# Showing Greens First Theorem Integral is 0

• EngageEngage
In summary, the conversation is about solving a problem using Greens first theorem. The problem involves finding the integral of two vectors, given certain conditions. The solution involves setting the integral to equal the requested answer and eliminating unnecessary terms. The conversation also includes a helpful suggestion to switch the roles of the two vectors.

#### EngageEngage

[SOLVED] Greens Functions

## Homework Statement

show:
$$\int\int\int_{D}\vec{F}\cdot\vec{G}dV = 0$$
where:
$$\vec{F}=\nabla\phi$$
$$\vec{G}=\nabla\psi$$
$$\nabla\cdot\vec{F}=0$$
$$\psi|_{\partial D}=0$$

## The Attempt at a Solution

This looks like a problem for Greens first theorem:

$$\int\int\int_{D}\phi\nabla^{2}\psi dV = \int\int_{\partial D}\phi\nabla\psi dS - \int\int\int_{D}\nabla\psi\cdot\nabla\phi dV$$

The very right term is clearly the integral that I'm looking for. So, i will set it to look like the requested answer. Also, I know that
$$\psi|_{\partial D}=0$$
meaning that I can also throw out the second term because that term wants me to integrate the gradient of psi over the surface, while I know that psi is 0 over the surface. So, I am left with this:

$$\int\int\int_{D}\phi\nabla^{2}\psi dV = - \int\int\int_{D}\vec{F}\cdot\vec{G} dV$$

So, this means that the term on the left mus equal zero. Does anyone know how I can show this? Psi is not zero through the domain, and the problem doesn't specify that it is a harmonic potential (although I suppose it could be). Could someone please help me with this step? Any help at all is greatly appreciated.

Uh, phi is given to be harmonic, not psi. Since the divergence of it's gradient is zero. Why don't you move the laplacian operator over to phi?

Last edited:
I.e. just switch the roles of psi and phi?

Thanks for the help! This will also eliminate the middle term anyways too, since psi will be zero there on the boundary, right?

EngageEngage said:
Thanks for the help! This will also eliminate the middle term anyways too, since psi will be zero there on the boundary, right?

Right.

thanks a lot for the help Dick