 So, in the last class, we did FTCS, FTFS, FTBS, we saw that the first two were unconditionally unstable. We saw the third one was conditionally stable and we sort of quickly did BTCS which I will look at again one more time and we saw that BTCS was unconditionally stable. So, basically where we are, so backward time central space, BTCS, backward time central space, I will just so that you know on applied to a greater than 0, okay, that is the equation that we are looking at and the stencil just to be clear, the stencil looks like that, okay. So, this would be delta t that would be delta x and we decided to call the new time level q plus 1 staying consistent with what we are doing before. So, this is pq plus 1, I will just go through it quickly, okay, that is fine, this p plus 1, q plus 1, p minus 1, q plus 1, pq, okay, the difference between what we did FTCS and so on and this is that the equation is being now represented at the time level q plus 1, that is the difference, okay. Please remember, you have to keep that in mind, after a point once you are starting to do these difference discretization and different schemes, easy to forget where you are representing the differential equation, so the differential equation is being represented at the point pq plus 1, okay. So then it is straight forward as I had indicated dou u dou t is upq plus 1 minus upq divided by delta t dou u dou x is up plus 1, q plus 1 minus up minus 1, q plus 1 divided by 2 delta x, okay. Substituting back into our differential equation gives us up plus 1, q plus 1 or maybe I do not skip a step here one point. So substituting back and what we are going to do is we are going to keep all the new time level q plus 1 on the left hand side of the equation and everything else is going to go on to the right hand side. So I get up plus 1, pq plus 1, okay, plus and a delta t by delta x or 2 delta x up plus 1, q plus 1 minus up minus 1, q plus 1 equals upq, fine. Again we identify this sorry wrong one as sigma. So we have gone through, we have gone through the stability analysis for this and we have shown that it is unconditionally stable for all sigma because the gain mod g or mod g squared I am not going to go through the whole thing I just wanted to get to this point, okay. Mod g squared was 1 over 1 plus sigma squared sin squared theta, right and we wanted that to be less than 1, is that fine, okay, at theta equals 0 it is 1 but otherwise it is strictly less than 1, is that fine, okay. And as always I am not really concerned about the DC component right now, okay. Are there any questions, yeah. This is for one particular n, right, this g or we can say this is like from g n. No, this is like yeah it should be, it should have, it should be a g n, this. Say that sigma mod of g n squared should be less than 1, right because we cannot just take this one wave number and I mean take, we have to take the general. No, no, the system of equations is linear, the system of equations is linear. So essentially we are looking at the, it is like saying that if you say sigma f equals 0 it is like saying does the x component grow. The system of equations is linear, right. So you can just look at, you can look at the individual components, look at the individual components. Finite dimensional vector space, so just having each component being finite does not necessarily mean the norm of the whole vector is finite. Well, no, there you have, see there is an issue here, we are just basically saying that, no there are more serious issues there, right. I have just basically if you are going now go to the fact that it is an infinite dimensional vector space, the original space is an infinite because we are not gone to the limit, right. This is definitely finite dimensional because I cannot represent anything over a certain largest wave number. So in this case for the numerical scheme that question does not mean, it is not an issue, right. I am going to take 10 grid points, I know 5 is the largest wave number, I take 10 end rules, 5 is the largest wave number I can represent. So that is it, I mean it is finished with that. However, if you are talking, there are other issues that if you are talking about it from an analysis point of view, there even you know like if I say take the sine theta component, if you actually go through the process you will see that you will be trading summations infinite sums with, there are two limit, two infinite processes that you will be exchanging. So then the issue of uniform convergence and so on, all those also become issues. So even before you get to this there are other issues that you have to handle. Am I making sense? Okay, right. So if you want to look at it from a mathematical analysis point of view there is a problem there which I am not paying attention to any of that, right because I am never going to go to that infinite process, right. So really you will not hear me, typically if I say uniform convergence yes when I am talking about SOR I will say something about oh it is not uniform convergence, it is indeed interesting, right. But in this case you want uniform convergence, if you are actually going to an infinite sum and you are going to say that I am going to exchange the differentiation, right. If you substitute for you, you expand you in terms of Fourier series, right, you expand you in terms of Fourier series and you want to substitute it into the derivative. You expand you in terms of Fourier series and you want to substitute it here. If I want to trade, I want to exchange the derivative with the infinite sum then there are certain properties that I need, uniform convergence being one critical property that I need, okay, right. But we are not going to do that. But it is an important consideration, I am not saying it is not, it is an important consideration but we are sort of in a finite dimensional space, okay. So in that sense it does not matter, right, okay. And there is this has a consequence to something that I am going to say later in class, this has a consequence but we are in a finite dimensional space, is that fine, okay, right. So in this case and like we always do in, like we always do in mathematics or in any of our mathematical physics or, right, deriving governing equation, the wave number was chosen arbitrarily. That is we are just basically saying for an arbitrary wave number n, for an arbitrary wave number n what does this do? And then we ask is there any particular wave number for which it is very bad, right. So and it turns out that there is no sigma, see that is basically what it comes because the n is, what do you call it, the n is embedded in here, there is no sigma for which this thing is going to blow up. There is no combination of parameters, physical parameters that we have as a choice, right, either delta x, delta t or of course given this speed of propagation for which this is going to, any of the wave numbers will diverge, okay, is that fine, that is basically what we have shown. So it is unconditionally stable, okay. Now we have got that, you know, so in fact yesterday I said, yeah great, we finally got, we finally got something that is unconditionally stable because it is a bit disappointing, you do FTCS, you expected to work and it did work, right. So it is very nice that it is unconditionally stable. But at what price? What is the price? If I look at FTCS and I look at this equation, I sort of messed it up so I can rewrite it. So if I say that now I am going to write it in a slightly different form so that we get it. Okay, I have written it in this form so that you can see that this actually forms entries in a tridiagonal matrix. The diagonal term is UPQ plus 1, the sub-diagonal is P minus 1, the super-diagonal is P plus 1, okay. And when you come to the, again if you take the, right, this is at time level Q plus 1. So when you come to, when you come to this point, clearly you need the boundary condition there, okay that will go over to the right hand side. And when you come to that point, when you come to computing the value, P is this value, then you need P plus 1, you need P plus 1, okay. So this is a boundary condition that is required because of our numerical scheme, choice of numerical scheme. I chose central difference because I wanted more accuracy, so we need to do something here. So this is a boundary condition that needs to be specified at the right hand side, which is not required as part of the, either the physics or the mathematics, okay. More we get a system of equations, okay. So you may say what is the big deal, so you get a system of equations. But if you solve the system of equations, what do you get? You have taken one time step, I want you to understand this. You solve the system of equations, you have taken one time step, now you get ready to take the second time step and you have a whole system of equations again, okay. As it turns out, because these coefficients are constant, it is not a nonlinear equation, right. This is a setup where something like LU decomposition will do well, right. So you pre-decompose, then you are left with only forward substitutions and back substitutions, okay. This is a situation where a scheme like LU decomposition works well, okay. You do some work, any work that you do beforehand on the matrix is work that you do not have to do at a later date, right because the matrix has not changed as long as it does not depend on these variables. That is one way to do it, it is a direct method. Direct methods have their own problems and the matrices grow very large, okay. Because if you do look at the process of forward substitution and backward substitution, you see that the last quantity that you determine is based on a series of computations that you have made on the previous n-1 quantities, okay. And therefore, it is prone to error. If you are solving a 1000 by 1000 system and you have 1 million unknowns or in 3 dimensions you are solving, right, 100 by 100 by 100 system which is still 1 million unknowns, then the last quantity that you calculate, the 1 millionth quantity that you calculate comes as a consequence of calculating all the previous 99, you know 99 lakh, whatever, 99,000, 999,000, am I making sense, right? So it is too much, right? So the error, the round off error tends to accumulate. So for that reason, I typically do not use direct methods, this is a personal opinion now. I typically do not use direct methods for matrices that are, if they are banded matrices like this may be up to about 1000 by 1000, right? And if they are not, if it is a dense matrix, may be about 200 by 200, this is a personal opinion. You understand what I am saying? If you give me a matrix, when I say dense matrix, all the entries, essentially all the entries are non-zero, right? You are doing panel method or something of that sort, okay, 200 by 200, that is about the largest. Otherwise, I immediately switch to iterative schemes, which is why I started you off on Gauss-Seidel, okay. If it is a sparse matrix like this, maybe I will go up to about 1000 by 1000, 1000 unknowns. But once it goes beyond that, I will always go to an iterative scheme. I may use this LU decomposition as an initial guess, fine? In this case, however, if you are wondering why the heck is this guy harping so much over? In this case, however, with iterative schemes, there is an advantage. What is a good guess? What is a good guess for UPQ plus 1? UPQ, right? You have a good guess. If your delta t is reasonable size, right? You have a good guess already. You have UPQ as an initial guess, am I making sense? So you already have something with which you can start, right? So it may not be that bad, it may not be, but the fact of the matter is that you have to solve a system of equations and when you are done solving it, you have taken only one time step. I want you to remember that. So there may be some effort that has to be put in to solve this other than just the schemes that I have told you, fine? There are ways by which you can accelerate these schemes. I am not going to spend time on that. There are ways by which, so this region, this gives you a system of equations. That is the key. The point that I want to make is this gives you a system of equations and because these equations appear together, these expressions, these terms appear, the quantities that you want appear together, packed together, it is called an implicit method, fine? It is called an implicit scheme. I will say a little more about implicit schemes, explicit schemes. As we go along, you will see that I will illustrate various kinds of situations that occur. So this is called an implicit scheme because you get a system of equations where the unknowns are tied together in a linear fashion and you cannot, you have to solve the system of equations to get it, okay? It is the reason why right in the beginning I said some of the mathematics that you want to learn implicit function theorem is one. So it will tell you when you can, in the case of a linear system, of course, the system just has to be invertible, right? But implicit function theorem will tell you in general, if you have an implicit expression, when can you invert it, okay? Implicit scheme as opposed to, I will just write it just for the sake of, this is FTCS. So this is BTCS, this is FTCS. This is an explicit scheme. Explicit because I have the expression that I want explicitly occurring on the left hand side, no further work need to be done, okay? Fine? Are there questions? Okay. So now we get to this point that we are talking about earlier. So I have 10 intervals, we have already shown that the largest wave number that we can represent is basically 5, right? Okay. So bear that in mind. There is, so if I, if I have something that has, if I have sin 10x, we have already seen in the demo that sin 10x, right, sin 20x, sin, all of these, sin 15x, all of them are basically going to look the same. You have only 10 grid, 10, 11 grid points on which you are sampling the function, okay? So bearing that, bearing that point in mind, I ask the question now, these equations approximate the original wave equation, linear wave equation. I go through the effort of solving the system of equations and I end up with a solution. The solution is an approximation to the solution to that equation that we want to solve, okay? So I am going to ask an ill-posed question basically. To what equation is it an exact solution? The solution that I am calculating here is an approximate solution to the problem that I am trying to solve. To what problem is it the exact solution? Put it another way. What is the problem that we are solving? I want to solve the linear wave equation, but I am making an approximation, I am solving some other equation actually. When I say I am making an approximation, so I am sort of turning around and saying well I have this u, I have this function, right? And if this function is, if this solution, this approximate solution is close to the exact solution, then I must be solving a problem that is close to the exact problem. I must be solving a differential equation that is close to the exact differential equation. What is that equation? Does that make sense? Does the question make sense? I say it is ill-posed because I have just told you, if I have only 11 grid points, I have these values at 11 grid points, we already know that there are so many, an infinity of functions that it represents, okay? But though it is ill-posed, it is still a worthy question to ask because the answer is interesting, okay? The answer is interesting. Am I making, is that clear, right? So the question we are asking is, I have an approximate solution to the equation that I want to solve. To what equation is it the exact solution? We want that equation, okay? Is that fine, right? So what we do is, we do a very simple thing, maybe I will start off, since I have FTCS here, I will start off with FTCS, right? And we will go back to our friend Taylor series, we will use Taylor series, expand these about the point, FTCS, we were representing the equation at the point PQ, that is important, okay? This is what I say, you have to constantly remind yourself, so in FTCS we are representing the equation at the point PQ. So we will expand, Taylor series always says where are we representing the function about which point are we expanding, right? So we are going to expand about the point PQ, okay? So let us do UPQ plus 1. So UPQ plus 1 is UPQ plus dou u dou t times delta t plus dou squared u dou t squared times delta t squared by 2 factorial dou cubed u dou t cube delta t cube by 3 factorial, how far do we want to go? Maybe we will add one more, is that fine? This supposedly equals UPQ, which is the first term, minus sigma by 2, this is going to get a little messy, UP plus 1Q. So this is UPQ plus dou u dou x delta x plus dou squared u dou x squared delta x squared by 2 plus dou cubed u dou x cubed delta x cubed by 3 factorial plus the fourth derivative minus, this part is okay, minus UPQ minus minus plus, be careful, there is a minus sign here, P minus 1, so it is minus delta x, okay, plus dou u dou x delta x minus dou squared u dou x squared delta x squared by 2 plus dou cubed u dou x cubed delta x cubed by 3 minus the fourth derivative. It is a mess but lots of things will cancel, well enough things will cancel or lots of things will cancel but enough things will cancel. First of all the UPQ goes away, the second derivative goes away, you may ask why are you doing this, we did it for, right, we did it when we are doing the derivation of the different scheme but anyway humor me, fourth derivative goes away, is that fine anything else? The UPQ here goes away, okay, I think that is it, we are stuck, we cannot do anything else here. So I will rewrite that, that tells me that, right, I will rewrite that and if I divide through by delta t, it gives me a dou u dou t plus dou squared u dou t squared delta t by 2 dou cubed u dou t cubed delta t squared by 3 factorial dou to the fourth u dou t to the fourth delta t cubed by 4 factorial equals minus a dou u dou x, that is the first term, there are two of them, so minus a dou u dou x, okay, then what else do we have? Minus sigma by 2, minus sigma by 2 delta x cubed, 2 will go away because you will get two of these by 3 factorial times delta t dou cubed u by dou x cubed, you can please check to make sure there are no issues and I do not think we have, right, the fourth derivative goes away, there are no other, there are higher order terms, so I should say plus dot dot dot, right, everyone. So I can easily see that I have a dou u dou t minus a dou u dou x, I mean you should expect that I am going to get that. What do I do with the time derivatives? You know, I would really like to have this as, because we have already seen, I mean we have gone to forward time or backward time, right, because we have higher derivatives in time are clearly not that easy to handle, okay. So as far as possible, we want to retain this as a first derivative in time, I do not really want, I do not like these higher derivatives in time for that reason. At a given time level, I have, right, special derivative, I can, I know, I am comfortable with that. At a given time level, I can evaluate special derivatives, right, whatever order derivative that you want, that is, right, second derivative, third derivative, no problem. Temporal derivative is a bit shaky, because when I start off, I only have one time level at which the initial data, right, so somehow I do not want this. So what I propose to do is, I propose to replace this derivative, time derivative, by a space derivative, right. So the game that we play is this. So if you say that dou u dou t equals –A dou u dou x, this is our differential equation, okay. Then dou squared u dou t squared equals –A dou squared u dou t dou x equals – or plus A squared dou squared by dou x squared of u. Is that okay? Yes. At this point, someone should have a complaint. Does anybody have a complaint? I am proposing to replace the second derivatives with this. I am proposing to replace the second time derivative with this second spatial derivative, fine. So if I go to dou cubed u dou t cubed, which is the time derivative of this, that should give me, okay. See, if you ignore that u, this basically says that dou by dou t is –A dou by dou x. It is essentially what it is saying, right, dou by dou t is –A dou by dou x, fine, okay. So fourth derivative would be something similar. Fourth derivative will be something similar. Fourth derivative would be, okay. So we actually end up with something like this. So I am proposing to substitute this here. See, okay, I will tell you. The complaint that you could have, the actual complaint that you can have is this equation. What is this equation? This is the equation that we are actually solving. When you do FTCS, when you do FTCS, right, when you use FTCS, FTCS in this case happens to be unstable, but if you are, or FTBS or whatever, when you are trying to solve using FTCS, right, you are getting, you are trying to get an approximate solution to the original equation. You are getting an exact solution to this equation. This is the equation that you are actually solving, okay. This is the equation. So this is, since this equation, dou u dou t equals –A dou u dou, it looks very similar to the original equation, right. This is called the modified equation. This equation is called the modified equation. But there is a point here. The point is, if I am going to substitute for the derivative, time derivative, for the second derivative to eliminate the terms there, should I not use the modified equation instead of the original equation? You understand what I am saying? If this is the modified equation, this is the equation that I am actually solving, if I am going to make the substitution, should I not use the modified equation to do it instead of the original equation? The answer is yes, you should, right. I am going to use the original equation for two reasons. One, doing this and eliminating the terms is a lot more difficult, right. So that, if you want, you can look it up and my book comes out, you can look at the book, but whatever, you know, when it, when it, you can look at, you can look it up, you can work it, work through it. So you are, because the trouble is you will get a lot of mixed derivatives. It is a very messy equation, right. So you can make the decision that you will only keep terms up to the fourth derivative and every time you differentiate this, you know that for instance, this term will go away. Once you differentiate with respect to time, this will become a fifth derivative, you can drop it, right. It is, you can, there are things that you can do to simplify, but still using this equation is a messy process, okay. And there is a second issue that will become clear later that using this actually by itself has its, has a certain advantage, okay. You will see, you will see what I mean when I, when I will explain that when I come to the end. Using this by itself has an advantage. So instead of using the modified equation to eliminate the time derivative terms, which is what you should technically do, I am going to use the original equation to eliminate that, right, those terms. Am I making sense? Is that okay? Simply because A, on the blackboard, it is easier to do, right. B, because that equation that you get is still a useful equation. Are there any questions? Yeah. We can write this Taylor series expansion only if the function is analytic rate. But here we, yeah, no, we have, we have, we have an issue. That is, how many terms of the Taylor series, you see, you get to that, you cannot in a sense get to even any of our finite difference, you understand, or the differential, original differential equation. You have to ask the question, right, if the Euler equations support shocks discontinuities, then how did you write? So that goes into a different realm, right, that goes with slightly different realm. So maybe we do not, we do not, we will assume that we can do this and get away with it. In fact, even for the Euler equations, if you do quasi 1D flow also, you can, you will get shocks, right, we will look at this, you will get shocks. And we will use finite difference methods and we will use a differential equation representation to actually go through and solve, okay, right. So then the issue is why does it work? So that theory is something that is sort of outside the scope of this course, right. Because if you get to, if you get to, if you look at it from a point of view of theory of differential, the way I usually handle it is I always go with integral equations. I do not usually solve them in the differential equation form, right. But even otherwise, there is a whole theory of weak solutions and so on, right. So when we say a solution, we are saying a solution is something that you can substitute into the original differential equation to verify whether it satisfies that equation or not. But if the solution has a discontinuity, how do you substitute it into a differential equation, okay. So you have to do something to weaken that differential equation. So that is a whole, there is a whole mathematical machinery that goes with it. So we are not really going to, maybe at the end of the semester if we have time, we will get to it, okay, right. So at this point, we will just leave it at this and basically say that, you know, I will sort of repeat something I said right in the beginning then. I can take as many derivatives as I want, right, right, okay. I can take as many derivatives as I want and typically I will stick to forth, right, okay. So I will, I will substitute this back there and we will see what we get. We will see what we get. So I get my original differential equation dou u dou t minus plus a dou u dou x equals and what does it equal? Is there a second derivative term? A minus a delta t by 2 dou squared u by dou x squared, okay. Then what else do we get? Do I have a third derivative term? A minus and that is a plus. So I get a a cubed delta t squared by 3 factorial. I think I got that right. Minus sigma delta x cubed by delta t and then there is a fourth derivative term. A minus a to the fourth delta t by 4 factorial. Is that fine? Okay, you can check whether I have, in fact in this case this is something that I prefer to do for my modified equation. In this case you can actually factor out a delta x cubed, right. This will turn out to be a, you will get, you will get, you can write this expression in terms of sigmas in the parenthesis. Did I miss a sign somewhere? The first one is minus a squared, a squared, a squared delta t, okay. That is fine and there are, right, there are infinite such number of terms. We are not going to look at, as I said, I am not really bothered, right. We may be even able to write a general expression after looking at this for the even terms and the odd terms. If you write this in terms of sigma that pattern may become more obvious but it does not matter. It is not, it is material for me right now. What matters to me right now? What matters to me right now? Right, what matters to me right now is that we have this modified equation, okay, modified equation. I have it in terms of, I have it in terms of the original expression equals all the extra terms and we ask the question what happens when delta t and delta x go to 0, okay. And again when you have two such things going to 0 there is always an issue as to whether something is going fast. So we want delta t and delta x going to 0 in a similar fashion, okay. Delta x and delta t go to 0, all of these individual terms go to 0. The right hand side, all the terms go to 0. Is that fine? Everyone, as delta x and delta t go to 0 in a comparable fashion, right, in a comparable fashion. Basically I am saying that the ratio of delta x and delta t does not change for instance. In a comparable fashion as delta x and delta t go to 0, this equation, the modified equation goes to the original equation. As this goes to, this becomes dou u dou t plus a dou u dou x equals 0. The modified equation goes to the original equation, okay. So as the number of points that you are using to represent the solution increases which is what delta x going to 0 means. And as the time steps that you take in evaluating it go to 0, right. So as your representation gets more and more accurate, as your representation gets more and more accurate, the modified equation goes to the original equation, which is a good thing. I mean we are happy that that happens. This property is called consistency. The scheme is said to be consistent. So you say I took a differential equation, I discretized the equation. I used Taylor series in this case. I discretized the equation. I end up with a modified equation. If in the limit my delta axis, say when I did the finite difference scheme, I said I am not going with the infinite process. If in the limit of doing the infinite process, I actually let delta x and delta t go to 0, if my discrete equation goes to the original equation, the scheme is said to be consistent, okay. You may be under the impression, you can say, wait a minute, we substituted finite difference method, individual derivatives, the definitions work. How can you come up with a scheme that is inconsistent? It is possible. As someone who has invented schemes that are inconsistent, it is actually possible. Terms can cancel. You put them together, there are terms that can cancel and you can be left with it. You can be left with the term that just refuses to go to 0. You understand, right? So it is possible for you to invent a scheme. It is possible for you to invent a scheme that is inconsistent, right? And you should suspect that I mean we invented a scheme which would not work. FTCS, which is modified, which is consistent. It is consistent and it would not work, right? It is possible actually for you to invent a scheme which is inconsistent. It is possible. So this consistency is very important, okay. So as delta x and delta t go to 0, modified equation goes to the original equation. The scheme is said to be consistent. As delta x and delta t go to 0, if u, I will write h where delta x, delta t are like h, they are like h. And going back to that, remember the notation where I put a superscript h means that it is a discrete solution. If in the limit h going to 0, u of h goes to u, then we have convergence. Am I making sense? As h goes to 0, our candidate solution, approximate solution converges to the exact solution. A solution converges to the exact solution. I do not want you to get this confused with our iterations converging. That is a different convergence. That is a sequence of solutions that are being generated which converge. This is a sequence of solutions with different h's, different grid sizes. And as I change the grid sizes, it converges. Am I making sense? So you have consistency, you have convergence, we have stability of the scheme. There are three things that are there. And there is a famous theorem, you can go look it up called the Lax equivalence theorem which basically says if you have two of them, you have the third. Consistency, convergence, stability. There are three properties for the scheme that we have seen so far. And they tell us how is our scheme going to behave? They tell us if you generate a solution, are you generating, does the solution converge to the original, right? Does it converge to the solution? Does the approximate solution converge to the solution? Does the equation that you are solving for that scheme? Does the equation converge to the actual equation? Okay. We are to a large extent we are satisfied with this, okay. But if you are doing mathematics formally, you would want to know not just that the u's converge, you would also want to know that the derivatives converge. The dou u, dou x converges and dou t, dou x also converge, right? Right? We are as engineers we are a little more coarse. We say okay u converges, we are very happy, right? And when you are doing your actual calculations, once you have a scheme, you say well somebody has looked at consistency, convergence, stability and all that and you tend to just take the u for what it is, right? But this is an issue, okay. Is that fine? Consistency, convergence, please remember in this context, convergence, what it means, okay? As h goes to 0, u h goes to the actual solution. And of course we have seen stability, fine? That is all very nice. Something else comes out of this, something else comes out of this. We had a form of a solution for this equation. It looks like if you do this, you get these extra terms. If you look at the modified equation, you get these extra terms on the right hand side. Is there a way for us to infer an analytic solution? We had an analytic solution, a proposed analytic solution for this in terms of Fourier series and so on. Is there a way for us to get an analytic solution with something on the right hand side? Am I making sense? Is it possible for us somehow to figure out a way that we say I have this, I know the solution to this. If I add something to the right hand side, is it possible for me to get a solution? Is that fine? Okay. So we can, I will just basically start us off on that. What we are going to do is, these are called semi-inverse techniques in the sense that we are going to guess the form of a solution and try to find some disposable coefficient that we have thrown in so that it becomes a solution, right? Because we already know that for dou u dou t plus a dou u dou x equals 0, we already know a solution of the form an exponent, I am writing only one term, I am writing only un, in 2 pi, we will drop the 2 pi and l, okay. We will just, for now, if l is 2 pi, it takes care of that, right? We will drop the i and x minus at, right? Just to keep the chalk dust to a minimum, we will drop the 2 pi and l, we will just assume l is 2 pi. We know that if we substitute this in there, it works, okay. So if instead of being 0, what if it is, pick one of those terms. What if it is, I chose mu 2 as a coefficient dou square d o dou x square, what is with that form, okay, fine? Can we guess the solution for it? What did Laplace, this looks like, second derivative is like, you know, so Laplace was smoothing, I would expect, I would expect some kind of, and from our physical intuition, which is why to trigger that intuition, I called it mu, viscosity, I would expect that it is, right? So anyway we have oscillation, we will see if something can happen, see there are different ways by which we can argue this. Remember we are, we are only trying this out, we are only trying this out, right? That is basically it, if it does not work, it does not work, we are only trying this out. So I would say that I have thrown an extra parameter b and look at the candidate solution u, I mean I should write n but I am not going to write n, u sub n, right, of that form. And ask the question, if I substitute this into this equation, I say consider a solution, see this is how you do it now, know that we have guessed this, you say consider a solution of this form, substitute that in there and see if you can actually get a value for b and does that value mean anything, is that fine, okay. So what is dou u, dou t? a n exponent i n x-at, you want to do it in one shot or you want to split it, e power bt times the derivative of the thing, right? I am basically using chain rule, minus i n a plus b, okay, this is the derivative of time minus i n a plus b. What is dou u, dou x? Instead of writing out the whole thing, I will just write it as u, now that we know it is u, u times i n, substitute it back there, so you get u times minus i n a plus bt, oh, sorry, I need one more. What is dou squared u, dou x squared? The u of course cancels, the i n a cancels with the i n a, I have a, take this, it is not a b. So b is, is that fine? If you want, okay, this is called a semi-inverse method, you sort of guess the form, you have a few disposable coefficients, you substitute it and see if you can fit the coefficient so that it turns out to be the answer, okay, is that fine? We will come back to this, we will come back, it is possible, now we know that it is possible, what does this mean, how does it, what does it mean for FTCS, what does it mean for FTBS, can we explain what we got, what we are going to look at, okay, in the next class, fine, okay, thank you.