 Hello, so last time we were talking about boundary value problems for a differential equation with the two point boundary conditions. We looked at Dirichlet boundary conditions, we looked at Neumann boundary conditions and in the exercises we looked at a couple of other boundary conditions. We talked about eigenvalues and eigenfunctions and I mentioned to you that this should be thought of as a generalization of the eigenvalue problem in linear algebra except that the matrices that correspond to these kinds of problems are real symmetric matrix or complex Hermitian matrix. And before we continue with boundary value problems for differential equations, we should be looking at this linear algebra notions of eigenvalues and eigenvectors in a slightly different perspective. So, that is exactly what we want to do. Let us look at this slides, let us go to problems that is displayed in the slide. A rigid body is rotated with uniform and fixed angular speed. So, the angular velocity is omega, omega is a vector, omega 1, omega 2, omega 3 and it is rotating that mod omega is fixed. So, let us assume that mod omega is 1 for simplicity, it is a unit vector. But and let us fix a point in the rigid body and let us call that point O the origin. Now, so this point is fixed, the speed of rotation is fixed. What can be varied is the direction of the axis of rotation. How should it choose the direction of omega vector, the angular velocity vector so that the rotational kinetic energy is maximized. Well, if you are looked at some elementary books on physics, then you will know that the rotational kinetic energy is given by 1 half, omega transpose i omega. But what is this matrix i, it is known as the inertia matrix, the 3 by 3 matrix consisting of moments of inertia and products of inertia. It is a real symmetric 3 by 3 matrix and the kinetic energy apart from a factor of half is omega transpose i omega or it is an inner product of i omega and omega. So, you see an inner product AXX, where A is a general real symmetric n cross n matrix and you are looking at a unit vector x1 square plus x2 square plus zeta plus xn squared equal to 1 and the problem is to optimize, maximize or minimize this inner product AXX. When A is the inertia matrix, the matrix of moments and products of inertia and when x is the angular velocity vector, which I am going to assume is a unit vector, I must maximize this. This is the twice the rotational kinetic energy and I want to maximize the rotational kinetic energy. So, how should we go about the problem? So, now that I told you the answer to the question. So, if you do question number 7, then automatically you are done question number 6. So, now let us go on to question number 7. Well, one thing that might come to you is the idea of Lagrange multipliers. So, this AX comma X that is a quadratic form and if you are right A as Aij and this quadratic form will be A11 x1 square plus A22 x2 square plus dot dot dot plus Ann xn square and then the cross terms 2Aij xi xj i less than j and that quadratic form has to be maximized subject to this constraint. So, you can think of this as a constrained optimization problem and you can try to apply the method of Lagrange multipliers. But let us just go at a more elementary and a more fundamental level. The quadratic form is V transpose AV the inner product of AV with V simply V transpose AV and this is a continuous function and the unit sphere is a compact space. So, continuous function on a compact space has a infimum and the infimum is attained. So, it has a minimum and also has a maximum. So, the maximum and the minimum are both going to be attained. So, the minimum value is attained at some point on the unit sphere V1. I repeat the quadratic form restricted to the unit sphere. We are not looking at the quadratic form in Rn we are restricting it to the unit sphere unit sphere is compact. Now, what we do is we perturb this unit vector V1 and we just push it away. So, V1 plus epsilon h and h is a any vector unfortunately this may not be a unit vector anymore. So, I need to divide with a length and this w is again a unit vector, but remember that V1 is the absolute minimum on this unit sphere. So, q V1 will be less than or equal to q w for all epsilon small enough. In fact, epsilon can be either positive or negative in we will need both. Let us find this expression. So, q w q w will be q of V1 plus epsilon h. What is that V1 plus epsilon h transpose a V1 plus epsilon h and the denominator will have V1 plus epsilon h norm squared. So, I have to multiply through by norm V1 plus epsilon h squared to clear the denominators and we will get this inequality here that is displayed q V1 norm V1 plus epsilon h squared less than or equal to V1 plus epsilon h transpose a V1 plus epsilon h. Now, we must write down the this matrix product. You must write it as V1 transpose a V1 and then there will be epsilon h transpose a V1 and we will have V1 transpose a epsilon h and then there will be a term which is epsilon squared h transpose a h the epsilon squared term has been written out here. And now what happens is that V1 transpose a V1 what is V1 transpose a V1 that is exactly q V1 and here we will expand this again you are going to get a norm V1 squared norm V1 squared is 1. So, you are again going to get a q V1 the q V1 is going to cancel from both sides. So, from both sides that constant term namely terms that do not depend on epsilon cancel out. Then there will be epsilon squared terms the epsilon squared terms have been combined in one particular term here. And then there is a middle term here 2 epsilon h transpose V1 when you expand this or the 2 epsilon times in the product of V1 and h. So, that is going to be this. So, again you are going to get 2 epsilon q V1 h transpose V1 less than or equal to from here there will be two terms containing epsilon's and what are the two terms h transpose a V1 V1 transpose a h. These two terms are going to be equal because h transpose a V1 is a 1 cross 1 matrix V1 transpose a h is also a 1 cross 1 matrix. If you write down these two 1 cross 1 matrices you will realize that they are the same. Here we are critically using the fact that a is a symmetric matrix. Please check the correctness of what I just said. So, that is why those two have been combined into one term 2 epsilon. Of course, divide by 2 epsilon and let epsilon go to 0. That is the next thing to do. So, let us do that. So, divide by epsilon let epsilon go to 0 and epsilon may have either sign either it could be positive or negative. So, when I divide by epsilon if epsilon is positive the inequality is going to be preserved. If epsilon is going to be negative inequality is going to get reversed. So, epsilon can go to 0 through positive values or through negative values accordingly I will get 2 inequalities. Now, remember that h is an arbitrary vector and arbitrary vector transpose this object is less than or equal to 0 and the same arbitrary vector transposed with the same vector is greater than or equal to 0. So, we get h transpose a V1 minus q V1 V1 must be equal to 0 because these two inequalities simultaneously hold. But h is arbitrary what is h transpose times x it is the inner product of x with h. But now I am saying that x inner product with h is 0 for every h that means that this x itself must be 0 or written out a V1 must be q V1 V1 remember q V1 is a real number it is a value of a quadratic form. And so this equation will tell you that q V1 is an eigenvalue of a and the point at which the quadratic form assumes its minimum is an eigenvector. So, we saw that the smallest eigenvalue is going to be the infimum of the value of the quadratic form restricted to the unit sphere and the point at which it attains its infimum is the eigenvector corresponding to that eigenvalue. Now, we want to proceed further. Let S be the intersection of the unit sphere with a hyperplane what is this hyperplane V dot V1 equal to 0 those vectors which are perpendicular to V1. So, you have got a unit vector V1 and you are looking at the orthocomplement of this unit vector V1 you are looking at those vectors which are all perpendicular to V1 and that is this locus V dot V1 equal to 0. And you are looking at this intersection of this hyperplane with the unit sphere and the intersection is being called S remember this is a closed set and the unit sphere is compact. So, S is also compact. So, this quadratic form restricted to S is again going to attain a minimum at say V2 and the minimum value q V2 is going to be greater than or equal to q V1. I asked why remember you got a function f a real valued function f it has a minimum on say a and you are looking at its minimum on a subset V. If you are minimizing it over a subset the value of the minimum will go up it will not come down all right. So, now we do the same perturbation argument we perturb V2 and we make it V2 plus epsilon h divided by norm V2 plus epsilon h and that is again become a unit vector. And unfortunately this W is a unit vector, but it may not be in S namely it may not be orthogonal to V1. So, we need to do something else to make it orthogonal to V1. So, what we need to do is that we need to select this h in such a way that h dot V1 equal to 0. Once you make such a selection V2 is already orthogonal to V1 V2 is in my set S. So, V2 dot V1 is 0 certainly holds I want this W to be orthogonal to V1 and that can be secured by making sure that this h satisfies this equation h dot V1 equal to 0. I am going to restrict h to those vectors only. So, now again since q attains its minimum on S at the point V2 and since W is also on the same set S q V2 is going to be less than or equal to q W. Again I want to multiply the inequality by norm V2 plus epsilon h the whole squared expanding the whole thing and cancelling out q V2 exactly as we did before. We get again the same kind of an inequality to epsilon h transpose q V2 V2 minus a V2 less than or equal to epsilon squared times h transpose h minus q V2 norm h squared. Again we was divided by epsilon and again we must allow the epsilon to go to 0 through positive values and negative values will get 2 inequalities and combining these 2 inequalities we get one equality what is that equality h transpose q V2 V2 minus a V2 0. So, now this q V2 V2 minus a V2 is some vector x. So, we say h transpose x is 0 that is dot product of h and x is 0 and I would like to conclude that h is 0, but for that I need this equation to hold for all values of h, but right now what have we proved we have proved this holds for all those values of h for which h dot V1 is 0 that the restriction is there. So, this holds only for those h which satisfy h dot V1 is 0, but now remember that this equation is trivially true for V1 also if I instead of h if I take V1 then I will get q V2 which is a scalar V1 dot V2, but V1 dot V2 is 0 by definition and V1 transpose a V2 is 0, but what is V1 transpose a V1 transpose a is lambda 1 V1 transpose remember V1 was an eigenvector of a and a is a real symmetric matrix. So, you get lambda 1 the first eigenvalue times V1 transpose V2 V1 transpose V2 is again 0 because the dot product of V1 and V2 is 0. So, this holds for all those h which are orthogonal to V1 it also holds for V1. Now any vector in Rn can be written as a vector which is orthogonal to V1 and a vector which is lying along V1 and therefore this equality that we see in the middle of the slide h transpose q V2 V2 minus a V2 equal to 0 holds for all h and therefore q V2 V2 minus a V2 must be 0. So, again V2 is also an eigenvector corresponding to eigenvalue q V2. So, the quadratic form q attains its minimum on the sphere at a point which happens to be an eigenvector corresponding to the smallest eigenvalue that was V1. Now I took the ortho complement of V1 repeated the process and quadratic form restricted to S attains its minimum at V2 V2 is an eigenvector corresponding to eigenvalue q V2 and there is a second eigenvalue. Then we can keep going further now what we do is that we take all those vectors V which are perpendicular to both V1 and V2 simultaneously and this is a closed subspace and I intersect it with the unit sphere in Rn call that set T is again a compact set and restrict the quadratic form to the compact set T it will attain its minimum at say V3 again we will prove that V3 is an eigenvector with eigenvalue q V3. Proceeding thus we would have constructed a orthonormal basis of eigenvectors and that gives you a proof of the spectral theorem a real symmetric matrix A has an orthonormal basis of eigenvectors and we are given a proof of the spectral theorem via a variational approach. Now you might ask why you prove the spectral theorem like this because in the next chapter we are going to prove the spectral theorem for a compact self adjoint operator on a Hilbert space and the proof more or less is the same spirit as this particular proof and that is the reason for giving this proof. Okay the analogous result for self adjoint differential equations with Dirichlet boundary conditions is a serious matter that has led to a huge corpus of mathematical research. So there is a beautiful account of these developments in this classic book by Richard Courant the same Courant that I referred to before Courant and Hilbert methods of mathematical physics. Richard Courant was a student of David Hilbert. He wrote a book called the Dirichlet's principle conformal mappings and minimal surfaces which was reprinted by Dover in 2005. Now some preview of the first few pages are available freely in the internet but now let us consider the problem of minimizing the energy. Look at this display 6.4 integral 0 to 1 y prime t the whole square dt. Remember I mentioned to you the problem of minimizing the kinetic energy omega transpose i omega and omega is a angular velocity. So it is a velocity vector of some sort and that was a kinetic energy rotational kinetic energy. So we have taken this vibrational problem and we have taken the solution yt and we are differentiating it with respect to t and you are squaring it and you are integrating it from 0 to 1 with respect to t calling it energy certainly makes sense. So now let us consider the problem of minimizing this energy but what will be the analog of the unit vector that we talked about. Remember in the linear algebra problem we took a quadratic form and we restrict the quadratic form on the unit sphere. So we have to look at unit vectors in my space but which is my space remember that the space that we are talking about is l2 of closed interval 0 1 not the Lebesgue measure but the weighted Lebesgue measure rho t dt. So the measure is rho t dt. So the unit vectors will be integral 0 to 1 yt square rho t dt equation 6.5. So the constraining equation is 6.5 and the objective functional that you are talking about is 6.4. So again we are looking at a constrained optimization problem. Of course now this has to make sense. First of all 6.5 has to make sense. So y must be in l2 of 0 1 with respect to the measure rho t dt and I have to differentiate this function y unfortunately l2 functions may not be differentiable every continuous function will be in l2 but every continuous function is not differentiable in fact it may be everywhere not differentiable. So we need to put a restriction. So let us restrict it to piecewise smooth functions continuous functions which are piecewise smooth. So the function is continuous and I can chop up the interval into finitely many pieces such that on each piece the function is smooth that is it is differentiable. And at the junction points the left hand derivative and the right hand derivative exist but they may not be equal. Also the boundary conditions have to be satisfied I am going to work with Dirichlet boundary conditions only y of 0 is 0 and y of 1 is also 0 and rho of x as always is a positive continuous function on a closed interval 0 1. So this is the problem the optimization problem that we are looking at. Now the Dirichlet principle it is called a principle it is a weighty term because it is indeed a highly non-trivial principle. So the minimization problem that I talked about does have a solution and that solution is twice continuously differentiable and it satisfies the differential equation y double prime plus lambda rho x y equal to 0 y of 0 equal to 0 y of 1 will be equal to 0. In other words the solution to the minimization problem is actually an eigenfunction for this particular boundary value problem. How do I know that this eigenfunction is not identically 0 remember it is a constrained optimization problem it is a unit vector integral y t square rho t dt is 1 not 0. So the 0 function is ruled out also it says something more the eigenvalue that I get is a smallest eigenvalue of the stem level problem. So this is the Dirichlet principle in one variable there are number of difficulties in trying to make this as a theorem the principle difficulty is as follows we have to prove that the minimizer exists remember we are talking about an infinite dimensional problem L 2 of 0 1 with respect to the measure rho t dt is an infinite dimensional space and the unit ball in an infinite dimensional space is not going to be compact we are going to see these things the next chapter. Chapter 7 will be functional analytic ideas in Fourier analysis. So the unit ball the unit sphere is not going to be compact. So unlike the linear algebra problem that we talked about we cannot be sure that the infimum is attained infimum exists because it is non-negative the the objective function is non-negative that you are trying to minimize the infimum may not be the minimum the infimum may not be attained you do not have compactness or even if the minimizer exists even if the infimum is attained it may only be once differentiable it may not be twice differentiable and so demanding that the differential equation 6.6 is satisfied would require the minimizer that is the minimum that is attained that should be a function that is twice continuously differentiable that is a lot to demand these are rather deep waters. So what happens is that motivated by potential theoretic considerations Bernard Riemann uses ideas of classical potential theory to prove his celebrated theorem in complex analysis today known as the Riemann mapping theorem and the Riemann mapping theorem has led to a huge corpus of results in analysis. In fact it is fair to say that the Riemann's ideas on the Riemann's method for proving his celebrated theorem has given much impetus to the development of modern analysis in the decades following the appearance of this result. I think it is a good place to stop this capsule and we shall continue from here. Thank you very much.