 Hello, so we did the isoperimetric theorem last time and there are certain small things that I had left for you to figure out, probably you have figured it out yourself but let us just tie up these loose ends before taking up the next application. So couple of things we had left out, first the area formula, when you take a closed curve C, a simple closed curve, what are the area enclosed by the closed curve? Last time I wrote the formula integral x dy, integral x dy, the line integral x dy where the curve is traced counterclockwise gives you the area enclosed by the curve. This is a consequence of the Green's theorem, I mentioned it is a consequence of Green's theorem, so let me formally tell you how to get that. So let us recall the Green's theorem, okay, so you see in the slide the statement of Green's theorem, integral over the curve C p dx plus q dy is a double integral del q by del x minus del p by del y dx dy over the interior of the curve. So now you take q equal to x and p equal to 0, p equal to 0, so this term goes away, q equal to x, so this becomes 1, dx dy is going to give you the area enclosed by the curve. Right hand side gives you the area enclosed by the curve, left hand side gives you integral x dy, integral x dy and that completes the formula for the area enclosed by the curve. The other thing I left out for you to think about is the stuff about arc length. Now let us recall that if you have a regular smooth curve gamma parameterized by the interval AB, so gamma from AB to R2 is a regular smooth curve, what is a regular smooth curve? We say that a curve is a regular smooth curve, the derivative is not equal to 0, I can always make that assumption that the derivative is not 0. So now let us look at the arc length function, the arc length function measured from a specific point. Let us call that arc length function S of t, then in elementary calculus courses you would have seen the formula dS by dt equal to dS by dt equal to mod gamma dot t that you see on the slide. So square both sides dS by dt the whole square equal to mod gamma dot t the whole square which is dx by dt the whole square plus dy by dt the whole square, this is the equation that you get for the arc length. Now we need to use the relation between t and S, that is this relation that you see over here, t equal to 2 pi S upon L, so dt by dS is 2 pi by L, so dS by dt is L upon 2 pi, so we get this dx by dt the whole square plus dy by dt the whole square equal to dS by dt the whole square that is L squared upon 4 pi squared. And you integrate both sides from minus pi to pi and divide by 1 upon 2 pi, you get exactly L squared by 4 pi squared that is 2.10, so that point has now been clarified. The next thing that we have to clarify is the thing about the Parseval formula. Now first of all let us take f and g to be in L2 of minus pi pi, I am going to assume that f and g are real valued, otherwise you will have to put the complex conjugates in the appropriate places, it is not difficult for you to figure out what to do, but let us take two reals for now and let us assume that f has Fourier coefficients alpha 0, alpha n's and beta n's, so alpha 0 plus summation alpha n cos nx plus beta n sin nx and for g it is gamma 0 plus summation n from 1 to infinity gamma n cos nx plus delta n sin nx ok. So, with that out of the way what is the Fourier series for f plus g and f minus g? It will be the alphas plus gammas respective with the betas plus deltas, for f minus g it will be alpha minus gamma and the respectively beta minus deltas. So, you can write down the Fourier coefficients for f plus g and f minus g, we can write down the Parseval formula for f plus g and f minus g, let us write down the Parseval formula for f plus g and f minus g. So, I am combining the two in one formula 1 upon 2 pi integral minus pi to pi mod f t plus minus gt whole square dt equal to the coefficient of the constant term square, remember we are in the real domain, so no need to put absolute values. So, alpha 0 plus minus gamma 0 squared plus 1 half summation n from 1 to infinity alpha n plus minus gamma n squared plus beta n plus minus delta n the whole square. So, this is the Parseval formula for f plus g and f minus g subtract, subtract and you can get we can get the next equation 1 upon 2 pi integral minus pi to pi f t gt equal to alpha 0 gamma 0 plus 1 half summation n from 1 to infinity alpha n gamma n plus beta n delta n. So, this is the formula that we are using and we are using this formula where f of t is x of t and g of t is y prime of t, remember the area formula line integral x dy, but you are going to parameterize the curve right. So, it is going to be integral x t dy by dt dt and the parameter t goes from minus pi to pi and so you are applying the Parseval formula for the case when f of t equal to x of t and g of t equal to y prime of t. Now, you have the Fourier series for y and you want to find the Fourier series for y prime. So, once you write down the Fourier series for y, how do you write down the Fourier series for y prime? Simply do the term by term differentiation, but how do you know that the term by term differentiation is a valid operation? Remember that our curve is assumed to be a smooth curve which means that x t and y t are smooth 2 pi periodic functions. When you have smooth 2 pi periodic functions the Fourier series can be differentiated term by term because the series of derivatives will converge uniformly. So, you get the Fourier series for x of t you get the Fourier series for y prime of t and then you have got to put them in the appropriate places. That is how we got the n factor over here over here, we got this we picked up a n here and the minus sign came because the cosine term had to be differentiated. So, we put all these things together and we get the result and so those 3 small things which had left for you to verify have not been completed. Now, we will go to the next application of Parseval formula that is a maximum modulus theorem in complex analysis. This is one of the most important theorems in complex analysis and let me state the theorem first. Suppose f is a non-constant holomorphic function on a connected domain omega in the complex plane and mod f is bounded in omega and the supremum of f cannot be attained at any point of D. Supremum of f cannot be attained at any point of D. Assume that the maximum modulus is attained at a point in omega proved by contradiction. Assume the contrary that is you see what I written in red. So, we may assume that this point at which the mod f attains its supremum is at the origin take a closed disc D take a closed disc D of radius r centered at the origin and contained in omega. The power series for f converges absolutely and uniformly in D and so you see in the display the power series for f of z. f of z is a naught plus a 1 z plus a 2 z square plus da da da put z equal to r e to the power i theta a naught plus r e to the power i theta plus r squared e to the power 2 i theta plus da da little r varies from 0 to capital R theta varies from 0 to 2 pi of course and we recognize here a Fourier series for the 2 pi periodic function theta going to f of e to the power i theta. We see a Fourier series appearing here and then let us now fix an little r and let us use the Parseval formula for this particular Fourier series. It says for a fixed r with 0 less than r less than capital R we compute integral 0 to 2 pi mod f of r e to the power i theta the whole squared d theta. We simply expand this thing write mod f is f f bar write it as f f bar f is a naught plus a 1 r e to the power i theta plus etcetera write the corresponding series with complex conjugates and we pick up a m a n bar e to the power i n theta e to the power minus i m theta minus because of the complex conjugation and you integrate from 0 to 2 pi you take the integration under the summation term by term integration is valid because for a fixed r with 0 less than r less than or equal to r remember that the closed disk is contained in my domain of holomorphy omega. So, the series will converge uniformly and so I am allowed to take the integration inside the summation and the only term that survives is when n equal to m and so we get 2 pi times summation n from 0 to infinity mod a n squared I already answered this question how to justify the exchange of summation and integration. What we see in the displayed equation is exactly the Parseval formula for the function theta going to f of r e to the power i theta. Now by our very assumption that this function attains its maximum modulus at the origin but at the origin the value of the function is a naught so we get mod f of r e to the power i theta less than or equal to mod f of 0 there is mod a naught. So, we infer that 2 pi times summation n from 0 to infinity mod a n squared from the top display which is equal to integral 0 to 2 pi mod f of r e to the power i theta the whole squared d theta but this entire thing the integrand is less than or equal to a naught. So, it is less than or equal to 2 pi times mod a naught squared and so this forces a 1 to be 0, a 2 to be 0, a 3 to be 0 etcetera. So, the it means that the function f is constant on the disc of radius capital R centered at the origin the power series collapses to a constant. So, the function f is constant on the disc of radius capital R centered at the origin. Now how do we proceed from there? How do you how do you infer that the function f must be constant throughout omega? It is a simple connectedness argument that I will leave it to you because the connectedness argument that I am leaving out is not relevant for Fourier analysis but it is a routine argument that you have seen in a complex analysis courses several times. So, I will leave that part. So, that proves the maximum modulus theorem. So, I have given you now two applications of Parseval formula one to complex analysis and other to a famous geometrical problem. Now let us proceed a little further and let us look at some applications to partial differential equations. Let us look at the Laplace's equation on a disc. So, we apply the theory of Fourier series for studying some classical differential equations. The most basic differential equation is a Laplace's equation and we shall do it for the two dimensional Laplace's equation del 2 by del x squared plus del 2 by del y squared and we will work with the unit disc x squared plus y squared less than or equal to 1. The problem is to find a twice continuously differentiable function u such that Laplacian of u is 0 on the disc and the value of u on the boundary cos theta sin theta is f of theta where f is a 2 pi periodic function of theta. We are going to assume that f is lip sheets and 2 pi periodic. Why would you assume that f is lip sheets? We want the Fourier series to converge point wise. First let us write this Laplace's equation in polar coordinates. Here is a little exercise for you to derive equation 2.12. Del 2 by del x squared plus del 2 by del y squared is del 2 by del r squared plus 1 upon r del del r plus 1 upon r squared del 2 by del theta squared. This is a standard thing in elementary calculus so I will not derive this. So our PDE in polar coordinates become del 2 u by del r squared plus 1 upon r del u by del r plus 1 upon r squared del 2 u by del theta squared 2.13. So in this equation 2.13 we shall look for a solution of the form u of x y as a function of r and a function of theta. So we use the method of separation of variables where g theta is a 2 pi periodic function. So let us substitute this particular object in the differential equation 2.13. So u equal to v of r g of theta. So the first term will become v double prime r g theta. The second term del u by del r will be v prime r g theta. The third term will be v of r g double prime theta. So when you do that we get this equation r squared v double prime plus r v prime by v equal to minus g double prime by g. Left hand side is a function of r alone right hand side is a function of theta alone. So both sides must be constant and let us call this constant k squared. And so we get 2 ODE's r squared v double prime plus r v prime minus k squared v equal to 0 g double prime theta plus k squared g theta equal to 0. The first one is a Cauchy Euler equation which is familiar to you and its solutions are a r to the power k plus b r to the power minus k. And the second differential equation is the differential equation for harmonic oscillator g theta equal to c cosine k theta plus d sine k theta. So far so good. Now since g is a 2 pi periodic function we must have that k must be an integer because if k is not an integer the right hand side will not be periodic with period 2 pi. We want at least one or the c or the d to be non-zero because if both of them are 0 we are looking at the 0 solution which is not interesting. So in order to get 2 pi periodic solutions this k must necessarily be an integer and the solution is continuous at the origin and so this k must be non-negative and then the b must be 0. So we get so this b must be 0 and so this v r into g theta what is v r into g theta r to the power k into Cauchy k theta there is a c here sine k theta there is a d and this a can be clubbed with a c and the a can be clubbed with a d you can call ac as c theta ad as d theta there is no need to put the d theta and c theta again there is some other constant I call the constant a k and b k if you like. Now you have got a special solution what is the special solution r to the power k ak cos k theta r to the power k bk sine k theta. Now you can have more general solutions by taking linear combinations or super positions of course when k is 0 that also has to be considered the k is 0 means it is a constant if you get the a naught term. So the most general solution is displayed here as equation 2.14 the last displayed in the slide. So u of r cos theta r sine theta equal to a naught plus summation n from 1 to infinity r to the power n a n cos n theta plus b n sine n theta. Now we got the problem of computing a naught a n and b n how to do that we simply put r equal to 1 simply put r equal to 1 and recall that when r equal to 1 u of cos theta sine theta is f of theta. Now you have to bring in the Fourier series so so we put r equal to 1 and we get f of theta equal to a naught plus summation n from 1 to infinity a n cos n theta plus b n sine n theta that is displayed 2.15 in the slide and from this equation we can at once deduce a a naught a 1 etcetera and b b 1 b 2 etcetera. Simple a naught is simply what 1 upon 2 pi integral f of theta d theta from minus pi to pi what is a n 1 upon pi integral minus pi to pi f of theta cos theta d theta etcetera the the formulas for the Fourier coefficients. So the a naught so the a's and the b's the coefficients can be determined uniquely. So here are some exercises simple instance f of theta is mod sine theta f of theta is mod sine theta it is obviously lip sheets. Now if I take the function to be lip sheets then we know that the Fourier series will actually converge in fact we can even prove that the series will converge uniformly. So mod sine theta is certainly a lip sheets function and we can calculate the a's and the b's of course the b will be 0 because that is an even function so the sine term will not be there. Another exercise show that if u is harmonic then the average value 1 upon 2 pi integral minus pi to pi u of r cos theta r sine theta d theta is u of 0 0 the mean value of the function along a circle you are integrating u along a circle r cos theta r sine theta and you are dividing by 1 upon 2 pi and you are going to get the value at the center. This is called the mean value theorem for harmonic functions simply integrate 2.15 term by term you will get the result. So we have proved two important theorems. So let us continue with the formula that we obtained in the last slide. So now we got u of r e to the power i theta we got a naught plus summation n from 1 to infinity r to the power n a n cos n theta plus b n sine n theta but we know the formula for these coefficients. And let us put the formulas for these coefficients and we will get u of r e to the power i theta equal to 1 upon 2 pi integral minus pi to pi f of t dt plus 1 upon pi summation n from 1 to infinity r to the power n cos n theta times what is the formula for a n minus pi to pi integral f of t cos n t dt. Similarly plus sine n theta integral minus pi to pi f of t sine n t dt. Now what we are going to do is that we are going to club these two integrals together and we are going to bring the integration outside the summation. So the integrals decay to 0 by Riemann Lebesgue Lemma and by Riemann Lebesgue Lemma's and the coefficients are decaying to 0 and R is less than 1 it is easy to justify the exchange of summation and integration. So when you do the exchange of summation and integration what you are going to get is outside you got 1 upon 2 pi integral minus pi to pi and everything else is inside the integral sign all the terms are f t in common. So f t dt so what is the other factor 1 plus twice summation n from 1 to infinity r to the power n cos n theta cos n t plus sine n theta sine n t okay. Now cos a cos b plus sine a sine b is cos of a minus b as you all know. We use that formula and we get what we get 1 plus twice r to the power n cos ns is 1 minus r squared upon 1 plus r squared minus 2r cosine s the middle exercise. How do you do this exercise 1 plus summation n from 1 to infinity to r to the power n cos nx? How do you find the sum of that? Write cos ns as 1 half of e to the power i ns plus e to the power minus i ns. What you get are simply two geometric series to individual geometric series and r is strictly less than 1. So these two geometric series individually converge write down the sums of those two geometric series do the simple algebra and you get this 1 minus r squared by 1 plus r squared minus 2r cosine s. It is a very easy exercise you please do it and we get the result u r e to the power i theta equal to 1 upon 2 pi integral minus pi 2 pi 1 minus r squared upon 1 plus r squared minus 2r cos theta minus t and there is an ft dt of course 2.18 this expression 1 minus r squared upon 1 plus r squared minus 2r cos theta minus t is called the Poisson kernel with a 1 upon 2 pi thrown in it is called the Poisson kernel. So this Poisson kernel is going to play a very important role in what is going to come. We are going to use this to discuss Abel's summability and many other things. So the solution of our problem so the solution to our problem 2.11 has been expressed as an integral. Remember right at the first lecture I mentioned to you that integral representations were preferred in classical in classical mathematics it is it is preferred because it is easy to do estimates or using integrals. So the solution to this Laplace's equation with boundary conditions f of theta can be written as a integral with respect to the Poisson kernel that solution has been expressed as equation 2.18. I think it is a good time to stop this lecture here and we will continue with this in the next capsule. Thank you very much.