Laplace equation for circular symmetry: given a potential $\phi (z,0)$, how to find $\phi (0,r)$? The potential function $\phi(z,r)$ has circular symmetry with respect to $r$. It also satisfies the Laplace equation $\partial \phi^2/\partial z^2 + \partial \phi^2/\partial r^2 + (1/r)\partial \phi/\partial r = 0 $ in the source-free half space $z>0$.  Assume it is also well-behaved, i.e. differentiable and fades to zero at large $r$ or $z$. 
It is often easy to derive the potential along the axis of symmetry, $\phi (z,0)$. For example, deriving the potential or field on the axis of a charged circular loop is straightforward. Given the constraint imposed by the Laplace equation and the circular symmetry, this knowledge should be sufficient to determine the potential along a radius, $\phi(0,r)$, or, more generally, the potential in the entire half-space, $\phi (z,r)$.
3D illustration showing circular symmetry
The very simplest example might be $\phi (z,0) = 1/z$ which maps to $\phi (0,r) = 1/r$, corresponding to the potential from a point charge.  I'm interested in functions that decay more slowly, especially the integrals of 1/z like $\phi (z,0) = -log(1/z)$ and $\phi (z,0) = x-x.log(1/z)$.
Is there a simple method or mapping that allows one to go from the potential along the axis of circular symmetry, $r=0$, to the potential on the surface $z=0$, or, indeed, to anywhere in the half-space?
[Edit] Another Example
If I consider a line charge extending along the z<0 axis, the potential ends up with unmanageable infinities. However, the vertical field $\partial \phi/\partial z$ for z>0 is a little better behaved and decays as 1/z. More or less by trial and error, I have found a solution for $\partial \phi/\partial z$ and $\partial \phi/\partial r$ that I think satisfies the Laplace equation. 
$\partial \phi/\partial z = 1/\sqrt(z^2+r^2)$ 
$\partial \phi/\partial r = (1-z/\sqrt(r^2+z^2))/r$ 
Subtituting:
$\partial/\partial z [1/\sqrt(z^2+r^2)] 
+ (1/r)\partial /\partial r[r.(1-z/\sqrt(r^2+z^2))/r] = 0 $ 
I'd like to be able to do this systematically rather than by guesswork.
 A: $\underline{\text{Is the problem well posed?}} $
Specifying the potential $\phi(z)$ on $z>0$ is not enough to uniquely determine $V(r,z)$ in the space $z>0$. This is because the boundary data on the $z=0$ plane is missing. For example, consider the two potentials defined along the entire $z$ axis
$$
\phi_1(z)=1 \\
\phi_2(z)=\operatorname{sgn}(z)
$$
Where $\operatorname{sgn}$ is the sign function. For $z>0$ we have $\phi_1=\phi_2$, yet they lead to different potentials in all space (and consequently different potentials in the space $z>0$). To demonstrate the non-uniqueness, it is sufficient to look at the $z=0$ plane. $\phi_1$ leads to $V(r,0)\sim\ln(r)$ while symmetry dictates that $\phi_2$ leads to $V(r,0)=0$.
Laplace boundary value problems require boundary data on the full surface enclosing the volume of interest. As we are concerned with the half space $z>0$ here, part of that boundary is the $z=0$ plane. The problem is not well posed unless:

*

*We specify boundary data on the $z=0$ plane.

*We work in the full space and specify $V$ everywhere along the $z$ axis.

*We specify a potential on $z>0$ that is a complex analytic function of $z \in \mathbb{C}$ (see below). This fails even for the point charge potential which is $1/|z| \neq 1/z$.

$\underline{\text{An integral representation}} $
If the specified potential $\phi(z)$ may be analytically continued to complex $z$, we have the following representation due to Poisson (see eg. here)
$$ \tag{1}
V(r,z)=\frac{1}{\pi}\int\limits_0^\pi d\eta \ \phi(z+ir\cos\eta) 
$$
Which does what you are asking for$^\dagger$: generate $V$ in all space given $V$ along the $z$ axis. Let us verify that this representation satisfies the axisymmetric Laplace's equation $\nabla^2V=\partial_{zz}V+\partial_{rr}V+r^{-1}\partial_rV=0$. Upon taking derivatives we have
$$
\nabla^2V=\frac{1}{\pi}\int\limits_0^\pi d\eta \ \left[\phi''-\phi''\cos^2\eta+\frac{i \phi'}{r}\cos\eta \right]
$$
Integrating the third term by parts yields $-\phi'' \sin^2\eta$ (the boundary term vanishes as long $\phi'$ is finite at the integral bounds) and we are left with
$$
\nabla^2V=\frac{1}{\pi}\int\limits_0^\pi d\eta \ \phi'' \left[1-(\cos^2\eta+ \sin^2\eta) \right]=0
$$
By setting $r=0$ in (1) we may also verify that the candidate $V$ satisfies the boundary condition $V(0,z)=\phi(z)$
$$
V(0,z)=\frac{1}{\pi}\int\limits_0^\pi d\eta \ \phi(z) = \phi(z)
$$
Thus (1) is the unique solution to $V(r,z)$.
For the type of functions you are interested in, such as $\phi(z)=\ln(z) \neq \ln(|z|)$, you may need to worry about branch cuts while performing the integral. Finding $V(r,0)$ by setting $z=0$ in (1) simplifies things somewhat
$$
V(r,0)=\frac{1}{\pi}\int\limits_0^\pi d\eta \ \phi(ir\cos\eta)
$$
For $\phi(z)=\ln(z)$, performing the integral yields
$$
V(r,0)=\ln(r/2)
$$
$\dagger$ Almost. The form of potential you specify on $z>0$ is continued to all $z$ when doing this.
$\underline{\text{The Neumann problem}} $
An alternative is to motivate physically plausible boundary data on the $z=0$ plane. Since you want to find $V(r,0)$, it wouldn't be useful to consider a Dirichlet condition on the $z=0$ plane. The examples you give, such as that of a point charge or disk suggest we apply the Neumann boundary condition $\partial_zV(r,0)=0$.
The simplest way to enforce this boundary condition is: given $\phi(z)$ for $z>0$, let $\psi(z)$ be the even continuation of $\phi$ to all $z$. Now we can work in the whole space and forget about the boundary condition at $z=0$.
Separation of variables for the axisymmetric Laplacian yields
$$ \tag{2}
V(r,z)=\int\limits_{-\infty}^\infty \frac{dk}{2\pi} \ A(k) I_0(|k|r)e^{ikz}
$$
Where $I_0$ is a modified Bessel function. The function $A(k)$ is determined by the boundary data at $r=0$
$$
V(0,z)=\psi(z)=\int\limits_{-\infty}^\infty \frac{dk}{2\pi} \ A(k) e^{ikz}
$$
Which we recognize as the Fourier transform of $\psi$
$$
A(k)=\int\limits_{-\infty}^\infty dz \ \psi(z) e^{-ikz}
$$
The $A(k)$ determine $V$ everywhere in space via (2). Since $\psi$ is even by construction, we have
$$
A(k)=2\int\limits_{0}^\infty dz \ \psi(z) \cos(kz)=2\int\limits_{0}^\infty dz \ \phi(z) \cos(kz)
$$
Finally, on the $z=0$ plane, we can write $V$ as
$$
V(r,0)=2\int\limits_{-\infty}^\infty \frac{dk}{2\pi} \int\limits_0^\infty  dz' \ \phi(z') I_0(|k|r) \cos(kz')
$$
Which is expressed only in terms of your specified potential $\phi(z)$ on $z>0$. Actually performing the integrals in closed form for various $\phi$ may be difficult.
$\underline{\text{Multipole expansion}} $
Outside of a finite charge distribution, we have the multipole expansion of the potential in inverse powers of $r$
$$
V(r)=\frac{q}{r} + \frac{p}{r^2} + \frac{Q}{r^3} + \dots 
$$
The coefficients $q$, $p$, $Q$, will not concern us here. What is important is that the potential decays algebraically in $r$ outside of a finite charge distribution. If you are studying potentials that scale like $\ln r$, then either your region is not source free, or your charge distribution is infinite in extent (eg on or below the $z=0$ plane the charge density is non-vanishing as $r \to \infty$).
A: 
Starting over from the beginning ...
$$\frac{\partial \phi^2 }{\partial z^2} + \frac{\partial \phi^2}{\partial r^2 }+ \frac{1}{r} \frac{\partial \phi}{\partial r }= 0 $$
Letting $\phi(z,r)=\Phi(z) \Psi(r),$
$$
\Phi^{\prime \prime}(z)\Psi(r)+\Phi(z) \Psi^{\prime \prime}(r)+ \frac{1}{r} \Phi(z) \Psi^{\prime}(r)=0\\
%
\frac{\Phi^{\prime \prime}(z)}{\Phi(z)}+\frac{ \Psi^{\prime \prime}(r)+ \frac{1}{r}  \Psi^{\prime}(r)}{\Psi(r)}=0\\
%
$$
We can separate variables and write
$$
\frac{ \Psi^{\prime \prime}(r)+ \frac{1}{r}  \Psi^{\prime}(r)}{\Psi(r)}=-\frac{\Phi^{\prime \prime}(z)}{\Phi(z)}=-\lambda^2
$$
Resulting in two ordinary differential equations:
$$\Psi^{\prime \prime}(r)+ \frac{1}{r}  \Psi^{\prime}(r)+\lambda^2 \Psi(r)= 0 \\
 \Phi^{\prime \prime}(z)-\lambda^2 \Phi(z) =0
$$
To meet the boundary conditions, you can know express the solution as a Fourier series.
