= Solution
Let
$$
s_m=a\exp\left(\frac{P_m-P_c}{4\tau_Y}\right),\qquad
\Delta=P_m-P,\qquad0\leq\Delta\leq P_m.
$$
At maximum <pressure>, the <stress> is the field from part (c) with $P=P_m$ and $s=s_m$. Under the assumed <elastic unloading>, previously accumulated plastic <strain> is frozen. Therefore it is the <stress> increment, rather than the total <stress>, that satisfies homogeneous <linear elasticity> and <radial elastic compatibility>. Applying the <superposition principle>, a <pressure> change $-\Delta$ produces the elastic increments
$$
\delta q=\Delta\frac{a^3}{r^3},\qquad
\delta t=-\frac{\Delta}{2}\frac{a^3}{r^3}.
$$
They have zero far-field <stress> and change the wall <traction> from $-P_m$ to $-P$. These increments are the same in the formerly plastic and formerly elastic regions because their elastic <Lamé parameters> agree. Adding them to the maximum-load field gives
$$
\boxed{q(r;P)=
\begin{cases}
-P_m+4\tau_Y\log(r/a)+\Delta(a/r)^3,&a\leq r\leq s_m,\\
-P_c(s_m/r)^3+\Delta(a/r)^3,&r\geq s_m,
\end{cases}}
$$
and
$$
\boxed{t(r;P)=\tau_{\phi\phi}(r;P)=
\begin{cases}
-P_m+4\tau_Y\log(r/a)+2\tau_Y-\tfrac12\Delta(a/r)^3,&a\leq r\leq s_m,\\
\tfrac12P_c(s_m/r)^3-\tfrac12\Delta(a/r)^3,&r\geq s_m.
\end{cases}}
$$
These formulae retain $s_m$: it records the plastic loading history and is not the instantaneous loading-front formula with $P$ substituted for $P_m$. If the entire return is elastic, setting $\Delta=P_m$ gives the <residual stress> after removal of the applied <pressure>.
To check admissibility, examine the signed difference of the <principal stresses> in the formerly plastic shell:
$$
D(r)=q-t=-2\tau_Y+\frac{3\Delta}{2}\left(\frac ar\right)^3.
$$
It immediately moves into the interior of the <Tresca yield criterion> when unloading starts. It can next reach the opposite limit $D=+2\tau_Y$. For fixed $\Delta>0$, $D$ is largest at $r=a$ and stays above $-2\tau_Y$. At the wall,
$$
-2\tau_Y+\frac32\Delta=2\tau_Y
\quad\Longleftrightarrow\quad
\Delta=\frac83\tau_Y=2P_c.
$$
There is no earlier violation in the outer region. Indeed,
$$
D(r)=D(s_m)\left(\frac{s_m}{r}\right)^3,\qquad r\geq s_m,
$$
so its magnitude is no larger than $|D(s_m)|$, which has already been controlled by the shell calculation. Hence the first <reverse yielding> occurs at the cavity wall, with
$$
\boxed{\Delta_{\mathrm{reverse}}=2P_c=\frac83\tau_Y,\qquad
P_{\mathrm{reverse}}=P_m-2P_c.}
$$
If $P_c<P_m<2P_c$, the <stress> stays within the <yield surface> throughout unloading to zero. If $P_m=2P_c$, it reaches reverse yield exactly at $P=0$; further unloading would produce reverse plastic deformation. If $P_m>2P_c$, reverse plastic deformation starts while the applied <pressure> is still positive. Thus
$$
\boxed{\text{reverse yielding before zero pressure occurs iff }P_m>\frac83\tau_Y.}
$$
If “yield again while unloaded” includes first touching the <yield surface> at the final endpoint, the corresponding condition is $P_m\geq8\tau_Y/3$. The displayed <elastic unloading> fields are physically valid only for $0\leq\Delta\leq\min(P_m,2P_c)$; beyond that interval a new reverse-plastic region must be solved for. Continuing the elastic formula to $P=0$ when $P_m>2P_c$ would violate the given <Tresca yield criterion>.
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