= Solution
The radial equilibrium equation gives $dt_r/dr=(t_\theta-t_r)/r$. Set $C=a^2-A^2$, so that $R^2=r^2-C$. Substitution of the <neo-Hookean solid> <stress> difference eliminates both $q$ and $t_\theta$:
$$
\boxed{\frac{dt_r}{dr}=\frac\mu r\left[\frac{r^2}{r^2-C}-\frac{r^2-C}{r^2}\right]
=\frac{\mu C(2r^2-C)}{r^3(r^2-C)},\qquad a<r<b.}
$$
This is a first-order <ordinary differential equation> for the radial <stress>, parametrized by the unknown inner radius. For inflation $C>0$, its right side is positive, so $t_r$ increases from the negative inner <pressure> to zero at the outer surface. The denominator stays positive because $r^2-C=R^2\geq A^2>0$.
Back to article page
