 So, in the last class we saw that Laplace's equation discretized could be written as a system of equations. In fact, I just wrote that as Af equals some b to keep the conversation general, keep the conversation general, right. I will switch to the standard notation that you are used to, which is Ax equals b. I will put the tilde under the x to indicate that it is a vector, fine, and also so that we do not confuse it. I mean, after all we were doing x, y coordinates and so on, so that we do not confuse it with the x, y there. So, we are basically solving the system Ax equals b, okay, and there were two schemes that we saw. One was the Jacobi iteration and the other was basically Gauss-Seidel. You may not have realized it. The other scheme that we basically used to solve that system of equations was Gauss-Seidel, okay. Now, what I want to do is both of these schemes are iterative schemes as opposed to direct schemes like Gaussian elimination, right, or an equivalent which is LU decomposition, okay. So, we are not talking about using a direct method. Direct methods like we would not do this like either Gaussian elimination, we are not going to do this, or equivalently LU decomposition. You can actually show that these two are equivalent to each other, okay. The elimination in the back substitution part turns out to be same as doing the LU decomposition. So, we are going with iterative schemes, okay. We are going with iterative schemes to solve this. In the context of Laplace equation, it is called a relaxation scheme, okay. Now, what we will do is we will see, we will try to write these in the iterative form that we have actually written for phi, okay, in a recursive form that we have actually written for phi. In order to do that, I need to partition A. I want to partition A. That means I want to write A as the sum of various parts as the sum of three parts, D, L and U. This is basically where we entered in the last class. So, if the entry is in A or Aij, the entry is in A or Aij, the entry is in D or Dij. Correspondingly, you have Lij and Uij. I am defining my D as the diagonal terms D, how am I defining D? Dij equals Aij if i equals j equals 0 otherwise. So, those of you who have already seen upper triangular and lower triangular matrices, this may just be a repeat, but let me just for completion, let me just write this out. So, for L, Lij is Aij if i less than j, okay, equals 0 otherwise. And U, which is Uij is Aij if the opposite of that, i greater than j and 0 otherwise. So, basically this will give me the diagonal elements, this will give me the elements below the diagonal, this will give me the elements above the diagonal, okay. So, I have partitioned my matrix A in this fashion. Let us just take one equation that we did from Jacobi and see what we are talking about. So, with the Laplace's equation, when we were doing the regular Jacobi iteration, basically was phi ij at n plus 1 is 1 quarter. I will write a 0 here, which was the right hand side of the equation, del square, nabla square phi equals 0 plus phi i plus 1jn plus phi ij plus 1n plus phi i minus 1jn plus phi ij minus 1n. This would basically be, the 0 here is basically the right hand side, right hand side happen to be 0, okay. That is the entry in B, okay. So, if I look at this Jacobi iteration, basically this, what is this 0.25? That is 1 divided by 4, this is 1 divided by 4, which is essentially D inverse, corresponds to D inverse. So, looking at this by inspection, I say that my x at n plus 1 is D inverse B minus l plus u times x at n, is that fine, everybody? So, I can rearrange terms. I get xn plus 1 is some p times xn plus c, right. And what is p? What is the matrix p? Minus D inverse l plus u, okay. This is called an iteration matrix. So, p is a iteration matrix. We can write it in this form, okay. So, this is p, that would be Jacobi. I will put a j there, p Jacobi, okay. What about Gauss-Seidel? Gauss-Seidel is going to have, if we did Gauss-Seidel instead, we had phi ij n plus 1 equals 0.25 times, my right hand side happens to be 0. I am just picking some arbitrary entry, plus phi i plus 1 j n plus phi ij plus 1 n plus phi i minus 1 j n plus 1 plus phi ij minus 1 n plus 1. This is of course assuming that I am going from the lower left hand corner, right, left to right bottom to top. There is an implicit assumption on the sweep direction, okay. So, these basically come from the n plus 1 side. In a sense, that is like saying that we have, we are solving, let me just, so let me take these over to that side, right. Let me take these over to that side, okay. So, that would be like saying, so I am undoing whatever we did, I am undoing it, right, to just get the matrix form, okay. So, what do we have? So, we have phi i minus phi i minus 1 j n plus 1 minus phi ij minus 1 n plus 1. You can either multiply it by 0.25 or we can take that, get that 4 out, 4 times phi ij n plus 1 and this equals 0 plus phi i plus 1 j n plus phi ij plus 1 n. What is this? This is the diagonal term. This is d. This is l, right. So, this is l plus d minus l d minus l into phi at n plus 1 equals d plus u times, using x, x tilde, u times xn. Is that fine? So Gauss-Seidel turns out to be xn plus 1 equals d minus l inverse b plus u xn tilde. We have to be a bit careful with the signs because really the way we had written a, this would have been minus 4 and those would have been plus 1. You have to be a bit careful with the signs. You have to be a bit careful with the signs. Should it be d plus l? Have I made a mistake? They are opposite signs. You are right, d plus l because the signs are right. D plus l. Fine, fine. Yeah. The diagonals and the off diagonal terms are opposite signs in the original matrix. So, actually in the original matrix, from the original matrix, this is actually minus d, minus off d plus l. From the original matrix, this is minus off d plus l. And this is, okay. You have to be a bit careful. Okay, that is fine. Right? Yet again, we are able to write, we are able to write that matrix for Gauss-Seidel, Gauss-Seidel as xn plus 1 is p Gauss-Seidel xn plus a c. The c is of course also a c Gauss-Seidel. Right? The c's are not, c's are also, c's are also dependent. Is that fine? Everybody. Okay? And the p is called the iteration matrix. We can actually figure out now that you can ask the question if you are going through, we have the iteration matrix. If you are going through the iteration, you are generating given an x0, you are generating a x1, given the x1, you are generating an x2, you are generating a sequence. Now we have a form that relates one to the other in a simple, right, equation. Something at least that the chalk dust looks as though it can be written simply. We can ask ourselves the question, is there a way for us to do an analysis of this? Is there a way for us to figure out to analyze this, to analyze this problem? So does it converge, does the sequence that we get, does it converge? In the last class, we saw that we could use the Cauchy test to decide whether convergence occurs or not. Here we will see what else we can do. So we start with this equation. And so I will just write this in this general form, either whether it is Gauss-Seidel or Jacobi iteration, whatever iteration. We will see how to do the analysis for that. So coming back here, what we basically do is, so the general form that we have is x at n plus 1 is some p times x at n plus a c. This is the general form that we have. Is that okay? So you will guess an x0 and get a, and get an x1. That is what you are going to do. And then get an x2. From there you are going to get an x2 and so on. You are going to generate the sequence xn. The actual solution to this equation we will represent by x. So the solution that we want is x. Is that fine? What do I mean by that? That means that if x is the solution, this implies x equals px plus c. So from this equation, if I subtract out the equation with the solution, that satisfies the solution. From this, if I subtract out that, okay, if I subtract out that, I will get an error. Essentially what I am saying is, if at any given time I have a candidate solution xn, right, I know my actual solution x, the difference between them I will define as the error en. It also happens to be a vector. Is that okay? That is fine. So if I subtract from this equation, I subtract this equation. This is, both of them are linear. Fortunately I am using that fact. Then I get the iteration equation en plus 1 equals p times en and by converting to en, what I have achieved, what is the difference? There is a difference. The c is gone. The c is gone. And now I can ask the question, do the sequence of en go to 0. I am generating a sequence. So my iterations are basically generating e0, e1, e2, en and the question is, does the sequence converge and you want it to converge to 0? In this case, we know, we know earlier it had to converge to a solution. We do not know what the solution is. In this case, we know that this has to go to 0. Fine. So in order to do that, if this were a scalar equation, you do a ratio test. If this were a scalar equation, you do a ratio test. So let us look at how we go about analyzing this. So please bear this in mind. We are going to come back to this equation now. I am going to do, I am not actually going to do fixed point theory, but I am going to talk about something called a fixed point theory. We are not actually going to do fixed point theory, but we are actually going to talk about fixed points. So in general, an equation of this form, so what is a fixed point? I have suddenly introduced this idea of a fixed point. So x is a fixed point or e here is a fixed point. If it turns out that you actually get psi equals p times psi. Psi is a fixed point. So if you substitute psi back into the equation on the right hand side, you get back psi. So when do we do this? In general, this equation in general looks something of the form xn plus 1 equals g of xn. g is some arbitrary function. So most iterations will look like this. You will have some function. You give me an xn. You give me a guess. We perform some magic and you get a new xn plus 1. So and this is a particular case of that. So the question is if you have something of this nature, if you have something of this nature, when does it have a fixed point? That is when will I have a psi so that psi equals g of psi. I am not going to talk about it in this general context. I will still talk about it only in this context, but I want you to see that. So that is a question that you can ask. And what do g in my p? What do they do? What do they do? Given a psi, so what g and p do is if you take xi in the domain of definition of g, you understand what I am saying. So those are the set of values for which g will give you an meaningful answer. It is defined. g is defined on those set of values. It will return xi that you can plug back in. That means g maps psi basically back into the domain. The domain and the range are the same. It is important. So this equation, the way we have set it up, the way we have set it up for it to work, g has to map any argument of its own back into the domain. Am I making sense? So g or p, for instance, if you say that I have some interval on which g is defined, so what g has to basically do is let us take x constrained, if we constrained x to 0 to pi by 2 and I am looking at x equals sin of x, xn plus 1 equals sin of xn. Am I making sense? In fact instead of pi by 2 I could even taken 1, 0 to 1. Then it turns out that if I even taken x is in 0 to 1 then this will constrained. This will map back not constrained. Sin x maps back into the interval 0, 1. If I start off here and guaranteed that I am going to be in that interval. Sin x is not going to take me out of that interval. Does that make sense? So first that it maps back into itself. The second question that we have is, is that enough for us to be able to guarantee that there is a xi which is g of xi and that we are going to get to it. So the fixed point theory basically tells you when you have a fixed point. As I said I am not going to really sit down and state and prove the theorem but we will just look at the essentials of fixed point theory that we require because it is pretty intuitive. Let us take an example similar to that but let us take a scalar. So I will remove the tilde. I have an x, no tilde underneath. This is a scalar. x equals alpha xn plus 1 is alpha xn. These are just numbers. What will this generate for a given alpha? What is the kind of series this will generate? What is the sequence it generates? It will generate a geometric sequence. So when does that converge? So you want mod alpha less than 1. So mod alpha less than 1 will guarantee that this will converge to, there is a fixed point. It will converge to 0. So there is another situation where there are fixed points. If alpha equals 1, then every point is a fixed point. If alpha equals 1, every point is a fixed point. So mod alpha less than 1 and alpha equals 1. Both of them will work. We will see where this takes us. So it is possible that if mod alpha less than 1, you will generate a sequence that converges. And basically there in a sense you did a ratio test. Actually doing a ratio test, not the Cauchy test. What if I had a second sequence? yn plus 1, now I will make this alpha x, alpha y, yn. So you can ask me the question where are these coming from? Where are these sequences coming from? So these sequences correspond, I am not going to give you an actual problem but let us in our mind imagine that we are solving a problem. And because of knowledge that I have, I pick a coordinate system and I pick a convenient coordinate system. I pick a convenient coordinate system. I look at my problem and I pick a coordinate system. I have some knowledge in my fluid mechanics or whatever it is. I look at the problem and say this is a convenient coordinate system and in this coordinate system I get these two equations. Which if I iterate gives me some answer to a fluid mechanics problem. It does not matter what is the problem. What is important is that I pick the convenient coordinate system and that I deliberately set the convenient coordinate system at an angle here obviously. The coordinate system is not, is convenient to the problem. So I picked a convenient coordinate system and I have these two equations that come out of my problem just like we got a system of equations from Laplace equation. In a similar fashion I do some magic in my fluid mechanics or whatever and I end up with these two equations and if I solve these two equations I solve my problem. That is all we need to know. We do not need to know what the problem is. So I can now generate iterates. So I will generate a sequence xn, x0, y0, x1, y1, x2, y2, xn, yn. Am I making sense? And when do these converge? When do these converge? You want mod alpha x, you want mod alpha x less than 1 and you want mod alpha y less than 1. Is that fine? Or in general, I can write it in a different way because I have an AND and it is a max of mod alpha x, mod alpha y less than 1. Maximum of those two is less than 1. Then that x0, y0 that sequence will converge. Fine? Any questions? So if I have three equations instead of two equations then it will be max of alpha x, alpha y, alpha z and so on. The largest one has to be, the magnitude has to be less than 1. But see in real life this does not always happen this way. Normally what happens is we have a problem we may not know, we may not know of a convenient coordinate system to start with. So this is basic, this coordinate system here, I have clearly drawn it this way, it is a setup. In reality what you would have done and what you have seen we always do is I will draw a coordinate system this way. I will have an x prime, I do not, I would not call it something else. Xi eta coordinate system this way. In real life that is what would happen. You draw a coordinate system typically parallel to your page because you have no other reason to pick a coordinate system. If you had knowledge about the solution you would pick a convenient coordinate system. But sometimes we do not do, it does not happen. Sometimes we think deep, think of a coordinate system, draw it and it turns out it is not really the convenient coordinate system. We try our best. So let us say you have picked this coordinate system instead. But this is the underlying sequence that you are going to generate. So how do I transform this into the new coordinate system? I just have to rotate the coordinates. In order to rotate the coordinates, I will first write that xn plus 1, I will write those as in a matrix form because I am going to do some matrix manipulation here. Is that okay? This is the same iteration matrix and these converge, these that is if I call this xn plus 1 is some, let me make it capital lambda. This is this matrix. It contains alphas, right capital Y and xn. So the matrix form of this equation can be written in this fashion. And it looks suspiciously like what we were talking about earlier. This looks, the reason why I am doing this, it looks very much like en plus 1 is p times en. That is where we are going. And this sequence of x is x0, x1, x2, xn converges if the maximum of alpha x and alpha y are modulus alpha x, alpha y are less than 1. That is what we have. Now if we had picked a different coordinate system which was at an angle theta with respect to the convenient coordinate system xy, if we had actually picked a different coordinate system, then we can go from one coordinate system to another coordinate system by performing a rotation, okay. So rotation matrix R basically is cos theta sin theta, these are entries in a unit vector, right, minus sin theta cos theta, orthogonal matrix. So if I take this, if I take this equation, if I take this equation and pre-multiply it by r, so I pre-multiply it by r, r xn plus 1 will give me psi, okay, the vector psi which consists of the entries. So r xn plus 1, let me see, I will write that here. So this equation turns out to be r xn plus 1, I am sorry, r xn plus 1, we need to give it some nice symbol. I will use, I will use, the capital psi does not look that good. So I will use psi with a tilde underneath, okay, just bear with the notation, right. So that if I take this equation r xn plus 1 equals psi n plus 1 equals r lambda, so I stick an r, r inverse, r inverse, r there xn. So this tells me psi n plus 1 equals, if I call this matrix A, A times psi n, okay. These are the vectors that we get in the second set of coordinates, right. So depending on your coordinate system, the equations that you get will change. The iteration goes on but if you could rotate from one coordinate system to another coordinate system, you may actually find that there is a convenient coordinate system in which the equations decouple, right. So all of you know where I am headed now, right. So this matrix r matrix is often called the modal matrix. We will come back to this later. We will need this, okay. And the alpha x and alpha y, what are these called? Alpha x and alpha y are eigenvalues, okay. And these of course will correspond to eigenvectors, right. These are eigenvalues. So if you have a matrix A, it has eigenvalues which are alpha x and alpha y. So if you have to do this iteration psi n plus 1 equals A x psi n, if you have to do this iteration, right, or you have to do the iteration En plus 1 is PEN, what we are going to say is if this sequence of psi that you get will converge, max of the eigenvalue is less than 1, modulus of the eigenvalue is less than 1, the largest eigenvalue is less than 1, is that fine? Everyone, okay, right. So En plus 1 equals PEN, this iteration matrix generates a sequence and that sequence converges if the largest, the modulus of the largest eigenvalue is less than 1, okay. So now we have reduced finding that fixed point to finding out what is the largest eigenvalue. And if you know that the largest eigenvalue is less than 1, you are set in the sense that at least you know that you have, if you are going to get to the fixed point. So how does this work? This is, this by the way, this mapping, right, I forgot to mention this earlier, this mapping or that xn plus 1 equals g of xn, see this maps into itself but this mapping when mod alpha x is less than 1, right, basically what is happening is there is a shrinking that is happening, it is called a contraction mapping. That is if you take values in a certain interval, it will map into a smaller interval, it is called a contraction mapping. I will just write that here, contraction mapping. So the mapping is not only into itself, it is a contraction mapping. The typical example that I would give for this would be, you imagine, you can imagine a steel rod but it is easier to think of a piece of sponge, right. So you take a prismatic piece of foam or something of that sort that you can squeeze with your hand easily. So you sit down and make markings on it at equivalent rules. So if I take that foam and the contraction mapping that I do is that is easy. I squeeze, right. So I hold the foam in my hand or I hold it in a vice, something of that sort and I squeeze. So when I squeeze, the right extreme moves in, the left extreme moves in, all of the points move. There is one point that does not change. You understand what there is a fixed point. Even in that example there is a fixed point. So if I go through a contraction mapping, there is still a fixed point. There is one point that does not move. Am I making sense, right? I mean I do not have the foam in my hand but I think you can imagine it, right, okay. So there is one point that does not change. So if you have a contraction mapping, right, well yeah, you could have a translation which is why I got rid of the C, okay. So you see that is the issue. So if I were to translate, if I did not, if I did not push them together, if I did not push them together in the proper fashion, it is possible that I could actually have a translation. Now you will know why I moved to the En plus 1, okay, because I had got rid of the C. I want something that is linear. I want something that is linear. Remember, remember, this is a standard confusion. The standard equation, this is outside the full normal discussion, Mx plus C, y equals Mx plus C is not linear, right. This is not linear. It is a straight line. Unfortunately, linear also means line, right. So curvy linear does not mean, then you will get confused. If you assume linear means straight line, curvy linear does not make sense, right. Curvy linear means the curved line, right. So y equals Mx plus C is not linear in the sense of a function is linear, okay. Is that fine? Remember that. So we got rid of that C by going, switching to the error, right. It is a little game that we played. Now, so if you have a, so the fixed point theorem basically says, right, you have this map. I am not going to go through all the details, but you have a map back into the domain and it is a contraction mapping, right. Then there is a fixed point. And you can get to that fixed point in this case because we are generating a sequence that will help us converge. So our automaton will actually converge because of this, because of the fact that we have a contraction mapping. Is that fine? Are there any questions? Okay, no, that is fine. So now the thing is how do we deal with this? We have reduced ourselves going from En plus 1 is P times En. How do we find out what are the largest eigenvalues and eigenvectors? Instead of dealing with this directly, I will deal with Laplace's equation itself directly. So in Laplace's equation if phi is the solution and phi tilde or phi capital phi is a candidate, right. And this of course will give us a corresponding E. This will give us an E, okay. And we have seen, when we talked about uniqueness, what are the boundary conditions that E satisfies? Well, this is, well let us call it En. What does the boundary condition that En satisfies? It will be 0, right. The boundary condition will be 0 because we have subtracted from the original which satisfies the boundary condition. See the candidate has to satisfy the boundary condition exactly. You cannot violate, the boundary conditions are inviolate. You cannot violate the boundary condition. You can do whatever you want on the interior. You cannot violate the boundary conditions, okay. So the candidate solution also satisfies the boundary condition. Therefore this is 0. This has homogeneous boundary conditions. It is 0 on the boundaries, okay. So how are we going to go about solving this? I cook up, I do use my Fourier series. I will write Fourier series in two dimensions. If you have not seen this, you will see how this works. So I look at this function Exy, Alm, okay, Alm. You know that these are Fourier coefficients are not the entries of the matrix A. You understand what I am saying? Do not confuse them with the entries of the matrix A. Exponent I pi Lx by L, L is the length of the domain. In our case it is 1 but I will just leave it as capital L. Exponent I pi My by capital L, right. So instead of a 1 by 1 unit square I have an L by L. This is summed over M, summed over L. I know I am going to use a grid of a certain size, right. If I have n intervals, then what is the highest frequency that I can represent? So here I do not have a 2 pi. It should have been 2 pi, right. It is only pi. So this goes from M equals 1 to n-1, L equals 1 to n-1. I have dropped the M equals 0 and L equals 0 because my boundary conditions are 0, right. My boundary conditions are 0, Laplace equation maximum minimum occurs on the boundary block, okay, fine, right. You know that it is going to go to 0 anyway. So I have dropped the DC complex, M equals 0 complex. So I have M equals 1. So what is the relationship between ex y and Laplace equation? If I substitute this into Laplace equation what will happen? If I substitute this into Laplace equation what will happen? Nabla squared, what is Nabla squared of ex y? Minus pi, pi squared L squared by L squared minus pi squared M squared by L squared, is that right? You can just check to make sure that I am not. So if I take any one of these for M equals 1, 2 or 3 or what if I take any one of these and this is what I am going to get, is that fine? Just substitute it and try it. Just substitute it and try it. So these are basically, now look at what we have just done. These are basically Eigen functions. These are Eigen values. These are Eigen values and these are Eigen vectors or Eigen functions, okay. They are Eigen vectors or Eigen functions, fine. So these are Eigen functions or Eigen vectors. Why am I doing this? Why is it important that they are Eigen vectors or Eigen functions? Because I know that I want to get my Eigen values in the largest Eigen value. Am I making sense? I know from my iteration matrix somewhere along the line, I want to do the ratio test. Let me not just say that I want to get the largest Eigen value. The largest Eigen value apparently comes from the ratio test. So I will do the ratio test in some fashion, then ask the question for what value will the ratio test, right? Give me the smallest change, the smallest ratio or I should say the least change, right, okay. So ex y, so it turns out that, it turns out that any one of these, any one of these, right, any one of these, which we call ELM or whatever, or any one of these can be now substituted into our differential equation or finite difference scheme and we will see what it is that we get there, okay. So what do we have? So we have, let us pick Jacobi, do Jacobi iteration, do Jacobi iteration. So Jacobi iteration, so I will write this n plus 1. I have done this so many times. These i, i plus 1 are spatial coordinates, the spatial coordinates. So we can actually substitute. We can sample the function ex y, we can sample the function ex y, right and substitute in here. Am I making sense? In a similar fashion, I guess I should not have written it this way because this will cause you to think that there is a b also. Maybe I should not have written it in terms of phi, I should have written it in terms of, call En also satisfies, E also satisfies del squared E. Does it satisfy Laplace's equation? Does del squared, is del squared E equals 0? So we want, this will be, this will be for, this will be for any n, okay. Del squared E equals 0, what is this going to give us? Let me see if we go through this. En plus 1 equals, I am going to substitute now. The function phi I am going to represent in terms of, the function E I am going to represent in terms of these exponentials. So what do I have? En plus 1 at ij is 0.25 times En at ij, En i plus 1j plus En at i minus 1j plus En at ij plus 1 plus En at ij minus 1. They are all equivalent rules. So what is this going to give me? i plus 1j, i minus 1j. So what is the relationship between E a y plus 1j at n and E ij? Is there a relationship between these two? One will be exponent, some constant times n delta x or n h, right? I am sorry, not n. i plus 1 h, x equals x i equals i times h because we are taking equivalent rules, right? I am slowly getting ahead of myself here. And yj equals j times h, we are taking equivalent rules, okay. So the relationship and E ij would be of the form exponent. Everything else is the same i h, okay. So the relationship between E i plus 1j and E ij would be just one, exponent of all of this coefficient times h, times h. I have made a mistake here, you guys have not corrected me, okay. What is the mistake that I have made? I am using i to be square root of minus 1, I am also using i as the subscript. I should not do that. You have to be very careful. So henceforth, henceforth, henceforth I will switch to the notation pq, p plus 1q, p minus 1q, pq plus 1, pq minus 1. You have to be very careful, right? p plus 1q, pq, p plus 1q, pq. Is that fine? Okay. Because right now, all of a sudden when I introduce Fourier series, I have decided that i equals square root of minus 1. When I could use a different i but I am likely to forget. So we will just switch the notation, fine? And in your electrical sciences, you may have used the j, you may have used the j instead of an i but it does not matter. We will stick to i equals square root of minus 1. So this equation using this information then becomes e pq at n plus 1 equals 0.25 times exponent of i pi l by capital L plus exponent of minus i pi l by capital L plus exponent of i pi m by capital L plus exponent of minus i pi m by capital L whole into e pq at n. Is that fine? Did I make a mistake? i pi l into h, you are right. Thank you. i pi l into h, h is very important here. i pi l into h, h is very critical for me. Okay. So what we will do is in the next class, we will see where we can take this, right? So in the next class what we are going to do is we will take the ratio, we will take the ratio of the two and find out what is the, what is the rate at which, right? What is the growth, the geometric growth that we are getting over the decay? What does Gauss Seidel, what does Gauss Jordan do? We will see if we can do something similar to, similar by way of convergence to Gauss Seidel and so on. Is that fine? Okay. Thank you.