 Now we saw some properties of the logistic map last time and we also saw that there was an intricate sequence of bifurcations followed by the onset of chaos interrupted I mentioned by periodic windows and finally fully developed chaos. Now one of the interesting questions that arises immediately is when you have a chaotic attractor how frequently two different parts of the attractor get visited by a typical initial condition by a trajectory and is there some kind of stationary distribution or invariant probability distribution which tells you how these attractors are occupied by a typical trajectory. This is a question of great interest because as you know once you have the property of ergodicity then time averages which are not computable in general can be replaced by averages over some invariant distributions probability distribution so we need to compute these distributions. Now this the history of distributions is very very long goes back a long way and just to recall to you even in statistical physics in ordinary equilibrium statistical physics the primary task of the theory is to give you the invariant distribution. Now once you have this distribution then you can compute averages physical averages using the property of ergodicity. Let me recall to you what happens in thermodynamics and equilibrium statistical physics for a minute if you have a system a macroscopic system isolated from the surroundings in thermodynamic equilibrium then we know that the energy of the system is constant it doesn't change at all and you could ask what's the probability with which different states of the phase space different portions of the phase space of the system get occupied and this is called the micro canonical ensemble in equilibrium statistical mechanics and you could ask what's the probability distribution on the micro canonical ensemble and that's easy to see because it says if the system is described by a Hamiltonian q, p then for an isolated system in thermodynamic equilibrium the energy is constant. So this minus the constant value of the energy is equal to 0 and if I insist on describing my system by a probability density in phase space in the space of the p's and q's that density rho as a function of all the q's and p's is equal to something which ensures that this Hamiltonian always have that numerical value. So it's essentially a Dirac delta function at that point on the energy surface. So it is this thing that's the equilibrium or stationary distribution in a micro canonical ensemble. So this is for micro the same system if it's put in thermal equilibrium with a heat path at some temperature T absolute temperature T then we know that the ensemble we use is the canonical ensemble it's called the canonical ensemble and that's the one in which we do normal usual thermodynamics because normal physical systems cannot don't generally remain isolated they are in interaction with their surroundings and they are in general closed systems where the system can exchange energy with its surroundings with a heat bath but not matter in such a case we have the canonical ensemble and the corresponding density operator rho of q, p is something which ensures that the value e of the energy if e is one of the possible values of the energy of the system the relative probability with which the energy e is taken is equal to the exponential of minus e over kT where k is Boltzmann's constant this is the standard rule you have in the canonical ensemble. So the relative probability is e to the minus the energy over kT then the absolute probability is the same number divided by a normalization constant which is the sum over all values of e so it would look something like e to the power minus beta h of q, p whatever value this takes divided by a sum over all possible values or an integral in this case e to the minus beta h of q, p dq, dp this is an integration over all the degrees of freedom and the corresponding generalized momenta and it serves to normalize this probability density. This is the thing which you normally call the Boltzmann factor because you might have come across this in the form e to the minus e over kBT I have used the symbol beta for one over k Boltzmann times T if this is the relative probability that the energy is e if you sum over all possible values of e the denominator that gives you the normalization constant and it is called the partition function in equilibrium thermodynamics statistical mechanics. If you have other physical conditions such as an open system which can exchange both energy and matter with its surroundings then you use not the canonical ensemble but something called the grand canonical ensemble and the grand canonical ensemble again is of this form the density the probability density or the phase space density the equilibrium density is of the form e to the minus beta times h minus chemical potential times the number of particles in the system and that is the grand canonical ensemble depending on what experimental conditions you are interested in you have different ensembles which specify for each of them you specify what the invariant measure is what the probability distribution is now of course we also know that in classical mechanics this probability distribution in general for a classical dynamical system obeys an equation called the Liouville equation which is not quite the equation of motion for a dynamical observable recall that if a of q,p is a dynamical observable we found out that the rate of change of this was equal to the Poisson bracket of this a,h we found that this was the case long ago on the other hand the density itself rho of q,p is equal to the Poisson bracket of h with rho it has an extra minus sign now if the system is stationary in other words the density is completely stationary steady state then this cannot change with time so the equation was this this was the more equation of motion for the density and if it is completely stationary this quantity has to be 0 if this is 0 it means that the Poisson bracket of h with rho is 0 identically pardon me the additional minus sign ensures that first of all it says that this rho is a probability density and is not an observable this these are observables this minus sign ensures that the average values of physical quantities are independent of which picture you use to describe them either you use an active picture in which the dynamical variables vary with time and the density does not or you use a passive picture in which the dynamical variables do not change with time but the density probability density changes with time and these are called the Heisenberg and Schrodinger pictures respectively they are related to each other they transform transformations of each other but yeah exact this is exactly the point by insisting that averages of physical quantities remain independent or be the same whatever picture I choose to describe this evolution in so it is a little bit like to give an analogy it is like saying that if you have a particle moving in space then two ways of doing this either you say the position of the particle changes in space and the axis remain fixed or I could equivalently say the particles really stays where it is but it is the axis that are changing in space and both these things should give you exactly the same results they just to two transformed ways of looking at the same thing one of them is an active picture and the other is a passive picture in quantum mechanics we would say one of them is the Heisenberg picture and the other is the Schrodinger picture and physical answer should be independent of which picture you choose that imposes this equation on the density matrix on the density operator itself now yeah by insisting on this namely the time average of any physical quantity now I take the average value of this quantity and when I do that then it turns out that if I write the formula down for the average value and look at its time derivative then depending on whether I take the average value of a and let us suppress the q and p for a moment if I take an active picture and say this is the average value of this physical quantity which changes with time with respect to a stationary distribution row this must be exactly the same as a picture in which the a doesn't have explicit time dependence doesn't have time dependence at all but it's with respect to a row which depends on time and when I insist that this be the case then automatically it says consistency demands that row itself obey an equation with an extra minus sign here and this follows from the cyclic invariance of the trace I'll demonstrate this after a little while okay so eventually the point I was trying to get across here is that since delta row over delta t equal to this if if the distribution is to be a stationary distribution then this should be equal to 0 and that implies that the Poisson bracket of row with the stationary distribution or the equilibrium distribution let me call it stationary be equal to 0 identically and what does that imply what does this imply for this distribution this is a function of all the phase space variables and this is the Hamiltonian of the system and you're saying the Poisson bracket of this should be identically equal to 0 so what's the most general solution you can think of what condition does it put on row stationary it's a constant of the motion but it's not a physical observable so what condition does this put on row if I say that the Poisson bracket of row with H row stationary with H must be identically 0 in general it implies that this row stationary should be some function of H that's the most general thing you can think of under all conditions if this is to be true this must be a function of that H and that's exactly what has happened as you can see the stationary or equilibrium distributions row equilibrium are functions of H but what functions is not specified by this equation here that is specified by your experimental conditions whether the system is isolated or the system is in close system or not and that's why invariably the in equilibrium statistical mechanics the probability density in phase space is a functional of the Hamiltonian of a system and not of anything else and that follows from here now what can we learn from this for our problem well in our problem unlike equilibrium statistical mechanics where this is all you can say you can say that this must be some functional of this but it doesn't say what functional at all that has to be put in from outside and the fundamental postulate of the equilibrium statistical mechanics I recall to you says that all accessible microstates of a system in thermodynamic equilibrium and isolated system are equally probable and that's this measure so it says on this energy hyper surface you have a uniform measure it's completely ergodic and you have a uniform measure because this immediately tells you that this measure this delta function says there's only one constraint that H be equal to E it doesn't preferentially say this portion of the energy hyper surface is better than that portion or anything like that on the other hand when you have a canonical ensemble when the system is in equilibrium with a heat path then it says this thing it's called the Gibbs measure this factor the Boltzmann factor appears here and this is the particular functional of H so in the case of statistical mechanics you need extra inputs it's not yet fully derivable from dynamics at all and that tells you what the form of this invariant or equilibrium distribution is but in our case we have some dynamical systems specified by set of equations and everything must follow from those equations there's no external input possible anymore so where does that leave us well let's see how to find invariant measures and this is our next task so let me call it the invariant measure or the invariant density again to make things specific we work with a one-dimensional map which is chaotic same map of some unit interval or something like that and then the equation the map is the following Xn plus 1 is equal to some non-linear function of Xn and Xn is an element of some interval that's the sort of situation we have in mind and now let's assume that you are in a region of the parameters where you have chaotic motion completely and this is the interval this is the attractor and you would like to find is there an invariant measure or not how do we do this well if I start with a point X in this interval and I iterate once then it goes to a point which is f of X therefore if I start with some initial point Y and I ask what is the density row after one time step as a function of X what is this equal to well clearly if I start with a point Y it goes to a point f of Y and therefore the support of row 1 of X is a delta function of X minus f of Y if Y is my starting point so this is clearly true what happens if I start with a distribution in Y what happens if I start with a distribution which is row 0 of Y a given Y goes to an X which is given by X minus f of Y therefore a distribution or a probability density in Y row 0 of Y each Y goes to its corresponding X according to this rule and if I sum over all possible Y's I end up with row 1 of X so different Y's would go to different X's but each Y goes to an X which is given by this delta function condition because it's completely deterministic to give me the Y and I know exactly which X it goes to and now the probability density in X is just the integral of the probability density in Y weighted with this kernel so it's an integral equation for row 1 of X given after one iteration a density row 1 of X what is row 2 of X what's row 2 of X going to be equal to well once again you take the density after one iteration and then you again weight it with the corresponding delta function kernel and you get row 2 of X so this is again equal to I dy delta of X minus f of Y but here you put row 1 of Y because everything that I want here is a function of X and I keep integrating this well if you like you can write this out in several ways you could write this out as I dy delta of X minus f of Y times an integral over I del some dz delta of Y minus f of Z row not of Z you could do this so you could start with Z that goes to Y then it goes to X so certainly this is also true you could also have written this as equal to I dy delta of X minus f 2 of Y row not of Y so I could have said well you have to have two iterations now in time and the function that takes you from a given Y to a given X after two iterations is f 2 of Y so if this is your kernel then you integrate the original density with this delta function kernel and you have the density after two iterations so these are more or less obvious statements so therefore in generalize this and say if you give me the density after n iterations then the density in the n plus 1th iteration is given by this which you could also write if you like you could write this as f n out here and then put a row 0 here but now suppose you have an invariant density in other words under iteration it does not change at all that would correspond to taking the limit n going to infinity in this n goes to infinity in this so what would that lead to so let us call limit n goes to infinity row n of X equal to row of X and let us call this the invariant density that is like your equilibrium or stationary density in the case of statistical mechanics but now we have a dynamical equation here like the Liouville equation we have a dynamical equation here and you need to take the limit n going to infinity in this and what would be the equation obeyed by this invariant density if it exists yeah so all we have to see is to see what happens to this equation when n tends to infinity so if you call that row of X that is it because as n tends to infinity so does n plus 1 so the invariant density obeys this equation what kind of equation is this is it a linear equation this is a horrible nonlinear function but now remember the equation we are talking about is an equation for this function so is it linear yes indeed of course because it is just the first power of it so it is a linear is it a homogenous equation or in a pardon me this is horrible this is some function this is some complicated some K of X comma Y in row that is the unknown that is the unknown quantity this is unknown right we supposed to be given this you give me the map function and you know this so it is very much a linear equation is it homogenous or inhomogeneous where is the inhomogeneous term here it is an integral equation definitely it is not a differential equation it is an integral equation but is it homogenous or inhomogeneous in other words by multiply row it by a constant would this still be true yes of course so it is a linear homogenous is it a differential equation or an integral equation or a difference equation it is an integral equation it is a very much an integral equation. Is it an Eigen value equation? In some function space, this is some function x defined over some interval. Yes indeed, it is an operator. It is an integral operator acting on this row and that gives you this row again. Therefore, it is an Eigen value equation. It is an Eigen value equation for an integral operator with kernel k of x, y. Is this kernel smooth or is it singular? It is singular because it is a delta function. So it is a singular integral equation. It is not an easy thing. It has got a singular kernel. If it is an Eigen value equation, then in some abstract notation, we have an equation of the form row. Some vector in some function space in the language of linear vectors is equal to some abstract operator k, an integral operator acting on row. That is the way this equation structure is. So it is very much an Eigen value equation. Operator on this Eigen vector is equal to the Eigen vector. What is the Eigen value? 1 in this case plus 1. So you want to solve a singular linear homogeneous integral equation such that the solution corresponds to Eigen value plus 1 for this integral operator with a delta function kernel. Is there any other condition on row based on the physical requirement that this should be a probability density? It should be normalizable. Can it be negative? It is a density. Therefore, it could be unbounded. It does not have to be between 0 and 1, but certainly it must be normalizable. So you have extra conditions on it. So we also require row of x to be greater than equal to 0 for all x in the interval. And you also would like to have integral dx row of x equal to 1. This is the normalization. So if this solution exists, that depends on whether this what sort of map function you have, then you have got yourself an invariant measure. But you have to satisfy these conditions. This kernel may not have an Eigen function which satisfies these properties. It may well not have one. But if it does, then we are in business. Provided we know that it is the only solution for sure. So you also need uniqueness because we need to be sure that this is the only solution. There is only one solution which is physically acceptable in this case. So it is not an altogether trivial matter even in this one dimensional map. You have a complicated situation. You really have to solve an integral equation. It is homogenous. So it is overall constant, overall multiple cannot be found from the equation itself. But that is fixed by this normalization condition. This will fix what the constant multiplying this row is, the overall normalization constant. It should be non-negative, could be unbounded. We do not care. It should be integrable so that this condition is satisfied. And it should correspond to an Eigen vector in function space of that singular kernel with Eigen value plus 1. There could be other Eigen vectors corresponding to other Eigen values. So the spectrum of this operator could be very complicated indeed. But the physically acceptable solution is one which satisfies these conditions. And there are theorems which tell you that if such a solution exists then it is unique. That is too intricate for me to prove. So I am not going to do that. But let me take it as an assertion here. Let me make this assertion that if a solution to such an equation exists which satisfies these conditions then it is unique. So if you grant me that let's try and see what happens in the simplest cases we know of like the Bernoulli shift for instance. So let's apply it right away to the Bernoulli shift and see where it takes us. So we need this equation. We retain that and let's apply it to the Bernoulli map. And recalled in this case the map function f of x was equal to 2x modulo 1 and the interval i was 0 to 1. So to write this out this was equal to what? To write it out explicitly it was equal to 2y for y less than a half between 0 and a half and it was equal to 2y minus 1 for half less than y less than unity. And the map was discontinuous at the point half. So to refresh your memory this is what the map function looked like. So what f of y was equal to. So let's apply it and find out what rho of x is. So rho of x therefore is equal to an integral over this function from 0 to 1. So there are 2 delta functions here one of which will be applicable as long as y is less than a half and the other between a half and one. So let's write them out separately. So it is 0 to a half dy a delta function of x minus 2y rho of y plus an integral from a half to 1 dy delta function of x minus 2y plus 1 rho of y. In order to do this integral we got to convert this to a delta function over y. So we have to write y minus something or the other and then make sure that this delta function actually fires when y is in this range. So it's these lines on which the delta function fires and what we have done is to break up the integration from 0 to this and remember the constraint is that this quantity is equal to x whatever be x and x runs from 0 to 1. So what would this become? I have delta of x minus a function of y but we use this formula for the delta function of a function of the argument of the integrand and if you recall a delta function of a function phi of y can be written down. First of all what is the delta function? There are certain properties of the Dirac delta function which we need for this business and I am not sure if you recall them right away but let's write them down. First of all delta of a constant times y minus some y naught is equal to 1 over modulus a delta of y minus y naught. Modulus is some real constant a and I need a modulus in the denominator. So this makes sure that the normalization is kept correct. We also know that delta of y minus y naught is equal to delta of y naught minus y. The delta function is an even function of its argument about the point where it fires. We also know that delta function of a phi of y where phi of y is a nice function of y. This contributes whenever phi of y vanishes because that's where a delta function contributes. So suppose y sub i is a root of phi of y equal to 0. So phi of y i equal to 0 and let's label the roots as y 1, y 2, y 3 etc. Then this fires whenever you have such a root but near one of these roots if it's a simple root then it's clear that phi of y could be written as phi of y i which is equal to 0 by definition plus y minus y i times phi prime at y i but that's like a constant. You bring that down so this is equal to modulus phi prime at y i and you have to sum over all the roots. So these are the properties of the delta function that we need here and it's very simple in this case because by symmetry I write this immediately as delta of 2 y minus x and then I take out the 2. So this becomes equal to one half because I take out the 2 here and then after that I can do the integral because this becomes delta of y minus x over 2 and you've got to weight it with rho of y and integrate over all values of y from 0 to half and please notice that going back to this graph whatever be the value of x between 0 and 1 when y is between 0 and a half there is a contribution from this delta function. It does fire at one unique point and fires at a point which is given by x over 2. Therefore this integral contributes for all x between 0 and 1 that's important to remember. So you have to draw a picture of this kind to find out where these delta functions contribute and in this case this is the picture we have to look at. So as a function of y this is a half this is 0 this is 1 this is 1 you have x on this axis and wherever x is between 0 and 1 as long as y is between 0 and a half there is a point where this graph is intersected and therefore there is a point where the delta function contributes. Therefore I can write this first integral down immediately as rho of x over 2 plus once again I write this as delta of 2 y minus 1 plus x and I bring out the 2 which comes out here and this again becomes rho of 1 plus x over 2 again as before for every y between a half and 1 no matter what the x is between 0 and 1 this delta function does contribute is one unique point where it contributes for every x and therefore for all x in this equation x element of 0 to 1 you have this equation we have therefore converted this equation here for the invariant measure from an integral equation to what kind of equation what kind of equation is that it's a recursive equation but it's not a difference equation it's for a function in which the arguments are different it's not a differential equation it's not a difference equation it's a functional equation this sort of thing is called a functional equation this integral equation here is very important this thing here for the invariant density this is called this is the Frobenius Perron equation for the invariant measure it's an integral equation with a singular kernel and we've reduced it in the case of this one dimensional map to this functional equation and for every x this must be true doesn't look like a very straight forward equation to solve because functional equations are not easy to solve there are no standard algorithms in most cases to solve these equations at all there are some techniques but they are very intricate and in general there are no systematic algorithms as you have for the case of integral equations or differential equations or recursion relations difference equations or integral differential equations and so on functional equations are difficult to solve but we have this uniqueness criterion that I mentioned earlier as a very big help in other words if you can find a solution which satisfies this functional equation then it's got to be the unique solution provided it's normalizable and provided it's non-negative now of course one can immediately see from this that if you put x equal to 0 then it says rho of 0 must be equal to rho of a half then you put x equal to 1 and it says rho of half must be equal to rho of 1 and then you put values in between and you discover that all those rows must be equal to each other so what would be a guess it's a constant right and what kind of guess constant should it be it should be normalized 0 to 1 rho of x dx must be equal to 1 1 itself 1 itself so if rho of x is a constant equal to 1 then it says 1 is equal to 1 plus 1 divided by 2 which is true right so in this case the unique solution to this density is in fact a constant so the unique normalizable normalized non-negative solution rho of x equal to 1 0 less than equal to x less than equal to kind of establishes why I said that in the case of the Bernoulli shift the system is ergodic and the invariant density is unity in other words all irrational points on this line their iterates will uniformly and densely fill up the entire interval uniformly with a very simple invariant measure just a constant the moment you have that under your belt then you ask what about the other maps that we looked at for instance let's look at the tent map at fully developed chaos and see what happens in that case so for the tent map at r equal to I use 2r so this must be equal to 1 tent map at r equal to 1 recall that this map itself was equal to 2rx for 0 less than equal to x less than equal to a half and it was 2 2r into 1 minus x for a half less than equal to x less than equal to 1 pardon me yeah it should be 1 is anticipated as so it should be 1 it turns out to be exactly equal to 1 once again we can check this out right away this is what the map looks like in this case right and this function here this f of y equal to when r is equal to 1 it's 2 minus 2 y and you put that equal to x so what happens what kind of functional equation do you get this portion remains the same what happens here so you don't get this so if 2 minus 2 y equal to x on this side right so what do you get so what's y equal to what do what do I get here plus rho of what yeah 2 minus x over 2 which is 1 minus x over 2 again so it says rho of x is equal to this quantity I put x equal to 0 and it says rho of 0 is equal to rho of 1 when I put x equal to 1 and it says rho of half equal to rho of 1 and so on yeah yeah there is sort of yeah because after all what I did was to take this map going up that way I cut and pasted it in one case and I folded it in the other case in some sense you expect the distributions would not change at all and indeed that's true the difference of course is that in one case you have a discontinuous map in the other case you have a continuous map so it's continuous here although its slope is not defined at this point really doesn't matter and you end up with once again rho equal to 1 again by inspection I put rho equal to 1 and it satisfies this equation on both sides so for both these maps so for the this is the 10th map the symmetric 10th map at fully developed chaos so in both these cases I have rho of x is equal to 1 for the Bernoulli map for the Bernoulli map or the Bernoulli shift not true for the 10th map for r less than to less than 1 between half and 1 that's not true because then it's not the unit interval at all it's much more complicated than that and the invariant measure is not findable by analytic means find it numerically because the equation is not easy to solve in that case but here for this fully developed cases for these onto maps this is certainly true now what's the great advantage of finding the invariant measure well any physical quantity whose average you want time average you want can now be replaced by an ensemble average so all we have to do once you have an invariant measure is to say that if I have the iterate xn and I take any function of it here and I'd like to compute the time average of this quantity then I would write the following phi of x the time average of this observable is equal to limit n tends to infinity 1 over n a summation from j equal to 0 to n minus 1 it's the phi of xj where xj plus 1 equal to f of xj that's the rule for iteration I iterate each time I get a new variable each time I get a new value of the argument and I put that in here and I sum over this take the arithmetic average and I've got a time but if the system is ergodic on some interval I can replace this by an ensemble average weighted with the invariant measure therefore this can also be written as equal to and integral over the interval dx rho of x where this is the invariant measure or the invariant density rather times phi of x itself if I know rho of x this is just an integral now what was our definition of the Lyapunov exponent remember the Lyapunov exponent lambda was defined in these cases as equal to limit n tending to infinity 1 over n a summation from j equal to 0 to n minus 1 the log of the modulus of f prime right so instead of iterating the map and finding the slope at all the iterates and taking the modulus and then the log and summing over the whole thing you can write down now where god is it you can also write this down as equal to an integral over the interval dx invariant density rho of x log of the local stretch factor which is log of mod f prime of x so you give me the map and I compute this number for you and what is it for the Bernoulli map log 2 because rho of x is 1 yeah we implicitly use this we actually implicitly use this although in that case it was trivial because I know this thing is independent of xj it was piecewise linear with the constant slope so I this was rigorously equal to log 2 times 0 to n minus 1 of 1 which was n and it cancelled on both sides and you ended up with log 2 explicitly but now it is consistent with the fact that rho of x is equal to 1 so this immediately implies that rho of x is equal to 1 and lambda equal to log 2 independent of the starting point so the system is uniformly hyperbolic everywhere there is a stretch factor which is essentially log 2 the log of the stretch factor is essentially the Lyapunov exponent and it is uniform everywhere it is log 2 in this case we are not going to be able to say the same thing for the logistic map at fully developed chaos because for one thing this is no longer constant so you certainly cannot do this and we do not know what the invariant density is we need to discover what the invariant density is for the logistic map before we can assert that the Lyapunov exponent is log 2 it turns out to be log 2 for fully developed chaos I mentioned this last time but then it is not obvious unless you know what to put in here and this quantity here of course is x dependent because the map is 4 times x times 1 minus x and its slope changes from 4 at the 2 end points right to 0 in the middle and you have to do this integral in order to compute what the Lyapunov exponent is but before that you need to know what rho of x itself is but what is the Frobenes param equation in that case for the logistic map what you have to do is to take this function this is f of y and this is y and we know that f of y in this case equal to 4 y times 1 minus y you must set that equal to x and solve now for y to find the 2 roots of this equation and certainly there are 2 roots because for every given x there are 2 values of y and one of them is y 1 of x and the other is y 2 of x at every level x so you need to find that and then this equation changes and becomes a functional equation and it looks like rho of x is equal to delta of sorry equal to rho of the delta function fires at 2 points now in each case so for a given x you have a y 1 of x which is this root and you have a y 2 of x which is that root and therefore the density is evaluated at y 1 of x divided by the slope f prime at y 1 of x modulus plus rho of y 2 of x divided by the modulus of the slope of the map at the point y 2 of x where this is y 1 and this is y 2 or the left branch and the right branch of the inverse function you have to compute this by solving y 1 and y 2 are the roots of this equal to x and it is fairly messy it is a functional equation which is fairly messy because you have some complicated function of x here which involves square roots and so on because this is a quadratic function of y and you solve it there are 2 roots which you have to write down explicitly so let us do that so it says 4 y squared take it on the right hand side minus 4 y plus x equal to 0 so y 1, 2 equal to plus or minus equal to 4 plus or minus square root of 16 minus 16 x divided by 8 which we can simplify so this becomes 1 plus or minus the square root of 1 minus x divided by 2 that is y 1 and y 2 as functions of x you have to plug that in here put in the values of the slopes of the map at those points and then you have a functional equation whose solution is by no means obvious but it is immediately clear that because of these guys sitting here that a constant is not a solution you put x rho of x equal to 1 it is not a solution to this equation so guess work does not seem to work here so what does one do no I am not no I am not going to find a row there is no reason why it should be locked to yes we know I said that it is locked to so now you want to work backwards but it is not so easy guessing solutions to this because as you can see immediately you are going to have something which is rho of and look at it it is not trivial 1 plus root 1 minus x over 2 when that is horrendous rho of 1 minus root 1 minus x over 2 and then there are these factors as well so it is not by any means trivial this itself involves square roots and so on by the way since it is modulus you expect both of them will come out because it is clear that the map is symmetric so the magnitude of the slope here is the same as the magnitude of the slope here so this factor will come out but it is a function of x and constant is not a solution so what we need is a clever trick in this case this clever trick was found by von Neumann and Ulam long ago in 1947 and they solved for the invariant measure invariant density of this map and this turns out to be a very well known solution turns out to be very interesting solution also with many many ramifications but the idea is the following and let me give you a hint as to how one goes about solving this problem at fully developed chaos at parameter value 4 this map is solvable in the sense same sense in which the Bernoulli map was also solvable we could write the solution xn equal to 2 to the n x 0 modulo 1 that is formally in exactly the same way if you have something which says xn plus 1 equal to 4 xn into 1 minus xn and xn is between 0 and 1 so it suggests very strongly it is like sign of something or the other but the sign runs from minus 1 to 1 perhaps the sign squared would be a good idea so suppose I put xn equal to sign squared theta sub n what happens to this map so it says sign squared theta n plus 1 equal to 4 sign squared theta n cos squared theta n but that is equal to sign squared 2 theta n so as long as we define the range of theta properly it looks like theta n plus 1 is twice theta n we could put in a pi to make sure everything runs from 0 to 1 instead of 0 to whatever right but that is solvable that is the Bernoulli shift if you put your put in a factor of pi so let us put sign squared pi some beta n let us say sign squared pi beta n plus 1 is beta n plus 1 is pi beta n pi beta n and this whole thing becomes n squared 2 pi then simply says beta n plus 1 is twice beta n modulo 1 but that is just the Bernoulli shift for which we know the solution so it suggests strongly that I can actually write down the solution to this guy and in fact it is of the form xn equal to sign squared pi times beta n which is equal to we also know x0 equal to sign squared pi beta 0 so this becomes square root of x0 here and then a sign inverse gives you this and then a 1 over pi gives you this so this times 2 to the n over pi that cancels off sign inverse root x0 please check the factors of pi and so on what is something like this is the solution to this map at fully developed chaos so at the parameter value 4 things become a little simpler and it looks like it is essentially the Bernoulli shift in some transformed variables that is the reason why it turns out to be log 2 and incidentally that also tells us that if the measure in beta n is uniform this has a Bernoulli shift so it is constant then the measure in x is just the acrobian of the transformation that takes you from beta to x once again and I will write this down and we will verify this subsequently for the logistic map row of x for the map the fully developed chaotic map so for f of x equal to 4x times 1 minus x the solution to this equation is row of x is equal to 1 over the normalize solution x times 1 minus x this is due to Ulam this map is called the Ulam map very frequently this was used this thing was used as a random number generator by them originally by for reasons which I will explain later on it is like the Bernoulli shift in transformed variables and you actually have this as the invariant density notice it is not bounded at x equal to 0 or x equal to 1 but it is integrable just where roots in the denominator and gets integrable so if you draw a picture of this invariant density then it looks like this here is 0 here is 1 it is a very symmetric thing which looks like this and this curve is 1 over pi root x times 1 minus x in the total area under the curve is unity so the system predominantly likes to stay near the origin or at the other end and it is very rare that it comes in here the total area is actually 1 it is completely symmetric the cumulative distribution is some sign inverse as you can easily see you have to do an integral over this this is the density the probability density so you could ask how much time does this iterate spend in any sub interval thus just just the ratio you want this sub interval the ratio of this area to unity which is the total area by ergodicity once again so you can calculate the fraction of the time a very long trajectory a typical chaotic trajectory spends in any given region sub region of the tool full interval now I leave it to you as an exercise to verify that for this map the integral from 0 to 1 dx over pi root 1 minus x times 1 minus x times the log of the modulus of the slope of this map but the slope of this map is 4 minus 8x this integral can be done for too much difficulty and I leave it to you to verify that this is equal to log 2 so that it rigorously establishes that the Lyapunov exponent is log 2 for the chaotic for the logistic map at fully developed chaos at parameter value 4 this trick does not work for any value less than 4 of course when there are fixed points which are stable or periodic orbits we know the Lyapunov exponent explicitly but in the chaotic region in general one one only knows it numerically but it reaches a maximum value at fully developed chaos for all mu between 0 and 4 the largest value of the Lyapunov exponent is log 2 and it occurs at 4 parameter value 4 this transformation is part of a more general relation called topological conjugacy the the maps we are relating to each other they are related by transformations of variables that is the reason why this trick works in this case I will come back and talk about topological conjugacy little more because it says given one map you are able to find things for other maps to which this initial map is related but right now the point I wanted to make was that even though you do not have a uniform density the Lyapunov is still exponent is still log 2 and that these points are preferred for this map rather than anywhere in the middle in this case there is one more value of the parameter where this map is explicitly solvable in the sense that you can find xn as a function of x0 explicitly and that case does not have any chaos unlike this case and that is the case when you have a super stable fixed point so in that case it was twice in between and recall this was the case where the fixed point was at x equal to a half and the slope there was equal to 0 so the Lyapunov exponent went to minus infinity at that point but you can also solve this you can solve the equation xn plus 1 is twice xn 1 minus xn by a similar trick you can write down xn as a function of x0 explicitly what would you do what would you what does it suggest you do let me give you a hint suppose you multiply both sides by 2 what happens what does it suggest for the right hand side what should one do let us write this out so 4 xn minus 4 xn square what does it suggest yeah it suggests something very strong I mean does not this suggest that you complete squares what should I do to complete squares so I do a minus 1 there and I get a minus 1 here so 2 xn is this so 2 xn plus 1 minus 1 is equal to that but what is this equal to minus yeah 2 xn minus 1 the whole square isn't it so let us write this right so what does it suggest we do next bring the minus to this side so it is a minus this is equal to that so this 1 minus 2 xn plus 1 and this of course is 1 minus that is now trivial because it immediately says define a new variable u n equal to 1 minus 2 xn so it says u n plus 1 equal to minus goes away so what is the solution what is the solution to this recursion relation yeah each time you square right so what is u u n as a function of u 0 sorry sorry I am sorry u 0 to the power so now go back to x and the problem is solved these are the only two cases which can be solved analytically the only cases where the map new x times 1 minus x for x in the unit interval can be solved in closed form explicitly the recursion relation can be solved are a parameter value 2 and 4 it is lucky for us 2 is not very interesting because it is just a fixed point which is very stable no problem there but 4 is fully chaotic so we have like the Bernoulli shift we have a simple model it is not so simple it is called intricate properties as we saw but at that value at 4 things become analytically tractable you can solve things you can write down the Lyapunov exponent explicitly and we can form a relationship with the tent map and the Bernoulli shift so that is why this map again is used as a kind of paradigm for chaos many cases so let me stop here today and resume from this point onwards we are going to look at a few other maps which have some other interesting properties and then I come back and talk about the properties some more properties of the logistic map.