 We were looking at the damped harmonic oscillator and we had found that the solution to this equation, to this linear equation for weak damping using a regular perturbation series produces similar secular terms as we had encountered in our solution to the non-linear pendulum while using the regular perturbation method. We had understood this that the exact solution of the damped harmonic oscillator contains an oscillatory term which is cosine of t with a certain frequency and that frequency is a function of the small parameter epsilon. So, this was the origin of those secular terms. In particular we had mentioned that a generalization we will introduce a more general technique which is the method of multiple scales using which we will be able to eliminate such secular terms up to any given order. We are also motivated this technique by saying that if you look at the exact solution of the damped harmonic oscillator you can observe different processes showing up at different time scales. So, at early times you can just describe it as a harmonic oscillator with unit frequency and no damping. At slightly longer time scales we had seen that the effect of damping starts showing up at even longer time scales you can see that the frequency of the oscillator is slightly different from unity. So, in order to reflect that description we said that we are going to convert from a single time to at least 3 different time scales. So, T0, T1 and T2 you can go higher up but we will have to do more algebra in order to obtain solutions up to a given order in epsilon. So, we are going to now do this up to order epsilon square and we will see what we get as a consequence. So, let us now continue. So, as I had said we are going to convert this equation from an ordinary differential equation to a partial differential equation. The independent variables will become T0, T1 and T2. So, our derivative d by dt becomes del by del T0 into del T0 by del T and then this is just 1. I will replace this derivative del by del Tn with the symbol dn. This is just a shorthand notation. So, del by del T0 which just becomes d0. We have to just remember that all the d's are derivatives and the subscript indicates which variable it is being derived with respect to. So, this is epsilon d1 plus epsilon square d2. We want the second derivative in our equation. That term, the first term is a second derivative with respect to time. So, we want, this is an operator operating on itself. So, this is the square of this and if you open it up you get d0 square plus twice epsilon d0 d1 plus epsilon square d1 square plus twice epsilon square d0 d2 plus dot dot dot. So, I am not writing any further because these will be higher powers of epsilon. If we now, we also have to do an expansion for our equation. So, x is now x0 which is a function of T0, T1 and T2 plus epsilon times x1 which is also a function of the same 3 variables plus epsilon square x2. So, let us find the solution. So, now we will have to go back and substitute this expansion into the governing equation. So, let us do that. So, we have the operator in the first term and as a coefficient of epsilon square we had 2 terms d1 square and twice d0 d2 and this term, this operator operates on x0 plus epsilon x1 plus epsilon square x2 plus the second term is the damping term that has a first derivative. So, the first derivative is just d0 plus epsilon d1 plus epsilon square d2 operating on again the same thing and the last term is just x which is and this is our equation. Now, like usual we have to collect terms at various orders. So, you can readily see that the term at order 1 is d0 square. So, that is this term operating on x0. From the second term we do not get an order 1 contribution because there is a prefactor epsilon overall and then we have this term which contributes an x0. So, I can write it as d0 square plus 1 x0 equal to 0. Remember that d0 is defined as del by del t0. So, d0 square is nothing but del square by del t0 square. Now, let us collect the terms at order epsilon. So, we will get, so we will get the structure of the left hand side will remain the same. So, we will obviously have d0 square plus 1 operating on x1. So, that is d0 square operating on x1 that is an order epsilon term and then we have an additional x1. So, that is the, let me put it this in color. So, this operating on this and then this and on the right hand side we will have, so we are collecting terms of the order epsilon. So, we will have minus 2 d0 d1 on x0 and minus 2 d0 of x0. How do we get those terms? So, we will have 2 epsilon d0 d1 operating on x0. You can see that there is that is an order epsilon term. We will also have d0 operating on x0 and that is an order epsilon term because of the epsilon pre-factor in this term. We have already taken into account the order epsilon term from the last equation that I have indicated in green and that is on the left hand side of the equation. So, this is my order epsilon equation. Note the structure of the equation. On the left hand side we always have d0 square plus 1 operating on the variable which reflects the order of the order at which we are operating. So, this is order epsilon to the power 1. So, that is why d0 square plus 1 is operating on x1. At order 1 which was epsilon to the power 0, it was d0 square plus 1 operating on x0 and on the right hand side at the lowest order it is always 0. At higher and higher orders we have more and more terms but whatever appears on the right hand side should not depend on a quantity which is at the same order. So, you can see that all these terms operate on something which is already known. So, order epsilon, we already know the order 1 solution. We cannot go to order epsilon and solve the equations unless we already know the order 1 solution. So, on the right hand side should only be things which we know and on the left hand side should be things which we are going to find out at the given order. So, this is the structure. Let us write one more order and that is because so we are really going up to order epsilon square. What we are going to do is we are not going to do a very long tedious calculation where we go all the way up to order epsilon square and find out all the corrections. We will just find the order epsilon square correction in the first term or in x0 and that you will see will contain the essential aspects of the calculation. So, we are let us go to one more order. So, order epsilon square. So, we are going to order epsilon square because as you can see the damping is going to show up at order epsilon as we will see. But the change in frequency, the fact that the frequency is not 1 but slightly different from 1 will appear only at order epsilon square. So, in order to see that we will really have to go all the way up to order epsilon square. So, with that in mind we are writing down terms up to order epsilon square. So, again the same structure we will have d0 square plus 1 now at operating on x2. So, that is coming from operating on that and then that. So, the left hand side is coming from the terms I have indicated in red. What do we have on the so, on the right hand side we will have let me write down the terms and then I will explain. So, the first term is twice d0, d2, x0. So, let me use another colour. So, twice d0, d2, x0 will be coming from this term operating on that term. You can see that there is a prefactor epsilon square. Then we have d1 square x0. So, that is going to come from this operating on this. Again there is a coefficient x0 square. Then we have twice d0, d1, x1. So, we have twice d0, d1 operating on x1. So, you can see the product of d0, d1 contains an epsilon and then there is another epsilon. So, the whole thing will be epsilon square. So, then we have twice d1, x0. So, this is going to come from the product of this with that. You can see that there is a epsilon sitting before. So, it will multiply and then we will get twice d1, x0. And then at the end we have 2 d0, x1. So, that will be once again this into that. So, that is the origin of various terms. I have listed them out one by one. So, this, this, this, this and that. Note that all of them will have a minus sign because they occur on the left hand side of the equation, but I am going to shift them to the right hand side to show that they are really inhomogeneities for the differential equation. So, now the procedure is exactly like before. It proceeds by solving the equations at every order. We have to start at the lowest order and then go step by step. At every order, we will have to find the complementary function and the particular integral. In particular, we will have to be careful that the particular integral should not contain any resonant forcing term. If it contains resonant forcing terms, we will have to set those to 0. When we do that, you will find that we actually obtain something called amplitude equations. And the quantity which will appear as a constant at a given order will actually be a function of a slower variable or a longer timescale variable at the next order. And the resonant elimination of the resonant forcing term will give us the equation which governs that quantity. So, let us learn how does this work. So, at order 1, that is the simplest problem. This is just a linear homogeneous equation. x0, which is a function of t0, t1, t2 is a function of a0 e to the power i t0 plus complex conjugate. x0 is a real quantity. In this case, x is a displacement and x is being written as a sum of x0 plus epsilon x1 plus epsilon x1 square x2 and so on. All the epsilon's are non-dimensional. So, x1, x2, x3 everything represents a displacement. So, it is a real quantity. So, I have to, if I am using complex notation, I have to add its complex conjugate to make it real like before. Why did I give a gap here? Because I know that x0 now is not just a function of t0, it is also a function of t1 and t2. We are integrating this equation as if it is an ordinary differential equation. So, we are going to integrate it. However, when we integrate it, we have to remember that x0 is a function of t1 and t2. And so, what I call as my constant of integration, actually here is not a constant, but is a function of t1, t2. Similarly, you will get a complex conjugate term like before. This is going to be a complex quantity, which means that if you give it a real t1 and t2, it will return a complex number to you. So, a0 is a complex function of t1 and t2. So, this is slightly more complicated than what we had encountered earlier. Earlier we were encountering complex constants. Now, we are encountering things which are functions of t1, t2 and are going to return complex numbers. Let us see how to determine these unknown functions of t1 and t2. So, up to this order, this is the only thing that we can write. If y a0 is a function of t1, t2 is not clear to you, I encourage you to differentiate this equation with respect to t0 and substitute back at the order 1 equation and convince yourself that this equation satisfies the differential equation that we have written at order 1. So, now with this, so our order 1 problem is completely determined. There is nothing more to be done at this order. It is also clear that a0 remains undetermined at this order. Now, before we go further, you can see that a0 is a function of the longer time scales. t1, t2 are long and longer time scales. So, you can immediately see that on a short time scale, so order 1 is the shortest time scale. So, on the shortest time scale, a0 is going to behave. So, if you think of a0 as some function of t1 and t2, the dependence of a, and the fact that a0 depends on time is not going to show up unless you go to times as large as t1 and t2. So, at very early times, a0 is effectively a constant. So, if we stop the solution to the problem at this order, it is just telling us that this is a constant into e to the power i t0. What does that mean? That means that if you convert this into real notation, you have to add the complex conjugate part, you will just get a constant into cos t. t0 is just small t. So, a constant into cos t. This is consistent with what we have argued earlier, that at very early times, it just behaves as a harmonic oscillator, no damping and with unit frequency. So, this is the description at this order. Suppose I want a more detailed description, which is consistent with what I observe if I go to longer times, at slightly longer time, I will start seeing the fact that this is a damped harmonic oscillator. The effect of damping will start to be seen. So, we have to proceed to the next order. So, at order epsilon, we have the left hand side remains the same operating now on x1, but now it is an inhomogeneous equation. And we had found that the right hand side is twice d0, d1, x0 and then there is one more term twice d0, x0. We have just found that x0 is a0, which is a function of t1, t2 into e to the power i t0 plus complex conjugate. So, what is, so let me work out these terms on the right hand side, this and that. So, the red term is twice d0, d1, x0. And so, you can see that this term is just twice i, the d0 of x0 will bring, will differentiate e to the power i t0 and it will just pull out an i. And then the d1 will differentiate just a0 because e to the power i t0 is not a function of d1. Once again I remind you that d0 is del by del t0 and d1 is del by del t1. So, d0 into d1 is nothing but del square by del t0 t1. And if this operates let us say on x0, then it is just del square by del t0 del t1 operating on x0, which is a0 e to the power i t0. And when you do this derivative, suppose you do the t1 derivative first, then this becomes del by del t0. So, when you do the t1 derivative first it is only a0 which gets whose derivative gets taken because this part does not depend on t1. So, it just becomes del a0 by del t1 e to the power i t0. And then when you do the next derivative which is with respect to t0, then it is only this part which will get derived because this part is just a function of t1 and t2. So, this part will behave as if it is a constant and then you will have i e to the power i t0. So, this is what I am writing here. So, it is twice i del a0 by del t1 e to the power i t0. Now, that is not enough because I also had a complex conjugate. So, I need to take the complex conjugate also. I leave it to you to show that if you took the complex conjugate, then what you will obtain will actually be just the complex conjugate of this. So, this is just equivalent to saying that you can do the derivative first and the complex conjugation later or vice versa. So, I will get a complex conjugate of this part. So, all the i's will become minus i's and remember that a0 is a in general a complex function. So, you will have to put a bar on the top of the derivative del a0 by del t1 also. In this way, the one advantage of the complex notation is that that we can only take one half of the part into account while doing our algebra. However, when we multiply things, we will have to be more careful. We will encounter that later. So, now let us take the next term. The next term is twice d0 x0 and twice d0 x0 is just i, so 2i a0 e to the power i d0 and again like before it is cc. Whenever I am writing cc, it only means that it is cc of this. So, this cc means cc of the term in red. This cc in black means cc of the term in black. When I add them up, I will have a cc of the term in red plus a cc of the term in black. So, I will write a single cc, which means the complex conjugate of the whole expression, the first term plus the second term. So, let us write the expression. So, I will have, so I am now just writing the equation. It is d0 square plus 1 into x1 is equal to minus twice i. There is a minus because both the terms on the right hand side have a minus. And then, so you can see that this term also has a twice i, this term also has a twice i. I am going to add them up. So, I can put take them 2i common and then I have a del a0 by del t1 plus a0. e to the power i t0 is common in both the terms. So, I am writing it outside the bracket. And as I said before, the sum of this complex conjugate plus that complex conjugate gives me another complex conjugate, which is now the complex conjugate of this entire expression. So, this complex conjugate is the complex conjugate of this entire expression. I hope it is not confusing to you that I am using cc, cc. So, this cc is not the same. So, if you want to prevent confusion, then you can write cc1, cc2, cc3 and this is the cc of the whole thing. So, just to ensure that this cc, cc1, cc2, cc3 is not the same as this cc. So, now I need to solve the equation at order epsilon, until now we have just written down the equation and we have figured out the explicit form of the right hand terms. Now, you can immediately see that although we are using complex notation, when you add the complex conjugate e to the power i t0 is basically cos t0. So, this is going to give me a cosine term with unit frequency. So, it is cos t0. So, now look at the solution to the homogeneous part. If you just look at the left hand side and if you set it equal to 0, you will see the expression for the left hand side is the solution to the left hand side is sum alpha e to the power i t0. So, some constant into e to the power i t0 plus minus i t0 is a solution. The plus is what comes here, the minus will come in the complex conjugate. So, once again we have the familiar situation where we have a right hand side term which is a solution to the homogeneous equation. And so, this term unless we eliminate this term, this is going to cause secular terms. So, I am going to call this a resonant forcing term. It will give me a term of the form in expressed in real notation. It will give me a term of the form cos t0. And if you look at the left hand side, cos t0 is a solution to the homogeneous equation for the left hand side. So, if you have to guess a particular integral, you will have to take t0 cos t0. And so, that will give me secular term which will grow in time. I do not want that because that was the problem we had encountered earlier when we were doing regular perturbation. Now, this method gives me a way of getting rid of such kind of secular terms. What do we do? We say that we want to set this entire term, this entire term to 0. You can pay attention that if you and set this entire term to 0, then you are also setting the complex conjugate to 0 because if you have some function for complex number and if you are setting it to 0, then it is equivalent to setting the complex conjugate also to 0. So, this is one advantage of the complex notation that if you just set this to 0, this part automatically, the second part automatically gets taken care of and it is also set to 0. So, what I am going to do is I am going to set the coefficient of e to the power i t0 to 0. Notice that it gives us an equation for a0. This is exactly what it should be because as recall that in the earlier calculation, in my order first order calculation, I had said that if you stop the calculation at the lowest order, then a0 is just a constant. a0 is not really a constant. We already know that a0 is a function of the longer variables t1 and t2. So, at this order, so if you look at very short times, you will find that a0 is just behaving like a constant because you cannot see its variation. You need to look at a longer time window to actually realize that a0 is varying and varying slowly as a function of time. So, when you go to the next higher order, that correction appears. Going to the next higher order is equivalent to asking what happens on longer timescales. So, on longer timescales, a0 actually varies as a function of time and the process of elimination of resonant forcing term at the next order tells us what is the equation for a0. This is why I had mentioned that this is like an amplitude equation. If you think of a0 at the lowest order, if you think of a0 as an amplitude of the oscillator, the e to the power i t0 term is an oscillatory term and a a0 was its amplitude. Its amplitude is at the lowest order, its amplitude is just a constant. But if you go to a longer time window, you actually realize that its amplitude is not a constant, it is slowly varying. And what is the equation which governs that slow variation? That equation is determined by eliminating the resonant forcing term at the next order. So, the equation that we will find is just this, del. So, I ignore the constants. The minus 2i is just a constant because I am going to set this term to 0. So, that constant is not relevant plus a0 is equal to 0. So, this is my equation governing a0. Note that a0 actually is a function of two variables t1 and t2. But this equation will not tell me the dependence of a0 on t2. This will only tell me what is the dependence on t1. We will still have yet another unknown amplitude which is function now of t2. And one has to proceed once again to the next order in calculation to determine that unknown amplitude. We will continue in the next video.