 So, what we will do in today's lecture is continue the discussion on Turing patterns which we started off yesterday. And towards the end of the last class, this is what we had done. We had found that the Jacobian matrix for the reaction diffusion system was given by this, okay. This is when a finite diffusion coefficient is actually added to the system, okay. And the determinant of this particular matrix is given by this expression. It is fourth order in alpha, okay. And what we had established was that the trace of this particular matrix would always be negative because it is already known to us that f u plus g v is negative, okay. So, there were 2 conditions which could possibly be violated for getting an instability. So, what we established was that the trace condition cannot be violated. The only condition which could possibly be violated was the determinant condition. And in order for the determinant condition to be violated, we know that this is positive because the system without diffusion is stable. This is positive and the only way the determinant can be negative is if this particular quantity is positive, okay. So, in order for it to be unstable and this is a necessary condition for instability. So, this is for instability, this is a necessary condition. What we already know also is that f u plus g v is less than 0. So, in order for you to possibly have an instability, there are 2 options. One is that both of them are negative. If both f u and g v are negative, then clearly this is always going to be negative and you will not have a violation. So, the only option available is one of them is positive and the other is negative, okay. And the negative guy is dominating the positive guy, okay. And so, what we had done towards the end of the last class was we had assumed that f u was greater than 0 and g v was negative. And because of the nature of this dependency, what we said was look the reaction rate increases with u. That is what it means. f is a reaction rate. So, as you increase u, the reaction rate increases and so this is u acts like an activator, okay. And as far as the other reaction rate, g, the reaction rate decreases with an increase in u. That means v acts like an inhibitor, okay. So, basically what you need to have is in your chemical kinetic expression, chemical reaction rate expression, you need to have this kind of a dependency. That is one species is actually going to slow down the reaction as you increase the concentration and if you have a regular first order kinetics, then you do not have this action of inhibition, okay. And you will not be able to get these Turing patterns for example, okay. So, this is what we had concluded. But these are all qualitative arguments and what we need to do is see if we can get some more quantitative information. So, if you want f u plus g v to be negative, if we have f u plus g v to be negative and if g v is negative, you can assume the other way but we are just assuming this, okay and we proceed. That means what the absolute value of g v must be greater than the absolute value of f u, right. Therefore, the absolute value of g v must be greater than the absolute value of f u. Only then this is going to be negative. If the absolute value of g v is less, then this is going to be positive, okay. And now I come back to this expression d u g v or d v f u rather, d v f u must be greater than minus of d u g v. g v is negative, so minus g v is positive, okay. So, basically what this means is d v divided by d u, d v divided by d u must be greater than minus of g v by f u. And we already know that the absolute value of g v is greater than the absolute value of f u, okay. That means d v must be, this must be greater than 1. In other words, d v by d u must be greater than 1, okay. So, all I am doing is I am using the fact that g v is the mod of g v is greater than the mod of f u and I go back and I am trying to find conditions on the diffusion coefficients, okay. Because clearly the diffusion coefficients are, you can think of as parameters which I can to possibly control my system, okay. And if I can do that, so I am just manipulating the second thing. I am saying that this is going to be positive and that means I can possibly have instability. And if I have an instability, the resulting solution is going to be a periodic pattern because that is what we assume it will be of the form of sin alpha x, okay. And this is greater than minus d u g v. So, now I am just dividing by d u, f u is positive, take it down here and this is positive by positive, okay. And from this, this is greater than 1. In other words, what it means is that the inhibitor has to diffuse faster than the activator. That is the conclusion I have. That is, you know, if this is a necessary condition, remember, this is only a necessary condition for instability just because it is diffusing faster does not mean that you will have a spatial pattern, okay. Because this is only guaranteeing that the middle term is positive, but this is going to dominate the first and the third term in the determinant that we do not know. See, I want this to dominate these two terms. I am only looking at the sign of this term, okay. In order for you to actually have an instability, you have to make sure that this is dominating over these two terms. In this case, the determinant will actually be negative, okay. Otherwise, this guy can be positive but a small value and these guys can dominate and still determinant will be positive, okay. So, again, I want to emphasize that this is a necessary condition. This is a necessary condition for spatial patterns. So, now, how can I make sure that you indeed have spatial patterns? In order for you to have spatial patterns, you need to have the determinant negative, right. So, in order to obtain spatial patterns, we must necessarily have the determinant of, I have added a subscript D here just to tell you that this is the Jacobian of the system with diffusion. When I do not have D, that means there is no diffusion, okay. Determinant of JD must be negative. That is, this is system with diffusion, okay. So, basically what I am going to do now is make a plot of this determinant of JD versus alpha squared, okay. Clearly, this is a quadratic, okay. For lower values, this is going to dominate. As alpha goes to minus infinity, this is going to dominate. It is going to be positive. For large values, this is going to dominate. It is going to be positive. So, what I know is this is going to be of a shape of a parabola, right. So, the question is this is the general shape that I have. If this is the shape of the parabola, that means my determinant is always positive, okay. And what decides the shape? This shape is going to be decided by the values of the elements in my Jacobian evaluated at the steady state and the diffusion coefficients, okay. The diffusion coefficients and Gv evaluated at the steady state decides the shape. It is possible that this guy may start dominating. The middle term may be large in a small interval. So, that happens then. So, this implies steady state is stable. Which steady state? The one which is having a uniform concentration. Because we have done the linearization about that steady state, okay. That is the homogenous, especially homogenous, especially homogenous solution. This means no matter what the alpha square is, no matter what the, if you give an arbitrary disturbance, you can decompose it into different wave numbers. And for all wave numbers, this alpha, the determinant is positive. And remember trace is negative. So, trace is, what is trace represent for a 2 dimensional system? The sum of the eigenvalues. The determinant represents the product of the eigenvalues. So, now this basically means that then the real parts are going to be negative, okay. And because the product of the 2 eigenvalues is negative, so both are negative and trace is negative, okay. And so, if eigenvalues are negative, the eigenvalue was my growth rate e power sigma t. When I did the linearization, I had sigma multiplied by u star v star equal to the Jacobian multiplied by u star v star. So, the eigenvalues basically are my growth rate, okay. So, however, if you have a situation of this kind, if your determinant can be like this and that is going to be decided by the diffusion coefficients, then you are necessarily guaranteed that that is an interval of alpha over which the determinant is negative, okay. So, this is stable here, this is stable here. And this is exactly similar to what you had done earlier, your neutral stability curves, etc., when you had your really been a problem, okay. So, the point I am trying to make here is that in this interval, let us say alpha 1 to alpha 2, when alpha belongs to alpha 1, alpha 2, the determinant is negative. If the determinant is negative, it means one eigenvalue has to be positive, one eigenvalue has to be negative, okay. That means the stability is going to be decided by the guy which is positive and the eigenvalue remember is the growth rate. That means one eigenvalue is going to make it unstable, the one which is positive, okay. So, here sigma 1, let us say is positive and sigma 2 is negative and sigma 1, sigma 2, I use sigma 1 last in the class, right. So, sigma 1, sigma 2 are the eigenvalues of JD and these are the growth rates. So, what is it that is going to guarantee you that this fellow is going to be negative. So, so far we have just made sure that we had a necessary condition but what is it that is going to guarantee you that the determinant can be negative. You have to look at the discriminant of this particular thing and if the discriminant is positive, if the discriminant is positive, I mean you treat this as a quadratic in alpha square, if the discriminant is positive that means you are guaranteed 2 real roots and these will be the 2 real roots. So, in addition to dv by du must be being greater than 1, you need to make sure that the discriminant is of this equation is positive, okay. So, the discriminant of this equation being positive implies, so we are guaranteed that the determinant of JD is less than 0 if the discriminant of JD is greater than 0 because this guarantees me 2 roots, okay. So, that is my condition which I need b square-4ac. I need to make sure that du gv plus dv f u whole square-4 multiplied by du dv, okay. If the discriminant of that determinant expression is positive, I am guaranteed 2 real roots and these will be the 2 real roots. If the discriminant is negative, then it is going to be of this kind, no real roots, okay. So, that is what I have written here. So, now if you have a steady state, if you have a steady state for a particular system, you have your reaction rate expression f and g. You know what your Uss is, Vss is, you can find out by just solving those algebraic equations, right and using Newton Raphson or something. Then you can also calculate the partial derivatives of the reaction rate expressions and if it happens, you need to satisfy this, okay. That is a necessary condition but if you satisfy this, then you know for sure that the determinant is going to be having a negative value and one of the growth rates is going to be positive, okay. If one of the growth rates is positive, then from our form of the expression that we have assumed for the disturbance, the steady state solution which we are going to get is going to be a periodic solution because you assume sin alpha x, okay. So, what I am saying is, these are quantities that are known to you from your reaction rate kinetics. Du and dv, you can possibly think of as experimentally controlled parameters. Actually, they are not, they depend upon the system but if you actually have this condition satisfied, then you can actually have a spatial pattern, okay. So, this is the necessary and sufficient condition in some sense because this is going to give you the thing. Now, the spatial pattern. I just want to spend a little bit of time on that expression. So, there was a neat explanation given about this dv by du. So, I think I will mention that dv by du being greater than 1, okay. So, we were looking at these two species, one as an activator and one as an inhibitor, okay, talked in terms of chemical reactions. So, rather than talk in terms of chemical reactions, I would like to give a slightly broader, more general argument. Now, many places where you would have seen, you would have read the newspapers about forest fires, okay, spreading and you know people especially when the U.S. in Australia, the weather becomes very dry and at one place, the forest, the wood catches fire and then it spreads and it causes a lot of devastation, right. So, and of course, how do you control it? You control it by having some kind of a firefighter who goes is spraying some chemical from the top or you know there are these fire engines who are going to go and douse the fire, right. So, you can look at this particular thing as if that is one of them is an activator and one is an inhibitor. The fire by itself has a rate by which it is spreading, okay. So, the fire is an activator because it just spreads whereas the fire engine or however you are spraying the aeroplane with which you are spraying your chemicals to control the fire, that is an inhibitor that is actually trying to quench the spread. Now, supposing the fire engine is moving very slowly, it does not move fast enough. The fire engine remember is the inhibitor, right. Supposing the guy does not spread fast enough, the fire is going to spread everywhere and the entire forest is going to be burnt. You will have a specially uniform homogeneous steady state, no forest anymore, a lot of ash, right. But if this guy is able to move faster, if the fire engine can move faster, then you will have pockets which are burnt, pockets which are not burnt, okay and then you are going to get a special pattern. So, that is just one way for you to possibly think of this particular thing. So, I mean I just thought this is a nice way to explain this. So, if in a forest fire, the fire spreads fast, that means the diffusion coefficient, that is the rate of spreading. Diffusion is basically a rate of spreading. Diffusion of the activator is large. Then the fire fighter moves, I mean in comparison slowly, that means D of the inhibitor is low, okay. Then we have uniformly charred forest, okay. But if D is a fire fighter moves fast relatively, I mean these are all relative things, right. That is D inhibitor is high. Then we will have pockets of green and you know burnt portions giving a special pattern. Of course the theory for this particular Turing pattern is whatever I have discussed here, okay. But then experimentally established because you know diffusion coefficient is not something which you can actually control. It is not a flow rate that you can come open your valve or it is not a temperature that you can actually increase. So, to actually experimentally establish the existence of these Turing patterns, that took a lot of time, okay. And then finally it was established and then some people started actually believing Turing patterns actually exist. So I think that is something which you people can read because there was a huge gap between the theoretical prediction of how diffusion can actually cause an instability, cause a spatial pattern and by the time in this particular theoretical prediction was actually verified. It took experimentally, it took time because you need to have the right system with the activator, with the inhibitor so that the kinetics was having this kind of a mechanism. Then you need to make sure that the diffusion coefficients are also right, okay. So that would depend upon the concentrations. So I mean with lot of persistence people finally managed to get this. The point I want to I want to just make one more point here and that is when the determinant condition is violated, when the determinant condition is violated, the eigenvalue which is actually causing the instability is a real eigenvalue, okay, that is the eigenvalue causing the instability is a real eigenvalue because now you can have a complex eigenvalue of the real part. What I am trying to tell you here is that the, see there are 2 ways in which you can have instability. One is the real part of a complex eigenvalue is positive or the real eigenvalue is positive, okay. So the point I am trying to make here is when the determinant condition is violated that means what? I mean when determinant is negative, when determinant is negative, one eigenvalue, one of the eigenvalues is real and negative, the other eigenvalue is real and positive, okay, that is sigma 1 is real and negative say and sigma 2 is real and positive for instance. Just before the violation, just before the guy cuts the x axis, both of them are going to be negative, okay. This means what? The linear stability indicates an exponential growth and to another steady state which is spatially periodic, okay. This is like same as what we had for our Rayleigh-Bernard problem. You had a stationary solution which was not moving at all and then when you had instability, you got a periodic pattern which was steady, okay. You have a periodic pattern spatially varying but that was also a steady state. So this is exactly the same thing happening here, okay. So I mean the new solution that you are going to get when the homogeneous solution becomes unstable by changing the diffusion coefficient is going to be a steady state pattern but it is going to be periodic. The earlier solution was a steady state pattern but it was spatially uniform, okay. So what is the other situation possible? So let me just explain what I mean by this. I am just going to draw the determinant versus alpha squared, okay. It is like this. So this is the situation where sigma 1 and sigma 2 are both real and negative. For this case, sigma 1, sigma 2 is real and negative whereas here, sigma 1 I wrote as negative and sigma 2 is positive. What is the difference between these two? I have changed something like a diffusion coefficient, okay. That is, let us say I am changing only diffusion coefficient of u. So for some value of du, I have this business. I keep changing du, this curve is going to change and it is going to come below, okay. When it comes below, just at that point, here you had both eigenvalues real and negative whereas here you have one eigenvalue negative, one negative positive which means a real eigenvalue has crossed the imaginary axis. A real eigenvalue has changed from negative to positive, okay. And that is what I want to say. The sigma 1 remains negative but sigma 2 has changed from negative to positive making it unstable. There is another scenario which is possible. I just want to mention that. Suppose we proved in this particular problem that the trace was always going to be negative, okay but suppose you have a situation where the trace can possibly get violated and that is not applicable to our problem here because we proved that trace is always negative with the diffusion coefficient. Suppose hypothetically or for another problem, the trace condition is violated which means what? Trace was earlier. Trace is negative without diffusion coefficient, without the d, okay. Trace is positive with the d. Suppose then if you actually calculate the eigenvalues, it will turn out that now the eigenvalue is causing the instability are a complex conjugate pair. So the real part will change from negative to positive, okay. The real part changes from negative to positive. So supposing this happens, what can you say about the nature of the new solution or the new state? The new state is not going to be a steady state. The new state is going to be a time dependent, time periodic solution because it is a complex conjugate pair. You will have e power a plus i b, okay. You have e power i b t. The e power i b t is going to be periodic in time and so what you will actually see is a periodic solution in time, okay. So here the new solution is periodic in that is earlier there is no imaginary part. Now I have an imaginary part that means my growth rate which is of the form e power sigma t is going to be of the form e power a plus i b t, okay and this is e power a t times e power i b t and e power i b t tells you this is periodic in time. So you actually now put a probe and you have to measure I do not know concentration or velocity or temperature. You will actually find if this situation arises time periodic behavior, okay whereas in the other case but in our Turing pattern problem that does not happen because the trace condition is not violated. Trace condition improved is always going to be negative, okay. So I mean I just wanted to mention this that this is also a possibility so that you not always think that whenever there is an instability you always get a steady state. You can also get a time dependent solution, okay and when this happens when you have a time dependent solution that means the new solution is periodic in time this arises at what is called a half bifurcation. So I am just dropping a couple of keywords for those of you are interested to actually read a bit more on this, okay. I think so what we have seen here is in order for you to actually have a Turing pattern or a spatially uniform spatially periodic solution when you have a system with a reaction and a diffusion added. You need to your kinetics should be such that there should be something which is some chemical which is activating, some chemical which is inhibiting and the diffusion coefficient of the inhibitor has to be faster than that of the activator, okay. So that is the prediction and then there are some necessary conditions and sufficient conditions which we derived. So I think the very fact that we have taken a very simple 2 dimensional example with 2 variables has actually given us a lot of insight into this particular problem. If you had taken actually more complicated system then you possibly would not have been able to get this kind of an understanding, okay. The other important point which I want to emphasize here is that look what we have been doing so far is using this framework of linear stability analysis which is a very general framework which can be used for understanding any system. It is not necessarily in fluid mechanics, it could be in reaction diffusion, it could be in some other system. So as long as you write your governing equations, okay and there is some kind of a nonlinearity which is embedded in the system which is going to be describing the system behavior. The procedure is very, very clear, okay. So I think the important point what I want to say here is linear stability analysis gives the general mathematical framework. You can apply it to any system and I think that was one of the things which I wanted to do that is you not necessarily hydrodynamic problems, fluid mechanics, it can be reaction diffusion, it can be something else just only chemical reaction, okay. And the idea is we have to find a steady state, okay. I mean they are always trying to find the stability of a particular steady state. So you need to find what the steady state is, only then you can find out whether it is stable or not and that is what we did. We had a spatially uniform solution. In the fluid mechanics problems it was velocity was 0, that was my steady state. In this problem uss vs is some constant value, okay. Then we give a perturbation, a small perturbation, okay and do a see how this evolves, how this actually is evolving in time. And sometimes if you have diffusion coefficient you will have evolution in space as well. And if the growth rates are positive then steady state is unstable as it is stable. And one of the things is the nature of these curves is such that the system automatically the intrinsic features of the system it decides what the periodicity is spatially. It is not something which you know depending on how you do the experiment. So for example in the Rayleigh-Binard problem the wavelength of the periodic solutions came was decided by the properties of the fluid, okay. The space between the things. Here the alpha where it is just going to cross the imaginary axis, the x axis sorry is going to be the one which decides the periodic spacing, right. In the determinant when I plot it vs alpha square the point where it is just tangential when I actually had the determinant of JD vs alpha square and some value is just going to be tangential. The periodicity that you are going to see is that one, okay. And that is going to be decided by the kinetics gv fu is going to be decided by the diffusion coefficients. So whatever is there in the system that is going to decide what the spatial pattern is. So what you can of course do is in a very simple system one dimensional equation in x you have a reaction kinetics you can just do a finite difference method and you can possibly solve the system and see when the determinant condition is violated when it is not violated what kind of a solution you get. You can get a spatially uniform solution when you have the determinant is positive, when the determinant is negative then depending on the choice of the parameters you might be able to get a spatially periodic solution. It is a very simple experiment you can do on the computer. It is more difficult to do an experiment in the lab with the test tube, okay. So I think that is something which you guys can do. What we will do is I do not want to start something new at this stage. So we will stop as far as this particular class is concerned and in the next class what we will do is we will derive the boundary conditions for the tangential stress wherein the gradient of the surface tension will show up, okay and we will have the formal derivation then we will do a problem on Marangoni convection and see how Marangoni convection sometimes is the cause for natural convection and not the buoyancy term, okay. Then we will work on 2 or 3 more problems in fluid mechanics but just to take you slightly away from fluid mechanics and kinematic boundary conditions we just threw in this example. We will go back to kinematic boundary condition and fluid mechanics next week, okay. Thanks.