 We were looking at the solution to the damped harmonic oscillator using the method of multiple scales. We had chosen 3 times scales T0, T1, T2 and then we had expanded and expressed everything in terms of 3 independent variables T0, T1 and T2. In particular we had found that at the lowest order of description our system just behaves as a undamped oscillator with unit frequency. At the next order where we expect some effect of damping to show up we had found that we have an inhomogeneous ordinary differential equation. The right hand side has a term which is proportional to e to the power i T0 that is a resonant forcing term and in order to in order to prevent appearance of secular terms we need to eliminate that term. That in turn gave us an equation for the amplitude A0. The A0 was actually a function of T1 and T2 and was appearing had appeared earlier as a coefficient of the lowest order solution. So now by eliminating the resonant forcing term we have an equation governing A0. If we solve this equation this is easy to solve this is just a first order ODE. This is actually a PDE but we can again solve it as an ODE. So you can see that this equation just has the solution A0 which is a function of T1 and T2 as we had indicated earlier is equal to some small A0 into e to the power minus T1. Once again this A small A0 is not a constant but it is a function of the other unknown T2 of the second independent variable T2. So now we know the T1 dependence of capital A0. The T2 dependence is still unknown. So this part is still unknown. And so now we have found out the T1 dependency of A0 the T2 is still unknown and that helps us in simplifying the equation governing X1. So this is the equation governing X1. The right hand side is now completely 0 because we have said the only term which appeared was e to the power i T0 and e to the power minus i T0 by setting the coefficient of e to the power i T0 to 0. We have also set the coefficient of e to the power minus i T0 to 0 because the coefficient is just the complex conjugate of the expression that we have just set to 0. So if this expression in the red box is 0 its complex conjugate is also 0 automatically. So now this tells us that the equation governing X1 is also a homogeneous equation because the right hand side is now completely 0. So I can write the solution to X1 as so the equation after this simplification governing X1 is just this. So this is after elimination of resonant forcing on RHS on the right hand side. So once we have eliminated that that gave us the T1 dependency of the variable capital A0 and this is in turn leading us to this equation for X1. This turns out to be the same equation which governs X0 also. Again this is easy to solve. So X1 which is a function of T0, T1 and T2 can be once again written as some coefficient like before. At the lowest order I was using the symbol A0 now I will use the symbol A1 because this is order epsilon calculation A1 is a function of T1 and T2 into e to the power i T0 plus the complex conjugate of this part. Like before like A0 A1 is also a complex function of T1, T2 you give it two real numbers it may give you a complex number. So this is in general a complex quantity and because this is complex we have to add its complex conjugate part to give us a real X1. So now we have determined X1, we have determined A1, A1 is not completely determined because in the expression that we have for A1 there is a small A0 sitting here. So you can see that there is a small A0 sitting here this is a function of T2 and we do not know what function of T2 is this. So for this we need to go to higher orders. So let us proceed now to the third order of calculation. So order epsilon square here the left hand side is the same now operating on X2 and we had written earlier that there were a few terms in the right hand side we had written that earlier that was minus twice d0 d2 operating on X0 minus d1 square X0. As I said before you can see that the right hand side is getting lengthier and lengthier at the lowest order we had 0 in the right hand side. At the next order we had we had two terms on the right hand side the term indicated in the red bracket box and the term indicated in black box both of them produced a resonant forcing term we eliminated it and so our right hand side was once again 0. Now at the or at order epsilon square I have 5 terms and so you can see that the amount of algebra that we have to do keeps becoming more and more as we go to higher orders. So let us do this so we now have to express so you can see that all the terms on the right hand side are known because we know now what is X0 and what is X1 we only have to take the derivatives. So let us do them one by one so I will call this 1, term 2, term 3, 4, 5. So 1 is minus twice d0 d2 of X0. So for reference let me write let me raise this complex thing and let me write the solution for X0 also so that we have both the expressions for X0 and X1 at one place. So we had found that X0 plus complex conjugate again if to eliminate confusion I will put a C C1 and C C2 to indicate that this C C1 is an complex conjugate of the term on its left whereas the C C2 is a complex conjugate of the term on its left. And we had also found A0 so I am just summarizing what we have found so far we had also found that A0 is small A0 this is unknown yet e to the power minus t1. So up to now this is an unknown and that is an unknown. So because small A0 is unknown so you can also think that capital A0 is also an unknown but it is partially known because we know that A0 depends on t1 as e to the power minus t1. However the t2 dependence is not known so that is why I am not putting A0 capital A0 in a red box it is partially known and then partially unknown. So now let us calculate this term this is equal to minus twice d0 d2 plus C C again I am not going to write C C1 C C2 it should be understood that whenever I write a C C it indicates that it is a complex conjugate of everything that appears on the left of the C C. So every C C is not the same so if I have multiple C C is in my in my in what I am writing every C C represents a different expression that should be remembered. And this is just minus twice i del A0 by del t2 into e to the power i t0. We know A0 from this expression we will do the derivative later but let me write it let me keep it in this form right now. The second term is minus d1 square x0 this is equal to minus del square A0 so by del t1 square e to the power i t0 plus again the complex conjugate of this term. The third term is twice d0 d1 x1. We need all of these terms before we can work on the equation at order epsilon square because each of these terms can potentially produce a particular integral first we will have to see whether it produces a resonant forcing term or not if so we will have to eliminate the resonant forcing term then what is left we will have to work out the particular integrals of those terms and then we will have to write the general solution as a linear combination of the homogeneous solution plus the particular integral. So this is the general structure that we will follow. So minus twice d0 d1 of x1 is just minus twice i del A1 by del t1 e to the power i t0 plus cc x0 and that is minus 2 del A0 by del t1 e to the power i t0 plus its complex conjugate and the fifth term is just minus 2 d0 x1 and this is equal to minus 2 i A1 e to the power i t0 plus its complex conjugate. So, we have now calculated all the five terms let us put them together so the left hand side is just putting on x2 and then it is 1 plus 2 plus 3 plus 4 plus 5 so then that is minus I will put minus common so then we have twice i del A0 by del t2 that is my first term then my second term is a second derivative plus del square A0 by del t1 square. Note that all the terms have a e to the power i t0 all the five terms have a e to the power i t0 so I am going to pull that out and keep it common you can also see that it will produce all of these terms will produce a resonant forcing term. The third term is twice i del A1 by del t1 you can check that and then the fourth and the fifth term. So, you can take a 2 also common there is in the second term there is no 2 so we cannot take 2 common and then the fifth term is twice i A1 plus complex conjugate of everything on the left. So, the complex conjugate of this whole thing in square brackets multiplied by e to the power i t0. So, now we can simplify this a little bit we have seen earlier that A0 is small A0 which was a function of t2 an unknown function as yet and then a known t1 dependence. So, wherever we have derivatives with respect to A0 we can work out those derivatives. So, I have a derivative with respect to t0 A0 here a derivative here and a derivative there. So, we have so del A0 so let me write it on the side del A0 comma del t2 is just del A0 by del t2 e to the power minus t1 and then del square A0 by del t1 square it will take the derivative of e to the power minus t1 twice. So, I will get back the same expression and then we also have del A0 by del t1 which is just minus of this. So, if I plug in this on the right hand side I have a minus this is twice i del A0 by del t2 to the power minus t1 plus A0 e to the power minus t1 that term it remains intact. Then the fourth term gives me and the fifth term is just twice i A1 plus the complex conjugate. If we remember that we have to take the complex conjugate we do not have to keep writing it all the time. So, now you can see that all the terms so there are five terms and you can see that all the terms have a pre factor which is this which is this or a post factor in this case e to the power i t0. As I told you earlier e to the power i t0 is a solution to the homogeneous equation. If I set the left hand side is equal to 0 any alpha into e to the power i t0 any constant into e to the power i t0 is a solution to the homogeneous equation. Once again we will have to set the entire coefficient of e to the power i t0 to 0. So, everything in square bracket here goes to 0. So everything in square bracket goes to 0 and so we obtain. So, I am just going to write twice i A1 plus twice i del A1 by del t1. So, I am writing this term and this term and I am shifting others to the right hand side. So, this and this can be combined and it just gives me minus a0 e to the power minus t1. If I shift it to the right hand side it just becomes plus a0 e to the power minus t1. Then we have minus twice i del A0 by del t2. Please remember what are we trying to do here? We have obtained this equation by setting the coefficient of e to the power i t0 on the right hand side to 0. So, this minus that is there is not relevant for our calculation. We do not have to worry about it because I am taking the entire square bracket and setting it equal to 0. Even if I had a minus it does not matter because I am going to set it equal to 0. So, I took the entire square bracket without the minus set it equal to 0. The two terms that I have encircled they are the same. So, I can add them and it gives me a minus a0 e to the power minus t1. I shift it to the right hand side and then this the first term here the first term here this is also shifted to the right hand side. So, it comes here. So, I have missed e to the power minus t1 here and then the terms that I have indicated by tick marks. So, let me put them in color. So, this tick mark and this tick mark. These two terms stay on the left because these terms depend on capital A1 and the terms on the right depend on small a0. This is the structure in which I have written down things. So, this now is telling me something interesting. Like before it seems to be giving us an equation governing A1. Recall that A1 was an unknown at order epsilon. A1 was just some unknown function of t1 and t2. We have seen this before. So, A1 in square brackets in the red rectangular bracket at the top of the slide you can see is an unknown function yet. And now by eliminating resonant forcing at second order order epsilon square we seem to be getting an equation for A1. However, on the right hand side is small a0 and we do not yet know what is small a0. So, let us see how to proceed with this. So, we have to solve this equation. Let us do it. It is easier if we shift the 2i on the left hand side. Both terms have a 2i. If I shift it to the right hand side it becomes a half. There is an i in the denominator but I can multiply numerator and denominator by i and bring i to the numerator and it becomes minus i because of i square in the denominator. So, then the equation becomes del A1 by del t1 plus A1 is equal to minus i as I told you earlier and multiplied by half into whatever we had a0 minus twice i del a0 by del t2 and I am taking the e to the power minus t1 out. So, this is also again this is actually a partial differential equation but can be solved like an ordinary differential equation. There will be a Cf and there will be a Pi. So, the Cf is just taking the left hand side and setting it equal to 0. So, Cf is just del A1 by del t1 plus A1 is equal to 0 and so this implies A1 which recall is a function of t1 and t2 is some constant it is not really going to be a constant into e to the power minus t1 and because A1 is a function of t1 and t2 and we are integrating in t1. So, this small a1 becomes a function of t2. So, this small a1 is the like the small a0 which we had earlier and which is appearing in the right hand side of this equation. So, a1 a0 small a1 small a0 are still unknowns but we have figured out the t1 dependence of a1 from the homogeneous part. So, this is the complementary function. The particular integral is easy you can see that this entire thing is just a function of t2 there is no t1 here why because small a0 is a function of t2. We have said that earlier that small a0 is a function of t2 alone. So, this part that I have underlined here is a function of t2 because it involves a0 and derivatives of a0 with respect to t2. So, this is just a function of t2 it is an unknown function but it is a function of t2. So, the entire term on the right hand side. So, this entire term on the right hand side has e to the power minus t1 with a coefficient which is just a function of t2. So, I can find the particular integral for this very easily. I say that the P i is going to be alpha which is a function of t2 t1 e to the power minus t1. Why did I take t1 e to the power minus t1? Because some constant or some function of t2 into e to the power minus t1 is a solution to the homogeneous part. So, this is like a resonant forcing term but you will see that this will not produce a secular term but we will still have to eliminate it. So, I have taken t e to the power because this right hand side is e to the power minus t1 is a solution on the left hand side. So, we have to take t e to the power minus t1. So, that is why alpha t2 into t1 e to the power minus t1. There has to be a function of t2 here because the right hand side has a function of t2 where small a0 is a function of t2. So, I hope this part is clear. So, now we have to in order to determine P i, we only need to find out what is alpha. So, we have to go back and substitute into this equation. This is the usual way in which we find particular integrals. We will just get alpha into I take the derivative of t1 first. So, it is just t to the power minus t1 plus alpha. Alpha is just a function of t2. So, it does not participate in the differentiation. And then this and there should be a minus sign here e to the power minus t1. That is the first term del a1 by del t1 plus a1. A1 we have said is alpha t1 e to the power minus t1. So, alpha t1 e to the power minus t1. And this is equal to minus I by 2 a0 minus twice I am just writing the right hand side of that equation. This term and this term cancel each other. This tells me that alpha is just minus I by 2 a0 minus twice I del a0 by del t2. It may appear a little bit tedious. It may appear more complicated than what we had done in the Linstead Poincare method. But this is a much more powerful technique. And we will see use of it when we do interfacial waves. So, this and then I can push the minus sign inside if you want minus a0 plus. So, alpha as we had expected is just a function of t2. This a0 is a function of t2 del a0 by del t is also a function of t2. So, that is alpha. So, we now know the complementary function. The complementary function was this. We also know the particular integral. The particular integral was alpha times t1 times e to the power minus t1. And we now know alpha. So, we can write the most general solution of this equation. So, this is the equation whose general solution we are trying to write. So, let me write. Say a1 which is a function of a small a1 e to the power minus t1. That is my complementary function. And then a particular integral which is i by 2. I am just writing the value of alpha into t1 e to the power minus t1. So, this is the expression for a1. Now, what do we do next? So, let us summarize what we have obtained so far. So far we have found that x is equal to x0 plus epsilon x1 plus epsilon square x2. We have not yet solved for x2, but we have solved for x0 and x1. x0 turned out to be a0 t1 t2 e to the power i t0. I am not writing the complex conjugate, I will write it at the end. Epsilon times x1 was a1 which is also a function of t1 t2 e to the power i t0 plus we have not solved for x2 yet. So, let us put a dot dot. We then found out that a0 is given by small a0 which is a function of t2. Then it is e to the power minus t1 into e to the power i t0. Similarly, epsilon times a1 we have found a small a1 which is some unknown function of t2 plus i by 2 minus a0 plus twice i. This is the alpha that we had found earlier and then we had a common e to the power minus t1 and then this gets multiplied by e to the power i t0. Plus we have higher order quantities and then we have to add the complex conjugate of everything to the left. So, this is what we have found so far. Now, if you look at x0, then x0 is of the form a0 e to the power minus t1 e to the power i t0. If I go back to my small t variable then this will turn out to be e to the power minus epsilon t. So, there is the damping that you can see that we had anticipated and then this would be i t. This is fine. Epsilon x1 the second term, the second term is epsilon x1. What did we find this to be? This is epsilon a1 of t2 plus i by 2 into what is there in the bracket. T1 is epsilon t into e to the power minus epsilon t into e to the power i t. This is what we would find if we went back to small t variables. Now, look at the term inside the square bracket, look at the term because the second term. Now, what is here is a0 plus twice i del a0 by del t2. You can see that this epsilon will multiply that epsilon and this will be a term of the form epsilon square t into e to the power minus epsilon t into e to the power i t with some coefficient. We can prevent this by setting the coefficient of this term to 0. So, we can prevent this disordering. By ordering I mean every term in the perturbation series is smaller than all the terms on its left. So, we can prevent this disordering by setting minus a0 plus twice i del a0 by del t2 equal to 0. You can immediately see that this gives you the equation that we had sought for small a0. So, we are going to solve this equation and we will find that there are no more unknowns. There will only be an unknown constant when we integrate this equation. There will be no more unknown functions which appear here and that unknown constant can only be obtained by from initial conditions. We will continue this in the next video.