 In our previous lecture, we started with a brief introduction to lubrication theory. So we will have a brief recapitulation of the physical situation, what we are trying to address. So let us say that there is a passage where the lateral length scale is much less than the axial length scale and let us consider this example where the top wall moves with a velocity relative to the bottom wall and that velocity may be arbitrary with the velocity components along x and y. So the situation is 2 dimensional and one of the important hallmarks of the particular arrangement is that the existence of a physical small parameter. The small parameter is in this particular case, the ratio of the transverse length scale to the axial length scale. So with this arrangement, let us try to see that how we can physically describe this arrangement through governing equations. Let us come to the board and try to explain that. So the physical situation, let us draw a schematic that you have 2 walls and you set up your x and y axis like this, the gap h is a function of position and time, you have an arbitrary velocity of the top wall relative to the bottom wall. The axial length is lc and the scale that we consider in the transverse direction. So I mean because h is continuously varying, you may consider the scale to be any local height provided that the variation in the height is not strong. That means there is a weak variation or slow variation of height as we move along x. If there is a slow variation in height, there is not much difference between this height, this height, this height and this height. So any particular height you can choose as the characteristic scale. Just as an example, if this height is Ho, then we can consider that as a characteristic length scale. I mean if this is the maximum height, then if the maximum height by lc is small, then all the other heights are also much smaller than the corresponding axial length scales. So with this particular arrangement and we have to keep in mind also that the top wall moves with a velocity v. And as we discussed yesterday that we will indicate the dimensional quantities with prime and non-dimensional quantities without prime. So that we will consider here and we will begin with the continuity equation. So let us assume that it is an incompressible flow. So the continuity equation for 2-dimensional incompressible flow. So this is x prime and this is y prime in terms of dimensional coordinate. Dimensionless there will be x and y. Let us describe the orders of magnitude of various terms. So what is the order of magnitude of this? Let us say that the characteristic scale is uc, characteristic velocity scale is uc. Now what is uc? We do not know because it depends on the physics of the problem that what is the characteristic length scale. If the characteristic length scale is, sorry what is the characteristic velocity scale? Characteristic length scale is described by the geometry. But characteristic velocity scale is described by the physics of the problem. So if the top plate velocity is the dominating physical influence then its velocity along x will be the characteristic scale. However if there is some body force which is dominating towards dictating what is the velocity then there will be a velocity scale dependent on that body force. We will see that how that can be obtained. So at this stage uc is totally arbitrary. It depends on the boundary condition or the physics of the problem for us to decide that what is the appropriate choice of uc, okay. So this scales as uc by lc where you do not know a priori what is uc. This scales as say this is vc by h0. So as we have discussed earlier that these 2 terms they must have the same scale so that the net effect is 0. So that means vc by h0 is you can say is of the order of uc by lc. That means vc is of the order of uc into h0 by lc. H0 by lc we denote by epsilon. So vc is equal to epsilon into uc. So that if epsilon is small that means there is one dominant flow direction. That is also another hallmark of the lubrication theory that you have a dominant flow direction because the dominant flow direction is dictated by u and not by v. However you cannot trivially neglect the momentum transport along y because of may be possibility of existence of a pressure gradient may be due to body force. Some body force may be there and also I mean you may have some sort of motion along y because the plate is being pulled or pushed vertically upwards or downwards that might be a possibility. So either the kinematics or the kinetics may decide that there may be some significance in the momentum transport along y although the velocity component along y is one is significantly less as compared to the velocity component along x okay. So this is what the continuity equation gives then let us write the x momentum equation put dash in all the parameters just to indicate that they are dimensional parameters. Find out we will now non-dimensionalize various terms. So we will write u is equal to u dash by uc, v is equal to v dash by vc, x equal to x dash by lc, y is equal to y dash by h0, p is equal to p dash by pc, fx is equal to fx dash by fxc. I am trying to derive a very general formulation where whatever body force may be there is present and so the normalization factor for the body force depends on what physical it is. Now certain scales we have already discussed like uc, vc, lc, h0 but we have not discussed about what is pc, what could be fxc. What could be fxc depends on what is actually acting as a body force along x. So in a general treatment it is difficult to prescribe what could be fxc but it is possible to prescribe what is pc and we will try now to find out what is an appropriate pressure scale that is one of the key parameters in the lubrication theory. So now let us one thing we have not yet done the time t is equal to t dash by tc. So there are different possibilities of choosing the time scale like effective time scale, diffusive time scale, forcing based time scale we have seen that these are there. Now here an appropriate choice would be an advection based time scale. There is no forcing here and the advective time scale will naturally have this term of similar effect as that of these terms. So if these terms are not important, this term will not be important. So the advection time scale so what is tc then tc is lc by uc okay. Now let us write the various terms in a dimensional form and certain non-dimensional parameters will come out of that. So rho what will be this term uc by tc into del u del t plus uc square by lc del u del x plus vc uc by h0 that was u del u del x this is v del u del y is equal to minus pc by lc del p del x plus uc by lc square del 2 u del x2 plus uc by h0 square del 2 u del y2 plus fx into fxc. Next we will make some simplifications. So tc is lc by uc so this becomes uc square by lc alright uc square by lc this is uc square by lc now vc by h0 is same as uc by lc from the continuity equation. So this also becomes uc square by lc so the left hand side let us write the next step you have rho uc square by lc as common into del u del t plus u del u del x plus v del u del y is equal to minus pc by lc del p del x plus mu uc by h0 square next maybe let us utilize the top part of the board. When we are writing these equations what appears to be a small parameter to me is the size of the board because I mean whatever you write it does not fit either in the left or in the right or in the top in the bottom so that is at least one small parameter. Now let us continue with this equation in the top little bit down we write so that it is covered by the camera. So what we do now is let us try to find out what is the appropriate pressure scale what is pc okay so we have to keep in mind that the pressure scale see the dominant pressure gradient term first of all forget about the pressure scale first of all what we do is that we multiply both sides by h0 square by mu uc okay this h0 so basically to get rid of this we multiply both sides by h0 square by mu uc so rho uc square by lc into h0 square by mu uc. Now let us look into various terms see one has to keep in mind that the order of magnitude of various terms depend on how the terms are non-dimensionalized for example see there is a key difference between normalizing with an appropriate scale and any non-dimensionalization for example in the problem that we have we have we have been discussing you could non-dimensionalize y dash by lc as well right because lc is also a length but that is not a proper normalization that is not a proper scaling based normalization since y dash it is range is from 0 to h0 not 0 to lc. So if you normalize a variable with its range then the dimensionless parameters are constrained to be within 0 to 1 in terms of order when the dimensionless parameters in terms of order are constrained to be within 0 and 1 then all their derivatives are also constrained to be within the order of 0 to 1. So that means you can safely say that this is of the order of 1 if this is of the order of 1 you can now quickly convert to a conclusion that this is of the order of epsilon, epsilon square sorry not epsilon of the order of epsilon square. So because this is of the order of epsilon square we can conclude that this term is asymptotically smaller than this term. So whenever you are considering an asymptotic expansion formally this term will not come out to be important but from physical arguments you can tell it without doing any asymptotic expansion into h0 square by mu uc right into h0 square by mu uc. The next observation is that in a situation where this term is of the order of 1 and there is a competition between viscous force and force due to pressure gradient and appropriate choice of the pressure scale should be such that this is of the order of 1. So that you will get a term like just dp dx. So this gives a choice for PC. So what is PC? So PC by LC is equal to mu uc by h0 square. So that means PC is equal to mu uc into LC by h0 square right. So this we can write as multiply both numerator and denominator by LC and h0 by LC is epsilon. So this is mu uc by epsilon square LC. Now what is this term? Let us calculate this, this is rho uc into LC by mu into h0 by LC square. 1 uc gets cancelled so 1 uc into LC by mu into h0 by LC square. So what is this? This is nothing but the Reynolds number, Reynolds number based on LC and this is epsilon. So this is epsilon square finally the last term the multiplying factor what is this? In place of h0 you can write LC square by mu uc epsilon, epsilon square. So the final equation in terms of epsilon is Reynolds number into epsilon square del u del t plus u del u del x plus v del u del y is equal to minus del p del x plus epsilon square del 2 u del x2 plus del 2 u del y2 plus fx fxc into LC square by mu uc epsilon square. Which one? Last one yes yes right. So this will be this will be LC square and this will be epsilon square right. Oh sorry sorry yes epsilon square LC square right. So this is epsilon square will be in the numerator right epsilon square LC square. So now can you tell from this equation just by looking without making any mathematical treatment or anything that which terms have potential importance relative to the other terms or which terms are potentially less significant as compared to the other terms. See here you have certain terms which are of the order of epsilon square and epsilon being small order of epsilon square will make it significantly smaller as compared to order of 1 right. So this term and this term you may well argue that well this term has also epsilon square but that is a dangerous conclusion to talk about because you really do not know what is the scale of fx fxc okay. So it might so happen that adjustment with the scale of the fxc will not eventually make it of the order of epsilon square although elusively epsilon square is happening to be there in the numerator. So we cannot make any conclusion of the relative dominance of this force or not because simply we do not know what is fxc. Not only that we do not know what is uc. So uc depends on the physics of the problem as we discussed. So we do not know what is fxc we do not know what is uc. So just by looking into this term although it contains epsilon square that does not mean that eventually when all the expressions are substituted finally it will be also of the order of epsilon square. Therefore we cannot conclude that this term is trivially negligible but we can conclude that these two terms are trivially negligible. That means if you make an expansion if you write an expansion u equal to u0 plus epsilon u1 plus epsilon square u2 this is called as an asymptotic expansion. In this way you expand u like this v is equal to v0 plus epsilon v1 plus epsilon square v2 like this p is equal to p0 plus epsilon p1 plus epsilon square p2 like this. You may well argue that if we make this asymptotic expansion how will we ensure that this expansion is justified. This expansion is justified only if there are 2 possible I mean 2 important constraints that are obeyed. One epsilon is small that we have ensured the other is that this parameter u0 u1 u2 these are constrained to be within 1 and that is ensured by considering appropriate scales for normalization and that is why we are actually doing it in a dimensionless form otherwise we could work on this with a dimensional form as well but the dimensionless form gives us this possibility that we are dealing with individual parameters which are of the order of 1. Dimensionally if we are handling these parameters that we cannot ensure. Dimensionlessly we can ensure by choosing suitable scaling parameters okay. So now if you substitute you will see just let us do it mentally because if we I mean this board does not give me enough space of writing all these expansions so let us just do it mentally and from that we can figure it out. So here you will get u0 plus epsilon u1 plus epsilon square u2 just substitute in place of u u0 plus epsilon u1 plus epsilon square u2 for v also like that for p p0 plus epsilon square plus epsilon p1 plus epsilon square p2 for u also like that okay. So when you do that you will see that the dominant terms of the order of 0 will be what of the order of epsilon to the power 0 that is what I am calling of the order of 0 of the order of 0 means of the order of epsilon to the power 0. So of the order of epsilon to the power 0 which terms will remain. This term will definitely not be there because there is a epsilon square multiplier. So you will get 0 is equal to minus del p0 del x plus del square u0 del x square and the body force u it is better to retain because you do not know what is the scale what is the appropriate scale. So plus fx fxc epsilon square lc square by mu uc y square right del y square if no body force is present then the last term is not important. So then we are left with an equation which is del p0 del x is equal to del square u0 del y square that is the leading order equation and in lubrication theory we do all the calculations using the leading order equations only because that gives the dominant effect. Now there may be an interesting yes no no no that is what I told that it is this is illusion you do not know what is uc you do not know what is fxc you do not know that what is the net effect. So in the leading order the body force may be dominant okay. So we should have a provision for keeping it and if the body force is not dominant in the leading order that has to be that has to be considered from the physics of the problem not from anything else then you drop this term. Now a very interesting point before we move on to the y momentum equation. The interesting point is that out of pressure and viscous stress which is more dominant in lubrication theory let us see. So when you are writing the pressure okay the viscous stress the viscous stress is what viscous stress is of the order of like mu del u del y right. So mu uc by h0 right that is the viscous stress. So h0 is what epsilon into lc right. So for a given lc the viscous stress scales with 1 by epsilon. The pressure scales with 1 by epsilon square right. So for small epsilon the pressure is what is dominating and not the viscous stress right. So see this I mean these are I mean certain things which are intuitive and certain things which come out of the smallness of the parameter epsilon. So that is one very important observation. So that is why in lubrication theory at the end to calculate the force we use the pressure distribution only. I mean people may argue that there are viscous stresses which are present then why do you just calculate the force only on the basis of pressure. The reason is that in the leading order the pressure is what is dominating being scaling with 1 by epsilon square whereas the viscous stress scales with 1 by epsilon. So to sum up the discussion on the x momentum equation we can see that the x momentum equation boils down to this. We will come to the boundary conditions of this equation later on but now let us move on to the y momentum equation. y momentum del v del t plus u del v del x plus v del v del y is equal to minus del p del y plus we will add a dash to all the terms this is the dimensional form. Next what we will do we will use the suitable scaling parameters to non-dimensionalize this. So as usual we have v is equal to v dash by vc u equal to u dash by uc then x equal to x dash by xc y is equal to y dash x is lc right y is equal to y dash by h naught t is equal to t dash by tc p equal to p dash by pc and f y is equal to f y dash by f yc. One very important point is that the pressure scale whatever is established to the x momentum equation same pressure scale has to be considered for the y momentum equation pressure is just a scalar. So its scale will not change as you go from one direction to the other okay. So let us quickly write the various terms in their dimensionless forms from our previous experience of writing it for the x momentum equation we can do similar thing for the y momentum equation. So rho see how do you do these things quickly just it is very important to learn how to do these things quickly without getting into like writing all the derivatives and all. See this term will be u scale into v scale by x scale right. So and by using the continuity equation and the advective time scale all the terms will have the same coefficient at the end. So we write that uc into vc by lc into del v del t plus u del v del x plus v del v del y minus pc by what h naught del p del y plus mu vc by h naught square. Now what is pc just tell me from your notes what is pc mu by epsilon square lc. So the next step what we will do we will multiply both sides by h naught square by mu vc right just the same way as we did for the x momentum equation. So it becomes rho uc vc by lc into h naught square by mu vc. Now let us just simplify this this term. So rho uc vc is uc into epsilon. So uc square by epsilon lc anyway one vc gets cancelled out right. So we do not have to care about this vc this vc gets cancelled out with this. So rho uc into epsilon square lc by mu. So that is equal to what Reynolds number into epsilon square alright this is then let us come to this term oh we have not written del p del y right del p del y was missing here. So this term mu uc one h naught gets cancelled one h naught by epsilon is okay mu also gets cancelled uc by vc into h naught by lc into 1 by epsilon square right. So uc by vc gets cancelled with h naught by epsilon. So this becomes 1 by epsilon square and the last term if you want it you can simplify it it will not make much of a difference so far as what we are doing currently because we have not assumed any body force but just for the sake of completeness this h naught square is lc square into epsilon square and vc is lc into epsilon sorry uc into epsilon. So the last term becomes fy fyc into lc epsilon by mu sorry lc epsilon square by mu uc lc square epsilon lc square epsilon right there is a epsilon in the denominator so fy fyc lc square epsilon by mu uc. Now you multiply both sides by epsilon square okay let us do it at the top you multiply both sides by epsilon square the reason is 1 by epsilon square has come here so just to bring it in the left hand side so you have Reynolds number into epsilon to the power 4 del v del t plus u del v del x plus v del v del y is equal to minus del p del y plus epsilon 4 del 2 v del x 2 plus del 2 v del y 2 epsilon square del 2 v del y 2 plus fy fyc lc square epsilon cube by mu uc. Next what we do we make an asymptotic expansion u equal to u0 plus epsilon u1 plus epsilon square u2 like that v equal to v0 plus epsilon v1 plus epsilon square v2 like that p equal to p0 plus epsilon p1 plus epsilon square p2 like that. If you do that what will come out as a dominant term forget about the body force keep it separate what will come out as a dominant term only the pressure gradient term. So the consequence of this if you neglect the effect of this term is simply this one because other terms are of the order of epsilon square epsilon 4 like that body force term of course what is the order we do not know but we are not keeping the body force in per view for this particular discussion. Now there is one subtlety here we are considering that this term is of the order of epsilon 4 so this is small but what if the Reynolds number is large but fortunately we are dealing with microfluidic problems where the typical Reynolds number is small. So we are dealing with low Reynolds number hydrodynamics so there is no chance of blowing this term up despite the smallness of epsilon. So the assumption that we are making are very much consistent with microfluidics. So to sum it up let us write our x momentum and y momentum equations. So you have 0 is equal to minus del p0 del x plus del 2 u0 del x2 sorry del y2 and 0 is equal to minus del p0 del y if you are not considering body force that means that because this del p0 del y is equal to 0 that means p0 is not a function of y so this you could as well write as dp0 dx. So your governing equation actually boils down to this one in absence of body forces remember this what are the boundary conditions. Now let us just draw the sketch that you have 2 boundaries this boundary say has a velocity u prime star i plus v prime star j this boundary is having this general velocity. So to make one of these as 0 instead of giving both as non-zero components what we do is that we apply a transformation of coordinates where you are using a coordinate system which is translating with respect to a velocity u star prime along the x direction and we fit that coordinate system with the bottom plate. So with respect to this this will become 0 one of the components will become 0. Now if this be the case then what are the boundary conditions this is y equal to 0 this is y equal to h these are all these h is dimensionless h. So this if you want to put this as dimensional parameters then just get rid of the dash that will make dimensionless parameters get rid of the dash. So these are dimensionless parameters in terms of which the governing equation is written. So we will write the boundary conditions also in terms of dimensionless parameters at y equal to 0 what is the boundary condition u0 equal to minus u star right that is the no slip boundary condition relative to this moving reference frame which is drawn by the green color and y equal to h what is u0 0. See the description of see one of the boundary conditions we have made 0 by using this transformation and this does not depend on what is v at that boundary. See this is what is the very critical thing that whenever we have made this transformation we have not cared about what is v we have made the transformation with the translation along x nothing to do with y. Why because in the leading order equation v has never appeared in the leading order equation the net effect of the momentum equation along x momentum equation along y as well as the continuity equation the continuity equation we have forgotten its role but its important role is to set up the scales for u and v the relation between vc and uc that is set up by the continuity equation. So, the smallness of the parameter epsilon is established through the continuity equation now. So, you can see there is no v0 which is appearing so it does not matter what is v of the top boundary so far as description of this boundary conditions are concerned. So, once we get this equation along with this boundary conditions next we will make an attempt to solve this equation to get u0 and from that establish a governing equation for the pressure distribution. We will see that how do we establish a governing equation for pressure distribution for that we have to couple this equation with the continuity equation. We have not considered the continuity equation as yet because we have used only the x momentum and the y momentum. We have never used the continuity equation to close this system and as we have understood that to obtain a pressure distribution we must close the momentum equations in conjunction with the continuity equation. So, we will bring the continuity equation into the picture and couple that with this equation to get a pressure distribution and hence a force distribution on one of the plates say the top plate that we will take up in the next lecture. Thank you very much.