 Today and in the next few lectures we are going to study a very important class of continuous Markov processes called diffusion processes and there is a very large body of literature on these processes and very large number of applications both in physics and in other subjects chemical physics for instance chemistry and so on and so forth. But what we are going to focus on is not so much the detailed technical mathematical rigorously mathematical aspects of the subject as the possible applications to various physical situations. Let me begin by recalling to you that we had just started defining continuous Markov processes and in particular I said we will talk about stationary continuous Markov processes we can relax this assumption of stationarity a little bit and talk about stationary or non-stationary processes and as you will see as we go along the most important non-stationary random process is in fact the diffusion of a particle that itself its position is a non-stationary process the velocity turns out to be stationary but the position is non-stationary as you will see when we go along. So to start with recall that we said that these processes are described by the conditional density which satisfies an equation of this kind in the stationary case X is the random variable and a set of values taken by the random variable and the probability density function the conditional density function conditioned on this initial value satisfies an equation of the form an integral over all possible X primes dx primes and then inside you have a W of X X prime P of X prime t at X naught minus the loss term which is W of X prime X P of X t this is the continuum analog of the discrete equation that we wrote down in the case of a process which took on a discrete set of values or attained the discrete set of states. Now as in the other case this is an integral differential equation although it is linear in this P and therefore is not very trivial equation to solve certainly if it had been a matrix equation we could have written the solution as the exponential of a matrix multiplied by t and then try to look for methods of exponentiating this matrix but here that is not true this is some kernel so it is an integral differential equation this is some kernel function of X and X prime d t o here and it is not at all so obvious what the solution is here one approach would be to try to convert this to a differential equation but because it is an integral equation and you are integrating over all values of this here and no conditions have been put on these transition probabilities at all these transition rates it is immediately sort of intuitively clear that the order of the differential equation in the X variable will tend to become infinite in this case and in fact that is so and I will write that down without going through the intermediate steps except to indicate how to do it what you have to do is to treat this as a function of X prime and you put X for instance is X minus X prime you put it equal to some delta X or something like that does not have to be small and then you do a Taylor expansion in terms of this delta X and whenever you get derivative operator is you try to put it on the P by integration by parts and the result is that this becomes equal to also equal to and I am going to skip these steps as summation from n equal to 1 to infinity minus 1 to the n over n factorial delta n over delta X n a n of X P of X t for given X naught of course. So it becomes equal to that formally these two are equal to each other where these coefficients a n of X equal to an integral of moments of this guy X plus delta X starting with X for example and then delta X to the power n and D delta X this difference and the nth moment of this increment with this as the weight factor is the definition of a n of X here and delta X is over the range allowed range here so it is a definite integral this thing is a definite integral and it is a function of where you start namely X. So this is the exact formal equivalence now of course it immediately raises the question of when is this valid when is it convergent and so on and so forth I am not going to talk about those technical issues at the moment except to say that you can make specific conditions put specific conditions under which this equation reduces to this infinite order partial differential equation and this is called the Kramers-Moyal expansion does not serve much purpose except for formal purposes because it is an infinite order differential equation. So it is not any easier to solve than this but it gives you a little bit of physical insight as to what are the terms that are contributing out here and how do you interpret them and so on and so forth okay but there is one great simplification that occurs for a specific class of processes called diffusion processes and by this I do not mean I mean diffusion in a technical sense which is not restricted to the physical diffusion of a particle in space or anything like that but a mathematical term which says there is a class of processes called diffusion processes for which this equation simplifies enormously okay and there is a theorem a rigorous theorem which says remarkably enough if these moments of this increment delta x the amount by which x jumps to a new value if these moments vanish for any n greater than equal to 3 so happens that a n is identically 0 for any n greater than equal to 3 then a n is equal to 0 for all n greater than equal to 3 it is called Pavoula's theorem and it is a remarkable theorem it is not magic it is possible to derive it very very fairly straightforwardly but the statement is if a n equal to 0 for some n greater than equal to 3 and of course that immediately simplifies matters enormously and processes for which this happens are called diffusion processes because that would immediately imply so that this equation becomes the following delta p over delta let us leave out all the arguments of this p equal to minus delta over delta x a 1 of x p plus 1 half delta 2 over delta x 2 a 2 of x because the higher moments are 0 and this 2 factorial I have put in here as a half there this is called the Fokker-Planck equation they originally derived it in a different context altogether in a related context but in a physical context of a particle diffusing in space and this is was the velocity of this particle but today we call it the Fokker-Planck equation in general for any diffusion process and I will use the same terminology and you can go ahead and interpret what this means and it will turn out we will see the specific examples that this term represents the effect of noise on this variable x that is what causes x to fluctuate randomly whereas this term very often describes the effect of deterministic evolution in this x as we will see from the examples so this part is what x would do its distribution would do in the absence of any noise and this is what makes it random so very often this is called the drift term and that is called the diffusion term and we use we will use these terms in general even though they come from the physical application I am going to talk about so this is if you like this is the drift and this portion is the diffusion it is still not a trivial equation as you can see because it is got this term here it is a second order in the position variable I will frequently call this the spatial variables because for want of a better term although it need not be that at all in fact in the original context of the Fokker-Planck equation it was a velocity variable it is first order in time but second order in the other variable so this is technically not as simple an equation as say Laplace's equation or Poisson's equation because of this inhomogeneity you will recognize special cases of it for example if this A1 had not been present and if that A2 were a constant this looks like the diffusion equation the ordinary diffusion equation for particles diffusing on a line where which would be delta P over delta T is D times D2 P over DX2 we will see how that comes about now of course one could ask what happens can I solve this equation in general and so on the answer is no in general for arbitrary coefficients A1 and A2 it is not so trivial to solve at all can I incorporate non-stationary process in this yes indeed that has nothing to do with the vanishing of the moments or anything like that it is an independent statement if these were time dependent explicitly then of course you have a non-stationary Markov process and then you have to be careful you have to write X0, T0 etc keep track of that and then this would become time dependent here and it would be time dependent here because W would be time dependent explicitly so it is possible to incorporate non-stationarity into this business by looking at time dependent coefficients we are not going to do that all the cases we look at would be stationary in that sense but we will come across a non-stationary process we will see what happens in that case where you do not have explicit time dependence and yet you will have a non-stationary random process which will be the position for instance we will see how it comes about okay. Now to make it familiar with the diffusion equation one possibility is to derive this diffusion equation independently all together we already looked at diffusion on a linear lattice in the presence of a bias we looked at a random walk and the question was can the random walk be made into an equation of this kind then of course you begin to see immediately the connection between these two so we will do that now but one thing I want to point out is that there is a special case of even this Fokker-Planck equation and that is very very important it is a very basic process in fact it is the most basic Gaussian stationary Markov process it is just one of them it turns out and everything else can be mapped on to that and it is called the Onstein-Volenbeck process it corresponds to the case in which you have the following extra simplification so the case in which a1 of x equal to a1 x and a2 of x equal to a2 a1 to equal to constants so a case in which the drift is linear in x and the diffusion term is just a constant this coefficient a2 is a constant this is called this particular process then is called an Onstein-Volenbeck process so let us write it down a very important special case and we will spend some time solving this the density for the density function of this Onstein-Volenbeck process along with the physical example but in the meantime let us go back and see whether we can derive this kind of equation from the random walk model altogether so let us go back to the case of a biased random walk on a linear lattice in one dimension so if you go back and recall what the statement of this problem was we had a linear lattice and infinite lattice say label by the site index J some arbitrary site was the origin and then you had a probability if you are at the site J of jumping to the right with the probability alpha and to the left with the probability beta and this was true at every site you would toss this unfair coin and you either jump to the right or to the left. Now we did that in the discrete time case but we also did it in the continuum we said the time was continuous and the steps were given by a Poisson process with some mean rate lambda in which case the process that corresponded to right steps steps to the right had a mean rate lambda alpha and those to the left had mean rate lambda beta and if you recall the master equation in that case was dp J, t I suppressed the fact that we started the origin we keep that going so that just a matter of simplifying the notation this was equal to lambda times alpha p of J minus 1 t plus beta times p of J plus 1 t minus alpha plus beta p of J comma alpha plus beta is equal to 1 but I put that back here as alpha plus beta because I am going to recombine terms so do you recall this this was the master equation for this probability p of J comma t and the initial condition was p of J comma 0 is delta of J comma 0 the Kronecker delta we solve this and we discovered the distribution was a modified Vessel function ij of 2 lambda t square root of alpha beta etc right now we are not interested in that solution but we want to see what happens in the continuum limit when this J becomes a continuous index so what we do is to introduce a lattice constant this spacing a I am going to let a go to 0 and correspondingly let the rate of jumps lambda become infinite because the distance you have to jump is going to go to 0 and the rate becomes infinite in a specific manner so as to derive a finite limit on for the right hand side a proper limit for the right hand side so the first step is to write this as equal to lambda times let us choose this first beta of p J plus 1 t minus p of J t minus let us subtract the other difference also p of J t minus p of J minus 1 t so I have taken care of this term and this beta here in this portion here and then I added I subtracted that too so I need to add that back so this becomes plus lambda times beta minus alpha times p of J comma t minus p of J minus 1 so that is the other term did we go through this continuum approximation earlier have we explicitly done that okay so it is worth looking at it carefully to see what exactly is involved so what I have done is to add and subtract this thing here and I get this now it is clear what you should do in order to get the continuum limit because this looks like the second difference this looks like p J plus 1 minus twice p J minus p J plus p plus p J minus 1 so this looks like the double difference the second derivative and this looks like the first derivative if J were a continuous variable right so what we need to do is to multiply and divide by the lattice constant and take limits so out here I need I can rewrite this this thing can be rewritten as p of J plus 1 t minus twice p of J t plus p of J plus 1 t I divide the whole thing by a squared because it is a double difference here and multiply by a squared I multiply this by a and divide this by a and take the limit so the correct limit that we need to take the continuum limit is lambda tending to z infinity a tending to 0 and I want lambda a squared to become finite that can only happen if beta minus alpha also tends to 0 so that I get something which goes like an a squared here so we need a times such that and J tending to infinity because what I am going to do is to put J a tends to x so J also becomes infinite such that J times a is my x coordinate just like we went to the continuum limit in time by saying the time step n multiplied by the by tau the unit the unit time step was such that n tends to infinity tau goes to 0 such that n tau went to t the continuous variable exactly the same way J tends to infinity a tends to 0 such that J goes to the variable x such that what we need here is lambda a squared beta limit lambda a squared beta is finite equal to some number D by the way if alpha tends to beta this is the same as the limit half lambda a squared equal to D because beta is also going to go to alpha and a times alpha minus beta times lambda tends to limit equal to what would be the physical dimensions of this limit of this quantity that is a length and that is a rate velocity just a velocity so let us call it alpha minus beta equal to C then then this quantity P of J, t tends to the probability density P of x, t but you got to pay attention to the fact that there is a dimensional change here because this is dimensionless probability that is a density probability density it has dimensions 1 over a 1 over length right so you have to be careful about it there is an extra a factor there which you can put in but it is not so serious because it will appear on both sides out here and when you take that limit you end up with this equation becoming delta P over delta t of x, t starting from some x not you do not care we do not put it here equal to this quantity is D and now here here we have minus C delta P over delta x because that is the first derivative plus D and that is exactly in the form of this Fokker-Planck equation that we have written down so in this problem a 1 of x equal to C a 2 of x equal to D both of which are constants a very trivial example in which these coefficients have actually become constants here so the position in the case of bias diffusion the position variable is a see looks like it is the Markov process it obeys a Fokker-Planck equation with constant coefficients both the drift and the diffusion terms are constant looks exactly like that right this kind of equation for the positional probability density when you have diffusion in the presence of an external field this fellow here looks like it is a drift caused by some external field because you are saying systematically either alpha is bigger than beta or smaller than beta it drifts to one side whichever is larger and that is exactly what happens when you have a constant force on the particle so this looks like this equation is describing the diffusion of a position of a particle positional probability density of a particle subject to diffusion but under a constant external force of some kind and indeed it is so it is indeed so because if you recall the problem of sedimentation that we talked about this is exactly what happened you had an extra term exactly of this kind we even saw the solution we wrote the steady state solution I think in that case and what did we get in that case the problem we looked at was I said J equal to 0 here 1 here 2 here and we looked at a case where this part was bounded and then it turned out that P of J P stationary of J was proportional to the bias alpha or beta so you have bias such that these rates are alpha and these rates are better downwards and this was proportional to alpha or beta to the power J this is what we had we imposed a boundary condition on this we said it cannot go below 0 so the rate alpha minus 1 to J equal to 0 was 0 and the rate from J to minus 1 the beta 0 was also equal to 0 then we immediately got this as a steady state solution and we need to normalize this we need to normalize this guy over J from 0 to infinity should be equal to 1 and that of course you sum this geometric series is 1 over 1 minus alpha over beta which is beta over alpha minus beta correct so this whole thing is proportional to this guy so equal to whatever is normalization in this problem beta was greater than alpha this is the steady state solution we got but we will now let us try and take the continuum limit of it here and see what you want to get so I need to put all these guys in I need to put in all these fellows here back again so let us do that alpha is going to go to beta but I can write this fellow here in this problem beta is bigger than alpha in this case so we got to be a little careful here about the sign I define my drift velocity C as alpha minus beta so if C is positive it says alpha is bigger than beta otherwise C is going to be negative we have to remember that sign here so let us write this as e to the power J log alpha over beta or log beta over alpha with a minus sign and I am going to take the limit in which alpha is equal to beta so this becomes the limit that guy becomes one the log of one it is going to get so what should I write I write this as log beta minus alpha over alpha plus alpha I can write it like that surely which is one plus beta minus alpha over alpha I mean we can do this very rigorously but you can see what is happening and this is going to go to 0 beta minus alpha what is log 1 plus e as e goes to 0 the leading term the itself right so this is can be replaced as beta minus alpha by alpha this form apart from some normalization we will worry about that later and I want to make this J into x so I multiply by an a divide by an a but a times beta minus alpha is going to C right so let us multiply this by another a this is what I had and I multiply this by another a so I got to put another a here and let us put a lambda here and a lambda here we all said to take this limit because what does this whole thing go to this fellow goes to x that guy goes to C minus C and this fellow goes to D where did the x come from alpha this was an alpha which is the same as beta in the limit right so you are going to get something like e to the power minus e to the power C x over D where C is negative I probably use the symbol C for the downward velocity limiting velocity but I have defined this C as an upward positive in the upward direction so increasing J so that is why I change the sign here but this is exactly what we got earlier we interpreted this as the Peclet number and so on but that is exactly what this gives you this equation gives you because you go back to this equation and ask what is the stationary solution what is it going to be P stationary satisfies the equation D D to P over DX to stationary minus C D P stationary over DX equal to zero that is what this tells you but I can pull out a D D over DX from here and write this as D over DX DP stationary over DX minus C times P or C over D so equal to zero and C by the way is minus this is plus modulus that is the equation that is when the what is the solution what is the solution to this equation not quite not quite because what you can what you can say from this is that this quantity is independent of X so this guy must be equal to some constant independent of X whereas here we did not have any such problem we did the random walk problem and we immediately got the answer right away but here we are getting an equation which says this follows actually independent of X nothing more than that what would you have to do to match that to this and make sure that that constant is zero you have to put a boundary condition somewhere we already put a boundary condition on the floor we said it cannot go below the floor we already did that here in this case we have not yet imposed that condition there we need to impose that condition which will be precisely that this quantity is zero at X equal to zero because this is the flux remember that this equation can be written in the form of an equation of continuity because I can write this as equal to minus delta J over delta X where J of X equal to D times D P over DX in this case plus mod C P so it is in the form of a continuity equation in this case and that is the flux at any point because it is precisely a continuity equation for this probability density and then it says you cannot go through the floor so it means this quantity this current here must vanish at X equal to zero but in the stationary case and only in the stationary case this quantity is independent of X completely and since it vanishes at X equal to zero it must vanish for all X because it is independent of X okay so I emphasize again this quantity is not zero for X not equal to zero in general there is a current otherwise you would have you would not have any dynamics at all certainly P of X comma T in general is a function of T okay but when you go to the stationary distribution there is no T dependence anymore okay so the statement is that the boundary condition says that the current as a function of T vanishes at X equal to zero for all T what a partial differential equation I have to give you an initial condition and I have to give you a boundary condition the boundary condition must be valid for all T the initial condition is valid for all X for a given T right at T equal to zero so in this case this acts as a boundary condition and it says this quantity here vanishes at X equal to zero and that same boundary condition applies even in the stationary distribution but in the stationary distribution you discover that this quantity must be independent of X and since it is zero at X equal to zero it is zero everywhere identically and once you put that in this is the solution so you see our discrete model went exactly into that that is just a verification that these limits are all right that all these factors were right just right and it gives you this equation here in this special case in which you have this particle diffusing under a constant force field here you can apply to other cases it could be an electric field causing a steady drift or whatever but this is the exact continuum limit so the lesson is that the bias random walk with a constant bias the same bias at all sites is equivalent to the diffusion of a particle under a constant force field in the continuum limit but now we are approaching the whole thing from the continuous Markov process angle okay so we are going to write down although we did not have any differential equation for the position of the particle in the random walk problem but only difference equations for the probability density now that we have a continuous Markov process we could go back and ask one more thing which is to ask okay it is a random variable but does the variable itself satisfy a differential equation or not this is not a differential equation for the variable it is a differential equation for the probability density of this variable and that is a nice object but the variable itself will be very irregular will be random because it is being driven by some fluctuations in this case we are going to find out that this will satisfy a differential equation but it is what is called a stochastic differential equation a random differential equation and it should correspond and be consistent with the fact that the probability density satisfies this master equation here the general name for particle which is for the positional probability density of a particle under diffusing under an external force field it is called a Smolukowski equation so this is an example of the Smolukowski equation I will call it a Smolukowski equation because it is much more general than this you already saw that this has the effect of constant force field what would you say happens would happen if there was a force here explicit force which was position dependent how do you think this equation would change it would not be a constant this a 1 of x would not be a constant right a 1 of x in some sense would be the force would involve the force so if the force were due to a potential v of x I would expect that something like minus v prime of x appears in this drift term okay and then the diffusion the scattering would come from the d part here so this is something to keep in mind that the first term will be a drift due to deterministic forces and the second term would be the diffusion due to random forces okay I will will systematize that so let us go back to this Fokker-Planck equation and ask is there a correspondence between an equation of the form delta p over delta t I will continue to use the variable x which does not have necessarily the connotation of a position but a random continuous random variable Markov process so delta p over delta t equal to minus delta over delta x a 1 of x t plus 1 half this is the Fokker-Planck equation I turns out that this Fokker-Planck equation is entirely equivalent to a certain stochastic differential equation for this variable x random variable x which is now called a diffusion process in the mathematical sense and that equation is the following I will write it down but I am not going to prove this that this is entirely equivalent to a certain differential equation for x itself which reads in sort of physics notation it is not the most rigorous notation it reads x dot equal to some function of x perhaps even a function of t if this is a function of t but we are looking at stationary processes so let us just call it f of x times plus g of x times a white noise and let me call it by eta t and explain what this eta is and this is a stochastic differential equation where f and g are prescribed functions and they are related to a 1 and a 2 as I will write it down in an instant but this eta of t is called a Gaussian white noise a stationary Gaussian white noise and I will explain what that is separately where f of x is essentially a 1 of x and g of x g squared of x is a 2 of x eta of t is a stationary Gaussian white noise I have to say what this means okay eta of t is a random process in time such that all its probability distributions multiple time probability distributions are all Gaussian in shape so that is why it is called a Gaussian noise it is stationary so all its statistical its statistical average and higher moments are all time independent correlation functions or functions only of the time difference etc and it is a white noise in other words it is delta correlated in the following sense equal to 0, 0 mean and it is got a delta correlation delta correlated delta function as an autocorrelation it is clearly the limit the mathematical limit of some physical noise whose correlation time would not be 0 because this implies the correlation time is 0 whereas I would expect for a stationary process if t is bigger than t prime I would expect this correlation to look like this as a function of mod t minus t prime I would expect this correlation to come down in this fashion and this characteristic time scale would be the correlation time of this noise but that is now going to 0 and the amplitude is going to infinity such that in the limit it becomes a data function so it is a mathematical idealization clearly it would have to be justified on physical grounds each time okay for instance in the problem of the collisions of in a gas of particles in a fluid for instance this noise would be caused by all the other molecules colliding against some particular tag particle then the state of t would be the correlation time of that force the random force caused by the collisions all these other guys in the scale on which the particles motion itself is tracked the time scale would be much longer it would remember its memory for much longer than what the noise does so the correlation time of the noise typically would be of the order of nanosecond or a picosecond for instance whereas the correlation time of the velocity of the particle that is being tracked that could be of the order of microseconds so as far as a microsecond is concerned a nanosecond or smaller intervals essentially 0 intervals so in that sense one can justify this approximation okay but each time in any problem when you model this you have to ask whether there is a clear separation of time scales of this kind or not but at the moment from a mathematical point of view the formal point of view this is what this equivalence is so the statement is that a stochastic differential equation of this kind is entirely equivalent to this Fokker-Planck equation for the probability density of this random variable so please take this as a theorem I am not going to prove it here but take this as a theorem we are going to exploit it over and over again now you can see why I call this a drift term because if you did not have this noise at all this is deterministic evolution of this variable under some prescribed function f of x here it may be minus dv over dx we do not care what it is or anything else so this term is indeed describing deterministic dynamics and this is the noise is entirely here in this thing here and that showing up in the second term here okay what is interesting in this problem as opposed to even simpler problems is that this g has x dependence in general so it says given a current value of x of this random variable the way the noise affects it the amplitude of that noise depends on this random variable on this x on the other hand and that is why it shows up here inside here but in the example we looked at in the diffusion problem as a diffusion equation remember this a2 turned out to be a constant so in that case this would have been square root of 2d and that is it so now we can kind of identify what would be the stochastic equation corresponding to delta p over delta t equal to d d2 p over delta x2 this would be equivalent to a stochastic differential equation for x which would be of the form x dot equal to in this case there is no a1 so clearly there is no external force or anything of that kind no drift at all a1 is 0 identically and a2 well this d half d is a2 so a2 is square root of is 2d and therefore g is square root of 2d and that is it this is the stochastic differential equation corresponding to for the position corresponding to the simple diffusion equation in one dimension one can write a formal solution for this guy and that formal solution would be x of t you have to define these integrals minus x of 0 equal to square root of 2d times integral 0 to t dt prime 8 of t prime we can call this x of t not if you like and integrate from t not to t so in that sense it is a plane diffusing particle doing free diffusion the x variable corresponds to the integral of white noise this guy corresponds to the integral of white noise and it is called a Wiener process is it a stationary random process it is Markov because we wrote the mastery equation down I said satisfies a Fokker Planck equation and so on so it is clearly Markov process but is it a stationary Markov by the way it is Gaussian that is something else you have to recognize because we know by hindsight we know the solution of this guy although we did not derive it here explicitly we know the fundamental solution of this is that Gaussian e to the minus x squared over 4 dt which I will come back to talk about so what it is telling us is you are going to hit the particle with a Gaussian white noise that means the distribution of this eta is Gaussian and it is delta correlated stationary it is and Markov then what is the output variable the driven variable x after this integration what properties does it have well it remains Gaussian because its probability density is Gaussian so the shape remains Gaussian that is robust what else happens it is Markov it certainly Markov it is over is this Fokker Planck equation but is it stationary will the stationarity remain does this look like if I if I had x of t not here and this is t not to t does this guy look like a function of t minus t not in general no no certainly not it is not stationary and you already know this because given this diffusion equation what does it imply for this quantity x of t minus x of 0 whole squared what does this become it is the mean squared displacement from some given origin and what is that equal to it is diffusing and therefore what is it equal to 2 dt exactly exactly it is 2 dt it is a function of t it is a function of t so it cannot be stationary because if it is a stationary random process all these moments should be independent of t but here right away tells you it is not stationary immediately we have not computed what the correlation function of x is we have not found what is x of t prime x of t double x of t x of t prime average we have not found that yet but certainly we found what is x of t x squared of t average and that is 2 dt that is just a Gaussian integral so it is not stationary it is not it has stationary increments because you can write this guy obviously as dx equal to square root of 2d 8 of t dt you can write it like that and then of course this is stationary this has stationary increments but it is not a stationary random process by itself or more less rigorously it is derivative is stationary but this function is not the variable itself is not and that you can see directly when you take something which is got stationarity but you integrate it in this fashion integration makes it non-local in some sense so it is not stationary so in general that is a lesson that means you integrate white noise you may not retain the stationarity property but we are going to see that if you put a proper drift you will be able to do this that is what the velocity would do and then it would attain an equilibrium distribution and so on there is another way of saying that this guy is not stationary because it is probability density p of x comma t given an x naught this fellow is decreasing as a function of time and it does not tend to any stationary distribution as t tends to infinity it goes to 0 this Gaussian broadens out over in infinite range the total area under the curve remains 1 but the value at any point is tending to 0 so there is no stationary distribution in this problem the variable is not a stationary random variable either okay. So we will get back to all these things but at the moment I wanted to simply pattern remember the fact that the most general definition of diffusion process could be either this or this does not matter either way now mathematicians do not like to work with these delta correlated noises they rather work with the differential so there is something which smooths it out by integrating so weiner process can be handled more rigorously than this singular object here so you integrate it once to make it smooth and so on but we are not going to pay attention to these niceties we will be careful not to make any mistakes but at the same time we will do this rather heuristically here provided we know there are certain rules which we have to obey and one of them is this that there is this equivalence between the Fokker-Planck equation on the one hand and this on the other and we will see what happens with this now already you can begin to see whether there is going to be a stationary distribution or not by saying if so this thing must have a solution when you make this an ordinary differential with respect to x and if this guy has a solution which is normalizable and so on then you know there is a stationary distribution okay and if you do not have this you do not have this and a is constant then the stationary distribution is to reality itself in this case if at all there is a p stationary it must satisfy d2 p stationary over dx2 equal to 0 but what does that say about p stationary must be a linear function that certainly not normalizable right away you have gone it is finished in an infinite range in an infinite range or even in a semi infinite range it is still not normalizable what would happen if you had a finite range yes then indeed it can be so then you have you do not require integration up to infinity then yes indeed it can it can it is true suppose you are told that p stationary is between these two points nothing more than that I put a diffusing particle here it is like putting a drop of ink inside a beaker of water the ink does not go anywhere but it becomes uniform everywhere intuitively we know this my diffusion so that is what will happen this particles probability density will be uniform will be constant in this boundary provided there is no escape from the ends there is no leakage from the ends so in that case yes indeed because I then I would say p stationary equal to ax plus b and I put boundary conditions at the ends and you discover finally that is 0 and you have just b which is normalized okay so the boundary conditions also play crucial role in the whole thing okay we will talk about these aspects in the next next time we will start with the Fokker-Planck equation and see where we can go