 So, in the previous lecture, we obtained a general solution to the Chapman-Colmogorov equation as P carrot k t as e to the power some unknown function lambda k t, where all that we know is lambda 0 is 0 given lambda 0 equal to 0. Another property we have is property 2 lambda k is an even function of k. Let us write property 2 lambda k is even function. This emerges because we started with a symmetric random walk problem starting from the origin itself. Since it is a symmetric random walk, the probability of moving to the left or to the right should be exactly equal. So, once P x t is a symmetric, when we do it to the power i k x transform, basically it will be only the cosine kind of transform will survive the sin transform will go away. Once it is cosine transform k x comes as a product. So, the solution should be also symmetric with respect to k, if it is symmetric with respect to x. So, since P x t is a symmetric in x unbiased problem because of the unbiased case, it is a Fourier transform P tilde also should be symmetric. It is important. I argued why it should be same thing can very easily see that this should be so. Third, the property 3 is let us focus or let us try to explore the long time or the large displacement behavior of the probability density function. So, from the perspective of predicting the behavior at large x for predicting large x behavior which is of interest generally of the probability density function, we should focus on small k behavior of small k behavior P tilde k t. Why this statement? Because k and x are called conjugate variables, they always come in the product. When you take the Fourier transform, it is always e to the power i k x. So, the large x behavior and the small k behavior, they are connected. The contribution to the answer at large x will come from small k, then only the k x product will have some finite value, then only you will get an answer. Therefore, the small k behavior of the Fourier transform will be controlled by the correspondingly the small k behavior of lambda k. Hence, small k behavior of lambda k is important, important for predicting the large x behavior that is the idea. Thus, we assume this lambda k equal to 0 should be 0. Now, as we are considering the case of symmetric random walk lambda k should be an even function of k, even function of k. This is because we know that P x should be an even function of x for symmetric random walk, this is for symmetric random walk and its conjugate variable always multiplies x and hence it should be symmetric in k also. Now, we focus on the behavior of lambda k near k equal to 0, lambda 0 is 0. So, what is interesting would be to see how lambda k is expected to behave as k approaches 0. This is because our interest is primarily in knowing how the distribution behaves for large x. So, to understand or to understand the behavior of the distribution for large x, for say large x, let us say mod x tends to infinity it is necessary to understand the behavior of lambda k as k approaches 0. So, now that we are considering symmetric random walk and it should be an even function and it should approach to 0 the most likely way it would happen is that it should go as either k to the power 0 or k square or etcetera, if we want to consider only integer values of powers. But however, there is another way that it can approach 0 that is as a power of a modulus of k. So, we assume that lambda k has a behavior of the type mod k to the power alpha some power alpha where alpha is a real number k approaches 0 mod k ensures evenness and alpha should be greater than 0 that ensures that it is not a divergent quantity it will approach 0 lambda k will approach 0 as k goes to 0. At the same time we want the integrals of Fourier inversion to be finite or to exist which means convergence is necessary that is lambda k should be negative let us say some constant b mod k to the power alpha. So, we offer an option in which b is now greater than 0 alpha is greater than 0 that means, lambda k will always remain negative as k approaches 0. So, with this let us examine how the inversion integral of the probability distribution behaves. So, in a sense with this assumption we have the Fourier transform of the distribution function in the k variable k t is e to the power minus b t mod k to the power alpha where b is greater than 0 and alpha is also greater than 0. So, this is what we have derived and then we are now substituting basically to the function lambda k this form. Now, if we execute the inversion which by definition is p x t equal to 1 by 2 pi minus infinity to infinity e to the power minus i k x p carried k t d k is a definition Fourier inversion. So, this we can write it as upon substituting 1 by 2 pi minus infinity to infinity e to the power minus i k x minus b mod k to the power alpha t d k. As usual b is we should always remember b is greater than 0 and alpha is greater than 0 the real part at least. Now since we expect p x t to the even function symmetric random walk the e to the power i k x the sin part of that should should vanish because sin is an odd function at the rest is all even and hence this can be written as 1 by 2 pi minus infinity to infinity cos k x e to the power minus b mod k to the power alpha t d k. Now in order to get rid of that negative side where mod k you have to remember to ensure the sin change is taken care of we convert this integral to one sided integral that is possible because both the cos k x as well as mod k to the power alpha e 1. So, this is going to be 0 to infinity if I mean if we convert it to 0 to infinity then it will be twice of 1 by 2 pi. So, it will be 2 by 2 pi then it will be 0 to infinity cos k x e to the power minus b. Now you can forget the mod part because it is any only positive side t d k. So, here this 2 and 2 will cancel. So, we will be left essentially with 1 by pi. So, it is 1 by pi 0 to infinity e to the power minus b k to the power alpha t cos k x d k. Unfortunately there is no general solution available for this for all alpha greater than 0. Even numerical integration is somewhat formidable because of the highly oscillatory nature of the cos function. It is possible today, but it is it does not give you a general feature you will be able to get some plot of the behavior, but we want to understand the behavior of the distribution function for large x that is our aim. So, for that we do a small transformation. Let us define to obtain large x behavior we are focusing on large x behavior. Let us do some transformation that is we define a variable eta equal to b t by x to the power alpha. So, and also we put k x equal to u k x equal to u which means d k the integration variable d k will be d u by x x is now a parameter of the system. So, with this the integral you can easily see b x t first we will write down and confirm that it is correct pi x 0 to infinity e to the power eta is now a new parameter which is a function of x and t eta u to the power alpha cos u du. So, all that we have done here in arriving at this is in arriving at this is we in this k x equal to u we have put. So, d k is d u by x and then this k therefore, will have that 1 by x to the power alpha etcetera which we have taken as eta. So, that is that will then come to this integral with this definition and we always remember that eta is a similarity variable of the problem. It includes the time as well as space in a combined manner very often we would see many solving many differential equations similarity transformation is quite useful technique. So, in a way we are combining or variables into a single variable. So, this is something like a similarity variable. Now, let me just give you a general information that this integral has some analytical solution only for alpha equal to half of course, for alpha equal to 1 it is analytically solvable there is no that we will see for alpha equal to 0 of course, it is a trivial cosine function integral for alpha equal to half is one fractional case where it is solvable and that solution is special case. This is a special case because we have assumed mod k to the power alpha with the alpha as a parameter to be explored what kind of alphas are allowed. So, let us consider alpha equal to half case where the solution is given in terms of slightly complicated functions I will not solve for you this is available in books it will have eta by mod x root 8 pi into cos eta square by 4 1 minus twice there is something called a Fresnel integral Fresnel cos and Fresnel sin Fresnel c eta by root 2 pi plus sin eta square by 4 1 minus twice Fresnel s integral eta by root 2 pi square bracket closes where Fresnel function is Fresnel c of z is equal to 0 to z cos pi t square by 2 dt and Fresnel s of z is 0 to z sin pi t square by 2 dt. So, all functions are defined here and this is the one analytical solution just for information as an information that it is possible to solve some complex integrals and they exist. The way this distribution for alpha equal to half it would look is like this supposing I now plot as a function of x and this is p x t then for various t's for example, it will be a it will have form like this there is a little cusp like behavior, but it is not really cusp because we are talking of alpha equal to half case and it will be sharper and sharper and with the larger time this is a shorter bt this is a these are of course, not to scale. So, this is t small. So, let us say t equal to some arbitrary time say t 1 and this is t greater than t 1 where t 1 is an arbitrary time. So, as time progresses it becomes flatter and flatter this is the space and there is a symmetry along both sides and interestingly this distribution tends to have a behavior of x you can actually because it is analytical you can take the limit as x tends to infinity this shows that p x t will have behavior 1 by x to the power 3 by 2. We are not putting all the pre factors, but functional form why we are interested in the functional form is this supposing the function actually goes as 1 by x to the power some number n and if this n is less than 2 then this function is integrable its area under the curve exists, but its moments do not exist it does not have first it does not have second moment higher moments. To have at least a second moment this should have a behavior of x 1 by x to the power 3 or more. In fact, one more than 1 by x to the power 3 then only it will have a at least a second moment. So, we see that for alpha equal to half case obviously, the this is a kind of a distribution which does not have a second moment will not satisfy the central limit theorem will not go into a Gaussian, but will have a tail of a long tail falling slowly as 1 by x to the power 3 by 2 that is why. So, alpha equal to half case. So, this is alpha equal to half case clearly illustrates the existence of power loss to the tail which will not support the moments higher moments. So, of course, although this is one analytically soluble case we can obtain the analytical behavior for general alpha. This is by certain asymptotic expansion methods. So, behavior for general alpha greater than 0 of course. So, for this we visit back and note that the integral that we have written here for a general alpha you know that cos u is a very rapidly oscillating function when u approaches upper limit infinity. And as x becomes infinity eta tends to become 0 which means the behavior of the function when eta becomes 0 is what matters which means the behavior at a large u and small eta is what is important. As a first approximation we assume that cos u is the one which essentially governs the behavior of the integral. So, we give a Taylor expansion to this under the assumption of small eta. So, with that we can expand we expand e to the power minus eta u to the power alpha as equal to 1 minus eta u to the power alpha plus higher order terms. Under the assumption that for eta tends to 0 that is x tends to infinity. And let us keep it in mind our eta is nothing, but some coefficient b into time divided by x to the power alpha. So, as x tends to infinity eta goes to 0. So, with this the integral now takes the form p x t equal to 1 by pi x the first term will be just cosine integral 0 to infinity cos cos u du cos u du. And the second one will be minus of eta 0 to infinity u to the power alpha cos u du and so on. Let us see the dominant behavior for x tends to infinity will be the first two terms plus h o t of course. Now integral of cos u du will be sin u and at u equal to 0 sin u is 0. And because of the oscillatory nature of the sin function we assume that at infinity also its value will be 0 and hence we set this integral as 0. So, that the leading term p x t we expect it to approach minus of eta by pi x integral 0 to infinity u to the power alpha cos u du as x tends to infinity. Where we always remember that eta has a reciprocal dependence on alpha alpha is greater than 0. Now therefore, the problem is reduced to evaluating this integral u to the power alpha cos u du again this is a formidable integral is formidable because carefully if you see cos u is a very highly oscillating function. And its value always will be lying between plus 1 minus 1 and u to the power alpha for positive alpha will be slowly diverging as u tends to infinity which means we are going to sum positive and negative values of ever increasing magnitude as we approach infinity and we are not sure of adding and subtracting infinities rapidly whether we will lead to a finite result or not. So, these kind of integrals are understood in the limit of certain convergent integrals these are called integrability criteria. So, one such criteria I would like to tell you is supposing I define this integral i which you want to know right now u to the power alpha cos u du and I take it as a limit supposing such a thing is possible limit some epsilon tends to 0 0 to infinity e to the power minus epsilon u then u to the power alpha cos u du and this integral that is the integral before you take the limit of epsilon equal to 0 there is no worry here because e to the power minus epsilon u will ensure convergence no matter how large of value of u you go then for any finite epsilon e to the power minus epsilon u will tend to 0 which means this integral will exist and it does. So, before we take the limit you can actually write cos u as for example, e to the power i u plus e to the power minus i u etcetera actually this will become a gamma function integral you have to make sure there is a imaginary part also, but because epsilon is real here epsilon is going from 0 positive 0 plus it exists. So, the I will do I will only take the value of this integral from the book that says any handbook of integrals one of the most widely used handbook of integrals is that of grad stain and Resick contains all integrable all integrations which can be executed finite definite integrals and others also. So, from that you can see that this integral has the value gamma of 1 plus alpha divided by 1 plus epsilon square to the power 1 plus alpha by 2 cos of 1 plus alpha tan inverse 1 by epsilon this is the tan inverse term is the most critical function. Now, if you take that is equal to limit epsilon tends to 0 plus if we take the limit we can see that when epsilon go tends to 0 this will go to 0 and here tan inverse of 1 by infinity will be pi by 2. So, the limit now nicely exists thereby implying that the I integral exists. So, I integral therefore, will have the form gamma 1 plus alpha and 1 plus epsilon square to the power 1 plus alpha by 2 will be just 1. So, it will be cosine of 1 plus alpha pi by 2. So, which means we have our answer P x t minus eta by pi x gamma 1 plus alpha cosine of 1 plus alpha pi by 2. We can expand this cos function into a simpler form because this minus sign is slightly irksome we do not want negative concentrations or negative probabilities. So, but then that is will be cancelled by the cos when you expand cos pi by 2 plus pi by 2 alpha is basically is minus sign will come there. So, if you do that expansion and replace eta with all that. So, you will get b t by pi x to the power 1 plus alpha 1 plus alpha gamma of 1 plus alpha sin of pi by 2 alpha. So, we have an expression now for the asymptotic behavior this is valid for x tends to infinity and of course, alpha should be greater than 0 and this is what we want to understand what values of alpha allowed. So, one thing is obvious P x t has a behavior of 1 by x to the power 1 plus alpha providing its coefficient exists providing 1 the providing its pre factor is positive non 0. So, gamma 1 plus alpha for positive alpha has no problem for any alpha it is a positive quantity and exists. Let us look at sin pi alpha by 2 if you see sin pi alpha by 2 the plot we see that when alpha is 1 it is pi by 2 this is alpha equal to 1 and when alpha equal to 2 it is exactly 0 because it is sin pi. So, here we are plotting sin pi alpha by 2. So, we notice that for higher values of alpha more than 2 if you now extrapolate it will be negative for alpha greater than 2 negative. Thus we see that the allowed values of this alpha parameter is lying between 0 to 2 only probably what exclude 2 the theory is not valid for alpha equal to 2 because solution will be 0. So, hence the allowed value of alpha is 0 0 is also not cannot be equal. So, alpha less than 2 and not equal to alpha equal to half we saw in a analytical solution and this also tells you when alpha equal to half it will go as 1 x to the power 1 plus half x to the power 3 by 2 reciprocal behavior. So, that is fine and for alpha greater than 2 the unphysical solutions will emerge that is the negative concentration negative probabilities will come. So, we have to discard that the only region where alpha is allowed is lying between 0 to 2. So, we have general possibility some kind of a summary that is the domain 0 less than alpha less than 2 this is actually very interesting conclusion from this study. These flights where alpha is less than 2 are called levy flights they distributions emerging from levy flights have integrable they are integrable, but they do not have moments higher moments. So, as a result they are the their exceptions to a Gaussian eventual transition to a Gaussian distribution. So, we complete this study for example, by considering a case of alpha equal to 1 another special case case 2 let us say if alpha equal to 1 it is a case of the Laplace transform k t having just mod k behavior minus b mod k k to the power alpha we assumed. So, this is the form and we have studied this case earlier also while discussing exceptions to central limit theorem and this is called as the Cauchy's distribution Cauchy distribution or in physics literature it is called Lorenzian or Lorenzian. We can easily see that the distribution function evolves in the way we have written it then it will be e to the power minus 0 to infinity eta u cos u d u is a nicely integrable function you can try to causes e to the power i u plus e to the power minus i u it is just an exponential and you can complete the exercise and actually finally, its answer will be 1 by pi b t divided by x square plus b square t square. So, this is the Cauchy distribution well known this is a one example which we knew earlier, but from the method that we adopted as a general solution to the random work problem in the continuous time domain we established a family of solutions that is achievement of a for various alpha. The next case of practical interest is of course, the classical case we call it as case 3 which corresponds to k equal to 2 the alpha equal to 2. So, if alpha equal to 2 it is a basically a Gaussian fluctuation now because it is quadratic eta will take the form b t by x square and our p x t will is we have done it earlier it will be 0 to infinity e to the power minus eta u square cos u d u it is a Fourier transform of a Gaussian and Fourier transform of the Gaussian also remains Gaussian and its final answer you can easily obtain as 1 by square root of 4 pi d t e to the power minus x square by 4 d t where b is a parameter and we identify d as the diffusion coefficient d equal to diffusion coefficient. So, this is the well known Gaussian solution. So, to summarize our present study we have the following summary to summarize we have for alpha less than 1 we have cusps this is power law with the cusp this is for alpha more than 0 less than 1 for alpha equal to 1 we have the Lorentzian or Cauchy this is for alpha equal to 1 this is Cauchy and for alpha equal to 2 for between 1 and 2 the behavior will be this is for alpha lying between 1 to 2 it will still be cusp but well let us make it not let us look like this it will have a slightly sharp faults but not Gaussian still a power law for of course, alpha equal to 2 it will have this is alpha equal to 2 and for alpha greater than 2 it will have perhaps negative values let us draw it for alpha greater than 2 it may exhibit negative concentration negative probabilities hence ruled out. Thank you.