 The x momentum equation and the y momentum equation which are summarized here you will of course have these in your notes for the previous lecture. Then the boundary conditions I mean we have expressed both in a dimensional form and in a non-dimensional form. So the kinematic boundary condition at the interface the kinematic boundary condition this is the dimensional form, this is the dimensionless form. The shear stress balance this is the dimensional form and this is the dimensionless form expressed in terms of capillary number. Normal stress balance again dimensional form and dimensionless form just to help you to recapitulate how to recover the dimensional form from the dimensional form I will briefly say take up any one of these boundary conditions say the normal stress boundary condition for example. So you have for example p dash-p dash atmosphere we have seen that-tau dash dot n dot n that particular term in terms of order will be one order less important as compared to the p-p dash term. So we will just keep this as it is say this will be roughly sigma dash by r. We will discuss about this excess pressure extra term which we attributed mainly due to van der Waals forces of interaction when the film thickness becomes very small. So this is pc into p dash-p dash atmosphere sorry p-p atmosphere. This is what sigma 0 into sigma by r and 1 by r is this one. So in the slides I have not put delta dash delta dash but just to emphasize that it is a dimensional operator that we are talking about we are I mean I have put it in the form of dash here. So this is-epsilon del h del x sorry first let us write the del dash operator. So i del del x dash plus j del del y dash dot product with n. So what is n-epsilon del h del x i plus j right this was n. So now this becomes now del del x dash x dash is what? x into lc. So this will become I mean this total operator will become 1 by lc del del x. So this will become-epsilon by lc del square h del x square right. What is pc? Tell from your notes what is pc? Yes? Mu uc by epsilon square lc. So you multiply both sides by pc basically 1 by pc that is divide both sides by pc. So I mean p x dash we are talking about so p-p atmosphere is equal to sigma 0 epsilon square lc by mu uc into-epsilon by lc del square h del x square sigma plus px prime divided by pc right. So here lc and lc get cancelled. So this term becomes mu uc by sigma 0 is capillary number. So –epsilon cube by capillary number sigma del square h del x square. So we will get back to the slides see that that is what is that –epsilon cube by sigma capillary number del square h del x square that is there as that term and last term is p dash x by pc. So this is the normal stress balance. Now in this analysis we have assumed that the surface tension along the liquid gas interface may vary. Now question is how can it vary? These variation may come from various sources. What are the sources? For example temperature variation along the surface present of surfactant etc. Since the surface tension is a strong function of temperature and surfactant concentration non-uniformity of temperature and surface tension might cause surface tension gradients. As an example the surface tension may be expressed in terms of a function of temperature. So as I told you that many fluids will actually have surface tension coefficient strongly depending on temperature. For water the variation is typically linear but there are other fluids for which higher order variations are also there. So this is the typical form and then if temperature varies with x then surface tension will vary with x. Now what about the term p dash x or the excess pressure across the interface? For very thin films typically in the range of microns or nanometers there can be extra contribution from van der Waals forces in the form of disjoining pressure. This is simply a pressure jump across the interface just like a Laplace pressure but due to van der Waals forces. So these forces become important only when the film thickness becomes so small that intermolecular forces also play a dominating role. So one typical way of representing it mathematically is this particular form. I mentioned your reference book where you will get the details of this variation but to summarize you can have in a continuum calculation this as parameter A by scales as a parameter A by h prime cube. 6 pi is just a multiplying parameter. This parameter A is known as Hamaker's constant and I mean it is possible to get the Hamaker's constant for various interactions by considering intermolecular forces and therefore the normal stress balance. So in a dimensionless form we look like this if you are considering the excess pressure. So in many problems in microfluidics and nanofluidics this excess pressure term is very important. Now one very important task that is still left is what is u c and what is v c. So that depends on what is dominating. So if the surface tension gradient is dominating. So surface tension gradient means the tangential gradient of surface tension. So let us look into the equation from which we get the scale, this equation. What is this equation? This is a scenario when the surface tension gradient is dominating. So if the scenario is the surface tension gradient is dominating then the velocity scale will be decided by this balance. Now in this balance what is the order of del u0 del y of the order of 1 because these are scaled suitably with normalizing parameters designed to be between 0 to 1. The dimensionless if u0 is a dimensionless u and y, I mean 0th order u and y is a dimensionless y, both u0 and y are constrained to be between 0 and 1 that is how uc has to be ascertain. So that means this is of the order of 1. The surface gradient of sigma this is also dimensionless designed to be of the order of 1. Therefore epsilon beta by capillary number should be of the order of 1. So quickly let us do it in the board. So epsilon beta by mu uc by sigma 0 is of the order of 1 that means uc will be epsilon beta sigma 0 by mu. So we can get back to that slide where various scales are ascertain. Epsilon sigma 0 beta by mu that is how the surface tension gradient driven velocity scale can be ascertain. Then a scale based on normal stress that is if the Laplace pressure is dominant. So you have to see this is where physics of the problem comes into the picture. What is dominant? It is a based on your judgment like mathematics will not tell what is dominant but mathematics will tell that based on what is dominant what is the scale that will be dictated by the mathematics but physics will dictate what is the physical effect dominant. So if you get back to that equation see in this equation equation number 17 in the slide you see that p s-p atmosphere this is designed to be between 0 to 1. So sigma del square h del x square that is also designed to be between 0 to 1. So epsilon cube by capillary number should be of the order of 1 if the Laplace pressure is dominating. So let us come to the board epsilon cube by capillary number of the order of 1. So epsilon cube sigma 0 by mu uc is of the order of 1 that means uc is of the order of epsilon cube sigma 0 by mu. So you can see that the analysis of the problem it is sensitively dependent on the scale and if you ascertain the scale wrongly the entire analysis will be wrong. So you have to be very very critical and judicious about the choice of scale. If the excess pressure is dominant if the excess pressure is dominant so you can see that p x s so that is actually p dash x s divided by pc. So p dash so let us get back to that get back to equation 17. So if all the terms of the order of 1 that means p dash x s into epsilon square lc by mu uc that is of the order of 1. So that means p dash x s uc is of the order of p dash x s epsilon square lc by mu. So if this is dominating if this is not dominating this will not dictate what is the scale then its magnitude will depend on what scale you have ascertain based on other influencing parameters but if this is the influencing parameter that will dictate the scale. So let us say there are groups of people now whoever is the leader has to play the leading role and the others will follow that. So whatever is the leading or dominant factor that will dictate the scale and the relative magnitudes of the others will be decided by that scale. So they will not themselves influence the scale x body force is dominant x body force is dominant. So you can see that this is the body force term in the x momentum equation right whatever is written in the slide epsilon square lc square rho g sin theta by mu uc okay. So this is the body force term in the x momentum equation if all the terms I mean if the dominant term has to be of the order of 1 then this term being the dominant one also has to be of the order of 1 that will give you what is uc and similarly if the y body force is dominant then this is the body force term in the y momentum equation and that will give rise to a uc. So you can see that different expressions for uc may appear depending on what is dominant and that has to be taken into account. So now what we will do is we will try to make an attempt to derive the final version of the thin film equation which we will apply to solve a problem and that will give you a clear illustration of how these equations which appear to be bit cumbersome and interrelated how this can be used to solve a practical problem. So instead of getting back and forward to the slides all the time I will refer to my notes and start with the corresponding basic equations. We will start with so our objective is thin film equations. So objective is derivation of a governing differential equation for H as a function of xt. This is what is our objective of what we are going to do okay. So we will start with the y momentum equation. What is the physical scenario? So you have a solid boundary over which you have some interface thin film H as a function of xt and you have x and y axis set up in this way where the angle of inclination of the inclined plane with the horizontal is theta and g is acting vertically downwards. So y momentum equation del p0 del y these are there in your slides or notes plus epsilon cube lc square rho g cos theta by mu uc equal to 0. Next what we will try to do? We will integrate this equation with respect to y right. So p variation with y cannot be ruled out here because the gravity effect is there. You might argue see there are there are so many arguments. You might argue that yes it is alright but the film thickness is small. So how will pressure variation in that thickness play a decisive role right. That might be an intuitive cross argument that why should we bother about pressure variation with y because the dimension along y normally we take the gravity effects important for pressure variation when the depth change is significant. Now here we have already presumed the thin film. So the depth change is not significant despite that y momentum equation is important and we will see why. So the analysis will make it clear. So if you integrate it with respect to y then p0 is equal to minus epsilon cube lc square rho g cos theta by mu uc y plus what function of x and t. See the variables are x, y and t say c, x, t. So what is the boundary condition? So this requires one boundary condition. What is that? At y is equal to h which is a function of x, t p0 is equal to say ps, pressure at the free surface and we have a boundary condition which relates pressure at the free surface with the atmospheric pressure ps minus p atmosphere. So at y is equal to h, so ps is equal to minus epsilon cube lc square rho g cos theta by mu uc h plus c, x, t. So what is c, x, t? Ps plus epsilon cube lc square rho g cos theta by mu uc h. So we can write p0 is equal to ps plus epsilon cube lc square rho g cos theta by mu uc into h minus y where h is a function of x and t. This is the governing equation for p0. Now ps is not a constant, right. Ps is governed by the pressure normal stress jump boundary condition across the interface which is given by p minus p atmosphere that formula. So we will write that. So I am giving you the most general formulation so that any complicated problem you can solve even using this formulation. So I will retain all the terms including the excess pressure. Although for simplicity when we have demonstrate a problem in the class I will work out a problem which I can work out analytically in the board without much of calculation. But the formulation that I will give will be pretty general and you can take it up for solving many complicated problems. So what is ps? So ps is equal to again if you look into your notes you will get this equation p atmosphere minus epsilon cube by capillary number sigma del square h del x square plus epsilon square lc by mu uc this one, okay. So ps what is very important to keep in mind is that ps is itself also a function of x and t because h is a function of x and t. So this is implicit it will appear as if it is a constant but it is also a function of x and t because it is a function of h which in turn is a function of x and t, okay. So these are subtle things so it is important to bring it up. Now let us write the x momentum. So x momentum minus del p0 del x plus del square u0 del y square plus epsilon square lc square rho g sin theta by mu uc is equal to 0, yes, right. So now we will use this equation for getting what is del p0 del x. So what is del p0 del x? So this is equal to minus del ps del x minus epsilon cube lc square rho g cos theta by mu uc del h del x plus epsilon square lc square rho g sin theta by mu uc what we will do we will alter this sign and write it as del square u0 del y square. This gives an answer to the question that I posed that although the film thickness is small still we are considering the y momentum effect and that effect comes into the picture because the pressure variation is with h and h is a function of x. Had h not been a function of x that would not have been important but pressure variation is a function of h and h is a function of x. So that makes it coupled with the x momentum equation which one plus yes this is plus this is minus yes. In terms of a shorthand notation we can write del del x of ps plus epsilon cube lc square rho g cos theta by mu uc h del square u0 del y square. So the shorthand notation see the right hand side with respect to integration with respect to y is a constant because the right hand side is a function of x and t but not y. So this is function of xt so this entire thing let us call it capital A just for ease in writing. We will integrate it with respect to y. So del u del y is equal to ay plus we have not used c1 so we can use c1 u0 and u0 is ay square by 2 plus c1 into y plus c2 which is a function of x and t. c1 is also a function of x and t but already we have written in the previous steps so I have not written again. Now we need to apply the boundary conditions to get c1 and c2. So what are the boundary conditions at y is equal to 0 u0 equal to 0 this is the straight forward no slip boundary condition. If there is a slip boundary condition instead of this you apply the slip boundary condition straight forward. So all the problems that we have worked out with no slip boundary condition I will give you a homework of repeat those with slip boundary condition and this homework is a very important homework because I will try to relate this homework with some problem in the n semester examination that I will n semester questions that I will give. So this is supposed to be a important homework please do this carefully. So at y equal to 0 u0 equal to 0. So that means 0 equal to okay let us write the other boundary condition then we will apply that. What is the other boundary condition? See when you write the boundary condition the mathematical form you can retrieve from the notes. But try to first say what is the physical boundary condition that we want to write here. What is the physical condition? Not no shear I mean there could be shear because of surface tension gradient. So the tangential stress balance or tangential force balance better to say not stress balance is not a correct term you cannot balance stresses. So tangential force balance. So tangential force balance what it gives we have worked out that but just at y equal to h del u0 del y is equal to epsilon beta by capillary number surface gradient of sigma. What is this del tilde? If you recall this is del del x plus del del y to del h del del x. We have discussed all this this is just like every time it is not necessary that we again go back to where from we started discussing and so I am just giving you the summary. So we need to calculate C1 and C2 based on these conditions which we will do in a moment at y equal to 0 u0 equal to 0. So that means but before that at y equal to h del u0 del y is equal to epsilon beta by capillary number del tilde sigma from that we can find out what is C1 right. So we can find out what is C1 from the boundary condition 2 use boundary condition 2. So if you use the boundary condition 2 then epsilon beta by capillary number del tilde sigma is equal to AH plus C1 which means C1 is equal to epsilon beta by capillary number del tilde sigma minus AH. Then the for this one use boundary condition number 1 at y equal to 0 u0 equal to 0 that means C2 is equal to 0. So we have got C1 and C2. So we can write a compact expression for u0 which I will write in a moment. So let me reproduce the expression for u0 u0 is equal to A into y square by 2 minus yh plus epsilon beta by capillary number check whether substituting the boundary condition this comes this should come right you have checked it. So what is A where A is equal to del del x of epsilon cube lc rho g cos theta by mu uc h plus ps minus epsilon square lc rho g sin theta by mu uc. So A is also a function of x and time. So what equations we have used so far? Let us make an accounting we have to use all equations we have to use all boundary conditions we have seen that that is the system is perfectly closed. So what are the things that we had in hand and what we have used let us make an accounting. So x momentum y momentum before that continuity x momentum y momentum these 3 equations boundary conditions no slip at wall no penetration at wall then at the interface kinematic boundary condition then normal force consideration and tangential force consideration at interface these are at interface. So let us see out of these what we had made use of. So we have made use of the x momentum and y momentum no slip at the wall not yet no penetration at the wall we have not used at yet as yet. Then interface what boundary condition we have used tangential force so we are yet to use the continuity equation no penetration at the wall kinematic boundary condition normal force we have used where we have used you have written when we have written p0 as a function of ps then that ps is in turn dictated by the normal force balance condition that is ps-p atmosphere is a function of the surface tension and the excess pressure. So normal force we have used so this circle things we have not yet used continuity no penetration at wall and kinematic boundary condition. So let us try to do that now let us write the continuity equation this is the continuity equation. Now what we will do is we will integrate the continuity equation with respect to y we will use a integral form of the continuity equation this h is a function of x and t that we have to keep in mind. Next what we will try to do is instead of working out this we will try to bring this derivative outside the integral. But as we understand that because h is itself a function of x that cannot be done without making adjustment of terms and that is given by the Leibniz rule. So we will use the Leibniz rule here to evaluate the differential differentiation under integral sign. So let us write the Leibniz rule yes u0. So Leibniz rule so basically if you want to bring the derivative out of the integral sign these 2 terms you have to make as an adjustment I mean I do not remember whether I have mentioned this but let me try to point out here that this Leibniz rule what we are writing in mathematics is just like a manifestation of the Reynolds transport theorem. So if you see here that this is the rate of change for the system this is the like the rate of change for the control volume and this is like outflow-inflow okay. So this is the beauty of mathematics see this has not been derived by considering system control volume or whatever but these are like the boundary terms the rate of changes in the boundary. No it depends on this is a function of what so here I mean the case that what we are going to use see I mean it would have been better to write x, t I have not written here it is a function of multiple variables. So it is better to save to write this when this is a single variable function I mean I should have written here as b as a function of x and t b is like h here. So let us write that had it been a function of a single variable it could have written as like d dx but concept remains same. So here what is f? f is u0. So del del x of integral u0 dy from 0 to h is equal to integral del del x of del u0 del x dy 0 to h plus u0 x, h. So this is the value of the value of the value of the value of the value into del h del x next term is of course 0 because a itself is 0. So in place of this term we will write del del x of integral u0 dy 0 to h minus u0 at x, h del h del x plus to h – u0 at xh del h del x. So, this term we have written. Next term, integral of next term is very straight forward del v0 del y from 0 to h. So, v0 xh – v0 at 0. v0 at 0 is 0 because of no penetration boundary condition, right. So, we have now used the no penetration boundary condition. So, that is why. So, I mean these things are important. These subtle steps are important because now if we encounter a problem where there are holes in the wall, then instead of that being 0, that will be replaced by velocity at which fluid is flowing through the holes at the wall, right. So, no penetration boundary condition, you have to think in a general manner. When there is no hole, there is no penetration, but if there is hole, there will be penetration. So, this is equal to 0. Now, we have not used the kinematic boundary condition. What is the kinematic boundary condition? Kinematic boundary condition is v0 kinematic boundary condition is equal to del h del t plus u0 del h del x at interface. This is the total derivative of h with respect to t. So, we can write u0 del h del x or – u0 del h del x which one? This is alright. This is the kinematic boundary condition, right. So, – u0 del h del x plus v0 is equal to what? At the interface is equal to del h del t, right. So, that means you can replace this term by del h del t. That is how we have used the kinematic boundary condition. So, this integrating the continuity equation gives us a natural way of incorporating the kinematic boundary condition at the interface. So, kinematic boundary condition has also been utilized. So, we are left with del h del t plus del del x of integral u0 dy from 0 to h is equal to u0. And u0 as a function of h, you know. What is u0? So, let us write a into y square by 2 – yh plus epsilon beta by capillary number del tilde sigma y. So, integral u0 dy from 0 to h is equal to a. This will become h cube by 6 – this will become h cube by 2, right, plus epsilon beta by capillary number h square by 2. So, this is what? This is 1 – so, – a h cube by 3, right, plus epsilon beta by capillary number s, this is sigma. So, you are left with del h del t plus del del x of – a h cube by 3 plus epsilon beta by capillary number this is equal to 0. We have utilized all equations and all boundary conditions. This equation has only one unknown dependent variable that is h, right. In a also there is implicit h, the a is function of h and other terms explicitly have h. So, this is the governing equation for h as a function of x and t which we will solve. So, we have come up with one governing differential equation which is a PDE and that PDE we have to solve for h as a function of x and t and there are some nonlinearities in this equation as well. So, in the next lecture, we will take it up from here. What we will do is we will summarize the generalized algorithm that has led to the derivation of the final form of this equation. So, that you can apply this for any problem and then we will work out a specific problem which will be giving rise to a governing differential equation that is a special form of this and we will solve that special form and give you some physical insight on a very important problem that is spreading of a droplet on a surface. We will do that in the next lecture. Thank you very much.