= 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 $q'+3q/r=0$, whence $q=C/r^3$. The <pressure> <boundary condition> is $q(a)=-P$: positive applied <pressure> is compressive. Therefore
$$
\boxed{\tau_{rr}=-P\frac{a^3}{r^3},\qquad
\tau_{\theta\theta}=\tau_{\phi\phi}=\frac P2\frac{a^3}{r^3}.}
$$
For completeness, the corresponding radial <displacement field> is $u=Pa^3/(4\mu r^2)$, which gives these components through <Hooke's law> and decays at infinity.
The <Tresca yield criterion> becomes
$$
f(r)=\frac12|q-t|=\frac{3P}{4}\frac{a^3}{r^3}.
$$
For $P\geq0$ 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 $3P/4=\tau_Y$. Hence
$$
\boxed{P_c=\frac43\tau_Y,\qquad r_{\mathrm{first}}=a.}
$$
The normalization matters: $\tau_Y$ in this paper is a shear <yield stress>, rather than the uniaxial <yield strength>.
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