 So, let us continue our discussion from last time on Eigenvalue problems for second order differential equations or regular storm level problems as they are often called. What we have done last time is that we took a problem in linear algebra the problem of proving the spectral theorem or the finding the Eigenvalues and Eigenvectors of a real symmetric matrix and we gave it a variational formulation. We gave it a variational twist and the reason for doing so is we want to apply the same ideas to study of Eigenvalue problems for differential equations. We also pointed out that the principal difficulty lies in the fact that the unit sphere in an infinite dimensional space is not going to be compact and the lack of compactness is the main issue. So, what was the variational problem that we are talking about? We are looking at minimizing the energy integral 0 to 1 y prime t square dt subject to the normalization condition 6.5 namely y of t must be a unit vector in the L2 space L2 of 0 1 with respect to the weight rho t dt or with respect to the measure rho t dt. And we saw that there are the principal difficulty lies in proving that the minimizer exists and that the minimizer is twice continuously differentiable. I also mentioned that these are very deep waters that this is so is called the Dirichlet principle and how did it all arise? It arose when Riemann proved the mapping theorem that bears his name the very famous Riemann mapping theorem using ideas from potential theory. And motivated by potential theoretic considerations he translated the problem of the Riemann mapping theorem into a problem on minimizing an objective functional subject to certain constraints. What are the Riemann mapping theorem state? The Riemann mapping theorem states that a proper simply connected open subset of C can be mapped onto the disk by a bijective holomorphic function. Remember that when you have a holomorphic function f from omega 1 to omega 2 if it is bijective the inverse function is automatically holomorphic. Such a thing is called a biholomorphic map or a biholomorphism. So, the Riemann mapping theorem says that a proper simply connected domain in complex plane is biholomorphic to the open unit disk. It is important that the domain be proper simply connected. For example, the complex plane itself is simply connected but the complex plane cannot be mapped biholomorphically onto the disk because of Lieuwehl's theorem. So, what happens is that Riemann set up the objective functional namely minimizing the energy mod grad phi squared dx dy where phi varies over all once continuously differentiable functions phi with prescribed boundary conditions along omega. The solution of the variational problem is the Dirichlet problem for the Laplace Cn in omega. So, the Laplace's equation which we see arising in electrostatics and other places also arises in complex analysis from the problem of proving the mapping theorem. The existence of the minimizer and choice differentiability of the minimizer is the main stumbling block and here are some exercises. Verify 6.6 formally. What is 6.6? We come back to the one-dimensional vibrating string example with minimizing integral 0 to 1 y prime t square dt subject to the L 2 norm of y is 1. If the assume that the minimizer exists and assume that the minimizer is twice continuously differentiable then prove that 6.6 holds. How do you go about doing that? Suppose y naught is the minimizer. Suppose y naught is the minimizer then you exactly imitate what we did in the linear algebra situation. Look at y naught plus epsilon phi where phi is a smooth function with compact support, smooth function which vanishes at 0 and 1 and you look at the objective functional 6.4 instead of y you take y naught plus epsilon phi and you divide by the norm. So, as to make it a unit vector and you simply imitate the idea of linear algebra that we just did in the last capsule and prove that if the minimizer exists and the minimizer is twice continuously differentiable then it satisfies the differential equation. So, a formal verification can be done. Try to give a intuitive that is non-rigorous geometrical argument that the eigen function corresponding to the smallest eigen value has no zeros in the interval 0. What is the problem? y of t is the eigen function corresponding to the smallest eigen value that is y double prime plus lambda rho y equal to 0, lambda is the smallest eigen value and the boundary conditions y of 0 equal to 0 and y of 1 is also equal to 0. This is called the fundamental mode in physics of vibrations. This particular function y of t has no zeros in the interval 0, 1. Let me give you some hints. The function that you have vanishes at 0 and 1 correct. So, it is y of 0 is 0, y of 1 is also 0. Suppose it has a 0 in the middle. Suppose it has 0 say at say 1 third for example. Then what does the graph do? Think graphically the function starts at the origin and say it goes up and then it passes through the 0 comes down below the x axis and then joins back to the x axis because y of 1 is 0. Now, what you do is that this lobe of the graph below the x axis just flip it above, just flip it above, reflect the lobe which is below the x axis to a lobe above the x axis. You get arches from 0 to the first place where the function vanishes and from there on to the next one. So, it is a arches. If you do this then remember 6.4 is not going to change because when you differentiate when you flip it over the derivative is not going to change and the norm is also going to not change because only the sign of y changes. Of course by flipping it over the new function is not differentiable, but it fails to be differentiable only at one point. So, that is perfectly okay. It is after an integral that you are looking at and now I got 2 hills from 0 to the point where it vanishes and from there on to the point 1. Now, simply flatten this hill, take join the peak of the first hill and the peak of the second hill by line segment. Now, that line segment has smaller slope than the thing that it covers. The part where the graph comes down from the peak to the 0 and from 0 to the essence to the next peak that slope is much higher than the line segment that it joins. In other words, this expression integral 0 to 1 y prime t squared has gone down in value. On the other hand, that particular area under the curve has become larger. So, the denominator has become larger and the numerator has become smaller and so you are going to get a value which is less than the infimum, but that cannot happen because the first eigenvalue is the infimum of the energy. So, I have given you some ideas for you to carry out the program to give an intuitive that is non rigorous geometrical argument that the eigenfunction corresponding to the smallest eigenvalue has no zeros, but I am going to use the Dirichlet principle. I am going to assume the Dirichlet principle. Then the next thing is to imitate the linear algebra exercise. Remember we got, we took the matrix A and we minimized the quadratic form over the unit sphere and we got our V1 and we looked at the orthocomplement of V1. Orthocomplement of V1 means those vectors which are perpendicular to V1 and now we are going to do the analog of that in the differential equation setup. What is the meaning of perpendicularity in this context? yt and y0 must be orthogonal, but orthogonal with respect to the measure rho t dt that is 6.8. So, this is exactly the same arguments that we did in the linear algebra situation will carry forward and again a formal verification can be done then the next eigenvalue will be attained in the next step and one can keep going just as we proved the spectral theorem. Here we can keep going, we can get more and more eigenfunctions and by constructions they will be mutually orthogonal. Will they be complete? That is a non-trivial question, but these things assume that we take the Dirichlet principle for granted. We are going to give a proof of the spectral theorem for compact self-adjoint operators in the next chapter in a manner that is similar in spirit. Zeroes of non-trivial solutions are isolated. As a preparation for the next round of results we insert here a very simple result in differential equations. Let rho x be a continuous function on an open interval i. Suppose yx is a non-trivial solution, non-trivial means the solution is not identically 0 and the non-trivial solution is second order ODE y double prime plus rho x y equal to 0 then the zeros of y cannot have a limit point in L. We have seen that in the context of Bessel's functions, but there we could resort to the fact that the Bessel's functions Jpx is simply x to the power p times a holomorphic function and we can use the result from complex analysis that the zeros of a holomorphic function are isolated. But here we are not talking about holomorphic functions, yx is a non-trivial solution of a second order ODE. So, let us prove the theorem. It is very easy p1, p2, p3, da, da is a sequence of zeros converging to p, we are going to derive it a contradiction. Then by continuity y of p must be 0. Without loss of generality we may assume that the sequence is monotone. In real analysis every sequence has a monotone subsequence passing to a subsequence if necessary we may assume at the very outset that this sequence p1, p2, p3 monotone increases to p or monotone decreases to p whichever one. Then between p i and p i plus 1 the derivative must vanish because y vanishes at p i and y vanishes at p i plus 1. So, the derivative must vanish somewhere say in q i between p i and p i plus 1. So, the p i is converged to p, the q i is also must converge to p. So, q i must converge to p as i goes to infinity. So, by continuity the derivative we see that y prime of p must also be 0. So, y of x satisfies the initial value problem y double prime plus rho x y equal to 0, y of p is 0, y prime of p is 0. But remember that the solution to the initial value problem is unique, but we already have a solution to this problem namely the 0 solution. So, in looking the uniqueness clause in the fundamental existence uniqueness theorem from ODEs we conclude that y of x is identically 0 and that is a contradiction. So, now thanks to this result it makes sense to talk about the successive zeros of a solution of this differential equation which is not identically 0. So, now let us state and prove the important terms comparison theorem. So, in the theory of second order differential equation the terms comparison theorem and separation theorem are very important oscillation and comparison theorems. So, we shall use this terms comparison theorem to demonstrate the existence of eigenvalues and eigenfunctions for the Dirichlet problem theorem 69. Suppose rho and sigma are both continuous positive functions on the interval 0, 1 and rho x is bigger than sigma x throughout the interval 0, 1 and now I am looking at solutions y of x and z of x of these two differential equations y double prime plus rho x y equal to 0, z double prime plus sigma x z equal to 0 between two successive zeros of z x there is at least 1 0 of y of x. So, remember what is the condition rho and sigma are both continuous and rho x is bigger than sigma x. So, you should think of this rho of x as a frequency suppose if rho and sigma are both constants. So, you can think of it as y double prime plus capital M squared y equal to 0 z double prime plus little m squared z equal to 0 capital M is bigger than little m. So, y double prime plus capital M squared y equal to 0 the frequency of oscillation is more z double prime plus little m squared the frequency of oscillation is less. So, for example, you can think of sine 3 x and sine 16 x between two successive zeros of sine 3 x there is at least 1 0 of sine 16 x the sine 16 x oscillates more frequently that is all that there is to it. The proof is very easy suppose a and b are successive zeros of z x in the interval 0 1 we may assume that z x is bigger than 0 throughout a b and let us assume that y of x does not vanish in the open interval a b because y of x does not vanish in the open interval a b it is strictly positive there. Now, we imitate the proof of orthogonality of eigen functions, but only difference is that we work in the interval a b instead of working on the interval 0 1 no prices for guessing multiply the first equation by z second equation by y integrate by parts and subtracts that is exactly what I am going to get. So, what you need to do is that you go when you integrate by parts you are going to get four terms out of these four terms two terms are going to vanish of course there is an integral involving z prime y prime which will cancel out and we will get a term with a rho x minus sigma x that you see in the slide. Now, look at the boundary terms two terms are going to be 0 and two terms will survive the z was 0 at a and the z was 0 at b and the z is positive in between the z is positive in between. So, z prime a is positive z prime b is negative and so you will get the an equation which is 0 on the right hand side on the left hand side you will have sum of three positive terms and that is evidently a contradiction and that is all that there is to it and the proof of the Sturm's comparison theorem is over. Now, we can give a very different proof of the Sturm's comparison theorem there is a beautiful discussion on page 2 19 222 in Lord Rayleigh's theory of sound volume 1 a very charming the description is available. So, that completes the proof of the Sturm's comparison theorem. Now, let us come to the important result namely we come back to the Sturm level problem y double prime plus lambda rho x y equal to 0 and I am going to work with Dirichlet boundary conditions at 0 and 1 namely y of 0 is 0 and y of 1 is 1. Now, let us do the following let us look at a solution y of x, lambda let us look at the solution of this differential equation y of x, lambda where y of 0 is 0 and y prime of 0 is 1. So, I am looking at the initial value problem, but now there is a parameter lambda appearing in the differential equation. So, that is why I am calling the solution as a function of 2 variables x and lambda, but the initial conditions are fixed y of 0 is 0 and y prime of 0 is 1. Now, this solution will depend continuously on initial conditions and parameters the initial conditions are frozen. So, it will be continuous with respect to this parameter lambda. It will also be differentiable with respect to lambda. These are general theorems in ordinary differential equations. The Picard's theorem existence uniqueness theorem says that the solution that is guaranteed to exist and it is unique actually has these additional features namely if there are parameters present in the differential equation the solutions will be nice it will be continuous differentiable analytic etcetera with respect to these parameters. So, now what happens is that we are interested in not in y x lambda for a general lambda we are interested in locating those lambdas for which the solution satisfies the boundary condition. Our final destination is not the solution of the initial value problem our final goal is to solve the boundary value problem. So, prescribing these initial conditions y of 0 equal to 0 y prime 0 equal to 1 and calling the solution y x lambda is only an intermediate step to reach our final destination to go where we want to go. So, we must select this lambda in such a way that the boundary condition is satisfied. So, we want to look at the roots of the equation y x lambda equal to 0 when x equal to in other words we want to put x equal to 1 and look at y 1 comma lambda and must equate it to 0 and must locate the roots of that and those roots lambda will be the Eigen values of the Sturm level problem. We are going to use the Sturm's comparison theorem to do this exercise, but you can think of it in a very different perspective look at the differential equation how does lambda appear in this differential equation the dependence on lambda is an analytic dependence. So, I can even allow the lambda to be complex and the solution y of x comma lambda will become a complex valued function of a complex variable lambda. So, for each x if I think of it as a function of lambda as a function of lambda it will be holomorphic. So, when I put x equal to 1 I am looking at y of 1 comma lambda and that is a holomorphic function of lambda I am looking at the zeros of a holomorphic function and I want to show that this holomorphic function has infinitely many zeros that is another way of approaching the problem. If you look at for example theory of differential equations by Coddington Levinson they do take this approach using complex analysis for those who are interested you may want to consult that. So, in the next capsule we are going to use the Sturm's comparison theorem that we have proved to establish and locate these roots lambda such that y of 1 comma lambda is 0. I think this will be a very good place to stop this capsule and we will continue this in the next capsule.