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

In summary, the conversation discusses using appropriate conditions to prove Green's identities and the use of Gauss theorem in the process. The conversation also includes an explanation and proof of the divergence and gradient operators and their differences.
  • #1
mathmari
Gold Member
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Hey! :eek:

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)
 
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  • #2
Re: prove also the Gauss theorem: Prove equalities

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

Hey mathmari!

Yep. (Nod)
mathmari said:
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?
$$\nabla \cdot (\phi\mathbf{f})
=\sum \pd {}{x_i} (\phi\mathbf{f})
=\sum \phi\pd{\mathbf f}{x_i} + \pd{\phi}{x_i}\mathbf f
=\phi\sum \pd{\mathbf f}{x_i} + \mathbf f\sum\pd{\phi}{x_i}
=\phi\nabla\cdot \mathbf f + \mathbf f\nabla \phi
$$

mathmari said:
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.

mathmari said:
For the third equality:

Is everything correct?

Yup. (Nod)
 
  • #3
Re: prove also the Gauss theorem: Prove equalities

I like Serena said:
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)
I like Serena said:
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)
 
  • #4
Re: prove also the Gauss theorem: Prove equalities

mathmari said:
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)
mathmari said:
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)
 
  • #5
Re: prove also the Gauss theorem: Prove equalities

I like Serena said:
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)
 
  • #6
Re: prove also the Gauss theorem: Prove equalities

mathmari said:
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)
 
  • #7
Re: prove also the Gauss theorem: Prove equalities

I like Serena said:
Yep.
Great! (Happy)
I like Serena said:
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)
 
  • #8
Re: prove also the Gauss theorem: Prove equalities

mathmari said:
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)
 
  • #9
Re: prove also the Gauss theorem: Prove equalities

I like Serena said:
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)
 

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

1. What is Gauss theorem?

Gauss theorem is a fundamental theorem in mathematics that relates the surface integral of a vector field over a closed surface to the volume integral of the vector field inside the surface.

2. How is Gauss theorem used in physics?

Gauss theorem is used in physics to calculate the flux of a vector field through a closed surface, which is important in understanding the flow of electric and magnetic fields, gravity, and fluid dynamics.

3. How do you prove equalities using Gauss theorem?

To prove equalities using Gauss theorem, you first need to understand the problem at hand and identify the vector field and the closed surface that are involved. Then, you can use the mathematical formula of Gauss theorem to solve for the surface integral and compare it to the volume integral to prove the equality.

4. What are the limitations of Gauss theorem?

One limitation of Gauss theorem is that it can only be applied to closed surfaces, meaning surfaces that have no boundary or holes. It also requires the vector field to be continuous and differentiable in the region of interest.

5. Can Gauss theorem be generalized to higher dimensions?

Yes, Gauss theorem can be generalized to higher dimensions, such as 4-dimensional spacetimes in the theory of general relativity. This extension is known as the generalized Gauss theorem or the divergence theorem.

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