 Hello. So, let us continue from where we left off last time. Let us recall that last time we were discussing the Aries function. What is the Aries function? The Aries function is a solution of the Aries differential equation y double prime minus x y equal to 0. Now we took the Fourier transform of this differential equation and we got a first order ODE and we solved this first order ODE and the solution was exponential of i chi cube and we applied formally the inversion theorem and we got the function y of x equal to integral over r x of i x chi plus i chi cube. The problem with all this derivation was that it was completely formal. We do not know that the Aries differential equation admits a solution which can be Fourier transformed. We do not know whether the Fourier transform y hat which we obtained as x of i chi cube in what sense can we take the inverse transform because exponential of i chi cube does not decay or anything. So, there are certain issues with this derivation, but the positive aspect of it is that we got an integral. We got an integral representation y of x equal to integral over r cosine x chi plus chi cube dx. Let us get to the slides 4.18 that you see in the slide. So, this equation 4.18 is supposed to give us a solution of the Aries equation, but the derivation of 4.18 was suspect. There is one way out of it. One possible option is to take 4.18 and directly check that it satisfies the Aries equation by differentiation. But there again there is a certain problem because if you try to differentiate 4.18 with respect to x, y prime of x will be integral over r minus sin x chi plus chi cube and a chi will be produced. y double prime will be minus integral over r cosine x chi plus chi cube times chi squared. We have picked up factor of chi and chi squared in two successive derivations and the integrand not only is not bounded anymore, it is actually becoming infinite. So, it is oscillating with large amplitudes chi squared. So, direct verification is going to be problematic. So, let us make a transformation. First let us discuss whether the integral even converges conditionally. Integral 4.18 it is not at all clear that even this integral even conditionally convergent. First we will prove that it is conditionally convergent. After that we will make a transformation and convert this conditionally convergent integral 4.18 into an absolutely convergent integral and we will work with a transformed integral and check that that transformed integral satisfies the Aries equation and that is the plan of today's lecture. Okay. So, first let us make a change of variables x chi plus chi cube equal to s. The function x chi plus chi cube as a function of chi is monotone increasing beyond a certain stage or infinity. It is not monotone on the entire real line and why do we need monotonicity because when we make a change of variables the change of variables must be a bijective mapping. So, we want it to be strictly increasing or strictly decreasing here because of chi cube we expect it to be strictly increasing and it will be increasing beyond a certain stage r to infinity. The corresponding value of s will be capital R. When you make a change of variables in this integral we are going to put x chi plus chi cube equal to s and that is how we are going to cause s and x plus 3 chi squared d chi will be ds. So, d chi will be ds upon x plus 3 chi squared that is 4.19. Now, we have the unpleasant task of transforming this into a integrand involving s alone solving this cubic for chi and substituting it here is going to be a difficult task and it is actually not necessary. First let us make some simple observations. Let us look at this equation x chi plus chi cube equal to s on this large interval r to infinity. As chi goes to infinity s must also go to infinity and conversely if s goes to infinity then chi must go to infinity. That is the first little observation. So, now let us do the following. Let us try to integrate by parts in 4.19. So, let us write down cos of s equal to the derivative dds of sin of s with a minus sign and throw the derivative on the other factor and we have to apply the chain rule to differentiate 1 upon x plus 3 chi squared. You differentiate with respect to s you are going to pick a 1 upon x plus 3 chi squared the whole squared into d chi by ds. Now, of course, when you apply the chain rule you will pick up a 6 chi s also. d chi by ds you will have to calculate using the equation connecting s and chi. But now we will not need to do any further simplifications because we must understand the behaviour of this chi upon x plus 3 chi squared the whole cube. I have suppressed the boundary terms that arise from integration by parts. You can figure out what happens to the boundary terms because it is this integral that we are concerned with and we want to show that this integral 4.20 is convergent. So, as I said we have the unpleasant job of expressing chi by x plus 3 chi squared the whole cube in in terms of s. But that would be needed if you want an exact value of the integral we do not want an exact value of the integral we only want that the integral converges. And now what happens is that as chi becomes very large s upon chi cube from this equation we can see that s upon chi cube must go to 1 as chi goes to infinity. So, for all large values of chi s upon chi cube must lie between half and 2 for instance. And I simply use this inequality and raise it to the power 2 by 3 and I get another inequality s to the power 2 thirds must be sandwiched between c chi squared and chi squared by c where c is a certain positive constant. So, now we get 3 chi squared plus x that must be bigger than 3 by c s to the power 2 thirds plus x. But note that if chi is very large then s will also be very large and x is fixed. So, that 3 upon c s to the power 2 thirds plus x is ultimately going to be much larger than say s to the power half because 2 thirds is bigger than half and s is going to infinity. So, we get this estimate chi upon 3 chi squared plus x the whole cube will be less than c times s to the power 1 third upon s to the power 2 thirds for instance and that is basically c times s to the power minus 7 by 6. And this estimates suffices to show that this integral here converges absolutely chi by x plus 3 chi squared the whole cube is dominated by s to the power minus 7 by 6. And since 7 by 6 is bigger than 1 1 upon s to the power 7 by 6 integral converges and so this integral here in 4.20 converges absolutely. So, we have completed the proof that the integral 4.18 is a conditionally convergent integral. Now, that would not suffice for us we need to go a little further and we will have to now transform that integral into an absolutely convergent integral. In other words we must shift the contour of integration into the complex domain this idea of shifting the contour of integration into the complex domain is a very frequently used idea in theory of differential equations. So, now let us start with this entire function f of z equal to x of i x z plus i z cube the chi has been replaced by z chi was real z is complex this is an entire function Cauchy's theorem will tell you that the integral of an entire function over a rectangle is going to be 0. So, let us choose the rectangle to have vertices minus r r r plus i eta minus r plus i eta. So, the base of the rectangle is the interval minus r to r. So, along the base of the rectangle the integral f z d z as r goes to infinity will give you the integral that you want ok. Then you have got to look at the vertical side from r to r plus i eta the contribution from that vertical side will go to 0 and then we must look at the contribution from the other vertical side minus r to minus r plus i eta that contribution will also go to 0. I am going to leave this verification for you as an exercise and now we will have to look at what happens on the top of the rectangle from minus r plus i eta to r plus i eta the top edge of the rectangle let us call it l 2 the base of the rectangle is called l 1 Cauchy's theorem has to be applied integral over l 1 plus integral over l 2 plus integral over v 1 plus integral over v 2 is 0 these integrals along the vertical sides go to 0 integral along l 1 gives you the integral that you want an integral along l 2 is what we are going to now find out and we are going to finally, get that integral over l 1 f z d z is minus integral over l 2 f z d z I am going to allow the r to go to infinity. So, integral that we want is going to be written as limit as r goes to infinity integral f z d z along l 2 it is this transformed integral that is going to be absolutely convergent. So, let us now analyze this integral along l 2. So, along l 2 z is what z is chi plus i eta where chi varies from minus r to r and eta is fixed. Let us estimate this absolute value of x of i x z plus i z cube it is going to be put z equal to chi plus i eta x is real remember we are going to get x of minus x eta plus eta cube minus 3 chi squared eta that is the real part and then the imaginary part of course, x of i c is going to be unit complex number when c is a real number. So, what is the estimate mod integral over l 2 of x of i x z plus i z cube d z that is going to be less than or equal to this factor e to the power minus x eta plus eta cube that does not depend on chi. So, that comes out an integral from minus r to r e to the power minus 3 chi squared eta d chi remember that eta is positive and chi is real varying between minus r and r. So, this integral certainly converges and so, this integral over l 2 is a nice absolutely convergent integral and so, let us write that down. So, finally, in the limit as r tends to infinity these two vertical pieces go to 0 the integral over l 1 gives the integral that you want and integral over l 2 will be what we are going to write down on the right hand side of this displayed equation in detail that is over here integral minus infinity to infinity x of i x chi minus x eta plus i times chi plus i eta the whole cube d chi. The left hand side is of course, the integral 4.18 that is the solution y x which we are going to check that is a solution not yet a solution. We have heuristically arrived at that y of x by taking Fourier transforms. Now that we know that the y of x that we got is exactly this expression that is displayed on the right hand side. So, the integral on the right hand side is going to be absolutely convergent and we can differentiate under the integral sign in a number of times that we want e to the power minus x eta does not depend on chi at all it comes out of the integral. Here let us look at the real part of this integral e to the power i times 3 chi squared i eta and. So, I get i squared which is minus 1 and I get chi squared eta. So, that is exactly causes the integral to be absolutely convergent. So, I get an e to the power minus 3 chi squared eta that is the term that is going to help us. When you differentiate with respect to x when you differentiate with respect to x I am going to pick up factors like a chi because of e to the power i x chi here for instance I am going to pick up a eta here, but eta is going to be constant. So, though I pick up a factor of chi I have e to the power minus 3 chi squared eta that is going to help us to cope with the integral. So, we can differentiate under the integral sign without any problem. So, let us differentiate under the integral sign twice with respect to x and let us calculate y double prime x minus x y by 3 that happens to be minus 1 upon 3 i integral minus infinity to infinity we get 3 i z squared plus i x x of i x z plus i z cube this chi plus i eta combination that appears here and it appears here has been replaced by z and this expression that we see inside we can write it as d d chi of x of i x z plus i z cube d chi. Now this integral is going to be 0 by the fundamental theorem of calculus and the rapid decay of the integrand as you go to infinity along the chi direction. So, that shows that y double prime minus x y by 3 does indeed turn out to be 0 in other words y of x is indeed a solution of the Aries equation. So, the integral representation that we got gives you a valid solution of the Aries equation and as far as estimates are concerned you can use this transformed integral to get your estimates on the Aries function. I think we will close the discussion of Aries function here with just one small comment it is elementary to obtain Aries equations as a power series solution, but the power series solutions are easy to get they lend themselves to algebraic manipulations such as term by term differentiations and multiplication and stuff like that it is very difficult to get information concerning the asymptotic behavior zeros and such integral representations are much better. In fact, the integral representation of the Aries function gives you the asymptotic behavior of the Aries function we would not get into that and that is important in optics. Now, let us take up a few odds and ends before we take the next topic in this chapter on Fourier transforms the Riemann-Lebesgue-Lehmann revisited. So, let us start with an L 1 function we know that the Fourier transform of f decays to 0 as chi goes to plus infinity and chi goes to minus infinity that is the Riemann-Lebesgue-Lehmann. However, I stated theorem 52 I said Riemann-Lebesgue-Lehmann revisited not only is the Fourier transform decaying to 0 as chi goes to plus minus infinity the Fourier transform is actually a continuous function indeed it is uniformly continuous and bounded the proof will not be given in detail I will give it as a guided exercise. First of all f is in L 1 moment you say that f is in L 1 it means outside an interval the contribution of mod f x as an integral is vanishingly small in other words there exists an a bigger than 0 such that integral of mod f x dx outside the interval minus a a is less than epsilon by 4 where epsilon greater than 0 is arbitrarily chosen. So, now let us write down the difference f hat of chi minus f hat of eta I want to discuss the absolute value of this difference. So, let us write the integral of f hat of chi and f hat of eta and the integral is over mod x less than or equal to a plus another integral mod x greater than or equal to a the two integrals are displayed over here. Now let us do the following let us look at the second integral and let us take the absolute value and take the absolute value inside the integral you strangle inequality they are both unit complex numbers and so this modulus is less than or equal to 2. So, that is an innocent constant it comes out and then I am left with mod f x dx and that is less than or equal to epsilon by 4 the integral. So, I get the contribution from the second piece in all is less than epsilon by 2. Now let us concentrate on the first piece suppose I use the mean value theorem let us call this function g of t. So, g of chi minus g of eta right you got a function evaluated at chi and the same function evaluated eta mean value theorem will give you that the difference will be equal to chi minus eta times the derivative when you differentiate I am going to pick up a minus i x i is an innocent constant and x is a variable, but mod x is less than or equal to a. So, the x that you pick up is bounded by a and then we get the derivative will be e to the power minus i x times theta for some theta between chi and eta, but that again is going to be unit complex number. So, when you try to estimate the first piece you are going to get a factor because you are pulled out an x while differentiating and then you got this mod f x dx and you also have a chi minus eta. So, you must choose the delta. So, that delta should be less than epsilon by 4 times a then your chi minus eta will be less than epsilon by 4 times the a that you picked up will get cancelled out and you will be able to show that the first piece is also less than epsilon by 2. So, we have proved that given any epsilon greater than 0, there is a delta greater than 0 and I told you what the delta is just now. So, that the mod of f hat chi minus f hat of eta is less than epsilon as soon as chi minus eta is less than delta in absolute value that proves that the function is uniformly continuous. Now, let us take the series of exercises use Parseval formula to compute the integral from minus infinity to infinity 4 sin squared chi d chi upon chi squared. What is the Parseval formula give you the Parseval formula gives you this beautiful equation the L2 norm of f squared and the L2 norm of f hat squared are related just by multiplication by 1 upon 2 pi. Now, if you know a function whose Fourier transform is 2 sin chi by chi. If you look at the list of Fourier transforms that we have calculated you will find that there is a function whose Fourier transform is exactly 2 sin chi by chi. So, this integral that you see the displayed integral is exactly norm f hat squared and that is the integral that you are trying to calculate the original function was much simpler what was it it was nothing but the characteristic function of the interval. So, you just have to understand what is the integral of f squared where f is the characteristic function of the interval and you use the Parseval formula and you calculate this integral that shows the use of Parseval formula. The Parseval formula as many uses in the next capsule we will prove the equipartitioning of energy in the wave equation and again use the Parseval formula. Then a couple of exercises on convolution use the convolution theorem to determine the convolution f s star f of t where f s is the Cauchy distribution what is what is the formula for f s s upon chi squared plus s square to the 1 upon pi thrown in s is a parameter s is a parameter and chi is the variable of the function. So, this 1 upon pi is a factor which is a convenience factor because when you integrate this f of s over the real line you will get 1. So, this is called the Cauchy distribution in probability theory, but this Cauchy distribution turns out to be the Fourier transform of some known function capital f s. Again you should go back and look at the list of Fourier transforms in the early part of this chapter and you will be able to find which function capital f is it whose Fourier transform is this Cauchy distribution. If you know that then this convolution is easy to calculate if without that if you try to calculate the convolution it is going to be a messy integration, but instead of calculating the convolution directly from the definition try to calculate the Fourier transform of the convolution and then once you get the Fourier transform of the convolution then you try to use the inversion theorem that approach may be much simpler than directly computing the convolution. The next exercise is another Fourier transform computation Cauchy ax is a positive real number and we know that 1 upon Cauchy ax is a rapidly decreasing function it is an element of the Schwarz class. It is an element of the Schwarz class. So, how to calculate the Fourier transform is there a way to calculate the Fourier transform? Yes there is a way to calculate the Fourier transform Cauchy ax is an even function 1 upon Cauchy ax is an even function. So, when you write down the definition of the Fourier transform only the Cauchy ax upon Cauchy ax term will survive the sin Cauchy ax upon Cauchy ax will disappear it is an odd function. Again you must use complex analysis to do this. Usually when you have a hyperbolic function in the denominator what you will do is that you will employ a rectangular contour. Again the rectangular contour will have base as the interval from minus r r and the contribution from the vertical sides r to r plus i t minus r to minus r plus i t the two vertical sides the contribution will go to 0 and integral along the base is the integral that you want as r goes to infinity. What happens to the integral along the top edge of the rectangle? What is the top edge of the rectangle? How would you choose this t in such a way that at the top along the top edge you get a multiple or the integral that you want. So, along the top edge what is z? z is x plus i t where x goes from minus r to r. So, when you put Cauchy of a into x plus i t you must get a multiple of Cauchy it the t must be what? The t must be pi by a in that case you will get Cauchy of ax plus i pi Cauchy of ax plus i pi is minus Cauchy ax and you will get that the integral along the top of the rectangle is a multiple of the original rectangle. I think it is a good place to stop this lecture here we will continue in the next capsule. Thank you very much.