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

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Nov9-12, 12:46 PM
P: 1
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 [ ] 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 [ ]. So their product cannot be a scalar function that I need. What is wrong about it?
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Nov12-12, 10:58 AM
P: 350
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.

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