 We were looking at perturbations on a thin film coating a rod of radius r naught without proving it I had told you that one can show that if you put perturbations of wavelength lambda or wave number k then the dispersion relation is given by the expression that I have written here. Now just to check for consistency it is useful to take the limit r 0 going to 0. So, you can see that when r 0 goes to 0 r 0 is the radius of the rod. So, I am shrinking the rod you can think of this process as one which holds this entire distance constant and shrinks the radius of the rod. So, the space that is occupied by the rod gets now occupied by the fluid. Now as the rod thickness shrinks to 0 in that limit this should reduce to a fluid thread or a fluid cylinder whose radius is r 1 and so we should the dispersion relation should reduce to the case that we have already derived earlier. This is just a check on the consistency of this dispersion relation. Now this can be shown easily that when r naught goes to 0 then this part is finite and this term is finite. This goes to 0 and this goes to 0 and so we are left with just the first term here and here. Note that the first term on in both numerator and denominator has a k 1 and k 1 the argument if I set it to 0 k 1 diverges. However, there is a k 1 in the numerator and a k 1 in the denominator. So, this divergence will get cancelled out and so we will just recover omega square is equal to T by rho r 1 cube k r 1 into k square r 1 square minus 1 into I am canceling out the k 1 and the k 1. So, I am canceling out this k 1 and this k 1 both of them will diverge k 1 diverges at the origin. So, as I said r naught to 0 for a fixed small k the argument will go to 0 and both the numerator and the denominator will diverge but both of them will diverge in exactly the same way because they are the same functions. So, it can be cancelled out and so it remains what remains is I 1 of k r 1 divided by I naught of k r 1 and you can check that this is exactly the same dispersion relation that we have obtained earlier. Now, we have replaced r 0 in the previous relation with r 1 now and that is because as the keeping this length fixed if I shrink the radius of the rod then the space that was earlier occupied by the rod now gets occupied by the fluid and the limit when the rod shrinks to 0 or just a line then it just becomes a fluid cylinder and so I should recover the same dispersion relation that we had got earlier except that now we are writing r 1 because the radius of the fluid cylinder is now r 1. So, it is this, it is this number. So, this is consistent with what we have done so far and kind of generalizes what we have done for a fluid cylinder. So, now let us analyze this dispersion relation you can once again go to MATLAB or Mathematica and plot this quantity and you will find that this quantity is always positive for all k this quantity is always positive for all k this can be shown. So, once again this dispersion relation can have omega square can become negative but the only place which can become negative is once again our old familiar term k square r 1 square minus 1. So, we get omega square negative omega square is negative when k r 1 is less than 1 and allow us to what we had got earlier earlier it was k r not less than 1 now it is k r 1 because r 1 is the radius of the free surface of the film that coats the rod. Now of course, for k r 1 greater than 1 you get waves or oscillations. So, there is no growth let us analyze this a little bit more. So, we will do exactly what we had done earlier I am going to substitute omega is equal to i omega i where omega i is now real that will introduce a minus omega i square on the left hand side like before and I am going to reverse this part in red. So, that it is positive now because I am looking at this limit of looking at this range of k r 1 less than 1. So, if k r 1 is less than 1 then k square r 1 square minus 1 is negative and so I am going to reverse it and bring out an overall negative sign like before the negative sign on the left and the negative sign on the right will cancel each other and I can plot omega i square as a function of the right hand side. So, once again I will do the same thing I will define a omega i star square which is a non-dimensional omega i square which is basically just omega i square divided by t by rho r 1 cube. Note that I am working on the full dispersion relation and so this is equal to it is just k r 1 into I have reversed this now. So, I will write it as 1 minus k square r 1 square into this part it is a long one. So, I am not going to write it again and so now this is purely positive because I have reversed the part which is actually negative. So, I have reversed the part and now I am going to plot the left hand side as a function of right hand side. So, if I do that or rather if I plot it as a function of omega i square as a function of k r 1 then you will see once again the same thing. So, while doing this you will have to choose a value of r naught. So, you will have to choose a value of the radius of the rod and then plot it as a function of k r 1. If you do that once again you will get a very similar curve and the curve will look like this. Once again this will hit the x axis at 1, 1.0 and that is because beyond k r 1 when it is greater than 1 omega i square is negative or in other words we are going to the regime where there would be oscillations and not instability. So, this is the place where we are getting growth k r 1 is less than 1. Does the growth rate get affected by the presence of the rod? That depends on what is the location of this maxima. Earlier we had seen that the location of the maxima was 0.697. In this case this maxima is approximately at 0.706 or approximately 1 by square root 2. So, it gets slightly shifted. So, now this is the fastest growing mode and so if this film breaks up into droplets or rather cylindrical droplets which will coat the rod then this will give us the wavelength of those or the size of those droplets. This wavelength will be the fastest growing mode. As you can see like before the criteria for whether it will grow or not grow is independent of surface tension. It is purely a geometric criteria which is given by this quantity k r 1 less than 1. So, any wave number which is less than any wave number which satisfies k r 1 less than 1 will exhibit instability and will cause exponential growth. Among all such wave numbers the wave number which grows the fastest satisfies k r 1 is equal to 1 by square root 2. This is slightly different from what we had done earlier for a liquid cylinder. Now this is a solid cylinder coated with a liquid film still prone to the relay plateau instability and the criteria remains the same. Let us now move over to yet another problem where we see the same instability and you will find that yet in this problem also the criteria for the instability remains exactly the same. It is a very analogous criteria that we have now seen 2 times. So, let us the third problem is that of a cylindrical bubble or a cylindrical column of air you know in water let us say. Such columns of air often get trapped when waves break in the ocean and it leads to air entrapment trapping a cylindrical column of air inside just below the surface of water. Now we will simplify that problem and we will ask the question that suppose I have a cylinder of some radius r naught and there is gas inside it. So, let us say there is air inside it and outside there is some liquid let us say it is water. Now due to the density difference between air and water we are going to ignore the dynamics of the medium inside we will model this column as being infinitely long. So, it goes from minus infinity. So, the coordinate system is again like before this is z and this is r and so the interface between air and water we are going to impose perturbations on it and we are going to solve. But now we are going to solve for the fluid outside not for the fluid inside. So, this is a filament with air inside it we are going to impose axisymmetric perturbations on it and we are going to ask are those perturbations stable to the oscillate or do they grow in time. We will find that some of these perturbations are still unstable to the relay plateau instability. So, this is here the base state once again is quiescent. So, both air and water is quiescent. We are going to ignore air the density of air is much less than that of water. So, as a first approximation we are going to ignore air. We have ignored air in all the examples that we have done until now except that in all the examples the gas was present either above or outside now the gas is present inside. So, this is like a air bubble. So, we are not solving for the air inside the bubble we are going to solve for water outside the bubble that is still modeled by Laplace equation. So, when we have perturbations the perturbation velocity potential in the water is going to be still modeled by the Laplace equation. Let us do that. So, in the base state the pressure once again is. So, if I say that the pressure here air pressure here is 0. In the earlier example when we had done it for a liquid cylinder we had modeled the air pressure outside to be 0 and so the pressure inside was more and so that was t by r naught. Now we are saying the pressure inside is 0 the pressure inside because of the weight is curved is always more than the pressure outside. So, if I have said the pressure inside to be 0 then the pressure outside has to be less than the pressure inside. So, that is minus t by r naught. So, this is the pressure in the water outside it is a uniform pressure we are ignoring the presence of gravity and we are going to solve for perturbations on the surface. We have already done this geometry. So, I am just going to skip a few steps because they are identical to what we have done for a cylindrical column of liquid. So, like before using variable separation in cylindrical coordinates axisymmetric we will find that the velocity potential the perturbation velocity potential is once again given by and now for the radial dependence we are going to have k naught and i naught. So, for the radial part of the velocity potential like earlier we will have two choices k naught and i naught and it will be a linear combination of the two. Now our domain we are not solving for air inside. So, our domain extends from r is equal to r naught plus some eta that we will put on the surface to infinity. So, our domain goes from r naught plus eta to infinity. So, small r does not go to 0. Recall that k naught diverges at small r, i naught diverges at large r. So, we our domain actually goes to infinity. So, I cannot include i naught in my calculation because i naught will diverge as the radius becomes larger and larger and we are solving for. So, we are solving for the water and the water is unbounded it is radially unbounded. So, I will write this. So, phi is the velocity potential in water. So, we are solving only for water we are not solving for air we are assuming the air to be quiescent and the pressure in the air to be always 0. So, then we are going to get a k0 of kr here. I hope you understand why. Earlier it was i0 of kr because the domain included small r equal to 0. Now it is k0 of kr because the domain includes small r is equal to infinity or small r going to infinity and so i0 diverges as small r becomes larger and larger. So, we have to set the coefficient of i0 to 0 now and only k0 will survive. Similarly, eta would be e cos kz plus f sin kz into e to the power i omega t. Like before eta is only a function of z and t by definition eta does not depend on r and these are my normal mode approximations. And so like earlier we will write the Bernoulli equation the total pressure plus the total velocity potential in this case the total velocity potential is just the perturbation velocity potential because this quiescent fluid in the base state. And this is equal to the Bernoulli constant the Bernoulli constant is once again found by applying the same equation in the base state. And in the base state it is just p b by rho p b by rho in the base state is just it is just p b by and like before so the pressure jump condition now becomes p at r is equal to r0 plus eta is equal to minus t grad dot m. Earlier our n was pointing from the fluid where we were solving into the fluid where we were not solving. Now also our n continues to point but now we are not solving for air but we are solving for water. So the n is still radially outward and that brings in this minus sign. So this is at r is equal to r0 plus eta. So now so n points from so n is a unit vector pointing from gas to liquid. You can check that this minus sign is necessary because this equation is also true in the base state and without this minus sign you will not so if you apply this equation in the base state and if you calculate the curvature in the base state that is just a constant and you will find that it recovers this equation only if you take that minus sign. So the minus sign is necessary. So now we will once again like before we are going to work out what is divergence of n. So this is we have already done this before. So divergence of n at r is equal to r0 plus eta is just going to be. So I am just going to write down expressions that we have already written. So we have done this before so I am just going to straight away give you the expressions and you can check it for yourself. This is exactly the same procedure that we have followed. You have to define a function capital F whose value is constant on the surface and then you have to take the gradient of that the components of that so the denominator is not there in the linear approximation. So n approximately is just gradient of capital F and then you have to calculate the various components of grad F in cylindrical coordinates and then take the divergence of that quantity. Once you take the divergence then you will get this omega t plus k square. Please refer to the previous video where we have already done this and of course there is a complex conjugate which I am suppressing. And then the expression we have just written that the total pressure at r0 plus eta is equal to minus t and then I can write my left hand side as pb and pb in the base state there is no eta. So it is just at r is equal to r0 plus the perturbation pressure at r0 plus eta is whatever we have calculated from this. So I am just substituting. So there is a minus sign overall. So minus t by r0 plus t by r0 square into e cos kz plus f sin and then minus k square tk square into e cos kz plus f sin kz e i omega t and then I have to write it plus cc also but I am not writing that. And we know that pb in the base state is just minus t by r0. In the base state there is a uniform pressure in the water outside. That pressure is just minus t by r0. There has to be a pressure difference between inside and outside. If we say that the pressure inside in the air is 0 in the base state then the pressure outside has to be lower and that is minus t by r0. It is only the difference which matters. So then this quantity and this quantity get cancelled out using this and so like before we just recover an expression for perturbation pressure. So p at r is equal to r0 plus eta is just the same thing. So t so it is e cos kz plus f sin kz into e to the power i omega t t into 1 by r0 square minus k square. Very similar expressions. There will be a overall difference of minus sign because now in the pressure boundary condition there is a minus sign compared to what was there earlier. So that is the only difference. So once again you can go back and substitute it into the linearized Bernoulli equation. So the linearized Bernoulli equation is just once again like we have to find that the perturbation pressure has to be applied at r is equal to r0 and not r is equal to r0 plus eta because that will bring in a nonlinear term is equal to minus rho. Similarly del phi by del t the perturbation the time derivative of the perturbation velocity potential also has to be evaluated at r equal to r0. If you substitute whatever we have found in the last slide we will recover the equation rho i omega I am straight away writing the equation. So compared to the last example where we had done this for a liquid cylinder the only difference is there is a minus sign and instead of i0 there we have k0 here. i0 was the modified Bessel function here this is the modified Bessel function the second kind. So instead of i0 here we have k0 a plus t into 1 by r0 square minus k square into e and this whole thing gets multiplied by cos kz plus rho i omega k0 k r0 B plus t 1 by r0 square minus k square into f this gets multiplied by sin kz the whole thing is multiplied by e to the power i omega t plus cc is equal to 0. This will be my equation 1 coming from Bernoulli equation linearized Bernoulli equation. Second so Bernoulli equation the second thing will come from kinematic boundary condition and so we will have del eta by del t is equal to del phi by del r at r equal to r0. This will involve derivatives of k0 with respect to its argument. So this will lead to I am straight away writing the final answer this will give you i omega e plus k into the modified Bessel function plus i omega f plus k this whole thing multiplied by e to the power i omega t plus a complex conjugate is equal to 0 and this is equation 2. Like before we have to set the coefficients of cos and sin to 0, four equations in four unknowns a, b, e and f the determinant will determine the dispersion relation. I will write down the dispersion relation it just requires a little bit of algebra you can do it yourself and verify that this is correct. So the final dispersion relation looks like omega square is equal to t by r0 cube into k r0 k square r0 square minus 1 into k1 of kr0 divided by k0 small k r0. You might be wondering where did all the minus the extra minus sign that we got why did not that made a difference to the dispersion relation. You can see that this k1 earlier we had a i1 in the numerator and i0 in the denominator and we had seen that the derivative of i0 with respect to its argument is i1. Here there will be an extra minus sign overall but the derivative of k0 with respect to its argument will be minus k1. So that minus and the overall minus will cancel and lead to exactly the same dispersion relation except that now we will have k1 and k0 here. So this is the only difference the rest of it remains the same. You can see that this also has the same structure as before that this is something so this is the part which has the dimensions of frequency squared. So this is the dimensional part and the rest of it is non-dimensional. So I can once again write it as maybe some capital H some function of small k r0 and capital H of small k r0 is defined as it is a non-dimensional function 0 by k0 of k r0. Once again you can show this part is always positive. So if this dispersion relation has to admit instability that instability has to again come from this part. Once again we find the exactly is the same criteria that all wave numbers which satisfy k r0 less than 1 are unstable. We have now seen three examples one of a liquid filament one of a thin film coating a cylindrical rod and one of an air bubble. In all of them the criteria for instability is the same the growth rates will be different. In this case the growth rate if you plot the growth rate so you can non-dimensionalize omega square. So you can substitute omega is equal to i times omega i like before omega i is real. This will give you a minus sign and then you can reverse. So you can write it as omega i square is equal to t by rho r0 cube into k r0 and reverse this 1 minus k square r0 square into k1 of k r0 and k0 of k r0 and this whole thing will be positive in the range k r0 less than 1. So you can plot it as a function of k r0 and it will look something like this. Once again it will hit the x axis at 1 and the maximum growth rate will be about 0.4838. So k r0 or rather the mode which grows the fastest so k max into r0 is equal to 0.4838. So you can see that whether we have a denser fluid inside or whether we have a denser fluid outside whether we have a solid inside it makes no difference to the criteria of instability. It makes a difference only to the growth rates of the fastest growing mode. So we will find that we will try to understand the origin of this behavior that what is so special about k r0 less than 1 and why does this cause instability in each and every case that we have seen so far. In all the cases k r0 greater than 1 gives you stability at least it is linearly stable and so you expect small amplitude oscillations governed by the dispersion relation. The dispersion relation is actually different in all the three cases. So the frequencies are not the same. However the criteria the boundary between stability and instability remains the same in all the three cases. We will examine the reason for this in the next lecture.