 We will continue with discussion on structural stability analysis, I would like to introduce in this lecture some elementary notions about fixed points and bifurcations in non-linear dynamical systems. See what we will quickly recall what we have been doing, we considered beam columns and we considered three situations which were departure from ideal situation like beam carrying axial load plus transverse loads and this axial load being applied eccentrically or the axial load is applied on a beam which is not initially perfect, there is initial imperfection Y naught of X, we showed that all these three problems are mathematically equivalent and for certain values of P the response of all these three systems become large the structure loses its stability, then we ask this question all these three cases represented departure from an ideal situation why such things happen if to answer that we need to consider the study of ideal situation itself then we showed that if we consider a perfectly straight beam which is loaded truly axially and we investigated the condition under which an equilibrium position in the neighborhood of Y equal to 0 is possible, so assuming that such a solution is possible we searched for the value of P which can lead to that, so under the assumption that a neighborhood an equilibrium position in the neighborhood is possible we were able to write the equilibrium equation for determinate systems we could write in the terms of bending moments whereas for statically indeterminate systems we need to use the fourth order equation, so I pointed out couple of times that this term PY that is P into this distance Y this force is computed this bending moment is computed with respect to the deformed configuration of the structure, whereas while finding reactions and other terms associated with the equations of motion we consider the undeformed geometry, so that difference we have to notice and we need to see later on how does that manifest in a more generic formulation of such type of problems. Then we plotted the mid-span transverse deflection versus the applied loads and we showed that the load deflection path has infinitely many branches and this is the, as the load is increased at this point at newer equilibrium states become possible and this is the overall profile of the equilibrium load deflection path. Now what we will do now is we will start considering certain issues related to dynamical systems and then we will relate all these two problems of stability analysis of structures as we go on, so for purpose of discussion we will start with a free vibration analysis of a single degree freedom nonlinear dynamical system, which is governed by this equation X double dot plus 2 eta omega X dot plus omega square X plus epsilon into G of X, X dot this is a nonlinear term which is equal to 0, we assume that this nonlinear term is a function of instantaneous values of X and X dot, it could as well be that this X could be function of tau where tau varies from 0 to T which is a current time, in which case the system will have hereditary nonlinearity, but right now we need not consider that, we are not going to consider that, so what we will do is we will introduce a new set of dependent variables I will call X of T as Y1 of T and X dot of T as Y2 of T, now clearly Y1 dot of T is X dot of T, but X dot of T is Y2 of T, therefore Y1 dot is Y2, Y2 dot of T is X double dot of T, X double dot of T we derive from the governing equation, so for X double dot we write minus 2 eta omega X dot minus omega square X minus epsilon G X, X dot, so that is what we are writing and in terms of Y1, Y2 this becomes the equation for Y1 and Y2 thus can be written in a compact manner like this, this is a pair of first order equations which are mutually coupled, so Y1 has terms containing Y2 and Y2 has terms containing Y1, now in general we consider equations of the form, suppose X and Y are the two dependent variables and I consider a pair of first order equations X dot is some F of X, Y and Y dot is some H of X, Y, we define the system state as a vector as X which consist of X and Y, now we define what are known as fixed points for this system, the fixed points are the system states at which the system state becomes time invariant, by that I mean those values of X and Y for which X dot and Y dot are 0, so that would mean at the fixed points we have F of X, Y equal to 0, H of X, Y equal to 0, so fixed points are those roots of this pair of equations, what do they mean? So we will consider some simple examples, a linear undamped single degree freedom system, Y1 dot is Y2, Y2 dot is minus omega square Y1, so Y2 equal to 0 and Y1 equal to 0 are the fixed points, you consider a damped system X double dot plus 2 eta omega X dot plus omega square X equal to 0, I have Y1 dot equal to Y2, Y2 dot is equal to minus 2 eta omega Y2 minus omega square Y1, again fixed points are Y2 equal to 0 and Y1 equal to 0, so origin is a fixed point, now we will consider another system X double dot minus alpha X plus beta X cube equal to 0, so Y1 dot is Y2 and Y2 dot is alpha Y1 minus beta Y1 cube, so here Y1 can be pulled out I write it as alpha minus beta Y1 square, so the fixed points are Y2 equal to 0 and this term equal to 0 means there are two possibilities Y1 equal to 0 or Y1 is equal to plus or minus square root of alpha by beta, so this system has three fixed points right, 0, 0, 0 comma plus this and 0 comma minus this, now these are linear differential equations, this is a non-linear differential equation, so one thing that we can observe immediately is for systems governed by linear systems there is only one fixed point, whereas for a non-linear system there can be more than one fixed point, now what are these fixed points of a dynamical system, what do they mean, what is their significance and how many fixed points can a system have, what happens if motion in the neighborhood of a fixed point is perturbed, okay, you have X dot equal to 0, Y dot equal to 0, now I slightly perturbed the motion, will the motion also die out or it will grow or it will remain unbounded I mean neither decays to 0 nor becomes unbounded what happens, next what is the rule of system parameters in deciding upon the number of fixed points, for example what happens if values of system parameters are varied, do the number of fixed points remain unaltered that you can quickly see here for alpha equal to beta equal to 0 the system has becomes linear and there will be only one fixed point, but beta not equal to 0 there will be three fixed point, so by changing beta I am changing the number of fixed points that is quite possible, do the nature of the motion in the neighborhood of fixed points change because of changes in values of system parameters, okay, so these questions we need to understand and of course relate them to the larger issue of stability of systems, so let's consider a linear undamped single degree freedom system, so as we saw the governing equation is X double dot plus omega square X equal to 0 and we will assume that the system starts from some initial condition X naught and X naught dot, now again I will write Y1 is Y2, Y2 dot is minus omega square Y1, fixed point is 0, 0, now we can solve this equation, so X of T we know it is given by R cos omega T minus theta and X dot of T is given by this, where R and theta depend on these two initial conditions, if initial conditions are changed R and theta would change, now what I can do now is I can plot X of T and X dot of T as a pair of time histories, that is what we have been doing, but instead I can also plot X dot on one axis and X of T and another axis and eliminate time from these plots, so if you do that what happens if you can quickly see here, if you write X square plus X dot square by omega square you can quickly see that this is R square, so that means if I plot X dot and X or X dot by omega and X, this will be a closed curve and it will be a circle if this scale is on Y axis I plot X dot by omega instead of just X dot, so that means at T equal to 0 we start the motion somewhere here and as time advances this point moves in the phase, this plane is known as phase plane and this point moves in the phase plane and traces a closed curve, for another set of initial conditions it traces another closed curve, so there can exist by slight change of initial conditions in another closed curve in the vicinity of this if I change the initial conditions slightly, okay, this has some significance we will see in due course what it is. Now let's consider damped system, so what is happening here there is no dissipation in the system and both displacement and velocity are sinusoidals and they never decay to zero and in the phase plane it traces a closed curve, so in the phase plane when I plot different trajectories of the system with different initial conditions, different system parameters and things like that what we get here is known as a phase portrait of the system. Now if we consider now a damped single degree freedom system, again in this case we know the exact solution so 0, 0 is the fixed point and if I now again find X of T and X dot of D, X of T is given by E raise to minus 8 omega T into A cos omega DT plus B sin omega DT I am assuming under damped system and X dot of T will be given by this, where by using initial conditions I can find A and B, so now if the motion is initiated somewhere here in the phase plane, X dot versus X of T plane it spirals and decays to zero, so how does a time history look like? This is typical plot of X of T, so X dot of T will be something similar and when plotted in the phase plane it traces a spiral which eventually lands at the origin, because at T tends to infinity X of T goes to zero, X dot of T goes to zero and therefore this trajectory goes to zero. If you start from some another initial condition similar another spiral will emanate and it will again as T tends to infinity goes to zero. We can consider systems which are externally driven, for example if I take a damped single degree freedom system with an external excitation F of T I can write it as Y1 dot is Y2, Y2 dot is now minus 2 eta omega Y2 minus omega square Y1 plus F of T, so if I now consider system states as Y1, Y2 I can write the governing equation in the so called state space form where Y dot is given by, I will write this in the matrix form, call this matrix as A and this is A into Y plus some capital F, so the governing equations here are of the form this. On the other hand if the system has nonlinear terms this type of matrix representation will not be possible we will simply get a generic equation of the form A is a function of Y, T, whereas here A is a matrix whereas here it is a 2 cross 1 vector of functions whereas it is 2 cross 2 matrix here, so that's a difference. Now we say that system is autonomous if A is independent of T, so that would happen clearly when there are no external excitations, so autonomous systems have the autonomy to choose their own frequency and amplitude of oscillations, that is why called it is autonomous. On the other hand if the system is driven externally, for example a linear system is driven externally by harmonic force at say a frequency capital lambda, even if the system natural frequency is omega in the steady state the system will be obliged to oscillate at lambda, so it doesn't have the autonomy to decide its own frequency, so this in such systems in the representation that we have used A will be a function of time, so external forcing for example influences the frequency and amplitude of oscillation. Now we will again return to the question of equilibrium points or the fixed point they are used synonymously, so I consider now an autonomous system Y dot of T is equal to AY, with Y of 0 is Y naught, now we define fixed points of this system as points at which the system state is at rest, that is Y dot equal to 0, so these points are obtained as roots of the equation A of Y equal to C, so these are set of equations, so as many equations as the size of Y and you will be able to solve for the roots of these equations. Again we have seen all this for linear system we got 0 0 and un-damped system, for damped system again 0 0 and here as we saw there are 3 points, so I as already pointed out non-linear systems have more than 1 fixed points. We now consider the question what happens if we perturb the motion or the system states equilibrium system states by small motion, if you apply a small force to the system when the system is at one of its equilibrium states what happens, so let's consider X dot equal to F of XY, Y dot is G of XY, I am considering a 2 cross 1 system of equation this can be generalized for more larger size problems, the equilibrium points are given by the conditions F of X star, Y star equal to 0, G of X star, Y star equal to 0, so this X star, Y star is a set of equilibrium points. Now let us examine the nature of motion around each of the equilibrium points by perturbing the equilibrium states by small amounts, so what I will do, I will take X of T as X star plus eta of T, and X dot of T that is this is Y of T as Y star plus some XI of T, now let us substitute this into this equation, X dot is X star dot plus eta dot, but X star dot is 0, because X star is a fixed point, so Y star is also 0, so I get X dot of T as eta dot of T which is F of this, and Y dot is XI dot is equal to this. Now what I will do is I will perform a Taylor's expansion around the fixed point and retain only the first order terms because I am assuming that the perturbations that I am providing are small, and mind you my question, the type of question that I will answer is not to determine the time histories of eta of T and XI of T in all detail, the only question that I will ask is that T becomes large, will eta of T and XI of T grow or decay or remain neither growing nor decaying, so these are the questions that we want to answer, in which manner they decay to 0 or in which manner they grow is not of primary concern, so we are asking therefore a qualitative question about the system behavior, so we are not so much particular about performing a complete time history analysis and answering this question, so if we now use Taylor's expansion F of X star, Y star plus the two gradients with respect to X and Y for the functions F and G, now we know that for fixed point F and G are 0, F of X star, Y star is 0, G of X star, Y star is 0, so I am now left with this pair of equation eta dot is this gradient into eta of T and this gradient into XI of T, and similarly another equation for XI of T, now these two equations I can put in the matrix form, I write it as eta dot, XI dot into this matrix Jacobian or gradient matrix evaluated at the fixed points and this is a vector of eta and XI of T, so these are linear homogeneous set of ordinary differential equation, now for sake of simplicity in writing I will drop writing this subscript that we are evaluating at X equal to X star and Y equal to Y star, let's assume that we are doing we need not have to write explicitly, now that would mean this is really a constant, now we seek the solution in the form, this is a linear system homogeneous therefore an exponential is always a solution, so we seek an exponential form of solution where the parameter S is unknown, so we want to know if this solution is permissible, if so for what value of S it is permissible, so I substitute into this, so I get S alpha beta exponential ST into this matrix alpha beta exponential of ST, now exponential of ST cannot be 0, so I get this equation which is nothing but an eigenvalue problem associated with this matrix, so the eigenvalues are S and alpha and beta are the eigenvectors, so clearly alpha beta equal to 0 is a solution to this problem for all values of S, but we are interested in those values of S for which alpha and beta are nonzero, so we perform the eigenvalue analysis in this case we get two eigenvalues S1 and S2 they will be complex and therefore they will be complex conjugates, so I can write the roots as S equal to A plus IB, now the solution therefore can be written as some alpha into exponential AT plus IBT, and this I can write it as alpha into E raise to exponential AT plus into cos BT plus I sine BT, similarly I will write a solution for XI of T, now this modulus of this exponential function E raise to IBT is 1, so this also modulus is 1, therefore as time becomes large if you are interested in observing whether eta of T, X of T grow or not we should look at how this multiplier behaves, E raise to AT will grow to infinity if A is positive, so the condition if A is greater than 0 then this limit of eta and XI as T tending to infinity becomes infinity and then we say that the fixed point X star, Y star is unstable, mind you this matrix is evaluated at this X star, Y star, so if A is less than 0 on the other hand as T tends to infinity the fixed point is the perturbation die we say that fixed point is stable. Now depending on how the nature of these eigen solutions that's also a solution we have different nomenclatures for the fixed point, we say that fixed point is a node if both eigen values are real and are of the same sign, they can be positive or negative but both should be real that means B will be 0, the fixed points can be stable or unstable depending on the sign of these two roots, both roots are negative then the node is stable, if both one root is positive another root is negative we get another situation. Now if both roots are real with one root real and other root imaginary, no other root negative the fixed point is unstable, now on the other hand the roots are complex conjugates and the fixed point could be stable or unstable depending on the sign of the real part, so if on the other hand if roots are pure imaginary that means real part is 0 then the linearized stability analysis is inadequate to answer the question on whether the fixed point is stable or unstable, okay, we have to carry out further analysis. Now let's see what we learn from, I mean if we apply these ideas to simple oscillators what is that we can learn, so let's take the damped single degree freedom system X double dot plus 2 eta omega X dot plus omega square X equal to 0 and I introduce Y1 dot is Y2, Y2 dot is this and this is my F and this is G, the fixed point is 0 0 and dou F by dou Y1 is 0, dou F by dou Y2 is 1, dou G by dou Y1 is minus omega square, dou G by dou Y2 is minus 2 eta omega, so the eigenvalues of this matrix are given by this characteristic equation and if I write this I will get lambda as minus eta omega plus minus square root this. Now depending on the nature of this eta there are different possibilities now, if eta equal to 0 then lambda will be plus minus I omega, then we say that origin is a center because this is a pure imaginary number, on the other hand if eta lies between 0 and 1 lambda will be given by minus eta omega plus minus I omega 1 minus eta square and since eta is, we were assuming that eta is positive, damping is positive it lies between 0 and 1, therefore origin is a stable focus, if eta is greater than 1 this will be, both roots will be negative and we say that origin is a stable node, if eta is less than 0 I leave it as exercise, origin is unstable, if negatively damped system, origin will be unstable and you can verify whether what is the range of eta for which this would be a saddle and a node or a focus, okay. Now let's consider now the nonlinear system, a simple nonlinear system, I have X double dot plus X dot minus X plus X cube is equal to 0, now again I introduce Y1 dot as Y2, Y2 dot as minus Y2 plus Y1 minus Y1 cube, where Y1 is X and Y2 is X dot, so I get this equation, so my F is Y2 and G is minus Y2 plus Y1 into 1 minus Y1 square, the fixed points you can see Y2 equal to 0, if Y2 equal to 0 then Y1 into 1 minus Y1 square is 0, therefore Y1 equal to 0 is a fixed point, so 0, 0 becomes a fixed point and 1 minus Y1 square equal to 0 is also a fixed point, so the fixed points are origin and plus minus 1, 0. Now we can evaluate dou by dou Y1 and other functions and this will be now function of Y, therefore we have to assign the right values for the equilibrium points and then evaluate the eigenvalues, so at the origin we have the eigenvalues to be given by this equation and we get lambda to be this and we can conclude that origin is unstable and is a saddle, we now consider the other pair of equilibrium points, this is plus minus 1 and 0 and we see that there is a term Y1 square here, so it doesn't really matter whether we take plus or minus, so the Jacobian will be, this determinant will be the same, so again the characteristic equation is this and we get the lambda to be a pair of conjugate, complex conjugate numbers with negative real part, therefore we say origin is a stable focus. Now we can see here if I now plot this potential energy associated with this system, so that will be omega square X square by 2, so if I plot U of XS versus X you get this type of a valley, so the motion of the system can be perceived as a ball rolling in this potential, okay, so what we are concluding here is origin is stable, there is damping, therefore this oscillation will continue and again it will come to rest, that means this is equilibrium point I give a slight perturbation, the system oscillates and it decays to 0, whereas here origin is here and if you place an object here any slight perturbation will move the ball away from this point, so the origin is unstable, whereas if you are here small perturbations you know motion will be stable, okay, now we will consider a simple pendulum, the governing equation is theta double dot plus G by L sine theta equal to 0, so again theta 1 dot is theta 2, theta 2 dot is minus G by L sine theta 1, so this F is this and H is this, the fixed points are theta 2 equal to 0 and sine theta 1 equal to 0, sine theta 1 equal to 0 means I have N pi, comma 0 where N varies from 0 plus minus 1 plus minus 2 etc., so this system has an infinite set of fixed points. Now the eigenvalues you can evaluate all the fixed points I mean can be written as N pi and by assigning different values to N we can approach different fixed points, so we will keep it as N pi and if you perform the analysis we get this to be the characteristic equation and for E 1 N I get the fixed points are centers, okay, then for 0 and E 1 numbers, so for example origin is a center, for N equal to 1, 3, 5 etc. the fixed points are saddles, okay, so we can plot on this axis the fixed points suppose this is a origin, suppose this is origin, now the pi will be here, minus pi will be here, so if I now plot the trajectory passing through that it will be like this, so these are the pendulum if you see what we are saying is origin is a fixed point and if the pendulum occupies a vertical position it's an unstable equilibrium that this is this point, since the pendulum can rotate over several you know complete motion is also permitted there are several fixed points, okay, so these points that you are seeing are origin and after it completes one revolution there is a 2 pi, another revolution 4 pi so on and so forth they are the fixed points, whenever the pendulum returns to this vertical position we come across an unstable fixed point and they are here, so these are centers, these are saddles and this is known as separate tricks. Now we will consider, we will apply these ideas to a mechanical system, a simple mechanical system which is made up of a rigid bar AB and there is a sleeve here and it is supported on a spring to this wall or some fixed condition here, so the position of this spring doesn't change but the sleeve can slide on this, so as a system oscillate this slide, this sleeve will be sliding on this, so now the question that we are asking is will the rigid rod remain in the upright position under quasi-statically load P, okay, so for some value of P can a neighboring equilibrium position exist, it's almost like asking the questions on, the kind of questions we asked on beam columns we are asking here, but mind you AB is a rigid bar and the flexibility is accounted by this presence of this discrete spring, so what we will do is we will formulate this problem as a problem in dynamics and see what we learn from this, that means we will basically consider the oscillation of this bar around this point. Now before we do that we should understand that if this bar starts deflecting to the right, moment this sleeve crosses this edge, our interest in this system is this, right, because that bar would have escaped from the sleeves and we are not interested in that, so we limit our interest to this range, minus cos inverse A by L to cos inverse A by L, so this range of angle, so we'll consider one of the snapshot of the deformed geometry of the bar, suppose at some point in during its oscillation it occupies this position, the load P is acting here, then the sleeve would have, you know, it would be originally at this distance, now it would have moved up, okay, so now I want to construct the expressions for kinetic energy and strain energy and write the Lagrangian and derive the equation of motion, so what we will do is we'll consider a point X, Y, and along this bar I choose a coordinate XI, so I will call X as XI sine theta, so theta is this angle, the axis of the deformed rod to the vertical, XI sine theta and Y is XI cos theta, so X is here and Y is there, so now DT is the incremental strain energy stored in a chunk D psi here, at a located at a distance XI, so this is half M D psi, M is a mass per unit length, X dot square plus Y dot square, for X and Y I will substitute this, so I will get this equation and the kinetic energy stored in the D psi element is given by this, and the total energy will be I have to integrate from 0 to L, and I will get it at ML cube theta dot square. Now the strain energy is given by the energy stored in the spring and work done by this load, so the change in length of the spring if you include and compute this, this will be half KA square tan square theta, and this is contribution to the load is minus PL 1 minus cos theta, so I can now construct the Lagrangian T minus V, and if I now put it in the Lagrange's equation and perform the necessary differentiations I get this equation. Now you see here inertia term is linear, but thus stiffness terms is highly nonlinear, there is cos cube theta here, sine theta here, so it's a highly nonlinear system, so for sake of discussion we will also include that there is dissipation in the system which is notionally represented through a viscous damper, so the governing equation for free vibration of this system is given by this. Now I will again introduce the state space form theta 1 dot is equal to theta 2, theta 2 dot I will derive from this minus 3 C ML cube I will divide by ML cube by 3 and I get this equation. So what I wish to do is we will identify the fixed points of this system and study their stability, how many fixed points are there? We will determine that first and around each of the fixed point we will apply a small perturbation and see whether perturbation grows in time or not, so this can be written as theta 1 dot is theta 2 I call F theta 1 theta 2 and theta 2 dot is this and this is some G of theta 1 theta 2, the fixed points are roots of F equal to 0 and G equal to 0, if you do that theta 2 is 0, once theta 2 is 0 this term drops off and I will be getting KA square by cos cube theta 1 minus PL sine theta 1 equal to 0, so sine theta 1 equal to 0 is a possibility and theta 1 is root of this equation which is secant inverse of this to the power of 1 by 3, we are not taking N pi here because as I said already we are not interested in the bar rotating about this, our limitation our interest is only in range of theta lying in this, so that is why we are taking only the one of the roots of this equation, theta 2 equal to 0 and theta 1 is this. So now we will evaluate the matrix whose eigenvalues we need to study, so dou F by dou theta 1 is 0, dou F by dou theta 2 is this and dou G by dou theta 1 and dou G by dou theta 2 are given by this, now consider the fixed point 0, 0 first, so I get the gradients in this case to be dou F by dou theta 1 is 0, 1 and dou G by dou theta 1 will be this quantity and dou G by dou theta 2 is this, so the eigenvalues of interest are to be evaluated by finding roots of this equation, if you do this I get the eigenvalues as minus C by 2 plus minus half square root of this etc. Now suppose for sake of discussion damping is 0, next if we see that if KA square minus PL is less than 0, okay, then real lambda will be greater than 0, therefore fixed point is unstable, on the other hand if KA square minus PL is less than 0, real lambda will be less than 0 and fixed point is stable, so therefore condition for stability of 0, 0 is P must be less than this, so this is the condition on value of P for origin to be stable, now let us consider the other fixed points which are given by this, now when deriving this we use this condition to find the roots KA square by cos cube theta 1 minus PL equal to 0, and consequently dou G by dou theta 1 will be given by this quantity and you have to see here all these quantities are real, there is cos to the power of 4 sine to the power of 2A square etc. KA is positive, this is alpha square, minus alpha square however alpha square is a positive number and all other gradients are as before, so the eigenvalues are given by this where alpha square of course is this messy term, and I get now this equation for lambda and from this we can see that the fixed points are always stable, okay, so now we can plot the load deflection diagram, so on this axis I plot theta and load is plotted here and first we look at origin 0, 0, until P crosses this, this will be the only fixed point that system has, and beyond this value of P the origin becomes unstable, so this cross indicates unstable points and dot represents stable points, but at this point two branches emerge, and along each one of these branch the solution is stable, okay, that is given by this, that is what this analysis of the second branch of fixed point shows. So P star which is the value of the load at which origin becomes unstable is given by KA square by L, so this is the complete load deflection analysis of this system by including large deformation and dynamics, okay. Now what we will do subsequently is we will assume small deformation and we will do a static analysis and see what happens, and then ask the question what are the type of problems where we need to include dynamic effects, what are the type of problem where we need to include large deformation effects, okay. So now if I were to do a linear dynamic analysis, that means after formulating the equation of motion I will linearize this term, sin theta 1 is written as theta 1, cos cube theta 1 is written as 1, and this equation becomes linear, and I get this equation. So again I can find out fixed points for this linear system, and perform the stability analysis, we find that, we will now only find origin, the other fixed points do not figure in this analysis, so we get the origin to be stable if this condition is satisfied, afterwards the origin will be unstable, so this is stable, and later on this is unstable. Now we can ask the question does the linearized stability analysis give qualitatively the correct picture of the phase portrait, do we, for example if a linearized analysis shows that a point is a center irrespective of value of certain system parameters, can we believe that? The answer to this question is yes provided the fixed point for linearized system is not a borderline case as in a center, if the linearized system predicts a saddle node or a focus then the fixed point is really a saddle node or a focus, but if it predicts it as a center we are not sure, so I will present an example to substantiate this argument, and this is taken from this book by Strogatz, the system here is given by X dot is minus Y plus AX into X square plus Y square, and Y dot is X plus AY X square plus Y square, now we can easily see that origin is a fixed point by equating this to 0, and Eigenvalue problem analysis shows that the origin is a center, because there are pure imaginary roots. Now let us analyze, now what we will do is we will solve this equation, it so happens that in this case it is possible to construct this solution, so what we will do we will make the substitution X equal to R cos theta and Y equal to R sine theta, and X square plus Y square is R square, so if I differentiate this it becomes 2XX dot plus 2YY dot is equal to 2RR dot, so I get XX dot plus YY dot is RR dot, so for RR dot I will write X into X dot, for X dot I will write the governing term from the governing equation, and for Y dot similarly I will write the second term here I get this, now after simplifying this I get the equation for R as R dot is equal to ARQ, so this can be solved, similarly the equation for theta dot can also be obtained and we can show that theta dot is equal to 1, so the governing equation in polar coordinates corresponding to the given equations in Cartesian coordinates is given by R dot equal to AR cube and theta dot equal to 1, now we can analyze this problem and show that if A is less than 0 as limit T tends to infinity R of T goes to 0, that means origin is a stable focus, now if A equal to 0 origin is a center, if A is greater than 0 R of T tends to infinity and origin is a non-stable focus, now whereas the linearized system analysis showed that origin is a center for all values of A, so exact solution does not support this observation, so if you get a center as an answer to your question on stability then you need to, you cannot be decisive, so what is advisable is to carry out a non-linear analysis, now we use the term bifurcations, what does this word mean, now non-linear systems are or dynamical systems are characterized by fixed points and their stability characteristics and also the nature of this fixed point, for example nodes saddle or focus etc. and they being stable or not, so and this dynamical system is characterized by set of system parameters, so as a system parameters in the dynamical system are changed, one observes that the number of fixed points can change, the nature of fixed points can change and the stability of the fixed points can change, the subject of bifurcation theory deals with these changes due to changes in the values of system parameters, bifurcation therefore is a change in the number of fixed points, their nature and their stability as a parameter of the system is changed. Now the occurrence of bifurcation is accompanied by qualitative changes in the nature of system behavior, if suppose a fixed point becomes from being stable it becomes unstable the system will behave differently, the system parameter values at which system undergoes bifurcations are called the bifurcation points, so Euler buckling phenomena can be interpreted as a phenomenon of bifurcation, so at P less than P critical the system will have certain types of fixed points and P greater than P critical it will have different types of fixed points. So we can consider to illustrate that a system with two parameters alpha and beta X dot plus alpha X plus beta X cube equal to 0, now if alpha is greater than 0, beta greater than 0 we can show that origin is a center, if alpha is less than 0 and beta is less than 0 this is center and saddle, if alpha less than 0 beta greater than 0 there will be two centers and one saddle, so you can analyze this problem and see if these statements are right. Now there is another kind of system behavior that we should be aware of what are known as limit cycles, now let's consider an equation of the form X double dot minus epsilon X dot multiplied by 1 minus X square plus X equal to 0, so here the nonlinearity term is associated with dissipation characteristics, we have been considering nonlinearity associated with stiffness they can be associated with dissipation also, this type of nonlinearity occurs for example in fluid structure interaction problems. Now if we now consider this system if X square is less than 1 then this 1 minus X square will be positive and this system will have negative damping, for small amplitudes of oscillation the system behaves as if it has negative damping, on the other hand if X is larger than 1, X square is larger than 1 then this term will be negative and negative into a negative term becomes positive and the system behaves as if it is positively damped, so if we now look at that in the phase plane, suppose if I plot a motion that is originating near the origin that is small oscillation will tend to grow, whereas a motion which is originated away from the origin will decay, so in between that there will be a periodic orbit to which the response will move towards in the steady state, okay, so that these periodic orbits are known as limit cycles, so they are isolated periodic free vibration solutions of nonlinear systems that means I am looking at free vibration solutions and irrespective of the initial conditions the same steady state periodic solution is obtained, and these in the neighborhood of this closed curve there cannot be another closed curve, a system can have several limit cycles, for example if you consider a dynamical system as given here you can show that this system has infinite number of limit cycles, so for example if you have the origin is unstable then small motions will start growing so there will be a stable limit cycle, any motion which is initiated within this will be repelled to this, whereas if there is another stable limit cycle, suppose this is another stable limit cycle at another distance, now if a trajectory gets initiated here to which orbit will it move towards, if both are stable there will be a contradiction, so there will be an unstable limit cycle sandwiched between two stable limit cycles, that means any motion within this region will go here, any motion in this region will go to that, so in systems with multiple limit cycles the unstable and stable limit cycles alternate, now I said that limit cycles are isolated close trajectories, if we now consider for example X double dot undamped single degree freedom vibrations, if I plot the phase plane plot for this, this will be one orbit for a given set of initial conditions say X naught and X naught dot, if I now change X naught by some small initial small change and similarly make a similar change on this I can realize in the neighborhood of this solution another close trajectory, this is not a limit cycle, any closed orbit in the phase plane is not a limit cycle, in the neighborhood of limit cycle there cannot be another close trajectory, so whereas in linear systems it is possible, but systems exhibiting limit cycles it is not possible, because if for example this oscillator that I discussed in the Van der Poel's oscillator, it has one limit cycle here, so if a motion is initiated here close to this it has to be absorbed by this, if it is here it is absorbed by this, so in the neighborhood there cannot exist another closed orbit, that is a major feature of a limit cycle oscillation. Now in the next part of our lecture what we will do now is we will consider energy methods for stability analysis, this analysis will be based on two axioms, here I will simply state the axioms we will consider what it means in the subsequent next lecture, so according to the first axiom a stationary value of the total potential energy with respect to the generalized coordinates is necessary and sufficient condition for the equilibrium state of the system, this is first axiom tells you condition for getting an equilibrium state, the next axiom tells a complete relative minimum of the total potential energy with respect to the generalized coordinates is necessary and sufficient for the stability of an equilibrium state of the system, so we will use these two axioms and study dynamical system or I mean here we are not considering dynamical system, we are considering basically statical systems as we will see and we will apply these problems to some of the simpler system that we have studied so far and then we will see that this can form the basis for developing stability analysis of elastic systems to start with we can use Rayleigh-Ritz type of approximations and then it will graduate to the finite element method, so a discussion on this will be taken up in the next lecture.