 Welcome to lecture 15 and what we are going to do is continue the solution to the problem we were looking at yesterday okay and if you recall this was the differential equation which we had derived okay and remember this only represents the radial dependency of the velocity profile when you consider only the periodic part of the pressure gradient okay and this is in the limit of t tending to infinity. So now this is of course subject to the boundary condition that at r equals 1 we have h must be 0 and h at r equal to 0 must be bounded. This equation is a non-homogeneous equation because you have this minus 1 appearing here and therefore the solution is going to be in the form of a complementary function and a particular solution okay. So clearly we have h has 2 parts h particular and h homogeneous okay and now you just have to recall some of those things you did in mathematics when you were trying to solve differential equations without possibly having a physical basis you were doing it mathematically. Now you see an equation where you have a differential equation and then this has some physical meaning in the sense that it tells you something about a velocity okay. So let us go back and I think we will first calculate the particular integral this is the particular solution and clearly the particular solution is h equals 1 divided by i r omega because when I put h as 1 by i r omega r omega is a constant when I differentiate it I get 0 and I get minus 1 and that gives me minus 1 equals minus 1 everybody is happy okay but then I do not like to have this i in the denominator I want to put it in the numerator so I am going to multiply the numerator and the denominator by i and that gives me minus i by r omega okay. So this tells me the particular integral that is a particular solution. Now we need to look at the homogeneous version of the thing and the homogeneous version is 1 by r d by dr of r dh by dr and this is the homogeneous solution minus i r w h equals 0 this is the homogeneous equation corresponding to which I will get my 2 solutions my complementary function okay. What I am going to do is I am just going to write I am going to make a small transformation r is my independent variable here I am going to seek the solution in the form of I am going to define r squared multiplied by minus i r w equals r star squared okay. What am I trying to do? I am trying to define a new variable and get this equation in one of the standard forms that you have possibly come across earlier okay. The idea is when I do this transformation you will get a differential equation whose solution is the classical Bessel's function okay I mean that is just redefining of the scales. So how did I get this? I basically want a coefficient of plus 1 here. If I get a coefficient of plus 1 here this is going to collapse to my Bessel's function equation. So in order to get plus 1 here the denominator scale has r squared. So I am just saying r squared multiplied by this is a new variable. So when I do the write the differential equation in terms of r star I would get a plus 1 that is the objective. So since I do not like this minus i I am going to multiply throughout by i I get r squared equals r star squared times i divided by r w okay. So I just multiply throughout by i I get minus i squared is plus so r w comes here I guess that. So basically what I am saying is my r I am going to write as r star times i divided by r w and r star is a new variable the what is the motivation say whenever you do something you need to have a motivation. Motivation here is to reduce the ODE to a standard Bessel's function form okay and if I now instead of using r as the independent variable if I use r star as independent variable what would I get I would get 1 by r star d by dr star of r star dh by dr star plus h equals 0 okay this gets transformed to this okay you can work out the algebra it is not a problem. Now this has my classical solution which is the solutions to h is basically going to be some constant a multiplied by j0 the Bessel's function of r star plus some constant B times the other Bessel's function r star yeah there is an under root where yeah absolutely correct absolutely correct there is an under root there yeah thank you okay. Now this is your solution for the homogeneous part because I have just reduced it to the Bessel's function so I mean you guys have done Bessel's functions in cylindrical coordinates in your math course. So that basically I am just trying to tell you that the solution is the Bessel's function see since you are working in radial coordinates in polar geometry your solutions are in the form of Bessel's function if you have been working in the form of in a rectangular Cartesian coordinates you have got in sine and cosine your trigonometric functions okay okay what we are going to do is we are going to look at this function y0 y0 at r star equal to 0 is unbounded and what that means is if you were to retain y0 that means your solution is going to become infinitely large and since you have a physical problem you really cannot have an infinitely large solution or velocity. So this implies that the constant B must be 0 because B is non-zero then your velocity is going to be unbounded at the center point r star equal to 0 and you know that velocity has to be bounded in the middle okay. So we use this bounded conditions whenever we are actually seeking solutions analytically supposing you are actually seeking a solution numerically then you would use something like a derivative equal to 0 from a symmetry okay but you are actually getting a analytical solution then I have used this bounded argument okay. So they are kind of equivalent but not exactly but since when you are doing a numerical code you cannot say they cannot be infinity right you have to have some other condition. So what this means is I have h equals a times in fact this is remember the homogeneous part okay j0 of r star. So what is my actual this is the homogeneous part. So what is my actual h? My actual h is going to be the particular solution plus this solution which is –i divided by rw plus a times j0 of r star I am just going to go back to r now because I know r is going from 0 to 1 okay and instead of r star I am going to write this as square root of rw divided by i times r going back to this definition of this because I mean r is my physical quantity which I know goes from 0 to 1 okay. Our job is to evaluate a now and remember I have not yet used my other boundary condition I have already used up one boundary condition which is the bounded boundary condition and I have gotten rid of b. I have got to use the other boundary condition which is the fact that h at r equal to 1 is 0 and that implies h at r equals 1 equals 0 implies a equals i divided by r omega divided by j0 of square root of r omega by i r okay I have just moved it to the I put i is equal to 0 I put r equal to 1 and therefore that is going to be it okay because this is evaluated at r equal to 1 and that gives me what my constant a is. So I can now substitute the a value back here and get the h that I was actually interested in so the h is minus i by rw times 1 minus j0 of square root of r omega by i r divided by j0 of square root of r omega divided by i okay remember this only tells you this is my analytical solution which tells you how the radial dependency is now how do I but what I am interested in is actually my velocity right that is what we want to find out this h is only giving me a part of the thing how is my actual velocity defined the uz u1 I think that is what I used I am not sure u1 remember was the imaginary part of the exponential of times rwt times h of r is not this correct you need to check your notes and tell me if this is consistent with what I wrote earlier okay and so this is basically the time so this basically what you need to do in order to get the velocity is you need you found the radial dependency you have found the exponential dependency on time which we are assumed to begin with and now I have got h I am going to multiply that by to the i r omega t and I am going to calculate the imaginary part of it and that gives me my velocity and that gives me only one component of the velocity remember because there is the other component with the constant pressure gradient which gives you the Hagen-Boysel so I could add this component of velocity to that and then find my actual velocity okay this is u1 and the actual velocity is I mean u0 plus epsilon u1 is it something like this I am not sure what I used yesterday so I just want to make sure subscripts okay we will just use this as a subscript just to be consistent great okay so yesterday so this is just trying to be consistent with the thing so the actual velocity that you are going to get is going to be composed of two parts we found this as a result of the constant pressure gradient get your Hagen-Boysel parabolic velocity profile u1 is the imaginary part of this and then you multiply that by epsilon and you get your solution. So this is the analytical form of the solution what you can do is you can possibly go to one of the software packages like Matlab or Mathematica and plot the velocity remember it is a function of rw t and r okay so what you need to do is you need to decide what you are interested in. If you are interested in a specific value of rw you just fix it if you are interested in a specific position fix r and plot the velocity as a function of time and you can do this for different r's you can do it for different rw's that gives you your actual velocity now since this is a complicated expression one may want to say like can I get a simplified expression okay and I would not call it a simplified expression and an approximate expression and one way to do this approximation is to invoke the fact that rw can be either very large or very small okay. So now let us look at like we did yesterday look at the limit where rw is very small okay and now so this is the total solution this is the exact solution if you get u0 from yesterday and u1 from today and put it together but so yeah but what I have done is no this is with r equal to 1 it is a function of r yeah this is a function of r yes this is a function of r but this is a constant the denominator is independent of r the numerator has r and the way we define this is r not r star r star is the way we have defined it okay yeah so it is a function so this quantity here is a function of definitely r and the velocity will be a function of r, t and rw okay now okay can we approximate a solution which let us say maybe valid for low rw and when I say low rw I am talking about rw being very much less than 1 okay why would we be interested in okay one of the things I want you to do is sketch or plot plot okay this is a homework problem for you plot the velocity for fixed rw and r as a function of time so you have the expression now you just have to go and code it you need to go to the computer okay and then you need to go and solve and get the graph okay like this sketching a function you can do the same thing you can sketch the velocity for a fixed rw and t as a function of r so this I think there are many packages I if you have a fancy for a specific package you can use your package it does not matter okay but I need the result at the end of the day so I am comfortable with MATLAB some people are more comfortable with Mathematica some people say Maple anything is fine but the reason you need to do this exercise is we are going to approximate this result and then we want to check how good our approximation is okay clearly when I say low rw how low is low is it 10 power-5 is it 10 power-10 so that is the thing that you want to know right there are 2 ways of doing the of doing the approximation one is you can take this expression and you can possibly expand the Bessel's function in terms of a power series and then do the approximation that is one option or rather than work with the solution you start with a differential equation itself and depending on your choice you can do this okay so what we are going to do is we are going to go back to the differential equation and explain the method of finding the solution using a perturbation series okay because I think that is easier in some sense rather than you because you need to know the exact form of the expressions of the Bessel's function and then you have to apply it so for low rw we seek a solution in the form of power series in rw the idea is I am implicitly assuming the small changes in rw give me small changes in the velocity profile okay so what do I do I have h right h clearly is a function of rw and r okay h which is solution to this equation depends both on rw and r those are the 2 parameters what we are going to do here is we are going to seek this as a power series expansion in rw because rw is my small parameter and now the coefficients will all be functions of so this is like your Taylor series expansion or the power series expansion depending upon the level of accuracy that you are interested in you are going to keep the terms okay so now I am doing a power series expansion in terms of rw so the coefficients will be depending on the radial position r you understand so basically h remember the function of rw and r what I am doing is the rw is in the form of this power series the small r dependency on the independent variable is captured in these coefficients so our job is very simple if I am seeking a solution of this kind I expect that this series should satisfy my differential equation okay so I am going to have to substitute this series in my differential equation and like we did earlier we equate terms of the same order which means we equate terms of which are independent of rw rw to the power 0 which have rw to the power 1 which have rw to the power 2 okay and then I am going to get a sequence of differential equations and I use these differential equations I solve and I am going to get h0 h1 and h2 okay and once I get that I have a form and then I can go back and substitute it here your job is to verify how good is this approximation clearly if you take one term you will have an okay approximation if you take two terms you will have a better approximation to the actual value so what you are going to do is you are going to find the actual solution and you are going to find out how good this approximation is to the thing and that will clearly depend upon the number of terms you take okay. So for example maybe rw less than 10 bar-3 for example the approximation is good for first order but suppose you want to push it for larger rw value then you may have to take a second order term as well okay so just like you take more terms you get better accuracy. So let us just do this exercise of substituting this in the differential equation and I am going to retain this as it is d by dr multiplied by h0 plus rw h1 plus rw squared h2 okay remember h0 h1 h2 are all functions of r okay-irw times h0 plus rw h1 plus rw squared h2 higher order term which I am neglecting equals-1 okay. So our strategy is now to find out h0 h1 h2 if I find out h0 h1 h2 I can substitute it here and I can find h. How do you go about finding h0 h1 h2 by equating terms of the same order okay. So if I look at terms of the order of r to the power 0 that is terms are actually independent of rw what do I get from this one I get 1 by r d by dr of rdh0 by dr okay and this will contain rw so I do not use this but this is independent of rw so I get equals-1 okay that is the differential equation which I get what about this guy to the power 1 clearly this is going to be contributed by this so I am only interested in the terms without the rw what I mean is I am not going to write the rw explicitly this gives me 1 by r d by dr of rdh1 by dr equals-i h0 plus this guy is this plus yeah when I am taking it to the other side of course it becomes plus yeah you are right I am taking it to the other side yeah it becomes plus i as 0 I am happy when you guys correct me because then I do not have to redo the lecture and the second one gives me this okay I was going to be plus i h2 no plus i h1 and so on and so forth what I want you to observe here is that we are solving a sequence of problems I solve for h0 h0 is known okay then I am going to now the right hand side becomes a non-homogeneity I solve for h1 we have a bunch of linear equations I solve for h1 then h1 is known I substitute it back here and I find h2 so this way I am able to proceed sequentially in my calculation and so once I know h0 h1 h2 I can substitute it back here and I have my solution h which is an approximation okay just want to make sure I have not made any mistake yeah I think everything is fine. So now in order to solve this we need to have boundary conditions because they are all differential equations anyway so what is the boundary condition on capital H capital H the boundary condition is that this should be 0 at r equal to 1 and so I am going to write it here the boundary condition is h at r equal to 1 equals 0 this implies h0 and r equal to 1 plus rw of h1 and r equal to 1 plus rw square times h2 of r equal to 1 equals 0 plus I have all the terms which I do not write okay remember I want this equation to be satisfied for any rw okay that is the idea when I am doing my power series expansion this has to be satisfied for all rw or any rw and this can happen only if h0 at r equal to 1 equals 0 h I should be more smart I should just say hi at r equal to 1 is 0 for all i and this implies so now I am all set because that is a very straight forward differential equation which we can solve and now since it is straight forward I will be bold enough to make an attempt okay and get the solution what we will do is we will try to get h0 and h1 I suggest you guys work it out on your own and then we can compare so or you can just follow me whichever way you are comfortable I will leave you to calculate h2 so how do you this is you can just directly integrate and get the solution h0 okay 1 by r d by dr of r dh0 by dr equals minus 1 I can take the thing over there and integrate this out I get r dh0 by dr equals minus of r squared by 2 I have r dr I take the r here integrate this with respect to this thing I integrate this once I get a factor constant c1 okay integrating once once with respect to r so I take the r there and do this integrating one more time but then I before that I am going to bring my r below okay and I get dh0 by dr equals minus r by 2 plus c1 by r integrate again what do you get h0 equals minus r squared by 4 plus c1 log r plus c2 okay so these are the constants which I have to evaluate I know that at r equal to 0 my log term becomes unbounded so I use the same argument as last time I say c10 since log of r at r equal to 0 is unbounded and I only have to evaluate c2 and that comes from a condition at 1 okay and the c2 is therefore equal to at r equal to 0 I have h0 equals at r equal to 1 I have h0 equals 0 and I have c2 is therefore 1 by 4th and this implies that h0 is 1 by 4th of 1 minus r squared so in some sense you can h0 corresponds to what the solution when rw is 0 okay h0 corresponds to the solution when rw is 0 when there is no omega remember rw is omega something multiplied by r squared divided by nu so that means there is no periodic part you can think of and so the solution is that only due to the constant part and you get your parabolic velocity profile and basically this is what we always try to do whenever we are doing a perturbation series solution when I am trying to expand it in terms of a particular parameter I want to make sure that when the parameter is 0 I have a solution okay and then I am trying to improve on the solution for nonzero values of the parameter okay so when rw is 0 I am getting a solution when rw is not 0 is some small finite value I am going to tell you the solution is going to be different and that difference the correction is incorporated in that rw h1 term okay and h1 is what we have to calculate now so depending on your level of accuracy you take more terms so now that I know h0 which is here I am going to substitute it in this equation and I am going to solve for the h1 okay I have 1 by r d by dr of r dh1 by dr equals i multiplied by 1 fourth of 1-r squared this is the governing equation for h1 I am just going to put that I think I am fine okay so now I can make space clearly this is a function of r and you know how to integrate this and you can get h1 now okay so I am just going to integrate this with respect to r again and what do I get d of r dh1 by dr I am integrating that equals i by 4 just doing it a little bit of a stepwise manner to reduce my chances of making a mistake and so this gives me r times dh1 by dr equals 1 fourth of i I get r squared by 2 and I get-r to the power 4 by 4 okay plus a constant of integration c1 okay all I have done is just taken this r dr to the other side and I am just integrating it I am going to do exactly what I did last time divide throughout by r r by 2-r cube by 4 plus c1 by r integrate this one more time to get h1 equals i by 4 times r by 2-r cube by 4 plus c1 something is wrong and I get r squared by 4 and r to the power 4 by 16 plus c2 same argument as last time I knock off c1 okay and I need to calculate c2 and I am going to use the boundary condition that h equals 0 at r equals 1 and get c2 okay let me just do this and r equals 1 h1 is 0 okay this implies-i by 4 times 1 by 4-1 by 16 let me just do this because equals c2 right and I have what 3 by 4 3 by yeah 4 by 16 is that right there is a 4 here yeah so 4 by 48 right equal to c2 16 4 or 64 yeah right that is it so that is my h1 is i by 4 you can write it whichever way you want some people like to write it in a different way 16-3i by 64 you can possibly write it in a slightly better way this is slightly clumsy I called i occurring a couple of times you can take things common and you can simplify things okay. So now if you wanted to get a more accurate solution go and get h2 but you can see that this is a very simplified way of getting the coefficients of rw what are you going to do now you are going to actually go back and h is now a composite h0 plus rw h1 I want you to see that h0 is independent of i the imaginary number h1 has i in it so what is it going to do you go back to get the velocity you have to do the imaginary part of the exponential e power i rwt multiplied by the h so this guy is going to be exponent that is the real term the exponential of that is going to give you your cos theta plus i sin theta so the imaginary part is going to give you only the sin theta which means that particular component is actually in phase so when rw is 0 your velocity is actually in phase with the pressure gradient whereas when rw is not 0 a small amount comes in the i comes and now when you actually take the imaginary part instead of the sin term you will also get a cosine term because now it is exponential of i theta multiplied by i something so the imaginary part will have the cosine term now okay. So this is going to give you that out of phase component of the velocity so I mean that is one way for you to actually figure out why when you have a finite value of rw you have an out of phase component so I think this becomes very clear here so this is something which you should I will let me write this down so clearly u is the imaginary part of e power i rwt times h0 plus rw h1 okay I mean I am neglecting all the higher order terms so this is the imaginary part of what cosine rwt plus i sin rwt multiplied by h0 plus rw remember rw actually has no sorry rw does not have it h1 point I am trying to make here is h0 is real so this multiplied by this is going to give me real part I am not interested in that this multiplied by this gives me the imaginary part. So this is going to be of the form this is of the form sin rwt multiplied by h0 okay so this h0 is contributing to my in phase part so this is in phase this is in phase what about this guy when this remember has the i in it so i multiplied by this is the real part so I am not interested in that i multiplied by this gives me the imaginary part which is the cosine part so this is basically going to have a cosine component times something okay some function of r so this gives you the out of phase so what I am trying to tell you is and remember this is multiplied by rw so this is the out of phase part so in rw is very very low the out of phase part will go off the more the rw the more the out of phase component okay so that is basically for you to understand so what you people will be doing is actually finding out the solution using this approximation finding out the solution using the exact solution and making a comparison just like you know what we did for the quadratic equation the quadratic equation you have the exact solution in terms of the discriminant then you have the binomial series expansion all the all the power series that we actually did and then you can compare from how accurate is this thing for a different epsilons the only when you do that you will get an idea because I am saying epsilon is low I am saying rw is low but how low is low that is going to depend upon the problem upon the specific problem for some problems rw may be low as 10 for some problems rw low could be 10 power minus 5 so how do you figure that out only by doing this comparison so how good is your perturbation series solution and because if it is very very low then possibly this approach is not very good but supposing retaining the two terms I am able to push to rw equals 5, 10, 50 the more the better then I am happy with it okay but then this can only be done when you actually sit down and do an actual calculation so that is what you people are going to do do an actual calculation and then verify how low is low I think one last thing I want to do and then we will stop somewhere in the beginning I chose my time scale as r square divided by the kinematic viscosity okay I want to go back to making the equation dimensionless okay and I told you that there were 2 possible choices of making the equation dimensionless because we had 2 choices for the time scale okay what was the equation we had to t equals g0 1 plus epsilon sin omega t plus mu by r d by dr of r du by dr that is the equation we began with okay I mean assuming that they have only the actual component of velocity only varying r and t I am going to just I am going to keep the velocity scale and the length scale the same okay because length scale clearly r there is nothing else happening and velocity is going to be decided by the pressure okay but if we choose the time scale the characteristic time as 1 by omega instead of r square divided by nu this is what we did earlier r square divided by nu is what we did earlier what you are going to see and I am just going to write this here you will have and you can do this okay keep all other scales the same you would get in a dimensionless form t the following equation the following equation equals okay. So you know by choosing a different scale I just wanted you to understand that I told you earlier that the scales can be chosen in different ways I am choosing it this way now the advantage of so anyway there is one parameter which occurs it is not that this rw has disappeared one parameter will occur and this again the ratio of the same time scales okay this guy now is multiplying my inertial term now it is possibly easy for you to see that when rw is 0 when rw is very very low the inertial component is not going to be significant in my earlier formulation I could not see that now I can see that when rw is 0 this guy is going to get knocked off okay. So when rw is 0 this guy goes off and remember this was the guy who was creating a problem with the outer phase component if this goes off then I can actually solve for the velocity and I can get my velocity directly okay. So basically what I am saying is here in this formulation here rw equal to 0 knocks off the inertial term and so the velocity is in phase with the delta p. So possibly by doing this scaling you can instead of solving it and then trying to understand whether it is in phase or out of phase even by looking at the differential equation you can actually make this conclusion. So whenever you are solving any problem you need to be able to actually work out by choosing different time scales and see what kind of information you are getting okay because a lot of information by choosing the right scales or different scales and that will give you some insight into the problem okay. Yeah let us stop thanks.