HigherLin4 - HIGHER-ORDER LINEAR ORDINARY DIFFERENTIAL...

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Unformatted text preview: HIGHER-ORDER LINEAR ORDINARY DIFFERENTIAL EQUATIONS IV: Laplace Transform Method David Levermore Department of Mathematics University of Maryland 21 June 2009 Because the presentation of this material in class will differ from that in the book, I felt that notes that closely follow the class presentation might be appreciated. 8. Laplace Transform Method 8.1. Definition of the Transform 2 8.2. Properties of the Transform 3 8.3. Existence and Differentiability 6 8.4. Transform of Derivatives and Initial-Value Problems 9 8.5. Piecewise Defined Forcing 12 8.6. Inverse Transform 14 8.7. Green Functions 19 8.8. Convolutions (not covered) 20 8.9. Natural Fundamental Sets (not covered) 23 1 2 8. Laplace Transform Method The Laplace transform will allow us to transform an initial-value problem for a linear ordinary differential equation with constant coefficients into a linear algebaric equation that can be easily solved. The solution of an initial-value problem can then be obtained from the solution of the algebaric equation by taking its so-called inverse Laplace transform. 8.1: Definition of the Transform. The Laplace transform of a function f ( t ) defined over t 0 is another function L [ f ]( s ) that is formally defined by L [ f ]( s ) = integraldisplay e st f ( t ) d t . (8.1) You should recall from calculus that the above definite integral is improper because its upper endpoint is . The proper definition of the Laplace transform is therefore L [ f ]( s ) = lim T integraldisplay T e st f ( t ) d t , (8.2) provided that the definite integrals over [0 , T ] appearing in the above limit are proper. The Laplace transform L [ f ]( s ) is defined only at those s for which the limit in (8.2) exists. Example. Use definition (8.2) to compute L [ e at ]( s ) for any real a . From (8.2) you see that for any s negationslash = a one has L [ e at ]( s ) = lim T integraldisplay T e st e at d t = lim T integraldisplay T e ( a s ) t d t = lim T e ( a s ) t a s vextendsingle vextendsingle vextendsingle vextendsingle T t =0 = lim T bracketleftbigg 1 s a e ( a s ) T s a bracketrightbigg = 1 s a for s > a, for s < a, while for s = a one has L [ e at ]( s ) = lim T integraldisplay T e ( s a ) t d t = lim T integraldisplay T d t = lim T T = . Therefore L [ e at ]( s ) is only defined for s > a with L [ e at ]( s ) = 1 s a for s > a. 3 Example. Use definition (8.2) to compute L [ t e at ]( s ) for any real a . From (8.2) you see that for any s negationslash = a one has L [ e at ]( s ) = lim T integraldisplay T t e st e at d t = lim T integraldisplay T t e ( a s ) t d t = lim T parenleftbigg t a s 1 ( a s ) 2 parenrightbigg e ( a s ) t vextendsingle vextendsingle vextendsingle vextendsingle T t =0 = lim T bracketleftbigg 1 ( s a ) 2 parenleftbigg T s a + 1 ( s a ) 2...
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This note was uploaded on 01/24/2011 for the course MATH math246 taught by Professor Levermore during the Spring '10 term at Maryland.

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HigherLin4 - HIGHER-ORDER LINEAR ORDINARY DIFFERENTIAL...

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