 We had started looking at interfacial waves. Recall that our big state was one where the interface was flat and the fluid underneath was quiescent. We have started our analysis by making a number of simplifying assumptions. Among them was first we had assumed that the medium would be incompressible. This would be true when the maximum speed of gravity waves we were looking at surface gravity waves to start with would be much lower than the speed of sound. Then we also looked at the then we also made the assumption that the medium would be assumed to be inviscid, the motion would be irrotational and the domain would be horizontally and vertically unmounted. And as a first step we would neglect surface tension. We mentioned that we would revisit the horizontal and vertical unmounted assumption later on and we would also include surface tension later in our analysis. Now with all these approximations we were led to the Laplace equation for the velocity potential phi. The pressure field was governed by an unsteady Bernoulli equation. We also had boundary conditions. Among them was a new boundary condition that we learnt how to derive. It was called the kinematic boundary condition. It essentially is a statement that the interface or the free surface is a material surface and so it is an additional statement of mass conservation. We looked at the kinematic boundary condition and then we also looked at the pressure boundary condition. Here we said that we are ignore, we are going to ignore the gaseous medium above. So we are going to ignore the air as a first step. And so the motion in the air is negligible. So the air is quiescent and it only exerts a pressure on the fluid below it. That pressure can be assumed to be 0 to start with. Then we looked at the boundary conditions and then we non dimensionalized our system. Our choice of length scale was K inverse. K is a typical wave number of a interfacial perturbation that we would put on the system. Then we chose the time scale as square root gk to the power half or rather this was the frequency scale. And I said that we would revisit these scales later on and try to understand them meaningfully. So then we plug these scales into the governing equations and we obtained our non dimensional set of equations. Then our small parameter here was epsilon. The product of a typical amplitude a0 into the wave number k and we expanded as a regular perturbation in epsilon. We plugged this in into our governing equations and boundary conditions and we obtained a set of equations at order epsilon. We have to remember here that the order 1 problem here is trivial because the base state is quiescent. The only non trivial variable which has an order 1 contribution is pressure here. But in the way we have written down things we can first solve for velocity potential and the interface and from there once we have determined these two we can extract the pressure field because from the unsteady Bernoulli equation. So let us proceed with what we had obtained at order epsilon. So we had obtained these set of equations at order epsilon. In addition we had also seen that at order epsilon our boundary conditions were linearized and because in particular we had two boundary conditions at the interface. One was coming from the fact that the Bernoulli equation in the Bernoulli equation we said pressure is equal to 0 at the interface. This led to one boundary condition and the second was a kinematic boundary condition. Both of these boundary conditions were simplified by the linearized analysis by virtue of the fact that using a Taylor series approximation we applied the boundary conditions instead of applying it at the unknown boundary z is equal to eta. We applied it at the known boundary as the first term in the Taylor series approximation. So you can see that all the terms del phi 1 by del z is applied at z is equal to 0 and del phi 1 by del t is also applied at z is equal to 0. Eta 1 is not a function of z and so we do not have to worry about the z dependence there. In addition we also said that we need because our domain is horizontally unbounded it goes from minus infinity to plus infinity the variable x and the variable z goes from eta at the top to minus infinity as we go deeper and deeper in the fluid. So we have to ensure that when in an unbounded domain whatever functional dependence we find out that does not diverge as we go to arbitrarily large depths or arbitrarily large horizontal distances. So we have these finiteness conditions. Now with having set up all these equations let us now write down the solution to these set of equations. So our first set of equation is the Laplace equation and we are going to use once again variable separation. We have met this technique once when we looked at the 2D vibrations of a rectangular membrane. We have done this earlier in the course. We are going to do the same procedure here and we are also going to do a normal mode analysis. The base state is quiescent. So I am just going to look for oscillatory solutions about the quiescent base state. Let us first work on the Laplace equation. So phi 1 I am going to write it as some function capital X of small x some function capital Z of small z into e to the power i omega t. Note that this omega is non-dimensional. This is because t is non-dimensional. We are in a non-dimensional framework here. So now our main purpose is to go back and substitute this form of phi 1 into the Laplace equation. Once we do that we get back ordinary differential equations for capital X and capital Z. We have seen how to do this before. So I can write it like this x double prime by x all the x dependencies are collected on one side. Then we have minus z double prime by z. I am just substituting in the Laplace equation. The e to the power i omega t just cancels out. It will not appear anywhere. And then I have to put a separation constant. I am going to choose the separation constant to be minus 1. Why am I choosing it to be minus 1? Recall that we are doing this in non-dimensional framework. So it should be ideally minus some quantity squared. Here it would have been minus k squared. k is a typical wave number that we have introduced while doing our non-dimensionalization. We have non-dimensionalized all our length scales by k. So you can see that if we wrote down the corresponding dimensional version of this, it would just be equivalent to setting the separation constant to minus k square. Why in the negative sign? We have met this negative sign before. In this case, you can justify the negative sign from the fact that if you have a negative sign here, then the equation for x would have would be x double prime plus x is equal to 0. The corresponding equation for z would be z double prime minus z is equal to 0. You can see why we are taking a negative sign. If I put a negative sign here, then it ensures that the solutions for x are oscillatory. Recall that small x goes from minus infinity to plus infinity. If we choose a positive separation constant, let us say plus 1, then the solutions for x would be exponential with a plus x and a minus x. So e to the power plus x and e to the power minus x and it would be a linear combination of the two. Because our domain is unmounded on at both ends, so both the exponentials are going to diverge. So it is not possible to get a finite solution by eliminating any one of the constants of integration. In order to prevent that, we look for oscillatory solutions in the horizontal direction and in the vertical direction, this choice of separation constant will give us two exponentials. In the vertical direction, the choice of exponential is okay because our vertical domain is semi-unbounded. So we can eliminate one of the exponential which diverges as we go to minus infinity and keep the other exponential which will anyway decay to zero as we go deeper and deeper in the fluid. So with that argument, we can now solve x double prime and the solution for x is simple. So I will call it b1 cos x plus b2 sin x. So this is a function of small x and those are small x's. Similarly, b1 and b2 are constants of integration. Small z is equal to c1 e to the power kz plus c2 e to the power minus kz. As I argued before, we said c2 to 0 because this term goes to infinity as z goes to minus infinity. So as we go deeper and deeper in the fluid, recall that our coordinate system is like this, z goes to minus infinity. So this is x, this is positive direction of z and z goes to minus infinity. We are in the deep water approximation. Because k is greater than 0, k is a wave number. It is related to wavelength. So k is greater than 0. So e to the power minus kz for negative z as we make z more and more negative, e to the power minus kz will become larger and larger. So we have to set the constant of integration to 0. And so we are left with only a single constant in the z dependence. Now let us go further with these solutions. So we are writing phi1 is equal to b1. Let me put a, let me put a prime here and I will explain why am I putting a prime because I want to reserve b1 and b2 for something else. So with this approximation, we can write the expression for phi1 as and now you can see that I can multiply and then there is of course e to the power i omega t. And now you can see that this constant can be absorbed into b1 prime and b2 prime. So I am just going to say that b1 is equal to b1 prime into c1 and b2 is equal to b2 prime into c1. So this is why I introduced a prime so that I can use b1 and c1 from here onwards. Now let us, let us now go back to our equations and try to understand. We have now got a form of phi. We now need to anticipate the form of eta. You can see that both the boundary conditions which are indicated by horizontal arrows, one is the linearized kinematic boundary condition, the other is the Bernoulli equation linearized form applied at the interface, at the linearized interface which is z is equal to 0. Both the boundary conditions tell us there is no x derivative in both the boundary conditions. So they are telling us that the x dependence of phi1 is the same as the x dependence of eta1. Now with that observation, we can anticipate that eta1 is also going to have the same functional dependence as far as x is concerned. So eta1 is also going to be a linear combination of cos x and sin x. That is because the boundary conditions do not involve any derivatives with respect to x. So we can see this is a1 cos x. So I am introducing some instead of writing it is just b1 cos x and plus b2 sin x. I am introducing some new constants, but I am keeping the functional dependency the same. So it is a linear combination of cos x plus sin x, but the constants are now different a1 and a2 for eta, eta1. Remember that eta1 is only a function of x and t, there is no z dependence. So from the normal mode approximation, this e to the power i omega t. So these two are coming from normal modes. And our purpose is just like before, we have to go back and substitute this into the equations. We anticipate that they should lead us to an eigenvalue problem and when we plug it in, it should lead us to some kind of a matrix whose determinant is going to determine omega for us. Let us see how. So we will write, so our boundary conditions are del eta1 by del t. I am just rewriting the boundary conditions. This is the linearized kinematic boundary condition at z is equal to 0 is 0. And then we have, so I am going to call this equation 1. And then del phi1 by del t at z is equal to 0 plus eta1 is equal to 0. This is 2. If we now substitute these forms of phi1 and eta1 into equation 1 and 2, then we obtain. So 1 implies, so substituting the form of eta1 and phi1 into equation 1, we obtain i omega a1 cos x plus a2 sin x. That is the derivative of eta1 into e to the power i omega t plus its complex conjugate. We have to remember to add the complex conjugate because these a1, a2, b1, b2, these in general are complex constants as we have seen earlier. So I will call this complex conjugate 1 because there is one more term which is minus del phi1 by del z in equation 1. So when I do the second derivative, that just gives me the same because the z dependence is only e to the power z. So del phi1 by del z is exactly the same expression plus its complex conjugate. So this is Cc2. So Cc2 is the complex conjugate of the term which appears on its left hand side, Cc1 is also similarly. And this is equal to 0. I can put them all together and write them as a single equation. By collecting all the coefficients of cos x, this whole thing gets multiplied by e to the power i omega t plus the complex conjugate of whatever I have written in the bracket. Only the a1 and the b1 will typically be complex conjugated. You will see that omega in this case will turn out to be a purely real quantity. So let me call this equation 3. We are now done with equation 1 because we have substituted the forms of phi1 and eta1 into 1. Let us do the same for equation 2. If we do that by an analogous procedure, we substitute and we collect all the coefficients of cos x together and cos sin x together. So 2 implies it just gives us the following equation. So we will get i omega b1. It is del phi1 by del t. So i omega and the exponential part of z vanishes because there is no derivative with respect to z firstly and then it is applied at z is equal to 0. In the previous one also it was the same argument that we had del phi1 by del z applied at z is equal to 0. So the exponential part did not appear in equation 3. So similarly, we will write one more equation now. i omega b1 plus a1 into cos x plus its complex conjugate is equal to 0. You can obtain this very easily by just substituting the forms of phi1 and eta1 into equation 2 and then collecting all the terms together. So let us call this now equation 4. Now our task is to work on equation 3 and equation 4. Now if you look at the equation 3 and equation 4, firstly it is enough to just work on the term which appears and not worry about the complex conjugate part because that is just the cc of what is written here. So in both the expressions you will see that we have a linear combination of cos x and sin x multiplied by e to the power i omega t. We want this to be 0 at all times the entire expression. e to the power i omega t is not 0 at all times. So we have to look at the expression inside the square bracket. So you can readily see that what is inside the square bracket in both equation 3 and equation 4 is a linear combination of cos x and sin x because cos x and sin x are linearly independent. So we have to set the coefficients of this to 0 in order to satisfy equation 4. So we are led to 4 equations. Equation 3 implies, so each of the coefficients of cos x and sin x are 0. So equation 3 implies i omega a1 minus b1 is equal to 0. Equation 3 this is the coefficient of cos x that we wrote. Now let us write the coefficient of sin x that is also 0. Equation 4 the coefficient of cos x is just i omega b1 plus a1 is equal to 0. Equation 4 the coefficient of sin x is i omega b2 plus a2 is equal to 0. So we have 4 equations and it is really telling us something very interesting. So we can for example use this. So we have 4 unknowns a1, a2, b1, b2. You can make this into 2 unknowns for example by using this one or two of these equations to eliminate. So for example we can express b1 in terms of a1. You can pay attention that this the first and the second equation just tells us that b1 is equal to i omega a1 and b2 is equal to i omega a2. You can also see that the last 2 equations basically tell us. So for example if I take this equation then it tells us that i omega b1 is equal to minus a1. So this is really b1 is equal to minus, so if I multiply both sides by i omega then I get minus omega square here and minus i omega a1 and this tells me that omega b1 is equal to i a1 and so b1 is equal to i a1 by omega. So now you can immediately see that if you compare this with the first equation that I have written here then you can immediately see that the first equation tells us b1 is equal to i omega a1 and this equation tells us b1 is equal to i by omega into a1. If I substitute this then I obtain i omega a1 is equal to i omega a1 and this is telling me that omega square is equal to 1. This is the dispersion relation that we are basically finding. Now formally you can do this a little bit more formally by not taking any 2 of these equations but working on all 4 of them. Now formally you can do this by taking not 2 of these equations as we have seen but working on all 4 of them and convincing yourself that that essentially leads to the same dispersion relation. So what you can do is you can do this more formally by thinking of these 4 equations. So this equation, this equation, this equation and this equation as an equation in a1 a2 b1 b2. So I can write this as a set of homogeneous equations in a1 a2 b1 b2. So let us do that. So first we write the coefficient of a1 then a2 then b1 then b2. So for the first equation the coefficient of a1 is i omega a2 is 0, the coefficient of a2 is 0 then minus 1 and the coefficient of b2 is 0. Similarly in the second equation 0 i omega 0 minus 1. Third equation 1 0 i omega 0 and fourth equation 0 1 0 i omega and this is equal to because this is a homogeneous set. So everything is 0 on the right hand side. So that is our matrix whose determinant will determine the frequencies at which the system can oscillate. If you solve this matrix you will just find that this is just equivalent to omega 4 or if you write the determinant of this matrix rather this is just equal to omega 4 minus twice omega square plus 1 is equal to 0. This can be written as omega square minus 1 whole square is equal to 0 and so we have plus minus 1. So we find essentially the same answer as we had found by just looking at two of the equations. Let us now use these to write down the solutions. So we will have one of the frequencies is 1. Remember that this is a non-dimensional omega. So in scaled units this is 1 when we dimensionalize our expressions we will obtain the real dispersion relation. Let us now write it. So we have found now that eta 1 is equal to a 1 cos x plus a 2 sin x into e to the power i omega t. I am not yet writing that omega is 1 but it is understood that omega is 1. Phi 1 was b 1 cos x plus b 2 sin x e to the power i omega t plus cc. We can express using the four equations that we have found between a 1, a 2, b 1, b 2. We can express b 1 in terms of a 1 and b 2 in terms of a 2. If we do that then we get a i omega here and a 1 here. We are just using this equation b 1 is equal to i omega e 1 a 1 and b 2 is equal to i omega a 2 to express b 1 and b 2 in terms of a 1 and e 2. And so we have this. So I can write this further as we have now omega a 1 cos x plus a 2 sin x e to the power i omega t plus pi by 2 plus cc. I have absorbed the i here in the phi 1 as a phase e to the power i pi by 2 is alright. Now can we proceed in real notation? So let us try to make this into slightly more real notation. So if I remember that there are complex conjugates everywhere then this will become a 1 plus a 1 bar. This is familiar to you from our earlier exercise plus i times a 1 minus a 1 prime. This is cos t and this is sin t. I am now putting omega is equal to 1. So this we have recall that omega square is equal to 1 is our dispersion relation. So omega is equal to 1 and it is really plus minus 1 but I am already taking the minus into account in the complex conjugate. We have done this before and so I am writing e to the power i t as cos t plus i sin t. And then we will have a e to the power minus i omega t which is cos t minus i sin t. If you add up everything and collect the coefficient of cos x you will find that the coefficient of cos x is this and the coefficient of sin x is that. This is more complicated than the problems that we have done before because here we have a variable where there are three dependencies. There is a time dependency and there are two space dependencies x, z and then in addition there is a time dependency. In the earlier 2D vibration problem we also had a similar thing where the displacement of the membrane eta in that case I think we used as a function of x, y and t. Here eta is just a function of x, t phi 1 is a function of x, z and t. Again I am substituting omega is equal to 1 sin x. Once again you can see that this is completely real because a1 plus a1 bar is real, i times a1 minus a1 bar is also real. And so this is completely real. One can similarly write down an expression for phi 1, phi 1 would have a e to the power z. So I have missed e to the power z here and similarly a1 plus a1 bar and in this case this would be cos t plus pi by 2. That is because there is a pi by 2 here, omega is again 1. So it is just cos t plus pi by 2 plus i a1 minus a1 bar sin t plus pi by 2. And we can work on those cos t plus pi by 2 and sin t plus pi by 2 later on. This whole thing multiplies cos x plus again e to the power z into and we can simplify this t to the power of t plus pi by 2, t plus pi by 2 to obtain the actual answer in terms of real variables.