7 3 laplace transforms translations unit step functions
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7.3 Laplace Transforms: translations & unit step functions a - PowerPoint PPT Presentation

7.3 Laplace Transforms: translations & unit step functions a lesson for MATH F302 Differential Equations Ed Bueler, Dept. of Mathematics and Statistics, UAF April 5, 2019 for textbook: D. Zill, A First Course in Differential Equations with


  1. 7.3 Laplace Transforms: translations & unit step functions a lesson for MATH F302 Differential Equations Ed Bueler, Dept. of Mathematics and Statistics, UAF April 5, 2019 for textbook: D. Zill, A First Course in Differential Equations with Modeling Applications , 11th ed. 1 / 18

  2. the Laplace transform strategy direct ODE IVP for y ( t ) y ( t ) = . . . L L − 1 algebraic equation solve for Y Y ( s ) = . . . for Y ( s ) • § 7.3: “operational properties” regarding translations (shifts) ◦ including the unit step function U ( t ) • § 7.4 (next): “operational property” re convolution 2 / 18

  3. recall Laplace’s Transform • the definition: � ∞ e − st f ( t ) dt L { f ( t ) } = 0 • when applying L to an ODE use: � y ′ ( t ) � L = sY ( s ) − y (0) = s 2 Y ( s ) − sy (0) − y ′ (0) � y ′′ ( t ) � L • doing L − 1 is mostly use of a table, e.g.: � 1 � ◦ L − 1 = e 4 t s − 4 � � k = e at sin kt ◦ L − 1 ( s − a ) 2 + k 2 Pierre-Simon Laplace (1749–1827) 3 / 18

  4. we have a decent table 4 / 18

  5. noticable in the table • compare the left and right columns in this part of the table: L { t } = 1 1 � te at � L = s 2 ( s − a ) 2 n ! n ! L { t n } = � t n e at � L = s n +1 ( s − a ) n +1 k k e at sin( kt ) � � L { sin( kt ) } = L = s 2 + k 2 ( s − a ) 2 + k 2 s − a s e at cos( kt ) � � L { cos( kt ) } = L = s 2 + k 2 ( s − a ) 2 + k 2 • multiplying by e at causes: s → s − a • this is a rule!: multiplying by an exponential in t is translation in s : e at f ( t ) � � L = F ( s − a ) 5 / 18

  6. why? • why does multiplying by e at cause s → s − a ? • recall definition: � ∞ e − st f ( t ) dt L { f ( t ) } = F ( s ) = 0 • so: � ∞ � ∞ e at f ( t ) e − st e at f ( t ) dt = e − ( s − a ) t f ( t ) dt � � L = 0 0 = F ( s − a ) 6 / 18

  7. examples from § 7.3 • start by just going back and forth using the new rule • exercise 1. e 2 t sin(3 t ) � � L = • exercise 2. � � 1 L − 1 = s 2 − 6 s + 10 7 / 18

  8. example like § 7.3 #23 • exercise 3. use L to solve the ODE IVP: y ′′ +4 y ′ +4 y = 0 , y (0) = 1 , y ′ (0) = 1 8 / 18

  9. example like § 7.3 #30 • exercise 4. use L to solve the ODE IVP: y ′′ − 2 y ′ +5 y = t , y (0) = 0 , y ′ (0) = 7 9 / 18

  10. unit step function • definition. the unit step function is 1 � 0 , t < 0 U ( t ) = 1 , t ≥ 0 t • the book defines it with a translation, and only on [0 , ∞ ) � 0 , 0 ≤ t < a U ( t − a ) = 1 1 , t ≥ a • why? because we want to model t a “switching on” at time t = a 10 / 18

  11. U ( t − a ) helps with switching on/off write each function in terms of unit step function(s): • example A. � 0 , 0 ≤ t < 1 f ( t ) = t 2 , t ≥ 1 • example B. 11 / 18

  12. Laplace transform with U ( t − a ) • U ( t ) is also called the Heaviside function • easy-to-show: if F ( s ) = L { f ( t ) } then L { f ( t − a ) U ( t − a ) } = e − at F ( s ) Oliver Heaviside (1850–1925) • show it: 12 / 18

  13. #57 in § 7.3 • exercise 5. write the function in terms of U and then find the Laplace transform: � 0 , 0 ≤ t < 1 f ( t ) = t 2 , t ≥ 1 13 / 18

  14. second version • the book then says: We are frequently confronted with the problem of finding the Laplace transform of a product of a function g and a unit step function U ( t − a ) where the function g lacks the precise shifted form f ( t − a ) in Theorem 7.3.2. • yup, that’s our problem • 2nd form of the same rule: L { g ( t ) U ( t − a ) } = e − at L { g ( t + a ) } • it will be in the table also, when it is printed on quizzes/exams 14 / 18

  15. once again • exercise 5. write the function in terms of U and then find the Laplace transform: � 0 , 0 ≤ t < 1 f ( t ) = t 2 , t ≥ 1 15 / 18

  16. like #66 in § 7.3 • exercise 6. use Laplace transforms to solve the ODE IVP: y ′′ + 9 y = f ( t ) , y (0) = 0 , y ′ (0) = 0 � 1 , 0 ≤ t < 1 where f ( t ) = 0 , t ≥ 1 16 / 18

  17. summary • assume L { f ( t ) } = F ( s ) • 1st translation theorem . e at f ( t ) � � L = F ( s − a ) • 2nd translation theorem . if a > 0 then L { f ( t − a ) U ( t − a ) } = e − as F ( s ) L {U ( t − a ) } = e − as ◦ includes easy case: s ◦ second form L { g ( t ) U ( t − a ) } = e − as L { g ( t + a ) } • these are all in the table you will get on quizzes and exams, so: goal is understanding not memorizing 17 / 18

  18. expectations • just watching this video is not enough! ◦ see “found online” videos and stuff at bueler.github.io/math302/week12.html ◦ read sections 7.3 and 7.4 in the textbook • you can ignore “beams” and example 10 in § 7.3 • only 7.4.2 Transforms of Integrals in § 7.4 ◦ do the WebAssign exercises for section 7.3 • I will quiz on problems like these 18 / 18

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