Showing Greens First Theorem Integral is 0

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The discussion focuses on demonstrating that the integral of the dot product of two gradient fields, \(\vec{F}\) and \(\vec{G}\), over a domain \(D\) is zero, based on Green's first theorem. The user attempts to manipulate the theorem's equation by recognizing that the boundary condition \(\psi|_{\partial D}=0\) allows them to eliminate certain terms. They conclude that the left-hand side of the equation must equal zero, as the integral of \(\phi \nabla^2 \psi\) simplifies under the given conditions. A suggestion is made to switch the roles of \(\phi\) and \(\psi\) to further clarify the solution. The conversation highlights the importance of boundary conditions in solving integral equations in vector calculus.
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[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.
 
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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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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