- #1
B3NR4Y
Gold Member
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Homework Statement
Prove Green's theorem
[tex]
\int_{\tau} (\varphi \nabla^{2} \psi -\psi\nabla^{2}\varphi)d\tau = \int_{\sigma}(\varphi\nabla\psi
-\psi\nabla\varphi)\cdot d\vec{\sigma}[/tex]
Homework Equations
[tex] div (\vec{V})=\lim_{\Delta\tau\rightarrow 0} \frac{1}{\Delta\tau} \int_{\sigma} \vec{V} \cdot d\vec{\sigma} [/tex]
is the divergence, and Gauss's divergence theorem says
[tex]\int_{\tau} div(\vec{V}) d\tau=\int_{\sigma} V \cdot d\vec{\sigma} [/tex]
[tex] div(\vec{V})=\nabla \cdot \vec{V} [/tex]
The Attempt at a Solution
The first thing I'll do is apply Gauss's Divergence theorem to the vectors [itex] \vec{A} = \varphi\nabla\psi [/itex] and [itex]B=\psi\nabla\varphi[/itex]
Applied to A:
[itex]\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} [/itex]
Applied to B
[itex]\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}[/itex]
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.