Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 336 (Perturbation Methods)

The divide-and-conquer asymptotic expansion separates the $x=O(a)$ endpoint region from the $x=O(1)$ bulk, whose expansions individually contain terms that are nonuniform in the other region. Here the recombined answer can also be checked exactly. Put $y=\sqrt{a+x}$; thenAs $a\to 0^{+}$,Multiplication givesThe nonanalytic $\sqrt{a}$ term is the contribution that a naive fixed-$x$ expansion misses at the endpoint.

Write $w=z-1=X+iY$. The phase and its derivative areso the saddle points arewith $\varphi (z_{+})=2i$ and $\varphi (z_{-})=-2i$. The contour geometry can be drawn fromThe stationary-phase level $\mathrm{Re}\varphi =0$ consists of $X=0$ and $X^2-3Y^2+3=0$, meeting at the saddles. The steepest curves through $z_{\pm }$ are the levels $\mathrm{Im}\varphi =\pm 2$. Far away, sectors with $\cos (3\arg w)>0$ are exponential hills and those with $\cos (3\arg w)<0$ are valleys.

The stated contour deforms through the upper saddle. If $z=z_{+}+s$, thenThe descent tangent has $s=e^{i\pi /4}t$, because $3i{s}^2=-3t^2$. The simple-saddle contribution in steepest descent is therefore

When the contour begins at $z=1$, deform it first from the endpoint into the decaying negative-real direction and then onto the same upper-saddle descent path. Near the endpoint, $s=z-1$ and $\varphi =3s+O(s^3)$, so the endpoint contribution in steepest descent isAdding the saddle and endpoint pieces gives

Use the fast time $t$ and slow time $T={ϵ}^{2}t$, and writeThere is no $O(ϵ)$ resonant forcing. Solvinggives a convenient particular solutionAt $O({ϵ}^2)$, the coefficient of $e^{it}$ in the forcing must vanish by the solvability condition in the method of multiple scales. This givesorThe two slow exponents satisfyThe second instability tongue of a weak Mathieu oscillator has real $s$ when $-1/12<f<5/12$. At either endpoint the repeated zero exponent permits a linearly growing slow solution, so boundedness for every initial condition requires the strict stable ranges

Let $X=x/ϵ$, $a(X)=1/d(X)$, and seekwith periodic correctors. The leading cell equation makes $u_0$ independent of $X$. At the next order, continuity of microscopic heat flux givesAveraging over one period and using $⟨u_{1X}⟩=0$ yieldsThus the periodic homogenization of a diffusion equation has effective diffusivity $1/\overline{d}$:

For the sawtooth profile, the two triangular areas giveThe homogenized problem is therefore the unit-diffusivity heat equation. Its half-line step solution isIt has the required initial and boundary limits, and direct differentiation verifies $u_t=u_{xx}$.

With $X=ϵx$, the equation isThe leading WKB approximation for a slowly varying oscillator isIt requires smooth nonzero $k$, $ϵ|k^{\prime }|/|k{|}^2\ll 1$, and distance from every classical turning point large compared with its turning-point scale.

For $k(X)=\sqrt{1+X^2}$,Since $k(0)=1$ and $k^{\prime }(0)=0$, the initial data select the cosine branch without an $O(1)$ phase correction. Hence, for $x>0$ in the WKB regime,

The reduced first-order equation cannot satisfy both endpoint values. The given outer expansion satisfies $y(1)=2$ but has $y_0(0)=e$, so the boundary layer lies at $x=0$ and has stretched coordinate $X=x/ϵ$.

Write the inner expansion as $Y=Y_0+ϵY_1+\cdots$. After multiplying the differential equation by $ϵ$, it becomesAt leading order,The boundary condition $Y_0(0)=0$ and matching to $y_0(0)=e$ giveAt the next order,Matching to $y_1(0)=e(\ln 2-1)$ and imposing $Y_1(0)=0$ gives

The additive composite expansion is outer plus inner minus their common part. With $X=x/ϵ$, it isIt satisfies $y_{\mathrm{comp}}(0)=0$ exactly through the retained order, satisfies the right boundary condition up to exponentially small terms, and is uniformly accurate to $O(ϵ)$.