 Hello students, we continue with the Fokker Planck description of a stochastic process. This is the equation that describes Brownian motion. It is also known in the language of stochastic processes as Wiener process. This is a special name for unbiased Brownian motion because it has certain characteristics. The if for example, it is it describes the evolution of a random variable in a time ordered indexing. So, if let us say we describe the time axis from 0 to t and if x is the random variable and in our case x is the displacement. So, let us say there is any time s and then the remaining time t minus s this is t minus s. Then when we say x t, it is the random variable x value at time t. Similarly, therefore x t and x s are the random variables at time t and s where t is greater than s. Then the Wiener process has a certain characteristics. Some important characteristic is basically already contained in the Gaussian solution that we obtained. But in a in a general way of expressing the we can list the characteristics of a Wiener process or a Brownian process, these are as follows. First is the x is that is a displacement, but it is actually the random variable x t minus x x s it will be a actually eventually what matters this basically net displacement difference between t and s. That net displacement eventually is a function only of t minus s. So, there is a kind of a translational invariance in the time space. We always should note that it is valid for all s less than or equal to t greater than 0 let us say. This is one characteristic. We can start of course with the property that the value of the random variable is 0 when the time is 0. And the third important property is that the variable x t minus x s is Gaussian distributed distributed with the mean 0 and variance t minus s for all s less than or equal to t. That is always a constraint as I mentioned we it is a time ordered process you cannot have a probability distribution from the information obtained with a probability distribution at a later time. So, it is always a progressive you can obtain the probability distribution at a later time from the information provided from the position at a prior time that is the idea. And the next fourth important property is that the covariance of x s and x t minus x s this is 0 for all s less than or equal to t covariance is basically you know if covariance of just a definition that is just information covariance is defined for a distributed quantity for distributed quantities is defined as say cough of any two quantities x and y is basically an integral x minus x bar y minus y bar w x y dx dy. So, this is the way we define covariance. So, what the condition fourth states is that the distribution of position distribution of position in a disjoint time intervals thus for example, this is the s length and this is the t minus s length. So, the distribution at a later times in disjoint intervals do not depend on the distribution at the earlier time. So, as hence there is sort of a correlation there they are uncorrelated the distribution at later times position at a later time is uncorrelated with the position at a prior time. So, this is the important property that comes from a Markovian assumption that we have made. So, with this interesting properties one can obtain a very important result that the covariance of x s and x t covariance of the displacement at time s and displacement at time t this is always minimum of s and t for all minimum of s and t. So, now let us say s and t we do not know which one is larger, but it will select whichever is smaller this can be proved easily from one of the four criteria that we have set for a Wiener process that is we know that covariance of x s x t can be written as covariance of x s and x t can be written as x t minus x x s plus x s I am just adding and subtracting x s. Now therefore, the covariance this can be expanded this will be covariance of x s covariance of x s with the x s plus covariance of x s with the x t minus x s and now let us say that s is less than t let s be less than t then we know from the previous condition that is the condition 4 above that covariance of x s with the x t minus x s is 0 this is 0 because there belong to disjoint intervals s and t minus s will always be disjoint and therefore, this will be equal to 0 by condition 4 for a Wiener process. Then covariance of x s x t now s is smaller that is why this happened is is is equal to covariance of x s comma x s and this is just s plus 0 equal to s. So, this proves our statement that the covariance into intervals selects the smallest of them. So, this is just a kind of a introductory information I am giving you as a digression to a to the equivalence of the solution that we obtained as a solution of Brownian motion which in the literature sometimes is also called as a Wiener process. All these conditions for example, are satisfied if we write W of x t comma x s the distribution function as 1 by square root of 2 pi t minus s e to the power minus x t minus x s whole square by 4 t minus s for all for all s lying between 0 to t. We have made an assumption that set this result is true assuming d equal to 1 we have set d equal to 1 above it is just a constant diffusion coefficient equal to 1 unit above. So, the probability of obtaining a displacements x t given that it was x s. So, it is a conditional statement we can also write it in the conditional way given that it was x s at an earlier time s is just given by a Gaussian distribution x t minus x s whole square by 4 into t minus s and this is completely consistent with all the Wiener process conditions that we have put. We can go further and obtain many other results right now we stop here just having gotten a flavor of how from the Fokker-Planck explicit solution of the Fokker-Planck equation we have mapped it into the into a Wiener process. So, with this we further continue to finding the solution of the continuous formulation that is a Fokker-Planck formulation and compare with the limits that we obtained by the discrete formulations that was done earlier using the random walk model of Brownian motion. So, we now take the Fokker-Planck equation solution to absorbing barrier problem. Starting from the random walk model considering a bias either towards an absorber or away from the absorber we have arrived at both a discrete form of the solution as well as the continuum form. This was done in lectures 27 and lecture 28 deal with random walks solution to this problem solution to 1D bias absorber problem. Let us try to see how the form of solution that you will get if you go by the continuum formulation of the Fokker-Planck equation. So, accordingly we proceed as follows. Now proof let us consider this is the x coordinate and there is a barrier at x equal to 0 this is a barrier absorbing barrier and the random walker starts from some point x naught at t equal to 0. Let us say that there exists a certain bias either away is a bias either away or towards the absorber. So, we want to obtain an explicit time solution to this problem by using the differential equation formulation that we derived. A part of taking up this problem is also to illustrate method of solutions when you have problems expressed in a continuum form. The advantage is the problems that we could not have handled by discrete methods or would have been very difficult to handle by discrete methods can be addressed by the continuum methods and this is an illustration of that advantage. There are well known problems like for example, a particle diffusing particle or a Brownian particle settling under gravity. Now the question is whether it will eventually escape gravity or whether it will be trapped on the surface is a very valid question. Similarly, there was a question of random walk modeling of gambler's ruin. So, this is relevance one Brownian motion say under an electric field or a gravity under external fields it can be electric or gravitational electric or gravity near an absorber. This absorber can be an electrode or it can be a ceiling or a wall as the case may be. So, this another another relevance is gambler's ruin this we saw. When we did this problem under discrete random walk a bias parameter gamma was used. However, in the Fokker Planck formulation we do not have explicitly a bias parameter, but instead we have an equivalent of a velocity which introduces asymmetry in motion. So, accordingly we keep the velocity term and formulate the Fokker Planck equation. Bias is accounted for the term velocity term in the Fokker Planck equation or drift velocity or the drift term. The equation is the FPE is we have written dw by dt equal to minus of let us say the velocity is u a steady flow type velocity then it will be first order term multiplying the velocity and if it has a diffusion coefficient d then it would be d 2 w by dx square in the domain x 0 to infinity. We are talking of positive axis as the space available for diffusion. So, to solve this we need certain conditions. So, we need one condition that w is a boundary and initial condition boundary conditions and initial conditions. So, the boundary condition is that the since it is an absorber at x equal to 0 at all times the probability of finding the particle in the closest vicinity of the absorber is 0. That means, w 0 t is 0 and we have seen it from the random walk model also that the solution with an absorber is equivalent to finding solution with absorber concentration kept at 0. Then we need another condition at infinity and we presume that at infinity it should be bounded it should be finite it cannot diverge. Similarly, the initial condition is w of x at all at 0 at time equal to 0 should be some delta of x minus x 0 it has started from some point x 0. So, this is the last of them is the initial condition the first two are the boundary conditions and this is the initial condition these are sufficient for solving this partial differential equation. We eliminate the number of parameters let us say this we call as equation 1. So, divide equation 1 by d then we will have dw by I will write a one step I will skip now and explain why. So, dw by d tau will be minus of nu dw by dx plus d 2 w by dx square where I have defined tau equal to dt and nu equal to u by d. So, dimensionally dt has a time of meter square per second into second. So, it is meter square dimensionally tau and nu has the dimension of 1 by meter. So, that eventually there is a meter square in the denominator of all the 3 terms. So, with this we can we have to think of how to solve this equation we can rewrite the equation here as hence we have dw by d tau equal to minus nu dw by dx plus d 2 w by dx square. And of course, we will keep one in the important condition w x at the absorber is I mean the distribution at initially time 0 is delta x minus x naught and main condition is w 0 at all now tau that is going to be 0 this absorption we see. To solve this equation we need to make certain transformations one of the tricks that we can follow is to try to reduce the first order term and that is done by the following transformation we define a nu dependent variable define nu dependent variable y x tau by definition is w x tau e to the power minus of nu by 2 x. We just multiply w by e to the power minus nu by 2 x and that we call as y this basically implies wherever w is there we can substitute as y e to the power nu by 2 x. Now the advantage of this transformation is that it exactly cancels with minus nu w by x and you can see it ourselves why because we need a term called to dw by dx. So, dw by dx is going to be e to the power nu x by 2 into dy by dx plus nu by 2 e to the power nu x by 2 or you can write it as e to the power nu x by 2 dy by dx plus nu by 2 y bracket here you can take it out. Now we need a second derivative. So, if you do the second derivative we get dw by dx square now there are 3 terms 2 terms, but you have to do it again. So, if I keep this e to the power nu x by 2 out then here it will be d 2 y by dx square here it is nu by 2 dy by dx and the other term will be e to the again a nu by 2 factor will come. So, I am going to take it inside e to the power nu x by 2 and here it will be nu by 2 d by by dx and here it is going to be nu by 2 into nu by 2. So, nu square by 4 y this can be simplified for the before we do that. So, I take e to the power nu x by 2 out again and this will be d 2 y by dx square plus nu by 2 dy by dx plus nu by it should be nu dy by dx plus nu square by 4 y. Now we can see very quickly using the by comparing the last term here d 2 w by dx square as written here and if you substitute d 2 w by dx square minus nu d w by dx since nu d w by dx is going to also have minus nu dy by dx whereas, here you have a plus nu dy by dx the dy by dx exactly cancels. Hence the right hand side of the equation becomes r h s that is d 2 w by dx square minus nu d w by dx that will go over to d e to the power nu x by 2 into d 2 y by dx square minus nu square by 4 into y. We can say the term nu dy by dx cancels. So, as a result hence the equation becomes if you substitute the left hand side is d w by d tau if you go back here the d w by d term here will simply be if you do this substitution there is no tau dependent. So, it will be simply e to the power nu by 2 x dy by d tau e to the power nu by 2 x exists both on left hand the right hand side. So, it will cancel completely. So, one obtains the equation dy by d tau equal to d 2 y by dx square minus nu square by 4 y. You can see that the y being our nu variable it will satisfy the conditions y at the absorber pointed tau by definition will be equal to w at the absorber pointed tau e to the power from our definition of y it is e to the power minus nu x by 2. So, minus 0 it will be which is only w 0 tau and that was 0 because e to the power 0 is 1. So, it does not matter and y at initial time should be equal to delta x minus x naught e to the power minus nu x by 2. Since it matters delta function has a relevance only at x equal to x naught e to the power minus nu x by 2 can also be written as e to the power minus nu x naught by 2 it does not matter because it is a selector. So, one can also write this as e to the power minus nu x naught by 2 by virtue of the property of delta function. This is now a new equation we call it as equation 2 and we try to solve equation 2 under the specified conditions which we believe or we hope that is going to be simpler than the first earlier equation 1 by virtue of our having eliminated the first order term. Thank you.