 In the previous lecture, we took up the problem of biased random walk in the presence of an absorber. You continue to discuss one-dimensional situation and for this problem of a biased random walker who has slight asymmetry in moving to the left or to the right, we found an exact solution for the occupancy probability W n m b 4 the biased random walker in the presence of an absorber and that expression is 1 plus gamma divided by 1 minus gamma to the power m minus m naught by 2 into 1 minus gamma square to the power n by 2 multiplied by W n 0 evaluated at m minus m naught minus W n 0 evaluated at m plus m naught where this W n 0 the 0 specifically denoting a free walker starting from m naught that is why it is m minus m naught and the second expression is the free walker starting from from the point minus m naught it is actually a virtual free walker. Gamma is the asymmetry factor or the bias factor m naught is the starting point of the walker to the right of the bias. So, this is the expression we derived it is an exact expression valid for values of gamma lying between 0 to 1 and any m naught or any m of course, they are all positive here because we are talking of half side of the real line. We also set a set about to evaluate the asymptotics of this solution the reason being that it is easy to handle to the asymptotic equation and find some general features of the solution then work with the exact numbers. So, exact distribution. So, in that process we assumed the situation for large n and small gamma. So, that we keep gamma to the first or second order as the case may be, but not higher orders. So, we then easily can see that 1 plus gamma by 1 minus gamma to the power m minus m naught by 2 will go as e to the power gamma of m minus m naught and similarly 1 minus gamma square to the power n by 2 will vary as e to the power minus gamma square n by 2. So, and then the third asymptotic form which we already know is the W it is the W n the free particle free particle occupancy probability function W n m which we had seen has a form 2 by pi n e to the power minus m square by 2 n. We have done it several times. So, we can now combine all these 3 expressions by suitably replacing m with m minus m naught for the first term on the right hand side and m plus m naught for the second term on the right hand side and obtain the asymptotics for the case of a bias that is W n d plus absorber at site m will vary now as e to the power gamma of m minus m naught minus gamma square by 2 n. Then we have the difference terms where we should remember it will be square root of 2 by pi n and e to the power minus m minus m naught whole square divided by 2 n here minus e to the power minus m plus m naught whole square divided by 2 n here. So, this is a asymptotic form supposedly valid for n much larger than 1. One can of course, work with this and find the properties it satisfies all the main conditions that we had imposed such as when m equal to 0 when m equal to 0 here for example, you put m equal to 0 then this term of course, will evaluate easily, but here you can see that when m equal to 0 this is e to the power plus m naught square and this is e to the power minus m minus of again plus m naught square the terms will cancel satisfying our required condition that automatically satisfies W n 0 equal to 0 b plus a b s. And we also in discuss the implication of a condition called the absorption boundary condition. We can now go a step further rather than use discrete step lengths or step variables m and n, step as well as a site variables n and m we can go pass over to the continuous variables x and time. So, passing over to continuous variables x comma t with our usual conversion formulae that is x is nothing, but m into l where l is the mean inter lattice distance. Similarly, t equal to n into tau where tau is the jump time and defining the new quantity d equal to l square by 2 tau. Effectively characterizing a quantity called the diffusion coefficient and further now for gamma we introduce a new concept called the velocity of flow. This is very important we will use the notation capital U velocity that will be defined as it is of course, proportional to gamma, but in terms of the scales that we have used it will be 2 d by l dimensionally if d is meter square per second and l is meter the dimension this dimension is going to be meter per second. So, it stands as mean velocity of the biased random walk. The physical meaning is that while the jump process strictly the left and right jump if we probable process half represents a diffusion that slight bias factor gamma should therefore, represent a steady motion in a particular direction. If gamma is positive it is motion is in the positive direction gamma is negative that steady motion is in the negative direction. Either way therefore, it should correspond to velocity when we pass over to the continuous variables. Hence we introduce this concept U one more parameter into the problem. It is a very important transformation we have done. So, with this we can arrive at an expression now or further of course, W m n that passes over to a continuous distribution W xt by the same method of dividing by the inter particle distance inter lattice distance, but making the allowance for the fact that it is always a unit of 2 n plus m oddness and evenness being maintained. So, with that we can obtain an expression for W xt as follows because we can go back and see that wherever m is there we are going to use x by l here wherever m the quantity m will be replaced by x by l n will be replaced by t by tau etcetera then naturally diffusion coefficients and velocities will come out as a result of which we are going to have the following expression. That is hence W xt can be written as e to the power u into x minus x naught divided by 4 d minus u square t by 4 d that gamma square getting replaced as u square here and then here it will be as the diffusion equivalent term x minus x naught square by 4 d t minus e to the power minus of x plus x naught whole square by 4 d t valid for all x comma t greater than 0. This is a half sided problem. So, we have to remember that it is valid for x comma t greater than 0 and also of course, x naught also greater than 0. One can actually plot this function now very easily because it is now just in terms of exponentials. We have also have to insert the term square root of 4 pi d t the denominator which basically is the square root of 2 by pi n term gets replaced with this. One can evaluate the occupancy probabilities at various times with this and this satisfies once again the important boundary condition that at x equal to 0 the W is 0 and so is it when x when x naught equal to 0. It also implies a certain effect of a steady motion which modifies the distribution significantly we can always plot it. Now the quantity of interest as we mentioned is the survival probability. So, once we have an expression for the occupancy probability we can evaluate the survival probability that is the probability that the random walker this is equal to the probability that the random walker has not been absorbed till time t it is a function of time. Alternatively it is the cumulative probability of finding the particles is somewhere along the positive side of the lattice. Alternatively also it is equivalent to the cumulative probability of finding the random walker somewhere lattice x greater than 0. So, we use the second definition which means s t equal to 0 to infinity W x t dx that is the total probability of finding the particles anywhere on the positive side at any time t. So, this integral is somewhat one has to do it very carefully. We can write here considering this expression now we have e to the power minus u square t by 4 d term which has nothing to do with space. So, we can take it out and insert the remaining terms inside then it will have the form e to the power minus u square t by 4 d divided by 4 pi dt will stand out and then it will be an integral there will be 2 different integrals. So, first integral will be 0 to infinity e to the power u x minus x naught by 4 d minus x minus x naught whole square by 4 d t over dx and the second integral similarly will be 0 to infinity e to the power u x minus x this is a common thing divided by 4 d into e to the power minus here it will be x plus x naught whole square by 4 d t dx and the bracket will close. So, you have 2 integrals to be evaluated. Now we can see that this both the integrals here there can be converted into a complete Gaussian integrals because here also it is x minus x naught here also it is x minus x naught. So, you can just combine them some residual will be there which will factorize out it will not include Gaussian. So, basically you can write it as a integral of a x square plus b x plus c type. The same thing we carried out here also although it is x plus x naught eventually as a function of x it can be written in a quadratic form and we know the all integrals of the quadratic form are either in terms of error functions or in terms of probability integrals they are called differently or in terms of complementary error functions and these are all related quantities. We as far as possible we will stick to the convention of using error function which we have defined once and we continue to use it. With the little reworking of these integrals one can arrive at the following form of the solutions and it can easily be checked that is half of 1 plus error function of x naught plus ut divided by 4 dt minus e to the power minus u x naught by d into 1 plus again error function of ut minus x naught by 4 dt it closes here. So, carefully if you look at we have an error function whose argument is x naught plus ut divided by 4 dt. So, it is basically a point which is moving along with the average velocity. So, x naught plus ut increases as time increases and there is another function which includes basically minus x naught plus ut. So, it is like a mirror image also moving in the positive direction. But the second term however is modulated by the existence of a time which does not existence of a term which does not depend on time e to the power minus u x naught by d. So, all these have implications and then it is basically the whole thing is expressed as an error function and for completeness here we once again define our error function as 2 by root pi 0 to z e to the power minus let us use the word some some some other variable let us say y square dy where y is just a dummy variable. So, this is this expression for survival probability for the 1D random walker. Let us look at the properties if you put a t equal to 0 s 0 we should get 1 because to begin with the random walker is surviving. So, e to the power x naught by dt and here it will be this term for example, if you put t equal to 0 it will be minus infinity. Here it is plus infinity because see if you put as a t goes to 0 the denominator 1 by root t has a will tend to infinity. So, it will be r infinity is plus 1. So, 1 plus 1 is 2 whereas, this in the second term when you put t equal to 0 it will be minus x naught divided by t. So, it will go as 1 by root t which is r minus infinity r minus infinity is minus 1 error function is an antisymmetric function it is an odd function. So, it is negative of r f of minus x equal to minus of r f x. So, this will become 1 minus 1 0 the second one vanishes the first one becomes 1 plus 1 divided by 2 and hence s 0 becomes 1 as required. Now the interesting point is the limit t equal to infinity when t tends to infinity as we can see and if u is positive then the first term becomes r f of plus infinity. So, 1 remains. So, 1 plus 1 2 divided by 2 becomes 1. So, we have 1 and the second term if u is positive and t goes to infinity this becomes r f of plus infinity again it will be 1 plus 1 equal to 2 divided by 2 that will be 1, but this term will be left. So, 1 minus e to the power minus u x naught by d. So, there is a certain survival probability at infinity if u is greater than 0. Interesting to see what happens when u is less than 0 when u is less than 0 this becomes r f of minus infinity as time goes to infinity and then 1 minus of 1 is 0 same happens with the second term also r f of minus infinity 1 minus 1 is 0 and we get s infinity equal to 0. So, we obtain very interesting bifurcation result starting from a simple one dimensional biased random walker in the presence of an absorber. Two kinds of survival exist depending on whether u is greater than 0 or less than 0. Once when we go back we note that our velocity that we have introduced is actually a measure of the bias of the random walker gamma as we can see from this expression. u being positive means gamma is positive and gamma is p minus q from our earlier derivations that means, p is greater than q which means the walker tends to has a slight excess probability of jumping towards the right as opposed to the left which is equivalent to a steady velocity moving away from the absorber. So, when the random walker has a steady probability of moving away from the absorber then he has a survival probability which is finite 1 minus e to the power minus u x naught by d is that survival probability where u is positive. The closer x naught is to the absorber survival probability will be smaller and smaller because he will be he could be absorbed, but further away he starts from the absorber x naught more and larger and larger the survival probability will be closer to unity. However, when u is negative he is actually effectively moving towards the absorber on an average and is the probability of survival becomes 0. So, this is a very useful result it has actually many practical applications we will discuss it shortly. Thank you.