Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 336 (Perturbation Methods)

Let $L=\log (1/ϵ)$. Taking logarithms gives $x^2+\log x=L$. Equivalently, $2x^2=W(2/{ϵ}^2)$ in terms of the Lambert W function. Iteration for large $L$ gives $x^2=L-\frac{1}{2}\log L+O((\log L)/L)$ and hence

Write $e^x=1+xg(x)$, where $g=(e^x-1)/x$ is smooth at zero. Thenand a uniformly integrable expansion givesBy integration by parts,Therefore

For $F(t)=\mathrm{sech}\alpha \sinh t-t$, the relevant saddle point is $t=\alpha$, where $F(\alpha )=\tanh \alpha -\alpha$ and $F^{\prime \prime }(\alpha )=\tanh \alpha$. The method of steepest descent gives

At $\alpha =0$, the ordinary saddles coalesce at zero and $\sinh t-t=t^3/6+O(t^5)$. The contributing scale is $t=O({\nu }^{-1/3})$. The standard cubic saddle-point approximation givesEquivalently, the coefficient is $2^{1/3}\mathrm{Ai}(0)$.

Introduce $T=ϵt$ and write $x_0=A(T)\cos [t+\theta (T)]$. Averagingover one fast period gives $A{A}_T=-3A^4/8$ and ${\theta }_T=0$. The initial data therefore give the method of multiple scales resultthrough $t=O({ϵ}^{-1})$.

For the replacement damping, $\sin (\dot{x})\dot{x}={\dot{x}}^2-{\dot{x}}^4/6+\cdots$ is even in $\dot{x}$. Every term has zero resonant projection onto the fundamental over a complete orbit, so the $O(ϵ)$ slow amplitude and phase equations vanish. Thusthrough $t=O({ϵ}^{-1})$: there is no first-order secular damping, although bounded mean and higher-harmonic corrections occur.

For constant width, separation of variables with $\varphi =X(x)\sin (n\pi y/h_0)$ givesHence $X=A_n{e}^{i{k}_{n}x}+B_n{e}^{-i{k}_{n}x}$, with imaginary $k_n$ representing an evanescent mode.

For $X=ϵx$ and varying width,Projection of the next-order equation onto $\sin (n\pi y/h)$, whose squared norm is proportional to $h$, givesThus the WKB amplitude in a slowly varying duct isThis applies for $n\ge 1$ away from turning points $k_n=0$.

The reduced outer equation is $(1+x)y_0^{\prime }+y_0=0$. Imposing the right boundary condition givesThe condition at $x=0$ requires $X=x/ϵ$. The first two inner terms areThey match $2+ϵ(-2X+3/2)$. The additive composite expansion is

On $[-1,1]$, the coefficient $1+x$ vanishes at the left endpoint. The ordinary $O(ϵ)$ exponential layer is replaced by a turning-point endpoint region of width $O(\sqrt{ϵ})$, where all three terms in the equation enter the leading balance.

The outer expansion behaves as $f\sim x^{-1}$ near zero. The terms $x$ and $ϵf$ become comparable when $x^2=O(ϵ)$, soThe equation has the exact first integralwhere the constant follows from $f(1)=2$. ThusWriting $F=F_0+ϵF_1+\cdots$ and choosing the branch matching the positive outer solution givesHenceIts large-$X$ expansion matches the supplied outer series.