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in this lecture i will i will talk about ah we look at the phase plane but for us slightly more complicated problem ok and in this case ah we look at a problem where we dont actually solve the problem ok so in the previous case we saw that we could actually solve the problem and ah we could write and that was because the problem was ah linear problem it was a linear ode the either the harmonic oscillator or the damped harmonic oscillator both are linear equations now lets consider a nonlinear equation and show and show how we can analyze the nonlinear differential equation so the equation that i will be considering is that of a pendulum of ah arbitrary amplitude so the differential equation has this form theta double dot so theta theta is the angle that the pendulum makes ok so is so if you have a pendulum ok and ah this is the vertical so the pendulum is oscillating and as its oscillate it makes an angle theta ok so theta double dot plus omega square sin theta equal to zero so clearly he is a sin theta so it is not ah its not just ah theta so for small amplitude it becomes linear but for arbitrary amplitude this is non linear sin theta has theta theta cube etcetera so sin theta remember sin theta remember the taylor series for sin theta sin theta equal to theta minus theta cube by three factorial plus theta five by five factorial plus all these are nonlinear terms ok so ah so ah this is a nonlinear second order ode lets write this in the usual phase plane picture so we will say theta dot equal to omega and omega dot which is theta double dot is equal to minus omega square this omega is ah in these this is capital omega this is small omega ah omega square sin theta now lets ah look at the critical points ok so theta dot equal to omega dot equal to zero implies omega equal to zero and omega square sin theta equal to zero so ah now omega square sin theta equal to zero implies sin theta equal to zero ok or theta equal to plus minus n pi or ah i will just write it as n pi where n equal to zero plus minus one plus minus two three and so on so these are the critical points so critical points omega equal to zero and ah theta equal to plus minus n pi so it can be zero plus pi plus two pi plus three pi minus two pi minus minus three pi and so on now ah theta equal to zero omega equal to zero ok this is your critical point ok so this is one critical point ok so when theta becomes zero ok then ah then you can clearly see omega is zero omega dot is zero everything is zero one kind of critical point ok and ah we will see soon that ah this point has a certain feature ok whereas the other other ones will have certain other features lets take this point sin theta for ah theta close to theta equal to zero so close to the critical point i can write sin theta as ah as ah sin theta approximately equal to theta i get omega dot equal to minus omega square theta ok and this is basically a simple harmonic oscillator and its a simple harmonic oscillator so ah so now if i look at this if i look at if i look at this omega theta plane the phase plane ok and i will deliberately show it a little long and the theta axis so so theta equal to zero is one ah critical point and around this critical point your trajectories look like they ellipses ok this is what the trajectories look like about this point this is theta equal to zero ok then you have ah another point where lets say theta equal to pi this is another critical point lets look at what the trajectories look like around theta equal to pi so in this case your trajectories look like periodic solutions this is a center theta equal to omega equal to zero is the center ok and ah around this critical point ok the trajectories look like ah ellipses what about

ah theta equal to pi ok so now in this case we will also look at two pi ok just to just for completeness ok what happens is that we have to look at what the solutions look like when you make small displacements around theta equal to pi theta equal to pi if you look at your pendulum ok that is ah this this is theta equal to zero ok so this is theta equal to zero and theta equal to pi is the pendulum pointing straight up so clearly if you make a small displacement around this ok if you make any small displacement ok you will you you can you can easily show that it will it will ah it will it will keep going away and away if you make a small displacement around this the solution is oscillatory if you make a small displacement around this the solution is not oscillatory it is ah it will completely go away from that ok so ah so clearly ah theta equal to pi three pi minus pi minus three pi so ah so pi is same as three pi ok same as five pi etcetera so ah are unstable critical points whereas theta equal to zero two pi minus two pi four pi minus four pi etcetera are stable critical points so so now lets look near unstable critical point pi theta equal to pi omega equal to zero so when theta is close to pi pi we define theta theta equal to theta ar e theta one equal to theta minus pi ok so then ah i will just so see theta one is a difference of theta from pi and now i can write my equations as ah d theta one by dt or theta one dot equal to omega c theta one dot is same as theta dot which is show omega omega but what i will get is that omega dot ok now now omega dot was was ah was basically given as ah was given by minus omega square sin theta ok this is so close too close to this ok so this is approximately equal to minus omega square it is it is equal to sin of ah pi plus theta one which is close to closed so for small theta one ok so this goes as plus omega square theta one ok so for small for small theta one so theta one is the difference of theta from pi and is since this is small i can do this ah sort of expression and now and now i can write my now i have a linear equation so i have theta one dot omega one dot is equal to zero one omega square zero theta one omega one so this procedure is called linearizing this equation so so what we did is we linearized nonlinear ode near critical point ok so near the critical point we linearized and we wrote it in this form this is the very important step ok so this this step linearization or of a non linear ode near the critical point is a is a very crucial step in analyzing of ah of nonlinear odes ok so so what we did was first we found the first we found the critical points ok so we had a nonlinear ode so first we found the critical points and then we linearize the ode so so so we identified ah the stable critical point so we know what the solution looks near the stable critical point ok now now we are going to linearize allow around the unstable critical point ok and we know what the solution looks like now now this ok you can you can you can easily write the solution theta one omega one this is a linear ode only thing now it is ah plus omega square so so this will have exponential solutions so it look like c one plus c two one minus omega e to the minus omega t you can do the analysis of this around this critical point so what you will find if you analyze the trajectories what you will find is that ah is that you

have lines of ah slope omega and minus omega ok so you have you have two lines ok so one of slope omega and one a slope minus omega so let me show it in this figure so so you have a line of slope omega and a line of slope minus omega ok so these two lines are there ok now the trajectories will ah what will they do ok so ah so in this case suppose you start some ah value of of ah theta and and some value of omega ok so theta one is this difference from here so ah so what you notice is that ah is that you have ah these two lines ok where ah whether are are basically asymptote for example if both c one and c two are not equal to zero ok then ah you will be starting somewhere lets say lets say you start here then what will happen to the trajectory is that ah is that ah as as time increases in this case what you will see is that is that as time increases this trajectory will go i will cross zero and then it will go in the other direction and it will keep it will keep going away so so theta one will keep increasing ah in the so so in this case theta one is negative ok so we started with a negative theta one and if theta one is negative then ah omega one will keep decreasing so so it will go in this direction ok so omega is decreasing so omega omega keeps decreasing ok theta one is negative now ah as omega keeps decreasing theta theta becomes less negative so so it follows ah so so theta becomes less negative it goes closer to pi when omega goes to zero then your theta one dot becomes zero so theta one so the slope of theta theta one goes to zero so in other other sense that when you are omega goes to zero essentially what you are what you are having is ah is your theta one just crossing this ah your your theta one is essentially constant your theta is not changing right at this point where omega goes to zero so the line becomes parallel to the omega axis and then and then it goes away now ah what is what is also important i didnt i didnt show this correctly here ok so this this will asymptotically approach these graphs so this graph will asymptotically approach approach these lines ok maybe may be i will show these lines in a different color so let me let me show them in a slightly different color just to make it clear so you have this line you have this line and it will asymptotically approach this now if you started with a point somewhere here ok then what would happen ah well well ok lets do the easier part lets do this first so if we start with somewhere here you will go like this so and and in this case you will have we will have it moving in the other direction if you start somewhere here then ah what will happen is that is that in this case your omega will never go to zero so what we started was ah [wa/was] was ah in in important this case your theta one will go to zero so you will get something like this so this is a line where theta one equal to zero so it will cross this and it will go like this and again it will asymptotically go to these and ah and the last one is if you start here you will end up like this ok so this is the asymptotic behavior around ah so so asymptotically all the graphs will either approach e to the it will it will either go to this one or to this one because ah theta is greater than zero and as t goes to infinity ok then ah then ah it will it will it will approach this curve in one of these ways we have a picture of what the trajectories look like around the unstable critical point and this will be true for ah this will be true for theta equal to three pi ok now theta equal to two pi is a stable critical point so around this the trajectories will again look like this will look like ah ellipses so now what you can do is you can make the overall picture ok so you have your trajectories look like ellipses and then you have these asymptotes so what is happening is that ah is that ah your trajectories are now ah they look like ellipses was small ok then as you go away they will start looking like this asymptotically they will start approaching this ah this line ok and ah in this direction it will look it will look like this ok now similarly the next critical point ok three pi will also have the same feature so three pi will also have the same feature so that will have that will show ah you will have you will have these two lines and you will have will have graphs that look like this

now lets look ah what is happening here so ah so so suppose you take any one trajectory ok ah so this is where your theta is actually greater than greater than pi ok and you are starting with omega that is negative ok so you are starting you are starting ah at a value of theta that is greater than pi ok and a value of omega that is negative and ah what you will find is that so so you started ah somewhere up here ok and your ah omega is negative that means you are moving in this direction ok so then what you are what you are what you are imagining is that you are you are starting in this case with your pendulum ah something like this and your omega being negative so omega is pushing it in this direction what will happen is ah it will go it will go past theta equal to zero so you are sending it with a it will go it will go all the way to theta equal to zero it will cross that and then it will come it will come it will come this way exponentially towards this ok so that is what is happening so this is the force this is the phase portrait of this simple pendulum ok and now ah there is a very special curve ok so ah so ah we saw that here here you have you have the simple harmonic oscillator solutions ok and ah now ah if you look at the simple harmonic oscillator solutions ok so what is the simple harmonic oscillator solutions so simple harmonic oscillator solutions near stable critical points ok so this has the feature that ah theta so so so we can we can we can write this as ah half theta dot square minus omega square so so so ok ok will will we we what we remember is that ah we had something like this ah theta square was equal to zero or d by dt of this this was the was equal to zero or we got this was constant ok so so so so we got this was constant now in this case in ah so so this led to theta dot square ah plus plus omega square theta square equal to constant ok so this was the this was the simple harmonic oscillator solutions that is the simple pendulum solutions now in this case in this case you will get a slightly different equation what you will get is ah so so ah lets get back to our equation so ah we have d by dt of theta dot is equal to minus sin theta into omega square ok so ah now if i look at if i look at ah d by dt of theta dot square ok and in the other the equation was d by dt of ah of theta equal to theta dot so if i look at ah d by dt of theta dot square ok so this looks like ah two theta dot theta double dot ok and ah and if i look at d by d by dt of ah lets say cos theta ok so this is minus sin theta theta dot ok so so now ah what you can see is that ah is that these two are related ah these two are related to each other so ah so suppose i take ah suppose i take d by dt of of suppose i take of theta dot square by two plus cos

plus omega square cos theta theta or maybe yeah so if i if i take minus omega square cos theta ok so if i look at this quantity ok so this is equal to theta dot theta double dot plus omega square sin theta theta dot and this is equal to zero ok so the trajectories satisfy theta dot square by two minus omega square cos theta equal to constant ok equal to c now c can be either positive or negative ok so so c can be either positive or negative ok and ah what what we can get from this is that ah what i can write is that i can write theta dot equal to omega is equal to two c plus omega square cos theta raise to half so so we have this expression for omega now omega is supposed to be real ok so ah so you should have you should have ah the c plus omega square ah cos theta has to be greater than equal to zero ok now ah if c greater than omega square ok c is greater than omega square ok then ah then ah you can have ah then then omega square then omega square cos theta will always be implies c as will be greater than minus of absolute value of omega square cos theta ok and ah what it implies is that c plus omega square cos theta greater than equal to zero for all theta all theta ok so theta is under is not restricted if c is less than omega square ok then ah what you can see is that ah c plus omega square cos theta is greater than equal to zero implies cos theta ok should be should be strictly ah such that so ah so ah absolute value of cos theta should be less than equal to c by omega square ok so that means theta is so cos inverse of minus c by omega square ok and ah you know i will just i will just put a minus absolute value of cos inverse of this should be less than theta should be less than equal to cos inverse minus c by omega square so so what is important is that your theta is restricted so so in this case ah so your theta can take only certain values ok so this is this is actually your periodic motion ok so so ah and ah this is ah basically ah this is non periodic motion ok this is not periodic so ah so what is the implication of this so based on the value of c and omega square ok so what your equation will ok so if you take i will just take i will just take a few points just to illustrate so zero pi ok so ah so the case where ah c and omega square are ah this is the rotary case where the way the motion is ah given by rotations and two pi so so ah and what you have is ah is here

here you have these these motions showed it so so so when you approach this you will end with something like this ok and ah and you have you also have things like this ok similarly at this at this next point you will have the same thing so so so what will happen is that ah is that whenever you are c is less than omega square ok you will have you will have a rotatory motion when c is is ah not restricted then you will ah have ah you will have this motion that is not periodic ok now ah let me show this in the next page ok lets just put all the all of this together ok so so what the motion looks like for ah for certain values of c you will just have motion like this ok so ah so so lets consider around this critical point now when see becomes greater than certain value ok then your theta is allowed to vary theta can go to any values there is no restriction on theta so you have this periodic so so you might ah for for larger theta you might get something that is not exactly elliptical you might get something like this ok now when this is this is theta ah large but c is less than omega square ok so c is less than omega square so theta is still restricted ok and theta is large so i am showing it ah i am showing it as not exactly an ellipse it is say this elliptical when it is small ok so it is periodic but not ellipse ok now ah so so this is what it looks like and then and then you have pi and ah what you have is ah if if c is greater than if c is greater than omega square then theta can take any values and you will get solutions that look like this ok so the ah so so this is the next so let me show this in a different color ok so this is the unstable unstable parts ok ah this is the next next critical point that is next stable critical point where you will have again the same kind of solutions you have periodic then you have things like this so so what is important is that is that if you start with ah with something like this then you will just keep going your ah your trajectory will just your pendulum ok so so here you started with a very large amplitude and it just keeps going ok spinning and ah going like this so your theta can just keep ah your theta can go all the way to infinity it it can just keep going round and round ok so now there is a boundary the boundary of these trajectories when c equal to omega square this is called a separatrix ok now when c equal to omega square ok then the pendulum what it will do is ah wherever you start off you will exactly go up to the up to pi ok so you start with some motion and theta will just go all the way to pi and then it will it will not go beyond pi ok or ah it is right at this point which separates these two these two stable and unstable modes ok this graph where c equal to omega square is called the separatrix ok so ah this c equal to omega square this separates periodic and we will call this as ah unstable

motion and unstable motion ok so ah so so what i want so this curve this curve is what is called the separatrix this red curve is called the separatrix ok and ah what i what we [wa/what] what i want to show in this ah in this is how you can take a take an equation that is completely nonlinear ok so the the equation that we start off with ok it it doesnt look like a linear equation where we started off with a highly nonlinear equation ok ah i say highly nonlinear because it contains terms of all orders of all powers in theta ok and we took it and we do the stability analysis around the critical points ok and then and then just by using ideas that ah the curves should be smooth ok we could draw we could draw the entire phase plane picture so this is a very powerful powerful ah way of looking at this where this problem ok i can i should also emphasize that ah you could also have solutions that look like this in this direction you could have more other other graphs also you could have more graphs also like this ok and so and so this ah this picture is extremely powerful to get an idea of what the pendulum motion is and we got all this ideas without actually solving the equation ok but but ah there is one important thing to keep in mind you should know the solution of the corresponding linear equation only then you can apply these methods so in the next class we look at ah we look at ah some of these [cri/critical] critical points in more detail ok thank you

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