= Solution
{c}
At $\alpha=0$ the transverse coordinates are independent <Brownian motions> with <diffusion coefficient> $D=1$. The <cylinder> is infinitely long, so the axial coordinate cannot affect its first contact time. By <rotational invariance of Brownian motion>, the <mean first-passage time> $u$ depends only on the transverse <radius> $\rho=\sqrt{x^2+y^2}$.
The <Kolmogorov backward equation> for a <mean first-passage time> is $Lu=-1$. Here $L=\partial_x^2+\partial_y^2$, so the radial <boundary value problem> is
$$
u''(\rho)+\frac1\rho u'(\rho)=-1\quad(0<\rho<R),
\qquad u(R)=0.
$$
The <absorbing boundary condition for a diffusion> is imposed on the circular cross-section. At $\rho=0$, the solution must extend smoothly over the axis. Multiplying by $\rho$ and integrating gives
$$
(\rho u')'=-\rho,
\qquad u'=-\frac\rho2+\frac C\rho,
\qquad u=-\frac{\rho^2}4+C\log\rho+C_0.
$$
Smoothness at the axis excludes the <logarithm> term, and $u(R)=0$ fixes $C_0=R^2/4$. Consequently
$$
\boxed{\tau(\widehat r_0)=\frac{R^2-\widehat r_0^{\,2}}4.}
$$
This also verifies the finiteness assumption behind the <Kolmogorov backward equation>. Let $T$ be the first contact time and apply the stopped <Itô formula> to the bounded nonnegative function $u$ just obtained. Its bounded derivatives on the disk make the stopped <stochastic integral> a <martingale>, giving
$$
\mathbb E(t\wedge T)=u(\widehat r_0)-\mathbb E u\bigl(\rho(t\wedge T)\bigr)\leq u(\widehat r_0).
$$
By <monotone convergence>, $T$ has finite <expected value>; continuity and <dominated convergence> then make the final term vanish as $t\to\infty$. Hence $u(\widehat r_0)=\mathbb ET$. In particular, the <mean first-passage time> is $R^2/4$ on the axis and tends to zero at the adsorbing wall.
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