 So, welcome to today's lecture. What we will do today is talk about a stability problem again but this is will be something which does not involve any fluid motion okay and in particular we are going to talk about Turing patterns. So, Turing patterns are basically patterns which arise when you have an interaction between reaction and diffusion okay and the motivation for this is would have seen many patterns on the animal skin. For example, you have the zebra which has stripes yellow and the white and black stripes, you have the leopard, you have the tiger which has a pattern and so the question is how do these patterns arise? Is there a way for you to actually describe the formation of these patterns okay. So, of course there is a theory which has been proposed and that is what we are going to discuss and what we will discuss at this stage is the basic theory which does not involve any fluid motion okay but the emphasis is in just understanding how reaction and diffusion and what are the kind of features reactions have to possess for you to actually see these kinds of patterns. So, with whatever we have done so far in the course we should be able to get some insight about this okay. The basic idea is when an organism is growing okay there are many cells which are going to start with one cell, it divides, proliferates, lots of chemical reactions going on, there is also transfer transport primarily by diffusion. So, if the chemicals are going to be such that there is going to be reaction and there is going to be diffusion, is it the possible that these can rearrange themselves which can finally give rise to some kind of a pattern okay. So, this is something which is not being forced by anybody outside but it is just that the intrinsic feature of the system is what is causing this. So, whether this can actually give any insight into the formation of patterns. So, what we are going to talk about today is tuning patterns and possibly another keyword for you to search is morphogenesis okay. So, what are we trying to do? We want to understand formation of facial patterns on animal skin. For example, the stripes on a zebra, maybe the spots on a leopard, etc. etc. okay. So, that is what we want to try and do. Whereas during the growth of the embryo, we have chemical reactions and cell proliferation okay. The cells are going to multiply and they are going to grow okay and the chemical reactions are also taking place simultaneously. So, is it possible that the interaction between these chemical reactions and as the thing grows, there is going to be some kind of a gradient in the concentrations okay. Is it possible that this gradient in the concentrations can actually give rise to some kind of a pattern formation okay. That is the question we are asking okay. Can gradients in concentration arise in the presence of reaction and diffusion? That is the question okay. So, the question which Alan Turing basically asked is whether this Turing patterns is of the name of the fellow who actually started this was let us consider a system where there is no diffusion taking place. So, to begin with we will keep the system very simple. We will talk about only 2 variables which are actually interacting with each other. So, let us say that U and V are concentrations of chemical species okay and the idea is there is a reaction going on between these 2 species and the reaction kinetics is going to be defined in terms of some kind of an expression okay. First order, second order whatever it is. So, we like to know if we have no diffusion okay. The concentrations of U and V change U to reaction okay and that is going to be given by du by dt equals f of u comma v and dv by dt equals g of u comma v. So, what does f and g represent? They represent the kinetics. Clearly the kinetics will depend upon the concentrations of species U and V and I am just writing it in a very abstract form instead of telling you this because one of the questions we are asking is what should be the property which f and g has to possess in order for you to actually see a spatial pattern. That is the question which we are trying to ask. It is not that if you put first order reaction here and second order reaction here it is going to happen. So, there are certain properties which f and g have to possess and that is the so what are these properties okay and we like to see whether our analysis in terms of this linear stability that we have been doing so far can actually give us some insight. This analysis is slightly different in the sense that we do not need to worry about things like kinematic boundary condition and stuff like that. U and V remember our concentration of species. That is the only similarity. U and V earlier were velocity components but here is concentration of species okay. So, f and g represent reaction rate expressions okay. Our question is what should be the property or features of f, g to get patterns that is what we are asking. So, what we are going to do is we are going to pose the problem as follows. We are going to find a steady state for the system okay and you all know how to find the steady state for the system. So, the steady state is given by f of Uss, Vss equals 0, g of Uss, Vss equals 0 that is the steady state. Clearly it is spatially uniform. There is no concentration gradient okay. So, it is spatially uniform. The variable is only a dynamic system. It does not change with space. Now, I am going to assume it is a stable steady state okay. Assume this is a stable steady state. Now, I am going to add diffusion to the system okay. I am going to say that earlier my steady state was spatially uniform. Now, I am going to add diffusion and when I add diffusion what is going to happen? I will have a spatial dependence right. Normally our understanding of diffusion is, diffusion is something which helps you to homogenize concentration. Supposing there are two in a vessel you have two regions one with high concentration, one with low concentration. What will diffusion do? As a result of diffusion everything becomes uniform. If you wait for a sufficiently long time, if diffusion is added to the system as a mass transfer mechanism then at the end of the day you will have concentration being uniform everywhere right. So, if I want to add diffusion you normally think diffusion has a tendency to homogenize concentration gradients. It will not create concentration gradients okay. So, what we are going to do now is we are going to add a diffusion to the system to the system here and say that actually that is reaction and diffusion both taking place and we are going to ask the question whether diffusion can actually destabilize the system. Whether adding diffusion can actually result in an instability which can give rise to a spatial pattern okay. The basic idea is the same as whatever we have done so far in our earlier examples but then so the methodology is going to be the same. We have a steady state, we do the linearization and then we can talk about this stability okay. So, what we are going to do now is we add diffusion to the system okay. Diffusion homogenizes concentration variations. It makes the concentration uniform in a system. Another question is can diffusion and reaction give rise to patterns and if the answer is yes what are the conditions, what are the properties which f and g have to satisfy okay. That is what we are trying to do okay. Let us go up to the problem where we have the spatially uniform solution, uniform system which is that the ODE. So, if I say that the steady state this is a 2 dimensional system right. If I tell you that the steady state is stable and that is the assumption which we are making. We are assuming that the steady state is stable, the spatially uniform steady state is stable. What do you infer from that? The USS, VSS is stable okay that is given to you. Now even stable that means if I had done the linearization and if I had calculated the Eigen values and this is what people have done earlier in the course then the Eigen values are going to be negative okay. And this will happen when your Jacobian matrix, the matrix which comes on linearization satisfies some properties right. So, what is the Jacobian matrix? This implies that the lambdas of the Jacobian matrix j have negative real part okay. Now what about the Jacobian matrix itself? It is the partial derivative of f with respect to u, partial derivative of f with respect to v, g with respect to u, g with respect to v okay. I mean if you do the linearization that is what you would get and write it in a vectorial form. We have done this earlier in the course. Now what about the Eigen values of the system? How do you find the Eigen values? The determinant of j-lambda i must be equal to 0 or the determinant of f u-lambda fv, gu and gv-lambda but the determinant must be equal to 0 okay. This implies f u-lambda times gv-lambda 0 or lambda squared-lambda times. Now I am not going to prove this but you can prove this to yourself. The condition was stability. This is stable. This has real part of lambda negative if f u-gv is negative and f u-gv-fv-gu is positive okay. This basically means the trace of j is negative and the determinant of j is positive. Remember I just want to emphasize that this is true only for a 2 dimensional system. What I have written here? If you have a higher order system this is not true okay. If you have a third order system you need additional conditions and in fact these conditions will change. Point I am trying to make here is basically if the trace is negative that means this particular polynomial will have no change in the signs of the coefficients of lambda. This is positive, this positive, this positive and if you go back one of the necessary conditions for roots to have negative real part there is no sign change. It is also a sufficient condition for a quadratic. For higher order systems you need additional conditions okay. So this no sign change is a necessary and sufficient condition for quadratic but this is just basic theory. You can you know analyze this in many different ways and you can come to this conclusion. So what I am trying to tell you here is that I have assumed that the spatially uniform state is stable. That means my f and g satisfy these conditions. Yeah. These are calculated at the steady state correct. This is evaluated at the steady state yes. The derivatives are evaluated at the steady state because what I have done is I have done the linearization around the steady state okay. So these are calculated at the steady state. So now what we want to do is this is given to me okay. You are right. This is at the steady state. I am talking about this at the steady state not in general. Now what I want to do is I want to go back and introduce my diffusion. So when I introduce my diffusion I introduce my spatial derivative. It also has this increases the mathematical complexity. So od have a partial differential equation right. So what I am going to do is I am going to solve look at this system. We have du by dt. When we add the diffusion I have du by dt equals du d square u by dx square plus f of u, v and db by dt equals. What I want to emphasize here is I am taking 2 different values for my diffusion coefficient okay. Basically I want to have some flexibility. I do not want to say that the 2 species are you know diffusing the same rate. That means I am constraining myself. I want to you know make this special pattern happen. So I want to have as much flexibility as I want to have this happen to make it happen. So I am keeping du and dv different okay. The diffusion coefficients are different. To keep my life simple I am just assuming transport only in one direction, only in x direction okay. Just like what we were doing earlier. If you want to complicate your life you can do other things. But I just want to introduce diffusion. We are looking at one direction and so as always what we will do is we will look for an infinite span in the x direction because then that helps me look for periodic solutions in the x direction okay. And then we can find what the critical wave number is, critical wavelength is if there is one okay. So du and dv and dv are different okay to get flexibility and then in the x direction we have an infinite extent. The point is even if you did not have infinite extent if you had a finite extent and you prevented the species from leaving the boundary okay. Which means the flux is 0, du by dx is 0. The infinite extent means I am not looking for any, there is no boundary condition I am imposing. But if you had a finite chamber and if the species cannot leave that means the flux is 0 that means the derivative is 0 okay. If we have a finite extent in the x direction and the species cannot leave then what is the boundary condition I am going to have du by dx equals dv by dx equals 0. Derivatives are 0, no flux okay. Why am I talking about this? Because I want to emphasize that if Uss, Vss is a solution to this system then Uss, Vss is also a solution to this system is also a steady state solution. For the infinite span it does not matter and goes off to infinity I am not talking about boundary conditions at all. For a finite span if I had the fluxes are 0 then constant means flux will be 0 okay. So if Uss, Vss is a solution the steady state for this system without diffusion is also a steady state for the system with diffusion okay. It satisfies the differential equations and it satisfies the boundary conditions okay. So it is a steady state. Now the question is it stable okay. So if it is stable that means that is the state you are going to see. If it turns out that it can be unstable now what I have done is I have introduced two additional parameters du and dv. So the rates of these diffusions can possibly make something which was stable unstable. So that is basically I want to try and get conditions for this instability and spatial patterns in terms of these diffusion coefficients and in terms of f and g that is the idea okay. So Uss, Vss is a steady state of the reaction diffusion system where the reaction where there is diffusion because Uss is constant so second derivative is 0 time derivative is 0 f is 0 already okay. Same thing for V it satisfies the boundary conditions. Question is is this steady state stable and clearly the stability of the steady state will depend upon the diffusion coefficients because these are the two additional parameters which have come into the picture okay. Depending upon the diffusion coefficients this can be stable or unstable. So what we want to do is we want to find out if at all it is possible to have conditions on du and dv the diffusion coefficients which can make this unstable okay. So what do we do? We do a linearization of the reaction diffusion system okay. We linearize the reaction diffusion system. So what we have? U tilde equals U-Uss V tilde equals V-Vss okay. These are my perturbation variables and these are small okay and I substitute it over there in that equation and what do I get? du tilde by dt equals du d square u tilde by dx square okay plus you do the linearization of f around the steady state you get f sub u u tilde plus f sub v v tilde okay and you get dv tilde by dt is this clear. So all I am doing since I mean you all are experts in doing linearization by now I am just linearizing okay plus du u tilde plus gv v tilde okay. I want to make sure I have not done anything silly great yeah. At the boundary at the boundary that is only at the boundary yeah correct this is at the boundary that is my boundary condition. So what I have to do is I will do what I have done earlier write this in a vectorial form okay and I am going to write okay. We will talk about the situation where it is infinite in extent okay. So what do I do? U tilde is a function of x and t and I am going to write this as u star e power sigma t times sin alpha x infinite x direction. So I am looking of periodic solutions in the x direction okay linear so I am talking about the linearized problem was exponentially in time like doing Laplace transform Fourier transform and that is the amplitude okay. We are looking for conditions under which I have nonzero u star. So now I am going to substitute this in an equation over there and when I put du tilde by dt I get sigma e power sigma t sin alpha x u star equals du. When I differentiate this twice I get minus alpha squared u star e power sigma t sin alpha x plus fu u tilde is u star e power sigma t sin alpha x then fv v star e power sigma t sin alpha x. So even we are basically kind of in phase spatially okay. What happens now? I can actually strike off the e power sigma t sin alpha x everywhere. That tells me that the assumed form of the solution is a possible solution okay. If I could not have done this striking off that means what I have assumed is wrong. I am assuming v and u tilde v tilde u tilde both are of a similar form okay. v tilde is also of the same form, same thing. So this gives me sigma u star equals minus alpha square du u star plus fu u star and then fv plus fv v star. If you did the same thing for the other equation, I have done this for one equation. I am going to do it for the other equation. I will get sigma v star equals minus alpha square dv v star okay plus du u star plus gv v star okay. That is what I would get. Now actually the sigma is my eigenvalue which tells me where is stable or unstable okay. I had lambda earlier but sigma is the growth rate. The real part of sigma being negative indicates stability. Remember that. So I am going to write this in a vectorial form. When I write this in a vectorial form, I get sigma u star v star equals fu minus alpha square du fv u star is du and I have gv minus alpha square dv okay. So what we have done is I have just added diffusion. The steady state is the same. I am doing the linearization to find out the stability of the steady state okay. When I do the linearization, I follow the usual method and take order of epsilon terms. These perturbations of order epsilon. So just take the first order terms okay and then I get these linearized equations since it is infinite in the x direction. I am looking for periodic solutions in the x exponential in time. It is first order and I proceed and when I proceed, this is what I get and this is exactly what you people did earlier in the other stability problems okay. Now I want you to understand that this is my new Jacobian matrix which has the diffusion included. So earlier when we did not have diffusion, these guys were 0 okay and you had a Jacobian matrix containing only these 4 elements. Now when the diffusion is included, I am getting on the diagonal elements 2 extra contributions okay and that is what is basically being reflected here. Now since this is again a 2 dimensional system, what are the conditions for stability? The conditions for stability are that the trace must be negative and the determinant must be positive correct. I mean because it is a 2 dimensional system, all I am saying is now that the Jacobian matrix is modified okay. We have a modified Jacobian matrix. We have a modified Jacobian matrix okay and the Jacobian matrix is given by f u-alpha square du, gu, fb and gv-alpha square dv. For the stability of the steady state or for rather okay for stability or for let us say instability, for instability of the steady state with the diffusion added, either trace must be negative. Let us do like this. Let us do like this. Let us talk about stability. For stability of the steady state, the trace of j must be negative okay. Trace of j must be negative means f u plus gv-alpha square times du plus dv must be negative. That is when diffusion is added, I want the steady state to be stable. So when will that happen? I am looking at the one condition first. I am looking at this condition. Now what do I already know? In the absence of diffusion, I have already said that the system is stable because if the system is unstable without diffusion then there is no point in talking about the stability with diffusion okay. So I already know that f u plus gv is negative. f u plus gv is the trace of the system. So given we know or we have f u plus gv is negative. That is what we started off with. So if f u plus gv is negative and alpha square of course is positive, diffusion conditions are positive that means this is always going to be negative. So what I am trying to say is that for the system with the diffusion, the trace is always going to be negative okay. So when we add diffusion, trace is always negative. That means the trace condition cannot be violated and it cannot give you an instability okay. Now we have to look for the other condition because both the conditions have to be satisfied simultaneously that trace and the determinant. In order for it to be stable, in order for this 2 dimensional system to be stable, the trace and the determinant both have to be satisfied. I am saying that the trace is satisfied. Now what about determinant? Let us look at the determinant of the system. So the determinant of that system tells me. Yeah? Yes? We should not be taking this argument from previous case right because this time are we not expecting steady state to be different. No, the steady state is the same. See the earlier problem, the diffusion was not there okay. So I had a particular steady state, USS, VSS. I have added diffusion but I am keeping the steady state the same. So question is when I add the diffusion to the system, can the diffusion induce an instability? That is what we are asking. So without diffusion, it is stable. The trace is negative. The steady state is the same. So FU and GV are all evaluated at the same steady state. So like he was asking, they are evaluated at the steady state. Now the steady state with the diffusion is also the same. So FU and GV, all this is being evaluated at the steady state remember. All these derivatives are evaluated at the steady state because my Jacobian matrix is evaluated at the steady state. So this FU and GV are the steady state which is the steady state of the earlier system. So that is the reason I am going to use the information from the earlier system. If this is another steady state, I cannot. The determinant of J, this guy is FU-alpha square du times GV-alpha square dV-FVGU as the determinant and I am going to multiply this out and get alpha power 4 du dV-alpha square times what? du GV-dVGU-FUGV-FVGU. That is my determinant of the system okay and remember all the FU, GV, etc. are at the steady state just to be making it clear one more time because I am done linearization. I am done linearization about the steady state okay. So the partial derivatives are evaluated at the steady state. Now what do we know? So this partial derivatives are also going to be evaluated at the same steady state. What do we know? The determinant of the system, the earlier system was positive. So this guy is positive right. Given from the earlier thing this piece is positive okay. This guy is also positive. Diffusion coefficients are positive. This is also positive and clearly the only way the determinant can be negative is if this is positive because this is associated with the minus sign then if this dominates these two terms. See look at this. This is positive. This guy is positive. There is a minus sign associated here. If DUGV-DVGU is negative, if this is negative, if something is wrong, this one, DVFU. Thank you. Thank you. So if this guy is negative then everything is positive because there is a minus sign here. That means the determinant is going to be positive. The determinant of the system with the diffusion okay. That means then the determinant of the RD system is positive and my steady state is stable okay. So in order for my, if at all you want to have instability, if at all you want to have instability of the spatially uniform solution then this has to be negative okay. So for instability, so for instability, FU must be positive and this is, is it a necessary condition or a sufficient condition? It is a necessary condition because just because it is positive you are not assured. It will be negative. It has to be significantly large only then. So this is a necessary condition. Just because this is positive, it will not guarantee you that you will have instability okay. This is a necessary condition. It does not give guarantee instability okay. So now you already know something about F and G. What do we know? Some trace condition. What do we know? We know FU plus GV is negative. That is given to me right. So now I want to see when looking at this condition and looking at this condition, when can this happen? When can I possibly have an instability? When can I possibly have a spatial pattern okay? If this is negative, I have 2 options. Both are negative individually or one is positive, one is negative okay. If both of these are negative then I cannot satisfy this. Agreed okay. If this FU is negative, GV is negative then this will always be negative. If therefore one has to be positive, one has to be negative then one is positive, one is negative and then maybe I can satisfy this condition. You understand? So both FU and GV cannot be negative since then TU GV plus DV FU will be negative always okay. So now you got a choice. You can choose whatever you want. I am going to choose FU to be positive and GV to be negative okay. So let us choose FU to be positive and GV to be negative. It does not matter because it is chosen in the other way. It really does not matter okay. So I am choosing FU to be positive and GV to be negative and if that is the case what exactly does this imply? Remember F occurs in the mass balance equation for U. DU by DT is F of U, V okay equals F of U, V and DV by DT is G of U, V right. If the partial derivative of F with respect to U is positive that means as the concentration of U increases the reaction rate is increasing okay and that is typically what you expect if you have a first order reaction or a second order reaction you say that the more the concentration the more the reaction rate. So this is a normal behaviour whereas for G, for here GV is negative that means as the concentration of V is increasing the reaction rate is actually decreasing okay. So here this is as the V as it increases actually is slowing down the reaction. So in some sense you can think of the species U as an activator that is the more the U the more the reaction rate. V is an inhibitor the more the V the lower is the reaction rate okay. So this is something which we have understood now that in order for U to possibly have an instability you need to have an activator and an inhibitor okay and we have to do this analysis some more but we will do that in the next class but all I want to emphasize here is U is an activator why because as the concentration of U increases the reaction rate increases V is an inhibitor why because as the concentration of V increases the reaction rate decreases. So basically this you need to have a combination of an activator and an inhibitor in your reaction system in order for U to actually have a pattern if at all but of course this is just a qualitative argument what we need to do is get conditions on the diffusion coefficient okay.