Presentation is loading. Please wait.

Presentation is loading. Please wait.

Laplace Transforms: Special Cases Derivative Rule, Shift Rule, Gamma Function & f(ct) Rule MAT 275.

Similar presentations


Presentation on theme: "Laplace Transforms: Special Cases Derivative Rule, Shift Rule, Gamma Function & f(ct) Rule MAT 275."β€” Presentation transcript:

1 Laplace Transforms: Special Cases Derivative Rule, Shift Rule, Gamma Function & f(ct) Rule
MAT 275

2 (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Derivative Rule: If 𝐿 𝑓 𝑑 =𝐻(𝑠), then 𝐿 βˆ’π‘‘π‘“ 𝑑 =𝐻′(𝑠). Proof: Using the definition of the Laplace Transform, we have 𝐿 𝑓 𝑑 = 0 ∞ 𝑓 𝑑 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 . Differentiate both sides with respect to 𝑠: 𝑑 𝑑𝑠 𝐻 𝑠 = 𝑑 𝑑𝑠 𝐿 𝑓 𝑑 = 𝑑 𝑑𝑠 0 ∞ 𝑓 𝑑 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 . The integrand can be differentiated β€œin place” with respect to 𝑠. In this step, 𝑓(𝑑) acts as a constant multiplier. Note that 𝑑 𝑑𝑠 𝑒 βˆ’π‘ π‘‘ =βˆ’π‘‘ 𝑒 βˆ’π‘ π‘‘ : 𝑑 𝑑𝑠 0 ∞ 𝑓 𝑑 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 = 0 ∞ 𝑑 𝑑𝑠 𝑓 𝑑 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 = 0 ∞ βˆ’π‘‘ 𝑒 βˆ’π‘ π‘‘ 𝑓 𝑑 𝑑𝑑. Thus, 𝐻 β€² 𝑠 = 0 ∞ βˆ’π‘‘ 𝑒 βˆ’π‘ π‘‘ 𝑓 𝑑 𝑑𝑑=𝐿 βˆ’π‘‘π‘“ 𝑑 , where 𝐿 𝑓 𝑑 =𝐻(𝑠). Corollary: 𝐻 β€²β€² 𝑠 =𝐿 βˆ’π‘‘ βˆ’π‘‘ 𝑓 𝑑 =𝐿 𝑑 2 𝑓 𝑑 . In general, 𝐻 𝑛 𝑠 =𝐿 βˆ’π‘‘ 𝑛 𝑓 𝑑 . (c) ASU - SoMSS Scott Surgent. Report errors to

3 (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Example: Find 𝐿 𝑑 cos 𝑑 . Solution: First, find 𝐿 cos 𝑑 , which is 𝐿 cos 𝑑 = 𝑠 𝑠 Thus, 𝐻 𝑠 = 𝑠 𝑠 Differentiating 𝐻, we have 𝐻 β€² 𝑠 = 1βˆ’ 𝑠 𝑠 Since 𝐿 βˆ’π‘‘π‘“ 𝑑 =𝐻′(𝑠), then moving the negative, we have 𝐿 𝑑𝑓 𝑑 =βˆ’ 𝐻 β€² 𝑠 . Therefore, 𝐿 𝑑 cos 𝑑 =βˆ’ 1βˆ’ 𝑠 𝑠 = 𝑠 2 βˆ’1 𝑠 (c) ASU - SoMSS Scott Surgent. Report errors to

4 (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Example: Find 𝐿{𝑑 𝑒 2𝑑 } and 𝐿 𝑑 2 𝑒 2𝑑 . Solution: Start with 𝐻 𝑠 =𝐿 𝑒 2𝑑 = 1 π‘ βˆ’2 . Differentiate twice, so that 𝐻 β€² 𝑠 =βˆ’ 1 π‘ βˆ’ and 𝐻 β€²β€² 𝑠 = 2 π‘ βˆ’ So we have 𝐿 βˆ’π‘‘ 𝑒 2𝑑 =𝐻′(𝑠) or equivalently, 𝐿 𝑑 𝑒 2𝑑 =βˆ’ 𝐻 β€² 𝑠 =βˆ’ βˆ’ 1 π‘ βˆ’ = 1 π‘ βˆ’ Similarly, we have 𝐿 βˆ’π‘‘ 2 𝑒 2𝑑 =𝐿 𝑑 2 𝑒 2𝑑 = 𝐻 β€²β€² 𝑠 = 2 π‘ βˆ’ Example: Find 𝑦= 𝐿 βˆ’ 𝑠 Solution: Integrate twice, so we have 𝑠 𝑑𝑠 =βˆ’ 2 𝑠 , and βˆ’2 𝑠 𝑑𝑠 = 2 𝑠+1 . Thus, 𝐿 βˆ’1 2 𝑠+1 =2 𝑒 βˆ’π‘‘ . Since 𝐿 𝑑 2 𝑓 𝑑 =𝐻′′(𝑠), we have 𝑦= 𝐿 βˆ’ 𝑠 =2 𝑑 2 𝑒 βˆ’π‘‘ . (c) ASU - SoMSS Scott Surgent. Report errors to

5 (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Example: Use Laplace Transforms to solve 𝑦 β€²β€² βˆ’6 𝑦 β€² +9𝑦=0, 𝑦 0 =1, 𝑦 β€² 0 =0. Solution: We have 𝐿 𝑦 β€²β€² βˆ’6𝐿 𝑦 β€² +9𝐿 𝑦 =𝐿 0 𝑠 2 𝐿 𝑦 βˆ’π‘ π‘¦ 0 βˆ’ 𝑦 β€² 0 βˆ’6 𝑠𝐿 𝑦 βˆ’π‘¦ 0 +9𝐿 𝑦 =0 𝑠 2 𝐿 𝑦 βˆ’π‘ βˆ’6𝑠𝐿 𝑦 +6+9𝐿 𝑦 =0 𝐿 𝑦 𝑠 2 βˆ’6𝑦+9 =π‘ βˆ’6 𝐿 𝑦 = π‘ βˆ’6 𝑠 2 βˆ’6𝑠+9 . Thus, 𝑦= 𝐿 βˆ’1 π‘ βˆ’6 𝑠 2 βˆ’6𝑠+9 . Using partial fractions, we have π‘ βˆ’6 𝑠 2 βˆ’6𝑠+9 = 𝐴 π‘ βˆ’3 + 𝐡 π‘ βˆ’ The denominator has a linear factor multiplicity 2. Solving for 𝐴 and 𝐡, we get π‘ βˆ’6 𝑠 2 βˆ’6𝑠+9 = 1 π‘ βˆ’3 βˆ’ 3 π‘ βˆ’3 2 (c) ASU - SoMSS Scott Surgent. Report errors to

6 (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
The solution is 𝑦= 𝐿 βˆ’1 1 π‘ βˆ’3 βˆ’ 3 π‘ βˆ’ = 𝐿 βˆ’1 1 π‘ βˆ’3 + 𝐿 βˆ’1 βˆ’3 π‘ βˆ’ The first term’s inversion is 𝐿 βˆ’1 1 π‘ βˆ’3 = 𝑒 3𝑑 . For the second, recognize that βˆ’3 π‘ βˆ’ is the derivative of 3 π‘ βˆ’3 . Using the rule 𝐿 βˆ’π‘‘ 𝑒 2𝑑 =𝐻′(𝑠), we have 𝐿 βˆ’1 3 π‘ βˆ’3 =3 𝑒 3𝑑 , so that 𝐿 βˆ’1 βˆ’3 π‘ βˆ’ =βˆ’3𝑑 𝑒 3𝑑 . Remember to attach a negative because of the leading negative in 𝐿 βˆ’π‘‘ 𝑒 2𝑑 . The solution of 𝑦 β€²β€² βˆ’6 𝑦 β€² +9𝑦=0, 𝑦 0 =1, 𝑦 β€² 0 =0 is 𝑦= 𝑒 3𝑑 βˆ’3𝑑 𝑒 3𝑑 . (c) ASU - SoMSS Scott Surgent. Report errors to

7 (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Shift Rule: If 𝐿 𝑓 𝑑 =𝐻(𝑠), then 𝐿 𝑒 π‘Žπ‘‘ 𝑓 𝑑 =𝐻(π‘ βˆ’π‘Ž). Proof: We have 𝐿 𝑓 𝑑 = 0 ∞ 𝑓 𝑑 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 , so that 𝐿 𝑒 π‘Žπ‘‘ 𝑓 𝑑 = 0 ∞ 𝑒 π‘Žπ‘‘ 𝑓 𝑑 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 = 0 ∞ 𝑓 𝑑 𝑒 (π‘Žβˆ’π‘ )𝑑 𝑑𝑑 = 0 ∞ 𝑓 𝑑 𝑒 βˆ’(π‘ βˆ’π‘Ž)𝑑 𝑑𝑑 =𝐻 π‘ βˆ’π‘Ž . Example: Find 𝐿{𝑑 𝑒 2𝑑 } and 𝐿 𝑑 2 𝑒 2𝑑 . Solution: Start with 𝐿 𝑑 = 1 𝑠 2 and 𝐿 𝑑 2 = 2 𝑠 3 . The presence of 𝑒 2𝑑 suggest to shift both results by 2 units. Thus, 𝐿 𝑑 𝑒 2𝑑 = 1 π‘ βˆ’ and 𝐿 𝑑 2 𝑒 2𝑑 = 2 π‘ βˆ’2 3 (c) ASU - SoMSS Scott Surgent. Report errors to

8 (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Example: Find 𝐿 𝑒 βˆ’3𝑑 cos 5𝑑 . Solution: We have 𝐿 cos 5𝑑 = 𝑠 𝑠 Now, shift the result by βˆ’3 units: 𝐿 𝑒 βˆ’3𝑑 cos 5𝑑 = π‘ βˆ’(βˆ’3) π‘ βˆ’ βˆ’ = 𝑠+3 𝑠 Example: Find 𝐿 𝑑 𝑒 3𝑑 cos 5𝑑 . Solution: Start with 𝐻 𝑠 =𝐿 cos 5𝑑 = 𝑠 𝑠 Differentiate 𝐻, so we get 𝐻 β€² 𝑠 = 25βˆ’ 𝑠 𝑠 Since 𝐿 βˆ’π‘‘π‘“ 𝑑 =𝐻′(𝑠), we have 𝐿 𝑑 cos 𝑑 = 𝑠 2 βˆ’25 𝑠 (The negative was moved to the right side and distributed into the numerator). Then, the 𝑒 3𝑑 suggests to shift the result 3 units: 𝐿 𝑑 𝑒 3𝑑 cos 5𝑑 = (π‘ βˆ’3) 2 βˆ’25 ( π‘ βˆ’3) = 𝑠 2 βˆ’6π‘ βˆ’16 𝑠 2 βˆ’6𝑠 (c) ASU - SoMSS Scott Surgent. Report errors to

9 (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
The Gamma Function: If 𝑦= 𝑑 𝑛 , where 𝑛>βˆ’1, then 𝐿 𝑑 𝑛 = Ξ“ 𝑛+1 𝑠 𝑛+1 , where Ξ“ represents the gamma function. Proof: Using the Laplace Transform, we have 𝐿 𝑑 𝑛 = 0 ∞ 𝑑 𝑛 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 . Using integration by parts where 𝑒= 𝑑 𝑛 and 𝑑𝑣= 𝑒 βˆ’π‘ π‘‘ , we have 𝑑𝑒=𝑛 𝑑 π‘›βˆ’1 and 𝑣=βˆ’ 1 𝑠 𝑒 βˆ’π‘ π‘‘ . Thus, 𝐿 𝑑 𝑛 = 0 ∞ 𝑑 𝑛 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 = βˆ’ 1 𝑠 𝑑 𝑛 𝑒 βˆ’π‘ π‘‘ 0 ∞ βˆ’ 0 ∞ βˆ’ 1 𝑠 𝑒 βˆ’π‘ π‘‘ 𝑛 𝑑 π‘›βˆ’1 𝑑𝑑 . The term βˆ’ 1 𝑠 𝑑 𝑛 𝑒 βˆ’π‘ π‘‘ 0 ∞ =0 after evaluation. Simplified, we get 0 ∞ 𝑑 𝑛 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 = 𝑛 𝑠 0 ∞ 𝑑 π‘›βˆ’1 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 . This is the same as 𝐿 𝑑 𝑛 = 𝑛 𝑠 𝐿 𝑑 π‘›βˆ’1 . This can be extended, e.g. 𝐿 𝑑 𝑛 = 𝑛 𝑠 𝐿 𝑑 π‘›βˆ’1 = 𝑛 𝑠 π‘›βˆ’1 𝑠 𝐿 𝑑 π‘›βˆ’2 , or 𝐿 𝑑 𝑛 = 𝑛(π‘›βˆ’1) 𝑠 2 𝐿 𝑑 π‘›βˆ’2 , and so on. (c) ASU - SoMSS Scott Surgent. Report errors to

10 (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
From the last slide, we showed that 𝐿 𝑑 𝑛 = 𝑛 𝑠 𝐿 𝑑 π‘›βˆ’1 = 𝑛 π‘›βˆ’1 𝑠 2 𝐿 𝑑 π‘›βˆ’2 and so on. If n is an integer such that 𝑛β‰₯0, then this formula is the same as 𝐿 𝑑 𝑛 = 𝑛! 𝑠 𝑛+1 (Try it for n = 3, for example). If n is not an integer such that 𝑛>βˆ’1, then the numerator is given by the gamma function Ξ“(𝑛+1), where Ξ“ 𝑛+1 = 0 ∞ 𝑑 𝑛 𝑒 βˆ’π‘‘ 𝑑𝑑 . The gamma function has the property that when n is a non-negative integer, then Ξ“ 𝑛+1 =𝑛!, but it β€œextends” the notion of factorial to include non-integers greater that –1. The integral 0 ∞ 𝑑 𝑛 𝑒 βˆ’π‘‘ 𝑑𝑑 is usually evaluated using numerical methods (e.g. a calculator). Example: Find 𝐿 𝑑 . Solution: We have 𝐿 𝑑 =𝐿 𝑑 = Ξ“ 𝑠 = Ξ“ 𝑠 β‰ˆ 𝑠 Using the relationship Ξ“ 𝑛+1 =𝑛!, then Ξ“ = !, which is found by evaluating 0 ∞ 𝑑 𝑒 βˆ’π‘‘ 𝑑𝑑 . A calculator shows that this is about Using substitutions and integrating in polar coordinates, it can be shown that != 0 ∞ 𝑑 𝑒 βˆ’π‘‘ 𝑑𝑑 = πœ‹ . Try it on your calculator! (c) ASU - SoMSS Scott Surgent. Report errors to

11 (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
The 𝒇 𝒄𝒕 Rule: If 𝐻 𝑠 =𝐿 𝑓 𝑑 , then 𝐿 𝑓 𝑐𝑑 = 1 𝑐 𝐻 𝑠 𝑐 . Proof: Start with 𝐿 𝑓 𝑐𝑑 = 0 ∞ 𝑓 𝑐𝑑 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 . Let 𝑒=𝑐𝑑 so that 𝑑= 𝑒 𝑐 and 𝑑𝑑= 𝑑𝑒 𝑐 . Make the substitutions: 𝐿 𝑓 𝑐𝑑 = 0 ∞ 𝑓 𝑐𝑑 𝑒 βˆ’π‘ π‘‘ 𝑑𝑑 = 0 ∞ 𝑓 𝑒 𝑒 βˆ’π‘  𝑒 𝑐 𝑑𝑒 𝑐 = 1 𝑐 0 ∞ 𝑓 𝑒 𝑒 βˆ’ 𝑠 𝑐 𝑒 𝑑𝑒 = 1 𝑐 𝐻 𝑠 𝑐 . Example: Find 𝐿 sin 2𝑑 using the above property. Solution: Starting with 𝐿 sin 𝑑 = 1 𝑠 and observing that 𝑐=2, we have 𝐿 sin 2𝑑 = 1 2 β‹… 1 𝑠 = 1 2 β‹… 1 𝑠 = 1 2 β‹… 4 𝑠 = 2 𝑠 (c) ASU - SoMSS Scott Surgent. Report errors to


Download ppt "Laplace Transforms: Special Cases Derivative Rule, Shift Rule, Gamma Function & f(ct) Rule MAT 275."

Similar presentations


Ads by Google