How to set up Neumann boundary condition for a PDE in a coordinate-invariant form?

In summary, the person is having trouble setting up Neumann boundary conditions for a diffusion-type PDE. They have a scalar function with a boundary region defined by another scalar function, and they want to specify a mixed boundary condition involving scalar numbers and functions. They are confused about the nature of the directional derivative in their example and the product of the gradient and normal vector. They mention that the gradient is a covariant vector and the normal vector is also a covector, but their product should be a scalar function. They are unsure of what is wrong and are seeking clarification.
  • #1
eugvas
1
0
I'm having trouble finding out how to set up Neumann (or, rather, "Robin") boundary conditions for a diffusion-type PDE. More specifically, I have a scalar function [itex]f(\boldsymbol{x}, t)[/itex] where [itex]\boldsymbol{x}[/itex] is n-dimensional vector space with some boundary region defined by [itex]A(\boldsymbol{x})=0[/itex] where A is another scalar function. Then I want to specify a mixed (Robin-type) boundary condition in the following way:
[itex]a f(\boldsymbol{x}) + b \frac{∂f}{∂\boldsymbol{n}} = c(\boldsymbol{x})[/itex],
where a and b are some scalar numbers and c is a scalar function. What I don't understand is what is the nature of the directional derivative which is multiplied by [itex]b[/itex] in my example: from Wikipedia [ http://en.wikipedia.org/wiki/Neumann_boundary_condition ] it follows that
[itex]\frac{∂f}{∂\boldsymbol{n}} = ∇f \cdot \boldsymbol{n}[/itex],
where the first term, gradient of the scalar function, is a covariant vector, and the second, the normal vector to the boundary, also seems to be not a "true" vector but a covector (covariant vector) given by ∇A [ http://en.wikipedia.org/wiki/Normal_vector#Hypersurfaces_in_n-dimensional_space ]. So their product cannot be a scalar function that I need. What is wrong about it?
 
Last edited:
Physics news on Phys.org
  • #2


It's not clear to me what is the problem you have. But perhaps this will help...

I'm never quite sure what people mean when they refer to the gradient, but in my book grad f is a vector field and df is a covector field (differential form). The coefficients of the gradient depend on a metric whereas the differential form components are simply the partial derivatives in whatever coordinate system you are using. With that convention the normal derivative is equal to df(n).

The vector n is a proper vector and is equal to the gradient of A divided by its own length. You need a metric tensor to define n. But you should already have one since you have a diffusion equation and the divergence operator depends on a metric.
 

Related to How to set up Neumann boundary condition for a PDE in a coordinate-invariant form?

1. What is a Neumann boundary condition for a PDE?

A Neumann boundary condition is a type of boundary condition that specifies the value of the derivative of a function at the boundary of a domain. In other words, it determines the behavior of the function at the boundary in terms of its slope or gradient.

2. What is the purpose of setting up Neumann boundary conditions in a coordinate-invariant form?

The purpose of setting up Neumann boundary conditions in a coordinate-invariant form is to ensure that the boundary conditions remain consistent regardless of the coordinate system used to describe the domain. This is important because different coordinate systems may have different orientations or scales, and using a coordinate-invariant form ensures that the boundary conditions are independent of these factors.

3. How do you set up Neumann boundary conditions in a coordinate-invariant form?

To set up Neumann boundary conditions in a coordinate-invariant form, one must first express the PDE in terms of a coordinate-invariant operator, such as the divergence or gradient operator. Then, the boundary conditions can be formulated using this operator, rather than specific coordinate directions.

4. What are some advantages of using coordinate-invariant Neumann boundary conditions?

Coordinate-invariant Neumann boundary conditions have several advantages, including improved numerical stability and accuracy, as well as the ability to easily adapt to different coordinate systems. They also allow for a more general and consistent treatment of boundary conditions, making it easier to apply them to a variety of PDEs.

5. Are there any limitations to using coordinate-invariant Neumann boundary conditions?

While coordinate-invariant Neumann boundary conditions have many advantages, they may not always be the most appropriate choice for a specific problem. In some cases, it may be more efficient or accurate to use boundary conditions that are tailored to the specific coordinate system or problem at hand. Additionally, setting up coordinate-invariant Neumann boundary conditions may require more advanced mathematical techniques and may not be as intuitive for some users.

Similar threads

Replies
4
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
2
Views
2K
Replies
8
Views
2K
  • Differential Equations
Replies
8
Views
4K
  • Differential Equations
Replies
13
Views
2K
  • Differential Equations
Replies
5
Views
2K
  • Differential Equations
Replies
4
Views
970
Back
Top