Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 31A / Solution

The lowest Liouville–Green approximation comes from $f=e^S$, for which $S^{\prime \prime }+(S^{\prime }{)}^2=Q$. First take $S^{\prime }=\pm \sqrt{Q}$; the next transport term is $-Q^{\prime }/(4Q)$. Integrating givesMultiplicative constants and the lower limit are arbitrary. The negative sign selects the recessive approximation in the usual WKB approximation regime, where the exponent dominates the amplitude and the derivative corrections are small. Positivity of $Q$ by itself does not prove that $f_{-}$ tends to zero or that the approximation is valid: for example $Q=e^{-x}$ makes this expression grow like $e^{x/4}$ with a bounded integral in its exponent. For $Q=x$, all the required large-$x$ conditions hold and $f_{-}=x^{-1/4}{e}^{-2x^{3/2}/3}\to 0$.

Put $a=f_{-}$. Direct differentiation givesSubstituting $\mathrm{Ai}=wa$ into the Airy ordinary differential equation ${\mathrm{Ai}}^{\prime \prime }=x\mathrm{Ai}$ and dividing by $a$ gives $w^{\prime \prime }+2(a^{\prime }/a)w^{\prime }+5w/(16x^2)=0$. Multiplication by $x^2$ provesThe recessive branch has a constant leading amplitude $c$. To find its first correction write $w=c(1+b{x}^{-\delta }+\cdots )$. Balancing $-2x^{5/2}{w}^{\prime }$ against the constant term $5c/16$ forces $3/2-\delta =0$, so $\delta =3/2$. The constant coefficients then give $3b+5/16=0$. The remaining terms are of order $x^{-3/2}$ and are cancelled by a next term of order $x^{-3}$. ThusThe multiplicative constant depends on the chosen normalization of the Airy function and of $f_{-}$. The calculation selects the recessive solution; a growing Airy solution would not have this constant-amplitude expansion.