Concerning Griffiths's book on electromagnetism I come more and more to the conclusion that it seems to be confusing sometimes and not always leads to correct ideas in the student's minds since it is well known that of course there are surface charges on conductors also for DC currents. The here critizized book by Assis is not as bad as you claim, because as far as I have looked at it, it gives correct solutions to the corresponding stationary problems. I'm not so sure concerning the didactics, because I don't see the sense of discussing old-fashioned action-at-a-distance models by Weber et al. The correct thing are the Maxwell Equations (in local form)!
So let's do the most simple example of an infinite coaxial cable. Let the inner cylinder of radius a be along the [itez]z[/itex] axis and the outer conductor a cylindrical shell with inner radius b and outer radius c.
In the stationary case the Maxwell equations separate completely into two equations for the electric and two equations for the magnetic field components. The electric part reads
\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=0.
Further we only need the non-relativistic approximation of Ohm's law, i.e.,
\vec{j}=\sigma \vec{E}.
We make the ansatz that in the inner wire and the outer shell we have homogeneous current densities, i.e.,
\vec{j}=\begin{cases}<br />
\frac{I}{\pi a^2} \vec{e}_z & \text{for} \quad 0 \leq \rho <a,\\<br />
-\frac{I}{\pi(c^2-b^2)} \vec{e}_z & \text{for} \quad b<\rho<c,\\<br />
0 & \text{elsewhere}.<br />
\end{cases}<br />
Here (\rho,\varphi,z) are the usual cylinder coordinates. Note that this ansatz takes into account global charge conservation. Local charge conservation is fulfilled anyway since \vec{\nabla} \cdot \vec{j}=0 everywhere, as it must be.
We shall now see that this ansatz admits an exact solution of the stationary Maxwell equations, satisfying all boundary conditions.
To that end we note that due to the first stationary Maxwell equations the electric field can be written as the gradient of a scalar potential, which due to symmetries should be of the form
\Phi(\vec{x})=z \phi(\rho),
where \phi is a still to be determined continuous function.
We further know that, according to Ohm's Law we must have
\vec{E}=\frac{I}{\pi a^2 \sigma} \vec{e}_z \quad \text{for} \quad 0 \leq \rho<a
and
\vec{E}=-\frac{I}{\pi (c^2-b^2)\sigma} \vec{e}_z \quad \text{for} \quad b < \rho<c.
The field is to be determined in the other region, i.e., in the gap between the conductors (for a<\rho<b) and outside of the coax cable, \rho>c.
To that end we use the ansatz for the potential to get
\vec{E}=-\vec{\nabla} \Phi=-\phi(\rho) \vec{e}_z - z \phi'(\rho) \vec{e}_{\rho}.
Ohm's Law implies that
\phi(\rho)=-\frac{I}{\pi a^2 \sigma} \quad \text{for} \quad 0 \leq \rho<a
and
\phi(\rho)=+\frac{I}{\pi (c^2-b^2) \sigma} \quad \text{for} \quad b<\rho<c.
For the missing regions we have to solve the potential equation, following from Gauß's Law,
\vec{\nabla} \cdot \vec{E}=-\Delta \Phi=0 \; \Rightarrow \; \frac{\mathrm{d}}{\mathrm{d} \rho} [\rho \phi'(\rho)]=0.
The integration of this equation leads to
\phi(\rho)=A_j \ln(\rho/a) + B_j
with integration constants A_{j}, B_{j} in
the regions 1 (a<\rho<b) and 2 (\rho > c). I've also written \ln(\rho/a) to avoid dimensionful quantities as the argument of a logarithm. Of course, one can choose any other constant instead of a, but changing it just adds a constant to the potential that is physically irrelevant.
To determine the integration constants we need the boundary conditions at the surfaces of the wire and the cylindrical shell. These are given by the continuity of the potential and of the tangential component, E_z of the electric field. With our ansatz the first condition already fixes also the second. So we have
\phi(a)=B_1=-\frac{I}{\pi a^2 \sigma}
and
\phi(b)=A_1 \ln(b/a)+B_1=+\frac{I}{\pi(c^2-b^2)\sigma} \; \Rightarrow\; A_1=\frac{I}{\sigma \pi \ln(b/a)} \left (\frac{1}{a^2}+\frac{1}{c^2+b^2} \right ).
For the region outside the coax cable we must demand that the field stays finite at infinity, leading to
A_2=0, \quad B_2=\frac{I}{\pi(c^2-b^2) \sigma}.
So finally we get
<br />
\phi(\rho)=\begin{cases}<br />
-I/(\pi a^2 \sigma) & \text{for} \quad 0 \leq \rho<a, \\<br />
I/[\pi \sigma \ln(b/a)] [1/a^2+1/(c^2-b^2)] \ln(\rho/a)-I/(\pi a^2 \sigma) & \text{for} \quad a \leq \rho <b,\\<br />
I/[\pi (c^2-b^2) \sigma] & \text{for} \quad \rho \geq b.<br />
\end{cases}<br />
At the surface of the wire \rho=a and the inner surface of the shell of the outer conductor are surface charges of a density given by the jump of E_{\rho}, i.e.,
\sigma_{Q}=-\frac{I}{\pi \sigma \ln(b/a)} \left (\frac{1}{a^2}+\frac{1}{c^2-b^2} \right ) \frac{z}{a}
and at \rho=b, it's
\sigma_{Q}=+\frac{I}{\pi \sigma \ln(b/a)} \left (\frac{1}{a^2}+\frac{1}{c^2-b^2} \right ) \frac{z}{b}.
The correct discussion is found in the good old book (for the case c \rightarrow \infty).
A. Sommerfeld, Lectures on Theoretical Physics, vol. 3 (electrodynamics)
There you also find a nice picture about the electric field lines and the energy flux (Poynting vector). One should note that the surface charges and the electric field outside the conductor are pretty small, so that we don't realize them in practice, but they are crucial for the entire conduction business and the energy flow.
