 Let us start with lecture 18. What we did in the previous 2 classes were basically talking about regular perturbation methods for finding approximate solutions to equations which had a small parameter and in both these systems the small parameter was in the differential equation itself okay. Now so far you are actually comfortable with solving problems where the boundary conditions are you know applied along surfaces which are surfaces of some coordinate being equal to a constant. For example if it is a spherical coordinate system you would say r equal to constant that is the surface of a sphere. If it is a rectangular condition coordinate system you would say y equals constant and you know how to apply the boundary conditions there right. But sometimes you could have a situation where the boundary the surface is not going to be defined as y being equal to a constant or r being equal to a constant it could be y is a function of x okay and when you write the boundary as y equals function of x maybe there is a small parameter occurring in this form of a surface then the definition of the surface okay. So that means this is actually occurring on the boundary and what we are going to do now is trying to find solutions to problems of this kind okay using the method of domain perturbation. Again we are going to follow Gary Leal very closely so this is also explained in Gary Leal the earlier 2 examples were also worked out in Gary Leal okay this is just for your reference. So now we are going to focus on domain perturbation we use this the boundary the domain in which we are solving is of the form you know y equals let us say f of x comma epsilon okay that is that is the surface of the actual problem I will give you an example to illustrate this what we are going to do is look at the surface this if we can visualize as a perturbation of y equals constant okay because you know how to solve problems where y is constant that is the when the surface is y equals 0 y equals 1 you know how to apply the boundary condition. So if this can be viewed as a perturbation of this kind of a surface then since we can possibly try to do a perturbation of the domain in terms of this and then seek a solution that is the idea okay. So then we get an approximate solution of the original problem whose surface is given by 1 okay you so the earlier problems that we talked about they were what we called regular perturbation problems because we have the difference the small parameter occurring in the differential equation here the small parameter is actually occurring in the definition of the surface which is actually defining my boundary okay so to give you an idea of this let us look at the flow in micro channel wavy wall now you are all used to pipes whose surfaces are rough okay but supposing there is a periodicity in the roughness then instead of just saying there is a random roughness that you have let us say that you have a periodically rough surface. So again to keep life simple what we are going to do is we are going to look at system like this okay so that is basically my wall with a wavy surface okay. Now let me get my coordinate system straight this is y and this is x and into the plane of the board is z okay so z is into the plane of the board into the board I am looking at a problem where we have flow in the z direction which is basically flow along the parallel to the grooves that we have okay it is not in the x direction it is in the z direction so flow so these grooves that you have they are actually in the z direction and the flow is in the z direction which is parallel to the grooves. Now why would anybody be interested in a problem of this kind sometimes what happens when you are talking about flows in micro channels the Reynolds numbers are so low that the mixing is very poor the flow is laminar okay so when you possibly give some induce some surface roughness then you can possibly induce not turbulence because your Reynolds number is still low but mixing because you can possibly have velocities because induce in directions perpendicular to the flow direction so you would have velocities induced in the x and y directions and so you could get better mixing okay so basically the surface is actually inducing the 2 other components of velocity or that is the idea okay. So now what we are going to do is we are going to so motivation is motivation for this is possible improvement in mixing in low Reynolds number flows okay so because of the surface being rough I may have some vortices induced here and so maybe better mixing. The other thing now to explain what is happening here how would I define the wall these are my walls okay actually rigid walls so this is supposing since this is periodic so I am assuming that this is periodic and the first trigonometric function which comes to my mind when I say periodic is sin okay. So I am going to say that this guy the mean value is let us say at a distance of y equals d by 2 this is d by 2 is the mean value okay so the way I am going to look at this actual surface is y is d by 2 multiplied by 1 plus epsilon sin 2 pi x by L where L is the wavelength in the x direction of this periodicity okay. So the actual is defined y equals plus d by 2 this is the top surface the bottom surface will be minus d by 2 the entire channel is of the thickness d okay. So epsilon basically it gives you some control over what is the amplitude of the periodic nature which is imposed. So the way I am looking at this problem there is a constant value d by 2 so this have been actually this is like again if epsilon is 0 it reduces to the problem of flow between 2 flat plates which we have seen earlier okay the top surface is given by plus d by 2 the bottom surface is going to be given by minus d by 2. So the channel in the y direction the gap between the plates is actually d it extends from minus d by 2 to plus d by 2 and L is L the wavelength of the periodic pattern epsilon is the amplitude you wanted very large amplitude thing epsilon controls the amplitude and of course what we are going to do is keep life simple and look for fully developed solutions that is we have a very long plate extent in the z direction and we look for fully developed solutions to the flow okay. So as you can see now if epsilon is 0 my surface it becomes y equals plus d by 2 y equals minus d by 2 those are the walls and then y equals constant you know how to solve the problem I mean you will just do whatever separation of variables and then you have to find those constants which come out when you do the integration so you put y equals d by 2 and then you are able to find the constant what is the problem now when you do the you can possibly solve your differential equation but when you are trying to find the arbitrary constants which are going to come you cannot put y is equal to this surface here because like you want a constant and what you are going to get is a function of x okay. So that is why the problem is going to arise so I just want to show you that you just can substitute y equal to this because you are trying to find a constant arbitrary constant of integration by applying the boundary conditions and since this is not a clean surface you got a problem okay. So I just want to show here that as epsilon tends to 0 the wall is given by y equals plus or minus d by 2 I mean the 2 walls 1 is plus d by 2 the other is minus d by 2 okay the walls I should write the walls are given by right and we know how to solve that so basically what I am going to do is I am going to do a perturbation analysis but now I am going to do a perturbation analysis keeping the fact that this guy is going to be perturbed about this okay that is the idea and since we are doing it on the boundary so you have a domain which is actually periodic I am going to look at this boundary as a perturbation of the constant wall that is the reason it is a domain perturbation problem the domain is being perturbed the domain of the solution is being perturbed okay. So we are looking at a steady state low Reynolds numbers flow and what is going to be the Navier's talks that the momentum equation this is going to be low Reynolds number flows so inertial terms are going off okay so we will just go back to what we did earlier so since we have low Reynolds numbers the inertial terms drop off and what I have is 0 equals minus dp by dz plus mu multiplied by d square w by dx square plus d square w by dy square because it does not depend on z it is fully developed it depends on y because clearly the boundary is there and it depends on x because clearly there is a periodic nature of the surface in the x direction okay so w is going to be now a function of x and y so what I am doing is I am basically looking at a 2 dimensional analog or extension of what do we did earlier in the class where we had only a 1 dimensional flow okay. So we are going to write this as d square w by dz square sorry x square plus d square w by dy square equals dp by dz and clearly the project gradient is negative so what I am going to do is write this as minus g where g is positive okay so here g equals minus dp by dz and I am going to keep the mu here okay we do the usual stuff which is try to make things dimensionless clearly what are the important scales that we need to look at we have a length scale in the x direction so what is the characteristic length scale in the x direction that is going to be l because that is the wavelength of your variation in the x direction so x characteristic is l what about y characteristic that is d that is the gap between the plates and the other thing that we need to do is worry about the velocity that is what we are trying to do find out and we need to have the characteristic velocity which is wch and since the flow now is driven by the pressure so we need to include the pressure gradient or g in this case g is positive in the definition of our characteristic velocity and that is going to give me g multiplied by d squared divided by mu okay the gap between the plates is being used to define because that is the one which is going to decide the average gap between the plates is also going to define your viscous resistance to the flow so let us use all this and make this equations dimensionless saying that I have w characteristic comes out I have g times d squared by mu okay times d squared w star by the x star squared and x characteristic is l squared plus d squared w star by d y star squared and y characteristic remember is d squared this must be equal to minus g and there is already a mu there which I like to keep here okay. So I have just made my equations dimensionless the star variables define my dimensionless quantity so w star is w by w characteristic and so on similarly for x star and y star so all the star follows are dimensionless okay. So you can clearly cancel of these mu's clearly cancel of the g's and what you have is and multiply throughout by d squared you get d squared by l squared times the partial derivative with respect to x star squared plus d squared w star by d y star squared equals minus 1 that is what I get. So this is my differential equation what I need to do is look at the boundary conditions the boundary condition is that on the surface since I have a rigid wall the velocity has to be 0 the no slip boundary it is not moving the no slip boundary condition basically tells me that w equals 0 at y equals plus minus d by 2 times 1 plus epsilon sin 2 pi x by l and I can define my characteristic use these definitions of ych and xch and write this equation in terms of the star variables which basically tells me that at w equal to 0 I mean w is equal to 0 at this tells me w equals 0 or w star equals 0 because I want to make w also dimensionless at y star equals plus or minus half 1 plus epsilon sin 2 pi x star okay that is my no slip boundary condition. So either I can solve the problem in the full domain that is from plus half to minus half okay or I can just say that look I am only interested the solution is going to be symmetric and I am going to be looking at only one half of the solution because we are looking at one half of the solution I can get the other half. So if you do that we can use these two boundary conditions and solve perfectly fine or we can also exploit the symmetry and use that at w star d w star by dy equals 0 by star equals 0 at y star equals 0 okay. So that means only solving for half domain and then you can extend what is happening in the other half domain by just extending it okay this allows us to find the solution in the half of the domain and we extend to the other half by symmetry arguments. No the center line unfortunately I have dropped the this thing so basically I am saying that the center line is the center of my domain everywhere. So at every point y star equal to 0 is going to be the center point of my channel. So when you have the two crests they are in phase the two troughs are in phase so y star equal to 0 is always my center line. See the way I have written this the top surface top wall and the bottom wall both are in phase the wavey pattern is in phase. So y star equal to 0 is always the center line and along this it is basically the pattern is symmetric across this okay if they were out of phase then you have to you may not be able to use this. So this basically tells me that y star equal to 0 is my center line throughout x for all as I go along x okay. So what this means is y star equal to 0 is the center line since the periodicity top and bottom walls are in phase that means the two peaks are coinciding I mean this guy is peak and this guy is the bottom thing will be coinciding and then the troughs will coincide okay. So now what we want to do is just extend what we have done earlier and which is seek w star okay we will now drop the stars we will drop the stars for convenience for my convenience. So I do not have to keep writing the stars. So w at y equals half times 1 plus epsilon sin 2 pi x so this is the function which I want to evaluate I want to evaluate w at along this surface at y equals half of 1 plus sin 2 pi x I like to write this as w at y equals half just x I am going to use the definition of the Taylor series of a function if I have f of x I am going to write it as f of x equals f of x naught plus f dash of x evaluated at x naught multiplied by x minus x naught right. So that is the whole idea that is what we are going to be using Taylor series expansion I am going to write this as f of x evaluated at x naught x naught is corresponding to epsilon equal to 0 okay dw by dy evaluated at y is equal to half multiplied by x minus x naught which is epsilon by 2 sin 2 pi x okay this is x so x naught is epsilon by 2 sin 2 pi x plus d square w by dy square times epsilon by 2 times sin 2 pi x the whole square right by 2 plus higher order terms. So all I am doing is we have obtained this using a Taylor series expansion which is f of x equals f of x naught plus f dash of x naught multiplied by x minus x naught plus etc okay the higher order terms which I am just not going to worry about. So what we are going to do now is now what am I done see what I have done is I have made a transformation this particular Taylor series expansion has allowed me to evaluate the w at y equals half and that is basically something which I am comfortable with that is in my comfort zone because if you tell me y equals half I know how to plug in the boundary conditions and get my arbitrary constants of integration because the boundary condition is basically used for getting my constants of integration okay when I have to evaluate w at y equals half I am fine I know how to do this. So basically what I have done is I have taken this w evaluated on this surface which is not one of the regular surfaces y equals constant and exploiting the fact that this is just a small deviation from a y equals constant surface y equals half and doing a Taylor series expansion okay. So I have w at y equals half plus the first derivative plus the second derivative term and of course all these derivatives are evaluated at y equals half okay. So now the problem that I am solving is the differential equation which is right here and that symmetric boundary condition which I am comfortable with because there is no epsilon there okay and this fellow here which has the epsilon in it. So what I am going to do is I am going to our problem is now reduced to so we now have to solve this problem equals minus 1 and let me write the easy boundary condition first which is the symmetric boundary condition 0 and the w equal to 0 on that surface I am going to write as 0 equals w at y equals half plus dw by dy at y equals half times epsilon by 2 sin square 2 pi x d square w by dy square evaluated at y equals half times epsilon by 2 sin 2 pi x the whole square okay. So w along the surface I am just using this w this w is 0 so 0 must satisfy this. So basically what I have done is I am trying to show you now I have just transformed the problem into something where the boundary condition is evaluated at y equals half the parameter epsilon is occurring and this is a small parameter assuming that this guy this wave is having a very small amplitude and I want to seek a solution to this problem right. So in order to seek a solution to this problem we know how to do this because my earlier irritant was my surface was having the function of x in it. So I have used the Taylor series expression and gotten rid of that problem I have all these conditions being evaluated at y equals half. So now I can hope to proceed further in the sense that use the same idea as what we had earlier seek w as w0 plus epsilon w1 plus epsilon square w2 etc because now w this guy is fine it does not have epsilon in it but this boundary condition has epsilon in it okay and so if epsilon were to be 0 I know what the solution is I mean this is a flat wall right. If epsilon is 0 that means the wall is flat and I can find the solution w at y equals half is 0 and so for epsilon equal to 0 I can possibly find the solution and for epsilon non-zero I will have to make corrections. So the same idea as what we had earlier I am just going to seek this form and now I am going to substitute this everywhere in the differential equation in this boundary condition and in this boundary condition and group terms of order epsilon to the power 0 epsilon to the power 1. Let us do the one which is challenging which is this and in the sense slightly more challenging than the other 2 they are not only challenging I mean 2 challenging I would not have done this in the class right. So what we are going to do is just substitute this. So shall I just say this is some number 3 and this is some number 2 substitute 3 in 2 and what do I get 0 equals w0 plus epsilon square w1 plus epsilon square w2 evaluated at y equals half that is my first term. I am just substituting this guy there okay plus epsilon by 2 sin 2 pi x that is this multiplied by the derivatives whole square and there is a by 2 also which I need to be careful about okay all I have done is substituted this in this expression here okay. So basically I am saying just look at this problem the first 3 lines that is my problem it has a small parameter epsilon in it and therefore that motivates me to seek my solution in the form of this power series okay in epsilon and I am just substituting this to find out w0, w1, w2 like we have done before only thing is now this epsilon was not in the differential equation but it was in the domain. So what we do we need to group terms of order epsilon to the power 0 epsilon to the power 1 etc. So what about order epsilon to the power 0 which is so this guy is clearly of order epsilon because there is already an epsilon here which multiplies everything. This has epsilon squared because it multiplies everything the only thing which is of order epsilon to the power 0 is this w0 term okay. So this tells me at order epsilon and this has to be valid remember for all epsilon any arbitrary epsilon that means each and every coefficient has to be 0 this implies w0 must be equal to 0 at y equals half because this is the only term which is of order epsilon to the power 0 or order 1 all these guys are whatever. What about order epsilon to the power 1 this contributes okay so I have w1 plus this will contribute only this term will contribute and that is w1 plus sin 2 pi x divided by 2 multiplied by dw0 by dy this equals 0 at y equals half consider equal to this that gives me this okay. So you can already see the sequence emerging I would have solved it for w0 first and once I know the solution I come back and I find w1 and so on and so forth and I am going to just do order epsilon squared and then we will stop okay. So order of epsilon squared gives me w2 at this term this multiplied by this gives me order epsilon squared okay which is plus half of sin 2 pi x times dw1 by dy okay and this multiplied by this gives me epsilon squared again all the other guys give me higher order terms so I have plus 1 by 8th of d squared w0 by dy squared multiplied by sin squared 2 pi x and this must be 0 okay and remember this is all evaluated at y equals half all these guys are evaluated at y equals half. So now I am in a good position because my boundaries are all being evaluated at y equals half and I can seek to proceed with finding my solution for w0, w1, w2 okay. Now that is the differential equation so these are the 3 boundary conditions which I have I mean it is the same boundary condition of different order terms. Let us do the same analysis for the differential equation as well as the other boundary condition that is I have to substitute w in terms of this here as well as here correct I go to do it everywhere. So let us do this easy fellow first dw by dy equal to 0 at y equals 0 implies d by dy of w0 plus epsilon w1 plus epsilon squared w2 plus etc equal to 0 at y equals 0 okay and here there is no expansion business because about I have already y equal to 0 so I am perfectly fine with just using it as it is so this implies dw not by dy is 0 and order epsilon to the power 0 implies dw i by dy equals 0 at order epsilon to the power 0 for all i okay that is what you are going to get because epsilon order epsilon will give me dw1 by dy is 0 and so on. What about the differential equation that is going to give me yeah it is fine yeah this is all evaluate y equal to 0 this is all at y equal to 0 substituting in the differential equation what would I get an order epsilon to the power 0 I would get d square by l square d square w0 by dx squared plus d square w0 by dy squared equals minus 1 and so on and so forth okay for order epsilon to the power 1 it would be the same thing okay similarly for other orders other orders other higher orders equal to orders would be equal to 0 you are right absolutely right thank you otherwise that would have been a bigger problem tomorrow so for n greater than or equal to 1 order epsilon to the power n d square by l square d square wn by dx squared plus d square wn by dy squared will be equal to 0 that is absolutely right because this is of order epsilon to the power 0 all the higher order terms do not exist okay so this would be 0 and that is true for all the higher order terms okay good so we all said to find out so we do the usual stuff find w0 find w1 w2 okay so how do you go about finding w0 so w0 a problem for w0 is d square by l square d square w by dx squared plus 1 okay subject to dw0 by dy equals 0 at y equals 0 and w0 equal to 0 at y equal to 1 half so now I want you to just look at the differential equation which tells me that w0 is a function of x and y okay and but when you look at the boundary condition you do not have anything in the boundary condition which is going to actually induce a variation in the x direction okay if at all the reason why we put a variation in the x direction was that the original problem had a variation in the x direction in the boundary but here I am saying look at the boundary condition there is nothing in the boundary condition to impose the flow the dp by dx is independent of dp by dz is independent of x okay so basically what I am saying is w0 is going to be independent of x since there is nothing to induce an x dependency and therefore I can come here and I can just neglect the dependency on x and I have my classical problem equals to minus 1 and w0 equals 0 and y equals half and w0 derivative equal to 0 at y equals 0 so that is basically saying that when epsilon is 0 you have flat plates okay and my w0 solves the remember this is all consistent because w0 solves the problem with epsilon being equal to 0 if epsilon is 0 my periodic perturbation that I have on the wall is not there I have flat plates and I have flat plates with the pressure drop impose in the z direction and it is sufficiently long in the x direction I can neglect the variation in the x direction and I can I just get my parabolic profile in the y direction with these boundary conditions so this of course you can solve and you should be able to get your parabolic profile okay so once you get the solution for w0 you go back and find the solution to the problem for w1 then find the solution for the problem w2 okay and one of the things which we would also like to ask is what is the effect of these perturbations on the wall on the flow rate which is through the channel okay if I am going to keep my pressure drop the same gradient the same for a fixed length and I have a flat channel and I have a wavy channel is it going to result in an equal flow rate or a higher flow rate or a lower flow rate so that way you can get an idea about whether you can process more chemicals in your channel or not.