Showing Greens First Theorem Integral is 0

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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.
  • #1
EngageEngage
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[SOLVED] Greens Functions

Homework Statement


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

The Attempt at a Solution


This looks like a problem for Greens first theorem:

[tex]\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[/tex]

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
[tex]\psi|_{\partial D}=0[/tex]
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:

[tex]\int\int\int_{D}\phi\nabla^{2}\psi dV = - \int\int\int_{D}\vec{F}\cdot\vec{G} dV[/tex]

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.
 
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  • #2
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:
  • #3
I.e. just switch the roles of psi and phi?
 
  • #4
Thanks for the help! This will also eliminate the middle term anyways too, since psi will be zero there on the boundary, right?
 
  • #5
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.
 
  • #6
thanks a lot for the help Dick
 

What is Greens First Theorem?

Green's First Theorem, also known as the Divergence Theorem or Gauss's Theorem, is a fundamental theorem in vector calculus that relates a surface integral over a closed surface to a volume integral over the region enclosed by the surface. It is named after the mathematician George Green.

What is the significance of showing that the integral is 0?

Showing that the integral is 0 means that the vector field being integrated over a closed surface is a conservative vector field. This means that the value of the integral is not dependent on the path taken, but only on the initial and final points. In other words, the vector field is "converging" or "diverging" at a point, rather than "circulating" around it.

Can you provide an example of using Greens First Theorem to show an integral is 0?

One example is using Green's First Theorem to calculate the flux of a vector field through a closed surface. If the flux is equal to 0, then the vector field is conservative. For instance, if the vector field is given by F(x,y,z) = (2xy, x^2, z), then the flux through a closed surface can be shown to be 0 using Green's First Theorem.

What are the applications of Greens First Theorem?

Green's First Theorem has many applications in physics and engineering, particularly in the study of electric and magnetic fields. It is also used in fluid mechanics to calculate the flow of a fluid through a closed surface. It is also a key tool in the study of partial differential equations.

Are there any limitations to Greens First Theorem?

Green's First Theorem is limited to vector fields that are continuous and differentiable within the region being integrated. It also requires the surface to be closed and smooth. Additionally, it only applies to three-dimensional spaces, so it cannot be used in higher dimensions. Finally, the region being integrated must be simply connected, meaning it has no holes or voids within it.

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