 We were looking at waves on a cylindrical base state geometry where this was a fluid cylinder in the base state, quiescent fluid with a pressure jump inside compared to outside due to surface tension. We had ignored gravity and we were looking at surface perturbations on the cylinder. So we had solved the Laplace equation. In this case we had found a modified Bessel function and we had expressed the r and the z dependence of the velocity potential, the perturbation velocity potential and the surface perturbation in terms of the modified Bessel function. So the I0 had appeared in phi and then there were these trigonometric functions which appeared in both phi and eta and then we did a normal mode analysis. Now we had worked out the linearized kinematic boundary condition and we had also looked at the expression for the divergence of the unit normal evaluated at the perturbed interface. Let us continue from there. So we had seen that the pressure boundary condition is just r0 plus eta is equal to d divergence of n also evaluated at r0 plus eta. And this pressure on the left hand side is a sum of base. In the base state the radius of the cylinder is constant and a perturbation pressure which has to be applied in general r0 plus eta is equal to the right hand side. We have already evaluated the expression for this earlier. So you can see that we have already evaluated the expression here. So we are going to use this expression in the equation that we have just written. So we write Pb is t by r0 we have seen this earlier plus the perturbation pressure is equal to t times divergence of n evaluated at r0 plus eta. That we have seen earlier is just this 1 by r0 minus 1 by r0 square plus k square the same thing. And of course there is a complex conjugate which has to be added which I am not writing explicitly. Now you can see that the first term here gets cancelled by the first term here that is just subtracting out the base state from our equation so that we are left with an equation for the perturbation. The perturbation pressure is just this expression. So it is e cos kz plus f sin kz that part is common and then we will have a e to the power i omega t and then the t can be pulled out. And what we have here is k square minus 1 by r0 square. So this is my expression for the perturbation pressure. We have seen the linearized Bernoulli equation earlier we have already written it down in the previous class. So the linearized Bernoulli equation is just this you can see that the quadratic term is missing and that is because we have linearized it. And in particular notice that the Bernoulli constant here is not 0 and this is because the this is the linearized Bernoulli equation applied at the free surface. And so the Bernoulli constant has been determined by applying the Bernoulli expression in the base state. In the base state the velocities are 0 and so we are just left with p b by rho. This quantity is not 0 in the base state. So using that equation we obtain so I can split the pressure into two parts. This term is also applied at r0 but this term is anyway a constant the first term is anyway a constant. So I can skip writing the r is equal to r0 is equal to p b by rho the Bernoulli constant. Now this and this will cancel each other p v is just a constant and so I obtain p at r is equal to r0 plus eta is minus rho del phi by del t at r is equal to r0 plus eta. Now while proceeding further we will have to remember that we have to do a Taylor series expansion in order to decide whether these expressions are to be evaluated at r0 plus eta or at r0. You can see that p small p is a perturbation pressure phi is also a perturbation velocity potential. So like before we will have to write these as so p at r0 plus eta can be written in a Taylor series as p at r0 plus del p by del r also evaluated at r0 into eta plus so on. You can see that this quantity is an order epsilon square quantity because p itself is order epsilon because this is a perturbation pressure and eta is also a surface perturbation. So if I had not done it properly with non-dimensionalization expansion would start as epsilon eta 1 plus epsilon square eta 2 epsilon p1 plus epsilon square p2 and so on. So this would give me an order epsilon term so which is neglected in a linear theory. Similarly you can see that this quantity also has to be expanded in a Taylor series and so it is clear that this equation in the linear approximation will be just applied at the undisturbed interface which in the base state is just small r is equal to capital r0. This is the same as what we had done earlier. So we obtain at r is equal to r0 this is the linearized version of the Bernoulli equation is minus rho del phi by del t at r is equal to r0. We already have the expressions for p and phi so we can plug them in and if you do that then you will get the expression cos kz plus f sin kz for i omega t into surface tension into k square minus 1 by r0 square is equal to we have to take a del phi by del t we already have the expression for phi earlier we have already written it earlier. So at the top of this slide you can see that we have already written the expression and so we have to just to take the time derivative of this expression it will lead to an i omega and then the rest of the part remains the same and we have to remember that we are applying this at small r is equal to capital r0. So with that we obtain minus rho i omega into a cos kz plus b sin kz into i0 now the small r has to be applied at capital r0 because of this into e to the power i omega t and of course we are suppressing the complex conjugate part. Now I like before we can combine all the all the terms which have a cos kz and all the terms which have a sin kz and later we will equate the coefficients to 0. If I do that then I obtain rho i omega i0 of kr0 into a plus t into k square minus 1 by r0 square e this whole thing multiplies cos kz plus a similar thing multiplying sin kz plus e into k square minus 1 by r0 square to f sin kz and this whole thing of course gets multiplied by e to the power i omega t plus complex conjugate is equal to 0. So this is my equation 1 which has come from the linearized Bernoulli equation. It was slightly more complicated than last time because in this problem my base state has a curvature. In the last problem that we have seen so far in the base state the interface was flat there was no curvature in the base state and so the calculations were a little bit easier. Here the base state has a curvature the natural coordinate system here is a cylindrical coordinate system so expressions get a bit longer and we have to deal with modified Bessel function in the radial direction. So this is my equation 1 we now will go back to the second equation which is our kinematic boundary condition. The kinematic boundary condition I have already written at the top of this page we have already written the kinematic boundary condition in this linearized form. Once again this Delphi by Dieler can be expanded in a Taylor series and you can convince yourself that this has to be applied at r is equal to r0 and not r equal to r0 plus eta because that would contain a non-linear contribution. So the equation for the kinematic boundary condition leads us to another equation. I am going to leave it for you to work out that equation and I am going to write down the final answer. The final answer is just this. I will tell you where I got the i1 from in a moment let me write it down and this whole thing gets multiplied by e to the power i omega t plus complex conjugate is equal to 0 and this is my equation 2. Now where did I get these i1s from? Note that the kinematic boundary condition contains a derivative of phi with respect to r. Phi, the radial part of phi contains an i0. So we have to do this derivative d by dr of i0 of k r. It is easy if you this derivative becomes easier if we express numerator and denominator in terms of k r. So I am multiplying and dividing by k r so that this actually becomes k d by dr bar of i0 which is a function of r bar. This actually turns out to be i1. The modified Bessel function, the first order modified Bessel function of the first kint. The zeroth order was i0. The first order is i1. You can see from the graphs that I had drawn earlier that i1 is just related to the slope of i0 because i0 is shaped like that i1 in that range is always going to be positive. So that is how I get my i1 and then we have to substitute r is equal to small r is equal to r0 which is why so this additional k that I am getting is why we have this additional factor of k here and here. So those are my two equations 1 and 2. Like before we will have to set the coefficients of sin and cos in both the equations to 0. We will be left with four equations in four unknowns A, B, E and F. I am going to write down those four equations and then straight away write the dispersion relation which will come by setting the determinant. Those will be linear homogeneous equations in A, B, E and F. We have done this a few times now. So it should be easy for you to follow this. So we will have the four equations r, r I have divided out by rho. So that becomes t by rho k square minus 1 by r naught square equal to 0. That is equation number 1 from the second part from equation 2. It has an i1. These are just the coefficients of sin kz and cos kz equal to 0. Once again we have to set the determinant equal to 0. The determinant you can write it down. The determinant you can so you can write it down as a matrix times the unknowns A, B, E, F and this is a homogeneous equation. So you can write this as 0, 0, 0, 0 and once again like before we do not want trivial solutions. So we do not want A, B, E, F all 0. For non-trivial solutions the determinant of this matrix has to be equal to 0 and that will lead us to the dispersion relation. You can see that is going to be a quartic in omega. So it will involve products of omegas but it can be factorized into a quadratic. So I am going to write down the dispersion relation and I leave the algebra to you. The algebra is not difficult. You just have to work it out. So the dispersion relation can be written in this form omega square i0 of kr0 minus pk by rho r0 square into k square r0 square minus into i1 of kr0 this whole square. So it factorizes into a quadratic and so this just tells me my dispersion relation which is tk by rho r0 square k square r0 square minus 1 into i1 of kr0 divided by i0 of kr0. So that is our dispersion relation. It can be written in a slightly more compact manner omega square is equal to. Note that kr0 is a non-dimensional quantity and so this I will write it as t by rho 0 r cube into so I am multiplying and dividing by r0. So that makes the denominator r0 cube and the numerator as kr0 and the rest of the expression remains the same. Notice the analogy of this. So I can write this as t by rho r0 cube. This is the part which has the dimensions of frequency squared. The rest of it is non-dimensional. So I can call it a sum function of kr0. Kr0 is a non-dimensional argument, f itself is a non-dimensional function. So you can see that this analogy, so this is our dispersion relation and here in this case f is defined as kr0 k square r0 square minus 1 i1 of kr0 divided by i0 of. Notice the analogy of this with what we had done earlier for pure capillary waves on a pool described by inner rectangular Cartesian geometry. There we had found rho r0 cube into some function g of kr0. If the pool was a finite depth or rather in that case it would be g of k h t k cube by rho we had found and this is some non-dimensional function and in this case g had turned out to be tan hyperbolic k h. Notice the analogy purely from dimensional arguments. You can see that whether we are solving for capillary waves on deep water or capillary waves on a cylindrical filament. We can always argue that omega square divided by something which has the dimensions of frequency squared. In this case it is t by rho r0 cube, t by rho into some quantity which has the dimensions of 1 by length cube. So it could be k cube in the Cartesian case or 1 by r0 cube in the cylindrical case. So this is a non-dimensional quantity. So this must be a function of another non-dimensional quantity. In the rectangular pool it was k h in the cylindrical case it is kr0. So this much you can anticipate from dimensional reasoning. What precisely is the functional form of f or functional form of g cannot be inferred from dimensional reasoning and one has to do a detailed calculation to figure out what is the functional dependency of f on kr0 or g the function g on the non-dimensional combination k into h. So now let us analyze this dispersion relation. So this is our dispersion relation for perturbations of wave number k. So this is the dispersion relation. Let us analyze this dispersion relation and we will find that there is something interesting about this dispersion relation which was not there in the earlier examples that we have seen so far. Analysis of the dispersion relation. So we have seen that omega square is equal to T by rho r0 cube into kr0 into case kr0 square minus 1 into i1 of kr0 divided by i0 of kr0. Recall that I had told earlier that the plot of i0 as a function of some non-dimensional argument x if we plot then it goes like this. We have also seen that di0 of by dx if i0 is a function of some non-dimensional variable x is just i1. The way this is curved you can see that i0 of x is positive and its slope is also positive. So i0 of kr0 is always going to be positive in the interval kr0 between 0 and infinity. i1 of kr0 which is just the derivative of i0 with respect to small r and then evaluated at kr0 is also going to be positive for any value of k. And so this part of the expression is always positive. These quantities are anyway positive. However, you can see that the quantity which in between can be negative. This can be negative. Why are we interested in negative values? We are interested in negative values because on the left hand side we have omega square. If we have the square of a quantity being negative then the quantity can become complex. In this case it will become purely imaginary when the right hand side becomes negative. So what is the criteria for the right hand side becoming negative? The right hand side becomes negative when so if kr0 is greater than 1 then omega square is greater than 0. You recall that we have done a normal mode analysis where we substituted e to the power i omega t. So I expect omega square to be real and omega square to be greater than 0. So I am going to get oscillations or waves when omega square is greater than 0. So when kr0 is greater than 1 then it leads to waves or oscillations. These are capillary waves. We have looked at the standing waveform. You can also put traveling waveform and you will recover exactly the same dispersion relation. Now let us ask the question what happens when kr0 is less than 1. If kr0 is less than 1 then omega square is clear is less than 0 because all the other parts are positive and this part in red here is the only part which becomes negative. So omega square is negative which implies that omega becomes a purely imaginary quantity. Now this has consequences. You can recall that we had done e to the power i omega t. If I set omega to be a purely imaginary quantity I would write it as i times some real quantity omega i. This implies omega square is minus omega i. So you can see that e to the power i omega t there is a square here. So this would give me e to the power i into i omega i into t. And so this is giving me e to the power minus omega i t. And so if I have an omega which is less than 0 this term is going to go to infinity as time goes to infinity. We will see that this is indeed what happens when this relation is satisfied. So this the fact that we have this quantity diverging to infinity implies that instead of oscillations about the base state we get an exponentially growing we put a perturbation and the amplitude of the perturbation grows exponentially in time. So this is what is known as an instability until now we have not looked at any instability. We have only looked at situations where the base state was stable. So if I introduced a perturbation about the base state it would oscillate about the base state. This is because the perturbation has a restoring force which wants it to bring it back to the base state. But when it arrives at the base state it arrives with a nonzero inertia. So there is an overshoot and the moment there is an overshoot there is again a nonzero restoring force which again tries to bring it back to the base state. So this interplay between inertia and restoring force leads to oscillations and we have seen a number of these examples so far both discrete mechanical systems as well as fluid interfaces where we found dispersion relation governing the frequencies of those oscillations. This is the first example where we are finding that the dispersion relation contains a particular case where which if satisfied can cause instead of oscillations can cause exponential growth in time. We will try to understand the meaning of this and including the physical reason why this exponential growth happens. But first let us analyze the growth in a little bit more detail. We are going to do right omega is equal to i times omega i in this dispersion relation. I am going to substitute it here and work out. So this omega i clearly will tell me something about the rate at which things are growing. We will see that it will give us a quadratic equation for omega i. There will be a positive value of omega i and a negative value of omega i. The negative value is of interest because as I said here the negative value of omega i will is what will give me a divergence in time. The positive value will decay to 0 exponentially fast. We will continue this in the next class.