Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 336 (Perturbation Methods)

The chain rule andgiveApplying the same rule to $L_1$ and $L_2$ in the Euler-Lagrange field equation yields the exact modulated equation

Multiply this equation by ${\Phi }_{\theta }$. Since $k$ and $\omega$ are independent of the fast phase,The product rule then rearranges the field equation into the exact modulated-wave first integralAverage this identity over one $2\pi$ period in $\theta$. Periodicity kills the first term, while differentiation of the averaged Lagrangian givesConsequently

The same two equations follow directly from the modulated variational principle. Varying $\Phi$ giveswhich is the exact field equation above. For a variation of $\Theta$, use $\delta k={\partial }_X\delta \Theta$ and $\delta \omega =-{\partial }_T\delta \Theta$. Integration by parts in $X$ and $T$ gives the averaged equation. Thus variation with respect to the periodic profile reproduces the local wave equation, whereas variation with respect to its slow phase gives the Whitham modulation equation for wave action.

Hereso the Euler-Lagrange field equation is the variable-coefficient wave equationAt leading order, $\Phi =A(X,T)\cos \theta$ hasand henceThe leading part of the modulated-wave first integral isA nonconstant periodic wave therefore requires the local dispersion relationThis is usually called the eikonal equation, or equivalently the leading geometric-optics dispersion relation.

Since ${\Phi }_{\theta }=-A\sin \theta$ and the average of ${\sin }^2\theta$ over one period is $1/2$, the leading averaged Lagrangian isIts Euler-Lagrange equation for the slowly varying amplitude is algebraic: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.

Define the wave-action densityThe local dispersion relation implieswhich is the group velocity. Moreover,Substitution into the phase-variation equation gives the wave-action conservation lawOn the two nondispersive branches $\omega =\pm ck$, the group velocities are $c_g=\pm c$.

LetThe inequality $|1-e^{-x^2}|\le x^2$ and a split into $0<x<{\epsilon }^2$, ${\epsilon }^2<x<\epsilon$, and $\epsilon <x<1$ show thatThus replacing the exponential by one does not affect any term through $O(1)$.

FactorwherePartial fraction decomposition givesThe required asymptotic expansion follows fromThereforeThe two small roots reveal the same nested scales $x=O({\epsilon }^2)$ and $x=O(\epsilon )$ that a divide-and-conquer asymptotic expansion would match explicitly.

Write the summand as $a_n=x^n/[\Gamma (n)n^n]$. By Stirling formula,SetThen the exponential phase isIt has a unique maximum at $t=1$, where $f(1)=2$ and $f^{\prime \prime }(1)=-2$. The contributing indices satisfy $n-N=O(\sqrt{N})$, so their width tends to infinity and the lattice sum may be replaced by a Riemann sum. The Discrete Laplace method therefore givesHence

Three distinguished limits are needed.

In the origin layer, put $x=\epsilon X$ and $y(x;\epsilon )=Y(X)$. Since $\mathrm{sech}(\epsilon X)=1+O({\epsilon }^2)$, the leading equation isIt has the first integralSolving this first-order linear differential equation givesRegularity at $X=0$ requires $D=-C$, and $Y(0)=1$ then fixes $C=1$. Thus Region I, $x=O(\epsilon )$, hasIts matching limit is $Y\sim X^{-1}$ as $X\to \infty$.

In Region II, $x=O(1)$ with $x\gg \epsilon$, setting $\epsilon =0$ reduces the second-order equation toHence $y_0=C{e}^{-x}/x$. Matching $C/x$ with the inner limit $\epsilon /x$ fixes $C=\epsilon$, so

The correction generated by $\epsilon x^{3}y$ changes the local decay rate when $x=O({\epsilon }^{-1/2})$. For Region III setThe terms of order ${\epsilon }^{-1/2}$ give the eikonal equationand the next balance gives $Z{G}^{\prime }+G=0$. Matching to Region II selectsThereforeThis solution is already beyond all algebraic orders in $\epsilon$ in the distant region. A leading multiplicative composite expansion that contains all three balances and satisfies the boundary value exactly iswith its value at $x=0$ understood by continuity.

Finally, the coefficient of $y^{\prime \prime }$ vanishes at the origin. Taking the regular limit of the original equation there givesso the prescribed value also fixes $y^{\prime }(0)=-1/(2\epsilon )$. Equivalently, the second inner solution behaves as $1/X$ and is excluded by regularity. This regular singular point is why one boundary condition determines the unique regular solution.