MHB Is Everything Correct in Applying Gauss's Theorem and Green's Identities?

[mathmari]

Hey!

With appropriate conditions, I want to show that $$\iiint_{\Omega}(\nabla \phi)\cdot \textbf{f}\ dV=\iint_{\Sigma}\phi\textbf{f}\cdot \textbf{N}\ dA-\iiint_{\Omega}\phi\nabla\cdot \textbf{f}\ dV$$ With appropriate conditions, I want to prove Green's identities: $$\iint_{\Sigma}\phi\nabla\psi\cdot \textbf{N}\ dA=\iiint_{\Omega}\left (\phi \Delta\psi+\nabla\phi\cdot \nabla\psi \right )\ dV\\ \iint_{\Sigma}\left (\phi\nabla \psi-\psi\nabla\phi \right )\cdot \textbf{N}\ dA=\iiint_{\Omega}\left (\phi\Delta\psi-\psi\Delta\phi \right )\ dV$$ Does $\phi\textbf{f}$ mean that $\phi$ is scalar and $\textbf{f}$ a vector? (Wondering) For the first equality we have:

Using Gauss theorem for $\phi\textbf{f}$ we get $$\iint_{\Sigma}\phi\textbf{f}\cdot \textbf{N}\ dA=\iiint_{\Omega}\nabla \cdot (\phi\textbf{f})\ dV$$ It holds that $\nabla \cdot (\phi\textbf{f})=\phi (\nabla \cdot \textbf{f})+\textbf{f}\cdot (\nabla \phi)$, right? How could we prove this? (Wondering)

Then we get $$\iint_{\Sigma}\phi\textbf{f}\cdot \textbf{N}\ dA=\iiint_{\Omega}\left [\phi (\nabla \cdot \textbf{f})+\textbf{f}\cdot (\nabla \phi)\right ]\ dV=\iiint_{\Omega}\phi (\nabla \cdot \textbf{f})\ dV+\iiint_{\Omega}\textbf{f}\cdot (\nabla \phi)\ dV \\ \Rightarrow \iiint_{\Omega}\textbf{f}\cdot (\nabla \phi)\ dV=\iint_{\Sigma}\phi\textbf{f}\cdot \textbf{N}\ dA-\iiint_{\Omega}\phi (\nabla \cdot \textbf{f})\ dV$$
For the second equality:

Using Gauss theorem for $\phi\nabla\psi$ we get $$\iint_{\Sigma}\phi\nabla\psi\cdot \textbf{N}\ dA=\iiint_{\Omega}\nabla \cdot (\phi\nabla\psi)\ dV$$ Does it hold that $\nabla \cdot (\phi\nabla\psi)=\nabla \phi\cdot \nabla\psi+\phi \nabla \cdot \nabla\psi$ ? If yes, how could we prove that? (Wondering)
If this is correct we get $\nabla \cdot (\phi\nabla\psi)=\nabla \phi\cdot \nabla\psi+\phi \Delta\psi$ and so $$\iint_{\Sigma}\phi\nabla\psi\cdot \textbf{N}\ dA=\iiint_{\Omega}\left (\nabla \phi\cdot \nabla\psi+\phi \Delta\psi\right ) dV$$
For the third equality:

Using Gauss theorem for $\phi\nabla \psi-\psi\nabla\phi $ we get $$\iint_{\Sigma}\left (\phi\nabla \psi-\psi\nabla\phi \right )\cdot \textbf{N}\ dA=\iiint_{\Omega}\nabla \cdot (\phi\nabla \psi-\psi\nabla\phi )\ dV=\iiint_{\Omega}\left [\nabla \cdot (\phi\nabla \psi)-\nabla \cdot(\psi\nabla\phi )\right ]\ dV$$ If the equality that I used previously was correct we get $\nabla \cdot (\phi\nabla\psi)=\nabla \phi\cdot \nabla\psi+\phi \nabla \cdot \nabla\psi=\nabla \phi\cdot \nabla\psi+\phi \Delta\psi$ and $\nabla \cdot (\psi\nabla\phi)=\nabla \psi\cdot \nabla\phi+\psi \nabla \cdot \nabla\phi=\nabla \psi\cdot \nabla\phi+\psi \Delta\phi$.
Then $\nabla \cdot (\phi\nabla \psi)-\nabla \cdot(\psi\nabla\phi )=(\nabla \phi\cdot \nabla\psi+\phi \Delta\psi)-(\nabla \psi\cdot \nabla\phi+\psi \Delta\phi)=\nabla \phi\cdot \nabla\psi+\phi \Delta\psi-\nabla \psi\cdot \nabla\phi-\psi \Delta\phi=\phi \Delta\psi-\psi \Delta\phi$.
Therefore we get $$\iint_{\Sigma}\left (\phi\nabla \psi-\psi\nabla\phi \right )\cdot \textbf{N}\ dA\iiint_{\Omega}\left [\phi \Delta\psi-\psi \Delta\phi\right ]\ dV$$ Is everything correct? (Wondering)

 

[I like Serena]

Re: prove also the Gauss theorem: Prove equalities

Does $\phi\textbf{f}$ mean that $\phi$ is scalar and $\textbf{f}$ a vector?

Hey mathmari!

Yep. (Nod)

For the first equality we have:

It holds that $\nabla \cdot (\phi\textbf{f})=\phi (\nabla \cdot \textbf{f})+\textbf{f}\cdot (\nabla \phi)$, right? How could we prove this?

How about writing it out in cartesian coordinates?
$$\nabla \cdot (\phi\mathbf{f})
=\sum \pd {}{x_i} (\phi\mathbf{f})
=\sum \phi\pd{\mathbf f}{x_i} + \pd{\phi}{x_i}\mathbf f
=\phi\sum \pd{\mathbf f}{x_i} + \mathbf f\sum\pd{\phi}{x_i}
=\phi\nabla\cdot \mathbf f + \mathbf f\nabla \phi
$$

For the second equality:

Does it hold that $\nabla \cdot (\phi\nabla\psi)=\nabla \phi\cdot \nabla\psi+\phi \nabla \cdot \nabla\psi$ ? If yes, how could we prove that?

Can we prove it by writing it out in cartesian coordinates? (Wondering)

Generally $\nabla$ does indeed behave as a partial derivative.

For the third equality:

Is everything correct?

Yup. (Nod)

 

[mathmari]

Re: prove also the Gauss theorem: Prove equalities

How about writing it out in cartesian coordinates?
$$\nabla \cdot (\phi\mathbf{f})
=\sum \pd {}{x_i} (\phi\mathbf{f})
=\sum \phi\pd{\mathbf f}{x_i} + \pd{\phi}{x_i}\mathbf f
=\phi\sum \pd{\mathbf f}{x_i} + \mathbf f\sum\pd{\phi}{x_i}
=\phi\nabla\cdot \mathbf f + \mathbf f\nabla \phi
$$

What is the difference between $\nabla \cdot $ and $\nabla$ ? Do we use the $\cdot$ when we have a vector after $\nabla$ ? (Wondering)

Can we prove it by writing it out in cartesian coordinates? (Wondering)

We have the following:
$$\nabla \cdot (\phi\nabla\psi)
=\sum \pd {}{x_i} (\phi\nabla\psi)
=\sum \left (\phi\pd{(\nabla\psi )}{x_i} + \pd{\phi}{x_i}\nabla\psi \right )
=\phi\sum \pd{(\nabla\psi )}{x_i} + \nabla\psi\sum\pd{\phi}{x_i}
=\phi\nabla\cdot \nabla\psi + \nabla\psi\nabla \phi$$ right? (Wondering)

 

[I like Serena]

Re: prove also the Gauss theorem: Prove equalities

What is the difference between $\nabla \cdot $ and $\nabla$ ? Do we use the $\cdot$ when we have a vector after $\nabla$ ?

The first is divergence and the second is gradient.
And they are different.
$$\operatorname{div} \mathbf f = \nabla\cdot \mathbf f = \sum \pd {f_i}{x_i} \\
\operatorname{grad} \mathbf f = \nabla \mathbf f = \sum \pd {\mathbf f}{x_i}\mathbf{\hat x}_i$$

Divergence is usually taken from a vector ($\nabla\cdot\mathbf f$), and yields a scalar.
Gradient is usually taken from a scalar ($\nabla\phi$), and yields a vector.
In the example above we have the gradient of a vector ($\nabla\mathbf f$), which is a vector of vectors.

Oh wait! (Wait)

That means that it should be:
$$\nabla\cdot(\phi\mathbf f) = \sum \pd{}{x_i}(\phi f_i) = \sum\phi\pd{f_i}{x_i} + \sum \pd\phi{x_i}f_i
=\phi\sum\pd{f_i}{x_i} + \sum \pd\phi{x_i}f_i
=\phi\nabla\cdot \mathbf f + \nabla\phi\cdot \mathbf f
$$
Sorry for that! (Blush)

We have the following:
$$\nabla \cdot (\phi\nabla\psi)
=\sum \pd {}{x_i} (\phi\nabla\psi)
=\sum \left (\phi\pd{(\nabla\psi )}{x_i} + \pd{\phi}{x_i}\nabla\psi \right )
=\phi\sum \pd{(\nabla\psi )}{x_i} + \nabla\psi\sum\pd{\phi}{x_i}
=\phi\nabla\cdot \nabla\psi + \nabla\psi\nabla \phi$$ right?

So this is not correct, since the divergence operator is not expanded correctly. (Worried)

 

[mathmari]

Re: prove also the Gauss theorem: Prove equalities

The first is divergence and the second is gradient.
And they are different.
$$\operatorname{div} \mathbf f = \nabla\cdot \mathbf f = \sum \pd {f_i}{x_i} \\
\operatorname{grad} \mathbf f = \nabla \mathbf f = \sum \pd {\mathbf f}{x_i}$$

Divergence is usually taken from a vector ($\nabla\cdot\mathbf f$), and yields a scalar.
Gradient is usually taken from a scalar ($\nabla\phi$), and yields a vector.
In the example above we have the gradient of a vector ($\nabla\mathbf f$), which is a vector of vectors.

That means that it should be:
$$\nabla\cdot(\phi\mathbf f) = \sum \pd{}{x_i}(\phi f_i) = \sum\phi\pd{f_i}{x_i} + \sum \pd\phi{x_i}f_i
=\phi\sum\pd{f_i}{x_i} + \sum \pd\phi{x_i}f_i
=\phi\nabla\cdot \mathbf f + \nabla\phi\cdot \mathbf f
$$

I got! (Nerd) So, we have that $\nabla\psi$ is a vector as $\mathbf{f}$ previously. So, we get:

$$\nabla\cdot(\phi\nabla\psi) = \sum \pd{}{x_i}(\phi [\nabla\psi]_i) = \sum\phi\pd{[\nabla\psi]_i}{x_i} + \sum \pd\phi{x_i}[\nabla\psi]_i
=\phi\sum\pd{[\nabla\psi]_i}{x_i} + \sum \pd\phi{x_i}[\nabla\psi]_i
=\phi\nabla\cdot \nabla\psi + \nabla\phi\cdot \nabla\psi$$

Is this correct? (Wondering)

 

[I like Serena]

Re: prove also the Gauss theorem: Prove equalities

So, we have that $\nabla\psi$ is a vector as $\mathbf{f}$ previously. So, we get:

$$\nabla\cdot(\phi\nabla\psi) = \sum \pd{}{x_i}(\phi [\nabla\psi]_i) = \sum\phi\pd{[\nabla\psi]_i}{x_i} + \sum \pd\phi{x_i}[\nabla\psi]_i
=\phi\sum\pd{[\nabla\psi]_i}{x_i} + \sum \pd\phi{x_i}[\nabla\psi]_i
=\phi\nabla\cdot \nabla\psi + \nabla\phi\cdot \nabla\psi$$

Is this correct? (Wondering)

Yep.

Oh, and I fixed the gradient definition in my previous post, since it should include the unit vector $\mathbf{\hat x}_i$. (Lipssealed)

 

[mathmari]

Re: prove also the Gauss theorem: Prove equalities

Yep.

Great! (Happy)

Oh, and I fixed the gradient definition in my previous post, since it should include the unit vector $\mathbf{\hat x}_i$. (Lipssealed)

Ah ok! At the exercise statement what is meant by "by appropriate conditions" ? Do we have to assume that something has to hold? (Wondering)

 

[I like Serena]

Re: prove also the Gauss theorem: Prove equalities

At the exercise statement what is meant by "by appropriate conditions" ? Do we have to assume that something has to hold?

We're applying Gauss's theorem.
We can only do that if its conditions are satisfied.
From wiki:

Suppose V is a subset of $\mathbb{R}^n$ (in the case of n = 3, V represents a volume in 3D space) which is compact and has a piecewise smooth boundary S (also indicated with ∂V = S ). If F is a continuously differentiable vector field defined on a neighborhood of V, then we have:​

So your $\Omega$, $\Sigma$, $\phi$, and $\mathbf f$ have to satisfy those conditions.
We assume that "appropriate conditions" hold, which are those.
Oh and it will also include that $\psi$ is differentiable twice, since otherwise we can't take its laplacian $\Delta$. (Thinking)

 

[mathmari]

Re: prove also the Gauss theorem: Prove equalities

We're applying Gauss's theorem.
We can only do that if its conditions are satisfied.
From wiki:

Suppose V is a subset of $\mathbb{R}^n$ (in the case of n = 3, V represents a volume in 3D space) which is compact and has a piecewise smooth boundary S (also indicated with ∂V = S ). If F is a continuously differentiable vector field defined on a neighborhood of V, then we have:​

So your $\Omega$, $\Sigma$, $\phi$, and $\mathbf f$ have to satisfy those conditions.
We assume that "appropriate conditions" hold, which are those.
Oh and it will also include that $\psi$ is differentiable twice, since otherwise we can't take its laplacian $\Delta$. (Thinking)

Ah ok! Thank you so much! (Smirk)