 Welcome again back to the lecture, especially the lecture on the first module. In the last lecture, we have seen few interesting examples starting with the population growth model both its linear and non-linear versions. And then we have seen the atomic waste disposal problem and finally, two problems from fundamentals of mechanical engineering they use it extensively the spring mass system and the electrical circuit simple electrical circuit. What surprisingly we have seen is that both this model leads to the same second order equation with constant coefficients. The another interesting observation from our first two examples is that even after getting solution especially the population model a getting solutions whether it is in an implicit or explicit form is not sufficient you have to do more analysis on it. Because the solutions in the explicit or implicit form will not reveal much about the physical model. And in the second example atomic waste disposal problem we have first model it as a linear problem we have obtained the solution, but it did not resolved the exactly what we wanted to. And then we remodeled it as a non-linear problem, but the non-linear problem was difficult we could solve it and get the equation in a implicit relation we could not make it an explicit solution, but that was enough for us to conclude the that disperse loss of the whether we can do it the question raised by the environmentalist. So, this see that how you model how you solve it and even after solving doing it in the analysis are very very important. And that is the purpose as I said in the beginning the purpose of the course is not just to solve it which normally is difficult, but very quite often explicit or implicit solutions are not the important aspects of the theory it is how do you understand its trajectory and how its behavior and what is the physical demands the at what we do it. Now, what we are going to do it we are continue with the last two problems the electrical circuit problem and mechanical thing and we have obtained the equation and we want to interpret the or rather analyze the equation. So, let me recall the second order equation which we have derived second order equation what was the second order equation m into d square y by d t square plus c into d y by d t plus k into y is equal to f. So, this was the second order very simple equation linear and what the other interesting thing is that with constant coefficients. So, you will learn more about this first order second order equation later, but so I am not introducing to you how to solve this equation in this particular lecture I will write down the solution and my aim is to give you the interpretation of this equation where the thing is that where m c k m c k are greater than in fact let me tell you m and k are greater than 0 then only you have known trivial mass and thing and c can also be. So, this is c is the this is the mass which you have let me recall mass this is the spring constant spring constant and this is the damping. So, I want to analyze this equation extensively now and give you an interpretation. So, I will be introducing few notions like that you will understand why this is I call it omega naught is equal to square root of k by m this is called the natural frequency of the system. You will understand why it is called a natural frequency when you analyze it this is called the natural frequency. So, when you develop a mechanical system it has natural frequency natural frequency of the system you see. So, this will introduce more and more notion. So, I am going to analyze it case by case. So, we will start with the case one and this in this case I assume. So, I am going to do it more and more assume c equal to 0 that means no damping and then I am also assuming that f equal to 0 that means this is no force no external force no external force. So, this is the situation free and damped motion free and damped motion then you know immediately this equation in this case equation is what is the equation m I will c is equal to c is equal to 0 f is equal to 0 and divide by m this equation I can write this equation as d square y by d t square is equal to plus omega naught square y equal to 0. This is a simple harmonic type motion using very equation nice equation there is no external force no damping then you know that the second law the first law of motion Newton's law of motion should obey let us see we can interpret that equation. So, let me write down the solution I am not going to tell you you will learn this when we as I said when we study the second order equation I will just write the solution y t those who are learned a course a preliminary course an ordinary differential equation know how to get the solution those who do not know will learn soon, but let me write the solution. So, second order equation there will be two equations. So, you can the solution will be a cos omega naught t plus b sin omega naught t. So, there where there will be a and b are constants arbitrary constants we call it. So, it is a general solution. So, I want to give you an exercise here it is nothing to do with a PDE. So, I will keep on giving you should do all when you learn this course the solution can be written in the form the solution can be written in the form you just do a bit of computations you write the solution in the form this is a better understanding in physics you will get a better understanding with the this way of writing you can write R cos omega naught t minus delta you do the a bit of a computation in the trigonometry. And what is R? R is nothing but square root of a square plus b square and your delta is equal to tan inverse of b by a. This is called this R is called the amplitude and this is called the phase angle you will see why it is important. You see whenever there is a pair motions you will see and it is also a cos is and sin a periodic function. So, you see the solution is periodic and the if you see that if you look at this case you see cos will take values between minus 1 and plus and hence this y t oscillates between minus r and plus r and you see the Newton's first law obeys it will never stop the only thing is this gives you phase angle. So, when t equal to 0 it starts from here. So, if there is no damping you will have a problem with the system you see. So, if you plot this curve y t and it will oscillate. So, if you mark here minus r and minus r and plus r it may start with a something somewhere. So, that is what is a phase it shows it may not start with 1 or something. So, you will have phase angle it will go. So, you see it is a periodic motion and the oscillation the conclusion oscillation never stops and that is what expect oscillation never stops. And as expected that is what it says because there is no damping and no external force. So, the moment if there is no external force acting on a system the motion has to continue and you have your complete thing. So, let me do to the next one case 2 you see you from the solution you understood case 2 you are still taking the situation with f equal to 0, but c not equal to 0 you see. So, how does the equation look like there is no external force. So, you have a system no external force coming to the system, but then you are putting the damping you are putting it in the oil for that particular example or in other mechanical system you add damping. So, you will have this equation m into d square y by d t square plus c into d y by d t plus k y equal to 0. So, this is your equation and let me write down again the solution I told you you will know how to find the solution. So, let me write down the solution in different cases. It depends on various depending on the there is what is called an auxiliary equation and you look at the roots of this auxiliary. So, what is the auxiliary equation here auxiliary equation that is equal to m into r square plus c r plus k equal to 0 and you have your solution r is equal to minus c plus or minus square root of c square minus 4 m k by 2 m. So, you have 2 roots r 1 and r 2. So, if c square minus 4 m k is positive you have 2 real roots and when it is 0 there is a double root and all of them you know. Accordingly you can write down the solutions now a e power r 1 t plus b e power r 2 t if c square minus 4 m k is positive and if it is a double root again you know how to find solutions which we will see later a and b are arbitrary constant b t e power r t. Because there is a if c square minus 4 m k is equal to 0 when c square minus 4 m k is equal to 0 when c square minus 4 m k is equal to 0 you have this r is nothing but in this case r is equal to nothing but minus c by 2 m. So, you have that you see and then the last case when c square minus 4 m k is negative in that is a complex roots in that case you can write down c is equal to your solution minus c t by 2 m into a cos mu t plus b sin mu t. This is the case if c square minus 4 m k is negative and what is mu if this is negative you define your mu is equal to in this case the mu is equal to square root of 4 m k. So, that is positive minus c square by 2 m that is what you can write down that solution c square minus 4 m k into 2. So, now the situation is different because c is always non zero in fact c is positive that means this implies c is positive because damping is always non negative. So, you have the case c is positive and look at the first two cases when the roots are zero in this two cases first two cases first two cases first case within this first case mean this situation. That means r 1 is different from r 2 and real that is the situation in that case it is a real you can analyze this equation if you analyze this equation c square minus 4 m k is positive and hence this will be smaller than c. So, it will retain the sign of this one that means in this case is actually r 1 and r 2 are negative you can see that because this is a even though if you take plus if it is minus no problem both are minus even if you take a plus it will be a quantity less than c an absolute value and hence this will retain the sign. When r 1 and r 2 are negative look at these terms e power r 1 t and e power r 2 t r 1 and r 2 are negative that means this will go to infinity zero. So, you will have e power r 1 t e power r 2 t goes to zero that implies y t goes to zero tends to zero. So, you have no problem. So, the case two this you know how to plot it. So, it is a it is all vanishes the solution. So, the moment you add some damping the earlier when there is no damping the solution were oscillating between minus r and plus r. The moment you add damping the solution goes to zero it may work in whatever depending on the sign of a and b whatever you take it in initially it behaves the way whatever way it is, but eventually every solution decay in an exponential rate to the zero level and what about the case two case two case two here within this case two this is the situation. So, you have the case two situation this is the thing I am analyzing it in this case c square minus 4 m k is equal to zero. So, your y t is has a term the first term no problem a into this one again will go to zero the second term there is a term term there is a term e power r t r is negative there is no problem even here r is negative because r is nothing, but minus 2 by m, but here is a term which goes to infinity and here is a term which goes there is a term which goes to infinity here is a term which goes to zero. So, there is an indirect terminated form, but let me give another exercise you have to keep on doing this exercise this is a nice exercise a small exercise from your analysis show that show that t e power r t goes to zero as t tends to infinity you will see. So, I have a nice exercise against the solution d k. So, both these cases case one and case two the solution d k's and there is no oscillation. So, when you add damping damping is when this will become zero suppose m and r fixed for a particular system m is fixed k is fixed and this sign of this is determined based on how much damping you add. So, if you add sufficient damping for m k is a positive quantity and if you add sufficient damping in such a way that c square is greater than or equal to 4 m k there is no oscillation there is no oscillation phenomena and the oscillation goes to zero. So, these two cases these two cases are called over damped system over. So, you are giving so much damping. So, that the entire over damped system. So, you have add a so much damping. So, that complete oscillation has vanished and the solution goes to zero exponentially, but now to the third case third case is this situation when you look at that term this is a cosine term this is a sin term both are bounded. So, the cos and hence this entire term is a bounded quantity, but this term is a quantity which to to zero. So, let me take to the third case. So, case c square minus 4 m k is negative in this case your solution y t is equal to e power minus e t by 2 m into a cos mu t plus b sin mu t. What I said is that this is a bounded quantity bounded and this term goes to zero. So, y t still goes to zero and this is a situation you have damping, but then damping is not too much. So, this is called the under damped system damping is there, but under damped system. So, you see all this when there is no damping complete oscillations oscillation does not decay. So, these two quantities shows that there are still oscillations. So, this will oscillate between some quantity. So, let me write down the solution in the another form the other form you can understand this oscillation much better way. So, that is what I say you can rewrite this equation in another form minus c t by 2 m into cos of mu t minus delta. So, you see it is a solution. So, this shows the oscillation is not removed completely, but then there is an amplitude and that amplitude basically goes to zero. So, it oscillates, but eventually goes to zero. So, if you plot the graph of this function it oscillates cos takes value between minus 1 and plus 1 and hence it will also takes the value between r this thing. So, if you have this if you plot this curve this is an exponentially decaying curve with r. So, if you plot that curve you will have a curve which goes to here this is nothing but minus r e power minus r I think and this minus r and here you have a plus r and you have a this curve it is a symmetric curve. So, though it does not look symmetric it is a you can plot it. So, this is the curve minus r e power minus c t by 2 m and this is the curve r e power minus c t by 2 m and it oscillates. So, it starts here it will go and maximum it will go. So, it will oscillate something like that. So, you see. So, you have this complete analysis in the this is a case 2. So, what is so far we have concluded? You have a system there is no external force we have considered only two cases. The case 1 with an no external force no damping full oscillations or oscillation continues in the same pattern then there are no external force, but then there is a damping it depends on how much damping you give either you can kill the entire oscillations or the oscillations will be if the damping is less the oscillation retains and then you have, but is still goes to 0 and that is the moment you have damping eventually the solution goes to 0. Now, let me go to the next case where you have the situation you have the case where you have force f not equal to 0, but you take c equal to 0. So, we analyze the case when f equal to 0 both c equal to 0 c not equal to 0 and now we are trying to analyze the case when there is an external trouble or external advantage. You do not know the external force is advantageous or external forces. In mechanical system quite often the external disturbance is advantageous, but we will make a remark that the same external force is advantageous in the electrical circuits. So, the interpretations will be different. So, we are trying to do anything in the set of of mechanical system. So, this is nothing. So, it is a forced, but undamped system. So, there is no damping what will happen is that one. So, this is so maybe I will this is a more difficult case. So, let me do the difficult case little later. So, let me do first because when already seen that when c equal to 0 when f equal to 0 there is vibrations, but now there is an external force coming into picture and still there is no damping we expect more trouble in the mechanical systems. So, let me do the easier case first and then we will go to the other case. This is a standard case. So, when f equal to 0 how do you write down the solution? So, you may not be able to write down the solutions with general f because it is a non homogeneous second order equation. So, first you have to solve the homogeneous linear equation and then with certain right hand side you will be able to write down the explicit solution. So, you have to find what are called particular solutions which you will study later. So, but certain f you will be able to study obtain the explicit solution. So, we will assume f is of the form with a single frequency coming at external force coming f f naught into cos omega t. So, this is f t where f naught is a constant this is a constant and then omega is the frequency coming from the external one omega naught is the corresponding frequency corresponding to your mechanical system. So, I can write down. So, already as I told you to write down the solution y t you need a solution of homogeneous equation which you know already homogeneous equation plus one particular solution. You will understand this more when you study the second order particular solution and let me write down the solution completely how to do it you will learn later. So, let me write down the homogeneous solution which you already seen whatever I have written down is the homogeneous solution plus you will have a particular solution. The particular solution let me denoted by y p t and what is a particular solution y p t? I can write down the particular solution I have something f naught this is cos of omega t minus delta omega t minus delta by is a little complicated, but you can see that omega square whole square plus c square omega square power whole power half. I hope the computation is correct anyway you should do that where delta you can also write down delta here is c by k m omega square. So, you can exactly calculate your phase coming using your external force here. So, there is nothing specific in this case the moment you have damping it should work when there is no damping this term was not when there is no external force this term was not there due to the, but this was there this is without exact homogeneous solution that means it is a solution with f equal to 0 you understand the behavior. So, that behavior will be retained and plus you have an extra term here and the extra term also have. So, there is no additional feature here the only additional thing is also the behavior everything more or less is the similar thing and we have seen that in all cases with f equal to 0 with f equal to 0 you have seen that the solution goes to 0. So, this is a term phi t which goes to 0 this will oscillate. So, what will happen is that y t eventually will go to this particular solution y p t let me not write it will go to 0 it will behave like y p t. So, you can understand the behavior of y p t and so that. So, as t tends to infinity. So, the behavior at large time y p will be of y p t which is coming due to the external force and nothing more interesting. So, this is called the steady state this y p if this is called this is the steady state solution. You can understand why it is called a steady state because eventually steady state and this phi the solution to the homogenous system is called the transient state transient state. You can understand this term terminology is very well it matches because in the beginning of the time initial time you will have the contributor transient state and eventually the transient state will go to 0 and the entire solution will behave like a steady state solution. So, now we come to the last part the next case the more interesting and difficult case. This is the case where f not equal to 0, but c equal to 0 you see this is no damping external force is there external force, but no damping you see y damping is important you will see here where is any system. So, that is it is the forced undamped vibrations this is the situation. So, let me here itself there are two cases. So, you have to study separately two cases now you can see. So, you have to split because it is a complicated thing two cases behaves differently. So, what is your f? f i am again assuming two of the form f not into something like cos omega t I have writing this one. So, what are the case in this case first let me tell you the sub case with that one this is called you understand why without resonance I am going to introduce a new terminology without resonance. That mean you have a natural frequency omega not coming from the system and you have an external frequency coming from the external force. So, the without resonance is the case the external frequency is different from the natural frequency. So, omega not is different from the frequency coming here then this is not the most interesting case you can just write down the solution y t you have a solution of the homogeneous system this will look like a into cos omega not t plus b sin omega not t very easy. So, this is the solution corresponding to your homogeneous part and then there will be a particular solution it may look like m in front into m into omega not square minus omega square into cos omega t. So, you see you will see that when omega equal not equal to omega this is not the particular solution the particular solution can have problem and why that problems mathematically you will understand when you solve second order linear equations with constant coefficient which we are going to do it anyway in future. But you see that immediately trouble we cannot write this equation is clear from here. So, this is again a natural thing. So, when you have c equal to 0 you have the your normal vibrations it is happening because then there is no damping naturally that full vibrations will be there plus you have a term here and it will vibrate as it is. So, this is the case which you have there is nothing new whatever we have discussed. The interesting another interesting case is with resonance why call it with the resonance this is the case when omega not equal to omega. This means you have a natural frequency to your mechanical system and external trouble external force is coming with the same frequency. When these things are there to solve this equation the finding the solution and writing the solution is in a different way and let me again write down the solution here. Just write down the solution here the equation in this case looks like this d y by d t plus omega not square y is equal to f not by m cos omega not t you see you already seen that cos omega not t is a solution to this one. So, whenever you want to solve an equation if you want to find a solution to this one anything like cos omega t when out work like a particular solution cos omega not t is a solution to your homogeneous equation. So, the solution is slightly you have to write it in a entirely different way. The solution let me write again here the solution will look like this y t is equal to a cos omega not t plus b sin omega not t plus this is the difference here how things behave to m omega not into t sin omega not t. So, the behavior of this term is fine it oscillates between a and b if you write it r e power you already seen that it will oscillate between fully. But, look at this term this oscillates this sin part oscillates the sin part oscillates between minus 1 and 1, but you see as t increases this increases to infinity. So, what happens is that even though this oscillates if this term this such a term such an unbounded term as time goes the amplitude is increasing that is what happens when t becomes bigger and bigger the amplitude is increasing like that. So, it will start from here the amplitude increases that means the vibrations become unbounded. So, if you plot this curve here the curve may you can look like that as time goes you see it goes to become bigger and bigger. So, you have a mechanical system you are meeting with an accident or something some vibrations coming in your motor bike or something and the vibrations comes into you that and the person who is riding the thing you do not want the vibrations to on you, but here if this is the situation an external frequency comes with that one and this vibrations becomes larger and larger and you are going to meet catastrophes. And this is what when you design any mechanical system like a bridge or anything you have to take into account in such a way that you have to put enough damping in the system in such a way that the external frequency it may be may not be the external frequency may not be due to your under your control. For example, there will be lot of vehicles travelling through your bridge. So, when you are vehicles travelling through the bridge it will create automatically its own external frequency and your system will have a certain frequency. So, if your bridge or whatever it is do not have the frequency either damping enough and it is a possible that the frequency of your bridge may coincide with the external frequency coming and the bridge will start oscillating and bridge will collapse. This is the famous you should go and read the famous pack comma bridge collapse. This is probably happened in 1940s or something is a famous bridge in USA where what happens is that in the early morning the bridge starts the vibrating with smaller and smaller vibrations by the time it and the vibration became bigger and bigger and by the time it 11 o clock or something the vibrations were so big. So, the amplitude was so huge and the bridge got collapsed. The interesting tale of that one is that nobody died in that huge disaster except a dog and this is also the reason probably you have seen that in the military or army marching in the cadence marching through the bridge the army marches not like what we walk we create different types of vibration, but army marching will be with a particular frequency and it will be very high that is why when the army marches through the bridge they would not allow the continuous marching of the army. They will break into small pieces a small set of people will cross the bridge and then only the other seat will. So, they will break because they do not want to create a frequency from the external side which marches with this one. So, as this example before going to quickly I will not go into quickly the other examples I will just explain few examples which you will see details in the in future other lectures in the later part of this course, but I will explain few important examples before going to that one when you analyze the same system in the electrical circuit the analysis is the same you look at it, but the interpretations are now different where here the exactly what is undesirable in a mechanical system is used to good use in electrical circuits. Because whatever happen is that when you want to tune a radio or TV what you are doing is that you are setting a frequency natural frequency to the system. So, when you are saying changing your channels or changing the all radios when you are changing the frequency means you are changing the frequency of natural frequency of the system and externally all that telecasting all the channels with the different frequencies what you are doing trying to do is that you are trying to put a resonance situation in such a way that when you are setting a particular frequency to your natural frequency to your system there will be all the frequencies around your radio or TV it will only match one of the frequencies of the particular channel or a particular broadcasting thing when these two matches these two will create a resonance and hence it to that is what is called an amplifying. So, basically a situation of resonance are happening even though whatever coming is well lit in it matches and then it amplifies and that is why you get a particular channel. So, you see the same thing the interpretations are different, but you have to it is coming from the same equation entirely two different fundamental problems from two fundamental ideas of engineering. So, this much details I will not do it any more in the next 10 minutes or something I do not know how many examples I can do it anyway I am not going to do too many examples I will be writing down few interesting already wrote down one equation is called the prepredator model called a Lotka-Walterer model and another equation you will be studying here in later is called the Duffing equation Duffing equation. The equation looks like this x double dot minus alpha x plus beta x cube plus delta x dot and you can also have the driving force if you want. So, that you have a homogeneous equation this is non-linear due to this is non-linear and you have the you can put something like gamma cos omega t if you want a driving force external driving force you see of course, if this beta and delta r 0 is nothing, but this part is the linear part of course, this is also linear, but the linear simple linear oscillator. So, you can think it as the perturbed model of the linear thing perturbed model if you want to see and alpha is if you want something thing alpha is nothing, but your stiffness you want it in a mechanical model stiffness you can interpret that one and beta may be the amplitude and delta whatever it is the damping. So, this is something a similar thing if you look at your mechanical model it is a similar thing. So, you have this term this term similar terms only this is a more term this is the damping you can see that one. This model was probably introduced by Joe Duffing as I said if you want to understand the more if you want a linear system you will be restricted to only thing if you want a more general understanding of the system you have to have your components the non-linear perturbations and perturbations and this is the thing. So, you may see say for example, your stiffness of the spring may not obey the Hooke's law exactly and such kind of things it will not follow precisely the things like that. So, you will see this equation and probably more about the derivation not too much, but we are going to study this equation in detail in our future lectures. You can write down this as a system you will see that also you can write down this as a system with x dot is equal to y c c to write when you have an x dot is equal to y you have your y dot is equal to you put that all the alpha x minus beta x cube plus minus delta y. So, you see you can write this as a system. So, you have a very beautiful equation as I said that the a reasonable extensive study will be done here at the end of it and this is one of the model which we want to understand it in the qualitative analysis. So, to understand your if you exact if you want to know how these equations are derived you can get into the references which we have given or you can also search through internets if you want to understand more very one of the all right. So, another model which you may see is called the van der poel equation. So, it is a van der poel equation these are all oscillator equation earlier it was an electrical mechanical model. This is also an interesting model coming from electrical circuits is again a call it oscillator you can also call it oscillator everything is a that form x square minus 1 into x dot plus x here also if you want can put it 0 or you can also put it something an external a sin omega t if you want a forced. So, this is called the unforced van der poel equation. So, this you can call it forced what you are putting a force external force there you can write down this again as a there are different ways possibly you can. So, when I am writing it as a system it does not mean that it is a unique way of writing it you can write down the systems in different ways. So, I am what I have done here is the easiest way of writing it and different ways of writing it will have different effects plus mu into x square minus 1 into x dot this is known by the Dutch van der poel was a Dutch engineer Dutch engineer van der poel and introduced in 19 probably 27 when he was working as in the Philips company that is the time he introduced, but it is also seems that this equation is derived a little earlier than van der poel due to Rayleigh load Rayleigh, but somehow it is not in that name it is a introduced probably 1896 itself 30 years before 30 years before that one, but it is a he is the one at least van der poel is the one who extensively studied this thing. So, you will see the behavior probably in this in this study in our course you will see the chaotic behavior probably periodic solutions and chaotic behavior all that analysis probably you will see that there is also further equation other people like Cartwright I think I am right Cartwright and Litchfield would they are further whole mathematical study later more detailed mathematical study here is some engineer he is more like a done like some mathematical theoretically and experimentally a complete study more detailed mathematical analysis probably a little later these people. So, you will see about this type of equations also in our this course you are going to see that of course, one of the easiest this equation the one more thing this equations have a not only application this kind of thing it also has applications in biological science that is in. So, as I told you as is as you already seen the same equation may have different effects in different thing just like the second order equation which we have derived and analyzed comes from a mechanical system it also comes from the thing. So, here in the study of the neurons the action potential of neurons and the same equations have extensively studied and its chaotic behavior it is also seismology is used where you want to understand the geological flow two plates. So, this has same equations has applications in other areas also. So, when you study so what you will be seeing here our concentration is not to a particular problem, but our concentration will be how this equations. So, our analysis is going to be the qualitative analysis of this general equation interpretations has to be done according to a problem according to a physical problem. The other equation of course, this would have been come here basically the pendulum equation this you also may see a pendulum equation and this looks like this more complicated let me write it in a slightly general way plus B L theta dot plus m g sin theta. It is a standard pendulum you have a string at least length of the string and mass is attached all of you know that one those are gone through your plus 1 plus 2 let the class and how the pendulum swings. So, this is the force due to friction force due to friction the term the this is something like looks like a damping part. So, if this term is not there and if you write it in a simple form say m l theta dot plus m g sin theta this is basically an ideal model more of an ideal model. So, you will see these things ideal model and which you are going to study here whatever it and the other case when you study B positive. So, there is a that is a more general case where there is a force due to friction. So, let me more or less finish by writing to you one more equation which is the probably the first to the chaotic behavior is observed in the Lorentz system I will write down the system. So, you can why I am writing this system is that by the time you come to the end of this course you should be familiar with the system. So, that you will understand its behavior the qualitative analysis in a much better way. So, if you go through this one do little more study on these systems how these systems are coming and then you understand. So, let me just write down the thing I will not again tell much about it this is minus sigma x plus sigma y and then y dot is equal to r x minus y minus x z and z dot is equal to minus b z this is again extensively studied to understand the chaotic behavior chaotic. And yeah this is actually developed in the study of meteorological problem with weather prediction. And you will see again a little bit analysis of these problems in future. So, there are some other interesting model we are not sure about we will be able to cover all these models because then over one semester course may not be good enough to cover that one. For example, some nice models if you get time I am not a thing for satellite motion. So, important satellite motion examples from satellite motion. So, is typically you have a satellite. So, you have a typically a satellite. So, you have a satellite is in this position p with x y as the coordinate system x y and then you have this has the thrusting capacity. So, you can you have the radial thrust if you want u 1 and you will also have a tangential thrust u 2. So, you can derive the modeling is much easier here. So, you can easily model these equations. So, the modeling is not the important thing the in the satellite thing in addition to model you have to also give a kind of study the control problems. So, if I want to write down the equation the equations may look like this x dot is equal to f of x and m poly dot is equal to f of y. So, this looks easier model, but then since this is a satellites a better model better way of writing is not in the Cartesian coordinate system the polar coordinates writing in polar coordinates is much more better. So, you can do the very simple analysis you do not need much things to be done. So, let me just write down the equation. So, it will be have something like that radial. So, you have r and theta. So, you can write down the radial and the. So, you can do very simple analysis not too much complicated we have time we could have done it, but we will not do it I am not sure we will be able to do a much better analysis also in this problem, but let me write down for your familiarity this equation can be converted with very simple analysis to the equation of the system of 2 second order equations. So, let me do that one r theta dot square equal to minus k by some r square and m theta dot double dot plus 2 again it is a non-linear system theta dot m 2 m into equal to 0. So, you see. So, you have a eventually. So, analysis for this kind of things is better to do it in polar coordinates not in our Cartesian coordinate. So, this is a system of 2 equations a system of 2 equations second order system that is what it is. So, it is not a first order system. So, when you go convert into a first order system it will a system of 4 equations in 4 r nodes and there are plenty of other examples where you can study and it is available and with this we will complete the module 1 giving examples that in the next module we will be giving the preliminaries. We will give some preliminaries in the case of linear algebra we will also present some preliminaries of analysis we will just do it which we require for our course and. So, thank you.