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MathGroup Archive 2007

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Re: DSolve with DiracDelta [CORRECTION]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75807] Re: [mg75764] DSolve with DiracDelta [CORRECTION]
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Fri, 11 May 2007 05:22:43 -0400 (EDT)
  • Reply-to: hanlonr at cox.net

Actually, the problem was that the original boundary condition should have been 

y[0]==1 

rather than 

y[0]==0


Bob Hanlon

---- Bob Hanlon <hanlonr at cox.net> wrote: 
> It appears that you need a boundary condition at a time other than zero.
> 
> Clear[soln];
> soln[t_]=y[t]/.
>     DSolve[{y'[t] + a y[t] == DiracDelta[t], 
>           y[1] == Exp[-a]}, y[t], t][[1]]
> 
> UnitStep[t]/E^(a*t)
> 
> 
> Bob Hanlon
> 
> ---- Steffen Paul <steffen.paul at item.uni-bremen.de> wrote: 
> > Hi
> > I tried to solve
> > DSolve[{y'[t] + \[Alpha] y[t] == DiracDelta[t], y[0] == 0}, y, t]
> > 
> > and got
> > 
> > -\[ExponentialE]^(-t \[Alpha]) (1 - HeavisideTheta[t])
> > 
> > which is zero for t >0.
> > 
> > The solution is correct but I expected somthing else:
> > 
> > exp( - alpha t) UnitStep(t)
> > 
> > which is zero for t <0 and which is also a solution.
> > 
> > In engineering, these solutions are called impulse responses.
> > 
> > The last solution is physically more usefull , because the system responds 
> > after the excitation (DiracDelta).
> > 
> > 
> > 
> > How can I force Mathematica to give only solutions with nonzero values for t 
> >  >0 ?
> > 
> > 
> > 
> > Regards,
> > 
> > Steffen
> > 
> > 
> > 
> > 
> > 



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