 Hello students. In the previous lecture, we had set out to solve the problem of Brownian motion in the presence of an absorber. We formulated the Fokker-Planck equation including a drift term. This is one level of complexity higher than the free particle Brownian motion in the presence of an absorber which we read earlier than that. To repeat the purpose of solving these equations which are basically the same problems done in biased random walk problem is basically to not to arrive at any dramatically different solutions, but to learn the methods of solving differential formulations of Fokker-Planck equation. We will get more or less the same results. In fact, the asymptotic results will turn out to be exactly the same. However, the approach of solving the Fokker-Planck equation teaches us to handle certain linear differential equations. Advantages we can then apply it somewhat more complex problems, more time dependent problems which cannot be done by the which cannot be easily done by the random walk method. So, accordingly we had set up the equation. We made some transformations of from the W function to the Y function and that is written here. The original equation as you can see includes a drift term nu is a dimensionless velocity scaled with respect to true velocity by diffusion coefficient. So, that has that formulation and then you have scaling of tau as a measure of time. It does it as a dimension of time it is actually tau equal to d into t. So, notations become simpler, then we transformed it to get rid of the first order term by defining the quantity y equal to w e to the power minus nu x by 2. So, then y becomes an equation continues to remain a second order, but the first order term gets removed instead we get a term which is proportional to y itself. So, it is more like a decay term with the boundary conditions which are slightly changed now delta x minus x naught becomes e to the power minus nu x naught by 2. Now how do we proceed further? So, this second order equation with the is basically in the domain x greater than 0. So, we can specify this entire solution is valid for all x greater than 0. So, we note that this is not like a free particle case the motion is in the finite domain. If it is an infinite domain we could have directly applied Fourier transforms to the x variable because we know that this Fourier transform of transform of the second derivative becomes an algebraic function. However, we cannot apply Fourier transforms because the function does not exist in the domain x less than 0. However, there is a good fortune the good fortune is that there is a another transform very closely related to Fourier transform. In fact, there are two such transforms known as the Fourier sine and Fourier cosine transforms. Let us talk of Fourier sine transform. Fourier sine transform in the convention we use we can define for example, if f t is a function or f x is a function then we can say f carrot denote by s for sine of let us say variable k will be simply defined as 0 to infinity sine k x e to the power of sine k x f x dx. So, it is just multiply the function by sine k x and then integrate and this is valid of course, our function f x is defined f x defined for x greater than 0. So, once a function is defined only on the positive interval and this transform also refers to only that interval. This can be derived actually by through the Fourier transform only by defining a function which is identically 0 on the negative side and all. We do not go into the derivation let us assume that this transformation law exists. Only thing we should know is how do we invert it? So, we can come back to f x say inverse sine transform inverse inverse Fourier sine transform that is simply 1 by 2 pi here it comes to 2 by pi. Since it is half domain the inversion comes to 2 by pi 0 to infinity again sine k x and here it is now f carrot sine transform that we adopted k dk. So, this is both forward as well as the inverse is simply multiply by sine k x first time you integrate over x and the second time you integrate over k and get back to the original. We can prove that this is completely consistent in fact, the double integrals turn out to be delta function and give us the results. The advantage of using a sine transform and correspondingly there is a cosine transform also are this. If your boundary condition of the function at x equal to 0 is 0 the case that we are having here then sine transform guarantees it automatically when we invert it. As you can see when we are multiplying by sine k x and inverting automatically f x is guaranteed to be 0 at x equal to 0 which is what we want. One can show that if your boundary condition at x equal to 0 has its derivative 0 then cosine transform way more appropriate. So, here since our boundary condition is 0 at x equal to 0 we decide to use the sine transform. So, accordingly we multiply equation the differential equation here is differential equation. We multiply this differential equation by sine k x and integrate over x and note that the Fourier transform of is Fourier sine transform of let us note the following result note Fourier sine transform of d 2 of d 2 y by dx square x to k x to k is simply minus k square y carat sine transform with respect to k. So, if you make use of this rule the differential equation here just reduces to the following. We get for the transformed equation carat s time derivative remains the same equal to minus you get k square here and there was already a nu square by 4 and that remains as such. So, it will be carat s k. Now we can we can denote these arguments as k tau it is a function of 2 variables here also it is not t we decided to use tau where tau equal to d into t. What happens to the boundary conditions? Now the second order differential equation as partial has become almost an ordinary type differential equation that to first order that is because when we did the Fourier sine transform we have to do the Fourier sine transform of the boundary condition also. So, you can see that the y carat sine transform of k at t equal to 0 or tau equal to 0 will be multiply 0 to infinity where rule is that we will multiply sine k x to the initial condition if you go back the initial condition is delta x minus x naught e to the power minus nu x naught by 2. So, that will be delta x minus x naught e to the power nu x or nu x naught by 2 dx is the definition. So, it simply will become sine k x naught e to the power minus nu x naught by 2 because of the selection property of the delta function it the value of the multiplying functions at the point x equal to x naught is realized. So, you get this. So, this being a first order equation has a very simple solution. Solution is y carat k tau sine transform will be we can take out the other terms e to the power minus nu x naught by 2 sine k x naught e to the power minus nu square by 4 plus k square tau. How we wrote this? Let us go back. So, this is the differential equation this is a dy by dt equal to some constant into y whose solution is y equal to some a unknown into e to the power minus that constant into time. So, that is what it is and unknown is nothing, but the initial value ys k 0 which is sine k x naught e to the power minus nu x naught by 2. So, when we plug all that in we will get the solution the initial value multiplied by e to the power minus nu square by 4 plus k square into tau. So, the formally the problem is completely solved in the domain of the transformed space. Now, we have to go back to the original solution. Hence by inversion we have y x tau will be 2 by pi 0 to infinity again sine k x now if the integration is going to be with respect to k. So, all the rest will wherever k comes we will keep it out. So, we will keep it as a sine k x naught e to the power minus k square tau and the other terms will stay out. So, we will write them as minus nu x naught by 2 minus nu square tau by 4 just stay out of the integral. Here of course, we should not forget to put d k the integration is with respect to k. So, we can write by taking out all other terms as 2 by tau 2 by pi e to the power minus nu x naught by 2 minus nu square by 4 tau. Now, to save a step we realize that sine a sine b can be expressed as a cosine function. Sine a sine b is cos of a plus b minus cos of a minus b divided by 2. So, half I can take out first and the whole thing can be written as cos of k x minus x naught minus cos of k x plus x naught e to the power minus k square tau d k. So, this now each of this integrals cos k x e to the power minus k square tau d k is actually a Fourier transform once again of the Gaussian function. We have done it several times over just for recalling we just use the result here which can be proved by recall that integral 0 to infinity cos k z some z e to the power minus k square tau d k is going to be half root pi by root tau e to the power minus z square by 4 tau. You can actually convert it into a full integral from minus infinity to infinity since cos is a even function and k square is an even function it is a totally an even function. So, you can make it actually a complete Fourier transform and then we will go get this result. As a result if we apply this rule using the above we get y x tau we have to note the various numerals that come. So, while the half there is a 2 by pi and there is going to be 1 more half. So, we are going to get 1 by if we put together all of them then e to the power minus nu x naught by 2 minus nu square tau by 4 and since these are 2 different Gaussian integrals here in the first one the z is nothing, but x minus x naught. So, we are going to have e to the power x minus x naught square by 4 tau and the second one e to the power x plus x naught whole square by 4 tau. We can see that the 2 terms emerge in the case of an absorber problem 2 Gaussian terms not a single Gaussian term. If we go back to our lectures L 27 and L 28 we had discussed the absorber problem by using the random walk picture by the method of images where we introduced a reflective a mirror or image walker on the other side of the absorber and subtracted trajectories which would be otherwise killed because of the absorption point absorption behavior. Here we get those 2 terms e to the power minus x minus x naught square by 4 tau and e to the power minus x plus x naught square by 4 tau apparently without using any image principle by just as a mathematical solution to a differential equation. However, certainly speaking both are equivalent in some way when we decided to impose the absorption boundary condition and follow through the sin Fourier transform we have implicitly assumed an equivalent of an image absorber they are mathematically equivalent. That is why we will get exactly the same result as we got earlier, but in the former case we had to make a asymptotic approximation using the Stirling's approximation from discrete to a continuous case. So, accordingly we the solution why now that we have can be converted to the full solution w and that solution is going to be upon using upon going back to the w variable and using the transformation between y and w we will get this whole thing as square root of 4 pi tau e to the power minus e to the power plus let us say e to the power minus nu x minus x naught by 2 minus nu square tau by 4 e to the power minus x minus x naught whole square by 4 tau minus e to the power minus x plus x naught square by 4 tau. You can check by inspection that at x equal to 0 automatically the solution becomes 0 because these 2 terms cancel each other x naught square and x naught square. So, this is this solution is the same same as the asymptotic solution obtained in lecture 28 we might have to appropriately identify nu with gamma etcetera. There we had at a bias factor called gamma and here that bias is represented by the velocity and velocity is nothing, but is related to nu because nu equal to u by d. So, there is a connection. We can proceed in the same manner to obtain now further results w is the full distribution. So, we can work with it to obtain all other quantities of interest. The first quantity of interest is the survival probability. What is the probability that the random walker continues to survive up to time tau. So, that survival probability for example, s tau would be the probability that the walker has not been absorbed that is his occupancy exists. So, the occupancy probability wherever he is along the positive line the existence is guaranteed by the value of w and integrated over x will give you the survival probability. And you can integrate this Gaussials and we have we have seen this seen this earlier it comes out to be half into 1 plus r f of error function of nu tau plus x naught divided by square root of 4 tau close minus e to the power nu x naught here it will be 1 plus r f of nu tau minus x naught divided by square root of 4 tau. In deriving this we are skipping this result we have done these integrals before in the previous lecture. In any case they are Gaussian integrals and the integrals of Gaussians turn out to be some error functions. So, it is only that we have to choose the correct representation. So, just for your reminder let us define where the error function of any quantity z is defined as 2 by root pi integral 0 to z e to the power minus u square d. So, this is the definition of the error function. From this we can ask the question asymptotically is there a chance of survival at all is it possible that the random walker or the diffuser or a Brownian particle is not absorbed in the boundary. Because there is a drift sort of perception tells us that if the drift is towards away from the wall not towards the drift is away from the wall there may be a possibility of absorption. But if the drift is towards the wall there is no chance because even without the drift the Brownian particle is going to be absorbed eventually. So, with the drift towards the wall chance of survival is 0. Hence, accordingly just like earlier we have the result that S infinity probability that the walker survives is 1 minus e to the power minus nu x naught if nu greater than 0 this is 0 if nu less than 0. So, if nu is greater than 0 the further the walker is from the further the walker starts from the absorber higher will his probability of survival. If he starts from very very far then his survival chance is almost 1 whereas, if he starts closer to the absorber his survival chance is going to be less and less and will eventually will increase only as he moves away from the absorber. So, this brings us to the end of the one dimensional absorber problem one dimensional single absorber problem. We can even calculate the distribution of contact probabilities the f function we called it as the first contact time distribution function all that was done previously and it shows it t to the power minus 3 by 2 behavior and all that. Just to summarize this this talk. So, you would sometimes wonder what is it that we achieved by deriving everything from differential equation approach when we already knew these results by the discrete jump method by the random walk method. The solution we are presenting is only an illustration of the methods such as Fourier transforms Fourier trying to sign transforms and many others which we are going to do. Supposing I have a problem in which the random walker of course, he has a diffusion he has also let us say a drift and on top of it let us say he has a decay property that he can disappear randomly in the course of his motion. You can imagine that the my random walker is a radioactive particle which as it is walking can decay. Now, I want to capture this particle before it decays onto a surface it has an advantage for me. Then I can genuinely ask a question what is the probability that ultimately the walker will be absorbed onto my surface before it decays. Now, this is somewhat difficult to do by random walk methods, but much easier to do by the differential equation methods. I do not propose to solve that problem for you because this is now a very area specific formulation. But let me assure you that the differential equation method that we proceeded same can be applied wherever the term new square comes the new decay terms get added. So, we will see that the ultimate solution will have almost the same form as what we got, but with the decay correction included. The ultimate result of the whole thing is there could be a non finite there will be no asymptotic survival of course, particle has to decay, but here you can ask the question of the probability of being absorbed onto the surface and that would be quite an interesting question to ask and answers out of this study could be quite interesting. From this point we move over to a description of Brownian motion based on velocity fluctuations. This will help us establish a mechanistic framework of positional fluctuations based on what is known as Langevin dynamics. Thank you.