Prof. Erwan Faou | Weak backward error analysis for stochastic differential equations

pleasure to introduce Eron Pow who talk about from India who talk about the weak backward error analysis for SD. Yes. So

do you hear me like this?

No. Okay. So um so thank you very much and thank you very much for this uh invitation. So as you might maybe um see

invitation. So as you might maybe um see in a few minutes I'm not specialist for for from of stoastic PDEs right but

um but I would like to talk you about something that is so backward error analysis this is some something well known in the geometric numerical integration theory. So, uh, Andrew was

integration theory. So, uh, Andrew was talking about in the in the last talk in the in the morning. And so, uh, um, here I would like to to to try to show you

how we can adapt some ideas that are now well known in this deterministic case to the stoastic case in finite dimension. So, um, so I will first in

dimension. So, um, so I will first in the first part I will I will talk I will try to describe you what is uh, what is uh, backward error analysis, right?

And so to do this I will consider a very special case of stoastic PTE. So maybe

from the school G or Z. So without noise and project it onto a finite dimensional subspace. Right? So essentially this is

subspace. Right? So essentially this is an OD an OD. So I will write it this way. So I

OD. So I will write it this way. So I

will not divide by dt because of the audience but so it's an od where here f is some vector field from ld to ld

right and associated with this uh od I will consider the flow so the flow I write it

tf of y which sends y to along along the flow of of this od to to after a time t right and if I do some numerical

approximation. So if I want to have

approximation. So if I want to have a to have uh to calculate a numerical approximation of

this then um I will so I take a numerical method for for ODEs and it's to with this numerical method is associated the

numerical flow so which I I will denote by fi big fi h of y. So here h is the

step is a time step. So this is a small parameter the time step and five h big five h of y will be an approximation of

the exact flow up to some error. So o of hp + one where p here is the order the order

of the in integrator. So if if you have such a local uh such a local estimate then you can accumulate the errors and then show that you have an a global

error of order h to the p if you if you calculate the solution over uh over a finite time

interval. So for example uh if if you

interval. So for example uh if if you take the oiler method 5 h of y will be just y

+ h * f of y and if you can if the midpoint rule will be which has the advantage of being uh

simplectic I will talk about that later which is defined by the solution of an implicit system f of

okay. And so if if you look at this

okay. And so if if you look at this system here and if you want if you try to to to show um to to see how how this uh y1 depends on y0 then you can see

that you have an expansion in powers of uh of h. So it start with f h f of y zero which is some kind of consistence of the numerical scheme and then you

have other terms etc. So it's essentially it's I mean here in this case it's analytic if if it is analytic right if etc. So you have an expansion

in powers of h so locally and other examples are given by splitting methods for example where you have the de composition of the of f into two parts that you can calculate exactly then you

solve alternatively one part and the other part and this these are splitting methods right.

Okay. And so uh so the goal of geometric what we call geometric numerical integration is to uh to try to have to try to understand the the qualitative

behavior of uh of uh of the numerical solution. With this you associate a

solution. With this you associate a numerical solution numerical

solution yn which is defined by induction yn equals to h of y n + one n

minus one sorry. So by induction you have a numerical solution that uh that approximate or is supposed to approximate the exact solution phi. So

the flow uh at time n times the step size for a zero. Okay. So now you we are not interested here in the in the

approximation of a single trajectory but more on the on the qualitative behavior of of this yn particularly in the case where uh where we have some strong structure in the in the in the vector

field. So in f which for example in the

field. So in f which for example in the case where it is Hamiltonian this is one of the major case of of applications. So

we had already this morning some uh some Hamiltonian systems. So in the case where um so if we consider Newton node

right where u is a is a potential and then you can write this system by in introduction in introducing the

velocity and you can write the system under the following way following form. Okay, so Q dot= P dot equals this

form. Okay, so Q dot= P dot equals this is the same and this you can write it as

this simple matrix times uh the gradient of H and here this is a simple matrix J Andrew was talking about this morning and in this situation I mean because

this J is a skew symmetric then you can show that along the flow along the flow of along the exact solution of this system then you have the preservation of the energy

and this is constant. So a natural question is

constant. So a natural question is uh do we have the same I mean so can we have an estimate for the discrete

solution if it's smaller than something and uh and for how long because this is for all time and now for how long do we have a preservation of

the energy so this is a kind of question with this so it's extremely important in um for the applications to so to Hamiltonian system because if you have preservation of the energy for the

numerical scheme. Then you also have the

numerical scheme. Then you also have the preservation of uh of the microchanical measure for example you also volume preservation and then you can use your numerical to sample also exactly like

this morning. So in molecular dynamic

this morning. So in molecular dynamic and but also in in astronomy uh uh people um are really interested in deriving numerical scaling energy. So

this is this is a natural question and uh okay so so how do we how

do we attack this? So um let me consider first so the the exact flow right. So the exact flow um I will

right. So the exact flow um I will define the lead derivative LF which is defined as

essentially the derivative of some function G. So G here goes from LD to LD

function G. So G here goes from LD to LD to LD.

Okay. And it's it's defined as the derivative of G uh in the direction F.

So this is a lead derivative along the flow along the exact flow and if if you want to write down this in coordinates you can write it this way

right which d fi okay then if you look at the exact flow t of y0 and you do uh you take you take

the tailor expansion of of this exact flow. So this is a sum for small t.

flow. So this is a sum for small t.

Okay. So for for for for t small here this is a sum of tk / k of and you take you take here k times the time derivative of the

flow. Okay maybe in s here in s equals

flow. Okay maybe in s here in s equals to zero. You write this is in the sense

to zero. You write this is in the sense of asynthetic expansion. Okay. So you if if you want to give a meaning of this then you truncate here to to some to some power n and then you have a small a

small remainder term. And if you look at this, if you look at the the equation, this is just L * the derivative of the flow. So this is L to the

flow. So this is L to the K in the direction F applied to the identity function and evaluated in

Y. So you just when you take a a time

Y. So you just when you take a a time derivative is nothing but taking a le derivative of something.

Um so so in some sense we can we can say that uh this flow this we are in the in the nonlinear case but we can say in some sense that this flow is an

exponential for small time exponential okay and it's it's you can

give a a correct meaning of this if you look at in the sense of asmtotic expansions. So now what is what is the

expansions. So now what is what is the backward error analysis uh uh problem? This is given my my my

uh problem? This is given my my my integrator f well h so it's a it's a it's a it's a small so it's a numerical integrator so

it's a it's a small kick in the in in it's a perturbation of the identity right because h is small so it begins with the identity plus essentially it

will be h* f this will be the oiler scheme and then you might have also some some other powers of something is it possible to find is it

possible to find some modified uh vector field f which uh which we expect to be a

pertubation of the original one at the order p where p is the order of the integrator right so something like this

so I mean something which will have an expansion in powers of h that will depend on h and such that this ph is

equal to the expon exponial of H * L. So

the lead derivative of this modified vector field which me which mean is it possible to find to express the numerical scheme

which is a discrete map as the exact as a solution of the exact flow at time H. And the answer is okay. I just take

H. And the answer is okay. I just take the log. You can do this. You you write

the log. You can do this. You you write fh equal to 1 / h the log of f h which

is identity plus hf etc. So in the case of imagine that you have a linear system. Imagine that you have a linear system. So f is just a multiplication by a matrix. Then 5 h is

also a multiplication by your matrix close to the identity. And this is this will be this this okay I I speak like this now. So this will be just a

now. So this will be just a multiplication by a matrix and then this will be the log of identity plus a small perturbation. Okay. So in the case in

perturbation. Okay. So in the case in the in the linear case uh it gives you a matrix and actually this converges in the nonlinear case it's much more

difficult but here you have you have to take um here in this if you expand this for example then if you expand this exponential then you will have a derivative of f in all the directions

and with this derivative you can associate some trees in the hop algebra and you can solve this equation in the hop algebra it's called uh And um

okay so okay that there exists f but in the nonlinear case here for this for this formal series there is no convergence in in

general there is no convergence. So we

have to to to make some uh some truncation. So imagine that imagine that

truncation. So imagine that imagine that this converges for just for a moment.

Then you will have ph uh compose n times. So here you you have n. So this

times. So here you you have n. So this

this will be really y n. Okay. If we if we manage to have this

n. Okay. If we if we manage to have this then this will be the exponential of and because it's the same the exponential commute with itself etc. So you can you

could put the n inside right imagine that imagine that we we have matrices here. So this would be n h times l of h

here. So this would be n h times l of h applied to uh to identity apply to y which means that this will be the

exact solution of a modified a modified a modified uh ordinary differential equation.

And it means that this solution will coincide here. This solution y evaluate

coincide here. This solution y evaluate it at the time tn. So tn you imagine that this is um this is n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n * n * h the

time the times where where we are after n step this will be close to uh to yn up to some error and here here I said that

this series does not converge in general. So we have to to to make some

general. So we have to to to make some some truncations and here uh we can make some truncation at any arbitrary order so that we can have hn plus one for all

n where here obviously is a constant dependent I will come back to that later okay so this is the general strategy uh and so now it's time it's time for me

to give you uh the the main theorem in fact in this in this theory I will come back to the origin and the people who worked on this

ator. So the main

ator. So the main theorem is that if we assume that f is

analytic analytic so over of over a ball some ball of radius m okay we are in finite dimension so all the nodes are the same and analytic

means that you are complex analytic in a small strip over the over the um and if uh then if you take then there exist

FH there exists some FH which is not which is a a truncation of this of this series such that if you take Y in the

ball then you will have that the flow of at time H of FH of Y minus

uh minus the the mod the the numerical flow this will be exponentially all with respect to uh to the time to

the time step. Here I put an h over you.

You will see why in a minute. So um so we can we can in fact construct construct a modified vector field such that at least at locally in time I I have only an exponentially small error.

So something extremely small. Uh where

does this does this term come from? Uh

actually uh you have to truncate this fh. So you truncate it to some power n

fh. So you truncate it to some power n which is what you think and uh if you work a lot. So here here you have a if you do this for this n you will have

some constant times h n + one. Okay, if

you work a lot and if you work if you work more then by analytic estimate you can prove that this constant is n to the n and this is this is the hard part of

the of the job and uh and then you take and I love this I really love this calculation you take n 1 / e * h so and so if you look at

this then it's then this is equal to exponential of minus n which is equal to exponential of minus okay a constant divided by h I cannot get tired of

this. Uh okay. So this is the basic uh

this. Uh okay. So this is the basic uh this is something standard if you want to make some divergent series converges right. Uh okay so application this

right. Uh okay so application this because this is only local this is for local local time step. If you want if you want to have some uh to know

something about the long time dynamics then you have to know something on FH on on this modified flow. And there is a

major case of applications which is the case of where if f is hamiltonian is

hamiltonian and if if uh ph is simplectic. So the midpoint rule any

simplectic. So the midpoint rule any splitting method with with uh with any splitting methods using Hamiltonian

Hamiltonian part then then uh FH is also Hamiltonian and associated with some

some H of H that is a modification of the original Hamilton etc. Yeah, that this is

something structural and now now obviously this H of H will be preserved along this flow and if you look at the accumulation of the error then you will

only accumulate linearly as long as you remain bounded in in BM so that you can make the the analytic estimates. So

there is a corary of this if if y belongs to some uh belongs to the ball. So this is kind of a posterior

ball. So this is kind of a posterior reestimate in some sense. Then I have that uh

h of yn minus h of y z. There there are a lot of h here. So it's it's a problem.

So uh this is smaller than exponential of minus 1 / 2 h for n h smaller than

exponential of 1 / 2 h. Yeah. So I have preservation of this

h. Yeah. So I have preservation of this modified energy and because the modified energy close to the original one uh on the on the same time scale I have the

preservation of of of the exact original energy.

Okay over the same time scale. So and once you have this

scale. So and once you have this modified energy you can you can do a lot of things. You can you can imagine that

of things. You can you can imagine that h is completely integraable and consider that this is a is an analytic pertubation of something completely integraable. So you can do pertubation

integraable. So you can do pertubation theory km theory stability analysis you can work a lot over exponentially long

time. Uh let me tell you a few words

time. Uh let me tell you a few words about the origin of this. So I I think that at least in the

this. So I I think that at least in the in the case of simplectic method this is something that is new since since uh since

1968 in a paper of mosa everything comes from motor in this uh in this in this in this kind of of works uh in a paper of mozzar lecture on

harmonian dynamics so it's written on with the typewriting machine he said essentially it works so just you do do the calculation. It works. If I have a

the calculation. It works. If I have a map that is simplectic and that is a perturbation of the of the identity, then it can be realized as the exact flow of a modified Hamiltonian. And he

called this Hamiltonian interpolation Hamiltonian interpolation. And it was only in the

interpolation. And it was only in the middle of of the '9s that this uh exponential exponential estimate were

rigorously uh defined. So in

1994 there was a paper by Benitin and George Giddi. So about

George Giddi. So about this sorry how many L. Okay. Uh and uh almost at the same

L. Okay. Uh and uh almost at the same time the same kind of result was were divided by Hyra and Lubish but not not this one right the father

[Laughter] up. Uh okay and um and also in 99 by by

up. Uh okay and um and also in 99 by by this so it was only in the middle of the 90s that this exponential estimate were

derived and um and um and then it really came because because people were using simplectic integration integrators more and more and observed that it would behave much more uh in in a better way

than you than the oiler method for example and now there are recent development concerning honor

directions and uh so made by u so by contribution by myself Arnold de with sufficiently many s's and

hes and benag extension of these two hamon pds okay so now uh I mean I will try to take

the same procedure and try to apply this to sdes okay and try to see what we can prove also I know. So I will consider here now an

know. So I will consider here now an SDE. Uh so maybe

SDE. Uh so maybe here f of xt dt plus sigma of

xt dwtd and u well so you in order to make some uh some uh analytic estimate or at least to to easily do some some estimates I will consider here that I

I'm on the on a dimensional taurus which means that f and sigma have are periodic in space okay and Um we can study the oiler

approximation defined by x n + 1 p + one okay is xp plus h with a small time

h f of xp plus sigma of xb times the burning increment.

Okay. So now if if we if you try to do the same So if you try to do the same I'm sorry but so if you ask the same question can

we find can we find some some y so of course you start here with x0 deterministic given

d can we find some y um XT. So that depend on H. Okay. Such that

XT. So that depend on H. Okay. Such that

D Y Y XT is equal to uh some modified f of Y

XT DT plus some modified sigma H of Y XT DW. Yeah. and such that you have the

DW. Yeah. and such that you have the same kind of things which means that x uh at time n time h will coincide at a

higher order with the numerical solution x and the answer to this question is no it's no and it's known I think since the

word of Tony Shardau right in 2003 so charlo 2003 I mean you can do the construction but but you will you will have to top n uh so so the the the

degree of approximation is smaller than two or three I don't I will write three okay so it's uh in some sense here the

the space uh where we want to take the log so the formal series space where we want to take the log is not big enough there is something missing in the and

and I hope that I will explain why in a minute So, so to go

um to to go further into this direction then I will consider the the weak error.

So I will work at the level of the of the generator the cologor equation because now it's deterministic equation and um we might hope to to do the same

kind of analysis in this situation. So along the exact flow here

situation. So along the exact flow here if I consider f that is smooth. Yeah. And if I con if if I look

smooth. Yeah. And if I con if if I look at the expectation of f of tx which I call sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry v of

xxt which I call u dx then you know that it solve the partial differential equation du = l0 l0 because it will be

perturbed of u uh u of 0 and x = to v and l0 is the colograph operator fi I d I so I write it with the summation

convention in the coordinates plus a i jig j. So here depends on x right

jig j. So here depends on x right depends on x and um and where this this term

corresponds to sigma sigma transpose. Okay. So the second order

transpose. Okay. So the second order operator and um and and now so if I consider the exact solution here. So the exact

solution u of tx is just the exponential of dl t0 of five. So this is not the exact exponential but uh so we have to define

it with with a semigroup. But okay, if T is smaller than if F is smooth, you should take the derivative. Then you can write down an asmtotic expansion of this

right TK plus LK in the sense of asmtotic expansion because F smooth. So this is this correspond to the expansion of the flow

right. So we are we are here now. Okay,

right. So we are we are here now. Okay,

let us continue and um let us try to expand the the the numerical

um to expand um the numerical flow. So

to do this I will consider x uh tilda xt which will solve the

following equation f of little x dt plus sigma of little x dw. So this is the oiler process you can say. So if you

if you start if if you calculate the solution of this process given XP then you at time t= h you have xp +

one and uh let us try to expand the uh uh the expectation of ph of

x tilda t for a small for a small time t uh well not the not the expectation for the moment. So is pi of x plus and here

the moment. So is pi of x plus and here we have the integral of from 0 to t of l

zero but now evaluated in little x dx 5 of x ds okay plus plus

am okay and this term will be of expectation zero so I don't I I keep I just keep it in mind but I don't touch it anymore and now I expand this term

again by using the same formula. So it

will give me that this is equal to 5x plus so the first term will be l applied to f evaluated in x integrate. So this

is tl of x dx uh tl of f of f dx evaluated in v plus and here there will be there will be

another integral another integral right integral of x integral of some operator a2 of x dx values v

etc plus the martingle term and another term that is that will be integral from 0 to t integral from 0 to s of something

right depending on xs and um and uh dws and this will be again of

expectation zero of zero expectation and if you if you push again if you push this uh this methodology again you will have

that you will have that the expectation of of X tilda x t for small t it can be

expounded as the sum of some tk a k of x dx uh plus o of t to the let's say n + one. So you

have an asmtotic expansion which is which correspond to the asytoic expansion of ph in the deterministic case which was the identity plus hf. So here and we have

the identity plus HL L etc. Okay. Uh and so now the here all these

Okay. Uh and so now the here all these operators if you look at the way they're constructed so you have L then you take the L of the L more or less. So the

these are operators of order 2k plus two just a calculation 2k plus two but always applied to some file that is completely

smooth. So now the question is this

smooth. So now the question is this operator here can I write that this

is exponential of so here at time h little time h exponential of h times

some modified lh applied to uh two and the answer is yes this is exactly the same

algebra this is exactly the same algebra and you will have that lh will be L + H * L1 + HK *

LK where all these operators are of order 2k + 2 I just take the log in the same way imagine that they are matrices for example then this is the same except that these are operators so it's it's a

more complicated algebra so here the algebra is power series in H and operators yeah in No, no, no. She didn't.

Ah, okay. Okay. Okay, so we can take the

okay. Okay. Okay, so we can take the log and um and now we have our modified equation, right? We have our modified

equation, right? We have our modified equation because we can say we can consider the the equation

dth equals to LH VH and uh VH in zero equals to five at least formally in in H. LH is a

modified generator if you uh so uh so now this is formal right now this is formal we have

only formal series the VH here the H of T and X will be a formal series in HK

VK T and X and uh and this is here the Koshi product just so this is in some sense this is

equivalent to to uh for all kt dt vk minus l0 vk this is the first term here apply to

vk is equal to something so to the sum from j = 1 to k of lj vk minus j

x okay this is this is uh in fact this has to be understood as as a collection of of equation. And now if you fix v and if you take a small h and if you truncate so which means that you define

some vn vn that is um that is uh just the sum for k equals 0 to uh to so obviously v 0

here is v 0 will be the exact solution right this is 80 l 0. So if you take uh if you truncate

0. So if you take uh if you truncate here VK HK so HK V can be

better then you can prove that for small H and X if you consider this truncated

solution that you constructed by just by Taylor just by in the sense of asytotic expansion for this equation minus the expectation of PH evaluated in

in H. So this will be the this will be

in H. So this will be the this will be the one one click of oiler process in some norm. So okay here we are on the to

some norm. So okay here we are on the to so we use infinity norm then this will be smaller than h to the n + one. Okay because I solve in the sense

one. Okay because I solve in the sense of asmtotic expansion I solve I solve the exact problem but here I will have

some um some derivative involved. So

here it will depends on five but in some higher norm. So here to be maybe to n

higher norm. So here to be maybe to n plus2 this time.

Okay. So this is and this correspond now to this corresponds now to obviously I have eras it right. This correspond to the local estimate of the numerical

flow. It's close to some it's close to

flow. It's close to some it's close to some modified modified equation to the solution of the modified equation. So

now if we want to have uh more um if you want to go um if if we want to really understand the dynamics over a long time we have to understand the dynamics of uh

of of of this modified equation exactly like in the deterministic case you can say things when you are Hamiltonian in general you can say nothing but if you're Hamiltonian then you have preservation of the energy under certain

circumstance. So now we have to look at

circumstance. So now we have to look at the longtime behavior of this modified equation.

And uh so to do this I will do some geometric assumption I can say some structural assumptions on L0 which is the the reference flow right so the

exact solution. So first let's uh it has

exact solution. So first let's uh it has some invariant uh measure uh given by

some C infinity density right uh X it has it has an invariant measure I

can solve I can solve the uh uh if I consider this equation L0 star so this is the transposer operator equal J And

if g is orthogonal to the constant uh and c infinity then there exists mu satisfying this. Okay. So you take the transpose of

this. Okay. So you take the transpose of this equation. Sorry. Uh and satisfying

this equation. Sorry. Uh and satisfying mud dx for example. I mean you can impose something equal z. And the third one which will be

crucial. If I consider the solution of

crucial. If I consider the solution of dtv= l0 v such that in zero is equal to

5. Then I have that v of t x minus the

5. Then I have that v of t x minus the integral of v row

dx in in uh in ck. So I take k derivatives then it's smaller than some polomial depending on k t exponential of

minus lambda t and here uh v in ck. Okay. So it will be satisfied in if

ck. Okay. So it will be satisfied in if uh if the SDE is elliptic right if sigma if if r is elliptic on the to

okay so when we have this when we have this let us go back here at this level then I claim that we can solve we can solve uh ah I don't want to erate

this Then I can solve the wait what is my aim aim you have you have the modified equation what time behavior do

you want to characterize ah my aim uh you will see in a minute u and then I want to impose that vk0 is equal to zero I I just said that you

need more structure if you want to have some more longtime behavior. Uh to

understand the longtime behavior here this is only a local thing. If you

accumulate this then you use gral and it gives you nothing. I'm just really curious. I mean is it going to tell you

curious. I mean is it going to tell you something?

Yes. In some sense yes because I mean you could use but you you want to construct an area. Okay.

more then then you can solve this by writing v of x this I call it gk as the integral from 0 to t of

exponential of t minus s l0 t minus s l0 of gk of dx

okay and now now uh with with the assumption here with the assumption here I can construct uh there exist some at least it's it's a formal series so row

plus h1 etc such that l h which is the modified the modified equation times row h star

is equal to zero so there exist a modified there exist a modified invariance measure at least formally and if you look at these equations If you look at these

equations then this will tell you the this this will be equivalent to L0 K. Ro K is just the function at the

L0 K. Ro K is just the function at the level K which is equal to minus the sum from LJ K minus J= 1 to K and this is orthogonal to the

constant to the constant because because LJ this is a star here. So you can construct some row with integral zero.

So that integral of row h is equal to to one again and which is at least formally a modified a modified measure for this.

when you truncate that series uh for sufficiently small h and u well maybe not no no no I'm sorry because row is row can row you can have zero

in row can be zero some places so now no so you're not sure not sure non- negative I think maybe is what you're asking yes I mean if non- negative I never understand

Please assign you mean this if this is this then okay it's positive if this is this then you can it's it's not negative then you

cannot ensure anything okay uh yes so there exist this modified measure that I mean at least it satisfies this and now if you calculate

if you look at this equation defining VK and if you calculate the the the part of GK that is uh the the component of GK

along row then it will correspond exactly to the component of VK along along row H along the along the modified

measure. So you will not be surprised

measure. So you will not be surprised that VK of TX

minus V dro K. Yeah. So in some in some so CN you

K. Yeah. So in some in some so CN you can derive this and this will be small a polomial that depends on k and n

evaluated t exponential of minus lambda t and here some uh some uh

some in some huge norm 2 + 2 and here if if you look at this this is because because because um because vk0

Here vk0 is equal to zero. This is

integral from row h vh dx evaluated. And now of course uh this is this is I mean at at the level k

at the level k at the level k so the k term of this formal series. So you have this and now you can sum you can take the sum of this from k= 0 to 1 and

you've got exactly the same uh the same estimate which means that if you truncate vn of t and x minus the

integral of v 0 k this will be smaller than some polomial depending on n of t and here exponential of minus

X time V okay to some m okay and my time is running so I will just uh conclude by

giving you the the main results uh when we have this so here we say locally I'm close to locally I'm this is for all time right this is for all time

so this is a this is something global on my modified equation this is not local this is like there exist a modified energy in the deterministic case. This

is something global on my modified energy. So now I use this is local. So

energy. So now I use this is local. So

and now I use tubaro technique for the weak convergence. So essentially you

weak convergence. So essentially you eval you have to evaluate at some point something like vn

uh t uh t tj + one minus s okay and and here x x uh okay so I don't remember exactly but

s you have to evaluate it something like this and this is very small and because because here you use because here you have this exponential decay

What you get in the end is that if you look at

the exponent expectation of XP so for all P minus the integral of five here uh

here it's not zero K this is the truncation of so the modified the modified measure dx X this will

be something that is uh a polomial here qn of t times uh exponential of minus lambda t

so something exponentially decaying plus and here uh a constant n time hn and this does not depend on

time because I have the exponential decay here and because always I use this local estimate I will have here something that decay exponentially. So

exactly like in in tally techniques. So uh okay so now you see

techniques. So uh okay so now you see that so it means that there exist a modified there exist a modified measure and but you might uh so you are on on the to so you know that this one has a

has a has an invariant measure but maybe maybe you don't have uniqueness and it means that uh once um if you have if you have two invariant measure they're chain

close very so this is can be something I do not prefer to really imagine what it can be Um and uh okay I will stop here just I

want to say that of of course this cen you really want to uh to optimize it to obtain maybe some exponential convergence. This is something I like.

convergence. This is something I like.

uh in the case of LD and if you impose some polomial growth for F and sigma then you will have something um if you

have uh bounded moments for for the for the numerical solution something like XP to the to the M is

noted uniform in B then you will then you can do more or less the same same kind of things Uh, and uh, okay, I think I'm I'm done. So, thank you very much.

Thank you.

So, just to make sure I understand something. This So, this this P here is

something. This So, this this P here is the P order method, right?

Uhhuh. This is the P of the time. I'm

sorry. Let's write this way. PH. Thank

you. Okay, fine. Fantastic.

This is the time and it's always for oiler scheme. Yes.

So it's for oiler scheme but it can be also for the implicit implicit explicit for splitting. I I don't see really

for splitting. I I don't see really structure. Let's just let it you just

structure. Let's just let it you just need to have an expansion the very first expansion but and you see why it does it does not work if you if you try to do

this. So with the modified FH and sigma

this. So with the modified FH and sigma H it's because the modified guy um has is not an operator of order two it's of order three and

four if I put a higher order method would I expect that expansion to would I expect to shift the exponent there by the higher order method I mean if I kept

two terms in your row expansion again oh yes exactly at all. No, you would get here uh so if

at all. No, you would get here uh so if I kept two terms there then you would get and then if I had a k order method then I would have plus a you write this something something that has to be that

has to be precise this is that here row zero in some normal okay so let's take it this the modified minus the exact one here is

h p where p is the number of term you you manage to to to get so for the order it's p equal

Could you ask a question? You modified

equation then is not process.

It has this uh it has this many this higher derivative, right? So it's a bron of the bro. So what do you know? Part of

the usefulness of backward error analysis is that you characterize the problem that you merely solve. So here

you've characterized that problem in some sense but it's you've gone out of the class of problems where you started is it can you do anything? I I don't know I

mean can you can you give in some sense can you give a probabilistic interpretation of this guy and it's probably it involves

operator of order four five six so it's iterated B motions maybe something like this iterated B the B of the

B why not so it's not it's not ah I don't know I would guess probably but there might be some cases where it is marked off and that would be

interesting to understand.

What happens in a linear additive case?

a linear additive case. Um,

well, I think you're You mean are you exactly?

No, but maybe it kills the high the high order operators. That's what you mean.

order operators. That's what you mean.

That's a good question. new

case in the construction here.

I can't see exactly. No, I don't I don't No, no, no, no. You really have a even if you take the the log maybe you have some kind of magic things that happen, but I don't

think so because you have this flow you you want to take the log go to the exponential and then then you do this at the level of operators. So if

even if you have something linear here, I don't think it changes anything. No,

no, it's not.

[Applause]