= Solution
The in-plane <equilibrium> equations are $\tau_{xx,x}+\tau_{xy,y}=0$ and $\tau_{xy,x}+\tau_{yy,y}=0$. On a <simply connected> patch, each <divergence-free> row has a <streamfunction>: write $\tau_{xx}=\phi_y$, $\tau_{xy}=-\phi_x=-\psi_y$, $\tau_{yy}=\psi_x$. Symmetry gives $\phi_x=\psi_y$, so another potential exists with $A_y=\phi$, $A_x=\psi$. Hence
$$
\boxed{\tau_{xx}=A_{yy},\qquad\tau_{yy}=A_{xx},\qquad\tau_{xy}=-A_{xy}.}
$$
If two such <Airy stress functions> give the same <stress>, their difference has zero <Hessian> and is therefore an arbitrary <affine function> $a+bx+cy$ on a connected domain. This is the full nonuniqueness for a fixed <stress>. The existence argument is local, or global on a <simply connected> domain; domains that are <multiply connected regions> require the associated force and moment period conditions for a single-valued global potential. <Equilibrium> alone should not silently erase that topological qualification.
For <plane strain> put $s=e_{xx}+e_{yy}$. The <constitutive equation> gives $\tau_{xx}+\tau_{yy}=2(\lambda+\mu)s$. Consequently
$$
e_{xx}=\frac{\tau_{xx}-\lambda s}{2\mu},\quad
e_{yy}=\frac{\tau_{yy}-\lambda s}{2\mu},\quad
e_{xy}=\frac{\tau_{xy}}{2\mu},\quad
s=\frac{\Delta A}{2(\lambda+\mu)}.
$$
The <displacement> <Saint-Venant compatibility equations> $e_{xx,yy}+e_{yy,xx}-2e_{xy,xy}=0$ then becomes
$$
0=2\mu(e_{xx,yy}+e_{yy,xx}-2e_{xy,xy})
=\frac{\lambda+2\mu}{2(\lambda+\mu)}\Delta^2A.
$$
For stable elastic moduli the coefficient is nonzero, so $\boxed{\Delta^2A=0}$. If $D$ is bounded and smooth and $A=\partial_nA=0$ on its boundary, integrating twice by parts gives $0=\int_DA\Delta^2A\,dV=\int_D(\Delta A)^2\,dV$. Thus $A$ is harmonic with zero boundary value and $\boxed{A=0}$ by uniqueness of the <Dirichlet problem>. The normal-<derivative> condition is what removes the extra fourth-order <boundary terms>.
For the pressurized <annulus>, substituting the radial <constitutive equations> into <equilibrium> gives $u''+u'/r-u/r^2=0$, so $u=Cr+D/r$. The stresses are
$$
\tau_{rr}=2(\lambda+\mu)C-\frac{2\mu D}{r^2},\qquad
\tau_{\theta\theta}=2(\lambda+\mu)C+\frac{2\mu D}{r^2}.
$$
Using gauge stresses relative to ambient, the boundaries are $\tau_{rr}(a)=-P$, $\tau_{rr}(b)=0$. Therefore
$$
2(\lambda+\mu)C=\frac{Pa^2}{b^2-a^2},\qquad
2\mu D=\frac{Pa^2b^2}{b^2-a^2},
$$
and
$$
\boxed{|\tau_{rr}-\tau_{\theta\theta}|=\frac{2Pa^2b^2}{(b^2-a^2)r^2},\qquad
\max_{a\leq r\leq b}|\tau_{rr}-\tau_{\theta\theta}|=\frac{2Pb^2}{b^2-a^2}.}
$$
The maximum occurs at the bore. Once the chosen <yield criterion> is reached, <Hooke's law> cannot describe the total <strain> alone: write total <strain> as elastic plus <plastic strain> and supplement the elastic <stress> law with a yield condition, a <plastic flow rule> and any required <work hardening> law. A plastic region starts near the bore and must be matched to the remaining elastic region. The exact conversion between a principal-<stress> difference and the quoted <yield stress> depends on whether the material criterion is Tresca or von Mises. This is the <elastic stress concentration in a pressurized thick cylinder>.
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