Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 74 / 2

Introduce the slow time $T=\epsilon t$ and treat $t,T$ independently in the method of multiple scales. WriteAt the next order,For fixed $R$ define the period average byHere $f$ inside the averages is evaluated at the leading position and velocity $R\cos \varphi ,-R\sin \varphi$. Orthogonality to both fundamental harmonics is the solvability condition in the method of multiple scales: it removes the resonant forcing that would otherwise generate a secular term. Since the squared sine and cosine averages are $1/2$, the amplitude-phase equations areEquivalently $\dot{R}=-\epsilon ⟨f\sin (t+\theta )⟩$ and $\dot{\theta }=-\epsilon ⟨f\cos (t+\theta )⟩/R$ at first order. These equations describe a bounded, weakly perturbed oscillation over $t=O({\epsilon }^{-1})$ while the amplitude remains in a range where the expansion is ordered. At zero amplitude the oscillation phase coordinate is singular; Cartesian harmonic coefficients or an equilibrium analysis should replace it.

For conservative forcing in averaged oscillator equations, $f=f(y)$, the average determining $R_T$ vanishes. If $F^{\prime }=f$, thenwhose period average is zero. Thus $R=R_0$ to this order, while a position-dependent force generally shifts the oscillation phase and frequency. The exact equation also conserves the energy $\frac{1}{2}{\dot{y}}^2+\frac{1}{2}{y}^2-\epsilon F(y)$, consistent with the absence of amplitude drift for bounded orbits.

For the cubic position force, $⟨f\cos \varphi ⟩=R^3⟨{\cos }^4\varphi ⟩=3R^3/8$. ThereforeThe leading oscillation is $R_0\cos [(1-3\epsilon R_0^2/8)t+{\theta }_0]$. For positive $\epsilon$ it is a softening cubic oscillator, obtained by weakly perturbing a harmonic oscillator. The weak-force condition is $\epsilon R_0^2\ll 1$; the constant $R$ is the leading harmonic amplitude, not a claim that the exact waveform remains sinusoidal.

For velocity-only forcing in averaged oscillator equations, $f=f(\dot{y})$, the oscillation phase average vanishes: $f(-R\sin \varphi )\cos \varphi$ is a periodic total derivative, using an antiderivative of $f$. Hence ${\theta }_T=0$ to first order, although the amplitude may change.

In the cubic case, $f=-R^3{\sin }^3\varphi$, soWith $R(0)=R_0>0$, integration givesFor positive $\epsilon$ this is cubic velocity anti-damping, not damping: the exact energy derivative is $d({\dot{y}}^2+y^2)/(2dt)=\epsilon {\dot{y}}^4\ge 0$. The slow solution grows and formally diverges at $T=4/(3R_0^2)$. The weakly nonlinear approximation loses validity before its amplitude becomes so large that $\epsilon R^2$ is not small; its divergence is not a controlled prediction of the exact late-time solution. If $R_0=0$, the exact zero solution should be used separately.

Let $\delta =1-m\downarrow 0$. This is the complete elliptic integral of the first kind in parameter notation; the modulus used in some definitions is $\sqrt{m}$. Near the endpoint, $s=\pi /2-\vartheta$ givesUse matched asymptotic expansion with an intermediate cutoff $b$ satisfying $\sqrt{\delta }\ll b\ll 1$. Away from the endpoint the leading integral iswhile the endpoint integral isAdding them removes the arbitrary cutoff and gives the logarithmic endpoint asymptotic of the complete elliptic integralThe order of the next term is therefore $(1-m)\log (1/(1-m))$, rather than merely $1-m$. More explicitly, if $L=\log (4/\sqrt{\delta })$,The logarithmic correction arises from the next terms integrated through the overlap. Its coefficient can also be found by substituting $L+\delta (aL+b)$ into the Gauss hypergeometric equation satisfied here, $m(1-m)K_{mm}+(1-2m)K_m-K/4=0$, giving $a=1/4$, $b=-1/4$.