 We were looking at the non-linear pendulum where we use the Linsted-Poyncare technique to eliminate the secular terms and obtain an expression which was bounded at all times. We determine the first non-linear correction and we found an approximation to the periodic solution of the non-linear pendulum. We also saw that as angles as large as 90 degree the two term expansion that we had found in the Linsted-Poyncare method was a very good approximation to the exact solution which was a more complicated formula expressed in terms of elliptic Sn and sine inverse functions. Now we will come to another more general technique which is called the method of multiple scales. Up till now we have been looking at the effect of non-linearity and we have looked at the non-linear pendulum and what it does, how does perturbation help us in obtaining approximations to the solution to the non-linear pendulum. Now I will show you that very similar problems arise even in linear equations. So let us look at a problem which is a slight modification to our simple harmonic oscillator problem. We are just going to put in a damping term. So you can imagine that this is earlier we had a mass connected to a spring now we have a mass connected to a spring and a dashpot. So this is the dissipative term. You can also assume that this has already been non-dimensionalized. So that introduces a small parameter in the problem. In this case the small parameter is a measure of the strength of damping. So epsilon is less than 1 and so it indicates that damping is weak. So this is a weakly damped system. This also implies so we are typically going to get underdamped oscillations. Recall that without when epsilon is 0 this is just a simple harmonic oscillator that we have already solved. By the addition of this term we have added damping. So we cannot get periodic solutions because this is like a friction term and so our oscillator is going to damp out with time because epsilon is much less we are going to typically get underdamped oscillations. Let us now try to so notice that this is a still a linear equation. We can of course solve it this is a linear constant coefficient equation. So we do know how to solve it exactly. Let us try to solve this perturbatively and you will find that there is an interesting feature which also comes out of this equation which is similar to what was seen in the non-linear pendulum. So once again because the equation has a term of the order epsilon. So I will say that the answer of the problem of the unperturbed problem which is just a undamped oscillator, undamped harmonic oscillator will also have to be perturbed by order epsilon. So x is equal to x naught plus epsilon x1 plus epsilon square x2 plus dot, if we substitute it into the governing equation then we will obtain x0 double dot plus epsilon x1 double dot as epsilon square x2 to 0. Once again like before we put together all the terms at various orders. So at the lowest order we just get the unperturbed oscillator. The solution to this is the general solution to this is sum a cos t plus 5. I have just chosen to write it like this. Note that you can also write it as a cos t plus b sin t, where a and b are constants of integration. Here a and phi are constants of integration. You can relate small a and phi to capital A and capital B. Now I am not going to solve an initial value problem here. So I do not specify the initial conditions. One can do that but the main point that we want to demonstrate here can be demonstrated without having to solve the initial without having to account for the initial conditions. So now let us collect order epsilon terms and we will have. So this left hand side is exactly the same operator but now for x1 and now it will be inhomogeneous. So I will have a order epsilon term from here. So that is minus 2 x0 dot. And that is my only order epsilon term for x0. For x1 I have already shifted it to the left hand side. And if I substitute what I know about x0 from here then this term just becomes plus 2 a sin t plus phi. Now you can already see an interesting feature which you have encountered before. This formula sin t plus phi is a solution to that homogeneous equation. If you take sin t plus phi, so if you take x1 is equal to some alpha sin t plus phi and substitute in the equation x1 double dot plus x1 you will see that it exactly satisfies the equation. So this is an exact solution to that equation. Now what does this imply? Whenever we have a right hand side which happens to be a solution to the homogeneous part of the left hand side or in other words is a multiple of the complementary function then the particular integral has to be multiplied by some power of t. In this case it is t sin t plus phi. So the particular integral in this case will take the form. So we will have to determine the exact coefficient but so the particular integral in this case has to be written as a linear combination of alpha times t sin t plus phi plus beta t cos t plus phi and we have to go back to this equation, the inhomogeneous equation and substitute it and find out what is the value of alpha and beta which satisfies the inhomogeneous equation. Alpha is equal to 0, beta is equal to minus a. So the solution for x1 of t is just minus a t cos t plus phi. I am not writing the most general solution. If I have to write the most general solution then I will have to add a solution to the homogeneous part which in this case is some c1 times cos t plus c2 times sin t. That part is necessary only if I am accounting for the initial conditions because initial conditions will determine the unknown constants of integration. So let me just not write the complementary function. Let me just write x1 in terms of the particular integral and this is the particular integral. The basic point that I want to emphasize here is that your x0 was an oscillatory function and x1 has turned out to be a function which has an oscillatory part but also a part which is going to grow with time. We have seen this behavior before. This is a secular term. This is arising in exactly the same manner. At the first order of correction we had a resonant forcing term and the resonant forcing term forced the left hand side and caused a secular term to be produced. Note that we have found this now in a linear equation. Earlier we had encountered this while we had solved for the non-linear pendulum. However, this feature is turning up or showing up even in a linear equation. Why is this so? The reason actually remains the same. Recall that by adding damping we have added a twice epsilon x dot. What is the exact solution to this equation? Because this is a constant coefficient solution you can go back and substitute e to the power lambda t. This will convert this equation into an algebraic equation for lambda. You will determine a complex conjugate pair and then write the answer in terms of a linear combination of e to the power lambda 1t plus e to the power lambda 2t. Lambda will have a real part and an imaginary part and if you express the final answer in terms of real things then you can show this is very easy to show that the exact solution to this problem is of the form a e to the power minus epsilon t cos square root 1 minus epsilon square t plus phi. Once again I have two constants of integration in my final answer an amplitude a and a phase angle phi. You could have we could have written this in terms of a cos and a sign in which case there would be an a 1 and an a 2. The two forms are exactly equivalent. Now you can see that why is there a secular term in my regular perturbative expansion. This is a linear pendulum but by add this is a linear oscillator. But by adding a dissipative term the exact solution is telling me that this is going to this is going to have an oscillatory part which oscillates at a certain frequency and a part which causes the amplitude to decay with time. If you look at the oscillatory part which is this cosine function you can see that the frequency of the oscillatory part the frequency of the oscillatory part which is 1 minus epsilon square depends on epsilon. It is not independent of epsilon. So, as you change epsilon the frequency will keep changing or in other words this is not exactly a time periodic system because the amplitude does not return back to its value but it is an oscillatory system and the 0 crossings actually the gap between the 0 crossings actually depend on epsilon. This is the same feature that we had encountered earlier in the nonlinear pendulum also. There it was an exactly periodic solution. So, the amplitude would return if it started from 1 it would again return to 1. Here because of dissipation if you start at 1 so we start at 1 we do not return to 1 after 1 oscillation. However, we one can still define a time period here the gap between things. So, the gap between here and here. And so, this is the frequency associated with the oscillatory behavior and the frequency depends on epsilon. Whenever the frequency depends on epsilon as I told you before you have a function whose you have an oscillatory function whose frequency depends on epsilon then the regular perturbation will typically lead to secular terms and you can get rid of the secular terms if you have a way of summing all these secular terms up to infinity. Now, the regular perturbation cannot do better than this. So, if you take a two-term expansion using the regular perturbation then provided t is fixed you can make it a good approximation by taking your small parameter to 0 by making a small parameter smaller and smaller. However, given a fixed number of terms in the expression we cannot take time to be bigger and bigger. So, this is a plot of x versus t. So, the two-term expansion which would be a sum of x 0 and x 1 would clearly show growth in time because of this part and this is a very unphysical behavior because our system does not admit any growth it does not there is no energy source all the energy was injected into the system at time t equal to 0 either through an initial displacement or initial velocity or both. Once that initial energy packet has been injected the system can only lose energy through dissipation. So, we expect an oscillatory behavior but the amplitude of oscillation is going to come down and it is going to come down exponentially as our exact solution tells us this is in the under damped limit. So, clearly the two-term expansion produces an unphysical behavior and we need to do better than this. Can we do an improvement just like we had improved the non-linear pendulum solution? So, for that we will introduce a new technique that technique is called the method of multiple scales it is more general than the Linsted point correct technique. Now, here the in order to understand the basic idea of the technique we I have plotted here the exact solution to this problem which is basically just this then I have plotted the linearized solution to this problem which is just this and I have plotted another approximation wherein I have an exponential damped part but I do not have the frequency correction in the cosine part. As I told you the exact expression has an exponential damping and its frequency of the cosine part also depends on epsilon. So, in this third approximation that I am indicating in red color there is an exponential part but the frequency of the cosine part is still the same as that of the linear part there is that square root 1 minus epsilon square is not there. Now, the reason why I am plotting this theory is to help you understand that we can think of the exact expression we can approximate the exact expression which is this so, in here it is plotted in brown as having different parts to it as time progresses. So, at very early times so, say let us say we look at only within this time window we can clearly see that the linearized approximation is good enough I do not need to worry about damping I do not need to worry about the fact that my oscillatory part has a frequency which is dependent on epsilon this part does not have any damping the cos t does not have any damping it also does not have any frequency correction beside t. So, the frequency is just 1. If we go to larger times so, let us say we went up to this time. So, up to here we were so, this is a good approximation. So, a good approximation at early time. Now, if I go further in time so, let us say I go up to here you can clearly see that this is no longer a good approximation the orange line is showing an amplitude which grows up to minus 1 whereas, the exact answer in brown has decayed. So, by the time you reach time which is about here about 3 you can see that you can start seeing an exponential decay. So, at very early times it is just cos t at slightly longer times you start seeing the exponential decay. If you go to even longer times then you will see that this brown curve and this red curve show slight deviation from each other. Now, what is the difference between the this brown curve and the red curve as I told you earlier the brown curve and the red curve are exactly the same they both have the exponential decay factor. But the brown curve which is the exact solution has the frequency correction inside cosine. The red curve does not have the frequency correction it is cos of just t. Do not worry about these extra factors that you are seeing here tan inverse of minus EPS and there is something else here all that has been put in so that all the 3 curves satisfy the same initial conditions. In this case that I have chosen the initial condition to be x of 0 is equal to some x naught which has been chosen to be 1 and then x dot of 0 is 0 because the formulas are different. So, their constants of integration in general turn out to be different. But if we adjust the constant of integration such that all the 3 formulas satisfy the same initial conditions then those will turn out to be the constant of integration. So, in this case this will be the one constant and that will be the other constant. In this case this will be the one constant and the amplitude is just 1. So, you can clearly see that at early times cos t is a good approximation. At even longer times you start seeing the fact that there is an exponential decay. So, the decay of amplitude starts becoming apparent. If you go to even longer times on this time scale you can see that the brown curve and the red curve show a slight difference. That difference can only be because the brown curve and the red curve have a cosine which are at slightly different frequencies. One is at cos of square root 1 minus epsilon square and the other is at cos of t. So, the frequency difference also starts showing up. So, what I am trying to say is that this problem can be broken up into pieces. So, at early times we only see an undamped oscillator which is oscillating with unit frequency. At slightly longer times we see the decaying amplitude of oscillation of the oscillator. At even longer times we start feeling the fact that the frequency of oscillation of the oscillator is slightly different from the exact problem. So, this suggests that the problem can be broken up into parts by treating each physical process. So, at early times it is just an undamped oscillator. So, that is an approximation that is the first level of approximation. At slightly longer times you will feel the effect of damping. So, it is not just an oscillator it is a damped oscillator. Then at even longer times it is a damped oscillator which is oscillating at a frequency which has a epsilon correction. So, that is the basic idea of the method of multiple scales. On different time scales different processes will different physical features of the problem will show up. And the basic idea is that that you can treat each of those time scales as it being independent of each other. So, let us solve this problem using the method of multiple scales. We will separate time scales, we will introduce multiple times in the problem. This will convert our ordinary differential equation into a partial differential equation, but that will not complicate the mathematics. It is still an if the resultant partial differential equations can still be solved analytically. So, let us understand the basic idea behind the method. So, the basic idea behind this method so, multiple scales is to correct the behavior of the regular perturbation technique. The regular perturbation technique had secular terms. And so, what we will do is up to now our independent variable in the problem is time. Now, we will say that there are multiple timescales in the problem. So, one time scale is t another time scale. So, these are all definitions epsilon t another time scale is epsilon square t. Where do we get the these time scales from? Let us look at the exact problem. So, the exact solution we have seen is A e to the power minus epsilon t cos 1 minus epsilon square to the power half into t. If I keep the e to the power epsilon t intact and if I just express this using an expansion, then you can see that at very early times, this is just A cos t roughly. At slightly later times, this is approximately A minus epsilon t into cos t. So, the frequency still remains unity, but we start seeing damping. At even later times, we start seeing damping and we also start seeing the fact that the oscillations are happening at a slightly different frequency. This effect is always there you know, but it starts it takes some time to show up ok. So, that is the basic idea that at order one time it is just a damp undamped oscillator at order epsilon t. So, when time becomes slightly larger, so then we start seeing a damped oscillator and at an even larger time we start seeing the effect which is due to the fact that this is oscillating at a slightly different frequency. So, this effect ok. So, this is where those time scales are coming from. You can see that we will call these t0, t1, t2 as so this is early time. So, when t becomes so all the t ends will be order one. What does that mean? That means that when t0 is order one we are at early time. When t1 is at order one that means that small t is very large only then can so when t1 is order one then it implies t must be of the size 1 by epsilon. Since epsilon is a small number so 1 by epsilon is a large number ok. So, this is at long time. When t2 is of the order one then time is of the order 1 by epsilon square. This is even bigger than this. If epsilon is a small number then 1 by epsilon is a big number, but 1 by epsilon square is an even bigger number. So, this is an even longer time. So, this is a longer. So, we will find so what we will do is we will convert this problem from small t to a problem where the independent variables are derivatives with respect to or with the independent variables are t0, t1 and t2 and we will have derivatives with respect to t0, t1 and t2. We will our dependent variable x will now be expressible as x0 plus epsilon x1 plus epsilon square x2. But each of these quantities themselves will become functions of t0, t1 and t2. We can keep going further we need not stop here, but we are going to do this process only up to order epsilon square as you can see this part is not 0. So, there will be higher and higher corrections to the frequency and the method of multiple scales is going to precisely give me that approximation if I solve it properly. So, we are going to learn how to do this using the method of multiple scales for this particular problem and we will see that the method of multiple scales will correct the behavior of the regular perturbation approach. In particular, it will give us a systematic way of keeping or eliminating secular terms up to any given order. So, if you say that we do not want secular terms up to order t2 then it will give us a way of getting an expression which has no secular terms up to order t2. This is a very important technique the method of multiple scales and is frequently used in the analysis of interfacial waves. Later when we do interfacial waves we will look at examples.