 So, in the last capsule we were trying to solve the Dirichlet problem for the Laplace's equation on a disc x squared plus y squared less than or equal to 1. The function was given on the boundary x squared plus y squared equal to 1 namely u of cos theta comma sin theta equal to f of theta, but f of theta was a Lipschitz continuous 2 pi periodic function so that the Fourier series converges etc. We also saw that the solution can be obtained as an integral with the Poisson kernel. What is the Poisson kernel? Let us look at the Poisson kernel. It is 1 minus r squared upon 1 plus r squared minus 2r cos theta minus t that is Poisson kernel of theta minus t. So, we got this integral 2.18. Now, we got to look at the properties of Poisson kernel. One of the properties of Poisson kernel is that it is a non-negative kernel unlike the Dirichlet kernel unlike the Dirichlet kernel the Poisson kernel is a good kernel it is a positive kernel. The Dirichlet kernel is not a positive kernel. It takes both values positive and negative values and it oscillates very rapidly Dirichlet kernel is not convenient because it was not a positive kernel because it oscillated very rapidly we had trouble when we try to prove point wise convergence. Here the situation is much better the Poisson kernel is a non-negative kernel and 1 upon 2 pi integral minus pi 2 pi pi r of theta minus t will be 1. How to prove this 2.19? Go back to the formula for the Poisson kernel namely 1 minus r squared upon 1 plus r squared minus 2r cos s is 1 plus summation n from 1 to infinity 2 r to the power n cos n s. Integrate this equation term by term with respect to s from minus pi to pi. You can use uniform convergence because r is strictly less than 1 and that is how you prove equation 2.19. So, what you can do is that you can imitate the same thing that we did for Dirichlet kernel you can put theta minus t equal to s in equation 2.18 and you will get 1 minus r squared upon 1 plus r squared minus 2r cos s except that over here this will become theta minus s theta minus s and the integral will be with respect to ds. So, do this small change of variables. Now, once you do this change of variables then what you do is that you can prove the next exercise show that limit as r tends to 1 minus more u r e to the power i theta minus f of theta equal to 0. Remember we got the solution u of r e to the power i theta. Now, we want to prove that this solution does indeed assume the correct boundary values namely f of theta that is what 2.20 is trying to tell you will carry out this program later on, but here I would like you to think about it. Now, what about the three dimensional Laplace equation? We could try to carry out a similar program for the Laplace's equation for a ball in r 3, but the problem is going to be more complicated because the analog of a Fourier series is going to be more complicated you have to develop the theory of spherical harmonics. The corresponding Fourier analysis in r 3 by writing the Laplace's equation in spherical polar coordinates that will lead you to the Legendre functions, associated Legendre equations and the spherical harmonics. So, the analysis of spherical harmonics is the higher dimensional analog of Fourier series. So, suppose you are given a continuous function f of x on the unit ball b centered at the origin then the solution of the boundary value problem Laplacian of u equal to 0 on b u on the boundaries f is given by this integral. So, you recognize the Poisson kernel for higher dimension and this d capital S x is the area element on the surface. You can derive this by other methods also, but the proof using spherical harmonics you will see in G. B. Folland's book Fourier analysis and its applications and page 180. It is a very delightfully written book and you must definitely consult this book and G. B. Folland is an authority in Fourier analysis. So, now we will come to the heat equation, how to solve the heat equation using Fourier series. This will be developed through a series of exercises. So, let u of xt be a smooth solution of the heat equation in the upper half plane. We are going to work in two variables t and x and the variable t is going to be considered as a time variable and the variable x is going to be considered as a spatial variable. We are going to work in the upper half plane real line Cartesian product with the positive half line. So, time is going to be positive t greater than or equal to 0 and we are interested in 2 pi periodic functions. We are going to postulate that the solution is 2 pi periodic u of x comma t equal to u of x plus 2 pi comma t for all x in R and t greater than or equal to 0. The first exercise is for you to show that the energy integral minus pi to pi u of xt the whole squared dx is a monotone decreasing function of time. Prove that the same result holds if the integral is over any interval of length 2 pi and then deduce that a 2 pi periodic solution of the heat equation with given initial condition if it exists is unique. Let us look at this exercise and let us try to see how to do it call the energy this integral Et energy integrated with respect to x and so the energy is a function of time call it Et. So, let us differentiate with respect to t I am going to differentiate under the integral sign I am going to get twice u of x comma t del u by del t. So, it is twice integral from minus pi to pi u of xt u t of xt dx. But remember u is a solution to the heat equation. So, ut equal to uxx. So, here in the last equation I am going to integrate by parts I am going to throw one of the derivatives on the other factor and I am going to pick up minus sign. So, when you integrate by parts I am going to get minus 2 times integral minus pi to pi u sub x x of t the whole squared. What happens to the boundary terms the boundary terms will cancel out because a function is 2 pi periodic u of xt equal to u of x plus 2 pi t. Remember that if the function is 2 pi periodic its derivative is also going to be 2 pi periodic. So, use the 2 pi periodicity to check that the boundary terms cancel out when I perform the integration by parts. And so, what you are left out is e prime of t equal to minus 2 times integral minus pi to pi a squared factor and everything here is real value. So, it is always less than or equal to 0. So, you have finished proving that the energy is monotone decreasing with respect to time. So, we get energy is obviously non-negative function. So, 0 less than or equal to e of t less than or equal to e of 0. So, the initial value u of x comma 0 were to be 0. So, the initial energy is also 0 and so, e of t will be identically 0 for all t and if e of t is 0 remember that e of t is a integral of a non-negative function. So, u of xt must be identically 0. So, we have now drawn the conclusion that the initial data is 0 and the solution is also going to be 0. So, for 0 data the 2 pi periodic solution is unique namely 0. So, from this uniqueness followed now we prove the existence uniqueness of the 2 pi periodic solution has been established we now derive the existence. Because f of x is continuous 2 pi periodic function the data the initial condition u of x comma 0 is f of x. Further conditions will be imposed as we go along we begin with the assumption that the initial distribution of heat f of x is simply continuous. Now, we are looking for the solution of u t equal to u xx we will now look for a solution of the form u of xt equal to a naught plus summation n from 1 to infinity a n cos nx plus b n sin nx. Remember that the solution that we are looking for is a 2 pi periodic solution because it is a 2 pi periodic solution it will be a Fourier series. But this time the coefficients will depend upon t because u is a function of 2 variables x and t. So, I can think of t as an auxiliary parameter if you like. So, that is how we get equation 2.22. So, let us substitute this ansatz 2.22 into the differential equation u t equal to u xx and see what happens. What are the what are the time derivative del u by del t will be a naught prime plus summation n from 1 to infinity a n prime t cos nx plus b n prime t sin nx and then we have to differentiate twice with respect to x. So, a naught disappears summation n from 1 to infinity a n t cos nx and then n squared will come plus b n t sin nx and n squared will come and this whole thing will have a negative sign in front. So, compare the two expressions u t equal to u xx we got two Fourier series and compare the coefficients we get a naught prime t is 0 a n prime t equal to minus n squared a n t b n prime t equal to minus n squared b n t equation 2.23 in the display. The first equation simply says that a naught of t must be constant. Then the second equation says that a n of t must be an exponential function. There will be some initial conditions a a naught of t is a constant alpha naught a n t is constant times e to the power minus n square t b n t is constant times e to the power minus n square t these constants are alpha n and beta n. Let us put t equal to 0 in equation 2.22 let us put t equal to 0. So, I get a naught of 0 will be alpha naught a n of 0 will be alpha n b n of 0 will be beta n u of x comma 0 will be f of x. So, f of x equal to alpha naught plus summation n from 1 to infinity alpha n cos nx plus beta n sin nx equation 2.24. But f of x is a given function and it is a 2 pi periodic continuous function and I can write out its Fourier series and I have my alpha naught alpha n and beta n with that I have my solution u of x t which assumes the form u of x t equal to alpha naught plus summation n from 1 to infinity e to the power minus n square t alpha n cos nx plus beta n sin nx equation 2.25 in the display. Now comes the task for us to show that as t goes to 0 this equation 2.25 does indeed converge to 2.24 the initial condition f of x is attained as t approaches 0 plus that has to be established the rapid decay of those exponentials help us. So, the problem amounts to a passage to limit under the summation sign what we have to do is you have to show that limit as t tends to 0 plus summation n from 1 to infinity that is equal to summation n from 1 to infinity limit as t tends to 0 plus that is what it really amounts to because when you allow the t to go to 0 you get the thing under the summation. So, to check that this approaches this if the limit as t goes to 0 is the same as checking this. As you know exchange of 2 limiting operations is a tricky business one should do it carefully it is not difficult to carry it out and it has been carried out in this book by Bachman and Naricci on page 203 I already given a reference to this book earlier in this course. So, I will just refer to this book and we will not carry out the analysis in this case. Instead I have given exercise determine the solution of the heat equation with initial condition f of x equal to pi squared minus x squared on the interval mod x less than or equal to pi that is an even function on the interval minus pi to pi extend it as a 2 pi periodic function on the entire real line and for that initial condition solve the heat equation using Fourier series. So, now let us continue with the last part of this chapter Abel's summability. Remember the case of the Poisson kernel I left it as an exercise for you to check that the solution that we obtained u of r e to the power i theta converges to f of theta as r goes to 1 we will discuss that in detail under the broader context of Abel's summability. So, let us recall what Abel's summability means a n is a sequence of complex numbers. Okay now let us recall Abel's limit theorem from elementary analysis for example you can look at Rudin's principles of mathematical analysis and you will see the proof of Abel's limit theorem. So, theorem 25 summation n from 0 to infinity a n converges suppose that the infinite series converges then cook up the function f of t equal to summation n from 0 to infinity a n t to the power n this function is obviously holomorphic on mod t less than 1. But this holomorphic function will satisfy the equation 2.26 limit of f of t as t tends to 1 along the real axis and from inside the disk of convergence that will be exactly equal to summation n from 0 to infinity a n. It is as if you can put t equal to 1 directly in the defining equation but you might wonder is not that obvious simply put t equal to 1 in this equation and then you should get this that is not quite the case the result is non-trivial the theorem is quite non-trivial and let us try to understand why is this theorem of Abel non-trivial because the left hand side of 2.26 what is the left hand side of 2.26 it is limit as t tends to 1 minus f of t that is limit as t tends to 1 minus summation n from 0 to infinity a n t to the power n but what is this infinite series this infinite series is limit as capital n tends to infinity sum from 0 to capital N. Let us look at the right hand side what are the right hand side right hand side is summation n from 0 to infinity a n well shall we write this says limit as capital N tends to infinity summation a n little n from 0 to capital N. But this thing can be written as limit as t tends to 1 minus summation n from 0 to capital N a n t to the power n. So, what we see here is an exchange of two limit operations. So, Abel's limit theorem is non-trivial because it asserts that a certain exchange of limits is valid. Now, let us say that a series summation a n n from 0 to infinity 2.27 is said to be Abel's summable. If the associated series summation n from 0 to infinity a n t to the power n has radius of convergence 1 and further the sum function f of t has a limit as t tends to 1 minus. Again t approaches 1 along the real axis and from within and the limit is then said to be the Abel's sum of the infinite series 2.27. It is important that the limit is taken as t varies along the real axis. I can relax the condition and I can take I can take the circle and I can take the point 1 and I can take a sector I can take a sector with vertex 1 and opening angle and some acute angle as opening angle that is called a Stoll's region. I will not get into the details of that you will see this being discussed in classical books. So, let us discuss Abel's summability of Fourier series. We know that if you take a continuous 2 pi periodic function on the real line the Fourier series may not converge to the given function point twice, but a nice analog of Abel's summability holds. So, let us introduce an r to the power n in the nth terms in the Fourier series. So, we pass from 2.29 to 2.30 exactly in analogy with the Abel's summability that we talked about a few slides ago. The series 2.30 will certainly converge for 0 less than or equal to r less than 1 simply because the a n's and the b n's decay to 0 by Riemann Lebesgue Lemma. So, now that we know that 2.30 converges for this range 0 less than or equal to r less than 1, we denote its sum by U R e to the power i t. It is of interest to know whether for a fixed angle theta limit of U R e to the power i t equals f of theta where R t approaches 1 theta. So, e to the power i theta is a point on the boundary of the disc mod z less than 1 and I want to know whether when R e to the power i t converges to e to the power i theta from inside whether do I recover or capture the boundary data f of theta that is Abel's summability. Let us look into the proof of this Abel's summability. First we begin by recalling the Poisson kernel which we studied a few slides ago. Pi r s is 1 minus r squared upon 1 plus r squared minus 2 r cos s the thing written in blue. We already have a formula for the sum of the series 2.30. We simply put in the definition of a n, b n and a naught as integrals from minus pi to pi and we find the sum of the series U R e to the power i t and we get it as an integral 1 upon 2 pi integral minus pi to pi pi r s minus t f of s ds or by the usual change of variables put s minus t equal to U and we get it is 1 upon 2 pi integral minus pi to pi f of t minus s pi r of s ds. Now remember that when I integrate pi r s ds from minus pi to pi I exactly get 2 pi so 1 upon 2 pi integral minus pi to pi pi r s ds is 1. So integral minus pi to pi f of theta pi r s ds is f of theta after dividing by 1 upon 2 pi. So now I subtract this expression from the previous one we get U of r e to the power i t minus f of theta equals 1 upon 2 pi integral minus pi to pi f of t minus s minus f of theta times pi r s ds. Now we exploit the fact that the Poisson kernel is a positive kernel. So the last expression that we got in the previous slide we have to take the absolute value and we need to take the absolute value under the integral sign nothing will happen to the Poisson kernel because the Poisson kernel is a positive kernel. So we get 2.32 mod U r e to the power i t minus f of theta less than or equal to 1 upon 2 pi integral minus pi to pi mod f of t minus s minus f of theta times pi r s ds no need to put the absolute value on pi r. Now it is time to bring in the epsilon's and delta's. So let epsilon greater than 0 be arbitrary. First we invoke the uniform continuity of the function f that will give us a delta such that mod f x minus f y less than epsilon whenever mod x minus y is less than 2 delta. Now we break this integral 2.32 into 2 pieces integral from minus delta to delta and on the complementary piece namely mod s greater than or equal to delta. So first let us deal with this piece. Now in this piece what happens mod s is less than delta. So if you take t minus theta absolute less than delta the difference t minus s minus theta absolute value will be less than 2 delta. So I can say that this factor mod f of t minus s minus f of theta is less than epsilon by 2 and then I am left with the integral from minus delta to delta pi r s upon 2 pi. But because this Poisson kernel is a positive kernel the integral from minus delta to delta is less than or equal to the integral from minus pi to pi and this last one is 1. So I get epsilon by 2 equation 2.32 prime in the display. Now let us proceed further and look at the other piece where mod s is larger than delta. What is the integral 1 upon 2 pi integral mod s larger than delta mod f of theta minus s minus f of theta times the Poisson kernel pi r s d s. Now let us assume that m is the supremum of mod f in which case mod of f of theta minus s minus f of theta will be less than or equal to 2 m 1 2 cancels out and we get m upon pi integral mod s bigger than delta pi r s. Again go to the definition of pi r s the expression for pi r s that I gave you. Look at the expression and you convince yourself that when mod s is bigger than delta pi r s is going to be less than or equal to 1 minus r squared upon 1 plus r squared minus 2 r cos delta. The cos s in the expression for pi r s has been replaced by cos delta and I get something larger. Now remember that this right hand side 1 minus r squared upon 1 plus r squared minus 2 r cos delta approaches 0 as r goes to 1. So there exists eta bigger than 0 such that when r is between 1 minus eta and 1 we have that this thing is less than epsilon by 2 into pi upon n. So pi r of s is also less than epsilon by 2 into pi by s. The integral that you see on the right hand side of 2.32 double prime is less than or equal to what m upon pi into pi upon m into epsilon by 2 into the range of the integral is mod s bigger than delta and so that is less than 2 pi. And so you all in all you will get that this contribution from the right hand side of 2.32 double prime is less than epsilon by 2. All this drama is taking place when provided 1 minus eta less than r less than 1 and remember mod t minus theta less than delta. Both these conditions must hold and then we get that mod of u of r e to the power i t minus f of theta will be less than epsilon when mod t minus theta less than delta and r is between 1 minus eta and 1. In other words I have estimated this difference to be less than epsilon and that completes the proof of theorem 26. So suppose f of t is a continuous 2 pi periodic function of the real line with Fourier series given by equation 2.29 then the function u r e to the power i t which is a sum of 2.30 converges to f of theta as r t converges to 1 minus comma theta. So this shows that the Fourier series of a continuous function converges to f in the sense of a bell summability not in the sense of point wise convergence. I think with this we should close this capsule and we shall close this chapter on mean convergence and a bell summability some applications to partial differential equations. In the next chapter we will begin with the fairer's theorem. Cesaro summability that is the next item. Thank you very much.