1 00:00:00,120 --> 00:00:06,420 So now we know how we can propagate the solution for our differential equation of First Order. 2 00:00:06,720 --> 00:00:10,320 And we will now discuss an example, which is the radioactive decay. 3 00:00:11,280 --> 00:00:19,290 So the differential equation for the radioactive decay is that the change of the mass of a radioactive 4 00:00:19,290 --> 00:00:23,370 substance is given by the amount of the substance itself. 5 00:00:23,520 --> 00:00:28,140 Because the decay, you will of course, be faster, the more substance you have. 6 00:00:29,040 --> 00:00:35,190 So the change of Y is given by a negative value times y. 7 00:00:35,940 --> 00:00:41,280 So in general, there is, of course, actually here also a constant which describes two different elements. 8 00:00:41,640 --> 00:00:48,330 But I would say for simplicity, we just write Y dot is equal to minus y and we'll leave out the constant. 9 00:00:48,330 --> 00:00:49,440 We just set it to one. 10 00:00:50,700 --> 00:00:55,680 So this is how you would write that down as a physicist, but as a mathematician, you would write the 11 00:00:55,680 --> 00:00:58,830 Y by D T, which is per definition. 12 00:00:58,830 --> 00:01:02,640 This function f of T and Y is equal to minus Y. 13 00:01:03,780 --> 00:01:10,890 So we know now to propagate this using this solution, we have to just use here to value minus y. 14 00:01:11,040 --> 00:01:13,740 And since it's your Y and we have to write minus point. 15 00:01:13,740 --> 00:01:21,870 And so y and plus one is y n plus minus y end times h. 16 00:01:23,460 --> 00:01:31,140 And before we do this, we want to briefly discuss the analytical solution because then we can compare 17 00:01:31,140 --> 00:01:32,760 later on if our solution is correct. 18 00:01:33,390 --> 00:01:38,520 Well, this differential equation, you have to calculate the first derivative with respect to time, 19 00:01:38,850 --> 00:01:42,180 and the function is essentially the same as just a minus sign here. 20 00:01:42,450 --> 00:01:47,130 So the solution will, of course, be an exponential function, and due to this minus sign here, it 21 00:01:47,130 --> 00:01:50,310 will be its potential function of minus T. 22 00:01:51,300 --> 00:01:55,260 So this is also the typical solution which describes radioactive decay. 23 00:01:56,670 --> 00:01:58,410 So now let's program this. 24 00:01:58,920 --> 00:02:04,380 And first of all, we must define it here some, or we must load some modules. 25 00:02:04,770 --> 00:02:08,639 And so far, we will not really use many of those. 26 00:02:08,639 --> 00:02:13,890 We will probably use a plot loop and we will use numpy, but not separate yet. 27 00:02:14,220 --> 00:02:18,140 But these are all the modules that we are going to need throughout this notebook. 28 00:02:18,150 --> 00:02:20,400 So let's just load all of them at once here. 29 00:02:21,600 --> 00:02:22,040 OK. 30 00:02:22,620 --> 00:02:25,300 So please write this code in your notebook and run it. 31 00:02:25,320 --> 00:02:29,970 And since this is done now, we can start calculating and start programming. 32 00:02:31,170 --> 00:02:39,090 So basically, what we want to do is we want to write a loop because we want to iterate this process 33 00:02:39,090 --> 00:02:39,360 here. 34 00:02:40,110 --> 00:02:45,270 So I write for I in range and then we can write, for example. 35 00:02:45,650 --> 00:02:47,570 Um yeah. 36 00:02:48,120 --> 00:02:53,010 And Max, for example, where and Max is the number of steps. 37 00:02:53,490 --> 00:02:55,320 So let's use 24 now. 38 00:02:56,040 --> 00:03:06,900 This would be the number of iterations, and I will loop from zero to 19 in this case. 39 00:03:08,190 --> 00:03:09,330 So what will happen here? 40 00:03:10,560 --> 00:03:17,880 Well, the first thing that we must do is we must define F because we know that y m plus one is y and 41 00:03:17,880 --> 00:03:19,620 plus f times h. 42 00:03:20,580 --> 00:03:26,130 So we write F is equal to minus y in this case because this is our function. 43 00:03:26,730 --> 00:03:30,060 This will, of course, change when we consider a different differential equation. 44 00:03:31,500 --> 00:03:41,100 And now the value of Y, which we will call Y so y at the step and plus one will be given by the old 45 00:03:41,100 --> 00:03:42,210 value of Y. 46 00:03:42,960 --> 00:03:47,910 And then we update it by the value of F and multiply by the step size of each. 47 00:03:49,680 --> 00:03:53,970 So of course, what we have to do now is we have to define these values here. 48 00:03:53,970 --> 00:03:58,320 We have to define H and we have to define some starting value off-line. 49 00:03:59,580 --> 00:04:00,300 So let me do this. 50 00:04:00,300 --> 00:04:06,920 We will define the step size, which is h is equal to zero point one. 51 00:04:06,930 --> 00:04:14,760 And also we will define the starting value, which is why. 52 00:04:15,120 --> 00:04:18,089 And here we can take basically any number doesn't really matter. 53 00:04:18,480 --> 00:04:20,010 It will just scale the function. 54 00:04:20,910 --> 00:04:22,680 So I just take one here. 55 00:04:23,850 --> 00:04:24,260 OK. 56 00:04:24,420 --> 00:04:27,360 So I think at this point, it should work already. 57 00:04:27,690 --> 00:04:29,430 And let's see what's the result. 58 00:04:30,030 --> 00:04:38,940 So we run this algorithm 20 times and then at the end, the result will be Y and it will be zero point 59 00:04:38,940 --> 00:04:40,380 one to one something. 60 00:04:41,400 --> 00:04:43,830 So we have started from one. 61 00:04:44,250 --> 00:04:49,830 So the amount or do you you could also see the mass of the radioactive substance, for example, was 62 00:04:49,830 --> 00:04:56,430 one kilogram and then 20 seconds later, the mass has reduced to 0.1 to one. 63 00:04:57,660 --> 00:04:59,940 So we don't really know yet if this is correct. 64 00:05:00,040 --> 00:05:08,650 Well, of course, we could just calculate and p e p of minus T, which is minus 20 in this case or 65 00:05:08,650 --> 00:05:09,460 minus and max. 66 00:05:10,640 --> 00:05:11,350 So. 67 00:05:15,620 --> 00:05:17,750 Obviously, you know this this is, of course, a mistake. 68 00:05:17,960 --> 00:05:24,350 So the time is, of course given by and max times, each so minus and max times each. 69 00:05:25,100 --> 00:05:31,850 And now we see the amount of substance, according to the analytical solution, is pretty similar to 70 00:05:31,850 --> 00:05:32,930 the numerical solution. 71 00:05:32,930 --> 00:05:35,240 But there is a small difference here. 72 00:05:36,080 --> 00:05:42,620 We can, of course, you know, increase the number of iterations or change to step size, for example, 73 00:05:42,620 --> 00:05:43,100 like this. 74 00:05:45,800 --> 00:05:48,410 And though you see the solution will be much closer. 75 00:05:49,160 --> 00:05:55,370 But we want to analyze the result, of course, in more detail and having just the final result is probably 76 00:05:55,370 --> 00:05:56,150 not sufficient. 77 00:05:56,390 --> 00:06:06,110 So what I will do here is I will collect data and I will define some lists and I will write the T values 78 00:06:06,110 --> 00:06:06,410 here. 79 00:06:06,930 --> 00:06:11,360 We could, of course, start from an emptiness, but I want to actually consider you two starting value 80 00:06:11,360 --> 00:06:11,810 of tea. 81 00:06:12,470 --> 00:06:17,660 And I want to also collect, of course, to my values. 82 00:06:18,020 --> 00:06:24,350 And here I also want to consider the starting value, which is one that we have to right. 83 00:06:24,350 --> 00:06:26,660 Here are tea values. 84 00:06:27,680 --> 00:06:40,280 Dots append is equal to PI times H and the Y values top per pound of Y. 85 00:06:41,210 --> 00:06:50,900 So if we run this, no, then we will see that tea values is a list starting from zero and then going 86 00:06:50,900 --> 00:06:53,570 to OK, maybe that's reduced as a bit. 87 00:06:53,930 --> 00:06:55,940 Let's go back to the initial parameters. 88 00:06:56,720 --> 00:07:03,590 So it starts from zero goes to one point nine and we want until we have twice the value zero and we 89 00:07:03,590 --> 00:07:05,930 want to give it instead to two. 90 00:07:06,710 --> 00:07:12,780 So we can fix this by just writing here range starting at one, two and max plus one. 91 00:07:12,800 --> 00:07:16,010 This is basically just because Python starts counting at zero. 92 00:07:16,010 --> 00:07:21,740 And when you write it like this, Python starts counting at one and this is what we want in this case. 93 00:07:22,370 --> 00:07:27,260 So that's why we run this and you know, you see the T values list is correct. 94 00:07:28,760 --> 00:07:37,790 So let me add here a plot because now we have our tea values and we have our Y values and we can just 95 00:07:37,790 --> 00:07:38,540 write Pulte. 96 00:07:38,570 --> 00:07:48,440 Total scatter tea values come our way values, and you see, this is an exponential function. 97 00:07:49,620 --> 00:07:58,370 Now, when we want to compare this with the analytical solution, we can write a pen pen total in space. 98 00:08:01,120 --> 00:08:02,860 And we have to start at zero. 99 00:08:03,340 --> 00:08:18,310 Go to and max times each and go instead of and max and for the Y values we write, Test Y is equal to 100 00:08:18,490 --> 00:08:20,250 and NPR Dot XP. 101 00:08:20,950 --> 00:08:21,850 We could also write. 102 00:08:23,020 --> 00:08:23,320 Yeah. 103 00:08:24,730 --> 00:08:29,630 Basically, the starting value of Hawaii, which in our case is one so MP. 104 00:08:29,630 --> 00:08:32,650 He thought he XP times oops. 105 00:08:32,770 --> 00:08:33,159 Sorry. 106 00:08:35,409 --> 00:08:37,710 Minus Test T.. 107 00:08:39,640 --> 00:08:41,110 And now we can plop this 108 00:08:43,600 --> 00:08:45,190 TLT dot plot. 109 00:08:47,900 --> 00:08:56,810 Test t come up test y and I will put it in red. 110 00:08:58,550 --> 00:09:05,780 So of course, it's always good to add some label just to make sure that you know what's plotted here. 111 00:09:06,050 --> 00:09:11,750 Also, when you come back to the notebook later on, it's often very helpful to label the boxes. 112 00:09:12,560 --> 00:09:13,970 And you see now finally, 113 00:09:16,910 --> 00:09:23,240 that we plot y over T and the blue points are a numerical solution and the red one is the analytical 114 00:09:23,240 --> 00:09:23,780 solution. 115 00:09:24,470 --> 00:09:30,770 And you see, it is kind of OK, but here at the larger the time gets, there will be an arrow, which 116 00:09:30,770 --> 00:09:33,200 we have also seen here by comparing these values. 117 00:09:33,800 --> 00:09:39,170 And we have also seen that we can reduce the arrow by decreasing to step size. 118 00:09:39,980 --> 00:09:43,790 And then in turn, we could, for example, increase the number of iterations. 119 00:09:44,930 --> 00:09:50,610 So let's run this and you see now it looks much better, I think. 120 00:09:51,980 --> 00:09:58,430 So this was an example where we have considered the radioactive decay, where the differential equation 121 00:09:58,430 --> 00:10:04,520 is characterized by the fact that the change in the radioactive substance is given by the amount of 122 00:10:04,520 --> 00:10:05,540 the substance itself. 123 00:10:05,960 --> 00:10:10,090 So why not is given by minus constant times y. 124 00:10:10,670 --> 00:10:15,050 And the analytical solution is, of course, in declining exponential function. 125 00:10:15,440 --> 00:10:22,190 And we have just iterated this equation here and have reproduced the result pretty well. 126 00:10:23,120 --> 00:10:29,480 So in the next lecture, we will define a function ourselves called Euler Odie for ordinary differential 127 00:10:29,480 --> 00:10:29,960 equation. 128 00:10:30,500 --> 00:10:37,040 And with this function, we will generally localize this cell here, which in our case was specific 129 00:10:37,040 --> 00:10:38,990 to this particular differential equation. 130 00:10:38,990 --> 00:10:43,670 But we will generalize this for a general first order differential equation.