Green's 2nd identity in two dimension is,

$\iint _{S}(u\nabla ^{2}v-v\nabla ^{2}u)dxdy=\oint _{\partial S}(u\nabla v-v\nabla u)\cdot {\textbf {n}}dl$

where $u$ and $v$ are twice continuously differentiable scalar function of $x$ and $y$ on a 2D domain $S$. $\partial S$ and ${\textbf {n}}$ are boundary and normal unit vector of $S$, respectively.

## Prove

Green's theorem in integral calculus is as below,

$\iint _{S}(\partial _{x}Q-\partial _{y}P)dxdy=\oint _{\partial S}Pdx+Qdy$

Define a vector ${\textbf {f}}=(f_{1},f_{2})$ where $f_{1}=Q$ and $f_{2}=-P$, there is

$\iint _{S}\nabla \cdot {\textbf {n}}dxdy=\iint _{S}(\partial _{x}f_{1}+\partial _{y}f_{2})dxdy=\iint _{S}(\partial _{x}Q-\partial _{y}P)dxdy=\oint _{\partial S}Pdx+Qdy=\oint _{\partial S}-f_{2}dx+f_{1}dy=\oint _{\partial S}(f_{1},f_{2})\cdot (dy,-dx)=\oint _{\partial S}{\textbf {f}}\cdot {\textbf {n}}dl$

which is the 2D divergence theorem; then choose ${\textbf {f}}$ to be $u\nabla v=(u\partial _{x}v,u\partial _{y}v)$ instead. There is

$\iint _{S}\nabla \cdot (u\nabla v)dxdy=\oint _{\partial S}u\nabla v\cdot {\textbf {n}}dl$

Therefore we can get Green's 1st identity in 2D

$\iint _{S}(\nabla u\cdot \nabla v+u\nabla ^{2}v)dxdy=\oint _{\partial S}u\nabla v\cdot {\textbf {n}}dl$

and similarly,

$\iint _{S}(\nabla v\cdot \nabla u+v\nabla ^{2}u)dxdy=\oint _{\partial S}v\nabla u\cdot {\textbf {n}}dl$

Finally the 2D Green's 2nd identity is available by subtracting the two equations above.

## 1D Green's 2nd identity

For two functions $\Psi$ and $\Phi$, there is an identity according to fundamental theorem of calculus (Newton-leibniz formula),

$\int _{a}^{b}[\Psi (x'){\frac {d^{2}}{dx'^{2}}}\Phi (x')-\Phi (x'){\frac {d^{2}}{dx'^{2}}}\Psi (x')]dx'$
$=\int _{a}^{b}\{{\frac {d}{dx'}}[\Psi (x'){\frac {d}{dx'}}\Phi (x')]-{\frac {d}{dx'}}[\Phi (x'){\frac {d}{dx'}}\Psi (x')]\}dx'$
$=[\Psi (x'){\frac {d}{dx'}}\Phi (x')-\Phi (x'){\frac {d}{dx'}}\Psi (x')]|_{a}^{b}$