 So, we continue with this theme and this is the last capsule in this chapter which also happens to be the last chapter in this part of the course and here we are going to prove the existence of eigenvalues and eigenfunctions using the Sturm's comparison theorem that we did in the last capsule. Also, we are going to prove that the Bessel's function Jpx has infinitely many zeros in the interval 0 infinity. You could use the Sturm's comparison theorem or you could imitate the proof of Sturm's comparison theorems but either way the circle of ideas is the same. So, these are the things with which we are going to finish this chapter. So, let us return to the proof of the existence of eigenvalues and eigenfunctions. So, let us recapitulate what we did last time. We got the differential equation y double prime plus lambda rho xy equal to 0, Dirichlet boundary conditions are going to be considered but before we take up eigenfunctions, we take the initial value problem y of 0 equal to 0, y prime 0 equal to 1 and that solutions is unique yx lambda. It will depend continuously, it will depend smoothly with respect to lambda, it will depend analytically with respect to lambda and so forth. We are going to put x equal to 1 because we are interested in the boundary value problem. y of 0 is already 0, we also want y of 1 to be 0 and we need to select the lambda in such a way that y of 1, lambda is 0. So, rho of x remember is continuous and non-negative on the interval 0 and we are actually going to assume that it is strictly positive. We are going to assume that rho x is strictly positive and little m squared and capital M squared of the infimum and supremum respectively. So, how are you going to apply this terms comparison theorem? Now, we got three differential equations y double prime plus lambda little m squared by equal to 0, the last line which is in blue y double prime plus lambda capital M squared y equal to 0 and the differential equation y double prime plus lambda rho xy equal to 0. This rho x is sandwiched between little m squared and capital M squared and so now we can use the Sturm's comparison theorem. Now, for each lambda the equation yx lambda equal to 0 has a discrete set of 0s. Remember, we proved the 0s are isolated. It is a discrete set of 0s and the 0s are simple that also we proved using the uniqueness clause in Picard's theorem. Here, let me just recapitulate the simplicity part. Suppose x naught is a double 0, suppose x naught is a double 0 then in addition to 6.11 we will also have y prime x naught lambda equal to 0 and y of x naught lambda equal to 0. So, the uniqueness clause will tell you that y is identically 0 and that is a contradiction. Now, let us show that each of these 0s zeta k lambda varies continuously with respect to lambda. Now, to show that zeta k of lambda is a continuous function of lambda, we must employ the implicit function theorem. Let us now prove the continuity at an arbitrary value lambda. Let us freeze the value of lambda, call it lambda naught and let us call the corresponding the kth 0 of yx lambda naught is zeta naught. Then by definition y of zeta naught lambda naught is 0 and we also know that y prime of zeta naught lambda lambda naught is not 0 because the 0s are simple. The implicit function theorem now immediately tells you that there are intervals containing lambda naught and zeta naught such that we can solve this equation y of x lambda equal to 0 uniquely for each choice of lambda naught in J 1 such that the corresponding root lies in J 2 and this unique solution is also continuously differentiable with respect to lambda. Remember when you want to apply the implicit function theorem you need two things you need the series solution. What are the implicit function theorem? You want to solve the equation f of xy equal to 0 for y as a function of x. Now first thing you need is you need a seed solution f of a comma b equal to 0 and you need del f by del y at a comma b should be nonzero. Both these conditions are being met with here zeta naught lambda naught is a seed solution and this derivative with respect to the first variable at zeta naught lambda naught is nonzero. So, implicit function theorem is applicable and we know that each of the 0s varies continuously. So, the 0s are discrete. So, there is a first 0, there is a second 0, there is a third 0, etc. I can arrange the 0s in increasing order. So, there is a there is a ordering among the 0s. So, everything is well defined. Now we are ready to prove the existence of an infinite sequence of 0s namely an infinite sequence of eigenvalues for the Sturm level problem. Let us do the following. Let us start with this function yx lambda. It satisfies this differential equation y double prime plus lambda rho xy equal to 0. Remember this rho x is less than or equal to m squared. So, I must compare that differential equation with y double prime plus lambda capital M squared y equal to 0 whose solution is sin m root lambda x. We know the 0s of sin m root lambda x between 2 0s of y of x comma lambda there must be a 0 of sin m root lambda x by the Sturm's comparison theorem. Now suppose for example the lambda is chosen very small then sin m root lambda x has no 0s on this interval 0 1 0 excluded. So, obviously since sin m root lambda x must have more 0s then y of x lambda we conclude that y of x lambda also cannot have 0s in this interval. So, the first 0 of y x lambda must be larger than 1 if lambda is very small. So, the first 0 of y x lambda must be larger than 1. Now look at a large value of lambda. Now let us look at a large value of lambda but let us apply Sturm's comparison theorem with y double prime plus lambda little m squared y equal to 0. But this time it has solution sin little m root lambda x between 2 0s of sin root lambda x there must be a 0 of y x lambda. Now if lambda is large then we know that sin m root lambda x must have lots of 0s here. So, if lambda is large it follows that y x lambda must have 0s here. So, for large lambda there must be a 0 for small lambda there is no 0 which means that there must be a specific value lambda 1 such that y of 1 comma lambda 1 must be 0 whether intermediate value theorem the 0s vary continuously for lambda is small the first 0 zeta 1 lambda is bigger than 1 and lambda is large the first 0 zeta 1 lambda must be less than 1. So, there must be a specific choice lambda equal to lambda 1 by intermediate value theorem such that zeta 1 lambda 1 must be exactly 0 which means y of 1 comma lambda 1 must be 0 and thus we got the first eigenvalue of the term level problem. Now what about the second eigenvalue the second eigenvalue proceeds along similar line looking at zeta 2 lambda the second 0 of y x lambda and look at it for small values of lambda it must be larger than 1 for large values of lambda it must be smaller than you must compare it with sin capital M root lambda x and sin little m root lambda x. And the proof of the existence of infinitely many eigenvalues is completed because it is simply go by induction. So, we have proved that there are infinitely many 0s now that let us look at the 0s of these eigenfunctions a few words about the 0s of these various eigenfunctions it is evident from the construction that the first eigenfunction has no 0s in the open interval 0 1 our proof just now that we are given evidently shows that fundamental mode of vibration has no 0s in the interval 0 1. All other eigenfunctions must have at least 1 0 here because if the second eigenfunction also does not have any 0 here then both eigenfunctions maintain the same sign in the open interval 0 1. So, the integral y 1 x y 2 x rho x dx cannot be 0 remember the density function rho x is assumed to be strictly positive. And so, since by orthogonality it means that other than the fundamental mode all other eigenfunctions must have 0s in my fundamental interval these 0s are called the 0s of the eigenfunctions and it is not difficult to show that nth eigenfunction has n minus 1 0s in the fundamental interval. In fact, our argument will precisely show that now you can also define these terms for storm level problems in higher dimensions such as the vibrating membrane for example or the vibrating ellipse elliptical membrane for example. In higher dimensions the study of the nodes of these eigenfunctions they assume a very spectacular aspect and again Lord Rayleigh's theory of sound contains lots of information concerning the geometry of these nodes of eigenfunctions. Proofs of some of these results can be found in Courant Hilbert's methods of mathematical physics. So, I think I would like to close this discussion here except that we still have one more task to complete to prove that the Bessel's function has infinitely many 0s and the open interval 0 infinity. Remember that the Bessel's function is a problem on the infinite interval 0 infinity whereas what we have been doing so far are regular storm level problem where we worked on a compact interval 0 1 and the density function was continuous non-negative and on the interval 0 1 and we looked at the Dirichlet boundary conditions. So, that is a regular storm level but the Bessel's function has a complication that the domain in question is infinite it is a singular storm level problem. So, as a preparation for that we need to first prove the mean value theorem for integrals. Suppose if f and g are continuous functions on the closed interval a, b and this function g is strictly positive on the open interval a, b then integral a to b fx gx dx is f of c time integral a to b gx dx. Let us see how to prove this first of all let us just work with one function let us assume that gx is 1 let us assume that gx is 1 then what does this equation read integral a to b fx dx is f of c times b minus a can we prove this is this easy to prove it is very easy to prove the function f of x is continuous on a closed interval a, b and so it must have its supremum and infimum and you divide by b minus a and use an intermediate value theorem that is one way to do the problem. So, I am going to assume that this has this easier problem has first been established or you can apply the mean value theorem for differential calculus that is also possible. So, you look at this integral from a to t fx dx call it phi t, phi t is integral from a to t fx dx and then you apply the Lagrange's mean value theorem to phi and that will do it. So, the easier case has been done the general case is very simple you simply use the integral of g as a variable of integration look at gx dx you want gx dx to be du and so look at integral g of lambda d lambda from a to t and call that as the new variable of integration. I will leave the details for you this mean value theorem for integral calculus is what we are going to use. Now, let us take the Bessel's function Jn of kx where k is a parameter assume that k is positive if you like now define ux equal to root kx into Jn of kx. Now, it is very easy for you to check but directly differentiating that ux satisfies this differential equation u double prime plus k squared u equal to n squared minus one fourth upon x squared u. Now, when you look at this kind of thing what you need to understand is that you think of what happens when x is large think of what happens beyond a certain stage think of what happens on the interval a to infinity where a is very very large. In that case since x is bigger than or equal to a this term n squared minus one fourth upon x squared is going to be vanishingly small. So, our intuition should tell you that this differential equation can be approximated by u double prime plus k squared u equal to 0 and you know that u double prime plus k squared u equal to 0 has a nice solution sin k times x minus a sin k times x minus a and we know the location of zeros of that and now we must apply this terms comparison theorem or imitate the proof of terms comparison theorem that is exactly what we are going to do. So, now let us look at sin of x minus a. So, let us compare it with sin of x minus a. So, let us call v of x is sin of x minus a and the u is the previous thing u satisfies this differential equation and sin of x minus a satisfies another differential equation v double prime plus v equal to 0. Now calculate d dx of minus v u prime plus u v prime you are going to get this expression the u prime v prime term will disappear and then u double prime is given by this differential equation and v double prime is basically minus v. So, you will get u v times k squared minus 1 minus n squared minus 1 fourth upon x squared. Now we are going to select the a to be so large that this expression k squared minus 1 minus n squared minus 1 fourth upon x squared is positive on the interval a to a plus pi. So, that this is going to be pretty small k is going to be larger than 1 just assume k larger than 1 to begin with just a matter of convenience. So, now we are going to look at this interval a to a plus pi. Now integrate this expression that you see on the slide on the interval a to a plus pi. Left hand side of course you are going to use the fundamental theorem of calculus. When you use the fundamental theorem of calculus and you know that the sine function vanishes this at a and a plus pi. The two of the boundary terms will be 0 and two of them will survive surviving ones are written here and for the right hand side integral a to a plus pi dada dx. But now we want to use the mean value theorem but are we sure that the functions and question are positive sine of x minus a is certainly positive on the open interval a to a plus pi and this parenthesis is also a positive quantity in the interval a to a plus pi. So, this expression written in red which is gx, gx is strictly positive on the open interval a to a plus pi. So, with the mean value theorem there is a c in the open interval a to a plus pi integral of the right hand side is given by uc times integral a to a plus pi this vx into k squared minus 1 minus n squared minus 1 fourth upon x squared dx. So, this is the expression that we should be focusing on. Now what happens is that this integral that you see here is positive because v is positive, this parenthesis is positive. So, this integral is positive. Look at these three numbers u of a, u of a plus pi and u of c. These three numbers cannot all have the same sign. Suppose if all three are negative, suppose u of c, u of a and u of a plus pi are all three negative. Then what happens the left hand shall be positive, the right hand shall be negative. If all three are positive, the right hand shall be positive, the left hand shall be negative. So, the three numbers u a, u c and u a plus pi cannot all have the same sign. So, by intermediate value theorem there must be a 0 of u anyway in the open interval a to a plus pi and that completes the job. In every interval of length pi, the function jn of kx has a 0 and so jn of kx has infinitely many 0s in the positive real line 0 to infinity. Now of course you want to remove this condition that k is bigger than 1, I will leave it to you to tinker with it and see how k bigger than 1 can be removed and can be replaced by k equal to 1. So, jn x has infinitely many 0s on the positive real line 0 infinity and we already seen an application of this in the theory of wave propagations. We needed this to construct our complete orthonormal system of eigen functions for the circular membrane. Another interesting proof can be given using the integral representation for the Bessel's functions and that is given in Dunham Jackson's book Fourier series and orthogonal polynomials Dover, New York 2004. You also want to consult the monumental treaties of G. N. Watson on the treaties on the theory of Bessel's function page 500 and following for a discussion of techniques used by Euler and Lord Rayleigh to actually compute the 0s of this Bessel's function JPX. I think that was the last bit that was remaining we had completed the bit. There is still one more small item left which I would like to simply mention without getting into the details. If you have been looking at various vibrational problems vibrating strings, circular membranes, you must be wondering what about membranes of other shapes. For example, the case of an elliptical membrane, the vibrations of an elliptical membrane were considered by Emil Leonhard Matthew in 1868 and the result in ODE today is known as the Matthew's equation that you see on the slide, y double prime plus a plus b cosine 2xy equal to 0 where b is given and a is a eigen parameter. It is like an eigen value parameter. This equation has led to a long and rich chapter in the theory of analytic ODE's and this equation of Matthew has been generalized and studied at great length by G. W. Hill about 20 years later in his researches on Lunar theory, specifically perturbations theory in Lunar motion and the work of Hill has been very influential in astronomy. Unfortunately we are not in a position to say anything about these exciting theory in this course. There is a whole book on Hill's equation by Magnus. Hill's equation and Matthew's equation play a very important role in certain problems in modern dynamical systems. So, I shall not say anything more about this. So, with this I would like to close this capsule which also happens to be the end of this particular chapter and also ending the third part of this course. Now we shall proceed to the next part namely study of functional analytic techniques in Fourier analysis. So, thank you very much.