 We were looking at the damped harmonic oscillator using the method of multiple scales. We had expanded up to some order and we had found that at various orders one has to eliminate the resonant forcing terms. In particular at order epsilon we had discussed that the coefficient of the term that I have indicated here this term in the box. The coefficient of this term is something which will get disordered. This is an order epsilon term and so at large times time of the order 1 by epsilon square this term becomes as large as the first term in the expansion. So, to prevent that we would have to set the term in that rectangular box to 0. I had shown that this leads to an equation for a naught. Recall that a naught is a function of t2. So, we now have an equation governing a naught. Let us solve that equation and determine the functional dependence of a naught on t2. So, one can solve this equation easily and it leads to the solution a naught as a function of t2 is now some constant of integration. Now the constant is no longer a function of any other variable because a naught is just a function of t2. So, this is really a constant into e to the power minus i t2 by 2. This is just the solution to the equation that I had indicated in the previous page at the bottom of the page. If you integrate that equation you will find this solution. Now this is a constant of integration and in general it is a complex constant. So, now let us put things together. So, now we have found that x0 which was a function of t naught t1 and t2 is now a00 e to the power minus i t2 over 2 then it is into e to the power minus t1 into e to the power i t naught. This is essentially rewriting this expression in the yellow box. x naught is a naught e to the power i t naught and a naught in turn we have found is equal to small a naught into. So, we have found earlier that a naught is small a naught into e to the power minus t1 and using the equation for small a naught we have now determined x naught completely. So, that is our x naught. Of course, this has to have a complex conjugate added to it to give us a real x naught. Similarly, one can find x1 and this is equal to a1 and this is a function of t2 e to the power minus t1 e to the power i t naught plus complex conjugate. Once again we have found this part before. We had found this part earlier we had written that it is. So, this is basically just a1 it is small a1 into e to the power minus t1 into e to the power i t naught. So, it is a1 into e to the power minus t1 into e to the power i t naught. The second part is not there because we have set it equal to 0 in order to determine small a naught. I hope this is clear. So, we need to determine this a1 as a function of t2. It is clear that we will have to go to the next order in order to do this. I am just going to write down the answer for you because the procedure is now straight forward. So, it may be shown that a1 of t2 is equal to some a11. I am following the same notation as earlier into e to the power minus i t2 over 2 where a11 is once again a complex constant. This I will leave it to you to show yourself. It is not difficult. Once you do this then we can write down the answer up to order epsilon we have found that this is a naught naught into we can combine all the exponentials. So, I can write so e to the power minus t1 does not have i. So, I will write it separately and then I will write i t naught minus t2 over 2 and then plus epsilon times a1 and we have seen that a1 is basically a11 and the same thing here e to the power minus t1 into e to the power i into t naught minus t2 over 2 plus of course the complex conjugate of the entire expression. So, there will be 2 parts to it. Now let us shift to real notation before we do that. Let us you can see immediately that we will have a term which is like this. So, we can write this as a naught naught plus epsilon a11 e to the power minus t1 into e to the power i t naught minus t2 over 2 plus complex conjugate. Now you can see that this is a constant, this is a complex constant because a naught a double 0 is complex constant, a double 1 is also complex constant and epsilon is real. So, the whole thing is a is another complex constant. I can write that a double 0 plus epsilon a11 is equal to some complex number whose amplitude is L and whose phase is some beta with the understanding that L is now real and beta is also a real number, both L and beta are real numbers. If we do that then the expression that I had written earlier may be written as L e to the power minus t1. I would also have e to the power i beta, but I will absorb the beta in the exponential which was already there with the i. So, I am going to just adding it up to this part and then plus complex conjugate of whatever is on the left. Remember that now L and beta are real by definition and so if we now shift to complex exponential to real notation using e to the power i theta is equal to cos theta plus i sin theta you can immediately see that this just becomes L e to the power minus t1 cos of t naught minus t2 by 2 plus beta and we do not have the sin part because the sin cancels out and so this if we now go back to instead of writing 3 different scales t1, t2, t0 we go back to our original scale then this gives you e to the power minus t cos of this is t, this is half epsilon square t and this is plus beta. This is equal to L e to the power minus t cos of 1 minus epsilon square by 2 into t plus beta. So, our final answer now of course there are higher order terms here. Now notice that so I have missed epsilon here, t1 is epsilon t. So, t1, t0 is equal to t1 is epsilon t, t2 is epsilon square. Now notice that this is not completely the solution up to order epsilon square we would have to add in general I will have to add another part here which would be epsilon square into x2. We have not done that step you can go ahead and do the step and make it completely consistent but the basic purpose of this exercise was just to demonstrate that what is multiple scales doing. So, first thing that you notice is that the multiple scales is giving you an answer which will not give become disordered in time. It has also eliminated the process of doing multiple scales itself eliminates all the resonant forcing terms and consequently there are no secular terms. Let us compare this solution with the exact solution that we had written at the starting of this example. So, we had seen that this so we were studying the damped harmonic oscillator and the this was the equation whose solution we have obtained perturbatively using the method of multiple scales. And this equation we have seen earlier has the solution this is the exact solution this is a linear equation. So, it can be solved exactly very easily it will be some a e to the power minus epsilon t cos square root 1 minus epsilon square into t plus phi. If I replace the a by l a and phi are real numbers so I can use any other symbol and I am replacing phi by beta just so that we can compare with this expression. So, you can see that these are this is nothing what we have recovered is nothing but the expansion of 1 minus half epsilon square t plus dot dot dot plus plus beta. So, you can see that you can guess what will be the higher order corrections the regular perturbation what it would have done is it would have expanded this term also and this term also and consequently it would have taken the product of the two expansion. So, consequently we would have found secular terms as long as we truncated the expansion up to any finite number of terms and took the limit t going becoming larger and larger. Here what it does is it keeps the e to the power minus epsilon t outside it recovers the e to the power minus epsilon t at the first order correction and then it expands the frequency. So, it basically says that square root 1 minus epsilon square t is approximately 1 minus half epsilon square into t when epsilon is sufficiently small. You can take this to higher orders of epsilon and you will recover more and more corrections. So, you can go to more time scales. So, you can go to t 3 which will be epsilon cube into t. So, essentially if you go to even longer time scales you will discover that the frequency is not exactly 1 minus half epsilon square but it is actually 1 minus half epsilon square plus some correction. So, this term will have a small mismatch with the theoretical prediction. So, that is how the basic method works. We will continue our discussion of the method of multiple scales particularly because this is a very important technique which is frequently used in analyzing interfacial waves. So, we will go on now we have just completed our discussion of using multiple scales to obtain an approximate solution to the damped harmonic equation. We now go on to another equation the damped harmonic equation was a linear equation. We now go on to a non-linear equation. So, this equation that we are going to now do using the method of multiple scales will not have any damping. So, the energy so this equation will have something like an energy which will be a conserved quantity. However, this equation will be slightly different from the non-linear pendulum in the sense that instead of having time theta which is a non-linear function of theta it will have a just a cubic term in theta. So, this equation was one of the early equations which was used for understanding an harmonic behavior of oscillators. So, the equation is like this. So, you can immediately see that up to here this is a harmonic oscillator omega 0 square is just some square of some characteristic frequency, but this term actually is a non-linear term epsilon is small but positive and so you can see that this is a kind of a so you can think that this is the equation of motion of some mass which is connected to a non-linear spring. So, if it was a linear spring it would just go as minus so this would be just this, but then I am adding some delta into some x cube term or rather this and then I am non-dimensionalizing it which is leading me to the Duffing equation. Now, you can immediately see qualitatively what is going to happen if I did not have the second term. The second term you can see that if x is positive if it was a linear spring the force would be from right to left it would try to the restoring force would try to pull it back to the equilibrium position. Here too the restoring force tries to pull it back and it tries to pull it back harder than a linear spring. So, we do expect periodic solutions we also expect that because the restoring force depends on epsilon. So, we do expect that the time period of motion will be dependent on epsilon like the non-linear pendulum. Note however that the first correction in the non-linear pendulum is a minus term approximate sin theta by its Taylor series about theta equal to 0 then it is theta minus theta cube by factorial 3. Here it is a plus epsilon into u cube. Let us look at this equation. So, I will just erase this. So, now before we do so this is we are going to do a multiple scale on this. This can also be solved by the Linsted-Poincare technique but this is just to show you that the multiple method of multiple scales is very general. It works it worked on the linear equation that I that we discussed earlier and it is going to work on this equation as well. So, let us draw the phase portrait of this oscillator before we attack it using the method of multiple scales. So, as before we convert this this is a second order ordinary differential equation. We convert this into two first order ordinary differential equations. They are expected to be coupled and non-linear. So, we define u to be x and we define du by dt to be y. So, this immediately gives me dx by dt is du by dt which is y and dy by dt which is basically the equation itself the Duffing equation itself. dy by dt if I shift the omega square u and minus epsilon u cube on the left hand side then I get minus omega 0 square into x minus epsilon x cube. So, this is we have we have learned to do this process before and I can write this as minus x into omega 0 square plus epsilon x square. Now, I can take the ratio of these two terms and if I take if I divide the second term by the first term then I get dy by dx is equal to minus omega 0 square x minus epsilon x cube divided by y. This equation can be easily integrated and this is telling me that y square by 2 plus omega 0 square x square by 2 plus epsilon x 4 by 4 is equal to some constant of integration. It is convenient if I choose the constant to be C by 2 so that I can cancel out a factor of 2 everywhere. If I do this then I get y square is equal to C minus omega 0 square x square minus epsilon x 4 by 2 and so like before we will have to plot this equations of y as a function of x for various values of C. Now, for convenience what in I have plotted it in my next slide I have chosen w 0 square is equal to 1. This is just for convenience it does not need to be unity. So, we will have two parts to it one part will be the positive square root and one part will be the negative square root. This is very similar to what we did for the non-linear pendulum. So, let us look at the phase portrait of the Duffing equation. Now, before we draw the phase portrait let us understand what are the fixed points of this equation. So, if you look at this equation it is clear that x dot is equal to 0 when y is equal to 0. Similarly, from this equation y dot is equal to 0 when x is 0 this part omega 0 square and epsilon are positive and so the only value of x at which y dot is equal to 0 is x equal to 0. So, consequently x comma y so let us call it y star and x star. So, x star y star being 0 comma 0 are the fixed points of the system. This is the only fixed point of the system. So, now let us draw the phase portrait. So, we are going to draw this curve. As I told earlier we will choose omega 0 square is equal to 1 for convenience and we have chosen an epsilon which is equal to 0.2. So, I am going to plot various values of this curve because this is a y square here. So, there will be a plus square root and a minus square root and for each value of c I will get one phase-phase trajectory. Let us look at those trajectories. So, this is what the phase portrait looks like. So, this is the fixed point of the system. So, fixed point this is corresponding to x is equal to 0 and y equal to 0. Now, note that I am plotting each curve as I said earlier for a different value of c. So, the curve recall are basically y square plus x square plus epsilon is 0.2 divided by 2 into x to the power 4 is equal to constant. So, in other words I have y square plus x square plus 0.1 times x4 is equal to c and I am choosing various values of c here on the right hand side. So, the value of c ranges from 0.1 to about 20. You can see that as c gets smaller the phase-phase trajectory also gets smaller. So, this is the smallest value of c for which I have plotted and that is the largest value of c for which I have plotted. Once again you can split this curves into two parts the upper part and the lower part the upper part will correspond to the positive square root and the lower part will correspond to the negative square. So, we notice that there is one fixed point in the system and there are these periodic oscillatory solutions about the fixed point. One can go to higher and higher values of c recall that c is related to the energy of the system and so higher and higher values of c just give me larger and larger phase-phase trajectories. So, now we are it is obvious from this figure that there are oscillatory periodic solutions to this equation. Let us find those solutions. Now, how do we find out these periodic solutions? Let us use once again the method of multiple scales. Once again what are the time scales in the problem? So, you can see that at very early times we know that the period of the non-linear oscillator is different from the period of the linear oscillator. Now, at very early times we are not going to see that period. We are not going to see the effect of that small difference in frequency which is there between a non-linear oscillator and a linear oscillator. So, as a first approximation we will find there is just a harmonic oscillator with time period omega 0. Now, at the next order approximation we will find corrections to that. Let us work out those corrections. So, we have we are not going to use the Linsted-Poynck technique. We will use the method of multiple scales. So, we will define like before. We do not intend to solve the problem up to order epsilon square but we have to go to order epsilon square in order to determine all the unknowns up to order epsilon. So, one has to do a little bit more algebra for here. So, we have this and then we have epsilon square t. So, like before d by dt is just d0 plus epsilon d1 epsilon square d2 and t square by dt square. If I take this and then I have to do an expansion which is x0 sorry u0 is equal to u is equal to u0 which is a function of t0 t1 t2 plus epsilon u1 t0 plus epsilon square. Now, let us do that expansion. So, we have to substitute all of this. So, the procedure is pretty standard now. I am sure all of you are familiar with this. So, I will straight away jump to collecting terms at various orders. So, all this is obtained once you substitute this into this equation and this also into this equation and then start collecting terms at various orders. The left hand side is the same. The right hand side now you are familiar with this. One has to be careful while doing the right hand side. If you miss even one term your answers will be incorrect. So, one has to be careful that you have not missed any term nor have you included a term which is not supposed to be there at that given order. Once again I would like to reiterate that we are not going to solve the problem up to order epsilon square. But to determine one unknown at order epsilon one has to go up to order epsilon square. So, that is why I am having to write up to order epsilon square. The algebra is slightly lengthy here. But since we have already learned how to do this and we are familiar with this from the damp harmonic oscillator. I am going to skip some steps. It is easy for you if you work out the algebra to do those steps. It is exactly similar to what we had done earlier. The only difference is here we are dealing with a non-linear equation. In the earlier example we dealt with a linear equation. So, these are our solutions at various orders. Now, let us solve these. So, let us write down the solution at order 1. So, at order 1 the solution is the simplest. So, u0 is just a which is a function of like before t1, t2 into e to the power i omega 0 t0 plus cc. Now, at order epsilon we will have to plug this in. The left hand side remains the same and we had written in the last slide that the right hand side consists of these terms which depend on the previous order. So, if I apply now I know the solution at the previous order and so this just becomes twice i omega 0 del a by del t1 minus a e to the power i omega 0 t0 plus its complex conjugate whole cube and there will be one more complex conjugate which is the complex conjugate of this term. So, you can put cc1, cc2 if you want. So, this is cc2 and this is cc1 for example. Now, we will have to work out the right hand side. So, this term is obviously there, then we have a cube e to the power psi psi omega 0 t0. Now, there will be one more term which will be the product of 3 times a square into b and that will give me. So, there is a minus 3 a square e to the power twice i omega 0 t0 into b and b is just a bar e e to the power minus i omega 0 t0. So, I will have a a bar here and e to the power i omega 0 t0. I hope this is clear what we are doing is we are just taking this and adding its complex conjugate part and cubing the thing. So, the cube of this thing I have already written plus 3 a square b is what I have written. You can see that e to the power twice i omega 0 t0 minus. So, this will give me e to the power i omega 0 t0 plus I will have thrice ab square and that will have e with a negative exponent, but that will exactly be the complex conjugate of this part. And then I will have one more term which is a bar cube e to the power minus thrice i omega 0 t0 and that is the complex conjugate of this part. So, I will just write the first two terms and then put a cc at the end to indicate the other two terms. We will continue this in the next video.