{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Laplace - Mathematics 421 Essay 1 Using the Laplace...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Mathematics 421 Essay 1 Using the Laplace transform Spring 2008 0. Introduction The Laplace transform of a function of t is a function of a new variable s defined by ˇ f f .t/ g D Z 1 0 f .t/e st dt This is an improper integral, so convergence must be considered. Typically, the integral will exist only for sufficiently large s , but explicit consideration of this restriction is usually not necessary. The functions that we will transform are covered by an existence theorem that guarantees that the integral exists for s > a for a piecewise continuous function f .t/ with j f .t/ j < Ke at . In section 4.1 of the textbook, an example was given in which this definition was easy to use: ˇ ˚ e at D 1 s a : The special case with a D 0 should be noted. However, it will turn out that even these examples are consequences of general properties of the transform, so that the definition will only be used to derive general properties. In stating these properties, only the simplest version will be shown; repeated application will be done as needed rather than used to state results with proofs requiring mathematical induction . Since the main applications involve very few steps , nothing is gained in most cases by pretending that there is a general formula. 1. Linearity The most important property of the Laplace transform is linearity . This is a direct consequence of the linearity of integration. The basic statements are ˇ f f .t/ g D F.s/ ń ˇ f c f .t/ g D c F.s/ ˇ f f .t/ g D F.s/ and ˇ f g.t/ g D G.s/ ń ˇ f f .t/ C g.t/ g D F.s/ C G.s/ Repeated use of this rule deals with a sum of arbitrarily many terms, each of which is a product of a constant and a known function. The generalization to such expressions has been common since the first course in algebra. Such general expressions are called linear combinations of the known functions. In addition to determining transforms, it will be necessary to find inverse transforms. Thus, any func- tion that can be written as a linear combination of 1=.s a/ can be recognized as the Laplace transform of a linear combination of e at . The method of partial fractions produces such an expression from some quotients of polynomials. Quotients of polynomials are called rational functions ; and a rational function is called proper if the degree of the numerator is strictly smaller than the degree of the denominator. The functions that are Laplace transforms of linear combinations of exponentials are proper rational functions whose denominator is a product of distinct factors of the form x a . In the first course on Differential Equations, solutions of linear differential equations with constant coefficients were found by assuming a solution of the form y D e at . Some equations had solutions that were trigonometric functions, and these could be found using Euler’s identity e it D cos t C i sin t . This leads to cos t D e it C e it 2 and sin t D e it e it 2i
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Mathematics 421 Essay 1, p. 2 If we accept these formulas, then ˇ f cos t g D 1 2 1 s i C 1 s C i D s s 2 C 1 ˇ f sin t g D 1 2i 1 s i 1 s C i D 1 s 2 C 1
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern