 So, let us continue with the third part of this course which we started out namely generalized Fourier expansions with respect to various complete orthogonal systems in Hilbert spaces. In this chapter there is one last bit was missing there is a 0's of the Bessel's function we still need to show that if P is real positive JPX has infinitely many 0's in the positive real line. What we have proved is that these 0's are simple, but we still need to show that there are infinitely many of them. The proof will be given in the next chapter after we do the Sturm's comparison theorem and now we turn to this chapter 6 where we discuss the vibrating string and regular Sturm-level problems. Suppose we have a string of length L and linear density rho x stretched along a line segment 0 to L where assuming that the length of the string is L and the string is clamped and the 2 ends and a string is set into vibrations then what are the governing differential equations? The governing differential equations can be written as y double prime plus lambda rho xy equal to 0. You are going to perform the same kind of calculations at the start of the linear wave equation and it is separate the variables as we did for the membrane and here you will get 6.1 instead of the reduced wave equation over there we get equation 6.1 y double prime plus lambda rho xy equal to 0 lambda is a parameter. The string is clamped at the 2 ends. So, we have the boundary conditions BC y of 0 equal to 0 and y of L equal to 0 display 6.2. A lambda for which there is a non-trivial solution of course if I take y of x identically 0 then surely the differential equation is satisfied and the boundary conditions are satisfied but such a solution is called a trivial solution I want a non-trivial solution I want a solution which is not identically 0 such that the differential equation is being satisfied and the boundary conditions are being satisfied the corresponding parameter value lambda will be called an eigen value of the boundary value problem and the corresponding solution is called a eigen function of the boundary value problem. So this particular boundary condition 6.2 is usually referred to as the Dirichlet boundary condition other types of physical problems will lead to other boundary conditions a more popular one is the Neumann boundary condition y prime of 0 equal to 0 y prime of L equal to 0. We shall discuss the Dirichlet boundary conditions and the Neumann boundary conditions these are the only two boundary conditions that we shall be discussing there are yet others such as periodic boundary condition that leads to Fourier series back to chapter 1 and there are Robin boundary conditions that we shall not touch upon. Of course we need to make some assumptions about this density function rho x the density function rho x is going to be assumed to be positive everywhere and it is going to be continuous we are going to make these assumptions about the function rho x. So, let us proceed further. So, a simple example is when rho x is 1. So, our problem simply reads y double prime plus lambda y equal to 0 and for simplicity I am going to assume that the string has length 1. So, the boundary conditions are y of 0 equal to 0 y of 1 equal to 0 they are Dirichlet boundary condition. Suppose if lambda is 0 suppose if lambda is 0 then what happens we get the differential equation y double prime equal to 0 what are the solutions of the differential equation y double prime equal to 0 a plus b x put x equal to 0 and I am supposed to get 0. So, a must be 0. So, solution is simply b x, but when I put x equal to 1 I must get 0. So, b must also be 0. So, a 0 b 0 the function is identically 0 it is not interesting lambda equal to 0 is not an eigenvalue. This condition that the solution should not be identically 0 that is very important remember in linear algebra when you look at a n cross n matrix a when would you say lambda is an eigenvalue if there is a non-zero vector v such that a v equal to lambda v. So, the eigenvector is not supposed to be 0. So, the analog of that is that the solution must not be identically 0. So, now let us take lambda to be non-zero if lambda is non-zero then we know the solutions of the harmonic oscillator y of x equal to a cos root lambda x plus b sin root lambda x put x equal to 0 and the condition y of 0 equal to 0 will immediately give you that a must be 0 because the sin term anyway disappears and cos root lambda x becomes 1 and so that forces a to be 0. So, a is gone from this discussion. Now, you put x equal to 1 we get that 0 equal to b sin root lambda. Now, b cannot be 0 because we are looking for a non-trivial solution. So, b cannot be 0. So, the only other option is sin root lambda must be 0. If sin root lambda is 0 then the possibilities of lambda are root lambda equal to 0 is out because we seen lambda equal to 0 is not an eigenvalue. So, root lambda must be plus minus pi plus minus 2 pi dot dot dot or the values of lambda are pi squared 4 pi squared 9 pi squared dot dot dot. In other words, a non-trivial solution for the boundary value problem exists only for a discrete set of values of the parameter. In short, the eigenvalues form a discrete set. Now, let us discuss the orthogonality of eigenfunctions and once we have orthogonality of eigenfunctions they got a sequence of eigenvalues lambda 1 lambda 2 lambda 3 will have the corresponding eigenfunctions y 1 x y 2 x y 3 x and these eigenfunctions will be mutually orthogonal. And again we will be looking at whether it is complete or not and if it is complete then we can do a generalized Fourier expansions with these. So, that is the idea. First and the most basic is the orthogonality and yes for the boundary value problem on the interval 0 L, y double prime plus lambda rho x y equal to 0 either with Dirichlet boundary conditions or Neumann boundary conditions. The eigenfunctions corresponding to distinct eigenvalues are orthogonal, orthogonal with respect to what? With respect to which space? I will be looking at L 2 of 0 1, but L 2 not with respect to the Lebesgue measure, but with respect to the weighted Lebesgue measure. The measure is rho x dx not the standard Lebesgue measure, the weighted Lebesgue measure. So, integral 0 to L u x v x rho x dx must be 0. This theorem 64 must be compared with the elementary theorem that you study in linear algebra when you have a real symmetric matrix and you take distinct eigenvalues lambda and mu. The corresponding eigenvectors are mutually perpendicular. A similar result holds for Hermitian matrices except that you will have to take a Hermitian product in C n instead of the usual inner product in R n. So, this is the analog of that. Suppose lambda and mu are distinct eigenvalues and let the eigenfunctions be u and v. So, u is an eigenfunction with eigenvalue lambda. So, u double prime plus lambda rho x u equal to 0. v is an eigenfunction with eigenvalue mu. So, v double prime plus mu rho x v equal to 0 and the Dirichlet boundary conditions are written down. u of 0 equal to 0, u of L equal to 0, v of 0 equal to 0 and v of L equal to 0. So, what do you do? We multiply the first equation by v and the second equation by u integrate from 0 to L and perform an integration by parts. By now this kind of argument has repeated itself several times and so this is a predictable line of proof. So, when you integrate by parts what is going to happen? Integral v u double prime, integral v u double prime you are going to get two terms one is the derivative will fall on v and you are going to get minus integral u prime v prime. Similarly, in the second equation I multiply by u and do an integration by parts I am again going to get minus integral 0 to L u prime v prime I am going to subtract and this term will cancel out. Then what is going to be left over? The boundary terms we get when we perform an integration by parts. Where are the boundary terms being evaluated at 0 and at L? What are the boundary terms for the first one for instance u prime 0 v of 0 with a negative sign but v of 0 is 0. At the other end you are going to get u prime of L v of L but v of L is 0. So, the boundary terms drop out. The same thing is going to happen with the Neumann boundary conditions again the boundary terms will drop out. So, what is left after you go through this rigmarole what is left over is integral from 0 to L u x v x rho x dx lambda minus mu is a constant comes out of the integral. Lambda is not equal to mu remember we are taking distinct eigenvalues. So, what must be 0 integral 0 to L u x v x rho x must be 0 and that completes the proof of the orthogonality assertion theorem 64 has been established. And the case of the Neumann boundary condition is left for you to work out. Now comes the question of simplicity of eigenvalues. To each eigenvalue or the boundary value problem with either Dirichlet boundary conditions or Neumann boundary conditions there is only one eigenfunction up to scalar multiples. Now go back to linear algebra. Go back to linear algebra and you take a matrix A. You want to discuss diagonalizability of the matrix. Now you know that if you have n cross n matrix and all the eigenvalues are simple and they are distinct and if the eigenvalues are distinct they are simple and that is the easiest situation. The eigenvectors will be linearly independent and they will form an eigen basis. The problem comes when the matrix has a certain eigenvalue and it has got certain multiplicities. In linear algebra you talk about algebraic multiplicity and geometric multiplicity. What is algebraic multiplicity? Algebraic multiplicity means that you take the matrix and look at its characteristic equation and take the eigenvalue lambda and you see what is the multiplicity of this eigenvalue lambda as a root of the characteristic equation. And that number is called the algebraic multiplicity. What are the geometric multiplicity? The geometric multiplicity means how many different eigenvectors are there which are linearly independent. Suppose you got an eigenvalue lambda and I can find three linearly independent eigenvectors and no more then the geometric multiplicity will be 3. So, in other words the geometric multiplicity is the nullity of a minus lambda i that number is a geometric multiplicity. The geometric multiplicity will be less than or equal to the algebraic multiplicity and the inequality can be strict. The inequality is strict then you know that in linear algebra there is a lot of problems then you have to look for a Jordan canonical form. But when you take a real symmetric matrix that does not happen for a real symmetric matrix the geometric multiplicity will exactly match the algebraic multiplicity. So, what we are talking about here are analogs of real symmetric matrix or a complex Hermitian matrix. These boundary value problems are analogs of those problems. But here we got something even better for the Dirichlet boundary conditions or for the Neumann boundary condition for each eigenvalue the geometric multiplicity is one. There is only one eigenfunction up to multiples and as I said there is no concept of algebraic multiplicity there is no concept of characteristic equation it is an infinite dimensional problem. So, let us prove theorem 65 explain to you the significance of theorem 65. Assume that u and v are linearly independent eigenfunctions linearly independent have been written in red and they are both eigenfunctions with the same eigenvalue lambda that means ux and vx both satisfy this differential equation y double prime plus lambda rho x y equal to 0. But they are both solutions of the same boundary value problem right. So, they must satisfy the initial condition. So, u of 0 must be 0 and v of 0 must also be 0 and if u and v vanish at 0 every linear combination of u and v vanish at 0. But u and v are two linearly independent solutions of a second order differential equation. What is the space of solutions of a second order ODE? It is a two dimensional space how do you prove that use Picard's theorem. So, let us do the following let us take initial conditions let us appeal to Picard's theorem let us take y of 0 equal to 1 y prime 0 equal to 0. So, with these initial conditions there must be a solution of the differential equation call it yx. But that solution does not vanish at that origin. But that solution must be a linear combination of u and v because u and v form a basis of solutions. So, that is a contradiction because y of x on the one hand must be a linear combination of u and v. On the other hand a linear combination of u and v vanishes at the origin but y does not vanish at the origin. So, that shows that you cannot have two linearly independent eigen functions with the same eigen value. So, each eigen value is a simple eigen value that is why I call it simplicity of eigen values. So, now we have got eigen functions to each eigen value lambda there is only one eigen function up to multiples. And so, we can take this set of eigen functions it is an orthogonal set. Is it complete in this L 2 space with measure rho x dx? Is it a complete orthogonal system? The answer is yes. The theorem is often stated under a stronger hypothesis. For example, you can look at Courant and Hilbert's methods of mathematical physics volume 1 page 293. However, if you look at E.C. Titchmarsh's book eigen functions expansion associated with second order ODE's volume 1 it proves this theorem of completeness with substantially weaker hypothesis I am giving you a page number. One can also discuss mean convergence but we shall not do so here. So, that concerns the completeness of eigen functions. Let me state this theorem 66 the possibility of expanding a reasonably nice function f of x as an infinite linear combination of these eigen functions. This is a complete statement for the term level problems in this question that is you are looking at y double prime plus lambda rho x y equal to 0, rho x is positive, rho x is continuous and we are looking at either the Dirichlet boundary conditions or Neumann boundary conditions. Then several things happen you have an infinite sequence of eigen values lambda 1 less than lambda 2 less than lambda 3 and these eigen values will tend to infinity and the set of eigen functions will be complete in the following sense. Each Lipschitz function on 0 L can be expanded as a four-year series f of x equal to summation j from 1 to infinity c j phi j x and the convergence is point wise on the open interval 0 L and the coefficients are given by the usual formula. Multiply both sides by phi j and take the inner product with respect to rho x dx. When you take the integral integral with respect to the measure rho x dx so when you right hand side you get c j integral phi j squared rho x left hand side you are going to get integral fx phi j x rho x dx so the c j will be given by integral 0 to L fx phi j x rho x dx divided by integral 0 to L phi j x squared rho x dx so the four-year expansion with respect to a eigen functions corresponding to a Sturm level problem that is a boundary value problem for a second order od y double prime plus lambda rho x y equal to 0 with Dirichlet or Neumann boundary conditions. Here are some exercises for you show that the eigen values of y double prime plus lambda rho y equal to 0 with y of 0 equal to 0 and y of 1 equal to 0 are positive real numbers that is this eigen values must be positive how do you check that the eigen values are positive let us do that let us multiply the differential equation by y and perform an integration by parts perform an integration with parts what do you get y double prime y integral from 0 to 1 the derivative will shift from y double prime to the other factor you are going to be left with minus integral y prime squared boundary terms will drop out because I am looking at Dirichlet boundary condition what was the second term lambda integral 0 to 1 rho x y x squared and that the first term had a negative sign so we get lambda times integral rho x y x squared equals integral 0 to 1 y prime of x squared so lambda must be positive that completes the problem. Determine the eigen values and eigen functions of the stem level problem y double prime minus 2 y prime plus 1 plus lambda y equal to 0 with boundary conditions y of 0 equal to 0 y of 1 equal to 0. The third problem is an interesting problem determine all the eigen values and eigen functions of the stem level problem x squared y double prime plus x y prime plus lambda y equal to 0 how do you do these exercises second one you will try the solution to be of the form c1 e to the power mx plus c2 e to the power m2x and you will try to fit the boundary conditions and so on and m of course has to satisfy the characteristic equation m squared minus 2m plus 1 plus lambda equal to 0 so you get the characteristic equation and you will try to fit the boundary conditions the second one here you see a Cauchy Euler equation x squared y double prime plus x y prime plus lambda y equal to 0 and the indicial equation is m into m minus 1 plus m plus lambda and you write the general solution and try to fit the boundary condition this boundary condition is different from the Dirichlet or the Neumann boundary conditions these are called periodic boundary conditions so there are some variants in the exercises show that the eigen value is the boundary value problem y double prime plus lambda y equal to 0 y of 0 equal to 0 and the boundary conditions are slightly different y of 1 plus y prime of 1 equal to 0 they are given by the equation lambda equal to k squared the eigen values are given by lambda equal to k squared where k satisfies the transcendental equation tan k plus k equal to 0 so again the general solution of this will be a cause root lambda x plus b sin root lambda x equal to 0 put x equal to 0 straight away a becomes 0 and then put the other condition and you will get this equation tan k plus k equal to 0 now this tan k plus k equal to 0 has infinitely many roots remember how does the tan function look like look at the graph of the tan function tan function is periodic with period pi it is an odd function so first draw the graph of tan x from minus pi to pi it starts from below and it passes through the origin and goes off to infinity it has two vertical asymptotes x equal to pi and x equal to minus pi two vertical asymptote and this picture repeats itself periodically with period pi so you draw this graph of tan k and then you take this k the diagonal y equal to x in the graph in the in the graph paper just draw the line sloping at an angle of 45 degrees with the x axis passing through the origin this line and the graph of tan x will intersect at infinitely many points there will be exactly one point of intersection in each vertical strip of width pi so that is how you will prove that there are infinitely many roots now this kind of boundary value problem appears in certain physical problems and this is discussed in Byerley's text on page 117 in section 67 the problem as such will appear in many textbooks a very popular exercise that you will see in almost all books in undergraduate text on boundary value problems but only Byerley gives you what is the origin of this boundary value problem the next two exercises are interesting a rigid body is rotated with uniform and fixed angular speed about an axis that is not specified how would you choose this axis of rotation so as to maximize the rotational kinetic energy formulate the problem mathematically and the other thing is the variational characterization of eigenvalues and eigenmeters these two exercises are basically items from linear algebra why is it that we are bringing in ideas from basic linear algebra because I want to first discuss the problem in the linear algebra setup which is more familiar and then take over to the differential equation setup where some highly non-trivial complications proper so I think this is a very good place to stop this capsule and we shall continue this next time with a discussion of these two problems we will begin by discussing the rigid body problem and this problem on variational principles of eigenvalues and eigenvectors maybe you want to think about this problem number 18 you probably want to consult some elementary physics books on rigid body mechanics and find the formula for the rotational kinetic energy of a rigid body maybe you want to take a cube and ask yourself whether the axis of rotation should be a diagonal for instance or should be an axis passing through one of the sides of the cube how should you choose this axis of rotation so we will stop this I have given you some interesting questions to think about thank you very much