Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 335 / 2 / i / Solution

Use the sign convention in the question,Then the total field satisfiesThe outgoing free-space Green function iswith $({\nabla }^2+k_0^2)G_0=-\delta$. Hence the Lippmann-Schwinger equation isReplacing the unknown interior total field by the incident field gives the first Born approximation

For the Rytov approximation, put $\psi ={\psi }_i{e}^{\varphi }$. After division by $\psi$, the wave equation givesNeglecting the quadratic term $(\nabla \varphi {)}^2$ makes ${\psi }_i\varphi$ obey the same inhomogeneous equation as the first Born scattered field. ThusandIts power series beginsso the two approximations agree through first order in the scattering potential.