 In the previous lecture, we set out to solve the problem of reflector a random walker in the presence of a reflecting barrier. So, by definition the reflecting barrier was a case of a restricted random walk in which the walker is not allowed to cross the barrier and move on to the lattice points on the other side of the barrier. So, we then found an exact solution using the idea of virtual walker. So, you put an image walker assuming it was a symmetric random walk. So, then there is a complete symmetry between the walks executed by an image walker and the true walker in their respective domains and then we use that property to establish a solution for the occupancy probabilities of a restricted walker using the free occupancy probability solutions for the infinite random walk. So, we said it is actually just the sum of the two random walks of the virtual walker and the true walker. So, then we obtained an asymptotic approximation, a Gaussian approximation using the property of the free walker and we arrived at the probability of the reflecting probability of occupancy for the walker in the presence of a reflector. So, we can call it as R indicating that it is a reflection. This was given by actually just recapitulate it is a sum of the probabilities of a free walker starting from M minus M naught. We will denote the free walker by a superscript 0 or free plus walker at free walker starting with the virtual point which is at minus M naught. So, asymptotically we showed that this boils down to a Gaussian 4 pi dt e to the power minus x minus x naught whole square by 4 dt plus e to the power x plus x naught whole square by 4 dt, where we identified all x and x naught and time in terms of the step length and the site distance, intersite distance. So, the advantage of the Gaussian approximation is it makes it easy for us to compute various quantities which are otherwise a bit too cumbersome to obtain from the discrete steps. Although they are very exact and many sums are available, but it makes it easier to operate with these exponential integrals and one hopes that for most cases the accuracy is a good. For example, we can show that this walker that is now we write it as W x t it is no longer probability it is a probability density for the reflected case when you use the continuous approximation. So, we can show that the walker probability total probability is conserved on the positive side which is what is expected. So, that is we can show that integral 0 to infinity W r x t dx will remain as unity. We expect from physical considerations since the walker is getting reflected at the reflection point. So, just recapitulate. So, this is the reflector and now in the continuous formalism he starts from a point x naught corresponding to m naught in the discrete case and we are asking for this probability for being found at a distance x subject to the constraint that he cannot cross the point 0 that is the problem. So, we can show therefore, that in the domain where he is allowed to walk he will be somewhere that is all this conservation laws. It is interesting to prove this. So, we take this let us call it as expression 1. So, if you carry out the integration. So, the left hand side will be W r x t dx and the right hand side will be 1 by 4 pi dt under root then here 0 to infinity e to the power minus x minus x naught whole square by 4 dt dx and on this side it will again be an integral 0 to infinity e to the power x plus x naught whole square by 4 dt dx and the square back it is closed. Now, we do a small trick in the second integral on the right hand side we make a change of variable new variable x prime equal to minus x in the second integral which is right hand side. So, then this becomes 0 to infinity the LHS integral of course, will remain the same is what we want to evaluate dt will be equal to 1 by 4 pi dt the first integral let us keep it like this 0 to infinity e to the power x minus x naught square by 4 dt dx and now x becomes dx becomes minus dx prime and the limits if you see when 0 will remain 0, but infinity will become minus infinity. So, it will be 0 to minus infinity, but the dx becomes minus dx prime. So, we will absorb that minus sign by changing the limits instead of 0 to minus infinity we make it minus infinity to 0. So, very easily you can see that it can be written in terms of the new dummy variable integration variable x prime. So, it will be minus x prime plus x naught minus x prime plus x naught let us leave it right now in the same form. So, that you can see the advantage is transformation divided by minus x prime plus x naught 4 dt dx prime. So, the second integral I can take this minus out of the square then it will again become x prime minus x naught whole square. So, now the integrants on both these integrals have the same form. The first one it is e to the power minus x minus x naught whole square by 4 dt and in the second one again it is x prime minus x naught whole square by 4 dt dx prime. x prime is just a dummy integral in dummy index integration variable you can as well replace again by x. So, one can then write the left hand side what we want to calculate dx is going to be 1 by 4 pi dt and this whole thing now will become minus infinity to infinity e to the power x minus x naught whole square by 4 dt dx. I replaced x prime once again with x and it was a broken integral 0 to infinity and minus infinity to 0 of the same integrants. So, I can just combine it make it a single integral from minus infinity to infinity and we know that this is a Gaussian integral. So, it is this integral is just going to be unity. We have we have made use of the general property that integral minus infinity to infinity e to the power sum u square divided by 2 sigma square du is going to be sigma root 2 pi. So, here we should remember that sigma square is nothing, but 2 dt then you are going to get the same expression and it will be virginity. Thus the particles are the particles are conserved. The solution is consistent with the conservation law. So, we basically checking our solution from various angles. So, particle are confined in the region 0 x less than infinity and probability is conserved. We can ask many questions with this and one of them of course, we showed that the solution is also consistent with the flux being 0 at the barriers level and that was a new boundary condition we sort of derived from the discrete equation. And so, it has a certain certain additional scientific value because we did not make such assumption when we started with the discrete case. We merely assumed a virtual walker and and added the paths which resulted in your solution which in its asymptotic limit led to a boundary condition saying that the flux is 0 at the surface. It is a great utility when we solve differential equations corresponding to particle diffusion in the presence of a reflecting barrier then you need a boundary condition which is supplied by this. So, that that was a new thing that we found. We can find many things for example, like we can ask a question what happens to the particle at the barrier what is the probability of finding the particle at the barrier. So, if you go back to our expression here equation 1. So, if you put x equal to 0 it will be e to the power minus of x naught square by 4 dt this also will be e to the power minus x naught square by 4 dt. So, it becomes a twice. So, you will have w 0 t will be 2 by square root of 4 pi dt e to the power minus x naught square by 4 dt. So, of course, this says that at t equal to 0 the probability was 0 that is quite true because our walker has to start from x naught and it would take certain time to reach the boundary. So, we do not expect the walker to exist at the reflecting boundary. However, as the time proceeds the probability increases because as t increases e to the power minus x naught by t the value will go on increasing. However, as t increases the 1 by square root t term will tend to bring it down. So, we expect that at t very very large as goes to infinity this e to the power term will become unity because e to the power minus 1 by t will be e to the power 0 and t equal to infinity which is 1, but 1 by root t will remain. So, we expect the probability to decay. Basically, this function therefore, will have a form if you plot as a function of t and here it is w 0 t for the reflecting case at t equal to 0 it is 0 it will remain very sharply 0 then it will rise have a peak, but then have a very slow decay. This functional form of decay will be of the order of t to the power minus half. So, what is the physical meaning of this? After a long time the walker's probability of being found at the reflector points tends to 0 very slowly though which is because and the walker is still conserved though total walker he is in the real line somewhere, but he moves on an average farther and farther by diffusion diffusive mechanics and his see he reaches farther and farther points tends to become sort of a uniform distribution all along the line, but you cannot have a uniform distribution at t equal to infinity because it will be 0 into infinity kind of a problem total area has to be conserved. So, you will have a certain uniformity up to some distance and then the probability will fall, but this distance will go on increasing with the time. So, that is why he will be found less and less often with less and less probability at x equal to 0 point. So, this is of course, is a function of t. So, at x equal to 0. So, this is w 0 t r. There are many other probabilities you can calculate once you have an analytical expression, but these are some of these essential features. Now, we go over to another you know opposite case of a restricted random walk that is an absorbing barrier instead of reflecting case 2 let us call it absorbing barrier. So, an absorbing barrier is a very useful situation in many transport problems. Most of the particle transport is about the behavior of the particle at certain interfaces. So, it random walks hits the interface and then the question of interest is whether the particle is going to be absorbed or whether the particle is going to be reflected. And if it is absorbed then there is a question of how much is the mean time of absorption, how much one has to wait typically for particles to be absorbed or what is the distribution of residence time before being absorbed etcetera. So, many questions can be asked. So, to answer this questions it is necessary that we find the probability distribution of a walker in the presence of absorbing barrier. Let us go back to this discrete case. So, we have now a barrier it is an absorbing barrier at m equal to 0. And the walker as earlier starts from some point m naught and it is desired to find his occupancy probability at some site m. So, always a question how does the barrier affect his probability at a point which could be quite far. In fact, these kind of transport problems they have some level of non locality that is what it is said because the presence of near presence of a barrier with some properties can affect the distribution at far off points. So, that is why these questions have certain other implications in physics. Now, first of all how do we formulate or how do we specify this property of absorption. Obviously, from a jump of probability principles we are talking that he starts with the probability half and probability half to the right. So, one step at a time. So, at 0 his transition probability p if you call it as p upward that is 0 p at m equal to 0 equal to 0. And q also we say is 0 because he is a we are not really interested in the negative side, but he should be interested in slightly different class of problems then you could specify q also as 0 or q is something it he does not allow it to come back, but he transmits it to negative side. We are not discussing that. So, let us also say that q at m equal to 0 also is 0. We are not in fact, discussing the point 0 at all. So, all that is true is that both p and q and of course, p plus q will be 1 elsewhere for all m excluding the 0 point 2 etcetera. So, strictly speaking this is the problem that one has to solve with these conditions. Like earlier the technique of using a virtual walker simplifies the problem greatly and that is specifically suited to the symmetric random walk problem. So, absorbing barrier symmetric random walk or unbiased as we said. As in the case of a reflecting barrier we can now extend the negative line put a virtual walker m naught. In view of the fact that it is symmetric random walk the walkers behaviour or walkers paths to the left of the barrier are completely symmetrical left of the barrier including the barrier are completely symmetrical with respect to the walkers paths to the right of the barrier. So, the virtual walker in the left side and the real walker on the right side have complete symmetry. Now, if you consider a trajectory we will make another picture. So, if you consider this situation 0 and m naught and we are interested in m and we have now a minus m naught here. Now, a path at this point which you could have he would have he had a probability half of coming back in the normal walk. If it is a free space he would have had a 50 percent probability of returning and that path originating from some point of contact at some level of step let us say nth step could have led him to m let us say this is the path by which he could have come. Since now the absorber kills that path any path the and since the path from the virtual absorber will be very symmetric in his region with respect to the path of the real absorber in his region. So, this path starting from the absorber which has been killed we can account for it by the subtracting the free walk of the virtual walker in the right side of the plane. So, if we subtract the path that minus m naught walker would have contributed to m after crossing the barrier and since crossing the barrier is not allowed it is killed. So, all trajectories that would have crossed from here from minus m naught they if you subtract those from the free walk that would account for the process of absorption of the walker m naught. So, this principle makes use of the symmetry of the walker in their respective domains including the barrier and the fact that all free walks starting from the barrier from the virtual absorber will have to be now subtracted. So, that results in the expression for the probability of finding the walker at m given that he has started from m naught in the nth step under the condition of absorption that should therefore be equal to the free walks of the all the free walks of the walker at a positive m naught. So, all the 0 now represents free. So, m and we know because of translational symmetry it will be m minus m naught. So, all the free walks that we discussed starting from m naught minus the free walks that the virtual walker would have had starting from minus m naught. So, m minus minus m naught is plus m naught. So, this solves the problem we merely therefore have to have an expression for the free walk. So, here we say the superscript 0 0 represents free walks barrier free walks free means barrier free walks or unbounded walks. This solves the problem it can be more explicit and right inexplicitly this amounts to saying w n with the absorption starting from m naught equal to n factorial by 2 to the power n into 1 divided by n minus m minus m naught divided by 2 factorial then n plus m minus m naught divided by 2 factorial maybe we can put brackets to separate the minus most important is minus divided by 1 divided by n minus m minus m naught by 2 factorial into n plus m plus m naught divided by 2 factorial again. If you just put m equal to 0 you can see that both the terms become identical if you put m equal to 0 this becomes n plus m naught here and n minus m naught here. Here n plus m naught also exists n minus m naught also exists they just exactly cancel each other thereby implying that automatically it implies a very interesting boundary condition. We note that w n with the absorption 0 m naught equal to 0 this is a very important boundary condition again derived by without assuming by merely using the property of symmetry and killing the paths and accounting for the killed paths by a virtual walker that is all we have done. So, this is often called as absorption boundary condition and has a great role to play in diffusion theory absorption boundary condition BC we will expand it for the first time. In other words the problem of absorption is equivalent to assigning a priori a probability 0 of the walker being found at the absorption site. This may raise a certain questions because all that we have done is to not allow the walker to jump back into the region of interest. We really did not have any idea he could have been trapped there and could have been present concentration need not necessarily be 0, but it is consistent. So, that may not be a unique boundary condition, but the boundary condition is consistent or the solution is consistent with this assumption and it is it m allows us to solve wide class of problems in a continuum approximation. Thank you.