 What we will do today is that we will work out the conditions for this patent formation in a reaction diffusion system. So, so last class we saw examples of how you can generate patents. The idea being that you you have some cell where you are undergoing some sort of chemical reactions between multiple species. So, we saw an example with two species let us say A and B and the idea being that the chemical reactions themselves can lead to some sort of a steady state. They can lead to some sort of a steady state, but if you now allow and if this reaction we consider is taking place in multiples of cells the same reaction. If you now allow these cells to communicate with each other specifically through diffusion then what we showed was that in certain cases you can destabilize this steady state of the non interacting system by introducing this sort of diffusive interactions between the different cells right and by destabilizing the steady state you can generate patents which are you can generate spatial patents which are different from this steady state that we saw in this isolated model ok. So, what I will try to do today is that I will try to sort of formalize this what are the conditions on this reaction system or this diffusion prop what are the constraints on this reaction diffusion system that will allow us to see the formation of a patent ok. So, this is generic it is true for any reaction diffusion system and you can analyze any reaction diffusion system using this sort of a framework to say when will a patent form or when will it not form ok. So, the basic I will stick with this two species model. So, the basic idea being that I have some sort of a chemical reaction between two species. So, I can write down the time evolution of the concentrations of the two species. Let us say del a del t is some reaction term f of a comma b just write capital F f of a comma b and del b del t equal to some g of a comma b. So, this is my reaction part of the system it encodes the chemical kinetics that is going on between these two species a and b ok. In addition now I want to say that these species can diffuse in space. So, I say a diffuses with some diffusion coefficient b diffuses with some other diffusion coefficient. So, this is in some sense the basic equation that I want to analyze ok. The specifics of the equation will come of course, in this f and the g terms the specifics of the strength of the diffusion will come in the d a and the d b. Before I start analyzing I will just rewrite these I will just rewrite these equations. So, the first thing to do is to non-dimensionalize these equations. So, let us say I have non-dimensionalized them and I write them now in terms of this non-dimensional concentrations I write in terms of u and v ok. What was a and b I just non-dimensionalize and I call it u and v and I will just write it in this form f u comma v plus. So, it is essentially the same thing as here except what have I done I have replace this d a by 1 and this d b by small d. So, you can think of this small d as the ratio of the two diffusion coefficients ok. Similarly, this capital F and capital G after read after non-dimensionalizing I have pulled out a factor gamma ok quite generically. What this gamma factor I mean I could have called this as f prime I just pulled one term out because I want to sort of encode. So, you can think of this gamma as sort of saying how strong is your reaction term in comparison to the diffusion term right or another way of interpreting gamma is that it contains information about the length scale of the system ok. So, in a finite system for example, what is the size of the box that you are going to form this we are going to do this reaction systems. So, it encodes for example, the relative strength of reaction versus diffusion and also it has encodes information about length scales. So, this is non-dimensionalization. So, these are the non-dimensionalized equations that I want to start quite generally again I have not really changed anything from here to here it is a different way of writing. Now, given that this is of course, a differential equation in order to solve it you need boundary conditions and you need initial conditions. So, let us specify some boundary conditions let us say no flux boundary conditions. So, let us say I have some boundary conditions no flux boundary conditions which is n dot grad u is equal n being the sort of surface normal n dot grad v is equal to 0. And let us say that I specified for you the initial conditions initial conditions something. So, u of x comma 0 and v of x comma 0. So, this is a time t equal to 0 what were these concentrations u and v ok x or whatever you know generally r you can write it is in 3D. So, these are also given. So, given the boundary conditions and the initial conditions together with this differential equation I now have everything in principle that I need to solve this problem. So, that is what we will try to. So, the first thing is of course, to find out remember the initial idea is to say that well there is a steady state in the absence of diffusion and there is a stable steady state in the absence of diffusion and we want to see whether diffusion can destabilize the steady state ok. So, first then we will consider this problem the absence of diffusion and then find out what is the requirement on these reaction terms in order to have a stable stable steady state. So, let us say I have a steady state. So, in the absence of in the absence of diffusion so, no diffusion no diffusion let us say I have a homogeneous steady state which I call some u naught v naught. So, these are the concentrations of the steady state if there was no cross talk between the cells and so, these are my equations now del u del t is equal to gamma times f u comma v and del v del t is gamma times g of u comma v. So, how do I find the steady state of this equation? I just say that del u del t is 0 del v del t is 0. So, this u naught v naught our solutions of so, u naught comma v naught our solutions of f u v equal to g u v equal to 0 right. So, this is the homogeneous steady state in the absence of diffusion. Now I want so, whatever it is right as long as I do not specify f and g you cannot find the particular solution, but in general given an f and a g you can equate them to 0 you can find out what is the stable steady state ok. So, you can find out the steady state now we must look at the stability of the steady state right. So, how do I look at the stability of the steady state? So, what I can do I will do a linear stability analysis. So, I will define perturbations away from this steady state. So, let me say let me define a quantity w. So, I will just write in vector form. So, which is perturbations away from this steady state. So, w is u minus u naught and v minus v naught right. So, I can now substitute this w into this equation right. So, what will I get? I will get a del w del t. So, by introducing w instead of having to write two equations for u and v I am just writing it in vector form. So, I will just write it together . So, del w del t is going to be what? So, if I do an Taylor series expansion of this f and the g terms around the steady state right. So, the first term will be I do not. So, let me just write del. So, if I do a Taylor series expansion the first term will be f at u naught comma v naught right plus delta into del f del u delta into del f del v and so on right. The f at u naught comma v naught is of course, 0 because I am expanding about the steady state which means that if I just write it in the matrix form what will I get? I will get a gamma times del f del u del f del v del g del u del g del v right times this w itself right. So, because I am doing a linear stability analysis I do not keep the higher order terms I just keep the first order terms ok. So, this this matrix let me call it as my stability matrix. So, let me define it as my stability matrix let me define it as a stability matrix a let us say and in short form I will just write f u f v g u g v these being the partial these partial derivatives f u is del f del u and so on ok. Evaluate it at this evaluated around at this steady state evaluated at u naught comma v ok. So, next in order to say whether the steady state is stable or unstable we can look for solutions of this type we can look for solutions of the form w is equal to sum e to the power of lambda times t right. And then I will say that this steady state u naught v naught is stable if these lambdas are negative if the real parts of these lambdas are negative right. So, u naught v naught is a stable steady state is a stable steady state if the real parts of the lambda is less than 0 right let us make a straight because if this is less if the real part is less than 0 then as time goes to infinity these perturbations will die out we will come back to the u naught comma v naught state ok. So, what I need to then find is the eigenvalues of this matrix. So, I need to solve this characteristic equation I need to solve this characteristic equation this is gamma times the matrix a. So, I need to solve this gamma times this matrix a minus lambda times identity matrix is equal to 0. So, this is my characteristic equation. What does this mean? This means basically gamma f u minus lambda gamma f v gamma g u gamma g v minus lambda is equal to 0 right. So, what does this give me? If I just write it out so, gamma square f u g v minus lambda gamma f u minus lambda gamma g v plus lambda square minus gamma square f f v g u is equal to 0 ok. So, in order to find out these eigenvalues lambda I need to solve this equation which is just a quadratic equation. So, this is lambda square minus lambda into gamma into f u plus g v right and then plus gamma square f u g v minus f v g u. So, here is my quadratic equation that I need to solve. So, this is of course, I know what is the solution. So, lambda is equal to lambda equal to minus b. So, gamma into f u plus g v minus b plus minus square root of b square. So, gamma square into f u plus g v whole square minus 4 is c minus 4 gamma square into f u g v minus f v g u. So, this is under square root divided by 2 a ok. So, these are my eigenvalues ok lambda times f u plus g v plus minus this quantity divided by 2 ok. What I want is that the real parts of these lambdas are negative then I know that this the solution that I have written down the steady state solution u naught comma v naught that is linearly stable. So, it is a stable steady state ok. So, what can I say? So, what are the conditions for this lambdas to be negative? So, let me see this clear. So, that is my lambda. So, let me say that if f u plus g v this quantity if this was less than 0 ok. So, if this quantity was less than 0 and if and so, let us say this is case a and then let me say that f u g v minus f v g u this was greater than 0 ok. Let me say this is greater than 0 and what else and say that is is less than f u plus g v I think I have specified everything. So, if this was the condition right that this term is negative this quantity in the bracket is positive and is less than this quantity. Then other what are the roots are the roots positive or negative under these conditions anyone? Negative. Are they real? So, if ah yes I should have included the factor 4 somewhere then these are real and negative right. So, in this case under this condition both roots are actually real and negative or real and negative because anyway when this sign because this minus when this sign is minus anyway it is minus even when this is plus this is less than this. So, therefore, it is it will not be greater than that f u plus u ok. What would happen if I said that ok this is still positive f u g v minus f v g u is still positive it is greater than 0, but this is delta. So, f u g v minus f v g u 4 times is greater than f u plus g v whole square then then the real. So, then the roots would become complex, but the real part would still be negative right. So, these would be complex roots these would be complex roots, but real part is negative. What if I now relax? So, what if I now say that f u g v minus f v g u is less than 0. So, if this is a negative quantity then what happens? So, one would be positive then and another would be negative which means you would have this blowing up right this state would no longer be stable. So, so one positive one negative. Now, what would I say if I said that f u plus g v which I had taken to be negative if this was positive then can you ever have both roots the real parts negative no right. There will always be at least one which is positive. So, here you will always have at least regardless of whatever on the other condition at least one positive. So, quite generally then having written down a steady state u naught comma v naught the condition for that steady state to be a stable steady state of this non diffusive non interacting system is that f u plus g v must be negative and what is f u plus g v? f u plus g v is the trace of this matrix that is the stability matrix that I have written down and this other quantity f u g v minus f v g u which is the determinant that must be positive right because in both these cases this was positive and I had real parts were negative ok. So, the condition for the steady state. So, then let me rub out somewhere. So, if I solve this characteristic equation what it tells me is that the condition for the steady state for a stable steady state condition for a stable steady state is that f u plus g v which is nothing, but trace of this stability matrix a that must be negative and f u g v minus f v g u which is nothing, but determinant of this matrix a that must be positive. So, if I satisfy these two conditions then these are sufficient for me to guarantee that this steady state that I have written down u naught comma v naught is a stable steady state fine. So, this is as far as this non diffusing part goes where I am just talking about the reaction parts of this equation have not yet brought in diffusion actually I should have written it here. So, that I can rub that out. So, let me just if you will allow me let me just rub this out. So, my stability matrix a let me just write is f u f v g u g v and condition for stable steady state is that trace of a equal to f u plus g v is negative and determinant of a is f u g v minus f v g u that is positive. So, these are my two conditions.