 Now, let me repeat this point about Lyapunov stability and the theorems in the simplest form in which we looked at it yesterday last time. If this is a critical point, say the origin is a critical point of a system and you have some neighborhood of this origin and I would like to understand the stability or otherwise of the system in this neighborhood. Then the statement was for the dynamical system given by x dot equal to f of x, one constructs or considers or pulls out of the hat an auxiliary function v of x called the Lyapunov function which has the following properties. The properties are if v of x is such that v of 0 is 0 and v of x is greater than 0 not equal to 0 anywhere but strictly positive for all x not equal to 0 in this neighborhood. So, let us call this neighborhood u then v of x is a positive definite function if it is less than 0 for all x not equal to 0 in u it is a negative definite function if it is greater than or equal to 0 you can find other 0s in the neighborhood u then it is semi definite and if it is less than or equal to 0 and you can find other 0s then it is negative semi definite with these statements with this with these definitions the statements we made were the following. So, this is positive then v is positive definite and so on. Now the statements made were the following the rate of change of this function v v dot turns out to be my very simple piece of algebra this turns out to be grad v dot f and the statements were if you have a positive definite Lyapunov function and v dot is strictly negative in this neighborhood namely its negative definite then the critical point is asymptotically stable on the other hand if you have a positive definite Lyapunov function and v dot is only negative semi definite in other words there exist other points other than the origin where v dot vanishes then what you can assert is that the critical point is stable you cannot assert that it is asymptotically stable and if you have a positive definite v and v dot happens to be positive also then it is unstable or if v is negative definite and v dot is negative then it is unstable this is for sure. Now in the examples we looked at we came across instances we came across one instance where in the neighborhood of the origin for a positive definite Lyapunov function we discovered that in the presence of damping v dot was minus gamma y squared where gamma was the damping coefficient and I commented that this only enables us to deduce that this particular critical point at the origin is not necessarily asymptotically stable but merely stable because we know that minus gamma y squared is vanishes everywhere on the x axis so we can assert that it is stable but we cannot go on to assert that it is asymptotically stable which is what the actual critical point was in the presence of damping. So it is not that now now it might turn out it might turn out that you have a better Lyapunov function not the one that we chose but a better Lyapunov function which is positive definite for which you can show that v dot is actually negative definite in which case you can assert that this is asymptotically stable as well it is not that this thing exhausts all possibilities it is just that you could choose a bad Lyapunov function in which case you cannot make any statement at all could change sign or you could choose one which turns out to be only negative or positive semi definite in which case you can assert something about stability but not about asymptotic stability but the point is if you can find a Lyapunov function which is such that it is positive definite and v dot is negative definite in the neighborhood then you can definitely assert that the critical point is asymptotically stable now we have seen there are critical points which are asymptotically stable but not stable and the critical points which are stable but not asymptotically stable and certainly there are critical points which are both stable as well as asymptotically stable and the corresponding Lyapunov functions would reflect these possibilities in various cases. So these are not mutually exclusive categories in that sense something can be stable as well as asymptotically stable but something which is stable does not have to be asymptotically stable and something which is asymptotically stable does not have to be stable it is true that you have proved something is asymptotically stable you made a very powerful statement you said that points going to fall into this a tractor eventually so it is a very strong statement in that sense stability is weaker than asymptotic stability it says things do not go away from the neighborhood even if you wait long enough once they have entered the neighborhood but it does not say they are going to fall into the critical point at all. So I hope this clarifies to some extent the question that you had so let us take an example now go back a little bit and look at a mechanical example of a bifurcation and I will work this problem out fully because it is a standard problem so a mechanical we will see there is a bifurcation and I am going to ask you at the end what sort of bifurcation this system displays and it is an extremely simple one but it is an illustrative one because it says something about Lagrangian mechanics as well which we have not really discussed in detail in this course at all but this would give us an opportunity to in a free capitulate or ideas about Lagrangian mechanics it is a following problem I have a hoop of wire which is say in the shape of a circle and I have a bead on this hoop of wire a bead of mass m which moves frictionlessly on this ring of wire and let us say this hoop is in a vertical plane and the bead is constrained to move on this circle and let us choose for convenience coordinates such that the origin of coordinates is at the bottom position of this hoop which is in a vertical plane and let us suppose this plane is the YZ plane for instance and this hoop has a radius R what sort of motion does this constrained system have this point mass under gravity if I take it up here and let go it is going to oscillate without any dissipation back and forth about this point here and it is a constrained system because it is constrained to move on this hoop as it stands how many degrees of freedom does this particle have now how many independent degrees of freedom does it have well I fixed the plane of this hoop and that is the YZ plane so to start with this particle has only Y and Z coordinates the X coordinate is always 0 moreover I am saying that it moves on the circle so it is clear on the circle Z is a function of Y and therefore it has one independent degree of freedom which you can choose in many ways a convenient one would be to choose the angle it makes with the vertical axis its instantaneous position makes with the vertical axis that would be convenient degree of freedom but now I make the problem a little more interesting by saying that this hoop is set rotating about the Z axis with uniform angular speed omega so the entire hoop rotates about this diameter with uniform angular speed omega and so does the bead also rotate and the question is can we do the dynamics can we solve for the dynamics of this point particle of this mass with this time dependent constraint namely it is sitting on this hoop but the hoop itself is rotating at a constant angular speed omega so this is the problem to start with so what do we do we start by writing down the Lagrangian of the particle which is in this case the kinetic energy minus the potential energy of the particle and let us do this by saying L which would initially be a function of the XYZ coordinates of this particle because now that it is set rotating there is also an X axis coming out of the board you have a right handed coordinate system here and this L would be a function of all the coordinates as well as the generalized velocities corresponding velocities and what would this be well to start with it would be equal to one half m times the kinetic energy which I could write as X dot squared plus Y dot squared plus Z dot squared in Cartesian coordinates minus the potential energy and what is the potential energy of this hoop well it when it is here at Z equal to 0 potential energy could be taken to be 0 and then when it is at a height Z above the XY plane its potential energy is just mg Z so this is minus mg Z to start with but then the hoop is the bead is constrained to move on this hoop and what sort of coordinate system should I choose when I do that should I choose spherical polar coordinates there is an axial symmetry here it is rotating about the Z axis so what would be a good set of coordinates to choose cylindrical polar coordinates so let us choose cylindrical polar coordinates and what are these coordinates well the radial distance the axial distance of this particle from the Z axis which is equal to the square of that distance is X squared plus Y squared that would be one of them so let us choose rho there let us choose the azimuthal angle phi so X is rho cos phi and Y is rho sin phi a plane polar coordinates in the XY plane and of course Z itself these would be the cylindrical polar coordinates and what is the kinetic energy in such a case so L becomes equal to one half m what does X dot squared plus X dot squared plus Y dot squared become in cylindrical in polar coordinates rho and phi exactly there is an radial velocity rho dot squared plus yes indeed rho squared phi squared because that is what X dot squared plus Y dot squared become when you have when you go from X and Y to plane polar coordinates rho and phi plus Z dot squared is always present this is there minus mg Z this is what the Lagrangian would be but now we have to impose the constraint that Z is a function of the other coordinates because it has to remain on the circle and we need an equation for this circle itself at any instant of time now what is the equation to this circle the circle has a center at Z equal to R and rho equal to 0 and in the static case when it was not moving what would have been the equation to this circle it is a circle in the YZ plane with the center at Y equal to 0 Z equal to R and its radius is R so indeed this circle would have been for example it would have been Y squared plus Z minus R whole squared equal to R squared that would have been the equation of this circle when it is static when it is in the YZ plane the center of the circle is at Y equal to 0 and Z equal to R now when it starts rotating there is this circular action symmetry about the Z axis in the XY plane and therefore what would this equation become instead of Y squared you simply have to replace it by the actual distance of this particle from this axis it becomes rho squared indeed so this is the constraint therefore if you solve for Z it says Z minus R whole squared equal to R squared minus rho squared or Z equal to R plus or minus square root of R squared minus rho squared which root should I choose the two roots for Z and what do they correspond to here yeah if you fix a value of the other coordinate you could either be here or here and we are interested in oscillations about this point here so we choose the plus root or the minus root we choose the minus root so this is the root we choose that is the physical root because when rho is 0 this is 0 Z is 0 sitting here so let us get rid of this plus and write minus for the branch we are interested in and now we are ready to eliminate out here this is phi dot squared by the way that is got to have dimensions of angle of velocity squared so that is the kinetic energy term and when I rotate it then what does phi become what does phi dot become if I rotate it at constant angular speed omega phi dot is omega itself and phi dot is identically equal to omega therefore we are now ready to write our Lagrangian down but we need to use this in order to eliminate Z dot from the problem and Z all together so if Z is this then it implies that Z dot is equal to minus 1 over twice square root of r squared minus rho squared it is rho that changes with time r is a constant and I am differentiating this with respect to time so you differentiate this and then you differentiate minus rho squared which is equal to minus 2 rho rho dot so that gives us equal to rho rho dot divided by square root of r squared minus rho squared we can put all that in here and we finally have our Lagrangian which is a function of rho and rho dot alone just the radial coordinate rho and it is corresponding generalized velocity rho dot and this becomes equal to 1 half m rho dot squared plus rho squared omega squared because I get rid of phi dot squared plus Z dot squared which as you can see is rho squared rho dot squared divided by r squared minus rho squared this term minus mg times Z which is mg r plus mg times square root of r squared minus rho squared and we can simplify this a little bit and get a Lagrangian which is equal to 1 half m we combine these two terms here and it is clear that you get just m r squared rho dot squared divided by r squared minus rho squared out here plus 1 half m rho squared omega squared that is this term plus mg square root of r squared minus rho squared minus mg r this is just a constant in a relevant constant that is how Lagrangian with 1 degree of freedom independent degree of freedom rho and it is a function of rho and the corresponding velocity rho dot the constraint has been taken into account automatically and notice that in the so-called kinetic energy term you have a coefficient which depends on the coordinate this is typical of a constraint problem you eliminated the constraint and it has become a function it is true that it is quadratic in the generalized velocity but it is got a coefficient which depends on the coordinate itself and there is this extra term which has emerged this looks like part of the potential what would this be due to physically what is happening what is that going to give us this gives us the effect of the quote unquote centrifugal acceleration that is exactly what it is as you can see because it is a rotating coordinate you end up automatically with the term which will give us this pseudo force once you write the equation of motion down and that is one of the advantages of the Lagrangian formalism where constraints can be taken into account easily and non inertial forces are automatically included it is emerged automatically once we wrote down what this thing was the kinetic energy was now what is the equation of motion we know that in a Lagrangian system of this kind the equation of motion if you have L as a function of q and q dot then delta L over delta rho the partial derivative of L with respect to the coordinate is equal to d over dt delta L over delta rho dot this is the Euler Lagrange equation for this simple system that is the equation of motion which is supposed to give you the correct counterpart of Newton's equation of motion well all we need to do is to apply that here so this would imply if I differentiate I get delta L over delta rho there is a term here it sits here differentiate this you have to differentiate this you have to differentiate this and equate it to d over dt delta L by delta rho dot which means twice or whatever rho dot divided by this and you get the equation of motion but we are interested in writing things down in the Hamiltonian framework because we would like to write everything as first order equations in this case and so we make a change from the Lagrangian to the Hamiltonian formalism and if you recall if you start with an L which is a function of q and q dot then the way to go to the Hamiltonian formalism is to go from this function to a Hamiltonian which is a function of q and p by making the following transformation you first define a generalized momentum which is delta L over delta q and that is the definition of the generalized momentum given a Lagrangian the momentum conjugate to a particular coordinate q is the partial derivative of L with respect to q then once you have done that with respect to q dot sorry with respect to the velocity then this Hamiltonian is equal to p q dot minus L and it is a transformation which takes you from the Lagrangian to the Hamiltonian. However you have to be careful that you can that you invert this equation this in general gives you p as a function of q and q dot because L is a function of q and q dot but in going to the Hamiltonian which is a function of q and p you must eliminate q dot and you do that by solving this equation for q dot as a function of q and p so this gives you p as a function of q and q dot you must now write q dot as a function of p and q substitute that here and wherever q dot appears in L substitute for q dot in terms of p and q and you have a Hamiltonian and you are guaranteed after that Hamilton's equations of motion. So this is the procedure for going to a Hamiltonian let us do that here let us apply that here so what we need to do is to first find the momentum conjugate to the generalized coordinate row I should really write p sub row it is a radial kind of momentum but since there is only one degree of freedom let me just call it p and this is equal to delta L over delta rho dot and what is that equal to it says differentiate this pretending everything else is a constant except rho dot the two cancels and you get m r squared rho dot divided by r squared minus rho squared please notice now that we are not turning out it is not turning out that the momentum is some mass times a velocity not at all it is a function of the coordinates times the corresponding generalized velocity this again is typical of a problem which involves constraints but that is the definition of the canonical momentum conjugate to rho once you have this of course you can invert this trivial in this case this implies that rho dot is p times r squared minus rho squared divided by m r squared and we can write down therefore the Hamiltonian as a function of rho and p which is equal to p rho dot minus L we are ready to write down then the Hamiltonian therefore h of p, rho is equal to p times rho dot and rho dot is given by this so it is p squared times r squared minus rho squared divided by m r squared minus the Lagrangian but in the Lagrangian you must replace rho dot in favor of p since it must be a function of rho and p alone if you did that you get p times rho dot squared so it is going to give you when I square this I get an m r squared whole squared that cancels against this and then r squared minus rho squared the whole squared one power cancels against this and as a factor a half so it is clear we have one of this minus half of the same therefore it is just this that takes care of the kinetic energy part minus one half m rho squared omega squared that is this quantity minus mg square root of r squared minus rho squared plus mg r I hope you kept track of all the minus signs because if not we are going to get wrong answers so let us see what happens that is our Hamiltonian so we can get rid of this now we are ready to write down Hamilton's equation some motion now it is a non-linear Hamiltonian it certainly does not look like simple harmonic motion or anything like that and notice the signs of these terms now what are the equations of motion well we know that rho dot is delta H over delta rho that is the first of Hamilton's equations and that of course is going to give us something we already knew because delta H over delta p and this is something we already knew because if I differentiate this with respect to p it gives p times r squared minus rho squared over m r squared but we already knew that p was rho dot times m r squared over r squared minus rho squared that is how we got p so this is not telling us very much nothing new in the case it is just the old relation between the generalized momentum and the generalized velocity but the other equation says something interesting and that is p dot is minus delta H over delta rho and what is this equal to now we have to differentiate everything I differentiate with respect to rho and I get minus p squared rho divided by m r squared it is a plus sign because I already have a minus sign here and I have another one here so it is a plus sign and then plus m rho omega squared this term here minus mg divided by twice square root r squared minus rho squared I am not very happy with what happened I hope we did not leave out any minus sign anywhere in between because if we have then we are going to have an unphysical result here so I need to differentiate this and it came from minus should be a plus should that be a plus we started off by writing L equal to the kinetic energy minus mg z that was the potential energy but if you recall z itself was r minus square root r squared minus rho squared and so it gives us this is equal to various terms plus mg square root of r squared minus rho squared in the Lagrangian so when I wrote the Hamiltonian I put a minus sign there so that gives me a minus similarly similarly the original term in the Lagrangian was of the form one half m rho squared omega squared the phi dot squared term in the Lagrangian with a plus sign and when we write a minus for the Hamiltonian it is pq dot minus L that to appears with a minus sign so we are okay all right now let us differentiate this so we were in the middle of that and then I need to differentiate it and change the sign so there is a minus here and a minus here both these things go away and you get a plus and then I have to differentiate this with respect to rho so that is equal to a minus twice rho and therefore the two goes away and the minus turns up and let us take the rho out of the bracket and the m as well so these are the two equations of motion they are the Hamilton equations of motion again you see they are badly non-linear equations because you have all kinds of powers here you have a cube here and then you have the third power here and then you have this one over the square root and so on so it is really a non-linear problem in this sense now the question is where are the critical points of the system they happen when the right hand side is equal to 0 of course you put r equal to rho you end up with an infinity here so the critical point would occur at p equal to 0 and let us write that down p equal to 0 and certainly rho equal to 0 is a critical point because if I put rho equal to 0 this vanishes and this vanishes as well rho equal to 0 that is always a critical point that corresponds to this point the system at rest here which we know is an equilibrium point whether it is stable or not we do not know is there any other critical point in the system so you said p equal to 0 and this term goes away and you see this could vanish this bracket here could vanish when does that happen so this would vanish with provided omega squared minus g over r squared minus rho squared is 0 or omega squared was equal to this or r squared minus rho squared is equal to g squared over omega 4 that is the root here or rho squared is r squared minus g squared over omega to the power 4 provided this is positive provided this quantity is positive otherwise it is not going to happen it is not a real root what would this imply when is this going to be positive so another solution non-trivial solution for rho at the square root of this value provided r squared was greater than g squared over omega 4 or since omega is the quantity that you have in control r and g are constants this implies that you have a second solution I need this equation so I am going to ask you to dictate it to me afterwards so you have another critical point provided omega to the 4 is greater than g squared over r squared or omega is greater than omega critical which is equal to square root of g over r then you have another solution it gives you something else altogether we need to compute what that is so that root is given by this let us write it in reasonable form so you have a second root at rho squared another cp at p equal to 0 and rho equal to the square root of this so let us take out an r completely r times square root of 1 minus g squared over r squared over omega 4 but g squared over r squared is omega critical to the power 4 right so this is equal to omega critical squared omega critical 4 omega c is some constant its square root of g over r now what kind of stability do we have here what can we say about the stability of these solutions so we have to go back and write the equations of motion down and see what it does for linear stability and if you recall we had p we had rho dot is equal to m r squared p over r squared my was this right what was what was rho dot p over mr squared that was the equation and the other equation was p dot is equal to now you have to tell me what it was I believe there was a p squared rho over mr squared plus m rho omega squared plus minus m g rho root of r squared minus rho squared and that was it so what we did was to take the rho out and the m out so we wrote it in this form so you could rewrite this as in terms of rho over r and the factor g over r you could write in terms of omega c but anyway let's leave it in this fashion so we have one critical point here and another one there and let's look at the stability of these critical points so let's linearize rho equal to 0 p equal to 0 and then the equations of motion are rho dot is approximately equal to the r squared cancels p over m which is what you'd expect because you'd expect very close to this for small oscillations you'd expect that rho dot is just p over m or p is m rho dot as usual the nonlinearity doesn't play a role there and then this equation this term is already third order we're going to look at it near rho equal to 0 this can be got rid of and then what you get p dot is approximately equal to so this can be dropped because it's second order in rho this can be dropped because it's got third order terms and then you have omega squared minus g over r times rho but what's g over r omega c squared so you have m times omega squared minus omega c squared times rho what kind of critical point is that well the linearized matrix l is of the form 0 1 over m m times omega squared minus omega c squared and then a 0 here what kind of critical point is that it depends on the value of omega whether omega is greater than omega c or less than omega c it depends on that completely this determines now the stability at the origin so let's look at the case omega less than omega c first so you start rotating at a small speed first and see what happens then so I think we need to retain this so let's do this therefore omega less than omega c the eigenvalues of l well what are the eigenvalues of this matrix omega is less than omega c plus or minus with an i there right because omega is less than omega c so what kind of roots are these pure imaginary so it's center absolutely so you have a center which is stable at the origin this means that if you rotate this hoop at sufficiently slow speed when as it rotates this point is a stable center so if you start here it oscillates about this point while rotating so it comes back and you have undamped oscillations about the bottom that's your critical point which is stable and you can show simultaneously that the other critical point is unstable on the other hand what happens if omega is greater than omega c so this is stable and if omega is greater than omega c then the eigenvalues of l plus or minus square root of omega squared minus omega c squared what kind of critical point is this it's a saddle point and therefore unstable the origin becomes unstable you would have to do the same thing about this point to discover that this becomes stable so this implies yes you're right second critical point doesn't exist at all if omega is less than omega c because this is pure imaginary absolutely right yeah so it says wherever you are it's going to oscillate about the bottom most position and these are stable oscillations on the other hand the moment you have omega crossing the value omega c the origin becomes unstable and a new stable critical point emerges so this is going to help us find what kind of bifurcation this is and this is the interesting question but before we do that let's see what happens let's draw this bifurcation diagram and see what happens so here's the bifurcation diagram so it's not an exchange of stability transition at all it's not an exchange of stability bifurcation because this one doesn't exist only if omega is greater than omega c okay bifurcation diagram is as follows we plot rho equilibrium the equilibrium distance radial distance axial distance from the z axis as a function of omega which takes on only positive values and here's omega c p is always 0 so I don't plot that it comes out of this axis but p is always 0 at equilibrium and then it says this is your stable root okay and we know it becomes a saddle point as soon as omega exceeds omega c so the rest of this diagram should really be a dotted line and the new root that comes out is this what's the graph here what kind of behavior does it have at omega equal to omega c it's 0 it's bifurcating out of this point that's clear what happens as omega tends to infinity what happens to that equilibrium value of rho it tends to r so it's very clear physically that this is what's happening it's clear that this hoop as you rotate it faster and faster with some angular frequency omega this is the furthest it can actually go it can't go beyond that it's constrained to remain on this hoop so once it zooms up to this point it has to remain there and this is a stable oscillatory point so here if this is an equilibrium value then if you displace the hoop about it as it's rotating it would actually do this about this point so it gets shot out a certain distance from the bottom as long as you didn't start exactly at this point then of course it remains at that saddle point it's an unstable equilibrium point but a little bit of perturbation puts you away and it zooms up crawls up to this point and oscillates about this point so this goes off and asymptotically hits the value r as omega becomes very large now what kind of behavior does it have at this threshold would the slope be infinite would the slope be finite would it do this would it do this would it do this what would it do that's crucial to know yes we indeed we need to know exactly what the behavior is like and that's easily seen because you can see that this row equilibrium is in fact equal to r times square root of omega 4 minus omega c 4 divided by omega squared take it out of the square root and near omega equal to omega c it's clear you can write this as omega squared plus omega c squared times omega squared minus omega c squared and near omega equal to omega c you can replace this with omega c that with omega c and you basically have this minus that which in turn you could write as omega minus omega c times omega plus omega c and this harmless factor you could write as twice omega c so the whole thing goes like the square root of omega minus omega c so what's the shape of the curve here what's the slope what's the slope of y equal to square root of x at x equal to 0 it's infinite so this is in fact the thing like this goes off and this thing here in this neighborhood it goes like omega minus omega c to the power of that's the exponent and this is the stable root what kind of bifurcation is this is it an exchange of stability bifurcation no because there isn't a critical point at all to start with the other one doesn't exist for omega less than omega c is it a saddle node bifurcation that means a spare a stable unstable pair of critical point emerges as you cross the bifurcation value but that's not true because you already had one here so it's not saddle node it's not exchange of stability it must be a pitchfork bifurcation but then a pitchfork bifurcation is one where a stable point bifurcates into an unstable point and a pair of stable equilibrium points where is the other pair one where's the other member of this pair pardon me why is it not physically realistic because we because row can't be negative we plotted a coordinate where it can't be negative but it's indeed a pitchfork bifurcation because there are indeed two possible states of this suppose you color this part of the hoop green and this part of the hoop red certainly something that sticks here and something that sticks here can be distinguished between depending on what kind of infinitesimal perturbation you had here it would either zoom up to the right hand side or the left hand side the green side or the red side and therefore it is in fact the situation where you have two different equilibrium points one on one side of the hoop and the other on the other side of the hoop and you don't see that in this picture because this is just the actual distance indeed there are two critical points that have emerged from a single one so this is very much a pitchfork bifurcation you could use this to measure little g to measure everything else and you're sure that you don't have friction you could use this actually to measure little g because you could measure where the equilibrium point is and this will tell you directly in terms of r and g and omega c tells you so that's one possible thing you could also put this in a ball in a spherical ball and rotate it rotate the ball you'd still have this problem but this is the way a constrained mechanical problem is simplified when you use the Lagrangian framework and its stability is easily analyzed when you use the Hamiltonian framework in this fashion you could now ask what happens if I had different shapes I don't necessarily have to have a circular hoop so let's see what happens quickly see what happens if you have a slightly different shape of hoop one possibility is to write is to use for instance a parabolic hoop so here's z and once I set it rotating in this fashion a parabolic hoop if this z is half ky squared for example then on this parabola z is equal to one half k rho squared because that's why squared is replaced by x squared plus y squared once you set it rotating by cylindrical symmetry and then you see that z dot is equal to k rho dot in this fashion we run through our steps quickly then the Lagrangian becomes equal to one half m they certainly a rho dot squared plus this is z dot squared but the z dot squared is that k squared rho squared if I pull the rho dot squared out this is one plus this times rho dot squared in this fashion plus in this case one half m rho squared omega squared that came from the rho squared phi dot squared term in the kinetic energy minus mg times z but this is over 2 k rho squared this is what the Lagrangian is and what's the Hamiltonian well to run through these steps quickly the conjugate momentum p is delta l over delta rho dot and that if I differentiate here is m times 1 plus k squared rho squared rho dot or rho dot is p over m 1 plus k squared rho squared that's the conjugate momentum again it depends on the coordinate so what's the Hamiltonian this is equal to p rho dot minus l p rho dot gives you p squared over m times 1 plus k squared rho squared here first portion minus l therefore you got to subtract this minus this term times the square of this so you have a p squared over m squared that gives you 2m and therefore it's clear that you just get twice as before plus one half m rho squared omega squared minus thank you minus this plus mg over twice k rho squared like so so we can simplify this a little bit take out the m rho squared over 2 and then write this as omega squared minus gk that's it and what are the equations of motion so again we have rho dot is delta h over delta p that's not going to give us anything new except this whole thing p over m into 1 plus k squared rho squared but then p dot is minus delta h over delta rho and that gives you something non trivial this is equal to minus p squared over m 1 plus k squared rho squared whole squared multiplied by the derivative of this quantity which is equal to k squared rho this term I differentiate this and then I differentiate that and that gives me minus m rho omega squared I need a minus out here by differentiate so this becomes a plus what do you conclude what kind of critical points do you have and what's your conclusion you have a bifurcation here at all what happens now clearly p equal to 0 rho equal to 0 is a critical point so the linear matrix about that point so near so what happens well I mean it's clear this is going to this is a linear term and it's going to change sign depending on omega is less than square root of gk or greater than square root of gk so the linear term the linear linearized problem is p dot equal to p over m to first order and rho p rho dot sorry rho dot is this and raise this problem write it properly rho dot and p dot is equal to m times omega squared minus gk times rho so what do you conclude from this so l is 0 1 over m m times omega squared minus gk and a 0 here yes if omega squared if omega is less than so center for omega less than square root of gk you could have guessed that by dimensional arguments square root of gk is the only quantity of dimensions frequencies so other than omega itself so for omega less than square root of gk you end up with a center which means you have stable oscillations what happens if omega is greater than gk the square root of gk becomes a saddle becomes unstable but you don't have an alternative after that and what does the system do goes off to infinity it goes off to infinity exactly this critical point just becomes unstable that's it it was because you had the curvature in the circle was appropriate that you ended up with another stable point so it deeply depends on the function that you had it depends on whatever was multiplying it it depended on the shape of the wire the constraint which gave you which depended on the shape of the wire told you whether you had another solution or not yes if omega square equal to if omega equal to root gk yes well if omega equal to root gk this goes away the problem is not linearizable anymore it's intrinsically nonlinear as you can see so that's the marginal point at which the center is losing its stability at this point infinitesimally greater value of omega and you're off it goes off so you can only have stable oscillations about the center as long as omega is less than square root of gk I leave you to play with this and figure out what happens if instead of this kind of situation you had for example you took a stick and you put it on a stick just a rigid rod you put a bead on that and you rotated this about the vertical axis so that's just a linear function so the circle is special of course there are other functions for which this would happen but the circle was a simplest case and we didn't look at what would happen in the hoop problem if you started with something not on the lower half but up here and I leave you to analyze that it's fairly straightforward the only thing is in the constraint equation you have to write Z is R plus the square root of R squared minus rho squared and go through the algebra once again to see what would happen to a particle sitting here when you start rotating but the result we got was completely physical it said that initially you have for small velocities small angular velocities you have a stable oscillations about this origin and once you exceed a critical point then this gets flung out and you have oscillations about a new point and the largest value that point can have is in fact when you write here at this stage which happens when you have an unbounded angular speed and then it gets flung out as far as it can go here and we also found out what was the nature of this bifurcation diagram we found out what kind of exponent you had and we found that there was a kind of square root behavior the function of square root of omega minus omega c was way the way the new function the new solution took off so it's a fairly complete analysis although we didn't solve the equation of motion in general that's complicated because a motion in general could be very complicated depending on the initial conditions and you could translate it back to the original Cartesian coordinates but there isn't much percentage in doing that in this problem it's obviously of interest to stay with the cylindrical coordinates that we had yes pardon me yes you have yes you are now we didn't draw the phase diagram because we didn't really write down we didn't really draw the phase diagram at all because we didn't write the solutions down we just looked at what happens in the critical points and so on but you're right there's obviously a hyperbolic point at a certain stage and then there would be separators is coming out of it and so on so I leave this as an exercise to try and generate this but in practice although because the expression is fairly complicated it might be messy to do this but you can do this numerically can write these solutions down can solve this yes it's clear that once the critical point becomes unstable and it becomes a saddle point then there is no question of oscillation so absolutely so things would diverge yes we didn't look at the time period of small oscillations about this point we didn't do that we didn't write down what would happen but that's straightforward because for sufficiently sufficiently close to the origin for small oscillations it was like a simple harmonic oscillator problem for which you can straight away write down the solution you can write down the time period yes would it diverge yeah presumably this would end up diverging yes because it becomes unstable beyond that point but one should check this out explicitly we didn't do the large oscillations we kept things very close to the origin and that was it but now you make the amplitude larger and larger and then you begin to see what would happen your question is what happens to the actual time period of small oscillations itself in the linear problem yes indeed it diverges that you can see directly because let's go back his question is what happens to the time period of oscillations as omega hits omega C in the previous problem the circular hoop if you recall in the parabolic case yes indeed it becomes infinite as you can see immediately because here's our problem here's our here's our linear matrix so what are the eigenvalues lambda 1, 2 is equal to what are the eigenvalues of this matrix right yes okay now we're going to look at small values of omega to start with and therefore omega is less than the square root of this in which case let's write it as I times square root of GK minus omega squared and let's call omega C squared equal to GK right therefore this is equal to plus or minus I times square root of omega C squared minus omega squared those are the eigenvalues which means that the small oscillations will have e to the power plus or minus these eigenvalues multiplied by T and the time period of oscillation T is therefore proportional to 1 over square root of omega C squared minus omega square which diverges as omega hits omega C it is exactly his conjecture so it's clear that as you hit this critical point the time period diverges it's about to take off and beyond that it becomes unstable you could also have deduce this by looking at the shape of the potential does this correspond to what kind of thing does this correspond to when I put this in here and you have an equation which says rho dot is P over M and P dot approximately equal to and let's write this as omega C squared and keep omega less than omega C squared so minus M omega C squared minus omega squared times rho what kind of potential are we talking about it's like a simple harmonic oscillator with a potential this is an equation of motion so you have to integrate this so you get a potential effective potential we effective which is proportional to one half M omega C squared minus omega squared rho squared isn't it if I integrate this and change the sign right this is like the force f f of rho so V is equal to integral f of rho d rho with a minus sign since f is minus d V over d rho this is certainly true what kind of shape is this as a function of rho because we are looking at positive rho this is a parabola upwards provided omega smaller than omega C but it's getting flatter and flatter as omega approaches omega C and therefore the potential is starting to look like this and once omega exceeds omega C the potential inverts and inverted parabola is unstable at the origin so this is exactly what happens the stiffness is going down as omega approaches omega C it's becoming like a spring with a smaller and smaller spring constant gets more and more flabby and therefore the time period is longer and longer and it diverges at omega equal to omega C beyond that it's not going to sustain oscillations because it's turned around the other way okay that yeah but we haven't really written the phase portrait down just some things near the origin but this is good enough it tells us all that we need to know okay.