Mathematical Biology- Neumann BCs, Turing Analysis

  • Thread starter binbagsss
  • Start date
  • #1
1,222
10
This is probably a stupid question but I have Neumann BC boundary : ## \nabla u . \vec{n} =0## (same for ##v##)conditions for the following reaction-diffusion system on a [0,L_1]x[0,L_2]x...x...[0,L_n] n times in n dimensional space so ##u=u(x_1,.....,x_n,t)## is a scalar I believe?

so that ## \nabla u . \vec{n} ## is a vector times a vector is a scalar,


my notes then say:

Untitled.png



and so, I'm confused how to argue, from, a summation of ##u_i## derivatives we conclude that each ##u_i## derivative must individually be zero? unless we are specifying ##n## different normal vectors, one for each surface (divided by two for the what would be a negative of this normal vector) ? so like ##(1,0....0)## ,...,(0,0,.....1) ##

ahh this must be the case actually and i have misinterpreted the boundary condition?? thanks
 

Attachments

Last edited by a moderator:

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,650
You only have one boundary condition per surface. The boundary conditions state that the derivative in the normal direction of the boundary is zero. This takes different forms on each boundary
 

Related Threads on Mathematical Biology- Neumann BCs, Turing Analysis

Replies
0
Views
1K
  • Last Post
Replies
0
Views
5K
  • Last Post
Replies
2
Views
16K
Replies
2
Views
892
  • Last Post
Replies
1
Views
1K
Replies
2
Views
4K
  • Last Post
Replies
2
Views
12K
Replies
2
Views
9K
Replies
26
Views
16K
Top