 So, far we discussed 1D random walk problems without bias and with bias, but these biases were constant. They did not depend on the sites. And our understanding is in such situations the system continuously evolves and the ultimate equilibrium solution is actually a trivial result it will be 0 everywhere because my space is infinity along the real line and it will spread everywhere and at any specific site if you ask for the occupancy probability it will tend to 0. Overall of course, it will be normalized it will be like product of infinity to 0 remaining finite. However, a matter of great importance in several engineering and scientific applications is the question of site dependent bias. So, if bias depends on site it is possible to attain non equilibrium possible to attain non 0 solutions in steady state some non trivial results are possible. Let us understand in this lecture how to arrive at such solutions, what are the context, what is the difference between steady state and equilibrium state. So, let us start with the random walk equation in 1D which we have written several times through the occupancy probability W step n n plus 1th step and m is the site index is equal to we write the forward transition probability from m minus 1 at the nth step plus the backward transition probability q m plus 1 W n from m plus 1 side. So, this is basically nearest neighbor transition problem. So, P m plus q m is unity they need not be half each locally there could be different then half values, but they have to add up to 1. Let us assume there is no waiting at each side every time at every step a transition occurs and this is 1D problem. Let us that is the my m extends it includes 0 plus minus 1 plus minus 2 etcetera. Let us postulate that somehow after sufficient number of steps W does not further change that is W n tends to some steady state value W infinity which further does not change. So, with this let us postulate this is a postulate we are doing. Let us assume the existence postulate a steady state. So, then W n plus 1 will also be W infinity. So, the left hand side is going to be W infinity at site m is going to be P m minus 1 W infinity at site m minus 1 plus backward transition from q m plus 1 again W infinity at m plus 1 for all m. So, we basically postulate it and seek if some some non trivial solution for W infinity exists. I can rewrite this equation by multiplying the left hand side with the P m plus q m because that is 1. So, I can write it as P m plus q m W infinity m equal to P m minus 1 W infinity at m minus 1 plus q m plus 1 W infinity at m plus 1. We can reorganize that equation by defining a current term. So, as we did earlier we define a current term J m at the mth side. So, we are let us superscript it with the infinity as as follows P m minus 1 W infinity m minus 1 minus q m W infinity m. That is I am defining a current term at some point m as the net that occurs with the forward transition and a return transition from m. So, forward flux on to the site m and the return flux from m to the backward site we define as a partial current there. So, with that definition very easily we can identify the previous equation if you go here P m minus 1 W infinity m minus 1 minus q m W infinity m is going to be J m by the definition and the other term is going to be J m plus 1 because here it is m plus 1 is involved in this. So, we can easily say that with this notation my steady state condition becomes G infinity m plus 1 equal to J infinity m for all plus minus 1 plus minus 2 etcetera. So, if I have a infinite lattice my current across every point should be the same. In fact, if you iterate with m equal to 0 1 2 etcetera we find that this is equal to J 0 equal to J 1 let us say on the positive side J 2 etcetera constant. So, it does not depend on m because this condition J m plus 1 equal to J m implies that J is constant. Let us call it as J say. So, there exists a possibility of a steady state which finally, says that each of the currents that is a constant, but there is a net current J which passes through all the sides and supposing there is a special situation when J is 0 that we call as an equilibrium state. Basically the existence of J in a system implies that is ultimately since J exists at all points it has to finally, get out of the system maybe at it to infinity, but it will be a loss eventually to the system. So, existence of a steady state implies that there must be some replenishment at some point in the site because there is finally, going to be removal via J at some other point in the state space. On the other hand an equilibrium situation is the one where there is no current net current at any point that means, there is no removal of the walker from the system. So, we say that specific specific case or special case J equal to 0 is referred to as is equilibrium state. So, from what we said this is a state in which there is no net current in the system, no removal of the walker from the state space or site space. So, this specific case we discuss now in the next few slides what are the implications for the occupancy probabilities. So, if we set J equal to 0 and go back to the definition since we know that J m for example, is 0 which was by definition P m minus 1 W infinity m minus 1 minus q m W infinity m that should be 0 for all m. So, that basically means I can write let us say now let us work on a positive space to be specific. So, let us consider a positive space situation then q m W infinity m equal to P m minus 1 W infinity m minus 1. Let us iterate it with let us iterate with m equal to 1, 2 etcetera that is a positive space. Then we get for example, if I put m equal to 1 q n W infinity 1 is going to be P 0 W infinity 0. Let us assume 0 is we do not know this, but some value exists. Similarly, q 2 W infinity 2 will be P 1 W infinity of 1 it will be that is going to be P 1 P naught it will be W infinity 1 we re substitute which is going to be P naught then there is this P 1 and here it is going to be q 1 W infinity 0 or W infinity 2 is going to be P naught P 1 q 1 q 2 W infinity 0. So, on if you iterate we can easily see that in general W infinity of let us say mth side for positive side is going to be P naught P 1 it will come to up to m minus 1 here because when m was 2 it stopped at P 1. So, we can easily foresee that it is going to be this way there will be m terms because it starts with 0. Similarly, the denominator will be q 1 q 2 it will be q m into W infinity. So, formally we have obtained an expression for this site occupancy probability on the positive side and 2 etcetera in terms of the value at the origin. We can do a similar exercise starting on the negative side if it exists again in terms of W infinity 0. Now, we have to evaluate it is required to evaluate W infinity 0. So, we need to now possess a specific model. So, we say that W infinity my space is now positive side 1 2 etcetera general mth side. So, consider the positive lattice positive side of the lattice assume that the entire matter is confined you can say it is a reflector here particle does not go back at 0. So, let 0 be a reflector then the particle should be confined from 0 to infinity. So, W 0 plus sigma m equal to 1 to infinity W m of course, at infinity. So, I should let us stick to our notation W infinity 0 plus m equal to 1 to infinity W infinity m should be equal to 1. So, already we have W infinity m expressed via W infinity 0. So, this leads to a solution W infinity 0 will be given by 1 by 1 plus m equal to 1 to infinity P naught to P m minus 1 divided by Q 1 to Q m. Let us. So, this is now formally the problem is solved we have evaluated this in terms of the P naughts P's and Q's. To understand the implication of this let us take the example of random walk width bias on the positive side only that is on positive lattice. That is let us say that my P the P is is less than half let us say divided minus gamma by 2 and Q therefore, necessarily has to be half plus gamma by 2. So, this is a situation in which I have transition to m minus 1 m m plus 1 there is a greater tendency to transit towards origin let us say my origin is here. So, P and this is Q sorry this is this way this is P and this is Q and P is less than Q that is the picture we are having here which means there is a higher tendency to return back towards the origin. So, some kind of a confinement problem it is a confinement problem a random walker is confined partially in a reflecting barrier basically. So, then we can work out that W infinity at 0 that is going to be 1 by 1 plus m equal to 1 to infinity and since half is common factor there are m terms we can write the formal expression is it was P naught to P m minus 1 and it is Q 1 to Q m and remember the reflecting barrier P naught is defined, but Q 1 is not defined. So, this will turn out to be 1 by 1 plus using the expression 1 minus gamma by 1 plus gamma to the power m for all m equal to 1 to infinity. Now, there is no harm in putting m equal to 0 and it can be extended to include m equal to 0 to infinity 1 minus gamma by 1 plus gamma to the power. So, this is now a geometric series which can be summed, but we make a small approximation for small gamma for gamma much less than 1 1 minus gamma by 1 plus gamma to the power m it approximates to e to the power minus 2 gamma m and how does one do that you write it as e to the power log of 1 minus gamma divided by log of divided by 1 plus gamma expand the log to the first order. So, you will get e to the power minus gamma m in both of them and it comes to e to the power minus 2 gamma m for small gamma that is gamma much less than unity is just an approximation you need not do that, but helps us to visualize the solution is more easily. So, with this my W infinity 0 can be explicitly calculated as the sum 1 by 1 plus we said 1 plus can be clubbed in the sum itself sigma e to the power minus 2 gamma m m equal to 0 to infinity which is nothing, but this sum will be 1 by 1 by 1 minus e to the power minus 2 gamma. So, geometric series in e to the power minus 2 gamma. So, z to the power m is 1 by 1 minus z and that is what you have done here. So, that solution is simply 1 minus e to the power minus 2 gamma. So, W infinity 0 exists is finite. Hence, we can go back and obtain the general solution W infinity therefore, we have hence the complete solution is W infinity m is going to be same argument p naught to p m p m minus 1 divided by q 1 to q m W infinity 0 this will be e to the power minus 2 gamma m into 1 minus e to the power minus 2 gamma which is a constant. So, thus we have a solution specific equilibrium solution for an infinite lattice which shows that the occupancy probability decreases exponentially as you move away from the confining point. It is the probability exponential probability it occupies decreases sort of exponential e to the power minus 2 gamma m as m increases this is W infinity m of course, the pre factor we can include 1 minus e to the power minus 2 gamma which is a pre factor. So, it is approximately of the order of 2 gamma itself. So, the existence of equilibrium is demonstrated even for a constant q's constant p and q system so, long as we are using a partial boundary problem. One can now actually extend this to include other transition probability expressions by invoking other dependencies and equilibrium solutions could exist even in the entire lattice. For example, one can visualize particles being confined in harmonic potentials which can be converted to appropriate transition probabilities of p and q and equilibrium solutions obtained. We revisit all these problems later when we move over to continuous approximations. This is only an illustration that even with the random walk step based random walk equation one can arrive at an understanding of existence of equilibrium. There are interesting problems of finite space equilibrium solutions and one very celebrated problem in this is what is called as the iron fests. In finite space equilibrium the model is what is called as iron fests dog flea model of course, may not like to use the word dog we like animals. So, we like dog especially so, we can even change it to the animal flea model. So, this model has been used to demonstrate equilibrium in a finite system by asking a question of some number of fleas total number of fleas which exist between two animals confined in a cage or in a shed nearby and these fleas keep jumping from one to another. There is no specific preference they both the animals have equal probabilities, but by sheer force of propensity of numbers the transition probabilities would depend on the existence of numbers on one animal as opposed to the other and then you can formulate a transition probability in a finite system. Proceed along the lines that we have already proved for equilibrium and then obtain an equilibrium solution which is a very exactly solvable problem. Overall this study with the finite random walks is quite vast there are very very many problems which can be covered under this, but we move over from the discrete lattice problem to continuum formulations. The advantage of continuum formulations is we are now in the world of we are going to enter the world of differential equations, but the transition from discrete to continuum has to be done very carefully and that requires us to revisit and go back to the Markovian property expressed in the continuum domain we did mention it once extend the Markovian theorem for continuous variables. So, in the subsequent lectures we will examine these aspects. Thank you.