 In our previous class we derived the governing equations for the thin film coupled with the boundary conditions. So we will summarize those today and then proceed with the illustrative example of how to solve the thin film equations. So the basic equations that we solved or that we derived are the continuity equation, the x momentum equation and the y momentum equation which are given in the slide, then the boundary conditions at the solid boundary at the interface between the solid and the fluid you have no slip, no penetration boundary condition and at the interface kinematic boundary condition the shear force balance and normal force balance. Then we assume the variation of the surface tension of a particular form and also introduce the concept of disjoining pressure so that we could write the pressure jump across the interface in terms of these parameters. I mean these we did in significant details so I am just summarizing. Now based on various forcing parameters we ascertained various velocity scales. So like for example if surface tension gradient is dominant then what is the velocity scale if the normal pressure gradient is, pressure difference is dominant then what is the velocity scale. If the excess pressure is dominant then what is the velocity scale all this, if the x body force is dominant what is the velocity scale, if the y body force is dominant what is the velocity scale. Now in the previous lecture we outlined a method of solution of these equations. See these equations are strongly coupled and the nature is non-linear and however it is possible to derive a single expression of h as a, I mean single expression of governing differential equation not expression for h directly but governing differential equation for h as a function of x and time. So that we outlined and the procedure is something like this we started with the y momentum equation then used the normal stress related boundary condition. Then we calculated the axial pressure gradient use the x momentum equation integrated the velocity profile twice and have used the no slip boundary condition at the wall and the tangential stress related condition and that u we finally substituted in the continuity equation where we use the Limnitz rule to bring the ddx out of the integral and in the process we use the kinematic boundary condition. So that eventually led to the equation number 30 which is the form with which we will start today that del h del t plus del del x of 0 to h u0 dy equal to 0. Starting from here we will go to the board and work out try to work out a simple problem. I will write the expression of u0 what we derived in the previous class and we will start from there. So you have del h del t plus del del x of u0 dy u0 is equal to a into y square by 2 minus yh then plus epsilon beta by capillary number del t del sigma y where the expression for a we will write a is equal to del del x of epsilon cube lc square rho g cos theta by mu uc h plus ps minus epsilon square lc rho g sin theta by mu uc. So integral of u0 dy we can calculate we will calculate it for a special problem and then we will substitute that integral here. So once that integral is substituted here you will get a governing equation which contains h as a function of x and t where are so this h I am not writing all the time that h is a function of x and t here you have h as a function of x and t this term when integrated these 2 terms you will have h as a function of x and t even within a you have parameter h which is a function of x and t even ps has the dependence of h as a function of x and t. So I mean the entire equation is flooded with the parameter h which is a function of x and t. So we will define a problem for problem definition and schematic I will refer to the slide but I will work out the problem in the board rather than using the slide. Example 1 gravitational spreading of a cylindrical drop okay this is a very very very classical problem. Consider the drop shown in the schematic with that blue color it is of cylindrical shape that is along the z axis the length is very large. So it is an infinitely long cylindrical drop so to say. So the z axis is not important because it is infinitely long along the z axis all the gradients along the z axis are not important. So eventually whatever is happening in the xy plane is what is important. This drop is spreading due to gravitational forces acting on it that is the situation that we want to mimic. This type of spreading is known as gravitational spreading. What is our objective? Our objective is to find out the height of the drop as a function of t and x. We would also find out the radius of the drop L t as a function of time. It can be assumed that we are only interested at a time which is long after the drop begins spreading. Why because initially when the drop starts spreading the inertia force may be important and in the thin film analysis that we have made remember that you can make thin film analysis even with inclusion of inertia force. But the analysis of thin film the special type of thin film analysis that we have made we have not and that is the traditional thin film analysis that is the classical thin film analysis that by utilizing the ratio of the length scales we have trivially neglected the inertial forces. But you can see that when the droplet spreads it is I mean you cannot say see you cannot always say that h is much much less than L and that is why during the initial stages the inertial effects may be important. Now however during the later stages when the droplet is almost towards the completion of the spreading process then the height will become small and the length will become large. So then the low inertia limit will work better. So we will assume that at the time of our investigation the height of the film is much smaller than its radius so that the thin film assumptions can be applied. Finally we would also assume that the effect of capillary forces are negligible I mean it is just for demonstrating a problem that is what I intend to tell you because if you include the effect of capillary force I mean the same thing but equations will become sometimes equations may become intractable so far as what we can do in the scope of a class course work. But I mean inclusion of surface tension forces surface tension gradients, disjoining pressures these are some of the very standard things that right we commonly do in research and even in not may not be in research even in advanced text that is what is commonly used but just purely to demonstrate you that how to solve the thin film equation I will consider that the gravity is the dominating parameter. So as we have mentioned in the problem statement we are only interested in a time which is long after the drop starts to spread clearly the drop starts to spread from an initial shape which is unknown. See this is this problem setting of the length scale is very challenging because the droplet is its dimensions are continuously evolving you do not have a fixed l you do not have a fixed h and these are continuously changing after sufficiently long time there is no memory in the drop of its initial shape. The droplet has lost its memory that what was its initial shape completely we are interested in precisely this time frame that is the physics that we are interested to capture. Looking at the gravitational forces in the thin film equation we find that these forces are proportional to sin theta and cos theta here theta is 0 because like the plane is horizontal and gravity is vertically downwards. So here theta is 0 and only active component of gravity is the cos theta component right. So that means so basically g the in a y momentum g is acting right. So the main reason this is the main dominant force here and this force is the main reason for the drop to spread. So we have our velocity scale from the y momentum body force right because the y momentum in the y momentum there is a body force which is proportional to g cos theta and here theta is 0 right. So because the gravity is vertically downwards that is where the y axis is that is how the y axis is aligned. That means that your uc will be decided by the body force in the y momentum equation. See this is the first thing that we have to decide what will be your scale. So that is the first point. Second point we can easily infer that the radius of the drop will keep on changing with time and the height will keep on getting shorter right. This is obvious height will decrease radius will increase because volume has to be conserved. We note that in this problem there is no natural time scale to speak of that is why we cannot speak of any natural length scales also because natural because the length scales along x and y directions will keep on changing forever. Thus one can conclude that in the present case in absence of any natural scales we are likely to have a similarity solution. So I will show you that what kind of similarity solution we can think of for this particular problem okay. So we will start with the velocity scales maybe I will go to the board to explain this. So in the y momentum equation I will not erase this part because we will require this soon. uc what is uc? Epsilon cube lc square rho g by mu uc this is of the order of 1. So if you look into your notes epsilon cube lc square rho g cos theta by mu uc is of the order of 1 for y momentum body force to be dictating the uc from there we are getting this. So this means you have uc is of the order of mu by epsilon cube lc square rho g is the other way right. uc is of the order of epsilon cube lc square rho g by mu. Look at the expression for PS I will try to get back to the slide these equations are so I mean involved that it is better to use the slide to do it quickly PS. So P- so PS look at the equation number c in the right column PS is equal to P atmosphere- epsilon cube by capillary number into something plus the disjoining pressure contribution. So first assumption is we are neglecting the disjoining pressure contribution for this particular problem. Many research problems it is very important I am telling you but if you know how to do the algebra you can do it with the disjoining pressure term also that is not that is just a mathematical exercise. I am trying to demonstrate the mathematical exercise. The other term epsilon cube by capillary number that is what so where from that term has come epsilon cube by capillary number sigma del 2 u del 2 h del x square that is that has come from basically sigma the into 1 by r the radius of curvature. So that is the surface tension term now how important that term is we will now assess. So what is the capillary number mu uc by sigma so that is equal to mu epsilon cube lc square rho g by mu sigma. So this is epsilon cube rho g by mu sigma. So this is epsilon cube rho g lc by sigma by lc. What is this? This is gravitational pressure right and this is the Laplace pressure due to surface tension. So this ratio this is an indicator of the gravitational force by surface tension force this is called as bond number, bond number Bo. So recall that in the Laplace pressure term we had a term epsilon cube by capillary number right. So epsilon cube by capillary number is equal to 1 by bond number. So we will so our assumption that Laplace pressure term is not important can be justified only if bond number is large right. So we will assume that the bond number is large that means epsilon cube by capillary number will be small. So if you look at this expression for a if you look at this expression for a now p s is equal to p atmosphere plus the Laplace term plus the disjoining pressure term. Laplace pressure term is negligible and the disjoining pressure we have already neglected. So this for this present problem for only this problem right do not generalize it for only the example problem that we are working out now this is p atmosphere okay. Now what are the other parameters for this problem? This is cos theta is 1 and sin theta is 0. Not only that epsilon cube L c square rho g by mu u c this is equal to 1 right. The reason is obvious that is how we set the velocity scale. By looking into that body force only we set the velocity scale u c. So that will be 1. So based on this we will proceed further not only that we will assume that there are no tangential gradients of surface tension. For example you have not applied any temperature gradient. So that it creates a surface tension gradient. Integral u 0 dy between 0 to h is equal to a integral y square by 2 minus y h dy 0 to h. So this is a this becomes h cube by 6 minus h cube by 2. So minus a h cube by 3 right. So it becomes del del t del h del t plus del del x of minus a h square by 2 h cube by 3 equal to 0. So del h del t is equal to del x of h cube by 3 del h del x. This is our governing differential equation. What are the boundary conditions at x equal to 0? What is at x equal to 0? Look at the diagram. Let us go to the diagram. x equal to 0 in this diagram is what? y axis. At x equal to 0 what should be the boundary condition? del h del x should be equal to 0 right. This is because of symmetry. Gravity is not breaking the symmetry. So the droplet is spreading maintaining the symmetry with respect to y axis. So at x equal to 0 del h del x equal to 0 at x equal to l which is a function of t what will be it? h equal to 0. So this is our problem definition. We can now forget about the previous derivation. This is the equation that we want to solve. Now you assume is equal to t to the power alpha into f eta where eta is equal to some constant into x by t to the power beta. Question is why do we why can I mean why are we attempting to assume this form? Now if self similarity does not exist then we will see that this form will not be satisfied. So there is not always a guarantee that this form will be satisfied. Now what it is saying? It is saying that h is varying with t to the power an exponent. From your intuition what will be this exponent? Positive negative 0. Yes. So how do you see think about it? It is a very simple answer. With time as t increases what does happen to h? h decreases. So you expect a negative exponent right and this eta see this is like and when h decreases l should increase. l is the horizontal footprint of the droplet which is called the droplet radius. So if h decreases l should increase so l scales with t to the power beta. So we expect beta to be positive. So this is like x by l which is like a similarity parameter. So whether this will be actually a similarity parameter or not it depends on whether in the final simplified version of this governing differential equation this t will be eliminated. If this t is eliminated it will be a function of a single variable eta which contains x by l okay. So we will see that whether the condition of getting a similarity solution may be satisfied because this problem physically hints self-similarity. Droplet height decreases and in proportion the stretching by stretching the droplet footprint increases right. So now let us treat that governing differential equation. So del h del t we will do this term very carefully. If I make any algebraic mistake please correct me I have forgotten to bring that notes with me. So I will do it in the board but in case I make a mistake you please let me know. So del h del t is equal to alpha t to the power alpha-1 f I am just writing short f means f eta. So this is alpha t to the power alpha-1 f plus t to the power alpha df d eta, df d eta we write f dash short hand notation del eta del t del eta del t is Ax. This is basically t to the power-beta. So minus beta t to the power-beta-1 alright. So this is alpha t to the power alpha-1 f plus t to the power alpha f dash Ax means eta into t to the power beta Ax is eta into t to the power beta. So that t to the power beta and t to the power-beta cancels. So minus beta into t to the power-1 just check this is alright. Then there is a relationship between alpha and beta. See this similarity will hold with a constraint of volume conservation right. The h decreases and the l will increase in consistency with that that we can say from intuition because volume of the droplet is conserved. If that is not conserved you cannot tell anything about that. So volume of the droplet is conserved means integral h dx from x equal to 0 to x equal to l that is constant. This is volume conservation. So let us substitute integral of t to the power alpha dx is t to the power beta by A d eta is equal to constant. Can you tell that this is constant? That the final result should it be sensitively dependent on what constant we choose? Actually not because the constant because ultimately the prefactor in the solution will be scaled with this constant. So you can normalize this and you can choose this as 1 for example. Not a must. I mean whatever you choose the final result will be scaled with that okay. So this integral eta 0 to let us say eta 0. What is eta 0? Eta 0 is eta when x equal to l. When x equal to l the corresponding eta we call as eta 0. So t to the power alpha plus beta integral f d eta is equal to A. When will this result in a self-similar deformation of the droplet? When there is no t variation that means alpha plus beta is equal to 0. So beta is equal to minus alpha okay. So here in place of beta we will write minus alpha. So this will be alpha. Alpha plus 1 f plus f dash eta alpha t to the power alpha minus 1 is common. This is del h del t. Then the right hand side let us calculate h cube by 3 del h del x. So del h del x. Yes this is mass conservation if density is constant that is true. I mean that is quite obvious. Del h del x del h del x t to the power alpha df d eta into del eta del x. So t to the power alpha df d eta is f dash del eta del x is A by t to the power beta. Beta is equal to minus alpha. So t to the power 2 alpha A is equal to F dash. Then h cube by 3 del h del x. So h cube means this will become t to the power 3 alpha. That 3 alpha and 2 alpha will make it 5 alpha. So t to the power 5 alpha f cube then another del del x is there. So that means this is d d eta into del eta del x. So that means del eta del x is A by t to the power beta. So d d eta of t to the power 5 alpha f cube A f dash by 3 del eta del x is A by t to the power beta. So beta is minus alpha. So it becomes t to the power 6 alpha f cube sorry d d eta is there. A we can take out of the eta. d d eta of f cube A by 3 let us take out. A square by 3 basically. A square by 3 d d eta of f cube f dash. So we have to basically equate this term with this term. Let us write that. Clean up some part of the board to write that. A square by 3 t to the power 6 alpha d d eta of f cube f dash. So I will erase the remaining part. Not equate it in that what is the condition under which we will get a similarity solution. When the function of t gets cancelled from both sides that means t to the power 6 alpha is equal to t to the power alpha minus 1. So that means for similarity solution to exist you must have 6 alpha is equal to alpha minus 1. So alpha is equal to minus 1 by 5. See a good check is that alpha has come out to be negative. Many times I have seen that I mean it is not just true for students. For anybody I mean for us as teachers or researchers it is possible that we make calculation mistakes. So some of the calculation mistakes are very difficult to detect because those are like sort of I mean very difficult to be resolved by physical argument whether that is a calculation mistake or not. But there are certain types of calculation mistakes which you can resolve by physical argument and those things we should check. So instead of minus 1 by 5 had your calculation given you minus 1 by 6 it would have been very difficult to check I mean whether that is a mistake or not. But instead of minus 1 by 5 had it come plus 1 by 5 you should get a caution. No this cannot be correct because physically as time progresses the height of the droplet decreases. So alpha is equal to minus 1 by 5. So here we can substitute minus 1 by 5. So now we can so these terms are eliminated. We can make a further simplification of the equation very nicely by choosing s square is equal to 3 by 5. If we choose s square equal to 3 by 5 you will get nice numbers. I mean your answer is independent of choice of s square because these are just scales. So this will scale you the final whatever you assume as a that will scale like I mean a good analogy is that in school when you are dealing with ratio and proportion a by b equal to c by d you assume some k. Now you can take k 1 2 3 4 5 whatever your significant point is whether a by b equal to c by d is maintained and not what is k. According to k the answer will be just scale. So similarly this a is like k s square is equal to 3 by 5 plus f dash eta is equal to d d eta minus minus d d eta of f cube f dash. I have this final equation in my notes at least let me check whether this is all right before we can proceed. Yes this is fine so up to this at least it is fine. Now this is what this is d d eta of f eta right simple and trivial d d eta of f eta then we can integrate both sides. So f eta equal to minus f cube f dash or better take away we can do this much better take minus f cube f dash here we put a minus sign algebra will be bit f cube f dash is equal to minus f eta plus a constant c 1 right. It does not matter I mean you can just play with whether you can put in the left hand side or right hand side where I have put it in this way because f dash if we keep in one side x basically we integrate it to get it that is the objective of writing it. What we will do is integrate this so at eta is equal to 0 you have f dash equal to 0 right. So that means you have c 1 equal to 0. So you have this is all right so d f d eta f square d f d eta is equal to minus eta so f square d f f cube by 3 f is equal to minus eta square by 2 plus c 2 f cube by 3 right. So f is equal to minus 3 eta square by 2 plus 3 c 2 to the power one third. Now the next so at eta is equal to eta 0 you have f is equal to what at eta is equal to eta 0 f is equal to 0 because h is 0 h is equal to t to the power alpha into f. So 0 is equal to minus 3 eta 0 square by 2 plus 3 c 2 that means c 2 is equal to 3 eta 0 square sorry eta 0 square by 2. So f is equal to 3 by 2 eta 0 square minus eta square to the power one third. So this completes the solution but you have to know what is eta 0 that you do not know. So for that we get back to the volume conservation. So integral f d eta 0 to eta 0 is equal to a what is the value of a that we have chosen no no a is not 1 root 3 by 5. So then f you substitute so f will 3 by 2 eta 0 square minus eta square to the power one third d eta 0 to eta 0 is equal to square root of 3 by 5. Now you can make a substitution eta is equal to eta 0 sin theta I mean that is the standard substitution sin theta or cos theta whatever and then you can evaluate this integral numerically to get the value of eta 0. I will tell you the final answer please do that do it as a small exercise you will get eta 0 is equal to 2.23. Once you get eta 0 you get the that means you have got the complete solution. So what is your complete solution h is equal to t to the power minus 1 by 5 into 3 by 2 into 2.23 whole square 2.23 square minus eta square to the power one third. So this and eta is equal to x by t to the power 1 by 5 square root of 3 by 5. Now what is our objective our objective is to get the droplet radius as a function of x. So how will you get l how will you get l in this expression when eta is equal to eta 0 x equal to l right that means 2.23 is equal to l by t to the power 1 by 5 square root of 3 by 5 this l is a function of time. So this will give you what is l t. So through this problem I have tried to demonstrate you that how to make use of a thin film equation and this is a very very classical problem that I have demonstrated that how it is possible to solve the thin film equation to get h as a function of time and x. So we will close our discussion on thin film dynamics with this and from the next lecture we will start with a new topic. Thank you very much.