Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 76 / 2

Apply the method of multiple scales with independent fast time $t$ and slow time $T=\epsilon t$, and write $x=x_0(t,T)+\epsilon x_1(t,T)+\cdots$. The chain rule gives $d/dt={\partial }_t+\epsilon {\partial }_T$. At leading order,At first order,A forcing component at the natural frequency produces a secular term in $x_1$. Thus the solvability condition in the method of multiple scales is orthogonality to both $\sin \theta$ and $\cos \theta$. Define the fast-phase average, holding $R,\varphi ,T$ fixed, bySince $⟨{\sin }^2\theta ⟩=⟨{\cos }^2\theta ⟩=1/2$ and $⟨\sin \theta \cos \theta ⟩=0$, the amplitude-phase equations areThey are leading averaged equations, not exact equations for a purely sinusoidal solution of a general nonlinear oscillator.

For the cubic conservative force, $f=-R^3{\cos }^3\theta$, odd symmetry gives $⟨{\cos }^3\theta \sin \theta ⟩=0$, while $⟨{\cos }^4\theta ⟩=3/8$. HenceIf $x(0)=x_{\ast }$ and $x^{\prime }(0)=v_{\ast }$, leading initial matching givesand thereforeon bounded slow time intervals for fixed bounded initial data. The positive frequency shift agrees with the increased restoring force of $x^{\prime \prime }+x+\epsilon x^3=0$. The zero initial state is the exact zero solution, with irrelevant phase. If a first waveform correction is desired, the remaining first-order forcing is $-R_0^3\cos 3\theta /4$, so $x_1=R_0^3\cos 3\theta /32$ plus a homogeneous correction chosen to match the initial data at the next order. This exhibits explicitly why the leading harmonic ansatz is approximate.

For the oscillator coupled to slow damping feedback, expand $y=Y(T)+\epsilon y_1(t,T)+\cdots$, alongside the method of multiple scales oscillator expansion. The leading equation ${\partial }_{t}Y=0$ makes $Y$ independent of fast time. The oscillator force evaluated on $x_0=R\cos \theta$ is $2YR\sin \theta$. The amplitude-phase equations therefore giveThe next equation for the damping variable isA bounded periodic $y_1$ exists only when the mean of its right side vanishes. Thus the full leading averaged system isSet $R_0=\sqrt{x_{\ast }^2+v_{\ast }^2}$, ${\varphi }_0=atan2(-v_{\ast },x_{\ast })$, and $Y_0=y_{\ast }$. These specify the arbitrary leading initial conditions.

For $g(x)=x$, the phase average is zero. ConsequentlyThus $x(t)=R_0{e}^{-Y_0\epsilon t}\cos (t+{\varphi }_0)+O(\epsilon )$ and $y(t)=Y_0+O(\epsilon )$. Positive $Y_0$ damps the oscillator, negative $Y_0$ amplifies it, and zero $Y_0$ leaves the leading amplitude unchanged. The fast correction $y_1=R\sin \theta$ plus an initial-matching constant confirms that $y$ can oscillate slightly even though $Y$ is constant. For $R_0=0$, the exact solution is $x=0$, $y=y_{\ast }$.

For the logarithmic feedback, assume $R_0>0$. Its phase singularities are integrable, and the supplied integral givesWriting $L=\log (R/2)$ turns the nonlinear amplitude-phase equations into the linear pairso $L_{TT}+L=0$. With $L_0=\log (R_0/2)$,The first integral $L^2+Y^2=L_0^2+Y_0^2$ provides a direct check, and the amplitude remains positive on every finite slow time interval. At the retained averaged order, the physical approximation is $x\simeq R(\epsilon t)\cos (t+{\varphi }_0)$, $y\simeq Y(\epsilon t)$.

The logarithmic case needs an interpretation at oscillator zeros. The original right side is undefined at $x=0$, so a globally classical solution through each crossing, or a solution from $x_{\ast }=v_{\ast }=0$, cannot be asserted. At a transverse crossing, $\log |x|\sim \log |t-t_{\ast }|$ is locally integrable; $y$ can be interpreted as absolutely continuous, with its equation holding away from the crossings and almost everywhere. In that integrable interpretation the displayed averaged formulas apply formally for nonzero leading amplitude. The standard smooth-forcing averaging theorem cannot be invoked without addressing this singularity. The phrase arbitrary initial conditions must exclude the identically zero state for this logarithmic problem.