Mathematical Biology- Neumann BCs, Turing Analysis

Click For Summary
SUMMARY

The discussion centers on Neumann boundary conditions (BCs) in a reaction-diffusion system defined in n-dimensional space, specifically the condition ## \nabla u . \vec{n} = 0 ## for the scalar function ##u=u(x_1,...,x_n,t)##. Participants clarify that each component of the gradient must be treated individually, leading to the conclusion that each derivative ##u_i## must be zero at the boundary. The confusion arises from the interpretation of normal vectors, which must be specified for each surface of the boundary, confirming that only one boundary condition applies per surface.

PREREQUISITES
  • Understanding of Neumann boundary conditions in mathematical biology
  • Familiarity with reaction-diffusion systems
  • Knowledge of vector calculus, particularly gradients and normal vectors
  • Basic concepts of n-dimensional space
NEXT STEPS
  • Study the implications of Neumann boundary conditions in reaction-diffusion models
  • Explore vector calculus applications in mathematical biology
  • Research specific examples of n-dimensional reaction-diffusion systems
  • Learn about the role of normal vectors in boundary value problems
USEFUL FOR

Mathematicians, biologists, and researchers working on mathematical modeling of biological systems, particularly those focused on reaction-diffusion processes and boundary value problems.

binbagsss
Messages
1,291
Reaction score
12
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

  • Untitled.png
    Untitled.png
    3.9 KB · Views: 526
Last edited by a moderator:
Physics news on Phys.org
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
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K