We consider a wave propagation problem in a medium ${\displaystyle D}$ bounded by a surface ${\displaystyle \partial D}$. By the Green's function method, we can represent the field at a distinguished location ${\displaystyle {\boldsymbol {x}}_{0}}$ using the Green's function method integral equation of the field

${\displaystyle u(\mathbf {x_{0}} )=\int _{D}f(\mathbf {x} )g^{\star }(\mathbf {x} ,\mathbf {x_{0}} )\;dV+\int _{\partial D}\left\{g^{\star }(\mathbf {x} ,\mathbf {x_{0}} ){\frac {\partial u(\mathbf {x} )}{\partial n}}-u(\mathbf {x} ){\frac {\partial g^{\star }(\mathbf {x} ,\mathbf {x_{0}} )}{\partial n}}\right\}\;dS.}$

We can pose boundary value problems by specifying values of either the field ${\displaystyle u(\mathbf {x} )}$ (called a Dirichlet condition) or derivatives of the field ${\displaystyle {\frac {\partial u(\mathbf {x} )}{\partial n}}}$, (called a Neumann condition.

## Unbounded medium

For this discussion, we consider the boundary condition of an unbounded medium. In this case the distance between the support of the source function ${\displaystyle f}$ and the boundary ${\displaystyle \partial D}$ is taken to tend to infinity.

In this case the boundary condition is a radiation condition. The basic notion of the radiation condition is that the field is outward propagating from the region of interest described by the the volume integral, and that the wavefield falls off sufficiently rapidly that the surface integral contribution tends to zero.

A mistake that is often made it to imagine that this means that the terms in the integrand cancel. This is not the case. This an asymptotic analysis.

### The Helmholtz equation and the Sommerfeld radiation condition

For the problem, in three dimensions, where the Helmholtz equation is the governing equation of the problem, the Sommerfeld radiation condition takes the form that any field ${\displaystyle U({\boldsymbol {x}},\omega )}$ that is a slolution to the Helmholtz equation has the following asymptotic behavior

${\displaystyle U({\boldsymbol {x}},\omega )=O(1/r)\qquad }$ as ${\displaystyle \qquad r\rightarrow \infty }$

and that

${\displaystyle {\frac {\partial U({\boldsymbol {x}},\omega )}{\partial r}}-{\frac {i\omega }{V(x)}}U({\boldsymbol {x}},\omega )=o(1/r)\qquad }$ as ${\displaystyle \qquad r\rightarrow \infty }$,

where ${\displaystyle r}$ is the radial distance from the source, and any scatterers of interest.