Welcome back to this course on Advanced Process Dynamics. As we enter the fifth week of instructions. Let us now switch gears and let us learn non-linear dynamics.
As I might have mentioned before, most processes around us are dynamical in nature and most systems are non-linear if that be the case, why study linear dynamics. So, the reason is quite simple that, in fact, there are two three reasons why we study linear dynamics and all the techniques that we have been studying right from the beginning of this course. The first thing being that, well, it is quite possible that the system is really linear in nature for which you need to know the techniques for handling such linear systems.
But more importantly, when most processes are non-linear, it is possible that over a small range of parameters or time the processes may behave linearly. Or in other words, there may not be a large difference between the behavior of a non-linear system and a linear system. And in such a case, the study of linear systems would make a good case for the study of non-linear systems.
Further, as we know that for linear systems as well as non-linear systems, which we are we can identify equilibrium states. Now, equilibrium is a term which is used for equilibrium solutions rather is a term which is used by mathematicians, for engineers more appropriate term is steady state. So, at steady state is the equilibrium solution exists and therefore, nothing changes with time.
And if nothing changes with time, it is basically in material whether your system is linear or non-linear, but about this steady state, we can actually do a linear analysis of a non-linear system. In other words, we can convert a non-linear system two possibly a linear system and this would be more prominent around the steady state. So, this is what we are going to do, we are going to take up examples of non-linear systems and there would be two approaches of studying non-linear systems.
The first approach would be to linearise a non-linear system. And the second approach would be to in fact, use the techniques of non-linear dynamics or in other words, do the analysis in a non-linear domain itself and understand the characteristics of the system. So, before we do any of such things, let us first try to understand the basics of non-linear systems and how do they differ from linear systems.
So, in our previous lectures, we defined in detail what a linear system is. So, for ensuring that a system is a linear system, what you need to do is you need to test the principle of linearity whether the principle of linearity holds true or not. So, what you would do is you would identify the operator corresponding to your model equation and if L is an operator for your system.
And your solution space contains two vectors u and v, they can be two solution functions, then the solution then the system is called linear if L operating on u plus v is same as L operating on u and L operating on v and L operating on alpha times u where alpha is the element and the field over which your solution space is defined, then it should be equal to alpha times L of u or L operating on u. And then we give this definition that those systems which do not follow this particular definition, are in fact, non-linear systems. So, let us take an example of both of these systems we have been using this equation dx by dt is equal to ax now, for quite some time and we know that this is a linear first order autonomous equation, it is first order because there is only one equation one ODE first order ODE, its autonomous because the right hand side has only x there is no t, but, we really took it as granted that this is a linear system, let us formally verify whether the system is indeed linear or not.
So, my equation is dx by dt is equal to ax. So, if x1 and x2 are the two solutions then, what I will do is I will identify the operator and operator in this case can be identified as d by dx, d by dt, d by dt minus a and if x1 and x2 are the two solutions, then L operating on x1 would be equal to dx1 by dt minus ax and L operating on x2 would be dx2 by dt. So here it should be x1, here it should be x2.
And according to my first condition L operating on x1 plus x2 has to be determined and this would be equal to d by dt operating on x1 plus x2 plus minus ax1 plus x2 I will expand it this would be dx1 by dt plus dx2 by dt minus ax1 minus ax2 and then I would do some rearrangements and put some brackets this would be dx1 by dt minus ax1 plus dx2 by dt minus ax2. Now, from equation two I see that dx1 by dt minus ax1 is nothing but L operating on x1 and from equation three I see that dx2 by dt minus ax2 is L operating on x2. So, therefore, what I see is that the first condition of linearity holds true for this equation rather for this operator L defined as dt, d by dt minus a.
Now, the second condition L operating on alpha times x1 would be equal to d by dt of alpha x1 minus a times alpha x1 and let me do further simplification. Since alpha is a constant this will be alpha times dx1 by dt, a as well as alpha constant. So, I can write this as minus alpha times ax1 which means, this is alpha times dx1 by dt minus ax1 and again from equation two dx1 by dt minus ax1 is alpha L operating on x1.
So, the second condition of linearity is also satisfied by the operator and therefore, equation one dx by dt is equal to ax is a linear equation. And the system whose dynamics is given by equation one dx but dt is equal to ax is, in fact, a linear system. Now, I will make a small change in equation one, and let us write equation one now as dx by dt is equal to ax squared, this is my equation, I can write this as dx by dt minus ax squared is equal to 0.
And now I need to identify an operator identifying an operator for the first term is easy at simply d by dt minus for the second term ax squared, what you basically do is you take the solution, you square it and multiply it with a, so there is no standard notation for it and therefore, we would have to resort to a notation what we will write here as this. So, in an analogous manner, let me write for the first operator derivative operator as well. So, this is my L operator.
So, therefore, L operating on x1 would be equal to dx1 by dt minus ax1 squared. And similarly, L operating on x2 would be dx2 by dt minus ax2 squared. So, let me write this as equation two and equation three.
So, therefore, I can write L operating on x1 plus x2, this would be equal to instead of this dot here, I would write x1 plus x2. So, therefore, I will have d by dt of x1 plus x2 minus a and in place of x, I will write x1 plus x2, so x1 plus x2 squared, I will simplify this. So, this will become equal to dx1 by dt plus dx2 by dt minus ax1 squared minus ax2 squared minus 2ax1x2 and from here, I can write this as dx1 by dt minus ax1 squared plus dx2 by dt minus ax2 squared minus 2ax1x2 from where I can write L operating on x1 plus x2 is equal to L operating on x1 plus L operating on x2 minus 2ax1x2 and I have this additional term minus 2ax1x2.
So, therefore, I can see here that for a naught equal to 0 for a naught equal to 0, L operating on x1 plus x2 is not equal to L operating on x1 plus L operating on x2 and when x and when a is in fact equal to 0 then you simply have the equation dx by dt is equal to 0 and that can be very easily proved to be a linear operator. But, as long as a is a non-zero quantity, you do not satisfy the first condition for linearity. So, this is equation four.
The first condition for linearity is not satisfied. We do not need to go further, because the first condition is not satisfied, but we can still look into the second condition L operating on alpha times x1 would be equal to d by dt of alpha x1 minus a alpha x whole squared. Which means this is equal to alpha dx1 by dt minus a alpha squared, x squared which can be written as alpha times dx1 by dt minus ax, but this is multiple but this is.
So, there should not be a bracket here ax squared and this is multiplied by alpha and you can see that this is not equal to alpha times L operating on x1 equation five. So, you can see here that neither the first condition nor the second condition for linearity is satisfied and therefore, a system described by the equation dx by dt is equal to ax squared equation one is a system which follows non-linear dynamics. So, can we have some physical examples, perhaps examples from the previous lectures that we studied, so, we had previously studied the dynamics of the liquid level in a tank our model equation was the dh by dt is equal to 1 over A q1 minus q2 we took several examples and one of the specific example was that when the input flow rate q1 is 0 and output for it q2 is such that you have a valve and q2 is equal to A times h then your system becomes linear first order autonomous system.
Then imagine that you have a gravity driven flow, gravity driven flow with no inlet rather than using inlet let us say no inflow there is no inflow. So, therefore, q1 is equal to 0 and I have a gravity driven flow. So, for gravity driven flow I have the expression for q2 as the area of the pipe multiplied by the discharge coefficient of the pipe multiplied by root over 2gh this is going to be my expression.
So, therefore, I can write this as dh by dt is equal to minus Ap Cd root over 2g upon capital A times root h and if I denote this entire quantity as some quantity alpha in fact, we will use alpha later in our notation. So, instead of alpha let me use the term simply a if this is equal to a then I have dh by dt plus a the root h is equal to 0. Now, this is my model equation for a gravity driven flow.
So, now let us see if equation one is a linear equation or a non-linear equation. The system is linear or not on linear, so, I have the equation dh by dt plus a root h. So, dh by dt plus a root h is equal to 0.
So, I will identify the operator as d by dt plus a square root this will become my operator the process of taking the square root and multiplying it with a. So, if h1 and h2 are the two liquid levels, then L operating on h1 is equal to dh1 by dt plus a square root of h1. L operating on h2 is dh2 by dt plus a operating on h2.
This is two this is three from where I can write L operating on h1 plus h2 will be what d by dt operating on h1 plus h2 plus a operating on square root of h1 plus h2, I can simplify this and this would be dh1 by dt plus dh2 by dt plus a square root of h1 plus h2 and I know that this is not equal to L operating on h1 plus L operating on h2. So, therefore, my condition for linearity is not satisfied. Can I take the case of second property?
L operating on alpha times h1 would be equal to d alpha times h1 by dt plus a square root of alpha times h1 and this is equal to alpha times d h1 by dt plus square root of alpha multiplied by a root of h1, which is not equal to alpha L of h1. So, we took the case of liquid level problem and what we saw was that under various conditions the same system can act in different manners in all the previous cases whenever we took this example, the system was linear, today when we have a system where there is no inlet and the outlet is now gravity driven flow then we saw that the system becomes non linear. Let us take one more example the example of cooling of a body, we took this example previously and we saw that you can rearrange this equation by considering all of these quantities in the bracket as constants this is what we did previously.
So, when hAs upon rho Vc is a constant say a then what happened. We saw this previously that we have the equation which would be dT by dt is equal to minus a T minus T infinity and when you add T infinity in fact also was a constant and when all of these conditions are satisfied, then you could write dT star by dt is equal to minus a T star where T Star was T minus T infinity. And this equation in the box which you got was a linear first order autonomous equation, but when we did this analysis and in fact, we did this analysis pretty thoroughly to determine the time evolution of temperature and we determined all the face portraits in all of those conditions, the results were true when this condition of constant values of this entire multiplication and division giving rise to a holds true and constant T infinity holds true.
But when we see these properties, heat transfer coefficient surface area density volume specific heat, how sure we can be that we in fact have a system which in which the properties are perfectly constant, we cannot be sure, if you do this experiment over a very large range of temperature, then certain certainly the density of the body is susceptible to change that can happen changes in the volume of the body et cetera. In fact, this is true for all of these properties, which are shown here. And therefore, the moment you introduce any of these properties with a function of temperature, when you introduce temperature in any of these, your system will become highly non-linear.
And therefore, all of the analysis which we did previously was true only under certain assumptions. And as I mentioned before that you can have the linear dynamics as an approximation of non-linear dynamics only under certain conditions or under specific ranges in this particular case, when the temperature range of operation is small, then you can assume that the properties are independent of temperature of properties do not vary considerably with temperature. So, the since all the properties are constant, our linear analysis holds true otherwise, it is not.
If that is the case, then how do I analyze a non-linear dynamical system? Well, to analyze a linear dynamical system, we always used to write these dynamical equations and the output equations the vector x1 x2 up to xN is the dynamical vector the vector y1 y2 y2 yP is the output vector and the vector u1, u2 up to uM is the input vector. So, this was the typical representation of a of an Nth order linear system.
Now, what would happen if I do not have linearity in my system? Then what would happen is and in fact, you have all of these dimensions for these equations in front of you, which we have come across before. Please make sure that you still remember the dimensions and how do they come from where do they come from.
But if I have a non-linear dynamical system, then what is going to happen is that instead of the right hand side being very beautiful matrices multiplied by vectors, now, you will have to write individual equations. So, for example, you may have dx1 by dt is equal to ax1 and dx2 by dt is equal to bx2 as a dynamical system in this particular case autonomous so, you did write this as d by dt of x1 x2 is equal to a 0, 0 b x1 x2 we did this previously. But, if this is not the case and you have in fact a non linear system.
Let us take an example dx1 by dt is equal to x1 x1 plus x2 and dx2 by dt is equal to x2 x1 plus x2 if this is the case then but you cannot in a straightforward manner write this system of equations as a matrix equation and we saw several advantages of converting a system of equations to matrix equations and then by then we could do an Eigen value analysis and comment upon the stability of the system and so, so on and so forth. So, if that is not the case that means, you have a non-linear system then in general you would have a set of equations which would be your dynamical equations and the dynamical equations would be functions of the individual dynamical variables and also the individual forcing functions. So, the functions here go from f1 f2 up to fN.
Similarly, the output equations go from y1 individual locations from y1 to yP and now, the right hand side some other function and the right hand side goes from g1 g2 up to gP and all of them again are individual functions of the dynamical variables and input functions. But, suppose if I have this, if I have these equations. Can I do anything with these equations to convert them to matrix equations?
And that is basically the meaning of linearization, linearization. So, what I will do, what I will do is, for example, when I have dx1 by dt is equal to x1 x1plus x2 and dx2 by dt is equal to x2 x1 plus x2, I will determine the steady state solution in the language of mathematicians, I will determine the equilibrium solutions, as an engineer I know that it is not very appropriate to call steady state as equilibrium, but we will follow the general convention. So, I will determine the equilibrium solutions by setting up this as 0, this as 0.
And if those equilibrium solutions are x1s and x2s then at x1s and x2s, my system dynamics ceases to occur which means I have 0 gradients nothing changes with time and therefore, it does not matter whether I have a linear system or a non-linear system and therefore, about that point x1s x2s, I can do an analysis which matches my analysis with the analysis of linear systems. That is the meaning of linearization. I am converting my non-linear system of equations or non-linear system to a linear system about the steady state or equilibrium solution.
So, let us see how would they look like. So, if the steady state solution of the non-linear system is described with this vector, a few moments back I wrote the steady state solution the procedure to drive the steady state solution for a two by two system. So, if this entire vector is the steady state solution, which is basically obtained by determining the equilibrium solution.
Then what I can do is I can write that at any location x1, x2,… xn, u1, u2,… um in proximity of the steady state solution as the value of the function at the steady state plus the derivative with respect to the first component multiplied by the difference between that that particular location and the steady state plus the same thing for the second component and so on. Similarly, I can do this analysis for the function g as well and what basically am I doing this is nothing but Taylor series expansion, in fact multivariable Taylor series expansion. So, currently I have how many variables for f I have n plus m number of variables and for g also I have n plus m number of variables.
So, if I have these many variables then for first, for just one variable I know that I can drag that illustrates much of it easily in higher dimensions instead of just derivative I will have to take partial derivative and in fact all the partial derivatives and then what I have done is I have truncated the series and I have just gotten rid of the second and higher order derivatives. So, this is basically the Taylor series expansion. And when I do this, when I do this, I can write now my equation as a matrix equation.
So, now, I define the deviation vector I define my deviation vector x1 minus x1 steady state x2 minus x2 steady state and so on and similarly, deviation vector for u, deviation vector for y and when you look at this equation when you look at this equation take it at this side then you will find that you can rearrange this entire equation to equations in the form of matrices matrix equation further. So, what would happen in the deviation variable form in the deviation variable form you will you can now write dx star by dt in fact, there has to be an under bar here there has to be an under bar here please make a correction here. So, in the deviation variable form you have Ax star plus Bu star and y star under bar is equal to Cx star plus Du star.
And this is basically the same equation exact same equation as what you wrote previously, the x under bar by dt is equal to A double under bar x single under bar plus B double under bar u under bar and y under bar is equal to C double under bar, x single under bar plus D double under bar u single under bar this is for linear system. And what you have here is not for non-linear system this is for linear rise the system remember, this is for linearized the system which means this will hold true in the proximity of your steady state, you cannot guarantee that this would be the most generic behavior of the non-linear system over the entire possible state. So, this is what we learned today that when you have a system, which is non-linear, there is one particular way to handle such system by linearizing this.
So, lean non-linear set of equations in general would not be elegant enough to be put in matrix form, but what you can do is you can determine the steady state and about the steady state you can do a Taylor series expansion and then you will realize that you get the exact same form of equations which is dx by dt is equal to Ax plus Bu. And y is equal to Cx plus Du, which will be the linearized form except that you need to realize that here x, u and y are in the deviation variable form. So, we will stop here today and continue our discussion on non-linear systems in the next lecture.
Thank you.
