Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 74 / 5

The positive drift places the boundary layer at the left endpoint. Set $y_{\mathrm{out}}=y_0+ϵy_1+\cdots$ and impose the right boundary condition order by order. The leading ordinary differential equation is $[(1+x)y_0{]}^{\prime }=0$, so $y_0=2/(1+x)$. At the next order,which integrates toThe outer limit at the left endpoint is two, so it cannot satisfy the left boundary value without a rapidly varying correction.

For the inner expansion use $X=x/ϵ$ and $y=Y_0(X)+ϵY_1(X)+\cdots$. Its equations and boundary values areMatching the leading constant gives $Y_0=2-e^{-X}$. The first-order forcing is then $-2+(1-X)e^{-X}$. A particular solution is $-2X+\frac{1}{2}{X}^2{e}^{-X}$, and matching determines the constant to be $3/2$. Enforcing $Y_1(0)=0$ givesIn the overlap region the common expansion is $2+ϵ(3/2-2X)$. Adding the inner and outer approximations and subtracting that common part gives the uniform asymptotic approximationIt has error $O({ϵ}^2)$ uniformly for $0\le x\le 1$, apart from an exponentially small endpoint adjustment of the same asymptotic irrelevance. At $x=0$ it equals one exactly, and at $x=1$ it differs from one only exponentially. The polynomial multiplying the exponential remains bounded on the inner scale.

There is an independent check from a first integral. The equation is $[ϵy^{\prime }+(1+x)y{]}^{\prime }=0$, so its exact solution has the formwhere $C$ is selected by the right boundary value. Endpoint expansion of this integral gives $C=2-ϵ/2+O({ϵ}^2)$, the same outer coefficients and the same inner expansion. This check also establishes the uniform order of the composite approximation without treating an inner residual as an outer estimate.

The negative drift moves the outflow boundary layer to $x=1$. The first-order outer equation is $-(1+x)y_0^{\prime }+y_0=0$. It must satisfy the left boundary condition, giving $y_0=1+x$. This tends to two at the right endpoint.

Use $X=(1-x)/ϵ$. The leading inner equation is $Y_0^{\prime \prime }+2Y_0^{\prime }=0$, and impose $Y_0(0)=1$ and $Y_0\to 2$ on matching. Thus $Y_0=2-e^{-2X}$, and a leading uniform asymptotic approximation isFor higher accuracy, expand the drift as $2-ϵX$ in the inner equation, solve the successive inhomogeneous equations, and match to outer corrections satisfying zero left endpoint data. There is no need for a left endpoint layer. Choosing the growing exponential at the wrong endpoint would prevent matching and is the reason the layer location matters.

Here the drift vanishes at the left endpoint. The outer equation is $x{y}_0^{\prime }+y_0=0$, giving $y_0=1/x$ after applying the right boundary value. The ordinary $x=O(ϵ)$ layer does not balance diffusion and drift: their balance selects $x=O(\sqrt{ϵ})$ instead.

For the square-root boundary layer at a vanishing drift, put $X=x/\sqrt{ϵ}$. The inner equation isMatching $1/x$ requires a leading inner amplitude of order ${ϵ}^{-1/2}$, so an inner expansion with an everywhere order-one leading term would be incorrect. This can be seen directly by integrating the original equation once:The right boundary value determinesIn the inner variable the exact representation becomesThe first term enforces $Y(0)=1$, while the second is zero there and matches $C/(\sqrt{ϵ}X)=C/x$ for large $X$. One would expand this integral in the overlap region, determine successive outer terms from the right boundary, and subtract their common expansion. The endpoint layer has width $\sqrt{ϵ}$ and an $O({ϵ}^{-1/2})$ interior peak, while its value at the endpoint remains one. This distinguishes it from the ordinary exponential layers in the two preceding cases.

Introduce the slow time $T=ϵt$ and expand $y=y_0(t,T)+ϵy_1(t,T)+\cdots$. At leading order writeThe first-order equation isIts resonant sine and cosine components must vanish by the solvability condition in the method of multiple scales. Let angle brackets denote a full-period average. The cosine projection of the nonlinear term is zero, since $⟨{\sin }^n\psi \cos \psi ⟩=0$. The sine projection givesThis is the averaged oscillator damping by a power of velocity.

If $n$ is odd, defineThen $R_T=-C_n{R}^n$, givingFor every fixed $n$, these leading multiple-scale expansions are uniform on bounded slow-time intervals $0\le ϵt\le T_0$. The unforced phase remains $t$ to this order. A bounded first-order correction supplies the small adjustment to the initial derivative and initial displacement; it does not change this leading result.

If $n$ is even, the average of the odd power ${\sin }^{n+1}\psi$ vanishes. Thus $R_T=0$ andThe first-order forcing has only a constant and even harmonics, so it produces no fundamental resonance. Its bounded particular solution can be supplemented by homogeneous sine and cosine terms to satisfy the initial conditions. There can be an $O({ϵ}^2)$ frequency correction, but its phase accumulation is only $O(ϵ)$ on the present time scale. In contrast to the odd case, the even-power force is invariant under velocity reversal and the equation is reversible; calling it positive damping for both signs of velocity would be wrong. The mechanical energy identity makes the distinction clear:This is nonpositive for odd $n$ and changes sign for even $n$; its cycle average reproduces the amplitude equation above.

For large odd $n$, the central binomial coefficient givesThis form reveals several time regimes. For $L_nϵt\ll 1$, $R=1-C_nϵt+\cdots$; at $t=O(1/(ϵ\sqrt{n}))$, the amplitude change is only $O(1/n)$. For $L_nϵt\gg 1$ but $\log (L_nϵt)\ll n$, the leading change is $R\simeq 1-\log (L_nϵt)/n$. In particular, at $t=O(1/ϵ)$ it is only $O(\log n/n)$, despite the long elapsed time. High powers damp mainly near velocity maxima, and even a small amplitude reduction strongly suppresses subsequent damping.

An order-one amplitude reduction in the averaged law requires $\log (L_nϵt)=O(n)$. Formally, $t=e^{(n-1)\sigma }/(ϵL_n)$ gives $R\sim e^{-\sigma }$, and for fixed odd $n>1$ its late envelope is an algebraic power $t^{-1/(n-1)}$. These exponential-in-$n$ time scales are predictions of extending the averaged law; the fixed-$n$, bounded-$ϵt$ multiple-scale error estimate alone does not establish uniform accuracy that far. A joint large-$n$, small-$ϵ$ approximation also needs control of the increasingly sharp nonlinear peaks, rather than assuming fixed-$n$ error constants are uniform. Taking $n$ fixed first and then examining the large-$n$ envelope avoids that unsupported interchange of limits.