Green's Theorem and Laplace's equation

[Gear.0]

Homework Statement

Show that for a solution w of Laplace's equation in a region R with boundary curve C and outer unit normal vector N,
$$\int_{R}\left\| \nabla w\right\| dxdy = \oint_{C}w\frac{\partial w}{\partial N}ds$$

Homework Equations

The book goes through the steps to show that the following is true (which is very similar to the problem I am doing so I thought it would be relevant):
$$eq(1): \int_{R}\left( \nabla^{2}w \right) dxdy = \oint_{C}\frac{\partial w}{\partial N}ds$$
Also, Green's theorem, which is just Stokes theorem in the plain:
$$eq(2): \int_{R}\left( \frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y}\right) dx dy = \oint_{C}\left( F_{1}dx + F_{2}dy\right)$$
Also:
$$eq(3): (\nabla w)\cdot \hat{N} = \frac{\partial w}{\partial N}$$

The Attempt at a Solution

I have been trying to follow the steps in the book that they use to get eq(1) but it just doesn't seem the same.
What they do is to define $F_{2} = \frac{\partial w}{\partial x}$ and $F_{1} = -\frac{\partial w}{\partial y}$ so that when those F's are plugged into eq(2) you end up with the laplacian of w. That is how they get the left side, and using that this laplacian can now be written in the form of eq(2) they get the relation on the right side.

But in my case the integrand on the left side is now:
$$\left\| \nabla w \right\| = \left( \frac{\partial w}{\partial x} \right)^{2} + \left( \frac{\partial w}{\partial y} \right)^{2}$$
I don't see how I can write this in the form of eq(2), so I don't know if the book's steps will help me.

 

[hunt_mat]

Use:
$$\nabla^{2}w=\nabla\cdot\nabla w=\textrm{div}(\textrm{grad}w)$$
And use equation (1)

 

[Gear.0]

I'm sorry, I'm really trying to get this but I just don't see how writing eq(1) as
$$\int_{R}\left( \nabla\cdot\nabla w\right) dxdy = \oint_{C}\frac{\partial w}{\partial N}ds$$
will help. I tried doing it and writing this next to the form I want to get:
$$\int_{R}\left\| \nabla w\right\| dxdy = \oint_{C}w\frac{\partial w}{\partial N}ds$$
but nothing came to me.

All I can think of is that you can pull the divergence out to get
$$\nabla\cdot\int_{R}\left( \nabla w\right) dxdy = \oint_{C}\frac{\partial w}{\partial N}ds$$
is that right? Then that looks a little closer to the final answer, but I can't see what to do next.

 

[gabbagabbahey]

All I can think of is that you can pull the divergence out to get
$$\nabla\cdot\int_{R}\left( \nabla w\right) dxdy = \oint_{C}\frac{\partial w}{\partial N}ds$$
is that right?

No, you can't pull the divergence out...it is analogous to claiming that

$$\int \frac{df}{dx}dx = \frac{d}{dx} \int f(x)dx$$

...which is clearly wrong...for example, try it with $f(x)=2$.

Instead, try calculating $\int_{R}\matbf{\nabla}\cdot\left(w\matbf{\nabla}w\right)dxdy$ using an appropriate vector calculus identity (the divergence of a scalar times a vector).

 

[Gear.0]

I must sincerely apologize, the original equation should have had a square in there on the first integrand:
$$\int_{R}\left\| \nabla w\right\|^{2} dxdy = \oint_{C}w\frac{\partial w}{\partial N}ds$$

But thanks for the help, I will try your suggestions with this corrected equation.

 

[Gear.0]

Instead, try calculating $\int_{R}\matbf{\nabla}\cdot\left(w\matbf{\nabla}w\right)dxdy$ using an appropriate vector calculus identity (the divergence of a scalar times a vector).

But I don't have anything with in the equation
$$\int_{R}\nabla}\cdot\left(w\matbf{\nabla}w\right)dxdy$$

I tried it anyway and I got that it can be rewritten as:
$$\int_{R}\left\| \nabla w\right\|^{2}dxdy + \int_{R}w\nabla^{2}w dxdy$$

The term on the left is exactly what I want but I don't know what to do with it because the term on the right, and term that is equal to the whole thing don't resemble anything else from my problem.

 

[gabbagabbahey]

But I don't have anything with in the equation
$$\int_{R}\nabla}\cdot\left(w\matbf{\nabla}w\right)dxdy$$

I tried it anyway and I got that it can be rewritten as:
$$\int_{R}\left\| \nabla w\right\|^{2}dxdy + \int_{R}w\nabla^{2}w dxdy$$

The term on the left is exactly what I want but I don't know what to do with it because the term on the right, and term that is equal to the whole thing don't resemble anything else from my problem.

What is $\int_{R}w\nabla^{2}w dxdy$ going to be equal to if $w$ satisfies the Laplace equation?

What do you get when you apply Green's theroem to $\int_{R}\mathbf{\nabla}\cdot\left(w\mathbf{\nabla}w\right)dxdy$?

 

[Gear.0]

That makes perfect sense, thanks! It is zero so that the two integrals are equal in this case.

My problem was not really knowing how to apply green's theorem to something that doesn't resemble a curl.
But instead of completely simplifying $\nabla\cdot\left( w\nabla w\right)$ if I do it up to this point:
$$\frac{\partial}{\partial x}\left( w\frac{\partial w}{\partial x}\right)+\frac{\partial}{\partial y}\left( w\frac{\partial w}{\partial y}\right)$$
Then I can define:
$$F_{2}=w\frac{\partial w}{\partial x}$$
$$F_{1}=-w\frac{\partial w}{\partial y}$$

and now it resembles green's theorem. When I apply it I end up with the answer I needed.
Thanks for your patience :D