 Good morning, we are going to continue our discussion of linear instability analysis towards the end of the last class, we had derived a set of linearized governing equations. So let us start by looking at the linearized governing equations and our first job is going to be to define a set of boundary conditions that can go with these linearized governing equations. So, we will make that our job for today. We took the complete two dimensional Navier Stokes equations and sorry Euler's equations and after scratching out the terms that are going to be order epsilon and epsilon after scratching out the terms that are order epsilon squared and only keeping the terms that are order epsilon, epsilon being the order of the perturbation quantity which are the primed quantity. So, essentially only the part that is linear in the primed quantity, we now have a set of equations that will tell us what the behavior of the primed quantities is in time and space. Now from these, so this is basically like saying these are the equations that govern the growth or evolution of the disturbance as the evolution grows in time. But as what we want is always go back to our analogy of ball in a cup. So, if I take the ball and move it slightly to the left from its bottom most equilibrium position, I can write a set of mechanics equations that describe the movement of that ball from this perturbed position. So, that is essentially saying the force imbalance due to the slight perturbation in the position of the ball in that cup is going to result in an acceleration. And if that acceleration happens to point towards the previous equilibrium position that is a stable equilibrium that is our essential understanding of stability. These equations 8, 9 and 10 here essentially govern the growth of those perturbations or really not even the growth the change of those perturbations in time and space x and y. Now what we want to do is impose a sinusoidal perturbation on the meniscus itself on the interface between the two fluids. And we started to say that we are going to make this meniscus have a function of the form cosine 2 pi x over lambda times e power omega t times eta 0. The 2 pi x over lambda essentially tells us the functional form of the of a wave that we will impose on the free surface and watch it either grow or decay in time. And the growth or decay is going to be determined by the real part of omega. If omega r is greater than 0 then the wave amplitude grows in time because the amplitude is this eta 0 times e power omega t. And the imaginary part of this wave number of the growth rate essentially creates another sinusoidal motion in time. So, if I take a wave and if the wave is say for example, simply going up and down in time just like that that would involve no imaginary part to the omega. If the wave is also doing this like all waves do there is a motion of the crest in time to either left or to the right there is a wave speed associated with that crest that wave speed is given by this omega i and k the wave number put together. So, as far as the actual stability whether the thing grows in amplitude or decays in amplitude is only determined by the real part of omega. And then we made one small distinction that said say for example, k in this equation here is real meaning k is a real wave number. And omega is allowed to be complex and that in turn gives us the real part of that complex number gives us the actual growth of the amplitude in space. So, this kind of an instability analysis is called a temporal linear instability analysis where we impose real spatial waves I will even put that will impose real spatial waves and watch their behavior in time. The other possibility is to impose a wave in time say for example, at some inlet location I have a small perturbation that I can add to the time component and watch what that small time perturbation does over space. We will leave that out of our argument for now, but essentially this is our discussion. Now, we have chosen to impose a disturbance on the meniscus remember just for clarity I will say y equal to 0 is the unperturbed free surface. So, on top of that unperturbed free surface we have added this disturbance. So, essentially you might think of this as 0 plus this just as everything else that we have done was the mean flow plus the small perturbation. In this case the mean interface location is y equal to 0. So, it is 0 is an order epsilon quantity. So, we will need to be aware of that. So, now we need the next order of business is to define a set of boundary conditions. Before that we will quickly make a list of all the variables that we have u i prime v i prime p i prime each of this for i equal to 1 or 2 and eta 0, eta 0 is the amplitude of the disturbance So, as far as u i prime and p i prime is concerned we have 3 equations in the form of 8 9 and 10. Now, remember these are still subscripted in i which means each one each of these for i equal to 1 and 2 gives us a total of 6 6 equations for the 6 6 variables the eta 0 is an interface condition it is not necessarily an unknown quantity it is what we are imposing and we just want to watch its growth in time or this is like an initial amplitude of the disturbance let us be more specific. The first set of boundary conditions is what we will call kinematic boundary conditions. So, let us quickly see what that is I will redraw our sketch from the earlier page here just take the condition the mean flow condition that u equals capital U 2. So, now I have imposed these surface waves on the free surface and the liquid if I take a liquid particle right on that free surface. So, this is a particle that is on the perturbed free surface as this fluid flows this way the wave itself is going to be moving with the fluid because of that the fluid has because a fluid particle let us say this is the next time position of the same wave what was here is now here just because the wave moved this is the previous time position this is the next time position now as the wave reaches its crest and comes back down this particle which is at that free surface is only going to go up and down. So, the first set of boundary conditions which are called the kinematic boundary conditions basically say that the y direction velocity of the particle is given by the rate of change of eta plus u d eta d x. So, this part here is a co-moving derivative of eta it is a co-moving derivative of eta and what we say is even if eta is not a function of time. So, think for a moment remember we said eta is this is our eta as a function of this whole thing as a function of x and t. So, even if eta even if let us say this omega was 0 just by a wave moving now at a velocity u i you get an up and down motion which is v i. So, if I ignore the first term here called d eta d t the partial derivative of eta with respect to time if I ignore that for a moment and only take the second term on the right hand side here u i times d eta d x is saying that the rate of movement of a particle in the upward direction is essentially governed by the x direction movement of the velocity and the slope of the interface. So, if the interface has a certain slope and it is moving this way then a point on that meniscus is essentially like it is being pushed upwards with a velocity u i times d eta d x d eta d x is the local slope of that interface. If on top of this if the wave amplitude itself is growing that is further additive y direction velocity. So, I get two kinds of y direction velocity at that free surface one is no growth in the amplitude just movement of a slope meniscus in this case a wavy meniscus and the second kind is where the growth there is you know like this is the meniscus it is just growing in time. So, the trophies coming down and the crest is reaching for higher values of y that in itself means that there is a y direction velocity that is that part of the y direction velocity is represented by the first term on the right hand side. If I had this wave on the free surface and if I was simply observing a point on the free surface that point as this wave moves that spatial location shows an upward y velocity. That is represented by the second term on the right hand side the total y velocity at any point at any x location on the free surface is given by the sum of these two. Now, there is still one issue if I take the u i and do our perturbation plus the mean flow plus the perturbation substitution that part there again happens to be order epsilon squared. So, what we end up is v i prime at y equal to 0. So, the actual velocity now we will come to this in just a moment. Now, remember we are talking of the fluid velocity on the meniscus right there this fluid particle used to be there now it is here and now it is down here. So, at this x location there is a negative y velocity because the slope is negative with no d eta d t being positive. Now, ideally I am evaluating the v i velocity on the meniscus. So, this velocity in this equation is the upward velocity on the free surface which is if I was to represent that mathematically v i at y equal to eta, eta is the free surface itself v i at y equal to eta can be written as v i at y equal to 0 plus eta times d v i d y evaluated at y equal to 0 plus this is simply a Taylor series expansion. If you notice again this part here is order epsilon squared. So, just as we simplified the right hand side of this equation to only have order epsilon terms. If I simplify the left hand side to only have order epsilon terms then v i at y equal to eta can be replaced by v i at y equal to 0 or v i prime at y equal to 0. So, ideally we want v i prime because v i prime is the only upward velocity there is no mean flow that is in the upward direction. So, therefore, this is again ideally I should have v i equal to d eta d t plus u i d x and v i is equal to 0 plus v i prime. So, we can drop the 0 plus v i prime because we have now sort of looked at this enough to know that the mean flow is 0. So, if I take the last part this v i prime is actually the perturbation free surface velocity at the free surface at y equal to eta, but that can be substituted by taking v i prime at y equal to eta is equal to v i prime at y equal to 0 to order epsilon. If I have to include the higher order term you know sometime later that is up to order epsilon squared, but as far as a linear instability analysis is concerned I am only going to keep terms that are order epsilon. So, we have the governing equations 8 9 10 to go with I have this boundary condition 8 9 10 to go with that I need this boundary condition let us get to that. I need this free surface condition 11 and this condition 12. Now, mind you equation 12 has two possibilities for i equal to 1 and i equal to 2 meaning I have the same miniscous d eta d t, but depending on the fluid I am in the upward velocity of that fluid is equal is determined by the balance from whichever direction I take that blue particle argument. So, whether this blue particle was in the i equal to 2 fluid or i equal to 1 fluid makes no difference as long as the free surface I am able to imagine the free surface is composed of two fluids one on top of the other and whichever fluid I look at it does not matter this condition has to hold. Now, the second set of conditions is called a dynamic condition what do we mean by that. If I go back to the same diagram if I imagine for a moment that the fluid let us look at the mean pressures capital P i or will be specific capital P 1 is equal to capital P 2 equal to 0 by definition that is where we left off. So, we found that for a flat free surface capital P 1 is the mean pressure in the entire fluid fluid one capital P 2 is the mean pressure in the entire fluid P 2 this pressure cannot vary with x and y we found that from the equations. And then once I know they are constants and they are both equal to each other I have the freedom to set that to any value we set that equal to 0. So, now that I have a curved interface the red line here in this figure the perturbed versions of these pressures do not necessarily have to be equal. So, if I take P i prime P 1 prime is the pressure on the fluid above P 2 prime is the pressure in the fluid below right that is what already appeared in the governing equations in the governing equations 8 9 and 10. So, if this P 1 prime and P 2 prime are now functions of x y and time if I take an arbitrarily small element of fluid just like that and draw a free body diagram of it. So, I will expand that out there is pressure P 1 prime on there P 2 prime on the other side and there is a surface tension force. So, because I cut out a small element of the fluid the rest of the fluid exerts of a force on this small element due to surface tension. So, if I write a force balance here we can see that P 1 minus P 2 sigma times kappa. So, this is this kappa is basically the fluid is the meniscus curvature this free surface curvature is essentially dictates the difference in the forces between P 1 and P 2. So, just to understand this we will go back to just a simple curve free surface if I have a pressure P 1 acting here and the pressure P 2 acting on the other side you can see that this infinitesimal piece so I am now looking at a little infinitesimal piece of fluid if I take this infinitesimal piece of fluid then let us say it exerts a small angle delta theta. This infinitesimal piece of fluid exerts a net there is a net force at this point in the magnitude P 1 times r delta theta that is the outward positive minus P 2 times r delta theta which is the inward pressure force we are assuming the meniscus is one unit width in and out of the plane of the chalkboard this equal to sigma by r times the total meniscus so essentially sigma over r is equal to sigma which is a surface tension force itself times acting over an angular this thing one times delta theta which is the width of the which is the length of the free surface. So, over which so essentially the this sigma force acts on this end here and this end that is one unit long this delta theta comes from the fact that if I take a vector sum of this in this direction and this direction likewise in this direction and in this direction the horizontal forces cancel out the vertical component of this surface tension force is the only force that contributes to the pressure differential especially as this delta theta becomes smaller and smaller in the limit of delta theta going to 0 the horizontal components of these surface tension forces becomes smaller and smaller and the vertical components are what add up to the pressure differential P 1 minus P 2. So, in this note we can see that sigma by r in a more generalized notation if r is if I have a circular boundary or a spherical bubble r is simply the radius of curvature so we will generalize it to say P 1 prime minus P 2 prime equal to sigma times kappa which is the radius of curvature. So, this is my force balance conditions the right hand side having surface tension pressure and this is static pressure difference if I take the free surface itself I know the equation of the free surface that is I will rewrite this equation one more time here just for our referral y equal to eta 0 e power omega t plus i k x. This is the equation of the free surface itself which means for a given curve if I know y equal to f of x curvature kappa is given by d square y dx squared by 1 prime minus 1 prime plus d y by dx the squared raise to the power 3 halves with a negative sign only to account for the direction of these curve of the curvature. So, this negative sign says wherever your free surface is concave make the curvature negative wherever the free surface is convex concave and convex viewed from fluid one let us say. So, wherever the free surface is concave as viewed from fluid one fluid one is on top wherever it is concave as viewed from fluid one will call this curvature negative wherever it is convex as viewed from fluid one will call it a positive curvature it is simply a matter of convention. If this negative sign was not there we will essentially invert this p 1 prime and p 2 prime. I am going to show you how I do not ever remember this I want to show you how you can not remember it either, but get to the correct answer. So, now given this if I want to calculate the curvature kappa from this equation I need this derivative. So, let me go ahead compute the derivatives dy dx is i k eta 0 times e power omega t plus i k x another derivative on that is minus k because i k times i k gives me minus k. So, if I make these substitutions kappa then becomes minus k eta 0 e power omega t plus i k x divided by omega t plus i k x divided by 1 plus i k eta 0 e power omega t plus i k x the squared this whole thing raise to the power 3. If you look at this term this is order epsilon squared. So, I can use 1 over 1 plus epsilon squared I will just call make the simplification on this. So, this I can approximate it to 1 minus 3 halves epsilon squared plus higher order terms. So, to order epsilon the denominator is equal to 1 because the next term after 1 is order epsilon squared. So, I can come here and say to order epsilon kappa equals minus k eta 0 e power omega t plus i k x. So, for the given free surface wavey shape this is the curvature at every point. So, I told you how I do not remember this I want to show you how I end up not remembering and still getting the right answer. It is this fact that if just as a check at x equal to 0 the curvature is minus k eta 0 e power omega t. So, that is a negative number which means as viewed from and the shape of the meniscus here the real part would end up looking like a cosine wave we saw that. So, as viewed from fluid 1 this part at x equal to 0 is concave meaning you have a negative curvature which means kappa is if d squared y d x squared is minus k which means this was wrong this should be plus. The fact that I tried to put a negative sign there if I did take that negative sign into account this would end up being positive here because d squared y d x squared itself is minus k eta 0. So, if I want kappa to be minus k eta 0 then this is plus d squared y d x squared. So, this sign here gives me viewed from curvature viewed from fluid 1. If I want to write the same exact equation for curvature viewed from fluid 2 I would end up putting making that a negative sign. So, we now have another boundary condition here if I take. So, now if I simplify our dynamic balance pressure condition p 1 minus p 1 prime minus p 2 prime is minus sigma k eta 0 times e power omega t plus i k x. So, just to recap one more time these are our variables I have u i prime v i prime p i prime for all i equal to 1 and 2 and eta 0 is the initial amplitude. Now if I impose an initial amplitude that is sinusoidal in the on the interface then as long as the equations are all linear the only spatial variation that you can expect in the x direction. So, the interface is sinusoidal in the x direction the only variation of all other quantities in the problem that you can expect in the x direction will be the same function sinusoidal will be the same will be of the same sinusoidal form. So, we can make us we can make what is called a normal mode assumption that all u i prime which is now a function of x y and t is of the form u i 0 times e power omega t plus i k x where the pre multiplicative factor to the exponential u i 0 is purely a function of y and e power omega t plus i k x is basically saying the functional form of the variation of u i both in x and y x and time are exactly the same as the functional form for the impose disturbance. This is a quality of only linear problems. So, if I have a linear governing equation partial or ordinary the frequency whether it is in space or time is not altered in the response. So, I have an impose disturbance which is my input disturbance the output disturbance let us just call which is the impose disturbances manifestation in the other fluid mechanic quantities is also in exactly the same frequency. I do not want to go into the details of why this happens, but let us just take that to be the case, but the function does still have a variation in y likewise I can do the same thing with this. This is not v i to the power 0 u i 0 y is simply I will put this in parenthesis or let us just switch notation just to avoid any confusion will call this u i double prime which is only a function of y these are called normal mode assumptions. So, all the modes of the of u i v i and p i are assumed to be of the same there of the same mode as the imposed wave. So, if I now substitute will call this 14, 15 and 16 into 8 to 13. So, first of all 8 becomes if I go back to equation 8 d u i d t plus u i d u i prime d x equals minus 1 over rho i d p i d x. So, just for the sake of clarity I will copy these will paste them on the side here, but then we will delete them d u i d t d u i prime d t is omega times u i double prime times e power omega t plus i k x plus u i d u i prime d x is u i times u i prime d x plus u i prime i k times u i double prime times e power omega t plus i k x this equals minus 1 over rho i d p i d x which is i k p i double prime e power omega t plus i k x. What I want you to notice is that what used to be a partial differential equation is now become an algebraic relationship between u i double prime and p i double prime. So, if I simplify this I get omega plus i k u i u i double prime equals minus i k p i double prime i k over rho i p i double prime. The exponentials all cancel out and what I have is this we will call this equation 17. If I continue on and write down the equations for the other two without going into this same level of mathematics substituting the same process 9 becomes. So, omega plus i k u i v i double prime minus 1 over rho i d p i double prime d y. So, we have now a set of ordinary differential equation system for u i v i p i in the form of 17, 18 and 19. Look at these are ordinary differential equations because I only have p i double prime quantities are only functions of y. All the variation in x and time has already been absorbed. So, what I have between 17, 18 and 19 is three quantities u i v i p i and three equations this forms a system of ordinary differential equations. We will take up from here in the next class.