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Laplace Transforms: Special Cases Derivative Rule, Shift Rule, Gamma Function & f(ct) Rule
MAT 275
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(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Derivative Rule: If πΏ π π‘ =π»(π ), then πΏ βπ‘π π‘ =π»β²(π ). Proof: Using the definition of the Laplace Transform, we have πΏ π π‘ = 0 β π π‘ π βπ π‘ ππ‘ . Differentiate both sides with respect to π : π ππ π» π = π ππ πΏ π π‘ = π ππ 0 β π π‘ π βπ π‘ ππ‘ . The integrand can be differentiated βin placeβ with respect to π . In this step, π(π‘) acts as a constant multiplier. Note that π ππ π βπ π‘ =βπ‘ π βπ π‘ : π ππ 0 β π π‘ π βπ π‘ ππ‘ = 0 β π ππ π π‘ π βπ π‘ ππ‘ = 0 β βπ‘ π βπ π‘ π π‘ ππ‘. Thus, π» β² π = 0 β βπ‘ π βπ π‘ π π‘ ππ‘=πΏ βπ‘π π‘ , where πΏ π π‘ =π»(π ). Corollary: π» β²β² π =πΏ βπ‘ βπ‘ π π‘ =πΏ π‘ 2 π π‘ . In general, π» π π =πΏ βπ‘ π π π‘ . (c) ASU - SoMSS Scott Surgent. Report errors to
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(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Example: Find πΏ π‘ cos π‘ . Solution: First, find πΏ cos π‘ , which is πΏ cos π‘ = π π Thus, π» π = π π Differentiating π», we have π» β² π = 1β π π Since πΏ βπ‘π π‘ =π»β²(π ), then moving the negative, we have πΏ π‘π π‘ =β π» β² π . Therefore, πΏ π‘ cos π‘ =β 1β π π = π 2 β1 π (c) ASU - SoMSS Scott Surgent. Report errors to
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(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Example: Find πΏ{π‘ π 2π‘ } and πΏ π‘ 2 π 2π‘ . Solution: Start with π» π =πΏ π 2π‘ = 1 π β2 . Differentiate twice, so that π» β² π =β 1 π β and π» β²β² π = 2 π β So we have πΏ βπ‘ π 2π‘ =π»β²(π ) or equivalently, πΏ π‘ π 2π‘ =β π» β² π =β β 1 π β = 1 π β Similarly, we have πΏ βπ‘ 2 π 2π‘ =πΏ π‘ 2 π 2π‘ = π» β²β² π = 2 π β Example: Find π¦= πΏ β π Solution: Integrate twice, so we have π ππ =β 2 π , and β2 π ππ = 2 π +1 . Thus, πΏ β1 2 π +1 =2 π βπ‘ . Since πΏ π‘ 2 π π‘ =π»β²β²(π ), we have π¦= πΏ β π =2 π‘ 2 π βπ‘ . (c) ASU - SoMSS Scott Surgent. Report errors to
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(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Example: Use Laplace Transforms to solve π¦ β²β² β6 π¦ β² +9π¦=0, π¦ 0 =1, π¦ β² 0 =0. Solution: We have πΏ π¦ β²β² β6πΏ π¦ β² +9πΏ π¦ =πΏ 0 π 2 πΏ π¦ βπ π¦ 0 β π¦ β² 0 β6 π πΏ π¦ βπ¦ 0 +9πΏ π¦ =0 π 2 πΏ π¦ βπ β6π πΏ π¦ +6+9πΏ π¦ =0 πΏ π¦ π 2 β6π¦+9 =π β6 πΏ π¦ = π β6 π 2 β6π +9 . Thus, π¦= πΏ β1 π β6 π 2 β6π +9 . Using partial fractions, we have π β6 π 2 β6π +9 = π΄ π β3 + π΅ π β The denominator has a linear factor multiplicity 2. Solving for π΄ and π΅, we get π β6 π 2 β6π +9 = 1 π β3 β 3 π β3 2 (c) ASU - SoMSS Scott Surgent. Report errors to
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(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
The solution is π¦= πΏ β1 1 π β3 β 3 π β = πΏ β1 1 π β3 + πΏ β1 β3 π β The first termβs inversion is πΏ β1 1 π β3 = π 3π‘ . For the second, recognize that β3 π β is the derivative of 3 π β3 . Using the rule πΏ βπ‘ π 2π‘ =π»β²(π ), we have πΏ β1 3 π β3 =3 π 3π‘ , so that πΏ β1 β3 π β =β3π‘ π 3π‘ . Remember to attach a negative because of the leading negative in πΏ βπ‘ π 2π‘ . The solution of π¦ β²β² β6 π¦ β² +9π¦=0, π¦ 0 =1, π¦ β² 0 =0 is π¦= π 3π‘ β3π‘ π 3π‘ . (c) ASU - SoMSS Scott Surgent. Report errors to
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(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Shift Rule: If πΏ π π‘ =π»(π ), then πΏ π ππ‘ π π‘ =π»(π βπ). Proof: We have πΏ π π‘ = 0 β π π‘ π βπ π‘ ππ‘ , so that πΏ π ππ‘ π π‘ = 0 β π ππ‘ π π‘ π βπ π‘ ππ‘ = 0 β π π‘ π (πβπ )π‘ ππ‘ = 0 β π π‘ π β(π βπ)π‘ ππ‘ =π» π βπ . Example: Find πΏ{π‘ π 2π‘ } and πΏ π‘ 2 π 2π‘ . Solution: Start with πΏ π‘ = 1 π 2 and πΏ π‘ 2 = 2 π 3 . The presence of π 2π‘ suggest to shift both results by 2 units. Thus, πΏ π‘ π 2π‘ = 1 π β and πΏ π‘ 2 π 2π‘ = 2 π β2 3 (c) ASU - SoMSS Scott Surgent. Report errors to
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(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
Example: Find πΏ π β3π‘ cos 5π‘ . Solution: We have πΏ cos 5π‘ = π π Now, shift the result by β3 units: πΏ π β3π‘ cos 5π‘ = π β(β3) π β β = π +3 π Example: Find πΏ π‘ π 3π‘ cos 5π‘ . Solution: Start with π» π =πΏ cos 5π‘ = π π Differentiate π», so we get π» β² π = 25β π π Since πΏ βπ‘π π‘ =π»β²(π ), we have πΏ π‘ cos π‘ = π 2 β25 π (The negative was moved to the right side and distributed into the numerator). Then, the π 3π‘ suggests to shift the result 3 units: πΏ π‘ π 3π‘ cos 5π‘ = (π β3) 2 β25 ( π β3) = π 2 β6π β16 π 2 β6π (c) ASU - SoMSS Scott Surgent. Report errors to
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(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
The Gamma Function: If π¦= π‘ π , where π>β1, then πΏ π‘ π = Ξ π+1 π π+1 , where Ξ represents the gamma function. Proof: Using the Laplace Transform, we have πΏ π‘ π = 0 β π‘ π π βπ π‘ ππ‘ . Using integration by parts where π’= π‘ π and ππ£= π βπ π‘ , we have ππ’=π π‘ πβ1 and π£=β 1 π π βπ π‘ . Thus, πΏ π‘ π = 0 β π‘ π π βπ π‘ ππ‘ = β 1 π π‘ π π βπ π‘ 0 β β 0 β β 1 π π βπ π‘ π π‘ πβ1 ππ‘ . The term β 1 π π‘ π π βπ π‘ 0 β =0 after evaluation. Simplified, we get 0 β π‘ π π βπ π‘ ππ‘ = π π 0 β π‘ πβ1 π βπ π‘ ππ‘ . This is the same as πΏ π‘ π = π π πΏ π‘ πβ1 . This can be extended, e.g. πΏ π‘ π = π π πΏ π‘ πβ1 = π π πβ1 π πΏ π‘ πβ2 , or πΏ π‘ π = π(πβ1) π 2 πΏ π‘ πβ2 , and so on. (c) ASU - SoMSS Scott Surgent. Report errors to
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(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
From the last slide, we showed that πΏ π‘ π = π π πΏ π‘ πβ1 = π πβ1 π 2 πΏ π‘ πβ2 and so on. If n is an integer such that πβ₯0, then this formula is the same as πΏ π‘ π = π! π π+1 (Try it for n = 3, for example). If n is not an integer such that π>β1, then the numerator is given by the gamma function Ξ(π+1), where Ξ π+1 = 0 β π‘ π π βπ‘ ππ‘ . The gamma function has the property that when n is a non-negative integer, then Ξ π+1 =π!, but it βextendsβ the notion of factorial to include non-integers greater that β1. The integral 0 β π‘ π π βπ‘ ππ‘ is usually evaluated using numerical methods (e.g. a calculator). Example: Find πΏ π‘ . Solution: We have πΏ π‘ =πΏ π‘ = Ξ π = Ξ π β π Using the relationship Ξ π+1 =π!, then Ξ = !, which is found by evaluating 0 β π‘ π βπ‘ ππ‘ . A calculator shows that this is about Using substitutions and integrating in polar coordinates, it can be shown that != 0 β π‘ π βπ‘ ππ‘ = π . Try it on your calculator! (c) ASU - SoMSS Scott Surgent. Report errors to
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(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu
The π ππ Rule: If π» π =πΏ π π‘ , then πΏ π ππ‘ = 1 π π» π π . Proof: Start with πΏ π ππ‘ = 0 β π ππ‘ π βπ π‘ ππ‘ . Let π’=ππ‘ so that π‘= π’ π and ππ‘= ππ’ π . Make the substitutions: πΏ π ππ‘ = 0 β π ππ‘ π βπ π‘ ππ‘ = 0 β π π’ π βπ π’ π ππ’ π = 1 π 0 β π π’ π β π π π’ ππ’ = 1 π π» π π . Example: Find πΏ sin 2π‘ using the above property. Solution: Starting with πΏ sin π‘ = 1 π and observing that π=2, we have πΏ sin 2π‘ = 1 2 β
1 π = 1 2 β
1 π = 1 2 β
4 π = 2 π (c) ASU - SoMSS Scott Surgent. Report errors to
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