Lecture 21 : Compact Schemes II

[Music] [Music] hello everyone welcome back uh in the last class uh we looked at the the compact schemes and we saw that uh for a given stencil uh the compact scheme gives you better accuracy compared to the regular schemes for example Central defensing and we derived the compact scheme for the threo stensil and that was the fourth order Equity scheme compared to the second order equ scheme that you see in central finishing and we also noted uh that uh for certain derivatives the compact scheme give you better results better approximation of the derivative compared to the central differencing scheme of course like we have this one-sided formulation you can also have the one-sided formulation for compact scheme so the approach Remains the Same you write your the left hand side in terms of the implicit points your five I primes 5 IUS one primes 5 I + 1 one primes and the right hand side are explicit points where you have the functions available basically create a matrix do the Matrix inversion and solve it uh to get your the value of all the derivatives at the same time okay not sequentially but all the derivatives uh in a grid will be calculated at the same time okay that is why it is called the compact scheme so so today we'll see that uh we can basically increase the stensil size and we can basically get uh more High compact schemes okay this is also consistent with the the the thought that we have in the last class that uh recall that we have discussed that if you want to get the N derivative you need n plus one exential point and at least n tell Series right so for example if you want to get the first order derivative you need at least two stencil point and uh at least one t expansion right but we also said that you can uh basically uh make use of bigger stencil and compare more derivative terms in your t expension uh and get more High Rec system so what we're doing in the compact scheme is that we are comparing uh the higher order derivatives in the T Series on the left and right hand side and we are Computing the coefficients so for the same penil we are getting uh basically more accuracy right by comparing the homs for example for the compact four uh scheme which is the three-point stanil system we compared the first derivative and the third derivative on the left hand side and right hand side and we got the value of Alpha and a okay so uh we can increase the exential size apply more higher order derivatives and get uh more accurate schemes so for example today we are going to look at the compact 6 scheme okay so this is the 6,000 compact scheme so in this particular case uh you can have a grid so this is your grid okay use a black color this is I I + 1 i - 1 I + 2 and I -2 right so what we going to do we are going to use all these five points to compute our derivatives at these three points okay so you can use the Five Points also I mean in the derivative on the left side also but that will create a pentagonal system and that will become even more expensive to compute right so what we going to do on the left hand side keep only the I -1 I and I + 1 on right hand side we can use the five point stensil so that will become Alpha 5 i - 1 Prime + 5 I prime + Alpha 5 I + 1 Prime will be now we'll have two unknowns so when we say a this is 5 I + 1 - 5 i - 1 by 2 H and this uh you can quickly recognize by I'm using two and then you have B 5 I + 2 - 5 i - 2 by 4H right because uh when you are using this i-1 and I + 1 the distance between uh among these is 2 H and this is basically 4 so that will Anyway come out so you can uh use it like this and uh then uh what you need to do you need to use the T expansion for all these points I + 1 i - one I + 2 i- 2 I I prime i - one and Pi Prime I + 1 okay and then you have to compare the coefficients on both hand side of Pi I prime Pi I Tri Prime and 5 fifth Prime okay which is basically d f by DX i d Q5 by DX q i and D 55 by d x i by at I okay so if you compare this you'll get three equations okay one correspond to this another to this to this and you have three unknowns which are alpha A and B okay and uh you can make the computations and you will get the value which is basically 1X 3 4 by 9 and 1 by 9 right so this you can see that is quickly becomes very uh tedious to make the hand calculation but you can use Mathematica or any uh symbolic software or even the math lab to get these values so you just write these three equations compare the left hand side and right hand side and you'll get this three values you can basically write a code in mat lab also okay uh it's it's very easy okay the other thing that I want to tell you is that you can have a trivial uh thing which is the explicit system so set Alpha equal to Z right if you set Alpha equal to Z these terms Will D vanish and then you can use your regular General disc approach okay and uh you can compare your uh basically right now need one A and B so you can compare your the first derivative and the third derivative okay and then you'll get this uh for a = to 4x3 and Bal to- 1 by 3 so this is your explicit fourth order scheme and this this is your impessive 6 compact schem so you can see that using the same stanil i - 2 i-1 I I + 1 and I +2 you can get a high accurate scheme uh let's say you want even more accuracy so what you can do you can consider this I + 3 and IUS 3 as well okay and then your equations will change like plus C 5 I + 3 - 5 i - 3 by 6 h okay and then you have to compare this uh Del 75 by DX 5 okay at I okay so you have four unknowns now A B C and Alpha and you have four equations pertaining to all these four things and you can get this values of alpha a b and c right and you have the a. comp scheme right similarly you can keep on increasing stensil sides and you can have the 10th order competive scheme 12th order competive scheme right so uh this is how you can increase your order of accuracy but remember it comes at a cost if you increase the stensil size the the competition cost increases right so typically in the computations you I mean the fourth order and the sixth order comput schemes are very very popular okay uh one thing that I've not discussed here that uh compact schemes by itself uh despite being accurate uh they are not table so you need to have a filter okay they are called compact filters so we'll probably discuss these compact filters at a later point but for now I want to leave you with a thought that uh for a given uh stencil you can increase the order of accuracy by making the system implicit okay and these schemes are called the uh the compact schemes so uh this is just a comparison this is three these two are three point stals and these are fiveo stals okay and these are the coefficients that you have got for Alpha A and B okay we have discussed earlier and you can see this is the the trailing age error for the cd2 you have this H square and for the compact you have H4 right so this this is the second order accurate system and this is the fourth system but exential Point Remains the Same the number of points in the exential remain the same and if you compare the CD4 and compact 6 they both you uh require Five Point stencil okay okay uh but the order of accuracy in the CD4 is uh 4 and for the compact it is six right so that's why compact schemes are more accurate okay and U some finishing thought that the compact schemes provide better accuracy than the expit schemes on the same stencil but are computationally more expensive uh the reason for that is uh it that lead to a banded system and which needs to be inverted or solved using a TDM solver tal Matrix solver or some some you might have read about the thos algorithms okay basically you use the forward substitution and backward substitution and calculate it okay these uh schemes are not very uh good with gpus for example right so gpus require your system to be uh highly Parable but these systems are not highly Parable so you need basically special techniques although there are some uh sh bu DDM Sol bu already implemented in the ca labion Nidia but generally these are uh not as good compared to let's say other schemes which are not implicit okay precisely because you have a matrix here which needs to be inverted and it's a very costly Affair let's say if your grid size is very very large your Matrix size will be very very large and inverting that Matrix is not that easy okay and the last thing is that at the boundary so let's say uh this is your boundary Point okay 1 to n and you want to calculate this uh the compact scheme at two okay these are the bond Val are known but you want to get the value at two right so you have only two points 1 and three available so right there's nothing on the left here and nothing on the right here so maximum that you can use is basically compact 4 here okay even if you might be using compact 8 or compact C at the other points at point two you can maximum use your compact four schemes right and uh the other way to do it is basically use the one-sided stencil that means like you want to calculate the value at two but use the points at 2 3 and 4 for example or basically maybe 2 3 4 and 1 okay so that will give you uh uh one-sided uh I mean you need to use the one-sided formulation for the compact scheme but practically speaking um even if you use a Fed compactor scheme or you lower the order near the boundary uh it does not make much of a difference so uh more more practical way is that as you come towards the boundary from your internal points you decrease the order okay so you might want to use the fourth order here sixth order here another places you can might want to use the 8 scheme okay doce the out of the scheme or use the one-sided compact scheme I've not discussed the one-sided compact scheme derivation but you can do it like we have been doing uh for the two-sided scheme okay only thing is that the Symmetry argument that you have been making uh to reduce the number of coefficients will not be applicable when you use the uh one-sided scheme okay so this finishes our discussion on the compact scheme uh one last thing I want to touch upon uh in terms of discretization is the non-informed discretization so very often your grid will not be uniform for example let's say if you want to calculate some something on a flat plate okay so you will have the bond layer which is developing okay like this and you want to capture the events very close to the wall uh in the bond deer very very well and outside anyway you know that uh the flow is not that vertical uh you might use the iser equation also even the B equation will be fine but here there are a lot of things which is happening uh your boundary your velocity is changing very quickly okay so all these things are happening here right so uh in order to uh capture this Behavior well what you need to do you need to make very fine mes close to the bond layer so this is called the bonday missing right and then as you go up basically you can uh reduce the number of points right this is a very common approach in cfd so this is called the bonday maching right so you have one deer machine so you have very close points near the wall and as you go away from the wall uh the points basically uh become uh course right so uh in order to handle this kind of situation let's say if you're going to use the finite difference formulation so this Delta X is different from let's say this Delta X okay so you have one on this distance and another let's say here okay and that's the ratio between the two is called the stretching ratio so you can call it if this is I and this is I + 1 and this is IUS one right so you can call this as a Delta x i and this is your Delta x i + 1 and similarly Delta x i i + 2 will be different Delta x i + 3 will be different right so Delta x i is = to x i - x i - 1 and Delta x i + 1 is will be x i + 1 - x i okay and the task is to find this gradient at I point I so how do we do that okay we will use our good old old approach that we had is so you have f i + 1 the Tor R approach so remember this I I + 1 and IUS one okay so this is Delta x i + 1 and this is Delta x i okay so you will have fi plus Delta x i + 1 fime I plus Delta x i + 1 sare by factorial 2 F Prime I Delta x i + 1 Q factorial 3 fi triple prime and so on so forth fi minus one will be fi minus Delta x i right here is I + 1 but here it is Delta x i okay- fi prime plus Delta X i² by factorial 2 f i primeus Delta x i + 1 CU by factorial 3 f i Tri Prime okay this is your 1 and this is your 2 so do 1 - 2 that will be fi + 1 - fi -1 this will be cancel will be equal to Delta x i + 1 + Delta x i right f i Prime and earlier when we had uniform M this used to get cancelled but this time it will not get cancelled and this will become Delta x i + 1 s - Delta X i² by 2 f i double Prime and then we'll have other terms right so you rearrange it take f i Prime will become fi + 1 - fi - 1 by Delta x i + 1 Delta x i okay minus so this is a square - b square and this is A + B so that will become a minus B that's minus Delta x i - I + 1- Delta x i by 2 fi Prime plus other terms and this Delta x i + 1 and Delta x i the summation of these two so this is Delta x i + 1 is equal to x i + 1 - x i and Delta x i is basically uh x i - x i - 1 if you add these two that will become x i + 1 - x i - 1 right so this will become fi + 1 - f i - 1 by x i + 1 - x i - 1 and minus Delta x i + 1 - Delta x i by 2 fi Prime and plus other terms right so you can uh calculate the derivative like uh this but you can quickly notice that here you have used the central defening approach you consider this uh I +1 and i1 S1 both but despite that our the the first the accuracy is here goes as Delta X or H right if you like that so that means it is first order accurate so despite using 3o stencil and the two sided formulation uh we um got the scheme which is the first order of crate so because of the non-uniform mass uh these terms do not cancel out okay and that's why your order of requency goes down so you this is a very uh common thing you'll be hearing that when you use a non-uniform mess uh your order frequeny goes down compared to when you use a uniform M right but sometimes you don't have a choice as I told you that uh uh let's say near the bond layer you have to use stretching okay because the things are changing very very fast and Away uh from the bond layer the things uh the events are not that interesting so you're not that interested in that so if you use uniform m in the entire thing your G SI will be very very huge because the minimum grid size that you require near the bond layer is very very small right and you don't want to use the same grid size away from the B dealer also okay so despite of the fact that uh your Ry goes down you need to basically use the uh the non-uniform mess right so uh this is the same thing is on here okay so uh now one question comes that the compact scheme that we have derived is for the uniform M so how do we use the nonuniform m i mean compact for the nonuniform M okay or even for let's say for messes which are not very uh uh nicely aligned for example if you want to do uh flow over an arrow file right so this is your arror file your mes will look like this okay like this okay and then you have M will some look something like this right the center one is your arrow file okay so can we use uh basically our compact scheme or whatever fining difference that we have this that we have discussed can we use on these kind of systems right but because all of the things that we have discussed basically for the uniform mes okay on a cartisian system let's say XY or X Y and G okay where things are uh changing very smoothly and uniform right but we can use those formulation all those ideas on a nonuniform mes or even mess like this by doing something called curv linear transformation okay that means B basically even if your things are changing uh very rapidly uh right you can con basically convert it into a uniform mes okay uh by using a Jacobian right so from this mes which is changing basically uh rapidly and this is your uniform mess so this is your non-uniform mess and this is your uniform M so between that you can use a transformation you'll get a Jacobian which basically Maps uh each grid point to the grid point in the uniform mess but uh the advantage that you can use all derivations and the compact schemes on this uniform M uh calculate the metric GD metric jacobians do the calculation uniform M and do the back transformation and get back to your non uniform M right so you can use your hyro schemes by doing this transformation similarly if you have a uh complicated grid like this not very complicated but uh a grid which is closed okay uh like in a cur system we can use the transformation and use the same ideas that we discussed for the discreation okay so we'll be discussing the cinear transformation uh in the later classes so for uh today uh we will finish with the thought that uh uh we can approximate a function derivative using a numerical approach and you have many different ways of getting ration you can have the central defensing or the compact scheme or the one-sided formulation and uh uh they all all approximate uh the derivative in a different way and you have different uh orders of accuracy for those schemes and we have seen some examples where we saw that how the compact schemes are doing better than the regular schemes right uh but we need to discuss about the stability issues also that we'll be discussing in the next classes so first we'll start with the model equations like the convection equation diffusion equation and conenction diffusion equation and solve very very simple problems on that okay using the disc as appro that we have discussed okay after that we'll go to the stability and uh the accuracy consistency uh robustness and all those kind of things how do you know that the solution is going to converge or not so all these things we are going to discuss okay and then we'll also look at the uh the Cur linear transformation uh at a later point of time so with that uh we'll finish the uh discussion today I'll see you in the next class thank you [Music] [Music] [Music]