My name is Alexander FufaeV and here I will explain the following topic:

First and Second Green's Identity

Here the two formulas, called Green's identities, are derived using the Divergence theorem. Green's identities are useful identities for converting integrals with gradients and divergences into integrals with normal derivatives. They are used, for example, in electrostatics to calculate electric potentials.

Consider a vector field $\boldsymbol{F}(\boldsymbol{r})$ (for example, it could be an electric field) that depends on the spatial coordinate $\boldsymbol{r} = (x,y,z)$ and is defined as follows:

$$ \begin{align} \boldsymbol{F}(\boldsymbol{r}) ~=~ \Phi(\boldsymbol{r}) \, \nabla \, \Psi(\boldsymbol{r}) \end{align} $$

Here $\Phi(\boldsymbol{r})$ and $ \Psi(\boldsymbol{r}) $ are smooth scalar functions (e.g. they could be two electric potential functions). And the symbol $\nabla$ is the nabla operator applied to the scalar function $ \Psi(\boldsymbol{r}) $. The result is the gradient of the scalar function: $\nabla \, \Psi(\boldsymbol{r})$. In the following we do not write the dependence on $\boldsymbol{r}$ to make the equations a bit more compact. Of course, the corresponding functions still depend on $\boldsymbol{r}$.

We want to apply the Divergence theorem to the vector field $\boldsymbol{F}$. The Divergence theorem states the following:

$$ \begin{align} \int_{V} \left(\nabla \cdot \boldsymbol{F}\right) \, \text{d}v ~=~ \oint_{A}\boldsymbol{F} \cdot \text{d}\boldsymbol{a} \end{align} $$

Divergence theorem illustrated.

Here $V$ is any volume and $A$ is the associated closed surface of the enclosed volume. Using the Divergence theorem, we can convert a volume integral (left side of 2) to a surface integral (right side of 2) and vice versa.

As you can see, the divergence of vector field $ \nabla \cdot \boldsymbol{F} $ occurs on the left side of the Divergence theorem. Therefore, the first step we want to take is to form the divergence of our vector field $\boldsymbol{F}$ defined in Eq. 1:

$$ \begin{align} \nabla ~\cdot~ \boldsymbol{F} &~=~ \nabla ~\cdot~ \left( \Phi \, \nabla \, \Psi \right)\\\\
&~=~ \Phi \, \nabla^2 \Psi ~+~ \nabla \Psi ~\cdot~ \nabla \Phi \end{align} $$

Here $\nabla^2$ is the Laplace operator, that is the sum of the second derivatives with respect to the spatial coordinates $x$, $y$ and $z$. In the second step, we have exploited the following identity for divergence:

$$ \begin{align} \nabla ~\cdot~ (u \, \boldsymbol{v}) ~=~ u \, \nabla \cdot \boldsymbol{v} ~+~ \boldsymbol{v} \cdot \nabla u \end{align} $$

Now we insert the divergence field 3 into the left side of the Divergence theorem 2:

$$ \begin{align} \int_{V} \left(\Phi \, \nabla^2 \Psi ~+~ \nabla \Psi ~\cdot~ \nabla \Phi\right) \, \text{d}v ~=~ \oint_{A}\boldsymbol{F} \cdot \text{d}\boldsymbol{a} \end{align} $$

Next, we express the area element $\text{d}\boldsymbol{a}$ with the area normal vector $\boldsymbol{n} $. This normal vector is perpendicular to the area element and points in the same direction as $\text{d}\boldsymbol{a}$ element:

$$ \begin{align} \int_{V} \left(\Phi \nabla^2 \Psi ~+~ \nabla \Psi ~\cdot~ \nabla \Phi\right) \, \text{d}v ~=~ \oint_{A}\boldsymbol{F} \cdot \boldsymbol{n} \, \text{d}a \end{align} $$

Then we insert the vector field 1 we defined into the right-hand side of the Divergence theorem 6:

$$ \begin{align} \int_{V} \left(\Phi \nabla^2 \Psi ~+~ \nabla \Psi ~\cdot~ \nabla \Phi\right) \, \text{d}v ~=~ \oint_{A} \Phi \, \nabla \, \Psi \cdot \boldsymbol{n} \, \text{d}a \end{align} $$

Here $\nabla \, \Psi \cdot \boldsymbol{n}$ is the directional derivative of $\Psi$. It is the component of the gradient field $\nabla \, \Psi$ in the direction of $\boldsymbol{n}$ on the surface $A$ (also called normal derivative). The normal derivative is also notated as follows:

$$ \begin{align} \nabla \, \Psi \cdot \boldsymbol{n} ~:=~ \frac{\partial \Psi}{\partial \boldsymbol{n}} \end{align} $$

Let us use this notation in Eq. 7. The we obtain:

$$ \begin{align} \int_{V} \left(\Phi \nabla^2 \Psi ~+~ \nabla \Psi ~\cdot~ \nabla \Phi\right) \, \text{d}v ~=~ \oint_{A} \Phi \, \frac{\partial \Psi}{\partial \boldsymbol{n}} \, \text{d}a \end{align} $$

If you set $ \Phi \equiv 1 $, then you get a useful special case ( recall that the gradient $\nabla \Phi$ of a constant function is zero):

$$ \begin{align} \int_{V} \nabla^2 \Psi \, \text{d}v ~=~ \oint_{A} \frac{\partial \Psi}{\partial \boldsymbol{n}} \, \text{d}a \end{align} $$

To get the second Green's identity, we first swap the scalar functions $\Psi$ and $\Phi$ in the first Green's identity:

$$ \begin{align} \int_{V} \left(\Psi \nabla^2 \Phi ~+~ \nabla \Phi ~\cdot~ \nabla \Psi\right) \, \text{d}v ~=~ \oint_{A} \Psi \, \frac{\partial \Phi}{\partial \boldsymbol{n}} \, \text{d}a \end{align} $$

Then we subtract from the 1st Green's identity the swapped version 11. Thus $ \nabla \Phi ~\cdot~ \nabla \Psi $ is eliminated, since divergence operation $~\cdot~$ is commutative. What remains is:

$$ \begin{align} \int_{V} \left(\Phi \nabla^2 \Psi ~-~ \Psi \nabla^2 \Phi \right) \, \text{d}v ~=~ \oint_{A} \left( \Phi \, \frac{\partial \Psi}{\partial \boldsymbol{n}} ~-~ \Psi \, \frac{\partial \Phi}{\partial \boldsymbol{n}} \right) \, \text{d}a \end{align} $$