 So, good morning everyone and welcome to today's lecture. So, today we will be looking at a short introduction to the stability of dynamical systems. So, in the previous lectures so far in this course we have seen how to formulate problems in fluid mechanics including two phase flows and how to obtain a solution for the case of a steady state in most of the problems. Ultimately we are building up to a position where we want to be able to study some complex two phase flows which have time dynamics. So, how it fits in with the previous part of the course is that so far we have seen how to find one particular solution under certain assumptions to the Navier-Stokes equations. But the problem is that that solution is not always stable which is what we discussed at the end of the previous lecture that often there will be multiple solutions to the Navier-Stokes equations because they are highly non-linear. And what will happen physically is that for some set of parameter and Reynolds numbers and other parameters you will have one solution stable and that is what you will see in experiment. But on the other set of conditions when you increase the Reynolds number or in terms of an experimental parameter if you turn up the flow rate at some stage you will have a transition to a new type of behavior and spatially you will see a new kind of pattern. So, what has happened is a solution has become unstable and a new solution has become stable. So, to understand these transitions we can draw on the theory of dynamical systems and the stability of these different steady states. So, in this class I will take a very simple second order two dimensional dynamical system and we will track go through how we can understand the stability of a steady state and how that transition happens. So, before I start how many of you all are familiar with the idea of linearization and then stability of systems in ODE's you can have a show of hands. So, I can gauge where to pitch the lecture. So, in for the audience watching this lecture also this is the basics of what we would do in a course like process control as well and I will just focus on the simple ideas here. For those of you all who already know some of this what I will do is try to give it a spin that allows me to generalize what I am going to do today to the partial differential equation setting which is what we are really interested in this course. So, let us begin. So, to let us consider very simple system. So, I have two variables x1 and x2 which are dependent on time and system is governed by two equations that tell me about the time dynamics are going to be and in general this first derivative will be some function f1 of the two dependent variables and since the right hand side does not have time in it it tells us that the system is basically the derivatives are time independent or how the system evolves depends only on x1 and x2 it is current state. I can rewrite this these two ODE's in a more condensed form where the dot represents the derivative f is a vector function and this capital X is a vector of the two quantities x1 and x2 and capital F is. So, this is a standard representation of the system in a condensed notation. So, whatever try to do today is lay out the steps that one would need to follow to analyze a system for steady states and stability. So, the very first step is to identify the steady states and generally it will be plural for non-linear systems. So, in order to identify the steady state what we need to do is we are looking for x1 and x2 values of the system such that the left hand side derivatives go to 0 or there is no evolution in time. So, these states are simply given by setting x1 so x1 and x1 star x2 star rather steady states to the system which must satisfy these two equations because only then my evolution in time will be 0 or have a time independent steady state. Again in vectorial form I can write this as so this first step is what we have already covered in the course so far. So, if these were if you are looking at the Navier-Stokes equations we have already seen how to calculate steady states for some condition and generally the idea will be that this x star this steady state is something that is quite simple I mean it is the it is an obvious steady state solution of the problem for some parameter values. So, for example, it could be a situation where there is no flow or for example, Poiseuille flow in a pipe just a simple Poiseuille for solution at steady state and then we want to understand how that solution undergoes a transition. So, in the ultimate case transition to turbulence in some simpler cases that we are looking at this course maybe some convection cells will set in and different features may arise. So, this will be a simple base state that we are going to start with and now we want to see whether that x star is stable or not. So, that brings us to step 2. So, step 2 is called linearization and the idea here is that because we want to see whether the system is stable around x 1 star and x 2 star we give small perturbations to the current position and see whether the system returns back to the original steady state or whether it deviates off and this ties in very well with the perturbation theory we have just completed where we have seen how to derive equations when you have small changes epsilon order changes. So, that is what we will do here. So, the variable x 1 which is my dependent variable I am going to give it a small perturbation about x 1 star. So, I will have x 1 is equal to x 1 star plus epsilon which has the amplitude of my perturbation into x 1 tilde. So, here tilde represents the perturbation that I am giving but this x 1 tilde is bounded by 1. So, that is magnitude is an epsilon. So, this entire quantity is very small compared to x 1 star. So, this quantity is a small quantity similarly I can do that for x 2. So, x 1 tilde x 2 tilde tell me in this 2D case the direction in which I am pushing the system or the what kind of deviation I am going to give to the system and epsilon just reminds me that that has to be a small deviation because I am looking at small perturbations. So, again I can write this in vectorial form as x which was the vector of my 2 dependent variables as the steady state plus epsilon times x tilde over x tilde as the deviation variables. So, this is the expansion and now like any perturbations calculation that we have done we need to substitute this into the original equations and then obtain simplified equations for the evolution of x 1 tilde and x 2 tilde because that is what we want to identify with stability. We are giving perturbations of this form and we want to see whether these how these grow in time. So, to be explicit I have x 1 of t and x 2 of t they are both time dependent and for small deviations about the steady state this time dependence is simply going to be of this form. So, you will have the steady state part and to the steady state there is some time dependent deviation and the idea is to find out whether this time dependent deviation is growing in time or whether it is going to decay back down to 0 and give me my steady state. So, to do that we come back here and substitute it in these equations. So, if you put it into the left hand side we will get simply the derivatives of the deviation variables to order epsilon naturally because these are of course time independent. Now, coming to the left hand right hand sides here f 1 and f 2 is any general function. So, while I cannot make the substitution mechanically what I can do is use Taylor series and that is what we have been doing in the course when we do not have simple functions. So, we can expand f 1 as a Taylor series. So, let me write it down here explicitly. So, x 1 is simply x 1 star which is some constant plus epsilon x 1 tilde. So, that is f 1 and now if I expand it in a Taylor series about x 1 tilde and x 2 tilde because it is a 2 dimensional function. Then I will have the function f 1 at x 1 star the value of the function at the steady state plus the deviation into the derivative of the function with respect to x 1. Of course evaluated at x 1 star x 2 star same thing for the variable x 2 and I can write the same over here. So, that is pretty clear I just use the Taylor series up to first order epsilon. So, to be exact I should add the fact that there will be order of epsilon square terms which I have neglected to write down at the moment. So, now the first thing you can see is that the function value f 1 x 1 star x 2 star is naturally going to be 0 because that was how we got x 1 star x 2 star in the first place. In other words these two are steady states. So, naturally they will evaluate to 0 on the functions f 1 and f 2. So, these guys are just knocked out and I mean that is by construction it is not a miracle that is going to happen every time naturally. So, what we are left with if you see is just a linearized system that is why it is called linearization. So, the derivative of x 1 tilde which is now the growth of the deviation is going to be related linearly to x 1 because this is simply a constant evaluated at the steady state. So, this is a function the derivative but we have to evaluated at the steady state. So, it is a constant factor multiplying x 1 tilde. Now of course, you will ask me that there are order of epsilon square terms as well. So, if I go to epsilon square I will get high order corrections which will involve x 1 tilde square and so on. And then we know from the perturbation theory that we have already looked at that since the left hand side is order epsilon. So, naturally we will equate it to the right hand side terms of order epsilon itself because these order epsilon square terms will be much smaller. So, if we look at it as a perturbation problem we just equate the coefficients of epsilon on both sides and what will be left with well again the dots represent derivatives is simply I just put the star here to indicate I am evaluating at steady state. So, that is what I get after I equate terms to order epsilon. And now I can write this in my vectorized notation. So, then I will get the growth of x tilde is simply equal to a matrix which I will call j and use the double under bar to signify a matrix times x tilde itself which is a classic first order linear algebraic system. And this j is called the Jacobian matrix and it is simply given by the partial derivatives. So, the matrix j has got the 4 partial derivatives all evaluated at the steady state and note that it is f 1 in the row and f 2 in the second row. Of course, if you ever forget that you just have to come back and re derive this which is quite straight forward. So, ultimately what we have seen here is I mean we should not get too caught up in the formal notation. I want you to remember from the beginning that we have some dynamical system this could for all you know be a partial differential equation also on the right hand side. But the point is we found a steady state and then we linearized about the steady state by the basic idea is giving small perturbations. Now, in the case of this simple 2D system dot small perturbation procedure ultimately led to this very simple x dot equal to j x form. This same form may not arise in an immediately obvious fashion when we are doing PDEs. But the idea will be the same we will have to give small perturbations and we know how to deal with those perturbations using perturbation theory. So, ultimately we will land up in some linearized description of the system. So, that is the point about step 2. So, moving on to step 3 and step 3 is where we now want to ask the question given my linearized equation how is x tilde going to grow in time. So, imagining that we have absolutely no clue how to solve this system by any formal rules what we would do is assume a solution for the time dependence. So, if you look back there you can see the full matrix form and since it is linear I can assume that x tilde which is the function of time is going to be sorry x 1 tilde is going to be some constant u 1 which is a number like 2, 3, minus 5 some number which I have yet to determine and the time dependence can grow as e to the power sigma t or exponential growth. The reason I would do e to the power sigma t is quick note that suppose I have a system d y by d t equal to say lambda say sigma y. This is your classic first order single equation O d right and what is the solution to this problem? This is e to the power sigma t into the value at t equal to 0 that is just an exponential growth. So, you can see that the time dependence has e to the power sigma t. So, in analogy to this we are looking for a similar solution when we have more than one equation and I am postulating beforehand that this growth rate for all the variables that is your x 1 tilde and x 2 tilde have the same sigma. So, right now just look at it like a hypothesis and let us proceed and then if things work out in the end we can look back and think about why it happened. So, right now I am saying that this has e to the power sigma t and so does x 2 all right. So, what I have done now is made a statement about how the time dependence is going to be and what this tells me now is that suppose my sigma is positive then I will have this exponential term growing unboundedly which means that x 1 tilde and x 2 tilde will both grow. So, if sigma is positive I will naturally have an unstable system right because my initial perturbations here x 1 tilde x 2 tilde are going. So, I give some small initial perturbation that initial perturbation you will see here is just u 1 and u 2 correct in the form that I have shown if time t is equal to 0 the initial condition I will just have u 1 and u 2. So, u 1 and u 2 represents that small perturbation I am giving now the question is whether this factor will grow. So, if sigma is positive that initial u 1 and u 2 is going to grow without bound if sigma is positive and this x 1 and x 2 of t will very rapidly move away from the steady state because of these terms. On the other hand if sigma is negative then e to the power sigma t dies off because sigma is negative. So, then the system will be stable because this portion will rapidly decay to 0 with time and I will return back to simply x 1 star and x 2 star. So, in this form of my perturbation sigma is crucial and that is why sigma is often called the growth So, typically again in these problems at step 3 we look at the linearized equations and these equations as you will notice will be homogeneous which means that if I say x tilde 0 it will satisfy the equation and that is it has to happen because saying x tilde 0 basically is saying that I am on the steady state and I know that the steady state satisfies the original equation. So, I will have a linearized homogeneous system for the evolution of the disturbance variables and the next step is always to assume a form e to the power sigma t for the growth where sigma is the growth rate. So, now for this particular system we will see where this leads us. So, if I substitute this again into the equation that I have here. So, let us look at the expanded notation and let us see what we are going to get. So, what will the right hand side be x 1 tilde derivative is simply d by dt of this and e to the power sigma t gives me back e to the power sigma t. So, I will have u 1 but with a sigma and my right hand sides this is of course a constant I have x 1 tilde which I will substitute from here. So, I will get u 1 e to the power sigma t back right do the same thing for the second equation. And now you see that because the equations were linear I have this e to the power sigma t term common everywhere right and so naturally I can knock that off. So, what am I left with I am left with an equation for u 1 and u 2 which of course are unknown at this point I mean that is one way of looking at it sigma is also on that is what we are trying to find. So, this is what I have. So, now if we again look at this in terms of the matrix notation remembering that those 4 partial derivatives are j I will just have sigma times u capital U where capital U is I will write that at the end. So, sigma times u where capital U is a vector of u 1 and u 2 is simply the Jacobian matrix times u itself where here u is u 1 and u 2. So, this process of assuming this form e to the power sigma t has led us to this equation in this particular system. And now the question is how do we use this to evaluate sigma what does this tell us about sigma what does it tell us about u. So, suppose sigma has some random value maybe 5 then I need to find a u such that 5 u is the same as Jacobian times u 1 sure solution is if I take u equal to 0. So, u equal to 0 will always solve the system that is because it is a homogeneous set of equations and saying u is equal to 0 means I have not given any disturbance. If u is non-zero this equation may not be solved there is no guarantee that for any non-zero values of u I will have this equation satisfied. But there could be some values of sigma such that a non-zero value of u will satisfy this equation. So, my objective really is to find those values of sigma and those non-zero disturbances such that this equation is satisfied. And this is the classic Eigen value problem that we have solved in general dynamical systems theory this is what an Eigen value problem looks like where you have a homogeneous system of equations which always has the 0 solution. But there will be some parameter like sigma and in this case the growth rate it is. So, there will be some values of the growth rate such that you will have non-zero disturbances propagating to the system and that is what we want to solve. Now in this particular 2D system it takes this simple matrix Eigen value form which we can now solve with the standard theory we already know. But in the PDE system we will also reach the stage we will have an Eigen value problem to solve and then we need to proceed with that solution either analytically or numerically. So, that brings us to so the end of step 3 is the Eigen value problem. So, in this case sigma is the Eigen value and the disturbance is the Eigen vector. So, that brings us to step 4. So, step 4 is to solve the Eigen value problem and determine the growth rate and determine the disturbance. And how do we solve classic matrix Eigen value problem? We realize that this is equivalent to saying. So, this itself is a matrix and for this to have non-zero solutions u this matrix has to be singular which means that the determinant should be equal to 0. So, for a second order system like the one we are looking at this will lead me to a quadratic equation and I will get 2 values of sigma. In general this equation can be some polynomial nth order if my system is nth order if you have a partial differential equation then this can be some extremely complicated function more complex than we can imagine sometimes. But it is always solvable if you cannot do it analytically we can do it numerically. But this is the equation is called the characteristic equation that leads us I do not know where lambda pop into my head but it should be the identity matrix naturally. So, this is the characteristic equation that gives me the characteristic values or Eigen values which in this case is simply the growth rate and you will have many values of sigma in general. For a 2D system you will get 2 growth rates for a partial differential equation you potentially have an infinite number of sigma or infinite number of growth rates. So, what that means is that there are various key directions in which you can disturb the system and if sigma is positive along even one of those directions or if even one value of sigma from here is greater than 0 then the system is unstable and the Eigen vector corresponding to that Eigen value sigma the positive guy will be the direction along which the system will grow and that Eigen vector will contain information of the new pattern or the new state. So, this equation can usually be solved and we can obtain the values of sigma and once we know sigma we can come back here and calculate the Eigen vectors knowing sigma we then calculate u which is the Eigen vector and that u will contain information u contains the information of my new state if sigma is greater than 0. So, if sigma is greater than 0 the previous state has become unstable and I am the system has to go somewhere. So, where it is going to go what will be the new pattern that I see will that if some of that information at least is there in you and we can get a lot of information if you do a slightly non-linear calculation. But from linear calculations we can tell if it is stable or not to very small perturbations because that was the initial hypothesis it has to be infinitely small. So, only to those types of perturbations we can tell whether the system is stable or unstable. So, that was step 4 where we solve the Eigen value problem and that will immediately tell us and answer all our questions about stability. Are there any questions at this point? Yes. So, then what you do the same thing that you do in the case of solving these linear algebraic equations because they are linear you will recognize that each of those sigmas and each of those u's correspond to linearly independent solutions. So, then the general solution will be given by the sum of them. So, because it is a second order system you will have 2 solutions and you will have 2 different directions in which the system can grow and in the general case your actual growth will be some linear combination of the 2. So, I will write that down formally. Yes, the question is that after solving this problem I have a value of sigma 1 and a value of u 1 right and I have also calculated another sigma 2 and another u 2 that is not a good idea. Let me call it sigma 1 sigma 2 u 1. So, u is Eigen vector corresponding to sigma 1 the first Eigen value. Second Eigen value has another Eigen vector phi and you will have only 2 for a second order for a 2 dimensional system and if I have a higher order high dimensional system I will get more of these. So, then I can represent x 1 of t for example as the value u 1 that comes from here e to the power sigma 1 t right and multiplying with some constant alpha some linear combination or in terms of my so all this is some linear combination of the 2 fundamental solutions I got from the Eigen value problem and you see that makes sense because it is a second order system. So, I need 2 I mean I will have 2 unknown constants because I have 2 initial conditions 1 for the initial value of x 1 1 for the initial value of x 2. So, there are 2 constants here which I can determine from my initial condition. Now, if I substitute so then this is definitely a solution because if I put it back in the equations they will each satisfy the equation on each half. So, the first part will be 0 second part will be 0 independently and there are solutions. So, a standard superposition of solutions. So, this is why if sigma 1 if either of sigma 1 or sigma 2 are positive it is unstable. So, even if sigma 1 is negative this part will go to 0, but this guy will blow up if sigma 2 is positive. So, for stability I need all the sigmas to be negative and even if one of them is positive it will grow. So, if everything else is reasonably clear at this point what we can do is look at a simple example of a 2 dimensional problem where we will get a feel for the actual mechanics and see how it plays out in a real system. So, there are a lot of subtleties that I have left out here. For example, what would happen if you get sigma equal to 0. So, that would mean that my Jacobian matrix is I mean it has a 0 eigenvalue. So, then I have some problems I cannot calculate the eigenvector and the eigenvector is can be anything. So, in such systems you actually read a you will end up at a contradiction in the mechanics because the fundamental theorem that lets me do this linearization fails if you have a steady state where you have a 0 eigenvalue. And those points are very special actually and that is usually what happens the problematic points are the ones that should be paid special attention to. So, in such situations honestly this theory completely fails. So, if the sigma is equal to 0 I cannot tell whether the what will the dynamics be whether it will be stable or unstable I have absolutely no idea. So, to find that out we need to go to the next order of epsilon. So, here we stopped at first order in epsilon we need to go to the epsilon square order to get more information. And the formal theory is called center manifold theory and we discuss more of these higher order dynamics in the course that we do in the alternate semester which is steady state and dynamical systems. So, next semester probably we will be having that course again. But for our purpose in this course we will be mainly wanting to identify the transition point. So, sigma equal to 0 is the transition point of the system. So, we will be varying some parameter like the Reynolds number and for small Reynolds numbers say the sigma will be negative system is stable. For higher Reynolds numbers suddenly one of those sigmas in a calculation will become positive. So, that point where it is 0 marks the transition and that is very important for us. Because that will tell us the critical parameter values for which my I am going to have a change in my flow. And how to investigate the dynamics over there we need to go to high order in our perturbation. So, to read more about the ODE systems theory dynamical systems theory and go through some of the things I have said you can refer to professor's book and its mathematical methods for chemical engineering. I think it is a Prentice Hall India. So, this book is available in the library and so I think in his last chapter he goes through this stability theory in some detail with a lot of practice problems and even answer some of the questions about high order dynamics. So, that is the place you can refer for what we have done today. So, we will now look at a specific example to which we can apply the past steps that we have just discussed and analyze the stability of a dynamical system. So, now I will work in x, y, z variables rather than subscripts. So, the change in x where the dot denotes the derivative is minus x, y and the change in y is minus y plus x square minus lambda. Here lambda is a parameter, write that clearly. So, if you look a bit carefully at these equations you will see that for positive values of both x and y the change in x pulls it back towards 0 as thus this term in y, but the x square term takes y further away from 0. So, there is a chance for some instability and some dynamics in this system and we will get right to the very first step of our analysis which is to find the steady states. So, to find the steady states we put the right hand sides of both these equations to 0. So, from the first equation we can get that say x is equal to 0 then from the second equation I would immediately find that y is equal to minus lambda. So, this is my first steady state and the other option from here is that y is equal to 0 in which case I am going to get x square is equal to lambda or x is equal to square root of plus minus square root of lambda. So, you see that I actually have 3 different possibilities one is 0, minus lambda and the other 2 are and you see another interesting thing here that these 2 steady states are possible only if lambda is positive because if lambda is negative naturally both of these would be imaginary and our vector field to begin with is real. So, these exist only for lambda greater than 0 whereas this steady state exists for all lambda right. So, this is the picture that we have and clearly at lambda equal to 0 I have a transition from one steady state to 3 steady states and this is something that we should keep in mind as we go on and we will get back to this at the end after we analyze stability. So, now that we have looked at the steady states we will move on to step 2 and that was of course linearization. So, in this step we write the variables as the steady state value plus a perturbation of order epsilon and substituting this into our original system of dynamic equations and linearizing we already know that we will get equations of the following form right where the Jacobian matrix contains the partial derivatives of the 2 right hand side vector fields. So, why do not we compute the Jacobian matrix now? So, here I will have derivative of the first function with respect to x moving on to the second term right and now that we have the Jacobian we want to evaluate the Jacobian at the base solution which in this case the base solution that we want to study the stability is the one that exists for all values of lambda and that is when x is equal to 0 and y is minus lambda right and we store this need result. Okay, so you might ask the question why are we picking this particular steady state to look at of course we should study the stability of these as well. But for this lecture I am going to focus on this for 2 reasons firstly this is the steady state that exists for all values of this parameter lambda and so in some sense it is the trivial steady state that is always a solution and in this sense which comes to the second point this steady state is very similar to the kind of simple steady states that we find in the fluid systems that we will be studying in this course. So for example Rayleigh-Bernard convection experiment when the fluid is at rest without any convection that is the simplest possible steady state of the system which exists for all values of the parameter like the Rayleigh number or the amount of temperature difference of the plates and then after a certain amount of heat input you get new steady states arising which is the convection cells which correspond in this analogy to these new states. So that is why right now we are going to focus on this trivial simple steady state which exists for all lambda and we can see that the Jacobian at that point at 0,-lambda is simply this diagonal matrix. So having completed the first 2 steps we now move on to the crucial step 3 which involves the Eigen value problem. So if you remember at this stage we make a substitution for our variables some constant which is governed by its initial condition or at this point itself I can say x0 e to the power sigma t. So we assume a exponential growth having the same growth parameter sigma for both variables and we can make this substitution into a linearized system where the Jacobian about a base state is lambda and minus 1 the diagonal matrix. So after doing this as we know we are going to get sigma times after simply substituting this in this expression we will get sigma times the vector is equal to Jacobian times the vector and this has the classic Eigen value form matrix times vector gives me the Eigen value which in this case is sigma times the vector. So directly as we had already said sigma is going to take the values of the Eigen values of the matrix j of course evaluated at 0,-lambda at the steady state of interest. So in this case sigma can take 2 values plus lambda and minus 1. So in a 2 dimensional system the 2 Eigen values or growth rates are lambda and minus 1. So we can see here directly that this minus 1 term this Eigen value is only going to lead to a decay and exponential decay of the solution. Lambda on the other hand can make the system unstable and lead to exponential deviation away from the base steady state 0,-lambda if the values of lambda are positive. So at this stage now we can conclude about the stability of the steady state of interest 0,-lambda and say that it is stable when lambda is less than 0 and it is unstable when lambda is greater than 0. And this is now very interesting result because you see it fits in very well with the multiplicity of steady states that we had observed right at the beginning. So if you come here you will see that for negative values of lambda only one solution exists because these 2 solutions will become complex. So when lambda is less than 0 I have only my base steady state and in that case the base steady state is stable. However when lambda is greater than 0 my base steady state becomes unstable and that is exactly when the 2 new steady states arise with plus minus square root of lambda,0 and now these come in so to speak to take the place of the original steady state. So I can represent all of this in terms of a diagram of the steady states plotted against the parameter lambda which in the literature has been called a bifurcation diagram. So in the bifurcation diagram now I will plot so on this axis I am going to plot values of lambda and on this axis I am going to plot x steady state and here lambda is 0. So on this side lambda is negative and here lambda is positive and if I plot the steady states now I will get a situation where I have only one solution all the way up to lambda equal to 0 and then exactly at this point this original solution which has simply x equal to 0 becomes unstable which I have denoted by the dashed line and instead of this I will get the 2 new solutions which grow as the square root of lambda and you can see that they will be symmetric because they are plus minus square root of lambda this is 0 and positive and negative. Now I will leave it up as a homework exercise to do the linear stability analysis about these 2 steady states which are plus minus square root of lambda, 0. So you can linearize the system about these 2 steady states calculate the Jacobian it is Eigen values and find whether these are stable or unstable. So that is the homework problem but I can tell you now that in fact these will turn out to be stable and this is a classic exchange of stability concept where one system one state becomes unstable and 2 new steady state solutions emerge which are actually stable. So for lambda less than 0 the system will be at this steady state globally and as soon as lambda crosses 0 it is going to leave this steady state now which is become unstable and move to either the one above or the one below and this bifurcation diagram is called a pitchfork bifurcation and is in fact a classic one of the classic bifurcations in one dimensional systems and is in fact seen in high dimensions as well as in pattern forming systems involving partial differential equations and there is special reason for that and that is the inherent symmetry of this bifurcation diagram and the reason for the symmetry you can trace back to the vector field which if you see is symmetric for values of plus or minus x. So what that means is if I substitute x equal to minus x I will get a minus sign out from here as well as from here which will leave the equation unchanged and you can see the same thing happens here because of x square. So this equation also remains unchanged if I make the substitution x goes to minus x and that is the reason why along the x axis I have this symmetry about 0. So you will find that in our physical systems this is quite common where there is the inversion symmetry about positive and negative values whole and that is why the pitchfork bifurcation is quite common in many of our systems. So we will be seeing this later on in the course again in a more complex context but it is good to remember this right here. So with this we have analyzed today the key stages in a stability analysis and I have shown it to you in ODEs but the same ideas propagate forward when we look at partial differential equations and therefore very often even though we may not be able to find out these steady states because in this case we have a simple vector field. So we could directly obtain the solutions analytically but in complex partial differential equations which have many solutions it is very difficult to find out to solve those nonlinear PDEs and get all the steady states in the system. However we will usually be able to find a simple steady state which exists for all values which in this case is this base steady state x equal to 0. So we would know this and by doing a stability analysis we can find when this becomes unstable and possibly leads to new solutions. And so during the next the latter part of this course we will be focusing on analyzing these systems and studying their part information.