 So, good morning. In this class what we will do is to look at the computational assignment, it is actually problems at 3, but I have called it problems at 4 online. And what I will do is I will try to work through a couple of the problems and as I go along feel free to ask me any doubts that you have about the assignment and then later on about the course until this point of time we can discuss some of those things. Then I will talk about the new assignment I uploaded online yesterday. The computational one and explain how you can write a code to simulate that Swift-Henberg equation so that we can test some of the things that we have done in the theory. So, this is only work if you ask me your doubts so feel free to ask and you do not have to worry about the video going on that we can edit anyways. Alright, so the first question was the 1D Swift-Henberg problem, right so u is a function of x and t and x extends in this problem from minus infinity to plus infinity. And this equation is a simplified you could say averaged version of the Rayleigh-Benard equations. So, using the method of multiple scales what has been done here is essentially they looked at the set of Rayleigh-Benard and using a suitable averaging condensed everything into this simple equation. So, quantitatively this will not cannot be expected to give you the same results, but what has been done is all the physics of the Rayleigh-Benard problem have been captured with the simple equation. The idea being that by studying this equation we can understand a lot of basic features about pattern forming systems. And this has been derived in a book by Cross and Green Side which is the book that I have used to put together this assignment and the next one as well and that is a really good book for the course. So, I will give the reference a bit later. So, they derive it from a heuristic perspective and then from a rigorous mathematical point of view also. So, keeping that in mind this parameter r is similar to the Rayleigh parameter. So, we want to find the value of r for which the system is unstable and there is a fourth order term which will come from the coupling of I think temperature and velocity and then you have a second order term. So, everything fits together similar to the Rayleigh-Benard if you just look at it from a far. So, you can be interpreted as the vertical component of velocity. So, for example, if you say I mean say you wanted to compare this with some experimental data then if I have my Rayleigh-Benard cell and it extends to infinity in both directions then as you already know at some point I will have maybe some kind of role something will be happened and if it is stable there will be no convection at all. So, u is almost the cutout along the z direction as h by 2. So, the and it is the v z the z component of velocity that I have cut out through the centre of the channel. So, now it is only of and I have considered to be only a function of x. So, I have ignored the y direction variation say that cell is quite narrow. So, I am looking at the x direction. So, with this interpretation we will move on to the analysis which is actually quite simple because it is just one equation. So, the first step is of course to linearize the problem. So, by that we would write u as u steady state plus u let us call it pot up and u steady state is basically 0 here. So, homogeneous state and that makes sense with the fact that there is no flow in the flow. So, u steady state is 0 and if you were to substitute here and retain terms I will call multiply this by epsilon for magnitude retain terms about a epsilon then we would get the and the u cube which is the nonlinearity does not survive in the linearization. Now, the way this equation has translated the entire nonlinear term has simply been knocked off but this would not always happen such a kind of thing happens only if your base state is identically 0. If I had a nonzero USS then I would have a term like USS square times u per term and other terms as well. So, you need to be careful when you are linearizing only if the base state is 0 you can write it down by inspection, but on other cases is better to substitute always and make sure. So, this is the perturbed equation. Now, the next step is to take a form for the perturbation or the normal mode analysis. So, I will say a few things about that. So, typically what we have been doing throughout the course is at this stage writing up as some constant a which is the initial condition of the perturbation. We take an exponential form in time because it is the first order system and then this is the most important part where we write down something like e to the power ikx and this means we are considering the Fourier modes. Okay, so first of all I will just go through it because some people have asked me this question. So, the first thing is that this is a complex factor but then up has to be real. So how does it resolve? Well, it resolves with the Euler identity that relates I mean complex exponential terms to cos h and sin h. So, when we look at the Fourier series in this manner we actually need to consider k's from okay because we are looking at we are not looking at a system where k would be discrete right now. If we have the entire if x belongs to the real number line so if x is a continuous variable from minus infinity to plus infinity k should take all values from minus infinity to plus infinity. So, we are looking at both k's negative k and positive k even though the dispersion curve we always plot positively it is understood that we are also considering the negative k. Now, what happens is the positive k term combines with the negative k term to give you a real function and the real function will have the form of cos and sin. So, eventually this complex guy will manifest itself manifest itself via the Euler formula as something like I mean this will be the dependence finally you will get cos kx and or depending on the initial condition sin kx. So, if you want to go plot your variation you cannot plot this you can plot cos kx and sin kx. So, this coefficient a is also complex that has to be kept in mind. So, this a is not a real thing it is complex because this is also complex. So, when this a with the positive k combines with the a with a negative k it gives you these real parts and you will find that the a corresponding to the negative k is actually a star it will be the complex conjugate of a. So, in other words I have a mode this will be one mode this mode will have to be added with a star where a star is the complex conjugate of a and that will be obvious when you look at the equation because this when you write instead of k if you take minus k it is basically looking at this same equation and replacing i by minus i which means that any unknowns will just be the conjugates of what you get in any unknowns here will be the conjugates of what you get here that is clear. So, this is how even though we are looking at the complex modes ultimately they are going to manifest in a real fashion and this will always happen in all these problems it has nothing to do with the physics and all this is just the way the Fourier modes work right. Sir. A is the combination of both real and imaginary part of a plus b i that one e power i k x can be written as cos k x plus i sin k x. So, there will be some when you multiply this is a complex term this is a complex term this is its complex conjugate right if I add it I want to get a real. Just the real part of those both. No this is some complex number I can write it as a plus i b this is the same complex number conjugated I will write it as a minus i b and then I will add it and I will get 2 a and the sin part of it I mean if you what you are saying you will add the cos parts the sin parts will also get added and ultimately give you real and if you are still have a doubt just go back to some of your old books on the Euler identity and how you solve the equation I mean this all of this business comes in this equation. So, you see how the solution here is done you here you can write it directly write sin x cos x sin y cos y solutions or you can proceed with complex exponentials and then see how the 2 are resolved. So, that should clear any doubts here. So, ultimately when you write this it means you are looking at both cos and sin and that which of them come depend on initial condition boundary condition that is one important point the other important point is that here k will k is real when this is the case. So if x goes from minus infinity to plus infinity k takes all the values on the real number line from minus infinity to plus infinity. But if x belongs to some interval like say minus pi to plus pi with periodic boundary conditions. So, when would this come up suppose I am doing a numerical simulation. So, if I want to simulate this numerically which is what we are going to do in the next problem I cannot very well have a infinite long domain I will be waiting forever for the problem to get solved. So, one way to do it is to take a very large domain. So, may be minus 100 to plus 100 where the characteristic wavelength is of order 1 then that might work. Another way to look at it is I will take a small domain like minus pi to plus pi and impose periodic boundary conditions to simulate what would happen in the infinite case because when it is infinite our argument is it has to be periodic if it is not periodic then it has to blow off infinity at one of the extents. So, realizing that it must be periodic we could take a periodic domain and of course taking a periodic domain restricts the modes that we would get. So, that is my point here if we take a domain like this minus pi to plus pi then k will belong to a discrete set and so on. So, here k takes the integral values. So, because minus pi to plus pi so it will be sin x sin 2 x sin 3 x. So, it has the wave has to fit properly in the periodic domain it cannot you know fit half of it. If you had minus l to plus l then k will belong to 2n pi by l actually sorry I think this you will have minus n pi by l yeah it is 2n pi by l 2n pi by 2l I guess because the 2l is the length of the domain. So, again the n takes values and will belong to the integers. So, the idea is that if the domain is infinite the k will vary over all the real numbers if the domain is finite and periodic then k takes discrete special values. So, this is the same thing that happens in Fourier series and Fourier integral. So, the Fourier series is how you split up a function on a finite domain. So, when you write the Fourier series you sum over the series and the series sums over integer values of n that is what is happening here when you do Fourier integral you do the Fourier integral for a function on an infinite domain then you have to integrate across all the k values because k is not discrete anymore it is the whole real number line. So, with this understanding we are writing e to the power ikx. So, if x is infinite k takes all real numbers and your dispersion curve is continuous if x is finite in a finite periodic domain k takes only discrete values along that dispersion curve. So, what it means is if I take a domain minus pi to plus pi my k can only be 1 and then 2 k cannot be 1.5 because 1.5 that mode will not fit in that minus pi to plus pi. So, by controlling the size of the domain you can exclude modes or include modes in the periodic case if you take the infinite case all the modes will come. So, that is something that you should remember and thoroughly understand before you try to simulate the problem because when you take a periodic domain you will actually be restricting the modes that are entering the problem and then that should that can lead to different results unless you are aware that was one point the other point is with someone asked me is now when we do this let us carry on and I will come back to it. So, we have got this form now. So, we will go back and substitute it here and get the equation for the dispersion curve you are agree those of you all have done it. So, this problem is very simple because you start off with an PDE but with only one spatial direction because we are knocking off time and space we will just get the algebraic equation which is the dispersion curve it gives us the dispersion curve immediately. Now, at this point some of you all asked me the question as you know what happened to the boundary conditions because we had a PDE and then we certainly I mean reached a solution and we did not seem to have said anything about the boundaries. But the factor is that once again when you are writing e to the power ikx it implies certain boundary conditions. So, what it implies is either that you have minus infinity to plus infinity in the domain is basically going to be periodic. If your thing was a finite domain then taking e to the power ikx again forces it to be periodic boundary conditions where this minus l to plus l will control which are which modes come. So, the length 2l of the domain controls which of the k modes are allowed and what are those periodic boundary conditions let me just I will just write. So, periodic boundary conditions are simply u at is u at plus l and all its derivatives. So, n is n is 0 means just the values are equal when n is 1 it means the first derivative. So, it is 0, 1, 2, 3. So, you equate the values and all the higher derivatives until you get 4 boundary conditions because I need 4 boundary conditions in the x direction that clear. So, it is with respect to these boundary conditions periodic boundary conditions that we are using the form e to the power ikx when x is on a finite domain minus l to plus l. And if it is on an infinite domain then the idea is like we are splitting it up into the Fourier modes. So, that is fine and now we can you will have the dispersion curve. So, when k is 0 what will happen I get r minus 1. So, let us look at the case of r equal to 0 because you already know that that is going to be the critical k. So, when r is 0 basically k will be when k is 0 and r is 0 sigma will be equal to minus 1. So, it will be somewhere here and then it does something like that. So, it becomes critical here at any higher value it will cross right. So, basically you have a region of unstable modes right this is for r equal to 0 this is for r greater than 0 and r less than 0 it would have all been negative. And this peak it remains the same throughout which you can find out what it is and I will call it I mean this portion is called k max. So, the k max gives me the maximum value of sigma for all values of r which means that the peak of the curve does not change with r. That is not always the case as you would know when you do the assignment in this problem it is true. So, what it tells me is that this mode corresponding to k max is the fastest growing mode but there is also a range of modes along with it all of which are unstable and the number of modes which are unstable keeps increasing as my supercriticality keeps increasing as r becomes larger and larger than the critical value 0. So, that is as far as the first few points of the assignment work any doubts about this. So, essentially what we get finally is that at r equal to 0 or for r close to 0 you have the critical k or the k max basically equal to I think you will find 1 because that is where sigma becomes 0 when r is 0. So, k max is this is basically 1. So, the idea now is if you try to interpret pattern formation what will happen is if you have a r just above 0 maybe 0.1 or 0.5 also then you will see a mode manifesting which will be the mode k equal to 1. If you do the non-linear simulation of the equations you will find that from the base state 0 you will get a new steady state which has the same wavelength as this k equal to 1 which means the solution will look very similar to sin kx with some magnitude like a sin kx something like that the new non-linear state will close to r equal to 0. For r much greater than 0 higher non-linearities come in and then the steady state can change, but close to this onset the kind of wavelength you will see in the pattern in this case it is just one dimension will be a wavelength of k equal to 1 means lambda will be 2 pi. So, if my domain is minus pi to plus pi for example if I look at the thing in minus pi to plus pi whether it extends to infinity or not if I just look at the thing minus pi to plus pi I will just get one sin way and this is basically sin I mean this could this is sin x I could also get cos x because that is also equally valid. So, then I could have got I am not very good at drawing this, but basically got the idea this sin is 0 at both ends you would get cos where the derivatives are 0 at both ends is just the 90 degree phase shift. Let us try to do this sin cuts whatever but we will get cos x. So, that is fine now the second part of the question was what happens when you go to the 2 dimensional Swift-Henberg. So, in the 2D Swift-Henberg basically you will have these partial derivatives, but now extended in x and y. So, if you go through the analysis what you will find ultimately is that we will get the same story except instead of k square I will have to replace it by kx square plus ky square and this should not be a surprise because we have done this and seen this in many problems now in the course. So, this kx I mean this comes because I take e to the power i kx x e to the power i ky y. So, that is the kx and that is the ky. So, the interesting thing now is that in the 2D version of the problem my critical kc or whatever is actually given by kx square plus ky square equal to 1 and that is very different from what I had before that just k is equal to 1 in the 1D case now I have kx or plus ky square equal to 1. This is actually k square equal to 1 and then I took the square root. So, the point now is that the first question that I ask is is the mode therefore that will manifest is that mode unique is the wavelength that you will see unique is the kind of pattern that you are going to see unique in this case it was I mean okay there was it could be sinx it could be cos x but in an infinite domain or a periodic domain those things do not matter it is basically the same physical pattern you are seeing it just phase shifted. But here I can actually get an infinity of possible modes which look differently in and just look different in space mathematically they have a different representation in kx and ky apart from the phase shifting. So, what does that mean you understand why that is so if I you tell me a value of kx say 0.5 I can always find a value of ky to go along with that. So, in that sense I can get what actually happens is in the plane of kx ky I have an entire contour where it is maximum. So, all these points in the plane of kx ky are critical. So, physically what does it mean if I have a part of the domain just a piece of the infinite plane and I look at some manifestation of the mode. So, I do the nonlinear simulation give it a perturbation close to the onset suppose I take kx equal to 0 and ky equal to 1 that is one possibility or let me do kx equal to 1 because that is this basically falls back to the 1D case. It is almost like you had a 1D problem and I added a y direction and said nothing will change in y that is this case and that is also equally unstable. So, what we will get here is that there is a periodic variation in the x direction of wavelength 2 pi and in the y direction nothing is changing. So, you will just get stripes where this thing is means it is greater than 0 and this means it is less than 0. So, if you take any cross section you will just get the sin x or the cos x and you will get this kind of stripe pattern that is clear how that came. Now, if I looked at the opposite case where kx is 0 and ky is 0 then it is just a rotation of this pattern you will just have horizontal stripe sorry ky is you have horizontal stripe this is also equally valid as that if I took the intermediate case which is also equally valid I would get and the perpendicular vector will be oriented along the 45 that is right. So, now what you see is that and I can go on with this which I want thankfully. So, what you will see is that you have a whole range of patterns all of which are equally possible in terms of the stripes the basic thing is a stripes. Now, this is where the idea of symmetry breaking which is more often discussed by physicists but which is relevant to all of us is particularly important. So, the reason why you have all these different possibilities is actually very simple and obvious once you think about and that is that suppose I were to say that I am only going to consider the one dimensional problem. So, I will only have kx equal to 1 and ky equal to 0 means I am only going to get this pattern okay. So, fine but now the question is if I look at the physical system what is there to tell me what the x direction is. So, if you look at this table and that is my experimental system which is the x direction. So, you do not know that you can choose the x direction as you please because the system is isotropic and we were discussing a system like that in infinite fluid where nothing is particularly special about the fluid or about the boundary conditions or about the governing equation in any particular direction x and y. So, that was totally I mean it was just our choice of the axis and coordinate system to allow us to proceed mathematically but when we come back to look at the physical system there is no x direction. So, because of that you would immediately realize that it has to be that all orientations of this should be equally unstable because I could choose any direction as the x direction. So, any rotation of this pattern is equally valid because I could have taken that as the x direction and this is simply saying the same thing in a mathematical way. So, all of these patterns are just rotations of this pattern and the fact that all the rotations are equally unstable tells you that the original problem had rotational symmetry which means that I could if the x y axis was there and I rotated it any way it does not change the problem or if I took the x y axis I rotated the coordinate system by theta and I made that substitution. So, I replaced x by x star y by y star the rotated axis I would get exactly the same equations the problem would not change at all. So, that invariance to a rotational transformation is what we mean by rotational symmetry of the not of the pattern rotational symmetry of the base equations and the base state. So, because of that what happens is that if I have a pattern like this I can confidently expect that all rotations of this pattern will be equally unstable or equally possible in my new steady state. So, that is an important thing. However, you must realize that this pattern is not rotationally symmetric because this is the pattern that has a clearly well defined orientation in the y direction and the x direction. If I rotate it and make it this these two patterns are not identical. So, there is the rotational symmetry is been broken when I went from the base state to the new state. So, the base state was flat homogeneous that is rotationally symmetric is just everything is just 0. But when it came to a new state the new state broke that symmetry. So, because this is no longer invariant to rotation all right. But because there is no specific directions in my problem because the original equations and base state were rotationally symmetric the new state all its rotation should be equally valid that is the idea. None of those states are rotationally symmetric but any rotations the whole family of them should be equally unstable. So, that is the idea of broken symmetry which I have written down in the solution and given some references you can read more about it. The same idea comes when we look at the phase shifts that is also a manifestation actually of broken symmetry. Because let me look at the 1D case I remember I said if you look at minus pi to plus pi in the infinite problem I could get a sine wave. But in truth because the problem is infinite I could get any phase shift of that sine wave. So, basically if sine x is a possible pattern sine x plus any phi will be possible where phi is a phase shift or basically the pattern of sine x translated by any amount is an equally valid pattern to emerge in the new study state. Why? Because again if you go back to the base problem there is no reason to there is no place which I can fix as the origin. I can take anything as the origin. So, if I shift my origin along the plane basically that is translating the sine x curve. So, in the 1D problem I did not have this question of rotation what I had was translational symmetry because I can translate the origin on the x axis and it is not going to change the problem. So, because the base state and the base equations were translationally symmetric when I got the new pattern all translations of the new pattern were equally valid as a new pattern as an unstable mode. Again once again you will see that this sine mode which is the unstable mode is not translationally symmetric. Translations will give you a different pattern but all those different patterns are equally possible because the base state was translationally invariant because the origin can be placed anywhere on the x axis. So, there is a very good book a popular account of these questions in physics, chemistry, biology, crystallography and so on. It is called fearful symmetry by Gulubitski and Stuart. Stuart is a very good writer and Gulubitski has done a lot of work in this area for many years and that it is a popular science book. So, you can just pick it up and read it and it tells you a lot of different patterns in nature how symmetry plays a role and you will be able to understand those things very easily now that we have done some of the math in the course. Along with this I would recommend for not only this problem but in fact the entire assignment is based on that book and the next assignment also and it explains many of the ideas and maths really well in that book. Just get the reference. I have given it in the solution. So, you can look at the full thing. It is pattern formation and dynamics in non-equivalent. It is by Hengie Cross and Green side, Michael Cross and Hengie Green side, Cambridge. That is a very good book for all the ideas that I have spoken about. So, now when you look at the non-linear simulations of this problem or an experiment, the idea is you should be ready to expect all these different orientations of patterns. That brings us quickly to the final point I want to make about the patterns is that not only are all orientations of these stripes state possible but even superpositions of different stripes can come up and there is nothing to stop that. So, if I tell you that all the modes on this circle are equally possible then either any one of them can emerge individually or maybe three or four of them can or six of them can superpose together and lead to some pattern or they can all grow together and once they grow initially the non-linear terms will come in and modify the solution it is called non-linear saturation and give you a new steady state. So, that new steady state could bear resemblance to any orientation of this stripe set. It could also be given by some combination of those stripes. So, just as an exercise if you look at these stripes and they are 90 degree rotation what you can do is you can go to MATLAB and plot the contours of cos kxx into cos kyy. Take that function plot its contours when you take kx equal to 1, ky equal to 0 you will get these stripes black white black white then you plot the other case then you plot their combination and see what you get. So, in some portions the say if I look at this and this is a in this portion everything is positive right in this portion everything will be negative here things will be sort of cancelled out. So, you will have almost a square checkerboard kind of pattern emerging simply out of these two combinations and those types of patterns are known to be seen like a checkerboard new steady state has known to be seen in these kinds of non-equilibrium systems after the symmetry breaks after the instability. A similar combination also gives rise to other lattices like the hexagonal lattice or the Rayleigh-Bennard cells. So, that is how you get so the hexagon seem to be nowhere in the whole problem either in the 1D problem of Rayleigh-Bennard nor in the 2D problem you found hexagons you just had these stripes basically. But it is these combinations of those stripes that lead you to those lattice states that you see in many non-equilibrium systems that you will get hexagons and different kinds of cells. And the question as to which of these patterns will come how do I know the hexagons will come or how do I know that these stripes will come that depends on very often on the boundary conditions which are not in this thing right now it depends on the initial condition also. But very strongly it also depends on the non-linear terms in the equation which we have totally neglected from the linear stability. So, in that sense it is similar to the case where you get you know the grow the eigenvalue to be 0 and then you cannot say anything about stability you need to go to the non-linear terms. So, to understand how it is going to grow. So, in a similar way here all these modes have zero growth rate in fact zero eigenvalue. So, now the question you are asking me is which of them will come actually is asking me which of them is going to grow and dominate. So, to understand that we need to look at these different modes the different stripes and look at the effect of non-linear terms on their growth. So, that is called weekly non-linear theory and that will tell you which of these patterns will come. So, those of you are interested in that can again look at this book and he has a discussion about those things and that those are the more advanced topics in the area of you know non-equilibrium dynamics pattern formation. So, now you can go back to your other problems that you have done the Rayleigh Taylor problem the Rayleigh jet Rayleigh plateau jet instability capillary instability and look at the same thing over there look at the base state identified symmetries and then see that in fact your new steady state or that the modes that you get should all obey this idea of broken symmetry that all they are the whole family of translations rotations and so on should be equally unstable and then it will not be surprising to you anymore. Do you all have any other doubts in the assignment I can address that now and then I want to talk a bit about the computation assignment I put up. So, if you will have any if you do not have any doubts in the assignment I will directly do that otherwise if you will have a doubt I will continue with the assignment. So, when you solve the non-linear equation you will get a graph for u1 and u2 like in this case why does it go like that and then we can. It is oscillating right. Yeah. Why I mean that is the system is like that you know the dynamics are oscillatory so it is a 2D system so it can oscillate and understand like in the sense that there is a there is a base steady state you solve the non-linear equations. So, now the base steady state was actually unstable possibly at that case the eigenvalues were had a complex part. So, it was complex positive and then you had an oscillatory kind of behavior if it is stable it will oscillate and die out to 0 if it is unstable it will oscillate and you know reach a new steady dynamic maybe it be a limit cycle maybe it will be an oscillation I mean it will be a steady state. The main idea is that it all of the things should have died out the steady state which we are looking at should have been stable under all conditions because the eigenvalues were minus epsilon and minus 2 epsilon but that does not happen because of non-normal growth and I have given in the solution I worked out a bit and given some references where you can read more about or any problems with that reaction diffusion problem. So, that is clear in that problem you need to look at the rate r of c I have not specified any form you should continue with the unspecified form. So, that means where I have written so, when you linearize it about USS you do not know what is r of USS plus epsilon you put up I mean you do not know it in the algebraic expression form but that is the whole idea is you anyway doing a Taylor series. So, if you want to approximate the value of this function about USS for small epsilon this will just be r of USS plus r prime of USS into epsilon UP. So, this is a constant depending on the USS. So, you do not know what USS is it does not matter. So, this is some constant number you can replace it by a if you want and then the rest is just epsilon UP. So, you will get a UP basically from this term into RSS basically a UP you will get from this term in the linear equation and then you will find that if r prime of USS is positive the system will be unstable and if r prime or negative the system would be stable. So, diffusion also if you look at it diffusion would already be stabilizing. So, this value of sigma will be given by simply r prime of USS itself and the diffusion just stabilizes it because sigma will be r prime of USS minus k square or something like that. So, in this problem diffusion is stabilizing. We might look later at a situation where diffusion is actually destabilizing and that happens when you have two chemical species and gives rise to what are called Turing patterns but I will not talk about them now because I think we will do it in the course. If you have any other doubts I have written the solution set down. So, you can go through that and then we can discuss them later especially about that classification of the different states you can just read that part. All right. So, now the next thing that we need to do is will be on a computers. So, again we are going to look at the 1D Swift to Hamburg and the idea is to now simulate the non-linear equation all right and then see whether in fact if I perturb the system for r greater than 0 whether that I will get a new steady state will the new steady states pattern be similar to sin x or cos x then how the effect of boundary conditions will change the instability and so on. So, all these questions that we are trying to answer analytically in this 1D problem it is quite simple to go and see for yourself numerically. So, second numerical experiment you can see whether the theory matches the computation. So, the equation is this similar thing I have put it up the assignment online with the list of questions and exercises and there I have given a short description of how you can write a numerical code to simulate it. So, I will just discuss some of those things here. So, that might make it easier all right. So, that is the governing non-linear equation. Now there are the next now here the important question comes of boundary conditions because I cannot simulate an infinite domain. So, what I recommend is you go from minus l to plus l say and then you apply periodic boundary conditions which are those conditions they are just u n of minus l of plus l right. So, u the values of u and all the derivatives will be equal at both ends of the domain okay. Now this is where the important thing about that mathematics business comes in which I was talking to you about. So, if your domain is this minus r to plus l you have a length of 2l. So, the possible case that I would get under this periodic boundary conditions there will be 2 pi by 2l n okay where n will take values of and the negative values. If I look at only sine if I look e to the pi kx will be plus and minus otherwise you will have cos and sine with only the positive. But the idea is that the case that you will get on the dispersion curve won't be all the case the case will only be these discrete values all right. So, now because we know from theory that the most unstable mode is k equal to 1 right. So, I will want to allow that to manifest otherwise I cannot compare the thing with the theory. So, therefore if I want to allow the k critical equal to 1 to manifest it should be one of these possible modes one of these discrete set. So, therefore I should take l equal to pi or in general l equal to n pi. So, if I go from minus n pi to plus n pi any integral multiple pi then I can be sure that that mode this kc equal to 1 which came from the linear stability theory that that mode will be allowed to come in if my l is say you can go minus pi to plus pi okay. If my l is less than this then that mode will not be allowed only k equal to 0.5 or something will be will come in this k equal to 1 will not because that domain will be too small to allow sin x to come in because it needs a wavelength of 2 pi if you are not going to give it a wavelength of 2 pi how can it come. So, you are not allowing it only. So, that is an important thing you should keep in mind that the numerical simulation at some point will start failing if you do not observe that thing. This is more important when you do not have periodic boundary conditions but even in this case you can try it out and see. So, the important thing is you can keep this in mind and then keep it at domains of n pi. So, if you take n very large ultimately you will reach minus infinity to plus infinity then it will l will not matter anymore if you take it large enough but then you want to avoid that much of computation. So, that is the one point then about the numerical simulation one method of doing it is to do what is called method of lines. So, in a method of lines or mol the idea is to discretize the spatial variables and leave the partial derivatives and time as total derivatives. And then what you can do is instead of a set of PDEs you will get a set of ODE's in time. And using Matlab or Mathematica you can integrate those ODE's very easily using the inbuilt functions with robust numerical methods. So, you do not have to worry about the time stepping you just need to take care of the spatial dependence. So, I will do a simple example suppose I just had dou u by dou t equal to dou square u by dou x square diffusion equation. So, that is some like this is a more complex version let us look at this. So, I just had this simple diffusion equation then what I would do is discretize in space which means that if I have my x domain I will break it up into nodes. So, node 0, node 1, node 2 this is node n, node n minus 1, n minus 2 normal spatial finite difference discretization. And what I will do is I will replace the derivative in x by its central difference finite I mean central finite difference. So, I will get du i by dt where u i is the value at the node i. So, du i by dt because I have discretized in space and the second derivative at i is basically u i plus 1 plus u i minus 1 minus 2 u i upon delta x where delta x is this difference between 0 and 1. So, this is the central finite difference of second order. So, in the same way you can find central difference formulas for the fourth order derivative also and substitute here and u cube will just be u i cube. So, now what we have is we have a bunch of ODs. So, if I runs from the other question now you have this now where do I apply this equation I apply this equation for all the interior nodes. So, in this problem I would have 2 boundary conditions 1 boundary condition would be u 0 equal to u n let me write it here boundary conditions would give me u 0 equal to u n that would be the I mean equality of the values and the derivatives would be the derivative of u equal to derivative of n because I have a domain like this I can use a backward difference here and a forward difference from the other side because delta x is the same I will get u 1 minus u 0 should be equal to u n minus u n minus 1. So, actually here I have used a forward difference and there I have used a backward difference. So, there are other ways of doing this but this is a simple way you could also use central difference here and then you will have a dummy node and whatnot. But if you just start this is a simple thing you can do. So, use appropriate differences forward here and backward from the other side. So, then you have got a boundary conditions and this problem you will have to repeat that for the second derivative and the third derivative using again forward difference backward difference of the appropriate formulas. So, what you will see is that these equations allow you to relate the values of the exterior nodes u 0 and u n to the interior nodes. So, you can do that. So, basically ultimately what you need to do is you need to apply this equation for i going from 1 to n minus 1 okay. You apply this equation for 1 to n minus 1 those are the interior nodes you do not have equations for 0 and n but that gets provided by the 2 boundary conditions. So, you will have n minus 2 ods plus 2 boundary conditions which are basically algebraic equations and you can use them to describe the evolution of these n nodes. So, more on this is there in the assignment. So, with this basics you should be able to simulate it. So, we will discuss more later. Thanks.