Using the Divergence Theorem to Prove Green's Theorem

[B3NR4Y]

Homework Statement

Prove Green's theorem
$$\int_{\tau} (\varphi \nabla^{2} \psi -\psi\nabla^{2}\varphi)d\tau = \int_{\sigma}(\varphi\nabla\psi -\psi\nabla\varphi)\cdot d\vec{\sigma}$$

Homework Equations

$$div (\vec{V})=\lim_{\Delta\tau\rightarrow 0} \frac{1}{\Delta\tau} \int_{\sigma} \vec{V} \cdot d\vec{\sigma}$$
is the divergence, and Gauss's divergence theorem says
$$\int_{\tau} div(\vec{V}) d\tau=\int_{\sigma} V \cdot d\vec{\sigma}$$
$$div(\vec{V})=\nabla \cdot \vec{V}$$

The Attempt at a Solution

The first thing I'll do is apply Gauss's Divergence theorem to the vectors $\vec{A} = \varphi\nabla\psi$ and $B=\psi\nabla\varphi$
Applied to A:
$\nabla \cdot \vec{A}=\nabla \cdot \varphi\nabla\psi =\varphi \nabla^{2} \psi \rightarrow \int_{\tau} div(\vec{A}) d\tau= \int_{\tau} \varphi \nabla^{2} \psi \, d\tau = \int_{\sigma} \varphi \nabla\psi \,\cdot d\vec{\sigma}$

Applied to B
$\nabla \cdot \vec{B}=\nabla \cdot \psi\nabla\varphi =\psi \nabla^{2} \varphi \rightarrow \int_{\tau} div(\vec{B})\, d\tau= \int_{\tau} \psi \nabla^{2} \varphi \, d\tau = \int_{\sigma} \psi \nabla\varphi \,\cdot d\vec{\sigma}$

And there is where I am stuck. These look like they're supposed to, but I am not sure if I'm allowed to subtract them to get Green's Theorem or if I am supposed to do something else.

 

[Ray Vickson]

Homework Statement

Prove Green's theorem
$$\int_{\tau} (\varphi \nabla^{2} \psi -\psi\nabla^{2}\varphi)d\tau = \int_{\sigma}(\varphi\nabla\psi -\psi\nabla\varphi)\cdot d\vec{\sigma}$$

Homework Equations

$$div (\vec{V})=\lim_{\Delta\tau\rightarrow 0} \frac{1}{\Delta\tau} \int_{\sigma} \vec{V} \cdot d\vec{\sigma}$$
is the divergence, and Gauss's divergence theorem says
$$\int_{\tau} div(\vec{V}) d\tau=\int_{\sigma} V \cdot d\vec{\sigma}$$
$$div(\vec{V})=\nabla \cdot \vec{V}$$

The Attempt at a Solution

The first thing I'll do is apply Gauss's Divergence theorem to the vectors $\vec{A} = \varphi\nabla\psi$ and $B=\psi\nabla\varphi$
Applied to A:
$\nabla \cdot \vec{A}=\nabla \cdot \varphi\nabla\psi =\varphi \nabla^{2} \psi \rightarrow \int_{\tau} div(\vec{A}) d\tau= \int_{\tau} \varphi \nabla^{2} \psi \, d\tau = \int_{\sigma} \varphi \nabla\psi \,\cdot d\vec{\sigma}$

Applied to B
$\nabla \cdot \vec{B}=\nabla \cdot \psi\nabla\varphi =\psi \nabla^{2} \varphi \rightarrow \int_{\tau} div(\vec{B})\, d\tau= \int_{\tau} \psi \nabla^{2} \varphi \, d\tau = \int_{\sigma} \psi \nabla\varphi \,\cdot d\vec{\sigma}$

And there is where I am stuck. These look like they're supposed to, but I am not sure if I'm allowed to subtract them to get Green's Theorem or if I am supposed to do something else.

Why do you claim that $\nabla \cdot (\varphi \nabla \psi) = \psi \nabla^2 \varphi$? Have you proved it?

 

[B3NR4Y]

Why do you claim that $\nabla \cdot (\varphi \nabla \psi) = \psi \nabla^2 \varphi$? Have you proved it?

$$\begin{align*} \nabla \cdot \psi \nabla \varphi =\\ \nabla \cdot \psi <\partial_1 \varphi, \partial_2 \varphi, \partial_3 \varphi> =\\ \nabla \cdot <\psi \partial_1 \varphi,\psi \partial_2 \varphi, \psi\partial_3 \varphi> =\\ \psi\partial_{1}^{2}\varphi + \psi\partial_{2}^{2}\varphi + \psi \partial_{3}^{2}\varphi=\\ \psi (\partial_{1}^{2}\varphi + \partial_{2}^{2}\varphi + \partial_{3}^{2}\varphi)=\\ \psi\nabla^{2}\varphi \end{align*}$$

My logic may be wrong, but here's what I had written. I also "proved" the dot product is distributive, so I could use that for this problem.

$$\begin{align*} (\vec{A}-\vec{B})\cdot \vec{C}=\vec{A}\cdot\vec{C}-\vec{B}\cdot\vec{C} \rightarrow \\ \vec{A}-\vec{B}=a_{i}-b_{i} \rightarrow https://www.physicsforums.com/file://\\(\vec{A}-\vec{B})\cdot \vec{C}= (a_{i}-b_{i})c_{i} \, ; \, \, \\\vec{A}\cdot\vec{C}-\vec{B}\cdot\vec{C} = a_{i}c_{i}-b_{i}c_{i} = (a_{i}-b_{i})c_{i} = (\vec{A}-\vec{B})\cdot \vec{C} \end{align*}$$

Therefore I can use this to subtract the two integrals I got and boom, green's theorem?

 

[Dick]

$$\begin{align*} \nabla \cdot \psi \nabla \varphi =\\ \nabla \cdot \psi <\partial_1 \varphi, \partial_2 \varphi, \partial_3 \varphi> =\\ \nabla \cdot <\psi \partial_1 \varphi,\psi \partial_2 \varphi, \psi\partial_3 \varphi> =\\ \psi\partial_{1}^{2}\varphi + \psi\partial_{2}^{2}\varphi + \psi \partial_{3}^{2}\varphi=\\ \psi (\partial_{1}^{2}\varphi + \partial_{2}^{2}\varphi + \partial_{3}^{2}\varphi)=\\ \psi\nabla^{2}\varphi \end{align*}$$

My logic may be wrong, but here's what I had written.

Are you assuming $\psi$ is a constant? If not don't you have to differentiate it too? Use the product rule.

 

[Ray Vickson]

$$\begin{align*} \nabla \cdot \psi \nabla \varphi =\\ \nabla \cdot \psi <\partial_1 \varphi, \partial_2 \varphi, \partial_3 \varphi> =\\ \nabla \cdot <\psi \partial_1 \varphi,\psi \partial_2 \varphi, \psi\partial_3 \varphi> =\\ \psi\partial_{1}^{2}\varphi + \psi\partial_{2}^{2}\varphi + \psi \partial_{3}^{2}\varphi=\\ \psi (\partial_{1}^{2}\varphi + \partial_{2}^{2}\varphi + \partial_{3}^{2}\varphi)=\\ \psi\nabla^{2}\varphi \end{align*}$$

My logic may be wrong, but here's what I had written. I also "proved" the dot product is distributive, so I could use that for this problem.

$$\begin{align*} (\vec{A}-\vec{B})\cdot \vec{C}=\vec{A}\cdot\vec{C}-\vec{B}\cdot\vec{C} \rightarrow \\ \vec{A}-\vec{B}=a_{i}-b_{i} \rightarrow https://www.physicsforums.com/file://\\(\vec{A}-\vec{B})\cdot \vec{C}= (a_{i}-b_{i})c_{i} \, ; \, \, \\\vec{A}\cdot\vec{C}-\vec{B}\cdot\vec{C} = a_{i}c_{i}-b_{i}c_{i} = (a_{i}-b_{i})c_{i} = (\vec{A}-\vec{B})\cdot \vec{C} \end{align*}$$

Therefore I can use this to subtract the two integrals I got and boom, green's theorem?

Essentially, part of what you claim is that
$$\frac{d}{dx} \left( f(x) \frac{dg(x)}{dx} \right) = f(x) \frac{d^2 g(x)}{dx^2}$$
Do you honestly believe that?

 

[B3NR4Y]

Oh, I'm dumb.
$$\begin{align*} \nabla \cdot \vec{A} &= \nabla \cdot \psi \nabla\varphi\\ &=\nabla \cdot \psi <\partial_{1}\varphi \, , \, \partial_{2} \varphi \, , \, \partial_3\varphi>\\ &=\nabla \cdot <\psi\partial_{1}\varphi \, , \, \psi\partial_{2} \varphi \, , \, \psi\partial_3\varphi>\\ &=(\partial_{1}\psi \, \partial_{1}\varphi + \psi \partial_{1}^2 \varphi)+(\partial_{2}\psi \, \partial_{2}\varphi + \psi \partial_{2}^2 \varphi)+(\partial_{3}\psi \, \partial_{3}\varphi + \psi \partial_{3}^2 \varphi)\\ &=\partial_{i}\psi\partial_{i}\varphi+\psi\partial_{i}^2 \varphi \\ &=\nabla\psi \cdot \nabla \psi +\psi\nabla^2\varphi \end{align*}$$

And for B:
$$\begin{align*} \nabla \cdot \vec{B} &= \nabla \cdot \varphi \nabla\psi\\ &=\nabla \cdot \varphi <\partial_{1}\psi \, , \, \partial_{2} \psi \, , \, \partial_3\psi>\\ &=\nabla \cdot <\varphi\partial_{1}\psi \, , \, \varphi\partial_{2} \psi \, , \, \varphi\partial_3\psi>\\ &=(\partial_{1}\varphi \, \partial_{1}\psi + \varphi \partial_{1}^2 \psi)+(\partial_{2}\psi \, \partial_{2}\psi + \varphi \partial_{2}^2 \psi)+(\partial_{3}\varphi \, \partial_{3}\psi + \varphi \partial_{3}^2 \psi)\\ &=\partial_{i}\varphi\partial_{i}\psi+\varphi\partial_{i}^2 \psi \\ &=\nabla\varphi \cdot \nabla \psi +\varphi\nabla^2\psi \end{align*}$$

And therefore the integrals I had before from Gauss's Divergence Theorem are
$$\begin{align*} A\rightarrow &\int_\tau (\nabla\psi \cdot \nabla \varphi +\psi\nabla^2\varphi) \, d\tau = \int_{\sigma} \psi\nabla\varphi \cdot d \vec{\sigma} \\ B\rightarrow &\int_\tau (\nabla\varphi \cdot \nabla \psi +\varphi\nabla^2\psi) \, d\tau = \int_{\sigma} \varphi\nabla\psi \cdot d \vec{\sigma} \end{align*}$$

and THEN I subtract, the dot product part at the beginning cancels and bam I have green's theorem?

 

[Dick]

Oh, I'm dumb.
$$\begin{align*} \nabla \cdot \vec{A} &= \nabla \cdot \psi \nabla\varphi\\ &=\nabla \cdot \psi <\partial_{1}\varphi \, , \, \partial_{2} \varphi \, , \, \partial_3\varphi>\\ &=\nabla \cdot <\psi\partial_{1}\varphi \, , \, \psi\partial_{2} \varphi \, , \, \psi\partial_3\varphi>\\ &=(\partial_{1}\psi \, \partial_{1}\varphi + \psi \partial_{1}^2 \varphi)+(\partial_{2}\psi \, \partial_{2}\varphi + \psi \partial_{2}^2 \varphi)+(\partial_{3}\psi \, \partial_{3}\varphi + \psi \partial_{3}^2 \varphi)\\ &=\partial_{i}\psi\partial_{i}\varphi+\psi\partial_{i}^2 \varphi \\ &=\nabla\psi \cdot \nabla \psi +\psi\nabla^2\varphi \end{align*}$$

And for B:
$$\begin{align*} \nabla \cdot \vec{B} &= \nabla \cdot \varphi \nabla\psi\\ &=\nabla \cdot \varphi <\partial_{1}\psi \, , \, \partial_{2} \psi \, , \, \partial_3\psi>\\ &=\nabla \cdot <\varphi\partial_{1}\psi \, , \, \varphi\partial_{2} \psi \, , \, \varphi\partial_3\psi>\\ &=(\partial_{1}\varphi \, \partial_{1}\psi + \varphi \partial_{1}^2 \psi)+(\partial_{2}\psi \, \partial_{2}\psi + \varphi \partial_{2}^2 \psi)+(\partial_{3}\varphi \, \partial_{3}\psi + \varphi \partial_{3}^2 \psi)\\ &=\partial_{i}\varphi\partial_{i}\psi+\varphi\partial_{i}^2 \psi \\ &=\nabla\varphi \cdot \nabla \psi +\varphi\nabla^2\psi \end{align*}$$

And therefore the integrals I had before from Gauss's Divergence Theorem are
$$\begin{align*} A\rightarrow &\int_\tau (\nabla\psi \cdot \nabla \varphi +\psi\nabla^2\varphi) \, d\tau = \int_{\sigma} \psi\nabla\varphi \cdot d \vec{\sigma} \\ B\rightarrow &\int_\tau (\nabla\varphi \cdot \nabla \psi +\varphi\nabla^2\psi) \, d\tau = \int_{\sigma} \varphi\nabla\psi \cdot d \vec{\sigma} \end{align*}$$

and THEN I subtract, the dot product part at the beginning cancels and bam I have green's theorem?

Sounds much better.

 

[B3NR4Y]

Thank you both for your help, if it weren't for you guys I would have just subtracted my two original equations and gotten what looks right, but is a wrong answer. And would have been befuddled when I got a bad grade (most of the time we don\t get our assignments back despite the class only having about 13 students.)