Proof of Green's identity
Can anyone explain to me how to prove Green's identity by integrating the divergence theorem?
I don't understand how divergence, total derivative, and Laplace are related to each other.
Why is this true:
$$\nabla \cdot (u\nabla v) = u\Delta v + \nabla u \cdot \nabla v?$$
How do we integrate both parts? Thanks for answering.
$\endgroup$1 Answer
$\begingroup$The identity follows from the product rule $$ \frac{d}{dx}(f(x)\cdot g(x))=\frac{df}{dx}(x)g(x)+f(x)\frac{dg}{dx}(x). $$ for two functions $f$ and $g$. Noting that $\nabla\cdot\nabla=\Delta$ we get $$ \nabla u\cdot \nabla v+u\nabla\cdot\nabla v =\nabla u\cdot \nabla v+ u\Delta v. $$ Applying the divergence theorem $$ \int_{V}(\nabla\cdot \underline{F})dV=\int_{S}\underline{F}\cdot \underline{n}dS $$ (where $V\subset\mathbb{R}^n,~S$ is its boundary, $\underline{F}$ is a vector field and $\underline{n}$ is the outward unit normal from the surface) and inserting it into the above identity gives $$ \int_S u(\nabla v).\underline{n}dS=\int_V u\Delta v+(\nabla u)\cdot (\nabla v) dV, $$ ie, Green's first identity.
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