Mathematica for Dirac delta functions and Green functions

Transcription

1 Mathematica for Dirac delta functions and Green functions DiracDelta function Mathematic has Dirac s delta function built in for use in integrals and solving differential equations. If you evaluate it directly you get unless the argument is in which case it gives you the function back---it is not evaluated and does not evaluate to infinity. 8DiracDelta@D, DiracDelta@D, DiracDelta@- D< 8, DiracDelta@D, < So there is no peak in a plot of the function Plot@DiracDelta@D, 8, -, <, Eclusions None, PlotStyle 8Red, Thick<D However, it integrates to give the theta function: Integrate@DiracDelta@D, D HeavisideTheta@D Plot@HeavisideTheta@D, 8, -, <, Eclusions None, PlotStyle 8Red, Thick<D.. - -

2 And differentiating the theta function returns the Dirac delta function D Further derivatives are just denoted symbolically D Note that numerical integration over a delta function does not give the correct result because the routines cannot find the infinitely narrow peak. So, beware of this---use DiracDelta only in analytical integrations or solves. NIntegrate@DiracDelta@D, 8, -, <D NIntegrate::izero : Integral and error estimates are on all integration subregions. Try increasing the value of the MinRecursion option. If value of integral may be, specify a finite value for the AccuracyGoal option.. Integrals using Dirac delta function Defining properties Integrate@f@D DiracDelta@ - D, 8, - Infinity, Infinity<D f@d Integral vanishes if integration range does not include position of delta function Integrate@f@D DiracDelta@ - D, 8, - Infinity, <D. Dirac delta is an even function Integrate@f@D DiracDelta@- D, 8, -, <D f@d Correct Jacobians Integrate@f@D DiracDelta@ H - LD, 8, -, <D f@d Eample from lecture Integrate@ Ep@D DiracDelta@Sinh@ DD, 8, - Infinity, Infinity<D Checking eplicitly:

3 D, D Cosh@ D Cosh@D IntegrateB Ep@D DiracDelta@D, 8, - Infinity, Infinity<F Correctly picks up multiple crossings of zero by argument of delta function Integrate@f@D DiracDelta@Sin@ DD, 8, -, Pi + <D Kf@D + fb F + f@do More eamples from lecture Integrate@Cos@ D DiracDelta@Sin@DD, 8, - Pi, Pi <D Integrate@Cos@ D DiracDelta@Sin@DD, 8, - Pi, - Pi <D Integrals of derivatives of delta functions Integrate@f@D DiracDelta'@D, 8, -, <D - A few cautions about Mathematica s DiracDelta[ ]: Only HeavisideTheta[] s derivative gives DiracDelta[ ], functions with similar behavior do not testfunctions = 8HeavisideTheta@D, UnitStep@D, HSqrt@ ^ D + L, HAbs@D + L < :HeavisideTheta@D, UnitStep@D, + Abs@D +, >

4 8, -, <D & testfunctions :, >.. -,. -, The limits from each side have the appropirate values Limit@testfunctions,, Direction D 8,,, < Limit@testfunctions,, Direction - D 8,,, < At the discontinuous point they have different interpretations testfunctions. Power::infy : Infinite epression encountered. Infinity::indet : Indeterminate epression CompleInfinity encountered. Power::infy : Infinite epression encountered. Infinity::indet : Indeterminate epression CompleInfinity encountered. 8HeavisideTheta@D,, Indeterminate, Indeterminate< Mathematica will only correctly identify the derivative of HeavisideTheta[] as DiracDelta[] D@testfunctions, D :DiracDelta@D, Indeterminate,, True DiracDelta[ ] is not defined for comple arguments: DiracDelta@ + ID DiracDelta@ + äd Abs@D - + >

5 5 - ID, 8, - I, I<D à ä DiracDelta@- ä + D â - ä Limits which can be used to define a delta function will not return DeltaFunction[ ]: Limit@ H ^ + ^ L, D Integrals will not return a delta function: Integrate@Ep@I k D, 8, -, <, Assumptions -> k Î RealsD Integrate::idiv : Integral of ãä k does not converge on 8-, <. IntegrateAãä k, 8, -, <, Assumptions k Î RealsE But FourierTransform[ ] will: FourierTransform@Ep@I k D,, y, FourierParameters 8-, - <D DiracDelta@k - yd Sec. 8. #9---eample worked in class Mathematica can deal with general t (which it assumes is real, as DiracDelta is only defined for reals): Clear@t, yd soln9 = Flatten@DSolve@8y ''@td + y '@td + y@td DiracDelta@t - td, y '@D, y@d <, y@td, tdd Simplify :y@td ã-t+t HHeavisideTheta@ t - td - HeavisideTheta@- tdl Sin@ Ht - tld> Make into a function: yy@t_d = FullSimplify@Flatten@y@tD. soln9 DD ã-t+t HHeavisideTheta@t - td - HeavisideTheta@- tdl Sin@ Ht - tld Net consider t= for definiteness. This gives the answer obtained in class in various ways: t = ; yy@td ã-t HeavisideTheta@- + td Sin@ H- + tld Recall this is a damped oscillator at rest given a unit impulse at t=, so that its slope has a discontinuity, although the function is continuous.

6 6 8t, -, <, PlotStyle 8Thick, Red<, PlotRange 88-, <, 8-., <<D Finding Green function obtained in Boas using DiracDelta soln = DSolve@8y ''@td + y@td DiracDelta@t - tpd, y@d, y@pi D <, y@td, td ::y@td - Cos@tpD HeavisideThetaB - tpf Sin@tD + Cos@tpD HeavisideTheta@t - tpd Sin@tD Cos@tD HeavisideTheta@t - tpd Sin@tpD + Cos@tD HeavisideTheta@- tpd Sin@tpD>> To simplify, need to let Mathematica know the range of tp. This result agrees with that we found. soln = Simplify@soln, Assumptions 8 < tp < Pi <D 88y@tD - Cos@tpD Sin@tD + HeavisideTheta@t - tpd Sin@t - tpd<< Green function is thus GG@t_, tp_d = - Sin@tD Cos@tpD + HeavisideTheta@t - tpd Sin@t - tpd; Sec. 8. #---eample worked in class Forcing function. ff@_d = Piecewise@88, < Pi <, 8Pi -, Pi < Pi <<D < - < True

7 8,, Pi <, PlotStyle 8Thick, Red<D...5 Here s the result for the solution using the Green function from above applied to the forcing function: fullyy@_d = Integrate@GG@, pd ff@pd, 8p,, Pi <D Simplify K- J- + N Sin@D - HeavisideThetaB- H - - Cos@DL HeavisideThetaB- K H - Sin@DL + HeavisideThetaB- + F K- + + CosB + F + + F + SinB + F K- - CosB + FO + HeavisideTheta@D + F + SinB + FOOO To get simple form, need to tell Mathematica the range of : yyless@_d = Integrate@GG@, pd ff@pd, 8p,, Pi <, Assumptions 8 < < Pi <D - Sin@D yymore@_d = Integrate@GG@, pd ff@pd, 8p,, Pi <, Assumptions 8Pi < < Pi <D J - - Cos@DN Subsequent three plots show resulting y, y and y, so that one can see the nature of the solution, whose third derivative is discontinuous. 7

8 8 8,, Pi <, PlotStyle 8Thick, Blue<, AesLabel 8"", "y"<d y Plot@Evaluate@D@fullyy@D, DD, 8,, Pi <, PlotStyle 8Thick, Blue<, AesLabel 8"", "y'"<d y' Plot@Evaluate@D@fullyy@D, 8, <DD, 8,, Pi <, PlotStyle 8Thick, Blue<, AesLabel 8"", "y''"<d y''...5 Mathematica can also solve the differential equation including the piecewise function:

9 + ff@d, y@pi D <, y@d, DD ::y@d - J- + - J- + N Cos@D N Sin@D J Sin@D Cos@DN > < True >> Keep in mind when Mathematica says True for a piecewise function, it just means Otherwise 9