 So, welcome to today's lecture, what we will do today is continue our discussion on the Rayleigh Taylor problem from where we left off yesterday okay and this remember is the problem where we have two liquids one on top of the other okay and what we had done was we had found the steady state which was the stationary state and then we did the linearization, we had the linearized equations, we assumed a normal form for the disturbance in terms of some periodic disturbance in x and y and then we reduce it to an ordinary differential equation in z and what we finally found was that the solution in the first liquid is given by this expression which is valid for in the range minus infinity less than z less than 0 and this is valid in the range or the domain 0 less than z less than infinity okay and on the way we got this from the equation of continuity and this is something which we have used to get this expression, we did the elimination okay, I would be needing this later so I have just written this down. Now our objective is to find these constants and go ahead with the solution and remember what we want to do is find out what kind of wave numbers are going to grow, what kind of wave numbers are going to decay, so that is the idea. So, we want to get a relationship between the growth rate which is sigma and alpha okay and that is the dispersion curve but that tells me that this wavelength is going to grow, this wavelength is not going to grow and how is it that the disturbance is going to manifest itself, so for that I need to solve this w1 and w2 and there are 4 constants and we are going to use the fact that w1 at star at z equals-infinity should tend to 0 as z goes to-infinity okay. This implies that b equals 0 okay, similarly w2 star at z equals-infinity should tend to 0 and this implies c is 0 okay, so basically what this means is w1 star equals a e power alpha z okay and w2 star is d e power-alpha z okay, so I have used 2 boundary conditions to get rid of 2 constants, I still have 2 unknowns which I have to determine the a and the d and what we are going to do now is use the, now that we are allowing the interface to deform, we need to use the kinematic boundary condition and we also need to use the normal stress boundary condition. Like I was mentioning yesterday, when we are looking at the invested limit what we have to do is let go of one of the boundary conditions okay, so when because the viscosity is 0, I have 2 conditions which we need to satisfy which is the normal stress and the shear stress. The shear stress condition is not invoked, it is not used because you know it is in some sense trivially satisfied because viscosity is 0 on both sides, so we get 0 equals 0 okay. The normal stress boundary condition is what we have to use and that is what we will be using. So let us look at the kinematic boundary condition and this is a small recap of whatever we did earlier. So when this is a flat interface and this is given by z equals 0 and what we need to do is worry about the situation where the interface can possibly get deflected okay and so here the interface is going to be given by a function of this kind okay. We write this in an implicit form f of x, y, z, t as z-h of x, y, t equal to 0 okay and the kinematic boundary condition follows from the fact that df by dt is 0. This is the kinematic boundary condition okay and when I use this what do I get df by dt plus the velocity vector times gradient of f equals 0 dotted with gradient of f equals 0 okay. That is the form for the substantial derivative, df by dt is nothing but –dx by dt but what I want to do is since I want to keep in mind that this perturbation is infinitesimal okay, I am going to write this z as epsilon times h of x, y, t but then it is makes it easier for me to do this order of epsilon analysis okay. So remember this is an infinitesimal perturbation and I have forgotten to put this epsilon there so I am just saying that it is a small deviation from the z equals 0 okay and that is what I have done. So now –df by dt and order epsilon, this is going to be multiplied by epsilon okay plus u partial derivative of f with respect to x which is –epsilon dx by dx plus v times this is all multiplied by okay, dx by dy plus w times df by dz okay. Remember the u and the v and the w are the actual velocities okay, the u, v and w, u, v, w are the actual velocities. I have not done any breaking this up in the form of a steady state plus a perturbation. So what is u? u is actually uss plus epsilon u tilde okay and h remember is a perturbation itself. So actually u is uss plus epsilon u tilde, v is vss plus epsilon v tilde and so on and so forth. So when I now substitute for u, uss plus epsilon u tilde, I will get epsilon u tilde the disturbance I will get epsilon v tilde the disturbance here okay. So then this becomes of order epsilon squared, this becomes of order epsilon squared, w will be epsilon w tilde. So what this means is at order epsilon this equation reduces to w tilde equal to dx by dt okay. So I proceed further and I write this as –epsilon dx by dt plus epsilon times u tilde with a minus sign, epsilon squared times ds dx minus epsilon squared times v tilde ds dy okay plus epsilon w tilde equal to 0 and this is order epsilon squared these 2 terms of order epsilon squared and therefore we get w tilde equals dh by dt that is your kinematic boundary condition at order epsilon that makes sense the rate at which the h is changing with time is my vertical component of velocity that is what it says okay. Now remember this is going to be valid for both the phases. So w1 tilde is going to be equal to dh by dt if I write for the first phase if I write for the second phase it is going to be w2 tilde is equal to dh by dt okay. So we have w1 tilde equals ds by dt which is equal to w2 tilde for each phase. This is at the interface so in other words w1 tilde will be equal to w2 tilde okay. Now I have assumed that the perturbations and remember h is also a perturbation all the perturbations are of the form periodic in x and y and growing in time okay. So now how am I going to assume this h? h is the function of x, y and t correct and so this is going to be of the form h a constant multiplied by e power sigma t e power i alpha xx plus alpha yy okay so I am assuming this is going to be the same in both the phases. See the 2 phases are actually coupled to each other and they are coupled to each other through this boundary condition okay. The h is going to vary as alpha x and alpha y and that is going to decide how the velocity in one liquid is changing, how the velocity in the other liquid is changing. So the coupling of these velocities is actually occurring through this boundary condition. So this is what is going to make sure that the wave numbers are the same in both the 2 liquids in both the liquids okay. So what I am going to do now is I am going to substitute this h here and you already know what is w1 tilde and w2 tilde it is and this remember is evaluated at w1 tilde we already know what it is a e power alpha z so w1 tilde is a e power alpha z and I am going to evaluate this at z equal to 0 the interface okay and e power alpha z multiplied by e power i alpha xx plus i alpha yy times e power sigma t equals h times e power sigma t alpha yy okay. So when I look at this I am going to be looking at evaluating this at z equal to 0 I take z equal to 0 because when I evaluate this at z equal to epsilon h I would do a Taylor series expansion okay. If I want to calculate z at epsilon h I will get the value of e to the power alpha z at 0 plus the next term which will be order epsilon lower okay. So what I am saying is this guy this is this here at z equal to 0 yields a equals h okay. So this cancels off with that and that is what I get and I can use the other one w2 tilde and I would get a and d. So my job is now reduced to what I have just found out is that a equals d and my job now is reduced to finding out this either a or d whatever you want and for that I go back to using the boundary condition and anyway I am going to be able to find the solution only to within an arbitrary constant okay. So what I am going to do is you find the or use the normal stress boundary condition yeah it should be h sigma you are absolutely right I think it should be h sigma is right I want to differentiate this with respect to time I would get a sigma here yeah otherwise I would have been in trouble very soon. So sigma is important yeah let us keep that now the normal stress boundary condition is going to be a balance of the stresses right. So I am going to write this as p1 minus n dot t dot n minus p2 minus n dot p2 dot n equals gamma times del dot n the boundary condition which we have derived long time ago telling you that the difference in the pressures is going to be balanced by the surface tension and the curvature okay. Now since we assume things to be inviscid these 2 terms are going to drop off okay and what this basically reduces to is p1 minus p2 equals gamma del dot n since t1 equals t2 equals 0 for an inviscid liquid. Now what do you want to do is we want to evaluate del dot n okay. Now how do you evaluate del dot n you already know how to do this n is written as gradient of f divided by the absolute value of the gradient of f and remember f is z minus epsilon h of x comma y comma t okay. So the gradient of f is minus epsilon h subscript x that is a partial derivative with respect to x times e x the unit vector in x direction minus epsilon h y times the unit vector in the y direction plus e z when I differentiate with respect to z I get 1 okay that is my gradient of f and so n will turn out to be what? Minus epsilon h x minus e x minus epsilon h y e y plus e z divided by square root of 1 plus epsilon squared h x squared plus epsilon squared h y squared okay that is my n I need to get del dot n. So what is del dot n? It is e x d by dx plus e y d by dy plus e z d by dz dotted with this n which I have just found out minus epsilon h x e x minus epsilon h y e y plus e z with h x squared plus h y squared that is what I need to do okay. Now since I am taking the dot product only the terms and since the unit vectors e x e y e z do not change with x y z okay this basically reduces to calculating d by dx of minus epsilon h x divided by square root of 1 plus h x squared plus h y squared okay minus or plus d by dy of minus epsilon h y divided by square root of 1 plus h x squared plus h y squared that is a epsilon squared in the denominator yes that is an epsilon squared in the denominator and that is important yes I have it there and here also there is an epsilon squared okay. Now this guy is not going to make any contribution because the derivative with respect to z this guy is independent of z okay. So only these 2 terms are going to give contribution because this is independent of the z variable it only depends upon x and y okay. I am just going to illustrate one thing and then you people can differentiate this and verify for yourself how it is done for the 2 dimensional problem. I am going to just do a little bit of algebra just to show you that this reduces to a simplified form containing only the second derivative of h with respect to x. So for this to just reduce the math on the board we will neglect changes in the y direction I mean just so that when I differentiate I do not make mistakes okay you guys can afford to make mistakes and correct ourselves right. So hopefully this will reduce the mistakes I make. So what is this changes in the y direction means I am going to put h y equal to 0 and if I put h y equal to 0 I just want to show that this reduces to a simplified form at order epsilon that is the whole idea okay. I am interested in what is this term at order epsilon okay we have already done this. So anyway I need to do this for the sake of other people who may have not done the assignment okay. So I will just do this once and then we will stop I will not bore you too much. So now here we are this is h y is 0 and I need to use the quotient rule for the differentiation squared times the derivative of the numerator which is minus epsilon h xx then minus of the derivative of the denominator times derivative of the numerator minus the numerator times the derivative of the denominator squared divided by 1 plus epsilon squared h x squared okay. So now when I differentiate that what do I get this equals square root of 1 plus epsilon squared h x squared times minus epsilon h xx plus epsilon h x times the derivative of this thing is nothing but half of divided by square root of 1 plus epsilon squared h x squared right multiplied by that is what it is yeah and then that thing is divided by 1 plus epsilon squared h x squared into 2 should come here yeah that is right. So that gets rid of that now I take the LCM of the numerator and what do I get 1 plus epsilon squared h x squared times minus epsilon h xx divided by 1 plus to the power 3 by 2 and you will see that this guy multiplied by this knocks off that and what I am left with is minus epsilon h xx divided by 1 plus epsilon squared h x squared to the power 3 by 2 okay that is your actual curvature and this is a form which you are possibly familiar with when you did your course in calculus you must have come across this form in the second derivative on the top and to the power 3 by 2 in the denominator okay. Now if you want to do an order of epsilon analysis you would do a binomial theorem expansion of this you would get 1 minus something which is of order epsilon which contains epsilon in it and so at order epsilon so using binomial theorem this is minus epsilon h xx times 1 plus epsilon squared h x squared to the power minus 3 by 2 and this is nothing but minus epsilon h xx okay. So the whole idea I did that little bit of algebra to tell you that order epsilon this particular term gives me epsilon h xx if you now retain that h y also you would get a minus epsilon h y y the curvature in the other direction. So right now I have assumed no changes in the y direction things are flat in the y direction changes only in the x direction. So changes only in the x direction give me this as the curvature changes in the y direction will give me analogously minus epsilon h y y okay. So that was the idea similarly in the y direction if h y is not equal to 0 we get the curvature is minus epsilon h y y okay. So that is my del dot n term at order epsilon and now I can go and write the boundary condition. The boundary condition was p1-p2 equals gamma times del dot n that is my full force boundary condition okay p1-p2 is gamma del dot n that is valid for actual variables okay. Now I am going to write this is valid at z equals epsilon h okay the boundary condition has to be applied as the interface which is z equals epsilon h what did I just find out del dot n is at order epsilon it is minus gamma epsilon times h xx plus h yy that is what we found for the curvature okay. And this is now being evaluated at order epsilon what about p1 and p2 this is the actual pressure so p1 remember is going to be written as p1ss plus epsilon p1 tilde okay because this is the actual pressure I am writing it in terms of the base state plus the deviation from the base state. So I am going to substitute this here p1ss plus epsilon p1 tilde minus p2ss plus epsilon p2 tilde equals minus gamma epsilon times h xx plus h yy okay I want to group p1 and this is evaluated at z equals epsilon h. So I am going to write this as p1ss minus p2ss evaluated at epsilon h plus epsilon times p1 tilde minus p2 tilde minus gamma epsilon h xx plus h yy this is my normal stress boundary condition okay I am evaluating this particular boundary condition at z equals epsilon h everything is at z equals epsilon h. Now what is p1ss p1ss was minus rho 1 gz right from what we got p1ss remember yeah so now I am going to substitute this as minus rho 1 gz substitute this as minus rho 2 gz and evaluate this at epsilon h okay because this is a boundary condition so now I write this as so this gives me rho 2 minus rho 1 g epsilon small h plus epsilon times p1 tilde minus p2 tilde equals minus gamma epsilon h xx plus h yy so that is my normal stress boundary condition at order epsilon okay. Now these variables contain both x, y and time dependency right and I want to write this in terms of star variables in fact that is what I am going to do. So h is going to be written as rho 2 minus rho 1 times so at order epsilon I have g and small h remember h times e power i alpha xx plus alpha yy plus sigma t okay that is the form we assume for small h for pressure it is going to be p1 star of z minus p2 star of z and this is going to be evaluated at z equal to 0 e power i alpha xx plus alpha yy plus sigma t okay and must be equal to gamma times I have the epsilon here so this is at order epsilon okay. I have gamma times when I differentiate with respect to x I would get alpha squared so I will get minus is already there so I will get alpha x squared plus alpha y squared times h plus sigma t okay. So my point is this is canceling everywhere and what I am left with is this condition h plus h okay that is the equation which I get from the normal stress boundary condition. So now ideally what I want to do is I want to get conditions under which I have a non-zero solution okay. So I want to get an equation which contains only h this p1 star for example contains this particular term is independent of h. So I want to use my earlier relationships to see if I can find p1 star in terms of some h variable. So the idea is I am going to use p1 star and p2 star I am going to use this relationship here to relate p1 star in terms of the derivative of w1. I already know what the solution for w1 is okay and so I can get w1 and if you remember a and d we have obtained in terms of h. So the idea is I am going to write this particular term in terms of capital H okay. Then I have an equation which contains something multiplied by h plus something multiplied by h equals something multiplied by h and I want a non-zero solution so I can knock off h and get a relationship between sigma which has to occur somewhere and alpha squared okay that is the idea. So that is the strategy. So what is p1 star? From this equation of continuity that we got p1 star is –rho1 sigma dw1 star by dz divided by alpha squared okay and that we got yesterday by doing some elimination. So I am just using that relationship p2 star is going to be –rho1 sigma by alpha square over 2 sigma by alpha squared dw2 star by dz similarly. Is that a problem? I do not think so because there is already a negative sign there is not it. So when I differentiate this with respect to x I will get –alpha squared –alpha x squared e power i alpha x I differentiate this I get –plus i squared alpha x squared so I will get this okay. Now what is dw1 star by dz? It is nothing but a alpha e power alpha z I can use this okay. So this is nothing but –rho1 sigma a alpha e power alpha z divided by alpha squared and this is –rho2 sigma d and I am differentiating this with respect to z so I am going to get a –alpha here divided by alpha squared e power –alpha z. The point is these 2 terms are going to be evaluated at z equal to 0 and this in fact is the basis of this domain perturbation method which we saw earlier. Now remember I harped on this being evaluated at the interface okay. Now this is a base state and I want this equation to be valid at order epsilon since it is to be valid at order epsilon this is the base state so I have used z equals epsilon h here. This is already a perturbation okay. So this is already of order epsilon. So now if I would so this has to be multiplied by something which is evaluated at the base state so the z has to be 0 okay. If I would evaluate this at epsilon h then this would be a higher order term okay. I have not done the formal derivation here but maybe the next problem we will do it more formally. So the idea is that this is going to be evaluated at z equal to 0 and now I can substitute this here and remember we already have a relationship between a, d and h. We derived that a and d are a equals d equals sigma h. So I think I want everything now what I wanted I am going to substitute all this back here and get a relationship for h or get a relationship between alpha and sigma square rho 2-rho 1 g h plus p1 star is this-p2 star so I have again plus here. So p1 star is the minus sign rho 1 okay this is –rho 1. When I substitute sigma h here I get sigma square h and this is evaluated at 0 and divided by alpha okay that is that-p2 star-minus is plus again- okay equals gamma alpha square h okay. I do not think I made any mistake okay. So now what do I have? I want to get my growth rate sigma square sigma remember is my growth rate gamma remember is my surface tension and alpha is my wave number it tells me something about the periodicity okay and this I will just write here is my surface tension and clearly we seek h to be not equal to 0 that is what we want and that is the condition which gives you your dispersion curve and what I am going to do is this is a negative sign I am going to move this to that side and move that to this side and I get rho 2-rho 1 g h okay I keep this here I am bringing that here-gamma alpha square this goes off square equals rho 1 plus rho 2 by alpha sigma square okay. So in other words sigma square equals and that is my final expression g-gamma alpha square times alpha divided by rho 1 plus rho 2 okay that is what you should get. Now if you do not have any surface tension okay if gamma is 0 then your relationship reduces to sigma square equals rho 2-rho 1 g divided by rho 1 plus rho 2 times alpha okay. What this means is this remember is the growth rate of the disturbance if rho 2 is greater than rho 1 this is positive and you would have the growth rate you will have a positive value for the growth rate when you take the square root one will be positive one will be negative okay. So if rho 2 is greater than rho 1 sigma is positive the system is unstable okay and that is perfectly fine you possibly did not have to do all these analysis to find this because you know that if the denser liquid is on top it is going to be unstable okay but some information that the growth rate varies linearly with the wave number the more the wave number the more is the growth rate okay. And if rho 2 is less than rho 1 what happens this is negative okay and sigma square is negative that means what is the real part the real part is 0 because you purely imagine a plus or minus i multiplied by something. So sigma so the linear stability analysis cannot really tell you anything because you are on the boundary okay. So you really cannot conclude that it is stable just because by doing this linear stability analysis but if you can if you have a rho 2 greater than rho 1 you know for sure it is unstable okay. So this real part is 0 that means the real part of sigma is 0 and we are on the boundary of stable and unstable region. And the other thing you observe is all wave numbers alphas grow if rho 2 is greater than rho 1 all the wave numbers are growing to grow. Whereas if you now have a finite value of gamma if gamma is not equal to 0 and if I remember correct sigma squared is the numerator is this gamma alpha squared okay. When will the growth rate be positive sigma squared is positive if rho 2-rho 1g is greater than gamma alpha squared or alpha squared is less than rho 2-rho 1g divided by gamma okay. So what does this mean sigma squared is positive when your alpha squared is going to be low okay. So low wave numbers are large wavelengths means are unstable correct wave number and wavelength are reciprocal. So low wave numbers that is what we said right sigma squared is positive if this is positive because this guy is anyway positive the wave number is positive this guy is positive and so only if this is positive if this for this to be positive your wave number should be low than a threshold okay. If this is lower then you have instability or the wavelength is large. So if the wavelength is large means it is very slowly and if I write it the other way large wave numbers or low wavelengths are stable that means if the periodicity is very sharp that means the curvature is going to be very high then the surface tension is dominating okay because the curvature multiplied by the surface tension is the one which is contributing to your normal state boundary condition right. So that dominates when the wavelength is low that means there is a very sharp curvature then sigma comes into the picture not sigma the gamma the surface tension and that has a stabilizing influence. So point here is that gamma has a stabilizing influence okay the point I am trying to make here is that surface tension has a stabilizing influence because this is associated with the minus sign okay and surface tension is going to dominate when alpha is going to be large the wave number is going to be large or the wavelength is going to be low okay. So surface tension dominates and stabilizes for large wavelengths sorry for low wavelengths or large wave numbers okay I think that is the message from this analysis.