1 00:00:17,070 --> 00:00:23,610 Hello, in the last class we have discussed some general methods for stability aspects 2 00:00:23,610 --> 00:00:27,740 of multi step method. So, for example schur criterion 3 00:00:27,740 --> 00:00:32,480 we have discussed. So, given a multi step method, how to find out the stability 4 00:00:32,480 --> 00:00:35,450 regions using the schur criterion? . 5 00:00:35,450 --> 00:00:41,930 So, let us continue ahead with stability aspects of multi step method. So, let us discuss 6 00:00:41,930 --> 00:00:54,350 some further methods. So, one of the popular method is Routh-Harwitz criterion. So, 7 00:00:54,350 --> 00:01:29,479 consider the mapping of the open 8 00:01:29,479 --> 00:01:57,670 unit disk of the complex plane to the open left half 9 00:01:57,670 --> 00:02:14,460 plane, that is real z less than 0 of the complex z plane. So, what this mapping does? This 10 00:02:14,460 --> 00:02:33,750 maps unit disc mod r less than 1 to the left of plane in z plane, then the inverse of the 11 00:02:33,750 --> 00:02:53,149 mapping is given by inverse of this mapping. So, this is the inverse of the mapping. 12 00:02:53,149 --> 00:02:54,149 .. 13 00:02:54,149 --> 00:03:27,320 So, under the transformation, under this transformation the function. So, what is this? 14 00:03:27,320 --> 00:03:58,790 This is our stability polynomial. So, this is this becomes, so this polynomial becomes 15 00:03:58,790 --> 00:04:24,040 this. Now, say this is a star, multiply star by 1 minus z power k, we have. So, we are 16 00:04:24,040 --> 00:04:46,790 multiplying this by 1 minus z power k, where k is the order of multi step method. So, we 17 00:04:46,790 --> 00:05:09,200 get this and our aim is to identify this as a polynomial in z. So, what was our aim? So, 18 00:05:09,200 --> 00:05:11,420 this is order of the method. . 19 00:05:11,420 --> 00:05:28,790 .So, we are multiplying and this we are expressing as polynomial in z. So, then the roots 20 00:05:28,790 --> 00:06:13,620 of the stability polynomial lies inside the open unit disk, if and only if 21 00:06:13,620 --> 00:06:20,250 the roots of the 22 00:06:20,250 --> 00:06:56,300 polynomial double star, so that is of this lie in the open left half plane. So, the roots 23 00:06:56,300 --> 00:07:01,210 of the stability polynomial in the original variable 24 00:07:01,210 --> 00:07:07,270 r lie inside the open unit disc if and only the roots of the polynomial double start. 25 00:07:07,270 --> 00:07:08,270 . 26 00:07:08,270 --> 00:07:38,850 This is this lie in the open left half plane. So, then the criterion 27 00:07:38,850 --> 00:08:06,160 the roots of double star lie in the open left half plane if 28 00:08:06,160 --> 00:08:45,250 and only if the leading principal minors of the k cross 29 00:08:45,250 --> 00:09:21,130 k matrix. So, this is a, the roots of double 30 00:09:21,130 --> 00:09:38,050 star so that double star is of the form. So, lie in 31 00:09:38,050 --> 00:09:45,900 the open left half plane if and only if the leading principal minors of this matrix are 32 00:09:45,900 --> 00:10:04,610 positive, and a 0 greater than 0 are positive leading principle minors of this matrix are 33 00:10:04,610 --> 00:10:09,950 positive and a naught greater than 0. 34 00:10:09,950 --> 00:10:10,950 .. 35 00:10:10,950 --> 00:10:37,330 We assume that a j is 0, if j is greater than this, in particular for k equals to 2 the 36 00:10:37,330 --> 00:11:38,740 condition leads to this and k equals to 3 and k equals to 4. These represent the necessary 37 00:11:38,740 --> 00:12:19,820 and sufficient conditions for ensuring that all roots lie in the left half plane. Of course, 38 00:12:19,820 --> 00:12:34,709 open. So, this is these are the necessary and sufficient conditions. So, to start with, 39 00:12:34,709 --> 00:12:39,620 so we have used a transformation, what was the transformation? 40 00:12:39,620 --> 00:12:50,459 z equals to r minus 1 and the inverse of this is this. 41 00:12:50,459 --> 00:12:51,459 . 42 00:12:51,459 --> 00:12:56,160 .So, under this transformation the stability polynomial reduces to this, then we have to 43 00:12:56,160 --> 00:13:01,880 multiply by this factor. As a result of the multi step method, the linear multi step method 44 00:13:01,880 --> 00:13:09,180 of order k we are getting this factor. So, then multiplied by that we get a polynomial 45 00:13:09,180 --> 00:13:14,670 z. So, then the Routh-Hurwitz criterion states 46 00:13:14,670 --> 00:13:24,839 that the roots of this polynomial lie in the open left half plane if and only if the leading 47 00:13:24,839 --> 00:13:31,450 principal minors of this matrix are positive. And a 0 is greater than 0 and in particular 48 00:13:31,450 --> 00:13:37,270 we have the these being the necessary and sufficient conditions for ensuring that all 49 00:13:37,270 --> 00:13:41,860 the roots of the polynomial in z to lie in the 50 00:13:41,860 --> 00:13:44,430 open left half plane, ok? . 51 00:13:44,430 --> 00:14:15,050 So, let us see how it works out with a example. So, use the Routh-Hurwitz criterion 52 00:14:15,050 --> 00:14:21,899 to determine 53 00:14:21,899 --> 00:14:56,880 the interval of absolute stability of the linear multi step method. So, the given 54 00:14:56,880 --> 00:15:04,190 linear multi step method we need to determine the interval of absolute stability. So, we 55 00:15:04,190 --> 00:15:25,620 have r equals to this is a map, then for this method we have. So, this is given by r square 56 00:15:25,620 --> 00:15:51,910 minus 1 and this is given by accordingly, we have r square minus 1 is h bar. 57 00:15:51,910 --> 00:16:09,459 So, this will be the stability polynomial which is, so this is our stability polynomial 58 00:16:09,459 --> 00:16:16,910 and what is the order of the method 2. So, we 59 00:16:16,910 --> 00:16:32,110 need to consider this into variable change to 60 00:16:32,110 --> 00:16:53,399 this transformation. So, accordingly 61 00:16:53,399 --> 00:17:10,159 this should be written in the form, so we can 62 00:17:10,159 --> 00:17:19,740 simplify this. 63 00:17:19,740 --> 00:17:20,740 .. 64 00:17:20,740 --> 00:18:07,890 So, here this get cancelled minus 1 term there, so 1 minus z. So, this will be 1 minus z 65 00:18:07,890 --> 00:18:19,669 square. So, this will be 1 minus z square. So, minus h by 2 z square and this is usual 66 00:18:19,669 --> 00:18:43,279 expansion. So, we multiply, sorry this is minus, this must be minus. So, this will be 67 00:18:43,279 --> 00:19:49,740 these terms then so this is minus h bar z square 68 00:19:49,740 --> 00:20:03,270 plus 4 plus 3 h bar z. This is this 2 h. So, this is 69 00:20:03,270 --> 00:20:25,970 a 0 z square. Now, for k equals to this is the necessary 70 00:20:25,970 --> 00:20:32,509 and sufficient condition accordingly. So, using 71 00:20:32,509 --> 00:21:05,840 part a, we get h bar to be is the region of absolute stability. So, this is Routh-Hurwitz 72 00:21:05,840 --> 00:21:17,919 criteria, right? So far these stability methods, these two general methods, one is schur 73 00:21:17,919 --> 00:21:24,820 criterion, other is Routh-Hurwitz criterion. So, they give you using some kind of 74 00:21:24,820 --> 00:21:34,110 approach, the Schur criterion via Schur polynomial and then Routh-Hurwtiz criterion via 75 00:21:34,110 --> 00:21:40,621 in kind of a transformation. So, but there are another general methods, 76 00:21:40,621 --> 00:21:47,110 in the sense you try to after all it is a difference equation, so you try to find the 77 00:21:47,110 --> 00:21:50,190 characteristic polynomial and find the roots so 78 00:21:50,190 --> 00:21:55,780 that like a general recurrence relation you get the solution. See once we have a 79 00:21:55,780 --> 00:22:01,269 recurrence relation which is also a finite difference method, right? So, we can get the 80 00:22:01,269 --> 00:22:12,050 solution using the roots and hence analyze what happens to this solution, ok? So, let 81 00:22:12,050 --> 00:22:26,500 us see one such method. 82 00:22:26,500 --> 00:22:27,500 .. 83 00:22:27,500 --> 00:22:46,139 Stability of multi step methods parasitic term. So, we will see what is this parasitic 84 00:22:46,139 --> 00:22:53,470 term? So, let us see with an example, so consider 85 00:22:53,470 --> 00:23:07,460 the multi step method. So, obviously this is 86 00:23:07,460 --> 00:23:24,529 for n greater than equals to 1 so then so with reference to absolute stability the reference 87 00:23:24,529 --> 00:23:52,639 equation is lambda y. So, accordingly these become 2 h bar y n, where h bar is, so the 88 00:23:52,639 --> 00:24:16,210 characteristic polynomial r square minus this, right? Now, the 89 00:24:16,210 --> 00:24:42,049 roots are given by 2 h, so this is given by, so these are the roots. 90 00:24:42,049 --> 00:24:43,049 . 91 00:24:43,049 --> 00:25:01,050 .So, let us analyze, so what were the roots r equals to. Now, consider first root, call 92 00:25:01,050 --> 00:25:17,769 it r 0 so this one plus 93 00:25:17,769 --> 00:25:28,849 so this is h bar plus. So, having power series expansion we get 1 3. So, I 94 00:25:28,849 --> 00:25:43,450 write that first, then h here, then h square by 2 plus the higher order terms. So, if you 95 00:25:43,450 --> 00:25:52,279 try to approximate see what was our equation and 96 00:25:52,279 --> 00:26:00,220 naturally the exact solution must be of this form. 97 00:26:00,220 --> 00:26:07,509 So, if we approximate and try to identify with the exact solution y a of course, this 98 00:26:07,509 --> 00:26:20,759 is indeed e power lambda h bar. Now, consider 99 00:26:20,759 --> 00:26:38,869 r 1 h bar minus, so this is h bar minus, so this can be approximated as minus 1 minus 100 00:26:38,869 --> 00:27:00,009 h bar plus h bar by 2 plus third order terms. Now, this can be approximated as minus e power 101 00:27:00,009 --> 00:27:14,249 lambda h bar, I am sorry this is lambda which is e power h and this is e power minus 102 00:27:14,249 --> 00:27:16,889 lambda h, ok? . 103 00:27:16,889 --> 00:27:29,380 So, one you have this type and other is this type. However, once we have the roots the 104 00:27:29,380 --> 00:27:45,169 general solution y n is given by c 0 r 0 power n c 1 r power n, right? So, there we have 105 00:27:45,169 --> 00:27:51,279 r 0 is behaving like e power lambda h and r 106 00:27:51,279 --> 00:28:02,440 1 is behaving like this. Now, let us take real 107 00:28:02,440 --> 00:28:10,289 positive then what happens, lambda real positive. So, we need to compare because the 108 00:28:10,289 --> 00:28:18,359 general solution is this. Now, we need to identify which one is dominating 109 00:28:18,359 --> 00:28:26,419 and really the solution of the difference equation gives stable solutions, 110 00:28:26,419 --> 00:28:29,919 right? So, then what happens if lambda is real 111 00:28:29,919 --> 00:28:47,799 and positive r 0 is greater than this. Then what happens? This grows faster than this. 112 00:28:47,799 --> 00:28:48,799 So, 113 00:28:48,799 --> 00:29:21,649 .what happens r 1 power n increases less rapidly then r 0 power n, implies c 0 r 0 power n 114 00:29:21,649 --> 00:29:33,799 dominates, this dominates. So, that means the other term, so out of your 115 00:29:33,799 --> 00:29:49,450 solution c 0 e power lambda h n e power lambda h n. This is dominating this for this 116 00:29:49,450 --> 00:29:58,419 is a case which is the exact solution and this 117 00:29:58,419 --> 00:30:24,729 is the parasitic term. However, in this case does not create any trouble. So, why do we 118 00:30:24,729 --> 00:30:29,669 call this parasitic term and then why we conclude all this? Of course, this is exact 119 00:30:29,669 --> 00:30:42,269 solution for the given solution, but this is an extra. So, this is also called extraneous 120 00:30:42,269 --> 00:30:48,179 part and this is the parasitic term, but however 121 00:30:48,179 --> 00:30:50,243 in this case does not create any problem. The 122 00:30:50,243 --> 00:30:57,310 overall solution grows as per the exact solution. So, that means for these values of 123 00:30:57,310 --> 00:31:01,089 lambda the solution is stable. . 124 00:31:01,089 --> 00:31:12,029 . So, let us consider now the other case, lambda 125 00:31:12,029 --> 00:31:36,239 is negative. Then for of course, h is positive. So, then c 1 r power n will dominate 126 00:31:36,239 --> 00:32:01,289 c 0 r power n, as n goes to infinity for a fixed h. Therefore, so the parasitic term 127 00:32:01,289 --> 00:32:25,599 which is h, so this parasitic term increases. So, 128 00:32:25,599 --> 00:32:36,710 this creates a problem, so that is how so for lambda real and positive it agrees with 129 00:32:36,710 --> 00:32:40,549 the true solution, whereas for lambda negative 130 00:32:40,549 --> 00:32:44,320 this parasitic term creates a problem. So, this 131 00:32:44,320 --> 00:32:48,579 is another kind of a stability analysis, ok? 132 00:32:48,579 --> 00:32:49,579 .. 133 00:32:49,579 --> 00:33:03,110 Let us see this little more from little theoretical point of view, relative and weak stability. 134 00:33:03,110 --> 00:33:15,809 So, relative and weak stability. So, with respect to what it is relative we will see. 135 00:33:15,809 --> 00:33:35,549 So, consider this, so general solution is given 136 00:33:35,549 --> 00:33:50,499 by. So, for a general method this is 0 to n c j r j 137 00:33:50,499 --> 00:34:16,060 lambda h, right? The parasitic terms power n goes to 0 y a, this is true because as h 138 00:34:16,060 --> 00:34:25,200 goes to 0 the parasitic terms goes to 0. However, 139 00:34:25,200 --> 00:34:42,929 for finite h for any x n 1 would expect it to 140 00:34:42,929 --> 00:34:56,260 be small compared to c. 141 00:34:56,260 --> 00:35:04,710 So, that means we are bringing out the concept of relative stability. So, relative with 142 00:35:04,710 --> 00:35:17,430 respect to, what? So, relative with respect to the a h, the principle exact r 0. So, this 143 00:35:17,430 --> 00:35:30,920 implies 144 00:35:30,920 --> 00:35:47,490 for sufficiently small h. So, this is relative 145 00:35:47,490 --> 00:35:56,180 is a comparison the parasitic terms with the exact, so extraneous terms with the 146 00:35:56,180 --> 00:36:03,660 exact, right? 147 00:36:03,660 --> 00:36:04,660 .. 148 00:36:04,660 --> 00:36:42,370 So, a method is relatively stable if the characteristic roots 149 00:36:42,370 --> 00:36:55,790 satisfying 150 00:36:55,790 --> 00:37:10,520 for all sufficiently small. So, that means let there be some parasitic 151 00:37:10,520 --> 00:37:20,450 terms, but compared to the exact, these extraneous must be small for sufficiently 152 00:37:20,450 --> 00:37:58,550 small lambda h. The impact satisfies the strong root condition if a h, these roots which are, 153 00:37:58,550 --> 00:38:02,470 which would contribute to the extraneous terms if the magnitude it. 154 00:38:02,470 --> 00:38:12,870 Of course, for a h 0 if the magnitude is less than 1 then we say the strong root condition, 155 00:38:12,870 --> 00:38:54,640 right? So, remark relative stability does not imply strong root condition and if a linear 156 00:38:54,640 --> 00:39:20,360 multi step method is stable, but not relatively stable. It is weakly stable, so relative 157 00:39:20,360 --> 00:39:25,820 stability does not imply strong root condition and if a method is stable, but not relatively 158 00:39:25,820 --> 00:39:33,130 stable, then we say it is weakly stable, ok? 159 00:39:33,130 --> 00:39:34,130 .. 160 00:39:34,130 --> 00:39:53,320 So for example, the case y n plus 1 plus 2 h f n, so far in this case we have seen. So, 161 00:39:53,320 --> 00:40:05,820 this does not hold. So, hence this method is weakly 162 00:40:05,820 --> 00:40:22,520 stable, so that is how we can categorize the relative stability and weakly stability. 163 00:40:22,520 --> 00:40:30,660 So, we have some criterions to systematically capture the interval of stability region of 164 00:40:30,660 --> 00:40:34,050 stability, but there is another concept to analyze 165 00:40:34,050 --> 00:40:38,690 the parasitic terms and then say how the method behaves, ok? 166 00:40:38,690 --> 00:40:39,690 . 167 00:40:39,690 --> 00:41:17,170 .So, let us analyze with some more examples. So, for example consider right. So, this 168 00:41:17,170 --> 00:42:29,250 implies on y dash we get. So, we have, so this will be the characteristic polynomial, 169 00:42:29,250 --> 00:43:03,940 then roots. So, if r is this we get these roots, 170 00:43:03,940 --> 00:43:05,660 ok? . 171 00:43:05,660 --> 00:44:13,130 Accordingly z 1, z 1 which is so this would be written as this can be written as. So, 172 00:44:13,130 --> 00:44:20,080 this is for what if h goes to 0, r goes to 0, r square 173 00:44:20,080 --> 00:44:25,690 goes to 0. So, with that we can linearize and 174 00:44:25,690 --> 00:44:41,690 try to get this. So, you can identify, so we have done it in the last example. So, 1 175 00:44:41,690 --> 00:45:09,210 plus lambda h equal e power lambda h. So, as h 176 00:45:09,210 --> 00:45:29,310 goes to 0, z 1 behaves like, it has to behave like this. Therefore, y n c 1 z 1 power n 177 00:45:29,310 --> 00:46:04,350 plus c 2, then so this is c 1. So, this agrees with 178 00:46:04,350 --> 00:46:25,100 exact solution and this is parasitic term. Now, if lambda is negative, 179 00:46:25,100 --> 00:46:40,650 this grows and creates trouble. So, this is another example where we could analyze. 180 00:46:40,650 --> 00:46:54,070 Now, for these multi step methods there is a small technique, given your characteristic 181 00:46:54,070 --> 00:46:59,580 first and second characteristic polynomials are r h o of zeta and then sigma of zeta. 182 00:46:59,580 --> 00:47:03,200 So, for a particular method if r h o is given, 183 00:47:03,200 --> 00:47:05,430 can we determine the corresponding sigma? 184 00:47:05,430 --> 00:47:06,430 .. 185 00:47:06,430 --> 00:47:28,010 So, this is another small trick involved a h. So, let us try to do that. So, given this 186 00:47:28,010 --> 00:47:37,200 how do we determine, so one may also refer the numerical 187 00:47:37,200 --> 00:48:15,430 solutions book by Jain. So, to know lot of examples, so consider, so these are 188 00:48:15,430 --> 00:48:58,890 general linear multi step method case, step method, then the error. So, this is the error, 189 00:48:58,890 --> 00:49:11,760 then recall this linear multi step method said 190 00:49:11,760 --> 00:49:40,570 to be of order p if C is 0. So, this gives a p th order method. So, accordingly T i plus 191 00:49:40,570 --> 00:50:03,530 1 reduces to right. So, this is identically 192 00:50:03,530 --> 00:50:23,090 0 when y x is a polynomial of degree less than r 193 00:50:23,090 --> 00:50:28,810 equals to p, right? . 194 00:50:28,810 --> 00:50:54,600 .Now, observe that the constants C i and p are independent of y of x. So, let us choose 195 00:50:54,600 --> 00:51:08,010 y of x equals to this, then double star, what 196 00:51:08,010 --> 00:51:58,670 was our double star this. So, this grows like, so 197 00:51:58,670 --> 00:52:56,160 this is e power k h minus a 1. So, this can be written as r h o of e h minus h, so it 198 00:52:56,160 --> 00:53:25,510 is right. So, T i plus 1 this must be behaving like 199 00:53:25,510 --> 00:53:32,500 this, ok? . 200 00:53:32,500 --> 00:53:51,700 So, this implies we can get some new constant because there are some terms left out after 201 00:53:51,700 --> 00:54:07,670 getting cancelled x i, so that I have put it to this constant. So, this is the relation 202 00:54:07,670 --> 00:54:11,370 using which when r h o is given. How to determine 203 00:54:11,370 --> 00:54:24,680 sigma? So, this will suggest us. So, let us work out some problems, how to determine exactly 204 00:54:24,680 --> 00:54:30,370 for a given example using this relation in the coming lecture. 205 00:54:30,370 --> 00:54:32,710 Until then thank you, bye. 206 00:54:32,710 --> 00:54:32,710 .