Boundary conditions

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ssothro
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Joined: Mon Feb 06, 2006 5:12 pm

Boundary conditions

Post by ssothro »

Hi all,
I would like to ask what type of boundary conditions NEURON applies. If for example, a single section is created, how are its 0 and 1 ends terminated?

Thank you in advance.
ted
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Post by ted »

See chapter 5 of The NEURON Book.
Sections are discretized into segments of uniform length, using a central difference
approximation that places internal nodes at ( 0.5 + i ) / nseg where i = 0 . . nseg-1.
ODEs are solved at each internal node to compute v etc.. Nodes at 0 and 1 have
no associated membrane or cytoplasm, and v at those locations is calculated
as the weighted average of v at adjacent internal nodes.
ssothro
Posts: 13
Joined: Mon Feb 06, 2006 5:12 pm

Post by ssothro »

"Nodes at 0 and 1 have no associated membrane or cytoplasm, and v at those locations is calculated as the weighted average of v at adjacent internal nodes."

What if a section does not have two adjacent nodes, e.g. last node of a cable. In that case what happens to the end? Its potential becomes just the potential of the one adjacent node? Are the ends sealed?
ted
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Post by ted »

ssothro wrote:What if a section does not have two adjacent nodes
Every section has at least three nodes. If you don't have The NEURON Book, then read
Hines, M.L. and Carnevale, N.T.
The NEURON simulation environment. Neural Computation 9:1179-1209, 1997
(see http://www.neuron.yale.edu/neuron/bib/nrnpubs.html), especially
3.2 Spatial discretization in a biological context: sections and segments
e.g. last node of a cable. In that case what happens to the end? Its potential becomes just the potential of the one adjacent node?
If nothing is attached to the 0 end, v(0) is the same as v(first internal node).
If nothing is attached to the 1 end, v(1) is the same as v(last internal node).
Are the ends sealed?
If nothing is attached to the 0 end, the 0 end is sealed. Ditto for the 1 end.
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