1 00:00:00,450 --> 00:00:06,150 So in the previous lectures, we have spent a lot of time with solving this particular example of considering 2 00:00:06,150 --> 00:00:12,810 three harmonic oscillators that are couples, and it turned out that the numerical solution was quite 3 00:00:13,020 --> 00:00:13,620 difficult. 4 00:00:13,650 --> 00:00:18,390 So at first we could just say, OK, this is the solution, but we don't really know what it means. 5 00:00:19,140 --> 00:00:25,830 But then we established the eigenvalue problem and solved the equation for the eigenvalues, which told 6 00:00:25,830 --> 00:00:27,900 us the ion frequencies. 7 00:00:28,560 --> 00:00:33,690 And we've seen that we can rediscover his ion frequencies in the four you transform. 8 00:00:34,560 --> 00:00:42,030 And even we can take these characteristic frequencies and build harmonic functions and then superimpose 9 00:00:42,030 --> 00:00:43,260 these three functions. 10 00:00:43,660 --> 00:00:50,070 We've seen that when we fit the parameters, then we can even restore the numerical solution. 11 00:00:50,910 --> 00:00:55,830 So everything, everything makes total sense and everything agrees well with each other. 12 00:00:55,830 --> 00:01:01,350 And we have practiced here a lot of the previous topics that we have discussed early on. 13 00:01:02,790 --> 00:01:09,360 So a bonus lecture that I want to show you here is that we can generalize the whole thing not only for 14 00:01:09,360 --> 00:01:13,050 this system with three oscillators, but four and oscillators. 15 00:01:13,680 --> 00:01:15,600 And here I want to show you this. 16 00:01:16,110 --> 00:01:17,840 So we will solve it numerically. 17 00:01:17,850 --> 00:01:21,810 We will calculate the ion frequencies and we will calculate the full year transform. 18 00:01:22,710 --> 00:01:26,280 And the nice thing is we don't really have to change any of the code. 19 00:01:26,730 --> 00:01:28,890 So just really tiny changes. 20 00:01:29,310 --> 00:01:36,000 And the only thing that we really have to change here is the definition, of course, of the equations 21 00:01:36,000 --> 00:01:36,600 of motion. 22 00:01:37,020 --> 00:01:38,320 So there's a we will do first. 23 00:01:38,340 --> 00:01:43,680 So I will write Def F and see if I would have written down here. 24 00:01:43,680 --> 00:01:45,690 Three O C, we wouldn't have to change anything. 25 00:01:46,350 --> 00:01:48,750 But I want to keep the old results. 26 00:01:48,750 --> 00:01:56,640 So please make sure that whenever we wrote three and a c right note and or C, and the other thing that 27 00:01:56,640 --> 00:02:00,570 we have to change for the eigen frequency is the definition of a. 28 00:02:01,590 --> 00:02:07,140 So this is what I will do first and then we can just run the whole thing and see if this works also 29 00:02:07,140 --> 00:02:09,120 for a general revised version. 30 00:02:09,750 --> 00:02:17,010 And as an example, I will choose here already quite a difficult set up here with 20 oscillators. 31 00:02:18,120 --> 00:02:22,080 So function will, of course, again have the arguments T and Y. 32 00:02:22,620 --> 00:02:25,470 And now we have to think, how can we generalize this? 33 00:02:26,340 --> 00:02:32,940 First of all, since we have now so many equations, I will write our is equal to y from zero to end 34 00:02:34,320 --> 00:02:45,030 and then the value will be y from end to to end. 35 00:02:47,570 --> 00:02:54,360 And the equations will I will, of course, not define explicitly because we don't know how many equations 36 00:02:54,360 --> 00:03:03,030 we will have, this depends on the value of and so I will start with and p zero of 10 so that we start 37 00:03:03,030 --> 00:03:04,470 with some blank equations. 38 00:03:04,980 --> 00:03:16,620 And then I will write equation the second until the second last equation will be equal to minus our. 39 00:03:20,270 --> 00:03:29,900 The first equation until the minus two, so basically leave out the last two and plus two times are 40 00:03:30,920 --> 00:03:33,920 one to minus one and then minus. 41 00:03:35,610 --> 00:03:39,510 Are two to the last one. 42 00:03:41,490 --> 00:03:50,130 So this is what we have also seen in The Matrix, basically all the entries, except for the first and 43 00:03:50,130 --> 00:03:54,030 last one, will have to form minus one to minus one. 44 00:03:54,150 --> 00:03:59,220 And then many zeros, if we have more of these oscillators than we get when we go to the next one, 45 00:03:59,220 --> 00:04:01,500 we shift these three numbers wants to the right. 46 00:04:02,530 --> 00:04:05,700 And this is what we have programmed here, exactly. 47 00:04:06,510 --> 00:04:11,670 So for every one of the middle equations, which means everyone except the first and the last one, 48 00:04:12,240 --> 00:04:13,800 the equation will look like this. 49 00:04:14,700 --> 00:04:20,579 And then we have to explicitly define the first and the last one, which will be equation zero, because 50 00:04:20,579 --> 00:04:21,930 here it will be a bit different. 51 00:04:22,320 --> 00:04:28,500 So there will be two times are zero minus are of one. 52 00:04:29,340 --> 00:04:40,320 And then for the last one equation and minus one will be minus are off and minus two to the left neighbor. 53 00:04:40,950 --> 00:04:47,550 And then the two later itself will be two times out of and minus one. 54 00:04:48,870 --> 00:04:51,240 And then we have no right neighbors, so that's it. 55 00:04:52,030 --> 00:04:53,160 But then we must return. 56 00:04:54,120 --> 00:05:02,390 First of all, the values and then the equations, and we are missing still the factor. 57 00:05:02,400 --> 00:05:05,880 So I wrote here minus K over m times equations. 58 00:05:07,110 --> 00:05:11,250 So since these are lists and these are lists, we have to merge these lists. 59 00:05:11,250 --> 00:05:14,790 So I write here and p dot concatenate. 60 00:05:15,330 --> 00:05:17,160 And then it should work, I think. 61 00:05:17,670 --> 00:05:21,120 But of course, here we are missing the brackets. 62 00:05:21,360 --> 00:05:27,700 OK, now let's run it and you see we get a nice solution. 63 00:05:27,720 --> 00:05:33,810 We have now 20 oscillators, and you see this motion looks very difficult. 64 00:05:34,350 --> 00:05:39,300 But we also know that we could calculate the eigen frequencies, which we will do next. 65 00:05:39,780 --> 00:05:47,910 And so we could recover each of these 20 oscillators by just superimposing of 20 harmonic functions 66 00:05:47,910 --> 00:05:49,680 with the 20 eigen frequencies. 67 00:05:50,640 --> 00:05:55,620 So let's go ahead and program the matrix and then calculate the EIGEN frequencies. 68 00:05:56,370 --> 00:05:59,360 So a would be it. 69 00:05:59,380 --> 00:06:03,000 That's something we will also encounter in a different lecture. 70 00:06:03,330 --> 00:06:11,100 There's always a nice trick when you have such and such matrices that almost the zero and only have 71 00:06:11,100 --> 00:06:16,380 values on the diagonal and then on the other diagonal right next to it. 72 00:06:17,580 --> 00:06:21,660 So basically, we will just define this diagonal, this diagonal on this one. 73 00:06:22,050 --> 00:06:32,070 And then we will use a command that is called and pitot diak and here will show you how this works. 74 00:06:33,100 --> 00:06:38,070 He write, for example, minus one times. 75 00:06:40,750 --> 00:06:41,710 And minus one. 76 00:06:42,490 --> 00:06:48,610 So this would be just by itself, a list of minus one. 77 00:06:48,700 --> 00:06:53,680 And this whole thing 19 times in this case, this would be not the main diagonal, but the one the right 78 00:06:53,680 --> 00:06:54,280 next to it. 79 00:06:54,820 --> 00:07:00,160 And since it's not the main diagonal, we right k is equal to one. 80 00:07:01,180 --> 00:07:11,230 So if I call this, you see we have here these this vector here, not on the main diagonal, but on 81 00:07:11,230 --> 00:07:12,670 the one right next to it. 82 00:07:13,690 --> 00:07:15,250 Then we can add another one. 83 00:07:16,270 --> 00:07:20,490 Copy this and now our right here, plus two times. 84 00:07:20,560 --> 00:07:26,140 And because this is not a main diagonal, so it has and trees and key is equal to zero, this is to 85 00:07:26,140 --> 00:07:26,650 shift. 86 00:07:27,700 --> 00:07:31,060 And so now we have on the main diagonal a lot of twos. 87 00:07:31,600 --> 00:07:33,400 And so only one thing is left. 88 00:07:33,770 --> 00:07:39,340 This is the other one of those minus one. 89 00:07:40,960 --> 00:07:43,480 And now it works. 90 00:07:43,480 --> 00:07:48,610 Yeah, because here in the second line, we start with minus one to minus one. 91 00:07:49,180 --> 00:07:54,940 And then it shifts the indices zero minus one to minus one until in the last line, we have a lot of 92 00:07:54,940 --> 00:07:56,740 zeros and then minus one and two. 93 00:07:57,580 --> 00:08:03,340 And we know that we get our ion frequencies by just calculating the square root of the eigenvalues and 94 00:08:03,340 --> 00:08:06,220 multiplying by square root of K over them. 95 00:08:07,420 --> 00:08:10,870 So you see, we have a whole lot of eigenvalues here. 96 00:08:13,200 --> 00:08:16,020 Now we can continue and calculate the Fourier transform. 97 00:08:16,470 --> 00:08:23,670 So here you only have to be careful because you have to change the solution three o c to solution and 98 00:08:23,670 --> 00:08:28,860 our C because this is what we have named it, and then we can just go ahead. 99 00:08:31,190 --> 00:08:36,450 And look at the frequencies, so here I've plotted several different things. 100 00:08:36,470 --> 00:08:43,760 The first one would be the four you transform only of the first oscillator, which has the peaks, which 101 00:08:43,760 --> 00:08:48,080 should correspond to these values that we have here. 102 00:08:48,680 --> 00:08:54,980 For example, we can check we have here several values below zero at zero point one five zero point 103 00:08:54,980 --> 00:08:57,800 three zero point four four five. 104 00:08:58,370 --> 00:09:04,400 And this one here, for example, would be 0.3 zero point one five zero point four five. 105 00:09:04,850 --> 00:09:07,940 The next one would be slightly above zero point five. 106 00:09:08,570 --> 00:09:10,230 This would probably be this one. 107 00:09:10,310 --> 00:09:14,900 So you see each of these eigen frequencies is indeed in the four you transform. 108 00:09:16,070 --> 00:09:22,800 And then the second plot for the second plot I've used here a loop where I loop over all of these oscillators. 109 00:09:22,850 --> 00:09:28,700 So here you see the different colors to different oscillators, and you see they have peaks at the same 110 00:09:28,700 --> 00:09:32,690 frequencies, of course, but the amplitudes can be a bit different. 111 00:09:33,020 --> 00:09:40,790 And this we have also seen already for the three example case and maybe even some of the some of the 112 00:09:40,910 --> 00:09:43,720 coefficients of the eigenvectors on zero. 113 00:09:43,730 --> 00:09:47,690 So then in this case, there would be no peak for this particular oscillator. 114 00:09:49,220 --> 00:09:55,940 And then the last thing would be the same thing as here, but zoomed in, and you can really see that 115 00:09:55,940 --> 00:10:02,690 we have these peaks at zero point fifteen zero point three zero point forty five and then another one 116 00:10:02,690 --> 00:10:10,040 here at zero point five eight, which are exactly these eigen frequencies that we see here. 117 00:10:11,240 --> 00:10:16,250 So you see the with the tools that we have, we can very quickly solve new problems. 118 00:10:16,580 --> 00:10:22,820 And you see, we have not generalized the problem to 20 oscillators, which is, of course, if you 119 00:10:22,820 --> 00:10:28,460 would want to solve it analytically by hand would be impossible to do. 120 00:10:29,180 --> 00:10:31,730 But with our methods that we have established, it's very easy. 121 00:10:31,730 --> 00:10:34,700 And you see, still, everything agrees very well. 122 00:10:35,180 --> 00:10:36,800 We have the miracle solution. 123 00:10:37,250 --> 00:10:38,660 We have the eigen frequencies. 124 00:10:38,660 --> 00:10:41,920 We have to fully transform and we could even do to fit. 125 00:10:41,930 --> 00:10:48,590 But I want I don't want to do this now because this was just be a time consuming and we know that it 126 00:10:48,590 --> 00:10:49,490 would work, of course.