 Let us continue with the problem of 1 day absorber, a symmetric random walk. We obtained that this problem can be solved by considering a virtual walker just as in the case of a reflector. We showed particularly that the occupancy probability of a site m starting from m naught. Let us say in the presence of absorber is given by the difference between the occupancy probabilities of a free walker. So, that is given as W n 0 representing the free walker. Let us say at m minus m naught minus W n 0 m plus m naught, where the superscript 0 represents the free free walker walker without any obstruction or in the entire line including the negative values whereas, here the solution is however valid for all m comma m naught 0 1 2 etcetera only on the positive line. So, this is the solution corresponding to the case where you have an absorber here at 0 and these are the lattice points on which the random walker moves. So, he starts from some m naught and this is some m this is absorber. We made no assumptions accepting to make use of the fact that the random walker crossing the barrier is killed and therefore, the virtual this process is actually executed by considering the virtual walker and subtracting his trajectory is crossing the barrier. Interestingly now if you put m equal to 0 in this we get result W n absorber 0 m naught equal to W n with free walk minus m naught I am putting m equal to 0 minus W n free plus m naught since the free random walk plus m naught is a symmetric is equal to W n 0 minus m naught that is why it is called symmetric. We obtain since this is true we get for the absorber the occupancy probability at the origin is 0. This is an important boundary condition boundary condition which has an implication and usefulness when we actually move over from discrete to continuous space from difference equations to differential equations where we need a boundary condition and this is obtained without actually imposing that in the random walk problem. Of course, it also says that W n absorber for any m, but starting from 0 that will be 0 because if the walker is starting from 0 is already on the absorber and can never execute a random walk and so, all occupancy probabilities will be 0. To complete this we must we may recall that the free probability or occupancy probability for the free walker is given by the expression which I derived several lectures ago is given by n plus m by 2 factorial and n minus m by 2 factorial and this is to be noted that if n is odd m should be odd and if n is even m should be even or basically n plus or minus m should be always even that constraint has to be met. Now, there are let us understand the implication of a random walker in the presence of an absorber. We know that every time he steps onto the point 0 there is a certain probability of him being lost from the system. So, the matter of importance is how does is the survival probability that is probability of not getting lost into the absorber change vary with the number of steps. For that we have to compute the overall survival probability of the random walker on all by summing up the occupancy probabilities over the overall the lattice points. So, we define survival probability s at nth step starting from the point m naught we can superscript is for consistency sake as in the presence of an absorber defined as m equal to 0 to infinity w occupancy probability in the presence of an absorber from over all sites starting from m naught. So, it is a sum over all sites. So, this means if it was a reflector this survival probability will be always 1 because you should be found in one of the sites. Here since m equal to 0 is a site we have included because we know anyway the probability is 0 there although strictly we can even exclude it and define 1 to infinity, but since the probability itself is 0 it should not matter. So, if you actually now substitute we are going to get this is equal to m equal to 0 there are 2 terms now involving the free random walker that is w nth step 0 m minus m naught equal to 0 to infinity same distribution function, but evaluated at m plus m naught this m. We can simplify it by simply replacing m minus m naught by m prime here and m plus m naught by another m double prime or something and in any case is easy we can write it as if you put m minus m naught as m prime then when n is 0 it is minus m naught to infinity. So, here w n 0 m prime now m prime is a dummy variable. Similarly here it is going to be m prime will vary from plus m naught to infinity w n 0 m prime. We can write it in a slightly more simplified form because we can take the sum up to plus m naught minus 1 here and then from m naught to infinity this term will cancel exactly with the m naught to infinity this of the second term. So, as a result we will be left with for the survival probability at the nth step in the presence of an absorber for a random walker starting from m naught is going to be left with simply minus m naught to m naught minus 1 of w n free m prime where now m prime is a dummy variable. So, basically it is a sum over number finite number of sites corresponding to minus m naught to m naught minus 1 and of course, minus m naught the value at that point is same as the value at plus m naught. So, you can further write the whole sum in terms of only positive indices positive sites. In any case one cannot avoid actually adding over all the sites and if m naught is large it will be formidable to sum over a large number of sites. There is no easy or simpler formula for this expression, but to understand the implication of the survival probability let us take the case of walker starting from let us say m naught equal to 1 itself. The random walker has just started adjacent to the absorber that is here is the absorber and he has started from 1 to etcetera and it is a symmetric random walk. So, probability half and then with the probability half he would have jumped on to 0. So, obviously, we know that in the very first step the probability of him surviving is only half. Interesting thing is how does this probability change now the other half has given him a chance to escape the absorber. So, the interesting question is whether he will eventually escape being trapped by the absorber or ultimately there is inexorably he will be trapped as time progresses and in case he is trapped what is the functional form of decay of his survival probability. So, these are very very important and practical questions you can easily visualize that I am actually not referring to a random walker. I could be referring to a capture by surfaces of particles or adsorption of atoms and gases. So, let us therefore, get a feeling for this considering a case of m naught equal to 1. So, if m naught equal to 1 the survival S n with the absorber for m naught equal to 1 turns out to be you are going if m naught is 1 the upper 1 is 0 and the minus 1 minus 1 is the same as the plus 1. So, it will be 0 let us take W n free walker at site 0 it is a free walker. So, free walker is a probability of being at site 0 plus W n free walker at site 1 there are only 2 turns and we may note that if now n is even probability of being at the odd site is 0. So, it will be only this term is valid. So, if n is even then W n 0 1 is 0. Similarly, if n is odd then W n 0 0 is 0 at an odd step he must be away from the starting point. So, essentially therefore, the survival probability of the absorber starting from site 1 is either W n 0 0 if n is even or W n 0 1 if n is odd. So, very simply we saw that it is a just occupancy probability. So, let us evaluate this and create a table this is for case m naught equal to 1. So, n S n. So, if at n equal to 0 his survival probability was 1 like one can easily see that when n n is 1 in the first step it is a W n 1 that matters and W n 1 now if we note down the or formula W n 0 for example, is going to be n factorial divided by 2 to the power n if you remember it is n plus m divided by 2 factorial. So, m is 0. So, it is n fact n by 2 factorial other term was n also will be n by 2 factorial whole square and W n 1 is going to be again n factorial divided by 2 to the power n here it will be n plus 1 by 2 factorial and here it will be n minus 1 by 2 factorial. Let us remember this. So, at a naught step it is a second one. So, when n is 1 1 plus 1 2 by 2 1 and here no contribution. So, it will be just half which is what we expect. The second step we use the formula of W n 0 here again it will turn out to be half it will be 2 1 4th. So, it is going to be half third step again you will use this. So, if you use the now 3 factorial here and 2 to the power 3 3 plus 1 by 2 and all those calculations if you do you will get 3 by 8 it turns out that the fourth step also will be 3 by 8 and the fifth step if you continue you will get 5 by 16. So, if we just actually calculate the values the survival probability is 0.5 it is 0.5 here 3 by 8 is going to be 0.375. So, it will be 0.375 5 by 16 is going to be 0.3125. So, likewise you can see that the probability decreases survival decreases and it decreases in steps we plot now steps here n equal to 1 2 3 4 and 5 this is n step the probability was 1 here. So, the first step it will be 0.5 is a significant reduction in the first step in the second step also it is 0.5. So, it will come here in the third step it comes to 0.375. So, somewhere here continues to be so at the fourth step then at fifth step again it comes to 0.31 likewise the probability continues. So, we see that the there is a decrease in probability successively, but it appears to be decreasing very very slowly. So, there is a question whether this will finally, lead to some stabilized result some small percentage of survival or then if it goes to 0 then how does it go to 0 to understand these tendencies or so called how persistence effects it is more useful to go to continue a formulations rather than discrete formulations. So, with this in mind we perform the asymptotics of the occupancy probability functions that we had derived for free random walk and examine how the solution behaves. Let us get back to this occupancy probability itself and then again come back to the survival probability. So, we now perform the asymptotic 1 D absorber symmetric or unbiased asymptotic behavior. We had derived earlier that the free random walker distribution W n m has a form for large n and m by using sterling x approximation square root of 2 by pi n e to the power minus m square by 2 n using sterling approximation. We apply this form to the W m with the absorber. So, we can just there are 2 terms of the similar type with the m replaced with m plus m naught and m minus m naught. So, the answer is pretty simple here and then of course, we also continue by replacing m with x. The occupancy probability will now will become probability density. So, that we call it as W absorber x. It will go over to 1 by root 4 pi D t of e to the power x minus x naught square by 4 D t minus e to the power x plus x naught square by 4 D t. So, we arrived at this form by using the following symbols x represented the actual position given by m into L where L was a jump length. Similarly, the time t was nothing but n tau. The steps were replaced with time where tau is the jump time and new parameter got introduced called the diffusion coefficient which is the ratio of L square and tau specifically it is L square by 2 tau. So, with these notations the form immediately we can see that the form will come to what is given in this equation. So, it has all the properties that we saw for example, when x equal to 0 this function will be e to the power minus of minus x naught whole square which is e to the power minus of x naught square divided by 4 D t and the second term also will be e to the power minus x naught square by 4 D t it will be 0. So, it meets the condition that W satisfies W absorber 0 t equal to 0 for all x naught also W absorber x t at x naught equal to 0 that also will be 0 because if you put x naught equal to 0 in this expression once again the terms cancel each other. So, it satisfies all these criteria. We use this to answer some of our questions on the survival probabilities of the random walker in the presence of an absorber. So, we define the survival probability in this case since it is a continuous distribution survival probability. The survival probability S t will be defined as we are talking of the solution in the positive space. So, it is 0 to infinity of W x t d x with the absorber. This is also let us superscript it. So, it involves the integration over 2 Gaussians. We can say that it is integral 0 to infinity of e to the power minus x minus x naught whole square by 4 D t minus e to the power minus x plus x naught whole square by 4 D t d x of course, divided by square root of 4 pi D t. Each of them is actually a Gaussian integral with the shifted peak. So, we make a transformation x minus x naught equal to x prime here. Here x plus x naught equal to x prime. So, this simply becomes an integral of the type 0 to infinity e to the power minus I call it as x prime square by 4 D t. Here now since x equal to x naught x minus x naught will be minus x naught. It will not be 0. You can write it as minus x naught to infinity d x prime divided by square root of 4 pi D t. And the second integral now it will be from x plus x naught is equal to x prime. So, when x equal to 0 it is x naught to infinity e to the power minus x prime square by 4 D t d x prime by square root of 4 pi D t. These are very similar integrals. So, eventually what will be left is a term the Gaussian integrated from minus x naught to plus x naught. And it can be written in a neat form in terms of a very well known function and that function is so called the error function. This whole thing now turns out to be an error function of an argument x naught by square root of 4 D t. The survival survival probability S t in the presence of an absorber where we define the error function for any general argument as 2 by 2 by root pi integral 0 to z e to the power minus u square d u. So, it is a very well known function it is well tabulated in hand books and comes in the theory of error. So, basically it is a cumulative of a Gaussian function say cumulative distribution of a Gaussian function suitably represented. So, the survival probability is proportional to the error function. The survival probability can be plotted as a function of x naught and it shows the behavior that for sufficiently large x x naught the survival probability remains one at a given time. As a time increases of course, the survival probability will become less and less let us say this is for t equal to 1 and this is for t equal to 2. So, particle slowly the survival probability decreases. You can have a plot as a function of time which shows that the survival probability as a function of time at t equal to 0 the survival probability is 1 because it is a r of infinity and as t goes to infinity the survival probability will approach 0 for any x naught. Say x naught equal to 1 for x naught equal to half that survival will happen quicker and quicker, decay will occur quicker and quicker. We use this survival probability to predict many other characteristics of the absorption probability. Thank you.