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RE: Re: NDSolve with boundary condition(s) at infinity?

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  • Subject: [mg68584] RE: [mg68531] Re: NDSolve with boundary condition(s) at infinity?
  • From: "Tony Harker" <a.harker at>
  • Date: Fri, 11 Aug 2006 04:39:46 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

 I'm guessing from your description of the problem that there is no
convenient change of variables which could move the boundary to a finite
value of some new independent variable.

 Once you're stuck with numerical solutions with one of the boundaries at
infinity, you probably have to resort to numerical experiments -- in other
words, put in your boundary condition for some large value of the
independent variable and see whether the solution changes significantly when
you increase that value. It sounds as if you are looking for an
exponentially decaying solution, so this should work reasonably well. 

  One possible problem is that you might pick up a spurious solution which
is still oscillatory in the region where the independent variable is large,
and so just looking for a zero of the function will not be enough --
consider modifying the boundary condition to force both the function and its
derivative to be very small (and hope that the special case of an
oscillatory function that just kisses the axis does not leap out and bite

   Tony Harker

Dr A.H. Harker
Department of Physics and Astronomy
University College London
Gower Street


]->-----Original Message-----
]->From: Narasimham [mailto:mathma18 at] 
To: mathgroup at
]->Subject: [mg68584] [mg68531] Re: NDSolve with boundary condition(s) at 
]->AES wrote:
]->> Can NDSolve be used for ODEs with one (or two) boundary 
]->> being that solution must tend toward zero at infinity?
]->> [Eigenmodes of complex slab waveguides with transversely 
]->varying index 
]->> and loss profiles being the case of interest.]
]->> Pointers to alternative Mathematica-based attacks?
]->> [Didn't find any answer to first query in extended 
]->documentation for 
]->> NDSolve.]
]->If function behaviour is a priori known to have symmetry 
]->somewhere, perhaps better to avoid asymmetric asymptotic 
]->boundary conditions in the ODE itself. Otherwise I do not 
]->know.. For example in Gaussian distribution y''[x]= - y[x] ( 
]->1- ( x/a)^2 ), a choice of symmetric case boundary 
]->conditions as initial value problem y[0] ==1 , y'[0] ==0 
]->includes the zero at infinity automatically.

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