 Hello students. In the previous lecture, we set out to develop a differential equation formulation for stochastic phenomena. We started with the Chapman-Colmogorov equation, then developed Taylor expansions, constructed moments and then arrived at a differential equation formulation and that differential equation is actually known as the Fokker-Planck equation. So, the equation we derived is if W is the occupancy probability at space point x and time t, then the equation read dw by dt equal to d by dx of minus u effective which is of course, a function of x could be a function of x, w which is a function of x and t, we are just omitting its arguments plus d which could be a function of x, dw by dx bracket close where u effective of x was defined as u x, where u is the sum drift velocity minus d prime x, where primes denote differentiation with respect to the argument and here the argument is only space point x. So, d prime is basically d by dx. So, this is the effective drift velocity and the dx is the space dependent diffusion coefficient and u x of course, is space dependent actual drift velocity. Just to point out that true or effective drift does include the derivative of diffusion coefficient if the diffusion coefficient is space dependent. So, this is a sometimes very useful when we discuss mass transfer in turbulent fluids etcetera, where near a wall or a boundary the diffusion coefficient in the medium could vary from point to point. However, when both u and d are constant when u equal to constant meaning not a function of x and d also is constant the equation becomes simpler dw by dt will be d by dx of minus uw plus d dw by dx, where now u and these are constant d prime is 0. So, u effective becomes the same as u. So, we can call this family of equation as the Fokker Planck same Planck who discovered the Planck's law of radiation Fokker Planck equation. So, what were the assumptions we made when we arrived at this equation? Basically, we assume that the transition probability in the derivation of FPE FPE means Fokker Planck equation following assumptions are remain. That is the transition probability p at a space point x making a transition by a length xi in a time interval tau had properties such that p d xi equal to 1 xi p d xi which is xi bar this had a property proportional to tau. So, we said this has a property of u which could be a function of x tau and for small tau at least and xi square bar had the property 2 d tau plus of the order of tau square or so or more than tau in any case and higher moments xi to the power n bar they were of the order of tau square or so and higher. With these assumptions we could take the limit we can say specifically for small tau take Taylor expansion limits of tau going to 0 and then we found that all that survives strictly are the only these two terms xi bar and xi square bar that to those containing the coefficients u x and d x. This was illustrated that is such function such transition probability is physical this was illustrated by constructing a Gaussian right. We took the example that this could be realized say p x xi tau for small tau had the property 1 by sigma root 2 pi e to the power minus xi minus let us say some u tau whole square divided by 2 sigma square where sigma square was 2 d tau we assume this form u and d could be functions of x. So, with that we completed the derivation I wish to point out here that the assumption of a Gaussian transition probability was only an illustration, but it is not a necessary situation. One can obtain Fokker Planck equation by considering transition probabilities other than Gaussian functions providing they have moments of at least second order and perhaps higher order also which is better, but at least for us to say a an equation up to second order we need its moment up to second order. Just to illustrate that such a possibility exists let us consider let us consider non Gaussian transition probability when you say Gaussian transition probability what we meant was a function xi having a form like this this is a Gaussian supposing with consider a function let us say p x xi tau through a function such that let us say 1 by sigma phi of xi minus a where phi has some properties we will come to that soon it is a general function where a we stipulate to be u x into tau like earlier it was something like a function which is linearly dependent on tau and sigma is twice square root of 2 d tau supposing we merely have to assume that there exists a scaling function phi where phi has the following properties function phi u has the following property that is phi u is symmetric in u or even function of u it can be a function of mod u for example, not necessarily differentiable at u equal to 0, but it should be integrable such that it should have first of all it should be normalizable let us say 1 then it should be let us keep the domain as minus infinity to infinity then since it is an even function phi u d u will be 0 and it should be possible for us also to insist that u square phi u d u is finite it should be possible to make that finite as unity by tuning the some some constants let us say and higher order moments might exist, but we can even stop at this point then we can proceed exactly through the steps that we have carried out and show that the mean of xi the mean of xi square etcetera have the required tau dependence the phi function in in for example, need not be Gaussian for example, it can be supposing it can be a function like this this is in the u variable it can have a mod u behavior like this maybe an exponential it could be e to the power minus u mod u here. So, same reflected on the left side or it could be even a function on symmetric which is 0 beyond some length it does not matter for moments because we our space which is of course, from minus infinity to infinity. So, we extend the function saying that beyond some length l the probability is 0 this is a function which will have infinite all the moments for example. So, the situation is somewhat analogous to when we started with central limit theorem there is a recurrent there is a recurrent story about central limit theorem in the derivation of the Fokker-Planck equation also. So, long as the function has all these properties we can show that the Chapman-Colmogorov equation necessarily leads to a Fokker-Planck equation of a second order higher order moments will necessarily go to 0 there will be of the order of tau cube and beyond. Just to make sure that we are on the right path let us understand as I mentioned the transition probability expressed in the following form P x xi tau equal to 1 by sigma phi of xi minus a divided by sigma where we note that a is proportional to tau where a equal to u tau and sigma equal to square root of 2 d tau we assume these two properties, but the function is not Gaussian then can very easily see that integral P d xi I am not writing the argument because we know that the argument for our purpose xi is sufficient others are like parameters. So, it is integral is going to be just to carry out 1 by sigma integral xi varying from minus infinity to infinity phi of xi minus a divided by sigma d xi a and sigma are functions of x and tau, but not of xi. So, I can make a transformation to a new variable transform u equal to xi minus a by sigma which implies that I can replace xi through the equation a plus u sigma which means d xi is equal to sigma d u and you can note that integral minus infinity to infinity P d xi is going to be 1 by sigma minus infinity to infinity for the u variable now xi has been replaced with the u variable and that will be phi of the whole thing is u and d xi is sigma d u sigma does not depend on u. So, it is a function of tau and xi, but not on u. So, it cancels out. So, which is integral minus infinity to infinity phi u d u and since the function phi is normalized this will be 1. So, the transition probability is normalized if you take a function generalized function phi similarly you can show that integral minus infinity to infinity xi bar as I mentioned by definition it is xi P which is a function of xi d xi and we work through the same transformation u then we note that it is going to be 1 by sigma which remains there minus infinity to infinity what is xi from the transformation here xi is a plus u sigma. So, we substitute that a plus u sigma here. So, it is going to be a plus u sigma d xi we write first that is a sigma d u and then the phi its argument is u it will be just this like earlier sigma cancels. So, you write there are two terms here. So, the first one is a is a constant in so far as u variable is concerned. So, a phi u d u plus sigma minus infinity to infinity u phi u d u in view of the fact that phi is an even function this will be 0 and this because it is normalized will be unity. Hence xi bar is equal to a only and a is nothing but u we get back the relationship that we obtained regardless of the function that we have chosen same way we can confirm ourselves that xi square bar which is by definition xi square p d xi minus infinity to infinity let us work through by substitution. So, we have to go back here. So, wherever xi square comes we are going to replace it with this a plus u sigma whole square and d xi will be sigma d u. So, it is going to be minus infinity to infinity xi square is a plus u sigma whole square p is going to be replaced with phi u now and d xi sigma d u. So, like earlier of course, first sigma cancels and if we expand this there will be three terms one will be a square second will be 2 a sigma and u and the third one will be u square sigma square and this is phi u d u. We can note that the first one will be simply a square here now the first integral is a square and then integral phi u d u we know its unity the second one is going to be 0 because it is multiplied by u and it is an even function multiplied by u integrated over the domain minus infinity to infinity is going to be 0. So, that will be 0 and the third function sigma square will stand out and it will be u square phi d u and we decided to construct phi in such a way that this is constant and we can take it as 1. Since sigma is having a pre factor we can always adjust it in sigma. So, this let us say is equal to 1. So, it will be sigma square plus a square and sigma square we know is 2 d tau 2 d tau and a square is going to be u square tau square. So, it has a term proportional to tau and of the order of tau square and we know that when we take finally, the limit of tau going to 0 the coefficient u square tau square is going to go to 0. So, we can always write it as 2 d tau and a term of the order of tau square which is going to go to 0 when we take the limit. So, the for the operative purposes what matters is the term 2 d tau we can now go like this and show that the higher order terms xi cube and xi to the power n for all n greater than or equal to 3 are going to be at least of the order of tau square and higher in any case may be may be tau to the power 3 by 2 or something, but more than tau. So, that in the limit when tau goes to 0 all those terms are going to vanish higher terms xi to the power n average will be of the order of tau square or so and not of tau. Hence, they will so in the limit tau going to 0 in the limit tau turns to 0 xi to the power n bar by tau will tend to 0 that is the point whereas, other coefficients will survive. The main learning point in this exercise is that as I indicated earlier it is not necessary to only have a Gaussian distribution for deriving the Schaplan Walker Planck equation from the Chapman Kolmogorov equation just like we made a statement in central limit theorem that regardless of the original distribution after certain number of samples the mean tends to a Gaussian. We have not arrived at a Gaussian solution yet, but we arrived at a differential equation formulation. Later we see why this is very important this statement because the differential equation formulation is a restrictive formulation which basically implies transition probabilities with the at least second moment. On the other hand the Chapman Kolmogorov equation is a much more general equation of transition probabilities which could address problems where these transition probabilities may not even have moments they may only be normalized. We will soon come to that clarification. So, having come to this point we now discuss certain solutions to the Fokker Planck equation. So, let us do some solutions simple solutions most of which basically we are revisiting the random walk problem straight away from the differential formulation solutions to Fokker Planck equation. When we discuss the random walk problem on a real line we started with the simplest problem of a walker starting from a point with the equal probabilities to the left or right that is called the unbiased or symmetric random walk and we had exact solutions and we even showed that in the asymptotic limit that goes to a Gaussian. We are basically now going to confirm that Gaussian distribution that we obtained directly by solving the differential equation that is the Fokker Planck equation. The side benefit of this exercise is that it also helps us to understand more about the method of Fourier transforms in handling linear differential equations partial differential equations. We do this exercise in the next lecture. Thank you.