Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 72 / 2

Introduce the slow time $T=ϵt$ and use the method of multiple scales with $x_0=A(T)\cos \psi$, $\psi =t+\theta (T)$. The initial data give $A(0)=\sqrt{2}$ and $\theta (0)=-\pi /4$. At first order the equation for the correction isThe averaged amplitude for position-dependent damping follows from the solvability condition in the method of multiple scales, which removes the resonant sine and cosine terms. With brackets denoting a full-period average, this givesThe wave phase average vanishes because $f(A\cos \psi )\sin \psi \cos \psi$ changes sign under $\psi \mapsto 2\pi -\psi$.

For the quadratic position-dependent damping, $⟨{\cos }^2\psi {\sin }^2\psi ⟩=1/8$ and $⟨{\sin }^2\psi ⟩=1/2$, soSolving this logistic equation with $A^2(0)=2$ gives $A(T)=2/\sqrt{1+e^{-T}}$. Thus the uniformly valid leading approximation isHere $T_{\ast }$ is any fixed finite positive constant. Bounded nonresonant corrections and the averaging estimate make the error uniform on this slow-time interval. The leading expression meets the position datum exactly and the velocity datum to leading order; the first-order correction supplies its $O(ϵ)$ adjustment.

Use the same method of multiple scales and its solvability condition in the method of multiple scales. The position-dependent damping function is odd, so the lemma that odd position-dependent damping has zero first-order amplitude drift applies. In its wave amplitude average, $\psi \mapsto \psi +\pi$ changes $\sin (A\cos \psi )$ to its negative while preserving ${\sin }^2\psi$, so the average is zero. Hence $A^{\prime }=0$; the wave phase average again gives ${\theta }^{\prime }=0$.

Equivalently, expand the position-dependent damping in its uniformly convergent power series for bounded $A$. Each term ${\cos }^{2m+1}\psi \sin \psi$ has only even sine Fourier harmonics, so none resonates with the unit-frequency oscillator. With the initial wave amplitude and wave phase unchanged, the leading uniform approximation isThere is no first-order slow drift. Effects at the next order accumulate only an $O(ϵ)$ correction on this time scale.

Let $q_0$ be the stationary point and let $H=D^2\beta (q_0)$ be its real symmetric Hessian matrix. First suppose the stationary point is interior and nondegenerate, the oscillatory integral amplitude and oscillatory integral phase are sufficiently smooth, and boundary contributions are absent or smaller than the stationary contribution. Diagonalize $H$ by an orthogonal change of coordinates, with eigenvalues $h_1,h_2$. Locally,Scaling each principal coordinate by ${\lambda }^{-1/2}$ and evaluating its Fresnel integral gives the two-dimensional stationary-phase formula:Here $\mathrm{sgn}H$ is the number of positive eigenvalues minus the number of negative ones: $2$, $0$ or $-2$. The formula gives the leading nonzero term when $\alpha (q_0)\ne 0$. A partition of unity isolates this neighbourhood; integration by parts away from stationary points makes the remaining interior contribution smaller. For a smooth compactly supported oscillatory integral amplitude the local error is $O({\lambda }^{-2})$.

The printed assumption of a single stationary point alone is insufficient for an unconditional answer. A degenerate oscillatory integral phase $\beta =x^4+y^2$ with a smooth oscillatory integral amplitude supported near the origin has only one stationary point, but scales as ${\lambda }^{-3/4}$. Boundary terms can also contribute at order ${\lambda }^{-1}$: on the unit disk with oscillatory integral amplitude one and $\beta =x^2+y^2$,whose stationary contribution is $i\pi /\lambda$ and whose boundary contribution is $-i\pi e^{i\lambda }/\lambda$. Thus the boxed formula is the standard nondegenerate interior stationary contribution, with the stated hypotheses; the data as printed do not determine all possible leading behaviours.