Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 48 / 3 / b

Here the zero-parameter solution is $y_0=1$, but stopping there erases a weakly excited growing mode. Expand $y=1+\epsilon y_1+\cdots$ at fixed $x$. The first correction satisfiesand thereforeThe $\epsilon x^2/2$ term eventually dominates the constant, so this is another singular perturbation on the unbounded domain.

Using the exact logarithmic-variable system from part (a), the initial values are now $z(0)=\epsilon$ and $w(0)=0$. Its linear growing/stable modes have matching coefficientsThus $w\sim Cx$ is order one only at $x=O({\epsilon }^{-2})$, rather than $O({\epsilon }^{-1})$. This is the nonlinear crossover delayed by a small unstable-mode coefficient. The first change in the size of $y$ occurs much earlier, at $x=O({\epsilon }^{-1/2})$, when its weak growing mode becomes comparable to the initial constant; that is not yet the derivative-nonlinearity crossover.

At $X=Cx=O(1)$ the same logistic crossover in a weakly nonlinear Euler equation gives $w=X/(1+X)$ and $H(X)=X-\log (1+X)$. The nonlinear-region solution is now $y\sim H(Cx)/(\epsilon C)$, or ${\epsilon }^{-3}H({\epsilon }^{2}x)$ at the lowest order. A composite asymptotic expansion covering all these regions isIt gives $y(1)=1$ exactly, and its initial derivative is $O({\epsilon }^3)$ rather than zero, within leading accuracy. At fixed $x$ its expansion is precisely $1+\epsilon [(x^2-1)/2-\log x]+O({\epsilon }^2)$; for $Cx\ll 1$ it retains the weak $A{x}^2$ growth, and for $Cx\gg 1$ it gives $y\sim x/\epsilon$. The positive equilibrium and boundedness argument from part (a) applies after the initially zero derivative immediately becomes positive. Thus the same eventual slope saturation occurs, but much farther downstream. Both the seeding calculation and the changed scale are essential to a leading approximation on the entire half-line.