Laplace Transforms: Special Cases - PowerPoint Presentation

Slide1

Laplace Transforms: Special CasesDerivative Rule, Shift Rule, Gamma Function & f(ct) Rule

MAT 275Slide2

Derivative Rule: If

, then

.Proof: Using the definition of the Laplace Transform, we have .Differentiate both sides with respect to : The integrand can be differentiated “in place” with respect to . In this step, acts as a constant multiplier. Note that :Thus, .Corollary: . In general, .

(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu

2Slide3

Example: Find

.

Solution: First, find , which is . Thus, .Differentiating , we have .Since , then moving the negative, we have .Therefore, (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu3Slide4

Example: Find

and

.Solution: Start with . Differentiate twice, so that and .So we have or equivalently, .Similarly, we have .Example: Find .Solution: Integrate twice, so we have , and . Thus, . Since , we have

.

(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu

4Slide5

Example: Use Laplace Transforms to solve

.

Solution: We haveThus, . Using partial fractions, we have . The denominator has a linear factor multiplicity 2. Solving for and , we get  (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu5Slide6

The solution is

.

The first term’s inversion is For the second, recognize that is the derivative of . Using the rule , we have , so that . Remember to attach a negative because of the leading negative in .The solution of is (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu6Slide7

Shift Rule: If

, then

.Proof: We have , so thatExample: Find and .Solution: Start with and .The presence of suggest to shift both results by 2 units. Thus, (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu7Slide8

Example: Find

.

Solution: We have . Now, shift the result by units:Example: Find .Solution: Start with . Differentiate , so we get . Since , we have . (The negative was moved to the right side and distributed into the numerator).Then, the suggests to shift the result 3 units: . (c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu8Slide9

(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu9

The Gamma Function:

If , where , then , where represents the gamma function.Proof: Using the Laplace Transform, we have . Using integration by parts where and , we have and . Thus,The term after evaluation. Simplified, we get . This is the same as . This can be extended, e.g.

, or

, and so on.

 Slide10

(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu10

From the last slide, we showed that

and so on. If n is an integer such that , then this formula is the same as (Try it for n = 3, for example).If n is not an integer such that , then the numerator is given by the gamma function , where . The gamma function has the property that when n is a non-negative integer, then , but it “extends” the notion of factorial to include non-integers greater that –1. The integral is usually evaluated using numerical methods (e.g. a calculator).Example: Find .Solution: We have . Using the relationship , then , which is found by evaluating . A calculator shows that this is about 0.88623. Using substitutions and integrating in polar coordinates, it can be shown that . Try it on your calculator!

 Slide11

(c) ASU - SoMSS Scott Surgent. Report errors to surgent@asu.edu11

The

Rule: If , then .Proof: Start with . Let so that and . Make the substitutions:Example: Find using the above property.Solution: Starting with and observing that , we have