 In the previous lecture, we developed a solution to the Fokker-Planck equation for the primary case of a free particle undergoing random walk, but random walk described by continuum mechanics along x direction. We arrived at a Gaussian solution to the probability of finding the particle at some location x at some time t. Today, we discuss other than Gaussian solutions which are in principle possible, but they will not come from the Fokker-Planck equation one has to take a step back and go back to the Chapman-Colmograph equation. So, today's topic is general non-Gaussian solution to continuous time random walk. So, we take a step back and rather than use the Fokker-Planck equation which will give you a Gaussian solution, let us see what general possibilities exist. For that, we go back to the Chapman-Colmograph equation, we go back Chapman-Colmograph equation which is written as the transition probability to a point x 2 from x 1 in a time t is given by via the Markov assumption by the Chapman-Colmograph equation which says it is the probability that it has come to x 2 from some point x in a time tau, where it had come to x from the original point x 1 in the time t minus tau integrated over all possible values of x. So, this is the Chapman-Colmograph way of writing the Markov transition process for continuous variables via integral formalism. So, picture here is particle has come to x 2 it started from x 1 and we say that it came from x 1 to x. So, x 1 let us say time 0 this is x 2 at time t. So, at some time t minus tau it was here and then in the remaining time tau it reached x 2. So, this is the meaning of this equation. Let us consider a special case consider spatial homogeneity of the system spatial homogeneity which means the transition probabilities from x 1 to x 2 do not really depend on from where it transits. In other words this basically implies the transition probability from any point x 2 from x 1 in a time t is given by is equal to identically the difference between the two positions only. So, it does not depend on the position from where the transition took place. So, if space is completely homogeneous it is completely invariant then then what matters is only the length by which the system has to undergo a transition. So, that length is x 2 minus x 1. So, it is a function only of that length of transition. So, we used this definition in our previous derivation of the Fokker-Planck equation also. So, with this assumption we can write with this the C k equation takes the form P of x 2 minus x 1 in time t equal to minus infinity to infinity P of x 2 minus x it will be true for all points in a time tau and P of x minus x 1 in time t minus tau for all points x. So, P is now a function of difference between the spatial positions only. Denote x 2 minus x 1 by a new variable say x 2 minus x 1 equal to xi. So, xi is a new variable and let and also we denote x minus x 1 x is our intermediary variable by xi prime. Then we can always write for example, x 2 minus x then x 2 minus x becomes x 2 minus x 1 plus x 1 minus x and this can be written then as xi minus xi prime. Hence we have the C k equation can be written as xi t which is integral minus infinity to infinity P xi minus xi prime tau P xi prime t minus tau d xi prime. So, this is almost like the formulation we made when we derived the Fokker-Planck equation we are expressing the transition probability by the jump length xi here. So, this simply says the probability that a jump xi has been made at a time t is given by the probability that it made a jump xi prime in a time t minus tau and in the remaining time tau it jumped by x xi minus xi prime integrated over all xi prime values. Now, let us why we did this will be apparent now let us Fourier transform this equation that is definitionally the Fourier transform is we use the notation P carrot say k t is defined as integral minus infinity to infinity e to the power i k xi P xi t d xi. So, Fourier transform with respect to the spatial variable or jump length variable xi. So, e to the power i k xi P xi t d xi we have been doing this and that is the definition. So, upon applying this to C k equation we have left hand side for example, will be LHS is simply e to the power i k xi P xi t d xi which is P tilde k t to be more precise P carrot k t and hence we get P carrot k t will be equal to now already there was one integral in the C k equation. So, this transform introduces another integral. So, there will be two integrals 1 over xi prime and 1 over xi. So, we can write e to the power i q xi. So, i q xi can be written as i q xi minus xi prime plus i q xi prime you can always do that because i q xi prime anyway is going to cancel I had multiplied only by e to the power i q i k xi. So, we decided it is k. So, i prime then the transition probabilities xi minus xi prime in time tau multiplied by xi prime in time t minus tau integrated over both d xi and d xi prime. So, this is now simply becomes minus infinity to infinity minus infinity to infinity. So, if we define xi minus xi prime as xi double prime for the xi integral it is e to the power i k xi double prime P xi double prime tau d xi double prime and then e to the power i k xi prime exists as such xi prime P xi prime t minus tau d xi prime where we use xi double prime equal to xi minus xi prime and d xi is nothing, but d xi double prime. So, now these two are two independent integrals there is no coupling between them. So, they simply become the products. So, we get that is we get k t will be 1 integral e to the power minus infinity to infinity i k double P xi double prime tau d xi double prime multiplied by another integral minus infinity to infinity e to the power i k xi prime T xi prime t minus tau d xi prime. Now going this is exactly the Fourier transforms of the respective quantities which leads us to a neat algebraic, but it is not just an algebraic equation it is an algebraic product difference type equation P tilde k t is P P carat k t is equal to P carat k tau multiplied by P carat k t minus tau of course, let us recall that tau should be lying between 0 to t with the end points included it cannot exceed t. So, this is a very neat reduction in the transformed space of the Chapman-Kolmogorov equation for the special case where there is a spatial invariance. So, this is like a this is a convolution theorem equivalent of a convolution theorem of Fourier transforms. For example, if we choose the classical Gaussian packet which we have been discussing quite often P carat k t equal to e to the power minus k square dt is satisfies the above equation. So, this is just to reassure ourselves that we have whatever we have done so far is consistent with this because we know that if it is exponential e to the power minus k square d into tau it will be e to the power k square dt minus tau the tau term cancels and you get back t which is again e to the power minus k square dt. But this is not the only solution that it is so, that this is not the only solution we can obtain by converting this into some kind of a differential equation. So, given this equation we differentiate with respect to tau only to find other possible solutions find other possible solutions we differentiate the convolution convolution solution we can say convolution with respect to P that will be d P carat by dt k t and on the right hand side the first term does not depend on t at all. So, it will remain just like this P carat k tau it will be and the second one will be d P carat by dt k t minus tau. Now, we are free to choose a tau tau is an intermediary variable only thing is it should be less than it should be more than 0 and less than or equal to t. So, we make use of this possibility that it can take any value and we decide that it can we will put tau equal to t. Now set tau equal to t since tau can take any values less than or equal to t positive values less than or equal to t. So, when we do that the left hand side becomes then we have d P carat by dt k t it is going to be on the right hand side we will we will see here. So, when we put the first second term will remain just the same, but here when we put it will be completely independent of any time. So, it will be 0. So, this will remain P carat k tau, but tau is t. So, we can replace this tau with t and the other one therefore, will be d P carat by dt k 0. Now this is an entirely an independent quantity which represents the gradient of P carat at time 0 and we have no prior information on this quantity. So, we can justifiably assume it to be just a function of k the quantity d P by dt k 0 is unknown a priori we do not have a priori information on this. Hence, we assume it to be just a function of k we assume that it is a general function of k that is d P carat by dt k 0 identically we say it is some lambda of k we will discuss what is allowed. Hence, we get the equation d P carat by dt k t will be lambda of k of P carat k t this is a first order equation in time. Hence, solution is P carat k t equal to some constant of integration and in general it it will not be a function of time, but it can be a function of k we cannot stop because k is just a parameter and e to the power lambda k into t. So, it comes as a product of lambda k. So, this is the most general solution for the Fourier transform of the probability density function in the spatial homogeneity case arising from as a solution of the C k equation. Now, let us see what how to determine a k to find a k or to determine a k we use the initial condition that is we know that P x at time 0 was some delta x it has to start from somewhere let us say started from the origin since it is spatial invariant it does not matter we can always put x equal to 0 as a starting point. Hence, if you take the Fourier transform P tilde k at t equal to 0 that should be the Fourier transform of delta x which is unity. Therefore, from the solution on top of this page we have 1 equal to a k e to the power lambda k into time was 0. So, that is a k. So, a k is unity hence a k is 1. So, we have the solution P carrot k t equal to e to the power lambda k into t as general solution to C k equation under translational invariance. Now, some properties of lambda k can be estimated for example, what should be the value of lambda 0 properties of lambda k let us say what is should be lambda 0. Now, as k tends to 0 if you go back to the from the definition we have P tilde k at t equal to 0 is integral minus infinity to infinity P k t e to the power i k x P x t dx. So, as k tends to 0 this will tend to integral P x t dx then k equal to 0 and this we know is equal to 1 normalization of the probability density. The probability should be normalized because the particles should exist somewhere in space all the time. Since you have started from some and there we have no loss mechanisms you should be normalized unity which basically means hence P tilde k t limit k tends to 0 should approach 1, but from our solution P carrot k t equal to e to the power lambda k t will approach 1 if lambda k approaches 0. Hence, one property is that lambda k lambda 0 equal to 0 we can say that lambda 0 equal to 0 is an essential property. We cannot have a constant there. There are now many other properties will emerge when we examine in detail the behavior of the solution for various situations. We will discuss in the next lecture. Thank you.