1 00:00:14,519 --> 00:00:23,000 Good morning, we have been discussing boundary value problems, so far we have 2 00:00:23,000 --> 00:00:28,050 discussed linear and non linear two point boundary value problems. So, in case of linear 3 00:00:28,050 --> 00:00:34,820 we have seen that the discretization leads to the linear system of equations in particular 4 00:00:34,820 --> 00:00:41,000 tri diagonal system which could be solved using Thomas algorithm. But in case of non 5 00:00:41,000 --> 00:00:46,289 linear, we have observed that the corresponding non linear system could be solved using 6 00:00:46,289 --> 00:00:52,390 this Newton Raphson method, but so far we have concerned only 7 00:00:52,390 --> 00:00:57,940 type boundary conditions, in the sense only the function values are given at the 8 00:00:57,940 --> 00:01:04,080 boundary points. So, now let us consider boundary value problems 9 00:01:04,080 --> 00:01:10,030 where the corresponding boundary conditions involve derivatives, so definitely 10 00:01:10,030 --> 00:01:14,380 these types of conditions play a vital role in 11 00:01:14,380 --> 00:01:21,639 the overall structure of the problem. So, let us see how far we can proceed with the 12 00:01:21,639 --> 00:01:25,200 derivative boundary conditions. . 13 00:01:25,200 --> 00:02:17,150 .So, BVP’s derivative going to conditions, so let us start with an example say suppose 14 00:02:17,150 --> 00:02:39,630 the interval is 0, 1 h 1 by 3. So, this means 15 00:02:39,630 --> 00:02:49,680 now if you discretize this equation, so we get 16 00:02:49,680 --> 00:03:23,880 usual central we have been using. We get this, and then we have to write this equation 17 00:03:23,880 --> 00:03:29,990 strictly speaking, so these are the boundary points, so i correspond to one and i 18 00:03:29,990 --> 00:03:38,620 correspond to 2 there. So, if you run for i equals to 1, then we 19 00:03:38,620 --> 00:03:57,430 get y 0 y 1 y 2 plus h square x 1 y 1, so then 20 00:03:57,430 --> 00:04:20,440 suppose we run for i equals to 2, we get y 1. So, in the earlier case, we have seen where 21 00:04:20,440 --> 00:04:28,280 the corresponding boundary conditions does not involve derivative terms in this case 22 00:04:28,280 --> 00:04:39,110 we use to have y 0 and y 3. There would have 23 00:04:39,110 --> 00:04:47,930 been known y 0 and y 3, we have two equations in two unknowns y 1 y 2, but now 24 00:04:47,930 --> 00:04:54,169 the situation is entirely different, so what is 25 00:04:54,169 --> 00:05:19,719 the difference we do not have boundary values. Explicitly, we do not have 26 00:05:19,719 --> 00:05:27,930 boundary values, so then we have a corresponding boundary 27 00:05:27,930 --> 00:05:35,159 condition which involves derivatives, so one simplest solution could be why do not 28 00:05:35,159 --> 00:05:44,939 we discretize the corresponding boundary conditions, for sure we can discretize. 29 00:05:44,939 --> 00:05:45,939 . 30 00:05:45,939 --> 00:06:08,879 So, for example, consider this, so this involve y 0, so this means y 0 minus y 0 prime 31 00:06:08,879 --> 00:06:16,919 equals to minus 1, so let us consider what could be the first order derivative 32 00:06:16,919 --> 00:06:35,099 approximation. So, the first order approximation let us use central, this is second order 33 00:06:35,099 --> 00:06:44,180 approximation, so it makes sense if you see this is second order approximation. So, it 34 00:06:44,180 --> 00:06:45,180 is 35 00:06:45,180 --> 00:06:54,409 .better to use for the boundary conditions as well so if this is the case then y 0 prime 36 00:06:54,409 --> 00:07:03,849 this will be 37 00:07:03,849 --> 00:07:23,419 y 1 minus y minus 1 y 2 h. So, what this is x 0, x 1, x 2, x 3 and y 38 00:07:23,419 --> 00:07:40,919 0, y 1, y 2, y 3, but we have term. So, we have this 39 00:07:40,919 --> 00:07:53,289 if you think of this x minus 1, so that means we have two equations i 1 2, but then we 40 00:07:53,289 --> 00:08:02,300 have we do not have explicitly y 0, y 3. So, then we have gone for the boundary 41 00:08:02,300 --> 00:08:09,969 conditions that involve derivative, then if we discretize we are getting an additional 42 00:08:09,969 --> 00:08:17,649 term. Similarly, if you discretize at the other 43 00:08:17,649 --> 00:08:23,999 boundary we may get another additional term, so 44 00:08:23,999 --> 00:08:38,990 let us go back to general framework and try to discuss what would happen in general. 45 00:08:38,990 --> 00:08:39,990 . 46 00:08:39,990 --> 00:09:50,640 So, consider more general derivative boundary conditions, now 47 00:09:50,640 --> 00:10:20,010 discretizing one we have we have done it before 48 00:10:20,010 --> 00:10:42,100 where 49 00:10:42,100 --> 00:10:56,550 we get this, so let us call this three, so we get now one, i 2 50 00:10:56,550 --> 00:10:57,820 N. 51 00:10:57,820 --> 00:10:58,820 .. 52 00:10:58,820 --> 00:11:13,590 So, if 53 00:11:13,590 --> 00:11:45,960 this is the case three for i 1 to N involve y 0 y 1 y 2, its very straight forward, we 54 00:11:45,960 --> 00:12:07,710 run i 1, it starts from 0, then we end up with N plus 1. So, this is a total of N plus 55 00:12:07,710 --> 00:12:27,910 2 unknowns, further y plus 1 are not explicitly 56 00:12:27,910 --> 00:12:55,200 available, this is due to derivative boundary condition, so however we expect we expect 57 00:12:55,200 --> 00:13:02,330 i 1 to n, so N equations, but involve N plus 2 58 00:13:02,330 --> 00:13:07,820 unknowns. So, still we have not used boundary conditions, 59 00:13:07,820 --> 00:13:53,470 so let us discretize the boundary conditions, so with this at x 0 we get this 60 00:13:53,470 --> 00:14:05,840 approximation. So, this implies y 0 y, y of a 61 00:14:05,840 --> 00:14:37,260 and y dashed of a is to gamma 1. So, this become this is a 0 a 0 y 0 gamma 1, so from 62 00:14:37,260 --> 00:14:56,149 this one could get y minus 1, so let us write down a number say this is four. 63 00:14:56,149 --> 00:14:57,149 .. 64 00:14:57,149 --> 00:16:08,779 So, before we do anything at x equals to y, so b 0 implies that these are different and 65 00:16:08,779 --> 00:16:24,480 correspondingly we have these different and correspondingly we have, so 66 00:16:24,480 --> 00:16:45,270 four and five involve y minus 1 which 67 00:16:45,270 --> 00:16:53,980 are outside the interval a b. I mean which are outside the 68 00:16:53,980 --> 00:17:10,611 interval which corresponds to nodes outside the interval because nodes are x minus 1 x 69 00:17:10,611 --> 00:17:27,140 N plus 2, so these are called fictitious values. 70 00:17:27,140 --> 00:17:28,140 . 71 00:17:28,140 --> 00:17:52,120 .So, let us from four and five, one can express the fictitious 72 00:17:52,120 --> 00:18:46,240 values, so the fictitious values have been expressed like this, but unless 73 00:18:46,240 --> 00:18:50,360 we get rid of these two, we do not end up with 74 00:18:50,360 --> 00:19:17,440 unknown equals to equations. So, what is next step, eliminating the fictitious values. 75 00:19:17,440 --> 00:19:18,440 . 76 00:19:18,440 --> 00:19:38,410 If you recall we have this equation which we were running from 1 to n, so we were 77 00:19:38,410 --> 00:19:58,190 running from 1 to N, so that introduced 1 to N we have a N equations. 78 00:19:58,190 --> 00:19:59,190 . 79 00:19:59,190 --> 00:20:17,070 .Discretizing Introduced two fictitious values, so we have to eliminate, how we eliminate, 80 00:20:17,070 --> 00:20:43,530 we assume that assume that the discretized equation three holds for i equals to 0 and 81 00:20:43,530 --> 00:20:55,220 N plus 1 that is at the 82 00:20:55,220 --> 00:21:04,290 boundary points x 0, so this is alternative. . 83 00:21:04,290 --> 00:22:07,410 So, accordingly i equals to 0 and we have y minus 1, from six 84 00:22:07,410 --> 00:22:14,940 we have to put it y 0, y 1 coefficient of y 0 is b 0 and there is a coefficient 85 00:22:14,940 --> 00:22:28,290 here 86 00:22:28,290 --> 00:23:01,870 coefficient of y 1 a 0. The rest are known quantities 87 00:23:01,870 --> 00:24:14,809 and i N plus 1, again we have six, from that y N plus 2 we had and 88 00:24:14,809 --> 00:24:27,160 from this we can collect the coefficients. . 89 00:24:27,160 --> 00:24:44,050 .So, one may ask what is the necessity of doing this general, one can solve any problem, 90 00:24:44,050 --> 00:24:55,040 well the answer is straight forward, any numerical competition we have to really setup 91 00:24:55,040 --> 00:24:59,620 thinking that we would implement on a machine. So, this kind of a general framework 92 00:24:59,620 --> 00:25:13,480 would help us so that we could write the algorithm systematically, so this is f 1 and say f 93 00:25:13,480 --> 00:25:38,290 2, now we have f 3 which is the system and we have this f 1 and we have f 2. So, this 94 00:25:38,290 --> 00:25:54,690 will set the scenario with a number of unknowns equal to number of equations, of course 95 00:25:54,690 --> 00:25:57,340 three i 1 to N. . 96 00:25:57,340 --> 00:27:03,710 This implies we get a with matrix so on so forth and here we get this matrix, this big 97 00:27:03,710 --> 00:27:29,870 matrix where N plus to unknowns, so this is also tri diagonal system and one can solve. 98 00:27:29,870 --> 00:27:40,179 So, when you have derivative boundary conditions, what is the rule, the rule is when you 99 00:27:40,179 --> 00:27:50,010 discretize a equation just leaving the end points. So, then we have only N equations, 100 00:27:50,010 --> 00:27:52,830 but then if you run the equations at end points, 101 00:27:52,830 --> 00:28:00,920 then we get fictitious values, so when we have fictitious values what do we do? 102 00:28:00,920 --> 00:28:05,150 We discretize the boundary conditions as well and the boundary conditions, also 103 00:28:05,150 --> 00:28:11,281 introduce the fictitious values, now how do we eliminate the fictitious values? We 104 00:28:11,281 --> 00:28:16,809 eliminate the fictitious values between the boundary conditions and between the set of 105 00:28:16,809 --> 00:28:24,010 equations obtained by running the discretized equation at the end points. So, the matrix 106 00:28:24,010 --> 00:28:27,240 is second with respect to the numbers of unknowns 107 00:28:27,240 --> 00:28:30,640 equals to the number of equations and 108 00:28:30,640 --> 00:28:36,480 .we obtain tri diagonal system which we can solve using Thomas algorithm, so let us 109 00:28:36,480 --> 00:28:44,130 solve some problems so that we understand much better. 110 00:28:44,130 --> 00:28:45,130 . 111 00:28:45,130 --> 00:28:59,070 So, consider very simple case and the boundary conditions derivative type only at one 112 00:28:59,070 --> 00:29:48,090 end point just for simplicity. So, let us discretize the equations, so this we need 113 00:29:48,090 --> 00:29:58,910 to find out y 0, y 1, y 2 in this case, y 3 is given, 114 00:29:58,910 --> 00:30:04,310 so we have y 0, y 1, y 2 are the unknowns. So, 115 00:30:04,310 --> 00:30:25,650 let this be the equation, say this is e, b 1, b 2, so this is discretization e i, then 116 00:30:25,650 --> 00:30:43,710 b 2, this then b 1 i, means discretized that would give 117 00:30:43,710 --> 00:30:58,400 y 0 this is y 0 plus y dash, this equals to 1. 118 00:30:58,400 --> 00:31:25,600 So, this implies 2 by 3 y 0 plus y 1 minus 2 by 3. 119 00:31:25,600 --> 00:31:26,600 .. 120 00:31:26,600 --> 00:31:43,840 So, this is f 1, now let us run the equation i equal to 121 00:31:43,840 --> 00:32:26,760 0, so y, so from here we are running this, we had our equation, we had this, so 122 00:32:26,760 --> 00:32:52,860 when i equal to 0, we get this. So, this implies 123 00:32:52,860 --> 00:33:07,320 plus h square x 0, so this is 0 because of x 0, so we get this is say this is e 0, so 124 00:33:07,320 --> 00:33:22,820 e 0 and f 1 eliminates y minus 1. So, this would lead 125 00:33:22,820 --> 00:34:03,630 to, we can simplify this, so this is 2 minus 2 126 00:34:03,630 --> 00:34:52,980 by 3, 4 by 3 y 0 minus 2 by 1, this 6 by 1 plus 2 equals to 0, so this is one equation, 127 00:34:52,980 --> 00:34:59,450 so this is let us say s one, now i equals to 128 00:34:59,450 --> 00:35:02,040 1. . 129 00:35:02,040 --> 00:35:35,910 .We get again i equals to 1, so this we have h square is 1 by 9, x 1 is this, so this will 130 00:35:35,910 --> 00:35:53,580 be 27, 54, so that will be minus say this is 131 00:35:53,580 --> 00:36:34,490 s 2. This implies 132 00:36:34,490 --> 00:36:48,340 from s one, s two, s three implies so we have used y 3, y 3 is also 1, 133 00:36:48,340 --> 00:36:52,200 so we have also used that this is gone, so then 134 00:36:52,200 --> 00:37:37,580 we have 4, minus 6, 0, then this is 1, 1, 0, 1. We have minus 2, 0, minus 1, so one 135 00:37:37,580 --> 00:37:41,770 can solve this, it is simple system, so we get 136 00:37:41,770 --> 00:37:59,500 y 0 is minus point, so the desired solution we get 137 00:37:59,500 --> 00:38:05,560 it. So, when we have the derivative boundary conditions 138 00:38:05,560 --> 00:38:11,099 we have seen, of course this example it is just one of the boundary conditions 139 00:38:11,099 --> 00:38:17,690 contained derivative, but if we have both as well similar procedure can be adopted. 140 00:38:17,690 --> 00:38:27,080 So, before we see that situation let us go for little bit of theory on the linear two 141 00:38:27,080 --> 00:38:29,780 point boundary value problems. . 142 00:38:29,780 --> 00:38:53,280 So, some theory we consider, we have seen already, so this is our equation and we have 143 00:38:53,280 --> 00:38:59,620 not talked about for IVP, we have talked about existence uniqueness. So, it would be 144 00:38:59,620 --> 00:39:07,880 better to talk about the similar situation in case of two point boundary value problems. 145 00:39:07,880 --> 00:39:31,950 Suppose, you say this, suppose f is continuous f means, of course f is given in star at 146 00:39:31,950 --> 00:40:26,270 suppose f is continuous on this set. That dou f, dou y, dou f y dashed are also continuous 147 00:40:26,270 --> 00:41:23,590 on d if d is greater than o for every this and 2 a constant m exists with less than or 148 00:41:23,590 --> 00:41:33,710 equals to m for every d, what is this? 149 00:41:33,710 --> 00:41:41,410 Suppose, f is continuous on this set and both the derivatives with respect to y and y 150 00:41:41,410 --> 00:41:51,109 prime are also continuous on this domain if this is positive and there exists a constant 151 00:41:51,109 --> 00:42:08,560 .that means if the magnitude of this derivative is bounded. So, then star has a unique 152 00:42:08,560 --> 00:42:21,200 solution, so if you recall we have for IVP’s, we have assumed records existence 153 00:42:21,200 --> 00:42:32,830 satisfying Lipschitz condition and similar thought of existence theorem for the two point 154 00:42:32,830 --> 00:42:34,780 value problems. . 155 00:42:34,780 --> 00:43:10,640 Now, let us consider in case of this form, then we will have 156 00:43:10,640 --> 00:43:29,830 similar to what we have considered in the last lectures, so then in 157 00:43:29,830 --> 00:44:00,630 case of this if p x, q x, r x, they are all continuous on this and 2 q x is positive on 158 00:44:00,630 --> 00:44:28,450 a b. Then, double star has unique solution, now with respect to this uniqueness, we get 159 00:44:28,450 --> 00:44:31,900 a little bit of idea, so that means if you have 160 00:44:31,900 --> 00:44:36,100 initial value problems we need Lipschitz type conditions. 161 00:44:36,100 --> 00:44:42,500 In case of two point boundary value problems, we have continuity as well as positiveness 162 00:44:42,500 --> 00:44:50,140 and some kind of boundaries of the derivative terms with y dash etcetera. Now, if you 163 00:44:50,140 --> 00:44:55,760 observe for the linear two point boundary value problems without derivative boundary 164 00:44:55,760 --> 00:45:03,750 conditions we end up with tri diagonal system. So, then under what conditions tri 165 00:45:03,750 --> 00:45:10,940 diagonal system can be solved and things like that let us see little bit on that. 166 00:45:10,940 --> 00:45:11,940 .. 167 00:45:11,940 --> 00:45:26,510 So, we have say special type just we consider x belongs to this with y 168 00:45:26,510 --> 00:46:11,330 of a is gamma of one y of b is gamma 2, so then when we discretize 169 00:46:11,330 --> 00:46:41,430 and the corresponding boundary conditions. So, star can be written as star 170 00:46:41,430 --> 00:46:58,580 can be written as j plus h square f 1 y equals to 171 00:46:58,580 --> 00:47:06,470 c, so we have already noticed that this produces a tri diagonal system the entire thing. 172 00:47:06,470 --> 00:47:29,150 So, here j is of the form minus 2, look at this the coefficient of y i, then coefficient 173 00:47:29,150 --> 00:47:38,109 of y i minus 1 coefficient of y i plus 1, so accordingly 174 00:47:38,109 --> 00:48:08,790 this we are giving it to f 0, then 1 minus 2, 1, 0, 0 then 1. So, we get this and f bar 175 00:48:08,790 --> 00:49:09,680 contains f 1 f n, then then c could contain gamma 1 of course, so this is we split it 176 00:49:09,680 --> 00:49:14,880 actually y, I could have plugged in and we could 177 00:49:14,880 --> 00:49:19,350 have written this matrix in a different way, but I just put it like this. 178 00:49:19,350 --> 00:49:20,350 .. 179 00:49:20,350 --> 00:49:42,520 So, this is a tri diagonal system if the determinant is non zero, then y equals to that 180 00:49:42,520 --> 00:50:04,359 means is invertible, so c and we get the solution from the tri diagonal system. 181 00:50:04,359 --> 00:50:05,359 . 182 00:50:05,359 --> 00:50:32,420 So, this with a linear case, so let us see little non linear case, so we consider there 183 00:50:32,420 --> 00:50:45,300 are several ones. So, I am just continuing, sometimes 184 00:50:45,300 --> 00:50:55,160 one sometimes star, so then the discretized, so what we use because of this 185 00:50:55,160 --> 00:51:24,070 non linearity we make a special remark. Suppose, approximate one by a different scheme 186 00:51:24,070 --> 00:52:01,750 of the form 187 00:52:01,750 --> 00:52:14,190 subject to 188 00:52:14,190 --> 00:52:20,470 that means these are some weights subject to these conditions. 189 00:52:20,470 --> 00:52:22,035 So, this entire thing has been approximated 190 00:52:22,035 --> 00:52:29,750 .by this, but with a specific condition, then the discretized equation can be put in this 191 00:52:29,750 --> 00:52:31,160 form. . 192 00:52:31,160 --> 00:53:07,900 So, the discretized equation can be putt in the form where again j would contain f of 193 00:53:07,900 --> 00:54:09,270 a and b and alpha gamma 1 beta 0 h squared 0. 194 00:54:09,270 --> 00:54:20,590 So, this system is because we have f contains this which is a non linear type, 195 00:54:20,590 --> 00:54:34,050 so again one can solve Newton’s method and similar conditions on this matrix holds corresponding 196 00:54:34,050 --> 00:54:41,650 to linear case. So, with this we have settled the matter for linear and non 197 00:54:41,650 --> 00:54:47,200 linear BVP’s. Of course there are other techniques as well, but within the context 198 00:54:47,200 --> 00:54:52,940 of finite difference methods. We discussed when we discretized the cases 199 00:54:52,940 --> 00:54:58,540 where we end up with tri diagonal system and even we have derivative boundary conditions, 200 00:54:58,540 --> 00:55:06,490 we end up with tri diagonal system. So, we will see we have settled the matter 201 00:55:06,490 --> 00:55:12,870 only for second order linear two point boundary value problems. So, when we consider 202 00:55:12,870 --> 00:55:16,150 higher order what happens we will see in the coming lectures. 203 00:55:16,150 --> 00:55:17,150 Thank you. 204 00:55:17,150 --> 00:55:17,150 .