 Okay, let us continue our discussion of instability theory specifically linear instability analysis. What we are going to do today is we are going to go through formally look at the stability of stratified fluids. So I have an interface between a fluid one and a fluid two, fluid one moving with a uniform velocity u1 fluid two moving with a uniform velocity u2 need to introduce a coordinate system to study this problem. So the velocity vector in one I will call this v uppercase is basically u1 ex and the velocity vector in two is u2 ex the y component of velocity in both these cases is 0. I am going to assume nu1 equal to nu2 equal to 0. So this is saying 0 viscosity. So let me formally write down the set of assumptions and initially the interface is flat. This is the unperturbed as we will see in just a moment at the position at the position y equal to 0. So this is my interface. Now in general each of these fluids one and two can have a velocity in the x and the y direction. So we will just also assume that the flow is two dimensional. So there is no flow in and out of this the plane of the paper neither is there a possibility of any flow. So at the moment there is no flow in the y direction but there is a possibility that things could start to flow in the y direction if they choose but there is no flow in the z direction right now then there is no possibility also of the flow to be initiated in the z direction. That is the meaning of two dimensional flow. So let us write down the governing equations and the boundary conditions for this flow problem. If V i is a velocity field and P i is a pressure field then this is the so i equal to 1 is the top fluid i equal to 2 is the bottom fluid. So I have one this is simply what we very often call the Euler's equations. I am going to write this out in full Cartesian form. So the x and of course I did not mention the continuity equation to go with this that says del dot V i equal to 0 which means let me go back add one more assumption that one fluids are incompressible. So row one and row two do not change with x and y or any other independent variable. So if I write these out in Cartesian form the vector V i is composed of components capital U i and capital V i. So V i being the y direction component in general. So this plus U i let me write this clearly. So this is the equation I will call this one this is two this is three. So let us first you know we have we started with a certain flow configuration that fluid one has a uniform x velocity of U one fluid two has a uniform x velocity of U two fluid one and two both have no y component of velocity. So in this so if I substitute to check if the flow described above is a solution what do we find if I set V i to 0 now I also know U i is constant in time. So from that so if D U i D t is 0 not changing in time this is 0 because it is not changing in x this is 0 because V i is 0. So let me write those down call this this is due to one two three reason one is nothing changing in time reason two. So we are solving each of these equations you have for each fluid you have one set of three equations describing the U i U 1 V 1 for that fluid U 2 V 2 for that fluid for the other fluid. So nothing changing with x and y in each fluid and the third is of course saying V i is 0 likewise V i is 0. So I can say this is due to three V i is 0 this is due to three. So this is 0 due to three this is 0 due to reason number two. So what have we what have these simplified to see I do not know what P i was supposed to do all I know is a flow field. So essentially what this now becomes is that D P i D x equal to 0 D P i D y equal to 0. So all that says is P i is a constant the pressure in each fluid everywhere in the domain is a constant it is not dependent on x and y that is basically what we are learning from saying D P i D x is 0 and D P i D y is also 0. Now I still have not talked of the boundary conditions let us see if what the boundary conditions are I will simply I do not want to go into the detailed boundary condition right now except to say that look I have a flat meniscus for a flat meniscus to remain flat for all times you can clearly see that the pressure in fluid one and the pressure in fluid two have to be equal. So the boundary condition at the interface can be written as this the simplest of boundary condition says that P 1 minus P 2 equal to sigma times kappa where sigma is a surface tension or more specifically we will call it interfacial tension and kappa is the curvature. Since the meniscus is flat everywhere since the unperturbed interface is flat it is easy to see that kappa has to be 0 everywhere therefore P 1 minus P 2 equal to 0 we now know P i is a constant the governing equation said P i in each phase is a constant the boundary condition says P 1 minus P 2 has to be 0 what this says is P i everywhere is equal to P 2 equal to some P. So physically it is just constant pressure everywhere ok so P is just a number this capital P is whatever is the actual magnitude of the pressure and that is the same pressure everywhere both in fluid one and in fluid two. So without loss of generality you will see this a lot so where I know that that particular number has no significance I can give it any convenient value of convenient value in this case I will assign P equal to 0. So if P capital P equal to 0 everywhere and V 1 equal to U 1 E x and V 2 the vector V 1 equal to U 1 E x V 1 equal to U 1 E x and V 2 equal to U 2 E x and P equal to 0. So V 1 equal to U 1 E x V 2 equal to U 2 E x and P 1 equal to P 2 equal to 0 is a solution to the governing equations ok that is just like saying that it is a system where forces are balanced in the ball in a trough example that we saw in the last class ok. So this particular flow configuration is an equilibrium solution is it a stable equilibrium or an unstable equilibrium that is the question that we now have to answer ok. How do we go about that now this interface is infinite in the x direction. So essentially imagine an interface of two super post fluids that has no start and an end that is the kind of interface that we are looking at ok. So if I now take this kind of an interface and super post a perturbation. So I have these two fluids that are co-flowing ok these two fluids are co-flowing and I want to understand if I perturb this interface slightly will the under the action of these co-flowing fluids top fluid moving at U 1 bottom fluid moving at U 2 and mind you we did not make any assumptions on whether U 1 is greater than U 2 or the other way around it does not matter in fact. So under the action of this co-flowing fluids does this meniscus amplitude decay or grow that is the only question that we are interested in asking ok. So if I keep everything else the same but perturb this interface slightly ok what that does is it introduces a perturbation velocity field ok. If I now perturb the interface as soon as I perturb the interface I have introduced a perturbation velocity field to the to both the fluids ok. So now if I look at V i this is still U i plus a little U i I will call this U i prime for perturbation times comma V i prime. So this is a vector I am denoting using ordered pair notation U i plus U i prime is the new velocity in the perturbed interface configuration. So I have introduced a small additional x component of velocity now mind you both the prime quantities U i prime and V i prime both for i equal to 1 and i equal to 2 are both what are called field variables. So these are velocity fields they are functions of x and y and time ok. So U i prime and V i prime are both velocity fields which were not there until I perturb the interface. Now because I have perturbed the interface I want to see what this velocity field would look like. Now whatever be U i prime and V i prime and let us even say the pressure P i is P plus some P i prime ok and P itself is 0 we know that ok. So the pressure in each fluid i equal to 1 is the top fluid i equal to 2 is the bottom fluid the pressure in each fluid is now given by some small quantity P i prime. The key point to note here is that U i prime V i prime and P i prime the magnitudes are much less than the mean flow quantities. So they are infinitesimally small ok this is our new velocity field and this is our new pressure field. These new velocity and pressure fields also have to be function have to be solutions to our unsteady Euler's equations. The reason the previous I showed that we showed that the we showed that the this is an equilibrium solution U 1 prime U i prime and U 2 being constant everywhere and P being 0 everywhere is an equilibrium solution because nothing is changing with time that is the idea of equilibrium it is steady in time. So if I now come to the perturbed case U i prime and V i prime could be functions of time all I want to know is does U i prime and V i prime do U i prime and V i prime decrease with time or do they increase with time that is what I am interested to find out. If the prime quantities increase with time from some small initial starting value that is not good for the stability of the interface ok. So first of all substituting I will call this equation 4 I will call this 4 A and B just to what do we have to do. So what do we find the first is the x momentum equation second is the y momentum and the third is the continuity equation we are quite familiar with that. So this is U i plus U i prime plus U i plus U i prime times d dx of U i plus U i prime plus V i prime d dy of U i plus U i prime equals minus 1 over rho i d pi prime dx this is the equivalent of equation 1 the y momentum equation becomes the last one is this. Now this capital U i is basically U 1 and U 2 those are not changing with x y or time the prime quantities are allowed to change with x y and time in fact ok. So we can simplify 5 6 and 7 using that what we find is first one I took this term here U i capital U i plus U i prime times d dx of U i plus U i prime U i itself is not varying with x. So d dx of capital U i is 0 but d dx of U i prime is not 0. So that is left in the calculation still but I have U i that term being multiplied by U i plus U i prime I just want to write it out as two separate terms. So I have U i times d U i prime d x plus U i prime d U i prime d x plus I also have the additional term V i prime d U i prime d y. So what do I find here I can do I can write the other one as well d V i prime plus U i d V i prime d x plus U i prime d V i prime d x plus V i prime d V i prime d y equals minus 1 over rho i d P i prime d y. The last term is simply do U i prime d x plus d V i prime d y equal to 0. Now remember U i prime is a small number. In comparison with capital U i so I want some way to compare the magnitudes of these four terms on the right hand side of the top equation on the left hand side of the top equation. So if I there are four terms on the left hand side of the top equation if U i capital U i is order 1. So it is like 1 meter per second let us just say then d U i d t is order U i prime. So that is order epsilon it is like a small number in comparison to 1. This the first term is order epsilon the second term is order epsilon because it is U i capital U i is order 1 U i prime is order epsilon. So the second term as a whole is order epsilon the third and fourth terms are order epsilon squared. So if U i if these quantities are all order epsilon, epsilon is some small number. Then what it automatically means is that these two terms have a two prime quantities multiplying each other which means that the magnitudes of these two terms that I have scratched out in this particular equation are very very small in comparison to the terms that I have left. So the error that I will end up making by completely scratching them out of my equation is very small and that smallness becomes smaller and smaller as epsilon goes further and further towards 0. This is the idea of an infinitesimal perturbation. It is so small that epsilon squared is you can ignore epsilon squared in favor of order epsilon and it is a small it is as small a number as possible essentially. Now if I give that much of a perturbation is that perturbation further going to grow or decay that is all I am interested in. I just want to know a direction to the growth that is all. So I can ignore these two terms which are now order epsilon squared in favor of the other three terms that I have remaining. I can do the same thing with this the third equation the both the terms in there are order epsilon. So I really cannot throw away anything. So what I can now do I can rewrite the equations without the terms that I scratched out. So completely throwing away the terms I scratched out in that fashion. I can number these equations now 8, 9, 10 have are important. Now how do I know that they are linearized versions of the governing equation. I want to make sure that you completely understand this idea of linearized version. If you look at these equations they are three equations the three unknowns are u i prime, v i prime and p i prime. I have one pair of three one set of three equations for the fluid one one pair one set of three equations for fluid two. There are no terms in these equations that are like u i prime squared or v i prime squared even though our original Euler's equations have terms like u d u d x. What we have done is we have linearized the complete Euler's equations about a state that I already know is a solution to the Euler's equation capital u i being constant both in x y and time. E is a solution to the Euler's equation we found that and we have written a linearized version of the Euler's equation around that ground state if you will of u i being some number like u 1 is a number u 2 is a number. So, the u i the capital u i in the equations 8 and 9 is just a number that you already know the only unknowns are the prime quantities u i prime, v i prime and p i prime and these are field variables their functions of x y and time. Therefore, these equations since they do not involve any terms that are non-linear in the prime quantities this set of governing equations are linear. So, I have a system of linear partial differential equations describing the prime quantities. So, now how do I go forward from here I am going to we started to say that the we are going to introduce a perturbation to the interface in the simplest form of perturbation is a sinusoidal perturbation. We will see may be later on why a sinusoidal perturbation is introduced, but essentially what we want is a set of harmonic functions a set of orthonormal basis functions of this of the governing equations of the linearized version of the governing equations that I can use to take any arbitrary interface shape and express as a series of those orthonormal basis functions. This is simply going back to solutions of linear ODE is in fact not even linear PDE's right. So, our idea of coming up with a basis function for this particular geometry that gives me a set of orthonormal basis functions that gives me a set of basis functions in which I can expand any arbitrary interface shape. For this particular case it happens to be a sinusoidal function. So, I will express this as a functional form y equal to eta 0 times e power omega t plus i k x. So, here is I will express this I will write this out a little later on. Now, where did I get this functional form? I want to make sure we understand that. If I expand this out I have eta 0 e power omega t times e power i k x which I can write as cos k x plus i sin k x. If y is a real number then essentially the real part of this is nothing but e power eta 0 e power omega t cos k x. So, I have allowed k k is like is a wave number. So, we will see in just a moment what that is. If I take essentially if lambda is the wavelength then this can be written as eta 0 e power omega t cosine of 2 pi x over lambda. So, when x equal to lambda the phase angle of this wave is equal to 2 pi when x equal to 0 the phase angle is 0. The phase angle goes smoothly as a linear function of the distance from this point here. So, this is a and if I define k to be equal to 2 pi by lambda k is a wave number and has units of per meter. So, we will work in this wave number space much it is easier to work in the wave number space then it is to work in the wavelength space. So, we will define this wave number k equal to 2 pi by lambda and 2 pi by lambda times x in the cosine is a way to describe a sinusoidal or a cosine sinusoidal interface sin and cosine are only off by a phase angle of pi by 2. And on an infinitely long sheet in the x direction it does not matter at the end of the day correct. So, here is a description of this interface now what is eta 0 eta 0 is that amplitude. Now, like I said the only feature of this feature I am interested in is if I impose a disturbance of wavelength lambda or now wave number k is that disturbance going to grow in time or decay in time that is the only part that I am interested in. And we use this function to capture that behavior if omega is a let us just say omega is a real positive number. If omega is real and positive then e power omega t is a number that grows as t increases. If omega is a real negative number then e power omega t is a number that decreases as t increases. So, for a very in the short period of time around t equal to 0 all I want to know is if I give a certain lambda does the omega for this system come out to be a positive real number or a negative real number. In fact, I will go back to now. So, this is the physics of what we are going to look for. So, or rather the mathematics of what we are going to look for, but I find it easy instead of sines and cosines to work in this exponential notation with imaginary with complex numbers. So, we know that I is this number square root of minus 1 and e power i k x is simply cos k x plus i sin k x. So, we are not going to really make the distinction between real and imaginary numbers. We will just say that omega is a complex number, omega could be a complex number, but all I care is whether the real part of omega is positive or negative. So, let us take this if omega is a complex number like that e power omega t is e power omega r t times cosine of omega i t times plus i sin omega i t. So, really speaking I can let omega be any kind of a it can be a full complex number. The real part of omega is going to basically determine if the disturbance is going to grow in time or decay in time. The imaginary part of that omega is only going to cause a sort of a ripple effect. Omega i is like a imaginary part it is going to cause a wave to be transported in time, but the amplitude of the wave is not affected by that. Amplitude is only affected by this e power omega r t. So, think of it this way k is a real number. I am going to force k to be a real number which means I am giving a real perturbation to the interface. I perturb this interface with some waves of total wavelength lambda or wave number k and I let go of this template that I used to create this wavy interface. As soon as I have let go of this interface is the interface amplitude this eta 0 going to decay or grow further. That is the only way that I care about and that is mathematically conveyed to me by whether omega r is greater than 0 or less than 0. So, essentially this idea of imposing spatial waves and watching is called a temporal linear instability analysis. So, just to quickly recap we started with the full Euler's equations showed that the mean state is a solution. Then we perturbed the interface using the prime quantities and wrote a set of linearized governing equations in the prime quantities which are these equations 8, 9 and 10. And then we said that in order for me to solve them I am going to make this assumption. This specific interface shape assumption and from there we are going to look for solutions of whether the real part of omega is positive or negative. We will continue this in the next class.