 So, we will start today's lecture, the 14th lecture of the course and what I intend to do today is take a problem in fluid mechanics and explain the complete solution to that problem okay and then explain how that same problem can be solved using the method of perturbations. So, this is the same thing as what we did earlier when we had this quadratic equation for which we had the exact solution okay because it is a quadratic equation even with the parameter meter epsilon in it we could find the solution in terms of the square root sign right the discriminant and all that then we said we will do a perturbation series solution. So, I want to basically take you through the same process okay and in the course of this we will mention a few things which I think are most of the time implicit in whatever is done in the classes. So, I just want to be explicit about certain things which people do not explicitly mention okay. So, the problem here is to a pulsatile flow in circular tube. So, you have a circular tube and the liquid flowing through it and what do I mean by pulsatile flow I mean that the pressure gradient that is being imposed is going to be varying periodically okay. So, pulsatile flow implies that the pressure gradient varies periodically in time. So, why would anybody be interested in this I mean very classical example of a pulsatile flow is that of the blood flow through the capillaries okay your heart is beating periodically the pressure changes fluctuates okay in a periodic manner and so this kind of problem just by way of motivation as to why anyone would really be interested in this is this can mimic blood flow in the arteries for example etc okay. The blood flow problem of course is more complicated in the sense that the rheology is non-Newtonian the walls are flexible okay. So, I mean those are additional complications but now we will keep life simple and we assume that the liquid is Newtonian we will assume that the wall is rigid okay and if those of you are interested in making further studies on more complicated things can pursue okay. So, what we will do is we will idealize the problem and restrict ourselves to Newtonian fluids with rigid walls over or inside cubes with rigid walls okay. So, that is just an idealization so that we can understand certain things about the flow. So, depending on if you are not happy with that then you go about relaxing some of these assumptions and then you proceed. So, clearly I mean this is an extension of a problem which you all have seen before what is the problem you have seen before the Heggen Poiseuille flow in a circular tube in Heggen Poiseuille flow what is the story the pressure drop imposes constant okay and then you talk about the fully developed flow where the velocity does not change in the axial direction and you only have the velocity varying in the radial direction but now the pressure drop is not a constant pressure drop is fluctuating with time. So, clearly what this means is the velocity that you are going to see in the channel is also going to be fluctuating with time okay. So, earlier we were in a position to talk about steady state we are now not going to be able to talk about steady state because the pressure drop is varying fluctuating with time the velocity is going to fluctuate with time and very importantly I need to retain the term containing the time derivative in the Navier Stokes equation okay but we will still keep life simple in the sense assume that the axial velocity is a function only of the radial position and time okay. Earlier for the steady state problem you had velocity function only of the radial position now because of the periodic pressure drop I am saying look time is also going to come in I am not going to complicate life because I just want to illustrate some ideas and then we will just say now that the velocity changes with r and t okay. So, what I mean by this is that the pressure drop which is given by –dp by dz I am going to write this as some constant remember as epsilon sin omega t what you are used to is if epsilon is 0 or omega is 0 that means it is a constant pressure drop. So, the way I am looking at it I have a the pressure drop is varying periodically g0 represents the mean value the average value of the pressure drop okay so g0 is the average value of the pressure drop okay and the way I have written it I got –dp by dz here so that g0 is positive and I should actually say this is the time average value of the pressure drop. So clearly uz which is the axial velocity is going to be a function of r and t now okay and what I am going to do is see I have a periodic pressure variation if I have this kind of a situation for a long time the velocity profile is going to be periodic as well. So, this is like the equivalent of my steady state. So, in order to reach my steady state I have to wait for a sufficiently long time finally I go to my steady state I have an initial profile supposing you have a fluid in a channel which is at rest but the liquid is not flowing okay you have filled it with fluid. So, your initial state is rest now you put a pressure drop a constant pressure drop you will get your parabolic velocity profile but to attain the parabolic velocity profile it takes some time and after you wait for a sufficiently long time you have the parabolic velocity profile. So, that is in the limit of t tending to infinity you have a steady state your parabolic velocity profile same thing here we today and tomorrow we are going to concentrate only in the limit of t tending to infinity I am not interested in how does the velocity change from the state of rest at t equal to 0 to the final t value okay how does it go from the state of rest to the parabolic velocity profile that I am not interested in I am only interested in what happens in the limit of t tending to infinity the final solution you expect it to be periodic. So, if you want to actually keep a probe in one of your arteries you will see that the velocity is varying periodically with time and a fixed point. So, that is the thing we are interested in finding out okay. So, we want to focus in the limit t tending to infinity okay and here we expect velocity to be periodic that is one thing the other thing which I want to talk about is the fact that I have assumed sin omega t. So, of course the periodic function does not have to be sinusoidal it can be some arbitrary periodic function but you know if any periodic function you can resolve it in terms of the Fourier sin series or Fourier sin series or Fourier cosine series or Fourier series. So, basically if you give me any periodic function I will basically resolve it using my Fourier series and I will have different components sin omega 1 t sin omega 2 t sin omega 3 t. So, that is the justification for using sin here okay. So, basically what I am saying is any periodic function can be written in terms of a Fourier series this justifies the use of sin omega t there and epsilon basically represents amplitude of the fluctuation that you have okay. So, this is the amplitude and this clearly is the frequency. So, what I am going to do is just go directly to the problem that our objective is to find the solution du by dt the axial velocity profile okay and this is going to be in terms of minus dp by dz plus my viscous term okay. The left hand side contains my inertial terms I am just simplifying it and I am neglecting all the u dot del u terms saying that they are very small but I need to retain this because I need my time derivative okay and the dp by dz term I am going to write as g naught multiplied by 1 plus epsilon sin omega t okay and plus my viscous term which is going to be times r d by dr of r du by dr that is my viscous term. So, that is my viscosity and that is my del square because this I have only remember the variation of uz with respect to r is captured here the variation of uz time is captured here and this tells me how uz varies with r and t. Now, I am not you know derived it from the Navier Stokes equations but I made some assumptions of simplifications basically I do not want to consider uz as a function of r, theta, z and all that. So, we are assuming fully developed in some sense but it is unsteady state okay. So, like I mentioned earlier we want to solve this problem in the limit of t tending to infinity okay and we also want to solve the problem exactly and we want to solve the problem using a perturbation series that is the idea. So, now the first thing that you have to do whenever you have a problem is to try and make it dimensionless okay. Clearly in this case I am imposing my pressure drop and the flow is going to be induced by my pressure drop. So, the velocity the characteristic velocity scale is going to be decided by the pressure drop which I am inducing which is basically given by g0 the average value is g0. So, I am saying that u characteristic is going to be measured in terms of the pressure drop and if I look at the terms here I am going to choose g0 r squared divided by mu as my characteristic velocity scale okay. What I have done I have looked at these 2 terms this term as units of g0 and I am looking at this term I have a length squared at the bottom I have viscosity here. So, I am just saying g0 and velocity is there. So, g0 r squared divided by mu is as units of velocity and therefore I am just saying that is my characteristic velocity okay. What about the length scale? The characteristic length scale is going to be clearly the diameter of the tube which is r or the radius of the tube which is r okay and the time scale in this case is going to be given by I can look at these 2 terms okay and what I am going to get I am looking at the term on the left here term on the second term on the right here and mu divided by rho is my kinematic viscosity and so T characteristic is going to be given by r squared divided by mu. So, in some sense what we are talking about let us understand the physical meaning of the characteristic. What does it represent? It represents the time required for momentum to diffuse in the radial direction okay. If you want r is the distance in the radial direction mu is the kinematic viscosity which basically is facilitating the transfer of momentum. Now how much time does it take for the momentum to diffuse in the radial direction? So, once the momentum is diffuse in the radial direction then you will have your fully developed velocity profile if you wait for that much time or longer than that. So, that is the idea. So, that is the choice of the scale. Now one thing I want to caution here is that the way you choose these scales are not necessarily unique. You can always choose different ways of scaling a problem and proceed. The idea is by choosing certain scales you want to try and get insight about what is dominating what is not dominating the problem and so that in some limiting cases you can actually do some analysis and get some idea about how the behavior of the system is okay. So, please understand that I have chosen this some of you for example you could have chosen an alternative time scale which is alternative time scale is coming from the omega okay. You can say look there is a frequency with which I am changing the thing and that is my time scale of my system. Sure you can proceed with that argument okay. But then what is important is not necessarily that the value of the omega what you will see is that there are 2 time scales in the system and the ratio of these 2 time scales is what is going to actually decide the behavior of the system okay. So, at the end of the day I think you have to be consistent. The point is you can do the problem in different ways okay. So, let us make this thing dimensionless now and what will that give me if I choose u bar, u tilde as u divided by u characteristic and r tilde as r divided by length characteristic and t tilde as t divided by t characteristic. Suppose and you make the equation dimensionless what do you get? Rho multiplied by u characteristic divided by t characteristic du tilde I am forgetting I am not going to write the z dependency you all know that is axial velocity dependency okay du tilde by d tilde equals g0 times 1 plus epsilon sin omega t plus mu times u characteristic by r squared times d by d r tilde d by of r tilde and that is the 1 by r okay. So, that is basically your dimensionless equation with the tilde here. Now, if you were to substitute u characteristic as g0 r squared by mu and put t characteristic as r squared by mu substitute the u characteristic and the t characteristic what do you get? u characteristic divided by t characteristic is just g0 equals you put tilde there okay. Now, the point I am trying to make here is this g0 cancels off anyway this is a very small issue in the sense you can sit down and make the substitutions and you can find out for yourself what is happening okay. Now, some of you are thinking possibly that look I have got this epsilon sitting here and this man is going to do a perturbation series about this epsilon. So, let me tell you I am not going to do a perturbation series about epsilon okay. Wait a second I got a problem right I got a problem here in the sense when I am writing this in terms of t tilde and I write t in terms of t tilde I need to write this as t characteristic multiplied by that I need to write this as r squared by mu sorry about that okay I need to include an r squared by mu omega here when I because I have t here I am writing t in terms of the character the dimensionless time t tilde so I need to get the characteristic time here so I can r squared omega by mu okay because that is important to me because that is the ratio of the time scales that I was talking about so remember I told you you can choose omega as your time scale you can choose r squared by mu as your time scale the but what is important is this particular parameter here actually represents the ratio of these 2 time scales okay and this is called a Strouhal number so whenever you have something like a periodically imposed pressure gradient you will get this okay so remember that the behavior of the system therefore is I am going to let me write it correctly now what I have done is basically rewritten those equations again and just want to make sure that the things are a bit more clear now the left hand side contains the time derivative of the dimensionless variables the dimensionless velocity u hat with t hat and it is multiplied by rho times u characteristic divided by t characteristic and that equals g0 times 1 plus epsilon sin r squared times omega times t tilde divided by mu so that was the term which I made a mistake and now I just want to emphasize that the argument of sin is r squared omega by mu t tilde plus the viscous term which you have and what I have done in the next step is substituted the expressions for the characteristic velocity and the characteristic time and you see that the parameter g0 occurs in all the 3 terms and so you can actually knock it off okay and the other change that I have done is call the term r squared omega by mu r subscript omega this r subscript omega is something like my Strouhal number okay and I am following Gary Lille for the notation the Strouhal number r omega I have written it at the very last step it is the ratio of 2 time scales it is the ratio of the time scale for momentum to diffuse across a distance r that is r squared divided by mu the kinematic viscosity to the time scale of the oscillation omega remember is the frequency of the oscillation. So the reciprocal of omega is the time period of the oscillation and therefore 1 by omega is the time scale of the oscillation and the Strouhal number r subscript w represents the time scale for momentum to diffuse over a distance r to the time scale of the oscillation this Strouhal number is the small parameter which we are going to be doing a perturbation series solution that is we are going to use r the Strouhal number as a small parameter to seek a perturbation series solution I have d u tilde by dt equals 1 plus epsilon sin omega t plus 1 plus omega sin t plus okay I am dropping the tilde but now you know that this is dimensionless okay we drop the tilde but now the variables are dimensionless why am I doing this this is to reduce my chances of making a mistake okay so now remember that this is a dimensionless equation all these are dimensionless this is a dimensionless time dimensionless velocity dimensionless position okay so I have put in the tilde everywhere I have just I am writing it like this point I am trying to tell you is there are 2 parameters here one is r omega which is the ratio of the time scales epsilon which is the amplitude of the perturbation I should not call it perturbation of the periodic component yeah pardon me now this is r subscript omega oh there is no omega here you are right there is no omega there thanks yeah so this is r subscript omega is the thing yeah that is right there is no omega there okay now what I want to make you observe is in order to solve this equation I need boundary conditions and initial conditions right so what are the boundary conditions the boundary conditions are at r equals 1 we have u equals 0 that is my mostly boundary condition at r equals 0 my u is bounded and at t equal to 0 u equals 0 that is I am just assuming that my fluid was initially at rest yeah everything is fine so the fluid is at rest initially and now I am suddenly imposing this periodic pressure drop I am trying to understand what is happening okay okay the point I want you to notice that this equation is a linear equation and because it is a linear equation I can be bold enough to look for an analytical solution okay now this particular term this is my source term okay my pressure gradient that is my source term and I am going to look at the solution for u as being made up of two parts one arising because of the constant component one and another arising because of the time dependent component the sine rw of t that is supposing you have a problem a linear problem x equals b1 plus b2 okay what you can solve that problem as x as a inverse of b1 plus b2 or you can solve a or you can just look at x as a inverse b1 plus a inverse b2 okay so that is basically what we are going to do let me just write the analog here see this is a linear equation correct okay now supposing I am just going to give you an analog of what we are trying to do so that you can relate to the solution strategy if we have x equals b and let us say this is my equivalent of my b okay if b equals b1 plus b2 what I am going to do is solve x is of course equal to a inverse b but x is also equal to a inverse b1 plus b2 which is a inverse b1 plus a inverse b2 okay matrix multiplication just gets carried over so that is essentially what I am doing what this means is if I have two sources of non-homogeneity I can find the solution find the response of the system to one find the response of the system to the other and then just add up because it is linear I can do this superposition okay this is the principle of linearity and superposition that you have are used to so that is exactly what I am going to do I am going to look at the solution to this problem as being composed of two parts the first part being coming from this constant pressure gradient the second part coming from the time dependent pressure gradient okay the idea is that you already know what the solution is for your constant pressure gradient you get your parabolic velocity profile for the time dependent thing is what we are going to do today okay so then once we do this so that that gives you your complete solution then we go back and look at the perturbation series approach for using in the limits of rw my straw wall number being very low straw wall number being very large so the perturbation is going to be done about rw okay epsilon is just a magnitude okay of course you could have done the other way also if you wanted to you could have done a power series about epsilon but that is not what we are doing today seek u as being made up of u0 plus epsilon u1 okay now what is this this I want to explain is not perturbation series solution okay this is just writing finding the solution for two different sources of not homogeneities but I am putting epsilon here just so that that takes into account that epsilon which is there so u1 will be a solution of only sin omega t that is all okay this essentially captures the two different so I say source terms sources of delta p okay the constant part in this so what I am going to do is I am going to seek the solution for u0 and I am going to seek the solution for u1 so u0 satisfies so u has to satisfy that equation so u0 is going to satisfy du0 by dt equals 1 plus 1 by r d by dr of r du0 by dr okay and du1 by dt will satisfy sin of rwt plus 1 by r d by dr of r du1 by dr okay. What I want you to do is I want you to just add up these two guys multiply this by epsilon suppose we multiply this by epsilon the second equation by epsilon and you add you will get d by dt of u0 plus epsilon du1 plus 1 plus epsilon that plus some second derivative operating on u0 plus epsilon du1 so that is a solution to my original equation okay. So I mean this is just to I am just decompose the solution just like I explained here instead of doing A inverse of 1 plus this I am just looking at 1 separately sin omega separately. Okay what are the boundary conditions to which this is going to be subjective same boundary conditions u0 equals 0 at r equals 1 u0 is bounded at r equals 0 u1 is 0 at r equals 1 u1 is bounded at r equals 0 at t equal to 0 u0 0 at t equal to 0 u1 is 0. So see the boundary conditions also have to be consistent with what the original problem was having okay. I mean you need to make sure that when you are doing this splitting up the boundary conditions are consistent the equation is also consistent. Okay so now life is simple in the in some sense in the sense that if you look at the first term for and remember I am focusing on the limit in the limit of t tending to infinity. In the limit of t tending to infinity I am going to have something like for the first problem a steady state because the first problem corresponds to the flow when you have a constant pressure gradient okay. So u0 is independent of r independent of t sorry independent of t since physically it corresponds to a constant pressure drop okay that is the physical thing. Mathematically you would have gone about solving it if you had been a mathematician you would have just let me solve this using separation of variables or something like that but now you see look this is a constant pressure drop. So the limit of t tending to infinity in the limit of t tending to infinity I expect steady state if you are not interested in t tending to infinity then you have to worry about the change with time okay as t tends to infinity and that is kind of important and then what you have is just your regular ordinary differential equation in r and so u0 will be a having a parabolic dependency. So u0 has a parabolic dependency and that you know how to calculate just use the boundary conditions you will get some 1-r square and you can calculate that. So I am not going to do this you will do that I want to talk about the calculation of u1. So if I look at u1 what happens is can the velocity or this term here represents my how my pressure is changing with time yeah is that a problem it is fine yeah the epsilon is not here because I have written my epsilon here. See I am looking for the solution as epsilon if I have not put epsilon here then I would have had to substitute u1 and then so whatever solution I am going to get u1 I am going to multiply that by epsilon and then I am going to add to get my original solution okay that is the reason the epsilon got cancelled basically when I am substituting epsilon u1 coming here there is already an epsilon and there is already an epsilon here when I substitute this epsilon gets cancelled okay I am just wanting to make sure I did not miss it over there okay. Now the pressure is varying periodically sinusoidally with time will the velocity be in phase with the pressure will it be out of phase with the pressure okay and that is going to be decided by the only parameter which seems to be occurring in this equation which is Rw okay. So this particular physical thing we will discuss later but I just want to point out that it is not the velocity in general will not be in phase with the pressure gradient. For example if assuming that the velocity is varying periodically with time and is in phase that what does it mean velocity will also be varying as sin Rwt only then the pressure is varying as sin Rwt velocity is also varying as sin Rwt they are going hand in hand. Suppose now that velocity okay the question is the velocity in phase with the pressure drop that is the question okay. Suppose it is in phase if it is in phase then what do you expect the time dependency is also going to be in the form of sin Rwt multiplied by something which is a function of R okay. Then u1 is going to be of the form sin Rwt multiplied by some function of R I mean that otherwise it would have been out of phase right. So now the point is this a likely solution to the partial differential equation this is your typical separation of variables approach which you would do you would assume that something is a function of time multiplied by a function of R okay. So here supposing this is a solution I am going to substitute this in my differential equation. When I substitute this in my differential equation when I do the derivative with respect to time I will get cosine. This of course has the sin term and when I substitute here since I am differentiating with respect to space I have my sin term and I only have the derivative of f. So what you would get is this yields something like cosine omega t cosine Rwt multiplied by f of r maybe multiplied by Rw because I am differentiating sin no with respect to time I get that equals sin Rwt plus 1 by r d by dr of r times df by dr times sin omega t. Remember Babita will tell me that this should be total derivative and not partial derivative okay. So since I am doing f and f is a function only of R this is a total derivative. So now the point is if it have been in phase then I would have had a sin omega t in all my terms and I could have actually cancelled out my sin omega t but the fact that when I am putting the sin omega t here I get a cosine omega t tells you clearly that the velocity in general is not going to be in phase okay. In general it is that is going to be a phase lag okay. So here in general there is going to be a phase difference I should not use the word lag because I do not know if it is a lag or a lead phase difference between velocity and the pressure drop. Since the sin omega t sorry does not occur in all the terms. If it has been sin omega t occur in all the terms then that is the possible solution that means the velocity is following the pressure gradient okay. The fact that this guy is not so I got a problem. So what do you do? Now this also gives you some clue. It gives you some clue in the sense for example when I have differentiated this thing with respect to time I call it Rw here. If Rw is very very low okay if Rw is very very low these guys are for all practical purpose this periodic is bounded between 0 and 1. This is only telling you something about the frequency of the change but the magnitude of the term this is also bounded between 0 and 1. If Rw is very low this guy is going to go to 0. So in the limit of very low Rw Rw must smaller than 1 only these 2 terms are going to be present and then you can solve the equation okay. So if Rw tends to 0 I expect my velocity to be in phase but if Rw is much larger than 1 then this guy will be present and my pressure gradient and my viscous term will be present and then I am expecting it to be out of phase. So that is the reason what we are going to do now is we are going to find the solution to this problem as it is and then we will you know try to find the solution the limit of low Rw using a perturbation series and then do the comparison okay. So I am just saying that if Rw is very much lower than 1 then LHS is approximately 0 and velocity is likely to be in phase. Now what is Rw? It is omega multiplied by R square divided by nu Rw is very low means the omega is very very small my frequency is very slow is very low my frequency is very low. So my pressure gradient is changing slowly if my pressure gradient is changing very very slowly that means my velocity is able to catch up with it. So I am not changing if I change it very rapidly then my velocity my system cannot respond but if my velocity is my pressure gradient is changing very slowly because my omega is low then the velocity can quickly respond to the change in the pressure and it is like is going to just follow the pressure gradient and that is the reason it is in phase. So that is the physical way to look at it okay. I mean mathematically you can look at something so but because if I am giving you like if I slowly increase the pressure on you you will be responding to the pressure slowly but and you will be able to keep up with me but if I keep you know doing some very fast changes I think you are going to go bonkers right. So that is basically what is going to happen. So that is the idea and in the limit of Rw being much greater than 1 you have the other situation and we will look at how to do the and now that is the motivation for doing the perturbation series solution. For Rw0 we know what is going to be in phase and when Rw is small it is going to be slightly out of phase and we will be able to capture that okay. So that is the reason we are going to be doing the perturbation series about Rw but let us go back to finding the actual solution because that is the idea we want to find the actual solution and then do the perturbation series. So the problem of finding a solution and since this is a linear equation partial differential equation we can definitely solve by separation of variables alright and whenever you want a separation of variables you want that term you know to be repeated so that you can then convert the partial differential equation to an ordinary differential equation remember that is the strategy. What is preventing us from doing that it is a time derivative the first order time derivative the first order time derivative when I differentiate sin and get cosine. So is that a function which on differentiating first time gives back itself the exponential function gives you that the derivative of the exponential of x with respect to x is exponential of x right and you also know from your complex variables that e power i theta is cosine theta plus i sin theta. So rather than solve this equation here what I am going to do is I am going to solve this equation du1 star by dt equals exponential of i rwt plus 1 by r d by dr of r du1 star by dr. Now this is of course not the same as that problem but this is remember e power i theta is cosine rwt plus i sin rwt okay from your knowledge of complex variables. So if I actually found out u1 star and if I got the imaginary part so this is u1 star response to this source okay if I find u1 star and why do I want u1 star because when I now put the first derivative I will get back my exponential so I am okay I am happy with that because then I can cancel out the time derivative and I have only my ode in r and which I can solve okay. So if after I finding out the u1 star I find out the imaginary part of it because the imaginary part is the one which corresponds to my source of sin omega t. So basically what I am trying to tell you is that u1 is nothing but the imaginary part of the solution of u1 star okay. The u1 star is going to be complex there is going to be a real part there is going to be a complex part okay u1 star is in general complex real part plus imaginary part. The imaginary part corresponds to the sin term and that is what we are interested in okay. The real part corresponds to the cosine term but that is what of interest to me. So clearly now since I am talking about the exponential term I have a cost component and a sin component. So this phase like business phase difference business will be incorporated because I am taking both into account. Now what I am going to do is I am going to solve this problem I am going to solve this problem get the imaginary part I get back what I wanted okay. But how do I solve this problem? Seek u1 star as exponential of i rwt multiplied by some function h of r. I am seeking this as the usual separation of variables that you do. Now I am going to substitute this back here. When I substitute this back here I get i rw I am differentiating with respect to time. So I get that i rw times e power i rwt times h of r equals e power i rwt plus e power i rwt times 1 by r d by dr of r dh by dr okay. Because h is a function only of r. Now is this choice a permissible choice is this a valid choice? It is a valid choice because now the time dependent term e power i rwt is present in all the terms and I can cancel off okay. That was the reason what motivated me to go from sin to the exponential term okay. And now I have my ordinary differential equation for h as a function of r. I should be able to solve this. If I am able to solve this clearly I am going to be able to solve this tomorrow because otherwise I would not have done this. I know h I can substitute this back here. I get u1 star. I get the imaginary part. I get my solution okay. And remember okay. So clearly h satisfies 1 by r d by dr of r dh by dr. I am going to bring that thing here minus i omega h equals I am going to take this minus 1 to the other side minus 1 okay. And now I need to have boundary conditions on this. I will just write them down and we will stop. h at r equal to 0 is bounded because only then my velocity is going to be bounded and my velocity has to be 0 at r equal to 1 is 0. That comes from my no slip boundary condition okay. Now we need to be able to solve this. This is a linear equation. It is an ordinary differential equation. I only think it is a complex number sitting here but we need to be able to take care of that. I just want to give you a hint that this is an equation which you guys have seen before in the calculus course when you are talking about Bessel's functions. So the solution to this will be in the form of Bessel's functions okay. And so that again is a nice closed form solution what we get. That is my total solution. Then what we do is this should be r omega. I think I need to stop because people are not correcting me. This is r omega. So now in the limit of rw being 0, I will know what the solution is. In the limit of rw being small, I will can do a perturbation series solution and then we will see. We will do that tomorrow okay. Thanks.