 Hello, welcome back, we will continue our discussion of the linear instability analysis of a two fluid interface. Let us quickly recap what we have done thus far. We are looking at the stability of an interface formed between two fluids say I have a fluid of density rho 1 superimposed over another fluid of density rho 2. The fluid of density rho 1 is moving with a velocity u 1. The fluid of density rho 2 is moving with a velocity u 2. Rho 1 rho 2 u 1 u 2 are all constants. We started with our basic governing equations and boundary conditions and we quickly substituted this base flow given by rho 1 rho 2 u 1 u 2 into the governing equations plus the boundary conditions to check that it satisfies the equations as it does. Then we introduce a small perturbation to this base flow field which we called we denoted by the primed quantities u i prime, v i prime, p i prime, etc. And we substituted the perturbed flow fields into the governing equation and boundary conditions and linearized the flow. So the first step was to linearize the flow field about this perturbation about this base flow field. So that gives us a set of linear partial differential equations that will describe the growth or decay of the perturbed quantity. The next step that we followed was to assume a particular form for the perturbed quantity called the normal mode form. We wrote these in equations 14 to 16 and we are now at a point where we know we have our linearized governing equations in equations 8 to 10 and the normal mode form in equations 14 to 16. So we begin by our solution process allowing us to substitute. When we perform the substitution what we get? We get these 3 equations. The first one coming from the continuity equation, the second one from the x momentum equation and the third one from the y momentum equation just for our reference. The double prime quantities are only functions of y just so we recall and that is the reason the derivatives of the double prime quantities with respect to y remain but with respect to either x or time are simply converted into an algebraic multiplication with either ik or omega or as the case with the advection term omega plus ik ui. So from these equations we are now able to we want to get a single ordinary differential equation if we can for pi double prime by eliminating ui double prime and vi double prime. So let us see what that process looks like. So we are able to write down ui and vi, ui double prime is minus ik over rho i times pi double prime divided by omega plus ik ui and vi double prime from 18 or from 19 is given by this. Now I can use these 2 equations and substitute into 17. So in order to do that I need to take the derivative of the equation for vi double prime with respect to y and that is what I have done here. Now when I substitute the equation substitute this equation and this equation into 17 here which is what we got from continuity. So for in short I can write this as this is our equation 20. Now remember i equals 1 or 2 we have maintained that notation all the way through from the beginning of this derivation. So essentially p1 and p2 are the static pressures in fluids 1 and fluid 2 and they are independent they are separate. So I am going to solve for them separately and from there I will find okay. So that is our location of the 2 fluids 1 and 2. One is located in the region y greater than 0 and 2 is located in the region y less than 0. So this is the most general form of the solution of equation 20. Now if you think of fluid 1 for a moment for all y greater than 0 and remember k is a positive number in our temporal linear instability analysis k is a positive real number. So if k is a positive real number then this first term here C11 e power ky is going to become increasingly larger as y increases but the static pressure in fluid 1 especially noting that these are all little p these are all lower case p meaning they are perturbation quantities. The perturbation quantities have to remain bounded. So for the perturbation quantities to remain bounded and this to be the solution the only possibility is that C11 has to be 0. C21 not being 0 is okay because the e power minus ky function decays away exponentially for all y greater than 0. So C21 does not have any constraint on it but C11 has to become 0. Similarly using similar arguments for p2 double prime y and noting that fluid 2 is located in the region y less than 0. So if y is less than 0 k still being a positive number e power ky is going to decay exponentially as y becomes an increasingly negative number that is y becomes y goes from minus 2 to minus 3 dot dot dot to minus 100 etc. As y becomes an increasingly negative number e power minus ky is already bounded but that is not the case with e power ky is bounded but that is not the case with e power minus ky. e power minus ky now becomes unbounded. So therefore in order for the whole static pressure to remain bounded C22 must be equal to 0. So now we are in a position to write down the general form of the static pressure in fluid 1 and 2 which is our single prime quantity. From here now we already have a relationship between ui and pi prime ui double prime equal to minus ik over rho i times pi double prime divided by the omega plus ik ui. So using that we will be able to write down what ui double prime and is supposed to be and similarly I can write u2 double prime I am not going to write it down but you can see that once I know u1, u2 and I can also use the other equation I can use this equation to get to what vi prime would have to be. So now that we have u1 and v1 it is also write down and v1 double prime would be analogous. So now we have the flow fields and we have two undetermined constants C12 and C21. So let us quickly take stock of that and we will use the two boundary conditions that we have the kinematic and the dynamic two sets of boundary conditions the kinematic and the dynamic boundary conditions to calculate the value for C12 and C21 such that we have a non-trivial solution for the problem that is our goal. We do not want C12 and C21 to become 0 because if they become 0 then we are left with the trivial case that the base flow alone persists. We want to introduce a perturbation and study whether the perturbation will grow or decay. For that purpose we do need C12 and C21 to remain non-zero. So let us now write down our kinematic boundary condition just to recall a kinematic boundary condition is basically a requirement that if I have a fluid particle on the interface. So if I have a fluid particle sitting on this interface that fluid particle remains on that interface at all points in time. So if I have a meniscus kind of like that and if this fluid particle is sitting here whether the fluid particle is translating in whether the movement of the fluid is to the left or right or up or down the fluid particle is not detached from the interface that is pretty much the physical meaning of what we call our kinematic boundary condition. So let us write that down what that says is v1 prime evaluated at y equal to 0 okay. We have the same normal mode perturbation introduced to eta. Eta happens to be the actual physical perturbation that you have introduced to the wave. So if you have eta to be of some having some amplitude eta not and some temporal frequency omega like we have said before and a wave number k then the requirement that the y direction velocity in fluid 1. v1 prime as we approach y equal to 0 from fluid 1 which is given which is basically denoted by this quantity v1 prime at y equal to 0 is equal to the physical movement of the interface or physical movement of the interface in an advected sense. So dou eta dou t plus u1 dou eta dou x is basically an advection of that interface with the velocity u1. u1 happens to be the fluid velocity the base flow velocity in fluid 1. So once we figure this out now substituting what we have for v1 into this we get k times c21. So simplifying we see k times c21 equal to rho 1 times omega plus ik u1 squared times eta not. Similarly we have another kinematic boundary condition requirement for fluid 2 and from there we have essentially the same requirement but written now from the sense of fluid 2. Now substituting what we have for v2 prime in here we get minus k c12 again simplifying k times c12 equal to minus rho 2. So we have c1 c2 and we have eta not which is the amplitude the initial amplitude of the perturbation that was introduced for nonzero eta not we want nonzero c1 and c2 that is our goal like we said before. So that we have one last boundary condition this is basically a force balance at the interface. We call this our dynamic boundary condition let us rewrite it quickly says p1 prime minus p2 prime equals minus sigma. So if I have p1 prime and p2 prime in terms of c1 c2 and if I make that substitution what do I get c21 minus c12 simplifying. So we have equations just as a quick recap 21, 21, 22 and 23 are homogeneous equations and we are looking for we have three equations 21, 22 and 23 in three variables c12, c21 and eta not and we want to find a solution that is non-trivial where neither of these quantities goes to 0. So there are two ways of doing this I can look at these three equations and any time I have three equations in three variables the only time you have a non-trivial solution is when the determinant of the coefficients goes to 0 that is one way of solving it and that is probably the most general way of solving it. I will restrict myself to a simpler case because I have essentially from 21 and 22 a closed form solution for c21 and c12 respectively. So I will just substitute for c21 and c12. So essentially we are able to eliminate c1 and c2 by a substitution and we find that we have an equation for eta not and the only time this equation in eta not has a non-trivial solution for eta not is when the rest of the coefficients are equated. So that gives us this equation here which we call our dispersion relation. This is important essentially what we have done is we found a condition that is written kind of nice and elegant here that rho1 omega plus iku1 squared plus rho2 omega plus iku2 squared equals minus sigma k cube. For a given k we can use this equation to find omega you notice that I have a quadratic equation in omega. So each for each k value of course given u1 and u2 rho1 and rho2 you have two roots in omega. Now in general if k is real omega equals omega12 are complex. So in general I have two roots omega1 and omega2 which are in general complex. So if I take if the real part of omega1 comma2 is greater than 0 if the real part of either omega1 or omega2 does not have to be both if real part of either omega1 or omega2 is greater than 0 then the introduced in perturbation of wave number k will grow. So if the converse being if the real part of either omega1 or omega2 is less than 0 then the instability corresponding to that particular wave number k decays. For the decaying condition the real part of both omega1 and omega2 must be negative. So if either one of them is positive then the instability is likely to grow. So we will now use this or base dispersion relations simplified to see if we can if we can understand this. As you see rho1 rho2 u1 u2 given we have a quadratic in omega. In general I can write this equation down in a after expanding out the squares in this form. So this is simply the dispersion relation written out as a quadratic of the form a omega squared plus b omega plus c equal to 0. Now we are able to solve this equation closed form because we have a formula and in here I know a equals rho1 plus rho2 b equals I am going to not do the algebra here we will just write down the solution you can check this at your convenience. So this is a closed form equation that would yield us two omegas omega1,2 the one corresponding to the positive sign and two corresponding to the negative sign and the term under the radical of course coming from our discriminant b square minus 4ac. Now you notice that you know we remarked that if the real part of omega is positive that is when you are likely to have an unstable perturbation. So in this situation if you notice the first part here has an imaginary is purely imaginary which means this part can never be real the all the real part can only come from the discriminant under the radical. So for omega to be greater than 0 or for real part of omega to be greater than 0 we know k squared times rho1 rho2 has to be greater than 0. So leaving out the case where k is which is trivial we are able to show that rho1 rho2 times u1 minus u2 the squared has to be greater than k times sigma times rho1 plus rho2. I want you to I want to draw your attention to two observations here. First thing is that if I interchange the subscripts 1 and 2 in this equation the equation does not change. So rho1 becoming rho2 and rho2 becoming rho1 with u1 becoming u2 and u2 becoming u1 equation is not changed. That basically means that the condition for instability is not governed by the absolute values of the of the densities or the velocities and that we have no preferential direction whether 1 is super post on 2 or 2 is super post on 1 which is which should be obvious. The second point I want to draw your attention to is that it is also symmetric in the velocity. So whether I simply replace u1 with u2 or u2 with u1 the only quantity that matters is u1 minus u2 the magnitude of the relative velocity u1 minus u2 not the absolute values u1 or u2. So based on this we can set either u1 or u2 to 0 without loss of generality essentially assuming that the frame of reference is moving with a velocity of with one of the fluids either u1 or u2 and the condition still is remains unaltered. So if I rewrite this so this is the condition for instability that all k for a given rho1 rho2 u1 u2 and sigma all k less than this would have a positive value of the real part of omega. So if this quantity is what we will call our k cut off it is always good to have some numbers. So let us just pick some set of realistic numbers all these are in SI units for this case I can quickly calculate k cut off to be about 13.93 or if I want to convert back to lambda so essentially if I have a breeze of 1 meter per second blowing on water I have assumed rho2 to be 1000 and rho1 to be 1. So if I have a breeze of 1 meter per second blowing on water at about water which is at rest u2 is 0 I am expecting waves whose wavelength is roughly about half a meter and you will notice that as u1 increases k cut off increases which means lambda cut off decreases. Now remember this is not this is the lambda cut off that is you will not see waves shorter than 0.45 meters that is the meaning of lambda cut off k cut off being 13.93. Now so this is essentially information that we get from linear instability analysis which is purely an analytical calculation but leads us to draw some inferences in the physical space of the kinds of waves that we would observe if we had a particular flow condition velocities and densities. We will stop here and continue in the next lecture.