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Unformatted text preview: Math 334 Lecture #27 § 6.5: Unit Impulse Functions and Dirac Delta “Functions” Example. How does the solution of the IVP y 00 + 4 y + 4 y = u 1 ( t ) u 1+ k ( t ) k , y (0) = 0 , y (0) = 0 , behave as k → + ? [For k = 1, the IVP is that of the example in the Lecture #26.] [See Maple worksheet for animation of solution as k → + .] An Idealized Impulse Force. The discontinuous function f k ( t ) = u 1 ( t ) u 1+ k ( t ) k describes the constant force of 1 /k applied on the interval [1 , 1 + k ]. [Sketch the graph of the discontinuous forcing function for several values of k , especial small values.] The impulse (or strength) of f k is its integral over [0 , ∞ ): I ( f k ) = Z ∞ u 1 ( t ) u 1+ k ( t ) k dt = 1 k Z 1+ k 1 dt = 1 for k > . What is the limit of the impulse of f k as k → + ? lim k → + I ( f k ) = 1 . For a fixed value of t ≥ 0, what is the limit of f k ( t ) as k → + ? It is the “function” δ ( t 1) = lim k → + f k ( t ) = if t 6 = 1 , ∞ if t = 1 ....
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This note was uploaded on 11/29/2011 for the course MATH 334 taught by Professor Dallon during the Fall '08 term at BYU.
 Fall '08
 DALLON
 Math, Differential Equations, Equations

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