Sturm-Liouville Problems
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[{"Name":"Sturm-Liouville Problems","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Introduction","Duration":"28m 28s","ChapterTopicVideoID":28210,"CourseChapterTopicPlaylistID":294448,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/28210.jpeg","UploadDate":"2021-12-26T15:45:35.3370000","DurationForVideoObject":"PT28M28S","Description":null,"MetaTitle":"Introduction: Video + Workbook | Proprep","MetaDescription":"Applications of Fourier Series on Differential Equations - Sturm-Liouville Problems. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/calculus-i%2c-ii-and-iii/applications-of-fourier-series-on-differential-equations/sturm_liouville-problems/vid29443","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.450","Text":"Now we come to a new kind of problem called,"},{"Start":"00:03.450 ","End":"00:07.449","Text":"the Sturm Liouville problem."},{"Start":"00:08.000 ","End":"00:14.460","Text":"Abbreviated S-L because it\u0027s the long names."},{"Start":"00:14.460 ","End":"00:15.960","Text":"As far as I know,"},{"Start":"00:15.960 ","End":"00:22.170","Text":"this is pronounced the German way Sturm and this is pronounced the French way Liouiville."},{"Start":"00:22.170 ","End":"00:25.050","Text":"But you\u0027re not going to be tested on pronunciation."},{"Start":"00:25.050 ","End":"00:28.635","Text":"Anyway, I don\u0027t know who these mathematicians were,"},{"Start":"00:28.635 ","End":"00:33.240","Text":"but we\u0027ll lead up to their theory bit at a time."},{"Start":"00:33.240 ","End":"00:37.905","Text":"We\u0027ll start off with an easy example and then get more involved gradually."},{"Start":"00:37.905 ","End":"00:43.490","Text":"The first example is a second-order differential equation,"},{"Start":"00:43.490 ","End":"00:49.460","Text":"linear constant coefficient homogeneous as it happens on a closed interval from"},{"Start":"00:49.460 ","End":"00:55.955","Text":"0 to 1 and instead of initial conditions which we usually give,"},{"Start":"00:55.955 ","End":"00:58.760","Text":"we have boundary conditions."},{"Start":"00:58.760 ","End":"01:02.690","Text":"Initial conditions usually relate to y and y\u0027 at"},{"Start":"01:02.690 ","End":"01:08.420","Text":"a given point and boundary conditions when we give the value of y or y\u0027."},{"Start":"01:08.420 ","End":"01:10.445","Text":"But at the end points,"},{"Start":"01:10.445 ","End":"01:12.789","Text":"condition for 2,"},{"Start":"01:12.789 ","End":"01:14.944","Text":"conditions relating to the 2 endpoints."},{"Start":"01:14.944 ","End":"01:20.135","Text":"In this case, we require that the function should be 0 here and 0 here."},{"Start":"01:20.135 ","End":"01:22.580","Text":"Now I can spot the solution right away,"},{"Start":"01:22.580 ","End":"01:24.410","Text":"what we call the trivial solution."},{"Start":"01:24.410 ","End":"01:26.840","Text":"If I let y equals 0,"},{"Start":"01:26.840 ","End":"01:29.104","Text":"that will solve the differential equation."},{"Start":"01:29.104 ","End":"01:31.325","Text":"They always solves the homogeneous equation."},{"Start":"01:31.325 ","End":"01:33.570","Text":"I also solves these because of the 0,"},{"Start":"01:33.570 ","End":"01:36.575","Text":"and this is called the trivial solution in general."},{"Start":"01:36.575 ","End":"01:39.814","Text":"But I\u0027d like to know if there\u0027s a non-trivial solution."},{"Start":"01:39.814 ","End":"01:43.730","Text":"First thing we would do is just solve this differential equation."},{"Start":"01:43.730 ","End":"01:48.215","Text":"Before that, I should\u0027ve mentioned this is called for short a BVP,"},{"Start":"01:48.215 ","End":"01:50.315","Text":"a boundary value problem."},{"Start":"01:50.315 ","End":"01:54.540","Text":"Just like we had an IVP initial value problem."},{"Start":"01:55.180 ","End":"01:59.630","Text":"We begin with the characteristic equation for y double-prime."},{"Start":"01:59.630 ","End":"02:02.570","Text":"We put k^2 for y we put 1."},{"Start":"02:02.570 ","End":"02:05.795","Text":"We solve this, it has complex solutions,"},{"Start":"02:05.795 ","End":"02:11.600","Text":"which are plus or minus i and we know from this that the solution to"},{"Start":"02:11.600 ","End":"02:15.170","Text":"the differential equation is"},{"Start":"02:15.170 ","End":"02:20.945","Text":"c_1 linear combination of cosine and sine and just as written."},{"Start":"02:20.945 ","End":"02:26.695","Text":"Now let\u0027s introduce the boundary conditions, there they are."},{"Start":"02:26.695 ","End":"02:33.380","Text":"What we do is we substitute 0 in here for x and we have"},{"Start":"02:33.380 ","End":"02:39.925","Text":"to get 0 for y. I get c_1 cosine 0 plus c_2 sine 0,"},{"Start":"02:39.925 ","End":"02:44.075","Text":"and for the other 1, we substitute 1 and in each case we have to get 0."},{"Start":"02:44.075 ","End":"02:49.665","Text":"Now remember that cosine 0 is 1 and sine 0 is 0,"},{"Start":"02:49.665 ","End":"02:55.755","Text":"so the first 1 just translates to c_1 equals 0."},{"Start":"02:55.755 ","End":"02:58.980","Text":"Then the second one, we get this."},{"Start":"02:58.980 ","End":"03:03.240","Text":"But if we plug in c_1 equals 0,"},{"Start":"03:03.240 ","End":"03:07.935","Text":"here, we get c_2 sine 1 is 0."},{"Start":"03:07.935 ","End":"03:09.960","Text":"But sine 1 is 0,"},{"Start":"03:09.960 ","End":"03:12.510","Text":"so that makes c_2 also 0,"},{"Start":"03:12.510 ","End":"03:14.655","Text":"so they\u0027re both 0."},{"Start":"03:14.655 ","End":"03:21.455","Text":"Really, there is no other solution other than the trivial 1 I mentioned before,"},{"Start":"03:21.455 ","End":"03:25.175","Text":"where y is the constant 0."},{"Start":"03:25.175 ","End":"03:27.815","Text":"That was the first example."},{"Start":"03:27.815 ","End":"03:32.300","Text":"Example 2 is almost the same as Example 1."},{"Start":"03:32.300 ","End":"03:37.760","Text":"If you look closely, the only difference is this Pi^2 here,"},{"Start":"03:37.760 ","End":"03:42.390","Text":"which we didn\u0027t have in the first example."},{"Start":"03:42.390 ","End":"03:50.430","Text":"Of course, the a trivial solution would still work here because it\u0027s homogeneous,"},{"Start":"03:50.430 ","End":"03:54.410","Text":"so y equals 0 would solve this and"},{"Start":"03:54.410 ","End":"03:59.150","Text":"also because the boundary values are both 0, this would work."},{"Start":"03:59.150 ","End":"04:00.860","Text":"Now in Example 1,"},{"Start":"04:00.860 ","End":"04:02.890","Text":"this was the only solution."},{"Start":"04:02.890 ","End":"04:05.360","Text":"I\u0027m going to give you a spoiler here."},{"Start":"04:05.360 ","End":"04:10.055","Text":"Example 2 will get solutions other than the trivial solution."},{"Start":"04:10.055 ","End":"04:12.640","Text":"Well, let\u0027s check."},{"Start":"04:12.640 ","End":"04:19.040","Text":"This time the characteristic equation of this equation is k^2 plus Pi^2,"},{"Start":"04:19.040 ","End":"04:21.215","Text":"note previously we had plus 1."},{"Start":"04:21.215 ","End":"04:26.435","Text":"That\u0027s similar another solutions are plus or minus Pi i and from here,"},{"Start":"04:26.435 ","End":"04:30.940","Text":"we get that the general solution of the homogeneous"},{"Start":"04:30.940 ","End":"04:36.400","Text":"is this linear combination of cosine Pi x and sine Pi x."},{"Start":"04:36.400 ","End":"04:39.710","Text":"Now we want to plug in the boundary values,"},{"Start":"04:39.710 ","End":"04:43.260","Text":"so we plug in for x 0,"},{"Start":"04:43.260 ","End":"04:47.340","Text":"1 time and then we plug in 1 instead of x and we get"},{"Start":"04:47.340 ","End":"04:54.370","Text":"this and this is what we get."},{"Start":"04:54.370 ","End":"05:04.380","Text":"I should have reminded you that of course the sine of 0 is 0 and also, sorry,"},{"Start":"05:04.380 ","End":"05:10.360","Text":"and to say, sine of Pi is also 0,"},{"Start":"05:10.360 ","End":"05:14.970","Text":"so this is 0 and this is 0."},{"Start":"05:14.970 ","End":"05:22.550","Text":"That gives us, in each case that c_1 is 0."},{"Start":"05:24.370 ","End":"05:27.830","Text":"Could\u0027ve mentioned that cosine 0 is 1,"},{"Start":"05:27.830 ","End":"05:29.690","Text":"but doesn\u0027t matter, it\u0027s not 0."},{"Start":"05:29.690 ","End":"05:32.030","Text":"We couldn\u0027t divide it by it, so the first equation gives us"},{"Start":"05:32.030 ","End":"05:36.715","Text":"directly that c_1 is 0 and for the second one,"},{"Start":"05:36.715 ","End":"05:39.585","Text":"cosine of Pi is minus 1."},{"Start":"05:39.585 ","End":"05:46.905","Text":"Well, write the whole thing that too and so we get minus c_1 is 0."},{"Start":"05:46.905 ","End":"05:50.240","Text":"The second 1 gives us the same as the first 1."},{"Start":"05:50.240 ","End":"05:55.250","Text":"Basically, all we can get out of it is that c_2 is 0,"},{"Start":"05:55.250 ","End":"06:04.420","Text":"but c_1, I wrote it backwards, please forgive me."},{"Start":"06:04.420 ","End":"06:05.860","Text":"Where it says c_1,"},{"Start":"06:05.860 ","End":"06:08.020","Text":"read c_2 and the other way round,"},{"Start":"06:08.020 ","End":"06:10.255","Text":"c_1 is the 1 that 0,"},{"Start":"06:10.255 ","End":"06:13.375","Text":"c_2 unrestricted could be anything."},{"Start":"06:13.375 ","End":"06:21.190","Text":"Let\u0027s just call it big C and then if we put c_1 is 0 here and c_2 is capital C,"},{"Start":"06:21.190 ","End":"06:26.785","Text":"then the general solution is y equals any constant c times sine Pi x."},{"Start":"06:26.785 ","End":"06:28.000","Text":"Because C can vary,"},{"Start":"06:28.000 ","End":"06:33.430","Text":"I call it a family of solutions is not 1 solution."},{"Start":"06:34.580 ","End":"06:37.525","Text":"That was Example 2."},{"Start":"06:37.525 ","End":"06:41.965","Text":"Now let\u0027s take another variant and jump to Example 3."},{"Start":"06:41.965 ","End":"06:46.130","Text":"Example 3 is also similar to Example 1,"},{"Start":"06:46.130 ","End":"06:47.660","Text":"except that instead of a plus,"},{"Start":"06:47.660 ","End":"06:49.310","Text":"we have a minus here,"},{"Start":"06:49.310 ","End":"06:53.270","Text":"so obviously, the boundary conditions are the same."},{"Start":"06:53.270 ","End":"06:57.315","Text":"We know we have a solution, y equals 0."},{"Start":"06:57.315 ","End":"07:00.770","Text":"Let\u0027s see if we have any non-trivial solutions."},{"Start":"07:00.770 ","End":"07:03.650","Text":"We started off by solving the differential equation."},{"Start":"07:03.650 ","End":"07:06.300","Text":"The characteristic equation."},{"Start":"07:06.670 ","End":"07:10.350","Text":"Characteristic is k^2 minus 1 is 0,"},{"Start":"07:10.350 ","End":"07:13.120","Text":"so k is plus or minus 1,"},{"Start":"07:13.120 ","End":"07:17.495","Text":"which gives us the general solution for the homogeneous is this."},{"Start":"07:17.495 ","End":"07:22.160","Text":"Now we want it to satisfy the boundary conditions. This is what we get."},{"Start":"07:22.160 ","End":"07:24.995","Text":"Remember when we plug in x equals 0,"},{"Start":"07:24.995 ","End":"07:27.350","Text":"that e^ 0 is 1."},{"Start":"07:27.350 ","End":"07:28.700","Text":"Here, just instead of x,"},{"Start":"07:28.700 ","End":"07:36.175","Text":"we just put 1 and e^F1 is e. From here,"},{"Start":"07:36.175 ","End":"07:39.655","Text":"let\u0027s see, I get some space here."},{"Start":"07:39.655 ","End":"07:42.940","Text":"The first equation, I\u0027ll just copy as is."},{"Start":"07:42.940 ","End":"07:44.350","Text":"In the second equation,"},{"Start":"07:44.350 ","End":"07:49.780","Text":"I\u0027ll multiply both sides by e,"},{"Start":"07:49.780 ","End":"07:53.665","Text":"and that will get rid of this e to the minus 1,"},{"Start":"07:53.665 ","End":"07:55.060","Text":"because that\u0027s 1 over e,"},{"Start":"07:55.060 ","End":"07:59.335","Text":"so I get c_1 plus c_2 e^2 is 0."},{"Start":"07:59.335 ","End":"08:01.915","Text":"Now we know what to do with this."},{"Start":"08:01.915 ","End":"08:05.530","Text":"There\u0027s many ways to do it. One way would be to subtract."},{"Start":"08:05.530 ","End":"08:07.960","Text":"Let\u0027s say the second from the first,"},{"Start":"08:07.960 ","End":"08:09.835","Text":"and then the c_1 disappears,"},{"Start":"08:09.835 ","End":"08:12.310","Text":"and we\u0027ve got c_2 minus c_2 e^2,"},{"Start":"08:12.310 ","End":"08:14.964","Text":"which is, this, is 0."},{"Start":"08:14.964 ","End":"08:17.785","Text":"Now this number is not 0,"},{"Start":"08:17.785 ","End":"08:20.170","Text":"e is not plus,"},{"Start":"08:20.170 ","End":"08:23.410","Text":"or minus 1 is 2.7 something."},{"Start":"08:23.410 ","End":"08:26.620","Text":"That means that the c_2 is 0,"},{"Start":"08:26.620 ","End":"08:28.180","Text":"and if c_2 is 0,"},{"Start":"08:28.180 ","End":"08:29.470","Text":"and you plug that in here,"},{"Start":"08:29.470 ","End":"08:35.150","Text":"say, then you get those so that c_1 is 0."},{"Start":"08:35.400 ","End":"08:41.990","Text":"Once again, we only get the trivial solution."},{"Start":"08:42.240 ","End":"08:44.950","Text":"There is the less interesting cases,"},{"Start":"08:44.950 ","End":"08:49.810","Text":"what we\u0027re going to be looking for cases when we have non trivial solutions."},{"Start":"08:49.810 ","End":"08:51.850","Text":"Here in Example 4,"},{"Start":"08:51.850 ","End":"08:54.940","Text":"we\u0027re going to generalize things a bit."},{"Start":"08:54.940 ","End":"08:59.515","Text":"All the 3 examples we\u0027ve had so far,"},{"Start":"08:59.515 ","End":"09:02.650","Text":"have all been in this form,"},{"Start":"09:02.650 ","End":"09:05.470","Text":"just with the different value of this."},{"Start":"09:05.470 ","End":"09:08.335","Text":"This is the Greek letter Lambda."},{"Start":"09:08.335 ","End":"09:13.730","Text":"Remind you how it\u0027s pronounced Greek letter."},{"Start":"09:14.880 ","End":"09:18.310","Text":"If you look back at Examples 1, 2, and 3,"},{"Start":"09:18.310 ","End":"09:22.000","Text":"you see that in the first example we have Lambda equals 1,"},{"Start":"09:22.000 ","End":"09:29.260","Text":"and then the second example Lambda was Pi^2 and in the third example Lambda was minus 1,"},{"Start":"09:29.260 ","End":"09:31.600","Text":"and what we\u0027re really looking for,"},{"Start":"09:31.600 ","End":"09:34.405","Text":"are interesting values of Lambda."},{"Start":"09:34.405 ","End":"09:36.460","Text":"When I say interesting, I mean,"},{"Start":"09:36.460 ","End":"09:40.270","Text":"once that will give us non trivial solutions."},{"Start":"09:40.270 ","End":"09:45.670","Text":"The only 1 so far that was interesting from our 3 examples was Pi^2,"},{"Start":"09:45.670 ","End":"09:52.270","Text":"1 and minus 1 both gave us just the trivial solution."},{"Start":"09:52.270 ","End":"09:54.235","Text":"Let\u0027s explore this,"},{"Start":"09:54.235 ","End":"09:59.620","Text":"and we\u0027re going to divide up into 3 cases."},{"Start":"09:59.620 ","End":"10:02.230","Text":"I\u0027ll just make a note to that."},{"Start":"10:02.230 ","End":"10:06.865","Text":"The cases basically to do with the characteristic equation."},{"Start":"10:06.865 ","End":"10:10.405","Text":"The characteristic equation is k^2,"},{"Start":"10:10.405 ","End":"10:14.230","Text":"plus Lambda equals 0."},{"Start":"10:14.230 ","End":"10:21.925","Text":"We\u0027re dividing into cases according to whether there\u0027s 2 real solutions,"},{"Start":"10:21.925 ","End":"10:26.770","Text":"a double root, and complex roots."},{"Start":"10:26.770 ","End":"10:29.920","Text":"That would depend on whether Lambda is bigger than 0,"},{"Start":"10:29.920 ","End":"10:32.080","Text":"or equal to 0, or less than 0."},{"Start":"10:32.080 ","End":"10:34.330","Text":"That\u0027s why we\u0027re dividing it into cases."},{"Start":"10:34.330 ","End":"10:36.535","Text":"With Case 1 will be Lambda equal 0,"},{"Start":"10:36.535 ","End":"10:39.355","Text":"Case 2 Lambda bigger than 0,"},{"Start":"10:39.355 ","End":"10:44.125","Text":"or vice versa with Case 3 less than 0, anyway."},{"Start":"10:44.125 ","End":"10:49.510","Text":"If Lambda equals 0, then we plug it in here,"},{"Start":"10:49.510 ","End":"10:53.840","Text":"and this is what we get, just this drops out."},{"Start":"10:57.420 ","End":"11:03.865","Text":"The characteristic equation is k^2 is 0,"},{"Start":"11:03.865 ","End":"11:06.685","Text":"and that gives us a double root."},{"Start":"11:06.685 ","End":"11:10.090","Text":"k_1 and k_2 are both 0,"},{"Start":"11:10.090 ","End":"11:15.715","Text":"we\u0027ll just say that there\u0027s only 1 double root, k equals 0."},{"Start":"11:15.715 ","End":"11:22.630","Text":"The solution of the homogeneous without the boundary conditions is this."},{"Start":"11:22.630 ","End":"11:28.670","Text":"Of course, we don\u0027t write it like this with e^0 is 1."},{"Start":"11:29.160 ","End":"11:31.780","Text":"It just comes down to this,"},{"Start":"11:31.780 ","End":"11:36.535","Text":"which is just an arbitrary linear equation."},{"Start":"11:36.535 ","End":"11:38.890","Text":"That makes sense because it\u0027s"},{"Start":"11:38.890 ","End":"11:43.045","Text":"only linear equations are the ones where the second derivative is 0."},{"Start":"11:43.045 ","End":"11:46.885","Text":"Now let\u0027s take care of the boundary conditions."},{"Start":"11:46.885 ","End":"11:49.420","Text":"We plug in x equals 0,"},{"Start":"11:49.420 ","End":"11:51.220","Text":"then we get c_1 is 0."},{"Start":"11:51.220 ","End":"11:52.990","Text":"Plugging x is 1,"},{"Start":"11:52.990 ","End":"11:55.854","Text":"c_1 plus c_2 is 0."},{"Start":"11:55.854 ","End":"11:59.155","Text":"I think from here it\u0027s clear,"},{"Start":"11:59.155 ","End":"12:03.055","Text":"both of c_1 and c_2 are 0."},{"Start":"12:03.055 ","End":"12:08.260","Text":"We only have the trivial solution."},{"Start":"12:08.260 ","End":"12:13.345","Text":"Lambda equals 0 tends out to be not so interesting."},{"Start":"12:13.345 ","End":"12:17.695","Text":"Now Case 2, Lambda bigger than 0,"},{"Start":"12:17.695 ","End":"12:21.965","Text":"and it\u0027s the same equation, just copied."},{"Start":"12:21.965 ","End":"12:26.310","Text":"The characteristic equation, k^2 plus Lambda is 0."},{"Start":"12:26.310 ","End":"12:28.420","Text":"That\u0027s what it always is."},{"Start":"12:28.640 ","End":"12:32.990","Text":"When Lambda is bigger than 0,"},{"Start":"12:32.990 ","End":"12:35.530","Text":"then when I bring it to the other side,"},{"Start":"12:35.530 ","End":"12:39.130","Text":"k^2 is minus Lambda, it\u0027s negative."},{"Start":"12:39.130 ","End":"12:41.830","Text":"I get the solutions are,"},{"Start":"12:41.830 ","End":"12:45.805","Text":"plus or minus the square root of Lambda times i,"},{"Start":"12:45.805 ","End":"12:49.120","Text":"to complex conjugate solutions."},{"Start":"12:49.120 ","End":"12:53.530","Text":"That gives us the general solution to the homogeneous as"},{"Start":"12:53.530 ","End":"12:59.485","Text":"this combination of cosine and sine."},{"Start":"12:59.485 ","End":"13:03.160","Text":"Silly me."},{"Start":"13:03.160 ","End":"13:07.705","Text":"I wrote i instead of x, there, fixed."},{"Start":"13:07.705 ","End":"13:10.255","Text":"That\u0027s the general solution."},{"Start":"13:10.255 ","End":"13:13.870","Text":"Now, the boundary conditions,"},{"Start":"13:13.870 ","End":"13:16.780","Text":"we plug in 0, this is what we get."},{"Start":"13:16.780 ","End":"13:21.535","Text":"We plug in x=1."},{"Start":"13:21.535 ","End":"13:24.380","Text":"Then this is what we get."},{"Start":"13:25.140 ","End":"13:31.808","Text":"Therefore, let\u0027s just see,"},{"Start":"13:31.808 ","End":"13:33.820","Text":"now I rewrote the first 1 cosine 0 is 0,"},{"Start":"13:33.820 ","End":"13:35.875","Text":"and cosine 0 is 1."},{"Start":"13:35.875 ","End":"13:39.100","Text":"Right away we get that c_1 is 0,"},{"Start":"13:39.100 ","End":"13:41.320","Text":"and we could plug that in here,"},{"Start":"13:41.320 ","End":"13:48.800","Text":"and then we\u0027d get c_2 times sine of something non-zero,"},{"Start":"13:49.290 ","End":"13:54.430","Text":"c_2 square root of Lambda is 0."},{"Start":"13:54.430 ","End":"14:00.235","Text":"Now of course, we could have c_2 equals 0."},{"Start":"14:00.235 ","End":"14:02.410","Text":"But then, it wouldn\u0027t be interesting."},{"Start":"14:02.410 ","End":"14:06.775","Text":"Then we just get the trivial solution c_2 is 0, c_1 is 0."},{"Start":"14:06.775 ","End":"14:10.360","Text":"Let\u0027s explore the possibility that c_2 isn\u0027t 0,"},{"Start":"14:10.360 ","End":"14:15.490","Text":"and see which interesting values of Lambda could provide this."},{"Start":"14:15.490 ","End":"14:18.535","Text":"Now in general, from trigonometry,"},{"Start":"14:18.535 ","End":"14:23.020","Text":"solution to sine Alpha is 0,"},{"Start":"14:23.020 ","End":"14:26.410","Text":"is known to be a multiple of Pi,"},{"Start":"14:26.410 ","End":"14:28.825","Text":"or multiple of a 180 degrees."},{"Start":"14:28.825 ","End":"14:35.125","Text":"Here just means that the square root of Lambda has to be Pi times some whole number."},{"Start":"14:35.125 ","End":"14:42.955","Text":"In other words, Lambda is going to be Pi^2 times some whole number squared."},{"Start":"14:42.955 ","End":"14:46.255","Text":"For each whole number n,"},{"Start":"14:46.255 ","End":"14:56.200","Text":"what we can do is call the y that\u0027s associated with this particular Lambda, call it y_n."},{"Start":"14:56.200 ","End":"15:00.235","Text":"I guess you could call this Lambda_n."},{"Start":"15:00.235 ","End":"15:04.720","Text":"Then for this particular Lambda_n solution will be this."},{"Start":"15:04.720 ","End":"15:07.600","Text":"Then we rename the c_2."},{"Start":"15:07.600 ","End":"15:10.745","Text":"C_2 can now be anything, because this is 0,"},{"Start":"15:10.745 ","End":"15:14.310","Text":"let me call it big C. We have a family of"},{"Start":"15:14.310 ","End":"15:21.630","Text":"solutions for each n. You might wonder which values of n?"},{"Start":"15:21.630 ","End":"15:25.160","Text":"Well, if n is 0,"},{"Start":"15:25.160 ","End":"15:27.245","Text":"it\u0027s not interesting,"},{"Start":"15:27.245 ","End":"15:31.255","Text":"because then sine of 0 is 0."},{"Start":"15:31.255 ","End":"15:32.815","Text":"If n is negative,"},{"Start":"15:32.815 ","End":"15:34.460","Text":"and n^2 is positive,"},{"Start":"15:34.460 ","End":"15:37.469","Text":"so no need to take the negative, so n is 1,"},{"Start":"15:37.469 ","End":"15:40.805","Text":"2, 3, 4, etc."},{"Start":"15:40.805 ","End":"15:49.775","Text":"For each n, we\u0027ve got a nice non trivial solution for this particular Lambda."},{"Start":"15:49.775 ","End":"15:52.475","Text":"Now there\u0027s a name for the Lambdas,"},{"Start":"15:52.475 ","End":"15:55.685","Text":"and the functions corresponding to them."},{"Start":"15:55.685 ","End":"16:00.120","Text":"I\u0027m going to give you 2 new technical terms."},{"Start":"16:00.120 ","End":"16:03.120","Text":"Now these functions,"},{"Start":"16:03.120 ","End":"16:05.480","Text":"I\u0027m going to not use the letter y,"},{"Start":"16:05.480 ","End":"16:10.410","Text":"we\u0027re going to use the Greek letter Phi."},{"Start":"16:10.420 ","End":"16:14.755","Text":"For each n, we have a function Phi_n,"},{"Start":"16:14.755 ","End":"16:19.570","Text":"which is just this sine of Pinx,"},{"Start":"16:19.570 ","End":"16:21.821","Text":"and just change the order."},{"Start":"16:21.821 ","End":"16:24.395","Text":"I prefer to write it as nPix,"},{"Start":"16:24.395 ","End":"16:28.220","Text":"and n equals natural number."},{"Start":"16:28.220 ","End":"16:31.475","Text":"These are called eigenfunctions."},{"Start":"16:31.475 ","End":"16:33.350","Text":"Don\u0027t ask me where the word comes from,"},{"Start":"16:33.350 ","End":"16:40.019","Text":"just take it as that\u0027s name for the boundary value problem."},{"Start":"16:40.210 ","End":"16:44.645","Text":"As for the Lambdas, these Pi^2,"},{"Start":"16:44.645 ","End":"16:46.940","Text":"n^2, in general,"},{"Start":"16:46.940 ","End":"16:50.300","Text":"the interesting values of Lambda are called the eigenvalues."},{"Start":"16:50.300 ","End":"16:52.250","Text":"Interesting in the sense that they produce"},{"Start":"16:52.250 ","End":"16:57.997","Text":"nontrivial solutions to the boundary value problem,"},{"Start":"16:57.997 ","End":"16:59.875","Text":"so 2 new terms."},{"Start":"16:59.875 ","End":"17:05.460","Text":"We still haven\u0027t covered the third case where Lambda is less than 0."},{"Start":"17:05.460 ","End":"17:07.780","Text":"Now the last case,"},{"Start":"17:07.780 ","End":"17:09.500","Text":"Lambda less than 0,"},{"Start":"17:09.500 ","End":"17:13.580","Text":"everything else is the same, the characteristic equation."},{"Start":"17:13.580 ","End":"17:16.190","Text":"But this time when I bring Lambda over to the other side"},{"Start":"17:16.190 ","End":"17:18.860","Text":"because it\u0027s negative minus Lambda is positive"},{"Start":"17:18.860 ","End":"17:26.885","Text":"and we get 2 real solutions plus or minus the square root of the positive minus Lambda."},{"Start":"17:26.885 ","End":"17:32.640","Text":"The general solution of the homogeneous is this,"},{"Start":"17:32.640 ","End":"17:37.040","Text":"and if we plug in x equals 0,"},{"Start":"17:37.040 ","End":"17:39.630","Text":"1 and then x equals 1,"},{"Start":"17:39.630 ","End":"17:42.515","Text":"for the boundary conditions,"},{"Start":"17:42.515 ","End":"17:45.335","Text":"so this is what we get for 0,"},{"Start":"17:45.335 ","End":"17:47.810","Text":"and then for x equals 1,"},{"Start":"17:47.810 ","End":"17:50.390","Text":"e^0 is 1 of course."},{"Start":"17:50.390 ","End":"17:52.400","Text":"In the second equation,"},{"Start":"17:52.400 ","End":"17:58.535","Text":"I\u0027ll multiply both sides by e to the square root of minus Lambda,"},{"Start":"17:58.535 ","End":"18:02.720","Text":"and that way it will cancel with this,"},{"Start":"18:02.720 ","End":"18:09.043","Text":"and the second gets more space."},{"Start":"18:09.043 ","End":"18:13.570","Text":"The first one just gives us c_1 plus c_2 is 0."},{"Start":"18:13.570 ","End":"18:16.090","Text":"Here, we multiply it by this,"},{"Start":"18:16.090 ","End":"18:18.160","Text":"so it becomes twice."},{"Start":"18:18.160 ","End":"18:20.365","Text":"This cancels with this."},{"Start":"18:20.365 ","End":"18:23.530","Text":"Now, if we subtract these 2 equations, the c_2,"},{"Start":"18:23.530 ","End":"18:30.355","Text":"c_2 are going to cancel and we get c_1 times 1 minus this is 0."},{"Start":"18:30.355 ","End":"18:33.085","Text":"Now, this quantity is not 0,"},{"Start":"18:33.085 ","End":"18:35.065","Text":"so c_1 is 0."},{"Start":"18:35.065 ","End":"18:37.945","Text":"If c_1 is 0 and I plug it into here,"},{"Start":"18:37.945 ","End":"18:41.334","Text":"then I also get c_2 is 0, ensured,"},{"Start":"18:41.334 ","End":"18:45.914","Text":"both of them, c_1 and c_2 are 0."},{"Start":"18:45.914 ","End":"18:48.760","Text":"In this Case 3,"},{"Start":"18:48.760 ","End":"18:52.620","Text":"we also only get the trivial solution,"},{"Start":"18:52.620 ","End":"18:56.659","Text":"y=0 to our boundary value problem."},{"Start":"18:56.659 ","End":"19:02.920","Text":"The example we just did this boundary value problem, and I copied it,"},{"Start":"19:02.920 ","End":"19:06.400","Text":"usually with some parameter,"},{"Start":"19:06.400 ","End":"19:13.810","Text":"Lambda is an example of a Sturm Louisville problem."},{"Start":"19:13.810 ","End":"19:14.890","Text":"It\u0027s just an example."},{"Start":"19:14.890 ","End":"19:22.280","Text":"Let\u0027s give a definition of what we mean by Sturm Louisville problem."},{"Start":"19:22.560 ","End":"19:26.450","Text":"Here it is. I unveiled it."},{"Start":"19:26.640 ","End":"19:29.500","Text":"The Sturm Louisville problem."},{"Start":"19:29.500 ","End":"19:32.440","Text":"It\u0027s called the regular Sturm Louisville problem."},{"Start":"19:32.440 ","End":"19:39.400","Text":"On an interval, we had the interval being 0,1,"},{"Start":"19:39.400 ","End":"19:41.110","Text":"instead of writing it as x between,"},{"Start":"19:41.110 ","End":"19:43.120","Text":"we can write it in interval form."},{"Start":"19:43.120 ","End":"19:48.190","Text":"In general, 0 and 1 will be replaced by a and b."},{"Start":"19:48.190 ","End":"19:51.880","Text":"This was just an example of an equation, but in general,"},{"Start":"19:51.880 ","End":"19:58.600","Text":"we\u0027ll have a second order differential equation of this form."},{"Start":"19:58.600 ","End":"20:05.920","Text":"In our case, P was 0 and Q was also 0,"},{"Start":"20:05.920 ","End":"20:09.865","Text":"and R was just the function 1."},{"Start":"20:09.865 ","End":"20:13.390","Text":"If you do that, if you write at P is 0,"},{"Start":"20:13.390 ","End":"20:18.535","Text":"Q is 0 and R is 1,"},{"Start":"20:18.535 ","End":"20:21.220","Text":"then you will get this equation."},{"Start":"20:21.220 ","End":"20:27.280","Text":"The boundary conditions relate to some condition that a,"},{"Start":"20:27.280 ","End":"20:32.290","Text":"let\u0027s use the colors that are in blue for a and red for b."},{"Start":"20:32.290 ","End":"20:37.015","Text":"But it doesn\u0027t have to just be y(a) and y(b),"},{"Start":"20:37.015 ","End":"20:38.440","Text":"like we had here."},{"Start":"20:38.440 ","End":"20:44.200","Text":"It could have a mixture of conditions of y and y\u0027 at a,"},{"Start":"20:44.200 ","End":"20:53.050","Text":"some constant Alpha well, as written."},{"Start":"20:53.050 ","End":"20:56.600","Text":"There are some conditions that apply here."},{"Start":"20:56.850 ","End":"21:00.010","Text":"Lambda is a parameter and as I said,"},{"Start":"21:00.010 ","End":"21:03.040","Text":"we will be looking for interesting values of Lambda,"},{"Start":"21:03.040 ","End":"21:06.685","Text":"which gives non trivial solutions."},{"Start":"21:06.685 ","End":"21:12.325","Text":"The y=0 will always work to solve a Sturm Louisville problem."},{"Start":"21:12.325 ","End":"21:15.505","Text":"Because not only is the equation homogeneous,"},{"Start":"21:15.505 ","End":"21:18.175","Text":"but you\u0027ve got zeros here and here also,"},{"Start":"21:18.175 ","End":"21:20.530","Text":"so y is z=0,"},{"Start":"21:20.530 ","End":"21:23.485","Text":"you\u0027ll always have the trivial solutions."},{"Start":"21:23.485 ","End":"21:27.220","Text":"Basically, we\u0027ll be exploring in a more general context"},{"Start":"21:27.220 ","End":"21:31.045","Text":"which values of Lambda will give us nontrivial solutions?"},{"Start":"21:31.045 ","End":"21:34.645","Text":"What are those nontrivial solutions?"},{"Start":"21:34.645 ","End":"21:37.210","Text":"The restriction is that the P,"},{"Start":"21:37.210 ","End":"21:40.345","Text":"Q and R are continuous on the interval."},{"Start":"21:40.345 ","End":"21:46.494","Text":"You can\u0027t have both Alpha and Beta being 0."},{"Start":"21:46.494 ","End":"21:50.395","Text":"Otherwise that would just not be a condition."},{"Start":"21:50.395 ","End":"21:55.790","Text":"Similarly, you can\u0027t both Gamma and Delta 0,"},{"Start":"21:56.100 ","End":"22:02.710","Text":"and for example, is what we just did already."},{"Start":"22:02.710 ","End":"22:06.520","Text":"We\u0027ve seen this one before."},{"Start":"22:06.520 ","End":"22:09.370","Text":"Here\u0027s another example."},{"Start":"22:09.370 ","End":"22:12.505","Text":"I just made a note that if you look back,"},{"Start":"22:12.505 ","End":"22:16.165","Text":"I want to know what a and b and Alpha, Beta, Gamma,"},{"Start":"22:16.165 ","End":"22:17.890","Text":"Delta are and in our case,"},{"Start":"22:17.890 ","End":"22:19.660","Text":"then this is what they are."},{"Start":"22:19.660 ","End":"22:22.540","Text":"We\u0027ve already mentioned the eigenvalues and"},{"Start":"22:22.540 ","End":"22:27.490","Text":"eigenfunctions of the Sturm Louisville problem."},{"Start":"22:27.490 ","End":"22:31.630","Text":"The eigenvalues are the interesting values of Lambda."},{"Start":"22:31.630 ","End":"22:36.100","Text":"Put in short, what it means is that Lambda is an eigenvalue,"},{"Start":"22:36.100 ","End":"22:42.070","Text":"if you plug-in that value of Lambda,"},{"Start":"22:42.070 ","End":"22:45.980","Text":"the problem has a non-trivial solution."},{"Start":"22:46.350 ","End":"22:52.630","Text":"Y=Phi(x), which is not the zeros solution."},{"Start":"22:52.630 ","End":"22:57.160","Text":"Phi(x) is called in eigenfunction if"},{"Start":"22:57.160 ","End":"23:03.230","Text":"it\u0027s a nontrivial solution associated with an eigenvalue Lambda."},{"Start":"23:03.330 ","End":"23:08.994","Text":"They\u0027re paired off an eigenvalue of Lambda and give us a non-trivial solution,"},{"Start":"23:08.994 ","End":"23:11.720","Text":"which is an eigenfunction."},{"Start":"23:11.850 ","End":"23:20.370","Text":"Just a minor note on this Greek letter has a slightly different form."},{"Start":"23:20.370 ","End":"23:22.350","Text":"At least this was typed,"},{"Start":"23:22.350 ","End":"23:24.465","Text":"for example, in Microsoft Word."},{"Start":"23:24.465 ","End":"23:26.040","Text":"There\u0027s a Phi like this,"},{"Start":"23:26.040 ","End":"23:32.910","Text":"and there\u0027s a Phi like this and so we consider them both to be the same letter."},{"Start":"23:32.910 ","End":"23:35.425","Text":"But some books like this print,"},{"Start":"23:35.425 ","End":"23:38.570","Text":"this font and some like this font."},{"Start":"23:40.560 ","End":"23:46.270","Text":"I\u0027d like to end this introduction with a few remarks."},{"Start":"23:46.270 ","End":"23:48.880","Text":"First one, I think I\u0027ve said it several times,"},{"Start":"23:48.880 ","End":"23:52.465","Text":"but I\u0027d like to have it written here."},{"Start":"23:52.465 ","End":"23:54.790","Text":"In any homogeneous problem,"},{"Start":"23:54.790 ","End":"23:59.500","Text":"are always guaranteed that it has the trivial solution,"},{"Start":"23:59.500 ","End":"24:06.400","Text":"y=0 but the question is whether it has other non trivial solutions."},{"Start":"24:06.400 ","End":"24:11.380","Text":"Remark number 2 relate to a given Lambda, not varying Lambdas."},{"Start":"24:11.380 ","End":"24:14.230","Text":"Suppose for some particular Lambda,"},{"Start":"24:14.230 ","End":"24:17.256","Text":"we have a non-trivial solution,"},{"Start":"24:17.256 ","End":"24:22.420","Text":"call it f(x), better than Phi."},{"Start":"24:22.420 ","End":"24:28.075","Text":"Now, because the problem is linear and homogeneous,"},{"Start":"24:28.075 ","End":"24:31.990","Text":"if we multiply a solution by a constant,"},{"Start":"24:31.990 ","End":"24:34.375","Text":"it\u0027s also a solution."},{"Start":"24:34.375 ","End":"24:37.795","Text":"If you add 2 solutions, it\u0027s a solution."},{"Start":"24:37.795 ","End":"24:39.924","Text":"In the language of linear algebra,"},{"Start":"24:39.924 ","End":"24:43.300","Text":"any linear combination of solutions is a solution."},{"Start":"24:43.300 ","End":"24:44.530","Text":"I wanted to say solutions."},{"Start":"24:44.530 ","End":"24:46.600","Text":"I mean with this same given Lambda,"},{"Start":"24:46.600 ","End":"24:49.460","Text":"you can\u0027t mix Lambdas."},{"Start":"24:50.310 ","End":"24:54.415","Text":"Like if f is a solution for a given Lambda,"},{"Start":"24:54.415 ","End":"24:59.215","Text":"and g is a solution for a given Lambda that I could say af"},{"Start":"24:59.215 ","End":"25:05.630","Text":"plus bg is also a solution for that same Lambda."},{"Start":"25:06.630 ","End":"25:13.720","Text":"Ties in with linear algebra and in fact the words eigenvalue and eigenfunction,"},{"Start":"25:13.720 ","End":"25:17.230","Text":"well, eigenvectors more belongs to linear algebra,"},{"Start":"25:17.230 ","End":"25:22.130","Text":"but eigen prefix belongs to linear algebra."},{"Start":"25:22.290 ","End":"25:26.740","Text":"Now, if you go looking up Sturm Louisville,"},{"Start":"25:26.740 ","End":"25:30.610","Text":"problems in books or on the Internet,"},{"Start":"25:30.610 ","End":"25:35.350","Text":"the form we gave it in is not the standard form."},{"Start":"25:35.350 ","End":"25:40.645","Text":"If we take our Sturm Louisville and multiply by,"},{"Start":"25:40.645 ","End":"25:42.820","Text":"you remember integration factors?"},{"Start":"25:42.820 ","End":"25:48.520","Text":"If you multiply by this integration factor where p is as in the definition,"},{"Start":"25:48.520 ","End":"25:50.140","Text":"then you can bring,"},{"Start":"25:50.140 ","End":"25:57.730","Text":"I guess this means our typo forgive,"},{"Start":"25:57.730 ","End":"26:01.430","Text":"our Sturm Louisville problems to the following form."},{"Start":"26:01.890 ","End":"26:05.710","Text":"This is the way it\u0027s commonly presented."},{"Start":"26:05.710 ","End":"26:10.330","Text":"You have a function multiplied by the first derivative,"},{"Start":"26:10.330 ","End":"26:13.915","Text":"and all this derived."},{"Start":"26:13.915 ","End":"26:16.210","Text":"Lambda\u0027s put inside the brackets,"},{"Start":"26:16.210 ","End":"26:19.480","Text":"it\u0027s packaged like this basically,"},{"Start":"26:19.480 ","End":"26:22.825","Text":"these conditions are the same as before."},{"Start":"26:22.825 ","End":"26:27.430","Text":"Just this equation is presented in a bit of a different form."},{"Start":"26:27.430 ","End":"26:34.645","Text":"Just so you\u0027ll encounter it else where you won\u0027t say that I\u0027ve given you the wrong one."},{"Start":"26:34.645 ","End":"26:36.715","Text":"It\u0027s a variant."},{"Start":"26:36.715 ","End":"26:39.730","Text":"But this form is the more customary form,"},{"Start":"26:39.730 ","End":"26:44.275","Text":"we\u0027ve been a bit unorthodox, let\u0027s say."},{"Start":"26:44.275 ","End":"26:51.760","Text":"I\u0027d like to show you an example of how to convert from our form to the customary form."},{"Start":"26:51.760 ","End":"26:55.645","Text":"Let\u0027s take this equation, for example."},{"Start":"26:55.645 ","End":"26:59.470","Text":"Never mind the boundary conditions because they are the same."},{"Start":"26:59.470 ","End":"27:07.945","Text":"Our p(x) is minus 2."},{"Start":"27:07.945 ","End":"27:13.075","Text":"The integration factor is e to the integral of minus 2dx,"},{"Start":"27:13.075 ","End":"27:15.955","Text":"which is just e to the minus 2x."},{"Start":"27:15.955 ","End":"27:19.670","Text":"I multiply all this by that."},{"Start":"27:19.670 ","End":"27:24.930","Text":"That gives us this, just straight multiplication."},{"Start":"27:24.930 ","End":"27:29.160","Text":"Now, the first two terms can be written like this."},{"Start":"27:29.160 ","End":"27:34.425","Text":"I suggest you just differentiate this to verify product rule,"},{"Start":"27:34.425 ","End":"27:40.095","Text":"the derivative of this is minus 2e to the minus 2xy prime."},{"Start":"27:40.095 ","End":"27:41.730","Text":"That\u0027s this term."},{"Start":"27:41.730 ","End":"27:46.340","Text":"Then keep this as it is and differentiate this and we get this."},{"Start":"27:46.340 ","End":"27:54.085","Text":"As for this, we just throw the e to the minus 2x inside."},{"Start":"27:54.085 ","End":"28:02.445","Text":"This now looks in the customary form p,"},{"Start":"28:02.445 ","End":"28:04.720","Text":"r, well,"},{"Start":"28:04.720 ","End":"28:07.609","Text":"this will be minus q."},{"Start":"28:07.770 ","End":"28:10.075","Text":"Here it is again."},{"Start":"28:10.075 ","End":"28:14.050","Text":"Anyway, this just shows you that that\u0027s how we get it"},{"Start":"28:14.050 ","End":"28:21.170","Text":"from the format that we presented to the customary format."},{"Start":"28:21.270 ","End":"28:25.224","Text":"I don\u0027t think it\u0027s anything more I went to add-in."},{"Start":"28:25.224 ","End":"28:29.180","Text":"Finally, that\u0027s the end of the introduction."}],"ID":29443},{"Watched":false,"Name":"Exercise 1","Duration":"10m 30s","ChapterTopicVideoID":28211,"CourseChapterTopicPlaylistID":294448,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.005","Text":"This exercise is a Sturm–Liouville boundary value problem."},{"Start":"00:07.005 ","End":"00:10.320","Text":"We have the homogeneous equation,"},{"Start":"00:10.320 ","End":"00:11.940","Text":"we have the interval,"},{"Start":"00:11.940 ","End":"00:15.165","Text":"and here we have the boundary conditions."},{"Start":"00:15.165 ","End":"00:21.720","Text":"As usual, we divide into 3 cases according to the parameter Lambda."},{"Start":"00:21.720 ","End":"00:25.109","Text":"First case, Lambda equals 0."},{"Start":"00:25.109 ","End":"00:28.230","Text":"This is what our equation becomes,"},{"Start":"00:28.230 ","End":"00:30.645","Text":"this term just drops out."},{"Start":"00:30.645 ","End":"00:37.065","Text":"Now the characteristic equation is k^2 equals 0 and it has a double root."},{"Start":"00:37.065 ","End":"00:42.029","Text":"Basically, it\u0027s like one root, but twice."},{"Start":"00:42.029 ","End":"00:50.741","Text":"In this case, we know that the solution of the homogeneous is a constant times e^0x,"},{"Start":"00:50.741 ","End":"00:54.840","Text":"and then we have the xe^0x part."},{"Start":"00:54.840 ","End":"00:58.123","Text":"Of course, the e^0 is 1,"},{"Start":"00:58.123 ","End":"01:02.075","Text":"so it comes down to just a linear function,"},{"Start":"01:02.075 ","End":"01:03.710","Text":"c_1 plus c_2 x."},{"Start":"01:03.710 ","End":"01:05.390","Text":"We\u0027re going to need the derivative because"},{"Start":"01:05.390 ","End":"01:09.965","Text":"the boundary condition contains the derivative."},{"Start":"01:09.965 ","End":"01:12.830","Text":"We didn\u0027t have an example like that in the tutorial,"},{"Start":"01:12.830 ","End":"01:14.990","Text":"but now we do."},{"Start":"01:14.990 ","End":"01:22.015","Text":"The derivative is the constant function c_2."},{"Start":"01:22.015 ","End":"01:28.870","Text":"I want to substitute 0 and 1 into the y prime,"},{"Start":"01:28.870 ","End":"01:31.540","Text":"but when it\u0027s a constant function,"},{"Start":"01:31.540 ","End":"01:34.420","Text":"it doesn\u0027t matter what your substitute is,"},{"Start":"01:34.420 ","End":"01:36.490","Text":"you always get c_2."},{"Start":"01:36.490 ","End":"01:40.570","Text":"We get the same equation twice, c_2 equals 0,"},{"Start":"01:40.570 ","End":"01:45.160","Text":"c_2 equals 0, conclusion: c_2 equals 0."},{"Start":"01:45.160 ","End":"01:48.400","Text":"Let me just get some more space here."},{"Start":"01:48.400 ","End":"01:51.730","Text":"There\u0027s actually no restriction on c_1, and anything,"},{"Start":"01:51.730 ","End":"02:00.760","Text":"we could call it big K or big c. We have a family of solutions,"},{"Start":"02:00.760 ","End":"02:03.620","Text":"y equals any constant."},{"Start":"02:05.640 ","End":"02:11.815","Text":"Now we\u0027re going to talk about eigenvalue and eigenfunction."},{"Start":"02:11.815 ","End":"02:15.260","Text":"The eigenvalue is the value of the Lambda."},{"Start":"02:15.260 ","End":"02:18.400","Text":"I want to give it a subscript because I want to have others later,"},{"Start":"02:18.400 ","End":"02:21.445","Text":"so I\u0027ll call that one Lambda naught is 0."},{"Start":"02:21.445 ","End":"02:24.417","Text":"For the eigenfunction, we choose one specific one,"},{"Start":"02:24.417 ","End":"02:26.695","Text":"we just don\u0027t choose the trivial one."},{"Start":"02:26.695 ","End":"02:31.420","Text":"Most obvious one to choose is y equals 1."},{"Start":"02:32.210 ","End":"02:37.120","Text":"Instead of y, we use the Greek letter Phi."},{"Start":"02:37.120 ","End":"02:39.635","Text":"This is the eigenfunction,"},{"Start":"02:39.635 ","End":"02:45.520","Text":"the constant function 1 associated with the eigenvalue 0."},{"Start":"02:45.520 ","End":"02:49.240","Text":"Let\u0027s move on to Case 2."},{"Start":"02:49.240 ","End":"02:54.510","Text":"This time we take Lambda as positive."},{"Start":"02:54.510 ","End":"02:57.290","Text":"Here\u0027s our equation."},{"Start":"02:57.290 ","End":"03:01.160","Text":"I can\u0027t really substitute anything like I did when Lambda equals 0,"},{"Start":"03:01.160 ","End":"03:05.750","Text":"it just looks the same except that we know that Lambda is positive."},{"Start":"03:05.750 ","End":"03:10.995","Text":"The characteristic equation is from here,"},{"Start":"03:10.995 ","End":"03:13.920","Text":"k^2 plus Lambda equals 0."},{"Start":"03:13.920 ","End":"03:17.885","Text":"For convenience, since Lambda is positive,"},{"Start":"03:17.885 ","End":"03:22.650","Text":"I can let Lambda be Omega squared."},{"Start":"03:22.650 ","End":"03:25.065","Text":"Let\u0027s use another Greek letter."},{"Start":"03:25.065 ","End":"03:29.655","Text":"We can take Omega is positive, of course."},{"Start":"03:29.655 ","End":"03:38.820","Text":"We can rewrite this as k^2 plus Omega ^2 equals 0. k^2 is"},{"Start":"03:38.820 ","End":"03:42.998","Text":"minus Omega squared and is negative so we need to"},{"Start":"03:42.998 ","End":"03:48.680","Text":"use complex numbers plus or minus Omega times i."},{"Start":"03:48.680 ","End":"03:52.875","Text":"We know what this means,"},{"Start":"03:52.875 ","End":"04:00.230","Text":"the solution of the homogeneous ordinary differential equation is this."},{"Start":"04:00.230 ","End":"04:01.850","Text":"Normally, there would"},{"Start":"04:01.850 ","End":"04:02.210","Text":"be"},{"Start":"04:02.210 ","End":"04:11.850","Text":"an e^0,"},{"Start":"04:11.850 ","End":"04:16.385","Text":"which is 1, because it\u0027s really 0 plus or minus Omega i,"},{"Start":"04:16.385 ","End":"04:20.165","Text":"and normally, we have e to the power of whatever it is here."},{"Start":"04:20.165 ","End":"04:23.030","Text":"But since it\u0027s 0, we don\u0027t have the e part,"},{"Start":"04:23.030 ","End":"04:25.460","Text":"we just have the cosine and the sine."},{"Start":"04:25.460 ","End":"04:28.255","Text":"Anyway, you know all this stuff."},{"Start":"04:28.255 ","End":"04:36.235","Text":"We\u0027re going to need the derivative because the boundary conditions require it."},{"Start":"04:36.235 ","End":"04:40.360","Text":"Here\u0027s y prime, it\u0027s straightforward enough."},{"Start":"04:40.520 ","End":"04:45.650","Text":"I need to substitute both x equals 0 and x equals 1"},{"Start":"04:45.650 ","End":"04:51.390","Text":"here and get y prime is 0 from these conditions."},{"Start":"04:51.620 ","End":"04:55.290","Text":"These conditions give us this,"},{"Start":"04:55.290 ","End":"05:01.205","Text":"and remember that sine of 0 is 0,"},{"Start":"05:01.205 ","End":"05:05.090","Text":"but cosine 0 is 1,"},{"Start":"05:05.090 ","End":"05:08.890","Text":"and if we substitute that,"},{"Start":"05:08.890 ","End":"05:12.170","Text":"then these just become this."},{"Start":"05:12.170 ","End":"05:15.950","Text":"Actually, I noticed that we can divide everything by Omega."},{"Start":"05:15.950 ","End":"05:18.320","Text":"Omega is bigger than 0, so it\u0027s not 0,"},{"Start":"05:18.320 ","End":"05:21.630","Text":"so I can cancel the Omega\u0027s."},{"Start":"05:21.630 ","End":"05:25.710","Text":"The first one gives me that c_2 is 0."},{"Start":"05:25.710 ","End":"05:30.270","Text":"Then I also have c_1 sine Omega is 0."},{"Start":"05:30.270 ","End":"05:35.075","Text":"Because c_2 is 0 and we are looking for non-trivial solutions,"},{"Start":"05:35.075 ","End":"05:39.125","Text":"I\u0027m going to look for c_1 not equal to 0."},{"Start":"05:39.125 ","End":"05:42.298","Text":"I always know that 00 will work,"},{"Start":"05:42.298 ","End":"05:44.465","Text":"but what we\u0027re looking for is something different."},{"Start":"05:44.465 ","End":"05:47.915","Text":"In that case, if c_1 is not 0,"},{"Start":"05:47.915 ","End":"05:54.305","Text":"then I get that sine Omega is 0. I just wrote that down."},{"Start":"05:54.305 ","End":"05:57.060","Text":"You were asked, what about c_1?"},{"Start":"05:57.160 ","End":"05:59.195","Text":"Well, c_1 could be anything."},{"Start":"05:59.195 ","End":"06:00.500","Text":"There\u0027s no restriction."},{"Start":"06:00.500 ","End":"06:03.790","Text":"One sine Omega is 0, c_1 could be anything."},{"Start":"06:03.790 ","End":"06:07.950","Text":"But there\u0027s a condition on Omega because"},{"Start":"06:07.950 ","End":"06:12.771","Text":"Omega has to satisfy this trigonometric equation,"},{"Start":"06:12.771 ","End":"06:21.335","Text":"and we know that the solutions where sine is 0 multiples of Pi where n is any integer."},{"Start":"06:21.335 ","End":"06:25.115","Text":"I guess I need some more space here."},{"Start":"06:25.115 ","End":"06:29.690","Text":"We get this family of solutions."},{"Start":"06:29.690 ","End":"06:37.700","Text":"The anything, I call that K. Now remember that Omega is bigger than 0,"},{"Start":"06:37.700 ","End":"06:41.588","Text":"so we can restrict our values of n to be 1,"},{"Start":"06:41.588 ","End":"06:44.720","Text":"2, 3, and so on."},{"Start":"06:44.720 ","End":"06:51.320","Text":"However, note that n equals 0 will give us,"},{"Start":"06:51.320 ","End":"06:57.120","Text":"cosine 0 is 1,"},{"Start":"06:57.120 ","End":"07:04.320","Text":"y naught equals K, and we already had that back in Case 1."},{"Start":"07:04.320 ","End":"07:06.420","Text":"Remember we had y is K,"},{"Start":"07:06.420 ","End":"07:14.740","Text":"so we took 1 as the eigenfunction and the eigenvalue was 0,"},{"Start":"07:14.740 ","End":"07:20.610","Text":"which it would be if n is 0,"},{"Start":"07:20.610 ","End":"07:23.505","Text":"then Omega is 0, then Lambda is 0."},{"Start":"07:23.505 ","End":"07:28.690","Text":"These are going to be our eigenfunctions except we don\u0027t need the K,"},{"Start":"07:28.690 ","End":"07:30.745","Text":"we just need a representative."},{"Start":"07:30.745 ","End":"07:33.580","Text":"I\u0027ll choose K equals 1 is the most obvious,"},{"Start":"07:33.580 ","End":"07:35.695","Text":"just anything but 0."},{"Start":"07:35.695 ","End":"07:37.660","Text":"We didn\u0027t want 0."},{"Start":"07:37.660 ","End":"07:43.730","Text":"But, yeah, cos(n Pi x) are going to be"},{"Start":"07:43.730 ","End":"07:52.280","Text":"the eigenfunctions and the corresponding eigenvalues,"},{"Start":"07:52.280 ","End":"07:53.930","Text":"I guess it\u0027s scrolled off screen,"},{"Start":"07:53.930 ","End":"07:59.040","Text":"but remember that Lambda was Omega squared."},{"Start":"07:59.590 ","End":"08:01.940","Text":"For each Omega n,"},{"Start":"08:01.940 ","End":"08:07.160","Text":"we have the corresponding Lambda n. Instead of n Pi,"},{"Start":"08:07.160 ","End":"08:09.535","Text":"we have n^2, Pi^2,"},{"Start":"08:09.535 ","End":"08:12.080","Text":"and n is any natural number,"},{"Start":"08:12.080 ","End":"08:14.960","Text":"and these are the eigenvalues corresponding."},{"Start":"08:14.960 ","End":"08:20.250","Text":"Each particular n, we have an eigenvalue and an eigenfunction."},{"Start":"08:20.330 ","End":"08:24.165","Text":"That does Case 2."},{"Start":"08:24.165 ","End":"08:27.720","Text":"Now let\u0027s move on to Case 3."},{"Start":"08:27.720 ","End":"08:33.030","Text":"This is the case where Lambda is negative."},{"Start":"08:33.030 ","End":"08:36.270","Text":"Here\u0027s the equation."},{"Start":"08:36.270 ","End":"08:40.990","Text":"I just copied it with the boundary values."},{"Start":"08:40.990 ","End":"08:45.465","Text":"The characteristic equation."},{"Start":"08:45.465 ","End":"08:49.355","Text":"This time because Lambda is negative,"},{"Start":"08:49.355 ","End":"08:55.720","Text":"we let Lambda be minus Omega squared."},{"Start":"08:55.720 ","End":"09:06.004","Text":"That would give us that k^2 minus Omega squared is 0. k is plus or minus Omega,"},{"Start":"09:06.004 ","End":"09:08.070","Text":"that\u0027s a k_1 plus Omega,"},{"Start":"09:08.070 ","End":"09:10.395","Text":"k_2 is minus Omega."},{"Start":"09:10.395 ","End":"09:14.315","Text":"The solution to the differential equation part is"},{"Start":"09:14.315 ","End":"09:20.130","Text":"this with the Omega here and the minus Omega here."},{"Start":"09:20.260 ","End":"09:23.330","Text":"Here\u0027s the derivative as before,"},{"Start":"09:23.330 ","End":"09:28.210","Text":"we need it because we\u0027re going to substitute the boundary values."},{"Start":"09:28.210 ","End":"09:33.050","Text":"We substitute, and now I need some more space."},{"Start":"09:33.050 ","End":"09:36.590","Text":"Use the fact that e^0 is 1,"},{"Start":"09:36.590 ","End":"09:37.985","Text":"it simplifies a bit."},{"Start":"09:37.985 ","End":"09:42.575","Text":"Now we can divide everything by Omega."},{"Start":"09:42.575 ","End":"09:45.755","Text":"Omega is not 0 because it\u0027s less than 0."},{"Start":"09:45.755 ","End":"09:48.200","Text":"This is what we get."},{"Start":"09:48.200 ","End":"09:51.020","Text":"But I also did something else."},{"Start":"09:51.020 ","End":"09:58.410","Text":"I multiplied this equation by e to the power of Omega, that\u0027s not w,"},{"Start":"09:58.410 ","End":"10:03.420","Text":"that\u0027s an Omega, and e to the minus Omega,"},{"Start":"10:03.420 ","End":"10:05.190","Text":"times e to the Omega is 1,"},{"Start":"10:05.190 ","End":"10:07.725","Text":"and this times this is e to the 2 Omega."},{"Start":"10:07.725 ","End":"10:09.945","Text":"This is the situation we have."},{"Start":"10:09.945 ","End":"10:12.033","Text":"Now, we have a minus c_2 and a minus c_2,"},{"Start":"10:12.033 ","End":"10:16.840","Text":"so it\u0027s natural to subtract both of these."},{"Start":"10:16.840 ","End":"10:23.325","Text":"The c_2 will cancel and I\u0027ll get c_1 times 1 minus e to the 2 Omega is 0,"},{"Start":"10:23.325 ","End":"10:25.065","Text":"but this is not 0,"},{"Start":"10:25.065 ","End":"10:26.790","Text":"so c_1 is 0,"},{"Start":"10:26.790 ","End":"10:28.050","Text":"and if c_1 is 0,"},{"Start":"10:28.050 ","End":"10:29.655","Text":"then c_2 is 0,"},{"Start":"10:29.655 ","End":"10:33.660","Text":"and if both c\u0027s are 0,"},{"Start":"10:33.660 ","End":"10:37.550","Text":"then in Case 3 where Lambda\u0027s negative,"},{"Start":"10:37.550 ","End":"10:41.190","Text":"we only have the trivial solution."},{"Start":"10:41.370 ","End":"10:44.290","Text":"That\u0027s Cases 1, 2, and 3,"},{"Start":"10:44.290 ","End":"10:49.745","Text":"and at the end usually collect all the results together."},{"Start":"10:49.745 ","End":"10:54.030","Text":"Let\u0027s just do a summary here."},{"Start":"10:54.300 ","End":"10:56.945","Text":"Well, Case 3, as you saw,"},{"Start":"10:56.945 ","End":"10:58.295","Text":"it doesn\u0027t give anything,"},{"Start":"10:58.295 ","End":"11:00.845","Text":"meaning only we got trivial solutions,"},{"Start":"11:00.845 ","End":"11:04.775","Text":"and we got a single solution,"},{"Start":"11:04.775 ","End":"11:08.090","Text":"a single eigenvalue in Case 1,"},{"Start":"11:08.090 ","End":"11:09.290","Text":"and in Case 2,"},{"Start":"11:09.290 ","End":"11:11.810","Text":"we got a whole series of them."},{"Start":"11:11.810 ","End":"11:16.170","Text":"Actually, Case 1 has part of Case 2."},{"Start":"11:16.400 ","End":"11:19.825","Text":"It\u0027s up to you how you want to organize it."},{"Start":"11:19.825 ","End":"11:23.740","Text":"We could completely incorporate Case 1 into Case 2,"},{"Start":"11:23.740 ","End":"11:26.425","Text":"because we had that when n equals 0,"},{"Start":"11:26.425 ","End":"11:28.930","Text":"we got that special function that"},{"Start":"11:28.930 ","End":"11:35.062","Text":"Phi-naught(x) is the constant function one which isn\u0027t really a cosine,"},{"Start":"11:35.062 ","End":"11:37.300","Text":"in the way it is."},{"Start":"11:37.300 ","End":"11:44.425","Text":"You can split it up and say we have a special eigenfunction,"},{"Start":"11:44.425 ","End":"11:46.990","Text":"Phi-naught, and all the rest of them,"},{"Start":"11:46.990 ","End":"11:48.250","Text":"n equals 1, 2,"},{"Start":"11:48.250 ","End":"11:50.395","Text":"3, etc., cosines."},{"Start":"11:50.395 ","End":"11:56.470","Text":"Or you could combine them and go for the non-negative integers."},{"Start":"11:56.470 ","End":"12:02.395","Text":"The eigenvalues, and remember that Lambda was equal to Omega squared."},{"Start":"12:02.395 ","End":"12:12.985","Text":"The Omegas equal to n Pi."},{"Start":"12:12.985 ","End":"12:15.870","Text":"The Lambda n was Pi^2, n^2."},{"Start":"12:15.870 ","End":"12:19.755","Text":"Again, the 0 just gives us 0."},{"Start":"12:19.755 ","End":"12:27.439","Text":"We could separate off Lambda 0 and Phi 0 and say that\u0027s an exceptional,"},{"Start":"12:27.439 ","End":"12:32.105","Text":"you could throw them in the list and say that one is a cosine."},{"Start":"12:32.105 ","End":"12:39.810","Text":"Anyway, these are the eigenvalues. We\u0027re done."}],"ID":29444},{"Watched":false,"Name":"Exercise 2","Duration":"12m 45s","ChapterTopicVideoID":28212,"CourseChapterTopicPlaylistID":294448,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"In this exercise,"},{"Start":"00:01.800 ","End":"00:05.670","Text":"we have a fairly familiar differential equation,"},{"Start":"00:05.670 ","End":"00:09.420","Text":"but the boundary conditions keep changing."},{"Start":"00:09.420 ","End":"00:15.135","Text":"This time, we have 0 and 1."},{"Start":"00:15.135 ","End":"00:21.890","Text":"At 0, we\u0027re given the value of y and at 1 we have a combination of y and y\u0027."},{"Start":"00:21.890 ","End":"00:24.300","Text":"That\u0027s a thing we haven\u0027t had before,"},{"Start":"00:24.300 ","End":"00:28.855","Text":"but not really much more involved because of that."},{"Start":"00:28.855 ","End":"00:33.650","Text":"Let\u0027s begin by solving the differential equation."},{"Start":"00:33.650 ","End":"00:36.350","Text":"As before, we divide into 3 cases."},{"Start":"00:36.350 ","End":"00:39.590","Text":"The first case being lambda equals 0."},{"Start":"00:39.590 ","End":"00:42.665","Text":"Plugging that in here gives us this."},{"Start":"00:42.665 ","End":"00:45.050","Text":"Now the equation part we\u0027ve done before,"},{"Start":"00:45.050 ","End":"00:48.445","Text":"so we can advance fairly quickly through that."},{"Start":"00:48.445 ","End":"00:51.080","Text":"Characteristic equation k^2 0,"},{"Start":"00:51.080 ","End":"00:54.170","Text":"we get twice the solution 0."},{"Start":"00:54.170 ","End":"01:00.650","Text":"The double solution gives us the exponent and x times the exponent,"},{"Start":"01:00.650 ","End":"01:05.000","Text":"and we just simplify because e^ 0 is 1."},{"Start":"01:05.000 ","End":"01:07.400","Text":"This is what we get and we got this before."},{"Start":"01:07.400 ","End":"01:11.575","Text":"It\u0027s just a general linear function."},{"Start":"01:11.575 ","End":"01:14.900","Text":"Now we need the boundary conditions and we\u0027ll"},{"Start":"01:14.900 ","End":"01:18.200","Text":"be needing the derivative because we have a y\u0027 here."},{"Start":"01:18.200 ","End":"01:20.900","Text":"This of course it\u0027s just the constant c_2."},{"Start":"01:20.900 ","End":"01:23.365","Text":"Let\u0027s substitute here."},{"Start":"01:23.365 ","End":"01:32.490","Text":"Y(0) is 0 gives us that c_1 is 0 because x is 0, then y is c_1."},{"Start":"01:32.490 ","End":"01:38.760","Text":"Then y(1) if we plug in 1 here,"},{"Start":"01:38.760 ","End":"01:40.935","Text":"we get c_1 plus c_2."},{"Start":"01:40.935 ","End":"01:42.540","Text":"If we plug in 1 here,"},{"Start":"01:42.540 ","End":"01:46.050","Text":"well there is nowhere to plug it\u0027s just the constant c_2."},{"Start":"01:46.050 ","End":"01:53.205","Text":"Now, we get c_1 is 0 from here and c_2 is 0."},{"Start":"01:53.205 ","End":"01:57.680","Text":"If a plugin c_1 is 0 twice c_2 is 0, so c_2 is 0,"},{"Start":"01:57.680 ","End":"01:59.540","Text":"and only the trivial solution,"},{"Start":"01:59.540 ","End":"02:03.170","Text":"so Lambda equals 0 didn\u0027t give us anything, it\u0027s interesting."},{"Start":"02:03.170 ","End":"02:07.800","Text":"Let\u0027s move on to the next 1 where Lambda is bigger than 0."},{"Start":"02:07.940 ","End":"02:11.010","Text":"Next lambda bigger than 0,"},{"Start":"02:11.010 ","End":"02:16.165","Text":"and same just copy the equation and the boundary conditions."},{"Start":"02:16.165 ","End":"02:21.650","Text":"Signal this before, so I\u0027m going quicker because Lambda is bigger than 0,"},{"Start":"02:21.650 ","End":"02:25.600","Text":"I can let Lambda be Omega squared,"},{"Start":"02:25.600 ","End":"02:30.110","Text":"and then we get this and we get plus or minus Omega i."},{"Start":"02:30.110 ","End":"02:39.920","Text":"That translates to the solution linear combination of cos(Omega)x and sin(Omega)x."},{"Start":"02:39.920 ","End":"02:42.260","Text":"We just have to figure out,"},{"Start":"02:42.260 ","End":"02:44.510","Text":"make it satisfy the boundary conditions."},{"Start":"02:44.510 ","End":"02:48.820","Text":"We\u0027re going to need the derivative, which is this."},{"Start":"02:48.820 ","End":"02:53.280","Text":"Now, to substitute both of these,"},{"Start":"02:53.280 ","End":"02:56.325","Text":"y(0) is 0,"},{"Start":"02:56.325 ","End":"02:58.290","Text":"so that gives us this."},{"Start":"02:58.290 ","End":"03:07.100","Text":"It\u0027s a bit of highlighting will make it clearer."},{"Start":"03:07.100 ","End":"03:15.510","Text":"Y(1) is what I get when I put x=1 in here,"},{"Start":"03:15.510 ","End":"03:20.160","Text":"that gives me the first 2 terms."},{"Start":"03:20.160 ","End":"03:28.125","Text":"Then y\u0027(1) means I put x equals 1 over here,"},{"Start":"03:28.125 ","End":"03:32.235","Text":"and that gives me these 2,"},{"Start":"03:32.235 ","End":"03:36.320","Text":"and altogether it\u0027s equal to 0."},{"Start":"03:36.320 ","End":"03:39.150","Text":"That\u0027s the 0 from here."},{"Start":"03:39.440 ","End":"03:44.030","Text":"I\u0027ll clear a bit of space."},{"Start":"03:44.030 ","End":"03:49.010","Text":"You probably remember that cosine 0 is 1 and sine 0 is 0,"},{"Start":"03:49.010 ","End":"03:51.210","Text":"we see this a lot."},{"Start":"03:59.750 ","End":"04:05.420","Text":"From the first 1, we immediately get c_1 equals 0."},{"Start":"04:05.420 ","End":"04:08.560","Text":"I put c_1 equals 0 in the second 1."},{"Start":"04:08.560 ","End":"04:12.210","Text":"This term drops out because c_1 is 0,"},{"Start":"04:12.210 ","End":"04:15.480","Text":"and this term drops out because c_1 is 0,"},{"Start":"04:15.480 ","End":"04:16.755","Text":"and in these 2,"},{"Start":"04:16.755 ","End":"04:19.745","Text":"I take c_2 outside the brackets."},{"Start":"04:19.745 ","End":"04:28.705","Text":"What I\u0027m left with is sin(Omega) plus omega cos(Omega) is 0."},{"Start":"04:28.705 ","End":"04:33.785","Text":"Now we\u0027re not looking for the trivial solution which is always available,"},{"Start":"04:33.785 ","End":"04:36.230","Text":"so c_1 is already 0,"},{"Start":"04:36.230 ","End":"04:40.055","Text":"so we\u0027re going to assume that c_2 is not 0."},{"Start":"04:40.055 ","End":"04:46.750","Text":"In that case, I can divide both sides by c_2 and I get this trigonometric equation,"},{"Start":"04:46.750 ","End":"04:53.250","Text":"sin(Omega) plus Omega cos(Omega) is 0."},{"Start":"04:53.250 ","End":"04:58.120","Text":"Now if I throw this term to the other side and divide by cos(Omega),"},{"Start":"04:58.120 ","End":"05:01.595","Text":"I get that tan(Omega) is minus Omega."},{"Start":"05:01.595 ","End":"05:09.155","Text":"You might ask, how do you know that cos(Omega) is not 0?"},{"Start":"05:09.155 ","End":"05:12.590","Text":"I didn\u0027t want to get into this level of technicality,"},{"Start":"05:12.590 ","End":"05:16.135","Text":"but if cos(Omega) is equal to 0,"},{"Start":"05:16.135 ","End":"05:21.985","Text":"then sin(Omega) is going to be plus or minus 1,"},{"Start":"05:21.985 ","End":"05:24.955","Text":"because sine^2 plus cosine^2 is 1."},{"Start":"05:24.955 ","End":"05:29.155","Text":"If cosine Omega is 0 and sin(Omega) is plus or minus 1,"},{"Start":"05:29.155 ","End":"05:33.045","Text":"that will give me c_2 plus or minus c_2 is 0."},{"Start":"05:33.045 ","End":"05:34.755","Text":"In any case c_2 is 0,"},{"Start":"05:34.755 ","End":"05:38.040","Text":"and we\u0027re avoiding the trivial solution."},{"Start":"05:38.080 ","End":"05:41.030","Text":"We can safely get to this now,"},{"Start":"05:41.030 ","End":"05:44.644","Text":"this is not easy to solve."},{"Start":"05:44.644 ","End":"05:46.970","Text":"I\u0027ll return to that in a moment."},{"Start":"05:46.970 ","End":"05:52.100","Text":"Meanwhile, because c_1 is 0,"},{"Start":"05:52.100 ","End":"05:55.445","Text":"our general solution which was this,"},{"Start":"05:55.445 ","End":"05:58.895","Text":"becomes and if you also let see 2,"},{"Start":"05:58.895 ","End":"06:05.555","Text":"which is unrestricted be k. Then we get y is k sine Omega x."},{"Start":"06:05.555 ","End":"06:08.280","Text":"But Omega has to be a special Omega,"},{"Start":"06:08.280 ","End":"06:10.335","Text":"has to satisfy this."},{"Start":"06:10.335 ","End":"06:13.625","Text":"Meanwhile, I\u0027m going to write that the eigenfunction,"},{"Start":"06:13.625 ","End":"06:18.805","Text":"the eigenfunctions are going to be y equals sin(Omega)x,"},{"Start":"06:18.805 ","End":"06:25.025","Text":"but only for those Omega where tangent Omega is minus Omega."},{"Start":"06:25.025 ","End":"06:29.045","Text":"Now, let me relate to this equation."},{"Start":"06:29.045 ","End":"06:32.855","Text":"No easy analytical way to do it, I don\u0027t know of any,"},{"Start":"06:32.855 ","End":"06:36.865","Text":"but we can also use a graphical method."},{"Start":"06:36.865 ","End":"06:40.260","Text":"If I draw the graph of tan(Omega) and I draw"},{"Start":"06:40.260 ","End":"06:43.895","Text":"the graph of minus Omega and see where they cross."},{"Start":"06:43.895 ","End":"06:47.380","Text":"Well, I\u0027ve already prepared a diagram."},{"Start":"06:47.380 ","End":"06:49.575","Text":"Now here we are."},{"Start":"06:49.575 ","End":"06:52.805","Text":"This is the Omega axis."},{"Start":"06:52.805 ","End":"06:57.350","Text":"This one is the tan(Omega),"},{"Start":"06:57.350 ","End":"06:59.495","Text":"which has several branches."},{"Start":"06:59.495 ","End":"07:02.600","Text":"Every Pi, like this would be 0,"},{"Start":"07:02.600 ","End":"07:06.570","Text":"this would be Pi, this would be 2 Pi,"},{"Start":"07:06.570 ","End":"07:09.130","Text":"3 Pi, and so on."},{"Start":"07:10.040 ","End":"07:15.395","Text":"This is the function minus Omega."},{"Start":"07:15.395 ","End":"07:19.130","Text":"It\u0027s not 45 degrees because there\u0027s not drawn to scale."},{"Start":"07:19.130 ","End":"07:25.715","Text":"But notice that it cuts every branch of the tangent somewhere."},{"Start":"07:25.715 ","End":"07:30.131","Text":"I\u0027ve got all these points here."},{"Start":"07:30.131 ","End":"07:35.930","Text":"Now, each of these has a value of Omega."},{"Start":"07:36.090 ","End":"07:39.820","Text":"Let\u0027s call this one Omega_naught."},{"Start":"07:39.820 ","End":"07:45.535","Text":"This one will be Omega_1, Omega_2, wherever."},{"Start":"07:45.535 ","End":"07:49.885","Text":"In each case, I just drop the perpendicular to the axis,"},{"Start":"07:49.885 ","End":"07:52.240","Text":"that would be Omega_3."},{"Start":"07:52.240 ","End":"07:56.875","Text":"Here we have an Omega_4 corresponding to this one,"},{"Start":"07:56.875 ","End":"07:59.740","Text":"Omega_5, and so on."},{"Start":"07:59.740 ","End":"08:04.060","Text":"We can actually ignore the first Omega."},{"Start":"08:04.060 ","End":"08:08.920","Text":"Omega_naught is 0 because Omega is positive."},{"Start":"08:08.920 ","End":"08:14.290","Text":"Now, let\u0027s just summarize with the eigenvalues and the eigenfunctions."},{"Start":"08:14.290 ","End":"08:16.060","Text":"For each n, 1,"},{"Start":"08:16.060 ","End":"08:17.305","Text":"2, 3, et cetera,"},{"Start":"08:17.305 ","End":"08:20.215","Text":"the eigenfunction we call it Phi_n,"},{"Start":"08:20.215 ","End":"08:24.670","Text":"is the sin(Omega_n times x)."},{"Start":"08:24.670 ","End":"08:29.470","Text":"We don\u0027t know numerically what the value of each of these Omega is,"},{"Start":"08:29.470 ","End":"08:32.995","Text":"but the diagram explains what they are."},{"Start":"08:32.995 ","End":"08:36.820","Text":"These are the eigenfunctions of the boundary value problem and"},{"Start":"08:36.820 ","End":"08:40.240","Text":"the eigenvalues they\u0027re not the Omegas,"},{"Start":"08:40.240 ","End":"08:46.120","Text":"they\u0027re Lambdas because Lambda_n is Omega_n squared."},{"Start":"08:46.120 ","End":"08:48.130","Text":"The squares of these,"},{"Start":"08:48.130 ","End":"08:50.590","Text":"those are the eigenvalues."},{"Start":"08:50.590 ","End":"08:54.640","Text":"This is Part 2,"},{"Start":"08:54.640 ","End":"08:57.550","Text":"and now we\u0027re going to move on to Case 3,"},{"Start":"08:57.550 ","End":"08:58.780","Text":"Lambda less than 0."},{"Start":"08:58.780 ","End":"09:03.340","Text":"Well, we\u0027ve done all this before until the problem where we have to"},{"Start":"09:03.340 ","End":"09:08.215","Text":"substitute in the boundary conditions the characteristic equations, k^2 plus Lambda."},{"Start":"09:08.215 ","End":"09:13.360","Text":"Because it\u0027s negative, we let Lambda be minus Omega squared."},{"Start":"09:13.360 ","End":"09:14.920","Text":"This is what we get."},{"Start":"09:14.920 ","End":"09:17.110","Text":"K is plus or minus Omega,"},{"Start":"09:17.110 ","End":"09:19.630","Text":"2 different real solutions."},{"Start":"09:19.630 ","End":"09:22.675","Text":"We know how to write the differential equation."},{"Start":"09:22.675 ","End":"09:25.825","Text":"Here\u0027s the Omega, here\u0027s the minus Omega."},{"Start":"09:25.825 ","End":"09:30.370","Text":"We also need the derivative because of this, which is that."},{"Start":"09:30.370 ","End":"09:33.085","Text":"This is seen before."},{"Start":"09:33.085 ","End":"09:37.435","Text":"Now let\u0027s get to the part where we substitute the boundary conditions."},{"Start":"09:37.435 ","End":"09:41.200","Text":"The first 1 is easy, y(naught) equals naught."},{"Start":"09:41.200 ","End":"09:45.205","Text":"Just plug in 2 here,"},{"Start":"09:45.205 ","End":"09:47.195","Text":"x equals naught,"},{"Start":"09:47.195 ","End":"09:49.695","Text":"and this is what we get."},{"Start":"09:49.695 ","End":"09:57.010","Text":"Then we plug in 1 both here and here."},{"Start":"09:57.630 ","End":"10:03.355","Text":"y(1), we put x=1 here,"},{"Start":"10:03.355 ","End":"10:05.920","Text":"and we get these 2."},{"Start":"10:05.920 ","End":"10:12.920","Text":"Then y\u0027(1), we put x=1 here and we get these."},{"Start":"10:13.950 ","End":"10:17.870","Text":"Let\u0027s see what we get."},{"Start":"10:18.450 ","End":"10:25.090","Text":"E^naught is 1 need I say so that c_1 plus c_2 is 0."},{"Start":"10:25.090 ","End":"10:29.530","Text":"1 of them 0, the other one\u0027s going to be 0 also."},{"Start":"10:29.530 ","End":"10:33.265","Text":"This one, if I collect together,"},{"Start":"10:33.265 ","End":"10:39.080","Text":"I collect together the bits with c_1 and I\u0027ve got this."},{"Start":"10:39.420 ","End":"10:45.220","Text":"It\u0027s e^Omega times 1 plus Omega."},{"Start":"10:45.220 ","End":"10:47.980","Text":"Then the parts with c_2,"},{"Start":"10:47.980 ","End":"10:50.185","Text":"we have a minus here."},{"Start":"10:50.185 ","End":"10:57.400","Text":"Now, we can multiply both sides by e^Omega."},{"Start":"10:57.400 ","End":"11:04.000","Text":"The multiplication by e^Omega makes this equal to 1, and here we get e^2 Omega."},{"Start":"11:04.000 ","End":"11:08.290","Text":"The other thing I\u0027ve done is that from the first equation,"},{"Start":"11:08.290 ","End":"11:11.755","Text":"we see that c_2 is minus c_1."},{"Start":"11:11.755 ","End":"11:15.400","Text":"If I replace c_2 by minus c_1, and this is what I get."},{"Start":"11:15.400 ","End":"11:17.680","Text":"I have c_1 twice."},{"Start":"11:17.680 ","End":"11:21.010","Text":"Now I claim that from this being 0,"},{"Start":"11:21.010 ","End":"11:23.260","Text":"it follows that c_1 is 0."},{"Start":"11:23.260 ","End":"11:26.110","Text":"The asterisk means I\u0027ll show you in a moment."},{"Start":"11:26.110 ","End":"11:30.715","Text":"If c_1 is 0 and c_2 is minus c_1 and c_2 is 0,"},{"Start":"11:30.715 ","End":"11:33.565","Text":"so we just get the trivial solution."},{"Start":"11:33.565 ","End":"11:37.705","Text":"But let me just show you how I got this."},{"Start":"11:37.705 ","End":"11:40.720","Text":"It\u0027s not that complicated,"},{"Start":"11:40.720 ","End":"11:42.835","Text":"but it\u0027s not trivial."},{"Start":"11:42.835 ","End":"11:45.429","Text":"This is something I\u0027ve said already."},{"Start":"11:45.429 ","End":"11:47.920","Text":"If c_1 is 0 then c_2 is 0,"},{"Start":"11:47.920 ","End":"11:51.925","Text":"then in fact, this turns out to be the case."},{"Start":"11:51.925 ","End":"11:55.660","Text":"If c_1 isn\u0027t 0,"},{"Start":"11:55.660 ","End":"11:58.630","Text":"we\u0027re going to get a contradiction because then,"},{"Start":"11:58.630 ","End":"12:02.335","Text":"we can divide by c_1."},{"Start":"12:02.335 ","End":"12:06.625","Text":"Bring this to the other side and divide by this."},{"Start":"12:06.625 ","End":"12:09.560","Text":"You\u0027ll get this equation."},{"Start":"12:09.720 ","End":"12:14.920","Text":"We\u0027ll get e^2Omega equals 1 minus Omega over 1 plus Omega."},{"Start":"12:14.920 ","End":"12:25.165","Text":"Now, Omega equals 0 satisfies this because then we just get e^0 equals 1 over 1,"},{"Start":"12:25.165 ","End":"12:28.675","Text":"but no other Omega will satisfy this."},{"Start":"12:28.675 ","End":"12:32.320","Text":"Because thinking about it,"},{"Start":"12:32.320 ","End":"12:36.820","Text":"I didn\u0027t even have to try putting in Omega equals 0 just for interest."},{"Start":"12:36.820 ","End":"12:39.835","Text":"I didn\u0027t know that Omega was bigger than 0,"},{"Start":"12:39.835 ","End":"12:44.170","Text":"if you look back when we said that Lambda equals Omega squared,"},{"Start":"12:44.170 ","End":"12:47.305","Text":"we also wrote that Omega bigger than 0."},{"Start":"12:47.305 ","End":"12:49.600","Text":"But anyway, 0 would have worked,"},{"Start":"12:49.600 ","End":"12:51.205","Text":"but it\u0027s not legal."},{"Start":"12:51.205 ","End":"12:54.145","Text":"Now if Omega is bigger than 0,"},{"Start":"12:54.145 ","End":"12:58.915","Text":"notice that the exponential function is increasing."},{"Start":"12:58.915 ","End":"13:00.595","Text":"If Omega is bigger than 0,"},{"Start":"13:00.595 ","End":"13:02.755","Text":"2 Omega is bigger than 0,"},{"Start":"13:02.755 ","End":"13:05.755","Text":"so e to the power of it is bigger than 1."},{"Start":"13:05.755 ","End":"13:07.255","Text":"On the other hand,"},{"Start":"13:07.255 ","End":"13:11.990","Text":"1 minus Omega is less than 1 plus Omega."},{"Start":"13:13.020 ","End":"13:19.135","Text":"This over this is going to be less than 1."},{"Start":"13:19.135 ","End":"13:27.835","Text":"The denominator is positive and i the numerator is less than the denominator,"},{"Start":"13:27.835 ","End":"13:33.295","Text":"then the fraction is going to be less than 1."},{"Start":"13:33.295 ","End":"13:35.560","Text":"On the 1 hand,"},{"Start":"13:35.560 ","End":"13:37.240","Text":"the left-hand side is bigger than 1,"},{"Start":"13:37.240 ","End":"13:39.130","Text":"the right-hand side is smaller than 1,"},{"Start":"13:39.130 ","End":"13:41.845","Text":"so this can\u0027t be."},{"Start":"13:41.845 ","End":"13:46.720","Text":"Basically, we ruled out, get a contradiction."},{"Start":"13:46.720 ","End":"13:52.300","Text":"If c_1 is not 0,"},{"Start":"13:52.300 ","End":"13:55.915","Text":"which means that c_1 is 0,"},{"Start":"13:55.915 ","End":"14:02.740","Text":"so we only have the trivial solution to our equation,"},{"Start":"14:02.740 ","End":"14:05.155","Text":"to our boundary value problem."},{"Start":"14:05.155 ","End":"14:08.860","Text":"Now it\u0027s time for the summary because we\u0027ve done Case 1,"},{"Start":"14:08.860 ","End":"14:11.480","Text":"2, and 3."},{"Start":"14:11.520 ","End":"14:14.065","Text":"See where it is."},{"Start":"14:14.065 ","End":"14:17.439","Text":"I\u0027m bringing the whole summary at once."},{"Start":"14:17.439 ","End":"14:21.415","Text":"In Cases 1, and 3, we got nothing."},{"Start":"14:21.415 ","End":"14:24.175","Text":"When I say nothing, I mean only the trivial solutions."},{"Start":"14:24.175 ","End":"14:26.440","Text":"We only got something interesting from Case 2."},{"Start":"14:26.440 ","End":"14:29.995","Text":"Remember that fancy graph of the tangent."},{"Start":"14:29.995 ","End":"14:34.310","Text":"We got a series sequence of functions,"},{"Start":"14:35.970 ","End":"14:39.880","Text":"sin(Omega_n x), n is 1, 2, 3, et cetera,"},{"Start":"14:39.880 ","End":"14:41.455","Text":"where these Omega_n,"},{"Start":"14:41.455 ","End":"14:46.210","Text":"where the positive solutions of this equation,"},{"Start":"14:46.210 ","End":"14:47.890","Text":"we don\u0027t know them numerically,"},{"Start":"14:47.890 ","End":"14:53.695","Text":"but we know that there are solutions of this and these are the eigenfunctions."},{"Start":"14:53.695 ","End":"14:56.455","Text":"Now the eigenvalues are not the Omega_n."},{"Start":"14:56.455 ","End":"15:01.765","Text":"Remember we need the Lambda which is Omega squared."},{"Start":"15:01.765 ","End":"15:05.600","Text":"The square of the solutions of this equation,"},{"Start":"15:05.600 ","End":"15:09.485","Text":"the squares are the eigenvalues corresponding."},{"Start":"15:09.485 ","End":"15:13.680","Text":"Now we are done with this exercise."}],"ID":29445},{"Watched":false,"Name":"Exercise 3","Duration":"14m 28s","ChapterTopicVideoID":28213,"CourseChapterTopicPlaylistID":294448,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.279","Text":"Here, we have yet another Sturm Louisville type of problem."},{"Start":"00:05.279 ","End":"00:11.230","Text":"This particular equation has appeared several times already,"},{"Start":"00:11.230 ","End":"00:14.654","Text":"so I\u0027ll go over the first part just quickly."},{"Start":"00:14.654 ","End":"00:16.800","Text":"As I said, we\u0027ve seen this before,"},{"Start":"00:16.800 ","End":"00:19.095","Text":"as usual 3 cases."},{"Start":"00:19.095 ","End":"00:24.030","Text":"The first case, we get this equation."},{"Start":"00:24.030 ","End":"00:31.080","Text":"The characteristic is this, double root, k=0."},{"Start":"00:31.080 ","End":"00:34.420","Text":"We get the general solution."},{"Start":"00:34.580 ","End":"00:43.250","Text":"Remember, we have the e^kx and then xe^kx only here k is 0,"},{"Start":"00:43.250 ","End":"00:45.230","Text":"so it simplifies to this."},{"Start":"00:45.230 ","End":"00:47.150","Text":"Because of the boundary conditions,"},{"Start":"00:47.150 ","End":"00:52.190","Text":"we also need the derivative c_2."},{"Start":"00:52.190 ","End":"00:54.500","Text":"You\u0027ve seen all this before."},{"Start":"00:54.500 ","End":"00:59.705","Text":"This is the point where we need to substitute our particular boundary conditions,"},{"Start":"00:59.705 ","End":"01:03.355","Text":"which are this and this."},{"Start":"01:03.355 ","End":"01:12.250","Text":"Let\u0027s see, y(0) is putting 0 here, we get c_1."},{"Start":"01:12.250 ","End":"01:15.815","Text":"y\u0027(0), y\u0027 is a constant,"},{"Start":"01:15.815 ","End":"01:17.720","Text":"it\u0027s everywhere, c_2."},{"Start":"01:17.720 ","End":"01:19.670","Text":"That\u0027s the first equation."},{"Start":"01:19.670 ","End":"01:22.180","Text":"The second equation is y(1),"},{"Start":"01:22.180 ","End":"01:28.160","Text":"so we put 1 into here where y is,"},{"Start":"01:28.160 ","End":"01:31.490","Text":"because we want to get c_1 plus c_2 is 0."},{"Start":"01:31.490 ","End":"01:37.295","Text":"Actually, we get the same equation twice,"},{"Start":"01:37.295 ","End":"01:39.590","Text":"c_1 plus c_2 is 0,"},{"Start":"01:39.590 ","End":"01:47.510","Text":"so we can just eliminate 1 of them and just say that c_1 plus c_2 is 0."},{"Start":"01:47.510 ","End":"01:51.870","Text":"That means that c_1 could be anything,"},{"Start":"01:51.870 ","End":"01:55.700","Text":"but then c_2 would have to be minus whatever that is,"},{"Start":"01:55.700 ","End":"02:00.620","Text":"so say K and minus K. Plug them both into here."},{"Start":"02:00.620 ","End":"02:03.930","Text":"y equals K minus Kx."},{"Start":"02:03.930 ","End":"02:05.945","Text":"Not the trivial solution,"},{"Start":"02:05.945 ","End":"02:09.409","Text":"unless K is 0, but K could be anything."},{"Start":"02:09.409 ","End":"02:11.240","Text":"Now for the eigenfunction,"},{"Start":"02:11.240 ","End":"02:14.330","Text":"you just choose a particular value of K, anything but 0."},{"Start":"02:14.330 ","End":"02:18.470","Text":"Usually, you choose something convenient like 1."},{"Start":"02:18.470 ","End":"02:22.805","Text":"Actually, I chose K equals minus 1."},{"Start":"02:22.805 ","End":"02:24.953","Text":"Could\u0027ve been okay for 1,"},{"Start":"02:24.953 ","End":"02:28.820","Text":"and I would have got 1 minus x. I prefer it as x minus 1."},{"Start":"02:28.820 ","End":"02:32.135","Text":"The eigenvalue is 0."},{"Start":"02:32.135 ","End":"02:35.164","Text":"I\u0027ll call that Lambda naught."},{"Start":"02:35.164 ","End":"02:38.615","Text":"The eigenfunction final, it\u0027s associated with it."},{"Start":"02:38.615 ","End":"02:40.850","Text":"That\u0027s so far, Case 1,"},{"Start":"02:40.850 ","End":"02:44.136","Text":"we\u0027ve still got 2 more chances to find more,"},{"Start":"02:44.136 ","End":"02:47.140","Text":"so let\u0027s go on to Case 2."},{"Start":"02:47.300 ","End":"02:50.750","Text":"Case 2, Lambda bigger than 0,"},{"Start":"02:50.750 ","End":"02:52.820","Text":"but this part\u0027s the same."},{"Start":"02:52.820 ","End":"02:55.130","Text":"We\u0027ve already done this before,"},{"Start":"02:55.130 ","End":"02:56.870","Text":"so I just copy pasted it."},{"Start":"02:56.870 ","End":"03:00.755","Text":"It\u0027s been in several previous exercises."},{"Start":"03:00.755 ","End":"03:04.070","Text":"That\u0027s the differential equation and the interval,"},{"Start":"03:04.070 ","End":"03:07.730","Text":"but the boundary conditions are what change each time."},{"Start":"03:07.730 ","End":"03:10.640","Text":"Now, let\u0027s take into account the boundary condition."},{"Start":"03:10.640 ","End":"03:15.038","Text":"Let us point out, this is the y that we found the general solution and"},{"Start":"03:15.038 ","End":"03:20.460","Text":"y\u0027 for the derivative we need because there\u0027s a derivative here."},{"Start":"03:21.040 ","End":"03:27.410","Text":"The boundary conditions, y(0),"},{"Start":"03:27.410 ","End":"03:29.780","Text":"where is y? Here it is."},{"Start":"03:29.780 ","End":"03:39.830","Text":"If x is 0, then we get c_1 cos(0), and c_2 sin(0)."},{"Start":"03:39.830 ","End":"03:43.199","Text":"Then plus y\u0027(0),"},{"Start":"03:43.199 ","End":"03:45.660","Text":"so we plug in 0 into here."},{"Start":"03:45.660 ","End":"03:51.505","Text":"There should be an Omega here."},{"Start":"03:51.505 ","End":"03:53.660","Text":"I\u0027ll just do a quick fix here,"},{"Start":"03:53.660 ","End":"03:55.145","Text":"say there\u0027s an Omega."},{"Start":"03:55.145 ","End":"03:58.730","Text":"As it turns out, sin(0) is 0 so it wouldn\u0027t have made any difference,"},{"Start":"03:58.730 ","End":"04:01.411","Text":"but it should be an Omega,"},{"Start":"04:01.411 ","End":"04:06.030","Text":"plus c_2 Omega cos(0)."},{"Start":"04:06.500 ","End":"04:09.230","Text":"I forgot to highlight this,"},{"Start":"04:09.230 ","End":"04:12.830","Text":"that goes with the y and the y\u0027(0) is what gives us"},{"Start":"04:12.830 ","End":"04:19.660","Text":"this bit with the Omega and the minus."},{"Start":"04:19.870 ","End":"04:29.516","Text":"Now, y(1) just need to read off from y and plug in x=0."},{"Start":"04:29.516 ","End":"04:35.401","Text":"I\u0027m using this color for the function and the other color for the derivative,"},{"Start":"04:35.401 ","End":"04:39.010","Text":"so y(1) is just c_1 cos(Omega 1),"},{"Start":"04:39.010 ","End":"04:43.880","Text":"which is Omega, and c_2 sin(Omega)."},{"Start":"04:43.900 ","End":"04:51.790","Text":"Now, I hope you remember some of your trigonometry at that"},{"Start":"04:51.790 ","End":"05:02.300","Text":"sin(0) equals 0 and cos(0) equals 1."},{"Start":"05:02.300 ","End":"05:05.310","Text":"When we substitute this,"},{"Start":"05:05.310 ","End":"05:08.530","Text":"the 2 signs here disappear and here,"},{"Start":"05:08.530 ","End":"05:11.680","Text":"we just get c_1 plus c_2 Omega."},{"Start":"05:11.680 ","End":"05:14.248","Text":"That\u0027s the first equation, 0."},{"Start":"05:14.248 ","End":"05:16.975","Text":"Then the second one,"},{"Start":"05:16.975 ","End":"05:23.710","Text":"well, we just copy it as is."},{"Start":"05:23.710 ","End":"05:25.870","Text":"There\u0027s nothing to simplify."},{"Start":"05:25.870 ","End":"05:28.120","Text":"It\u0027s cos(Omega), sin(Omega),"},{"Start":"05:28.120 ","End":"05:30.860","Text":"we don\u0027t know what these are."},{"Start":"05:30.930 ","End":"05:33.130","Text":"Now from the first equation,"},{"Start":"05:33.130 ","End":"05:42.299","Text":"I can get c_1 in terms of c_2 and then I can substitute c_1 here using this."},{"Start":"05:42.299 ","End":"05:46.290","Text":"This bit I underlined the c_1 just becomes minus"},{"Start":"05:46.290 ","End":"05:53.600","Text":"c_2 Omega and this is the equation that we have."},{"Start":"05:53.820 ","End":"05:57.790","Text":"We\u0027re looking for non-trivial solutions."},{"Start":"05:57.790 ","End":"06:01.510","Text":"If c_2 is 0, for example,"},{"Start":"06:01.510 ","End":"06:03.940","Text":"then c_1 will also be 0,"},{"Start":"06:03.940 ","End":"06:06.390","Text":"so we want to c_2 not 0."},{"Start":"06:06.390 ","End":"06:09.055","Text":"I just wrote that down."},{"Start":"06:09.055 ","End":"06:15.650","Text":"Then of course we can divide this equation by c_2 so we just have this."},{"Start":"06:15.650 ","End":"06:20.990","Text":"Now, what I want to do is bring this to the other side and divide by cos(Omega)."},{"Start":"06:20.990 ","End":"06:25.714","Text":"I\u0027ll explain in a moment why cos(Omega) is not going to be 0."},{"Start":"06:25.714 ","End":"06:29.780","Text":"Anyway, once we\u0027ve done that, it becomes plus Omega cos(Omega),"},{"Start":"06:29.780 ","End":"06:33.830","Text":"and then we divide by cos(Omega)."},{"Start":"06:33.830 ","End":"06:37.220","Text":"We get it the other way round, Omega equals tan(Omega),"},{"Start":"06:37.220 ","End":"06:38.945","Text":"but then flip sides."},{"Start":"06:38.945 ","End":"06:47.075","Text":"The reason cos(Omega) is not going to be 0 is if cos(Omega) was 0,"},{"Start":"06:47.075 ","End":"06:52.580","Text":"then what we would get if we plug it in here is that sin(Omega)"},{"Start":"06:52.580 ","End":"06:58.010","Text":"is 0 and the cosine and the sine are never 0 together for many reasons."},{"Start":"06:58.010 ","End":"07:00.995","Text":"For example, cosine^2 plus sine^2 is 1,"},{"Start":"07:00.995 ","End":"07:02.583","Text":"so they can\u0027t both be 0,"},{"Start":"07:02.583 ","End":"07:10.275","Text":"so that\u0027s why cosine is not going to be 0."},{"Start":"07:10.275 ","End":"07:18.445","Text":"This equation, there\u0027s no simple closed formula way of solving it."},{"Start":"07:18.445 ","End":"07:21.220","Text":"We\u0027re going to have to do it using graphs,"},{"Start":"07:21.220 ","End":"07:24.370","Text":"but let\u0027s say we have a solution Omega to this problem."},{"Start":"07:24.370 ","End":"07:25.540","Text":"There might be many, in fact,"},{"Start":"07:25.540 ","End":"07:29.950","Text":"they\u0027ll turn out to be infinitely many but whatever Omega solves this,"},{"Start":"07:29.950 ","End":"07:32.720","Text":"we can then say,"},{"Start":"07:33.510 ","End":"07:38.620","Text":"we plugin to our equation, it\u0027s above here."},{"Start":"07:38.620 ","End":"07:43.390","Text":"y is c_1 cos(Omega) x plus c_2 sin(Omega) x."},{"Start":"07:43.390 ","End":"07:51.115","Text":"That was the general solution and we let c_2 equal anything we want to say k,"},{"Start":"07:51.115 ","End":"07:57.970","Text":"c_2 is unrestricted from these equations as long as c_1 observes this equation."},{"Start":"07:57.970 ","End":"08:02.620","Text":"c_1 would be minus k Omega."},{"Start":"08:02.620 ","End":"08:08.005","Text":"Then if you do that and put this in terms of k,"},{"Start":"08:08.005 ","End":"08:13.240","Text":"then this is what we get c_2 is k and c_1 is minus k Omega."},{"Start":"08:13.240 ","End":"08:15.490","Text":"I should have maybe said this is this,"},{"Start":"08:15.490 ","End":"08:22.629","Text":"and c_2 is just k. We can get an eigenfunction by taking, for example,"},{"Start":"08:22.629 ","End":"08:27.055","Text":"k=1 corresponding to"},{"Start":"08:27.055 ","End":"08:34.765","Text":"whichever Omega\u0027s fit this."},{"Start":"08:34.765 ","End":"08:37.465","Text":"I\u0027m getting ahead of myself."},{"Start":"08:37.465 ","End":"08:40.450","Text":"Let\u0027s just talk about the eigenfunction."},{"Start":"08:40.450 ","End":"08:48.220","Text":"The eigenfunction is going to be equal to if I let k=1 here,"},{"Start":"08:48.220 ","End":"08:53.170","Text":"minus Omega cos(Omega)x plus sin(Omega)x but we"},{"Start":"08:53.170 ","End":"08:58.240","Text":"can put in any Omega such that tangent Omega equals Omega."},{"Start":"08:58.240 ","End":"09:00.595","Text":"That\u0027s the equation here."},{"Start":"09:00.595 ","End":"09:03.820","Text":"Now we\u0027re going to solve this and then we still have to get from Omega"},{"Start":"09:03.820 ","End":"09:07.375","Text":"back to Lambda because Lambda is the eigenvalue."},{"Start":"09:07.375 ","End":"09:10.705","Text":"Let me jump to the next page with the graph."},{"Start":"09:10.705 ","End":"09:13.840","Text":"Here\u0027s the graph of both functions."},{"Start":"09:13.840 ","End":"09:23.950","Text":"This is the function just the independent variable is Omega."},{"Start":"09:23.950 ","End":"09:25.450","Text":"Then we have 2 functions."},{"Start":"09:25.450 ","End":"09:31.605","Text":"We have Omega itself and we have a tangent of Omega,"},{"Start":"09:31.605 ","End":"09:36.335","Text":"all these together are tangent of Omega,"},{"Start":"09:36.335 ","End":"09:46.540","Text":"just in another word you call it some other Greek letter equals the function of Omega,"},{"Start":"09:46.540 ","End":"09:48.895","Text":"which is Omega, and the function of Omega which is tan(Omega),"},{"Start":"09:48.895 ","End":"09:51.970","Text":"and where they cross is where we have the equality."},{"Start":"09:51.970 ","End":"09:55.959","Text":"We have, for example here,"},{"Start":"09:55.959 ","End":"09:58.750","Text":"here, here, and here."},{"Start":"09:58.750 ","End":"10:02.680","Text":"Now if you recall the condition was that Omega is bigger than 0,"},{"Start":"10:02.680 ","End":"10:06.070","Text":"so this one I wouldn\u0027t take for example."},{"Start":"10:06.070 ","End":"10:11.350","Text":"Also, that\u0027s why I only took positive side of the graph and we can give them names,"},{"Start":"10:11.350 ","End":"10:13.570","Text":"the value of Omega that\u0027s below them,"},{"Start":"10:13.570 ","End":"10:15.505","Text":"let\u0027s say this is below this,"},{"Start":"10:15.505 ","End":"10:17.080","Text":"and this is below this."},{"Start":"10:17.080 ","End":"10:18.760","Text":"This is just below this."},{"Start":"10:18.760 ","End":"10:23.005","Text":"That would be to say Omega_1, Omega_2."},{"Start":"10:23.005 ","End":"10:25.420","Text":"I don\u0027t take this Omega,"},{"Start":"10:25.420 ","End":"10:28.195","Text":"Omega_3, Omega_4, and so on."},{"Start":"10:28.195 ","End":"10:30.100","Text":"For any positive value of n,"},{"Start":"10:30.100 ","End":"10:36.955","Text":"I have Omega_n infinitely many and they form a sequence Omega_ns,"},{"Start":"10:36.955 ","End":"10:42.500","Text":"where n is a positive integer"},{"Start":"10:43.650 ","End":"10:49.540","Text":"and Omega_n, they\u0027re all positive."},{"Start":"10:49.540 ","End":"10:54.100","Text":"But as I said they want to somehow get from Omega back to Lambda."},{"Start":"10:54.100 ","End":"10:55.900","Text":"Just 1 more moment,"},{"Start":"10:55.900 ","End":"10:57.174","Text":"we\u0027ll get to the eigenvalues."},{"Start":"10:57.174 ","End":"11:03.820","Text":"We can summarize the eigenfunctions of this series for Omega is Omega_n."},{"Start":"11:03.820 ","End":"11:05.845","Text":"We get infinitely many."},{"Start":"11:05.845 ","End":"11:10.150","Text":"We get for each n eigenfunction Phi_n,"},{"Start":"11:10.150 ","End":"11:12.680","Text":"which is sine of Omega_nx."},{"Start":"11:13.380 ","End":"11:17.275","Text":"The eigenfunctions of our problem, boundary value,"},{"Start":"11:17.275 ","End":"11:20.515","Text":"and the eigenvalues finally,"},{"Start":"11:20.515 ","End":"11:26.755","Text":"there is the whole sequence of Lambda_ns."},{"Start":"11:26.755 ","End":"11:36.760","Text":"Also, an infinite number of Lambda_ns also n is bigger or equal to 1,"},{"Start":"11:36.760 ","End":"11:41.004","Text":"or sometimes you say from 1 to infinity or whatever."},{"Start":"11:41.004 ","End":"11:44.140","Text":"Just the squares of the solutions of these."},{"Start":"11:44.140 ","End":"11:45.865","Text":"If we square these numbers,"},{"Start":"11:45.865 ","End":"11:48.925","Text":"we get Lambda_1, Lambda_2, and so on."},{"Start":"11:48.925 ","End":"11:59.540","Text":"If we take the same n, Lambda_5 eigenvalue corresponds to Phi_5 eigenfunction."},{"Start":"12:02.850 ","End":"12:05.830","Text":"That\u0027s what we found for Case 2."},{"Start":"12:05.830 ","End":"12:08.410","Text":"We still have Case 3."},{"Start":"12:08.410 ","End":"12:13.180","Text":"In Case I1, we found and finitely many in Case 2."},{"Start":"12:13.180 ","End":"12:14.398","Text":"Now let\u0027s see case III."},{"Start":"12:14.398 ","End":"12:16.250","Text":"I\u0027ll start a new page."},{"Start":"12:16.250 ","End":"12:19.575","Text":"Case 3, Lambda negative."},{"Start":"12:19.575 ","End":"12:22.140","Text":"It\u0027s the same differential equation as in"},{"Start":"12:22.140 ","End":"12:29.010","Text":"several previous exercises so the first part, I\u0027ll just skip."},{"Start":"12:29.010 ","End":"12:33.850","Text":"Here it is, but I won\u0027t go into it in any detail because we\u0027ve done it before."},{"Start":"12:33.850 ","End":"12:36.565","Text":"Here\u0027s y the general solution,"},{"Start":"12:36.565 ","End":"12:38.020","Text":"and here\u0027s y′,"},{"Start":"12:38.020 ","End":"12:42.400","Text":"which we need because there is a y′ in the boundary conditions."},{"Start":"12:42.400 ","End":"12:44.395","Text":"The thing that\u0027s changed,"},{"Start":"12:44.395 ","End":"12:47.830","Text":"changes from the last few exercises from 1 to the other"},{"Start":"12:47.830 ","End":"12:51.250","Text":"is the boundary conditions and here they are."},{"Start":"12:51.250 ","End":"12:56.920","Text":"Let me explain first of all the bits with y. y is this,"},{"Start":"12:56.920 ","End":"12:58.870","Text":"so if I want y(0),"},{"Start":"12:58.870 ","End":"13:05.980","Text":"I just put 0 in here and I get these two terms."},{"Start":"13:05.980 ","End":"13:14.455","Text":"Here, y(1) is going to be what I get when I put x=1 here so I get this."},{"Start":"13:14.455 ","End":"13:17.245","Text":"The y′ bit is this,"},{"Start":"13:17.245 ","End":"13:21.430","Text":"and all I have to do is y′(0) here so I put 0 in"},{"Start":"13:21.430 ","End":"13:26.500","Text":"here and I get these and each of these is equal to 0."},{"Start":"13:26.500 ","End":"13:30.985","Text":"Now, remember that e"},{"Start":"13:30.985 ","End":"13:38.080","Text":"to the 0 is 1 and so the first equation can be simplified."},{"Start":"13:38.080 ","End":"13:40.655","Text":"Here, all the e to the 0 and e to the minus"},{"Start":"13:40.655 ","End":"13:44.499","Text":"0 disappear and if I collect this and this together,"},{"Start":"13:44.499 ","End":"13:47.200","Text":"I get c_1 times 1 plus Omega."},{"Start":"13:47.200 ","End":"13:48.610","Text":"If I take these two,"},{"Start":"13:48.610 ","End":"13:51.010","Text":"I get c_2 times 1 minus Omega."},{"Start":"13:51.010 ","End":"13:55.480","Text":"What I did was I multiplied"},{"Start":"13:55.480 ","End":"14:02.480","Text":"both sides by e to the Omega and this gives me this."},{"Start":"14:03.120 ","End":"14:06.955","Text":"The second equation gives me c_2 in terms of"},{"Start":"14:06.955 ","End":"14:10.900","Text":"c_1 and so we get this because if c_2 is equal to this,"},{"Start":"14:10.900 ","End":"14:19.315","Text":"I can replace the c_2 here by minus c_1 e to the 2 Omega from this equation."},{"Start":"14:19.315 ","End":"14:25.550","Text":"Now I want to divide both sides by c_1 but before I can divide by c_1,"},{"Start":"14:25.550 ","End":"14:28.474","Text":"I have to check that it\u0027s not 0."},{"Start":"14:28.474 ","End":"14:31.775","Text":"Well, let\u0027s say that it was 0."},{"Start":"14:31.775 ","End":"14:34.070","Text":"If c_1 was 0,"},{"Start":"14:34.070 ","End":"14:40.130","Text":"then c_2 is 0 so we get the trivial solution."},{"Start":"14:40.710 ","End":"14:43.810","Text":"That\u0027s not useful to us."},{"Start":"14:43.810 ","End":"14:45.200","Text":"It could be, but it\u0027s not useful."},{"Start":"14:45.200 ","End":"14:48.780","Text":"We want a solution which is not trivial."},{"Start":"14:48.900 ","End":"14:54.875","Text":"We want to check if it\u0027s possible that c_1 is not 0."},{"Start":"14:54.875 ","End":"14:56.780","Text":"See what that gives."},{"Start":"14:56.780 ","End":"14:59.620","Text":"Well, in that case,"},{"Start":"14:59.620 ","End":"15:01.690","Text":"we cancel the c_1 here and here,"},{"Start":"15:01.690 ","End":"15:04.380","Text":"and then we just get this equation."},{"Start":"15:04.380 ","End":"15:10.085","Text":"Then if I bring this over and divide by 1 minus Omega and then switch sides,"},{"Start":"15:10.085 ","End":"15:13.155","Text":"I end up with this equation."},{"Start":"15:13.155 ","End":"15:22.270","Text":"Now, we can see that Omega equals 0 would solve it and it\u0027s not"},{"Start":"15:22.270 ","End":"15:31.444","Text":"that easy to show that the only solution to this equation is Omega equals 0."},{"Start":"15:31.444 ","End":"15:35.030","Text":"Basically, at 0, you see that they\u0027re equal"},{"Start":"15:35.030 ","End":"15:38.780","Text":"and this one increases faster than this one using the derivatives."},{"Start":"15:38.780 ","End":"15:46.510","Text":"I\u0027m not going to get into that but Omega_0 is not in our range so that\u0027s a contradiction."},{"Start":"15:46.510 ","End":"15:51.535","Text":"Remember Omega is positive so basically,"},{"Start":"15:51.535 ","End":"15:57.155","Text":"it doesn\u0027t give us anything either that c_1 is not 0."},{"Start":"15:57.155 ","End":"16:00.560","Text":"This one is not possible, so c_1 is 0,"},{"Start":"16:00.560 ","End":"16:04.109","Text":"so c_2 is 0, so it\u0027s just the trivial solution."},{"Start":"16:05.460 ","End":"16:08.140","Text":"Next, I think we could just have to"},{"Start":"16:08.140 ","End":"16:15.680","Text":"summarize what we have and I also just wanted to emphasize,"},{"Start":"16:15.680 ","End":"16:18.020","Text":"I put it in writing is the only solution in"},{"Start":"16:18.020 ","End":"16:21.190","Text":"Case 3 is boundary value problem is a trivial 1."},{"Start":"16:21.190 ","End":"16:24.010","Text":"Now summarize Cases 1, 2, and 3."},{"Start":"16:24.010 ","End":"16:27.740","Text":"Case 3, which we just did gave only the trivial solution."},{"Start":"16:27.740 ","End":"16:29.075","Text":"In Case 1,"},{"Start":"16:29.075 ","End":"16:36.850","Text":"we got just 1 eigenvalue 0 and an eigenfunction x minus 1,"},{"Start":"16:36.850 ","End":"16:39.710","Text":"and Case 2 gave us a lot."},{"Start":"16:39.710 ","End":"16:47.875","Text":"Case 2 gave us a whole sequence of eigenvalues, Lambda_n,"},{"Start":"16:47.875 ","End":"16:55.520","Text":"which were Omega_n^2 and Omega_n were the solutions which couldn\u0027t obtain exactly"},{"Start":"16:55.520 ","End":"16:59.440","Text":"only we demonstrated it graphically and"},{"Start":"16:59.440 ","End":"17:03.499","Text":"the eigenfunctions that correspond to each of these eigenvalues,"},{"Start":"17:03.499 ","End":"17:08.270","Text":"the corresponding n like Lambda_4 corresponds to 5,"},{"Start":"17:08.270 ","End":"17:10.950","Text":"4, and so on."},{"Start":"17:11.400 ","End":"17:17.270","Text":"sin(Omega)x for each of this Omegas."},{"Start":"17:17.970 ","End":"17:24.620","Text":"Together we have infinity plus 1 eigenvalue with eigenfunctions,"},{"Start":"17:24.620 ","End":"17:26.105","Text":"we have this infinite,"},{"Start":"17:26.105 ","End":"17:30.330","Text":"and this odd one and there we\u0027re done."}],"ID":29446},{"Watched":false,"Name":"Exercise 4","Duration":"7m 24s","ChapterTopicVideoID":28214,"CourseChapterTopicPlaylistID":294448,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.923","Text":"This exercise probably looks familiar."},{"Start":"00:02.923 ","End":"00:05.010","Text":"We certainly had this equation before,"},{"Start":"00:05.010 ","End":"00:07.545","Text":"but notice that the interval\u0027s changed."},{"Start":"00:07.545 ","End":"00:11.010","Text":"Previously, we had several examples from 0-1."},{"Start":"00:11.010 ","End":"00:16.545","Text":"Here, we generalize the 1 to sum l, generally."},{"Start":"00:16.545 ","End":"00:21.420","Text":"Here, we have the boundary conditions and we give them both endpoints at"},{"Start":"00:21.420 ","End":"00:26.610","Text":"0 and at l. One side y is 0 and the other end,"},{"Start":"00:26.610 ","End":"00:28.545","Text":"the derivative is 0."},{"Start":"00:28.545 ","End":"00:32.565","Text":"Now, the first part we\u0027ve done many times."},{"Start":"00:32.565 ","End":"00:34.800","Text":"We split into cases."},{"Start":"00:34.800 ","End":"00:37.950","Text":"We solve the equation."},{"Start":"00:37.950 ","End":"00:40.905","Text":"In the case where Lambda was 0,"},{"Start":"00:40.905 ","End":"00:48.440","Text":"we got that this was the general solution and this is its derivative."},{"Start":"00:48.440 ","End":"00:55.770","Text":"Now, we\u0027re going to introduce the boundary conditions."},{"Start":"00:55.780 ","End":"00:59.765","Text":"The first one relates to y at 0."},{"Start":"00:59.765 ","End":"01:02.660","Text":"If we look y(0)=c_1."},{"Start":"01:02.660 ","End":"01:08.030","Text":"The other condition relates to y prime, but the l doesn\u0027t matter."},{"Start":"01:08.030 ","End":"01:13.970","Text":"It\u0027s a constant everywhere at c_2=0."},{"Start":"01:13.970 ","End":"01:19.630","Text":"We\u0027ve just got that c_1 and c_2 are both 0,"},{"Start":"01:19.630 ","End":"01:22.130","Text":"so we only get the trivial solution,"},{"Start":"01:22.130 ","End":"01:24.530","Text":"that is that y is 0."},{"Start":"01:24.530 ","End":"01:27.035","Text":"Of course, this is just case 1."},{"Start":"01:27.035 ","End":"01:28.635","Text":"In case 2,"},{"Start":"01:28.635 ","End":"01:32.690","Text":"we\u0027ve also done this several times before when Lambda is positive,"},{"Start":"01:32.690 ","End":"01:35.180","Text":"whether that be Omega squared and we get"},{"Start":"01:35.180 ","End":"01:39.665","Text":"the general solution is this and the derivative is this."},{"Start":"01:39.665 ","End":"01:45.245","Text":"What we need is the boundary conditions y(0)."},{"Start":"01:45.245 ","End":"01:47.450","Text":"We plug that in here,"},{"Start":"01:47.450 ","End":"01:50.650","Text":"the x is 0 and we get this."},{"Start":"01:50.650 ","End":"01:55.910","Text":"In the other one, we plug in x=l and we get this."},{"Start":"01:55.910 ","End":"02:00.660","Text":"Remember that cosine 0 is 1 and sine 0 is 0."},{"Start":"02:01.850 ","End":"02:08.180","Text":"The first equation just gives us straightaway c_1 is 0."},{"Start":"02:08.180 ","End":"02:09.410","Text":"This bit is 1,"},{"Start":"02:09.410 ","End":"02:11.797","Text":"this bit is 0."},{"Start":"02:11.797 ","End":"02:19.790","Text":"There\u0027s a little typo here."},{"Start":"02:19.790 ","End":"02:22.415","Text":"This should be l also."},{"Start":"02:22.415 ","End":"02:26.585","Text":"But it makes no difference because if c_1 is 0,"},{"Start":"02:26.585 ","End":"02:28.280","Text":"if I put this as 0,"},{"Start":"02:28.280 ","End":"02:30.305","Text":"then this first term doesn\u0027t matter,"},{"Start":"02:30.305 ","End":"02:35.105","Text":"which is why I just copied the second term here and this is 0."},{"Start":"02:35.105 ","End":"02:36.920","Text":"Now if c_2 is 0,"},{"Start":"02:36.920 ","End":"02:39.740","Text":"we\u0027re going to get the trivial solution because c_1 is already 0."},{"Start":"02:39.740 ","End":"02:42.635","Text":"We\u0027re looking for c_2 not 0."},{"Start":"02:42.635 ","End":"02:45.370","Text":"Also, Omega is positive,"},{"Start":"02:45.370 ","End":"02:48.650","Text":"and so we can divide by c_2 Omega."},{"Start":"02:48.650 ","End":"02:55.410","Text":"They\u0027re both not 0 and we just get the cosine Omega l is 0."},{"Start":"02:56.440 ","End":"02:59.810","Text":"We want to know how to solve that."},{"Start":"02:59.810 ","End":"03:04.445","Text":"But let\u0027s leave the actual solution aside for the moment."},{"Start":"03:04.445 ","End":"03:06.560","Text":"There\u0027s going to be an infinite number of solutions."},{"Start":"03:06.560 ","End":"03:10.715","Text":"This kind of equation always has periodic infinite number of solutions."},{"Start":"03:10.715 ","End":"03:12.020","Text":"But for each solution,"},{"Start":"03:12.020 ","End":"03:13.715","Text":"let\u0027s just say what we have."},{"Start":"03:13.715 ","End":"03:15.949","Text":"We find these Omegas,"},{"Start":"03:15.949 ","End":"03:17.600","Text":"are going to be many of them."},{"Start":"03:17.600 ","End":"03:19.340","Text":"For each such Omega,"},{"Start":"03:19.340 ","End":"03:28.155","Text":"we plug in c_2 is basically anything."},{"Start":"03:28.155 ","End":"03:32.415","Text":"I prefer to use K and c_1 is 0."},{"Start":"03:32.415 ","End":"03:38.150","Text":"We\u0027ve got that y is K sine Omega x,"},{"Start":"03:38.150 ","End":"03:40.745","Text":"where we still have to find the solutions Omega."},{"Start":"03:40.745 ","End":"03:46.940","Text":"But Omega is such that subject to or such that cosine Omega l is 0."},{"Start":"03:46.940 ","End":"03:49.895","Text":"Now, let\u0027s get to the trigonometric equation."},{"Start":"03:49.895 ","End":"03:53.090","Text":"Now, the cosine is 0."},{"Start":"03:53.090 ","End":"03:59.135","Text":"It\u0027s 90 degrees and then plus multiples of 180 in radians."},{"Start":"03:59.135 ","End":"04:05.625","Text":"It\u0027s Pi over 2 plus whole multiples of Pi,"},{"Start":"04:05.625 ","End":"04:16.800","Text":"n could be 0 because then that would still be positive."},{"Start":"04:17.120 ","End":"04:22.890","Text":"It\u0027s worth writing n=0 or 1 or 2 or"},{"Start":"04:22.890 ","End":"04:29.215","Text":"3 or whatever because we want this to be positive,"},{"Start":"04:29.215 ","End":"04:30.440","Text":"Omega has to be positive,"},{"Start":"04:30.440 ","End":"04:32.930","Text":"and l is already positive."},{"Start":"04:32.930 ","End":"04:38.850","Text":"From here, we divide by l and then take Pi outside the brackets."},{"Start":"04:38.850 ","End":"04:40.320","Text":"We\u0027ll get Pi over l,"},{"Start":"04:40.320 ","End":"04:45.645","Text":"a half plus n which is the same as this."},{"Start":"04:45.645 ","End":"04:48.825","Text":"Again, n is from this range here."},{"Start":"04:48.825 ","End":"04:54.950","Text":"For each n, we can take the Omega n and we get this family of solutions."},{"Start":"04:54.950 ","End":"05:01.914","Text":"We say family because K varies and Omega n is this."},{"Start":"05:01.914 ","End":"05:05.640","Text":"This is what we get and of course,"},{"Start":"05:05.640 ","End":"05:08.000","Text":"if we let K=1,"},{"Start":"05:08.000 ","End":"05:10.490","Text":"we can get an eigenfunction."},{"Start":"05:10.490 ","End":"05:14.635","Text":"You let K be anything but 1 is the easiest."},{"Start":"05:14.635 ","End":"05:19.790","Text":"These are the eigenfunctions that have y and call it Phi n and loops."},{"Start":"05:19.790 ","End":"05:22.220","Text":"You should start at 0,"},{"Start":"05:22.220 ","End":"05:23.855","Text":"0, 1, 2, 3."},{"Start":"05:23.855 ","End":"05:29.135","Text":"0 counts also because it\u0027s still positive as the eigenfunctions and"},{"Start":"05:29.135 ","End":"05:33.210","Text":"the Lambdas are the eigenvalues when we"},{"Start":"05:33.210 ","End":"05:37.470","Text":"get to the Lambdas through the Omegas by squaring Lambdas Omega squared."},{"Start":"05:37.470 ","End":"05:45.650","Text":"These are the eigenvalues and these are the eigenfunctions corresponding for each n,"},{"Start":"05:45.650 ","End":"05:47.810","Text":"we get one of these."},{"Start":"05:47.810 ","End":"05:52.050","Text":"We still just on case 2."},{"Start":"05:52.480 ","End":"05:55.700","Text":"In case 3, when Lambda is less than 0,"},{"Start":"05:55.700 ","End":"05:59.605","Text":"we\u0027ve done all this stuff before."},{"Start":"05:59.605 ","End":"06:07.445","Text":"Lambda\u0027s minus Omega squared and general solution is this and the derivative is this."},{"Start":"06:07.445 ","End":"06:13.760","Text":"What\u0027s different from one exercise to the other is the boundary conditions,"},{"Start":"06:13.760 ","End":"06:15.590","Text":"and this time, if we plug in,"},{"Start":"06:15.590 ","End":"06:19.610","Text":"what we\u0027ll get is y of 0."},{"Start":"06:19.610 ","End":"06:22.420","Text":"We put in x=0 here,"},{"Start":"06:22.420 ","End":"06:25.200","Text":"so e^0 is 1,"},{"Start":"06:25.200 ","End":"06:26.360","Text":"so it\u0027s just c_1,"},{"Start":"06:26.360 ","End":"06:29.060","Text":"and the other one also just c_2."},{"Start":"06:29.060 ","End":"06:31.025","Text":"The sum of these two is 0."},{"Start":"06:31.025 ","End":"06:34.490","Text":"From here, when x is l,"},{"Start":"06:34.490 ","End":"06:42.630","Text":"we just get this expression with l instead of x. I\u0027ll get some space."},{"Start":"06:43.480 ","End":"06:47.495","Text":"I just copied it without this bit."},{"Start":"06:47.495 ","End":"06:50.135","Text":"From the first equation,"},{"Start":"06:50.135 ","End":"06:52.130","Text":"c_1 is minus c_2."},{"Start":"06:52.130 ","End":"06:54.760","Text":"Basically, I\u0027m going to let c_2 be anything."},{"Start":"06:54.760 ","End":"06:57.540","Text":"But c_1 is got to be minus c_2."},{"Start":"06:57.540 ","End":"07:00.880","Text":"If I plugin instead of c_1 minus c_2,"},{"Start":"07:00.880 ","End":"07:02.075","Text":"this is what I get."},{"Start":"07:02.075 ","End":"07:04.900","Text":"Now, we can take stuff out of the brackets."},{"Start":"07:04.900 ","End":"07:10.850","Text":"The c_2 and the Omega come out and the minus and what we\u0027re left with is this."},{"Start":"07:11.660 ","End":"07:16.130","Text":"I making a lot of typos today. This is equal to 0."},{"Start":"07:16.130 ","End":"07:20.215","Text":"Now notice, Omega is bigger than 0."},{"Start":"07:20.215 ","End":"07:22.900","Text":"It\u0027s not 0, so we can divide by it."},{"Start":"07:22.900 ","End":"07:27.249","Text":"The other term is also not 0 because e to the anything is positive,"},{"Start":"07:27.249 ","End":"07:29.110","Text":"so positive plus positive is positive."},{"Start":"07:29.110 ","End":"07:33.190","Text":"It\u0027s not 0. Both of these are positive."},{"Start":"07:33.190 ","End":"07:35.350","Text":"Therefore, not 0, so we can divide."},{"Start":"07:35.350 ","End":"07:43.445","Text":"You\u0027re left with minus c_2 is 0 and if c_2 is 0, c_1 is 0."},{"Start":"07:43.445 ","End":"07:49.400","Text":"We\u0027re just left with the trivial solution for the boundary value problem."},{"Start":"07:49.400 ","End":"07:53.330","Text":"The trivial solution being y=0."},{"Start":"07:53.330 ","End":"07:58.190","Text":"That\u0027s case 3. I need to summarize the three cases."},{"Start":"07:58.190 ","End":"08:00.110","Text":"Now in our summary,"},{"Start":"08:00.110 ","End":"08:05.585","Text":"you should remember that the case we just did is only the trivial solution."},{"Start":"08:05.585 ","End":"08:09.935","Text":"I gave just the trivial and case 1 also."},{"Start":"08:09.935 ","End":"08:12.130","Text":"We\u0027re really down to case 2."},{"Start":"08:12.130 ","End":"08:15.470","Text":"If you remember, we got this series of functions,"},{"Start":"08:15.470 ","End":"08:21.680","Text":"Phi n for different n. These are the eigenfunctions."},{"Start":"08:21.680 ","End":"08:24.395","Text":"We can start from 0."},{"Start":"08:24.395 ","End":"08:29.835","Text":"The corresponding eigenvalues were the Omega n squared,"},{"Start":"08:29.835 ","End":"08:32.979","Text":"which were these, and also n is 0,"},{"Start":"08:32.979 ","End":"08:34.070","Text":"1, 2, 3, etc."},{"Start":"08:34.070 ","End":"08:38.345","Text":"These are the eigenvalues for each n. Got an eigenvalue and an eigenfunctions."},{"Start":"08:38.345 ","End":"08:41.340","Text":"That concludes this exercise."}],"ID":29447},{"Watched":false,"Name":"Exercise 5","Duration":"6m 32s","ChapterTopicVideoID":28215,"CourseChapterTopicPlaylistID":294448,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.790","Text":"Here\u0027s another one of those familiar Sturm-Liouville exercises."},{"Start":"00:05.790 ","End":"00:08.070","Text":"It\u0027s the same equation again."},{"Start":"00:08.070 ","End":"00:11.070","Text":"The difference is the end point is Pi"},{"Start":"00:11.070 ","End":"00:14.130","Text":"this time and the boundary conditions are different each time."},{"Start":"00:14.130 ","End":"00:18.415","Text":"Remember we divide up into 3 cases."},{"Start":"00:18.415 ","End":"00:26.010","Text":"It always starts out the same so just copy paste in the previous exercises."},{"Start":"00:26.010 ","End":"00:33.030","Text":"What\u0027s different here is the boundary conditions and what we get the following."},{"Start":"00:33.030 ","End":"00:35.220","Text":"So here\u0027s y and y\u0027."},{"Start":"00:35.220 ","End":"00:37.755","Text":"If I put y\u0027(0),"},{"Start":"00:37.755 ","End":"00:42.285","Text":"it\u0027s just c_2 because y\u0027 is a constant function."},{"Start":"00:42.285 ","End":"00:45.240","Text":"If I put Pi in here,"},{"Start":"00:45.240 ","End":"00:47.940","Text":"I get c_1 plus c_2 Pi."},{"Start":"00:47.940 ","End":"00:54.705","Text":"Here\u0027s our 2 equations in 2 unknowns, c_1 and c_2."},{"Start":"00:54.705 ","End":"00:57.170","Text":"Of course, since c_2 is 0,"},{"Start":"00:57.170 ","End":"00:58.925","Text":"I can plug that in here,"},{"Start":"00:58.925 ","End":"01:01.490","Text":"and I get that c_1 is 0."},{"Start":"01:01.490 ","End":"01:06.170","Text":"In short, this only gives us the trivial solution."},{"Start":"01:06.170 ","End":"01:08.875","Text":"Y is constantly 0."},{"Start":"01:08.875 ","End":"01:12.380","Text":"Now Case 2 with Lambda bigger than 0,"},{"Start":"01:12.380 ","End":"01:14.390","Text":"once again, just copy paste."},{"Start":"01:14.390 ","End":"01:17.240","Text":"It\u0027s the same as all the previous exercises."},{"Start":"01:17.240 ","End":"01:19.865","Text":"Here\u0027s our y, here\u0027s our y\u0027."},{"Start":"01:19.865 ","End":"01:22.295","Text":"The boundary conditions are different."},{"Start":"01:22.295 ","End":"01:23.760","Text":"In this Case 2,"},{"Start":"01:23.760 ","End":"01:30.130","Text":"y\u0027(0) is plug-in 0 here."},{"Start":"01:30.130 ","End":"01:32.720","Text":"Since sine of 0 is 0,"},{"Start":"01:32.720 ","End":"01:34.295","Text":"we just get the second bit,"},{"Start":"01:34.295 ","End":"01:39.330","Text":"c_2 Omega cosine 0."},{"Start":"01:39.330 ","End":"01:40.830","Text":"Cosine 0 is 1,"},{"Start":"01:40.830 ","End":"01:43.480","Text":"so don\u0027t even need that."},{"Start":"01:43.640 ","End":"01:49.160","Text":"The second y of Pi means I put in Pi here,"},{"Start":"01:49.160 ","End":"01:51.535","Text":"and I\u0027ll just copy that here."},{"Start":"01:51.535 ","End":"01:54.240","Text":"Let\u0027s see what we get from that."},{"Start":"01:54.240 ","End":"01:56.600","Text":"Now, Omega is positive,"},{"Start":"01:56.600 ","End":"01:59.030","Text":"so it\u0027s not 0,"},{"Start":"01:59.030 ","End":"02:01.384","Text":"so c_2 is equal to 0."},{"Start":"02:01.384 ","End":"02:03.635","Text":"Put c_2 equals 0 here,"},{"Start":"02:03.635 ","End":"02:10.074","Text":"we just get c_1 cosine Pi Omega is 0."},{"Start":"02:10.074 ","End":"02:13.310","Text":"We\u0027re looking for a non-trivial solutions."},{"Start":"02:13.310 ","End":"02:16.220","Text":"I mean, always c_1 and c_2,"},{"Start":"02:16.220 ","End":"02:17.915","Text":"both zeros always works."},{"Start":"02:17.915 ","End":"02:19.370","Text":"But we want to try something else."},{"Start":"02:19.370 ","End":"02:21.545","Text":"c_2 is 0, so we have to,"},{"Start":"02:21.545 ","End":"02:23.880","Text":"for c_1 naught to be 0."},{"Start":"02:23.880 ","End":"02:26.555","Text":"See if we can do that, find something."},{"Start":"02:26.555 ","End":"02:28.580","Text":"In that case, I can divide."},{"Start":"02:28.580 ","End":"02:31.850","Text":"So I get that this part,"},{"Start":"02:31.850 ","End":"02:33.890","Text":"cosine Pi Omega,"},{"Start":"02:33.890 ","End":"02:35.570","Text":"Omega Pi is 0."},{"Start":"02:35.570 ","End":"02:37.850","Text":"That\u0027s a trigonometric equation."},{"Start":"02:37.850 ","End":"02:41.030","Text":"We know when the cosine is 0,"},{"Start":"02:41.030 ","End":"02:43.310","Text":"the cosine is 0."},{"Start":"02:43.310 ","End":"02:47.645","Text":"We\u0027ve had this in the previous exercise when that"},{"Start":"02:47.645 ","End":"02:53.180","Text":"something is Pi over 2 plus multiples of Pi."},{"Start":"02:53.180 ","End":"02:59.435","Text":"Then that gives us that Omega is divided by Pi."},{"Start":"02:59.435 ","End":"03:03.875","Text":"We get 2n plus 1 over 2,"},{"Start":"03:03.875 ","End":"03:06.760","Text":"which is the same as n plus a half."},{"Start":"03:06.760 ","End":"03:08.885","Text":"What I meant to say earlier,"},{"Start":"03:08.885 ","End":"03:13.670","Text":"that we have our general solution that looks like this."},{"Start":"03:13.670 ","End":"03:18.750","Text":"We know that c_2 is 0."},{"Start":"03:19.090 ","End":"03:24.420","Text":"I meant to say c_1 is 0, sorry."},{"Start":"03:24.420 ","End":"03:25.790","Text":"C_2 could be anything,"},{"Start":"03:25.790 ","End":"03:33.604","Text":"but I\u0027d like to call K. That gives us that the general solution is K sine Omega x,"},{"Start":"03:33.604 ","End":"03:35.700","Text":"but not any old Omega,"},{"Start":"03:35.700 ","End":"03:42.320","Text":"only the Omega for which subject to or such that cosine Omega Pi is 0."},{"Start":"03:42.320 ","End":"03:45.590","Text":"In other words, only for these Omegas."},{"Start":"03:45.590 ","End":"03:51.815","Text":"Now this is just 2n plus 1 over 2."},{"Start":"03:51.815 ","End":"03:54.770","Text":"I\u0027ll simplify it in a moment 2n plus a half."},{"Start":"03:54.770 ","End":"03:57.905","Text":"I\u0027ll just call this the Omega_n."},{"Start":"03:57.905 ","End":"03:59.960","Text":"For each n, we get the whole sequence of these."},{"Start":"03:59.960 ","End":"04:04.435","Text":"For each n, we get a different Omega_n."},{"Start":"04:04.435 ","End":"04:09.485","Text":"It\u0027s true that this is equal to n plus a half."},{"Start":"04:09.485 ","End":"04:13.740","Text":"Leave it like this, doesn\u0027t really matter."},{"Start":"04:13.810 ","End":"04:18.695","Text":"For each n, we have the family of solutions."},{"Start":"04:18.695 ","End":"04:21.540","Text":"Y_n is K cosine."},{"Start":"04:21.540 ","End":"04:23.070","Text":"The appropriate Omega,"},{"Start":"04:23.070 ","End":"04:27.045","Text":"Omega_n. Omega_n is this."},{"Start":"04:27.045 ","End":"04:31.980","Text":"So this is going to be the eigenfunction."},{"Start":"04:31.980 ","End":"04:35.360","Text":"We can take whatever value of K we want except 0,"},{"Start":"04:35.360 ","End":"04:37.565","Text":"I\u0027ll take K equals 1."},{"Start":"04:37.565 ","End":"04:40.130","Text":"I\u0027ll call it Phi_n."},{"Start":"04:40.130 ","End":"04:44.580","Text":"So this we can take as the eigenfunction."},{"Start":"04:44.890 ","End":"04:48.410","Text":"Actually could be 0 also,"},{"Start":"04:48.410 ","End":"04:51.200","Text":"because then Omega would still be positive."},{"Start":"04:51.200 ","End":"04:56.965","Text":"These are the eigenfunctions and there is always eigenvalues."},{"Start":"04:56.965 ","End":"04:59.600","Text":"These are the Lambdas."},{"Start":"04:59.600 ","End":"05:01.410","Text":"We don\u0027t want Omega, we want Lambda."},{"Start":"05:01.410 ","End":"05:03.545","Text":"Remember, Lambda is Omega squared."},{"Start":"05:03.545 ","End":"05:05.870","Text":"So it\u0027s this expression squared."},{"Start":"05:05.870 ","End":"05:09.350","Text":"These are the corresponding eigenvalues for each n, 0, 1, 2,"},{"Start":"05:09.350 ","End":"05:14.035","Text":"3 and so on we\u0027ve got Phi_n and we\u0027ve got Lambda_n."},{"Start":"05:14.035 ","End":"05:18.540","Text":"That is Case 2."},{"Start":"05:18.540 ","End":"05:24.830","Text":"On to Case 3, which also always starts the same because it\u0027s the same equation."},{"Start":"05:24.830 ","End":"05:29.205","Text":"We get the solution,"},{"Start":"05:29.205 ","End":"05:33.090","Text":"y is this and y\u0027 is this."},{"Start":"05:33.090 ","End":"05:36.390","Text":"Omega is such that because Lambda\u0027s negative,"},{"Start":"05:36.390 ","End":"05:39.055","Text":"Lambda\u0027s minus Omega squared with positive Omega,"},{"Start":"05:39.055 ","End":"05:44.075","Text":"what distinguishes each problem from the other is the boundary conditions."},{"Start":"05:44.075 ","End":"05:45.800","Text":"Here\u0027s what we have in our case."},{"Start":"05:45.800 ","End":"05:49.285","Text":"I\u0027m just scroll a bit."},{"Start":"05:49.285 ","End":"05:53.450","Text":"We got y\u0027(0) is 0 and y of Pi is 0."},{"Start":"05:53.450 ","End":"05:54.815","Text":"Let me just see, okay."},{"Start":"05:54.815 ","End":"05:59.930","Text":"Here we can see why just about if we plug in Pi here,"},{"Start":"05:59.930 ","End":"06:04.250","Text":"this is what you get and you plug in 0 here."},{"Start":"06:04.250 ","End":"06:09.890","Text":"Then e to the 0 is 1 and so is e to the minus is 0."},{"Start":"06:09.890 ","End":"06:14.975","Text":"So this is what we use to equations we get."},{"Start":"06:14.975 ","End":"06:20.905","Text":"Now Omega is not 0, so I can divide this first equation by Omega and get c_1 minus,"},{"Start":"06:20.905 ","End":"06:23.490","Text":"this should be c_2,"},{"Start":"06:23.490 ","End":"06:26.955","Text":"of course is 0."},{"Start":"06:26.955 ","End":"06:32.000","Text":"From here we just copy it as is."},{"Start":"06:32.000 ","End":"06:35.260","Text":"Now we see from here that c_1 equals c_2."},{"Start":"06:35.260 ","End":"06:43.465","Text":"If instead of c_2, I put c_1 and I can put c_1 outside the brackets and get this."},{"Start":"06:43.465 ","End":"06:47.750","Text":"This means that c_1 is 0 because this is positive and this is positive."},{"Start":"06:47.750 ","End":"06:49.535","Text":"E to the anything is positive."},{"Start":"06:49.535 ","End":"06:52.130","Text":"So this sum is positive, not 0."},{"Start":"06:52.130 ","End":"06:53.780","Text":"So c_1 is 0,"},{"Start":"06:53.780 ","End":"06:59.900","Text":"but c_2 equals c_1 and c_2 equals 0 also."},{"Start":"06:59.900 ","End":"07:05.045","Text":"We only get the trivial solution, y equals 0."},{"Start":"07:05.045 ","End":"07:10.045","Text":"Then finally, we\u0027re going to collect together all 3 cases."},{"Start":"07:10.045 ","End":"07:17.510","Text":"The last case, remember we got just a trivial and if you recall also in the first case,"},{"Start":"07:17.510 ","End":"07:19.700","Text":"we only got something in Case 2,"},{"Start":"07:19.700 ","End":"07:24.040","Text":"we got this series of eigenfunctions,"},{"Start":"07:24.040 ","End":"07:33.155","Text":"where n equals non-negative integers and the corresponding eigenvalues"},{"Start":"07:33.155 ","End":"07:37.085","Text":"where this expression for each n."},{"Start":"07:37.085 ","End":"07:44.450","Text":"Each Lambda_n corresponds to Phi_n,"},{"Start":"07:44.450 ","End":"07:46.650","Text":"and that is it."}],"ID":29448},{"Watched":false,"Name":"Exercise 6","Duration":"10m 8s","ChapterTopicVideoID":28207,"CourseChapterTopicPlaylistID":294448,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:08.025","Text":"This exercise is another Sturm-Louiville boundary value condition problem."},{"Start":"00:08.025 ","End":"00:10.650","Text":"It\u0027s different than the previous ones."},{"Start":"00:10.650 ","End":"00:12.300","Text":"The equation is different."},{"Start":"00:12.300 ","End":"00:15.900","Text":"All the ones up till now was something else."},{"Start":"00:15.900 ","End":"00:20.190","Text":"We\u0027ll do this one from the beginning."},{"Start":"00:20.190 ","End":"00:25.815","Text":"As always, we divide into 3 cases with the Lambda."},{"Start":"00:25.815 ","End":"00:28.278","Text":"If Lambda is 0,"},{"Start":"00:28.278 ","End":"00:31.800","Text":"then our problem is this,"},{"Start":"00:31.800 ","End":"00:34.560","Text":"instead of the 1 plus Lambda,"},{"Start":"00:34.560 ","End":"00:37.540","Text":"just throw that out like 1y."},{"Start":"00:38.510 ","End":"00:43.370","Text":"We get the characteristic equation, which is this."},{"Start":"00:43.370 ","End":"00:45.560","Text":"Notice that this is a perfect square."},{"Start":"00:45.560 ","End":"00:48.635","Text":"This is k minus 1^2."},{"Start":"00:48.635 ","End":"00:56.430","Text":"Therefore we get that k equals 1 is a double solution."},{"Start":"00:56.430 ","End":"00:58.670","Text":"When we have a double solution,"},{"Start":"00:58.670 ","End":"01:06.380","Text":"then the differential equation general solution is as follows."},{"Start":"01:06.380 ","End":"01:11.900","Text":"Where normally you have kx here and here,"},{"Start":"01:11.900 ","End":"01:13.010","Text":"but k is 1,"},{"Start":"01:13.010 ","End":"01:21.680","Text":"so that\u0027s why you don\u0027t see the k. Now let\u0027s look at the boundary conditions."},{"Start":"01:21.680 ","End":"01:27.515","Text":"By the way, at first glance so that y\u0027 doesn\u0027t appear in the boundary conditions,"},{"Start":"01:27.515 ","End":"01:30.500","Text":"because normally I would differentiate this to"},{"Start":"01:30.500 ","End":"01:33.725","Text":"get y\u0027 but if I don\u0027t need it, then I don\u0027t."},{"Start":"01:33.725 ","End":"01:39.170","Text":"We just need to plug in x equals 0 and x equals 1."},{"Start":"01:39.170 ","End":"01:43.100","Text":"In this case, if we put x equals 0,"},{"Start":"01:43.100 ","End":"01:47.675","Text":"then this bit becomes 0 and e^0 is 1."},{"Start":"01:47.675 ","End":"01:49.750","Text":"So we get c_1 is 0."},{"Start":"01:49.750 ","End":"01:53.880","Text":"If we let x equals 1,"},{"Start":"01:53.880 ","End":"02:00.210","Text":"then we get e^1 is just e We get this basically."},{"Start":"02:00.210 ","End":"02:04.960","Text":"I need some more space."},{"Start":"02:06.260 ","End":"02:10.040","Text":"Let\u0027s just explain how I get c_1 equals 0."},{"Start":"02:10.040 ","End":"02:18.560","Text":"This is e plus e times c_1 is 0."},{"Start":"02:18.560 ","End":"02:21.815","Text":"I wrote it wrongly. Here\u0027s the corrected version."},{"Start":"02:21.815 ","End":"02:27.890","Text":"That means that c_1 plus c_2 is 0 and c_1 is 0,"},{"Start":"02:27.890 ","End":"02:30.370","Text":"so c_1 is 0,"},{"Start":"02:30.370 ","End":"02:32.590","Text":"c_2 is also 0."},{"Start":"02:32.590 ","End":"02:36.965","Text":"So for case 1, we only get the trivial solution."},{"Start":"02:36.965 ","End":"02:40.475","Text":"The case Lambda bigger than 0."},{"Start":"02:40.475 ","End":"02:46.840","Text":"Here\u0027s the boundary value problem copied."},{"Start":"02:46.840 ","End":"02:54.748","Text":"We want the characteristic equation for this differential equation,"},{"Start":"02:54.748 ","End":"02:57.250","Text":"and here it is."},{"Start":"02:58.750 ","End":"03:01.715","Text":"Shouldn\u0027t need any explanation."},{"Start":"03:01.715 ","End":"03:03.530","Text":"We can simplify this."},{"Start":"03:03.530 ","End":"03:05.930","Text":"If I take k^2 minus 2k plus 1,"},{"Start":"03:05.930 ","End":"03:07.655","Text":"it\u0027s a perfect square."},{"Start":"03:07.655 ","End":"03:10.230","Text":"This is what I get."},{"Start":"03:10.280 ","End":"03:13.730","Text":"This time, because Lambda is bigger than 0,"},{"Start":"03:13.730 ","End":"03:17.910","Text":"we can write Lambda equals Omega^2."},{"Start":"03:18.370 ","End":"03:24.145","Text":"It\u0027s easier to work with Omega^2 than with Lambda."},{"Start":"03:24.145 ","End":"03:27.815","Text":"We can always take Omega to be positive."},{"Start":"03:27.815 ","End":"03:32.405","Text":"I mentioned this because we use this positivity later."},{"Start":"03:32.405 ","End":"03:34.505","Text":"Here are the solutions."},{"Start":"03:34.505 ","End":"03:42.298","Text":"Basically you take Omega^2 to the other side and then you would get that k minus 1 is,"},{"Start":"03:42.298 ","End":"03:45.560","Text":"because we have minus Omega^2,"},{"Start":"03:45.560 ","End":"03:49.700","Text":"the square root of that is plus or minus Omega i,"},{"Start":"03:49.700 ","End":"03:51.215","Text":"and then you add to 1."},{"Start":"03:51.215 ","End":"03:54.395","Text":"We get the complex conjugate pair,"},{"Start":"03:54.395 ","End":"03:58.370","Text":"1 plus Omega i and 1 minus Omega i."},{"Start":"03:58.370 ","End":"04:01.354","Text":"This is the general solution."},{"Start":"04:01.354 ","End":"04:07.820","Text":"We know that when it\u0027s a plus or minus bi, you get e^ax."},{"Start":"04:07.820 ","End":"04:11.375","Text":"I didn\u0027t write 1x obviously."},{"Start":"04:11.375 ","End":"04:14.430","Text":"The b is the Omega."},{"Start":"04:15.760 ","End":"04:18.200","Text":"Here are the boundary conditions."},{"Start":"04:18.200 ","End":"04:21.260","Text":"Notice that we didn\u0027t get y\u0027 in either of them,"},{"Start":"04:21.260 ","End":"04:23.690","Text":"which is why I didn\u0027t bother to differentiate here."},{"Start":"04:23.690 ","End":"04:25.775","Text":"When I see a y\u0027 then I do."},{"Start":"04:25.775 ","End":"04:34.260","Text":"We just need to substitute 0 and 1 into this general solution,"},{"Start":"04:34.260 ","End":"04:36.585","Text":"and this is what we get."},{"Start":"04:36.585 ","End":"04:47.960","Text":"We can simplify this if you remember that the cosine of 0 is 1 and the sine of 0 is 0,"},{"Start":"04:47.960 ","End":"04:52.940","Text":"and so the first one gives us right away that c_1 is 0."},{"Start":"04:52.940 ","End":"04:57.680","Text":"I already put 0 then into the c1_ here,"},{"Start":"04:57.680 ","End":"05:06.650","Text":"so we don\u0027t get the cosine part we only get the c_2 sine Omega times a is 0."},{"Start":"05:06.650 ","End":"05:11.210","Text":"You can divide both sides by 0 by removing the e. Now,"},{"Start":"05:11.210 ","End":"05:13.520","Text":"remember, we are looking for non-trivial solutions."},{"Start":"05:13.520 ","End":"05:17.190","Text":"If c_2 is 0, and then c_1 is also 0,"},{"Start":"05:17.190 ","End":"05:18.945","Text":"we get a trivial solution."},{"Start":"05:18.945 ","End":"05:23.210","Text":"We\u0027re searching for possibility that c_2 is not 0,"},{"Start":"05:23.210 ","End":"05:25.775","Text":"and then we can divide by it."},{"Start":"05:25.775 ","End":"05:31.170","Text":"That gives us that sine Omega is 0."},{"Start":"05:31.300 ","End":"05:37.670","Text":"We know from trigonometry that the sine of something is 0 when that something is"},{"Start":"05:37.670 ","End":"05:44.015","Text":"a multiple of 180 degrees or multiple of Pi, we get nPi."},{"Start":"05:44.015 ","End":"05:45.867","Text":"Since Omega has to be positive,"},{"Start":"05:45.867 ","End":"05:48.080","Text":"we start from 1,2,3,"},{"Start":"05:48.080 ","End":"05:52.595","Text":"etc, all sequence of Omegas."},{"Start":"05:52.595 ","End":"05:58.500","Text":"Now, the original equation is here, I rewrote it."},{"Start":"05:58.500 ","End":"06:05.307","Text":"C_1 is 0 and c_2 really is unrestricted."},{"Start":"06:05.307 ","End":"06:06.890","Text":"C_2 could be anything,"},{"Start":"06:06.890 ","End":"06:14.210","Text":"call it K. But this only works for special Omegas that are from this sequence."},{"Start":"06:14.210 ","End":"06:20.000","Text":"I\u0027ll call this one Omega n. Then y is just"},{"Start":"06:20.000 ","End":"06:28.995","Text":"K sine Omega n times x. Omega n is nPi."},{"Start":"06:28.995 ","End":"06:34.510","Text":"This is the family of solutions,"},{"Start":"06:34.510 ","End":"06:36.750","Text":"and we have for each n family."},{"Start":"06:36.750 ","End":"06:38.960","Text":"We have a sequence of families."},{"Start":"06:38.960 ","End":"06:42.360","Text":"Family because K can vary."},{"Start":"06:42.650 ","End":"06:47.540","Text":"For an eigenfunction, you can pick any non-zero K,"},{"Start":"06:47.540 ","End":"06:50.820","Text":"easiest to choose K equals 1."},{"Start":"06:50.980 ","End":"06:55.070","Text":"We\u0027ll let this, we denoted usually,"},{"Start":"06:55.070 ","End":"07:01.695","Text":"where letter Phi_n is going to be sine of nPix."},{"Start":"07:01.695 ","End":"07:04.790","Text":"We get such a function for each n 1, 2, 3,"},{"Start":"07:04.790 ","End":"07:08.765","Text":"etc, for the positive integers."},{"Start":"07:08.765 ","End":"07:11.330","Text":"These are the eigenfunctions."},{"Start":"07:11.330 ","End":"07:15.810","Text":"For eigenvalues we have to go back from Omega to Lambda."},{"Start":"07:15.940 ","End":"07:21.080","Text":"Lambda is Omega n^2 and Omega is nPi, so it\u0027s nPi^2."},{"Start":"07:21.080 ","End":"07:23.990","Text":"Again, n is 1,"},{"Start":"07:23.990 ","End":"07:25.160","Text":"2, 3, etc."},{"Start":"07:25.160 ","End":"07:28.015","Text":"These are the corresponding eigenvalues."},{"Start":"07:28.015 ","End":"07:30.775","Text":"That concludes Part 2."},{"Start":"07:30.775 ","End":"07:38.630","Text":"Case 3 is where Lambda is negative and it\u0027s a bit similar to case 2,"},{"Start":"07:38.630 ","End":"07:40.205","Text":"at least in the beginning."},{"Start":"07:40.205 ","End":"07:45.515","Text":"It\u0027s the same characteristic equation and the same simplification."},{"Start":"07:45.515 ","End":"07:54.905","Text":"The difference is that we let Lambda be minus Omega^2 and that gives us the minus here."},{"Start":"07:54.905 ","End":"07:56.300","Text":"When Lambda is negative,"},{"Start":"07:56.300 ","End":"08:01.485","Text":"we can do this and Omega positive."},{"Start":"08:01.485 ","End":"08:04.430","Text":"The 2 solutions of this is easy to see."},{"Start":"08:04.430 ","End":"08:06.500","Text":"Just bring Omega^2 to the other side."},{"Start":"08:06.500 ","End":"08:10.010","Text":"K minus 1 is plus or minus Omega,"},{"Start":"08:10.010 ","End":"08:12.635","Text":"and so we got k_1 and k_2."},{"Start":"08:12.635 ","End":"08:16.470","Text":"2 different real solutions."},{"Start":"08:16.470 ","End":"08:19.450","Text":"One plus Omega 1 minus Omega."},{"Start":"08:19.580 ","End":"08:24.140","Text":"Here\u0027s the solution of the differential equation."},{"Start":"08:24.140 ","End":"08:29.765","Text":"Notice that here we have 1 root and here the other root, this is standard."},{"Start":"08:29.765 ","End":"08:34.795","Text":"Next will be wanting to take a look at the boundary conditions."},{"Start":"08:34.795 ","End":"08:37.910","Text":"Notice that when x is 0,"},{"Start":"08:37.910 ","End":"08:41.870","Text":"both these exponents is 0 and e^0 is 1."},{"Start":"08:41.870 ","End":"08:43.985","Text":"We can basically just throw them out."},{"Start":"08:43.985 ","End":"08:47.770","Text":"We\u0027ve got c_1 plus c_2 is 0."},{"Start":"08:47.770 ","End":"08:51.690","Text":"From y of 0 is 0."},{"Start":"08:51.690 ","End":"08:54.210","Text":"When I plug in 1, I also have to get 0."},{"Start":"08:54.210 ","End":"08:56.960","Text":"I plug in x equals 1,"},{"Start":"08:56.960 ","End":"09:06.485","Text":"I can use the fact that e^1 plus Omega is e times e^Omega."},{"Start":"09:06.485 ","End":"09:11.235","Text":"Similarly, if it\u0027s 1 minus Omega,"},{"Start":"09:11.235 ","End":"09:15.760","Text":"it\u0027s going to be plus or minus."},{"Start":"09:16.040 ","End":"09:19.490","Text":"This is what we get. Of course e is not 0,"},{"Start":"09:19.490 ","End":"09:22.770","Text":"we can cancel it on both sides."},{"Start":"09:23.000 ","End":"09:26.620","Text":"I need more space."},{"Start":"09:27.200 ","End":"09:31.339","Text":"Just copy what we have above."},{"Start":"09:31.339 ","End":"09:35.641","Text":"The first equation gives us that c_2 equals minus c_1."},{"Start":"09:35.641 ","End":"09:38.050","Text":"C_1 can be anything,"},{"Start":"09:38.050 ","End":"09:41.785","Text":"it\u0027s not restricted, but then c_2 has to be minus c_1."},{"Start":"09:41.785 ","End":"09:47.645","Text":"Now I can replace this c_2 here by minus c_1."},{"Start":"09:47.645 ","End":"09:51.321","Text":"Then I can take c_1 outside the brackets,"},{"Start":"09:51.321 ","End":"09:53.915","Text":"so we get this equation from here."},{"Start":"09:53.915 ","End":"09:56.680","Text":"I\u0027m looking for non-trivial solutions."},{"Start":"09:56.680 ","End":"10:05.310","Text":"If c_1 is 0,"},{"Start":"10:05.310 ","End":"10:07.005","Text":"then c_2 is 0."},{"Start":"10:07.005 ","End":"10:08.745","Text":"Both ways, let me get the trivial."},{"Start":"10:08.745 ","End":"10:11.790","Text":"I really would c_1 to be not 0."},{"Start":"10:11.790 ","End":"10:15.395","Text":"If that\u0027s possible, that would give us"},{"Start":"10:15.395 ","End":"10:20.885","Text":"that we can cancel by it and e^Omega minus e^minus Omega 0."},{"Start":"10:20.885 ","End":"10:24.950","Text":"The only solution to this is Omega equals 0."},{"Start":"10:24.950 ","End":"10:34.680","Text":"That\u0027s fairly easy to see because if e^Omega equals e^minus Omega,"},{"Start":"10:35.650 ","End":"10:40.730","Text":"then when these things are equal the exponents have to be equal."},{"Start":"10:40.730 ","End":"10:43.340","Text":"You have to get Omega equals minus Omega."},{"Start":"10:43.340 ","End":"10:47.975","Text":"The only number which is equal to minus itself is 0."},{"Start":"10:47.975 ","End":"10:50.684","Text":"But we know that Omega is positive,"},{"Start":"10:50.684 ","End":"10:52.725","Text":"that\u0027s why this is false."},{"Start":"10:52.725 ","End":"10:58.770","Text":"The assumption that c_1 is non-zero gives us something false, so contradiction."},{"Start":"10:58.810 ","End":"11:03.060","Text":"So c_1 does have to be 0."},{"Start":"11:04.460 ","End":"11:07.640","Text":"We only get the trivial solution."},{"Start":"11:07.640 ","End":"11:10.820","Text":"Let me write that there, c_1 is 0."},{"Start":"11:10.820 ","End":"11:14.855","Text":"One of you said c_2 minus c_1 is also 0."},{"Start":"11:14.855 ","End":"11:19.910","Text":"These 2 are 0, so it gives us just the solution y equals 0,"},{"Start":"11:19.910 ","End":"11:21.485","Text":"which is always there."},{"Start":"11:21.485 ","End":"11:25.255","Text":"That\u0027s case 3. Now we want to summarize."},{"Start":"11:25.255 ","End":"11:28.260","Text":"I have the summary all prepared."},{"Start":"11:28.260 ","End":"11:32.630","Text":"As we said, we only got the trivial solutions in cases 1 and 3,"},{"Start":"11:32.630 ","End":"11:36.885","Text":"so we just have the solutions from case 2."},{"Start":"11:36.885 ","End":"11:43.670","Text":"Recall we got a whole sequence of eigenfunctions,"},{"Start":"11:43.670 ","End":"11:47.540","Text":"sine of nPix for n being"},{"Start":"11:47.540 ","End":"11:55.655","Text":"a natural number and the corresponding eigenvalues for each n nPi^2."},{"Start":"11:55.655 ","End":"12:02.420","Text":"The eigenvalue corresponds to the eigenfunction for"},{"Start":"12:02.420 ","End":"12:09.960","Text":"each n. That\u0027s it for this boundary value problem, we\u0027re done."}],"ID":29449},{"Watched":false,"Name":"Exercise 7","Duration":"6m 7s","ChapterTopicVideoID":28208,"CourseChapterTopicPlaylistID":294448,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.890","Text":"In this exercise, we have a Sturm-Liouville boundary value problem."},{"Start":"00:04.890 ","End":"00:06.644","Text":"This is a differential equation,"},{"Start":"00:06.644 ","End":"00:08.505","Text":"is the boundary values."},{"Start":"00:08.505 ","End":"00:10.245","Text":"Let\u0027s start solving it."},{"Start":"00:10.245 ","End":"00:16.484","Text":"The characteristic equation is k^2 plus Lambda equals 0."},{"Start":"00:16.484 ","End":"00:20.595","Text":"I will distinguish 3 cases according to the discriminant."},{"Start":"00:20.595 ","End":"00:25.710","Text":"The discriminant, e^2 minus 4ac is minus 4 Lambda."},{"Start":"00:25.710 ","End":"00:30.390","Text":"There were 3 cases there where there\u0027s discriminant equals 0,"},{"Start":"00:30.390 ","End":"00:32.265","Text":"less than 0 or bigger than 0."},{"Start":"00:32.265 ","End":"00:34.215","Text":"Since it\u0027s minus 4 Lambda,"},{"Start":"00:34.215 ","End":"00:36.855","Text":"Lambda correspondingly will be equal to 0,"},{"Start":"00:36.855 ","End":"00:39.420","Text":"bigger than 0, less than 0."},{"Start":"00:39.420 ","End":"00:42.135","Text":"This will be case 1, case 2, case 3."},{"Start":"00:42.135 ","End":"00:44.430","Text":"Case 1, Lambda equals 0."},{"Start":"00:44.430 ","End":"00:48.380","Text":"We just get y\u0027\u0027 equals 0."},{"Start":"00:48.380 ","End":"00:50.600","Text":"If you integrate, once you get a constant,"},{"Start":"00:50.600 ","End":"00:52.295","Text":"you integrate second time,"},{"Start":"00:52.295 ","End":"00:55.430","Text":"you get a constant plus a constant times x."},{"Start":"00:55.430 ","End":"01:00.100","Text":"We can find these 2 constants from the 2 boundary conditions."},{"Start":"01:00.100 ","End":"01:06.765","Text":"From this one, we get c_1 equals 0 because x is 0 so we just have c_1."},{"Start":"01:06.765 ","End":"01:09.030","Text":"From the second one, we put L in,"},{"Start":"01:09.030 ","End":"01:12.675","Text":"so we have c_1 plus c_2L equals 0."},{"Start":"01:12.675 ","End":"01:14.580","Text":"C_1 is 0."},{"Start":"01:14.580 ","End":"01:19.835","Text":"Plug that into this equation and we get c_2L is 0,"},{"Start":"01:19.835 ","End":"01:22.580","Text":"but L is not 0, so c_2 is 0."},{"Start":"01:22.580 ","End":"01:25.795","Text":"So both c_1 and c_2 are 0,"},{"Start":"01:25.795 ","End":"01:30.495","Text":"which means that y equals 0, the 0 function."},{"Start":"01:30.495 ","End":"01:34.640","Text":"We only have the trivial solution, no eigenfunctions here."},{"Start":"01:34.640 ","End":"01:36.885","Text":"Next case, case 2,"},{"Start":"01:36.885 ","End":"01:38.670","Text":"Lambda is bigger than 0."},{"Start":"01:38.670 ","End":"01:41.640","Text":"You can write Lambda as Omega ^2,"},{"Start":"01:41.640 ","End":"01:44.080","Text":"where Omega is the square root of Lambda."},{"Start":"01:44.080 ","End":"01:50.645","Text":"Take the characteristic equation and it\u0027s k^2 plus Omega^2 equals 0."},{"Start":"01:50.645 ","End":"01:52.625","Text":"Well, always k^2 plus Lambda,"},{"Start":"01:52.625 ","End":"01:55.425","Text":"which becomes k^2 plus Omega squared."},{"Start":"01:55.425 ","End":"01:59.930","Text":"We need to go into imaginary solutions plus or minus Omega i."},{"Start":"01:59.930 ","End":"02:01.985","Text":"We know that in this case,"},{"Start":"02:01.985 ","End":"02:06.475","Text":"the solution is a combination of cosine and sine."},{"Start":"02:06.475 ","End":"02:10.455","Text":"From the boundary conditions plug in 0 or plug in L,"},{"Start":"02:10.455 ","End":"02:12.485","Text":"we get this and this."},{"Start":"02:12.485 ","End":"02:16.675","Text":"Cosine 0 is 1 and sine 0 is 0,"},{"Start":"02:16.675 ","End":"02:21.110","Text":"so this one gives us that c_1 equals 0."},{"Start":"02:21.110 ","End":"02:23.850","Text":"Then put in that here,"},{"Start":"02:23.850 ","End":"02:25.290","Text":"so this term disappears."},{"Start":"02:25.290 ","End":"02:28.335","Text":"We have c_2 sine Omega L is 0."},{"Start":"02:28.335 ","End":"02:30.270","Text":"C_2 isn\u0027t 0,"},{"Start":"02:30.270 ","End":"02:32.685","Text":"so sine Omega L is 0."},{"Start":"02:32.685 ","End":"02:36.914","Text":"The sine is 0 when the arguments are multiple of Pi,"},{"Start":"02:36.914 ","End":"02:40.580","Text":"which means that Omega is nPi over L. Well,"},{"Start":"02:40.580 ","End":"02:42.470","Text":"we have a sequence of Omegas,"},{"Start":"02:42.470 ","End":"02:51.660","Text":"Omega n. There\u0027s no need to take negative n because if we take sine of minus nPi over L,"},{"Start":"02:51.660 ","End":"02:56.735","Text":"it\u0027s just minus sine nPi over L. In other words, a linear combination."},{"Start":"02:56.735 ","End":"02:58.460","Text":"Anyway, n just has to be 1,"},{"Start":"02:58.460 ","End":"02:59.930","Text":"2, 3, 4, 5, etc."},{"Start":"02:59.930 ","End":"03:03.185","Text":"No point taking 0 either because that\u0027s the 0 function."},{"Start":"03:03.185 ","End":"03:11.960","Text":"For each n we have a solution and altogether we have a family of solutions."},{"Start":"03:11.960 ","End":"03:16.790","Text":"We have some constant k times sine Omega_n x,"},{"Start":"03:16.790 ","End":"03:18.965","Text":"which is when you spell it out, this,"},{"Start":"03:18.965 ","End":"03:22.520","Text":"sine nPi over L times x."},{"Start":"03:22.520 ","End":"03:25.100","Text":"These are our eigenfunctions."},{"Start":"03:25.100 ","End":"03:28.718","Text":"Sine nPi over L times x,"},{"Start":"03:28.718 ","End":"03:33.910","Text":"the eigenfunctions of the boundary value problem, the Sturm-Liouville problem."},{"Start":"03:33.910 ","End":"03:37.810","Text":"The Lambda n, which is Pi^2 n^2 over L^2,"},{"Start":"03:37.810 ","End":"03:40.085","Text":"these are the corresponding eigenvalues."},{"Start":"03:40.085 ","End":"03:46.630","Text":"Each Phi n, the eigenfunction has an eigenvalue Lambda n. That\u0027s just case 2."},{"Start":"03:46.630 ","End":"03:50.030","Text":"We come to case 3, Lambda\u0027s negative,"},{"Start":"03:50.030 ","End":"03:53.870","Text":"so we can write it as Lambda equals minus Omega^2."},{"Start":"03:53.870 ","End":"04:00.560","Text":"The characteristic equation would be k^2 plus Lambda equals 0,"},{"Start":"04:00.560 ","End":"04:06.200","Text":"or k^2 minus Omega^2 is 0 because Lambda is minus Omega^2."},{"Start":"04:06.200 ","End":"04:10.710","Text":"K^2 is Omega^2, so k is plus or minus Omega."},{"Start":"04:10.710 ","End":"04:13.069","Text":"When we have 2 real roots,"},{"Start":"04:13.069 ","End":"04:18.410","Text":"this is the form that the solution takes for each of the roots here and here,"},{"Start":"04:18.410 ","End":"04:20.905","Text":"and constants here and here."},{"Start":"04:20.905 ","End":"04:26.895","Text":"We can find c_1 and c_2 by using these 2 equations."},{"Start":"04:26.895 ","End":"04:31.950","Text":"Plugging in 0, we get this plugging in L, we get this."},{"Start":"04:31.950 ","End":"04:34.125","Text":"From the first one,"},{"Start":"04:34.125 ","End":"04:37.410","Text":"we get c_1 plus c_2 equals 0,"},{"Start":"04:37.410 ","End":"04:38.700","Text":"and from the second one,"},{"Start":"04:38.700 ","End":"04:40.000","Text":"we get this,"},{"Start":"04:40.000 ","End":"04:44.975","Text":"c_1 e^Omega L plus c_2 e^minus Omega L is 0."},{"Start":"04:44.975 ","End":"04:48.830","Text":"From this equation, we get that c_2 is minus c_1,"},{"Start":"04:48.830 ","End":"04:51.370","Text":"so we can put minus c_1 here,"},{"Start":"04:51.370 ","End":"04:56.630","Text":"and then we get c_1 times this minus, this is 0."},{"Start":"04:56.630 ","End":"05:02.510","Text":"From this one, we can say that either c_1 is 0 or this is 0,"},{"Start":"05:02.510 ","End":"05:04.865","Text":"which means that this equals this."},{"Start":"05:04.865 ","End":"05:07.010","Text":"This turns out to be impossible."},{"Start":"05:07.010 ","End":"05:08.330","Text":"I\u0027ll show you why."},{"Start":"05:08.330 ","End":"05:09.950","Text":"Because if this is true,"},{"Start":"05:09.950 ","End":"05:13.310","Text":"multiply both sides by each of the Omega L,"},{"Start":"05:13.310 ","End":"05:16.415","Text":"we get this, one is e^0,"},{"Start":"05:16.415 ","End":"05:20.760","Text":"so 2 Omega L equals 0, but L is not 0,"},{"Start":"05:20.760 ","End":"05:22.560","Text":"so Omega has to be 0,"},{"Start":"05:22.560 ","End":"05:27.125","Text":"and that\u0027s a contradiction because we know that Omega is bigger than 0."},{"Start":"05:27.125 ","End":"05:30.610","Text":"That leaves us with the possibility that c_1 is 0,"},{"Start":"05:30.610 ","End":"05:33.120","Text":"and then c_2 is minus c_1,"},{"Start":"05:33.120 ","End":"05:35.310","Text":"so it\u0027s also 0."},{"Start":"05:35.310 ","End":"05:38.940","Text":"Y, which is this is identically 0,"},{"Start":"05:38.940 ","End":"05:40.535","Text":"y is the 0 function,"},{"Start":"05:40.535 ","End":"05:43.040","Text":"so it\u0027s not an eigenfunction."},{"Start":"05:43.040 ","End":"05:47.470","Text":"There are no eigenfunctions in this case 3."},{"Start":"05:47.470 ","End":"05:49.455","Text":"Let\u0027s summarize."},{"Start":"05:49.455 ","End":"05:51.890","Text":"In both cases 1 and case 3,"},{"Start":"05:51.890 ","End":"05:54.835","Text":"we just have the trivial solution, no eigenfunctions."},{"Start":"05:54.835 ","End":"05:56.995","Text":"All we got is from case 2."},{"Start":"05:56.995 ","End":"06:04.415","Text":"Well, this is the family of eigenfunctions and the corresponding eigenvalues are these."},{"Start":"06:04.415 ","End":"06:07.740","Text":"That concludes this exercise."}],"ID":29450},{"Watched":false,"Name":"Exercise 8","Duration":"7m 16s","ChapterTopicVideoID":28209,"CourseChapterTopicPlaylistID":294448,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.009","Text":"In this exercise, we have another Sturm–Liouville boundary value problem,"},{"Start":"00:05.009 ","End":"00:09.375","Text":"and we have to find the eigenvalues and eigenfunctions."},{"Start":"00:09.375 ","End":"00:14.130","Text":"This is the differential equation part and this is the boundary condition part."},{"Start":"00:14.130 ","End":"00:17.205","Text":"The characteristic equation for this,"},{"Start":"00:17.205 ","End":"00:19.170","Text":"because it\u0027s a second derivative,"},{"Start":"00:19.170 ","End":"00:23.190","Text":"it\u0027s k^2 plus Lambda equals 0,"},{"Start":"00:23.190 ","End":"00:28.710","Text":"the discriminant b^2 minus 4ac is minus 4 Lambda,"},{"Start":"00:28.710 ","End":"00:31.980","Text":"and this will be positive or 0 or negative"},{"Start":"00:31.980 ","End":"00:34.920","Text":"according to whether Lambda is positive, 0, or negative."},{"Start":"00:34.920 ","End":"00:36.207","Text":"The three cases."},{"Start":"00:36.207 ","End":"00:39.163","Text":"Let\u0027s take Lambda bigger than 0 first."},{"Start":"00:39.163 ","End":"00:43.825","Text":"We can write that as Lambda equals Omega squared and Omega is positive."},{"Start":"00:43.825 ","End":"00:47.525","Text":"This is what our boundary value problem becomes,"},{"Start":"00:47.525 ","End":"00:49.430","Text":"just an Omega squared here."},{"Start":"00:49.430 ","End":"00:53.600","Text":"Now, the characteristic equation is k^2 plus Omega"},{"Start":"00:53.600 ","End":"00:58.120","Text":"squared is 0 and the solutions are k equals plus or minus Omega i."},{"Start":"00:58.120 ","End":"01:00.310","Text":"When we have this solution,"},{"Start":"01:00.310 ","End":"01:02.930","Text":"plus or minus an imaginary value,"},{"Start":"01:02.930 ","End":"01:07.805","Text":"then the solution is the combination of cosine and sine."},{"Start":"01:07.805 ","End":"01:10.730","Text":"We need the derivative for this condition."},{"Start":"01:10.730 ","End":"01:12.920","Text":"The derivative just multiply by"},{"Start":"01:12.920 ","End":"01:16.310","Text":"the inner derivative Omega and derivative cosine is minus sine,"},{"Start":"01:16.310 ","End":"01:19.160","Text":"derivative of sine is cosine."},{"Start":"01:19.160 ","End":"01:23.065","Text":"If we plug in the boundary conditions, 0 and L,"},{"Start":"01:23.065 ","End":"01:25.900","Text":"these are the two equations we get."},{"Start":"01:25.900 ","End":"01:28.300","Text":"Omega cancels."},{"Start":"01:28.300 ","End":"01:33.910","Text":"From the first one we get minus c_1 times 0 plus c_2 times 1 is 0."},{"Start":"01:33.910 ","End":"01:37.175","Text":"That just says that c_2 is 0,"},{"Start":"01:37.175 ","End":"01:39.510","Text":"so this part drops out."},{"Start":"01:39.510 ","End":"01:43.515","Text":"Here we have c_1 times sine Omega L is 0."},{"Start":"01:43.515 ","End":"01:45.975","Text":"If we take c_1 equals 0,"},{"Start":"01:45.975 ","End":"01:48.030","Text":"and we have c_1 equals c_2 is 0,"},{"Start":"01:48.030 ","End":"01:49.150","Text":"so that\u0027s a trivial solution,"},{"Start":"01:49.150 ","End":"01:50.785","Text":"so we don\u0027t need that."},{"Start":"01:50.785 ","End":"01:52.720","Text":"We\u0027ll take c_1 not 0,"},{"Start":"01:52.720 ","End":"02:01.995","Text":"so sine Omega L is 0 and Omega L is a multiple of Pi."},{"Start":"02:01.995 ","End":"02:05.460","Text":"Note that there\u0027s no condition on c_1,"},{"Start":"02:05.460 ","End":"02:08.010","Text":"so c _1 could be anything."},{"Start":"02:08.010 ","End":"02:11.160","Text":"So c_1 is k, c_2 is 0,"},{"Start":"02:11.160 ","End":"02:14.100","Text":"and Omega L is a multiple of Pi."},{"Start":"02:14.100 ","End":"02:16.380","Text":"That means that Omega,"},{"Start":"02:16.380 ","End":"02:17.970","Text":"which is really Omega n,"},{"Start":"02:17.970 ","End":"02:23.560","Text":"it depends on n is nPi over L. That\u0027s a family of solutions."},{"Start":"02:23.560 ","End":"02:31.640","Text":"y_n is c_1 cosine Omega nx plus c_2 sine Omega nx, this is 0."},{"Start":"02:31.640 ","End":"02:35.510","Text":"We just get this and n is 1,"},{"Start":"02:35.510 ","End":"02:37.310","Text":"2, 3 and so on."},{"Start":"02:37.310 ","End":"02:43.945","Text":"The reason that we don\u0027t take 0 is that Omega has to be positive,"},{"Start":"02:43.945 ","End":"02:48.500","Text":"and the reason we don\u0027t take n negative is because"},{"Start":"02:48.500 ","End":"02:53.440","Text":"the cosine of minus something is the same as the cosine of the something,"},{"Start":"02:53.440 ","End":"02:55.655","Text":"because cosine is an even function,"},{"Start":"02:55.655 ","End":"02:58.775","Text":"so it doesn\u0027t add anything to take the negative values."},{"Start":"02:58.775 ","End":"03:01.190","Text":"We just have n equals 1, 2, 3, etc."},{"Start":"03:01.190 ","End":"03:03.680","Text":"We can say what the eigenfunctions are."},{"Start":"03:03.680 ","End":"03:07.010","Text":"We can take anything we want for k and usually take k equals 1."},{"Start":"03:07.010 ","End":"03:09.020","Text":"It\u0027s determined up to a multiple."},{"Start":"03:09.020 ","End":"03:13.280","Text":"We take Pi n to be cosine nPi over L,"},{"Start":"03:13.280 ","End":"03:15.380","Text":"n is 1, 2, 3, etc."},{"Start":"03:15.380 ","End":"03:17.440","Text":"These are the eigenfunctions."},{"Start":"03:17.440 ","End":"03:20.570","Text":"The eigenvalues are the Lambda n,"},{"Start":"03:20.570 ","End":"03:22.340","Text":"which is Omega n^2,"},{"Start":"03:22.340 ","End":"03:24.650","Text":"which is Pi squared n^2 over L^2."},{"Start":"03:24.650 ","End":"03:29.590","Text":"We got the eigenfunctions and eigenvalues for Case 1 where Lambda is positive."},{"Start":"03:29.590 ","End":"03:33.185","Text":"Now for Case 2, let us take Lambda equals 0."},{"Start":"03:33.185 ","End":"03:37.250","Text":"The equation just becomes y double prime equals 0."},{"Start":"03:37.250 ","End":"03:41.670","Text":"The solution for that is, c_1 plus c_2x."},{"Start":"03:41.670 ","End":"03:44.790","Text":"That gives us that y prime is c_2."},{"Start":"03:44.790 ","End":"03:50.655","Text":"Now we can plug in 0 or L. If we plug in 0,"},{"Start":"03:50.655 ","End":"03:53.895","Text":"we just get that c_2 is 0."},{"Start":"03:53.895 ","End":"04:00.030","Text":"If you plug in L, same thing because y prime of anything is c_2."},{"Start":"04:00.030 ","End":"04:04.770","Text":"It doesn\u0027t matter if you plug in 0 or L you still get c_2 equals 0."},{"Start":"04:04.770 ","End":"04:07.990","Text":"There\u0027s no restriction on c_1."},{"Start":"04:08.090 ","End":"04:14.600","Text":"Yeah, y equals c_1 is a general solution and we could just take"},{"Start":"04:14.600 ","End":"04:17.630","Text":"a basis or a representative is just y"},{"Start":"04:17.630 ","End":"04:21.754","Text":"equals 1 because an eigenfunction is determined up to a multiple."},{"Start":"04:21.754 ","End":"04:27.950","Text":"So y equals 1 is an eigenfunction and the eigenvalue is Lambda equals 0."},{"Start":"04:27.950 ","End":"04:30.640","Text":"I mean we took Lambda equals 0."},{"Start":"04:30.640 ","End":"04:32.340","Text":"That\u0027s Case 2."},{"Start":"04:32.340 ","End":"04:34.335","Text":"Now let\u0027s move on to Case 3."},{"Start":"04:34.335 ","End":"04:36.810","Text":"This time Lambda is negative,"},{"Start":"04:36.810 ","End":"04:40.230","Text":"so we can write Lambda as minus Omega squared."},{"Start":"04:40.230 ","End":"04:46.985","Text":"Then our equation becomes y double-prime minus Omega squared y equals 0."},{"Start":"04:46.985 ","End":"04:51.155","Text":"The characteristic equation is k^2 minus Omega squared is 0,"},{"Start":"04:51.155 ","End":"04:54.305","Text":"which gives us that k is plus or minus Omega."},{"Start":"04:54.305 ","End":"04:56.435","Text":"We have two real solutions."},{"Start":"04:56.435 ","End":"05:02.300","Text":"The general form of the solution is as follows."},{"Start":"05:02.300 ","End":"05:08.315","Text":"We\u0027ll need the derivative for the boundary value and that\u0027s as follows."},{"Start":"05:08.315 ","End":"05:11.315","Text":"From the boundary conditions, what we get,"},{"Start":"05:11.315 ","End":"05:12.860","Text":"plugging 0 and L,"},{"Start":"05:12.860 ","End":"05:14.795","Text":"we get these two equations."},{"Start":"05:14.795 ","End":"05:17.540","Text":"We can divide both sides by Omega,"},{"Start":"05:17.540 ","End":"05:19.375","Text":"so the Omega drops out."},{"Start":"05:19.375 ","End":"05:23.935","Text":"Now e^0 is 1 and e^minus 0 is also 1."},{"Start":"05:23.935 ","End":"05:28.405","Text":"The first one gives us c_1 minus c_2 is 0,"},{"Start":"05:28.405 ","End":"05:30.830","Text":"and here we get the following."},{"Start":"05:30.830 ","End":"05:34.905","Text":"From this one, we know that c_1 equals c_2."},{"Start":"05:34.905 ","End":"05:37.560","Text":"Put c_2 equals c_1 here,"},{"Start":"05:37.560 ","End":"05:39.825","Text":"multiplied by e to the Omega L,"},{"Start":"05:39.825 ","End":"05:43.605","Text":"and we get c_1 times e^2 Omega L minus 1 is 0."},{"Start":"05:43.605 ","End":"05:46.485","Text":"I claim that this can\u0027t be 0,"},{"Start":"05:46.485 ","End":"05:49.035","Text":"because if this is 0,"},{"Start":"05:49.035 ","End":"05:51.510","Text":"then each of the 2 Omega L is 1,"},{"Start":"05:51.510 ","End":"05:56.105","Text":"so 2 Omega L would be 0 which it isn\u0027t."},{"Start":"05:56.105 ","End":"05:59.225","Text":"I mean, Omega is not 0 and L is not 0."},{"Start":"05:59.225 ","End":"06:05.835","Text":"Because of that, this is non-zero and therefore c_1 is 0,"},{"Start":"06:05.835 ","End":"06:07.140","Text":"but c_1 equals c_2,"},{"Start":"06:07.140 ","End":"06:09.300","Text":"so c_2 is also 0."},{"Start":"06:09.300 ","End":"06:11.735","Text":"We just get y equals 0."},{"Start":"06:11.735 ","End":"06:17.135","Text":"I mean, we have y is c_1 times this plus c_2 times this and c_1 and c_2 are both 0."},{"Start":"06:17.135 ","End":"06:20.910","Text":"We just got the trivial solution for Case 3."},{"Start":"06:20.910 ","End":"06:23.285","Text":"Now it\u0027s time to summarize."},{"Start":"06:23.285 ","End":"06:27.290","Text":"Case 1 gave us a family of eigenfunctions,"},{"Start":"06:27.290 ","End":"06:32.465","Text":"cosine nPi over Lx with these eigenvalues."},{"Start":"06:32.465 ","End":"06:37.880","Text":"Case 2 gave us a single eigenfunction y equals 1 with eigenvalue 0."},{"Start":"06:37.880 ","End":"06:41.410","Text":"Case 3 gave no eigenfunctions."},{"Start":"06:41.410 ","End":"06:45.050","Text":"Now this would be the answer except we can tidy up a bit"},{"Start":"06:45.050 ","End":"06:48.650","Text":"because Case 1 and 2 can be combined."},{"Start":"06:48.650 ","End":"06:54.154","Text":"The y equals 1 is a special case of this with n equals 0."},{"Start":"06:54.154 ","End":"06:56.015","Text":"If you take n equals 0,"},{"Start":"06:56.015 ","End":"07:02.619","Text":"then cosine of 0 is 1 and Pi squared n^2 over L^2 is 0."},{"Start":"07:02.619 ","End":"07:06.030","Text":"Which is correct. If we just change the 1,"},{"Start":"07:06.030 ","End":"07:08.029","Text":"2, 3 to include 0,"},{"Start":"07:08.029 ","End":"07:10.100","Text":"then we can combine Cases 1 and 2,"},{"Start":"07:10.100 ","End":"07:14.135","Text":"and this is the answer to the exercise,"},{"Start":"07:14.135 ","End":"07:16.770","Text":"and we are done."}],"ID":29451}],"Thumbnail":null,"ID":294448}]

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