Is normal derivative a definition?

[yungman]

Is $\frac{ \partial f}{\partial n} = \nabla f \cdot \hat n$ a definition? No article that I found said it's a definition. The term $\frac{ \partial f}{\partial n}$ does not make sense to me, what is $\partial n$?

Also is this correct:
$$\int_v\nabla\cdot (v\nabla u)dV=\int_s (v\nabla u)\cdot \hat n dS=\int_s v(\nabla u\cdot \hat n) dS=\int_s v\frac {\partial u}{\partial n} dS$$Thanks

 

[HallsofIvy]

Is $\frac{ \partial f}{\partial n} = \nabla f \cdot \hat n$ a definition? No article that I found said it's a definition. The term $\frac{ \partial f}{\partial n}$ does not make sense to me, what is $\partial n$?

Also is this correct:
$$\int_v\nabla\cdot (v\nabla u)dV=\int_s (v\nabla u)\cdot \hat n dS=\int_s v(\nabla u\cdot \hat n) dS=\int_s v\frac {\partial u}{\partial n} dS$$


Thanks

Well, what is $\hat n$? I would interpret that as the normal vector to some curve (in two dimensions) or surface (in three dimensions), but you don't mention a curve or surface. Assuming there is such an object, then I would interpret the notation $\frac{\partial f}{\partial n}$ as the rate of change of f in the direction perpendicular to that curve or surface. That is, of course the same as $$\nabla f\cdot \hat n$$ but I would say you can reasonably take either as a definition of the other.

 

[yungman]

Thanks for the reply. Yes, $\hat n$ is the normal of the boundary. Here is an article contains normal derivative.

http://en.wikipedia.org/wiki/Directional_derivative

I am almost sure it's a definition, but this is math, it's black and white.

Thanks

 

[saminator910]

$\int_v\nabla\cdot (v\nabla u)dV=\int_s (v\nabla u)\cdot \hat n dS=\int_s v(\nabla u\cdot \hat n) dS=\int_s v\frac {\partial u}{\partial n} dS$

This makes sense, simply the divergence theorem, I assume v is some other function, but you say you integrate over it in your first expression, I think that is meant to be a capital V, correct?

$\frac{\partial f}{\partial n}=\nabla f \cdot \hat n$

Take this as you will, the last poster confirmed that it is true and it makes sense to me as well.

 

[yungman]

$\int_v\nabla\cdot (v\nabla u)dV=\int_s (v\nabla u)\cdot \hat n dS=\int_s v(\nabla u\cdot \hat n) dS=\int_s v\frac {\partial u}{\partial n} dS$

This makes sense, simply the divergence theorem, I assume v is some other function, but you say you integrate over it in your first expression, I think that is meant to be a capital V, correct?

$\frac{\partial f}{\partial n}=\nabla f \cdot \hat n$

Take this as you will, the last poster confirmed that it is true and it makes sense to me as well.

My bad, I should have use a different variable name. $v$ is a function of the same coordinates as the $\nabla$ and $dV$ is volume integral.

 

[arildno]

It is a definition of what the short-hand notation df/dn is meant to signify, i.e, an identity, not some equation or perceived relationship between (otherwhere defined) LHS and RHS.

 

[yungman]

It is a definition of what the short-hand notation df/dn is meant to signify, i.e, an identity, not some equation or perceived relationship between (otherwhere defined) LHS and RHS.

Thanks.