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Unformatted text preview: HIGHERORDER 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 InitialValue 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 initialvalue problem for a linear ordinary differential equation with constant coefficients into a linear algebaric equation that can be easily solved. The solution of an initialvalue problem can then be obtained from the solution of the algebaric equation by taking its socalled 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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 Spring '10
 LEVERMORE
 Laplace

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