Use the sign convention in the question,
$$
V(\mathbf r)=k_0^2[n^2(\mathbf r)-1].
$$
Then the total field satisfies
$$
(\nabla^2+k_0^2)\psi=-V\psi.
$$
The outgoing free-space <Green function> is
$$
G_0(\mathbf r-\mathbf r')
=\frac{e^{ik_0|\mathbf r-\mathbf r'|}}
{4\pi|\mathbf r-\mathbf r'|},
$$
with $(\nabla^2+k_0^2)G_0=-\delta$. Hence the <Lippmann-Schwinger equation> is
$$
\psi(\mathbf r)=\psi_i(\mathbf r)
+\int_DG_0(\mathbf r-\mathbf r')
V(\mathbf r')\psi(\mathbf r')\,d^3r'.
$$
Replacing the unknown interior total field by the incident field gives the first <Born approximation>
$$
\boxed{
\psi_B(\mathbf r)=\psi_i(\mathbf r)
+\int_DG_0(\mathbf r-\mathbf r')
V(\mathbf r')\psi_i(\mathbf r')\,d^3r'}.
$$

For the <Rytov approximation>, put $\psi=\psi_i e^\phi$. After division by $\psi$, the wave equation gives
$$
\nabla^2\phi+(\nabla\phi)^2
+2\nabla\log\psi_i\mathbin\cdot\nabla\phi
=-V.
$$
Neglecting the quadratic term $(\nabla\phi)^2$ makes $\psi_i\phi$ obey the same inhomogeneous equation as the first Born scattered field. Thus
$$
\phi_1(\mathbf r)
=\frac1{\psi_i(\mathbf r)}
\int_DG_0(\mathbf r-\mathbf r')
V(\mathbf r')\psi_i(\mathbf r')\,d^3r',
$$
and
$$
\boxed{
\psi_R(\mathbf r)=\psi_i(\mathbf r)e^{\phi_1(\mathbf r)}}.
$$
Its <power series> begins
$$
\psi_R=\psi_i(1+\phi_1+O(\phi_1^2))
=\psi_B+O(V^2),
$$
so the two approximations agree through first order in the <scattering potential>.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-335/2/i/solution