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

Substitute

m_{4} = e^{Ψ}

into the given equation. The chain rule gives

Ψ_{x} = \frac{i}{k} (Ψ_{r_{1} r_{2}} + P s i_{r_{1}} Ψ_{r_{2}}) - Q .

The incident reduced field is the constant one, so

m_{4} (- Δ x) = 1

and

Ψ (- Δ x) = 0

. During a thin-screen crossing,

Ψ = O (Δ x)

. Its mixed-derivative contribution integrates to

O (Δ x)

. More precisely, its contribution to

Ψ

O ((Δ x)^{2})

, while the product of its first derivatives contributes only at still higher order. Therefore

Ψ (0, r_{1}, r_{2}) = - Q (r_{1}, r_{2}) Δ x + O ((Δ x)^{2}),

and hence

m_{4} (0, r_{1}, r_{2}) = e^{- Q (r_{1}, r_{2}) Δ x} + O ((Δ x)^{2}) = 1 - Q (r_{1}, r_{2}) Δ x + O ((Δ x)^{2}) .

b

Q = 0

, and the fourth moment obeys

\partial_{x} m_{4} = \frac{i}{k} \partial_{r_{1}} \partial_{r_{2}} m_{4} .

For the two-dimensional Fourier transform

m_{4} (x, p_{1}, p_{2}) = \int_{R^{2}} m_{4} (x, r_{1}, r_{2}) e^{- i (p_{1} r_{1} + p_{2} r_{2})} d r_{1} d r_{2},

\partial_{x} m_{4} = - \frac{i}{k} p_{1} p_{2} m_{4} .

Thus

m_{4} (L, p_{1}, p_{2}) = e^{- i L p_{1} p_{2} / k} m_{4} (0, p_{1}, p_{2}) .

Using the screen-exit value from part (a), the inverse transform gives

m_{4} (L, r_{1}, r_{2}) = \frac{1}{( 2 π ) ^{2}} \int_{R^{2}} e^{i (p_{1} r_{1} + p_{2} r_{2}) - i L p_{1} p_{2} / k} e^{- Q Δ x} (p_{1}, p_{2}) d p_{1} d p_{2}

through first order in

Δ x

m_{4} (L, r_{1}, r_{2}) = \frac{k}{2 π L} \int_{R^{2}} exp [\frac{ik}{L} (r_{1} - s_{1}) (r_{2} - s_{2})] m_{4} (0, s_{1}, s_{2}) d s_{1} d s_{2} .

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