7.3 Laplace Transforms: translations & unit step functions a - - PowerPoint PPT Presentation

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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


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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.

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the Laplace transform strategy

ODE IVP for y(t) algebraic equation for Y (s) Y (s) = . . . y(t) = . . . L solve for Y L−1

direct

  • §7.3: “operational properties” regarding translations (shifts)
  • including the unit step function U(t)
  • §7.4 (next): “operational property” re convolution

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recall Laplace’s Transform

  • the definition:

L {f (t)} = ∞ e−stf (t) dt

  • when applying L to an ODE use:

L

  • y′(t)
  • = sY (s) − y(0)

L

  • y′′(t)
  • = s2Y (s) − sy(0) − y′(0)
  • doing L−1 is mostly use of a table, e.g.:
  • L−1
  • 1

s − 4

  • = e4t
  • L−1
  • k

(s − a)2 + k2

  • = eat sin kt

Pierre-Simon Laplace (1749–1827) 3 / 18

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we have a decent table

4 / 18

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noticable in the table

  • compare the left and right columns in this part of the table:

L {t} = 1 s2 L

  • teat

= 1 (s − a)2 L {tn} = n! sn+1 L

  • tneat

= n! (s − a)n+1 L {sin(kt)} = k s2 + k2 L

  • eat sin(kt)
  • =

k (s − a)2 + k2 L {cos(kt)} = s s2 + k2 L

  • eat cos(kt)
  • =

s − a (s − a)2 + k2

  • multiplying by eat causes:

s → s − a

  • this is a rule!: multiplying by an exponential in t is translation in s:

L

  • eatf (t)
  • = F(s − a)

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why?

  • why does multiplying by eat cause

s → s − a ?

  • recall definition:

L {f (t)} = F(s) = ∞ e−stf (t) dt

  • so:

L

  • eatf (t)
  • =

∞ e−steatf (t) dt = ∞ e−(s−a)tf (t) dt = F(s − a)

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examples from §7.3

  • start by just going back and forth using the new rule
  • exercise 1.

L

  • e2t sin(3t)
  • =
  • exercise 2.

L−1

  • 1

s2 − 6s + 10

  • =

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example like §7.3 #23

  • exercise 3. use L to solve the ODE IVP:

y′′+4y′+4y = 0, y(0) = 1, y′(0) = 1

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example like §7.3 #30

  • exercise 4. use L to solve the ODE IVP:

y′′−2y′+5y = t, y(0) = 0, y′(0) = 7

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unit step function

  • definition. the unit step function is

U(t) =

  • 0,

t < 0 1, t ≥ 0

  • the book defines it with a translation,

and only on [0, ∞) U(t − a) =

  • 0,

0 ≤ t < a 1, t ≥ a

  • why? because we want to model

“switching on” at time t = a

t 1 t 1

a

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U(t − a) helps with switching on/off

write each function in terms of unit step function(s):

  • example A.

f (t) =

  • 0,

0 ≤ t < 1 t2, t ≥ 1

  • example B.

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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−atF(s)

  • show it:

Oliver Heaviside (1850–1925) 12 / 18

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#57 in §7.3

  • exercise 5. write the function in terms of U and then find the

Laplace transform: f (t) =

  • 0,

0 ≤ t < 1 t2, t ≥ 1

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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−atL {g(t + a)}

  • it will be in the table also, when it is printed on quizzes/exams

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SLIDE 15
  • nce again
  • exercise 5. write the function in terms of U and then find the

Laplace transform: f (t) =

  • 0,

0 ≤ t < 1 t2, t ≥ 1

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like #66 in §7.3

  • exercise 6. use Laplace transforms to solve the ODE IVP:

y′′ + 9y = f (t), y(0) = 0, y′(0) = 0 where f (t) =

  • 1,

0 ≤ t < 1 0, t ≥ 1

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summary

  • assume L {f (t)} = F(s)
  • 1st translation theorem.

L

  • eatf (t)
  • = F(s − a)
  • 2nd translation theorem. if a > 0 then

L {f (t − a)U(t − a)} = e−asF(s)

  • includes easy case:

L {U(t − a)} = e−as

s

  • second form

L {g(t)U(t − a)} = e−asL {g(t + a)}

  • these are all in the table you will get on quizzes and exams, so:

goal is understanding not memorizing

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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

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