In free space $Q=0$, and the fourth moment obeys
$$
\partial_xm_4=\frac{i}{k}\partial_{r_1}\partial_{r_2}m_4.
$$
For the two-dimensional <Fourier transform>
$$
\widehat m_4(x,p_1,p_2)
=\int_{\mathbb R^2}m_4(x,r_1,r_2)
e^{-i(p_1r_1+p_2r_2)}\,dr_1dr_2,
$$
the <Fourier transform of a derivative> turns this equation into the <ordinary differential equation>
$$
\partial_x\widehat m_4
=-\frac{i}{k}p_1p_2\widehat m_4.
$$
Thus
$$
\boxed{
\widehat m_4(L,p_1,p_2)
=e^{-iLp_1p_2/k}
\widehat m_4(0,p_1,p_2)}.
$$
Using the screen-exit value from part (a), the inverse transform gives
$$
\boxed{
m_4(L,r_1,r_2)=\frac{1}{(2\pi)^2}
\int_{\mathbb R^2}e^{i(p_1r_1+p_2r_2)-iLp_1p_2/k}
\widehat{e^{-Q\Delta x}}(p_1,p_2)\,dp_1dp_2}
$$
through first order in $Δx$. Equivalently, the <free-space fourth-moment propagator> has kernel
$$
\boxed{
m_4(L,r_1,r_2)=\frac{k}{2\pi L}
\int_{\mathbb R^2}
\exp\left[\frac{ik}{L}(r_1-s_1)(r_2-s_2)\right]
m_4(0,s_1,s_2)\,ds_1ds_2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 335 / 2 / iii / b / Solution

For $m_x=(i/k)m_{r_1r_2}$, the free-space fourth-moment propagator multiplies the two-dimensional Fourier transform by $e^{-ixp_1p_2/k}$. In physical coordinates its oscillatory kernel is
$$
K_x(r_1,r_2)=\frac{k}{2\pi x}e^{ikr_1r_2/x}.
$$


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