 So, let us begin the last lecture. I thought I will call it the last lecture. So, I suppose you have had opportunity to look at the overall syllabus. My confession is, I still have not made the question paper. I would make it as late as possible, maybe during my flight tomorrow. I will have some time to think about, but today let us discuss about whatever residual doubts that we may have. Well, you know this title that you see here the last lecture is a book. I do not know how many of you have heard of this book called the last lecture. Any of you know? There was a professor of computer science in Carnegie Mellon University. So, he was diagnosed with pancreatic cancer. So, he decided to retire, resign and take his family, so that they could settle after he dies to some other city. So, the Carnegie Mellon University has a tradition of inviting some of the retiring faculty to come and give a so called last lecture. They pass on the information to everybody in the campus and it is quite popular. So, this professor's name was Randy Poush. So, Randy Poush gave a lecture and it became quite well known all over the world. You may actually even see some book stall here. You should be able to pick that book up. He eventually died, but he gave a very moving lecture. So, that the title of that book is called the last lecture. So, as I said that I will tell you about few things which are not related to academics today because we have finished this level. I am more here to answer your residual queries and any comments that you would like to make. I hope I did not make the course difficult, but if I have, so I said this is my excuse. So, I fell back on the philosophers of the past. Kierkegaard, you know the Danish philosopher. He was extremely brilliant in articulating himself. So, he said that if the task is difficult, there is a reason for it because he wants only the noble hearted people to come out. So, I am happy to see there are at least 40 noble hearted people, population in the IIT Kanpur, those who have brave the whole semester here with me. Well, the idea is not to really make anything easy or difficult, but there are certain things that I wanted to tell you. I wanted to tell you something about one path which was asked by one of the students. She is not here, but still I should explain to you. This the transfer function for the filters that we discussed about. See, it has a real path and it has an imaginary path. I think she is here. She asked me the question about these two points that what does the imaginary part of transfer function do. If you look at the way we have defined transfer function, transfer function after filtering, if I call that as G, which is defined in the K plane, would be equal to G of the basic numerical method times the transfer function. So, let us call that as T. That is also a function of K. Now, please do understand that G as well as T, both could be complex. So, if there are complex, then you can see how you can define this. We are more interested in what? We are more interested in mod G to begin with. To make it neutrally stable, we want mod G to be equal to 1. So, that is the whole idea. So, if I take a mod G, then it should be like this. If G and T are complex, you have to see what will happen. Now, you can see that imaginary part does play a role in deciding whether it is going to be stable or unstable or not. So, whether these two statements are definitive or not depends on the transfer function and depends on G real and G imaginary, because you will be able to work it out yourself and see what is the contribution of this imaginary part of this transfer function. But if you look at each should be strictly real, then you know what this is. G hat of real would be just simply nothing but G of real times that T real. So, we are talking about when T imaginary is identical equal to 0. This is what we get and G hat of imaginary would be simply G of imaginary times T of real. This is the first part of the task. The next thing that you really want to find out is the propagation property of disturbances and that is given by beta. Well, we do not need to really even say J because it is nothing but minus tan inverse of G imaginary by G real. Now, if we are talking about the filtered quantity that will be that and now since in this case for strictly real filter with central filter T imaginary is 0, then you can see this is nothing but equal to beta itself. So, dispersion relation does not change. So, what happens is when you have a purely central filter, then you can see that it does change the numerical amplification factor, but dispersion property unaltered. That was one of the reason that we had said that we would like to use those list ordered centered schemes so that we always retain central attribute of the filter so that we can we basically do not tamper with the dispersion property. So, this is one thing that I wanted to tell you. If there are any other questions on filters, you could ask now. Anybody else has any other question? I thought I would clarify the function. Let us see we have this you are asking about this. The figure that we have here that is without a filter what is this? See basically why do you need filter? When filters were originally introduced in late 90s, mid 90s it was thought that it will take care of numerical instability problem alone. So, it should not actually tamper with the physics of the problem. So, you really want to have a filter working for you in that mode where it will take care of any numerical instability while being not intruding upon the physics of the problem. It is for that reason that we talked about using the filters which will actually affect more on high wave numbers. Say I suppose this is quite revealing you can see that if you compare a second order filter with a sixth order filter, sixth order filter leaves almost up to this value of k h unaltered. So, basically that figure that you saw was something that would tell you that we have computed a case which goes on up to at least this time. This was shown at a time t equal to 47, but you also see that here that without filter solution actually breaks down here. It just drops off vertically and then it stops. So, without filter you cannot basically you cannot compute indefinitely. For the reason is this big vortex that you are seeing here that this starts growing and this creates a very large gradient, vorticity gradient. Large vorticity gradient means you have a very wide range of k's excited and you also know that as you are going away from the surface of the cylinder the grid becomes sparse. It is rarified. So, what happens is your ability to resolve k also degrades as you go away. So, if this vortex was somewhere here probably you would not have any difficulty. This breakdown in this case the solid line breakdown is because of this large vorticity gradient which is in a path of the domain where the grids are not fine enough. So, it is a kind of a numerical instability problem that you are seeing here. So, a sixth order filter would like to rectify the situation with in this case it did momentarily that is the last line the dotted line and you could see it went up to the trough of that curve, but then again it broke down. So, what happened is it could delay it could delay whereas, if you apply the second order filter well it shows you as if it can go on forever, but at what cost? The cost is that this vortex has been completely diffused. This contour value is probably an order or two lower than this vortices that you are seeing here. So, what happens is in a situation scenario like this you would be able to go on for a long long time because you have removed numerical instability, but at the same time you may have also tampered with the physics. That is what we saw in this figure something like this a second order filter starts altering the spectrum right from very small value of k. So, that figure that you saw there the sixth order filter and the unfiltered case look similar because the sixth order filter kept on removing high wave number values of k and that is why it could continue for little longer, but eventually that also was not adequate to take care. Yes, any other questions? Direction of the filters that also you can make out that in this particular case you have two directions to know. One is of course the radial direction and the other one is the azimuthal direction. One good thing about the azimuthal direction is in this direction the geometry is periodic. So, you could actually get by using the central stencil for all the points. So, you do not need any closure or anything. So, it would appear as if the azimuthal filter would be preferable compared to the radial filter right because in the radial filter near the boundary I have to create an additional filtering if I want to have a uniformly higher order filter. See, if I have a second order filter I do not have to do anything. I could straight away start doing the filter from the point two onwards. So, second order filter is no issue, but if I want to have a fourth order filter then at j equal to 2 I cannot apply. So, I will have to use some closure. So, that is one thing that you have to realize that the filtering that we are performing. We can do either in the theta direction or in the r direction or we can do both. In fact, some people tend to think that well it is now that is what we also see that look this is a attribute of a azimuthal filter. What did it do? It actually elongated the vortex in the theta direction. While it was a actually circular vortex it has become elongated in the theta direction. So, to counteract this tendency some people would tell you that you not only apply a filter in the theta direction you also apply it in r direction, but what you are doing? The very act of filtering it once itself has reduced the strength of the vortex. If you would have also added in the r direction you would have even more loss of the signal. So, in fact I did not want to discuss going on in the lab. We have been working and I think I just finished looking at the work. We have developed a two-dimensional filter now, multi-dimensional filter. So, instead of doing it once in theta direction and then in the r direction or vice versa we will just simply apply the filter only once. So, there are possibilities we have just worked it out and we hope that it would be acceptable by the scientific community. So, that is the question of directionality, but you also know that when we actually filter let us say in two-dimension, let us say this is k 1, h 1 and this is k 2, h 2 then again what I would be doing is I would be working in a space like this and if you recall in designing the filter what did we do? We actually use the consistency condition plus additional condition depending on the order. Furthermore, to close the system we also prescribe the transfer function at some point. In one dimension what was it that we did at the Nyquist limit we put the transfer function equal to 0 that gave us the additional equation. So, in two-dimension also you may have to do some. How do you do it? That is the thing. This is let us say in say one of the direction it goes from 0 to say k 2 max and here it goes to 0 to k 1 max. So, I could make the transfer function 0 here, 0 here and then work out the additional. So, there are certain things, directionality of the filters are a very live issue and in recent times people are using filters in a much more creative and imaginative way that you can use the filter to do what is called as a large eddy simulation. I have before also large eddy simulation basically tells you the following that if you are solving a flow problem let us say and if energy spectrum then it say typical energy spectrum would look like this. So, that means you have to resolve satisfactorily all the way to where the energy decays to 0. But suppose your computing power is limited you can only resolve up to this satisfactorily then what you would like to do? You would like to put in some kind of a artificial sink here, sink of energy what it would do is it will try to reproduce the energy spectrum as it is up to the result scale and then it would decay. So, then you will go back and tell your peer group that look I have been able to resolve the energy spectrum up to some range and you should trust only this part of the result. These are numerical attributes and this is what is called as a large eddy simulation you may have heard of that is what people call LES. This is a very standard technique for it is not really standard in the sense it is up and coming people are investing lot of thought to this. So, filters actually can be used for large eddy simulation because I can now see that if this is my original energy spectrum and then I pass the spectrum through a filter to force it to 0 here. So, that is the your Nyquist limit for your calculation. So, that is what this is where filters are becoming very very important I mean we are doing some of this work here also. So, does that answer your question of the directionality of filters this one next one. What we are showing this? This is the experimental data these are experimental flow visualization data by these are these are basically vortices these are vortices that we are seeing. So, this cylinder is doing this. So, you see this is a this appears to be a very simple geometry, but the flow here is very very complex because if this is oscillating in a rotary manner then for some time the flow on the surface you are introducing a sort of a induced velocity which is against the outer flow. The outer flow is going from let us say left to right. So, what happens is you are doing something like your separation I do not know if you are familiar with you may have done a course on fluid mechanics if you have done you may have heard of boundary layer separation. So, here let us say when this is going in the anticlockwise direction the top part what will do? We are actually inducing a separation because the flow wants to go from left to right and we are imposing a motion which takes it from right to left. So, flow actually separates and that causes this vortex to be released. If you look at your book you will find that if this was not rotating it was stationary then this separation point would be here past the midnight it should be about 108 degrees to 130, 40 degrees in that range, but here it is almost happening 90 degrees or even before you see the separation begins pretty early. So, what happens is half the cycle this releases and another half of the cycle this vortices are released. So, these are this looks like your Vodkarman vortex tree where you have a set of vortices of one side sign on top the other sign on the bottom. So, this is your experimental data this is what we have computed for this case the Reynolds number is pretty low 150 that is what the experimental data is we have also computed it for much higher Reynolds number, but this is perhaps more feasible because Reynolds number is low flow is two dimensional the calculation method is two dimensional. This basically tells you the amplitude of oscillation and this last factor is basically the frequency of shedding that is more than what it is for the normal shedding of a stationary cylinder. So, we could do a calculation without using any filter we get this then what we have done what we have done here is basically we have done the same calculation, but now we have used a filter which is sixth order it has been done in two phase we have used a kind of a azimuthal filter only for the 30 lines. Now you know why you did it for 30 lines we did not want to do it for full domain then this vortices would have been smeared in the theta direction we wanted to filter this azimuthal direction only very close to the surface. So, it is almost like only working inside the boundary layer that is what we did and while we have used a sixth order radial filter for the full domain. This was done specifically to show that your filter should not really again interfere with the physical nature of the flow that is why you will find that these two pictures are completely indistinguishable. So, that may mean that let us say if you get by taking let us say few 500 points by 500 points in the domain with the filter I should be able to do it in fewer times. So, that you can use the filter for now I never said that I said it will not do anything provided you use central filter where t imaginary is 0. If you have t imaginary nonzero then it changes the dispersion relation all. I completely did not comprehend what you said yes fine attenuate yes right. Convection equation is non dispersive because your group velocity is equal to the phase speed. Say any other convection equation other than linear this is what you are seeing here what you just now said that equation is there. So, this is the central filter this path and this is the upwind path please allow me to correct this. This is not a fourth order term this is a fifth order term. So, this is once again needs correction here. So, this is not a fourth order term it is basically a fifth order term. So, it actually alters some introduces t imaginary see by itself if I did not have this last term it would be central. So, I will have t imaginary 0, but by adding this asymmetric filtering here we have actually brought in t imaginary that is probably is shown here yeah this is the t imaginary path and it is very interesting that you can this is a fifth order filter. So, it modifies your j equal to 5 to n minus 4 point along this solid line right. So, what it does actually done it introduces a t imaginary changes the dispersion relation and this is what we seen this is this is what we see. Suppose I do not add that path this is my group velocity contours in k h n c plane and we can see this is that border line for which v g is 0. So, above which we have a spurious upstream propagating waves and this is almost like the rotary oscillation case you see the signal wants to go let us say from left to right, but numerically you are inducing something which is going from right to left and this could be a disaster if you are solving it flow equation like what we do in fluid dynamic then you do not really want this is a bad attribute of this method. So, what we did we added that upwinded path of the filter with the coefficient as your additional degree of freedom we have small quantum of that fifth derivative term and what we find that this line actually does not intersect the k h plane it actually rolls over. So, you have a strip means a very small range of n c here that you to circumvent that quave formation. So, shall we go over to something else if you wish to discuss anything here is Mansi you have some question. Now, Paul decide for yourself I thought if you may have some question on error analysis or compact scheme we could discuss and as I told you the begin error analysis you want me to open that page. This is the last slide, you have to read the paper we have defined there you see what happens here that will actually it is not difficult for you to work it out if you read the paper this is where it would be. See if I have a discontinuous solution well may be I should that would half if I look at this what drives error these are the three essential ingredients. One is the phase error path that means if c n is not equal to c then this will contribute but that is also dependent upon the solution gradient. So, this is this part that is due to both phase error as well as solution discontinuity if there are this will come into play. So, the second term we have talked about this is due to dispersion error. So, if you are v g n and c n are not coincident then you will get this term and this is our stability path. So, if you now go back to that picture we have purposely taken this conduction of this ramp because we wanted to create a solution discontinuity here at the foot and at the shoulder of this ramp and that is where you see from both this region errors actually appear. So, what happens this error appears in the sense that 1 minus c n by c into del u and del x provides the seed and once that provides the seed across a whole range of k then your dispersion error also comes into picture. So, these two peaks actually correspond to those two sources of error one corresponds to phase error another corresponds to dispersion error. But simply looking at in the physical plane you do not get the feel of it but if you go to the spectral plane k plane then you can see that these are two distinct peaks those are caused by those peaks. Yes, how do you get from there that well I do not have a ready made answer for you I do not think we can analytically show it it has to be worked out you have to compute it and plot it and check for yourself. But for sure if you can at least show it for explicit schemes see the problem I am not venturing to give a ready made answer to you is because c is itself A inverse B. So, if I at least take a explicit scheme A is identity. So, by looking at B matrix you can work it out. So, the B matrix is if it is symmetric then you should be able to show that if it is symmetric you will not have any imaginary contribution coming from that product. You followed what I said see basically you are saying that i k equivalent is summation of C l j and the C k l going from 1 to n. Now, C by itself is nothing but A inverse B. So, if I look at explicit schemes for that A is the identity matrix and then what you are going to get is i k equivalent would be simply equal to well what is P j l it is nothing but e to the power i k x l minus x j. Now, if I take this as to be a symmetric matrix and then I take a product of this you will see the imaginary part will be identity. For long because if this is i k equivalent you have to see the correspondingly what you have to see. So, this you can take it out now you can draw your conclusion. So, the real part will turn out to be 0 you will get that i sin k h by k type of term. If you think of let us say C d 2 that is what you would get well any other questions on this then I would like to close here by giving some of my thoughts to you. I have spent almost 40 hours in this room with you. So, I wanted to make the subject as simple as possible if it is not these are the excuses it must be made difficult you do not just do not want to sell it through the local pawn shop. It has to be a has to have some class it is true, but we try to understand what it is and I am sure if we claim I it is all like Harvard and Cambridge of India we should be able to do it. We should have the confidence to surmount the difficulties and enjoy the glory. So, these are the fellows of speaking all these are some common sense things. Let me make some clear observations these are taken mostly from Tom Paine you know the person who is responsible for I think one of the author of American constitution Tom Paine was great thinker. So, he actually talked about certain things about science. So, I thought they will be of interest to you every science has for its basis a system of principles has fixed whether you are there or not the science is there. So, it does not depend on the observer with the apology to the quantum mechanics guys we say the observer is not important here the things will go on happening. So, we cannot make principles we can only find it out that is what we do we whether your error equation is right or wrong we do not even fable about it it is there when you compute you see it is there. So, there are people who quibble with us I said to them this that we are not discovery we are just simply pointing out that this is possible right. Another thing is this modern tendency to Chalta high attitude a moderately good thing is ok lots of you actually spend a lifestyle which is reflective of that, but that is not what he said it is not really good enough. The third thing relates to this course is you know I have not only this time many a times I have students come up to me and say oh you say this Von Neumann analysis is wrong we have been taught all this years and that we have all along thought it is right. So, that is what it is you know it is a habit forming if you thinking something is right it appears to be right, but it is not right. So, that is the issue that a long habit of thinking thing is wrong if it not thinking it is wrong gives it a superficial appearance. So, I suppose all I am saying is that be reasonable be analytic do not accept anything just because I stand here and I say something and you will have to accept that that is right. I always encouraged I always invited you to ask probing questions right. So, that is that is what it will be a reasonable thing to do. Now, talking about algorithms right what we do in science it is trying to do something new right. So, this is how many of you know of this author no I think it is probably from the first one what is that breaking tipping point I think it is from the tipping point or may be in one of his article in New Yorker I do not know where it is, but he said it that you cannot really say how to invent things you know like our honorable minister says IITs are not producing noble laureates as if there is some formula and you pour in some central fund and noble laureate comes out from the other end right. So, this is not. So, we do not have the algorithm, but what we know what is needed are these elements in which you have good people right. So, they would have something to contribute and the people should have also the passion to do it. You should be almost obsessive about doing things I have to do it this obsession is there and this is the luck element serendipity. I have told you before that there are scientists who got Nobel Prize, but did not know for what they have got it this scientist from the luck the background radiation they found out. You know about this word serendipity, the serendipity is the modern day Sri Lanka right. So, that is called is to be called Surnadeep. So, from that the Britishers have coined this word called serendipity. So, they were looking for India they reached Sri Lanka and found it is full of emeralds right. So, when you actually unexpectedly discover something that is your serendipity ok. So, that is that and this is something that God gives you you know epiphany you are inspired by something. So, if you are not inspired then probably you would have one missing element of it and there you have it. So, I suppose we will edit this out, I wish I could send it to our minister to say that we have been talking about algorithms for 40 lectures almost, but we still have not found out how to come out with an algorithm for creating a Nobel Prize. So, that is that and this is the last word again by here to God. You have to live in the present, but you will only appreciate when you look at it in the distance ahead and then you will see whether anything that is done at present makes any sense or not. So, I think I have said enough it is a long semester like you can say that every nightmare comes to an end with the day break right. So, with the end semester probably this dose of course also should come to an end. So, I will stop here ok.