f with compute its partial derivatives since it is a function of several variables. So, del alpha f upon del x 1 to the power alpha 1 del x n to the power alpha and alpha is a multi-index alpha 1 alpha 2 alpha n of non-negative integers. So, we differentiate the function certain number of times arbitrary number of times and we multiply it by a arbitrary polynomial p x 1 x 2 dot x n. So, this product p of x 1 x 2 x n times the alpha derivative of f the supremum over R n should be finite. So, in particular if I take alpha to be 0 and the polynomial to be the constant polynomial that is I am not differentiating the function at all and I am taking the constant polynomial 1 the function itself must be bounded. Suppose I take the polynomial to be x 1 square plus x 2 square plus x 3 square plus dot dot dot plus x n square and I take alpha to be 0. So, I am not differentiating the function at all it means that x 1 square plus x 2 square plus x n square times f that supremum must be bounded it means that the function must decay to 0 as x goes mod x goes to infinity and. So, you get the idea I could take polynomial of high degree I could take 1 plus x 1 square plus x 2 square plus dot dot plus x n square to the power 100 and then I do not differentiate I take alpha equal to 0. So, this polynomial times f must be bounded. So, f must decay very rapidly it must decay faster than 1 upon 1 plus x square plus x 2 square plus dot dot plus x n square to the power 100 and the same must be the case for all the derivatives not only the function must decay very rapidly all the derivatives must decay very rapidly and. So, this is the class of Schwarz functions examples of Schwarz functions e to the power minus x 1 square minus x 2 square theta minus x n square take the individual Gaussian e to the power minus x 1 square e to the power minus x 2 square theta e to the power minus x n square and multiply them together you get the an element in the multidimensional Schwarz class. So, you take an element of the Schwarz class and we want to define the Fourier transform and the definition is given in the slide. The Fourier transform f hat of chi 1 chi 2 chi n is the integral over R n e to the power minus i x 1 chi 1 plus x 2 chi 2 plus dot dot dot x n chi n times f of x 1 x 2 x n dx 1 dx 2 dx n that is a definition of the Fourier transform in several variables. The analogy is very clear the vector chi 1 chi 2 chi n is going to be the frequency vector. Now, suppose that the function f of x 1 x 2 x n depends only on the distance from the origin. If f depends only on square root of x 1 square plus x 2 square plus dot dot plus x n square the same is the case with the Fourier transform. f hat depends only on root of chi 1 square plus dot dot plus chi n square. What happens if the function is a function of one variable? What happens if n is 1? What does it mean to say that the function depends only on mod x? For example, the function is an even function or a function is an odd function. So, this idea that a function depends only on root of x 1 square plus dot dot plus x n square is a generalization of the notion of an even function. So, we know that the Fourier transform of an even function is again an even function. A generalization of this is the theorem 53 that has been displayed in the slide. As I said the most important example of an element in the Schwarz class in several variables is this multidimensional Gaussian e to the power minus x 1 square minus x 2 square dot dot minus x n square. This is a radial function and its Fourier transform is going to be again a radial function and you can calculate the Fourier transform of this explicitly. Now, let us turn to the proof of the theorem. So, now x 1 chi 1 plus x 2 chi 2 plus dot dot dot plus x n chi and that is the inner product of x n chi. So, now let us take a chi and let us take a matrix A which maps the nth standard unit vector to chi upon norm chi. I have taken a non-zero chi and I divide by the length that is a unit vector and I am going to choose an orthogonal matrix that takes this unit vector E n and maps it to chi by norm chi. Now, let us look at the Fourier transform of chi that is by definition integral f of x 1 x 2 x n dx 1 dx 2 dx n and the exponential factor. x 1 chi 1 plus x 2 chi 2 plus dot dot x n chi n is x 1 times chi 1 upon norm chi 1 chi 2 upon norm multiply and divide by norm chi and chi upon norm chi is A E n and so we get A E n in a product with x and the norm chi is here. Push this matrix to the other factor when I push this matrix to the other factor I am going to get inner product of E n n A transpose x but remember that the matrix is a rotation matrix. So, A transpose is A inverse. So, I get A inverse of x in the second slot and I get x of minus i norm chi inner product E n with A inverse of x and f of x 1 x 2 x n dx 1 dx 2 dx n. Now, the next step is to put A inverse of x equal to y or x equal to A y. When I put x equal to A y this simply becomes E n times y E n times y is simply y n. So, the exponential factor becomes x of minus i norm chi times y n and of course, f is radial f depends only on the square root of x 1 square plus x 2 square plus x n square. Square root of x 1 square plus x 2 square plus x n square is the same as the square root of y 1 square plus y 2 square plus y n square because A is orthogonal. So, f of x 1 x 2 x n simply becomes f of y 1 y 2 y n. What happens to dx 1 dx 2 dx n? It is going to be determinant of A mod times dy 1 dy 2 dy n but A is a rotation matrix and so, it is determinant is going to be the plus 1 or minus 1 and I am going to take the absolute value. So, the Jacobian factor is simply 1 and so, I get this last display f hat of chi 1 chi 2 chi n equal to integral over R n x of minus i norm chi y n f of y 1 y 2 y n dy 1 dy 2 dy n. Now, let us write f of x 1 x 2 x n as capital f of R where the distance from the origin has been abbreviated to R. So, let us look at the case for n equal to 3. So, f hat of chi 1 chi 2 chi 3. What is f hat of chi 1 chi 2 chi 3? Let us go back to the previous slide. This little f of y 1 y 2 y n is capital f of R and e to the power minus i times norm chi into y n. So, what is y n here? y 3 is R cos theta. So, x of minus R norm chi cos theta that is the exponential factor. The function f of x 1 x 2 x 3 is capital f of R. Where did this come from? R squared dr sin phi d phi d theta. Well, dx 1 dx 2 dx 3 is going to be R squared sin phi dr d theta d phi polar coordinates in R 3 and the theta varies from 0 to 2 pi phi varies from 0 to pi and the integral 0 to 2 pi d theta picks up a factor of 2 pi. Now, what remains is the middle integral and to deal with the middle integral there is a very convenient sin phi d phi out here. So, obviously, I am going to put cos phi equal to s and the integral becomes the outer integral is 0 to infinity f of R R squared dr the inner integral is e to the power minus i R norm chi times s d s and when I put cos phi equal to s when phi runs from 0 to pi s will run from minus 1 to 1 and this integral can be computed directly it is sin norm chi by norm chi and it is an even function. So, it is going to be twice. So, you get a 4 pi by norm chi and then the integral from 0 to infinity remains f of R remains and 1 R goes away because when I integrated this I pick up R the denominator and then this is what is left over. So, the Fourier transform of a radial function is again a radial function as we can see from this integral, but it is also what is known as a sin transform. When n is even say when n equal to 2 something very interesting happens again the little f became capital F of R dx dy R dr d theta and the exponential factor yn is basically written as R sin theta. So, we get cos of R norm chi sin theta d theta that is what we get in the even case. So, let us go back the exponential factor the sin part drops out and the cos part remains. Now, we are not so lucky we can not put sin theta equal to s and hope to simplify this integral this integral would not simplify the reason why it will not simplify is because this is a Bessel function here cos is an even function. So, I can say twice integral 0 to pi and so you see a 2 factor and a multiplied divided by pi. So, write it as 2 pi times 1 upon pi integral 0 to pi cosine R norm chi sin theta d theta. If you go back to the previous chapters you will see that that is the integral definition of the Bessel's function which we derived in chapter 1 from the Schrodinger Milchitz formula remember. So, this is exactly j 0 of R norm chi. So, the Fourier transform is again a radial function, but it is a Bessel transform it is a transform whose kernel involves the Bessel's function g 0. In general when you take a radial function in even space dimensions the Fourier transform will be a Bessel transform in odd dimensions it will be an elementary sin transform. If you take it in 4 dimensional space then there will be j 1 if you take it in 6 dimensional space it will be j 2 and so on so forth. The appearance of Bessel's functions in these formulas is significant because you will see this appearing in various problems in optics. For example, if you look at diffraction problems through a circular aperture the Bessel's function makes its appearance the radii or the Newton's rings there are zeros of the Bessel's functions. In fact, now why does the Bessel function appear in these kinds of problems because these kinds of problems exhibit cylindrical symmetries when you write the Laplace's operator in cylindrical coordinates you will see the Bessel's function staring at you. Now let us go to the next part of this chapter the wave equation. Let us begin with the one-dimensional wave equation utt minus uxx equal to 0 equation 4.19 in the display and we look at this equation in the half space the x varies over the real line the time t varies in the positive real line r plus. It is a second-order equation in time so two initial conditions have to be prescribed u of x0 is fx and ut of x0 is gx 4.20 in the display. For simplicity let us assume that the fx and gx are in the Schwarz class you can relax these conditions tremendously but let us not bring in those technicalities now we will discuss those things later. So let us take the Fourier transform or the differential equation with respect to the x variable. So simply write the Fourier transform of u and differentiate with respect to t under the integral sign and you see that ut hat is you first take the Fourier transform and then differentiate it with respect to t because my function u is the Schwarz class differentiation under the integral sign is permissible. Likewise utt hat will be del 2 u hat upon del t squared but what about the Fourier transform of minus uxx it is chi squared u hat remember exactly as we did it in the heat equation. So the Fourier transform of the wave equation transforms to equation 4.21 namely d2 by dt squared u hat plus chi squared u hat equal to 0 and this is a second-order ODE for u hat chi should be thought of as a parameter but you need initial conditions. So what you do is that you take the initial conditions 4.20 and take the Fourier transform with the space variable x. So u hat of chi 0 is basically f hat ut hat at chi 0 equal to g hat. So the initial conditions are given to be f hat and g hat and we can solve this ordinary differential equation through very elementary methods which we teach in undergraduate courses and the solution can immediately be written down u hat of chi t is cos t chi f hat of chi plus sin t chi by chi g hat of chi put t equal to 0 and check this part disappears cosine becomes 1. Yes you get this initial condition here differentiate and put t equal to 0 you will get the second piece and surely this 4.23 satisfies equation 4.21. Now the next thing that you would believe we are going to do is to take the inverse Fourier transform and obtain an integral representation for u that can be done and you will get the D'Alembert's formula for the solution of the wave equation but that is not what we are going to do since that thing is discussed in most books I will leave it to you to figure it out what we are going to do instead is something far more interesting. We are going to look at equation 4.23 and we are going to work with 4.23 using the Parseval formula. So let us look at the principle of conservation of energy let us take the wave equation u t t minus u x x equal to 0 multiply by u t and integrate over the space variables not over the time variable and in the second integration integrate by parts through one of the derivatives with respect to x in the first factor it will become a plus sign and we will get integral u x u t x, u x u t x is the derivative of u x squared u t u t t is a time derivative of u t squared. So after you perform integration by parts with the x variable this left hand side combines to give you d dt integral over r u t squared dx plus integral over r u x squared dx equal to 0 4.24. Of course, we have assumed that the boundary terms goes to 0 which is easy to verify you can use the fact that the functions are the Schwarz class and so on there is no problem. So now you can consult any book for the statement and proof of the D'Alembert's formula and and and so on. But this equation 4.24 is what we are interested you see that this expression under the d dt let us look at the expression under the d dt integral over r u t squared dx plus integral over r u x squared dx 4.25 this combination is called ET I have integrated with respect to x so what is left over will depend on t but the remarkable fact is that it is also independent of t in other words this function ET given by 4.25 is constant in time it is constant in time this is called the law of conservation of energy the expression ET is really the energy of the wave. So now we will use the Parseval formula what we see here is a square of the l2 norm of u t and what we see here is a square of the l2 norm of u x so the square of the l2 norm is related to the square of the l2 norm of the Fourier transform so we must replace these things by their Fourier transform there will be a factor of 2 pi coming in from the Parseval formula that is what you see here in equation 4.26 on the right hand side you see 2 pi e but when you take the Fourier transform of u x squared you are going to pick up a chi squared times the Fourier transform and so that is what you see here and when you take the Fourier transform of this piece you simply take the Fourier transform of u and take its time derivative so you get 4.26 is completely equivalent to 4.25 that is a law of conservation of energy. Since energy is independent of time as we have seen we can put t equal to 0 in 4.26 when I put t equal to 0 u hat of chi comma 0 is f hat u t hat of chi comma 0 is g hat so this is the energy equation 4.27 computed at time t equal to 0. Now what we can do is that let us look at the two parts of the energy equation one was u t squared dx integral and the other was u x squared dx integral this part involving the time derivative is the kinetic energy let us call it k t and the Parseval formula it is the same as 1 upon 2 pi integral over r u hat t chi t the whole square d chi t. Now on the other hand we also have the explicit formula that we derived 4.23 namely u hat of chi t is cos t chi f hat plus sin t chi by chi g hat so let us use this to calculate the kinetic energy so differentiate this with respect to time what happens you pick a chi factor here so minus chi f hat of chi sin t chi differentiate here and when chi goes away g hat of chi into cosine t chi so use this expression differentiate with respect to t and put it in 4.28 we get the expression for kinetic energy in terms of g hat and f hat okay so now what do we do with this 4.29 now we expand this expression g hat cos t chi minus f hat sin t chi the whole square this expression that we got we can expand and what we get is that we will get cos square t chi chi squared sin square t chi and we are going to get the cross terms 2 cos t chi sin t chi and all these things can be written using the double angle formula cos 2 alpha is 2 cos squared alpha minus 1 sin 2 alpha is 2 sin alpha cos alpha use these double angle formulas and you will get terms or the form the kinetic energy is 1 upon 4 pi times you are going to get g hat chi the whole square plus chi squared f hat chi the whole the pure square terms have been taken out plus dot dot dot but these is basically e by 2 plus dot dot dot what is e by 2 what is e we have the formula for e here remember over here from 4.27 and so using 4.27 what we get over here is exactly e by 2 and what are the dot dot dot the terms that have been left out the terms that have been left out is the sum of 3 terms mod chi f hat squared cosine 2 t chi d chi mod g hat chi cos 2 t chi d chi and integral chi f hat chi g hat chi sin 2 t chi d chi each of these terms individually go to 0 as t goes to infinity why is that remember f and g are in the Schwarz class f hat and g hat are in the Schwarz class and if since f is in the Schwarz since f hat is in the Schwarz class g hat is in the Schwarz class they are in L 2. So, f hat is in L 2 g hat is in L 2 chi times f hat is also in the Schwarz class. So, chi times f hat is in L 2. So, mod g hat squared is in L1, chi f hat squared is in L1, chi f hat chi g hat chi is also in L1, product of 2L2 functions is in L1. And now you got an L1 function and you got cosine 2 t chi d chi integral over R, Riemann Lebesgue Lemma. Riemann Lebesgue Lemma says that they individually go to 0. So, the terms that are not being written out go to 0 as t goes to infinity. So, what it proves is that the kinetic energy converges to e by 2 as time goes to infinity. Then the potential energy likewise must go to e by 2 as t goes to infinity because some of the kinetic energy and the potential energy is e. So, it says that for the wave equation as time evolves the kinetic and potential energy, the total energy is constant e, but individually the kinetic and the potential energy they tend to become equal, they equalize, they redistribute themselves in equal measures and that is called the principle of equipartitioning of energy. So, I have given you a reference for these kinds of things. Robert Strichards book, A Guide to Distribution Theory and Fourier Transforms, CRC Press, Boca Raton 1994. It is a delightfully written book on Fourier transforms and distributions and that would be a very good book to consult after you have learnt the fundamental principles of this course. I think it is a good place to stop here. Thank you very much.