 So far we have discussed about some examples of steady flows in the paradigm of low Reynolds number flow. Now we will discuss something more about steady flows later on and we will discuss about a theory which somewhat generalizes the elementary low Reynolds number hydrodynamics that we have discussed and that theory is known as lubrication theory. So we will discuss about that in details later on but prior to that let us move into some cases of unsteady flows. So let me talk about little bit of motivation as we discuss about the unsteady flows. So when we discuss about unsteady flows remember that whatever we are discussing here is in the broad with the broad motivation of studying some examples related to microfluidics. So we are proceeding towards that direction in many ways. Now we will start with an example of an unsteady flow with a classical problem which is not a problem of microfluidics that is Stokes first problem. So the Stokes first problem is something like this. You have a solid plate and there is a fluid over the top of the solid boundary. This fluid is initially stationary. Now you suddenly pull the plate with a particular velocity say u equal to u0. Then because of the movement of this plate there will be a disturbance in momentum in the fluid which will be propagated by virtue of the fluid property viscosity. So you want to see that how does this momentum disturbance get propagated as a function of position and time. That is as a function of the y coordinate and the time coordinate. Now what is the relevance of this problem in microfluidics? See this is not a microfluidic system. In fact this is not a confined system. Now in microfluidics there are several effects which come into the picture. One possible effect is the effect of confinement. That is if you make a plate coming on the top of this plate with a small gap between the two plates then it becomes a confinement. So the question is that how does the confinement affect the flow? To understand that we will first solve this problem when we will consider that there is no confinement. See if we want to understand what is the effect of confinement first we have to study that what is the scenario when there is no confinement and then in the next problem that we will be solving we will put a plate at the top of this one. So it becomes a confinement and we will see what is the difference between the results in the two cases and that will give us an indication of what could be a possible effect of confinement in unsteady flows. That is one of the classical motivations. I will talk about some more motivation later on. So as we proceed so when we get a little bit bored with the mathematical exercise we will talk about some practical motivation to compensate for that. So let us start this working out this problem by writing the governing equations. So let us assume that the velocity u is a function of y and t. It is not a function of z because the plate is considered to be of infinite width. It is not considered to be a function of x that is not fully developed flow to be in a literal sense because fully developed flow we generally talk about in the context of internal flows that is flow through channels pipes and all this. But the notion is similar that there is a translational invariance that means if you want to look into the velocity field at this location the same velocity field will be repeated if you go to a different axial location. If you translate and move on to a different axial location that is called as translational invariance. So with that u is a function of t and y or y and t whatever. So this is equal to so the left hand side of the Navier-Stokes equation will have this unsteady term. Remaining terms will not be there because u del u del x plus v del u del y del u del x is equal to 0 and as a consequence from the continuity equation v is equal to 0 that we have already discussed. So that means the remaining terms if u is not a function of x the remaining terms are not there but unsteady term is there that is the basic difference between the previous problems and this problem that they are the unsteady terms were not there now the unsteady term is there. And because of the length scales involved the dominant viscous term is the viscous term which is mu del 2 u del y 2. The x length scale is being much greater than the y length scale this term is dominating as compared to the gradients along x. There is no pressure gradient because it is like an infinite flow medium where it is subjected to like uniform like atmospheric pressure from all sides. So it is an open problem. Now of course I mean you can artificially impose a pressure gradient on this but just like a flow over a flat plate you have the flow is acting flow is being subjected to a 0 pressure gradient that is also very similar to this. Now as we can clearly see that the problem is dictated by the kinematic viscosity. So mu by rho we can write this as nu and the problem is dictated by the kinematic viscosity. So here what will be the time scale? The time scale will be we have discussed about 3 possible time scales. The advection time scale, the diffusion based time scale and the forcing based time scale. So here what will be the time scale? It will be the diffusion based time scale because you can clearly see that these two are competing with each other. Now there is a time scale associated with the problem but there is no natural length scale. There is no natural length scale involved. So why is there no natural length scale involved? Because there is it is not in a confinement. If it had it been in a confinement the height of the confinement would have been the natural length scale but it is an unconfined medium this is like in this is extended up to infinity. So there is no natural length scale but it is important to figure out that what could be a possible natural length scale based on the time scale. So what is that length scale? Length scale means the distance up to which the effect of this movement of this plate will be felt. So question is that on which factor is that distance related? One of the critical factors on which it is related is time. So if you allow more and more time, more and more fluid in the outer layer will feel the effect of the plate. Theoretically if you allow infinite time all the fluid will feel the effect of the plate but in a very short time only a small extent or a short depth from the plate will feel the effect of the solid boundary okay. So time is an important parameter. Is there any other important parameter? Yes, kinematic viscosity we have discussed about that. But more is the kinematic viscosity more will be the effect or the penetration depth of the effect of the movement of this plate. So the length scale up to which the disturbance due to the motion of this plate will be felt should be increasing with time and should be increasing with kinematic viscosity. But exactly how? To understand that we will make an order of magnitude analysis of this equation very simple analysis. So what is the order of magnitude of this term? U reference is U0 by T reference. This is of the order of nu U0 by y reference square. So y reference scales with square root of nu T reference okay. That means although there is no natural length scale in this problem but the length scale is eventually decided by the kinematic viscosity and the time scale okay. So from here we can qualitatively expect that U by U0 will be a function of y by y ref. And this function is expected to be self-similar. This is very much analogous to the boundary layer theory where velocity profiles like if you draw an analogy with the boundary layer theory if you have this is the boundary layer velocity profiles at various locations are indeed very different. But if you non-dimensionalize or normalize the velocity profile U by U infinity as a function of y by delta this being delta this being the nu delta at the nu location. So U by U infinity will be a single valued function of y by delta. Here the delta is this one that is the penetration depth just like the boundary layer okay. So U by U infinity is like U0 here is a function of y by delta. Is this delta it is a function of t right this we could get from a physical intuition. So we can write that U by U U0 is a function of y into a function of t right this gt roughly scales with 1 by delta okay. So we can because it is a single valued function of this we can write replace this with a variable eta which we call as a similarity variable okay and seek for a solution and when you express the solution in terms of this similarity variable again just like method of separation of variables the similarity transformation the objective is to convert the partial differential equation into ordinary differential equation because if you are able to express this as a function of eta as a single variable then it will be a U by U0 will be a function of a single variable only. Question is is it always true I mean is it always possible to cast this type of PDE into in the form of an ODE like this depends on the physical situation on hand in this physical situation it is possible. So I mean there are situations when that may not be possible but here the physics is giving us or driving us towards this kind of variable separation. Now we will proceed with the solution of this problem so U by U0 is a function of eta where eta is equal to y into gt. So what is del u del t U0 df d eta into del eta del t right using the chain rule. So del eta del t is y into g dash right where g dash means dg dt then del u del y del u del y y we require del u del y because we need to evaluate this term. So del u del y U0 df d eta into del eta del y so U0 df d eta into g. So what is the second derivative? Second derivative is U0 so you basically make dd eta of this into del eta del y. So another g will come so it will be U0 d2f d eta 2 into g square. So let us then write this equation del u del t that is U0 df d eta to y g dash is equal to nu del 2 U del y to nu U0 d2f d eta 2 into g square and what is y? y is eta by g right this y is nothing but eta by g. U0 gets cancelled from both sides so g dash by nu g cube is equal to d2f d eta 2 by eta df d eta right. So we have clubbed all functions of t in on one side and all functions of eta on one side. So this is function of t only this is function of eta only. So these 2 are the same only if it is a constant. So let us say that the constant is c. So we can write dg dt, g dash is dg dt then 1 by nu g cube is equal to c. So g to the power minus 3 dg is equal to nu c dt. So g to the power minus 2 is equal to by minus 2 is equal to nu ct plus c1. Now we have to remember that what is g? So g is what? g is 1 by g is the penetration depth at a given time right. So at t tends to 0 what is g? At t tends to 0 the penetration depth is what? The penetration depth is what? 0. So g is 1 by the penetration depth that will tend to infinity. So at t tends to 0 g tends to infinity that means c1 will be 0. So you have g is equal to 1 by minus 2 nu ct square root. So you can see so this is the beauty of scaling analysis. Say a rigorous mathematical analysis has given rise to this form of g but you could tentatively get this form without going into this mathematical analysis simply by scaling the two terms that are appearing in the governing differential equation. So you will clearly get that g scales with 1 by square root of nu t. Remember that g is 1 by y reference. y reference is scales with square root of nu into t reference that is what is sort of supported or justified by this more detailed mathematical treatment. Now the question is what is the value of c that you may take? So only restriction is that c has to be negative because this g has to be a real number. So c must be negative. So what is the typical choice of c? So that depends on convenience. So you may choose for example c is equal to minus half or 2 minus 2 like that so that you know the square root business goes off. So as an example let c is equal to minus 2. So then remember this is not a must. This is just one example. So whatever c you will take it does not disturb this equality. So you recall that in school level ratio of proportion problems a by b equal to c by d equal to k whatever k you take that does not disturb the equality of a by b and c by t that is what is important. So this equality is never disturbed because of the presence of this c. So for different c the solution of g will be affected and the solution of f will be equally affected. So I mean different c's will eventually give rise to the same solution. So you will have g is equal to minus 2 as an example. So c is equal to minus 2 as an example. So g will be 1 by 2 square root of nu t. This is about g. Now let us concentrate on f. So you can see that we have now been successful in converting the pde into an od. We will write the boundary conditions immediately but before that we will just integrate these ones. So the boundary conditions maybe we can write first and then integrate. So what are the boundary conditions of this at eta equal to 0 what is f? f what is the definition of f? Definition of f is u by u0. u by u0 is f. So at eta is equal to 0 what is u? u is u0 no slip boundary condition. You have to be careful here the plate itself is moving. So no slip boundary condition does not mean that the fluid is stationary. It means the fluid is stationary relative to the solid boundary. So if the solid boundary is moving with the velocity u0 the fluid will also move with the same velocity u0. That is what is the no slip boundary condition. So if u is equal to u0 then what is u by u0? 1. So at eta is equal to 0 f is equal to 1 and at eta tends to infinity f will be 0 because far away from the plate effect of the disturbance of the plate will not be failed. So the velocity will be 0. So our problem definition boils down to this. Now to solve this problem I mean it is a very straight forward solution. Let dfd eta is equal to some h. So you have dhd eta plus 2 eta h is equal to 0. So dh by h is equal to minus 2 eta d eta. So lnh if you integrate is equal to minus eta square plus ln of c2. We have already used a constant c1 for this problem. So let us give it a name c2. So h is equal to c2 into e to the power minus eta square. What is h? h is dfd eta. So what is f? c2 e to the power minus eta square d eta plus c3. Now we will apply the boundary conditions. So these boundary conditions let us give names. This is boundary condition number 1. This is boundary condition number 2. So the boundary condition 1 what does it say? The boundary condition 1 says that at eta equal to 0 f is equal to 1. So that means c3 is equal to 1. So you have f is equal to 1 plus c2 integral 0 to eta e to the power minus eta square d eta. We apply the boundary condition that at eta tends to infinity f tends to 0. So the boundary condition 2 0 is equal to 1 plus c2 integral 0 to infinity e to the power minus eta square d eta. Then we can make a substitution that eta square is equal to z. So eta is equal to z to the power half that means d eta is equal to half z to the power half minus 1 dz. So 0 is equal to 1 plus c2 by 2 integral 0 to infinity e to the power minus z into z to the power half minus 1 dz. By definition 0 to infinity e to the power minus z into z to the power n minus 1 dz is gamma of n. So this is gamma of n is equal to half, gamma function of half and gamma half has a value of root pi. So c2 becomes minus 2 by root pi. So f becomes what? 1 minus 2 by root pi integral 0 to infinity e to the power minus eta square d eta. This is the error function of eta, erf eta, sorry 0 to eta, not 0 to infinity, yes, 0 to eta. So we write the solution here f is equal to u by u0 is equal to 1 minus error function of eta where eta is equal to 1 by 2 root nu t. So 1 minus error function eta is also called as complementary error function, erfc. So that completes the solution of the stokes first problem. Now how will the velocity fields vary? We will look into that but as we are doing for most other problems, we will recapitulate through the slides what we have done for the stokes first problem and then we will look into the graphs which will give the solution for this velocity field, basically a visualization or the plotting of this velocity field. So if you look into this, so we have used this similarity transformation and I am just browsing through it. So now we can see, you can, so in the similarity variable transformation, we have written f as a function of eta and g as a function of time, we have separated these variables and then we have integrated to find out that expression for g, c can be any negative real number we have chosen c equal to minus 2. Then we have considered the governing equation for f and we have integrated the governing equation for f by assuming, by letting f dash is equal to h and then that has given rise to the solution by substituting the necessary boundary conditions and these boundary conditions eventually give rise to the solution that f is equal to 1 minus error function of eta. Now it will be interesting to see how the solution looks graphically. So if you look into this graph, you will see that u by u0 as a function of y by 2 root nu t, you see that the effect of y and time, effect of position and time they are coupled. So that is the first important observation. See if you are asked that what is the value of velocity at a given y, then it depends on what is the time. So if you allow more and more time, the same velocity will be appearing at a greater depth from the plate. The simple reason is that at greater time more and more amount of fluid will now feel the influence of the solid boundary. So if you have higher value of t, you will get a higher value of y for which the same velocity effect is felt. That is why when you map the solution u by u0 as a function of y by square root of nu t that coefficient 2 is of course not that important from a concept actual point of view from a mathematical data point of view that is important but concept wise it is y by square root of nu t that dictates that what will be the solution of the velocity as a combined function of position and time. Now if you are interested to plot the velocity in a dimensional plane, not in a dimensionless plane. So that is shown in the second graph where you plot for some problem u as a function of y for different time. So you can see, so look at these extreme diagrams, extreme curves. So the first one say this is for a short time t is equal to 0.1 second. So only a small depth from the solid boundary feels the effect of the plate. But as you allow more and more time say t equal to 4 second for the particular data that has been assumed, you can see that now the penetration depth is much beyond what it was for 0.1 second. So the penetration depth becomes more as you allow more and more time and it depends critically on the kinematic viscosity. If the fluid has a greater kinematic viscosity at a given time the penetration depth will be more. So that is what we have learned from the Stokes first problem. Now the thing is that this problem is very interesting and classical. Now how can we relate this problem with microfluidics? So we will consider a problem which is a little bit deviated from the Stokes first problem but has some kind of notional relevance with that. So in the Stokes first problem you had a solid boundary moving towards a particular direction with a velocity. Then what we do is that to if consider the effect of confinement we put another plate which is on the top of this. So this is in the steady state the classical quet flow problem. You have flow between 2 parallel plates. One of the plates is moving relative to the other with a particular velocity that is the classical quet flow. Now we are interested in the unsteady solution of this problem not really the steady solution. Now question is many times this kind of question comes to our mind that when we study fluid mechanics these kinds of classical problems are always introduced and microfluidics is of no exception. So people might talk about that well in the introductory lectures in microfluidics on one hand you talked about cell, DNA these that and suddenly you are going to an abstract problem when a plate is moving relative to the other what is the physical relationship between these 2 and I think I have a responsibility to discuss about that before entering into this problem. So now in teaching pedagogy it is important that we discuss about certain things in the context of why we are studying that. So when we are studying quet flow for example forget about steady or unsteady. When we are studying quet flow the basic reason that should come to our mind the basic question that should come to our mind is that why are we studying quet flow that relevance should be brought in. When we are studying a poiseuille flow again poiseuille flow that motivation is very clear that you have a pressure gradient driven flow and you in reality in engineering systems you may have a pump to drive the flow. So that motivation is quite clear but when you are thinking of a quet flow that motivation is not very obvious. Where have you seen somebody pulling a plate on the top of a fluid and moving the fluid in that direction. I have not seen anybody doing that I do not know whether I will come across that example in the remaining part of my professional career but I have never seen that. But I have seen people pumping water through pipes and channels that I have seen. So somebody is suddenly pulling the plate in a particular direction and allowing the fluid to move of course you can make such actuators not by somebody literally manually doing it but doing it by a mechanism that is possible but there is actually a broader motivation towards this. So when we work out the steady state solution of this problem you will see that the shear rate like if you set up your x axis in this direction and y axis in this direction the shear rate will be u0 by h and the steady state velocity profile will be linear. I mean from your undergraduate studies you already know about that and we will revisit that quickly while working out the unsteady state solution. So then if the shear rate is u0 by h you can control the shear rate by controlling u0 or by controlling h. So that means you and this flow is acting this is a pure quet flow without pressure gradient. You can have a quet flow with a pressure gradient that is called as quet Poiseuille flow that is a combination of pressure gradient and the this effect. So this effect is giving what? It is giving rise to a shear rate which is a constant or the rate of deformation which is a constant. So you can have a design rate of deformation or the rate of shear on the flow and then study what is the response of the flow against that rate of deformation. So this is called as shear driven flow just like pressure driven flow this is a shear driven flow and this is just one manner in which you can impose the shear by moving a plate. But if you could apply the shear by whatever by a different mechanism and study the shear driven flow mechanism study the response of the system against that shear then if you have shear of this magnitude then the response to the system you can study as a function of shear of this magnitude which can be controlled independently by u0 and h. Not only that you can see like all of you have seen that you have a parabolic velocity profile for a fully developed pressure driven flow. Now let us say that this dimension is over a few millimeters. Let us say there is a biological cell which is sitting on the wall of this channel this is let us say this is few centimeter or millimeter this is like say this is what around 10 micron roughly. So over the span over which this cell is sitting you can approximately linearize this velocity profile. So a small part of the parabola is like a straight line right a very small part of the parabola this is actually this is exaggerated in a figure but this is actually very small. So that means if you have a cell sitting on the surface of a blood vessel and is subjected to shear of a given amount then what is the response of the cell against that shear if you want to study that this can be one of the basic mechanisms of probing the fluid mechanics in presence of that kind of physical situation okay. So you can see just I just wanted to I mean I can get into deeper discussions on this but we will not spend much more time on this before getting into the mathematical analysis but my whole objective of telling this is to convey this information to you that whenever we are doing mathematics for solving a problem never try to think that we are doing mathematics for the sake of doing mathematics. So we have to keep in mind that there is a broad picture in mind and the broad picture is a very complicated picture to get the essential physics of the broad picture to begin with we can simplify the problem to a considerable extent and study the essential simplified problem of relevance and that is one of the motivations of studying the quiet flow. Now the governing equations are still very similar as the previous problem right because putting a solid boundary at the top does not change the governing equation it changes the boundary condition. So the governing equation is still this one but what is the most fundamental difference between the previous problem and this problem right the length scale. In the previous problem there was no natural geometrical length scale the length scale was a function of the time scale. Now here there is a natural length scale which is the height h still one may argue see these are deeper and deeper discussions I am trying to get into these and to ignite ideas in your mind. Now you have seen the solution of the Stokes first problem for very short time for very short time the penetration depth is very short let us say this h is greater than the penetration depth then this plate is actually as if located at infinity although it is not literally infinity see when we say in mathematics something tending to infinity in physics we mean that it is far field that means it does not feel the influence of the forcing effect it does not feel the effect of the influence of the forcing effect that is what is far field or infinity. So infinity does not literally mean that it has to be dimensionally infinity. So it may so happen that the penetration depth at a given time is shorter than this h. But in a microfluidic situation we are not we are not going to practically encounter such a situation because for the practical times whatever we will be considering the penetration depth will be greater than this micrometer dimension typically this is a very short dimension. So in microfluidics that we are interested about studying this problem independent of the Stokes first problem provided this h is the natural length scale otherwise it becomes the Stokes first problem itself. So with that motivation that we want a separate grasp of this particular problem we consider that h is the natural length scale. So what is the so if we do an order of magnitude analysis for this particular problem what is the order of magnitude of this u0 by t ref. What is the order of magnitude of this nu u0 by h square y ref square is h square. So t ref is of the order of h square by nu. So this is what this is the diffusion base time scale. So we non-dimensionalize the problem we non-dimensionalize the problem and write u bar that is non-dimensional u is equal to u by u0 non-dimensional y is equal to y by h and non-dimensional t is equal to t by h square by nu. See it is possible to non-dimensionalize by using different parameters. You could use here an advection base time scale also to non-dimensionalize but if you use the non-dimensionalization parameter as the correct physical parameter then the solution gives the correct physical meaning directly otherwise it needs further interpretation. So now del u del t is what del u del u bar del u bar del t bar into del t bar del t. So u bar by this is t ref in short we are writing this is t ref del u del y what is del u del y del u del u bar into del u bar del y bar into del y bar del y this is what so this is u0 by h into del u bar del y bar del square u del y square is basically u0 by h square into this another differentiation with respect to y. So if you write this equation then del u del t is equal to u0 by t ref u0 gets cancelled from both sides and t ref is h square by nu. So you can cancel t ref h square by nu so we will get that equation del u del t non-dimensional is equal to del 2 u del y 2 non-dimensional both of these terms are of the order of 1 because the non-dimensional parameters are so chosen that non-dimensional u is what u by u0 that is of the order of 1 non-dimensional y y by h that is of the order of 1. So all the terms are of the order of 1 now this problem will have a steady state solution and an unsteady solution. When will the unsteady solution be irrelevant? The unsteady solution will be irrelevant theoretically when we go to t tends to infinity. So at t tends to infinity we will get the steady state solution which is the solution for the quiet flow. So the general solution will be sort of the unsteady part of the solution plus the steady part of the solution where the unsteady part of the solution should tend to 0 as t tends to infinity. So we will work out the steady state solution separate it from the unsteady solution and then separately work out for the unsteady solution which should asymptotically vanish as t tends to 0 infinity. So we will take this up in the next lecture for the time being we stop here today thank you very much.