1 00:00:17,630 --> 00:00:24,640 Hi. In the last class, we have discussed about root condition, which has direct implication 2 00:00:24,640 --> 00:00:33,430 on zero stability of the method. So, before we proceed further on some more things 3 00:00:33,430 --> 00:00:42,989 related to stability, so let us state root condition as a necessary and sufficient condition. 4 00:00:42,989 --> 00:00:43,989 . 5 00:00:43,989 --> 00:01:16,009 So, we have not stated that. So, the linear multi step method is zero stable, if 6 00:01:16,009 --> 00:01:39,200 given y dash equals to f of x y f of x y satisfy Lipchitz 7 00:01:39,200 --> 00:01:55,569 condition and root condition holds. So, that means the linear multi step method is 8 00:01:55,569 --> 00:02:03,840 zero stable if given this f satisfy Lipchitz condition and root condition holds. So, as 9 00:02:03,840 --> 00:02:27,780 I mentioned for a given method say 2 y n plus 2 minus 3 y n plus 1 minus 6 y n minus h 25 10 00:02:27,780 --> 00:02:57,230 f n plus 1 minus 40 f n equals to 0, we can write, so 2. So, this is n plus 2. So, 2 zeta 11 00:02:57,230 --> 00:03:05,820 cube zeta, this is n plus 2, so this is not, this 12 00:03:05,820 --> 00:03:22,440 will be 2 square minus 6 and this will be this will be… 13 00:03:22,440 --> 00:03:23,440 .. 14 00:03:23,440 --> 00:04:05,930 So, that is how we identify. So, now consider another example let us say 15 00:04:05,930 --> 00:04:18,660 so the first characteristic polynomial 11 and the roots 16 00:04:18,660 --> 00:04:38,120 if we compute. Now, what is the root condition says root condition the roots must 17 00:04:38,120 --> 00:04:41,770 be within the unit disk, but unfortunately for 18 00:04:41,770 --> 00:04:57,750 this greater than 1, therefore this is not zero stable. So, we can compute accordingly. 19 00:04:57,750 --> 00:05:18,130 So, for example, consider 20 00:05:18,130 --> 00:05:39,020 so for this is 21 00:05:39,020 --> 00:05:55,690 so this implies simple root hence zero stable. So, that 22 00:05:55,690 --> 00:05:57,720 is how we compute. . 23 00:05:57,720 --> 00:06:10,620 .So, let us proceed further. If you recall for convergence, we have defined two things, 24 00:06:10,620 --> 00:06:15,810 one is stability and another is consistency. So, 25 00:06:15,810 --> 00:06:38,300 the linear multi step method of the form star is 26 00:06:38,300 --> 00:07:01,740 said to be consistent. If it has order 3, greater than 1, so these two are together 27 00:07:01,740 --> 00:07:10,500 necessary for convergence. So, before we go for further 28 00:07:10,500 --> 00:07:37,800 concept of stability, let us do some example. Determine all values 29 00:07:37,800 --> 00:08:13,570 of the real parameter b for which linear multi step 30 00:08:13,570 --> 00:08:26,760 method. Determine all values of the real parameter b for which this method is zero 31 00:08:26,760 --> 00:08:45,699 stable. So, let us identify rho of zeta. So, rho of zeta for this method, look at the distance 32 00:08:45,699 --> 00:09:12,190 plus 1 and plus 2 and plus 3, so this is our first characteristic polynomial. So, this 33 00:09:12,190 --> 00:09:38,390 can be written as this. 34 00:09:38,390 --> 00:09:39,390 . 35 00:09:39,390 --> 00:10:00,500 This can be written as, so call this, now the method is zero stable if all the roots 36 00:10:00,500 --> 00:10:25,940 of rho one belong to unit disk, those on the circle 37 00:10:25,940 --> 00:10:39,550 are simple and differ from 1. Why? This is because already we have zeta equals to 1 as 38 00:10:39,550 --> 00:10:43,020 one root. If you have multiple roots, then one 39 00:10:43,020 --> 00:10:49,010 of the necessities you have seen of the modulus is greater than 1. Then it blows up and 40 00:10:49,010 --> 00:10:59,300 multiplicity, therefore for rho 1, these are the conditions. The method will be zero stable. 41 00:10:59,300 --> 00:11:27,230 If this happens, hence suppose that the method is zero stable, then b is not equal to 1 by 42 00:11:27,230 --> 00:11:43,350 2 or b is not equal to 3 by 2 because 43 00:11:43,350 --> 00:12:01,370 for these values of b, rho 1 zeta has double roots on 44 00:12:01,370 --> 00:12:19,670 the unit circle, that is zeta equals to 1, minus 1 for these values of this further product 45 00:12:19,670 --> 00:12:35,600 of roots of rho 1 zeta is 1, further product 46 00:12:35,600 --> 00:12:48,880 of is 1. Then but since these values are not equal, 47 00:12:48,880 --> 00:13:11,020 roots are none of them are 1 or minus 1, neither of them are plus or minus 1. 48 00:13:11,020 --> 00:13:12,020 .. 49 00:13:12,020 --> 00:13:41,960 Hence, they are strictly complex. Therefore, the discriminant negative, this implies, this 50 00:13:41,960 --> 00:14:08,610 implies b belongs to half, 3 by 2 to b belongs to, so this is this interval for zero stability. 51 00:14:08,610 --> 00:14:50,920 Conversely, suppose b belongs to this. Then the roots are 1, zeta 2, 3. These are, so 52 00:14:50,920 --> 00:15:19,680 this is 1 and mod 2, 3, this is 1, but zeta 2 or 53 00:15:19,680 --> 00:15:28,480 3 not equals to 1, zeta 2 not equals to zeta 3 and 54 00:15:28,480 --> 00:15:53,680 all roots lie on the unit circle and are simple. So, this is the analysis to identify a 55 00:15:53,680 --> 00:16:02,560 particular method is zero stable or not. In this particular case, we were trying to obtain 56 00:16:02,560 --> 00:16:09,990 the two bounds on the b, the range for which this method is absolutely zero stable. 57 00:16:09,990 --> 00:16:10,990 . 58 00:16:10,990 --> 00:16:20,800 .Now, let us proceed to something called absolute stability. If you recall for single step 59 00:16:20,800 --> 00:16:39,540 methods, we have discussed this concept. So, the stability that we discussed with respect 60 00:16:39,540 --> 00:16:57,389 to h goes to 0, n goes to infinity, nh fixed, so this was the stability. The zero stability 61 00:16:57,389 --> 00:17:01,899 h goes to 0, n goes to infinity, so that nh 62 00:17:01,899 --> 00:17:14,149 is fixed. Now, we need to analyze something different, how a particular linear multi step 63 00:17:14,149 --> 00:17:30,990 method behaves in the case h greater than 0 64 00:17:30,990 --> 00:17:43,250 fixed and n goes to infinity. So, how a particular linear multi step method 65 00:17:43,250 --> 00:17:51,880 in the case h goes to 0 fixed and n goes to infinity. So, absolute stability with reference 66 00:17:51,880 --> 00:17:56,640 to always a particular reference equation, so 67 00:17:56,640 --> 00:18:22,750 that is our y dash equals to lambda y. Now, consider our method equals to h multi step 68 00:18:22,750 --> 00:18:36,780 method in this form, let us say both i equals to 0 to k; suppose we considered in this, 69 00:18:36,780 --> 00:18:49,220 then with reference to the reference equation, 70 00:18:49,220 --> 00:19:10,669 we 71 00:19:10,669 --> 00:19:19,840 get similar difference equation in this form. . 72 00:19:19,840 --> 00:19:42,470 Now, the general solution of this, what is this? This is a homogeneous difference 73 00:19:42,470 --> 00:20:05,070 equation or a recurrence relation can be expressed as a linear 74 00:20:05,070 --> 00:20:30,500 combination of powers of roots of the associated 75 00:20:30,500 --> 00:20:55,640 stability polynomial. So, this is a stability polynomial, first 76 00:20:55,640 --> 00:21:00,370 characteristic, second characteristic, and then one can define stability polynomial. 77 00:21:00,370 --> 00:21:10,350 So, this is possible. Now, the definition of absolute 78 00:21:10,350 --> 00:21:21,850 stability the linear multi step method of 79 00:21:21,850 --> 00:21:49,990 the form star is called absolutely stable for a given h bar 80 00:21:49,990 --> 00:22:04,420 if and only if for that h bar, all 81 00:22:04,420 --> 00:22:53,880 the roots all the roots say r s of the stability polynomial satisfy this condition, otherwise 82 00:22:53,880 --> 00:23:12,010 this method is absolutely unstable, absolutely stable 83 00:23:12,010 --> 00:23:17,919 unstable. So, what it says this linear 84 00:23:17,919 --> 00:23:24,460 .multistep method is absolutely stable for a given h bar. See zero stability we defined 85 00:23:24,460 --> 00:23:31,200 with h goes to 0 as n goes to infinity. Now, absolute 86 00:23:31,200 --> 00:23:37,600 stability we are defining for fixed h. . 87 00:23:37,600 --> 00:23:45,700 So, similar to single step method, it is our duty to find out an interval, an interval 88 00:23:45,700 --> 00:24:17,919 alpha beta of the real line is called the interval 89 00:24:17,919 --> 00:24:43,169 of absolute stability. The method is 90 00:24:43,169 --> 00:24:55,570 absolutely stable for all h bar belongs to this interval. 91 00:24:55,570 --> 00:25:20,429 Further, the method is absolutely unstable for 92 00:25:20,429 --> 00:25:53,429 every h bar, then the method is said to have no interval of stability. So, it is a very 93 00:25:53,429 --> 00:25:57,470 much essential to find the interval of absolute 94 00:25:57,470 --> 00:26:09,759 stability. So, what is the procedure? 95 00:26:09,759 --> 00:26:10,759 . 96 00:26:10,759 --> 00:26:40,170 .So, general methods for locating the interval of absolute stability, so the 97 00:26:40,170 --> 00:26:47,250 first one, first there are several approaches. So, let us discuss 98 00:26:47,250 --> 00:27:08,419 couple of them. So, the first one is schur criterion. So, before we state that some preliminaries, 99 00:27:08,419 --> 00:27:21,080 consider the polynomial phi r of 100 00:27:21,080 --> 00:28:01,450 the form c 0, c k non zero, c 0, non zero with complex coefficients, complex coefficients. 101 00:28:01,450 --> 00:28:30,910 Then, the polynomial phi is said to be a schur polynomial, if each 102 00:28:30,910 --> 00:29:00,820 of its roots r s satisfy. So, the polynomial phi is said to, so first 103 00:29:00,820 --> 00:29:03,860 of all we consider polynomial where these are 104 00:29:03,860 --> 00:29:11,240 complex coefficients and remember c k non zero, c 0 non zero, then the polynomial is 105 00:29:11,240 --> 00:29:18,000 said to be a schur polynomial if each of its roots are satisfy. That means the magnitude 106 00:29:18,000 --> 00:29:29,090 must be less than 1. So, then we say this polynomial is schur polynomial. 107 00:29:29,090 --> 00:29:30,090 . 108 00:29:30,090 --> 00:29:44,291 Now, let us consider the following polynomial. What is that? We 109 00:29:44,291 --> 00:30:12,559 denote it by phi hat c 0 bar. What are these bars, one can guess easily 110 00:30:12,559 --> 00:30:25,539 because the coefficients were complex. Therefore, c j bars are complex conjugates 111 00:30:25,539 --> 00:30:41,289 of c j. Please try to recall c k r power k, c k 112 00:30:41,289 --> 00:30:47,090 minus 1 r power k minus 1, c 1 r 0. Now, what is the polynomial that we are 113 00:30:47,090 --> 00:30:59,549 considering? Look c 0 conjugate has gone to r power k, c 1 conjugate has gone to r 114 00:30:59,549 --> 00:31:15,600 power k minus 1, c k conjugate has gone to r power 0, and so it is backwards. Then let 115 00:31:15,600 --> 00:31:32,320 us define phi 1 of r 1 over r phi hat of 0 phi 116 00:31:32,320 --> 00:31:53,450 of r phi of 0 that has degree less than or equals 117 00:31:53,450 --> 00:32:04,669 to k minus 1 that has degree. 118 00:32:04,669 --> 00:32:05,669 .. 119 00:32:05,669 --> 00:32:50,860 Then, this is the theorem. The polynomial phi defined earlier is a schur polynomial 120 00:32:50,860 --> 00:32:57,740 if and only if modulus of phi hat of 0 is greater 121 00:32:57,740 --> 00:33:26,610 than modulus of phi of 0 and phi 1. So, what it 122 00:33:26,610 --> 00:33:39,990 says that phi which was defined earlier is a schur polynomial, if and only if this condition 123 00:33:39,990 --> 00:33:51,700 holds and phi 1 is a schur polynomial. When do you say it is a schur polynomial? That 124 00:33:51,700 --> 00:34:14,220 means roots of phi 1 must be less than 1 in magnitude. So, let us check with an example. 125 00:34:14,220 --> 00:34:15,220 . 126 00:34:15,220 --> 00:34:40,529 .Consider this example characteristic polynomial phi of r square minus 1 because n plus 1 127 00:34:40,529 --> 00:35:09,230 is missing, this is r minus 2. Hence, the stability, the stability polynomial, sorry 128 00:35:09,230 --> 00:35:29,119 r h is given by r square minus 1 minus 129 00:35:29,119 --> 00:35:39,410 the stability polynomial. See, we have defined like this. 130 00:35:39,410 --> 00:36:32,450 So, this is h bar, so this minus so equal to r square minus, this is our stability 131 00:36:32,450 --> 00:36:40,099 polynomial. Now, we would like to use the schur criterion. 132 00:36:40,099 --> 00:36:47,930 So, what it says? So, this is our polynomial. Now, consider this as our relative 133 00:36:47,930 --> 00:36:56,779 polynomial. Then we have to compute phi 134 00:36:56,779 --> 00:37:12,209 hat where the coefficients are conjugated and given backwards. Therefore, pi hat is 135 00:37:12,209 --> 00:37:35,219 given by so this is c 2, c 1, and c 0. 136 00:37:35,219 --> 00:38:10,019 Now, we have to consider the other way. So, c 0 r square, so this is, this is our phi 137 00:38:10,019 --> 00:38:17,751 hat. Now, what should be checked? The polynomial 138 00:38:17,751 --> 00:38:23,529 phi is a schur polynomial if and only if phi hat of 0 moduli is greater than this and 139 00:38:23,529 --> 00:38:50,519 phi 1 is a schur polynomial. So, let us try to 140 00:38:50,519 --> 00:39:01,309 do that. So, what was the, so we have pi like this, pi hat like this. Now, we have to check 141 00:39:01,309 --> 00:39:17,410 mod, mod pi of 0. . 142 00:39:17,410 --> 00:40:11,099 This is mod 1 plus, then so the condition mod is greater than this implies 143 00:40:11,099 --> 00:40:23,349 h bar belongs to, so this can be verified very easily. Now, 144 00:40:23,349 --> 00:40:29,269 what is the next step? So, this is the range that means the polynomial phi is a schur polynomial 145 00:40:29,269 --> 00:40:35,490 if and only if this happens. That means the polynomial is schur polynomial only 146 00:40:35,490 --> 00:40:43,170 when h is in this range. Further, we have to check phi 1 is a schur polynomial. So, 147 00:40:43,170 --> 00:40:44,359 let us check. 148 00:40:44,359 --> 00:41:17,079 .Now, pi 1 r is 1 over r. So, we are computing as defined pi 1 r, we are computing pi 1 r. 149 00:41:17,079 --> 00:41:45,469 So, pi hat 0 to pi hat 0 is 1, so times phi of r that is phi of r, this is exactly this 150 00:41:45,469 --> 00:42:07,260 minus pi of 0, this minus 1 plus 2 h bar by 3, 1 minus 151 00:42:07,260 --> 00:42:30,760 1 by 3 h bar r. So, this is 1 over r, r gets cancelled. So, we can just put r minus h by 152 00:42:30,760 --> 00:42:47,160 3 minus 1 over r, then plus 1 over r with 1 153 00:42:47,160 --> 00:43:09,779 there, then minus h by 3, 1 plus this. If it gets cancelled, then this 1 minus 1 154 00:43:09,779 --> 00:43:29,390 plus 2 h bar by 4 square r, so this is so r 1 there 155 00:43:29,390 --> 00:43:48,559 and this is the coefficient. Then these two get cancelled. Then minus h by 3, if I take 156 00:43:48,559 --> 00:44:26,170 common 1 plus, so this can be written as so this is pi 1 r. Now, the definition is 157 00:44:26,170 --> 00:44:33,559 polynomial pi is said to be schur polynomial if each of its roots satisfy modulus less 158 00:44:33,559 --> 00:44:34,559 than 1. 159 00:44:34,559 --> 00:44:46,559 Now, further the schur criteria says the polynomial phi is a schur polynomial if and only 160 00:44:46,559 --> 00:45:02,529 if this and this is only a, so r equals to half, r equals to half is our root, so which 161 00:45:02,529 --> 00:45:21,549 implies, hence pi 1 is a schur polynomial, pi 1 is 162 00:45:21,549 --> 00:45:30,609 a schur polynomial. Now, three a multi step method of the form is called absolutely stable 163 00:45:30,609 --> 00:45:37,559 for a given h bar if and only if for that h 164 00:45:37,559 --> 00:45:52,319 bar, all the roots for a stability polynomial satisfy this. So, we have verified this. So, 165 00:45:52,319 --> 00:45:57,299 if it is a schur polynomial if each of roots satisfies, 166 00:45:57,299 --> 00:46:08,209 this it is interrelated. We have shown that phi 1 is pi 1 has this; therefore pi 1 is 167 00:46:08,209 --> 00:46:13,369 a schur polynomial. Therefore, the roots of their 168 00:46:13,369 --> 00:46:34,419 characteristic polynomial are less than 1. Hence, the method is absolutely stable. 169 00:46:34,419 --> 00:46:35,419 . 170 00:46:35,419 --> 00:47:19,609 .So, let us look at one more example y n plus 2 minus y n equals h by 2. Now, this 171 00:47:19,609 --> 00:47:21,749 can be written as so let us do everything in one 172 00:47:21,749 --> 00:47:36,969 go, E square minus 1 y n h by 2 E plus 3 f n. 173 00:47:36,969 --> 00:48:24,420 This implies rho E equals E square minus 1, so therefore plus 3. Therefore, now the 174 00:48:24,420 --> 00:48:59,799 stability polynomial, this is given by, this is given by this equals r square 1 plus 3 175 00:48:59,799 --> 00:49:21,559 by 2. Therefore, pi hat of this 176 00:49:21,559 --> 00:49:33,130 r square r plus 1. . 177 00:49:33,130 --> 00:50:05,720 So, we have to compute this equal to 1. Then let us schur criterion says the polynomial 178 00:50:05,720 --> 00:50:08,949 pi is a schur polynomial if and only if this 179 00:50:08,949 --> 00:50:26,459 is 0. So, let us so 180 00:50:26,459 --> 00:50:44,829 implies this implies 181 00:50:44,829 --> 00:50:54,189 h bar belongs to this range. Now, on the next step 182 00:50:54,189 --> 00:50:59,910 is to construct this pi 1 and show that this is 183 00:50:59,910 --> 00:51:43,579 a schur polynomial. So, this is 1 over r. So, what is this polynomial pi 1? Pi 1 has 184 00:51:43,579 --> 00:52:00,459 to be constructed. So, we have to construct this, 185 00:52:00,459 --> 00:52:56,170 this is 1 and pi of r. So, this will be r minus 186 00:52:56,170 --> 00:54:33,779 this, this minus, so this gets cancelled. So, this implies minus half 3 r, so this implies 187 00:54:33,779 --> 00:54:50,699 r equals to minus 1 by 3. So, this implies mod of r is less than 1. So, schur criterion 188 00:54:50,699 --> 00:54:59,599 says the polynomial pi is said to be schur polynomial if each of its roots modulus is 189 00:54:59,599 --> 00:55:05,809 less than 1. So, pi 1 is a schur polynomial, then 190 00:55:05,809 --> 00:55:12,729 so the polynomials pi is a schur polynomial because pi 1 is a schur polynomial. The 191 00:55:12,729 --> 00:55:25,209 polynomial phi is a schur polynomial and accordingly what it says for absolute stability, 192 00:55:25,209 --> 00:55:34,279 these stability polynomial roots are within modulus unity. 193 00:55:34,279 --> 00:55:41,339 So, this is schur criterion, helps one to conclude the region of absolute stability. 194 00:55:41,339 --> 00:55:45,160 For example, when we say h bar that is lambda 195 00:55:45,160 --> 00:55:50,179 h is within the range minus 4 by 3 to 0, one 196 00:55:50,179 --> 00:55:56,809 .must choose a combination of lambda and h because for a fixed h, you have to choose 197 00:55:56,809 --> 00:55:59,730 a lambda such that the lambda h is within the 198 00:55:59,730 --> 00:56:06,119 range. So, this trade off, suppose lambda, suppose for a given lambda, one cannot choose 199 00:56:06,119 --> 00:56:12,789 h, which makes the lambda h beyond this interval. So, the schur criterion is very 200 00:56:12,789 --> 00:56:22,789 much useful. So, in the next lecture, we shall discuss some more criterions for absolute 201 00:56:22,789 --> 00:56:28,079 stability. So, you may try for some other methods with schur polynomial, until then 202 00:56:28,079 --> 00:56:29,079 good bye. 203 00:56:29,079 --> 00:56:29,079 .