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:
[tex]\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[/tex]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:
[tex]\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[/tex]


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 [tex]\nabla f\cdot \hat n[/tex] 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]

[itex]\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[/itex]

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?

[itex]\frac{\partial f}{\partial n}=\nabla f \cdot \hat n[/itex]

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

 

[yungman]

[itex]\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[/itex]

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?

[itex]\frac{\partial f}{\partial n}=\nabla f \cdot \hat n[/itex]

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.