 We had just started talking about the Fokker Planck equation associated with the stochastic differential equation. So let us do this systematically and let us proceed as follows. So first we have an equation which is like the Langevin equation, kind of generalization of it and let us continue to call it the Langevin equation for a random variable xi which is the first order differential equation of the form xi dot equal to some function of xi. Let us look at stationary or non-stationary processes but no explicit time dependency. We could include it if necessary but let us look at the case without it because all the cases we are going to look at will be without it. Plus a term g of xi times of a white noise, a Gaussian white noise which has got zero mean and a delta function correlation. So xi of t equal to 0, zeta of t, zeta of t, zeta of t prime equal to delta function of t minus t prime with unit strength and any strength in the noise is subsumed in this quantity g of xi which could be a constant for instance. Now technically this noise is a Markov process, assumed to be a Markov process. It is Gaussian in the sense that all its probability distributions, joint distributions and so on are all Gaussian functions and it is a stationary noise as you can see here at the level of the auto correlation and it is a delta correlated noise so that the power spectrum is flat in this case. Then this implies and in turn is implied by an equation for the conditional probability density of this variable xi. So for this quantity p of xi t, xi not 0 for this quantity here which is also guaranteed to be a Markov process in the sense that this two point, this conditional density determines all joint densities. This quantity p satisfies an equation called the Fokker Planck equation and the equation reads as follows. So this is the stochastic differential equation for which I will use the abbreviation SBE and that implies for this p a Fokker Planck equation or FPE for short for this quantity which is as follows. Delta p over delta t equal to on this side a term which is minus delta over delta xi f of xi times p itself that is the deterministic part the drift part without this you have an ordinary differential equation it is deterministic plus a term called the diffusion term which is one half g square times p, g square is the square of this function that is the Fokker Planck equation for this quantity. Of course it is got to be in this case since the initial conditions are at xi not at t equal to 0 the value of xi not the initial condition on this is simply a delta function this p at t equal to 0 is delta of xi minus xi not so p of xi 0 equal to delta of xi not. So very often I am going to suppress this initial condition just call it p of xi, t and p of xi, 0 is some prescribed initial condition and this is what you have to solve this equation with. We are not going to derive this it is not very hard to derive it is actually quite straightforward I have to introduce moments of something called a transition rate or probability per unit time and then using that make an expansion etc etc and show that this is what is obtained in the case where this is a Gaussian white noise pardon me I do not have to discretize time but it is convenient to do so when you are deriving the equation if I am going to start from what is called the chain equation for Markov processes go to what is called a master equation and then the Kramers-Moyal expansion and so on and finally you show that for Gaussian white noise it reduces to this equation okay and there is also the question of the interpretation of this equation because there is this noise here multiply this deterministic term multiplies this noise so this is called multiplicative noise whose amplitude depends on the value of the variable at that instant of time through this function here and for multiplicative noise you have different ways of interpreting this equation and we are using what is called the Ito interpretation there is a Stratanovich interpretation there are several other ways of interpreting this equation which gets us into stochastic calculus which I do not want to do. So we will take this equation for granted because we are going to actually look at it in the simplest of instances in the simplest cases and our focus is on very physical processes like the motion of a Brownian particle or 3-dimensional motion in a potential etc etc okay. So this equation is not very simple to start with because it is a second order equation in the random variable in the variable xi and it is a first order equation in time so it is a complicated partial differential equation the solution is also very complicated. Now the cases we have already looked at falls squarely in this you can see what is going to happen immediately for instance we considered the velocity process we said one Cartesian component of the velocity of Brownian particle was satisfied in equation minus gamma v plus square root of gamma over m times eta of t I call square root of gamma zeta of t is what I had called eta of t earlier the noise because I said the correlation function is gamma times the delta function I want now a unit strength delta function here for the 2 point correlation so therefore I could put the gamma outside in this form. So in this problem xi is v f of xi is a linear function it is minus gamma v and g of xi is a constant and that is called additive noise because it just says to the deterministic part you are adding noise whose strength is independent of the value of the variable of the random variable this thing is a pure constant. Now look at what is going to happen immediately here this will imply as we have seen if you apply the general theorem out here this implies that delta p over delta t equal to gamma times delta over delta v v times p it is a minus f of xi which is minus gamma v is minus sign cancels plus one half and then a g square which comes out as a constant so it is just gamma over 2 m square d 2 p over dv so without proof using this general correspondence I am asserting that given this Langevin equation this is the equation satisfied by the conditional density of the velocity. Now of course we know from the Langevin equation we already proved the fluctuation dissipation theorem we know that gamma is 2 m gamma k Boltzmann t we already know that for consistency saying that the system remains in equilibrium thermal equilibrium when you take the full average over all realizations of the noise and then over all initial velocities we know that you should get the Maxwellian distribution finding equilibrium distribution should remain Maxwellian that happens only if this is equal to that and that is what we argued was the fluctuation dissipation theorem in this instance I said the strength of the dissipation and the strength of the noise must be related to each other in this consistent fashion but that follows from here too immediately you can see that if you ask what happens as t tends to infinity because what we are doing is starting with a p of v t v not because I know this is a stationary process so I do not write the 0 there it does not matter in this case or just a side remark the fact that this equation has a noise which is a stationary random process does not guarantee that the driven process Xi is stationary it guarantees it is Markov but it does not guarantee that it is either stationary or Gaussian because this could be all kinds of functions out here what is guaranteed is that it remains Markov and in general we will have a finite correlation time not 0 as the noise so what is carried over is a Markov property neither the stationarity nor the Gaussianity is carried over to the driven process in general but we know we have already computed the velocity correlation function in this case and we know it is e to the minus gamma t apart from k t over m and we know that it is a stationary random process so I do not bother to write the 0 there whenever it is a stationary process I will not write the initial time explicitly this can always shift it out this quantity here we know that it should become as t tends to infinity it should go to the equilibrium distribution the Maxwellian distribution right so this should tend to p equilibrium of v because it should forget the memory of the initial velocity and go to the final velocity and be an equilibrium distribution as t tends to infinity it should be the stationary distribution but we know what this is we know this is m over 2 pi k Boltzmann t to the power of half it is a single component we are talking about e to the minus mb square over 2 k Boltzmann t we know that is what it should become right on the other hand if I examine the equation itself and ask what happens to this as t tends to infinity if there is a stationary equilibrium distribution this must become 0 because it cannot depend on time and then as we have already seen before this will imply that the equilibrium distribution will satisfy an ordinary differential equation no time dependence there which will look like this this will look like gamma it will look like d over dv of gamma over 2 m squared dp equilibrium over dv that is this term here plus gamma v times p equilibrium is equal to 0 because this side is 0 no therefore the bracket is a constant therefore that implies that this fellow is equal to a constant independent of v but we also want an equilibrium distribution which is normalizable so this means p equilibrium as v tends to infinity must go to 0 it must have finite moments mean velocity mean square and so on which means that these fellows should also go to 0 all derivatives should also go to 0 as v tends to plus or minus infinity which implies that the constant must be 0 because this thing is equal to constant and we say that at v equal to infinity this constant is 0 so it is since it is a constant it is 0 for all v but then that is a first order equation and we know what the solution of it is it is trivial so this thing here will imply that p equilibrium is equal to is proportional to apart from constant of proportionality e to the power minus 2 m squared gamma and multiply through by this guy you should be careful with all the 2s and so on yes 2 m squared gamma over capital gamma yes v squared over 2 because I integrate v dv and that gives me v squared over 2 and that is this so cancelling out these two fellows you get this so this matches this thing here if and only if m over 2 k Boltzmann t equal to m squared gamma over gamma so you are back to this you are back to this so that is one more way of seeing that this must be the fluctuation dissipation theorem if and only if gamma equal to 2 m gamma so it is just saying it in the in terms of the probability distribution or then probability density function rather than the stochastic equation itself it is the same fluctuation here dissipation theorem you are going to get now what is the general solution of it well now we need to examine this a little more carefully this equation is not too hard to solve it is not trivial though because this differential operator is not self adjoint this is first order term here it is a bit of a mess there are several ways of solving this it turns out that the solution is precisely the Onstein Lundbeck distribution this turns out to continue to be a Gaussian process in this case and it is the Onstein Lundbeck distribution so we know what that is we know that p of v v naught equal to an exponential apart from normalization minus m v minus v naught e to the minus gamma t whole square that is the mean value the peak shifts to the left divided by 2 m 2 k t and then the variance and then a normalization factor which is essentially m over 2 pi k t times this time square root of the total so the mean is this value which goes to 0 as t tends to infinity starting at v naught and the variance is a delta function at t equal to 0 variance vanishes at t equal to 0 and then broadens out till it hits the value given by the Maxwellian distribution depends on the temperature and this is the O u Onstein Lundbeck distribution next question what about the phase space distribution what about the distribution of x v and t you need to know is a joint distribution is it possible to get an answer for that and the answer is yes we can see how that is going to come about because I should really augment this equation with another equation which says x dot equal to v we are talking about one Cartesian component so it is just an x and a v so x dot is v and v dot is this okay in this case there is no external force and therefore there is no f of x which you might get from a potential may be but you do have this systematic part and then you have a random force here so one should be able to join to write down a distribution of p of x comma v comma t given x naught v naught and zero and an equation for it a corresponding equation for that you need a generalization of this to higher dimensions to more to two dimensions we will come to that in a minute but before that let us settle this other question we know there is a diffusion limit and in the diffusion regime we know that the mean square displacement goes like 2 dt and we know that d is kt over m gamma can we get that from this does it follow from this thing at all does it follow from this equation for instance and the answer is yes it indeed does because what we need is an equation which says that the velocity is delta correlated because when you are in the diffusion regime the velocity correlation time is gamma inverse and now you are saying you are a t much much greater than gamma inverse so one way of implementing that is to say I take gamma to be so large that gamma inverse is negligible and then at all t practically you are in the diffusion regime so one could look at the high friction limit of this equation and ask whether that is going to work several ways of implementing this but let us do it that way and see what we get so the high friction limit of this system is going to be gamma tends to infinite gamma high friction high compared to what well the statement is that gamma is so large that all times that you look at are such that t is much much greater than gamma inverse so I drop the inertia term this came from m times v dot mass times acceleration I drop that term I retain this term and look at it as a stochastic equation so now my stochastic equation says x dot that is v is equal to root gamma over just one second before I do that there is one thing I want you to bear in mind which is since we need this we have imposed this fluctuation dissipation theorem for consistency we can rewrite this equation a little bit and it is going to be useful to do so pardon me so that I can take the large gamma limit but it is also going to be needed in another context we will see in a minute so this is really gamma delta over delta v vp plus for this I substitute 2m kt gamma then the 2 cancels the so this is gamma k Boltzmann t over m d2p over delta so the diffusion coefficient if you like in velocity space is gamma kt over m in position space it is already what we know it is kt over m gamma but in velocity space it is turning out to be gamma kt over m we will keep that aside for a moment so let us look at the high friction limit of this and that reads so I drop the v dot term this term and retain this and bring it to the left so it says v equal to root gamma over m gamma zeta of t so you are really saying sorry v which is x dot which is what it is in the diffusion regime because it says the velocity is uncorrelated for all practical purposes it is a delta function correlation but we can simplify this right so this is equal to square root of I will put that inside so 2m gamma k Boltzmann t over m squared gamma squared zeta of t equal to the m cancels one of them gamma cancels twice kt over m gamma but remember we had set d equal to k Boltzmann t over m gamma we discovered that the mean square displacement in the diffusion limit was actually 2dt where capital D was given to be kt over m gamma that is what we found so let us put that in this says this is square root of 2d zeta of t and now I go back and appeal to this general Fokker Planck correspondence between the stochastic differential equation and the corresponding Fokker Planck equation I stare at this and I say look x is xi now xi is x now there is no f of xi of zeta or xi that term is missing there is only a noise term and g is square root of 2d it is a constant so this tells us that delta over delta t p of x comma t for some given x not which I will impose as an initial condition is equal to one half g squared but g squared is 2d and half of that gives me d so this becomes d d 2 p over d x that is the diffusion equation that is precisely the diffusion equation. So you see the origin of the diffusion equation from this language from this point of view it is precisely the fact that you are working in a regime where the velocity correlation time is essentially 0 and therefore the velocity is a white noise and when you integrate it you get an equation for take this equation and write it for as a density in x position you have this diffusion equation so this emerges from that correspondence it is part of that correspondence there is no drift term in this equation it would be there if I put an external force if I said I am looking at diffusion in a potential even gravity there be an extra term there be that first the f of xi term would be present but that is completely missing here in the velocity case there was a friction term which was proportional to the velocity it was linear in the velocity that made life a little easy in the diffusion case that is even that is missing now if I look at sedimentation namely diffusion of molecule in a vertical column under gravity for instance then there would be a constant force and a constant force would lead to just p there be no v here just p on this side so you would have a first order term plus a second order term on this side and that would lead to an extra contribution it would change the solution here considerably it would lead for instance if you asked in a finite column if you asked or even a infinite semi infinite column under gravity if you asked is there a steady distribution the answer is yes but is there a steady distribution for this on the infinite line is there an equilibrium distribution for this if there were then this should be 0 and then you get d2p equilibrium over dx2 equal to 0 and the solution to that is p equilibrium is ax plus b but that is not normalizable it is not normalizable therefore there is no such equation there is no such distribution and you know that in this case what happens is that if you start at delta function at x0 equal to impose a condition p of x0 equal to delta of x minus x0 if you start at this delta function distribution and you look at this p as t changes does this etc whatever it does not even drift it does not drift because there is no drift at all what it does is to start at x0 it broadens out and broadens further in the case to 0 such that the total area under the curve remains 1 it is normalized you know no material is going away so this thing does not have an equilibrium distribution if you put gravity we will do that later on we will do the sedimentation problem later on you will discover it tends to the bottom everything goes to the bottom so there is an equilibrium distribution under certain cases but in general there is in confers so this is how you get the diffusion equation but now you could ask what about the phase space distribution what does that look like so now we have to be a little more careful or stochastic equation is a pair of equations so we have x dot plus v equal to 0 sorry minus v equal to 0 and you have v dot plus gamma v equal to whatever this fellow was yeah root gamma over m let us let us change this thing here this is equal to root gamma over m equal to root 2 m gamma k Boltzmann t over m square so 2 gamma k t over m let us just put it that way this fellow here times z dot so it is a pair of coupled equations now let me introduce a vector x which is x put that the matrix column matrix here I am going to write this as a single equation in a vector form and let us introduce a matrix k a drift matrix which is equal to the minus 1 here so it is 0 minus 1 the plus gamma here so it is 0 gamma introduce that matrix then the left hand side for these two together becomes x dot plus k x this is the left hand side equal to on the right hand side a noise which is essentially this fellow here 0 root 2 gamma k t over m times zeta I can give it a symbol some vector noise it does not matter now the question is what is the corresponding Fokker-Plank equation corresponding to this and it turns out that what you have is a special case of a much more general case in a case where the noise is additive this is an additive noise because there is no x or v dependence here there is no g there is only constants moreover the drift is linear so this is a linear drift so you have additive noise and linear drift this makes life easier the general case is also something we are going to write down but the expression for the Fokker-Plank equation for the linear case with linear drift and additive noise is very straightforward and it is very natural let me write it down it says so now we are going to look at the density rho of x v given x naught v naught etc x naught v naught oh by the way when we looked at the high friction limit of the original Langevin equation and got to the diffusion equation for the positional probability distribution function was x of t a stationary process or not so we had this quantity p of x t delta over delta t equal to d d to p of x comma t over d x 2 with initial condition x naught etc so we had p of x 0 equal to delta function of x minus x naught and we know how to solve this equation you do Fourier transform with respect to space and the class with respect to time etc many ways of solving this there is a fundamental Gaussian solution with natural boundary conditions namely p 0 as x tends to plus minus infinity then the solution of this equation is worth remembering it is a fundamental property it is equal to 1 over square root of 4 pi d t e to the power minus x minus x naught whole square over 4 d t that is the fundamental Gaussian solution to the diffusion equation. The peak remains at x naught and the peak actually goes to the width of this peak goes to infinity linearly with time that is why you have diffusion so in this case the average value of x minus x naught squared it actually diverges while the probability density itself decreases at all points there is a 1 over root t it is normalized to unity the integral from minus infinity to infinity in the x for all time is finite is 1. Is this a stationary process no eminently no because the variance changes with time so it is not stationary it is Gaussian it is Markov but non stationary earlier the velocity process alone the noise is Gaussian stationary Markov and Delta correlated the velocity process is stationary Gaussian Markov but not Delta correlated exponentially correlated the position in the friction limit is not stationary it continues to be Gaussian Markov exactly so this is not yes I emphasize once again this is not the exact equation for the positional density at all how would you get that well that would get you would get from here you would get from this quantity so if you did this if you integrated minus infinity to infinity d v rho of x v t if you did this with those initial conditions of course x naught v naught 0 so you integrate over the other variable velocity you would then get a distribution in x the conditional density for x the exact conditional density for x that is not this because as he points out this is true only in the region where gamma t much much greater than 1 only there is this true you really have to go back here and do this similarly if you got the exact answer here and you did this minus infinity to infinity dx rho of x v t x naught v naught 0 you would expect to get what would you expect to get here you would expect to get this should be equal to p of v t v naught you would expect to get this because you would get the conditional density in the velocity now and you do and you do in this case the x naught dependence must somehow disappear and you should get this but what happens and I am anticipating myself a bit what happens if you did this you integrate over the velocity is that you get an answer for p of x v t which involves both x naught and v naught which immediately tells you that it is not a stationary process x alone is not a stationary process even worse we know that this quantity satisfies a Fokker Planck equation in the velocity variable alone this quantity does not satisfy anything of that kind it does not satisfy any simple master equation so that is the problem it is a highly non stationary process even in the diffusion limit there is not there is v naught dependence of course we saw that there is v naught dependence certainly there is v naught dependence but the point is when you integrate here this quantity there should be no v naught dependence there should be x naught yes but this v naught dependence as well so you cannot decouple the velocity completely showing that this is retaining the calculations memory etc we will see we will see what the solution looks like and then be able to examine this so that is a good point that this quantity is equal to this only in the limit so this as gamma t much greater than 1 goes to this p of x this p of x but not for all t okay so the question is what is the Fokker Planck equation here and the answer is the following you have delta rho over delta t equal to in such a case let me call momentarily let us call this equal to x 1 x 2 in index notation first component second component then this is equal to delta over delta x i sorry it is equal to k i j delta over delta x i x j rho where a summation over repeated indices is implied plus a term which is the diffusion term it will look like some generalized diffusion matrix here d i j times d 2 rho over d x i we already know this we already know what this k i j is it is this I have to write down what is d i j the diffusion matrix let me use another symbol for it let us use d so this d in this case is 0 0 0 that is not surprising because this is essentially root twice the diffusion constant in velocity space that is what this so that is the Fokker Planck equation in phase space all we have to do is to substitute for this case substitute for this d and we are done so let us write it out so this becomes equal to the first term that is going to come to contribute is k 1 2 okay and that is going to have a minus sign delta over delta x 1 that is x 2 is v rho and then k 2 2 is going to contribute so that is going to be plus gamma delta over delta v v rho that is it plus this fellow the only term that contributes is d 2 2 so plus gamma k Boltzmann t over m d 2 rho over delta v 2 pardon me well x 2 is v x 2 is v and we are looking at only d 2 2 because all the other terms are 0 but this is precisely we what we got in the Fokker Planck equation for the velocity and this 2 but we have an extra term here but you see this term can be simplified a bit because x and we are independent variables in phase space so this is equal to minus v delta rho over delta x plus gamma delta over delta v v rho plus gamma k Boltzmann t over m d 2 rho over d v 2 this is the phase space Fokker Planck equation for the phase space conditional density rho remember this is in equation for rho of x v t given x naught v naught at t equal to 0 we essentially have delta of v minus v naught delta of x minus x naught and with those boundary initial conditions you have to solve this does this remind you of something well if you bring it to this side what is going to happen yeah it looks like total derivative this looks like the convective derivative it is indeed that indeed that it is just the convective derivative just sitting here no external force present no velocity dependent force no magnetic field none of those things then this is looks like convective derivative it even has a right sign you bring it to this side it is convective precisely the convective derivative which is what you kind of expect physically so can we write the 3 dimensional generalization of this that will be horrible thing but anyway before we do that I should tell you what the solution to this equation is you can again solve it with delta function initial conditions you can give an exact solution it is a 2 dimensional Gaussian it is a joint Gaussian in both x or x minus x naught and v minus v naught e to the minus gamma t it is a joint Gaussian so it will have an exponential which will involve minus x squared minus v squared it will also involve an x v term in between such that asymptotically so the solution to this asymptotically what would you expect it to do for t tending to infinite well the solution will actually vanish because if you look at the ordinary diffusion equation the probability density vanishes but but we are not asking as mathematically t becoming infinite we are not saying that you are saying when t becomes much larger than all the time scales in the problem gamma t much much greater than 1 for instance what would happen so let us write that that is much more reasonable right what would you expect it to become the velocity thermalizes in other words it loses type track of its initial value and gets into the Maxwell indistribution so you would certainly expect that to happen the position in the diffusion limit will have a diffusion equation solution Gaussian so I expect that this is going to become e to the minus m v squared over 2 k Boltzmann t times e to the minus x minus x naught squared over 4 dt times a normalization factors so this fellow is divided by root 4 pi dt and this fellow is divided by m over 2 k Boltzmann 2 pi k Boltzmann t times square root so I expect that to happen and it that is what that is that should be a check that indeed does so but as I said if you integrate this exact expression over v alone you get a very complicated thing for x which will involve v naught and x naught but you integrate over x from minus infinity to infinity you will get the once you know in back distribution with initial value v naught with no reference to x naught for the velocity distribution alone what would you expect the would happen this three dimensional case so we could write a generalization of this you could make these things vectors here in three dimensions so what would happen to this quantity everything else remains the same but of course different Cartesian components of the noise would be uncorrelated to each other you have to assume that so you would certainly have to assume that you have a zeta i of t zeta j of t prime average value is equal to delta ij delta of t minus t prime you have to assume that that is certainly true and then what this quantity does row of r v t given r naught v naught at 0 we can write down a Fokker Planck equation for this for this row phase space density in three dimensions we can read it off from here just the vector form of it so we would again get delta row over delta t is equal to what would happen to the first term minus v naught grad row grad with respect to r with respect to the components of the coordinate plus what is the next term going to be so gamma times grad with respect to v dot v row plus gamma k Boltzmann t over m grad with respect to v squared this is the exact Fokker Planck equation for the phase space density in three dimensions just a straight forward extension of this and the solution is a generalized Gaussian in all six variables with the same sort of properties once again so once you have this correspondence between the stochastic differential equation and the Fokker Planck equation then the matters writing down the Fokker Planck equation is very straight forward now we have to go to the next stage where we look at cases where you have multiplicative noise and then the question is what happens if you have a higher dimensional case you should be able to write generalization there too and we will do that and we will apply it at least in the simplest instances we will apply a couple of examples so that you see what the use of this working but this is a fairly intricate problem at the same time it is amenable to exact solution in this case let me stop here let me take it up.