 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/2/c/solution

The first-order condition

Q = 0

, combined with

b_{0} = - i ψ_{0, y}

, gives

b_{1} = - i ψ_{1, y} - \frac{i}{ω} ψ_{0, x} + \frac{1}{ω} ψ_{0, yyy} .

Differentiate this expression with respect to

y

and insert it into

S = 0

. The

ψ_{1}

terms cancel, leaving

0 = - iω ψ_{1, yy} - ω [- i ψ_{1, yy} - \frac{i}{ω} ψ_{0, x y} + \frac{1}{ω} ψ_{0, yyyy}] + i ψ_{0, x y} = 2 i ψ_{0, x y} - ψ_{0, yyyy} .

2 ψ_{0, x y} + i ψ_{0, yyyy} = 0.

It is exactly

\partial_{y} (2 ψ_{0, x} + i ψ_{0, yyy}) = 0

. Integrating with respect to

y

therefore gives

2 ψ_{0, x} + i ψ_{0, yyy} = f (x),

where the integration constant may be any function of

x

. Conversely, differentiating this last equation with respect to

y

recovers the compatibility equation.

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