= Solution
\b[Constitutive law and Poisson ratio.] Use positive <stress> for tension. In an isotropic linearly elastic medium, <Hooke's law> gives
$$
\boxed{e_1=\frac{\tau_1-\sigma(\tau_2+\tau_3)}E.}
$$
The analogous equations follow by cyclic permutation. Under uniaxial tension in direction 1, $e_1=\tau_1/E$ and $e_2=e_3=-\sigma\tau_1/E$. Thus <Poisson's ratio> $\sigma=-e_2/e_1$ measures transverse contraction relative to longitudinal extension when the transverse stresses are zero. It is dimensionless; it is not the ratio of transverse <stress> to longitudinal <stress>.
\b[Radial and hoop strains.] Let a material point originally at radius $r$ move to $r+R(r)$, with small axisymmetric displacement. Two neighbouring radii separated by $dr$ move apart by $dr+dR$, so the radial fractional extension is $dR/dr$. A circle's circumference changes from $2\pi r$ to $2\pi(r+R)$, so its fractional extension is $R/r$. Consequently
$$
\boxed{e_r=R'(r),\qquad e_\phi=\frac{R(r)}r.}
$$
The absence of longitudinal <stress> is a <plane stress> condition, not a claim of zero longitudinal <strain>.
\b[Equilibrium of an annular wedge.] Consider a slice of unit axial length, radial thickness $dr$ and small angular width $d\phi$. The difference of the outward and inward radial face forces is $(r\tau_r)'dr\,d\phi$ to leading order. The two tangential face tensions have inward radial resultant $\tau_\phi dr\,d\phi$: each contributes $\tau_\phi dr\sin(d\phi/2)$. With no body <force>, their balance gives the <radial equilibrium of a thick elastic cylinder>,
$$
\boxed{\tau_\phi=\frac d{dr}(r\tau_r)
=\tau_r+r\tau_r'.}
$$
This hoop contribution is why the radial <stress> need not be spatially constant in a thick wall.
\b[Displacement equation.] With $\tau_z=0$, the two <Hooke's law> equations read
$$
R'=\frac{\tau_r-\sigma\tau_\phi}E,\qquad
\frac Rr=\frac{\tau_\phi-\sigma\tau_r}E.
$$
Invert them:
$$
\tau_r=\frac E{1-\sigma^2}\left(R'+\sigma\frac Rr\right),\qquad
\tau_\phi=\frac E{1-\sigma^2}\left(\frac Rr+\sigma R'\right).
$$
Substitute in $\tau_\phi=(r\tau_r)'$. After cancelling the common factor, the right-hand side is $rR''+(1+\sigma)R'$, so
$$
\frac Rr+\sigma R'=rR''+(1+\sigma)R',
\qquad
\boxed{r^2R''+rR'-R=0.}
$$
The cancellation explains why this displacement equation does not depend on the value of <Poisson's ratio>.
\b[Power laws and wall pressure.] Try $R\propto r^n$. The indicial equation is $n(n-1)+n-1=n^2-1=0$, so
$$
R=c_1r+\frac{c_2}r.
$$
The resulting <Lamé stress distribution in a pressurized cylindrical wall> can be written
$$
\tau_r=P-\frac Q{r^2},\qquad
\tau_\phi=P+\frac Q{r^2},
\quad P=\frac{Ec_1}{1-\sigma},\quad Q=\frac{Ec_2}{1+\sigma}.
$$
The <pressure> boundary tractions are $\tau_r(a)=-p_1$ and $\tau_r(b)=-p_0$. Solving them yields
$$
P=\frac{p_1a^2-p_0b^2}{b^2-a^2},\qquad
Q=\frac{(p_1-p_0)a^2b^2}{b^2-a^2}.
$$
Hence the positive compressive radial <pressure> in the wall is
$$
\boxed{p(r)=-\tau_r(r)=
\frac{p_0b^2-p_1a^2+(p_1-p_0)a^2b^2/r^2}{b^2-a^2},
\qquad a\leq r\leq b.}
$$
It equals $p_1$ at $a$ and $p_0$ at $b$. This directional radial <pressure> is not an isotropic fluid <pressure>: radial and tangential stresses differ. The associated displacement is $R=[(1-\sigma)Pr+(1+\sigma)Q/r]/E$. Also $e_z=-\sigma(\tau_r+\tau_\phi)/E=-2\sigma P/E$, generally nonzero even though $\tau_z=0$.
\b[Maximum absolute tangential stress.] For $p_1>p_0$, $Q>0$ and $\tau_\phi'= -2Q/r^3<0$. Therefore the maximum of its absolute value must be at one of the two surfaces. Keeping the specified external <pressure> gives the fully general answer
$$
\boxed{\max_{a\leq r\leq b}|\tau_\phi(r)|
=\frac1{b^2-a^2}\max\left\{
\left|p_1(a^2+b^2)-2p_0b^2\right|,
\left|2p_1a^2-p_0(a^2+b^2)\right|
\right\}.}
$$
For the usual high internal <pressure> with negligible external <pressure>, the largest magnitude is the tensile <stress> on the inner wall:
$$
\boxed{\tau_{\phi,\max}\simeq\tau_\phi(a)
=\frac{p_1(a^2+b^2)}{b^2-a^2}\quad(p_0\simeq0).}
$$
For finite $p_0$, the inner-wall expression is $[p_1(a^2+b^2)-2p_0b^2]/(b^2-a^2)$, but its absolute value is not always the larger endpoint value. For example, nearly equal internal and external pressures can leave the outer surface more compressive. With $p_1=p_0=p$, both stresses correctly reduce to $-p$ everywhere. These results use a homogeneous linear elastic wall without axial loading; a closed-ended pipe would require a different longitudinal-<stress> condition.
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