= Solution
{c}
Away from the boundary, <derivatives> of a smoothly varying solution are much smaller than $k^2f$ unless the <outer solution> vanishes. Nonzero boundary data therefore require a rapidly varying layer. If its thickness is $\delta$, the radial second <derivative> is of order $f/\delta^2$; balancing it with $k^2f$ gives $\delta=k^{-1}$.
Put $R=k(1-r)$ and write $f(r,\theta)=F(R,\theta)$. The equation becomes
$$
F_{RR}-\frac1{kr}F_R+\frac1{k^2r^2}F_{\theta\theta}=F.
$$
For fixed $R$ and boundary data with angular variation independent of $k$, the leading equation is $F_{RR}=F$. Matching to the zero <outer solution> removes the growing exponential, and the <boundary value> at $R=0$ fixes its coefficient:
$$
\boxed{F_0(R,\theta)=g(\theta)e^{-R}.}
$$
This approximation describes the layer; it is not a uniform expansion in polar coordinates at the origin.
<Laplace's method> localizes an exponential <integral> near maxima of its real phase. At an isolated smooth interior <maximum> $t_0$ with $G''(t_0)<0$ and $F(t_0)\neq0$, <Taylor expansion> and <Gaussian integral> give
$$
\int F(t)e^{kG(t)}dt\sim F(t_0)e^{kG(t_0)}
\sqrt{\frac{2\pi}{k|G''(t_0)|}}.
$$
An endpoint <maximum> with nonzero inward phase slope instead gives an inverse-$k$ factor. Degenerate maxima require their first nonzero local phase term; continuous differentiability alone does not guarantee the quadratic formula.
Here $\cos(ix\sin\phi)=\cosh(x\sin\phi)$. In its two exponentials, the positive exponential dominates at the interior <maximum> $\phi=\pi/2$ of $\sin\phi$. Since $\sin(\pi/2+s)=1-s^2/2+O(s^4)$,
$$
I_0(x)\sim\frac1{2\pi}e^x\int_{-\infty}^{\infty}e^{-xs^2/2}ds
=\frac{e^x}{\sqrt{2\pi x}}.
$$
The negative exponential has no comparable positive <maximum> and is negligible. Consequently, where $kr\to\infty$,
$$
\frac{I_0(kr)}{I_0(k)}\sim r^{-1/2}e^{-k(1-r)}.
$$
In the <boundary layer> $r=1-R/k$, $r^{-1/2}=1+O(k^{-1})$, so this reduces to \b[$e^{-R}$], agreeing with the formal result when $g=1$.
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