Chapter 4 Lecture Notes-Laplace Transforms 1

# Chapter 4 Lecture Notes-Laplace Transforms 1 - McGill...

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McGill University Math 263: Differential Equations and Linear Algebra CHAPTER 4: LAPLACE TRANSFORMS (I) 1 Introduction We begin our study of the Laplace Transform with a motivating example: Solve the differential equation y 00 + y = f ( t ) = 0 , 0 t < 10 , 1 , 10 t < 10 + 2 π, 0 , 10 + 2 π t. with IC’s: y (0) = 0 , y 0 (0) = 0 (1) Here, f ( t ) is piecewise continuous and any solution would also have y 00 piecewise continuous . To describe the motion of the ball using techniques previously developed we have to divide the problem into three parts: 0-0

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(I) 0 t < 10 ; (II) 10 t < 10 + 2 π ; (III) 10 + 2 π t . (I). The initial value problem determining the motion in part I is y 00 + y = 0 , y (0) = y 0 (0) = 0 . The solution is y ( t ) = 0 , 0 t < 10 . Taking limits as t 10 from the left, we find y (10) = y 0 (10) = 0 . (II). The initial value problem determining the motion in part II is y 00 + y = 1 , y (10) = y 0 (10) = 0 . The solution is y ( t ) = 1 - cos( t - 10) , 10 t < 2 π + 10 . Taking limits as t 10 + 2 π from the left, we get y (10 + 2 π ) = y 0 (10 + 2 π ) = 0 . 0-1
(III). The initial value problem for the last part is y 00 + y = 0 , y (10 + 2 π ) = y 0 (10 + 2 π ) = 0 which has the solution y ( t ) = 0 , 10 + 2 π t < . Putting all this together, we have y ( t ) = 0 , 0 t < 10 , 1 - cos( t - 10) , 10 t < 10 + 2 π, 0 , 10 + 2 π t. The function y ( t ) is continuous with continuous derivative y 0 ( t ) = 0 , 0 t < 10 , sin( t - 10) , 10 t < 10 + 2 π, 0 , 10 + 2 π t. However the function y 0 ( t ) is not differentiable at t = 10 and t = 10 + 2 π . In fact y 00 ( t ) = 0 , 0 t < 10 , cos( t - 10) , 10 < t < 10 + 2 π, 0 , 10 + 2 π < t. The left-hand and right-hand limits of y 00 ( t ) at t = 10 are 0 and 1 respectively. At t = 10 + 2 π they are 1 and 0 .

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