 Hi, so let us continue from where we stopped. We were discussing the vibrations of a circular membrane which is clamped along the rim and we derived the differential equation for the circular membrane. I gave you a reference to Krazyk's book and we did a separation of variables and we got the radial part of the solution Vr must satisfy this Bessel's differential equation 5.9 that you see in the slide and we also saw that the physically tenable solution is Jn of Kr and since the membrane is clamped on the boundary that is Jn of k must be 0. So, we saw that the k the relevant k parameter k must be a root of the Bessel's function or the zeros of the Bessel's functions. We are going to later prove that the Bessel's function Jn has infinitely many zeros. Now, let us assume that n is 0. Let us assume that this function z does not depend on the angular coordinates at all. We are only looking at the radial vibrations of the membrane. Now, of course, we understand this function a cos ck t plus b sin ck t. This is a function we understand. So, it is important to understand this other function Jn of Kr a little more carefully. So, now let us look at the radial vibrations. Now, we are given a second order differential equation. So, we must give the initial displacement and the initial velocity membrane. The initial displacement is z x 0 and the initial velocity is z t x 0. They are the radial functions because we assumed that the n is 0. So, cos n theta term will drop out and so, the this depends only on root of x squared plus y squared. So, solution is simply J0. Remember n is 0 J0 of Kr into a cos ck t plus b sin ck t. k runs through the list of zeros of J0 zeta 1 zeta 2 zeta 3 dot dot dot. The most general solution is a superposition of these things namely z r t equal to summation j from 1 to infinity j0 zeta j r correspondingly there will be a j cos c zeta j t plus b j sin c zeta j t. The c is a constant which appears in the wave equation itself. The c does not change and it is a zeta j that keeps changing and then these coefficients a j s and b j s have to be determined. Now, we put t equal to 0 in 5.11. When you put t equal to 0 in 5.11 the sin term disappears and we will just get a equation for determining a s. We must differentiate with respect to t and then put t equal to 0. This part involving a j will disappear and we will get an equation determining b j. What are these equations? Let us look at them z r 0 which is a known function summation j from 1 to infinity a j j0 zeta j r. So, left hand side is known and from this equation we have to determine these coefficients a j. Now, when you differentiate with respect to time this c zeta j will come out and they will be clubbed with these coefficients a j s and b j s. So, after clubbing them I am going to use a different name for these coefficients. So, z t of r comma 0 z t of r comma 0 is a known function the initial velocity of the wave that is going to be an infinite superposition of j 0 zeta j r, but the coefficients are c j. The b j has become a c j because when we differentiate we pick up some factors j c and all these things to in order to proceed further we need a result from analysis from classical analysis called the Bessel expansion theorem. We are not equipped to prove this general theorem on Bessel's functions the relevant theorem is available in chapter 18 or the monumental work of G and Watson. I already mentioned to you in the past the work of G and Watson treaties on the theory of Bessel's function. The second edition came Reginia's Tepris and you also want to look at the historical introduction on page 577 to 579. So, using the Bessel expansion theorem you can determine the a j s and c j s. Let us look at the formula for a j for example, suppose if f of r is a smooth function on 0 1 I may not assume that the function is smooth then this function f of r can be expanded as a series of Bessel's function f of r equal to summation j from 1 to infinity a j j naught zeta j r where what are these a j's these a j's are given by equation 5.12 2 by j 1 zeta j squared integral 0 to 1 r f of r j naught zeta j r dr. In the chapter 1 we try to write a general function from minus pi to pi as a summation of sines and cosines here we are try to write a general function on 0 to 1 we are assuming it is smooth as a linear combination of j 0 zeta j r and with certain coefficients and the coefficients are given by equation 5.12 it is exact analog of those coefficients formulas a n equal to 1 upon pi integral minus pi to pi f of t cos n t d t etcetera. There are analogs for the Fourier Bessel expansion. Okay, so let us proceed further. So, now we discuss the orthogonality properties of the Bessel's function. So, observe that in this particular expansion you got this j naught zeta 1 r j naught zeta 2 r j naught zeta 3 r. So, we got an infinite set of functions these Bessel's functions remember that the Bessel's function is an entire function j naught is an entire function and zeta j is a fixed number and so j naught zeta j r is a nice function on the closed interval 0 1. So, now the question is that if I take these family of functions j naught zeta j r and allow the j to run from 1 2 3 up to infinity and we get a sequence of functions in 0 1 will that have any kind of orthogonality properties will it be orthogonal with respect to L 2 of 0 1 well it will be provided you describe the measure properly it is not the Lebesgue measure it will be a weighted Lebesgue measure and so you can talk about L 2 of a measure space L 2 of a measure space and I can have a sequence of functions and I can ask for orthogonality and completeness of the sequence of functions completeness is what the Bessel's expansion theorem talks about completeness is difficult and we shall not discuss that orthogonality we shall discuss here. So, orthogonality with respect to the measure r dr so that is the objective of the next series of 5 or 6 exercises. So, first thing would be to write the Bessel's equation in self adjoint form. So, first of all notice that the Bessel's differential equation is x y double prime plus y prime plus x minus p squared by x y equal to 0 I have taken the standard Bessel's equation x squared y double prime plus x y prime plus x squared minus p squared y equal to 0 and I divided through by x and I get this. Now this first two terms can be combined and keep written as d dx of x y prime plus x minus p squared by x y equal to 0 this is called the self adjoint form of this differential equation. We shall be writing the self adjoint forms of many of the classical differential equations such as the Legendre equation and the Hermits equation as well and you will get a hang of it and why is the self adjoint form important. Let us make a change of variables. Let us look at phi u x equal to Jp of x u. So, let us find out what happens when I use Jp x u in the differential equation what kind of a differential equation do I get when I make a change of variables x u equal to t. In other words what is the differential equation satisfied by phi sub u it is here 5.13 equation 5.13 is the ODE satisfied by the scaled Bessel's functions Jp x u eventually the u will be a 0 of the Bessel's function. Now let us assume that p greater than or equal to 0. So, these two exercises of exercise 4 and exercise 5 are routine calculus exercises I will just leave it to you. Exercise 6 is more interesting and we shall dwell a little deeper in exercise 6. So, first I am going to assume p is greater than or equal to 0 because p squared appears and p is real. So, I may as well assume it is non-negative. Zeta 1, zeta 2, zeta 3 is a list of zeros of the Bessel's function. Show that the family Jp zeta x is orthogonal on the interval 0, 1 with respect to the weight function x. What do I mean by saying with respect to the weight function x let us call this phi j of x. So, I mean phi j and phi k are orthogonal with respect to the measure x dx in other words integral 0 to 1 phi jx phi kx x dx is 0 if j and k are distinct. That is the meaning of saying orthogonality with respect to the weight function x. So, let us prove this result which is given as exercise 6. Some easy verifications I will leave it to you to check. Having discussed the orthogonality of the function we must determine what is the normalizing factor. Remember that when you have c 1 phi 1 plus c 2 phi 2 plus c 3 phi 3 plus dot, dot, dot. What are the ci's? The ci's are inner product x inner product with phi i divided by norm phi i squared. What is the measure in the L2 space? x dx is the measure. So, what is the norm squared? The norm squared is 5.14 x times Jp zeta x the whole squared dx. So, these are the normalizing factors in the orthogonal system. So, we must compute these as well if we want to compute the Fourier coefficients. So, now let us look at the list of exercises a little more and then we will look at the solutions to some of these exercises. Let zeta be a 0 or the Bessel's function and you multiply the Bessel's differential equation by 2x zeta Jn prime zeta x and we have to deduce 5.15. The normalizing factors that I talked about in the previous one we need to compute these integrals and these integrals are computed here in the next exercise. And using this exercise we deduce the formula of Lommel. What is the formula of Lommel? The formula of Lommel appears in the Bessel expansion theorem here. So, in this formula of Lommel we can determine a J. Here we are not going to prove the validity of this expansion. What we are going to do is that we are going to assume that this holds and proceed formally with a calculation. That is what I mean when I say prove the formula of Lommel or deduce the formula of Lommel, proceed formally and use the two previous exercises. Then you need the two identities that we derived in chapter 1, the two identities for Bessel's functions. One involving p and the other involving minus p here. And there is a next exercise of expanding x to the power n as a Bessel series. And this I can leave it to you to work out. A very interesting proof of the orthogonality properties of the Bessel's functions suggested by physical considerations. We are going to give an analytical argument for the orthogonality of the Bessel's functions. But the Bessel's functions arise from physics, from vibration problems. And so it is of interest to understand does this orthogonality have any physical relevance? Yes, it does. In fact, we can deduce the orthogonality from physical considerations. And for this, you should definitely look at the two volumes of Lord Rayleigh theory of sound. It is an amazing two volume work. It contains lots of information about various aspects or the theory of vibrations. It is difficult to find many of these things in other standard texts. I have given you a complete reference to it with the page numbers to the relevant orthogonality property derived from physical considerations. Now let us prove this orthogonality that I talked about using mathematics using analysis. So let us fix p and let us look at the list of zeros of the Bessel's functions. Now these zeros are simple zeros. So if I have a function f of x, remember that our Bessel's functions are entire functions. And for an entire function, when you have a zero of an entire function, you understand what does it mean to say that a zero is a simple zero. Why can't it be a double zero, for example? Suppose for example, if zeta 1 would be a double zero of JP. Let us arrive at a contradiction. That means what JP zeta is zero, JP prime zeta is also zero. And zeta is not zero. These are the non-zero roots of the Bessel's function. If I take p positive, if p is 3 for instance, we know that JP z is z cubed times a power series with a non-zero constant. So the origin is definitely a zero of the Bessel's function if p is positive. But that is not what we are talking about. This in this list zeta 1, zeta 2, zeta 3, none of the zetas are zero. We are looking at the non-zero zeros of the Bessel's function, if you like. So why is it that these zeros are simple? If it were not simple, if it were a double zero, for example, or a triple zero, then both the function and its derivative will vanish. But JP satisfies an ODE, which is of second order. And I can divide by the leading coefficient x squared because I am looking at the point where it is not zero. And the uniqueness clause in the fundamental existence uniqueness theorem of ODE's will tell you why this cannot happen. If that happens, then JP will be identically zero. So the zeros are simple. And let us write them in ascending order. The zeros are real. The JP does not have any complex zeros. So let us write phi Jx as JP of zeta Jx. Now again a routine calculation will tell you that this phi J satisfies this ODE. It is just that the x squared gets replaced by x squared zeta J squared. Nothing else happens in the differential equation. First we will look at the case when P is strictly positive and later we will look at what happens when P is zero. The case P is zero is easier. The K P bigger than zero is slightly more complicated. But that is also pretty easy really. Multiply the differential equation by 1 upon x phi kx. So this phi J phi k and there is an x here after multiplication by x inverse because 1 x inverse will cancel with the 1 x here will be left with 1 x. So that is the idea. So multiply the differential equation by 1 upon x phi kx and integrate over zero one. The question that I want to ask you is why is this an integrable function? See phi kx upon x is it integrable over zero one. Again remember that since P is positive and phi kx is cooked up out of a Bessel's function. JPx behaves like x to the power p. So when I divide by x I am going to get 1 upon x to the power 1 minus p that certainly an integrable function. So the 1 upon x is not going to cause any problems as far as integrability is concerned. So this is integrable. So what happens to the differential equation? Integral 0 to 1 phi k ddx of x phi j prime x because 1 x cancelled out plus the middle term zeta j squared x phi jx phi kx minus the last term p squared integral 0 to 1 phi jx phi kx dx by x equal to 0. All right what we are going to do now is we are going to perform an integration by parts in the first term. When you perform an integration by parts in the first piece the derivative will shift to the other factor and will become minus sign. So you will get a minus sign. So minus integral 0 to 1 x phi k prime x phi j prime x dx plus the zeta squared integral 0 to 1 x phi jx phi kx dx I have not done anything to the second integral and the third integral. What we are going to do is we are going to switch the rules of j and k and you are going to subtract the two things. So the term involving p squared will cancel out and the first term will also cancel out this minus integral 0 to 1 x phi k prime phi j prime that will also cancel out. What we will be left with is the middle term zeta j squared minus zeta k squared integral 0 to 1 x phi jx phi kx dx equal to 0. Now remember that zeta j and zeta k are distinct zeros and so that cancels out and we are left with the orthogonality condition 0 to 1 integral x phi jx phi kx dx equal to 0 as advertised. There is one small thing though something is written in red in the slide, explain why the boundary terms will drop out. When you do an integration by parts you are going to get a boundary term. Let us examine the boundary term what will be the boundary term at 0 and at 1. Of course the presence of x here means that when x equal to 0 the thing will drop out. At the other end of the boundary phi k1 what is phi k1 or phi j1? Phi k1 will be jp zeta k remember that zeta k is a 0 or the Bessel's function so jp of zeta k will be 0. So the boundary terms cancel out the boundary terms drop out. So there are no boundary terms to worry about and so the orthogonality of phi j and phi k has been established for P positive. What about the case when P is 0? When P is 0 the situation is much easier because the third term in the differential equation is not there and the x drops out one factor of x drops out of the differential equation and the differential equation simplifies to x phi j prime prime plus x zeta j squared phi j x is 0 the differential equation simplifies. Now the same technique will work again you again multiply by phi kx and integrate by parts over 0 1 and the rest of the argument is completely similar to the previous argument and since this part is easier than the previous one I can safely leave it to you as an exercise. So we see here that from the point of view of vibration theory of a circular membrane we have obtained a remarkable family of orthogonal functions. We take the Bessel's function JP x and we take its zeros zeta 1 zeta 2 zeta 3 etc JP zeta kx where k runs from 1 to infinity that gives you an infinite family of mutually orthogonal system of functions orthogonal with respect to which weighting function x dx that is the measure. So L2 of 0 1 with respect to the measure x dx in that Hilbert space they got a complete orthonormal basis but we are not proved completeness completeness is difficult to establish we will not do that and we will just leave this over here. Other exercises I will discuss in the next capsule the second exercise that we talked about this deriving 5.15 is more tricky and we will do that in the next capsule. I think this is a very good place to stop this lecture here. Thank you very much.