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Retaining the nonlinear factor $f$ determines the order-$ϵ$ forcing and the matching constants. Write $\delta =\sqrt{ϵ}$. The inner leading equation and its boundary condition at $r=1$ give $f_0=C(1-r^{-1/2})$. Matching to the far-field value fixes $C=1$.

The tail $-r^{-1/2}$ becomes $-\delta x^{-1/2}$ under $x=ϵr$, so a regular expansion in integer powers of $ϵ$ cannot suffice. This is a matched expansion with square-root and logarithmic corrections. DefineThe exponential integral here is normalized by the displayed definite integral. Integration by parts also gives $E(x)=2e^{-x}/\sqrt{x}-2\sqrt{\pi }\mathrm{erfc}\sqrt{x}$, using the complementary error function.

For the outer asymptotic expansion, put $F(x)=f(x/ϵ)$ andThe transformed equation gives $L{F}_1=0$ and $L{F}_2=-F_1{F}_1^{\prime }$, with both corrections vanishing at infinity. Matching the leading inner tail determines $F_1=-E/2$. Multiplying the equation for $F_2$ by its integrating factor yieldsFix the particular solution $G$ unambiguously byThen $(x^{3/2}{e}^x{G}^{\prime }{)}^{\prime }=E$, $G(\infty )=0$, and $G$ has no $x^{-1/2}$ term at zero. The general outer correction is $F_2=\beta E+G/4$, whereThe logarithm in this asymptotic expansion explains why the inner expansion needs an $ϵ\log ϵ$ term.

For the inner asymptotic expansion, all corrections larger than order $ϵ$ solve the homogeneous leading equation and vanish at $r=1$. At order $ϵ$, the forcing isIntegration gives the particular solution $1-\sqrt{r}+\log r$, which vanishes at $r=1$. Thus writeExpress this inner expansion in $x=ϵr$ and compare the terms that survive in the overlap $ϵ\ll x\ll 1$. The order-$\delta$ constant gives $A=\sqrt{\pi }$. At order $ϵ$, the term $C\log ϵ+\log r$ becomes $\log x+(C-1)\log ϵ$, so $C=1$. The singular term $-ϵA/\sqrt{x}$ matches $2ϵ\beta /\sqrt{x}$, giving $\beta =-\sqrt{\pi }/2$. Finally, the regular constants give $B+1=\pi +\mu /4$.

The requested expansions, including every correction through order $ϵ$, are thereforeThe inner formula satisfies the boundary condition at $r=1$ term by term; the outer formula satisfies the condition at infinity. Each formula has its own stated region of validity, and the matched asymptotic expansion relates them in their overlap.

The constant can also be made explicit. Integration by parts in the definition of $G$ gives $G=-EI-{\int }_x^{\infty }E(t{)}^{2}dt$, and $E(x)I(x)\to 8$. Squaring the complementary error function expression for $E$ separates the integral into three elementary pieces:The first piece is $-4\log x-4\gamma -4\log 2+o(1)$, where $\gamma$ is the Euler--Mascheroni constant. Substitution $u=\sqrt{t}$ makes the second integral tend to $\sqrt{\pi }/2$, because $(\mathrm{erfc}u{)}^{\prime }=-2e^{-u^2}/\sqrt{\pi }$. Integration by parts in the third integral gives $1/2-1/\pi$. ConsequentlyEquivalently, defining $G$ by its convergent integral already determines the same matching constant without needing this closed form.