 We started last time by introducing the idea of what I call persistent diffusion. So we will take this up today, we will look at this problem today and I will point out what its connection to a dichotomous Markov process is and we will see how it differs from ordinary diffusion. So we will get back to the diffusion limit and so on and so forth. It is a very very popular model and a very useful model of continuous random process. So what is the problem? It is as follows and I called it persistent diffusion but let me give its proper name dichotomous diffusion is a better name dichotomous or persistent and I will explain why it is called persistent. We have in mind a particle diffusing along the x axis exactly as in the normal diffusion problem but this time I specify the velocity of the particle to be fixed to be some finite value but capable of reversing in direction. So it either moves to the right or to the left with some fixed velocity C some speed C. So if I call the process x, x dot of t is equal to a dichotomous Markov process in the sense that it takes values plus C or minus C so psi of t and this fellow is a dichotomous Markov process taking values plus C or minus C and the question is what are the statistical properties of the process x itself which is the integral if you like offer dichotomous Markov process. Remember that the normal the usual diffusion problem we had x dot equal to square root of 2d times 8 of t where this fellow here was a Gaussian white noise and therefore this x became a Wiener process and this process was not differentiable anywhere it had all these strange properties it satisfied the probability density function satisfied the normal diffusion equation. But now the question is what happens if I integrate dichotomous noise here with these values what answer would I get what sort of probability density would I get and so on. We can actually sketch this function without too much difficulty because you can see that schematically if this is a dichotomous Markov process what it does is to flip between plus C and minus C at random intervals of time. So this is C and this is minus C this is what the process x dot does or xi of t does and when it is plus C x is uniformly increasing with time linearly and when it is minus C it is decreasing with time in the other direction. So it is a it is immediately clear that if I plotted t versus x of t not in this case but in the other case then if it is going up in time it does that the function of time and then it decreases with speed minus C then goes up again and decreases and so on. So it does something like this it is a lot more smooth looking except for these points of direction reversal it is a lot more smooth looking than a very jagged curve that Brownian motion actually is or a Wiener processes. So in that sense this is expected to be a milder process in a certain sense more handleable more tractable and we would like to see what its statistical properties are okay. Now you can get to this in several ways you can get to the probability density for this guy in several ways but let me just write down so this implies that x of t equal to integral 0 is equal to x of t minus x of 0 is equal to integral 0 to t dt prime xi of t prime. So in some sense you want the integral of a dichotomous process we can find out what the correlation of this is etc etc we will do that shortly but the process itself in a typical realization would look like this in this fashion. So as you can see everywhere there is this basic idea of a Poisson sequence sitting everywhere these points where this thing undergoes reversal are supposed to form an uncorrelated Poisson sequence of points. And we have already studied what a Poisson pulse process does okay we can we know what its correlation is and so on and so forth just as a digression let me remind you of what this whole thing is if you took the several ways of defining a Poisson sequence of points but the way one convenient way is to say the following is to say that on the time axis you have a whole lot of epochs or instance of time which are completely uncorrelated with each other. So this is tj tj plus 1 this is tj minus 1 dot dot dot completely independent of each other and now let us the statement is that in any finite interval of time the probability that you have n of these crosses is Poisson distributed with some mean rate. So if you say for example if you start with the number in 0 to time t n of 0 t that is a random variable the number of such epochs occurring in a given time interval 0 to t is a random variable can take on any integer value any non-negative integer value and one can ask what is the distribution of this n here. So the probability that this is equal to some integer r some non-negative integer r is a Poisson distribution. So this guy is equal to e to the minus lambda t lambda t to the power r over r factorial and r equal to 0, 1, 2 etc this lambda is called the intensity of the process it is the mean rate at which these crosses occur okay and these t's are independent of each other so we can actually define a Poisson sequence in another way which is to say that the mean that the gap between two successive crosses is distributed exponentially the probability that if you have a cross at t equal to 0 the probability that you do not have a cross till time t is goes like e to the minus lambda t that is it right. So several ways of defining this we also defined it yet another way by saying in any interval delta t any infinitesimal interval delta t there are only two possibilities either there is a cross in it with probability lambda delta t or there is no cross in it with probability 1 minus lambda delta t the probability of two or more crosses occurring in a delta t is a higher order in infinitesimal. So the several equivalent ways of handling this yet another way is to say that this is Poisson distributed and different non-overlapping intervals are independent of each other completely independent of each other. So n of 0 t and n of 2 t, 3 t for example they are not overlapping intervals these are independent random variables completely. What would the correlation of this n be what would the autocorrelation of this n of 0 t be so let us let us call this fellow let us leave it like that. So what is n of 0 t prime n of 0 t what would this be let us be very definite let us say 0 less than t prime less than t. What is this expectation value likely to be this autocorrelation do you think it is a function of t minus t prime remember the statistics of these t's the sequence t's this a sequence tj it is a stationary sequence in the sense that the mean rate lambda is independent of time. Whether you look at this interval as 0 to t or whether you look at it as t naught to t naught plus t whatever t naught is it does not make a difference you get exactly the same distribution. So in that sense it is stationary so my question now is do you think n of 0 t prime which is a random variable is independent of n of 0 t if that were true then I can remove this average I can write this average of this product as a product of averages incidentally what is this equal to what is this guy equal to exactly it is lambda t it is lambda t it is a Poisson distribution the average value is lambda t. So of course it says that if the interval increases the number of points in it will increase with the mean rate lambda so it is proportional to lambda t what is the variance of n of 0 comma t also lambda t also lambda t right now let us look at what this fellow is do you think first of all it is a function of t minus t prime first of all do you think that these two are random variables which are independent of each other no because there is an overlap there is certainly an overlap t is bigger than t prime and 0 comma t prime is sitting inside 0 comma t so these are not independent random variables therefore you cannot factor it out immediately and there is a non trivial autocorrelation so one way to do this would be to say let us look at it like this let us look at n of 0 t prime n of t prime to t let us look at this one here now this is just for reference so you have 0 t prime and t on the x axis on the t axis so let us look at this the quantity and ask its expectation value are these independent random variables yes because they are not overlapping so they are definitely independent random variables so this is equal to n of 0 t prime n of t prime t and what is n of 0 t prime lambda t prime and what is this lambda of t minus t prime so this is equal to lambda squared t t prime minus lambda squared t prime squared that is certainly true right but you can also write this as n of 0 comma t times n of 0 t minus n of 0 t prime and then take the average by definition n of t prime t is this difference right so that is equal to sorry n of 0 t prime and then this so this is equal to n of 0 t prime n of 0 t minus n of 0 t prime squared and then the average n squared of this fellow here so I want to find this quantity that is the correlation I want to find auto correlation so I move this to the right hand side and I have shown that this is already equal to that out there so this quantity therefore this implies this is equal to this fellow plus lambda squared t t prime minus lambda squared t prime squared but what is this quantity because we already have this other result we have this result it says variance of n of 0 t equal to lambda t because for a person the variance is the same as the mean right that means that n squared average is lambda t plus the square of the average lambda squared t squared right so if you put that in this is equal to lambda t prime plus lambda squared t prime squared plus lambda squared t t prime minus lambda squared t prime squared right and this cancels this guy cancels so it says this auto correlation is lambda t prime plus lambda squared t t prime not a function of t minus t prime but the average is not 0 here n of 0 t or n of 0, t prime does not have a 0 average so the auto correlation should be defined as the covariance by subtracting out the mean so what we need is the following what we need so let us write these results down so what we have is expectation n of 0 t prime n of 0 t this fellow is lambda t prime plus lambda squared t t prime t but what we need is delta n of 0 t prime delta n of 0 t expectation and this stands by the way for n of 0 t prime minus expectation n of 0 t prime that is the definition of this delta so what is this quantity equal to all you have to do is to take this guy here and subtract out the product of the means right and what is the product of the means lambda t prime times lambda t prime t so this is equal to this lambda t prime okay and we took this to be t prime to be less than t what happens if I took t prime to be bigger than t it will get replaced so in general this guy here is equal to lambda times mean it is not stationary so this n of 0, t I mean it has a correlation this random variable has a correlation so the number of such epochs such instance of time in a Poisson sequence in any interval of time the correlation function of that is given by this and of course when this is equal to that t prime equal to t this is equal to t lambda t as you expect it is a variance that is it okay. Now this t this linearity is at the origin of the linear dependence with t in all these diffusive processes this is what is finally going to finally leads to this whole business of linear dependence it comes straight from here from this Poisson sequence because now if I attach a process to this by saying there is a velocity process which goes either plus c or minus c at these instance of time the statistics of that carries over signatures from this okay and so on but remember that this process here how is it correlated now it is the xi process what is the correlation of xi this is a stationary Markov process definitely a stationary Markov process a jump process and what is this equal to xi of t prime xi of t what is this equal to we have looked at the symmetric case was a plus c and a minus c not c 1 and a c 2 and the other symmetry is that this I said that it is going to reverse with the mean rate which is the same whether it is going from down to up or up to down if it is some lambda then what is the correlation here it is exponential it c squared e to the minus 2 lambda 2 lambda actually more t minus t prime it is stationary it is stationary okay. So you must distinguish between these processes they look very similar to each other they were very closely related to each other but this is in time this is just a set of points in time uncorrelated that is called its own statistics but now if I attach to it a process which goes up or down in this fashion this becomes a Markov process and then it is exponentially correlated in this fashion now we are asking the next question what is the integral of this process going to look like what is this that is our dichotomous diffusion all right and now this x in the x space is the whole of the real axis say the whole of the x axis we would like to know what its statistics are and it is a continuous process this is a jump process but the x itself the integral is a continuous process although it has those kinks whenever the velocity reverses and we are looking at the statistics of that now you could ask can I get this process so let us keep this aside for a moment and we can ask can I get this process from a discrete random walk after all what is being remembered is the velocity namely the direction of motion so can I get this from a linear lattice by doing a random walk on it with the following kinds of rules so once again I have a lattice whose sites are labeled by the integer j this is j plus 1 j minus 1 and I ask for the probability that you are at j at time step n so I take steps of tau time step tau and there is a lattice constant a which I will introduce subsequently but for the moment we just label the lattice points by the integers on an infinite linear lattice and I want to know p of j comma n but with the following proviso just as earlier I said in the case of the DMP I said we would like to remember the velocity whether it is plus c or minus c the analog here would be to say is it moving to the right or is it moving to the left so I really have two different probabilities I have a PR and a PL so it is at this point at time n moving rightwards or moving leftwards and there is a certain rate at which or a certain probability with which it will reverse direction so let us put a bias in it in the following sense instead of saying I toss a coin with probability alpha I move to the right with 1 minus alpha I move to the left if it is a biased coin I now use the same biased coin I say with probability alpha I continue in the same direction as a previous step and with probability 1 minus alpha I reverse my direction then what are the rate equations for this PR of j n must be equal to clearly alpha times PR of j minus 1 at time n minus 1 so I have come here and I am moving right and I continue rightward so I am in the right it is going to contribute to PR plus a beta which is 1 minus alpha PL of j minus 1 n namely I am here at time n minus 1 but now I reverse direction and move towards the right and therefore I am here in the right moving state those are the only two contributions what about PL well this is clearly beta times PR of j plus 1 n minus 1 because I am here at time n minus 1 and I want to contribute to the left moving guy whereas I am moving right there so I am a switch and then move left and the probability is beta right plus alpha times PL of j plus 1 n minus 1 so I am here moving left and I jump here with probability alpha and I continue to move left and therefore I contribute to PR so these are the rate equations and the point is to solve these now for any given alpha between 0 and 1 okay so the solution is not that trivial as you can see the two coupled unknowns here PR and PL but in principle this can be solved this that can be solved what we are interested in doing is taking a continuum limit of it so that we get some diffusion type equations and how would you take this limit well the first thing you do is to say that the lattice constant should go to 0 so let us put j a equal to x and this is the lattice spacing n tau equal to t this is the time step and it is clear yeah by no you are asking what is the contribution to one of these say PR to be in the right moving state at this point at time n right it can only come from here or here because we have allowed only nearest neighbor jumps but then it could have come from different states and these are the only states from which you can feed into this point here that is it okay we are not writing a rate equation we are writing an actual difference equation for finite n right. So we have a time step and we have a space step a and clearly the velocity C so you must write limit 8 tends to 0 tau tends to 0 such that a over tau equal to C that will tell me the speed now what about the rate of reversal well reversal is measured by the probability beta so clearly you also want limit beta tends to 0 tau tends to 0 such that beta over tau equal to new this is the rate of reversals okay or lambda in the case of the dichotomous process the parameter used I will use new because I want to distinguish this from that lambda do not want to confuse it with a dichotomous process all the time so let us call it new so I put those limits in here I put this in here I write j times a n times tau and then subtract from it the n minus 1 part and so on and write differential equations now and then with a little bit of manipulation exactly as we did in the case of the diffusion equation you end up with the following you end up with delta over delta t so let me call this let us use some shorthand notation del t plus C del x these are partial derivatives with respect to x and t of p r of x, t this is a probability density function whereas these are probabilities this is the probability density the positional probability density function at time t to be at the point at the point x in the right moving state okay what is this equal to what can this possibly be equal to now comes the gain and loss terms because we are writing a rate equation new times p l of x, t minus p r of x that is the only possibility remember this guy here is like the total derivative it is like the convective derivative this delta over delta x this is delta over delta t so d over dt of this guy must be equal to this is the gain term that is the loss term okay and the other one has a velocity minus C so it is minus C del x p l of x, t new times now this is p r those are the two equations rate equations and they are coupled they couple to each other so we need to solve these equations and these are the solutions these are the equations for dichotomous diffusion the total probability p of x, t where you have integrated or summed over the both the velocity states plus C and minus C that when you integrate over x must be equal to 1 at all times right so we want this condition normalization condition integral minus infinity to infinity dx p of x, t to be 1 for all t greater than 0 where this quantity by definition is p r of x, t plus p l that should be normalized what are the initial conditions well this depends on what conditions you would like to specify so let us you do our usual kind of specification we say that the particle starts at x equal to 0 at t equal to 0 right now you could put whatever initial conditions you like but the most symmetric ones would be to say it starts from 0 and it starts with equal probability moving right or left you do not care right. So the initial conditions this would give us symmetric nice symmetric solutions incident you can solve this for any you can solve this set of equations for suitable initial conditions and boundary conditions could be quite general but let us look at the simplest cases initial conditions would be p r of x 0 equal to p l of x 0 equal to delta of x half of the half a delta function okay half delta of x half delta of x half in the left state half in the right equally proper right. So when you add the two you get a delta of x and you integrate it you get one but we need conditions on this because these are equations which involve derivatives you need conditions on delta deltas as well right because you see what you can do to solve these equations is to eliminate one of them you can eliminate p r or p l and get another equation for it right. So let us do that and you see what the equation looks like so by the way this guy here starts looking suspiciously like the wave equation is going to emerge right it looks like one factor of this wave equation. So what should I do I want to eliminate p l so the thing to do is to I know what this operator on p l is. So let us put this operate on both sides of this equation here. So delt minus C del x delt plus C del x p r equal to nu times this guy on p l that is this guy which is this so there is a nu squared p r minus nu squared p l that takes care of this fellow here and then I have to take this operator and act on this. So that is equal to minus nu delt minus C del x on p r now what does this give you that tells you the second derivative partial derivative and then there is a delt C times delt del x and a del x del t but it does not matter which order you do this and those cross terms cancel and then you have C squared with the minus sign del x x second derivative not the first derivative squared but second derivative of the derivative. So it says delt t minus C squared del x x of p r equal to nu squared p r minus nu del t p r plus nu C del x p r minus nu squared p l now what should I do for that p l what should I do to get rid of that p l I want an equation to eliminate p l I use this I use this equation for p l right. So I have a formula for nu times p l so let me write this to be equal to minus nu times whatever is there for nu p l and that is equal to delt plus C del x p r plus nu p r and if there is any justice in the world this says nu squared p r cancels against this then a nu C del x p r cancels this fellow cancels against this and you get this 2 nu del t p r. So we finally get an equation it says delt t minus plus plus 2 nu del t minus C squared del x x p r equal to 0 you get a second order hyperbolic equation now this looks like the wave equation except it has got that reversal term and nu what would be the equation for p l p l differs from p r in my very symmetric thing the C becomes a minus C in that state right but this equation here is square in C. So what would you expect is the equation for p l same equation I expect exactly and therefore for p and therefore for this guy too I expect them to obey the same equation how do then differentiate this between these solutions but the initial condition is the same for both of them the initial condition is the same that looks like it says p r equal to p l that is not true that is not true right I agree with you that this is going to be p r comma l equal to 0 exactly the same equation pardon me why does it give me two solutions no once I solve that equation I get solution is unique once I specify the initial conditions what is the order of the differential equation in time it is second order so for either p r or for p l I need one more initial condition on the derivative the time derivative right I need to know what is p r dot at 0 how do I find that how will I find that how will I find that initial condition I have I have all the data here I have this here feed it into the differential equation feed it into the differential equation at t equal to 0 after all the differential equation should start be satisfied even at t equal to 0 right so feed it in and what do you get you get p r dot of x comma 0 plus C del x and this is half delta of x and each of these is half delta of x so at t equal to 0 they cancel so what is p r dot at x comma 0 move this to the right hand side so minus C over 2 delta prime of x why not I define I am not defining the square of a delta function you find it is derivative it is a very singular object yes but I define it by integration by parts always after all how do I define an integral of delta function I select the value at the delta function fires for a derivative I do integration by parts on the test function I think of it physically what is a delta function it is a function which looks like this goes up very sharply and comes down in the limit what is the derivative of this function look like this is an even function so the derivative has got to be an odd function the slope at the origin is 0 for this so the derivative vanishes at the origin right and what does it look like well the slope is increasing very rapidly going to 0 and then decreasing very rapidly and then coming back to 0 so this function looks like this very sharp what would the second derivative look like goes up and down a few more times and then goes to 0 so it is getting more and more singular I agree but it is defined it is defined after all one way of defining a delta function or any distribution is through it is Fourier transform so the inverse Fourier transform of a constant is a delta function and differentiation is equal to multiplication by K so the inverse Fourier transform of I K is the delta prime apart from some sign and then I K squared is the second derivative and so on so I can define those things so it is minus C over 2 delta prime X on this side and what is PL dot of X, 0 what is this equal to plus C over 2 so I expect the solutions to be different PR and PL because the initial conditions are different but they both obey this equation and the initial condition on P of XT itself is even more different right so P of X, 0 equal to delta of X P dot of X, 0 equal to 0 it is not going anywhere with these two add up so these are three different solutions to the same differential equation now you are faced with a thing like this what would you do next it is not the wave equation because there is this guy sitting here in between now what is the effect of this new what is it physically doing actually what is this new comes from the dichotomous Markov process the velocity is reversing all the time right so it is preventing this guy from becoming ballistic motion every time it is moving fast this is reversed it is not getting anywhere right so it is acting like some kind of dissipation in that sense okay so you would expect that if I get rid of that dissipative factor then you would have something like the wave equation possibly now that can be easily arranged incidentally does it start looking like the diffusion equation at all is it like the diffusion equation ever well if you get rid of this it is exactly like the diffusion equation so we keep that in the back of our mind it looks like if you let C tending to infinity which is very reasonable because I remember I said that the formal velocity of a Brownian particle is actually infinite and C came from A over tau but the diffusion constant in the diffusion equation came from the limit A squared over tau A over tau is infinite and A squared you put in one more A which goes to 0 and then you get a finite limit right so it is very reasonable that C tending to infinity and nu tending to infinity such that C squared over 2 nu tends to diffusion constant D gives you the diffusion equation back again so we should keep this in the back of our mind that the moment you let this limit you take this limit in the solution to that equation we should be able to recover the diffusion equation solution the Gaussian solution but right now the solution to this the envelope P is R and L these are not like the Gaussian at all but the central limit theorem still operates and you will in the long time limit you should be able to get back the diffusion equation so we will come back we will come back to this meanwhile since we have a guess that this nu is creating this problem this term here and there is dissipation in the problem so let us set P R of X t equal to e to the minus nu t some Phi R of X t and do the same thing for L there has got to be some damping out so let us remove that and then see what happens right now this is not hard to show plug that in that this term will get cancelled out this goes away but you will have terms which come from the second derivative of acting on this guy here and that pulls down a factor nu squared so this is a matter of algebra all you have to do is to put this in here and see what happens to this then let us write the answer down and then this will imply that del t t minus c squared del x x minus nu squared ends up with that it does not change the initial conditions because as you can see this this guy here you can put in find out what are the new initial conditions etc write it down what does this equation remind you of it is not the wave equation but it is a Klein Gordon equation it is exactly like the Klein Gordon this is like the box operator in 3 2 3 plus 1 dimensions is a wave operator and then there is a constant square so if you can solve the Klein Gordon equation you can write the solutions for this you have to specify moment to give me anything this operator acting on some function of X t equal to 0 and if you tell me what the function does at t equal to 0 and its first derivative does at t equal to 0 arbitrary functions of X I can write the explicit solution but we need it in this very special case these things here let me write the solution down the solutions look like this so let us define sorry to use this variable but it is a standard symbol defines I to be equal to square root of c squared t squared minus x you expect that factor to appear all the time because this is this operator has appeared so this is going to appear now we immediately can see intuitively that since the velocity is never going is always the speed is always c it is clear that if you start at t equal to 0 at x equal to 0 you are never going to go beyond the point c t in time t similarly you are never going to go to the left of the point minus c t unlike the normal diffusion equation where the velocity was infinite and you started with a delta function at the origin at t equal to 0 and it immediately spread out into a Gaussian which had an infinite extent for arbitrary small t very narrow Gaussian it is true but still it was arbitrarily extended here that is not the case the envelope the diffusion envelope will end at plus minus c t so the graph will actually look like this so you start with the delta function here but then as a function of x for a fixed t p of x, t p r and p l will have a symmetry in the following sense it is not hard to see that p r of minus x t equal to p l of x t you have that symmetry at every instant of time with those symmetric initial conditions then it is cut off between minus c t and plus c t and what you really have is in p r you have a delta function spike in p l you have a delta function spike and in between you have some function which do it as this kind of thing which we will write down so the solution it turns out looks like this let me write the solution down the half remains except this delta function which was half of delta of x at t equal to 0 is really delta of x minus c t as you would expect it propagates this is the ballistic part propagates right that is the contribution which comes from no reversal of sign just keeps going and then there is a portion which has the rest of it which is plus let me see if I remember this right an envelope which is theta of x plus c t minus theta of x minus c t this restricts you to this region here that is a square pulse of unit height which tells you that this solution is restricted to that region times times nu over 4 c of i naught of nu xi over c what you need is a dimensionless quantity here and because of the symmetry of the differential operator remember that xi was defined to be equal to c squared t squared minus x squared to the power of half positive square root of that this has dimensions of t inverse that has dimensions of length so it is a velocity and it cancels here plus c t minus x over xi i 1 of nu xi over c these are modified Bessel functions which we have seen many are very often appear very often in random of problems we know what the properties are their entire functions of their arguments and so on okay and similarly P L of x t is one half delta of x plus c t that is this delta function here plus the same thing theta of x plus c t x minus c t nu over 4 c i naught of c plus c t plus x over xi these are the solutions which satisfy the initial conditions on the two on the two functions PR and PR and of course the whole thing is a piece is some of and the whole thing is multiplied so e to the nu t times that e to the nu t times that because there is that damping factor sitting there all the time so what happens when you sum the two is that this fellow becomes nu over 2 c times i naught but then this x and that minus x cancel out and then you have 2 c t the c cancels here and you have nu t over c etc etc okay and you are guaranteed due to the presence of these quantities these theta functions here that it is restricted to this region here and of course P of x comma t is a symmetric function of x because it does not distinguish between left and right okay now we can check if it will go to the correct diffusion limit or not that is not very hard to do but before that we need to also find out does this diffuse what is the mean square displacement going to look like etc we do not know what the mean square displacement looks like we need to compute that explicitly so we need to compute x squared average but it is sensible to compute x squared average by not by solving the differential equation but I am going to leave this as an exercise by solving a much simpler equation you know what each of them does PR and PL and what P does we know the differential equation for P right so all you have to do is to argue that x squared of t by the way what is the average value of x of t this is equal to integral minus infinity to infinity dx x squared P of x comma t by definition and it is obvious that in this case this fellow vanishes outside plus or minus ct on either side so the integral runs from minus ct to plus ct times some vessel functions it is too messy to do it that way what you should do is to say I start with the differential equation so I have del t t minus c squared del xx plus 2 nu del t P equal to 0 multiply by x squared and integrate from minus infinity to infinity or from minus ct to plus ct integrate this equation here get rid of these by integrating by parts and then you will get a differential equation for x squared average as a function of t and initial value is 0 right so you can compute what the actual average is explicitly so I leave it to show that this quantity turns out to be c squared over 2 nu squared t minus 1 plus e to the minus 2 write a differential equation for this guy in time ordinary differential equation and solve it with appropriate boundary conditions to get this it is linear in t at long times what would you expect it will do at as t tends to 0 it should be ballistic it should go like c squared t squared finished and indeed it does because you can see that the for one cancels the nu t cancels the next term is of order nu squared t squared the nu squared cancels and you have c squared t squared over 2 factorial the 2 cancels and so on so it is precisely c squared t squared for very small t for very large t what do you expect what is meant by large t nu t much much bigger than 1 right the velocity correlation time what is the velocity correlation time in this problem in the original diffusion problem there was a gamma there and I said the velocity correlation time was gamma inverse what is the velocity correlation time here because we know this is a dichotomous Markov process we know it is exponentially correlated what is the velocity correlation time 2 nu inverse 1 over 2 nu so when t is much much bigger than nu nu t much much bigger than 1 can drop this you can drop this you are left with this at c squared over this goes away so this goes to c squared remember in the diffusion limit we said that the d the diffusion constant was c squared over 2 nu so this says that it is equal to 2 dt which is exactly what the diffusion limit is so this thing here what do you think happens to it when t becomes very very large we really saying that mod x is much much less than c t or t becomes very very large t becomes much much larger than mod x over c right so the condition the diffusion limit is that t much much bigger than mod x over c but when c becomes infinite this is simply saying that t is positive that is what we are looking at in any case right what happens to this profile it moves further and further right but it is multiplied by e to the minus nu t because this guy goes to the right hand side so after a longer time it likes like this finally in the diffusion limit this goes away and you end up with a Gaussian you are back to that and we should be able to see that explicitly so let us see we can see that explicitly you see you look at so you agree that if I write pr plus pl and I put the e to the minus nu t on the right hand side and let nu t tend to infinity the e to the minus nu t the delta function goes away goes to 0 height and I have only got to worry about these guys so let us see what that does I have to make t very very large in it so I have an e to the minus nu t multiplied by I naught or I 1 does not matter and what is it going to look like look at this function here for instance look at I naught what does the argument t is becoming very large that is the only thing that is getting very large so Xi is getting very large because I square root of c square t square etc and what happens to I naught when the argument becomes very large let us write this out properly I naught of z goes when mod z becomes very large e to the z over square root of 2 pi z in fact I j of z does that for all positive integer j non-negative integer j and I minus j is the same as I j what does I do as t becomes very large c t you take it out and then you have 1 minus x squared over c squared t squared to the power of half so it is roughly this guy here right sorry yeah so this is equal to c t minus x squared over 2 c t this is equal to so Xi goes to this for very very large c so it says nu Xi over c becomes nu t nu t so our solution looks like e to the minus nu t and then e to this guy so it is e to the power nu t minus nu x squared over 2 c squared t divided by square root of whatever it is this is the leading term as t becomes very large so apart from some constants it goes like this this cancels against that and you are back to the Gaussian solution e to the minus nu x squared over 2 c squared c but remember that c squared over 2 nu was the diffusion constant so this is twice the diffusion constant here and there is another 2 here so it is 4 dt so this guy really is going to e to the minus x squared over 4 dt over square root of apart from some constants okay so that is why I said that that envelope becomes a Gaussian goes to the Gaussian so we have a very fairly complicated process but the fact remains that it finally at the end of the day reduces to the diffusion equation in the diffusion limit in a well defined limit right but it is a much smoother process than Brownian motion is much more tractable and so on in particular we can now answer the following question we have this thing going up and down we have it is a sample space it is sample parts look like this and then the same slope over again and we really have when we found pr and pl we have actually found the joint distribution in position and velocity because the velocity is either plus or minus c so we have really found p of x comma v comma t and once we have done that and it is not a Gaussian some crazy Bessel functions right once we have done that we know that the variances of these guys are finite at any time t so we can find what is the rate of level crossing we can ask if I put a threshold here how often does it cross what is the mean rate in particular how often does it cross 0 the origin itself and all those formulas we derive for level crossing can be applied immediately this right so do that and we will write down the solution next time so find let us take a simpler case let us take the simplest instance so find the average value of the number of times it crosses the threshold 0 in a time interval 0 to t number of 0 crossings so if you look at PR it is going to give you the up crossings and PL will give you the down crossings and you can compute it explicitly compute what is the exact contribution in this case I put x equal to 0 as a threshold because it simplifies the formulas because you can see that if you put x equal to 0 this xi becomes just CT and then it is much easier to handle this case so use those level crossing formulas to discover what this is and then we would be able to say on the average how many crossings will there be in a time interval t how will it go with t what is your guess so the question I am asking is if this was a 0 if this was a 0 level in x here as a function of t the question asked is in a time interval 0 to capital T how many on the average crossings are there of this up and down both together it will of course be an increasing function of capital T the question is how is it going to increase what would be your guess will go like t will go like e to the power t will go like square root of t what would you think so work this out work this out of course as t becomes very large it makes a huge difference what power of t is a crossing but it is explicit here you can actually get an explicit formula and we want the large t limit find this thing for as a matter of fact you can even find the variance of this number that is not really very much harder to do even compute the variance of this number although that is a little more intricate we did not talk about it in level crossing but you can compute even the variance in this case so I want you to find out what is the power law increases like a power law not clearly not exponential but increases like a power of t and the question is what power is it going to be so that is a useful thing to know because whenever you model something by like a Thomas diffusive process of this kind by the way this random walk is only a paradigm I mean this x is not really does not have to be positioned could be anything could be a large number of things just like a Poisson sequence will modify or model all kinds of things models the rate at which telephone calls arrive rate at which people arrive in a public place and it is a very large number of possible applications similarly here this process this dichotom integral of the dichotomous process it is also the continuous process models a large number of random phenomena and it would be interesting to find out what are the threshold crossings what is it so do this and then we will take it from here and I want to emphasize again that P x in this case is not a Markov process definitely not a Markov process on the other hand x, v together is a Markov process okay so it is jointly a Markov process and we alone is a Markov process it is a dichotomous Markov process okay. So this is exactly as it was in the case of ordinary diffusion as well but there x was also the integral of white noise was also a Markov process in this case the integral of a dichotomous Markov noise process is not a Markov process but together the two former vector process definitely is alright so let me stop here by the way I forgot to mention that the dichotomous Markov process is also called a random telegraph signal with some modifications of rates and things like that and these two equations we wrote down the coupled equations for PR and PL they called the telegraphers equations because they also appear in the equations for old transmission lines we are not interested in that aspect of it here but you might see in the literature the telegraphers equations the equation with the two new del T in between that is a telegraphers equation okay alright so let me stop here today.