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Handout: The Laplace Transform - First Year Interest Group Seminar | N 1, Study notes of Health sciences

Material Type: Notes; Class: FIRST-YEAR INTEREST GROUP SMNR; Subject: Nursing; University: University of Texas - Austin; Term: Spring 2006;

Typology: Study notes

Pre 2010

Uploaded on 08/30/2009

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Download Handout: The Laplace Transform - First Year Interest Group Seminar | N 1 and more Study notes Health sciences in PDF only on Docsity! M427K Handout: The Laplace Transform Salman Butt March 7, 2006 Notes 1. If you have not received your midterm back, come see me in office hours to pick it up. 2. Do not procrastinate about dropping by office hours. If too many people show up at office hours during the last week of class and I do not get a chance to talk to you, you will not be marked as having come to office hours. 3. Please pickup your old homework assignments from the box attached to my office door (RLM 11.114). 4. Office hours this Friday are cancelled. Series Solutions Let’s do an example to stretch a bit: determine the indicial equation, its roots, and two recur- rence relations for the following differential equation: 2xy′′ + y′ + x2y = 0 1 The Laplace Transform The Laplace transform is a technique used to solve second-order ODEs (it actually generalizes to higher order) by recasting the problem into a new variable. The central idea is to convert your differential equation into an algebraic one using an integral, solving this resulting (rather elementary) algebraic equation, and finally inverting your algebraic solution to get a solution to the differential equation. There are some technical details involved with using the Laplace transform. I will cover these mostly in lecture tomorrow. Today I want to instead talk about the critical properties to know about Laplace transforms – the properties you will need to solve differential equations. We have the following defintion: Definition 1. If the following integral ∫ ∞ 0 estf(t)dt (1) exists (i.e. it converges), we call this integral the Laplace transform of f , denoted L{f(t)} (s) = F (s). The Laplace transform is a special case of the more general construction of integral transforms. Generally, and integral transform of a function f(t) is gotten by integrating f against some function K(s, t), sometimes referred to as a kernel, across some interval from α to β:∫ β α K(s, t)f(t)dt Another such transform is the Fourier transform which we may be seeing later in this class. The convergence of (1) is a matter of concern and is discussed in Theorem 6.1.2. on p. 310 of the text. We will not be directly concerned with the existence of the integral today, but this is not a trivial matter and should not be overlooked. One of the critical properties of the Laplace transform is that it is a linear functional (or a linear operator), so The Laplace transform also obeys a simple identity when it comes to derivatives of a given function: Proposition 1. Suppose f, f ′, . . . , f (n−1) are continuous and f (n) is piecewise continuous on any interval 0 ≤ t ≤ A and |f(t)| ≤ Keat, |f ′(t)| ≤ Keat, . . . , |f (n−1)(t)| ≤ Keat for t ≥ M where K, a, M are constants and K, M are positive. Then L { f (n)(t) } (s) exists for s > a and is given by L { f (n)(t) } (s) = . In particular, we have the following identities: L { f ′(t) } (s) = , L { f ′′(t) } (s) = . A key idea to keep in mind is that there is a one-to-one correspondence between continuous functions f(t) and their Laplace transforms L{f(t)} (s) = F (s). Thus we are confident that our 2
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