 We were discussing about some examples of unsteady flows and while discussing unsteady flows we first discussed about a classical flow where there was no confinement and there was an oscillating boundary because of which fluid motion was actuated and that is known as I mean these kinds of problems where the boundary is set into motion or the boundary is set into oscillation these are classical problems in fluid mechanics. We studied one such problem in detail where the plate was moved suddenly in one particular direction and that allowed the fluid in the vicinity to move in the same direction. So that was the classical Stokes first problem and then we discussed about the motivation of studying these kinds of problems which are like shear driven flows and then we try to figure out that what would be the change in the scenario if there instead of a free instead of an open environment in one side there is a confinement. That means that you have a plate like this which is moving towards the right say with the velocity u1 and you have another plate which is stationary and the gap between the two plates is h. When h tends to infinity then the problem is like the classical Stokes first problem. So this can be thought of or imagine as a Stokes first problem with confinement right. It is a modified version of the Stokes first problem or in terms of the classical quet flow also it is like a quet flow we are it is basically a quet flow because you have two parallel plates one plate is moving relative to the other it is a quet flow but we are interested in the unsteady state solution of the quet flow problem. Now let us try to see let us try to see that what is the steady state behavior of this it is well known to all of you but just to recapitulate. So the first the governing differential equation for the general flow problem is what this is consistent with all the assumptions we have discussed in details earlier. Now before moving on to the steady state solution what we will do is we will non-dimensionalize this problem. So we define u bar as u by u reference there is a natural reference velocity scale which is u1 y bar is y by h unlike the Stokes classical Stokes first problem here the natural length scale is h and then there is a natural time scale. The time scale is a function of the nu and the h so that can be worked out by noting that the order of magnitude of this is u reference by t reference and the order of magnitude of this is nu u reference by h square since these 2 terms are of the same order then t reference this implies that t reference is equal to h square by nu. So you can define accordingly one t bar is equal to t by t reference where t reference is h square by nu. So what is this h square by nu? h square by nu is the characteristic time by which this characteristic change diffuses over a distance h that characteristic this what is what creates the characteristic change the motion of the boundary at the bottom creates the characteristic change and that motion gets propagated into the fluid because of viscous effects and the characteristic time that it takes to get transmitted up to a distance h is h square by nu. So that is the characteristic time scale based on h as a characteristic length scale. So with all these now a big question comes that why are we taking h as a characteristic length scale right. So we are taking it with a very important assumption and the assumption is that this h is smaller than like if you are considering that instead of h there is a y reference then t reference scales with y reference square by nu that means y reference is of the order of square root of nu t reference right. If instead of h it is y reference something arbitrary now we are considering the characteristic time scales to be such that this corresponding y reference is of the order of h if the corresponding y reference is much much less than h that means this plate is like as if like at infinity that is not the case which is of interest to us because then the situation will boil down to the stokes first problem. In fact I will ask you to work out a homework where you have to show that under what limiting condition the solution of this problem boils down to the solution of the stokes first problem effectively but that you can do after we complete the solution of this problem. So the understanding is pretty clear that our reference time scale is not such that this y reference is much much less than h and that in fact the case that we are considering that h is very small so that with a little bit of like time elapsed this will come to the order of h because it is a micro channel or a nano channel that we are considering that will be the practical scenario. So our assumption of y reference is equal to h is not obvious you have to keep in mind is not obvious but very practical from the consideration of many microfluidics and nano fluidics scenarios. So with that assumption your governing differential equation will be now we are interested to get the unsteady part of the solution. So we will separate that out from the steady part of the solution so let us be the steady state solution to this problem. What is USS? USS bar means non-dimensional steady state version of the solution to this problem. So the steady state version of the solution to this problem will be governed by this equation with the unsteady term equal to 0 same equation but with unsteady term equal to 0. So if we put the unsteady term equal to 0 then this if we integrate it twice this is what we get at the solution. The boundary condition at y is equal to 0 see the steady state boundary conditions should be the actual boundary conditions y. You might argue that I will dump the actual boundary conditions on to the unsteady state and make the steady state boundary condition something different from the actual boundary condition but that you cannot the reason is that the problem will eventually come to a steady state and when it comes to the steady state the whatever is the physical boundary condition that has to be satisfied. So you have to account for the boundary condition assuming that once the steady state has been attained that steady state is dictated by the actual boundary conditions which are physical to the problem. So at y equal to 0 what is the boundary condition? This is your y axis USS equal to what is the boundary condition at y equal to 0? This is 1 and at y equal to 1 non-dimensional y equal to 1 what is u? 0 this is no slip boundary both are no slip boundary condition. So the solution will be very straight forward 1-y just plug in C1 and C2 C1 is C2 is 1 and C1 is –1. So this is the steady state solution and this is the solution typically for the steady state quet flow that you learn in undergraduate fluid mechanics. You have 2 parallel plates one plate moves relative to the other with the velocity you have a linear velocity profile in between if the velocity profile is translationally invariant along the x axis that is the function of y only. Now the interesting part what is the unsteady part of the solution? See we have to keep in mind that many problems we solve at steady state many problems we solve at unsteady state. In the world no problem is actually a steady state problem because you think about a situation like say there is a water in a channel. Now there is some condition which makes it start from rest maybe there is a pressure gradient that is applied at time equal to 0 or maybe there is a shear that is applied at time equal to 0 maybe there is a electrical field that is applied at time equal to 0 whatever something is applied that creates a disturbance and there is a time that the system takes to adjust to the disturbance to come to a steady state that is the response time of the system. Typically if the response time of the system is much much shorter than the characteristic time scale let us say they are the characteristic time scale over which you are doing an experiment is hours and the response time is a few milliseconds that within a few milliseconds is reaches the steady state then we are mostly interested about the steady state solution we do not care about the unsteadyness but if the characteristic time scale over which you are doing the experiment is comparable with this time scale over which the system takes its time to relax or to come to adjust to the change then the unsteadyness of the solution is important and you have to look into the unsteadyness of the solution. So this is what I wanted to discuss very carefully because see in the class when we are giving lectures we are solving some problem a steady problem some problem an unsteady problem in engineering when you are solving a problem no problem in the real world is posed as steady or unsteady you have to give your judgment that whether your unsteady part of the solution is important or your unsteady part of the solution is not important over the time domain that you are considering and only the steady state solution is important provided that the system attains a steady state. If the system does never attains a steady state I mean that is that means it attains a steady state only at t tends to infinity that is a very special case that we are not considering here. Now we are interested about the unsteady part of the solution. So what we are writing is we are writing the total solution we are decomposing into a steady part and an unsteady part. How can we make this decomposition? We can make this decomposition because eventually if a steady state is attained the unsteady part should be 0 and there will be a steady state solution. So the unsteady solution should be decoupled from the steady state solution. So then you can write here u steady state we have already solved the solution is 1-y bar we will keep this in mind we will make use of this whenever we are going through the mathematical part of the solution. So you can see that the time derivative of u bar is same as time derivative of u hat because this is a function of position only this is not a function of time. Not only that the time derivative the second order special derivative of u bar is same as second order special derivative of u hat. The reason is that u steady state although is a function of y it is second derivative with respect to y is 0. So that means the unsteady state governing equation remains same if we write it in terms of the unsteady part of the solution. Boundary conditions at y equal to 0 what is u hat at y equal to 0 u is equal to u steady state right. So u hat must be 0 and similarly at y equal to 1 also u hat equal to 0 this is for time greater than 0 and initial condition is everything is stagnant at time equal to 0. So you can see that by decoupling the steady state solution from the unsteady solution we have achieved another interesting feat what is that we have made all the boundary conditions homogeneous. So that is a mathematical way of looking into the scenario but that is what it has made. See had we considered the complete solution for the complete solution the boundary condition at y equal to 0 is not homogeneous but for the unsteady part of the solution the boundary condition at y equal to 0 is homogeneous that is at u u u hat is equal to 0 at y equal to 0. Now because of this nice homogeneity in the boundary conditions we can use the separation of variables. So let us write u hat as a function of y into a function of time right that is the method of separation of variables. So its time derivative is f into g dash right and second order special derivative f double dash into g. So if you substitute that in the governing equation you will get f g dash is equal to f double dash g. So f double dash by f is equal to g dash by g question is so this is I mean by this time you have now appreciated that why we are interested to write it in this way this is a function of y only this is a function of t only that means we claim that each is equal to a constant. Now what constant positive constant or negative constant it is a negative constant minus lambda square say. Now why are we interested to or why are we writing this as a negative constant the reason is pretty clear like when you write dg dt by g is equal to something this will give a exponential solution of g as a function of time. If there is eventually a steady state solution then the unsteady state solution should decay exponentially with time. So that at t tends to infinity the unsteady state solution will be 0 and you will get the steady state solution. So that means there must be an exponential decay so the coefficient coming with the exponentiation at the top that should be a negative constant so that it is a decay and not arise that is why this minus lambda square. So let us complete this 1 by g dg dt is equal to minus lambda square so l n g is equal to minus lambda square t plus l n c 1. So g is equal to c 1 e to the power minus lambda square t. t means this is non-dimensional time t bar I have used c 1 before for this problem okay then c 1 c 2 we have already used before so c 3. So g is equal to c 3 e to the power minus lambda square t the functions of y. So f double dash plus lambda square f is equal to 0. So what is the solution of this in terms of sin and cos right. So f is equal to c 4 cos lambda y plus c 5 sin lambda y y means y bar. Now let us apply the boundary conditions at y equal to 0 u equal to 0. So boundary condition number 1 at y equal to 0 u equal not u u hat basically u hat equal to 0 means f is equal to 0. So f is equal to 0 at y equal to 0 means c 4 must be equal to 0 and boundary condition 2 at y is equal to 1 u hat equal to 0 right. So sin lambda equal to 0 and that means lambda is equal to n pi. I just want to reiterate that instead of sin lambda is equal to 0 had we considered c 5 equal to 0 that would also have satisfied this boundary condition but that would have given rise to a trivial solution that u hat equal to 0. So here we are seeking values of lambda that will give us non-trivial solution for this problem and that makes it an eigenvalue problem okay. So how many such possible values of lambda are there? There are infinite number of such possible values of lambda. So we call this lambda n lambda with subscript n which is equal to n pi. Now let us write the solution u hat is equal to into g f is c 4 sorry c 5 into sin lambda y and g is c 3 e to the power lambda square minus lambda square t. So c 5 c 3 e to the power minus lambda n square t into sin lambda n sin lambda n y and because so first we can replace this c 5 c 3 with a constant or just a name a and give it a subscript n because that constant is different for different n's and because it is a linear problem the resultant solution is a linear superposition of solution for all possible values of n that makes it sigma n e to the power minus lambda n square t sin lambda n y. How do we find out n? That is the only question that remains to be addressed to solve this problem for that we have to use the initial condition. We have not yet used the initial condition right this kind of problem is called as initial boundary value problem that means it has an initial condition and it also has boundary conditions. So we have used the boundary conditions but we have not yet used the initial conditions. Now to find a n see we have to somehow bring down this series to a single term. And that may be possible so this is Fn by using the orthogonality of Fn. So we now have to derive the orthogonality condition for Fn. I have already discussed how to derive the orthogonality condition in the context of one problem earlier and we will use a similar technique and we will try to do it little bit faster than before because we have already discussed the principle. So the Eigen function for the Eigen function Fn is given by this equation which is d2 Fn dn2 plus lambda n square Fn is equal to 0. What we do is we multiply this by Fm and integrate it by parts sorry d y. So we multiply this and integrate it by parts from y is equal to 0 to y is equal to 1 with Fm as the first function and the second derivative as the second function. So what we have is the first function into integral of the second minus integral of derivative of the first function into integral of the second d y bar. Now the boundary term clearly see F is an indicator of what u hat? u hat is 0 at both y equal to 0 and y equal to 1. So that makes this term 0. Now if you so there will be a d y bar here let us write this. If you swap m and n then just replace m with n and n with m and subtract whatever is that new equation by swapping m and n that equation you subtract from this equation. So what you will get? Lambda n square minus lambda m square integral of Fn Fm d y is equal to 0. For lambda n is equal to lambda m there is no necessity that this integral is 0 to generalize but when lambda m is not equal to lambda n this integral must be 0 to make the right hand side 0. Therefore we can conclude that integral of Fn Fm d y from 0 to 1 is equal to 0 if m not equal to n and is not equal to 0 if m is equal to n. This is the orthogonality condition. So now come back to this expression for u hat. In the expression for u hat you already Fn so you multiply it with Fm and integrate both sides that you do. So you can write u hat into Fm is equal to summation of An e to the power minus lambda n square t Fn Fm. Now you apply the initial condition. What is the initial condition? At t equal to 0 what is u hat? u hat is equal to minus uss because u is 0. So u is u hat plus uss so that is equal to minus of 1 minus y. So that if you apply minus of 1 minus y Fm is sin lambda m y is equal to summation of An sin lambda m y sin lambda n y and then we integrate this both sides with respect to y from 0 to 1. I am not writing each and every step just takes more space in the board but so next when you integrate this from 0 to 1 y equal to 0 to y equal to 1 then in that integral in this summation only one term will remain for which m is equal to n other terms will go away. So that will mean that minus of 1 minus y bar sin lambda m y is equal to minus 1 minus y bar sin lambda m y is equal to minus 1 minus y well in this summation only one term will remain for which m is equal to n other terms will go away. So that will mean that minus of 1 minus y bar sin lambda n y dy from 0 to 1 this integral evaluation is straight forward. Let us do maybe one or two more steps I mean it is very straight forward. So this is integral of 0 to 1 sin lambda n y remember lambda n is equal to n pi this we derived earlier. So sin lambda n y dy plus integral of y sin this is the left hand side y sin lambda n y dy 0 to 1. So sin lambda n y this will be so this becomes cos lambda n y by lambda n 0 to 1 then next one we can do integration by parts first function y into integral of the second minus cos lambda n y by lambda n minus integral of derivative of first into integral of the second that will make a plus sign. Now tell quickly what will be this terms. So cos lambda n by lambda n then minus 1 by lambda n then this term again minus cos lambda n by lambda n the subtracted term is 0 because y is 0 then this one plus sin lambda n y by lambda n y by lambda n square that term becomes 0. So eventually you are left with minus 1 by lambda n in the left hand side if I make any algebraic mistake please point out. So this will be minus 1 by lambda n right hand side see for each and every problem I will not work out all these details but initially at least for 1 or 2 problems I want to show you that how you can arrive at a very compact solution. The right hand side a n into integral of sin square is half into 1 minus cos to theta cos 2 lambda n y dy 0 to 1. So a n by 2 y minus sin lambda n y minus sin 2 lambda n y by 2 lambda n right. So this term will be 0 considering lambda n is equal to n pi. So this will be a n by 2. So left hand side is equal to right hand side this implies a n by 2 is equal to minus 1 by lambda n that means a n is equal to minus 2 by lambda n. So u hat will be equal to minus 2 by lambda n summation e to the power minus lambda n square t sin lambda n y and the general solution is u hat plus u steady state. So minus summation of 2 by lambda n e to the power minus lambda n square t sin lambda n y plus 1 minus y bar where lambda n is equal to n pi. Now you can try to understand the physics very deeply by referring to this solution. So the first question is that how quickly the system will attain steady state. So how quickly will the system attain steady state will depend on this exponent. See if this exponent is large if this is large this exponent is large then what will happen? This term will asymptotically approach to e to the power minus infinity pretty quickly. So you figure out that after what non-dimensional time your solution almost becomes a steady state solution. Theoretically it is infinity practically it is a finite time after which it is as good as a steady state solution. So the homework from this is you find out the threshold time below which you have to consider the unsteady part of the solution and beyond which the steady state solution is good enough and that you can simply do by making this function plotted as a function of time and y for different values of time and y on a single plot. You will see that after some time you will get the solution which is like the steady state solution which is the linear solution linear velocity variation between the two boundaries. So we have worked out two different problems of a suddenly driven boundary. In one case there was no confinement and in another case there is a confinement. Now we will do a similar exercise but now instead of driving the solid boundary in one direction suddenly we will oscillate the solid boundary and if it is done in an infinite fluid medium that is called as the classical stokes second problem. So we will discuss about the stokes second problem first and then we will discuss about that how the solution of this problem is modified if the similar forcing parameter you put with a confinement. So first stokes so you have a solid boundary like this and an infinite fluid medium there is no natural length scale in the y direction and this plate is oscillated let us say with a velocity u is equal to u0 sin omega t just as an example. I mean you could make it cos or combination of sin and cos whatever many things you could do but just as an example see any arbitrary oscillation if you could decompose into a Fourier series we will do with a large number of sin and cosine terms then one such sinusoidal variation can give a signature to the total characteristics of the system. So in reality one may not have classical sinusoidal like the typical pure sin or pure cos type of time variation of the velocity but this actually will give a signature even in situations when the boundary motion is much more complicated. So that is why these are very classical problems which can give you a strong classical basis of approaching more and more involved problems by using the same concept. Omega is the frequency at which the plate is being oscillated. So this type of flow is called as oscillatory shear flow like you have a shear like you have a shear that is created but the shear is created by the movement of the plate movement of the solid boundary that creates a velocity gradient but that shear is an oscillatory function of time it is not a time independent shear that is imposed. So that is called as oscillatory shear flow just like the previous scenario we will start analyzing the problem by writing the governing equation and describing the time scales and velocity scales and length scales and so on. So the governing differential equation see governing differential equation will not depend on whether the boundary is pushed in one direction or is oscillated governing differential equation will still remain the same. Now let us try to non-dimensionalize the scenario u bar is equal to u by u reference. What is a good choice of u reference? u0 that is the amplitude of oscillation velocity t bar is equal to t by t reference. Now this is a problem where you have a natural t reference. See your problem in this problem the oscillation of the plate is important provided the time scale is set up by the frequency of oscillation otherwise you may give oscillation but that oscillation will not be important. So to make the oscillation important you must have a time scale which is 1 by omega. So the non-dimensional t is omega t that will give a length scale that evolves as a function of this time scale. See you do not have a natural length scale in this direction but the length scale will be decided by the combination of the velocity scale and the time scale and that you can simply arrive at by doing an order of magnitude analysis of this equation. So you have u reference divided by t reference is of the order of nu u reference by y reference square. t reference is 1 by omega. So y reference scales with square root of nu by omega. So that means you can clearly see that what is the length scale up to which the oscillation of the plate is failed. So this is like a boundary layer. So you know the kinematic viscosity you know the frequency of oscillation. So the ratio of that square with a under a square root that becomes a natural length scale for that problem. So within that length scale only the effect of oscillation of the plate is failed. If you go beyond this length scale then the effect of oscillation of the plate is not failed. So if you consider these scales then we will set up the problem. So you have nu omega everything is absorbed in the scales. So this is a solution this is the governing equation and this governing equation should have a solution which is spatio temporally recurring over both position and time. So it is something where you may have a function of time you may have a function of position. The function of position may be decoupled from the function of time and the function of time is oscillated in nature. These are the physical signatures of the solution. So there are many ways of getting a solution of this. One very simple way is like you substitute u is equal to imaginary of e to the power i t into a function of y. The simple tricky substitution makes the solution of this problem very simple and today we will stop by discussing about the rational behind the choice of such a solution. So first of all you see e to the power i t will have sin and e to the power i theta is what cos theta plus i sin theta. So you have both a sin and cosine term and that will mean that you can represent a sinusoidal variation with a phase. So that is the time variation positional variation is not like that. So you have a spatio temporal variation where the temporal thing is sort of time periodic. Time periodic but in addition so here you have a pure sin term but you can also have a cosine and sin combination associated with that to account for a phase. The phase there can be a phase difference between the way in which you oscillate and the way in which the fluid responds to that oscillation as a function of time. Frequency will remain the same t bar this is t bar omega this is like this is omega t actually okay. So i omega t omega t is t bar. So then why this imaginary? The imaginary is because the boundary condition you are given you have given sin omega t. So this solution has to satisfy the boundary condition and at the boundary so you know the imaginary part of e to the power i theta is sin theta okay. So had i given this plate and oscillation u0 cos omega t this would have been just real. So real or imaginary that is being pushed by the boundary condition. So it will take a few more steps to complete the solution we are running out of time today. So we will stop here by considering that this is the solution that we are suggesting. So we will do this same operation from both sides and then we will get an ordinary differential equation because this is essentially separation of variables. I mean done in a little bit of different way to accommodate for the oscillatory time dependence of the solution. And then we will work out the complete solution and then we will go for a problem where we will not give an oscillation to the boundary but we will put a confinement and we will put a oscillatory pressure gradient that will like be now closer to model or to mimic the blood flow dynamics in a in a microfluidic confinement. So for simplicity we will not consider the flexibility of the blood vessel and we will consider still a Newtonian fluid but the forcing parameter we will consider to be pulsatile in nature. So that is the agenda that we keep for the next class for the time being thank you very much.