 So, welcome to today's lecture where what we are going to do is continue from where we left off in the last lecture where we were analyzing the linear stability of the Rayleigh-Bernard problem. So just for a brief recollect, we had taken the stationary solution and we had imposed small perturbations on this stationary solution and these perturbations are actually being denoted by the tilde variable. So, t tilde represents the perturbation on temperature, v tilde represents the perturbation on velocity and we started with the original non-linear equations and we linearized the equations about the steady state. So what this basically means is we assume the perturbations to be of order epsilon and retain terms only of order epsilon to the power 1. We do not consider terms of order epsilon squared because they are higher order terms and they are negligibly small. So when you do this, you get a bunch of linearized equations. So what we have done is if you remember you had the perturbations on u, the x component of velocity, the perturbation on pressure p tilde and we have eliminated those 2 variables and you had 4 perturbations on the 2 velocity components on the pressure and on temperature. What we did is we eliminated 2 and we reduced it to a system of 2 equations, one on the temperature perturbation and one on the vertical velocity component perturbation and this is what we had derived in the last class. What we want to do now is take the analysis further and in our quest for getting an analytical solution, okay. You observe that these 2 equations are linear and they are also coupled. The temperature equation has velocity in it, the velocity equation has temperature in it. So you have to solve them simultaneously and you also observe that they are homogeneous that is every term present in each of the equation contains the perturbation variable or the derivative of it to the first power, okay. So they are linear and there is no non-homogeneity and these are important characteristics of a linearized problem. So what we want to do now is remember that the temperature and the velocity perturbations are actually functions of are dependent on x, y and t, the time. They are independent on z where they assume the span to be infinity and which basically means that these are partial differential equations. So what we will do is, we would like to convert these to a set of ordinary differential equations. So we are going to seek the solution as t tilde of x, y, t as being equal to t star of y times e power sigma t times sin alpha x and v tilde of x, y, t as v star of y times e power sigma t times sin alpha x. What have I done here? I am going to give 2 interpretations to this functional form that I am seeking. One is the physical interpretation. The physical interpretation is I am going to seek solutions which are periodic in the x direction. Now the fact that the x direction extends to infinity allows me to actually seek periodic solutions in the x direction. I do not have to worry about boundary conditions in the x direction and the fact that this is a linear equation which is first order in time allows me to seek the time dependency to be exponential and sigma is the growth rate of the disturbance in time. The y dependency is captured in p star and v star. So clearly if you have a system which is stable then the real part of sigma is going to be negative and that implies stability and the real part of sigma is going to be positive. This implies the steady state is unstable. So this is the physical interpretation. The other interpretation is a mathematical interpretation. The mathematical interpretation comes from the techniques you have learnt in your mathematical courses where you have been talking about taking Laplace transforms and Fourier transforms. So when you are taking a Fourier transform or a Fourier sine transform or a Fourier cosine transform you are essentially seeking a periodic solution in the x direction. When you take a Laplace transform you are essentially seeking time dependency in the Laplace domain and which is also going to be exponential. Our objective is to seek solutions of this kind and get the t star of y and v star of y. That is what we are going to do is we are going to convert this bunch of partial differential equations here to a bunch of ordinary differential equations which are going to describe t star and v star. So remember that is the objective. Convert partial differential equations to ordinary differential equations. We can do that by using a physical argument or we can use a mathematical argument. So let us write this down. We want to convert PDEs to ODEs, ordinary differential equations. First is physical argument where we seek periodic solutions in space, exponentially varying solutions in time. The second is more of a mathematical argument, mathematical interpretation. We take a Fourier sine transform in the x direction and a Laplace transform in time. So the fact that we have actually assumed the x to extend to infinity is what is allowing us to do a Fourier sine transform okay and this is where the assumption of x being large comes in handy. The form assumed for the temperature and the velocity here implicitly assumes that the temperature and the velocity are in phase as far as the spatial dependency is concerned. That is both of them are varying as sinusoidal. It is not that one is varying as a sine and the other is varying as cosine okay and whether they are actually going to be in phase or not we can find out only by substituting these equations, these forms for the perturbation in these governing equations. So if it turns out that by substituting these forms in these equations the e power sigma t and sine alpha x occurs in every term then you can essentially cancel off the e power sigma t and sine alpha x okay from every term and which means that such a solution is possible and this has to be true for both the equations okay. So that is basically what we are going to do now. We are going to substitute these forms for the perturbation in these equations and find out how this equation can be reduced to an ordinary differential equation. So let us do that but before we proceed I am going to make some small calculations so that it will allow me to proceed faster. So let us see what is the time derivative of the temperature. So supposing we were interested in calculating d by dt of t tilde I need to get d by dt of t star of y e power sigma t sine alpha x. This is a function of y that is a function of x. This depends on time so when I differentiate I am going to just have to differentiate only this term and this is going to give me sigma multiplied by e power sigma t times sine alpha x times t star of y okay that is how the time derivative is found. Now let us look at how one can calculate del square of t tilde. Remember del square is 2 dimensional it only has variations in x and y and so del square is essentially d square by dx square plus d square by dy square of t tilde which is t star of y e power sigma t sine alpha x. Now when I am going to differentiate this term with respect to x these are for all practical purposes constant and the derivative of sine alpha x the first derivative is going to be alpha times cosine alpha x and when I differentiate it one more time I am going to get minus alpha square and I get back sine. So please understand that this is going to be the same as the first term it is going to give me. If I look at the first term I am going to get minus alpha square sine alpha x times t star of y times e power sigma t and now when I am going to differentiate this with respect to y these are going to be constant and what I have is essentially the second derivative of t star but since t star is only a function of y it is not going to be a partial derivative anymore but it is going to be a total derivative and this is going to be written as d square by dy square of t star times e power sigma t times sine alpha x. I can take out a power sigma t and sine alpha x common from these two terms and I can write this in a slightly more compact way as d square by dy square minus alpha square of t star divided by e power sigma t sine alpha x okay. The derivative operator is only in the y direction is alpha square is of course the wave number of the periodic disturbance okay it is the reciprocal of a wave number of the wave length sorry this is the wave number which is the reciprocal of the spatial wave length and these are the time dependency. So what is normally done is to make things a bit more compact I am going to write this as t star times e power sigma t times sine alpha x where d square is nothing but d square by dy square. So remember this is the Laplacian of t tilde you see that there is also del to the power 4 operator occurring in the velocity equation. So the del power 4 operator of the velocity equation is going to be of the velocity variable is going to be nothing but del square of velocity okay. And you already know that del square of velocity is nothing but d square minus alpha square of v star e power sigma t sine alpha x because now its velocity instead of t star I am going to have v star and so this is going to basically reduce to d square minus alpha square the whole square of v star e power sigma t sine alpha x okay. Again this operator is operating only on v star remember v star is a function only of y and this derivative d capital D here is a derivative only with respect to y okay. So that is basically what we have done we have just made sure that the partial differential operator del power 4 can be converted to an ordinary differential operator d square. Now it may appear to you that I have actually jumped a step but if you understood how you have done this I think what you would do is just go through the algebra and you can verify for yourself that this is indeed correct that is one when you apply del square you get the d square minus alpha square when you apply it again you get another d square minus alpha square and that basically gives you this okay. So if you are not comfortable with this I suggest you work this out in your home and make sure that this is indeed right. Our job now is to basically substitute these expressions in my partial differential equation and convert it to an ordinary differential equation. So let us do that. So the first equation here on temperature now when I want to look at the time derivative with respect to temperature I would get only sigma okay and this is got v tilde I am going to write this as v star of th minus t0 divided by h and what I have done is replace all my tilde variables in terms of my star variables okay. So this is going to be sigma times t star because remember the temperature derivative with time is nothing but sigma times t star okay and you also have the e power sigma t times sin alpha x okay and the right hand side is nothing but d square minus alpha square of t star times e power sigma t times sin alpha x and that is what we just did. We showed that del square can be reduced to d square minus alpha square of t star e power sigma d sin alpha x and v tilde is v star times that when you differentiate this I get a sigma times t star and this. Now remember we have made this assumption of things being in phase okay. Now the fact that we are on the right track that the velocity and the temperature are indeed in phase is going to be confirmed by the fact that every term here has this the exponential term and the sin term okay. So this is basically an indication that indeed those variables are in phase. So this now reduces to rho naught cp times sigma t star plus v star times t h minus t 0 divided by h equals d square minus alpha square of t star. So this is again a linear equation but now it is a linear ordinary differential equation. What I like to do next is take the second equation here this equation and convert it to a ordinary differential equation. So when I do that I get rho 0 times the time derivative gives me a sigma and the del square operator gives me at d square minus alpha square okay and there is already a minus sign. So that gives me this minus sign. So what I am doing is I am looking at this term here. I am looking at this term and I am getting a sigma because of this time derivative this del square gives me the d square minus alpha square and v tilde remember is now going to get converted to v star okay and this gives me v star of e power sigma t times sin alpha x that is my left hand side. On the right hand side I have 2 terms the first term is the del power 4 operator which is nothing but d square minus alpha square whole square of v star times e power sigma t times sin alpha x okay. So this is the viscous term which goes with the fourth order del or del power 4 and that gives me my d square minus alpha square square times v star times e power sigma t sin alpha x and then you have the body force term which remember is this term here. This term here is the body force term and this is associated with the second derivative with respect to x. Now this is something which I had not done earlier but remember the x dependency is sinusoidal. So when I am differentiating it twice I am going to get a minus alpha squared multiplying this and so this term now is going to be reduced to minus rho naught beta g alpha squared times t star times e power sigma t times sin alpha x. All I have done is said that invoked the fact that the second derivative of t tilde is nothing but minus alpha squared of t star. Again what we see is that the exponential term and the sinusoidal term cancels off because they are present in all the terms. They cannot be 0 because if they were 0 then my perturbation itself is 0 because I have assumed the perturbation to be of the form e power sigma t sin alpha x. So they are non-zero and that basically justifies and allows me to cancel them and the fact that they are occurring in all the terms tells me that the assumed form of the spatial dependence for temperature and velocity in the x direction the periodic sin alpha x in phase is indeed right okay. If they are not cancelled off then it means that those velocity components are actually out of phase with the temperature component and there are situations where variables can be out of phase. So this equation now simplifies to minus sigma times rho naught times d squared minus alpha squared times v star equals minus mu times v squared minus alpha squared whole squared v star minus rho naught beta g alpha squared t star okay. So what I have done is converted partial differential equations to ordinary differential equations and the idea is I know how to solve ordinary differential equations especially because these equations are linear. What I am going to do now is tell you that the objective we have is to find this point of onset of natural convection okay. When exactly is natural convection going to start that is this critical value of the temperature gradient for a fixed fluid and the geometry. So when the temperature gradient is less than this critical value if you were to impose any disturbance this disturbance would actually decay and you would have the system going back to the stationary solution where the fluid does not move okay. If the temperature difference is more than this critical value when you are going to be giving a disturbance the disturbance will get amplified such that you would actually see convection. So the transition between the stable to the unstable is going to take place by looking at the real part of the growth constant in time sigma. If the real part is negative I have a stable system if the real part is positive I have an unstable system. So the critical point where you have the change from stable to unstable is going to be given by the condition that the real part of sigma is 0 okay. So for the onset of natural convection the critical condition is given by the real part of sigma equals 0 and one of the things which we can establish and I am not going to do that in this course is show that for this particular problem the sigma is real that is the sigma is not complex there is no imaginary component okay. And so rather than talk about the real part of sigma being 0 I am going to talk about the sigma being 0 okay. We can show that sigma equals is real so transition occurs at sigma equal to 0. Now I just give you an inkling of how we can go about proving that sigma is indeed real and this arises because you know matrices which are real symmetric okay they have the property that their eigenvalues are real and what we can do is generalize this idea of a real symmetric matrix to that of a Hermitian matrix to that of what is known as a self-adjoint operator. So rather than talk about matrices we can look about this as an operator a matrix takes a vector and converts it to another vector this is an operator which takes a function converts it to another function and we can talk in terms of eigenvalues of this operator and for this particular system we can look at the fact that whether it is self-adjoint or not and establish that sigma is 0. So this is just some piece of information I am giving you for those of you who are interested in pursuing this otherwise you just accept what I am saying sigma is indeed real and so the transition occurs at sigma being 0. So since I am interested only in the onset of natural convection what I am going to do is I am going to further simplify my equations by putting sigma equal to 0 in these ordinary differential equations that I have just derived which describe my V star and T star okay. So we put sigma equal to 0 to find the transition and that basically means this particular equation reduces to mu times d squared minus alpha squared whole squared times V star equals minus rho not beta g alpha squared T star. I just want to make sure that I am not messing up negative sign anywhere because I might get into trouble later and the other equation becomes rho not Cp times T h minus T not divided by h times V star equals d squared minus alpha square of T star. So you have 2 equations now but these are ordinary differential equations and they have velocity and temperature and you can see that the velocity and the temperature again couple to each other okay. Remember all these are constants which I know for a given experimental system. What I like to do now is write down the boundary conditions for this system of equations that we have okay. The condition on temperature is going to be obtained from the boundary conditions on temperature that is going to be telling me what the boundary conditions are for the perturbation. So remember T equals T not at y equals 0 okay and you know that T was written as Tss plus epsilon times T tilde okay and what we want to do is we know that Tss is equal to T0 at y equal to 0. Therefore you know Tss equals T0 at y equal to 0 which basically implies that T tilde equals 0 at y equals 0 okay. So basically since the steady state satisfies the boundary condition of the original problem the perturbation is going to vanish at y equal to 0. Similarly and what we have done is we have decomposed T tilde to T star of y and so this is actually the condition on T star of y. T star equals 0 at y equal to 0 because T tilde is nothing but T star multiplied by e power sigma T sin alpha x and those are independent of y. So the only way T tilde can vanish is if T star can vanish. You can similarly establish that T star equals 0 at y equals capital H. These are the boundary conditions on the temperature perturbation. Look at the equation for velocity. This is a fourth order equation for velocity and therefore I need 4 conditions 2 on the upper plate and 2 on the lower plate. Remember v star is my vertical component of velocity. So since my plate is impermeable the liquid cannot penetrate my upper plate and so the velocity component is going to be 0, v is going to be 0 at both y equal to 0 and H okay. And so I am just going to write this here that v tilde equals 0 at v equals 0 at y equals 0 and H. This follows from the fact that the liquid cannot penetrate the wall and v remember is nothing but vss plus epsilon times v1 tilde v tilde and this is 0 and this is 0 at y is equal to 0 and H and therefore this implies p tilde is 0 at y equals 0 comma H and in other words v star is 0 at y equals 0 comma H okay. So I have v star also being 0. So I have Dirichlet conditions on temperature and on velocity v star but do I have enough conditions to solve the problem? The answer is no. Since I actually have a fourth order equation remember in v star. So I need 4 boundary conditions and what I have is only 2 boundary conditions. So I need 2 more boundary conditions. Where am I going to get this from? I am going to get this from the conditions on the x component of velocity u remember what we have done is we have converted this problem simplified it by eliminating the x component of velocity. In this process we have not used the boundary conditions on the x component of velocity okay. So we have to figure out a way for converting the boundary conditions on the x component of velocity to conditions on the y component of velocity v star. So let us see how we can do that and let me give you a clue we are going to use the equation of continuity to accomplish this okay. So what I want to do is I want to get 2 more boundary conditions for v star by using the boundary conditions on the x component of velocity u star okay. So the no slip boundary condition implies that u equals 0 at y equals 0 and h and you can make the same argument but the steady state velocity is 0 at 0 and h and so u tilde which is equal to u star equals 0 at y equals 0 and h and this is my no slip. So this is the boundary condition on my original velocity this is the boundary condition on my x component of the perturbation velocity. So this is for original and this for perturbation. I want to convert this to a boundary condition on v star remember du by dx plus dv by dy equals 0 that is my equation of continuity and I am going to write this again in terms of my perturbation variables. So I get du tilde by dx plus dv tilde by dy equals 0 you can convert this to star variables but remember u tilde is nothing but 0 and y equal to 0 and h. So that means all along for all x this is proof at y equals 0 and h no sorry this is not true at y is equal to 0 and h. This is the equation of continuity which is always valid but u tilde is 0 at y equal to 0 and h which means that u tilde does not change with x for no matter what the position is in the horizontal direction no matter what the x position is u tilde is 0. So not only is u tilde is 0 but du tilde by dx is also 0 u tilde equals 0 for all x therefore du tilde by dx equals 0 at y equals 0 and h. So that means if du tilde by dx is 0 that means the first derivative of v tilde is 0 at y equal to 0 and h and this I can use to say that dv star equals 0 at y equals 0 and h. So this is the boundary condition which I have on velocity okay. So what I have done is basically these are the extra 2 boundary conditions which I was talking about earlier which I need to solve my problem and this basically tells me that the first derivative of the velocity of perturbation v star is 0 at the 2 walls. So now I have 6 boundary conditions and I am all set to solve the problem okay. However what we will now do is do a further simplification and this simplification is going to come by converting the system of 2 equations that we have to only 1 equation 1 variable. So what we have here is a system of 2 equations which are coupled to each other in 2 variables v star and t star. I like to write this as a system of equations or only a 1 equation in only 1 unknown v star that is I want to eliminate my temperature perturbation between these 2 equations okay. I like to keep my equation as if it is an equation which describes only the velocity perturbation without bringing into account the temperature perturbation. So let us do that by operating on both sides by d square-alpha square. So if I do that I would these are all constants I have d square-alpha square of t star. I can use this equation and substitute for that expression from here and that way I can eliminate t star okay. So operating on this equation by d square-alpha square what I get is mu times d square-alpha square whole cube of v star equals-rho not beta g alpha square of d square-alpha square t star. So that is what I have done I have just operated on that using d square-alpha square and now I am going to use the fact that d square-alpha square of t star is given by my velocity perturbation from the second equation to write this as-rho not beta g alpha square and d square-alpha square t star is nothing but rho not cp times th-t not divided by h. You know I think I have missed thermal conductivity somewhere I have missed a thermal conductivity here I have missed a thermal conductivity in that equation. So this thermal conductivity is important okay because remember the del square goes comes with the thermal conductivity and so I would have a k at the bottom here and I can write this equation now as d square-alpha square whole cube of v star equals-rho not beta rho not by mu times beta times g times rho not cp by k times th-t0 divided by h and I am going to write I have missed a v star here and that is going to be a v star here okay. So all I have done is rewritten this as it is here and that is an alpha square which is important. I have brought the mu down here in the denominator I am going to remember that this is nothing but my kinematic viscosity and this is nothing but my thermal diffusivity alpha t and I am going to write this as-beta g alpha square divided by mu the kinematic viscosity times the thermal diffusivity times th-t0 divided by h times v star okay. So this is my sixth order equation and like I said I need 6 boundary conditions I have found 4 boundary conditions on velocity that is the velocity perturbation and the first derivative must be 0 but remember my other 2 boundary conditions are on temperature. So again what I want to do is I want to convert my temperature boundary condition a velocity boundary condition okay and how do I do that? I should not have erased that sorry yeah. I am going to use the fact that t star is 0 at y equal to 0 and h see if t star is equal to 0 at y equal to 0 and h that means this term has to be 0 at y equal to 0 and h and so from since t star equals 0 at y equals 0 and h this implies b square-alpha square whole square of v star equals 0 at y equals 0 and h. So what I have done is I have converted my boundary condition on temperature to my boundary condition on velocity okay. So this sixth order equation that I have just written is going to basically need 6 boundary conditions and the 6 boundary conditions are v star equals dv star equals d square-alpha square whole square of v star equals 0 at y equals 0 and h and the differential equation is d square-alpha square whole cube of v star equals-beta g alpha square times nu by alpha t times th-t0 by h v star. So this is the differential equation these are the boundary conditions and what we have to do is see how we can solve this. We will do this in the next class. Thank you.