= Solution
Put $\phi=R(r)\Theta(\theta)Z(z)$ in the <Laplace equation in cylindrical coordinates>. <Separation of variables> first gives $\Theta''+m^2\Theta=0$; $2\pi$-periodicity forces $m=0,1,2,\ldots$, with angular factors $\cos m\theta,\sin m\theta$ (only the constant for $m=0$). Taking $Z''=\kappa^2Z$ gives
$$
r^2R''+rR'+(\kappa^2r^2-m^2)R=0.
$$
For $\kappa>0$ the separated modes are therefore
$$
\boxed{[A J_m(\kappa r)+B Y_m(\kappa r)]
[C e^{\kappa z}+D e^{-\kappa z}]
[E\cos m\theta+F\sin m\theta].}
$$
Here $J_m$ and $Y_m$ are the <Bessel function of the first kind> and the <Bessel function of the second kind>. For the opposite sign $Z''=-\kappa^2 Z$, the modes instead involve $I_m(\kappa r),K_m(\kappa r)$, the <modified Bessel functions>, paired with $\cos\kappa z,\sin\kappa z$. At zero separation constant use $Z=C+Dz$ and $R=Ar^m+Br^{-m}$ for $m\ge1$, or $R=A+B\log r$ for $m=0$. Superpositions over the integer angular orders and appropriate sums or integrals over the separation parameter give the general separated expansion; boundary conditions select its spectrum and coefficients. Regularity at the axis excludes $Y_m,K_m,r^{-m},\log r$ terms.
For the specified decaying side data, bounded separated modes use $e^{-\kappa z}J_m(\kappa r)$. A particular solution is
$$
\boxed{\phi_p=e^{-4z}\left[\frac{J_1(4r)}{J_1(4)}\cos\theta+\frac{J_2(4r)}{J_2(4)}\sin2\theta\right]
+2e^{-z}\frac{J_2(r)}{J_2(1)}\sin2\theta.}
$$
All three denominators are nonzero. Each term satisfies the radial <Bessel differential equation>, is regular at the axis and bounded for $0\le r\le1,z\ge0$, and substitution at $r=1$ gives the stated side values. The modes behave as $r^m$ at the axis, so their apparent angular dependence there causes no singularity.
There is an actual nonuniqueness in the printed problem: no values are specified at $z=0$. If $j_{0,1}$ is the first positive zero of $J_0$, then for every real $a$,
$$
\phi_p+a e^{-j_{0,1}z}J_0(j_{0,1}r)
$$
is another bounded solution with the same side values. It remains smooth at the axis and even decays as $z\to\infty$. This explicitly proves that <side boundary data do not determine a bounded harmonic function in a half-cylinder>. Thus the boxed expression supplies a bounded solution; the phrase “the bounded solution” is not justified without an additional base boundary condition. The PDF has $z\ge0$, whereas the TeX transcription displays $z>0$.
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