 Welcome to the lecture where we are discussing the stability of the solution to reaction diffusion equation okay and what we did till the last time was we found a solution which was spatially uniform which was basically saying that the steady state was 0 for all values of the parameter okay and we decided we asked the question whether such a solution is stable because if it is stable that means you will actually observe it. If it is not stable that means you are not going to observe it okay so that was the question which we wanted to find out the answer to and in the process what we did was we found the steady state, we did the linearization and this was the linearized equation. Now what you have to remember is that Uss equals 0 is a steady state for all values of A, D and L it is always a steady state okay that is something which I want you to remember now. Because recall that Uss equals 0 is a steady state for all A, A is something like reaction rate constant okay, D is diffusion coefficient and L is the length of the thickness of the catalyst so these are the 3 physical parameters no matter what you take for all combinations this is always a possible steady state okay. Now the question is will we observe this that is the question and answer is yes if it is stable that is the answer but the question is when is it stable and that is what we are trying to find out okay. So when is this stable that is the next question and that is what we are trying to find out now when is this particular steady state stable. So in the process what we have to do is we have to solve this linear equation which is subject to homogeneous boundary conditions okay the boundary conditions are U tilde equals 0 at x equals 0 and L okay and since it is a linear equation we decided to seek a solution in the form of variable separable form and we substituted this here and then we get this equation and with a little bit of rearrangement you find that the left hand side is a function of time the right hand side is a function only of x and the only way these 2 functions can be equal is if both of these are equal to a constant. So what I am going to do now since it is a constant there are basically 3 possibilities the constant can be positive it could be 0 or it could be negative okay. So this constant can be positive 0 or negative right what we are looking for is a nonzero solution to this differential equation I mean 0 is a solution always but if you get 0 as a solution that means what your perturbation is 0 but what we want is given a particular perturbation as time goes to infinity we want to find out how the system is going to behave. So we are interested in finding out a nonzero solution so I am going to claim and this is something if you people have to verify that if this constant is positive or if this constant is 0 the solution to this equation the second order equation x is going to be 0 okay what I am saying is if this constant is positive let us say lambda squared plus lambda squared okay or 0 I am saying plus lambda squared lambda squared just to tell you it is positive and a plus sign in front there is no way this guy can be negative right or 0 the solution to this equation actually yeah this equation is this d squared x by dx squared multiplied by d plus ax equals lambda squared x is 0 when it is subject to the boundary condition x of 0 is equal to x of l equals 0. So how do I get these boundary conditions on x these come from the boundary conditions on u, u is 0 at the 2 ends so only way you can be 0 at the 2 ends if x is 0 at the 2 ends. So x has to be 0 at the 2 ends and then if x is 0 at the 2 ends I am saying if you solve this equation the solution is 0 if you have plus lambda squared if you have d squared x by dx squared plus ax equals 0 subject to x of 0 equals x of l equal to 0 also has only the trivial solution x equals 0 you understand and I want you to verify this you can just proceed do the algebra and find out find the solution to this put the boundary conditions and see if you can get nonzero solution okay. However if the constant is negative and I am going to indicate that as minus lambda squared then this equation dx double prime for the second derivative plus ax equals minus lambda squared x admits a nonzero solution and this is basically what you have done all along in your separation of variables solution you have assumed this constant to be equal to minus lambda squared and you have proceeded okay. So the reason why you proceed assume is minus lambda squared is because that is the thing which gives you the nonzero solution and that is what you want if you assume plus lambda squared and you will find that gives you only the 0 solution we assume 0 you get only the 0 solution okay. So the reason I am just going to justify why I am putting minus lambda squared okay. Now which satisfies both the differential equation and the boundary condition so what is the nonzero solution which will satisfy the differential equation and the boundary condition. I am going to claim and this is something which you people have done whenever you have done the separation of variables for example sin n pi x by L this is in fact you will have sin as well as cosine but because of the boundary conditions you can prove that the cosine term does not exist only the sin term exists okay. In fact what I need to do is yeah a and sin n pi x by L satisfies the boundary conditions whether everybody understand this a and sin n pi x by L see I am saying this solution xn of x this satisfies the boundary conditions okay and you can possibly do this in a slightly more formal way but since I have done this problem so many times and since you have also done this problem in calculus you should be able to figure out that I am just jumping a few steps but although it satisfies the boundary conditions I have to make sure that it satisfies the differential equation right and what I do not know yet is the lambda squared I do not know what lambda squared is okay so I am going to substitute this here and find out that the lambda squared what it is in terms of this n pi and the other stuff that is the plan okay. So to find lambda what I do is I just substitute this the second derivative which gives me substitute the solution in the differential equation to find the lambdas okay. So let us do that what do you get you have the d you have the a n and when you differentiate it 2 times you get minus n squared pi squared by L squared okay multiplied by sin n pi x by L plus small a multiplied by capital A and sin n pi x by L equals minus lambda squared a n sin n pi x by L okay. So this is a typical eigenvalue problem which you people have come across before okay and you know that the solution is sin n pi x by L I am just exploiting that knowledge which I already have in proposing the solution. Only thing is normally your eigenvalue problem will not have this a x term now there is an extra a x term here you have solved problems where you have x double prime equals minus lambda squared. So yeah all I am doing now is I am going to realize that a n sin n pi x exists everywhere and so this equation I am interested in a nonzero value so a n sin n pi x is not 0 I can cancel it off and what I get is the diffusion coefficient multiplied by n squared pi squared by L squared plus a must equal minus lambda squared okay. So basically what I am trying to tell you is that the lambda squared and remember what is n? n goes from 1, 2, 3 etc it is an integer okay and correspond to different values of n I will have different lambdas so I am going to put a subscript n here. Now I put this constant as minus lambda squared so what will be the solution for the capital T? The capital T of T is going to be of the form e power minus lambda n squared t okay that will be the solution for capital T. Now so I am saying that the u tilde is if you remember your separation of variables it is going to be summed over all these n's and it is going to be of the form e power minus lambda n squared t sin multiplied by n sin n pi x by L okay. So this is basically how the perturbation is going to change with both space and time. Now just to this is the mathematical solution which we have obtained but I want you to think of this thing slightly in physical terms when you are giving a perturbation or a disturbance to the solution the solution is Uss equals 0 you are giving some perturbation what happens because of this perturbation in some points instead of 0 you will have a non-zero value for the concentration for the u okay and the perturbation is the deviation of the actual value from the steady state. So what we can do is we can think of the perturbation at time t equal to 0 which is when I am starting the experiment as going to be some function of x is a function of x in the interval from 0 to L. Now if you were to extend this function of x periodically okay you can basically represent this function of x in the form of a Fourier sine series. So what we have done is think of an arbitrary perturbation if you have expanded this in the form of a Fourier sine series at time t equal to 0 these coefficients would basically tell you how this particular function how this particular disturbance is resolved in along these components. So this is just like resolving a vector in terms of some basis vector you take an arbitrary vector you can write it in terms of some basis e1 e2 e3 1 0 0 0 1 0 0 0 1 okay. So the way you should look at this is look at some function which is your perturbation at time t equal to 0 is resolved in terms of these Eigen functions okay this sign. And now the question is our interest is to find out how do things behave as time t goes to infinity clearly that is going to be decided by the lambda n squared okay. There is already a negative sign here. So if the lambda n squared were negative then the negative and negative would be positive and you would have the thing blowing up becoming unstable. If lambda n squared is positive then it is going to decay okay. So let me just summarize what I have said and then how do you calculate lambda n squared that basically depends upon the diffusion coefficient the geometry the rate constant that is what you want okay. So that is basically the link in the loop which we are trying to close at time t equal to 0 we give a perturbation okay and this is u tilde of x, t equal to 0 okay. And what I am saying is when I write u tilde of x, t equal to 0 my solution is a n sin n pi x by l. I am just saying that what we are doing is this is a function of x I am resolving this function of x in terms of these basis functions just like you resolve a vector in terms of basis vectors okay. So this is the sin n pi x by l are my basis functions and we resolve the disturbance in a long components okay that is the way I would want you to look at this physically not mathematically of course you got the solution okay. Now so how does each of this like for example in the finite dimensional problem you had a vector disturbance you had 2 eigenvectors and you wrote it in terms of 2 eigenvectors instead of 2 eigenvectors I have an infinite sum now where this summation is going from n equals 1 to infinity. So from what was the finite dimensional problem we move to an infinite dimensional problem okay. So these are basis functions okay and the evolution of the disturbance is given by e power minus lambda n squared t okay the time dependency and what we are interested in is as theta is infinity the guy has to go to 0. So if lambda n square is positive for all n then as t tends to infinity u tilde tends to 0 and we have a stable steady state okay else the steady state is unstable. So now remember lambda n square equals d times n squared phi squared by l squared minus a okay. So I have different eigenvalues lambda 1 squared will be d times phi squared l squared minus a lambda 2 squared is so on and so forth right. So now what do I want if lambda n squared is positive for all n then we have a stable steady state okay. So lambda 1 squared if d is less than or d phi squared by l squared minus a is less than 0 okay. If d phi squared by l squared minus a is less than 0 that means this guy is positive I do not like this I am doing something wrong yeah I have done something wrong here. Now I want lambda to be negative for stability okay. So if the lambda square positive for stability right I want lambda squared to be positive if this is greater than 0 then all then we have stability for the steady state right because this guy is positive then all the other fellows are positive okay. The first one is positive so I want to take 4 phi squared if d phi squared l squared minus a is positive then d 4 phi squared l squared minus a is positive 9 phi squared l squared times d minus a is positive and so on and so forth okay because I am just increasing this guy the positive fellow this is of course a positive term diffusion coefficient number is positive okay. So this guy is positive and keeps on increasing then we have stability for the steady state since for all n lambda n squared is greater than 0 okay which means d must be greater than l squared a divided by phi squared that is the threshold value that is the critical value of the diffusion coefficient. If it was for a given slab for a given reaction which is decided by the rate constant a for a given reaction decided by which decides a for a given slab which decides what the thickness is l geometry is fixed the kinetics is fixed and the diffusion coefficient is the only other parameter. If the diffusion coefficient is greater than this number okay then you have a stable solution for USS equal to 0 okay and that is basically what we said earlier remember if the diffusion coefficient is very large then you would have the any concentration variation which is present in the slab will gets made out because diffusion will flatten it because that is what diffusion does if you have a room where there is a concentration gradient and you just leave it diffusion is going to make it all equal okay. If the diffusion is greater than this critical value you have a stable steady state if the diffusion coefficient is less than this critical value then this guy is going to be negative these guys we do not care these guys could be positive negative the first guy to become negative is going to be this the first lambda n squared which becomes negative will correspond to n equal to 1 this is the first guy all these guys will be positive this guy will become negative first then this guy will become negative as you keep lowering d okay so first guy to become negative will be this guy if this becomes negative then remember the solution is going to grow exponentially okay. So that is basically the onset of instability which means your USS equal to 0 is not going to be observed if the diffusion coefficient is lower than this value if the diffusion coefficient is less you will have a nonzero solution you will have some function you get some idea about how that function is by looking at the corresponding value of the Eigen function here sin n pi x by L sin n pi x by L tells you how the spatial variation is going to be of the nonzero solution so the nonzero solution that you are going to get is going to be of the form corresponding to n equal to 1 is going to be of the form sin pi x by L it is going to go to 0 at the 2 ends and it is going to have one kind of hump in the middle okay so that is basically what the information is which is present in the Eigen function. So now I am just going to write this thing down and so if P is greater than L squared a by pi squared diffusion is fast that means concentration get smeared out we have uniform solution which can be observed and that is stable okay. If D is greater than L squared a by pi squared then all the lambda n squared lambda n squared are greater than 0 okay and therefore the solution is stable once when D is less than L squared a by pi squared just slightly less than okay I should say is equal to L squared a by pi squared minus epsilon where epsilon is positive okay where epsilon is positive and small so slightly less than this then what happens only lambda 1 squared is negative all other lambda n squared are that is for n equals 2 to infinity are positive okay when it is slightly less only slightly less only lambda 1 will be negative lambda 1 squared lambda 2 squared is multiplied by 4 so that guy will still be positive okay all others are positive. So what that means is supposing you have a disturbance and you resolve it in terms of sin pi x by L sin 2 pi x by L sin 4 pi x by L the thing which is going to grow is the one corresponding to sin pi x by L corresponding to n equal to 1 that is the only mode which is going to grow the modes corresponding to n equal to 2, 3, 4, 5 they are going to decay so as a result what you are going to observe in your system is going to be the solution which corresponds to sin pi x by L okay so that is the inside which you are getting the solution observed as t tends to infinity when D is slightly less I do not know how to write slightly less but slightly less than L squared by pi squared will have a spatial dependency of the form sin pi x by L because only corresponding to n equal to 1 is going to grow okay since only n equal to 1 grows all other ends decay okay. So what this means is you will get something like a solution for u which has got some kind of nonzero value inside maybe the single maxima okay so I think this is just to illustrate to you that you have a critical value so there are 2 processes which are taking place one is a reaction process one is a diffusion process okay if the diffusion process is slow if the diffusion coefficient is very low then you will have a nonzero solution but the diffusion is very fast then you have spatial homogeneity and that solution is stable okay that is what physically you expect but what the theory what the mathematics allows you to do is try to get you what this critical value is of the diffusion coefficient okay and so what this means is if you actually had to plot so if we made a plot of let us say u versus D okay and let us say we are plotting u at the center point u at the u at x equals L by 2 no steady state solution as a function of D what do I have for high values of D I know that 0 is my solution which is stable what I am going to observe is 0 okay so this is the value of the solution that I am going to observe this is u ss equal to 0 so u ss equal to 0 is along this line my u ss is 0 means everywhere it is 0 therefore at the center point also it is going to be 0 okay so what I am doing is I am trying to represent how the steady state depends upon the parameter D so the point is u ss equal to 0 lies on this line so u ss equal to 0 lies on this line and this is true for all D all the values of diffusion coefficient of course we do not want to go to negative diffusion coefficient because that does not make sense the point I am trying to make here is this guy is this is let us say my critical value what is my critical value A L square by pi square and that is the number which I can calculate and I am just putting a plot there I am saying that when D is less than this I am not going to observe this whereas I will get some other nonzero value inside my pellet inside my catalyst slab and I want to represent that nonzero value so I am just taking the value at a particular point and so let us say it is going to be positive value it is going to be some nonzero value which may be keeping on increasing so as I keep decreasing the diffusion coefficient the value the magnitude of the concentration of the center of the pellet is going to keep on increasing okay the further I go away from this the more is going to be the value of the center. So what we normally do to represent this kind of pictures we want to represent the stability information also on this kind of a picture okay so the way this is classically done is this this is just to tell you that whenever I have a solid line that represent a stable steady state solution when I have a dashed line I have an unstable steady state solution. So this is a stable steady state okay this is an unstable steady state okay when this particular so when the steady state when the D is sufficiently large I get the stable steady state I can actually observe experimentally but if you keep on decreasing the diffusion coefficient calculation tells you that the steady state is unstable what does it mean whether it mean the catalyst is going to vaporize or does it mean something else is going to happen nothing crazy like that is going to happen right. So what is going to happen is there is going to have a nonzero solution okay and that means this nonzero solution now is going to be the steady solution because that is what you are going to actually observe experimentally so that is the reason I have drawn this by solid line this is a nonzero stable solution I am telling you that this particular branch which I am getting is a another solution okay and if you want you can calculate this other solution by doing a finite difference scheme for the partial differential equation or for the OD equation you can just solve it by using some method some numerical method because it is a nonlinear equation and you can get this branch but this is also a steady solution but this will be observed only when the diffusion coefficient is sufficiently low if the diffusion coefficient is sufficiently large this guy will not exist this guy will collapse to u equal to 0. So this kind of a diagram where I am trying to represent the behavior of a particular system or the solution versus a parameter is called a bifurcation diagram okay. So this bifurcation this is what I have drawn here is actually a bifurcation diagram. So this is a bifurcation diagram okay and what does a bifurcation diagram do? This diagram depicts how a state or a solution changes with a parameter okay and the stability information is also captured I am going to redraw that thing which I had drawn earlier and this is d equals l square a by pi this particular point where you have the change from one branch to another is called a bifurcation point. So the bifurcation point is the point where a new solution branches emerging okay. So I have this is a solution a steady state solution which always exists okay because you will put u s s equal to 0 it satisfies the equation always no matter for all values of a d and l you however want to know if you can actually experimentally actually observe that steady state and that is where the question of stability comes in. Then you do the linear stability analysis and then it tells you only if diffusion coefficient is sufficiently large more than this critical value you will actually experimentally observe and this is being shown by a solid line. So low values of diffusion coefficient I have this dash line but what happens I mean when the diffusion coefficient is low clearly there has to be some kind of reaction going on something has to be happening in the pellet right. So there has to be some other solution which is going to be present in the system and that other solution is going to be of the firm sin pi x by l and that is what the linear stability analysis tells you. So you get a non-zero value for the concentration inside the pellet and this non-zero value I am just depicting in the form of this kind of a parabola. Physically I expect that the further I go away the more is going to be the amplitude correct as I come closer it should collapse to 0 so I go away it should be more so I am just drawing it in the form of this kind of a parabola to tell you that this is the kind of a parabolic dependency this solution has okay. So that is basically it as far as this particular problem is concerned okay I do not know if I have jumped too fast when I wrote down this Eigen value thing but you can solve that linear equation for capital X in different ways you will get the same conclusion okay. If you get a different conclusion then what you have done is wrong you come to the same conclusion then what you have done is okay okay. So what we have seen is how varying a parameter of course experimentally you cannot really vary diffusion coefficient that is a very difficult parameter for you to vary and the only way you can vary a diffusion coefficient is by either making the pores bigger or by changing the gas so but this is just illustrate the concept. But maybe what you can do is think of varying the length of the slab okay so if the length is going to be sufficiently low you will have a specially uniform solution because the distance to which it has to diffuse is lower the thickness of the slab is very large that means the diffusional resistance is high. So you can just invert the problem and then say talk in terms of the thickness of the slab for which you want to get a uniform solution. So this may have some implications especially if you have a non-isothermal reactor where you may want to keep the temperature uniform for example because you may not want to have the temperature go too high inside the catalyst okay for other reasons like maybe undesired side reactions are taking place. So the essential idea is when you do a linear stability analysis what you can do is you can find these kind of transition points where a solution becomes unstable a new solution is emerging okay and this will correspond to some basic change in the physics of the process something which was dominating diffusion which was faster has now become slower and that is basically what is caused and it is a relative rate of diffusion and reaction rate that we always have to look at okay. So we will stop with this for now and we will move on to a problem in fluid mechanics which is again going to be a partial differential equation but with system of equations and then we will solve and that particular problem that we will be looking at is the problem of natural convection which is called the Rayleigh-Bernard problem that is we will have only one fluid but it will have 2 solid walls and therefore it will be part of the multi phase flows course right and then we will have actual 2 phase 2 liquids okay.