 Alright, so let us just look at a few of these so called exact solutions. As you saw these Navier-Stokes equations, now that I can use your words, they are non-linear and coupled and all those things. In general, there is no closed form solutions of these equations available. Under some seriously simplifying assumptions though, you can obtain some solutions by integrating the Navier-Stokes analytically. So there are a whole lot of such situations. I think some of the advanced textbooks on fluid mechanics will list number of such problems. I am just dealing with 3 or 4 right now here just to outline what is done in an exact solution situation. On the other hand, when it comes to the main workshop, we can always include a few more examples. However, let me honestly tell you that most of these situations will turn into something of a mathematical exercise beyond a limit and then one has to get into only the mathematics part because all that you are looking for is using some trick or the other trying to obtain an analytical solution. So in my opinion, I am not such a great fan of doing too many of these. It is just to give some sort of a flavor as to what is involved in a so-called analytical solution is what we are doing. But nonetheless, we can include a few more. So the few specific situations that I am going to talk very briefly here are what are called as plane Poiseuille and Couette flow. Then we will look at one more situation where we have a free surface flow flowing down an inclined plane. The last example is something that is slightly more interesting which is what we will call a transient flow between infinite parallel plates. So it is the transient plane Poiseuille Couette flow. So let us see what these are. So the first one is what we will call fully developed flow between infinite parallel plates. Remember this is what we began with our discussion where we obtained that tau equal to mu times du dy etc. So what we have is two very very large plates in their own plane separated by a distance of h which is much smaller than either their length or let us say the depth. And in that sense they are called infinite because their length and the depth into the plane of the paper are very large. What we will assume and this is what it boils down to that unless you come up with such assumptions you cannot really solve those never so. We will assume that these are steady constant density flow. We neglect the body forces. The infinitely large plates will essentially imply that we are dealing with a two dimensional situation. What this means is that whatever picture that you may generate in the x y plane as is shown on the board will be repeating exactly in any plane that you can think of along the z axis. In other words what we are talking about is there are no boundaries on the z dimension and because there are no boundaries the boundary cannot affect the flow situation along the z direction. That is what the idea behind a two dimensional flow is. In that sense there is no z component of the velocity and any variable that you can think of is not varying with respect to z at all. So this is what is written by partial derivative of z of whatever so I have not even bothered to say whatever that is it could be anything. Pressure density velocity nothing is equal to 0. Further we come up with this simplification called fully developed flow which basically means that the axial derivative or the stream wise derivative of the axial velocity is equal to 0. So any particular axial location that you look at it is here or here or here or here the axial velocity u is exactly the same. In some sense this fully developed flow also has a link between the infinitely large extent in the x direction as well. So what we are saying really is that the x direction extent is also extremely large so there are no real boundary conditions on the x direction which can affect the flow in that sense. So fully developed flow of course is observed experimentally it is not like it is not observed and the condition that is experimentally observed is exactly this namely that the axial velocity is unchanged as a function of the axial distance. So actually when it comes to solving these equations what it means is that the u velocity which is the axial velocity is not a function of x. It immediately implies that so that is what is written down. So these are our assumptions under which what we will do is we will try to simplify the governing equations starting with the continuity. For constant density we have du dx plus dv dy equal to 0 because of fully developed situation we get rid of du dx all together which means that it is simply a dv dy equal to 0 which we can partially integrate with respect to y and claim that v equal to then some function of x. Now the question is what should be that function of x. To obtain that function we come up with the boundary condition. So what we have in this case is a solid plate at y equal to 0 and a solid plate at y equal to h. It is not a porous plate it is a solid plate. So as far as solid plate is concerned pure physical situation will dictate that there cannot be any y component of the flow getting into the plate. In other words the relative normal velocity that is what formally you can say is 0 at a solid surface. Since the solid surface in our case itself is at rest this relative is really not relative with respect to that surface is the absolute velocity v equal to 0 at both y is equal to 0 as well as at y equal to h. The only way this boundary condition and the result that v is equal to f of x can be satisfied is if that f of x itself is identically equal to 0 otherwise the boundary condition cannot be satisfied. So what this means is that the entire y component of the velocity is identically equal to 0. In other words this is what we call a perfectly parallel flow. This is what we had taken right at the beginning where there was no y component of the velocity. But the formal argument as to why there should be no y component of velocity is actually coming through the simplification of this continuity equation and putting those boundary conditions on the v velocity at y equal to 0 and y equal to h either way it does not have to be both. Even if you look at only y equal to 0 that is sufficient or at y equal to h fine. So then we basically conclude that there is no y velocity at all in this flow. The y momentum equation is taken up later and now it is written since for analytical purpose as I was mentioning it does not matter whether you write it in conservative or non-conservative form. So it is written here in the non-conservative form. We have assumed steady flow so that time derivative simply goes away. Everything else you can see here will go away except the pressure gradient term. Why? Because either the v velocity is identically equal to 0 everywhere so that all these terms involving v will go away. Similarly the terms on the right hand side will go away as well. Only thing that is left is that the pressure gradient in the y direction then has to be equal to 0 is our conclusion based on this. So what does that mean? That really means that pressure is not a function of y. So then we say that pressure at the most is a function of x since the flow being steady. Then we come to the x momentum equation and we realize that du dx will go away because of fully developed flow. v will go away because it is identically 0. The first term inside this viscous part is also going to be 0. Why? Because it is v dx of du dx. Now du dx as a function itself is identically 0 everywhere so therefore its derivative is going to be 0. What is left then is only a partial derivative of pressure with respect to x which now need not be written as partial derivative at all. Why? Because we have already shown that it is not a function of y and similarly partial derivative of u with respect to y the second one. This also need not be written as a partial derivative. Why? Because right at the beginning we have seen that u is not a function of x. So the conclusions then are that I write the total derivative of pressure with respect to x equal to mu times d2u dy squared the total derivative. Now see what has happened though? Through our successive arguments what we are able to say is that dp dx is at the most a function of x at the most. It is definitely shown to be no function of y here. So it can be at the most a function of x. On the other hand this d2u dy squared can be at the most a function of y. But then these two guys are said to be equal to each other. So is that possible? Unless each of them has to be a constant. Unless each of them has to be equal to a constant this situation will not make any sense whatsoever. So then what we say is that in order to then solve this problem we basically assume that the pressure gradient which is imposed on the flow is assumed to be known. So under the known assumed pressure gradient dp dx then you can integrate this mu times d2u dy squared equal to say this constant 2 times with respect to y and you can come up with this form of the velocity profile. So even though this comes out as a velocity profile in the sense that this is u as a function of only y in reality for a situation under the assumptions that we have listed it is actually the velocity field. It is just so that u is only a function of y in this case. So it turns out to be a profile. So now the only part that is remaining is to determine what these two unknown constants are for which we go to the boundary conditions and depending on which boundary conditions we want to employ there are two different cases. First one is what is called as a plane poiseuille flow where both plates are held stationary. In other words the boundary condition on u then is u is equal to 0 at y equal to 0 as well as at y equal to h. This is because we have a so called no slip boundary condition that the u velocity, the axial velocity has to satisfy. So what is the no slip boundary condition? It is basically this no relative tangential velocity at a solid surface. So again our solid surface itself is at rest. So it is essentially the absolute velocity which has to be 0 at both y equal to 0 and y equal to h. So these are the boundary conditions that we will normally use when it comes to a solid fluid interface. Which ones that we have used? The one that earlier was used was 0 normal velocity at the solid surface and now we are using a 0 tangential velocity at the solid surface. So this is what is called as a no slip condition. Let me point out here though that the 0 normal velocity has a fairly strong physical basis in the sense that if it is an impervious solid surface we cannot expect that the flow will actually penetrate into it. If the surface was porous then it is possible. We looked at that example yesterday there was a leakage flow rate etc. from that pipe. So it was a porous wall. This is not. So this is fine. On the other hand this no slip has no theoretical basis or no physical basis either. What is the basis? You know for this no slip. Nothing more than repeated verification through experimental expression. That is it. So just because it appears that velocity in the tangential sense that is seems to be going to 0 at a stationary solid surface we have decided to accept it purely based on experiment. There is absolutely no theory and in fact people who do molecular level simulations will actually say that this is incorrect. They will in fact say that there is indeed a slip. It just so happens that in continuum situations the slip is very, very, very small essentially equal to 0 and therefore we utilize it. It is worthwhile pointing this out to students because otherwise there is always this question that you know on what basis we claim that there is no slip and there is no theoretical basis. The only basis is really as he correctly said is experimental. Anyway so having incorporated these in our general expression we immediately obtain the velocity profile in the case of a plane poiseuille flow. In a minute I will show how it looks. Obviously you know how it looks looking at it. Second case is what is called as a quet flow where the lower plate is held stationary and the upper plate is moving with a constant velocity of capital U. So if you incorporate these two boundary conditions along with the solution here you obtain a different velocity profile which is the quet flow velocity profile. So if you plot these two guys you should get something like this. On the left this is the plane poiseuille flow and looking at the expression here is essentially a parabolic profile that you obtain which is what I have shown here. So it is a completely analogous situation to the circular poiseuille flow or flow through a pipe of circular cross section that we already know and for completeness I have just written down the expression for the velocity that we obtain there. So the plane poiseuille flow and the circular poiseuille flow are essentially analogous to each other. One is just flow between two parallel plates and the other one is inside a circular cross section pipe. The quet flow depending on whether there is an imposed pressure gradient or not is going to be looking different. So remember that to have a flow in the plane poiseuille situation you must have an imposed pressure gradient otherwise there will not be any flow. So here in the first case there is a non-zero pressure gradient present. In the case of quet flow the pressure gradient need not be there. It can be there it does not have to be there. The reason is because the top plate which is anyway moving will transfer its momentum through the action of viscosity and the successive layers below it will get into motion. So if that axial pressure gradient imposed is 0 then we essentially get from the expression before the second part will simply go away and what you get is a linear profile which is plotted as this curve A and that is what we call usually a simple quet flow as many of you would know. On the other hand if I have a negative pressure gradient which means that the pressure is assisting the flow because it is falling in the direction of the flow then you actually get a fuller velocity profile roughly something like this and if you have an adverse pressure gradient which is given by a positive dpdf then you have a profile like this. So depending on how adverse it is you may actually even get a reverse flow here. That is something that I think you normally see as part of the discussion of this quet flow situation where there is an imposed pressure gradient. So this problem in some sense we will solve using that finite difference. This is a steady problem obviously so this is the steady state solution that you are seeing. What we are actually going to solve is a so called transient problem but in the time limit of time tending to a very large value eventually it should get into a situation like this. So we can verify that when we solve it with our finite difference code. So this is the first set of exact solution as it is called but remember that the exact solution works only with certain seriously simplifying assumptions. One of the other reasons I wanted to point out these two problems is because in fact the validity of in some sense validity of the Navier-Stokes equation was actually ascertained through the quet and positive flow situations. In the sense that people worked out these solutions analytically as they have been outlined here and then they actually rigged up these experiments and measured relevant quantities and the analytical solution as obtained here was then compared with the measurement data and they seem to be matching quite well and therefore the validity of Navier-Stokes is ascertained. Remember I have been always saying that it is a model, it is a model so unless you make sure that the model comes up with a description that is realizable in physics or real life nothing is going to be accepted. In that sense these two flow problems have a fairly high historical value that only after these were realized in the lab and pertinent measurements were made people said that okay if this is the case then the solution that is worked out in this fashion must be the correct one which means that the governing equations from which you start these have to have some sense in there. So that is the idea behind this. That is the first example. The other example is to demonstrate essentially a different type of boundary condition at a free surface of a liquid. So what we have here is again a fully developed flow down an inclined plane okay. So this is what we will call a steady constant density flow same as before two dimensional so no z component and all derivatives with respect to z are 0 and we will assume for the solution process that it is a fully developed flow just like it was before. So what we have really is that this is a liquid film essentially falling on an inclined plane and the film has a thickness of capital H. Now note that here there is a gravitational force acting so we cannot neglect the gravitational force in this case. So let us see how we take into account that. But I will still go one by one looking at continuity through the fully developed flow. We come up with the same conclusion as before that there cannot be any y component of the velocity. Again this is a fully parallel flow only u component of the velocity. If you look at the y momentum equation the coordinate system is chosen such that the x direction is along the plate and the inclined plate and the y direction is normal to it okay. So then accordingly this g which is acting vertically downward needs to be resolved into component parallel to the plate and normal to the plate okay. Is that fine? So that is pretty straight forward. So if you do that the y momentum equation then will become in the non conservative form everything the same except that we add this minus of rho times g cosine theta as the relevant body force term acting in the y direction here. So that y component here is what you have to compute. Using all these previous results what you can quickly show then is that dp dy the pressure gradient in the y direction will be then rho times g times cosine of theta which you integrate and say that with respect to y partially that is that p then has to be of this form where this f of x is something that needs to be decided. To do that we go back to our y equal to h situation and realize that the boundary condition there is that there is a constant atmospheric pressure acting. So you employ that boundary condition and if you do that the pressure expression then will come as p atmospheric plus rho g cosine theta times h minus y. One way or the other this says that dp dx is 0 because this is purely a function of y now. So then going with the previous results including this dp dx 0 you realize that all that is left in the x momentum equation is d2u dy squared with a total derivative because we had right at the beginning realized that it is not a function of x the u velocity and the relative sorry the relevant component of the gaudy forces plus rho g sine theta in the x direction. Again it is a second order ordinary differential equation which you integrate and obtain this form for the velocity profile. The boundary conditions in this case one is obviously that at the solid surface at y equal to 0 we have a solid plate and there we need to satisfy this so called no slip condition so that is fine. The other boundary condition is actually at the free surface at y equal to h and many of you would know this which is what continuity of shear stress at the liquid air interface. So when can we utilize this the way it is can I utilize this continuity of shear stress the way it is written out here in all situations where can I not use it perhaps in that form. So we are really talking about immiscible liquid in some sense. So here it is actually a free surface between a liquid and a gas but in principle we are talking about immiscible liquid. Well the shear is actually continuous what I am saying is whether you approach it from this fluid or that fluid the interfacial shear stress is the same that is precisely what this condition is. What about my point is what about curved interface so here because the interface is basically assumed to be absolutely flat we do not need to take into account a surface tension effect. If there was a curvature to the surface tension actually will come in and it becomes a much more complicated situation to handle. In this case because the free surface is flat all that we can really say is that there is a continuity of shear stress which I simply write as mu times du dy because there is no v component anyway on the liquid side equal to mu times du dy on the air side at y equal to h. Now how do I further simplify this by invoking this idea that mu on the liquid side the dynamic viscosity of the liquid is much larger usually than the dynamic viscosity of air. So if I divide this entire equation by mu of liquid what I will get on the right hand side is mu of air divided by mu of liquid that ratio can be argued to be fairly small. So in general the right hand side then here will be in general argued to be fairly small and in other words we are coming up with a simplified boundary condition based on a physical situation that the du dy which is the slope of the velocity profile at y equal to h on the liquid side is essentially equal to 0 and if you employ this boundary condition along with the no slip you actually obtain the velocity profile as is outlined here. So that is the solution process the purpose of this problem was really to bring about this boundary condition at a liquid air free surface and of course there was a body force incorporated because the flow was actually taking place because of the gravity force. So here if you want to now attach some sort of a physical interpretation to this equation what can it be rho g sin theta plus mu times du d2u d y squared equal to 0. So there is a dynamic balance between gravitational force and viscous force and under this dynamic balance there is a steady flow involved or established. What I have not been pointing out in this example and the previous example is what happens to the left hand side of our momentum equation these guys rho times capital D dt of v rho times capital D dt of u they are identically 0 which means what? No acceleration. No acceleration is absolutely right and no acceleration but the flow is still going which essentially means that we are talking about a dynamic equilibrium under certain forces. So in the first example it was the dynamic equilibrium between the pressure force and the viscous force. In the second example is the dynamic equilibrium between the gravity force and the viscous force. So immediately you can attach these physical interpretations on these governing differential equations. Is that fine? Okay now let me just point out I am only working out here the velocity profile. Usually if you see textbooks they will go ahead and do little more massaging of this result. In particular once you have the velocity profile you can always find out the shear stress profile by differentiating this with respect to y and multiplying that by the viscosity. Second thing that you will normally see is that you can obtain a volume flow rate by how will you find the volume flow rate for example in this case you have to integrate that u rho or if it is a volume flow rate it is just u times dy you integrate from 0 to h with mass flow rate you simply multiply by rho. Such auxiliary results then you are normally able to bring about and such results you can find out in these textbooks when they are solving these kinds of problems. One more example which is little more interesting than the previous one is the following. Here now we have a transient flow between infinite parallel plates. So the same problem setting as it was in the first case only thing is that now we have now we have made the problem transient or unsteady. So the physical setting could be like this that there is a pair of parallel plates at time equal to 0 this is filled with a viscous fluid in between these and everything is stationary. So all velocities are 0 at time equal to 0. Now you have a stop watch and started at time equal to 0 what is done is the top plate is suddenly set in motion with a value of u velocity in its plane. Lower plate is held still stationary and simultaneously at time equal to 0 an axial pressure gradient is imposed. So this is the way I would like to point out a physical description of this problem is that fine. So start at time equal to 0 minus that is before 0 everything stationary at time equal to 0 you set the top plate in motion with a constant velocity of capital U keep the lower plate stationary as it is and simultaneously you impose an axial pressure gradient. And now what we are interested is how the flow evolves in time eventually it will reach a steady state but what we are actually interested as part of the solution here is how it evolves from time equal to 0 until it reaches that steady state solution. So this is the problem setting and mathematically speaking these are the assumptions that are listed. So unsteady constant density flow we neglect the body for boundary conditions because now the problem is transient we say that the tangential velocity at y equal to 0 for all times starting from time equal to 0 is 0 that is the lower plate is held stationary. Tangential velocity at the upper plate for all times is equal to capital U of course starting from time equal to 0. Initial condition for all use sorry for all y's between y equal to 0 and h at time equal to 0 the velocity is 0. This is the formal mathematical expression for the initial condition and the remaining two are our previous assumptions namely infinitely large plates and fully developed flow. So let us first see what happens to the governing equation and then we will look at its solution continuity exactly the same. There is nothing different here you obtain the same result that there is no v component or the y component of the velocity same for the y momentum as well you come up with the conclusion that there is no y pressure gradient. What happens to the x momentum? Now we cannot get rid of that time derivative of U so that is maintained here. On the other hand these two guys will go away because first is the fully developed assumption and secondly the v velocity identically equal to 0. Same thing for the right hand side we have d dx of du dx, du dx being identically equal to 0 everywhere its derivative is also 0. So what is left is having divided through rho partial derivative of U with respect to time equal to minus 1 over rho dp dx the total derivative I do not need to use a partial now. On the other hand for the d2u dy squared it is first of all nu because mu divided by rho but I need to use the partial derivative here why? Because U is not only a function of y but it is also function of t. So now we come to this situation that we realize that U can be both function of y and t therefore the derivatives of U whether it is a time derivative or the y derivatives can be at most functions of either y or t whereas this fellow in between is a function of at most x. So again the same argument this cannot be true unless this is a constant the pressure gradient and let me simply denote that minus 1 over rho dp dx equal to s so that from an applied mathematicians point of view I bring about a so called diffusion equation as the governing equation partial derivative of U with respect to time equal to nu times second partial derivative of U with respect to y plus a so called source term which is essentially coming through that minus 1 over rho dp dx this we need to solve to obtain U as a function of y and t subjected to the two boundary conditions and the one initial condition. So this is the formal setting for the partial differential equation that needs to be solved to obtain our U as a function of y and t. So before I get on to this let me ask again how many of you have solved things like these in the applied mathematics type classes? So now everything is mathematics from now on all physics is over no analytical. So that separation of variables idea is what I was mentioning earlier some of you are indeed familiar through heat transfer or some other classes. So let us see what is done the slightly complicating factor in this particular situation is this presence of the source term. If the source term was missing it is a relatively relatively easier situation. So let us see how to handle that. So here is where still I will point out some physics before we get on to this. What I will point out is let me not show that slide yet. The idea is the following we started with time equal to 0 we set this thing in motion also impose some pressure gradient everything otherwise at time equal to 0 was at 0 velocity. So what is going to happen physically what do you expect will happen forget the impose pressure gradient for now let us say that the impose pressure gradient is exactly equal to 0 as a special case. What will happen how will the flow set initially this is the lower plate and this is the upper plate upper plate is set in motion lower plate is stationary. So the momentum of this plate will gradually get transferred through the action of viscosity is not it. So if you want to look at an initial time the velocity in the flow or in the fluid I should say close to the top surface will be essentially equal to the velocity of the plate. But then it will suddenly drop to 0 everywhere as some time goes by that velocity at the top is constant but some more portion of the flow will get also induced in the same direction. So there will be some sort of wave like this and so on. It will keep on penetrating more and more toward the bottom plate eventually what will happen in steady state impose pressure gradient is 0. Yes it will be essentially a linear profile as what we saw in the case of the quiet flow situation where dp dx was 0. This is what it has to finally end with. Let us look at this steady state solution final solution. The steady state solution has to satisfy what boundary conditions for the velocity same as what we had before namely velocity equal to 0 here velocity equal to u there. So this is the idea that we will use and claim that let the solution be proposed in the form of a transient part and a steady part such that as time tends to infinity the transient will essentially go away and what will be left is only the steady solution part. So in that sense now I can put up that slide that I propose the solution to be u of y, t equal to a steady part which I am simply calling as u1 of purely y no time dependence plus u2 of y, t which is the so called transient part and through our physical arguments we realize that this transient part as time goes to infinity in the theoretical sense should basically go to 0. So then what is left in that case is the steady part which must satisfy those two equations. The pressure gradient is always there it is imposed at time equal to 0 and it is kept. So the steady part also needs to take along with it thus the pressure gradient which comes through that source. So I need to basically solve for this u1 an equation which is given by nu times d2 u d2 dy squared of u1 plus s equal to 0 subjected to the boundary conditions u1 of 0 at the lower plate u1 of capital U at the top plate this is pretty straightforward since s is the constant you work it out and you get exactly the same form it better be otherwise there is something wrong as what there was in the case of the simple quiet flow here is that fine. So that is that is one good all right what about the transient part the transient part is going to be a partial differential equation now there is no s to be solved along with it. So is this something that is followable are you are you getting it why should the transient part not have the source with it okay fine. So what you can do also one thing is you can take this u as a function of other way of looking at it that is and then our governing differential equation was du dt partial equal to nu d2 u d y squared plus s. So take this form and actually substitute it here in this so what will happen meaning that I am taking this here and substituting in the governing equation partial derivative with respect to time this fellow will not contribute to it because there is no time dependence in u1 only this will come in okay. So I will have d2 u dt partial that is all equal to nu what about here second derivative of u with respect to y meaning that I want to differentiate the right hand side with respect to y twice obviously this will contribute because it is a function of y but it is going to be a total derivative because it is only a function of y correct. So I will simply write this as d2 u1 d y squared plus what should be there partial derivative of u2 the second one with respect to y that is that that plus s okay and now we say that let us force u1 to be the steady solution which means that since that source is always there it will always be there in even the steady solution as well and therefore I say that nu times d2 u1 d y squared plus s equal to 0 is what the steady part must satisfy then what is left then from here if nu times d2 u1 d y squared plus s is equal to 0 then what is left is partial derivative of u2 with time must be equal to that is how we can obtain the governing equation for this transient part separately okay. So this is what is written out here now this is a truly transient situation in the sense that it will have its own boundary conditions how many of them 2 because it is a second order in space and there will be one initial condition okay the question now is what should be the boundary conditions on u2 I have already written those here but where are those coming from I am right correct well formally what you can say is that you do the exact same thing as what you did here for the boundary conditions also. So the basic boundary conditions were what u equal to 0 at y equal to 0 I will just work out this so you substitute this entire form in here with y equal to 0 so therefore u1 of y equal to 0 that is this part plus u2 of y equal to 0 comma t is equal to 0 we have already taken along with the steady part u1 of y equal to 0 has been taken as 0 this implies then that u2 has a boundary condition at y equal to 0 which is given by is exactly what is written out in the same manner if you substitute the second boundary condition at y equal to h since the steady part has already taken up that u equal to capital u what is left nothing is left so therefore the u2 function at y equal to h at all times must satisfy a 0 boundary condition the utility of splitting this situation in this fashion is that the problem that you have come up with for the transient part is a diffusion equation which is fine but what can you say about the boundary condition mathematically speaking both are 0 means what there is a word for it if you remember from the applied math homogenous because both are 0 we have now homogenous boundary conditions on the function u2 the moment you have this you are in business for solving it using a separation of variable technique so for the separation of variable technique to work your problem should have homogenous boundary condition so in some sense the applied mathematician will say that this entire exercise has been done to obtain homogenous boundary conditions for the transient part from a physical point of view an engineer like hopefully us we will argue that we are thinking that there is a transient part there is a steady part the transient part will die out and the steady part will remain as the steady solution and then automatically the mathematic fits into it either way it is okay to look at one way or the other finally what you end up with is a homogenous boundary condition problem for the function u2 which then becomes a candidate for solving using separation of variables however what needs to be done is a careful monitoring of the initial condition for u2 that has been written out here u2 for all y's at time equal to 0 is now written as minus of u1 y which is this guy where that is coming from same thing you have the baseline initial condition as u of y, 0 equal to 0 you substitute that in here u of y, 0 equal to u1 y there is no time dependence in here nothing to talk about plus u2 of y, 0 but what is u of y, 0 so 0 equal to u1 y plus u2 of y, 0 which is what we want so this u1 y goes on to the left hand side and becomes minus of u1 y so the problem is simplified in the sense that it becomes a homogenous problem for the transient part but the initial condition slightly is troublesome it is not troublesome it is a little bit cumbersome let us say but it does not matter the initial condition can be whatever as long as I have a homogenous pair of boundary conditions I can use the separation of variables technique so separation of variables as you may remember from some of your heat transfer or other classes it involves a product function like y of purely y, capital T of purely time as what is the functional form for this u2 and I am not now going to get into the algebra right now it turns out that this is how it will become if you use the initial and boundary condition you realize that there is an Eigen value problem to be solved where all these constants you can obtain and then eventually the solution comes out in the form of a big series like this the important thing to me is the algebra you can always work out once you know how the separation of variable techniques work the issue really is the design of the solution that is what the important part is and hopefully I have tried to explain it both from a physical point of view and a purely mathematical manipulation point of view mathematics person will say I want a homogenous problem for my transient part and that is the way I am going to play around till I get that physically I expect that the transient will die out and what remains is the final steady solution one way or the other those who are willing to go through this they can actually solve it and finally the expression comes in the form of this infinite series now what you make of this infinite series in the sense that if I say that this is the analytical solution students will say what to we do with what can answer what answer can you give them because this is what we are going to do in the afternoon we are going to take this analytical solution and do something with it so what can that be you actually evaluate the series how will you evaluate the series correct so you actually keep on evaluating the series until it converges from the mathematics point of view you cannot keep on doing this by hand obviously the only way to do this is program this series and the way to then test the convergence is to keep on adding the term every time keep on looking at the last term and the previous to that keep on seeing the difference between those as long as the difference between successive last two terms is less than some tolerance you say that I am satisfied with my convergence and then you say that I have evaluated my series so this actually is going to be done in the afternoon session that the same problem here this solve du dt equal to nu d2u by square plus s subjected to all these conditions will be actually numerically solved using the finite difference method and on the finite difference solution we are actually going to superimpose this analytical solution by appropriately evaluating the series appropriately meaning whatever terms are going to be required the computer will churn out and say that ok 10,000 terms were required fine so be it you do not have to do anything as long as you just keep the tolerance in place that the successive terms have to be less than this tolerance it will keep on adding. So this is what you can tell to students that analytical solution when it comes in the form of a series has no value unless you are ready to sit down on the computer and program that series to find out at a given time and at a given while location what the value is going to be. So for example I say that now I am interested in knowing value of u at time equal to something say 20 second at y of h by 2 so remember that our problem setting was y 0 to y so in the middle I am interested to know what the value is at time equal to 20 so what will I do this first part is ok this is straight forward as far as the series evaluation is concerned here is the time where I will substitute that 20 or whatever I want clearly there is a eigen value to be evaluated along with it but this is not one eigen value it is a series of eigen value so corresponding to m equal to 1 there will be one term so I take that I add to that the term corresponding to m equal to 2 I add to that corresponding m equal to 3 I keep on doing it until the two terms the last two terms are not differing by more than your tolerance and then they say that then you say that that is a time. So this is exactly how the series has been programmed when you look at this in the afternoon today it is exact in the sense that mathematics person will say that you add to infinity I have no problem formally speaking you can say that there is a very fast convergence of most of these series no no no it is not like that it is not like that you just look at this particular expression that you are seeing on the on the screen and usually these problems which have a reasonable physical setting will come in the form of some series which actually converges relatively fast. So in that sense it will never converge to the absolute value of say 0 in the 0 in the sense the successive difference between two terms but as an engineer you you have to take a call that this is good enough so exact is to be taken with a pinch of salt in the sense that as good as you want it is here so if you are ready to add 10,000 terms that is fine but you need not add actually 10,000 terms it may it may turn up that if your tolerance is 0.001 for example for the successive term it may end up with only 100 or even sometimes the series are extremely fast converging may be even 10 terms are good enough I am not too sure do you know exactly on the top of your head how many terms have been added 90 so it is less than less than 100 in this case so it is a fairly fast converging term though by no means is exact so exact is to be taken as a pinch of salt just like Navier stokes equations are the equations of governing equations of fluid flow really there is no nothing exact about this just that it comes in the form of a series and from the mathematics point of view it is exact because he says that here she will say that you can evaluate the series actually some series not this one but some series actually have a analytical close form solution so in those cases you can perhaps really call it as an exact this fellow does not have an exact solution in that sense that there is no close form final result of the series but for a practical purpose we say that we add up to a certain number of terms alright so that is all I have as far as the exact solutions are concerned as I said to you so this example you will not find in any book this is something that I have come up with myself and normally we do these kinds of things with our applied mathematics classes we normally do not do this kind of stuff with fluid mechanics but for the purpose of this one finite difference let us say discussion I wanted to come up with something slightly different nothing more about that some of these analytical solutions you will see in chapter 5 and 8 of fox chapter 7 of potter and chapter 6 of gupta and gupta yeah so my point was that most of the more complicated exact solution so to say will come in these form so I am not too sure what is the utility of doing this beyond a certain limit for an engineering student I would rather have that student code the equation using either a finite difference or a finite volume and generate a numerical solution rather than going through this for one or two things it is okay to do but for 20 different types of solutions does not make any sense to me at least is that fine so these are just few sample exact solutions what I can do though is in the in the main workshop if you feel I can add a few more just perhaps a little different geometry maybe I can do it in a cylindrical geometry so that some cylindrical governing equations can also be used presently I am sticking to only Cartesian but I can do that in so there is a very troublesome problem analytical solution where there is a cylinder in cylinder concentric cylinders with a fluid in between those two and a transient problem of so essentially that the problem that we just solved you just do it in a cylindrical geometry without any pressure gradient impose so any idea what will be involved in that which functions may come up yes but which mathematical functions may show up as part of the solution yes vessel functions will come in so it is up to you if you want to get into things like those you can always do that students have absolutely no interest in it let me tell you that so they would rather say that I will code that equation with a finite difference or a finite volume and be done with it and things like that so maybe I will add one or two more relatively straightforward exact solutions in the cylindrical geometry when it comes to the main workshop but that is about it personally I have no interest in doing that unless you tell me that you do it