 So, we will continue with the one-dimensional flow. So, the governing equation that we had was this and we basically said that we could discretize it using FTCS, right, discretize it using FTCS as Qp at Q plus 1 as being Qp at Q minus delta T by 2 delta x, Qp, Q plus 1, Pp, Q, P plus 1, Q is E p minus 1 Q. I am using central difference, okay, just to recollect what we did in the last class and then we said that if you look at two grid lines, this is at Q, this is at Q plus 1, these were the boundary points, right, I will draw a few interior points just to, I am not going to draw a whole lot of them just for the sake of this discussion, I will draw a few interior points. So, it is clear that we can do FTCS, right, using that stencil, no problem. We have everything at Q and we can determine the value at Q plus 1. The difficulty is when we get to determining the value, right, one shot of the boundary on both ends, on both ends determining the value one shot of the boundary requires values at the boundary, right, it requires values at the boundary, fine. And this clearly in the way FTCS works since we are going to do this and we are going to use them to determine the interior points, this clearly is a problem we have to solve because when we take the next time step, when we go from Q plus 1 to Q plus 2, we are going to have the same difficulty, we are not determining these points, okay. We are not finding out what these values are and what do we need at these places? We need the Q, I need Q at this point, I need Q at that point, I need all, okay. So, because we were in the situation, we decided that we look at the physical problem, actually because so far we have just been dealing with the Choctas, we have been working only with the theory. So, we said the physical problem just to recollect because it is a one-dimensional flow, I remember, right, I took a pipe of length L, right and there was a valve on one end where you place the valve will only determine the initial condition, right. I should not say we will only determine, it will determine the initial condition. So, if you have the valve here, then whatever, this is open, right, on this side to the atmosphere, typically you would have P ambient described on that side. So, the initial condition would be whatever the ambient condition is, right, since the valve is on this side, valve is on the other side, it would be whatever is in the chamber, right. And here, we have total pressure and total temperature. These are typical conditions that you would use, right, if you are performing an experiment and making sense in a gas dynamics lab, okay. And it is not a very exciting, even that experiment is not a very exciting experiment, but still for our purposes, it is simple enough that we can play around with it. So, the physics of the problem requires, as I pointed out in the last class, three conditions, right. Physics of the problem requires three conditions. Describing T ambient here does not really do anything for you, unless of course you need it, the T ambient here you require in order to determine the initial condition, but as the boundary condition, it does not do anything for you, okay. You understand. It is a pressure difference that is going to drive. See, this is just like normally when you say that you have, let me think of a situation. So, if you think of all the sources of power that we have, the most sources of power that we have, where we compress the fuel, air mixture, and we add energy, right. We ignite it, there is combustion, and then there is expansion. The process always involves compression, expansion. You understand what I am saying? If I were just to create a fireball in the open and raise the temperature of all the air, we are ignoring gravity here, all that will happen is the temperature of all the air will go up, right. In order in this room, if I were to raise the temperature, the temperature in the room will go up. There is no change in pressure, nothing of that sort. You actually need, so I need a lower pressure at one end, a higher pressure here, and the pressure drives the air and we extract the energy from the fluid. That is why we always in thermodynamics deal with enthalpy and not directly with pressure or temperature, right, because you need the pressure gradient in order to drive to start the motion, so that you can extract the energy. Having the temperature alone through mechanical processes is not enough. Am I making sense? So, in that sense, it is the pressure difference that is going to drive the flow. It is the pressure difference that is driving the flow. Is that fine? So, in this side, we can measure the total temperature and this total temperature will basically do what? What is that total temperature? Does it determine anything? Do we need it? You go back to your gas dynamics and see why I need to prescribe the total temperature there. So, a P ambient, P not, T not and this problem can actually work using gas dynamics, fine, okay. From the mathematical point of view, we have already said that because we have a time derivative, we have a doh q, doh t term, because of the doh q, doh t term, at t, for some given t, we will take t equals 0, at t equals 0. I need a condition, so I need, I have an initial condition for all spatial locations and I have a doh e, doh x term, right. So, for all, I have a q at t equals 0 and I have a doh e, doh x term, I have first derivative. So, integration in my mind will give me 3 constants of integration, which have to be determined and P not, P not and P ambient are given. So, both physically and mathematically, well determined, okay. It is just that our FTCS requires that I need a q at this point and the q at that point, fine. And consequently, I need to determine something here and something there, is that okay, right. The experiment, thought experiment we are running yesterday was if P ambient was P not, P ambient was P not, as I just indicated, there would be no flow. And if I started to lower, there are some mechanism by which I could lower P ambient, right. If I have another pressure vessel here or something of that sort, I start to lower P ambient, then a flow field will be set up. So, the effect, if I look at the first, if I look at the, this interface, if I look at this interface, valve is open. So, if I lower the P not, P ambient, the effect will be that there will be a u set up here because of the lower P ambient, right. So, presumably this boundary condition communicates itself here through an increase in u, fine. I am just giving you a hand-waving argument. So, you want three quantities, you want three quantities here, you have only P not, P not. I am saying here is a way by which we can do it. So, what we do is, from the first interior grid point, maybe I go back here, from the first interior grid point, so whatever I have here, I copy the u to this point, I extrapolate the u to that point, is it okay. So, this is a very naive way of doing this, but this is the first cut. We will see whether, how we can improve, we can improve, right. I copy the u here. You could also copy rho u if you want it, but I will copy the u there. So, what does that give me? At this grid point, I have prescribed P not and T not and now what we need to do is a little gas dynamics, right. Given u that I have extrapolated u, I use energy equation, what does it tell me? One-dimensional energy equation, not the differential equation that we are using, Cpt plus u squared by 2 and if you give me the u and you give me the T not, this tells me that T equals T not minus u squared by 2 Cp, is that fine. And once I have T by T not, I can find P by P not, right. So, from T by T not, I can find P by P not and consequently, I can find P. And given P and T, I can use equation of state, I can use equation of state or I can, I guess I might as well write P equals rho over T for example and find rho. Now you have set, now you have everything, you have rho, you have rho u squared, right. You can calculate everything, you have q tilde and therefore you can find q, right. Is that okay, fine. What about the right-hand side? I have only prescribed at this point, I have only P ambient prescribed at this point, right. I only have P ambient prescribed at that point at time level q. Remember this is not the initial condition, initial condition we knew everything. This is at some time level q, just in case you are wondering what is the deal. That is why it is at time level q, right. So, we have only P ambient at that point. Clearly not enough, I need 3 quantities, right. So, I need to extrapolate 2, fine. What can I extrapolate? The different variations now that you can do, right. There are so many possibilities. Even upstream, there are so many possibilities. You could have extrapolated P instead of u. Am I making sense? So, these are things that one can try out. There are lots of scope for you to play around here, okay. So, one possibility is that for example, you say that there are no sources and heat sources, energy sources in between, right. Maybe I extrapolate T0, possibly. I could extrapolate T0 and u. I could extrapolate T0 and u, possibly. I am just making a suggestion, right. The point that I am trying to make is that you need to extrapolate 2 quantities. See, these are boundary conditions that I have to generate. They do not exist. The physics does not provide them. The mathematics is not right now giving me a clue. We will look at it to see whether we can use the differential equations somehow to generate these. Is that fine? So, if I extrapolate the T0, if I extrapolate the T0 and u, what do we get? Again, if I extrapolate the T0 and u, I can find the static temperature. And given the pressure and static temperature, I can find the density. It is the same game. Then you can go through, right. Is that fine? Everyone? Okay. It seems like a rather arbitrary thing to do. It seems like a rather arbitrary thing to do. So, this is something that you can try out. As I said, it leaves scope. So, you can try it, let us say, to extrapolate density and rho and rho u or rho and u. There are various quantities. The thing that you want to make sure you do not do, these are things that you have to be careful with. You already have T ambient here. You do not want to really extrapolate density and temperature, for instance. Because potentially you could violate equation of state at that point. Am I making sense? If the pressure here is different from the pressure here, by extrapolating the density and temperature, you could potentially violate equation. So, you have to be very careful what quantities you are extrapolating. But you could try out various quantities. Some of them may be totally unstable. You may not work at all. You could try out various quantities. But these are something that I thought, okay. We can just start with. Is that fine? Is that okay? Can I justify that? Is it possible for me to justify that? Maybe I will stick with this. Is it possible for me to justify that? Justify extrapolating these quantities. Okay. So, the question is, when can I do this? What are the conditions? Is there any inherent assumption that I have made? I am saying there are these quantities. Can I always extrapolate? Will it always work? Okay. So, let us go back. Let us go back. Remember that this could be written as dou q dou t plus a dou q dou x equals 0. And this could be further decoupled into three equations. I am going to write now component form dou q i carat dou t plus lambda i dou q i carat dou x equals 0, right, where q1 hat, q2 hat, q3 hat are the characteristic variables. We are not going to solve the equation in this form because multiple dimensions anyway it is not going to help us, right. So, that it does not make sense going there. But it is useful to look at. So, what are lambda i? Lambda 1 is u, lambda 2 is u plus c and lambda 3 was u minus c. This is what we found. Is that fine? Okay. Now, we will actually investigate these equations. So, if you look at one equation, if you look at one equation, you have dou q1 hat dou t plus u dou q1 hat dou x equals 0. That says that q1 hat is being propagated along a characteristic. I will just draw a small figure here first. So, at this point, q1 hat is being propagated along the characteristic whose slope is 1 by u. What about q2 hat? q2 hat is being propagated along that. And this is, this corresponds to, so I would not write the 1 by u and so on. I will just write out what characteristic this corresponds to. So, this corresponds to u. This corresponds to u plus c and q3 hat, u plus c or u minus c? u minus c. This is u minus c. That is u plus c. Remember, this is x. So, it is travelling faster. This is u plus c. So, if this is u minus c, u, u plus c and all the slopes are positive, what does that tell you about this slope? u minus c, u is greater than c. So, this actually corresponds to supersonic flow. So, this actually corresponds to supersonic flow. That is u greater than c. So, if you are talking about subsonic flow, if you are talking about subsonic flow on the xt plane, you would have u plus c there. You would have u here and you would have u minus c headed out in the other direction. Am I making sense? Because u minus c will be negative. This is for u less than c. This is for u less than c. So, at a subsonic inlet, at a boundary condition, go back to those boundary conditions. So, if this happens to be subsonic, if that boundary happens to be subsonic or if you look at what happens to the first interior grid points, if the flow is subsonic, these are the grid points. If the flow is subsonic, if the flow is subsonic, what happens at this point? So, you have one characteristic which is u minus c. Maybe I will use a different color here, u minus c. One characteristic which is u and one characteristic which is u plus c. Is that fine? Actually, a sort of deliberately drew this characteristic this way. This, if you go think about your stability condition, the grid size is too, the delta t is too large. All of these characteristics should be above that. Just go back, think about your stability condition. But we are going to look at the stability issues in this seriously anyway. But I thought I would just sort of stick that there. I could resist the temptation to have that. So, you could have a u plus c that is going there. Is that fine? So, what this basically says is there is one characteristic that is propagating q1 hat or q3 hat in this case. It is propagating q3 hat in that direction, outward direction. And if I look at the boundary itself, at the boundary itself, at the boundary itself, since the flow is subsonic, I have one characteristic that is propagating q3 hat out. And I have two characteristics that are propagating q1 hat and q2 hat in. Is that fine? And if it is a subsonic exit, I would have the same situation. So, I draw at the last but one point. So, I would have u plus c here. I have u there. I do not know why I am putting arrowheads. And I have u minus c here. I guess I am looking at it. I am putting arrowheads because instinctively I am looking at it as being propagated along that direction. So, in which case then you have q1 hat going out, q2 hat going out, q3 hat coming in. Now, when I look at the problem, when I look at the physical problem that we prescribed, we prescribed p0, p0, two incoming characteristics, two quantities prescribed. That is nice. And I prescribed p ambient, one incoming characteristic coming into the domain, one quantity prescribed. So, the mathematics and the physical, the experiment that they are the same. I am happy with that. We are happy with the way that worked out. And we are extrapolating two quantities. We should be extrapolating q1 hat and q2 hat. Obviously, that would be the way to do it. And we will see how we go about doing that. So, there are two quantities that we are extrapolating that correspond to in our mind now the characteristics u and u plus c. And the one quantity, say I am giving you a rational. It is a hand waving argument. I should what is being propagated or q1 hat, q2 hat, q3 hat. That is what are being propagated. But I sort of say, well, I will extrapolate the u. It is easier to extrapolate the u instead of going through all these characteristic variables. Fine. If it works well, I am willing to live with it. But as I said, we will also look at the legal way to, the proper way to do it. Is that fine? Are there any questions? Okay. So, this is one mechanism by which we can apply boundary conditions. As I said, we will come back to, we will revisit boundary conditions again. It does not end with this. But this is something that you can definitely implement and try out. But before you do that, the question that we have is, is it going to be stable? Is it going to work? Right? Or do we need to do something? It is FTCS. So, is this going to work or do we need to do something? So, we come back to our original equation. How can we, what is the, how do we do the analysis? How can we do the stability analysis for this? Now, we have a system of equation. Earlier, we had only one static situation. Now, we have system of equation. What do we do? Any suggestion? Well, there are different ways to do this. Maybe we will, I will cheat a little. There are different ways to do this. One, you could write it in terms of delta Q's and so on. But let me just, I will just do a little cheat here. So, first, we will use the shift operator, right, and get this E out. So, Q, Q, P, Q, P, Q plus 1 equals Q, P, Q minus delta T by 2 delta x E, P, Q into E power i n delta x minus E power minus i n delta x. Obviously, there has to be a 2 pi by L and so on, right. I am acting as though that I am going to wave my hands a little and say that, okay, L was 2 pi, okay. And I do not know whether you check this out, right. The reason why I said I will do a little cheat is because in this particular case, it turns out that E is, E can be actually written as AQ, right. In other words, you have tried it out but E can actually be written as AQ. It so happens. E can be written as AQ, right. So, consequently, this is going to turn out to be Q, P, Q plus 1 equals Q, P, Q, okay. So, what I will do is, I know I am going to replace an AQ here. I am going to have an AQ here, okay. I am going to have an AQ here. Oh, I mean, okay, minus delta T by 2 delta x. Already you see that A P Q, Q P Q. What is e power i n delta x minus e power minus i n delta x? 2i, 2i sin, I will call it theta where theta is n delta x. As earlier, the 2s go away. So, I can write this as Q P Q plus 1 equals i minus delta T by 2 delta x. So, I can write this as delta x i sin theta A P Q, Q P Q, which of course is a very familiar looking. The only difference is that they are all matrices. It looks familiar but they are all matrices. What now? There is one way to decouple it. We could pre-multiply the equation by x inverse, right, okay, where x is the matrix of eigenvectors. If I pre-multiply the equation by x inverse, and of course I can choose x inverse so that it is normalized in some sense, right. It does no stretching, only rotation. I can take unit vectors, right. x inverse Q P Q plus 1 equals x inverse minus delta T by delta x i, I have that right, x inverse A x x inverse. This is i minus delta T by delta x, little i lambda into x inverse Q P Q, sin theta, oh, sin theta. Very important. How can I forget my sin theta? Remember, x inverse Q is not, they are not the characteristic variables, right. X inverse Q is not, they are not the characteristic variables. Characteristic variables are related through the idea of a derivative. D Q character is x inverse D Q. Characteristic variables are related this fashion, right. This is some strange map. We are just doing some kind of a transformation. Am I making sense? So in a sense this gives me some, I will just give it a name S so that you do not confuse it with Q character, right. I will just give it some name, Q P Q plus 1. So I want you to be careful here, right. Yes, now I just take the, I take the norm. I take the norm. I want the norm of this divided by the norm of that, norm of S P Q plus 1 divided by norm of S P Q equals norm of, this we have seen as an iteration matrix before when we did Laplace's equation, right. System of equations, think back to the example that I did with Laplace's equation. I had a scalar equation. I performed a rotation, right. I had two scalar equations. I performed the rotation and showed that they became coupled, right. Two scalar equations. I made them into a coupled system by performing a rotation. I am doing the opposite here, right. I have a coupled system of equations and I am trying to undo the coupling. I am trying to undo the coupling. So we might as well take the spectral norm. What is the largest eigenvalue is what determined, right. The largest eigenvalue is what determined whether that sequence of iterations converged or not. I could have just used that argument directly but I just wanted to tie it up with this. We have actually done this before, right. We have actually done this before. So what is the largest eigenvalue? U plus C looks like the largest eigenvalue but you do not know the sign of U. You understand what I am saying. So the largest eigenvalue are U plus A. If I use C throughout it does not matter. C and A they are the same, okay, right, fine. U plus A it looks like or U plus C whatever. So mod U plus A or mod U plus C since I am using, right. In general mod U plus A would be the largest eigenvalue, okay. In this case it is clearly going to be unconditionally unstable, clearly going to be unconditionally unstable just like it was for FTCS applied to any one of the equations, scalar equations. So I could add, I could add, what do I want to add? I can add artificial dissipation. It will make it upwinding, right. What was the term that I wanted to add? What is the term that I want to add? I can add a mu2 dou square q dou x square, right, and subtract out a mu2 dou to the 4th q by dou x to the 4th. Is that fine? Okay, fine. I usually prefer to write this in terms of, because if you think about the term that we added to eliminate, I prefer to add a delta x square but it does not matter. You will see as you go along here. I rather make it the mu2 delta x square dou square q by dou x square or you can try this delta x to the 4th, right. See if you think there is any difference. And of course the minute I give you two options like this, one of the, one thing, one possibility that catches, I can see a few smiles out there. It is obviously that you can try a mu2 delta x. See what happens, right. See what happens. What are the, right now I am not going to give you any helpful hints on what possible values of mu2 may work, right. You try it out. You see what it is. You already have, remember what we have done for the one dimensional flow, right. You already have enough, you already have enough of a background to figure out what you need to add, fine. The only difference there is there you had only one propagation speed, a. Here you have three propagation speeds, right. That is the problem. You have u, u plus a and u minus a, right. And after adding these, this dissipation term, if you do get a stability condition that corresponds to the CFL less than, CFL condition that is sigma less than 1, which sigma do you use, right. Because there are three sigmas actually. So for the wave equation you only had sigma, right, one sigma but here you have sigma u plus c, sigma u minus c or u minus a, u minus a and sigma u. There are three possible sigma because there are three propagation speeds, right. And all of them have to be less than 1, even if you do this, even if you do this, right. And if you go back to up winding, you basically add the right amount, let us say that is the clue. You add the right amount, it goes back to up winding. Which one do you add? Which one do you, which is the most constraining one? Which of these is the most constraining one? That would seem to be u plus a, right. It would seem to be u plus a. So along these lines I will write again that you want sigma, you want that constraints to be mod u plus a delta t by delta x. Is that fine? But remember that you can actually define three of them. So there are three propagation speeds. You can actually define three of them. Are there any questions? Is this fine? Okay. So you can ask the question why are we bothered with FTCS when we know that it is unconditionally unstable, right. Why are you wasting our time when you know it is unconditionally unstable. Fine. I know that we can add this, right. We can do something with it. Let us go to a more sensible scheme that we had. But the only problem there was, what was the problem? BTCS. BTCS, the advantage was, BTCS you may also see it as backward Euler centered space, whatever. BTCS unconditionally stable, it gave you a system of equation, right. And here we already have a system. So now we have to be a bit careful. We have to see what does that mean. So what does we did a semi discretization? If you remember when we did this, delta q by delta t plus dou by dou x, I am going to write this in the delta form E at delta q by delta t at time level pq. And this is at time level pq plus 1. I want to write dou q dou t at time level q plus 1. This equals 0. Okay. And just like we did last time, Eq plus 1 is Eq plus dou E dou t at q by delta t. Times delta t plus higher order terms. I am going to truncate the series here. Anyway I am planning to do only backward, right, backward Euler or backward time, backward step. And again I will use chain rule. E is a function of q. So dou E dou t, I will write as dou E dou q dou q dou q dou t. Okay. So that this equation becomes delta q plus delta t times dou by dou x of a delta q equals minus delta t into dou E dou x. This is at q. We are all at q. Is that fine? Everyone? So dou q dou t plus dou E dou x equals 0. If I am looking only for the steady state, seeking steady state, I will tell you what we will do when we are going to look for a transient. But right now we will look for only the steady state, seeking steady state. So this tells me that dou E dou x equals 0 is the equation that I am actually trying to solve. Okay. So that equation can be written as i plus delta t dou by dou x a acting on delta q is minus delta t into some matrix R. Just like we got for the wave equation, the generalized wave equation. But there is little difference here. What is the difference? Each of the delta q's, each of the delta q's is a matrix, vector. So each of these entries is a vector, is a matrix. So what does this equation look like? If we write it for an arbitrary pq, if we write it for an arbitrary point p, what does this, and if we use central differences to discretize that, this is going to give me delta q pq. I would not write the q, it is a time q, delta p at delta q at p plus delta t times ap plus 1 delta q p plus 1 minus ap plus 1 ap minus 1 delta q p minus 1 equals R at p. Okay. So you will get a tridiagonal system. You get a tridiagonal system. But the point is that each entry in the tridiagonal system is a block, a matrix block. So the general equation will turn out, so you will get a tridiagonal system. What I mean by that is you will get something, so this will be delta q p. So this is the equation p going through. And what you would get here is, what is on the diagonal? There is an identity matrix on the diagonal. There is ap minus 1 delta t by 2 delta x on the sub-diagonal. And there is an ap plus 1 delta t by 2 delta x on the super diagonal. You understand what I am saying? That is the diagonal, that is the sub-diagonal, that is the super diagonal. Everyone, is that fine? And each of these entries, each of these entries is a matrix. Each of these entries is a matrix, ap minus 1 delta t by 2 delta x. Just to make it clear, ap plus 1 delta t by 2 delta x. Those are the entries. These are the entries in the matrix. So if you decide for example to do Gaussian elimination, if you decide for example that you want to do Gaussian elimination. So where in Gaussian elimination you will say divide through by the pivot element, here you have to pre-multiply by the inverse. Am I making sense? Pre-multiply by the inverse. If you are planning to do this, if you are planning to do that, it is better that you factor this ap minus 1 using LU decomposition. There is always this question when you do numerical linear algebra, you know that LU decomposition and Gaussian elimination are essentially the same. LU decomposition and Gaussian elimination, operation for operation you can actually show that they are essentially the same. Am I making sense? You do the elimination part of Gaussian elimination, you end up with a upper triangular matrix. You do LU decomposition, you have a lower triangular, upper triangular, you do a forward substitution of the L, you end up with an upper triangular matrix. You would expect that you can map the two operations together. You actually can. You can show that they are equivalent. You can show that they are operation for operation, you can actually identify that they are equivalent. Fine. So then the question always pops up, why would you choose one over the other? Well if you do LU decomposition, if you factor it as L and U, then you only have back substitutions and forward substitutions. If you pre-factor this as L and U, if you are going to multiply through by, right, the inverse, you could actually do back substitutions, forward substitutions and back substitutions. That is the idea. Fine. That is if you are using, if you are using Gaussian elimination. Even if you want to do Gauss-Seidel, you are doing an iterative method, you may still have to do something of that sort. But of course Gauss-Seidel, the big advantage is the diagonal matrix here is I, right. So Gauss-Seidel is not that bad. Gauss-Seidel it is not that bad. Am I making sense if you are using an iterative matrix, it is not that bad. There may be consequences but it is not that bad. Okay. Is that fine? Everyone? Okay. So then we have only one question left. How do you apply boundary conditions here? You have a system of equations. How do you apply boundary conditions here? In this problem, how do we apply boundary conditions? Let me see where is the best place for us to discuss that. Maybe I will do it here. How do you apply boundary conditions here? So in the application of boundary conditions, we need to maybe look at it a little more carefully. Again, I will draw two timelines Q and Q plus 1. Like I did the first time around, I am not going to choose, I am going to just choose a few grid points. I would not choose a lot of grid points. So it does not become too messy. And what I had indicated at that time is you need a value here that you can extrapolate some quantity from the interior. Say for instance t0 or u or whatever or we said that from here you can extrapolate to u. It need not be from the current time level to the current time level. Right now what we want is we want the value here. So actually you could instead of extrapolating from the current time level to the current time level, you could actually propagate. Now that looks more like, that looks more like our characteristics. You could actually propagate from the previous time level to the current time level. You could extrapolate both in space and in time. There are so many different ways by which you can do this. You can extrapolate both in space and in time. Another value goes from the last but one grid point at the previous time level to the time level to which you are stepping. It is possible. I mean if you are going to do it hand wavingly that is possible. So in a similar fashion even here you could do the same thing. By the way this is just, I am suggesting this suggestion comes to mind only because we are looking at an implicit scheme. But even in the case of FTCS we could have done that. Even in the case of FTCS, even in the case of FTCS you are going to use this stencil. Even in the case of FTCS you are going to use this stencil. You assume at the initial condition this is known. You find all the interior points then you can ask the question what do I do at the boundary and you can actually extrapolate from the current time level to the next time level even in FTCS. Am I making sense? Even in FTCS you could do that and then when you go Q to Q plus 1, when you go to the next step Q becomes Q plus 1 you have everything at the current time level. So the minute you start this, so it is this, we needed the conditions. The conditions were required because of the scheme that we came up with. We have no physical basis by which we can justify this. Right now the only mathematical basis that we have done is though the characteristics are propagating in that direction. So that rationalizes why I am extrapolating. That is all we have done. The question that we can ask is cannot we just directly use the characteristics themselves. Why not just use the characteristic equations? Am I making sense? Why not just use the characteristic equations directly and say that at all of these interior points we use the regular equations and at the boundaries we will do something with the equations. Say in characteristic forms we will allow, we will use the equation in characteristic form to determine what is the boundary condition, what is the equation that is solved at this point. Am I making sense? We will admit that only p0 and t0 is provided here which means that from the differential equation we need one differential equation for one parameter to be solved at the boundary. Am I making sense? And it is the same thing at the right hand end. That is you have p ambient and p ambient provided. You have only one condition given. So two of them should come from the interior. Is that okay? Right? So in the next class what we will try to do is we will try to determine how do we apply these boundary conditions using the equation governing equations themselves. Is that fine? Okay. Thank you.