 So, now what I want to do is that I want to see that if I introduce now diffusion, if I now say that these neighboring cells can all talk to each other by diffusing from one cell to the next, can I destabilize the steady state? Is there some choice of this f f g d whatever such that I can destabilize this previously stable steady state. So, again if I now again write that equation in the matrix form. So, what I have is del w del t right is gamma times a the stability matrix times w ok a being whatever it was plus let me write down since I am writing it in matrix form plus let me write down a diffusion matrix times Laplacian w ok. And what is this diffusion matrix then? The diffusion matrix is simply 1 0 0 small d. So, for the u part of this remember there is a vector equation that has u component and v component for the u component you will just get Laplacian of u 1 into Laplacian of u for the v component you will get small d into Laplacian of v which will give you back those equations. So, I have just written it in a matrix form nothing else. So, what I want to do is I want to look at this diffusion driven instability if I can generate a diffusion driven instability of this homogeneous steady state that I have written down. So, what do I do? I can start off by saying that let me define these space let me just define the spatial eigenvalue problem which is is equal to some k square times w with the same boundary conditions that n dot grad w. So, let me define. So, what am I doing basically just to summarize. So, I want to solve this equation what is the standard thing to do maybe I do a separation of variables right. So, let me say that I will write this let me say this is a small w I will write this as some capital W of space times some capital T of time right. I will plug it over here into I will plug it into this equation then I will get one term which depends only on the time terms I will get one term which depends only on the space terms right. What will the space term will look like? It will look like something as Laplacian of this capital W is equal to some k square times this w. So, you can find out the eigen let us say that w k are the eigen vectors of this equation. So, let me say eigen w k are the eigen functions of this equation satisfying this boundary condition whatever is the boundary condition and then for this actual w that I am interested in you can construct a general solution. Let me write a small w you can construct a general solution by doing a linear superposition of all of these eigen functions right. So, I can do a linear superposition of this w k of r the time part will we look like e to the power of some lambda t times some coefficients c k some lower all wave vectors k right. Is this clear? You do a separation of variables you get a set of basis functions from the expansion from this equation of this special part which are these w k which obey whatever boundary conditions that you have which means that you will have some allowed wave vectors right. For example, if you are doing in 1 d your eigen functions would be some cos let us say n pi x by l or sin n pi x by l right and your allowed wave vectors would be n pi over l right for various values of pi. The time part the time equation will look like something like del t del t is equal to some lambda t right if you if you use this separation of variables which will have some solution like e to the power of lambda t. The general solution you construct by doing a linear superposition of all of these allowed solutions and then you have to find out these coefficients the c k's by matching whatever initial condition is given right this u r comma 0 and v r comma 0. So, now, I take this I take this general solution and I put it back into this equation that I want to solve del w del t. So, what does this give me del w del t? So, the left hand side will del del t will bring me down a lambda a lambda times a w right. On the right hand side I have this times this is the capital W this make it clear. The left hand side on the right hand side I have gamma times the stability matrix times this w of k plus d. So, ideally you get this with a summation. So, if this holds true for every value of k then you know that the your original differential equation is going to be satisfied. What is this Laplacian of this capital W k that is nothing, but minus k square w because that is how this is defined right this is the. So, I can just write this is gamma times a times W k minus d k square W k. The Laplacian of W k is k square of W k these are the Eigen functions of this equation. So, I just use that. So, now what I need to do because I want to now evaluate these lambdas because these will now tell me that in the presence of diffusion how does this how do perturbations from this steady state behavior right. So, again I will solve the characteristic equation and the characteristic equation in this case is. So, again the characteristic equation in this case is lambda times i the identity matrix minus gamma times a the stability matrix plus d times k square this is equal to 0 lambda times i minus gamma times a plus d times k square ok. So, now I can write this down I know what all of this is. So, this is lambda minus gamma f u plus d k square. So, plus k square then here is minus gamma f d then 0 minus gamma g u and lambda minus gamma g v plus small d k square is equal to c separate ok. So, this is now my two new equation that I have to solve I know what is my a I know what is my d I know what is the identity matrix I can just write this down and again I now need to solve it for lambda ok. So, let me write this down. So, this gives me lambda square that is the first term plus lambda into minus gamma g v plus d k square lambda square minus lambda k square minus lambda gamma minus gamma f u plus k square then what else have I miss plus gamma square f u g v gamma square f u g v minus gamma d k square f u gamma k square g v plus d k to the So, let me just reorganize this. So, let me write it a little better lambda square plus lambda into k square 1 plus d minus gamma f u plus g v I have missed one term ok. So, right I have not written the last term. So, what is that minus gamma square f v g u ok all right. So, plus something which is let me say d k to the power of 4 minus gamma what gamma terms do I have gamma k square minus gamma d f u d f u plus g v into k square right and then this and that. So, plus gamma square f u g v minus f v g u. It is a horribly complicated well not horrible. So, it is a slightly complicated equation, but provided I have done everything correctly this is what you get. So, this is the equation for lambda again a quadratic equation, but at least it is a quadratic equation. So, I can just solve it. So, lambda is equal to minus of this thing minus of k square 1 plus d minus gamma into f u plus g v plus or minus this whole thing square. So, I am not writing it out again minus 4 h of k square. So, it is just to save me writing this let me call this entire thing as h, h as a function of k square ok divided by 2. So, what I note is that in the absence of diffusion the lambdas are less than 0 or at least the real parts of lambdas are less than 0. Accents of diffusion basically means that k square is equal to 0 right. So, I know that the real parts of lambdas lambda at k square equal to 0 these are less than 0. So, because it is a stable steady state. What I want to find out is that is there any value of this wave number k such that this real part of lambda as a function of k this becomes greater than 0 right. If I can make this eigenvalue positive at least one of these eigenvalues positive then I can destabilize what was previously a stable steady state and I have this possibility of patterns arising ok. So, how can I do this? So, maybe let me see I can have this quantity itself whatever is in the bracket if this was positive right then this will come with a negative sign. So, then I would have at least one positive root ok. So, I could say that well what is the possible. So, if I want this thing that real part of lambda is greater than 0 if if this thing k square into 1 plus d minus gamma into f u plus g v if this thing was greater than 0 then I can have this that real part of lambda is greater than 0 or I have another possibility which is that if even if this was positive if if let us say this was positive, but h of k square is a negative quantity then this negative negative would become positive I would in this negative term I would get something which was ultimately negative right. So, another possibility is that this h of k square this term is less than 0. So, let us look at these possibilities let us look at this first one k square is the wave vector yes which one is what do I want that should be less than 0. So, if this is what I want remember k square is the wave vector. So, that is positive d is the ratio of the diffusion coefficients. So, that is positive. So, this is a positive thing f u plus g v I know is negative because that was a condition of my stable steady state which means that minus of a negative quantity is always positive. So, this is positive this is positive therefore, this cannot be true this can never be negative ok. So, this condition is out of these two possibilities this first one is never possible. So, all I am left with is this term that h of k square is less than 0 for some k for some for some k it may not be for all wave vectors, but at least for some k this should be less than 0 ok. Yes provided you have a stable steady state for the reaction diffusion if you did not then yes of course, I mean if this was if this was positive then in general yes, but then you would not even see a stable steady state even in the absence of diffusion. Yes, yes of course. So, for example, if I had an equation if I had a chemical reaction like this I have two species a and b let us say they both up regulate themselves and they mutually down regulate each other ok. So, neither is a clear activator or inhibitor right. Remember last day we were talking about a pattern in the equation that I saw in that we looked at where a activated itself and activated b whereas, b auto inhibited itself and inhibited a right. But if I think of something like this where each species activates itself and mutually they mutually repress each other then f u which is like del f del a is positive f g v which is del f del v that is also positive. So, therefore, this would not be true and this is a very common pattern you will see in lot of in lot of genetic networks when you talk about transcription factors mutually interacting this is a very common motive that you come across. I mean simplified no motive is just two chemical reactions, but roughly yes such motifs are possible. So, what is h of k square oh I have rubbed it off sorry let me write down one small what is h of k square. So, h of k square is d k to the power of 4 minus gamma small d f u plus g v plus gamma square into f u g v minus f v g u ok. So, this was my h k square and I need this to be less than 0 in order to have some wave vector where this lambdas are positive ok. So, in order to destabilize the system ok. So, now, let us look at this d k to the power of 4 is always positive ok. So, this term is always positive f u g v minus f v g u I have said needs to be positive in order for it to have a stable steady state gamma square is positive. So, therefore, this term is also necessarily always positive. So, the only way you can have and this comes with the negative sign remember. So, the only way you can have this to be less than 0 is if this term is greater than 0 right. So, that the negative sign will make it less than 0. So, the only possibility is if d f u plus g v this is greater than 0. The only way you can make h k square less than 0 is if this term d f u plus g v that is a positive term. Now, let us think about what that means. So, I have this condition now d f u plus g v is greater than 0. Can you have f u and g v both to be of the same sign? Yes. Can you have f u and g v both to be of the same sign? No right because then you would violate this condition f u plus g v less than 0. So, let us say let us let me say that let us say f u greater let us say f u is positive and let us say that g v is negative I could take the other way also. If f u is positive and g v is negative then d must be then d must be greater than 1 in order for you to have this equality this inequality being satisfied which means that remember what was d, d was the ratio of the second species to the first. f u greater than 0 again if I draw my speed, if I draw my reaction that I have f u greater than 0 means a activates itself, g v less than 0 means d inhibits itself. So, a is my activator and b is my inhibitor. So, which says that and this ratio being greater than 1 says that whatever is my inhibiting species that my inhibitor must diffuse faster than the activator faster than the activator. If I had reversed this if I had said that f u was less than 0 and g v was greater than 0 then I would have the less than 1 which again physically would mean the same thing that whichever was the inhibitor species that then in that case in this case b is the inhibitor if I took the other assumption then a would be the inhibitor and the what I this would say is that a would then need to diffuse faster. So, this inhibitor must always diffuse faster than the activator that continues to hold true ok. So, let me write this condition here as well the d f u plus g v that is greater than 0. So, this was 1, this was 2, this is 3 ok, but ok is there anything else? Even if I had said that this is greater than 0 I need this to be greater than 0 in order for this h of k square to be negative. So, it is a necessary condition it is not sufficient right because to get what to be sufficient this term d f u must overcome this this and this the contribute the positive contributions of this and this. So, this is a necessary condition, but in order for it to be sufficient condition. So, let me get rid of this anyway this condition does not matter. What I need to say is that let us say that I have this h let us say I plotted this h as a function of k square right some arbitrary whatever depending on what f u is some curve h of k square. What I need is that at least yes yes the first condition is still satisfied, but let us say that both are negative right. Can you ever get f a positive number into a negative plus another negative number giving greater than 0 will you cannot destabilize the steady state. So, that is what we are looking for right. So, in that case you cannot destabilize the steady state. So, a thing like this a b which is I think what you are saying that both sort of repress themselves it can be a valid chemical reaction, but it will never have a pattern in this sense. In order to have a pattern you need this that if one is activating itself the other needs to replace itself. So, that is what 1 and 3 together imply. So, coming back to this. So, I need this h k square to be less than 0 for some value of k at least. So, what I can say is that this is a necessary condition for it to be sufficient what I can look at is I can look at the minimum point of this h curve ok wherever this h of k square has a minimum. And if at the minimum point let us say this is my minimum point if at the minimum point the value is less than 0 then I am guaranteed that at least 1 k exists where I have this h of k square less than 0 there might be more. So, the curve might look something like this in which case this whole range of k's are k's that will destabilize the steady state, but at least if I have 1 then at least I am guaranteed that I have 1 wave vector that will be destabilized. Yes, k square the middle term has a k square. Thank you. Yes, I should not draw a random graph is that what you are saying? I can still draw a nice draw something like this yeah I should not draw a completely random graph ok. So, then what I want to do is that in order to have a sufficient condition that this h of k square is less than 0 what I will find out is that I will find out the minimum of this h k square curve. So, I will do del h del k square equal to 0 ok and find out. So, this I will solve this this will give me some value k m where this curve h of k square has a minimum. I will find out the value of h at this k m point and see whether that is less than 0 on so del h del k square. So, del h del k square is 2 d k square minus gamma d f u plus g v is equal to 0 which means that this k m square is equal to gamma into d f u plus g v divided by 2 d and the value of h at this point the value of h at this k m square the value of h at this wave vector at this minimum wave vector is d into k to the power 4. So, gamma square into d f u plus g v whole square by 4 d square minus gamma into then is right at straight up there is a gamma square d f u plus g v whole square by 2 d and plus gamma square this is a determinant or if you just write determinant of a what do I have? So, this is actually the same term this d and so this is 4 d and that is 2 d so minus 4 d. So, this is I can just write gamma square determinant of a minus d f u plus g v whole square by 4 d. So, this is the value of this h this quantity h at this minimum point. So, what this so for this to be negative what this means then is that this term must be greater than this determinant of a obviously. So, for this to be negative what it means is that d f u plus g v whole square must be greater than 4 d into determinant of a or if another way say if I just write it out d f u plus g v whole square minus 4 d into f u g v minus f v g u that has to be greater than 0 ok. If this condition is also satisfied then at least I am guaranteed to have one wave vector at least one it can be more I am guaranteed to have at least one wave vector for which you can destabilize this stable steady state and of course, the critical wave vector will come when you equate this to 0 right. So, basically if I think about this h of k square plot. So, if I think about this h of k square plot h of k square versus k square plot some h looks like this for some parameter values. Let me say I am just tuning the diffusion coefficient for example, then at some value it will just hit the axis and then for some values like this. So, this is let us say some critical ratio of the diffusions this is some d greater than dc. So, act the critical whatever parameter let us say the diffusion ratio of diffusion coefficients you have this one single wave vector that will destabilize the steady state. If you increase this ratio even more you have this whole spread of wave vectors from this k 1 to this k 2 for which this stable steady state will become destabilized. So, correspondingly if you were to plot this lambda the actual eigenvalue you would get something like this and then something like this and then something like this. At the critical point there would be one wave vector for which this lambda is positive or just positive and then there will be a range of wave vectors. So, this is the fourth condition that I now have for pattern formation which is that whatever I wrote down. So, d fu plus gv whole square minus 4d into the determinant of a that must be greater than 0. So, provided so, these two conditions 1 and 2 are required to say that you have in fact, a stable steady state even in the absence of diffusion. Conditions 3 and 4 are what is required for this diffusion in order to destabilize this steady state and give rise to some spatially extended patterns. So, these four conditions and this is quite generic I have not recourse to any particular form of f and g or whatever. So, you can take any chemical kinetics that you wish and think about whether you can destabilize the steady state or what this says is that in order to form a pattern at least by this sort of a reaction diffusion mechanism you need to have one of the species which is an activator and another species which is an inhibitor and then something about here. So, you can try to construct patterns and see that what sort of chemical networks even if they are going into the rates and so on what sort of chemical what patterns of these chemical reactions will can give rise in principle to pattern formation through this reaction diffusion mechanism. For example, like you said if both of these were inhibiting auto inhibiting then you would never have a pattern.