 So, towards the end of the last lecture what we had derived was this particular relationship for Rayleigh number okay and this dimensionless number the Rayleigh number is given by n squared pi squared plus alpha squared cube divided by alpha squared and alpha here is a dimensionless wavelength sorry a wave number okay alpha represents a dimensionless wave number and what does this particular thing signify this when the Rayleigh number equals this then we have a nonzero solution to the linearized equations because that is the condition we invoked to get this result okay. So, when the above condition is satisfied we obtain a nonzero solution to the linearized equations okay and what we can do is we can make a plot of the left hand side and the right hand side and you would get if you were to plot as a function of alpha you get something like a curve which has a minima okay and remember this particular value was 657 approximately and this is for n equals 1 that is important okay. Now what I wanted to reiterate is that along this curve the sigma is 0 so along this sigma equals 0 and so this particular curve is called the neutral stability curve okay and this is the neutral stability curve. So, on one side you will have stable on the other side you have unstable and we just reasoned out on the basis of the physical system that for on the below this curve you will have a stable situation where sigma will be negative and above this you will have sigma is positive and unstable. So, basically this region corresponds to sigma negative and this region corresponds to sigma positive okay. Now the important point here is that we can plot Rayleigh number on the y axis and Rayleigh number is something which I control experimentally okay. For Rayleigh number less than 657 sigma is negative for all alpha okay and this means the disturbances dk and we have a stable steady state okay. So, this disturbance that we have see this disturbance is going to be some kind of a function okay because disturbance for example the disturbance could be a deviation of the lower plate temperature from t0 okay. We have assumed you know analysis of the lower plate is uniformly at t0 there can be some deviation from this uniform value of t0. So, what we the way you I want you to understand this is this particular disturbance this small deviation from t0 because experimentally you cannot really get exactly whatever value you are trying to control it at 80 Celsius it could be 80.1 somewhere it could be 79.9 somewhere else right. So, the point is this particular disturbance you can you know view it as being decomposed into different components each component is going to correspond to a particular wave number supposing your deviation is actually sinusoidal with sin alpha x sin alpha 1 x that means the only mode of the disturbance is corresponding to alpha 1 alpha equal to alpha 1 you understand supposing the disturbance of the lower plate temperature the deviation of the lower plate temperature from t0 is periodic and the spatial periodicity is given by sin alpha 1 x then the disturbance has only one wave number corresponding to alpha 1. But since you are having an arbitrary disturbance this arbitrary disturbance has going is going to be decomposed and resolved into different components just like a vector is being resolved into different components okay. A three dimensional vector you write in terms of three components if you have a function you can write it in terms of different wave numbers okay and the what we are doing is we are trying to what this curve tells you is how does a particular wave number grow or decay okay and what we are interested in is for low values of Rayleigh number this tells me for all so if you give any arbitrary disturbance I am going to resolve it along all these different alphas. So, every component is going to decay as I keep increasing my Rayleigh number okay just when it is slightly above 657 the wave number which is going to grow is going to correspond to this particular value of alpha which we saw last time as being equal to some pi divided by root 2 or something correct. So, if the Rayleigh number is greater than 657 the wave number which becomes unstable grows is given by alpha equals pi by root 2 okay all other wave numbers are going to decay okay. So, what this means is the spatial periodicity which you are going to observe when you keep at the point where it is just going to convict just beginning to convict is going to be given by pi by root 2 okay. So, at the point of initiation of the natural convection just when it is about to become unstable the stationary state is about to become unstable you will see of spatial periodicity given by pi by root 2 okay. So, at the onset of convection the spatial periodicity is given by pi by root 2 I should be careful this is not the wavelength the wavelength will be the reciprocal of this only proportional to the reciprocal of this. So, now if you keep on increasing okay the other thing we want to point out is so how exactly is this is going to behave how exactly and for purposes of illustrating this the simplest possible way for you to understand how this natural convection is going to occur is by looking at the convection in the form of a cylindrical roll okay. So, what exactly is a cylindrical roll I mean you have this guy that way oops so I think I would prove it much bigger than what I should have. So, what is the pattern which is repeating itself the pattern which repeats itself is this particular combination of 2 cells see this vortex is clockwise this vortex is anticlockwise I need to have this what clockwise anticlockwise vortex because I need to have continuity of velocity here. So, this pattern is basically repeating itself and this is my lambda the wavelength is I am going to observe and this wavelength is related to my wave number okay. So, what I am saying is the pattern which repeats is this guy okay and this should all have been drawn equal but then they do not look equal so do not worry about that okay that is my lambda and this is the one which repeats forever now that is when you go along in the vertical direction there is only one roll okay and this corresponds to the fact that n is equal to 1 is the one which is most critical okay. So, here when we go in the vertical direction we have only one roll and this is because the most critical disturbance corresponds to n equals 1 okay. In the vertical direction this n represents sin n pi y by h okay. So, at the bottom it is 0 and the top it is pi. So, going only one side from 0 to pi so when you look at the direction of the velocity you will just see that you are going to go from negative to a positive region as you go up for W okay. It is not negative to positive back to negative that is what would have happened if n is 2 if n is 2 you would have had 2 sign changes okay but here n is 1 and therefore when you go up here you just change the sign once from negative to positive okay. So, the other important thing which I want to emphasize here is the fact that this particular pattern that you can observe is going to be another steady state another stationary solution stationary in the sense being steady. That is if you were to now keep a probe here and put a pitot tube or any probe that you want and measure the velocity and the temperature you will find that it is constant it does not change with time okay. So, you have a situation earlier also you had a steady state which was linear the temperature was linear there was no motion but now you have another steady state where the liquid is actually moving. So, you have a non-zero velocity and you have a temperature which is different from the straight line that you had. So, the point is this convective state is also steady okay that is if we keep a thermocouple we will have a constant reading. Now, what is it that makes this thing constant the growth rate sigma is real see I have not proved it but I just told you that the growth rate sigma is real and then we said the critical value is given by sigma equal to 0 okay. Now, because the growth rate is real and what it means is when I come here when I am operating in this range this particular wave number is going to be the one which I am going to see because that is the one which grows fastest okay. Supposing I have my Rayleigh number given by that value these wave numbers are going to die these wave numbers disturbances of this wave number will decay all these wave numbers are going to grow right but which is the one which is going to grow fastest the one which is going to grow fastest is the one corresponding to approximately this particular wave number why because these points see the value of sigma here sigma I know is positive the value of sigma is going to be given by the distance from the neutral stability curve here sigma is 0. So, here is going to be less when I go further away sigma is going to increase. So, sigma has to be 0 here sigma has to increase and again it has to go back to 0 assuming it is symmetric assuming this dependency is symmetric sigma will be maximum here. So, the wave number which grows fastest is going to be corresponding to this minima you understand. So, by point here is that sigma is 0 at this point and at this point it is positive here so it has to increase and it has to decrease and if you think it is symmetric and this guy the middle is the one which is going to have the maximum sigma okay and that is the other wave numbers are trying to grow but this fellow has grown and it has dominated the other wave numbers and so what you are going to see is the dominant wave number which is given by this okay. So, point is that if the Rayleigh number is greater than 657 I can make a plot of sigma versus alpha I am fixing my Rayleigh number and everywhere here it is negative on the left is going to go that way that is going to be the nature of the curve right. This is for a fixed Rayleigh number the Rayleigh number is greater than 657 and fixed for low alpha it is negative everywhere for high alpha it is negative in between it is positive okay and this will be the point which corresponds to the minima. This corresponds to pi by root 2 it may be shifted this way that we are depending upon the actual problem but the point I am trying to make here is the maximum growth rate is this. So, the point is the disturbances which have alpha equals pi by root 2 grow fastest and dominate the behavior okay and this wave number or wavelength is observed experimentally. The so basically what I am saying is we have decomposed this into different wave numbers okay. So, the way I want you to I do not know if I wrote this earlier so any arbitrary disturbance can be written in terms of the Fourier modes okay and that is what we are doing this is like this is equivalent to resolving a vector in terms of its components okay and we are saying which component is going to grow maybe this is not going to grow that is not going to grow. But even if one wave number grows then the system is unstable okay the system is stable only if all the wave numbers decay disturbances which are components all the wave numbers if all of them decay only then I say it is stable even one guys although everywhere else is negative but if one guy is positive it is unstable okay. So, this is let us say the curve for 700 really number equals 700 when it is really number equals 658 what do you expect I expect and to be slight a small maxima near pi by root 2 because only a small region around this pi by root 2 will have a positive thing okay. So, just illustrate this is really number equals 658 these are basically equivalent representations of the same thing here I am plotting the growth rate versus the wave number and this is called in the literature a dispersion curve and that particular thing which tells you how a parameter is varying with the wave number and identifies the region of stability is called a neutral stability curve okay. But you have to understand that these are basically the same information present in both these curves and depending upon what you are interested in you may want to make an appropriate plot. The other question which is going to arises look we have sigma which is positive here the growth rate is positive from a really number is greater than 657 in this case I will have my disturbances which are growing exponentially in time. So, you expect that velocity is going to go keep on increasing and it is going to get unbounded temperature will keep on increasing and it is going to get unbounded if you just focus on the linear stability analysis if you look at the linear stability analysis what does the disturbance form look like e power sigma t multiplied by some function of x and some function of time remember that is what we are assumed right sigma tells you the growth with respect to time and we are assumed an exponential dependency because your system is linear. So, the question is if sigma is positive and if you now were to just substitute it in your linear equation or disturbance has to exponentially increase with time and become unbounded. So, what is it that is actually going to prevent but I told you that what you are going to see is this kind of a role clearly the temperature is not going to become infinity is velocity is not going to become infinity. So, what happens is that the linear stability analysis is based on assuming that the perturbations are small remember okay the linear stability analysis we made a fundamental assumption that we have only small perturbations but as that velocity increases as the temperature increases then the linear stability analysis cannot be used anymore you actually have to the actual real system is going to involve taking record all the higher order terms the point I am trying to make here is that these higher order terms are the ones which are actually preventing it from going to infinity okay. So, that is the weight for you to resolve this contradiction that sigma is positive. So, it would look like the velocity becomes unbounded temperature becomes unbounded but then as the velocity increases as the temperature increases you cannot use the linear stability analysis anymore because I got the linearized equations assuming that I have only infinite similar perturbations okay. So, what is it that prevents although sigma is greater than 0 the disturbances do not become infinite as t tends to infinity equation as y because the linear stability analysis assumes small disturbances okay and as the disturbances increase as the disturbances increase the higher order terms have to be considered okay and these prevent the amplitude from increasing indefinitely. So, the point is when you include the higher order terms you would get a nonlinear equation because you are having second order term third order term okay. So, rather than solve those nonlinear equations which are quadratic and cubic you might and which you possibly have to do it iteratively you might as well solve the original governing nonlinear equations directly because the nonlinear equations you know there is no approximation apart from what you have made physically. So, you can just directly solve the nonlinear equations and you can find out what the behavior is because when you are solving the actual nonlinear equation you are including all the terms all the Taylor series expansion including all the infinite orders terms. So, what people in CFD would do is they would just do that they just go to the program the problem and then write the code and then they would simulate it okay. What exactly does this come into the picture? So, for example when you write a code in CFD you want to possibly validate what you have done okay or when you are doing this linear stability analysis you also want to validate this result. So, what the way to you have to make sure that whatever theory you propose they are all consistent. So, what this theory linear stability analysis is doing is it is telling you that for really number less than 657 we have no convection for really number greater than 657 we have a convection pattern and there is some periodicity that is you know steady state. So, what you should do is you go to fluent or one of these packages or you should write your own code and simulate the governing nonlinear equations the Navier-Stokes equations okay and do it so that for 2 conditions for really number equal to 600 and for really number equal to 700 for really number equal to 600 you should get a steady state velocity 0 temperature profile is linear for really number 650 you should get the same story but when you go past 657 go to 658, 659 you will start seeing a convection okay. So, that is basically going to give you confidence in both the linear stability analysis as well as the numerical code which you have written and which are using for solving the actual natural convection problem. So, you understand the 2 different approaches one is take the equations write the code solve you will get a result but how do you know that those results are actually accurate. So, I am just saying you can use this information from this linear stability analysis for example do for a lower temperature gradient physically you expect that to be no convection see if your numerical code gives you that see if you are going just above 658 where you get convection okay and what are the source of the disturbance when you are doing these numerical calculations see your computer will only have some kind of a finite precision is not although you are saying you are keeping your temperature at the bottom later 80 it is going to have a small round of error in the calculations which are coming okay. So, that those small round of errors will act as a disturbance and that is what is going to decide. So, although you may say look I do not give any disturbance temperature is only 80 when the computer is because there is only a finite precision to which is going to make the calculations okay. So, that error in the 5th decimal place 6th decimal place is actually going to act as a disturbance okay. The other important point is that so, for a Rayleigh number okay. So, what I am saying is we need to cross check and validate these results with let us say computational fluid dynamics that serves as a way to actually benchmark both okay. See for example if your computational fluid dynamics tells you that up to 800 there is no convection and more than 800 you are having convection and this as 657 then you know that there is some problem somewhere you are going to fix it. So, an vice versa okay. So, that kind of thing will only give you some more confidence about what to do. The other important point is the linear stability analysis it actually cannot give you any idea about the amplitude of the solution that you are going to see. See what did the why because the linear stability analysis is going to give you a bunch of homogeneous equations. If you remember we got a system of equations Ax equal to 0 okay and then we put the determinant of A equal to 0 and then we found out the conditions for a non-zero solution. So, if you are solving Ax equals 0 and let us say x equals c is a vector which is a solution and clearly k times c is also a solution to this okay. Since the equation is linear and homogeneous you can determine the solution only to within a scalar multiple okay. So, as a result what the linear stability analysis does is it cannot give you any idea about what the amplitude of the solution is what is going to be the velocity. Because you are only able to determine the velocity to within an arbitrary constant multiplicative constant okay and but what it can do is it can give you information about some qualitative features about the flow like the spatial periodicity that is going to be spatially periodic it is going to be steady okay. So, how do you know it is going to be steady that is because the imaginary part of the growth is 0 of the growth rate sigma is 0. If the imaginary part of the growth rate is non-zero that is sigma can actually be imaginary what does that mean the disturbance is going to have a imaginary component e power i sigma t and that is a periodic component because it can be written as cosine t plus sine t. So, the fact that I do not have a periodic solution when the guy is convecting and I have a steady state is coming because of the fact that my sigma my growth rate is real okay. So, these are some small subtle things which I wanted to emphasize and mention. So, before we keep going on further once you understand this then when we are solving different examples I do not have to emphasize this again and again okay. So, the 2 things I wanted to mention one is the linear stability analysis helps identify the onset of instability okay that is the real number equal to 657. It cannot tell us the amplitude of the disturbance why is that since we have a linear homogeneous system if A x1 equals 0 then A times C x1 equals 0 okay and the disturbance can be determined to only within a arbitrary multiplicative constant okay. And that is the reason when you actually have to find the higher order terms we actually find the amplitude you have to include the higher order terms because that is the one which is stabilizing okay otherwise it will just appear like it is going to be going off to infinity if you look only at the growth rate sigma. I think the point I am trying to make is if sigma were imaginary and sigma is of the form sigma real plus i sigma i the real part and an imaginary part then the onset of instability is given by sigma r equals 0 okay and what does sigma i do sigma i gives you oscillations in time and since the problem sigma is 0 there is no imaginary part there is no oscillation in time and that is the reason why the natural convection pattern that you see is a steady one okay. So if sigma i equals 0 then as for the natural convection pattern we have no oscillations in time okay and we get a steady state. So is this clear? So I think there are small things I want to emphasize today and that is what I am trying to do that I assumed or I told you that I can prove sigma is real for this problem and so in our problems there is no imaginary part sigma is only sigma r okay. If sigma sigma r then the disturbance is going to grow exponentially it keeps going exponentially but as it keeps growing the higher order terms are going to make sure it does not become infinite keeps it bounded. What is preventing in the thing from oscillating with respect to time? It is the fact that sigma is 0 but if you had a situation where sigma is actually not 0 then you would have the amplitude increasing with time and there is going to be also an oscillatory component. If that is the case your velocity for example would have increased in a oscillatory manner the higher order terms would have prevented the things from way off to infinity but then what you would have actually observed would have been not a steady state but a oscillatory state okay. So that is what would have happened if sigma i is non-zero and that is basically what I am trying to write here okay. So if sigma i is non-zero is not equal to 0 then the new state shows periodic behaviour because the sigma associated with t time okay. Now if you put a probe the probe will show an oscillation in time although you are fixed in space and so you do not have a steady state okay. So now what I like to do is just make a plot of whatever we said in the form of a bifurcation diagram okay. So when you talk about a bifurcation diagram what are we plotting? We are plotting some parameter which I can vary experimentally on the x axis and on the y axis some dependent variable okay. So on the x axis the dimensionless parameter which I am going to vary is the Rayleigh number okay and on the y axis I need to have something which is an indication of the solution. So it could be the average velocity or it could be the average temperature okay or it could be the temperature at the midpoint. Let us say I am just going to represent the temperature of the midpoint. Now actually this is a lousy choice. So I do not want to put temperature of the midpoint. I am going to put velocity. Let us say you take velocity at some location y equals 1 fourth okay in dimensionless form because it is going from 0 to 1 or 0 to h h by 4. So you decided that is where your probe is and so one particular steady state is the one where there is no velocity at all. So u is 0 everywhere. So that is my steady state which corresponds to u equal to 0 okay and this steady state remember is going to be valid for all Rayleigh numbers in respect to what the Rayleigh number is u equal to 0 is satisfying the equation okay. So this is a steady state which is valid for all Rayleigh numbers. But what we know is up to 657 and we are doing this imaginary problem. After 657 this guy is stable. More than 657 it is unstable okay. So I am going to put 657 here and I need to put a dashed line. So I am just going to mark it like that. So this is unstable. So this is stable and this is unstable and we keep increasing the Rayleigh number beyond 657. We expect that the velocity and remember what I am trying to do is I am going to plot the magnitude of the velocity okay. If I plot the magnitude of the velocity it is always going to be positive. If I decided to plot only the velocity then it will be positive or negative okay. So let us just say that we are plotting the magnitude of the velocity at this point. If I am going to plot the magnitude of the velocity at this point clearly it is going to be always positive because there is going to be a non-zero value there okay. If I have a non-zero value there then I expect the magnitude to increase as I go further and further away from the critical point. And this is something which I am not going to prove but then we can show that this has a square root dependency okay. And that will come by how will you be able to establish this by considering the higher order terms. If you consider the higher order terms you can actually prove that and possibly we will do this towards the end of the course if time permits that the amplitude is actually going to saturate at some point okay. So this is the way the amplitude is behaving okay. So this is proportional to Ra-Rac to the power half okay. The point I am trying to make here is this is my steady state. This is also a steady state and that is what we found out. And that is the reason I am drawing this as a solid line because that was what I said. All stable steady states is run by a solid line. And when Rayleigh number is greater than 658 what you need to do is you need to the growth rate how it changes for this branch is useless because this steady state is already unstable you understand. What you need to do is the experimentally observed state is going to be this steady state which has a finite view with a periodic roll pattern okay or some other pattern which is periodic. You have to do a linear stability analysis of this steady state. See you are always going to do a linearized stability analysis of a possible steady state. If this guy is already unstable what is the point in doing a linear stability analysis about this steady state? You already know it is unstable. So but you know this is stable. So now when Rayleigh number is greater than 658 the question is when does this become unstable okay. So what this means is you have to do the linear stability analysis of this steady state which means your solution u will be of the form of uss plus u epsilon u tilde and uss will be the periodic roll state you understand. So what I am saying is you always are going to look at a steady state which is stable and then see when does it become unstable. There is no point in doing a linear stability analysis around this steady state here because this guy is unstable okay. So for Rayleigh number is greater than Rayleigh critical we use the natural convection solution of the periodic rolls okay as the base state. Linear stability analysis of this state that is what is the meaning u equals uss but I am going to put nc at the bottom to tell you it is the natural convection solution not the other one which was 0. So what I did in the class was uss equal to 0 but now I am saying uss nc which is the one which you have to find numerically okay plus epsilon u tilde. Then you do the same thing but only thing is now the thing is slightly more complicated because I had uss equal to 0 earlier many things become 0 and then I could actually solve it analytically and I could get this magic number 657 but now you actually have to write a code and find out the onset of when this guy becomes unstable. So what is going to happen is this is also going to become unstable because you just imagine the situation will keep on heating in more and more then you will not have a steady pattern because you will have all kinds of convection which is going chaotic okay and you put a temperature probe it is not going to show a constant value it is going to show some fluctuations. So after what I am saying is for a sufficiently large Rayleigh number this also becomes unstable but if you want to find out the point where this becomes unstable you need to do a linear stability analysis of this steady state okay and which is more difficult in the sense you cannot do it on the blackboard and then you can find this and now you will see this is a steady state and you will see time dependent oscillations here oscillations and I will just say maybe even turbulence okay. So if your temperature lower plate is sufficiently large you expect the motion to be turbulent so clearly it is not a steady state. So how do you get to that turbulent state? So this is the mechanism first it is not moving then it moves in a steady manner then this guy becomes unstable and then you have turbulence okay. So that is something which people are interested in studying. So I just wanted to show to you an illustration of basic things because tomorrow we will be solving other problems okay and right now we have not used the kinematic boundary condition and the normal state boundary condition because the interface was always flat. So from tomorrow when we start solving problems and the interface deflects you need to use those conditions but the idea is the same. So we are beginning of the tomorrow's class we will just do a summary of the approach and then we will just apply it to a bunch of problems okay. Thanks.