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26 laplace transform I

26 laplace transform I - 18.03 Lecture#26 Nov 6 2009 notes...

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18.03 Lecture #26 Nov. 6, 2009: notes Topics for today are (first) Duhamel’s principle (meant to be on Wednesday’s lecture, but not covered then) and (second) an introduction to the Laplace transform. For the first topic, here is an enlarged version of the last theorem introduced on Wednesday. Theorem. Suppose f and y are (possibly generalized) functions vanishing for t < 0 , and p is any polynomial. Then (writing D = d dt as usual) we have p ( D )( f y ) = f ( p ( D ) y ) . ( Differentiating convolutions ) Suppose in particular that y is a solution of a differential equation p ( D ) y = g, with g vanishing for t < 0 . Then f y is a solution of the differential equation p ( D )( f y ) = f g. The proof is (formally) very easy: to differentiate ( f y ) = integraldisplay −∞ f ( s ) y ( t s ) ds with respect to t , you just differentiate y under the integral sign; that’s the only place that t appears. Corollary (Duhamel’s principle—see EP 4.6). Suppose w is a (generalized) function satisfying the differential equation p ( D ) w = δ 0 , w ( t ) = 0 ( t < 0) . Then if f is any (generalized) function vanishing for t < 0 , the solution of the differential equation p ( D ) y = f, y ( t ) = 0 ( t < 0) ( Forced ODE for positive time ) is given by y ( t ) = ( f w )( t ) = integraldisplay t 0 f ( s ) w ( t s ) ds ( t 0) . ( Duhamel’s formula ) Definition. A weight function for the differential equation (Forced ODE for positive time) is a solution w of p ( D ) w = δ 0 , w ( t ) = 0 ( t < 0) . (Weight function for p ) Duhamel’s principle says that calculating weight functions is very important for solving forced differential equations, now with arbitrary input f that begins at time zero: the solution just requires computing the single definite integral in Duhamel’s formula. Here’s a way to find the weight function. (We’ll see another way to think about weight functions using the Laplace transform.) 1
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Theorem (fundamental solution of constant coefficient linear ODE). Suppose p is a polynomial of degree n with leading coefficient 1 . Then the solution w to the ODE (Weight function for p ) may be calculated as follows. For t < 0 , w ( t ) = 0 . For t 0 , w ( t ) is equal to the solution to the homogeneous differential equation p ( D ) y δ = 0 satisfying y δ (0) = 0 , y δ (0) = 0 , . . . , y ( n 1) δ (0) = 1 . ( Delta initial conditions ) The function w has n 2 continuous derivatives; the n 1 st derivative jumps from 0 to 1 at t = 0 . Proof. Let y 0 be the response to the unit step input discussed on Monday: p ( D ) y 0 = u 0 , y 0 ( t ) = 0 ( t < 0) . Differentiating this equation with respect to t gives δ 0 on the right (by (Derivative of Heaviside step function), so p ( D ) y 0 = δ 0 , y 0 ( t ) = 0 ( t < 0) . That is, y 0 is the weight function that we want. For t < 0, y 0 is certainly 0. For t 0, y 0 satisfies p ( D ) y 0 = 0, with initial conditions y 0 (0) = 0 , ( y 0 ) (0) = y (2) 0 (0) = 0 , · · · ( y 0 ) ( n 2) (0) = y ( n 1) 0 (0) = 0 because those were the initial conditions we had Monday for the step response y 0 . Finally, ( y 0 ) ( n 1) (0) = y ( n ) 0 (0) = 1 as we observed Monday in Theorem: response to unit step . This identifies all the initial conditions for y δ , and so completes the proof of the theorem. Q.E.D.
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