 Now, let us look at the solutions of this equation. Let us first look at the equilibrium solutions. Let us look at equilibrium solutions of this. If I first take this standard diffusion equation without the drift. So, no force or no drift whatever. At equilibrium what will happen? Whatever probability or whatever density or concentration that you have that is not going to change with time anymore right. So, this del rho del. So, equilibrium or steady state whatever equilibrium implies that del rho del t is equal to 0 which means what you need to solve is del 2 rho del x 2 is equal to 0 right. So, what is the solution of this? So, we are just doing in 1 d you can extend all of these two higher dimensions as you want. So, if del 2 rho del x 2 is equal to 0 I can write let me write rho star because this is what equilibrium. So, rho star of x is sum A x plus B right. I need something else. I need some sort of a boundary condition in order to fully solve a different specify a differential equation. So, let me say I use is no flux boundary condition. So, let me say I use a no flux boundary condition which basically means the current or therefore, equivalently del rho del x is equal to 0 at boundaries. So, j or equivalently del rho the derivative of the concentration or density that is equal to 0 at the boundaries. So, nothing flows in or out. If you do that then of course, what this says is that what does it say? It says that A must be equal to 0. So, if you impose this no flux boundary condition that implies that A equal to 0 and therefore, the equilibrium solution then becomes what you would expect that it is a constant right. If you wait long enough the diffusion equation will take whatever inhomogeneities you had in your profile and it will make it a flat profile. So, regardless of start off with whatever if you wait long enough it will go to the straight line it will go to an uniform line. So, it is a boring equation in that sense right. So, that is the no drift case, but we can also do the case in the presence of on the presence of forces or in the presence of drift. So, with force so, we will solve the drift diffusion equation. Now, if I have equilibrium I will have no net current in my system right everything is sort of stationary which means that. So, at equilibrium at equilibrium the current is equal to 0 which means what remember the current is minus current is v rho minus d del rho del x. So, the current is equal to 0 there is no flow then what I have need to solve is this equation v rho is equal to d del rho del x. Remember that the velocity that I had that was proportional to the force right. So, the velocity the drift velocity was proportional to the force. Let me assume for the sake of simplicity that this is a conservative force which means you can write it as the negative gradient of a potential right which means this I can write as some minus mu del u del x right. I have some potential u of x the gradient of that gives me the force the force causes a velocity. So, let me write the velocity in this form it is minus mu del u del x or if you want to write in terms of the friction minus 1 by gamma del u del x all right. So, now, you need to solve that equation that. So, del u del x is equal to gamma d let me write in terms of gamma gamma d 1 by rho del rho. What is the solution of this equation? What is? So, all of these are equilibrium. What is the solution of this equation? Here you have a del rho by rho here you have a d u right. So, therefore, yes someone exponential right. So, exponential of what let me write its proportion exponential of did I miss minus sign somewhere yes right that minus sign I missed. So, exponential of minus u by gamma d right. And what does the Einstein relation say about the relation between the diffusion coefficient and the friction d is k B T by gamma. So, this is nothing, but proportional to exponential of minus u by k B T. So, recover back the Boltzmann the Boltzmann distribution at equilibrium. So, you take this yes this is a del rho by rho. So, that will give you a log rho if you integrate right here you will get a u. So, then rho is e to the power of whatever you have right. So, at equilibrium for this drift diffusion equation unlike this simple diffusion equation where you get a constant profile here you recover back the Boltzmann distribution right. So, that was the easy part. So, this is the solutions at equilibrium we can also write the full time dependent solution of the diffusion equation. So, we can also write the full solution. So, that is what we will do next it is good to know how to solve the diffusion equation at least completely in all cases. All right. So, let me first solve this the equation without drift and we will come to the drift part later. So, I will solve this one del rho del T is equal to D del 2 rho del x 2 all right. So, I need to specify some sort of an initial condition. So, I need to specify an initial condition and let me take a delta function for that. So, it is the particle started off at some given position. So, rho at x at x comma T equal to 0 the initial condition is some delta of x let us say. So, start off at some specified position and you want to know how does this this particle which started off at this position which you know how does the probability density for that evolve in time if it follows the diffusion equation. This is a good probability density because this is automatically normalized right if you integrate the delta function over space you will get one. So, it is normalized. So, what will you can solve this in many many ways what we will do it we will do it using Fourier transforms. So, if I Fourier transform the density I can write. So, I will do a Fourier transform rho tilde of k that is rho x comma T e to the power of minus i k x dx minus infinity to infinity. You can also write the inverse Fourier transform of course, you can write the inverse Fourier transforms. So, rho of x comma T will be 1 over 2 pi integral rho tilde k of T e to the power of plus i k x d k. Everyone is familiar with Fourier transforms are the depart. Since I know the initial condition I can find out this Fourier transform of the initial condition right. So, this is not k this is T sorry. So, I can find out the Fourier transform rho k tilde at time 0. So, I just need to substitute the initial condition over here and that will give me the Fourier transform at 0 time. So, what is that what is rho k tilde at 0 what is delta x e to the power of minus i k x dx. So, I know that all right. So, what I will do is that I will try out wave solution that is because this equation is translationally invariant. If I write down if corresponding to this I write down a solution rho of x comma T then any other solution rho of let us say x minus delta comma T will also be a solution where this which corresponds to a shifted initial condition. So, it is a translationally invariant solution system which means that will admit plane wave solution. So, let me try you try a solution of this form. So, I will try out a solution of this form rho of x comma T rho of x comma T is rho tilde k of T e to the power of i k x. You can substitute this back into this equation. So, this is my answers you can substitute this back into this equation which means that del rho del T is del rho k tilde of T del T e to the power of i k x and on the other side del 2 rho del x 2 is what minus. So, there is a d over here. So, let me write that also. So, minus d k square rho k tilde of T e to the power of i k x. Each differentiation will bring down an i k i k into i k is minus k square. So, if I equate these two then the equation that I get is del rho k tilde del T is equal to minus d k minus d k square rho k tilde of T which I know how to solve which has the solution that rho k tilde of T is equal to rho k tilde of 0 e to the power of minus d k square T and this rho k tilde of 0 I have already found because I know what was the initial condition that is just 1 let me forget. So, I know what is rho k tilde of T. Once I know what is the this Fourier transform I can use the inverse Fourier transform to get back to my density. All I need to do is I need to substitute for that rho k tilde that I found in this equation. So, the rho x of T is going to be 1 by 2 pi integral e to the power of minus d k square T that is my rho k tilde then e to the power of plus i k x plus i k x d k from minus infinity to infinity. How would you do this integral? How does one do an integral like this? What is this integral? e to the power of minus a x square d x this is the Gaussian integral right. So, what is the solution of this? root pi by a this you should know by now. If you are not familiar with Gaussian integrals just go back and look at them it is fairly standard. So, if you have an integral of this form then what does one do? Complete the squares right you add and subtract to multiply and divide by a term. So, that you complete the squares and then you recast it in this form in the form of a Gaussian integral. So, please do this what will you get? Krishna what will you get? This is a diffusion equation what will you get is the full time dependent solution of the diffusion equation Gaussian. So, what is the Gaussian? What is the normalization? 1 by square root of 4 pi dt e to the power of minus x square by 4 dt. So, please do this you complete the square recast it in the form of a Gaussian integral do that Gaussian integration and you should get this answer. So, this is a simple Gaussian e to the power of minus x square by 4 dt with this normalization. So, what does that mean? It means that I started off with this if I plot this rho x of t rho x versus x at time t equal to 0 I started off with the delta function right this is delta of x as time goes on it will slowly smear out the probabilities and it will smear them out in the form of a Gaussian right. This is at some time t 1 greater than 0 if you wait longer it will get smeared out even more. So, this is some time t 2 greater than t 1 and so on. After infinite time of course, it will become flat because that we have already seen that at equilibrium it will go to a flat profile. But how it goes the full time dependence that it goes in the form of this Gaussian and the variance of this Gaussian of course, is sigma square is 2 dt which is exactly what we were doing earlier as well the variance grew with the variance grew with the number of steps of the time that the random walker has walked for. So, this plays the role of n in these earlier coin flip or drunkards walk examples that we took. So, that is the variance. All right. So, this was the simple diffusion equation what would you do for the drift diffusion equation? So, what would be rho x of t for the drift diffusion equation? Yes Can you just guess at the answer? If you have something which is travelling with a velocity v so, remember here in this case the mean stays at 0 right the peak is always at 0. So, if something is travelling with a velocity v how would this change x minus v t. So, this you should again check for yourself that if you solve the full drift diffusion equation then the time dependent solution the full time dependent solution is x minus v t whole square divided by 4 dt. So, here if you look at the mean that is going to be travelling with a velocity v ok. So, if this is what you started off with at some time after some time it will have shifted after some time it will have shifted even. So, that is the idea. Couple of things this is of course, in 1 b generally here you will have a coefficient which depends on the dimensionality the normalization will depend on the dimensionality and so on, but the technique the method or the at least the salient feature that it is going to be a Gaussian either a stationary or a travelling solution like that that is going to remain independent that is going to remain unchanged. So, this is this clear. So, this is a little bit about what I wanted to say formally before I come back to one more biological application. So, let me write this as well rho of x comma t is square root 1 by 4 pi