 Solution

In the completely elastic exterior, radial elastic compatibility makes $q+2t$ constant. The far-field stress vanishes, so $q+2t=0$ and $t=-q/2$. Radial force balance then gives $q^{\prime }+3q/r=0$, whence $q=C/r^3$. The pressure boundary condition is $q(a)=-P$: positive applied pressure is compressive. ThereforeFor completeness, the corresponding radial displacement field is $u=P{a}^3/(4\mu r^2)$, which gives these components through Hooke's law and decays at infinity.

The Tresca yield criterion becomesFor $P\ge 0$ this is largest at the cavity wall and decreases strictly with radius when $P>0$. The first yield point is therefore at $r=a$, where $3P/4={\tau }_Y$. HenceThe normalization matters: ${\tau }_Y$ in this paper is a shear yield stress, rather than the uniaxial yield strength.