 So, we have been looking at one D Euler equation, one dimensional Euler equations, dou Q dou t plus dou E dou x equals 0. We have seen FTCS, we have seen sort of FTCS applied, FTCS and BTCS applied to this and BTCS applied to this. We have seen the BTCS applied to it backward time central space applied to it in delta form. We have also seen that FTCS can also be interpreted as being in the delta form, right. That is basically, so this equation, we could write this as I plus, I just want to make one observation, delta t dou by dou x A acting on delta Q is minus delta t into R and we have noted, we have registered, right, we have noted that I delta Q equals minus delta t into R. So, this is backward time, that is backward time and this is power time. So, in fact, if you use central differences to discretize R, right and central differences to discretize that, then it becomes backward time central space and forward time central space. So, I delta Q, which is delta Q is nothing but R minus delta t into R, right. So, another way to look at it, I just want to just point out, another way to look at this would be that BTCS as I had indicated, you can delta Q and now I will say, I stick an I in there to say implicit equals delta Q explicit, am I making sense? It is like we are calculating the explicit, the correction, explicit correction and from there getting the implicit correction, that is a different way to look at. So, any algorithm, you should try to see the, look at the algorithm from different points of view. It may give you some insight, right. It is always a good idea to try to look at it from different points of view. I just wanted to point that out before we left this form. In the last class, we had looked at application of boundary conditions using characteristics, right and I promise that in this class, we would start off looking at a finite volume method. But before I leave the finite difference class of schemes, let me just do one last thing that comes from what we have been doing in the class so far. Now, instead of multiplying this by the x inverse, x being the matrix of eigenvectors, I am going to go back and multiply this by x inverse, which is what we had done right in the beginning, right. Because this business of splitting the waves based on which way they are travelling could give us a hint as to how to generate a new scheme. That is the idea. I just want to point out, so every little thing that we do, see typically what happens is you are trying to solve a problem. You have a local issue that you fix. You say, oh, I have to apply boundary conditions. I want to do it better. How do I do it? And you are thinking imagining ways by which you can fix that, solve that particular problem. But you should also realize that you may come up with something or there may be clues by which you can generate other techniques, right, other techniques. So for instance, if you take this and if you were to multiply it, dou q dou t plus, I will write it in the non-conservative form in conservative variables, a dou q dou x equals 0. And if I want to pre-multiply this by x inverse, x is the matrix of eigenvectors. This gives me dou q tilde dou q karat dou t plus lambda dou q karat dou x equals 0, right, where the matrix lambda is u 0 0 0 u plus a 0 0 0 u minus a. Is that fine? So we have already seen that in the problem that we were looking at, the problem that we are looking at, right, that these two are basically propagating left to right and this one is propagating q 1, q 3 hat right to left, okay. So this lambda can be partitioned into lambda plus plus lambda minus, okay. So lambda plus will have the right running characteristics and lambda minus as you would expect will have the left running characteristics. See, I just, so characteristics that propagate right to left, I am calling the left running characteristics, characteristics that propagate right to left, I mean left to right, I am sorry, no, I am getting my left and right confused, I see a few smiles in the, right. So characteristics that propagate from left to right, I will face the board, right, left to right as seen on the board, okay, which are like this or right running characteristics. Make sure that I do not get them, get that confused. So that is t, that is x, u, so that would correspond to u plus a and that would correspond to u, fine, right, so they will go from left to right. Thank you. See, the more I keep on explaining the left to right till you point out to me that I have made some other mistake, I mean, I see question marks on your face and that is lambda minus, okay, I mean, I have said fine, okay. So these have only the right running characteristics and these have only the left running characteristics, okay, sort of, okay, my left running characteristics, I mean characteristics that propagate right to left. So if I wanted to go back now, if I wanted to go back now to the original coordinate system, what I would basically do is I would pre-multiply this by x, you understand what I mean? So we transform from the primitive variables rho u, rho e t using x inverse to q1 hat, q2 hat, q3 hat, okay. Now I am transforming back, I am pre-multiplying by x, I am transforming back. So this lambda, I am going to see what happens to it if I pre-multiply this by x because matrix multiplication distributes over and if I pre-multiply it by x inverse, matrix multiplication distributes over addition. So this gives me a equals some a plus, plus a minus, is that okay, everyone. And in this particular case I have an easy way to get out for these equations. If I were to multiply this equation by, if I were to multiply this equation by q, what would I get? aq was e which is e plus, plus e minus, right, I have split the fluxes. Of course the other thing is you could go back there and you could multiply by delta q, okay, that is a different, that is a different scheme. So what you basically get here is therefore dou q dou t plus dou e plus dou x plus dou e minus dou x equals 0, okay, where I have basically split, I have basically split the fluxes into two parts. I will let you do the, I will let you do the algebra for this is not that difficult, right. So I will let you do the algebra for this, I am not going to do it. So what have we managed to do here, what is the consequence of this? We hope that you could, this corresponds to, what is the, this corresponds to eigenvectors that are, eigenvalues that are propagating in this direction, okay, right. So we should use, what kind of spatial derivatives? Backward space, right, we should use backward space here. So you use backward space here, you use forward space here, is that fine, okay, right. So you could in fact try instead of using just two point backward space, if you still want second order representation for your spatial derivatives, you can try using a higher order backward space, you can use a three point backward space, right. Last class when we were talking about applying boundary conditions more accurately, we have already talked about, that is what I am saying, you have all the bits and pieces now, right, it is just a matter of coming up with a scheme with all the, taking the combinations. So very often it is a matter of being alert, the individual ideas, all the ideas are there, now it is just a matter of being alert and making that match saying that oh, I can try to do this and see what happens, okay, right, is that fine. So you could use backward space here, forward space here and forward time if you want, right and forward time, is that fine, any questions, okay. So with this, this of course when I am saying backward space, forward space, I am talking still in terms of finite differences, right. So as promised in the last class, what we will do is we will try to look at finite volume methods, we have talked about it when we are doing the one-dimensional wave equation, quasi linear wave equation, we have talked about it when we are talking about the one-dimensional quasi linear wave equation. So since it is one-dimensional flow, if you want I can draw that tube again that I had, I had a pipe last time, so I will have, I will draw the pipe, so that is the pipe. So at this point we have the grid point P, at that point we have the grid point P, at this point we have P-1 and at this point we have P-1. These intermediate values are P-1.5, P-1.5 and P-1.5, is that fine, P-1.5 and P-1.5, okay, is that okay. Just like we did in our derivation, we have an outward normal n here, outward normal n there, so the control volume that we are talking about, that is the control volume we are talking about, that is the control volume we are talking about, is that fine. So if you say you have, what was the integral form of the equation, do you remember, integral form of the equation? Integral Q, if you want xi, tau or xi, t, dxi over xp-1.5 to xp-1.5, xi is some kind of a running coordinate, d by dt, time rate of change of the amount of Q that you have in, right. We will take the area to be unit area right now. That is a control volume. This is the amount of Q that is there in that volume. This is the rate at which it is changing, time rate at which it is changing, is the integral, what is it? E dot n, E is the flux, vector E dot n ds over s with a minus sign because n is an outward norm. S in this case goes from xp-1.5, right, over s. So s is a unit area. It has to be this minus that, right. So this turns out to be E p-1.5, E p-1.5, E, well, E is a vector but E is a vector that is either pointed this way, that is a vector E. It has only one component or has only the x component. Is that fine, right? E is a vector which has only one component. So E dot E here will be the E that we know, right, which is a matrix. Remember that E itself is a matrix. It is E i if you want it in Cartesian coordinates where this E is rho u, rho u squared plus p rho Et, right, and this is the unit vector i. Am I making sense? That is what I mean, okay. Is that fine? By doing this I am basically doing all the equations mass, momentum and energy conservation in one shot, okay. Yeah. So what do we get? I have a minus sign E p plus half minus E p minus half. Is that fine? Okay. I will replace this by an average value in the volume. This is what we did even there, right. I will replace this by an average value in the volume which I will call qp, okay. So this becomes d by dt qp delta x equals minus E p plus half minus E p minus half, right. And you can see that you can see one of the reasons why I want to go through that is you can see the equivalence between the finite volume method and finite difference method here, okay. That you get the same equation. That is it. We can get into all sorts of discussions about what scheme where but sometimes quite a few of these will come out with the same kinds of equations, right. But they are not quite the same, similar equations but not quite the same. What is the difference? Here we have the fluxes at, here we have the fluxes at E p plus half, we have p plus half and p minus half. Are we making sense? We have the fluxes at some intermediate value, right. So what does this have to do with, what was the motivation for doing finite volume method? Why did I suggest this yesterday? What does this have to do with application of boundary conditions? Well, what we said was instead of applying, we know that we can apply boundary conditions. We can apply boundary conditions. We now know how to apply boundary conditions on the whole domain, right. So you have two incoming characteristics, one outgoing characteristic. So for a large domain, we use this idea, right. These correspond to u, u plus c, u plus a, u minus a, u, oops, u, u plus a, u minus a, okay. These correspond to those characteristics. The proposal last time was why not shrink this down? Why not shrink this down to a small volume? Why not shrink this down to a small volume? This whole pipe that you have, we can shrink it down. So we break this big pipe into small pieces like this. I am just separating out the individual pieces. You break it down into small individual pieces like this, small volumes and look at the ideas that we generated for boundary conditions and see where that takes us, okay. Since here this material I am only going to be giving you the motivation. You understand what I am saying? That is we will, I will show you that yes, it is possible to use these characteristics. It is possible, there is a possibility that we can reconstruct, right. There is a need to find the value of the flux at the intermediate point which we have already seen. What is the way by which we can estimate it, okay. And there is a whole gamut of schemes, huge number of schemes on estimating the flux, right, at that interface. Am I making sense? So I am not, as usual, I am not going to get into it. I just want to present the basic ideas, okay. Is that fine? So let us get back. Let us look at an interface. So I have this, I have an interface. That is the interface. So that is P or P plus half or whatever it is. That is the interface. So the label itself does not matter. So at this interface what do I have? Or at this point, let us pick P. At the point P what do I have? We will look at the interface later. At point P what do I have? I have three characteristics. One corresponding to u plus a, one corresponding to u, one corresponding to u minus a. Is that fine? On the left hand side, on the left hand side I have the state. You can call it ul. So I have ul, rho l, Pl. These are the state variables that I have. Or you have q1 hat. This is going to get messy. L, q2 hat, l, q3 hat, l. Is that fine? So right here I have q1 hat r, q2 hat r, q3 hat r. Across this, across this I have a jump dq, u minus a. Across that I have a jump dq3 hat. Across this I have a jump dq3 hat. Across that line. Fine. So the question is what happens to dq3 hat. Dq3 hat is propagated along that. That is what our equations are. Q3 hat. So what we basically have here is when you cross the first characteristic, when you first cross the first characteristic, what is this characteristic say? Remember let me, let us go back to the linear wave equation. dou u dou t, dou u dou t plus a dou u dou x equals 0. So if you had a step, if you had a step, if your initial condition was a step. So this was ul, this is ur. The characteristic was 1 over, propagated with 1 over a. So to the left of this is the state ul. To the right of this is the state ur. That is ul. Am I making sense? So in a similar fashion, here when you cross the characteristic corresponding to u minus a, only q3l changes to q3r. Only q3 hat changes. Does that make sense? Only q3 hat changes. So in this in-between region, q3 hat becomes q3 hat r. In a similar fashion, so in this range from here, you have q3 hat r. What happens when you cross the u line? q1 hat will change. Am I making sense? So then beyond this, you get q1 hat r. And beyond here of course, the jump will be in q2 hat r. Am I making sense? So it is actually possible for us to use these characteristics to figure out what is this change, what is the change in state as you cross each characteristic. If you know the conditions on the left hand side. So it is like saying that if I have a small locally, if I have a small shock tube, small jump. So how is that jump going to propagate? How are those, what are the characteristics corresponding? I have a small shock tube, I have a small diaphragm, small pressure difference. I break the diaphragm. And we have a shock tube problem. And we are asking the question, the jump is very small. So I seem to have assumed here, whatever I assumed here in drawing these characteristics like this, I seem to have assumed A is constant. I have seemed to have assumed that capital A is constant. I seem to have assumed that capital A is constant. Because capital A is constant, propagation speed and the eigenvalues are constant. And therefore these curves have a constant slope. So locally around here or if A is constant or if the jumps are very small, so that A is constant, we can derive linearized equations. You have seen it in gas dynamics. I am not going to derive the equations here. But if A is comes from a system which is sort of a small perturbation system or A is essentially constant, then these characteristic lines turn out to be straight lines. These characteristic lines turn out to be even otherwise they are straight lines. And when you cross them, you have to think about this. And when you cross them, the q1 hat, q2 hat, q3 hat, depending on which characteristic you cross, the state will change. So it is actually possible for us to, it seems starting from here, moving into the future to predict what is going to happen. It looks like that. There is a possibility here. Now we go back to the volume. Now if you look at the volume, it is like you have the u-A characteristic that is coming in. This is at time q. This is at time q plus 1. You have the u-A characteristic coming in. And you have the u plus A characteristic. Again, as I said, just to give you trouble, I deliberately choose it so that it goes down, goes through. But anyway, it does not matter. It does not matter. So this is u plus A. That is u. So in this volume, we actually know what, we can actually take, you know what are the states at the various levels, we can actually reconstruct. We can actually reconstruct a function. Am I making sense? If I tell you, what is on this side, what is on that side, it is possible for me to actually reconstruct the function along this line. Because I know what is the value. I know what is the value through this, through past u. I know what is the value that is coming in, say, past u from there. So I know the value of, I know what is the value of q1 hat there. I know the q1 hat. I know what q1 hat is doing there. Am I making sense? In a similar fashion, I know what is the q1 hat. I know q1 hat, for instance, is doing something like that. I know q2 hat is doing something similar. q2 hat is basically just a constant. q2 hat, q2 hat are not to scale. q2 hat, these graphs are not to scale. q2 hat is just doing something like that. And q3 hat again is the same thing. q3 hat may be somewhere in between. q3 hat is, so in the volume, I know the variations. In fact, right through, I know the variation of q1 hat, q2 hat and q3 hat. I can actually reconstruct. So the whole class of schemes that are basically based on this kind of a reconstruction. Is that fine? So it is possible for us to reconstruct and get, so using, it looks like we are sort of using this idea of characteristics. We are sort of using this idea of characteristics. It is possible for us to come up with a class of schemes. As I said, what I want to do is I want to sort of bring this just to give you an idea of time where you are. We are most probably now by way of large volume of research somewhere in the mid-80s. By the time the large volume of research started, we are most probably somewhere in the mid-80s. Back here, is there a way, there is another issue, there is another factor that I want to talk about. So if I have p-1 and I have p, okay and we have the issue of finding E at p-1. We have the question of finding at E at p-1. We have the issue of finding E at p-1. Of course, we can look at these characteristics and try to estimate the state at that point using these characteristics. The other possibility is, and in order to do that, of course, there are other possibilities. What were the possibilities we looked at right in the beginning? We can just take averages. So we could say that E p-1. Let us get rid of the simple possibilities first. This is E p plus E p-1. We will just eliminate, first eliminate, just recollect what were the other possibilities we had. E p-1. These are possible definitions. I will say E p-1 for all of them. These are possible definitions. I am not going to use different symbols for E p-1. Could be E of q plus q p-1 where q p-1 is q p-1 plus q p. Am I making sense? q p-1, q p-1. Thank you. q p-1 plus q p. This is the other way to do it. These are two possible ways to do it. So basically what you say is you say that the E, the E, the state there is determined by the value of E there is determined by the state of q there. One possibility is that you say that actually I do not need the q at the interface. The value of E there is just the average of the E is at the q. These are what we have considered so far. This is what we have considered so far. What we have to realize is that we are trying to do a interpolation of some kind. What we are trying to do is an interpolation of some kind. We have seen that if we knew the characteristics and A was constant and A was constant, we could actually do an interpolation of some kind using characteristics. The question is what is the A? What is that propagation speed? You have a state A here, you have a state A there. You have an A, you have a capital A. You have a capital A p-1 here, you have a capital A p-1 here. What is the A that you use? One possibility is you take the average of A. See there is no limit to this. One possibility is you take the average of A. Am I making sense? One possibility is that you take average of A. What else can we do? In order to do what I indicated earlier, you say A p-1.5 is 1.5, A p-1 plus A p. So that you know the A. Then what? These are possibilities. But actually there are certain basic relationships that I want. The A is, the E is and the Q is to satisfy. And this satisfies one of them. This basically says that E at this point is E of Q at that point. I mean you are starting with the Q. Am I making sense? This is just like I said earlier if you extrapolate both pressure, if pressure is prescribed at the exit and you extrapolate both the density and temperature, then it is possible that you actually have an inconsistency. You have to be careful because there is an equation of state that ties all of them together. So in a similar fashion, we have governing equations and relationships. If you just start taking averages, then there is always this fear that there is going to be a mismatch. So what we would like to do is we would like to look at, get back here, we would like to look at, we would like to look at some relationship between A, E and Q and that we have in the definition. That is A equals dou E dou Q or A is a derivative. And therefore we know that D E is A dQ, right. This is the definition of the derivative. Definition of a directional derivative. I am going to draw this as though I am doing it in regular one-dimensional dealing with scalars. But actually we understand that this is a vector of 3 elements, that is a vector of 3 elements, right. And this is a 3 by 3 system which is 9 elements. Am I making sense? But I am going to do the explanation using just scalars for now. So what are the things that we have been doing so far? Okay, what are the things that we have been doing so far? Everything that we have done so far, this is the x axis. This is either Q or E, whatever dependent variable. This is what we have been doing so far. So we have P minus 1 P and we want the intermediate value somewhere, right. It is possible that this intermediate value, it is possible that the intermediate value because we are taking equivalent rules is actually in the middle. If you are taking unequal intervals, it may not even be in the middle, right. That brings up a whole other bunch of issues as to taking the averages and so on. So when we just take the average EP plus half, EP minus half, what are we saying? So if you have a variation of the, if you have a variation of E, if you have a variation of E or you have a variation of Q, by taking EP plus half, EP minus half, what are we saying? What is the value that we are taking in between? You are joining these two by a straight line and taking the mid-value, right. You are taking the mid-value. That is basically what you are doing. And of course depending on how this function varies, that mid-value may or may not be a value at that point, right. The function could essentially be, the function could essentially be flat and then drop rapidly. Am I making sense? Okay, right. So this is, it is a matter of representing. What is it that we are representing? Am I making sense? Okay. So if you look at, if you look at dou E dou x, if you look at dou E dou x, if you look at dou E dou x at this point, look at dou E dou x at that point, it is the tangent to the curve. I will draw a bigger picture. Let me just forget the coordinates. I will just draw a bigger picture. Okay. So you could have a nice smooth curve like that. You could have a strangely behaved curve like that, also a nice smooth curve. This is the chord. That is the chord for both of them. At the midpoint, this is the tangent. That is the tangent for that. At the midpoint of course, this is the tangent for that. Is that fine? And at this point, dou E dou x, dou E dou x, dou E dou x could either be that tangent or this tangent. If you were to take an average, you would actually be representing it by the chord. Am I making sense? Okay. So the chord in some sense, yeah, it is mean value theorem says, in fact for the white one, you can see that it is almost parallel. Mean value theorem basically tells us that somewhere along the line, it is going to be parallel to it. It just so happens that in this case, in the case of the red curve, it is somewhere there at some other point. The point can vary. It can be at different points. But it is possible for us to actually approximate this. In reality, what we want to do is we want to approximate this by a straight line. Fine? In reality, what we are trying to do, in reality what we are trying to do is we want to approximate the function by a straight line. You think about it. That is basically what we are trying to do. We are trying to approximate the function by a straight line. The normal two possibilities that you have is either the tangent or the chord. You either approximate the function using the chord or you use the approximate the function using the tangent. Fine? Because we we are actually performing an integration finally. So you are trying to approximate the function okay. The other way to look at it is we are trying to approximate this tangent using this chord. But typically we are trying to approximate the function using the chord. We are trying to find some linear approximation to our function. That is typically what we are trying to do okay. Then the question is what is that, what is the best possible approximation okay. The two things that we looked at is we do a linear interpolation for Q or we do a linear interpolation directly for E. The two possibilities we considered. The third possibility that we just threw out now is that we try a linear interpolation for A, which is the slope that we are looking at. Am I making sense? Which is the slope that we are looking at, fine. It is possible that the error that we are making, the reason why we have difficulties with this, the error that we are making is that we are interpolating these quantities, we are interpolating these quantities on the x coordinate axis right. But in reality E is a function of Q and this is really what we want to approximate. This is the tangent, this is the tangent. This is the tangent not dE dx, not dQ dx. The tangent that we are talking about is here. This is the tangent that we are trying to represent okay. Am I making sense? Or E as a function of Q is what we are trying to represent. And the question is, is the tangent better or is the chord better? Okay. So the chord typically, especially when you look at functions like this red curve here, the chord typically is on the average a better representation okay. So basically what you do is we convert these increments and we will say we look at delta E equals sum A delta Q, delta E is sum A delta Q. This A is some average A, this A is some average A. Am I making sense? This A is some average A that needs to be determined. It is some average A and not necessarily this A. It is not necessarily this A. This A is some average A that is not necessarily this A and it needs to be determined. So in the E Q, I write E and Q but I am only drawing one coordinate right. I am only doing it, still doing it only as a scalar. In the E Q space, we know what is delta Q and we know what is delta E. What do we want to find? We want to find the A. We want to find that average A. So we want to find the average A. As I said, I draw one dimensions, in one dimension, given two points, the line through it is unique. In two dimensions, if I have a surface right, E Q, E as a function of Q, let us not look at it 3 by 3, let us look at it in two dimensions right. So you have Q1, Q2, E1, E2. So we do not forget E1, E2. Let us look at this E1. You have Q1, Q2. You have a surface. I give you two points. Then infinity of planes that go through that point, those two points right. So we need some way, we need some way to make sure that we are picking right. We need some way to constrain. So what we basically say is, I want this A of A right. I want a Q. I want a Q P plus P minus half or P plus half or whatever. I want a Q P plus half, P minus half. I want an estimate of that. I want this A to equal A at Q P minus half. Am I making sense? See the geometrical, we will get back. Just in case, I mean just in case you did not get the why there are multiple planes. You look at it directly here. What do we have? How many components do we have here? This looks like an equation Ax equals B right. This looks like this is a vector matrix vector right. So normally what you do is, you are given this, you are given that and you are asked to find this. That is the usual problem that you are used to in linear algebra, matrix algebra. Here the problem is, you are given delta E, you are given delta Q. You want to find A. How many unknowns do you have in A? You have 9. Well, if A is going to have the structure of, is going to have this structure, you may not actually have 9. But in theory you have 9 right. But you have only 3 equations. There are lots of solutions. There are lots of solutions. But fortunately for us, this A, the first few entries of this A are something like 0 1 0 or whatever it is. There is at least one line of them that is just gone okay right. There is one set of them by placing this constraint, by placing this requirement which we can argue in a physically intuitive fashion right. We have sort of eliminated, we have eliminated right that it have this structure. We are saying something about what does the structure look like okay. By insisting on a certain structure, we have already eliminated a whole bunch of unknowns. But the difficulty that we have is that this, we need given this vector and given that vector, we need to determine A. That is the problem that we have. Is that okay? What is delta E? I will just set it up and see what difficulties we have. What is the problem? And then we will see whether we are able to okay. What is delta E? Delta rho U, delta rho U squared plus P delta rho E t, rho E t plus, I am thinking of delta Q, rho E t plus P times U. It looked a little too easy okay. And of course you can expand this out. But we leave the delta rho U as it is right now. It is delta rho U squared plus delta P and this is delta rho E t, delta rho E t times U plus delta P times U. Fine. What is delta Q? Delta Q is delta rho, delta rho U, delta rho E t. Do you remember the values of A? Does anyone have the values of A? Otherwise I will, we have the values of A, gamma minus 3 by 2 U, P minus gamma, gamma minus 1 into U. Yeah. That is what we have. Actually there is a reason why I expanded this. Normally I would not go through this headache. There is a reason why I expanded this. What do we have here? Just by inspection because this is, we are setting up for, we are setting up for the next class basically. What do we have here? We have E t, we have P. Really if I look at this in terms of the kinds of unknowns that I am going to have, I have E t's and I have a P. I have a rho and U. Is that it? Now you think back to your thermodynamics. You think back to your thermodynamics. There is a combination of E t and P that we have a term. If we switch instead of total energy, if we switch to total enthalpy, then we can absorb the P. You can potentially get rid of the P from the equation. Am I making sense? We can potentially, so we will retain that is, the reason why I am writing this out is, we will retain the A. Remember if you did a change of variables, the Jacobian equation changed. We are going to retain this equation. To this equation, to the Jacobian equation we are going to do a change of variables. Am I making sense? So what we propose to do now is, tomorrow I will write out the equations in terms of enthalpy. So that I will have a total enthalpy here instead of the pressure term. So we will try to write out the equation in terms of the total enthalpy. See if we can get rid of the pressure in some form. So that all the pressure terms will be written in terms, then we will have only the enthalpy rho and u. And see where we can go from there. Is that fine? So in tomorrow's class what I will do is, I will start off with that. I will start off with the enthalpy part and we will go back and we will try to solve this equation. We will try to solve this equation. We will see what this gives us when we try to get in terms of delta rho, delta rho, delta rho u and delta rho ht or something of that sort. Fine. And why are we looking at deltas? Because at one level those would be the jumps across any given interface. Across the, they would be basically, so we have looked or you can look at it as, if I have two points either specially distributed or distributed in our state space so to speak. So if you have u left and u right and you are trying to figure out what is the interpolant that I should use given that I have u left and u right, we can actually use it, do it here instead of doing it there. Especially if you are talking in terms of I have a small jump, then this kind of a discussion does not make sense. Even these graphs do not make sense. Whereas a discussion here in the function space still makes sense. Is that fine? Let us see idea. So in the next class we will basically derive the averaging process. After that on the subsequent class I will try to do a demo for you. And then we will move on to other, we will leave 1D flow equation at this point and go on to other schemes. Fine. Thank you.