Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2010 / A (Perturbation Methods)

Put $\delta =\sqrt{ϵ}$ and write the outer asymptotic expansion as $y_0+\delta y_{1/2}+{\delta }^2{y}_1+\cdots$. Equating powers in the differential equation givesThe first two equations give $y_0=A{e}^{x^2/2}$ and $y_{1/2}=B{e}^{x^2/2}$. Since $y_0^{(3)}=A(x^3+3x)e^{x^2/2}$, an integrating factor yieldsHere $A,B,C$ are matching constants; boundary conditions have not yet been imposed.

In a layer of thickness $\ell \ll 1$, the third-derivative and first-derivative terms have sizes $ϵ{\ell }^{-3}y$ and ${\ell }^{-1}y$. Their dominant balance requires ${\ell }^2\sim ϵ$, soThe local fast modes are exponentials with rates $\pm 1/\sqrt{ϵ}$. A mode decaying into the interval can occur at either endpoint: $e^{-x/\sqrt{ϵ}}$ at zero and $e^{-(1-x)/\sqrt{ϵ}}$ at one. The zero fast exponent represents the matching outer component. The term $xy$ is lower order on either fast scale, and even smaller near zero; it creates no further independent layer thickness.

For the left inner coordinate $X=x/\delta$, the equation becomes $Y_{XXX}-Y_X+{\delta }^{2}XY=0$. At leading order the solution that matches a bounded outer expansion is $Y_0=A+D{e}^{-X}$; the growing exponential is excluded. The boundary value gives $A+D=1$. Since $y^{\prime }(0)=1$ is finite, the possible $O({\delta }^{-1})$ derivative must vanish, so $D=0$ and $A=1$.

At the next order, $Y_{1/2}=B+E{e}^{-X}$. Its boundary value is zero, giving $E=-B$, while its physical derivative at zero is $-E=1$. Hence $B=1$, and the left inner expansion is $1+\delta (1-e^{-X})+O({\delta }^2)$.

At the right endpoint put $\xi =(1-x)/\delta$. The equation is $Y_{\xi \xi \xi }-Y_{\xi }=\delta (1-\delta \xi )Y$. Matching the outer expansion at one and imposing $Y(0)=1$ giveAt the next order, $Y_{1/2}^{(3)}-Y_{1/2}^{\prime }=Y_0$. The solution with no growing exponential, the correct matching polynomial, and $Y_{1/2}(0)=0$ isIndeed the operator $D^3-D$ sends $-\sqrt{e}\xi$ to $\sqrt{e}$ and sends $\xi e^{-\xi }$ to $2e^{-\xi }$.

Add the two decaying inner corrections to the outer solution, subtracting their common matching parts. Since $\delta \xi =1-x$, the composite asymptotic expansion isIt reproduces all three boundary conditions to the retained algebraic orders, with exponentially small interactions between endpoints. The right endpoint has an $O(1)$ boundary layer in $y$. The left correction has only amplitude $O(\sqrt{ϵ})$ in $y$, but its derivative is $O(1)$ and is necessary for the derivative boundary condition. Thus the left layer is a weak derivative boundary layer; if “genuine” refers specifically to an order-one change in the function itself, only the right layer is genuine in that sense. Both nonzero corrections belong in the expansion through $O(\sqrt{ϵ})$.

For an asymptotic sequence $a_{n+1}(ϵ)=o(a_n(ϵ))$, the asymptotic expansion condition is that every fixed truncation has remainder smaller than its last retained term:This limit is taken as $ϵ\to 0$ in the specified direction or sector. The number $N$ stays fixed while taking that limit. The infinite series need not converge for any nonzero parameter.

Let $g_x(t)=(1+xt{)}^{-1}$. Repeated integration by parts using the factor $e^{-t}$ givesBoundary terms at infinity vanish, and $g_x^{(n)}(t)=(-1{)}^{n}n!x^n(1+xt{)}^{-n-1}$. ThereforeFor every fixed $N$, this is $o(x^N)$ as $x\downarrow 0$, provingFor each fixed $x>0$ the displayed infinite series diverges, since successive term magnitudes have ratio $(n+1)x\to \infty$. Its usefulness is precisely that of an asymptotic expansion, not a convergent power series.

Fix $0<ϵ<\pi /2$. On $[ϵ,\pi /2]$, ${\sin }^{2}t\ge {\sin }^2ϵ>0$, so that part of the integral is bounded by $(\pi /2)e^{-x{\sin }^2ϵ}$, exponentially smaller than $x^{-1/2}$.

By the assumed local equivalence, the remaining part is asymptotic to ${\int }_0^{ϵ}{e}^{-x{t}^2}dt$. Substitute $s=\sqrt{x}t$:The last step uses the Gaussian integral and monotone convergence of the upper limit to infinity. Hence

A fixed small-angle approximation alone is insufficient because its error is multiplied by $x$. Introduce the shrinking cutoff $\eta =x^{-1/3}$, which satisfies $\eta \ll ϵ$ but $\sqrt{x}\eta \to \infty$.

On $[0,\eta ]$, the Taylor expansion ${\sin }^{2}t=t^2+O(t^4)$ gives $x({\sin }^{2}t-t^2)=O(x{\eta }^4)=O(x^{-1/3})$ uniformly. Thusuniformly there, and the two central integrals are asymptotically equal. On $[\eta ,ϵ]$, choose a fixed $c>0$ with $\sin t\ge ct$ for $0\le t\le ϵ$. The sine-integral tail is at most $ϵe^{-c^{2}x{\eta }^2}=ϵe^{-c^2{x}^{1/3}}$; the Gaussian tail has the same exponential bound with $c=1$. Both are $o(x^{-1/2})$, while the central Gaussian integral is asymptotic to $\sqrt{\pi }/(2\sqrt{x})$. HenceThis justifies the localized Laplace method rather than interchanging a nonuniform approximation with integration.

Matching powers of $ϵ$ gives $u_0^{\prime \prime }+u_0=0$, $u_1^{\prime \prime }+u_1=u_0^2$, and $u_2^{\prime \prime }+u_2=2u_0{u}_1$, with $u_0(0)=2$ and homogeneous initial data for every higher correction. ThusThere is no resonant fundamental forcing at first order. At second order,The first term resonates with the homogeneous oscillator. Solving with the initial data givesConsequently the regular perturbation approximation is $u=u_0+ϵu_1+{ϵ}^2{u}_2+O({ϵ}^3)$ for fixed $t$. The first secular term is $(10/3){ϵ}^{2}t\sin t$, at order ${ϵ}^2$. It makes the regular expansion nonuniform on times of order ${ϵ}^{-2}$.

Use the fast time $T_0=t$ and slow time $T={ϵ}^{2}t$. A possible intermediate scale $ϵt$ has no leading modulation: the absence of first-order resonance imposes independence of that scale. WriteAt first order choose the nonresonant particular solutionwith any needed fundamental homogeneous correction assigned consistently when matching the initial data. At second order,The coefficient of $e^{i{T}_0}$ on the right is $(10/3)|A{|}^{2}A-2i{A}_T$. A bounded oscillatory correction requires this resonance coefficient, and its conjugate, to vanish. The quadratic oscillator phase-modulation condition is thereforeIt resums the secular phase accumulation into a slow modulation instead of leaving it as a growing term in the correction.

Write $A=a{e}^{i\varphi }$. The modulation equation gives $a_T=0$ and ${\varphi }_T=-5a^2/3$. Matching the leading initial conditions gives $a=1$, $\varphi (0)=0$. ThereforeThe leading amplitude remains two and has no slow growth or decay. The frequency changes from one to $1-5{ϵ}^2/3+O({ϵ}^3)$; expanding this shifted cosine recovers the second-order secular term from part (i).

There are bounded first-order waveform changes as well. If they are retained while matching $u(0)=2$, one may write $2\cos \tau +ϵ[2-(4/3)\cos \tau -(2/3)\cos 2\tau ]+\cdots$, where $\tau =(1-5{ϵ}^2/3)t$. Thus the first harmonic's coefficient is $2-4ϵ/3$, with a mean shift and a second harmonic. These bounded shape corrections are distinct from the constant leading amplitude in the zeroth-order multiple-scales solution.