4 3 2 1 1 2, Figure 1 A plot of the unstable potential energy function blue curve for. k 1 The orange line indicates the amount of energy a particle would. need in order to be able to hop out of the potential minimum and travel. towards negative infinity, the somewhat strange shape near the origin a particle with an arbitrarily large. energy will be trapped inside of the potential minimum Thus we can consider. motion at an arbitrarily large energy without worrying about issues of stability. With this potential energy function my differential equation now becomes. y 2 y 2 y y 2 y 3 f t 7, For simplicity we will assume that there is no damping no driving force and. that the cubic term in the potential is zero so that the potential energy is. symmetric around zero In this case we find the Duffing equation. y 2 y y 3 0 9, It is a nonlinear differential equation that describes a simple harmonic oscillator. with an additional correction to its potential energy function This type of. oscillator is often known as an anharmonic oscillator How do we solve for. the motion of such a system, The most important thing to notice before embarking upon our search for. a solution is that in the presence of a nonlinear term the principle of super. position no longer applies That is if I have two solutions to my differential. Figure 2 A plot of the stable potential energy function for k 1. equation y1 t and y2 t then a linear combination of these two solutions is. NOT necessarily a solution because of course, y1 y2 6 y13 y23 10. Almost every solution technique we have used so far has at least in some way. involved the principle of superposition a property which we have now lost So. While we may no longer be able to use the principle of superposition we. do have one old tool which we can always fall back on perturbation theory. Our goal here is to understand how under a suitable approximation we can. think of the motion of the anharmonic oscillator as being a perturbation of. the harmonic oscillator s motion For nonlinear problems there will often be. many different ways to perform perturbation theory each with their advantages. and disadvantages We ll explore two techniques here although this list is far. from being exhaustive, Small Parameter Perturbation Theory. Let s imagine that the quantity describing the anharmonic term in our poten. tial is sufficiently small Actually more carefully we should say that. y 3 2 y 11, for the entire region in which the particle moves can you see why the cubic. term will always become important for large enough y no matter how small. is While we may not know how to solve for the motion of the particle exactly. we do know how to find the region in which it travels and thus we can check. whether this condition holds If the particle has total energy E then its turning. points must be the locations at which, U yM E kyM yM E 12. yM 1 1 2 E 3k 2 13, If the quartic part of the potential is indeed only a small correction in between. these two points then the nonlinear term in our differential equation should. only represent a small perturbation to the linear oscillator described by the. differential equation, y 2 y 0 14, whose solution we know to be. y0 t A cos t B sin t 15, Motivated by this thinking we might imagine that in some sense the solution. to the anharmonic oscillator is given by a small correction to the harmonic. solution a correction which depends on the small quantity Such a correction. might look something like, y t y0 t y1 t 16, The function y1 t is the correction term and we can think of it as the next term. in an expansion in powers of In this case however notice that the coefficients. in the expansion are functions of time That is not unlike the previous cases. we considered where for example the expansion was in powers of the drag. parameter but the coefficients in this expansion could depend on the other. parameters of the problem mass initial velocity etc in a more complicated. way If we plug this proposed solution into our full differential equation we find. y 0 2 y0 y 1 2 y1 y0 y1 0 17, If we expand this equation out to first order in epsilon we find. y 0 2 y0 y 1 2 y1 y03 0, Now in order for both sides of this equation to be equal for all values of. it must be the case that each parenthetical term vanishes In this case the. zero order parenthetical term yields, y 0 2 y0 0 19. which is nothing other than the equation for the linear oscillator which we know. how to solve If we also match the first order parenthetical term then we have. y 1 2 y1 y03 20, Since we already know what y0 is this represents a forced linear differential. equation for y1 which is something we do know how to solve In particular. this equation describes the function y1 as the coordinate of a simple harmonic. oscillator with frequency, For concreteness let s assume we ve chosen our initial conditions such that. y t 0 1 v t 0 0 21, where I ve avoided using y0 and v0 so as to not confuse them with the coefficients. in the expansion Applying these initial conditions to the zero order solution. y0 t cos t 22, Our perturbative equation then tells us. y 1 2 y1 cos3 t 23, Using a trigonometric identity we can rewrite the cubed term as. y 1 2 y1 cos t cos 3 t 24, This is just a linear undamped oscillator subject to two sinusoidal driving forces. As for its solution we can simply quote our results from the past lecture in. order to find, y1 t A cos t B t sin t cos 3 t 25, where the constants A and B arise from the homogeneous part of the solution. Thus the full motion through order is given by, y t A 1 cos t B t sin t cos 3 t 26. If we apply the same initial conditions as before a short calculation reveals. and so we find, y t 1 cos t cos 3 t t sin t 28, 32 2 32 2 8. Notice that the coefficient on cos t is slightly less than one by the same. quantity which multiples the cos 3 t term Thus some of the weight of. the solution has been transferred into an oscillatory term with a frequency. that is an integer multiple of the original frequency We say that the nonlinear. interaction has excited a higher harmonic of the oscillator which is a general. feature of nonlinear differential equations While the linear system required an. external driving force in order to excite higher harmonics the nonlinear system. is capable of doing so under the action of its own internal dynamics This type. of behaviour would also appear for example in a course on the Standard Model. in which a similar type of differential equation this time describing something. known as a Quantum Field would be used to describe how the interaction of. particles can create and destroy new particles, The Poincare Lindstedt Method. There is however an obvious problem with the answer we have found from using. perturbation theory Notice that the sine term has a factor of t it continues to. grow over time increasing without bound This is totally inconsistent with the. behaviour we expect on physical grounds the particle should simply oscillate. back and forth between the two turning points The appearance of this term is. actually quite general and not special to this case The term appears because. in the expansion of y0n for odd values of n there will always be a sinusoidal. term with the undamped frequency of the linear oscillator. cosn t cos t 29, involves a binomial coefficient Even powers also cause problems although this. only becomes clear at higher orders in perturbation theory Because the differ. ential equation for y1 has the same natural frequency as the linear oscillator. y 1 2 y1 y03 31, then there is an undamped resonance resulting in this diverging oscillation. amplitude This type of term one which arises in perturbation theory and. grows without bound over time is often known as a secular term. The resolution to this problem comes from realizing a second somewhat. more subtle problem with our solution it oscillates at the wrong frequency The. frequencies which appear in our solution are and also a harmonic multiple. 3 Thus our solution is still periodic with frequency However this is. inconsistent with the fact that in addition to changing the functional form of. the solution the quartic perturbation to the potential will in general also change. the oscillation frequency of the spring In some sense the reason that our naive. version of perturbation theory has failed is because we have not taken into. account the fact that the new frequency of oscillation in our system will no. longer be the same as the frequency of the linear oscillator So how do we take. this fact into account, As a fist step notice that if y0 were to oscillate at a frequency other than. we would no longer have our secular term problem the forcing function. would no longer be in resonance with the differential equation for y1 This. realization gives us the idea that maybe we can make a slightly better choice. for our unperturbed solution and instead choose, y0 t cos t 32. for some 6 This revised choice reflects an attempt to incorporate the. additional change in the frequency of the oscillator while still perturbing. away from the solution we already know So how should we choose One. guess might be to simply replace it with the actual frequency of oscillation in. the full potential which we can find from the period. where y and y are the two turning points However while this will give the. correct oscillation frequency perhaps it may seem as though our unperturbed. solution y0 should not necessarily incorporate the exact frequency of the oscil. lator it is after all only supposed to be approximately correct not exactly. correct Of course we can always perform a perturbative calculation of. writing the new frequency in terms of an expansion in. as you did on the homework However it s not clear exactly how many terms. we should take, In order to side step this thorny issue altogether we will make use of a new. tool sometimes known as a dual series expansion The idea is that we have. two objects the frequency and the trajectory y t which both need to be. expanded in terms of By expanding them simultaneously in just the right. way we can eliminate the secular term from our solution In order to do so we. will in fact not modify the solution y0 directly but instead define a new time. so that in terms of this new variable our differential equation becomes. 2 y 2 y y 3 0 36, The derivatives in the first term are now derivatives with respect to and so. the chain rule pulls out a factor of 2 since, Make sure to understand that this is exactly the same equation as before simply. written in terms of a new coordinate The difference will come when we conclude. later that is something other than, We now seek a solution of the form. y y0 y1 38, If we plug in this proposed solution along with the expansion for we arrive. at the equation, 1 y 0 2 y0 1 y 1 2 y1 y0 y1 0 39, where we used the fact that. which must be true in order to recover the correct value for the frequency when. 0 If we expand in and only keep terms up to first order we find. 2 y 0 y0 2 1 y 0 2 y 1 2 y1 y03 0, The vanishing of the first term tells us that. y 0 y0 0 42, which has the general solution, y0 A cos B sin A cos t B sin t 43. or to zero order in the frequency, y0 A cos 0 t B sin 0 t A cos t B sin t 44. This is indeed the correct solution when 0, So far it would not appear that our new technique has accomplished any. thing However things start to look different when we consider the first order. y 1 y1 2 y 0 2 y03 45, This is again a linear differential equation describing y1 with a forcing term that. depends on y0 although it has a slightly different appearance Again choosing. our initial conditions such that, y t 0 1 y t 0 0 46. we find according to the chain rule, y 0 1 y 0 0 y 0 0 47. so that our zero order solution is, Therefore our first order equation reads. y 1 y1 2 cos 2 cos3 49, or using the same trigonometric identity for the cosine cubed term. y 1 y1 1 cos cos 3 50, Again we find an equation for y1 which contains an undamped resonance. the natural frequency of y1 in this case simply 1 is matched on the right. by a sinusoidal forcing term with the same frequency This resonance would. cause y1 to contain an overall factor of which also goes to infinity for very. large times Thus it would seem that we still have the same problem as before. However notice that in this case the resonant forcing term is multiplied by a. factor which involves the expansion coefficient 1 If we were to set. then this problematic term would be gone and we would have a well behaved. solution Thus we see how our secular term can actually be turned into a. useful tool rather than a problem If I require that my solution be free of any. problematic divergent terms which I know must be the case then this forces me. to make a specific choice for 1 which helps me determine the series expansion. for This technique is known as the Poincare Lindstedt Method and it. is a very useful tool for studying periodic potion in a nonlinear potential It. can be continued to higher orders in and at each step in the expansion the. elimination of a resonant forcing function will fix another term in the expansion. Having made this choice for 1 we find, y 1 y1 cos 3 52. Quoting our result from before the solution to this equation is. Nonlinear Oscillation Up until now we ve been considering the di erential equation for the damped reasonably small oscillations around an equilibrium point For this reason we should add at least one more term to the Taylor series expansion of the potential U y 1 2 ky2 1 6 y3 1 24 y4 6 For gt 0 this now describes a stable potential energy function A plot of this improved

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