Lecture 25 : Numerical Schemes: MPDE II

[Music] hello everyone uh welcome back so we looked at some of the implicit schemes in the last class and we saw that uh the implicit scheme allows for a high Del T we'll look into the theory of why implicity schemes allows the high Del t uh but uh before that we I'd like to introduce some more numerical schemes to you and then we'll go into the theoretical part of uh uh something called LAX equivalence theorem where we say that uh uh there are some properties of the scheme uh or the solution uh which are basically the consistency stability and so on and so forth which basically tells you uh and what conditions you will have the uh stable and accurate solution okay so before uh going into that let's look at some of the more popular schemes uh that we introduced uh at the end of the last class so one is the Lea frogging scheme okay so Lea frogging scheme is also called the uh basically Central in time and Central in space okay so if you take the convection scheme for example C by delx equal to Z okay this is your conviction problem so use a Lea frogging scheme okay this is basically Central in time and Central in space so if you Central in time this is u i n + 1 minus UI n -1 by 2 delta T right plus C UI + 1 N - u i -1 n by Delta x equal to 0 right so if you reorganize it okay you can get u n + 1 will be equal to u y n - 1 uh minus C Delta X that 2 this will be 2 2 two will cancel out so minus Lambda UI + 1 n minus UI - 1 n right now if you look at this scheme closely you can uh basically derive any scheme that you want as we have discussed you have the general scheme you can take as many Sten as you want and derive your scheme but that scheme should have some meaning okay and it should serve some purpose so here the leap frogging scheme uh is if you look at the structure of this scheme okay so this is my nth point so this is let's say this is my uh X and this is my T right so if you're starting you know the value at n right so let's say this is my point I and this is I + 1 and this is IUS one okay but this is uh scheme requires you value at n minus one right so of course you don't know the if you are starting a simulation you don't know the value of n minus one right so this is not a selfish in scheme what does it mean that means that you have to start uh solving this problem using some other scheme maybe the forward in Central in space or forward in time and backward in space whatever and so that you can get the values of uh some time steps and after that you can start using this Li frog scheme because then you have the value at n minus one also so let's say you started at t equal to uh 1 2 3 and up to 10 you have solved the using the approach which is Ft uh let's say ftvs okay and then from time Step 11 you can switch to the ctcs or lip frogging scheme because you know the value of uh 10 as well as 9 okay to get the value of 11 right so you know the value you need the value at uh so if you want to calculate the value at 11 so let's say this is the value at 11 so you need the value at 9 right so n+1 if you want to calculate you need to get get the value of n minus1 right so this is not a selfish starting scheme but you can use another scheme to start your simulation and then you can switch to this scheme and you have to do a big bookkeeping so that you have the value that nus one available when you're trying to use this scheme right so if you want to calculate the value at n + 1 so this is your n and this is your n + one okay and this will be your value so this is my I at n + 1 I + 1 and I -1 okay and this is the value at n minus one right this I I + 1 and I - one right so what is the value that you want to get you want to get this value I'll use the red color here u n + 1 and what is the values that you need you are need UI n minus one and you are basically you have the unit value at at any time step IUS one and I + 1 right so if you connect this okay to get these values you need this and this and this okay so it's basically it looks like a frog and basically jumps the step right you can see that uh it is you want to get the value at n+ one it is jumping from n minus one right so that's why it's called the lip frogging scheme it lips okay so this is called Lea frogging scheme and there are some stability properties of this scheme which is uh uh superior to the explicit scheme and that's why these schemes are used in the simulation we'll look at the stability of criter of these schemes also perhaps solve some problem okay the another scheme that you have is basically The Lex Frederick scheme okay so this Lex fed scheme this is uh forward in time and Center in space okay but we take the average of I +1 and IUS one so why we need that see you know that ftcs is not stable right forward in time in certain space is not a stable that we have seen for a linear convection problem right not for the diffusion problem so we they are using this for the linear convection problem okay so uh let's look at that scheme let's fedick scheme okay we'll call it LF okay so your problem is d by DT plus c d by DX = to Z okay and uh you can write it as d by DT equal to minus c d by X okay and what Le does the usual approach let's see if you use ftcs right so this will be u i n + 1 - u i n by delta T will be equal to minus c u i + 1 n minus u i - 1 n by 2 Delta X right this is your ftcs scheme or you can write it as UI n + 1 will be equal to UI n and minus C delta T by 2x 2 Delta X okay UI + 1 n minus UI - 1 n okay so this is your ftcs scheme but if you use this scheme as we know we have seen that uh for example the problem that you have solved okay we had a lot of Wiggles here and this portion right and eventually if you keep on running this problem this will eventually blow up right so ftcs scheme is not suitable for the conviction problem so what the LAX and Frederick proposed they were to famous scientist they said that instead of using u n we use so now it is called The Lex fedick scheme so UI n + one so you use the average of uy n and u n + 1 right let me just see no so average of central not the time okay so the time is St Remains the Same okay so this is basically u i + 1 n plus u i -1 n minus C delta T by 2 Delta x u i + 1 N - UI - one n okay so that means so you are taking this value as u i n this is your I and you're taking the value of UI n so we replace it back with this is I + 1 this is i - one so you replace the u n value as half of UI + 1 and and plus UI -1 n okay and then you got a new scheme and this scheme is basically stable okay conditionally stable but it's stable it is not unconditionally uh unstable like the ftcs scheme okay so we'll look at uh the property of these schemes also so I'm just introducing a class of schemes which are very popular okay and we'll look at the stability uh properties of these schemes in future classes okay the last one that I want to introduce of course there are many more schemes than what is uh uh written here okay uh I'm just introducing a few to you okay which are very very popular so the third one is The Lax VR scheme okay and uh this again LAX and VR have proposed this scheme so this is also ftcs for when time Center in space but we use a second order numerical diffusion so these terms are basically additionally added okay so uh let's look at the ftcs scheme okay so I can work it here only so this is my ftcs scheme right so return that but add a second order term so that will become your LAX vandr scheme okay there are two kinds of LAX vandr scheme one is the two step Lex vandr okay we will uh look at that scheme also later so this is the one step Le scheme this is UI n + one will be u y n minus C delta T by Delta x u y + 1 N - UI - 1 n and then you add some basically numerical diffusion because this is not stable by itself okay so Delta t² by Delta x² okay and uh if you are adding a diffusion the diffusion terms are normally the second order accurate terms that is basically u xx okay or Delta U by DX squ okay this is nature we have seen in the diffusion problem also when you have dlu by DT equal to New D to U by D x² right so new is basically you can equivalent it as c² Delta d by delx square and that you have to use the central defening scheme for that that will become u i + 1 N - 2 u i n + U u i -1 n by Delta x² okay so this is your LAX V scheme okay so this is the uh ftcs forward in time and center and space but then you have added some defent ter here okay so that Delta x² I it twice okay so this is how it will look like okay you can have sign you can use other sign also okay so c² delta T okay there will be you can use the two term okay or you can replace this entire thing with a another diffusion term of your choice okay uh so the purpose of this term is to basically make the scheme uh stable okay the central defensing scheme is not stable because it does not have any diffusion okay for any scheme to be stable it should have some diffusion but if you have it has got too much of diffusion or too much of dissipation it will basically smear out any nonlinear phenomena okay which is also uh not something that you want so you want the scheme to be stable but at the same time it should be accurate enough to predict the flows accurately okay so uh we will discuss these schemes stability characteristic in the letter classes okay for now what I want to do I want to go to something called understanding these schemes and want to see that when we discretize the scheme what is the actual PD that you're solving okay we are trying to solve this PD okay but then uh when we discretize it we are actually not solving this PD but a modified PD okay so we'll look at uh those approaches I mean uh those equations and that will give us some understanding that why uh some schemes are stable and some schemes are not stable we have been talking about the diffusion term right what do you mean by diffusion term okay how does it add stability when we use the backward differencing backward differencing or forward differencing basically the upwinding scheme but when we use the center differencing that disappears and we get an unstable solution why that that happen right so let's look at that uh MPD scheme uh basically modified PD equations so we'll start with the the convection problem again uh d by DT plus c d by d x equal to0 okay because this is the most complicated problem the diffusion problem is stable because there's uh anyway it is just diffusing and as long as your viscosity or the numerical diffusion term is positive uh it will be stable but uh the more critical problem is your uh the linear conviction okay in the diffusion also you have to honor this uh the diffusion Quran number but that is normally not that restrictive in a flow problem okay so uh if you write this in ftcs okay so if you try to do this in ftcs work this out out okay so we have a scheme d by DT + C D by DX = to Zer okay and we are doing it using the forwarding time Center and space so this is basically UI n + 1 - UI n by Delta t + C UI + 1 N - u i - 1 n by 2 Delta x equal to 0 okay so there's something called consistency of the scheme okay so the any scheme that you we are using should be consistent okay uh what do you mean by consistent that means that if you reduce the delta T and Delta X to Zero basically if they tend to zero your uh the discretize equation okay should lead to your PD that you're trying to solve okay so uh that means uh in other words so when we we have discretized in this form we have basically let go of some of the tration error terms right remember there will be tration ER term in this also which will be order of delta T and there'll be tration ER this term the leading TR eror will be order of Delta x² right so we have basically let go of those terms right so we have to make sure that if we uh in those terms if we put Delta t ending to zero and Delta X tending to zero it should become zero the truncation error should go to zero so tration error should go to zero as delta T and Delta X goes to zero okay that is what uh mean by the consistency of the scheme okay so let's uh look at this problem okay and look at what is the tration error of this scheme okay so earlier we have taken the seen the tration error for let's say d by DT separately and for d by DX separately okay but now we looking at the tration error of the scheme itself the entire thing when we consider together what is the tration error that we are getting okay so how do you do that same our good old friend Tor series will use that and we'll find out what is the tration error of the scheme so this is UI n + 1 will become UI n okay and plus delta T so now you see when you are writing this uh Tor series we used U Prime and so on so forth but now we have derivative with respect to time also and with respect to uh space also you will have d by DT term and you'll have d by DX ter and all their higher orders and and so on and so forth right so what I'll do I'll use a notation so I'll say U that represent d by DT and when I say ux it represent d by DX and similarly if I said u xx then it delt U by DX Square u tt will become delt U by DT sare okay so I use this representation for my T expansion okay so this will become Del t u i okay and uh this will be at n Del t² factorial 2 okay U TTI I okay I is at I place at n plus delt Cube by factorial 3 u t t t i n and so on and so forth right so what we did we expanded uh our n + one around n okay and we know our delta T is basically the jump from n to n+1 that's our timeing St similarly these things we already know U + 1 n will be basically u n so everything I want to expand around u n okay plus Delta X this will be u x i n plus Delta x² factorial 2 u xx i n plus Delta x q by factorial 3 u x x x i n and so on and so forth and you have u i -1 n will be UI n minus Delta X ux i n plus Delta x² factorial 2 uxxx i n and minus Delta x q by factorial 3 u x x x i n and so on and so forth right so now use this and put it back in this one right so when you do that so what I'll do I will basically rearrange this term so I'll take this minus this and divide by delta T okay and that will give us the term so if I use that term so I'm trying to approximate this portion first so that will look like uh u t i n plus delta T by 2 okay u t t i n delta T Square will get cancel with delta T here because I'm dividing by delta T also okay and uh plus Delta t² by factorial 3 or 6 u t t t i n okay and plus some other term okay just let's return up to this third term only okay and then we'll have plus C and I'm now expanding this okay what I'm doing so I have let's if I call this a b and c okay so I'm doing b - c by 2 Delta X okay I'm using the value that gives us the this values okay so that will become C uxi okay and this thing will anyway get canceled out okay uh plus Delta X Cub by 3 u xx I and and then will be the next term will also get canceled out and you will have something like Delta X5 to Delta x 5 by factorial 5 right u x x x x x i n and so on and so forth right so uh this is our value okay now if you uh rearrange it this will become so I'm collecting this term and this term okay so this is U + CX I okay U plus c x is What U is plus CX is basically d by DT plus c d by DX okay is equal to this entire thing is zero okay so that will become minus delta T by 2 U TTI i n and uh minus Delta xqb by 3 u xx i n and then plus or minus doesn't matter okay order of Delta t² and Delta x to the power 4 okay so there is an error here because we are dividing by Delta X so this will become Delta x² and this will become Delta x 4 okay this is the Delta x square and Delta x 4 okay so this is what our ter looks like okay by just putting this value and putting our T expansion right so if you look at this is the equation that you want to solve and the right hand things that you see is basically our track error right so um for this particular scheme so when you take Delta T tends to Z and Delta x ts to Z you would see that the entire term on the right hand side goes to zero that means that equation is consistent okay so our forward in time and s in space is a consistent scheme it may not be stable but it is a consistent scheme because as we reduce our time step to zero and our spatial uh size to zero we recover our scheme which is Del UT plus c xal i equal to zero okay now when we are saying uh that we are using this scheme to approximate our uh linear convection equation right so basically we are solving this problem right so let me just write it in the next one so we are solving d by DT + C D by DX I is equal to minus delta T by 2 okay U DT i n okay and uh minus uh let's leave other terms okay so that will become ordered or just just let us let me just see a second okay so that will become Delta x² okay we have written Delta x² by 3 so just return that term so order Delta t² Delta x² right so now what we have done we have put all these THS together and we have just ring the first term here so uh when we are trying to solve uh using our ftcs scheme we're basically solving this problem okay and you can see that we wanted to solve d by dt+ c d x equal to Z okay but the right hand side actually not zero it has got certain terms right now what we'll do we will write everything in terms of the space okay so this is utti so UT is we want to put in in terms of space so you have U + c x = to0 that means U equal to minus CU X okay so I'll take another derivative of with respect to time u tt = to minus c u x t right and you can replace it minus C uh Del by D X U this is what it means that minus c d by D X and UT will become minus c d by DX right same thing so that become C c² uxx right so u u tt is equal to c² uxx okay so this equation can be written as d u by dt+ c d by DX I is equal to minus c² delta T by 2 u xx i n plus order Delta t² Delta x² okay so we wanted to solve a problem which was basically d by DT plus C D to Z but because of our choice of our disc scheme which is forward in time here and centrer in space here we got a scheme uh we got a uh we are solving an equation which is looking like this okay so this equation is called the modified or exact sometimes exact also modified partial differential equation modified PD or MPD what is it exactly this is basically when we are using a certain Des scheme it actually modifies the PD okay and instead of solving this problem we are solving this problem right now if you look at this equation closely okay ignore this uh the higher order terms but look at the just the the first term okay so this you will see it looks like d by DT plus c d by DX is equal to new Del to U by D x² right so u xx is this term this is U XT so this is not we are not solving a convection linear convection problem but we are solving a convection diffusion problem we cause our choice of the diffusion scheme not only that you will see this term the new here if you compare to this one new is equal to minus c² delta T by 2 c² is a positive number delta T is a positive number and there's two so this entire thing new is a new is negative so that means you have negative diffusion right your scheme has got a negative diffusion and earlier we have seen when we're trying to solve the problem of the diffusion we saw that when the new is negative the numerical differ term is negative it basically blows up very very fast and that is precisely the reason why you are not able to use the ftcs scheme in solving uh your problem your linear connection problem because the schemes gives you a negative diffusion okay and that leads to the blow up of the scheme in the next class what we'll do we'll look at the ftps forward in time and backward in space and we'll get this same modified PD okay for that uh particular problem okay and we'll see uh at what condition uh they become stable so we know that fbh is stable okay from our experience for a linear contion problem if the C delta T by DX is less than 1 right by just solving the problem we know that right but can we also notice in our modified PD okay so we'll look at that ttbs scheme in the next class okay for now you should uh go back with the message is that when we are trying to use a disc scheme on a given differential equation okay we basically when what we solving is a modified differential equation and we have to look at those terms closely to know about how the schemes would actually behave okay with that we'll stop here thank you very much [Music] [Music] [Music]