Since $\Phi_\theta=-A\sin\theta$ and the average of $\sin^2\theta$ over one period is $1/2$, the leading <averaged Lagrangian> is
$$
\boxed{
\overline L^{(0)}(k,\omega,A)
=\frac14A^2(\omega^2-c^2k^2).}
$$
Its <Euler-Lagrange equation> for the slowly varying amplitude is algebraic:
$$
\frac{\partial\overline L^{(0)}}{\partial A}
=\frac12A(\omega^2-c^2k^2)=0.
$$
For a nonzero wave this again gives the <dispersion relation>. It does not determine $A$ because the original wave equation is linear and homogeneous: the leading amplitude is fixed only by the next-order transport equation and by initial or boundary data.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 336 / 1 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-336/1/b/ii/solution