1 00:00:00,090 --> 00:00:02,490 The following content is provided under a Creative 2 00:00:02,490 --> 00:00:04,030 Commons license. 3 00:00:04,030 --> 00:00:06,330 Your support will help MIT OpenCourseWare 4 00:00:06,330 --> 00:00:10,690 continue to offer high-quality educational resources for free. 5 00:00:10,690 --> 00:00:13,320 To make a donation or view additional materials 6 00:00:13,320 --> 00:00:17,270 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,270 --> 00:00:18,200 at ocw.mit.edu. 8 00:00:35,070 --> 00:00:36,295 INSTRUCTOR: Hi. 9 00:00:36,295 --> 00:00:38,520 Today we're going to conclude our study of learning 10 00:00:38,520 --> 00:00:40,470 our differential equations. 11 00:00:40,470 --> 00:00:43,050 And we're going to use the occasion 12 00:00:43,050 --> 00:00:46,980 to introduce the concept of a Laplace transform. 13 00:00:46,980 --> 00:00:49,440 I should point out that the Laplace transform has 14 00:00:49,440 --> 00:00:52,770 much greater application than just two linear differential 15 00:00:52,770 --> 00:00:54,040 equations. 16 00:00:54,040 --> 00:00:56,480 It's related to the Fourier transform. 17 00:00:56,480 --> 00:00:59,160 It has applications in the subject known 18 00:00:59,160 --> 00:01:01,980 as the convolution integral-- many applications, which 19 00:01:01,980 --> 00:01:04,530 we'll talk about during the learning 20 00:01:04,530 --> 00:01:06,600 exercises in this unit. 21 00:01:06,600 --> 00:01:09,060 But within the framework of learning our differential 22 00:01:09,060 --> 00:01:11,850 equations, I thought this would be a good time 23 00:01:11,850 --> 00:01:14,280 to introduce this rather important concept. 24 00:01:14,280 --> 00:01:16,890 Consequently, our topic for today is called, 25 00:01:16,890 --> 00:01:19,290 quite simply, the Laplace transform. 26 00:01:19,290 --> 00:01:26,040 And it hinges on the fact that e to the t goes to infinity-- 27 00:01:26,040 --> 00:01:27,780 as t approaches infinity, you see-- 28 00:01:27,780 --> 00:01:31,590 much more rapidly than most functions of t. 29 00:01:31,590 --> 00:01:33,090 And by the way, I say t here instead 30 00:01:33,090 --> 00:01:35,940 of x, because the Laplace transform was introduced 31 00:01:35,940 --> 00:01:38,250 primarily for differential equations 32 00:01:38,250 --> 00:01:40,290 in which the independent variable was time. 33 00:01:40,290 --> 00:01:43,320 So traditionally, one usually talks about f of t 34 00:01:43,320 --> 00:01:47,370 when one is talking about Laplace transform concepts. 35 00:01:47,370 --> 00:01:49,740 But the entire idea is this. 36 00:01:49,740 --> 00:01:53,880 Remember, as we saw in part 1 of our course, that e to the t 37 00:01:53,880 --> 00:01:59,430 goes to infinity much faster than t to the n for any integer 38 00:01:59,430 --> 00:02:00,300 n. 39 00:02:00,300 --> 00:02:04,410 Consequently, what this means is, since most functions can 40 00:02:04,410 --> 00:02:06,180 be represented in the power series, 41 00:02:06,180 --> 00:02:08,639 one would expect that if one were 42 00:02:08,639 --> 00:02:12,210 to take a function like f of t and multiply it 43 00:02:12,210 --> 00:02:17,400 by e to the minus st where s was a constant greater than 0, 44 00:02:17,400 --> 00:02:21,600 one would expect that e to the minus st f of t 45 00:02:21,600 --> 00:02:26,430 would approach 0 rapidly as t approaches infinity. 46 00:02:26,430 --> 00:02:27,620 Now, this may not happen. 47 00:02:27,620 --> 00:02:29,940 We'll talk about that in more detail later. 48 00:02:29,940 --> 00:02:31,920 But at any rate, let's assume that we're 49 00:02:31,920 --> 00:02:34,770 dealing with a particular function f of t 50 00:02:34,770 --> 00:02:38,520 where not only does e to the minus f of t 51 00:02:38,520 --> 00:02:43,320 go to 0 rapidly for some value of s as t approaches infinity, 52 00:02:43,320 --> 00:02:49,200 but so rapidly that the integral of e to the minus st f of t dt 53 00:02:49,200 --> 00:02:51,690 from 0 to infinity converges. 54 00:02:51,690 --> 00:02:53,820 Let's just keep that in mind for the time being. 55 00:02:53,820 --> 00:02:57,000 And with this in mind, one defines the Laplace transform 56 00:02:57,000 --> 00:02:58,440 of a function f of t. 57 00:02:58,440 --> 00:03:02,550 Namely, given f of t, the Laplace transform of f of t, 58 00:03:02,550 --> 00:03:05,580 written l of f of t, is defined to be 59 00:03:05,580 --> 00:03:09,030 the integral from 0 to infinity e to the minus st 60 00:03:09,030 --> 00:03:13,620 f of t dt, provided that this integral converges. 61 00:03:13,620 --> 00:03:17,970 Notice, by the way, that since t is the variable of integration, 62 00:03:17,970 --> 00:03:21,920 when we evaluate this integral at t equals 0 63 00:03:21,920 --> 00:03:26,520 and t equals infinity, the resulting function 64 00:03:26,520 --> 00:03:28,440 becomes a function of s alone. 65 00:03:28,440 --> 00:03:30,120 In other words, we evaluate this. 66 00:03:30,120 --> 00:03:33,930 As t goes from 0 to infinity, the resulting integral 67 00:03:33,930 --> 00:03:35,790 is a function of s alone. 68 00:03:35,790 --> 00:03:39,820 Consequently, one often-- to identify the function with this 69 00:03:39,820 --> 00:03:41,520 Laplace transform-- 70 00:03:41,520 --> 00:03:44,040 indicates that, if the function was f of t, 71 00:03:44,040 --> 00:03:48,030 the Laplace transform is f bar of s. 72 00:03:48,030 --> 00:03:51,630 And notice, by the way, what this thing means pictorially. 73 00:03:51,630 --> 00:03:54,105 This is an area under a curve. 74 00:03:54,105 --> 00:03:55,980 And what you're really saying, if the Laplace 75 00:03:55,980 --> 00:03:58,230 transform of a function exists, is 76 00:03:58,230 --> 00:04:03,270 that this area under the curve as t goes from 0 to infinity 77 00:04:03,270 --> 00:04:05,105 is, indeed, finite. 78 00:04:05,105 --> 00:04:06,480 By the way, this should also lead 79 00:04:06,480 --> 00:04:10,050 you to believe that if this is finite for one value with of s, 80 00:04:10,050 --> 00:04:13,290 if s is replaced by something larger, 81 00:04:13,290 --> 00:04:15,900 this goes to 0 even faster. 82 00:04:15,900 --> 00:04:18,930 And consequently, the area under the curve 83 00:04:18,930 --> 00:04:23,790 will exist for all values of s beyond a certain value 84 00:04:23,790 --> 00:04:26,250 once the area exists for that one value. 85 00:04:26,250 --> 00:04:29,760 Maybe the easiest way to say that is in terms of a picture. 86 00:04:29,760 --> 00:04:32,880 Here, I've drawn two representative functions 87 00:04:32,880 --> 00:04:37,920 y equals f of t, one case being a pulsation type of thing 88 00:04:37,920 --> 00:04:41,880 where you have a irregular pulse repeated here, 89 00:04:41,880 --> 00:04:45,510 the other being a polynomial y equals t squared. 90 00:04:45,510 --> 00:04:48,600 What we're saying is that, if I multiply f of t by e 91 00:04:48,600 --> 00:04:49,950 to the minus st-- 92 00:04:49,950 --> 00:04:53,640 remember, e to the minus st for positive values of s 93 00:04:53,640 --> 00:04:56,890 is a curve that goes to 0 very rapidly. 94 00:04:56,890 --> 00:05:00,870 And what this tends to do is to pull this thing down so 95 00:05:00,870 --> 00:05:03,060 that you go to 0 very rapidly. 96 00:05:03,060 --> 00:05:05,130 And the fact that the integral exists 97 00:05:05,130 --> 00:05:08,190 says not only does this go to 0 very rapidly, 98 00:05:08,190 --> 00:05:11,850 but it goes to zero so fast that the area under the curve 99 00:05:11,850 --> 00:05:12,840 stays finite. 100 00:05:12,840 --> 00:05:15,510 Remember, that was our version of infinity times 0. 101 00:05:15,510 --> 00:05:19,020 It's not enough that this goes to 0, that this 102 00:05:19,020 --> 00:05:20,820 approaches the t axis. 103 00:05:20,820 --> 00:05:22,770 It has to approach fast enough so 104 00:05:22,770 --> 00:05:25,092 that the area under the curve is finite. 105 00:05:25,092 --> 00:05:26,550 And all we're saying is that if you 106 00:05:26,550 --> 00:05:29,580 can find one value of s that makes the area under the curve 107 00:05:29,580 --> 00:05:32,760 finite, any larger value of s will certainly 108 00:05:32,760 --> 00:05:34,990 make the area under the curve finite, 109 00:05:34,990 --> 00:05:37,660 because this will be pulled down even faster. 110 00:05:37,660 --> 00:05:40,150 And perhaps you can begin to see intuitively 111 00:05:40,150 --> 00:05:42,580 that, as I let s get larger and larger, 112 00:05:42,580 --> 00:05:45,130 not only does the area under the curve stay finite, 113 00:05:45,130 --> 00:05:48,280 but one would expect that it would also approach 0 114 00:05:48,280 --> 00:05:51,580 because, for large values of s, e to the minus 115 00:05:51,580 --> 00:05:56,110 st goes to 0 so rapidly that even large values up here would 116 00:05:56,110 --> 00:05:58,150 be pulled down very rapidly. 117 00:05:58,150 --> 00:06:01,730 At any rate, let me illustrate this by means of an example. 118 00:06:01,730 --> 00:06:04,630 Suppose I pick, for f of t, something as simple 119 00:06:04,630 --> 00:06:07,780 as e to the at where a is a given constant. 120 00:06:07,780 --> 00:06:10,900 What I mean by "simple" here is recalling that the Laplace 121 00:06:10,900 --> 00:06:15,940 transform involves multiplying f of t by e to the minus st. 122 00:06:15,940 --> 00:06:18,290 This will give me a simple algebraic manipulation 123 00:06:18,290 --> 00:06:19,150 to perform. 124 00:06:19,150 --> 00:06:23,140 By definition, how do I form the Laplace transform of e 125 00:06:23,140 --> 00:06:24,220 to the at? 126 00:06:24,220 --> 00:06:29,020 What I do is I multiply e to the at by e to the minus st 127 00:06:29,020 --> 00:06:31,400 and integrate that from 0 to infinity. 128 00:06:31,400 --> 00:06:32,680 And if that integral exists-- 129 00:06:32,680 --> 00:06:34,097 meaning if the integral converges, 130 00:06:34,097 --> 00:06:35,710 if this limit is finite-- 131 00:06:35,710 --> 00:06:37,600 I call that the Laplace transform. 132 00:06:37,600 --> 00:06:40,510 At any rate, just substituting n here then, f of t 133 00:06:40,510 --> 00:06:41,980 becomes e to the at. 134 00:06:41,980 --> 00:06:44,110 This says integral from 0 to infinity 135 00:06:44,110 --> 00:06:49,330 e to the a minus s times t power dt. 136 00:06:49,330 --> 00:06:56,850 Since a and s are constants, the integral e to the a minus st 137 00:06:56,850 --> 00:07:01,020 comes out to be 1 over a minus s, e to the a minus st, 138 00:07:01,020 --> 00:07:04,270 evaluated as t goes from 0 to infinity. 139 00:07:04,270 --> 00:07:05,940 Now, the key point to notice here 140 00:07:05,940 --> 00:07:09,600 is that, as soon as s is greater than a, 141 00:07:09,600 --> 00:07:11,820 this gives me a negative exponent-- 142 00:07:11,820 --> 00:07:13,350 as soon as s is greater than a. 143 00:07:13,350 --> 00:07:15,660 Consequently, as t goes to infinity, 144 00:07:15,660 --> 00:07:17,920 this essentially behaves like one over e 145 00:07:17,920 --> 00:07:19,780 to the infinity, which is 0. 146 00:07:19,780 --> 00:07:22,380 In other words, once s is greater than a, 147 00:07:22,380 --> 00:07:24,420 the upper limit drops out. 148 00:07:24,420 --> 00:07:29,940 My lower limit would just be e to the zero, you see. 149 00:07:29,940 --> 00:07:31,260 And I'm subtracting that. 150 00:07:31,260 --> 00:07:33,720 At any rate, to make a long story short 151 00:07:33,720 --> 00:07:35,490 and to summarize what we're saying, 152 00:07:35,490 --> 00:07:38,280 simply observe that, if s is greater 153 00:07:38,280 --> 00:07:40,890 than 0, the limit as t approaches infinity 154 00:07:40,890 --> 00:07:45,690 e to the a minus st is 0 and that, therefore, the Laplace 155 00:07:45,690 --> 00:07:49,980 transform of e to the at is simply 1 over s minus a, 156 00:07:49,980 --> 00:07:52,800 provided s is greater than a. 157 00:07:52,800 --> 00:07:55,110 In other words, again, just looking back here-- 158 00:07:55,110 --> 00:07:57,120 see-- I subtract the lower limit. 159 00:07:57,120 --> 00:07:59,767 When t is 0, this is 1. 160 00:07:59,767 --> 00:08:00,600 But I'm subtracting. 161 00:08:00,600 --> 00:08:02,100 It makes it minus 1. 162 00:08:02,100 --> 00:08:07,560 And minus 1 over a minus s is the same as 1 over s minus a. 163 00:08:07,560 --> 00:08:10,020 To state this then in other words, if f of t 164 00:08:10,020 --> 00:08:13,830 is e to the at, f bar of s is 1 over s 165 00:08:13,830 --> 00:08:17,400 minus a where the domain of f bar is a set of all 166 00:08:17,400 --> 00:08:19,560 s, such that s is greater than a. 167 00:08:19,560 --> 00:08:21,570 And what does this mean, geometrically? 168 00:08:21,570 --> 00:08:26,690 This is the area under the curve e to the minus st-- 169 00:08:26,690 --> 00:08:28,170 e to the at-- 170 00:08:28,170 --> 00:08:30,600 as t goes from 0 to infinity. 171 00:08:30,600 --> 00:08:35,460 That area will be finite as soon as s is greater than a. 172 00:08:35,460 --> 00:08:39,023 By way of illustration, if I replace a by 2, 173 00:08:39,023 --> 00:08:41,190 what we're saying is that the Laplace transform of e 174 00:08:41,190 --> 00:08:43,890 to the 2t is 1 over s minus 2. 175 00:08:43,890 --> 00:08:46,730 Notice that nice polynomial quotient idea, 176 00:08:46,730 --> 00:08:48,180 that rather simple expression-- 177 00:08:48,180 --> 00:08:51,210 1 over s minus 2 if s is greater than 2. 178 00:08:51,210 --> 00:08:55,210 If we replace a by minus 3 and use this recipe, 179 00:08:55,210 --> 00:08:58,290 we have that the Laplace transform of e to the minus 3t 180 00:08:58,290 --> 00:09:01,560 is 1 over s minus minus 3, 1 over s 181 00:09:01,560 --> 00:09:05,850 plus 3, where S must be greater than minus 3. 182 00:09:05,850 --> 00:09:10,142 Notice also here that, by the comparison test, so to speak-- 183 00:09:10,142 --> 00:09:12,600 remember, the same comparison test that we use for infinite 184 00:09:12,600 --> 00:09:13,480 series-- 185 00:09:13,480 --> 00:09:18,270 if the magnitude of f of t is less than some constant c 186 00:09:18,270 --> 00:09:21,870 times e to the at power for all values of t, 187 00:09:21,870 --> 00:09:25,470 then the integral from 0 to infinity e to the minus st 188 00:09:25,470 --> 00:09:28,680 f of t dt must also converge. 189 00:09:28,680 --> 00:09:32,190 In other words, we already know that this integral converges, 190 00:09:32,190 --> 00:09:36,788 shall we say, if the integrand were e to the at. 191 00:09:36,788 --> 00:09:39,330 Therefore, we're saying it will converge if you multiply that 192 00:09:39,330 --> 00:09:41,670 by a constant, because convergence isn't affected 193 00:09:41,670 --> 00:09:43,560 by multiplying by a constant. 194 00:09:43,560 --> 00:09:47,400 Consequently, if the integrand that we're investigating 195 00:09:47,400 --> 00:09:49,650 is smaller than this in magnitude, 196 00:09:49,650 --> 00:09:53,940 it must also have a Laplace transform-- that this thing 197 00:09:53,940 --> 00:09:55,600 here must converge. 198 00:09:55,600 --> 00:09:59,130 Now, because this is so obvious and so important, 199 00:09:59,130 --> 00:10:02,070 one defines a function with this property 200 00:10:02,070 --> 00:10:03,750 to have a very special name. 201 00:10:03,750 --> 00:10:06,390 Namely, the function f of t is set 202 00:10:06,390 --> 00:10:08,790 to have exponential order if and only 203 00:10:08,790 --> 00:10:11,630 if there exists constant c and a such 204 00:10:11,630 --> 00:10:14,520 that the magnitude of f of t is less than ce 205 00:10:14,520 --> 00:10:19,470 to the at for t, all t, greater than 0. 206 00:10:19,470 --> 00:10:21,870 What we are saying in summary is that, 207 00:10:21,870 --> 00:10:26,130 by this definition of exponential order 208 00:10:26,130 --> 00:10:29,910 and the comparison test, all functions of exponential order 209 00:10:29,910 --> 00:10:32,550 have Laplace transforms, because that integral from 0 210 00:10:32,550 --> 00:10:36,300 to infinity-- e to the minus st f of t dt-- 211 00:10:36,300 --> 00:10:40,530 will converge by comparison with the integral 0 to infinity 212 00:10:40,530 --> 00:10:45,242 e to the minus st e to the at dt for s greater than. 213 00:10:45,242 --> 00:10:47,700 In other words, this will all converge where the value of s 214 00:10:47,700 --> 00:10:49,800 has to be greater than a. 215 00:10:49,800 --> 00:10:52,770 Now, the idea is, what's so important about functions 216 00:10:52,770 --> 00:10:54,150 of exponential order? 217 00:10:54,150 --> 00:10:55,800 Why do I stress these? 218 00:10:55,800 --> 00:10:58,170 And the answer is that, in terms of linear differential 219 00:10:58,170 --> 00:11:00,900 equations with constant coefficients, 220 00:11:00,900 --> 00:11:03,720 the functions that we tend to deal with all 221 00:11:03,720 --> 00:11:05,350 have exponential order. 222 00:11:05,350 --> 00:11:07,950 In other words, they will all have Laplace transforms. 223 00:11:07,950 --> 00:11:11,010 For example, e to the at, trivially, 224 00:11:11,010 --> 00:11:14,490 is a function that has exponential order. 225 00:11:14,490 --> 00:11:16,530 That was our model for the definition. 226 00:11:16,530 --> 00:11:19,920 In fact, e to the at is 1e to the at. 227 00:11:19,920 --> 00:11:23,130 Simply take a to ba c to b1, and we 228 00:11:23,130 --> 00:11:27,000 have the criterion for exponential order being obeyed. 229 00:11:27,000 --> 00:11:30,390 The sine of at where a is any constant has exponential order. 230 00:11:30,390 --> 00:11:31,080 Why? 231 00:11:31,080 --> 00:11:35,400 Because the magnitude of sine at is certainly no greater than 1. 232 00:11:35,400 --> 00:11:37,770 And 1 is certainly no greater than e 233 00:11:37,770 --> 00:11:40,680 to the t of any positive value of t. 234 00:11:40,680 --> 00:11:43,060 See, e to the 0 is already 1. 235 00:11:43,060 --> 00:11:48,720 Therefore, sine at is less than 1e to the 1t 236 00:11:48,720 --> 00:11:50,340 if t is greater than 0. 237 00:11:50,340 --> 00:11:52,890 So sine at has exponential order. 238 00:11:52,890 --> 00:11:55,890 Finally, we dealt with polynomials t to the n 239 00:11:55,890 --> 00:11:57,820 where n was a positive integer. 240 00:11:57,820 --> 00:12:01,320 And in part 1 of our course, either using L'Hospital's rule, 241 00:12:01,320 --> 00:12:03,300 or power series, or what have you, 242 00:12:03,300 --> 00:12:07,455 we showed that for any fixed positive integer n, 243 00:12:07,455 --> 00:12:12,330 t to the n times e to the minus t went to 0 in the limit 244 00:12:12,330 --> 00:12:14,040 as t went to infinity. 245 00:12:14,040 --> 00:12:16,530 Consequently, for large values of t, 246 00:12:16,530 --> 00:12:19,440 the magnitude of t to the n is less than e to the t. 247 00:12:19,440 --> 00:12:21,240 I'll leave the details out here. 248 00:12:21,240 --> 00:12:23,530 I'm just trying to illustrate what's going on. 249 00:12:23,530 --> 00:12:27,120 The important point is that the usual functions encountered 250 00:12:27,120 --> 00:12:30,840 in linear differential equations with constant coefficients have 251 00:12:30,840 --> 00:12:32,780 Laplace transforms. 252 00:12:32,780 --> 00:12:35,232 And I would like to make a couple of notes 253 00:12:35,232 --> 00:12:35,940 about this thing. 254 00:12:35,940 --> 00:12:40,765 One I've already said, but I'd like to reinforce it. 255 00:12:40,765 --> 00:12:44,400 What we've already seen is that if s has exponential order, 256 00:12:44,400 --> 00:12:47,040 f bar of s, the Laplace transform of s 257 00:12:47,040 --> 00:12:50,760 must be less than or equal to the Laplace transform of e 258 00:12:50,760 --> 00:12:51,480 to the-- 259 00:12:51,480 --> 00:12:54,000 well, a constant times e to the at. 260 00:12:54,000 --> 00:12:58,800 But the Laplace transforms e to the at is 1 over s minus a 261 00:12:58,800 --> 00:13:02,435 so that the Laplace transform of f of t, 262 00:13:02,435 --> 00:13:04,890 if f has exponential order, can be no bigger 263 00:13:04,890 --> 00:13:06,900 than c over s minus a. 264 00:13:06,900 --> 00:13:09,270 Since c is a constant and a is a constant, 265 00:13:09,270 --> 00:13:13,650 as s goes to infinity, f bar of s goes to 0. 266 00:13:13,650 --> 00:13:16,310 In other words, for sufficiently large values of s, 267 00:13:16,310 --> 00:13:20,160 the Laplace transform, which names an area under a curve, 268 00:13:20,160 --> 00:13:22,380 goes to 0 meaning, as we would expect, 269 00:13:22,380 --> 00:13:24,630 the area under the curve does go to 0 270 00:13:24,630 --> 00:13:26,460 as s increases without bound. 271 00:13:26,460 --> 00:13:28,860 What that means in particular, you see, 272 00:13:28,860 --> 00:13:31,800 is that not any function can be a Laplace 273 00:13:31,800 --> 00:13:33,240 transform of something. 274 00:13:33,240 --> 00:13:34,920 You see, in particular, this says 275 00:13:34,920 --> 00:13:37,950 that for f bar to be the Laplace transform of something, 276 00:13:37,950 --> 00:13:41,950 f bar of s must go to 0 as s goes to infinity. 277 00:13:41,950 --> 00:13:48,120 For example, if I were to write down f bar of s to be s over s 278 00:13:48,120 --> 00:13:53,260 minus 1-- or s plus 1, even, f bar of s is s over s plus 1-- 279 00:13:53,260 --> 00:13:57,180 notice that the limit of f bar of s, as s goes to infinity, 280 00:13:57,180 --> 00:13:58,860 is 1. 281 00:13:58,860 --> 00:14:03,030 Consequently, this f bar cannot be the Laplace transform of any 282 00:14:03,030 --> 00:14:04,920 function, at least of exponential order, 283 00:14:04,920 --> 00:14:07,440 because we've already seen that the Laplace transform 284 00:14:07,440 --> 00:14:09,150 of a function-- 285 00:14:09,150 --> 00:14:10,950 at least of exponential order-- 286 00:14:10,950 --> 00:14:15,630 must go to 0 as s goes to infinity. 287 00:14:15,630 --> 00:14:19,090 So not every function can be a Laplace transform of something. 288 00:14:19,090 --> 00:14:23,250 And secondly, not every function has exponential order. 289 00:14:23,250 --> 00:14:25,590 It may be true, as we just showed, 290 00:14:25,590 --> 00:14:28,320 that every function that we're dealing with when we're dealing 291 00:14:28,320 --> 00:14:29,820 with linear differential equations 292 00:14:29,820 --> 00:14:32,130 with constant coefficients-- or most functions 293 00:14:32,130 --> 00:14:34,980 that we're dealing with linear differential equations 294 00:14:34,980 --> 00:14:37,830 with constant coefficients-- has our functions 295 00:14:37,830 --> 00:14:39,210 of exponential order. 296 00:14:39,210 --> 00:14:41,790 Not all functions have exponential order. 297 00:14:41,790 --> 00:14:44,220 For example, e to the t squared does not 298 00:14:44,220 --> 00:14:45,970 have exponential order-- 299 00:14:45,970 --> 00:14:49,710 not order order, order-- 300 00:14:49,710 --> 00:14:54,960 since, notice, that e to the t squared divided by ce to the at 301 00:14:54,960 --> 00:14:59,310 would be 1 over c, e to the t squared minus at, 302 00:14:59,310 --> 00:15:01,755 which is t times t minus a. 303 00:15:01,755 --> 00:15:05,820 And notice now that, if t is greater than a, 304 00:15:05,820 --> 00:15:08,760 this exponent will be positive. 305 00:15:08,760 --> 00:15:12,210 And as t goes to infinity-- you see, once t is greater than a, 306 00:15:12,210 --> 00:15:14,430 this thing increases without bound. 307 00:15:14,430 --> 00:15:16,800 And consequently, since c is a constant, 308 00:15:16,800 --> 00:15:20,610 you essentially have infinity divided by a non-zero constant. 309 00:15:20,610 --> 00:15:21,600 So you have, what? 310 00:15:21,600 --> 00:15:24,090 That e to the t squared over ce to 311 00:15:24,090 --> 00:15:27,720 the at goes to infinity as t approaches infinity. 312 00:15:27,720 --> 00:15:31,020 In particular, this says that e to the t squared 313 00:15:31,020 --> 00:15:34,800 dwarfs ce to the at for large values of t. 314 00:15:34,800 --> 00:15:40,170 In particular, you cannot have that e to the t squared is less 315 00:15:40,170 --> 00:15:44,220 than some constant times e to the at. 316 00:15:44,220 --> 00:15:45,840 Now the question that comes up is, 317 00:15:45,840 --> 00:15:48,990 what's so important now about the Laplace 318 00:15:48,990 --> 00:15:53,480 transform in terms of linear differential equations? 319 00:15:53,480 --> 00:15:59,190 And the answer is that the Laplace transform, amazingly 320 00:15:59,190 --> 00:16:03,030 enough, has properties of linearity. 321 00:16:03,030 --> 00:16:06,120 What do you mean by properties of linearity? 322 00:16:06,120 --> 00:16:07,400 You mean, what? 323 00:16:07,400 --> 00:16:10,380 That for a function to be linear, 324 00:16:10,380 --> 00:16:13,500 l of a sum must be the sum of the l's. 325 00:16:13,500 --> 00:16:15,570 And l of a constant time something 326 00:16:15,570 --> 00:16:17,660 must be the constant times l of something. 327 00:16:17,660 --> 00:16:22,210 Well, look at-- when you're dealing with integrals, 328 00:16:22,210 --> 00:16:24,700 given two definite integrals which exist, 329 00:16:24,700 --> 00:16:27,630 the sum of the integrals is the integral of the sum. 330 00:16:27,630 --> 00:16:30,550 The integral is a constant times an integrand 331 00:16:30,550 --> 00:16:35,030 is a constant times the integral of the integrand itself. 332 00:16:35,030 --> 00:16:37,120 In other words-- sparing you the details, 333 00:16:37,120 --> 00:16:39,550 because they follow right from the definition. 334 00:16:39,550 --> 00:16:41,980 And that is assuming that f and g have Laplace 335 00:16:41,980 --> 00:16:45,820 transforms-- in other words that that improper integral from 0 336 00:16:45,820 --> 00:16:50,530 to infinity, e to the minus st f of t dt, converges. 337 00:16:50,530 --> 00:16:55,645 Then what's true is that l of f plus g is l of f plus l of g. 338 00:16:55,645 --> 00:17:00,045 And l of c times f is c times l of f. 339 00:17:00,045 --> 00:17:02,020 See-- the linearity properties. 340 00:17:02,020 --> 00:17:03,910 Now, how does that help us? 341 00:17:03,910 --> 00:17:06,400 And why does that give us a hint that there 342 00:17:06,400 --> 00:17:10,180 will be some usage of the Laplace transform in solving 343 00:17:10,180 --> 00:17:13,006 linear differential equations with constant coefficients? 344 00:17:13,006 --> 00:17:14,589 Well, perhaps the best way to see this 345 00:17:14,589 --> 00:17:16,450 is by means of an illustration. 346 00:17:16,450 --> 00:17:18,010 Let's assume that we're given that y 347 00:17:18,010 --> 00:17:20,440 is a twice differentiable function of t 348 00:17:20,440 --> 00:17:22,869 and that y double prime plus 2a y 349 00:17:22,869 --> 00:17:26,890 prime plus by where a and b are constants is some function 350 00:17:26,890 --> 00:17:29,170 f of t. 351 00:17:29,170 --> 00:17:30,430 The idea is this. 352 00:17:30,430 --> 00:17:34,550 I simply take the Laplace transform of both sides. 353 00:17:34,550 --> 00:17:36,280 See? 354 00:17:36,280 --> 00:17:40,870 By linearity, since the Laplace transform is a linear function, 355 00:17:40,870 --> 00:17:46,360 a linear operator, l of y double prime plus to 2ay prime plus by 356 00:17:46,360 --> 00:17:51,430 is simply l of y double prime plus 2a l of y prime plus b 357 00:17:51,430 --> 00:17:52,720 l of y. 358 00:17:52,720 --> 00:17:55,900 And the Laplace transform of f of t, assuming that f of t 359 00:17:55,900 --> 00:17:59,350 has a Laplace transform-- in particular, if f of t 360 00:17:59,350 --> 00:18:01,865 happens to be a function of exponential order, 361 00:18:01,865 --> 00:18:04,240 we've already seen that it will have a Laplace transform. 362 00:18:04,240 --> 00:18:08,050 All we're saying is, let's call a Laplace transform of f of t, 363 00:18:08,050 --> 00:18:10,000 as usual, f bar of s. 364 00:18:10,000 --> 00:18:13,840 And now what we have is an equation involving the Laplace 365 00:18:13,840 --> 00:18:16,420 transform of y, the Laplace transform 366 00:18:16,420 --> 00:18:19,720 of y prime, the Laplace transform of y double prime. 367 00:18:19,720 --> 00:18:24,250 In terms of f bar of s, if somehow we could manipulate 368 00:18:24,250 --> 00:18:26,960 this to solve for l of y-- in other words, 369 00:18:26,960 --> 00:18:29,860 if we could find what the Laplace transform of y was 370 00:18:29,860 --> 00:18:31,120 in terms of s-- 371 00:18:31,120 --> 00:18:34,660 maybe we could then invert, so to speak, and find out 372 00:18:34,660 --> 00:18:37,360 what the function itself must have been. 373 00:18:37,360 --> 00:18:39,760 In other words, the idea of inverse functions 374 00:18:39,760 --> 00:18:43,960 comes up the same way here as it comes up every place. 375 00:18:43,960 --> 00:18:48,420 Namely, starting with a function that maps x into y, 376 00:18:48,420 --> 00:18:51,880 one often says, if I know what y is, can I determine 377 00:18:51,880 --> 00:18:53,230 what x must have been? 378 00:18:53,230 --> 00:18:55,728 And that's why 1 to 1-ness is so important. 379 00:18:55,728 --> 00:18:58,270 In other words, if it turns out that the Laplace transform is 380 00:18:58,270 --> 00:19:02,200 1 to 1, once I know the Laplace transform, 381 00:19:02,200 --> 00:19:04,720 I essentially know the function, because if two 382 00:19:04,720 --> 00:19:07,600 different functions can't have the same Laplace transform-- 383 00:19:07,600 --> 00:19:10,390 and by the way, as a finale for today's lecture, 384 00:19:10,390 --> 00:19:12,700 we will show that this is essentially 385 00:19:12,700 --> 00:19:15,610 the case that the Laplace transform is 1 to 1. 386 00:19:15,610 --> 00:19:17,545 And what I mean by essentially, it 387 00:19:17,545 --> 00:19:19,960 will be pointed out in the learning exercises. 388 00:19:19,960 --> 00:19:21,910 But don't worry about that now. 389 00:19:21,910 --> 00:19:23,710 The idea is simply this. 390 00:19:23,710 --> 00:19:27,310 Suppose there was some way of expressing l of y prime 391 00:19:27,310 --> 00:19:30,460 and l of y double prime in terms of l of y. 392 00:19:30,460 --> 00:19:33,040 I could then solve the resulting equation 393 00:19:33,040 --> 00:19:35,740 for l of y in terms of s. 394 00:19:35,740 --> 00:19:38,980 Having a table of Laplace transforms, 395 00:19:38,980 --> 00:19:42,850 I could then locate what that function of s 396 00:19:42,850 --> 00:19:45,250 is the Laplace transform of and then 397 00:19:45,250 --> 00:19:48,730 invert this to find what my y must have been. 398 00:19:48,730 --> 00:19:50,830 The problem is, how do I know that I 399 00:19:50,830 --> 00:19:54,670 can express l of y prime, l of y double prime, 400 00:19:54,670 --> 00:19:55,930 l of y triple prime? 401 00:19:55,930 --> 00:19:57,880 I'm only going up to the second derivative 402 00:19:57,880 --> 00:20:01,150 in terms of our usual convention of illustrating everything 403 00:20:01,150 --> 00:20:03,760 by second-order equations, even though the results 404 00:20:03,760 --> 00:20:05,050 hold the higher orders. 405 00:20:05,050 --> 00:20:06,490 The question though is, how do we 406 00:20:06,490 --> 00:20:08,380 know that we can express these Laplace 407 00:20:08,380 --> 00:20:12,520 transforms in terms of the Laplace transform of y itself? 408 00:20:12,520 --> 00:20:15,270 And this brings up the second reason why that fact-- 409 00:20:15,270 --> 00:20:18,760 that e to the minus st has a multiplying factor-- 410 00:20:18,760 --> 00:20:19,930 is so important. 411 00:20:19,930 --> 00:20:23,260 You see, aside from the fact that, for most functions, 412 00:20:23,260 --> 00:20:26,800 e to the minus st makes sure that the resulting integral 413 00:20:26,800 --> 00:20:30,220 will converge, the other key property is that when you 414 00:20:30,220 --> 00:20:33,700 integrate or differentiate e to the minus st, 415 00:20:33,700 --> 00:20:36,700 because s is a parameter-- meaning it's being treated 416 00:20:36,700 --> 00:20:40,340 as a constant at a given time-- 417 00:20:40,340 --> 00:20:43,390 what this thing means is that the integral or the derivative 418 00:20:43,390 --> 00:20:47,650 of e to the minus st is simply a constant times e to the minus 419 00:20:47,650 --> 00:20:50,470 st. Well, let me show you how this is used. 420 00:20:50,470 --> 00:20:51,970 Let's suppose, for example, that I 421 00:20:51,970 --> 00:20:54,840 wanted to find the Laplace transform into f prime of t. 422 00:20:54,840 --> 00:20:56,410 And I didn't know any tricks. 423 00:20:56,410 --> 00:20:59,230 And by the way, I hope that this is one of the fringe benefits 424 00:20:59,230 --> 00:21:01,750 that this course is teaching you-- that we do not have 425 00:21:01,750 --> 00:21:03,730 to be tricky in mathematics. 426 00:21:03,730 --> 00:21:07,000 All we have to know is the basic definitions 427 00:21:07,000 --> 00:21:08,740 and how to manipulate them. 428 00:21:08,740 --> 00:21:11,140 I do admit that it is a stroke of genius 429 00:21:11,140 --> 00:21:14,260 sometimes in inventing the basic definition that we're 430 00:21:14,260 --> 00:21:14,950 going to use. 431 00:21:14,950 --> 00:21:18,010 For example, it's my own belief that it's a lot easier 432 00:21:18,010 --> 00:21:21,470 to use the Laplace transform than it was to have invented 433 00:21:21,470 --> 00:21:23,220 the concept in the first place. 434 00:21:23,220 --> 00:21:26,210 But once I've defined what the Laplace transform is, what 435 00:21:26,210 --> 00:21:27,860 happens here is very simple. 436 00:21:27,860 --> 00:21:29,780 Namely, by definition, what is the Laplace 437 00:21:29,780 --> 00:21:31,970 transform of f prime of t? 438 00:21:31,970 --> 00:21:36,380 I simply multiply f prime of t by e to the minus st 439 00:21:36,380 --> 00:21:39,300 and integrate that from 0 to infinity. 440 00:21:39,300 --> 00:21:42,830 And if that integral converges, then the Laplace transform 441 00:21:42,830 --> 00:21:43,970 exists. 442 00:21:43,970 --> 00:21:45,620 Let me assume, for the sake of argument 443 00:21:45,620 --> 00:21:49,160 now, that the Laplace transform of f of t exists. 444 00:21:49,160 --> 00:21:51,410 And by the way, again remember, I'm not really 445 00:21:51,410 --> 00:21:52,850 assuming anything when I'm dealing 446 00:21:52,850 --> 00:21:55,790 with linear differential equations 447 00:21:55,790 --> 00:21:58,040 with constant coefficients, because the functions that 448 00:21:58,040 --> 00:22:02,867 come up as possible solutions there are of exponential order. 449 00:22:02,867 --> 00:22:04,700 And we have already seen that, for functions 450 00:22:04,700 --> 00:22:07,400 of exponential order, the Laplace transform does indeed 451 00:22:07,400 --> 00:22:08,180 exist. 452 00:22:08,180 --> 00:22:09,770 But anyway, the idea is this. 453 00:22:09,770 --> 00:22:12,050 By definition, this is the Laplace transform 454 00:22:12,050 --> 00:22:14,000 of f prime of t. 455 00:22:14,000 --> 00:22:16,680 I would like to be able to integrate this. 456 00:22:16,680 --> 00:22:20,570 Well, the idea is that the integral of f prime of t 457 00:22:20,570 --> 00:22:22,670 is certainly f of t. 458 00:22:22,670 --> 00:22:24,520 And if I-- well, to a constant. 459 00:22:24,520 --> 00:22:25,670 But that's not important. 460 00:22:25,670 --> 00:22:28,040 What I'm thinking of is integrating by parts. 461 00:22:28,040 --> 00:22:31,580 Notice that if I let u equal e to the minus st 462 00:22:31,580 --> 00:22:38,790 and dv be f prime of t dt, then du is minus se to the minus st. 463 00:22:38,790 --> 00:22:41,150 v would just be f of t. 464 00:22:41,150 --> 00:22:43,520 Remembering the recipe for integration by parts-- 465 00:22:43,520 --> 00:22:45,320 remember, just by way of quick review-- 466 00:22:45,320 --> 00:22:51,830 integral u dv is equal to uv minus the integral v du. 467 00:22:51,830 --> 00:22:55,340 What we're saying here is that to evaluate this integral, 468 00:22:55,340 --> 00:22:56,600 we simply take, what? 469 00:22:56,600 --> 00:23:00,380 u times v-- e to the minus st f of t. 470 00:23:00,380 --> 00:23:03,470 Evaluate that as t goes from 0 to infinity, 471 00:23:03,470 --> 00:23:05,600 minus the integral v du. 472 00:23:05,600 --> 00:23:08,900 du is already negative. s is a constant. 473 00:23:08,900 --> 00:23:15,530 So minus integral v du is just plus s integral 0 to infinity, 474 00:23:15,530 --> 00:23:19,280 e to the minus st f of t dt. 475 00:23:19,280 --> 00:23:22,370 And lo and behold, you see that's that beautiful property 476 00:23:22,370 --> 00:23:26,480 of e to the minus st. It's still in here. 477 00:23:26,480 --> 00:23:30,680 The integral is precisely what we mean by, what? 478 00:23:30,680 --> 00:23:34,010 l of f of t. 479 00:23:34,010 --> 00:23:37,700 In other words, the Laplace transform of f prime of t 480 00:23:37,700 --> 00:23:42,290 is this thing plus s times the Laplace transform of f of t. 481 00:23:42,290 --> 00:23:43,820 But let me not be so informal. 482 00:23:43,820 --> 00:23:46,700 Let me not refer to this as "this thing." 483 00:23:46,700 --> 00:23:49,250 Let's see what "this thing" really is. 484 00:23:49,250 --> 00:23:52,310 Notice that, because the Laplace transform of f of t 485 00:23:52,310 --> 00:23:55,640 is assumed to exist, the fact that the integral of this 486 00:23:55,640 --> 00:23:59,150 from 0 to infinity is finite means, in particular, 487 00:23:59,150 --> 00:24:02,960 that ad infinity, the integrand must be 0. 488 00:24:02,960 --> 00:24:05,420 You see, otherwise, how could the area under the curve 489 00:24:05,420 --> 00:24:09,260 be finite if the curve didn't at least asymptotically approach 490 00:24:09,260 --> 00:24:10,620 the t axis? 491 00:24:10,620 --> 00:24:12,440 So in other words, the assumption 492 00:24:12,440 --> 00:24:16,460 that f has a Laplace transform means that the upper limit 493 00:24:16,460 --> 00:24:18,230 gives me 0. 494 00:24:18,230 --> 00:24:20,120 The lower limit gives me, what? 495 00:24:20,120 --> 00:24:22,310 When I plug in t equals 0, this is 496 00:24:22,310 --> 00:24:24,620 e to the minus s 0, which is 1. 497 00:24:24,620 --> 00:24:26,460 This is f of 0. 498 00:24:26,460 --> 00:24:28,490 So the lower limit is f of 0. 499 00:24:28,490 --> 00:24:31,840 And since I'm subtracting the lower limit, I have what? 500 00:24:31,840 --> 00:24:34,280 That the thing that I call "this thing" is, 501 00:24:34,280 --> 00:24:37,680 more precisely, minus f of 0. 502 00:24:37,680 --> 00:24:40,790 In other words, to find the Laplace transform of f prime 503 00:24:40,790 --> 00:24:46,370 of t, it's simply minus f of 0 plus s times the Laplace 504 00:24:46,370 --> 00:24:48,140 transform of f of t. 505 00:24:48,140 --> 00:24:49,490 Well, look. 506 00:24:49,490 --> 00:24:51,740 If I happen to know what f of 0 is-- 507 00:24:51,740 --> 00:24:55,490 in other words, if I pick a particular member of the family 508 00:24:55,490 --> 00:24:58,520 f of t, given the curve y equals f of t-- 509 00:24:58,520 --> 00:25:00,500 I certainly know what f of 0 is. 510 00:25:00,500 --> 00:25:01,400 This says, what? 511 00:25:01,400 --> 00:25:06,260 The Laplace transform of f prime of t is some constant plus 512 00:25:06,260 --> 00:25:09,770 s times the Laplace transform of f of t itself. 513 00:25:09,770 --> 00:25:11,270 In other words, somehow or other, I 514 00:25:11,270 --> 00:25:14,390 have now managed to express the Laplace transform 515 00:25:14,390 --> 00:25:19,130 as f prime of t in terms of a polynomial involving s 516 00:25:19,130 --> 00:25:22,970 and the Laplace transform of f of t itself. 517 00:25:22,970 --> 00:25:25,730 By the way, notice that generically here 518 00:25:25,730 --> 00:25:29,170 there is nothing sacred about f of 0, or f. 519 00:25:29,170 --> 00:25:32,120 Think of f as being any function and f prime 520 00:25:32,120 --> 00:25:33,680 as being its derivatives. 521 00:25:33,680 --> 00:25:37,940 In particular, I could allow f prime to play the role of f. 522 00:25:37,940 --> 00:25:38,960 See-- this says, what? 523 00:25:38,960 --> 00:25:41,660 The Laplace transform of the derivative 524 00:25:41,660 --> 00:25:47,930 is minus the function evaluated at 0 plus s times the Laplace 525 00:25:47,930 --> 00:25:50,300 transform of that function. 526 00:25:50,300 --> 00:25:53,220 So if I now take my function to be f prime, 527 00:25:53,220 --> 00:25:55,610 its derivative is f double prime. 528 00:25:55,610 --> 00:25:58,190 The function is, itself, f prime. 529 00:25:58,190 --> 00:26:00,320 So all I really do is, what? 530 00:26:00,320 --> 00:26:02,570 Go into this recipe here. 531 00:26:02,570 --> 00:26:05,930 Every place I see a prime, replace it by a double prime. 532 00:26:05,930 --> 00:26:08,600 Every place I see no prime, put a prime in. 533 00:26:08,600 --> 00:26:11,570 And I have that the Laplace transform of f double prime 534 00:26:11,570 --> 00:26:15,830 of t is minus f prime of 0 plus s times the Laplace 535 00:26:15,830 --> 00:26:17,600 transform of f prime of t. 536 00:26:17,600 --> 00:26:19,970 By the way, to make sure you see this, notice-- 537 00:26:19,970 --> 00:26:23,160 what would the Laplace transform of f triple prime be? 538 00:26:23,160 --> 00:26:26,430 The Laplace transform of f triple prime of t 539 00:26:26,430 --> 00:26:29,580 would be minus f double prime of 0 540 00:26:29,580 --> 00:26:32,460 plus s times the Laplace transform of f double prime 541 00:26:32,460 --> 00:26:33,330 of t. 542 00:26:33,330 --> 00:26:38,400 So assuming that f prime also has a Laplace transform, 543 00:26:38,400 --> 00:26:39,980 we see what the Laplace transform 544 00:26:39,980 --> 00:26:43,850 of f double prime of t looks like in terms of f prime of t-- 545 00:26:43,850 --> 00:26:45,900 the Laplace transform of f prime of t. 546 00:26:45,900 --> 00:26:48,300 We know what the Laplace transform of f prime of t 547 00:26:48,300 --> 00:26:51,540 looks like in terms of the Laplace transform of t. 548 00:26:51,540 --> 00:26:53,580 In other words, from here, I simply 549 00:26:53,580 --> 00:26:57,840 replace l of f prime of t by its value in this equation. 550 00:26:57,840 --> 00:27:01,000 And I wind up with that the Laplace transform of f double 551 00:27:01,000 --> 00:27:04,710 prime of t is nothing more than minus f prime of 0 552 00:27:04,710 --> 00:27:09,270 minus s times f of 0 plus s squared l of f of t. 553 00:27:09,270 --> 00:27:11,310 And now, you see, I've solved the problem 554 00:27:11,310 --> 00:27:13,090 that I was discussing over here. 555 00:27:13,090 --> 00:27:16,500 Namely, once I know what y of 0 is-- 556 00:27:16,500 --> 00:27:19,830 by the way, don't confuse the f here with the f here. 557 00:27:19,830 --> 00:27:22,530 That's an unfortunate choice, which I've just noticed. 558 00:27:22,530 --> 00:27:25,050 The f that I'm referring to here names 559 00:27:25,050 --> 00:27:28,450 the function whose Laplace transform I'm trying to find. 560 00:27:28,450 --> 00:27:32,190 Notice that the f prime and f double prime are served by y 561 00:27:32,190 --> 00:27:33,990 prime and y double prime here. 562 00:27:33,990 --> 00:27:36,210 What we're saying in terms of this notation 563 00:27:36,210 --> 00:27:40,500 is that I can express l of y double prime and l of y prime 564 00:27:40,500 --> 00:27:42,720 in terms of the Laplace transform of l 565 00:27:42,720 --> 00:27:48,540 of y plus powers of s, provided I know only what y of 0 is 566 00:27:48,540 --> 00:27:50,430 and what y prime of 0 is. 567 00:27:50,430 --> 00:27:53,700 And by the way, again, notice that in a physical problem, 568 00:27:53,700 --> 00:27:56,880 it is very natural to look at y and y prime 569 00:27:56,880 --> 00:27:59,820 when t equals 0 because, in many instances, 570 00:27:59,820 --> 00:28:03,270 t equals 0 represents the time at which we've 571 00:28:03,270 --> 00:28:05,490 started the measurement at our experiment. 572 00:28:05,490 --> 00:28:07,980 And this becomes a very natural interpretation 573 00:28:07,980 --> 00:28:10,800 for what you mean by the initial conditions-- 574 00:28:10,800 --> 00:28:14,040 namely, what's going on when t equals 0. 575 00:28:14,040 --> 00:28:17,790 At any rate, we are now in a position to apply our results 576 00:28:17,790 --> 00:28:20,790 to an actual linear differential equation with constant 577 00:28:20,790 --> 00:28:24,510 coefficients given initial conditions 578 00:28:24,510 --> 00:28:29,040 as to what's happening at 0 in terms of the y-coordinate, 579 00:28:29,040 --> 00:28:31,170 and the y-prime coordinate, et cetera-- 580 00:28:31,170 --> 00:28:32,730 in other words, as an application 581 00:28:32,730 --> 00:28:34,170 to a linear differential equations 582 00:28:34,170 --> 00:28:35,670 with constant coefficients. 583 00:28:35,670 --> 00:28:37,650 Suppose we want to solve-- 584 00:28:37,650 --> 00:28:38,340 meaning, what? 585 00:28:38,340 --> 00:28:41,730 Find what function y is of t, if y 586 00:28:41,730 --> 00:28:44,790 double prime minus 4y prime plus 3y equals 587 00:28:44,790 --> 00:28:48,720 e to the 2t, given that my initial conditions-- namely, 588 00:28:48,720 --> 00:28:53,970 when t is 0, y is going to be 0, and y prime is going to be 1. 589 00:28:53,970 --> 00:28:56,490 This could be any conditions I want over here. 590 00:28:56,490 --> 00:28:58,650 I just chose these to give me a rather simple 591 00:28:58,650 --> 00:29:00,900 algebraic example. 592 00:29:00,900 --> 00:29:03,910 And I'll pick more complicated things in the exercises. 593 00:29:03,910 --> 00:29:05,190 But the idea is this. 594 00:29:05,190 --> 00:29:07,800 We have already seen by linearity 595 00:29:07,800 --> 00:29:10,650 that, if I take the Laplace transform of both sides here, 596 00:29:10,650 --> 00:29:13,290 I have the l of y double prime minus 597 00:29:13,290 --> 00:29:18,240 4l of y prime plus 3l of y is the Laplace transform of e 598 00:29:18,240 --> 00:29:19,680 to the 2t. 599 00:29:19,680 --> 00:29:23,520 Now, we just saw, right over here, 600 00:29:23,520 --> 00:29:26,220 that the Laplace transform of y double prime is 601 00:29:26,220 --> 00:29:31,500 minus y prime evaluated at 0 minus sy of 0 plus 602 00:29:31,500 --> 00:29:33,300 s squared l of y. 603 00:29:33,300 --> 00:29:37,440 The Laplace transform of y prime was minus, you see, 604 00:29:37,440 --> 00:29:42,990 y of 0 plus sl y. 605 00:29:42,990 --> 00:29:47,220 So minus 4 times that is just 4 times y of 0 606 00:29:47,220 --> 00:29:51,420 minus 4s l of y plus 3l of y. 607 00:29:51,420 --> 00:29:54,510 And we saw earlier that the Laplace transform of e 608 00:29:54,510 --> 00:29:57,810 of the at is 1 over s minus a where 609 00:29:57,810 --> 00:29:59,790 s is any number greater than 2. 610 00:29:59,790 --> 00:30:02,935 In particular, this is that formula with a equal-- 611 00:30:02,935 --> 00:30:04,620 s is any number greater than a. 612 00:30:04,620 --> 00:30:06,630 In particular, this is that formula 613 00:30:06,630 --> 00:30:09,170 with a equal to 2, so that the Laplace 614 00:30:09,170 --> 00:30:12,720 transform of e to the 2t is 1 over s minus 2, 615 00:30:12,720 --> 00:30:15,570 provided s is greater than 2. 616 00:30:15,570 --> 00:30:17,610 By the way, notice, up to this point, 617 00:30:17,610 --> 00:30:20,350 I have not used the initial conditions. 618 00:30:20,350 --> 00:30:24,600 I have not used the fact that y of 0 is 0 and that y prime of 0 619 00:30:24,600 --> 00:30:25,540 is 1. 620 00:30:25,540 --> 00:30:30,000 Notice that, even without this, the l of y is here by itself. 621 00:30:30,000 --> 00:30:34,410 It's going to be multiplied by s squared minus 4s plus 3. 622 00:30:34,410 --> 00:30:37,620 These terms can go over onto the other side of the equation. 623 00:30:37,620 --> 00:30:39,630 I will get a function as s alone. 624 00:30:39,630 --> 00:30:43,410 I then divide through by s squared minus 4s plus 3. 625 00:30:43,410 --> 00:30:46,410 That will give me l of y as a function of s. 626 00:30:46,410 --> 00:30:48,900 And if I can then find one function which 627 00:30:48,900 --> 00:30:52,350 has that function of s as its Laplace transform, 628 00:30:52,350 --> 00:30:53,790 I can conclude that that function 629 00:30:53,790 --> 00:30:56,370 is the one I'm looking for, provided only that l is 630 00:30:56,370 --> 00:31:00,300 a 1 to 1 operator-- that the Laplace transform is a 1 to 1 631 00:31:00,300 --> 00:31:01,200 function. 632 00:31:01,200 --> 00:31:03,240 At any rate, all I did in our problem 633 00:31:03,240 --> 00:31:08,130 was to simplify this by specifying that y of 0 was 0 634 00:31:08,130 --> 00:31:10,920 and that y prime of 0 was 1. 635 00:31:10,920 --> 00:31:12,630 These two terms drop out. 636 00:31:12,630 --> 00:31:14,070 This is minus 1. 637 00:31:14,070 --> 00:31:17,310 It comes over onto the other side as plus 1. 638 00:31:17,310 --> 00:31:21,460 And so I wind up with this particular equation. 639 00:31:21,460 --> 00:31:24,280 And by the way, notice that this simply says, what? 640 00:31:24,280 --> 00:31:28,390 s minus 2 plus 1 over s minus 2. 641 00:31:28,390 --> 00:31:31,990 In other words, this is s minus 1 over s minus 2. 642 00:31:31,990 --> 00:31:36,010 This factors into s minus 1 times s minus 3. 643 00:31:36,010 --> 00:31:39,850 So in other words, what I'm now faced with 644 00:31:39,850 --> 00:31:44,680 is that s minus 1 times s minus 3 times l of y 645 00:31:44,680 --> 00:31:48,070 is equal to s minus 1 over s minus 2. 646 00:31:48,070 --> 00:31:51,240 I have to be careful. 647 00:31:51,240 --> 00:31:53,070 You see, I wanted to divide-- see, 648 00:31:53,070 --> 00:31:56,460 I want to be able to cancel out the s minus 1's here. 649 00:31:56,460 --> 00:31:59,190 I want to be able to divide through by s minus 3. 650 00:31:59,190 --> 00:32:01,800 I have to be careful of 0 denominators. 651 00:32:01,800 --> 00:32:04,170 And the safest way to take care of everything 652 00:32:04,170 --> 00:32:06,582 is that all I really care about is 653 00:32:06,582 --> 00:32:08,790 whether the Laplace transform exists for sufficiently 654 00:32:08,790 --> 00:32:10,950 large values of s. 655 00:32:10,950 --> 00:32:12,690 To get rid of my headache, I'll simply 656 00:32:12,690 --> 00:32:14,790 assume that I'm not even looking at the Laplace 657 00:32:14,790 --> 00:32:18,560 transform until, shall we say, s is greater than 3. 658 00:32:18,560 --> 00:32:22,170 You see, once s is greater than 3, this factor can't be 0. 659 00:32:22,170 --> 00:32:23,730 This factor can't be 0. 660 00:32:23,730 --> 00:32:27,960 Consequently, I can solve for l of y and concluded that l of y 661 00:32:27,960 --> 00:32:31,530 is 1 over s minus 3 times s minus 2, 662 00:32:31,530 --> 00:32:35,070 provided that s is greater than 3. 663 00:32:35,070 --> 00:32:40,580 Now I resort to that same method of partial fractions 664 00:32:40,580 --> 00:32:43,130 that we used when we integrated back in part 1 665 00:32:43,130 --> 00:32:44,780 using partial fractions. 666 00:32:44,780 --> 00:32:48,020 I simply try to find constants a and b, 667 00:32:48,020 --> 00:32:51,530 such that I can write this as a over s minus 3 668 00:32:51,530 --> 00:32:53,120 plus b over s minus 2. 669 00:32:53,120 --> 00:32:55,310 In other words, break this down into simpler parts. 670 00:32:55,310 --> 00:32:58,160 Sparing you the details, it follows very simply 671 00:32:58,160 --> 00:33:01,550 in this case that this expression is simply 1 over s 672 00:33:01,550 --> 00:33:04,408 minus 3 minus 1 over s minus 2. 673 00:33:04,408 --> 00:33:06,950 In other words, that if I put this over a common denominator, 674 00:33:06,950 --> 00:33:10,520 I get 1 over s minus 3 times s minus 2. 675 00:33:10,520 --> 00:33:13,580 Now here's where the kicker comes in. 676 00:33:13,580 --> 00:33:18,560 Remember, we have already seen that 1 over s minus a 677 00:33:18,560 --> 00:33:21,620 is the Laplace transform of e to at, 678 00:33:21,620 --> 00:33:23,690 provided s is greater than a. 679 00:33:23,690 --> 00:33:27,590 In particular, the Laplace transform of e to the 3t 680 00:33:27,590 --> 00:33:31,940 would be 1 over s minus 3, provided s was greater than 3. 681 00:33:31,940 --> 00:33:35,510 Similarly, the Laplace transform into e to the 2t 682 00:33:35,510 --> 00:33:40,040 would be 1 over s minus 2, provided s was greater than 2. 683 00:33:40,040 --> 00:33:43,010 By the way, notice that both of these conditions 684 00:33:43,010 --> 00:33:45,890 are obeyed as soon as s is greater than 3. 685 00:33:45,890 --> 00:33:49,550 Obviously, if s is greater than 3, it must be greater than 2. 686 00:33:49,550 --> 00:33:51,320 So I'm sure that both of these results 687 00:33:51,320 --> 00:33:53,870 are true once s is greater than 3. 688 00:33:53,870 --> 00:33:56,750 You see the same 3 I have up here. 689 00:33:56,750 --> 00:34:01,910 Now, by linearity, I know that the Laplace transform 690 00:34:01,910 --> 00:34:04,490 of a difference is the difference of the Laplace 691 00:34:04,490 --> 00:34:05,570 transforms. 692 00:34:05,570 --> 00:34:07,490 You see, by equals added to equals, 693 00:34:07,490 --> 00:34:11,989 this tells me that l of e to the 3t minus l of e to the 2t 694 00:34:11,989 --> 00:34:15,830 is 1 over s minus 3 minus 1 over s minus 2. 695 00:34:15,830 --> 00:34:21,590 But by linearity, l of e to the 3t minus l of e to the 2t 696 00:34:21,590 --> 00:34:25,850 is l of the quantity e to the 3t minus e to the 2t. 697 00:34:25,850 --> 00:34:29,570 In particular then, do I know one function 698 00:34:29,570 --> 00:34:33,770 whose Laplace transform is 1 over s minus 3 minus 1 699 00:34:33,770 --> 00:34:35,120 over s minus 2? 700 00:34:35,120 --> 00:34:36,139 And the answer is, yes. 701 00:34:36,139 --> 00:34:37,460 We've just constructed it-- 702 00:34:37,460 --> 00:34:41,795 namely, the function y equals e to the 3t minus e to the 2t. 703 00:34:41,795 --> 00:34:44,659 In other words what we do know is that, in any event, 704 00:34:44,659 --> 00:34:48,620 whether y is the given function or not, whatever l of y is, 705 00:34:48,620 --> 00:34:51,889 it's l of e to the 3t minus e to the 2t. 706 00:34:51,889 --> 00:34:55,590 And therefore, we could conclude that y must equal e 707 00:34:55,590 --> 00:34:59,930 to the 3t minus e to the 2t, provided that l is 1 to 1-- 708 00:34:59,930 --> 00:35:02,780 provided that there was only one function that can have a given 709 00:35:02,780 --> 00:35:04,040 Laplace transform. 710 00:35:04,040 --> 00:35:06,620 You see, you cannot automatically conclude that, 711 00:35:06,620 --> 00:35:10,450 because these two things are equal-- 712 00:35:10,450 --> 00:35:12,020 that the whole expression's equal-- 713 00:35:12,020 --> 00:35:14,270 that the inputs must be equal. 714 00:35:14,270 --> 00:35:18,260 For example, you cannot conclude that x equals pi over 6, 715 00:35:18,260 --> 00:35:22,070 simply because you know that sine x equals sine pi over 6. 716 00:35:22,070 --> 00:35:24,740 You see, the sine function is not 1 to 1. 717 00:35:24,740 --> 00:35:28,160 Well, at any rate, I think you can sense intuitively 718 00:35:28,160 --> 00:35:30,680 that-- because we already know that, for linear differential 719 00:35:30,680 --> 00:35:32,552 equations with constant coefficients, 720 00:35:32,552 --> 00:35:34,760 there are no singular solutions and that, once you've 721 00:35:34,760 --> 00:35:37,550 found one solution, subject to given initial conditions, 722 00:35:37,550 --> 00:35:38,660 you've found them all. 723 00:35:38,660 --> 00:35:41,390 I think you can guess that, at least for functions 724 00:35:41,390 --> 00:35:45,170 of exponential order, this should be a true result. 725 00:35:45,170 --> 00:35:48,710 And the fact that it is true is known in the literature 726 00:35:48,710 --> 00:35:51,650 under the name of Lerch's theorem-- 727 00:35:51,650 --> 00:35:55,490 "Lerch," not L-U-R-CH, but L-E-R-CH. 728 00:35:55,490 --> 00:35:57,450 It's known under the name of Lerch's theorem. 729 00:35:57,450 --> 00:36:00,470 And I have taken the liberty of stating it 730 00:36:00,470 --> 00:36:03,800 in a less controversial form, saving the generalization 731 00:36:03,800 --> 00:36:05,380 for the learning exercises. 732 00:36:05,380 --> 00:36:08,690 But essentially, all Lerch's theorem says is this. 733 00:36:08,690 --> 00:36:14,990 If f and g are continuous and of exponential order, 734 00:36:14,990 --> 00:36:19,460 then if the Laplace transform of f equals the Laplace transform 735 00:36:19,460 --> 00:36:23,250 of g, for all sufficiently large values of s-- in other words, 736 00:36:23,250 --> 00:36:27,620 if beyond a certain s, f bar of s equals g bar of s, 737 00:36:27,620 --> 00:36:30,590 in other words, f and g have the same Laplace transform-- 738 00:36:30,590 --> 00:36:34,280 then f and g must be identical. 739 00:36:34,280 --> 00:36:36,020 And there is a slight modification 740 00:36:36,020 --> 00:36:40,810 of this that comes up if you leave out the word "continuous" 741 00:36:40,810 --> 00:36:41,310 here. 742 00:36:41,310 --> 00:36:44,650 Lerch's theorem is stated only in terms of f and g 743 00:36:44,650 --> 00:36:46,610 being piecewise continuous. 744 00:36:46,610 --> 00:36:49,120 In other words, there could be jump discontinuities 745 00:36:49,120 --> 00:36:51,610 in the curves and things like this, at which case 746 00:36:51,610 --> 00:36:55,090 this would hold, except possibly at the points at which you 747 00:36:55,090 --> 00:36:57,250 had jump discontinuities. 748 00:36:57,250 --> 00:37:00,490 I prefer not to get into this at this particular time, 749 00:37:00,490 --> 00:37:02,920 but rather to wrap up our lecture 750 00:37:02,920 --> 00:37:06,220 at this particular point, and to simply say 751 00:37:06,220 --> 00:37:07,780 that we have done-- what now? 752 00:37:07,780 --> 00:37:10,960 We have studied linear differential equations. 753 00:37:10,960 --> 00:37:13,900 We have used this last lecture to introduce 754 00:37:13,900 --> 00:37:17,170 the concept of the Laplace transform, 755 00:37:17,170 --> 00:37:21,670 which has secondary value in terms of solving differential 756 00:37:21,670 --> 00:37:24,130 equations, but which has other applications that 757 00:37:24,130 --> 00:37:27,610 go far beyond the scope of this particular course. 758 00:37:27,610 --> 00:37:31,300 And what I'm hoping is that this rather introductory lecture, 759 00:37:31,300 --> 00:37:34,150 coupled with well-chosen exercises, 760 00:37:34,150 --> 00:37:36,610 will give you enough insight to the Laplace 761 00:37:36,610 --> 00:37:39,100 transform so that you will be able to handle it 762 00:37:39,100 --> 00:37:42,250 not only in terms of solving linear differential 763 00:37:42,250 --> 00:37:44,230 equations with constant coefficients 764 00:37:44,230 --> 00:37:48,100 but in other contexts where they might occur 765 00:37:48,100 --> 00:37:51,550 in the particular line of work and research that you're doing. 766 00:37:51,550 --> 00:37:54,940 At any rate, this completes our work on block seven. 767 00:37:54,940 --> 00:37:57,970 And in our next lecture, what we shall do 768 00:37:57,970 --> 00:38:00,580 is start the final block of our course 769 00:38:00,580 --> 00:38:04,000 and revisit the concept of vector spaces 770 00:38:04,000 --> 00:38:07,120 from, hopefully, a more productive point of view 771 00:38:07,120 --> 00:38:08,310 than what we've had before. 772 00:38:08,310 --> 00:38:09,880 But more about that next time. 773 00:38:09,880 --> 00:38:13,700 Until next time, goodbye. 774 00:38:13,700 --> 00:38:16,100 Funding for the publication of this video 775 00:38:16,100 --> 00:38:20,960 was provided by the Gabriella and Paul Rosenbaum Foundation. 776 00:38:20,960 --> 00:38:25,130 Help OCW continue to provide free and open access to MIT 777 00:38:25,130 --> 00:38:30,565 courses by making a donation at ocw.mit.edu/donate.