 Let us use a short form for Fokker-Planck equation, let us call it f e e f p e from now on. So, we are going to obtain the case 1, first case is a free particle case free particle and node drift that is u equal to 0, also d is space independent position independent. So, let us say it is a constant. So, our equation is now going to be simple equation d w by d t is d d 2 w by d x square in one dimension, we are talking of one dimension, we will go to generalizations later. A few word about this equation, in our context we are discussing a single particle via the concept of probability density that is w. If you have a collection of particles which you do not interact, then w can be considered as instead of probability density kind of a concentration. And if these particles are virtually moving independent of each other, what is true for one is true for the other and true for the collection and then w has a separate interpretation what is called as a concentration, notation c or n is used. In that case this equation has a structure of what is known as a diffusion equation. For all practical purposes this equation can also be called a diffusion equation, although we must keep in mind that we are talking of a single walker or a single diffuser. Secondly, we can see that similar equations occur very often in physics in engineering. For example, the heat conduction equation has very same structure where the w will be replaced by temperature for example. So, it will be a conduction of heat. There are many other occasions where one obtains same partial differential equation, second order in space and first order in time. So, what the solution we are developing has wider implications. Although we are focusing on the case of a free random walker or a diffuser along the x axis along an x axis. So, we assume that the walker starts from the origin at t equal to 0. So, w x 0 is delta x where delta is the direct delta function. So, with this we now systematically solve this equation. As I mentioned when we were discussing mathematical preliminaries method of Fourier transforms comes very handy for solving free space diffusion equation. So, we define the Fourier transform like earlier define w let us say character of a conjugate variable k. So, our Fourier transform is now with respect to k space and this is also a point where we can note that Fourier transform as a characteristic function has a similar property as the generating function when we discuss discrete cases. So, we can always keep seeing that what we call as k would be something like z not exactly, but some measure some function of z in the case of discrete variables. So, we are basically mapping or transforming from the real space to a kind of a conjugate space through a characteristic function or a generating function or a Fourier transform and the definition for Fourier transform particularly is that an integral over ikx w xt dx this is the definition. So, multiply both sides by e to the power ikx both sides of equation 1 let us say if we call this as equation 1 by e to the power ikx and integrate over and integrate over dx. So, when we write integral dx which is we are integrating over x then the left hand side simply becomes d e pi dt of w carried k t because we have multiplied by e to the power ikx and integrated and the this is the left hand side the right hand side becomes we will write it explicitly minus infinity to infinity e to the power ikx d 2 w by dx square dx. Let us focus on the right hand side in particular we can see that integral e to the power ikx d 2 w by dx square integrated over x we can do it by parts. So, parts when we do we integrate that differential term. So, the first integral is going to be d w by dx and the second term e to the power ikx keep just that way and then minus infinity to infinity here also it is minus infinity to infinity then the second will be minus and we now differentiate the other term that is e to the power ikx. So, when we differentiate it will be ik the differentiating with respect to x. So, ik e to the power ikx d w by dx and this integral we use we can simplify the first part since the function w is expected to vanish at x equal to minus infinity to infinity the reason being that it is a probability density of finding the particle at x at any time. We know that at any time a particle which is starting from origin would not have reached minus infinity and infinity. Hence this will be always 0 and here ik does not depend on x. So, it can be taken out minus ik e to the power ikx d w by dx. Now we identify this is the Fourier transform of d w by dx. So, once again we can do it by parts and then if you notice this e to the power minus ik will remain and here we are going to integrate this. So, w will come this e to the power ikx w again minus infinity and infinity limits and the second term is going to be differential of e to the power ikx. So, straight away we will keep it outside here it will be e to the power ikx and w dx. The differential of e to the power ikx is with respect to x that is why we took out the coefficient ik out of the integral. So, like earlier this will go to 0 and so we are going to have minus ik then again minus ik and this back to the definition is going to be the Fourier transform of w or w caret of k t. So, this will be w caret k t. Now minus minus plus and i square. So, this simply minus k square that is the attraction of doing Fourier transforms it reduces the entire differential term to an algebraic term. Such this trick could work normally this kind of tricks are useful when you have differential equations which do not have space dependent diffusion coefficient and other things. If they have space dependent diffusion coefficient we would not have been able to do this. So, this is a very useful for very specific problems. So, in any case with this transformation we will now have hence after Fourier transforming we obtain d w caret by d t is equal to minus minus d k square w caret k t. Just to note a PDE has been reduced to a PDE of second order is reduced to an ordinary differential equation of first order which is much easier to solve. Now we note that it is integral is going to be simple. So, one can write down the solution w tilde w caret of k t is going to be some constant let us say some unknown constant with which will not be a function of time because we are integrating over time. So, it could be a function of k in general because there is a other parameter which exists and this is e to the power minus k square d t. If you integrate it is simply e to the power minus k square d t. Now to evaluate a k we use the initial condition. What is the initial condition? That is w cap k 0 at t equal to 0 which is nothing but the Fourier transform i k x w x 0 dx and we know this is delta x particle started from the origin and hence it is e to the power i k x delta x dx and we know that when delta x is integrated with another function it just selects the value of that function at the point x equal to 0 or wherever the argument becomes 0 and hence in this case when x equal to 0 e to the power i k x is 1. So, the answer is 1. So, as a result hence at t equal to 0 w tilde k 0 which is identical to a k which is 1. So, we have finally, w tilde k t is simply e to the power minus k square d t. So, in the Fourier space we have solved the problem into a simple exponential solution. To obtain the real distribution we have to invert it that is Fourier inversion. We know the inversion theorem that is by definition w x t is the inverse Fourier transform that is let us write it as inverse Fourier transform which is 1 by 2 pi minus infinity to infinity over k variable now, but with the sign change e to the power minus i k x w character k t d k is the most general definition. So, upon substituting it will have the form 1 by 2 pi minus infinity to infinity e to the power minus i k x minus k square d t d k. If we consider a general solution consider an integral we treat it as an integral now i equal to 1 by 2 pi minus infinity to infinity e to the power minus let us say a k square minus d k d k is closely analogous to the our integral, but we are replacing d t with a and i k with b where a equal to d t d equal to i x i x because that is a parameter we are integrating over k. Now, if you consider the integral i with these definitions of these definitions of a and b and if you consider this i integral it is the same as the w integral, but we will first do this integration. So, we can write the i integral which was by definition 1 by 2 pi minus infinity to infinity e to the power minus a k square minus b k d k if we make it as a perfect square first we make it as a perfect square in the argument minus infinity to infinity e to the power we can convince ourselves that this is correct k plus b by 2 root a whole square plus b square by 4 k. So, why I wrote like this you can see that if you actually expand this root a k plus b by 2 root a whole square the first term is a k square which is here minus sign is already taken out second term is 2 a b that is a 2 root a b and root a root a cancels. So, 2 2 also cancels. So, we are going to get with just b k and to make a perfect square you would have had now 1 term minus of b square by 4 a and you will have to compensate for it by adding b square by 4 a which does not depend on k. So, it will have a form e to the power b square by 4 a into 1 by 2 pi I could rewrite this e to the power some variable y square d y by root a where we substitute a new variable which is y new integration variable y equal to k root a plus constant b by 2 root a. What we have done is this perfect square term we put the whole thing as y we want to get rid of this perfect square. So, k will be replaced with a new variable y and that is how that y is defined here k root a plus b by 2 root a then this will be simply minus y square this is a constant which will come out and only difference will be d k when I am doing it will replace with d k equal to d y by root a that is why we have 1 by root a here. So, you can see that it is going to be 1 by 2 pi root a e to the power b square by 4 a and minus infinity to infinity e square e to the power minus y square d y this integral is a Gaussian integral with the unit variance with the not exactly unit 1 by root 2 variance or half variance 1 by root 2 standard deviation but half variance. So, in any case it is answer is root pi the well known Gaussian integral. So, then the answer becomes this root pi cancels with the pi here leaving behind root pi. So, we are going to have square root of 2 root pi a e to the power b square by 4 a. So, if you go back and see our a and b and substitute a and b here b square interestingly has i x. So, b square is going to be simply minus x square and a is. So, root a is going to be root d t. So, this will be we will complete it here which is going to be which is equal to 1 by 2 square root of pi d t e to the power minus x square by 4 d t. This is about the i integral and i integral is the same as w. Hence, the solution is solution to free particle diffusion is w x t equal to 1 by you can write it as all inside 4 pi d t e to the power minus x square by 4 d t. It is easy to see that this solution where we started with from the at the origin at t equal to 0. This can be generalized to diffuser starting from any point from a point x equal to x naught at located at some time t naught it did not be 0 all the time. We can easily see that the equation that we have started with the diffusion equation is invariant with respect to the transformation. Since the equation is since the differential equation is invariant with respect to transformation x prime equal to x minus x naught and t prime equal to t minus t naught. To see that we can just visit revisit the differential equation you can see in this equation supposing I simply replaced t by t minus t naught nothing would have changed because it is a differential over t. So, d by d t is the same as d by d of t minus t naught also since t naught is a constant shift. Same way d 2 by d x square would have been just x minus x naught also would have worked because x naught is a constant when you differentiate it does not matter only thing difference I would have had is I would have written W x t equal to t naught would have been delta x minus x naught. So, with that change and the transformation you can easily see that by mere translationally invariance the solution the above solution can be written as W x t given that it was at x naught t naught equal to 1 by square root of 4 pi d t minus t naught e to the power minus x minus x naught whole square by 4 d t minus t naught and we must be sure to assume that t is always greater than t naught. There is no negative time concept or you cannot go backwards in time and interpret the differential equation that is a unique part of entire subject of stochastic mechanics of stochastic processes is that we always march forward in time and assign transition probabilities for later events from a previous or time state. So, this is the most general solution to a free particle diffusion equation. So, it is actually a probability density which can also be called as the transition probability can interpret is the probability that the particle transits to a point x at time t given that it was at x naught at time t naught is given by a Gaussian distribution is of course, a time dependent normalization term multiplying it just to complete it will be if we plot a function of x say let us say this is x naught some point then for various times if you plot it will have for example, it will be very sharply peaked for short times and as the time progresses this will broaden for example, this will be a structure for sufficiently long time. So, we can call it as a short time this is after long time you must remember that these distributions are always normalized to unity that the as the peak comes down the particle moves drifts away from the center as a time progresses this distribution broadens however, the area remains conserved since the probability is conserved. So, the important properties we can see from this that the variance the variance sigma square is 2 D t because it is a Gaussian you can see that it is a Gaussian with variance as 2 D t minus t naught if we this is if let us say t naught equal to 0 otherwise it is t minus t naught mean is of course, at x naught the interestingly the probability of being at the origin and t which is W x 0 probability of being at the origin. So, x equal to x naught. So, W let us say 0 t let us keep x naught as 0 is going to be 1 divided by square root of 4 pi D t if we say t naught equal to 0 and let us say x naught also 0 which is varies as t to the power minus half exactly at all times. So, of course, this is not valid for t greater than 0 which should not include the point 0 because then we were not right in putting x equal to 0 there. If we look back at the derivations we carried out for the persistence at the origin W m n n step m side the probability of the origin had exactly the same square root of n behavior for large n. So, this of course, is expected we are discussing the same problem, but the power of a Fokker Planck formulation is that the problem can be solved by a common formalisms more importantly as we will see in later lectures we can generalize it to higher dimensions much more easily treat it as a partial differential problem in general dimensions obtain different power laws of the decay of concentrations or probability densities at the origin you get different power loss. And also we can address more complex problems of a finite boundaries even in 1D and in higher dimensions which almost are very very difficult to do with discrete step walks they are mathematically more involved and there is no uniform procedure available whereas, here we can obtain it by a fairly uniform methodology. In the coming lectures we explore such and very many interesting aspects of the continuum time continuum space formulations some problems in higher dimensions also and this will be a very useful learning to solve general random walk problems by using the continuous formulation and obtained solutions. Thank you.