 Hi, so let us continue from where we left off last time. Last time you recall that we discussed generalized Fourier series. The need for generalized Fourier series, we looked at some examples of complete orthogonal systems. We looked at the case of the vibrating membrane, circular vibrating membrane and we got the Bessel's function. So JP zeta jx call it phi j and we saw that they were orthogonal where the zeta j varies over the zeros of the Bessel's function JP. I mentioned to you that later we shall show that there are infinitely many zeros of the Bessel's function. We also looked at the orthogonality, but remember that it is not enough to just prove orthogonality of the Bessel's functions, we also have to compute the L2 norms of these functions with respect to which weight x dx that is the weight of the, that is the measure with respect to which we take the L2. Now let us go a little further, we want to calculate integral 0 to 1 JP zeta x the whole square dx. So observe that x dx is the measure and this is the L2 norm of the function phi jx with respect to this measure. So we begin again with the definition of phi ux, recall that phi ux is JP of ux and it satisfies the differential equation x squared phi u double prime plus x phi u prime plus x squared u squared minus p squared phi u equal to 0. At this juncture you need not be a 0 of the Bessel's function as far as this differential equation is concerned not yet. Later we will take it to be the 0, but we will also need information as to what happens when u is not a 0 of the Bessel's function. Multiply the differential equation by x inverse phi v and integrate over the interval 0 1. What do we get this x squared becomes an x phi v phi u double prime plus phi vx phi u prime x the x disappear plus x squared u squared minus p squared phi u phi v dx upon x equal to 0. Again I am going to assume that the p is not 0 and so this is an integrable business and when p is 0 it even gets better because this term disappears and straight away 1x gets cancelled out. Now what happens is that we are ultimately going to switch the roles of u and v and we are going to subtract. So if you do that then this term minus p squared phi u phi v dx upon x that term will cancel out and so I will not even bother to discuss this term in the future because this is going to cancel out. In the first piece we are going to integrate by parts. When you integrate by parts what happens one of the derivatives will shift from phi u double prime and it will come over here and there will be a negative sign. When it comes over here what happens there are two terms in one term the derivative will fall on x and we will get phi v phi u prime x dx and there is a negative sign and that exactly cancels with the middle term. And there is a other term where the derivative shifts from phi u double prime and falls on phi v. What is that term minus integral 0 to 1 x phi v prime phi u prime dx again when we switch the roles of u and v and we subtract that term also cancels out. So the terms arising by integration by parts in the first integral will cancel out except that we have to discuss the boundary terms when we perform integration by parts. Remember u is not a 0 of the Bessel's function. So with these things in mind let us proceed a little further and that is exactly what I got. The integration by parts gives you three terms two of them are integrals and the third one is a boundary term. And we explain to you why the middle integrals will drop out and the first integral will drop out of the middle term in the previous slide. So the only thing that we need to look at is the boundary term. What happens to the boundary term x phi v x phi u prime x at the lower end anyway is 0. What about the upper end at upper end it is phi v of 1 what is phi v of 1 it is going to be J p of v. What is phi u prime x what is the derivative of phi u x the derivative of phi u x is the derivative of J p x at u times remember phi u x is J p of x u so when you apply chain rule and that will produce a u outside so the boundary term has been written down. And then we do the switching of u and v and subtracting and what remains basically are these terms u squared minus v squared integral 0 to 1 x phi u x phi v x dx the right hand side comes from these boundary terms v J p u J p prime v minus u J p v J p prime u idea is to divide by u squared minus v squared and let v tend to u ok. So when v tends to u we have to understand what happens to the right hand side of course we have to divide by u squared minus v squared. So in the limit what we are going to get is integral 0 to 1 x phi u squared x dx equals limit as v tends to u minus u J p v J p prime u upon u squared minus v squared you have to apply the L'Hopital's rule and you are going to get exactly one half J p prime u the whole square and the proof of L'Hommel's formula is thereby complete. Now that we have discussed this we have obtained the solution or the vibrating membrane using the Fourier Bessel expansion and the L'Hommel's formula. Now we need to ask ourselves a very basic question how do you know that the solution to this problem is unique maybe there is a solution is not unique maybe this is only one of the solutions that we got. We have to discuss this matter very completely and we proceed to do so. We shall prove uniqueness we shall prove that this is the only solution and there are no others. Suppose Z 1 and Z 2 are two solutions of the same differential equation with the same initial conditions and the same boundary conditions. Then what happens is that take the difference Z 1 minus Z 2 call it capital Z then what differential equation this capital Z satisfy the same wave equation but what about the initial condition because I have taken Z 1 minus Z 2 the initial conditions will drop out and capital Z has 0 initial conditions the boundary conditions are anyway 0 for both Z 1 and Z 2. So, the boundary condition for capital Z also is 0. So, now we are going to look at this particular problem over here. So, we will show that the only possibility here is Z equal to 0. So, let us proceed to do that what we will do is we will use the energy method something that you have seen before in the context of the wave equation. So, what we can do is multiply the wave equation by Z t we will get del 2 Z by del t squared del Z by del t that is one half the del del t of del Z by del t the whole squared it is the derivative of Z t squared. Of course, there is a one half factor which I suppressed and then what happens is over here you will get del 2 Z by del x squared into del del t that is going to be the same as this. How would I get this? Remember that the Z x x here was here and Z t was here one derivative shifted from here to there I have done an integration by parts I have done a integration by parts in these 2 terms and there is a c squared and the minus sign became plus sign and I get this from this we infer that the energy E t is basically conserved what is the energy this quantity c squared Z x squared plus c squared Z y squared plus Z t squared. Remember this is again one half del del t of Z x squared this is one half del del t of Z y squared and there was one half here also which I so I have suppressed the one half in all the places there is 0 on the right hand side there is no problem. So, the time derivative of Z x squared if you calculate you are going to get exactly this. So, this equation says that d dt of this is 0 in other words the energy E t is constant in time. So, since the initial value of the energy what are the initial value of the energy the initial value of the energy is 0 because the initial conditions are 0 and the boundary terms are also 0. The energy is identically 0 means Z t squared must be 0 Z y squared must be 0 Z x squared must be 0 because the energy is an integral of squares sum of squares and so individually Z t Z x and Z y must be identically 0. So, Z must be constant what is Z is constant why should Z be 0 remember the boundary condition the membrane is clamped along its rim. So, the this constant must be 0 itself and the proof is complete the uniqueness has been established. Now, imitate this energy method to show that a twice continuously differentiable initial value problem for the heat equation is unique. So, do the same idea not with a wave equation for the heat equation. Again you will have to write the energy function E t this time E t is not going to be constant, but it is going to be monotone decreasing in time. So, there is a small difference between the two, but you will figure it out if you work out the problem. Show that the twice continuously differentiable solution to the boundary value problem Laplacian of u equal to 0 on D u of x y z equal to f of x y z on the boundary is unique in D where D is a region in R 3 with a smooth boundary del t and f is a function which is prescribed along the boundary del t. Again you can imitate the same idea, multiply the differential equation by u and integrate. Now, for more discussions on these kinds of wave phenomenon, we are just touched upon wave phenomenon. There are number of excellent books available which describe these kinds of things in very great detail. A classic reference is Courant and Hilbert, Richard Courant and David Hilbert's methods of mathematical physics volume 1. We already cited this earlier. The discussion of vibration of a circular plate is available on page 307 308. These involve Bessel's functions of imaginary orders IP. They are also known as a modified Bessel's function. The biharmonic equation will show up. This is something that you could look at. The other thing is the reference that I have already given you, Lord Rayleigh, theory of sound volume 1 volume 2. This is the most comprehensive account of the theory of vibrations. The discussion of the vibrating plate is also there. This is the best source on the physics of vibrations. Another application of Bessel's function is a skin effect. See Frank Bowman's introduction to Bessel's functions, Dover 1958. Bessel's function appeared in optics. The radii of the successive interference fringes formed by the diffraction of light by a circular aperture. They are given in terms of the zeros of the Bessel's function. Besides the authoritative treaties of G. N. Watson, you should also look at the book of Bayer-Lee which is available for free online. Now we go to the next example, Legendre polynomials. Legendre polynomials give you another orthonormal system on the interval minus 1 1. I mentioned to you the Legendre polynomials on L2 of minus 1 1 forming a complete orthogonal system. So, what is the Legendres differential equation? The Legendres differential equation is the display 5.3. 1 minus x squared y double prime minus 2 x y prime plus p into p plus 1 y equal to 0. The coefficient 1 minus x squared does not vanish at the origin and so the origin is a non-singular point. It is a regular point. So, we can seek a power series solution of this differential equation 5.3 as a power series 5.4. So, the solution that we are looking for, we take it as a power series y of x equal to summation n from 0 to infinity a n x to the power n. Remember that a power series can be differentiated term by term if it has positive radius of convergence. So, let us differentiate 5.4 term by term minus 2 x y prime because I have to differentiate and I am to multiply by minus 2 x. So, minus 2 x y prime what is it going to be term by term differentiation a n x to the power n is going to give you n a n x to the power n minus 1. So, multiply by minus 2 x it will go back to minus 2 n a n x to the power n 5.5 and for the x squared y double prime term we get x squared y double prime equal to summation n from 0 to infinity n into n minus 1 into a n x to the power n 5.6. So, now we have x squared y double prime 2 x y prime and p into p plus 1 y you simply multiply 5.4 by p into p plus 1. What are you going to get? You are going to get 3 equations 5.4, 5.5 and 5.6 all of them have x to the power n in them. Now, we have to deal with one more term of the differential equation y double prime. y double prime is summation n from 0 to infinity n into n minus 1 a n x to the power n minus 2 put n equal to 0 it is 0 put n equal to 1 it is 0. So, the summation starts from 2 to infinity. Now, there is a very important warning which is given in red it is very important to arrange it in such a way that all the exponents of x in the general term should be the same whereas, it is n in the 3 previous terms it is x to the power n x to the power x to the power n here we have x to the power n minus 2 it is at most important to make this also n. So, we make a change we put n minus 2 equal to capital N. So, when little n runs from 2 to infinity capital N runs from 0 to infinity and we get y double prime equal to n plus 2 into n plus 1 into a n plus 2 x to the power n n is a dummy index and so, we can write y double prime is summation little n from 0 to infinity n plus 2 n plus 1 a n plus 2 x to the power n 5.7. So, we combine everything together 1 minus x squared y double prime minus 2 x y prime plus p into p plus 1 y equal to all the summations are going from 0 to infinity all the x's have nth power. So, n plus 2 into n plus 1 a n plus 2 minus n into n minus 1 a n minus 2 n a n plus p into p plus 1 a n equal to 0. Now, this power series is identically 0. So, every coefficient must be 0. So, equating this coefficient to 0 we get what is called as a recurrence relation we collect the all the a n's together and then you simplify a n plus 2 equal to minus a n into p minus n into p plus 1 plus n upon n plus 2 n plus 1 n goes from 0 1 2 3 etcetera. So, now you put n equal to 0 1 2 3 in succession and we get a 2 put n equal to 0 and we get a 2 a 2 is what minus a 0 p into p plus 1 upon 2 factorial put n equal to 1 a 3 is minus a 1 into p minus 1 into p plus 2 upon 3 factorial put n equal to 2. When you put n equal to 2 we get a 4 equal to minus a 2 times something, but a 2 has just been computed in the previous step. So, put that. So, if you keep doing this you will get a 2 and a 3 a 4 is again getting in terms of a 0 a 5 will be in terms of a 1 and the law of formation is very clear the even ones are minus 1 to the power n a 0 p into p minus 2 into teta teta p minus 2 n plus 2 p plus 1 p plus 3 teta p plus 2 n minus 1 upon 2 n factorial for the odd ones a 2 n plus 1 equal to minus 1 to the power n a 1 all of them will have a 1 they will get p minus 1 into p minus 3 into teta p minus 2 n plus 1 and then p plus 2 p plus 4 teta p plus 2 n minus 1 exactly what we get. So, we get a 2 n and a 2 n plus 1. General solution is thus given by a 0 into 1 minus p into p plus 1 x squared upon 2 factorial plus p into p minus 2 into p plus 1 into p plus 3 x to the power 4 upon 4 factorial minus zeta plus a 1 into x minus p minus 1 into p plus 2 x cube upon 2 factorial plus p minus 1 into p minus 3 into p plus 2 into p plus 4 x to the power 5 1 4 factorial minus etcetera signs alternate in both the terms. Of course, the coefficients a 0 and a 1 can be given arbitrary values. So, we can assign a 0 equal to 1 and a 1 equal to 0. So, the second piece completely drops out and a 0 is 1 and I get 1 solution. On the other hand, I can put a 0 and I can put a 1 as 1 and I get the second solution. So, when the first one was an even powers, second one involved only odd powers. So, the first one is an even function and the second one is an odd function both of them are power series. Now, you check using the D'Alembert's ratio test that each of these two series will have unit radius of convergence except in one situation when p is an integer. For example, if p is 2 then look at the first series after the third term onwards everything is going to collapse and it will be just a quadratic polynomial. If p is 3 for example, look at the second parenthesis the first two terms will remain after that everything will collapse and you get a cubic polynomial. If p is 5 the first three terms will survive and the rest will be 0 and you get a fifth degree polynomial etcetera. So, these polynomials are called Legendre polynomial when p is an integer exactly one of these two series terminates and we have a polynomial solution and the other one is a non-terminating power series and that will have unit radius of convergence. So, check these things with suitable normalizations that we shall presently specify these polynomials are called Legendre polynomials. So, we have finally obtained the Legendre polynomials up to some normalization. Remember that the Legendre differential equation is a homogeneous differential equation. So, y of x is a solution then constant times y of x is also a solution. So, the normalization is necessary before we call them the Legendre polynomials. So, let us define them clearly. Assume that p is an integer assume that p equal to k and we have seen that one of the two series described above terminates into a polynomial solution f of x. Now, we have to show that this polynomial solution f of x has the property that f of 1 is not 0. This polynomial does not vanish at x equal to 1. Why are we interested in the non vanishing of this because the normalization that we are talking about is f of x upon f of 1. We have a polynomial solution already and we want to divide it by the value at 1 and that is how we normalize the Legendre polynomials. We do not take the L 2 norm and divide that is something that not done. So, now how do we know that f of 1 is not 0? If f of 1 is 0 then we cannot do this. So, let us prove that f of 1 is not 0 suppose not let us prove it by contradiction. Suppose that f of 1 is 0 then look at the differential equation what is the differential equation 1 minus x squared y double prime but 1 minus x squared y double prime will become 0 minus 2 x y prime plus p into p plus 1 y equal to 0. Now, we assume that y of 1 is 0 that is a solution f of x vanishes at 1. So, the third term in the Legendre differential equation also became 0 and the middle term therefore should give you f prime of 1 should be 0. So, now we have f of 1 is 0 and f prime of 1 is also 0. Let us proceed by induction. Let us assume that the nth derivative of f at 1 is 0. We will prove that the n plus first derivative is also 0. Take the Legendre differential equation and differentiate the differential equation n times using the Leibniz rule for a product. You know how to find the nth derivative of a product of two things. In the Legendre's differential equation you have got the 1 minus x squared y double prime you have to differentiate this n times and the middle one will be minus 2 x y prime you have to differentiate that n times. Obviously, you have to use Leibniz rule when you differentiate 1 minus x squared y double prime n times what happens at max two derivatives can fall on 1 minus x squared beyond that it will become 0. So, all derivatives falling on y double prime and none on 1 minus x squared but that will become 0 because 1 minus x squared vanishes when you put x equal to or 1 derivative falls on 1 minus x squared and n minus 1 derivatives fall on y double prime that is minus 2 x and the n plus first derivative of and you put x equal to 1 at the end of it and you keep doing this and deal with the middle term also in the same way. Use induction hypothesis that all the derivatives obtained up to and including the nth derivative are all 0. Then you will conclude from there that the n plus first derivative will also be 0 and therefore we have proved that all the derivatives are 0 and so f must be the 0 polynomial and that is a contradiction. So, this contradiction shows that f of 1 is non-zero and this normalization is a valid normalization. So, after this normalization the normalized polynomial solution is called the Legendre polynomial. So, since I have taken p equal to k this polynomial will have degree k and I am going to denote it by p k x. So, we record its three properties the three properties of Legendre polynomial p k x is a polynomial of degree k it satisfies the Legendre differential equation its normalization p k of 1 is 1 and it is clear that p k is an odd function if k is odd and it is an even function if k is even. So, and the 0th Legendre polynomial is a constant polynomial 1 I think it is a good place to stop this lecture here. Thank you very much we will continue it next time.