$\int_V \left( g\nabla^2f-f \nabla^2g \right) dV=\int_S \left( g\nabla f-f \nabla g \right) \cdot u_n \, dS$ not depend on time

Question

I' m wondering why the following relationship, known as Green's identity, doesn't depends on time.

Let

$f(x,y,z,t)$ and $g=\frac{e^{ikr}}{r}$

so

$$\int_V \left( g\nabla^2f-f \nabla^2g \right) dV=\int_S \left( g\nabla f-f \nabla g \right) \cdot u_n \, dS$$

Answer 1

The identity is easily shown using the divergence theorem: $$ \int_V \left( g \, \nabla^2 f - f \, \nabla^2 g \right) dV = \int_V \nabla \cdot \left( g \, \nabla f - f \, \nabla g \right) dV = \oint_S \left( g \, \nabla f - f \, \nabla g \right) \cdot u_n \, dS $$ Both $f$ and $g$ may depend on $t$, but this dependency does not enter the identity. There will be no derivatives w.r.t. $t$. That is related to the integral $\int \cdot \, dV$ being only over space, not over time.