Question

Help with Green's theorem/vector calculus identities

My notes say I can get Green's Theorem

\[\int_V \left(w \nabla^2 u - u \nabla^2w\right)dV = \int_S \left( w\nabla u - u\nabla w\right)\cdot d\mathbf{S}\]

by substituting \(\mathbf{A}=w\nabla u\) into the divergence theorem

\[\int_V \left(\nabla\cdot\mathbf{A}\right)dV = \int_S \mathbf{A}\cdot d\mathbf{S}\]

Answer 1

I am assuming that I am missing some vector calculus identity here.

Answer 2

if you are writing the div theorem as

\(\int_V \nabla \bullet \vec A ~ dV = \int_S \vec A \bullet d \vec S\)

and set \(\vec A = w \nabla u\)

then sub that into div theorem

\(\int_V \nabla \bullet (w \nabla u) ~ dV = \int_S (w \nabla u) \bullet d \vec S\)

\(\int_V \nabla w \bullet \nabla u + w \nabla^2 u ~ dV = \int_S (w \nabla u) \bullet d \vec S \qquad \star \)

then switch u and w around to get

\(\int_V \nabla u \bullet \nabla w + u \nabla^2 w ~ dV = \int_S (u \nabla w) \bullet d \vec S \qquad \circ\)

then \(\star - \circ \implies\)

\( \int_V \nabla w \bullet \nabla u + w \nabla^2 u - (\nabla u \bullet \nabla w + u \nabla^2 w) ~ dV = \int_S (w \nabla u - u \nabla w )\bullet d \vec S \qquad \)

\( \int_V w \nabla^2 u - u \nabla^2 w ~ dV = \int_S (w \nabla u - u \nabla v )\bullet d \vec S \qquad \)